Inquiry Into Life 15th Edition Mader Test Bank Chapter 01 The Study of Life Multiple Choice Questions 1. Organs are composed of tissues, which are composed of cells. This is an example of which characteristic of life? A. Living things grow and develop. B. Living things respond to stimuli. C. Living things maintain themselves by homeostasis. D. Living things have levels of hierarchical organization. E. Living things are adapted to the environment. Organs represent one level in the hierarchy of biological organization but they are a higher level of organization than either tissues or cells. The other answer choices are also characteristics of life; however, the example is concerned specifically with levels of organization.
Bloom's Level: 1. Remember Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Levels of Biological Organization
2. The smallest unit that has all of the characteristics of life is the A. cell. B. tissue. C. organ. D. organ system. E. organism. The cell is the basic unit of life. Some organisms are single-celled; however, others are comprised of cells arranged into tissues, organs, and organ systems.
Bloom's Level: 1. Remember Learning Outcome: 01.01.02 Distinguish between the levels of biological organization. Section: 01.01 Topic: Levels of Biological Organization
3. Which level of biological organization is composed of several tissues? A. organism B. organ system C. organ D. cell E. molecules Organs are composed of tissues working together in a common set of functions. Tissues are collections of cells, and cells are made up of various molecules. Organs make up organ systems, which comprise the organism.
Bloom's Level: 1. Remember Learning Outcome: 01.01.02 Distinguish between the levels of biological organization. Section: 01.01 Topic: Levels of Biological Organization
4. Which sequence correctly lists the different levels of biological organization, from the smallest and simplest to the largest and most complex? A. cells-organs-tissues-organ systems-organism B. cells-tissues-organ systems-organs-organism C. tissues-cells-organs-organ systems-organism D. tissues-organs-organ systems-organism-cells E. cells-tissues-organs-organ systems-organism The cell is the smallest, simplest unit of life. Cells are organized into tissues, which make up organs. A group of organs working together for a common set of functions comprises an organ system. An individual organism contains several organ systems.
Bloom's Level: 1. Remember Learning Outcome: 01.01.02 Distinguish between the levels of biological organization. Section: 01.01 Topic: Levels of Biological Organization
5. Which statement concerning reproduction is false? A. Hereditary information is passed on to the next generation. B. Hereditary information is in the form of genes. C. The offspring of asexual organisms have the same genes as the single parent. D. The offspring of sexual organisms have roughly one half of the genes from each parent. E. The offspring of multicellular organisms tend to be identical to the parent. Multicellular organisms generally use sexual reproduction, in which each parent contributes to the genetic makeup of the offspring; thus, the offspring is not genetically identical to either parent.
Bloom's Level: 1. Remember Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Characteristics of Life
6. Which sequence of classification categories is in the proper order from least to most inclusive? A. genus, class, kingdom, domain, order, phylum, species, family B. domain, class, genus, family, species, order, phylum, kingdom C. species, genus, family, order, class, phylum, kingdom, domain D. genus, species, order, class, family, kingdom, domain, phylum E. species, genus, family, class, order, phylum, kingdom, domain The correct sequence is species, genus, family, order, class, phylum, kingdom, domain. Species is the least inclusive category in the taxonomic hierarchy; domain is the most inclusive.
Bloom's Level: 1. Remember Learning Outcome: 01.02.01 Describe how living things are classified. Section: 01.02 Topic: Levels of Biological Organization
7. What is the correct format for the binomial name of human beings? A. Homo Sapiens B. homo Sapiens C. homo sapiens D. Homo sapiens E. Sapiens homo When using binomial naming, the genus name should come first, and its first letter is always capitalized. The species name follows, in all lowercase letters. Thus, the correct answer is Homo sapiens.
Bloom's Level: 1. Remember Learning Outcome: 01.02.01 Describe how living things are classified. Section: 01.02 Topic: Levels of Biological Organization
8. Corn belongs to the kingdom A. Plantae. B. Animalia. C. Fungi. D. Protista. E. Archaea. Corn is an example of a multicellular, photosynthetic organism belonging to kingdom Plantae.
Bloom's Level: 1. Remember Learning Outcome: 01.02.01 Describe how living things are classified. Section: 01.02 Topic: Levels of Biological Organization
9. The three major domains of life are A. plants, animals, and protists. B. bacteria, fungi, and eukaryotes. C. eukaryotes, prokaryotes, and animals. D. archaea, bacteria, and eukaryotes. E. eukaryotes, prokaryotes, and fungi. The three major domains of life include the bacteria, archaea, and eukaryotes. The rest of the living organisms listed are not classified at the domain level.
Bloom's Level: 1. Remember Learning Outcome: 01.02.02 Distinguish between the three domains of life. Section: 01.02 Topic: Levels of Biological Organization
10. A multicellular, photosynthetic organism with complex, specialized cells and tissues would most likely be assigned to A. kingdom Animalia. B. kingdom Fungi. C. domain Archaea. D. kingdom Protista. E. kingdom Plantae. Members of kingdom Plantae are multicellular, with specialized cells and tissues, and can form their own "food" by photosynthesis. Organisms in kingdom Animalia are also multicellular with specialized cells and tissues, but they are heterotrophs that ingest their food. Kingdom Fungi is comprised of unicellular and multicellular organisms, all heterotrophic by absorption. Kingdom Protista contains mostly unicellular organisms with a few multicellular forms, some of which are heterotrophic while others are photosynthetic. They lack the high degree of cellular and tissue specialization seen in plants. Domain Archaea contains only unicellular, prokaryotic organisms.
Bloom's Level: 1. Remember Learning Outcome: 01.02.01 Describe how living things are classified. Section: 01.02 Topic: Levels of Biological Organization
11. The layer where organisms can exist on the surface of the earth is the A. biosphere. B. ecosystem. C. population. D. homeostasis. E. community. The biosphere is the zone of air, land, and water on the earth where living organisms are found. An ecosystem is a community of organisms plus the non-living components of their environment. A community is a collection of populations of different species living in the same area. A population is a collection of members of the same species living in the same area. Homeostasis is the maintenance of a steady internal environment.
Bloom's Level: 1. Remember Learning Outcome: 01.01.02 Distinguish between the levels of biological organization. Section: 01.01 Topic: Levels of Biological Organization
12. All the organisms of various species living within a given area constitute a(n) A. biosphere. B. ecosystem. C. population. D. domain. E. community. A community is a collection of populations of different species living in the same area. All the members of a particular species living within a given area are known as a population. The biosphere is the zone of air, land, and water on the earth where living organisms are found. An ecosystem is a community of organisms plus the non-living components of their environment. Domain is the highest, most inclusive category for the classification of organisms.
Bloom's Level: 1. Remember Learning Outcome: 01.01.02 Distinguish between the levels of biological organization. Section: 01.01 Topic: Levels of Biological Organization
13. A community of organisms along with their physical environment constitutes a(n) A. biosphere. B. ecosystem. C. population. D. habitat. E. community. An ecosystem includes not only all the living organisms in a given area, but also the nonliving, physical environment. The biosphere is the zone of air, land, and water on the earth where living organisms are found. A population is a collection of members of the same species living in the same area. A habitat is where an organism lives. A community is a collection of populations of different species living in the same area.
Bloom's Level: 1. Remember Learning Outcome: 01.01.02 Distinguish between the levels of biological organization. Section: 01.01 Topic: Levels of Biological Organization
14. The most important factor that determines where major ecosystems are located on the globe is A. soil type. B. available vegetation. C. climate. D. oxygen levels. E. political boundaries. The main determinant of where the major ecosystems are located is climate. Climate exerts the most important influence over what vegetation is available, and vegetation contributes to soil formation. Oxygen levels are a concern only at the highest altitudes, and political boundaries have no inherent influence on ecosystem locations.
Bloom's Level: 1. Remember Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Characteristics of Life
15. The term "biodiversity" refers to A. a study of ecosystems without consideration of the living organisms. B. the similarities between species. C. the rate of extinction of species due to human activities. D. the total number of species, variation of species, and the ecosystems in which they live. E. the study of food chains within an ecosystem. The term "biodiversity" is quite broad, encompassing the total number of species, variation of species, and the ecosystems in which they live. All of the other answer choices are too narrow to fit the definition of biodiversity.
Bloom's Level: 1. Remember Learning Outcome: 01.04.02 Summarize some of the major challenges facing science. Section: 01.04 Topic: Biodiversity
16. A possible explanation for a natural event, based on observations and past knowledge, is a A. hypothesis. B. phenomenon. C. control. D. variable. E. theory. A hypothesis is defined as a possible explanation for a natural event (phenomenon) based on observations and past knowledge. When a hypothesis has withstood repeated testing over time, it may be elevated to a theory. In a controlled experiment, the control is the subject or group not subjected to the factor, or variable, being tested.
Bloom's Level: 1. Remember Learning Outcome: 01.03.02 Distinguish between a theory and a hypothesis. Section: 01.03 Topic: Scientific Method
17. Which statement regarding the scientific method is false? A. Hypotheses are tentative explanations of observed phenomena. B. A hypothesis may become a theory, which may become a law. C. Deductive reasoning is often used in the design of an experiment. D. A control group is always exposed to the factor being tested. E. Conclusions are formed after an experiment is conducted. An experimental group, not a control group, is always exposed to the factor being tested. All the other answer choices are accurate statements.
Bloom's Level: 1. Remember Learning Outcome: 01.03.01 Identify components of the scientific method. Section: 01.03 Topic: Scientific Method
18. Which answer choice lists the steps of the scientific method in the correct order? A. hypothesis, observation, experiment, conclusion, data collection B. conclusion, hypothesis, observation, experiment, data collection C. observation, hypothesis, experiment, data collection, conclusion D. observation, experiment, hypothesis, conclusion, data collection E. data collection, conclusion, hypothesis, experiment, observation In a typical application of the scientific method, observations are used to formulate a hypothesis. The hypothesis may then be tested, as in an experiment. Data is collected as part of the experimental process. Based on the results of the hypothesis testing, a conclusion is made.
Bloom's Level: 1. Remember Learning Outcome: 01.03.01 Identify components of the scientific method. Section: 01.03 Topic: Scientific Method
19. Which of the following is one of the domains of life? rev: 10_04_2013_QC_36723 A. Animalia B. Eukarya C. Molds D. Protista E. Fungi Eukarya is a domain of life. Animalia, Protista, and Fungi are all kingdoms of life. Molds is not a level of scientific classification.
Bloom's Level: 1. Remember Learning Outcome: 01.02.02 Distinguish between the three domains of life. Section: 01.02 Topic: Levels of Biological Organization
20. Technology is the . A. application of scientific knowledge for a practical purpose. B. study of living organisms. C. study of the interactions between living organisms and their environment. D. application of common knowledge for a practical purpose. E. application of laws to benefit society. Technology is the application of scientific knowledge for a practical purpose.
Bloom's Level: 1. Remember Learning Outcome: 01.04.01 Distinguish between science and technology. Section: 01.04 Topic: Experimental Design
True / False Questions 21. Using technology in the field of agriculture has enabled farmers to feed a growing human population. TRUE Advances in technology have enabled farmers to grow more food in order to feed a growing human population.
Bloom's Level: 1. Remember Learning Outcome: 01.04.01 Distinguish between science and technology. Section: 01.04 Topic: Scientific Method
Multiple Choice Questions 22. A physician specializes in surgery involving the following group of organs: mouth, esophagus, stomach, and intestines. Overall, what is the highest level of organization that this physician is specialized in? A. cell B. tissue C. organ D. organ system E. organism The mouth, esophagus, stomach, and intestines are all organs belonging to the digestive system. An organism is made up of multiple organ systems. The individual organs are made up of tissues, which are collections of cells.
Bloom's Level: 2. Understand Learning Outcome: 01.01.02 Distinguish between the levels of biological organization. Section: 01.01 Topic: Levels of Biological Organization
23. Which statement about living organisms is not correct? A. Living organisms create energy. B. Living organisms maintain homeostasis. C. Living organisms reproduce. D. Living organisms have adaptations. E. Living organisms grow and develop. Living organisms cannot create energy. Instead, they convert it from one form to another. It will then be used to maintain homeostasis, reproduce, and adapt, as well as grow and develop.
Bloom's Level: 2. Understand Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Characteristics of Life
24. When you are overheated, you perspire, and when you are too cold, you shiver to generate heat. Which property of life is best represented by this example? A. homeostasis B. development C. behavior D. organization E. adaptation Homeostasis is the body's physical and physiological responses to changing conditions over a short period of time.
Bloom's Level: 2. Understand Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Characteristics of Life
25. The body temperature in humans is maintained around 37C. Which characteristic of life does this information represent? A. Living things acquire materials and energy from the environment. B. Living things are homeostatic. C. Living things are adapted. D. Living things grow and develop. E. Living things respond to stimuli. Keeping the internal body temperature within a constant range is an important part of homeostasis. Living things acquire materials and energy from the environment and respond to stimuli in order to maintain homeostasis. Adaptations that organisms have for surviving in their environment enable them to maintain homeostasis. The maintenance of homeostasis allows organisms to grow and develop.
Bloom's Level: 2. Understand Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Characteristics of Life
26. The bones in a bird are hollow, reducing its weight for flight. This is an example of which characteristic of life? A. Living things grow and develop. B. Living things acquire materials and energy from the environment. C. Living things reproduce. D. Living things are adapted. E. Living things are homeostatic. Adaptations, such as the hollow bones of birds, are features that make organisms better suited to their environment. All the other answer choices are also characteristics of living things, but the example of the bird skeleton has to do specifically with adaptation.
Bloom's Level: 2. Understand Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Characteristics of Life
27. Which classification category includes the most species? A. family B. genus C. class D. phylum E. kingdom Placed in order from most inclusive (containing the greatest number of species) to least inclusive (containing the smallest number of species), the categories would be listed: kingdom, phylum, class, family, genus.
Bloom's Level: 2. Understand Learning Outcome: 01.02.01 Describe how living things are classified. Section: 01.02 Topic: Levels of Biological Organization
28. Which of these classification categories contains the closest related group of organisms? A. domain B. genus C. family D. phylum E. kingdom Members of closely related species are grouped into the same genus. The categories listed, from the greatest to least degree of relatedness of organisms, are: genus, family, phylum, kingdom, and domain.
Bloom's Level: 2. Understand Learning Outcome: 01.02.01 Describe how living things are classified. Section: 01.02 Topic: Levels of Biological Organization
29. The common name for "cat" in Spanish is "gato" and in Chinese is "mao." Which of the following statements pertaining to the use of scientific names instead of common names is false? A. Common names may include different sets of organisms. B. Scientific names can be known and recognized by all scientists throughout the world. C. To recognize an organism in the literature by its common name, a scientist would have to know multiple languages. D. The scientific name begins to tie the organism into related groups. E. The common name clearly identifies an organism as unique. The only false statement about the use of scientific names is that common names do not identify organisms as unique. In fact, different organisms are often called by the same common name.
Bloom's Level: 2. Understand Learning Outcome: 01.02.01 Describe how living things are classified. Section: 01.02 Topic: Levels of Biological Organization
30. All the banded sunfish () in a pond would comprise a(n) A. population. B. ecosystem. C. community. D. biosphere. E. species. A species is a category of organisms within the same genus that share very similar characteristics. Banded sunfish ( ) are a species. A population is all the members of a given species within a particular area, as in the example of all the banded sunfish in one pond. All the different populations in the same area make up a community. A community together with its physical environment makes up an ecosystem. The biosphere is the zone of air, land, and water on the earth where living organisms are found.
Bloom's Level: 2. Understand Learning Outcome: 01.01.02 Distinguish between the levels of biological organization. Section: 01.01 Topic: Levels of Biological Organization
31. When comparing energy and chemicals in an ecosystem, A. both chemicals and energy cycle over and over again. B. chemicals cycle over and over again but energy flows through once. C. both chemicals and energy flow through once. D. energy cycles over and over again but chemicals flow through once. E. neither cycle nor flow but both are simply present. Chemicals can be recycled in an ecosystem as they pass from one organism to the next. The same is not true for energy, because every time an energy conversion occurs some of the energy is "lost" in the form of heat.
Bloom's Level: 2. Understand Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Energy and Chemical Cycling
32. Which statement is false regarding science? A. Science helps us to understand the natural world. B. Science strives to be objective rather than subjective. C. Correct scientific conclusions are permanent and never subject to change or refinement. D. Information is gathered by scientific methods. E. Information is gained by observing and testing. Scientific conclusions are always subject to change and refinement. All of the other statements are accurate.
Bloom's Level: 2. Understand Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions. Section: 01.03 Topic: Experimental Design
33. Which statement regarding the scientific method is false? A. Inductive reasoning is used to form a hypothesis. B. Observations are used to form a hypothesis. C. Experiments need to be repeatable. D. Original hypotheses are formed after an experiment. E. The control and experimental group are identical except for one variable. Original hypotheses are formed prior to an experiment, through inductive reasoning based on observations. Inductive reasoning is the process of taking individual observations and weaving them into a cohesive explanation. In a controlled experiment, the control and experimental groups should be the same except for a single factor, called a variable. Experiments should be repeatable, even by different individuals in different places.
Bloom's Level: 2. Understand Learning Outcome: 01.03.01 Identify components of the scientific method. Section: 01.03 Topic: Scientific Method
34. Which statement regarding the scientific method is false? A. Experimental results are reported in scientific journals. B. Hypotheses are finally proven as absolutely true or false. C. The control is not subjected to the experimental variable. D. Statistical tests may be used to analyze experimental results. E. Experimental results are often summarized in a graph or table. Hypotheses are always subject to repeated testing because they are not absolutely provable. This is because, in experiments to test a hypothesis, there is always the possibility of inaccuracy. All the other answer choices are true statements.
Bloom's Level: 2. Understand Learning Outcome: 01.03.01 Identify components of the scientific method. Section: 01.03 Topic: Scientific Method
35. Which statement regarding the scientific method is false? A. A theory that has been repeatedly supported is sometimes called a law. B. Consistent observations about a hypothesis may lead to a theory. C. Theories formed in everyday life carry the same validity as a scientific theory. D. Only one experimental variable is used in a controlled experiment. E. Published research should provide enough detail to allow anyone to repeat the experiment in the same manner. In everyday life, the word theory is used as a synonym for speculation. But in science, a theory grows from a hypothesis that has withstood repeated testing and consistently explains many observations. Hypotheses can be testing in various ways, but one of the most common is controlled experiments, in which only one variable is studied. Scientists know about one another's studies through publication; this enables them to test hypotheses by repeating experiments.
Bloom's Level: 2. Understand Learning Outcome: 01.03.01 Identify components of the scientific method. Section: 01.03 Topic: Scientific Method
36. When researchers test a new human cancer drug using mice, the mice constitute the A. hypothesis. B. data. C. experimental design. D. model. E. control. An experimental model is a representation of an actual subject; in this case, the mice represent humans. The manner in which a scientist intends to conduct an experiment is called the experimental design. A hypothesis is a tentative explanation for a natural event. The results of an experiment are referred to as the data. A control is a subject or group of subjects that goes through all the steps of an experiment but lacks the factor, or is not exposed to the factor, being tested.
Bloom's Level: 2. Understand Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions. Section: 01.03 Topic: Experimental Design
37. Which of the following domains contains organisms that are adapted to life in harsh environments? A. domain Archaea B. domain Bacteria C. domain Eukarya D. domain Animalia E. domain Plantae The domain Archaea contains organisms that can survive in harsh environments. Domain Bacteria contains organisms that can be found everywhere but tend not to be in harsh environments. Domain Eukarya contains the plants, animals, fungi, and protists which are not adapted to harsh environments either. Animalia and Plantae are not domains.
Bloom's Level: 2. Understand Learning Outcome: 01.02.02 Distinguish between the three domains of life. Section: 01.02 Topic: Levels of Biological Organization
38. Which domains contain organisms that lack a membrane-bound nucleus? A. Archaea and Bacteria B. Archaea and Eukarya C. Bacteria and Eukarya D. Eukarya and Animalia E. Archaea and Animalia The domains Archaea and Bacteria both contain organisms that lack a membrane-bound nucleus. Domain Eukarya contains organisms that have a membrane-bound nucleus. Animalia is not a domain.
Bloom's Level: 2. Understand Learning Outcome: 01.02.02 Distinguish between the three domains of life. Section: 01.02 Topic: Levels of Biological Organization
True / False Questions 39. If the human population faces potential extinction, cloning of humans is the only technology that will save us. FALSE False, cloning is not a technology that will save the human population from extinction. Technologies in medicine, agriculture, etc. would play vital roles in the survival of the human species.
Bloom's Level: 2. Understand Learning Outcome: 01.04.02 Summarize some of the major challenges facing science. Section: 01.04 Topic: Experimental Design
Multiple Choice Questions 40. A piece of petrified wood was once part of a living organism that has had its tissues replaced by minerals. Which property of life will still be present in the wood? A. organization B. homeostasis C. growth and reproduction D. response to stimuli E. metabolism A piece of petrified wood is a fossil, a preserved remnant of a once-living organism. Although it lacks metabolism, can no longer maintain homeostasis, and does not grow, reproduce, or respond to stimuli, the original pattern of organization is still evident.
Bloom's Level: 3. Apply Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Characteristics of Life
41. An unmanned spacecraft has been sent to another planet to detect other life forms that might be quite different from those on Earth. If the probe could only send back one still picture, which property or properties of life would be observable in a picture? A. organization B. homeostasis and metabolism C. growth and reproduction D. response to stimuli E. evolution A still image would not be likely to show dynamic processes such as homeostasis, metabolism, evolution, growth, or reproduction. However, organization would be evident.
Bloom's Level: 3. Apply Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Characteristics of Life
42. Ever since the antibiotic drug penicillin was discovered in 1928, the incidence of resistant bacteria has steadily increased as a direct result of A. biodiversity. B. natural selection. C. homeostasis. D. development. E. reproduction. In natural selection, members of a species that are best adapted to survive and reproduce in their environment have more offspring than those that are not. Thus, more members of the population will come to have the successful adaptation. In this example, when people started using penicillin to fight bacterial infections, only susceptible bacteria were killed; resistant ones were able to survive and reproduce. Now there are many more resistant bacteria. Biodiversity is the total number of species, the variability of their genes, and the ecosystems they inhabit. Although genetic variability is the raw material for natural selection, it is not equivalent. Homeostasis is an organism's ability to maintain a stable internal environment. Development is the series of changes that occur throughout the lifespan of an organism. Reproduction is the ability of an organism to make more like itself; it is related to natural selection in that more successful organisms can pass their adaptations on to their offspring, but the terms are not equivalent.
Bloom's Level: 3. Apply Learning Outcome: 01.01.03 Recognize the importance of adaptation and evolution of life. Section: 01.01 Topic: Characteristics of Life
43. Nutrient molecules are used to build cellular structures or for energy. This best represents which characteristic of life? A. Living things acquire materials and energy from the environment. B. Living things are homeostatic. C. Living things are adapted. D. Living things grow and develop. E. Living things respond to stimuli. The ability to acquire materials and energy from the environment is crucial to all the other activities of living organisms, including homeostasis, adaptation, growth and development, and responses to stimuli.
Bloom's Level: 3. Apply Learning Outcome: 01.01.01 Identify the basic characteristics of life. Section: 01.01 Topic: Characteristics of Life
44. You have discovered a previously unknown organism. It is multicellular, with a filamentous form, and absorbs food from its environment. Upon microscopic examination, you see that the cells have nuclei. How would you classify this organism? A. domain Bacteria B. kingdom Animalia C. kingdom Protista D. kingdom Fungi E. kingdom Plantae Kingdom Fungi is comprised of unicellular and multicellular organisms, all heterotrophic by absorption. The multicellular forms are generally filamentous. Members of kingdom Plantae are multicellular, with specialized cells and tissues, and can form their own "food" by photosynthesis. Organisms in kingdom Animalia are also multicellular with specialized cells and tissues, but they are heterotrophs that ingest their food. Kingdom Protista contains mostly unicellular organisms with a few multicellular forms, some of which are heterotrophic by ingestion while others are photosynthetic. Domain Bacteria contains only unicellular, prokaryotic organisms. All four of the kingdoms listed as answer choices belong to domain Eukarya.
Bloom's Level: 3. Apply Learning Outcome: 01.02.01 Describe how living things are classified. Section: 01.02 Topic: Levels of Biological Organization
45. Which of these is an example of inductive reasoning used to form a hypothesis? A. Every fungus that has ever been studied absorbs its food; therefore, food absorption is characteristic of fungi. B. All fungi absorb their food; if mushrooms are fungi; then the mushroom absorbs its food. C. A mushroom is classified in the kingdom Fungi of the domain Eukarya. D. This cell from a mushroom has a nucleus. E. Fungi are not capable of photosynthesis. In order to formulate a hypothesis, a scientist could use inductive reasoning, in which a collection of facts or observations are woven together to support a statement. Inductive reasoning is exemplified by, "Every fungus that has ever been studied absorbs its food; therefore, food absorption is characteristic of fungi." To test a hypothesis, a scientist could use deductive reasoning, which is based on "if, then" statements such as, "If a mushroom is a fungus, then the mushroom absorbs its food." The other answer choices are accurate observations or statements, but not representative of either type of reasoning.
Bloom's Level: 3. Apply Learning Outcome: 01.03.02 Distinguish between a theory and a hypothesis. Section: 01.03 Topic: Scientific Method
46. You are conducting an experiment to determine which brand of fertilizer results in the greatest amount of fruit production by tomato plants. In this example, the response variable would be the A. brand of fertilizer. B. unfertilized tomato plants. C. fertilized tomato plants. D. amount of fruit produced by the tomato plants. E. variety of tomato plants. The response (dependent) variable would be the amount of fruit produced, because this is the response to the different fertilizer varieties used. The fertilizer varieties would constitute the experimental (independent) variable, because this is the factor that the researcher is manipulating. It would be a good experimental design to have two groups of plants: a test group (fertilized plants) and a control group (unfertilized plants). To minimize the number of variables in the experiment, only one variety of plant should be used.
Bloom's Level: 3. Apply Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions. Section: 01.03 Topic: Experimental Design
47. Which of the following might plausibly be said by a scientist? A. Although no one else gets the same results as I do, since I get the results in my experiments, it is still science. B. Since all the accumulated body of data support it, a hypothesis is definitely "true." C. Test methods are unimportant since anyone can run another test; it is the results that make the substance of science. D. The test's results and conclusion finish the scientist's job; it is enough that a scientist alone now knows the results. E. There is always a possibility that a more advanced experiment might falsify any hypothesis. Hypotheses are always subject to additional testing; they are only conditionally accepted or rejected. If two individuals performing the same experiment obtain different results, this would fail to support the hypothesis. Test methods are important because they may influence the results obtained. Scientific endeavors do not end with obtaining results and a conclusion; researchers share their results via publications and meetings.
Bloom's Level: 3. Apply Learning Outcome: 01.03.02 Distinguish between a theory and a hypothesis. Section: 01.03 Topic: Experimental Design
48. Which of the following is a potential consequence if people stop using technology in the agriculture industry? A. There could be a significant decrease in the amount of food produced. B. There would be a collapse of many of the terrestrial ecosystems on Earth. C. The extinction of the top level species that reside in the communities around the agricultural land. D. The amount of food produced would increase. E. None of these consequences will occur if people stop using technology in the agriculture industry. If the agriculture industry stops using technology the most likely consequence will be a significant decrease in the amount of food produced. Ecosystems will flourish with a decrease in technology which will then make it easier for the top level species to survive. Unfortunately, the amount of food produced would decrease.
Bloom's Level: 3. Apply Learning Outcome: 01.04.02 Summarize some of the major challenges facing science. Section: 01.04 Topic: Scientific Method
49. Which of the following is of least importance when concerning the preservation of ecosystems? A. Ecosystem preservation maintains the status of species in their current state. B. Many different organisms perform services that ensure our continued existence. C. Humans derive medicines, food, and many important materials from functioning ecosystems. D. Preservation of ecosystems allows humans to exploit as many resources as possible. E. It is estimated that possibly 400 species become extinct each day due to activities of humans. When humans exploit as many resources as possible, ecosystems are destroyed, not preserved. All the other answer choices are very important statements about why ecosystems should be protected.
Bloom's Level: 4. Analyze Learning Outcome: 01.04.02 Summarize some of the major challenges facing science. Section: 01.04 Topic: Ecosystem Ecology
50. Which of the following represents a potential threat to biodiversity? A. People construct artificial reefs to support marine life. B. Humans clear land for agriculture and housing. C. Energy flows through an ecosystem, with much lost as heat. D. Tropical rain forests and coral reefs are found where solar energy is the most abundant. E. In a food chain, one population feeds on another. When humans clear land, biodiversity is lost because we replace the former inhabitants with a much smaller number of species. The construction of artificial reefs can actually help maintain or restore biodiversity as species that have been forced to move from destroyed coral reefs find new homes. The feeding order of a food chain is an integral part of a community and does not generally have a negative impact on biodiversity. Tropical rain forests and coral reefs are the most diverse ecosystems on our planet. Energy flow is a normal aspect of ecosystems, and does not adversely affect biodiversity.
Bloom's Level: 4. Analyze Learning Outcome: 01.04.02 Summarize some of the major challenges facing science. Section: 01.04 Topic: Biodiversity
51. Which statement regarding science technology is false? A. Science technology has brought about life-improving discoveries, such as antibiotics. B. Science technology helps us to understand the causes of cancer. C. Science technology is a basis for all ethical or moral decisions. D. Science technology may ease the feeding of the world population by producing new plant strains. E. Scientific experimentation may make use of a model instead of an actual subject. Science technology is morally and ethically neutral, and can be used for benefit (as in understanding cancer, producing better food sources, and developing antibiotic drugs), or for harm (such as damage to the environment). Thus, science technology cannot serve as a basis for all ethical and moral decisions. Such decisions must be made by all people, not just scientists. Scientific experimentation may be guided by morals and ethics, as in using a cell culture or animal as a model for a new drug or procedure before testing it on humans.
Bloom's Level: 4. Analyze Learning Outcome: 01.04.01 Distinguish between science and technology. Section: 01.04 Topic: Scientific Method
52. If a new anti-cancer drug is found to be effective in initial tests with mice, what might researchers conclude? A. If the drug was effective in a large number of mice, it will therefore be effective in humans. B. If the drug was effective in a small proportion of mice, it will be effective in a small proportion of humans. C. The mice have provided a positive control in this experiment that proves the drug is effective in humans. D. The drug is effective in the mouse model; it must still be tested in humans. E. The effect of the drug on mice has no bearing on the effect of the drug on humans. Results obtained using models are considered a hypothesis until the experiment can be performed using the actual subject. In this case, the mice are the experimental model, and humans would be the actual subjects. The mice cannot be described as a control; a control would be a group of mice that go through all the steps of the experiment but are not exposed to the drug being tested.
Bloom's Level: 4. Analyze Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions. Section: 01.03 Topic: Experimental Design
53. Some students consume large amounts of coffee and so-called energy drinks to help them stay alert when studying. You notice that many who engage in this practice seem to do poorly on exams. Suppose you want to investigate the relationship between caffeine consumption and exam performance. Which of the following statements would be an appropriate hypothesis? A. Students who consume large amounts of caffeine while studying will have lower exam scores than those who consume less caffeine. B. One should avoid consuming too much caffeine while studying. C. Too much caffeine is harmful to your health. D. Many students consume large amounts of caffeine while studying. E. Caffeine increases alertness but also increases anxiety. A hypothesis is a tentative explanation for a phenomenon that can be tested and then supported or falsified based upon the results. To test the hypothesis, you would compare the exam scores of students who drink large amounts of caffeine to those who do not. If the students who consume more caffeine have significantly lower scores, then your hypothesis is supported. Otherwise, the hypothesis is falsified. The statement that one should avoid consuming too much caffeine while studying is a recommendation, not a hypothesis. The statements that caffeine increases alertness and many students consume large amounts of caffeine are observations, which may be used in formulating a hypothesis. Too much caffeine is harmful to your health is a conclusion that has been reached by some scientific studies; it is not, however, a hypothesis that is appropriate to this experimental design.
Bloom's Level: 4. Analyze Learning Outcome: 01.03.02 Distinguish between a theory and a hypothesis. Section: 01.03 Topic: Experimental Design
54. You are conducting an experiment to determine what concentration of disinfectant is most effective in killing bacteria. In this example, the concentration of disinfectant would represent the A. control. B. experimental variable. C. response variable. D. data. E. hypothesis. The experimental (independent) variable is the factor being tested—in this case, the concentration of disinfectant. The response (dependent) variable is the result or change seen in response to the experimental variable—in this case, the quantity of bacteria killed. A control is not subjected to the experimental variable; an example would be an object or surface contaminated with bacteria that is not treated with disinfectant. Data are experimental results. A hypothesis is a tentative, testable explanation for a natural phenomenon.
Bloom's Level: 4. Analyze Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions. Section: 01.03 Topic: Experimental Design
55. In the late 1800s, Louis Pasteur was searching for a vaccine for anthrax in livestock. One French veterinarian had a local reputation for being able to cure anthrax by applying oils and wrapping the animal in cloth to induce a fever. Pasteur also knew that some animals got well on their own when left untreated. Pasteur tested the effectiveness of the local veterinarian's methods by injecting four cattle with anthrax bacteria. He then directed the veterinarian to perform his procedures on two cattle. The other two cattle were left alone. What is the rationale for Pasteur's experimental design? A. Two cattle represent a test of the veterinarian's hypothesis; two represent a test of Pasteur's hypothesis. B. Two cattle represent a test of the treatment; two serve as a control to determine the likelihood of survival without treatment. C. Two cattle represent a test based on inductive reasoning; two serve to test deductive reasoning. D. The two cattle that are treated are the only test being conducted; the other two cattle serve no purpose beyond representing all the untreated cattle in France. E. The two cattle not being treated were just a whim on Pasteur's part. The two cattle left alone constitute a control group, since they were injected with anthrax but not subjected to the veterinarian's treatment; they are not intended to represent all the cattle in France. The other two cattle represent the test group, since they received both the anthrax injection and the veterinarian's treatment. The entire experimental design is based on deductive reasoning: If the veterinarian's treatment is more effective than lack of treatment, then the treated cattle should recover and the untreated ones should not.
Bloom's Level: 4. Analyze Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions. Section: 01.03 Topic: Experimental Design
56. As the human population size increases, A. ecosystems remain unaffected. B. fewer fossil fuels are burned and carbon dioxide levels remain constant. C. it becomes evident that preserving the biosphere has no benefit to humans. D. fewer ecosystems are destroyed, resulting in an abundance of biodiversity. E. biodiversity is adversely affected as humans have destructive effects on ecosystems. Human activities continue to disrupt and destroy ecosystems, reducing biodiversity. The more our population increases, the more fossil fuels are burned, causing a rise in carbon dioxide levels which can lead to a decrease in biodiversity.
Bloom's Level: 5. Evaluate Learning Outcome: 01.04.02 Summarize some of the major challenges facing science. Section: 01.04 Topic: Ecosystem Ecology
57. Which statement about an ecosystem is incorrect? A. Chemicals are recycled in an ecosystem. B. Energy flows through an ecosystem, with some being lost as heat. C. Some organisms produce food used by themselves and others. D. There is a community of organisms, but no non-living components. E. Climate is the main factor that determines where an ecosystem will be located. An ecosystem is comprised of the living things that inhabit it as well as the non-living physical components, such as soil, water, and air. All the other statements are accurate.
Bloom's Level: 5. Evaluate Learning Outcome: 01.04.02 Summarize some of the major challenges facing science. Section: 01.04 Topic: Ecosystem Ecology
58. Which of the following statements is based on scientific observation and reasoning? A. Some chemical compounds react differently in different human cultures. B. If a tree falls in a forest but this is never witnessed by a human, there are no physical or biological consequences. C. Many physical and biological phenomena lack any cause-and-effect; outcomes from one situation may be one way today and another tomorrow. D. Objects fall downward anywhere on the earth's surface, regardless of the observer. E. A scientist records only the data that are easiest to interpret.. Science seeks to formulate laws that objectively explain natural phenomena, as when objects fall due to gravity. Natural phenomena operate on a cause-and-effect basis and have consistency of outcome that is not influenced by human cultures or the presence of human observers. Scientists must be conscientious in recording all data, not just those which are easiest to interpret.
Bloom's Level: 5. Evaluate Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions. Section: 01.03 Topic: Scientific Method
59. In the late 1800s, Louis Pasteur was searching for a vaccine for anthrax in livestock. One French veterinarian had a local reputation for being able to cure anthrax by applying oils and wrapping the animal in cloth to induce a fever. Pasteur also knew that some animals got well on their own when left untreated. Pasteur tested the effectiveness of the local veterinarian's methods by injecting four cattle with anthrax bacteria. He then directed the veterinarian to perform his procedures on two cattle. The other two cattle were left alone. One cow from each group died and the other one in each group survived. Pasteur dismissed the veterinarian's procedure as ineffective, but he also considered this to be a poor test. Why would he consider this to be a poor test? A. The difference in survival between the two groups was not dependent on the treatment. B. In a small group with 50% survival, the treated cows would have survived anyway. C. The number of cows that survived was the same as the number of cows that died. D. With such a small group you can't determine if survival was dependent on the treatment or random chance. E. The production of a fever in the treated cattle interfered with the outcome. This was a poor test because of the small sample size. The larger the sample size (in this case, the number of cattle studied), the less likely that the result is due to chance alone. There was no difference in survival between the two groups—one of each survived and one died— but due to the small number of subjects it was not possible to determine if the veterinarian's treatment (inducing a fever) had any real effect.
Bloom's Level: 5. Evaluate Learning Outcome: 01.03.03 Analyze a scientific experiment and identify the hypothesis, experiment, control groups, and conclusions. Section: 01.03 Topic: Experimental Design
Short Answer Questions 60. Give examples of the costs and benefits of using technology. Answers will vary. Some of the potential costs will include: an increase in pollution, the accidental destruction of ecosystems or species that were not supposed to be destroyed. Some potential benefits include: an increase in the amount of food that can be produced, the ability to eliminate a pest species.
Bloom's Level: 6. Create Learning Outcome: 01.04.01 Distinguish between science and technology. Section: 01.04 Topic: Scientific Method
Chapter 02 The Molecules of Cells
Multiple Choice Questions 1. The two isotopes shown below differ in the number of
A. electrons. B. protons. C. neutrons. D. carbon atoms. E. ionic bonds. Carbon 12 and carbon 14 are different isotopes of the element carbon; this means they have the same number of protons and electrons, but they differ in the number of neutrons. The number of bonds an isotope can form is not determined by its neutrons.
Bloom's Level: 1. Remember Learning Outcome: 02.01.03 Describe how variations in an atomic nucleus account for its physical properties. Section: 02.01 Topic: Atomic Structure
2. If an element has an atomic number of 12, how many electrons are in its outermost shell? A. 1 B. 10 C. 8 D. 2 E. 12 Two electrons fill the innermost shell and eight fill the next, leaving two for the outermost shell.
Bloom's Level: 3. Apply Learning Outcome: 02.01.01 Describe how protons, neutrons, and electrons relate to atomic structure. Section: 02.01 Topic: Atomic Structure
3. If an element has an atomic number of 15, then A. the atomic mass must also be 15. B. the atom has 15 electrons. C. there are 7 electrons in the outermost shell. D. the atom has 15 neutrons. E. the atom must have only one orbital. In an electrically neutral atom, the number of protons (the atomic number) is the same as the number of electrons. The atom would need three orbitals to accommodate 15 electrons, and there would be 5 electrons in its outermost shell. The atomic mass includes the protons and the neutrons; the number of neutrons is not specified in this question.
Bloom's Level: 3. Apply Learning Outcome: 02.01.02 Understand how to interpret the periodic table of elements. Section: 02.01 Topic: Atomic Structure
4. The chemical reactivity of an element is dependent on A. the number of protons. B. the arrangement of neutrons. C. the number of electrons in the outermost shell. D. the number of protons and neutrons. E. the number of electrons in the inner shell. It is the outermost shell of an atom that can potentially react with electrons in the outermost shells of other atoms. The protons and neutrons remain in the nucleus and do not engage in chemical reactions.
Bloom's Level: 1. Remember Learning Outcome: 02.01.03 Describe how variations in an atomic nucleus account for its physical properties. Section: 02.01 Topic: Chemical Reactions
5. The atomic mass of an element A. is determined by the number of protons and neutrons it contains. B. equals the number of protons plus the number of electrons. C. equals the number of neutrons. D. changes after each reaction. E. depends on the number of electrons in the outermost shell. The atomic mass is essentially the sum of protons and neutrons in the nucleus; it is not changed by chemical reactions. The mass of electrons is negligible.
Bloom's Level: 1. Remember Learning Outcome: 02.01.02 Understand how to interpret the periodic table of elements. Section: 02.01
6. The nucleus of an atom contains A. neutrons and electrons. B. electrons only. C. protons, neutrons, and electrons. D. protons and neutrons. E. neutrons only. The nucleus contains protons and neutrons; electrons orbit the nucleus.
Bloom's Level: 1. Remember Learning Outcome: 02.01.01 Describe how protons, neutrons, and electrons relate to atomic structure. Section: 02.01 Topic: Atomic Structure
7. Isotopes of a given element have A. the same number of protons but differ in atomic mass. B. the same atomic mass but a different number of protons. C. a different number of electrons. D. the same number of protons and atomic mass. E. the same number of neutrons. Isotopes are atoms of the same element that differ in the number of neutrons only; thus, they have the same atomic number but different atomic masses.
Bloom's Level: 2. Understand Learning Outcome: 02.01.02 Understand how to interpret the periodic table of elements. Section: 02.01 Topic: Atomic Structure
8. The isotope shown below has A. 14 electrons. B. 6 neutrons. C. 8 protons. D. 8 neutrons. E. 8 electrons. Carbon 14 has eight neutrons, 6 protons (and thus 6 electrons), and an atomic mass of 14.
Bloom's Level: 2. Understand Learning Outcome: 02.01.02 Understand how to interpret the periodic table of elements. Section: 02.01 Topic: Atomic Structure
9. To measure the activity of the human brain during certain thought processes, a short-lived radioactive sugar is injected in the carotid artery and is utilized by those cells that are most active. This shows up on a PET scan and demonstrates the detection of A. ionic bonds. B. high levels of radiation. C. covalent bonds. D. neutrons. E. isotopes. The isotopes used in PET scans constitute a low level of radiation used for beneficial purposes. Types of chemical bonds are not registered by this procedure. Although isotopes of a given element differ in their numbers of neutrons, neutrons are not detected by PET.
Bloom's Level: 1. Remember Learning Outcome: 02.01.04 Identify the beneficial and harmful uses of radiation. Section: 02.01 Topic: Atomic Structure
10. Which of the following molecules is NOT a compound? A. H2O B. HCl C. H2 D. C6H12O6 E. NaOH A compound consists of atoms of two or more different elements bound together; H2 is two molecules of the same element (hydrogen) bound together.
Bloom's Level: 2. Understand Learning Outcome: 02.02.01 Describe how elements are combined into molecules and compounds. Section: 02.02 Topic: Chemical Reactions
11. Which of the following statements is NOT true of chemical bonds? A. Both ionic and covalent bonds involve electrons in the outer shell. B. Covalent bonds share electrons between two atoms. C. An atom involved in an ionic bond has an unequal number of electrons and protons. D. Salts are covalently bonded. E. The atoms in a molecule of water (H2O) are covalently bound together. Salts, such as sodium chloride (NaCl), are ionic compounds. All the other answer choices are accurate statements.
Bloom's Level: 1. Remember Learning Outcome: 02.02.02 List the different types of bonds that occur between elements. Section: 02.02 Topic: Chemical Bonds
12. An ion is an atom that A. exists in a gaseous state. B. has a net charge. C. does not have a net charge. D. shares electrons with other atoms. E. shares neutrons with other atoms. An ion has more or less electrons than a neutral atom of the same element, so it has a net positive or negative charge.
Bloom's Level: 1. Remember Learning Outcome: 02.02.02 List the different types of bonds that occur between elements. Section: 02.02 Topic: Chemical Bonds
13. If neutral atoms become positive ions, they A. gain electrons. B. lose electrons. C. gain protons. D. lose protons. E. do not change. The electron transfer is what will determine if an atom becomes a positive or negative ion. To become a positive ion an atom will need to lose electrons so there are more protons than electrons.
Bloom's Level: 2. Understand Learning Outcome: 02.02.01 Describe how elements are combined into molecules and compounds. Section: 02.02 Topic: Chemical Reactions
14. When an ionic bond forms, electrons are A. lost from both atoms. B. gained by both atoms. C. shared equally by both atoms. D. totally lost from the paired atoms. E. transferred from one atom to another. Ionic compounds form when one atom gives up an electron, which is accepted by the other member of the pair. Now that one atom is a positively charged ion and the other is a negatively charged ion, the two atoms are attracted to one another.
Bloom's Level: 1. Remember Learning Outcome: 02.02.02 List the different types of bonds that occur between elements. Section: 02.02 Topic: Chemical Bonds
15. Calcium chloride, CaCl2, is an ionic compound in which A. one chlorine atom transferred an electron to the other chlorine atom. B. each chlorine atom has lost electrons. C. calcium has two extra electrons in its innermost shell. D. calcium has gained two electrons. E. calcium has lost two electrons. In the ionic compound CaCl2, calcium has transferred two electrons from its outermost shell, becoming a calcium ion (Ca2+). One of the electrons has been accepted by each chlorine atom, so they become chloride ions (Cl-).
Bloom's Level: 1. Remember Learning Outcome: 02.02.02 List the different types of bonds that occur between elements. Section: 02.02 Topic: Chemical Bonds
16. A covalent bond is A. a type of bond that results in ionic compounds. B. the transfer of electrons from one atom to another. C. a sharing of electrons between two atoms. D. an attraction of charged atoms. E. a sharing of protons between two atoms. A covalent bond results when two atoms share electrons in such a way that each atom has eight electrons in its outermost shell. In contrast, ionic compounds result from the complete transfer of electrons between bonded atoms.
Bloom's Level: 1. Remember Learning Outcome: 02.02.02 List the different types of bonds that occur between elements. Section: 02.02 Topic: Chemical Bonds
17. Potassium, a metal with one electron in the outermost shell, will react with how many chlorine atoms? (Chlorine is a nonmetal with seven electrons in the outermost shell.) A. 5 B. 7 C. 1 D. 2 E. 3 Potassium will attain stability by transferring the one electron in its outermost shell to chlorine, which needs only one more electron in its outermost shell to become stable. The result will be a potassium ion (K+) and a chloride ion (Cl-). The two oppositely charged ions will be attracted to one another, thus forming an ionic compound.
Bloom's Level: 2. Understand Learning Outcome: 02.02.01 Describe how elements are combined into molecules and compounds. Section: 02.02 Topic: Chemical Bonds
18. Polar covalent bonds result from A. unequal sharing of electrons in a covalent bond. B. equal sharing of electrons in a covalent bond. C. equal sharing of electrons in an ionic bond. D. unequal sharing of electrons in an ionic bond. E. hydrogen bonding between molecules. Covalent bonds result from the sharing of electrons between bound atoms; when the sharing is unequal, it is a polar bond, and when the sharing is equal, it is a nonpolar bond. Ionic bonds are a different type of chemical bond from covalent bonds; in an ionic bond, one or more electrons are completely transferred from one member of the compound to the other(s). Hydrogen bonding is a relatively weak attraction between hydrogen in one molecule and a highly electronegative atom (such as O or N) in an adjacent molecule.
Bloom's Level: 1. Remember Learning Outcome: 02.02.02 List the different types of bonds that occur between elements. Section: 02.02 Topic: Chemical Bonds
19. Which of the following statements about hydrogen bonding is incorrect? A. Hydrogen bonding occurs only between water molecules. B. Hydrogen bonds are easily broken. C. Hydrogen bonding can occur between different molecules or within the same molecule. D. Most hydrogen bonds involve hydrogen and oxygen or nitrogen. E. The structure of a large, complex molecule can be influenced by hydrogen bonding. Hydrogen bonding is not limited to bonding between adjacent water molecules. For example, hydrogen bonds can form between hydrogen in water and nitrogen in ammonia, or between the two strands of a DNA molecule. Hydrogen bonds are important in determining the shape of large, complex molecules such as proteins. Even so, hydrogen bonds are relatively weak when compared to ionic or covalent bonds, and are easily broken.
Bloom's Level: 1. Remember Learning Outcome: 02.02.02 List the different types of bonds that occur between elements. Section: 02.02 Topic: Chemical Bonds
20. Which of the following is not a property of water that results from hydrogen bonding? A. The temperature of water changes very slowly. B. Many polar substances dissolve in water. C. Water molecules have cohesiveness. D. Ice melts at -100C. E. Water has a high surface tension. Due to its hydrogen bonding, water melts at 0C instead of -100C. All the other answer choices are accurate statements.
Bloom's Level: 1. Remember Learning Outcome: 02.03.01 Evaluate which properties of water are important for biological life. Section: 02.03 Topic: Properties of Water
21. Water is a liquid at room temperature. This is due to A. ionic bonding of the atoms in the water molecule. B. covalent bonding in the water molecule. C. covalent bonding between water molecules. D. hydrogen bonding within the water molecule. E. hydrogen bonding between water molecules. Hydrogen bonding between water molecules keeps water in a liquid state at temperatures typically found on the Earth's surface, including room temperature. Water molecules do not covalently bond to one another, and the water molecule is too small to permit intramolecular hydrogen bonds to form. Water molecules do ionize, but this does not influence the fluid nature of water.
Bloom's Level: 3. Apply Learning Outcome: 02.03.01 Evaluate which properties of water are important for biological life. Section: 02.03 Topic: Properties of Water
22. The moon lacks life and varies dramatically in temperature. If we could keep a layer of water spread on the surface of the moon, what effect would it have? A. Life would be possible but it would have to withstand these extremes in temperature. B. Water would absorb and hold heat and moderate the temperature extremes. C. The temperatures would drop to the lower extremes. D. Because water has a high heat of vaporization, the temperatures would rise to the upper extremes. E. Physical conditions would remain the same. Water has a high heat capacity; this serves to moderate temperature changes. Although water does have a high heat of vaporization, this also has a moderating effect, and would prevent temperatures from rising to the highest extremes.
Bloom's Level: 4. Analyze Learning Outcome: 02.03.01 Evaluate which properties of water are important for biological life. Section: 02.03 Topic: Properties of Water
23. In water, a weak hydrogen bond occurs between hydrogen in one molecule and A. an oxygen atom in the same molecule. B. an oxygen atom in a different molecule. C. a hydrogen atom in the same molecule. D. a hydrogen atom in a different molecule. E. either hydrogen and oxygen atoms of different molecules. Because water molecules are polar, and each oxygen has slight negative charge and the hydrogen a slight positive charge, hydrogen bonding occurs between a hydrogen atom of one water molecule and the oxygen of another.
Bloom's Level: 1. Remember Learning Outcome: 02.02.03 Compare the relative strengths of ionic, covalent, and hydrogen bonds. Section: 02.02 Topic: Properties of Water
24. You notice that rain water forms "beads" on your car. This is an example of what property of water? A. cohesion B. dissociation C. high heat of vaporization D. adhesion E. solvent The formation of water beads on the surface of a car is due to the cohesiveness of water molecules for one another, thanks to hydrogen bonding. Adhesion is attraction of water molecules for a surface—a property that is not demonstrated here, since the surface of the car (especially a freshly waxed car) repels the water. The high heat of vaporization and solvent capabilities of water are not apparent in this example. Water does dissociate into ions, but this does not manifest in the formation of beads. Bloom's Level: 1. Remember Learning Outcome: 02.03.01 Evaluate which properties of water are important for biological life. Section: 02.03 Topic: Properties of Water
25. Hydrogen bonding produces which of the following properties of water? A. Water boils at a lower temperature than expected. B. Water is less dense as ice than as liquid water. C. Water absorbs heat with a large change in temperature. D. Water releases heat with a large change in temperature. E. Water can be more dense as ice than as liquid water. Due to the increased stability of hydrogen bonding at lower temperatures, water is less dense as ice than as liquid water. Water can absorb and release heat, but with a relatively small change in temperature. Due to hydrogen bonding, water boils at 100C; without hydrogen bonding, it would boil at -91C.
Bloom's Level: 1. Remember Learning Outcome: 02.03.01 Evaluate which properties of water are important for biological life. Section: 02.03 Topic: Properties of Water
26. The water strider is an insect that skates across the water without sinking. The tips of its feet must be coated with molecules that are A. ions. B. hydrophilic. C. hydrophobic. D. basic. E. acidic. The water strider's feet should be coated with hydrophobic molecules to repel water; a hydrophilic coating would cause the insect to stick to the surface of the water and sink. Ions, acids, and bases are hydrophilic.
Bloom's Level: 3. Apply Learning Outcome: 02.03.01 Evaluate which properties of water are important for biological life. Section: 02.03 Topic: Properties of Water
27. The lower the pH A. the lower the hydrogen ion (H+) concentration. B. the more acidic the solution. C. the higher the pH. D. the greater the hydroxide ion (OH-) concentration. E. the closer the hydroxide ion (OH-) concentration comes to equaling the hydrogen ion (H+) concentration. The pH scale is based on hydrogen ion (H+) concentration. The higher the concentration of H+ (and the lower the concentration of OH-, hydroxide), the lower the pH, and the more acidic the solution.
Bloom's Level: 1. Remember Learning Outcome: 02.03.02 Identify common acidic and basic substances. Section: 02.03 Topic: Acids and Bases
28. Since pure water is neutral in pH, it contains A. no hydrogen ions (H+). B. no hydroxide ions (OH-). C. neither hydrogen ions (H+) nor hydroxide ions (OH-). D. an equal number of hydrogen ions (H+) and hydroxide ions (OH-). E. seven times more hydrogen ions (H+) than hydroxide ions (OH-). Pure water is neutral, with a pH of 7 (midway between 0 and 14 on the pH scale), meaning that it has equal concentrations of hydrogen ions (H+) and hydroxide ions (OH-).
Bloom's Level: 2. Understand Learning Outcome: 02.03.02 Identify common acidic and basic substances. Section: 02.03 Topic: Acids and Bases
29. The pH of blood is slightly basic. Which of the following would therefore be an expected pH for blood? A. 6.4 B. 4.6 C. 4.7 D. 7.4 E. 13.8 The pH scale ranges from 0 to 14, with 7 being neutral. Numbers lower than 7 are acidic, and those higher than 7 are basic. Thus, a pH of 7.4 would be slightly basic.
Bloom's Level: 3. Apply Learning Outcome: 02.03.02 Identify common acidic and basic substances. Section: 02.03 Topic: Acids and Bases
30. Potassium hydroxide (KOH) almost completely dissociates in aqueous solution into K+ and OH-, which means it is a A. strong acid. B. strong base. C. weak base. D. weak acid. E. nonpolar covalent molecule. Because potassium hydroxide dissociates completely and adds hydroxide ions (OH-) to an aqueous solution, it is a strong base. A weak base would not dissociate so completely. An acid would contribute hydrogen ions (H+) to an aqueous solution. Potassium hydroxide is an ionic compound; a nonpolar covalent molecule would not dissociate or dissolve in an aqueous solution.
Bloom's Level: 4. Analyze Learning Outcome: 02.03.02 Identify common acidic and basic substances. Section: 02.03 Topic: Acids and Bases
31. Which statement regarding acids and bases is correct? A. Acids increase the pH, and bases decrease the pH. B. Acids increase the proportion of hydrogen ions (H+), and bases reduce the proportion of H+. C. Acids are harmful, but bases are not harmful. D. Acids combine with bases to form buffers. E. Acids combine with bases to form sugars. Acids raise the hydrogen ion (H+) content of a solution, while bases reduce the proportion of H+. The lower the pH, the more acidic the solution, and the higher the pH, the more basic the solution. Strong acids and bases are both harmful. When acids combine with bases, salts result.
Bloom's Level: 1. Remember Learning Outcome: 02.03.02 Identify common acidic and basic substances. Section: 02.03 Topic: Acids and Bases
32. Buffers A. are strong acids or bases. B. keep the pH within normal limits. C. release large amounts of hydrogen ions (H+). D. will only lower the pH. E. will only increase the pH. Buffers are the chemicals or combinations of chemicals that keep pH within normal limits. Weak acids and bases may be used as buffers, not strong ones, which would greatly influence the H+ concentration of the solution and thereby raise or lower the pH.
Bloom's Level: 1. Remember Learning Outcome: 02.03.03 Describe how buffers are important to living organisms. Section: 02.03 Topic: Acids and Bases
33. Aspirin is acetyl salicylic acid and can therefore pose a problem to people who have ulcers. Bufferin is an alternative to aspirin that uses a buffer to neutralize this effect by A. substituting another ingredient for the acetyl salicylic acid. B. adding a drug to stimulate the immune system. C. adding salts to neutralize the acid. D. adding an equal amount of hydroxide (OH-) ions. E. adding chemicals that take up excess hydrogen (H+) ions. Bufferin contains a buffering system to bind the excess H+ from the aspirin (acetyl salicylic acid). This would not be a salt; salts form when acids and bases react.
Bloom's Level: 3. Apply Learning Outcome: 02.03.03 Describe how buffers are important to living organisms. Section: 02.03 Topic: Chemical Reactions
34. Rain falling in the northeastern U.S. has a pH between 5.0 and 4.0. Normally, rainwater has a pH of about 5.6. Which of the following statements is not correct? A. The pH of the rainwater has changed from neutral to acidic. B. The pH of the rainwater has become more acidic. C. The hydrogen ion (H+) content of the rainwater has increased. D. The proportion of hydroxide ions (OH-) in the rainwater has declined. E. The rainwater with a pH of 4.0 is ten times more acidic than the rainwater with a pH of 5.0. The pH of rainwater is normally acidic (5.6) not neutral (7). All the other answer choices are accurate statements.
Bloom's Level: 2. Understand Learning Outcome: 02.03.02 Identify common acidic and basic substances. Section: 02.03 Topic: Acids and Bases
35. Organic molecules A. always contain carbon. B. always contain hydrogen. C. always contain carbon and hydrogen. D. are found only in organisms, hence their name. E. are always food molecules. Organic molecules, by definition, must contain both carbon and hydrogen. They are found in organisms and in food, but can also be made in a laboratory.
Bloom's Level: 1. Remember Learning Outcome: 02.04.01 Compare inorganic molecules to organic molecules. Section: 02.04 Topic: Chemical Reactions
36. Which of the following molecules is inorganic? A. CH4 B. CO2 C. C6H12O6 D. C12H22O12 E. C6H6 Carbon dioxide (CO2) is inorganic because, although it contains carbon, it does not contain hydrogen. All the other molecules contain both carbon and hydrogen and are therefore organic.
Bloom's Level: 2. Understand Learning Outcome: 02.04.01 Compare inorganic molecules to organic molecules. Section: 02.04 Topic: Chemical Bonds
37. One carbon atom can form covalent bonds with up to molecule. A. 2 B. 3 C. 4 D. 6 E. 8
other atoms to form an organic
Carbon, with an atomic number of 6, has 4 electrons in its outermost shell. Thus, carbon can form 4 single covalent bonds with other atoms.
Bloom's Level: 1. Remember Learning Outcome: 02.04.01 Compare inorganic molecules to organic molecules. Section: 02.04 Topic: Chemical Bonds
38. Two molecules of glucose combine to form a disaccharide molecule during a(n) reaction. A. dehydration B. hydrolysis C. hydrogen bond D. ionic bond E. inert The glucose molecules are monomers; forming a covalent bond between them requires a dehydration reaction. A hydrolysis reaction could be used to break the disaccharide apart into individual glucose monomers. An inert material would not react at all.
Bloom's Level: 1. Remember Learning Outcome: 02.04.03 Recognize how monomers are joined to form polymers. Section: 02.04 Topic: Chemical Reactions
39. DNA codes for the sequence of amino acids in the primary structure of a protein, but not for sugars or lipids. This is because A. only proteins are involved in living metabolic reactions. B. sugars and lipids code for their own replication. C. sugars and lipids are ever present in the living environment and are not used in living structures. D. other hereditary molecules code for sugars and lipids. E. proteins are the main structural and functional components of cells. Proteins are the main structural and functional components of cells. As enzymes, proteins catalyze the synthesis and degradation of other biological molecules, including sugars and lipids.
Bloom's Level: 3. Apply Learning Outcome: 02.07.03 Compare the four levels of protein structure. Section: 02.07 Topic: Proteins
40. A genetic mutation can cause a change in the sequence of the 20 amino acids used to build proteins. Such a change is a change to the protein's A. primary structure only. B. secondary structure only. C. tertiary structure only. D. primary structure, but this will likely alter higher levels of structure as well. E. quaternary structure only. A mutation (a change in a DNA sequence) may directly alter the primary structure of a protein, since this is the sequence of amino acids in the chain. However, the primary level of structure dictates the higher levels of structure—secondary, tertiary, and even quaternary—so these may be indirectly affected as a result of the mutation.
Bloom's Level: 3. Apply Learning Outcome: 02.07.03 Compare the four levels of protein structure. Section: 02.07 Topic: Proteins
41. Glycogen is a A. monosaccharide used for quick energy. B. protein found in cell membranes. C. polysaccharide used to store glucose/energy. D. fat found in margarine. E. nucleic acid found in the nucleus of a cell. Glycogen (a polysaccharide) is the storage form of glucose (a monosaccharide). It is rich in chemical bond energy, found in animal tissues such as liver and skeletal muscle. DNA would be a nucleic acid found in the nucleus of a cell, and trans fats are found in margarine.
Bloom's Level: 1. Remember Learning Outcome: 02.05.02 List several examples of important monosaccharides and polysaccharides. Section: 02.05 Topic: Carbohydrates
42. Maltose is classified as a A. nucleic acid. B. fatty acid. C. protein. D. carbohydrate. E. lipid. Maltose is classified as a carbohydrate due to its carbon and hydrogen backbone.
Bloom's Level: 1. Remember Learning Outcome: 02.05.02 List several examples of important monosaccharides and polysaccharides. Section: 02.05 Topic: Carbohydrates
43. All carbohydrate molecules A. contain amino acids. B. contain nitrogen and phosphate. C. are organic acids. D. are composed of atoms of C, H, and the functional group -OH. E. are composed of atoms of C, H, O, and N. Carbohydrates are organic molecules, thus, they contain both carbon and hydrogen atoms. They are further characterized by the hydroxyl (-OH) functional group. Although some carbohydrates do contain nitrogen, this is not a requirement in order to be classified as a carbohydrate. Amino acids are the monomers of proteins, not carbohydrates. Carbohydrates do not release H+ in aqueous solutions, so they are not organic acids.
Bloom's Level: 2. Understand Learning Outcome: 02.05.01 Identify the structural components of a carbohydrate. Section: 02.05 Topic: Carbohydrates
44. When two glucose molecules combine they form a disaccharide molecule and A. another glucose molecule. B. another disaccharide molecule. C. a dipeptide molecule. D. a lipid molecule. E. a water molecule. Glucose is a monosaccharide (a carbohydrate monomer); two glucose molecules combine to form a disaccharide, not a dipeptide or a lipid. In the process, a water molecule is released, which is why this is called a dehydration reaction.
Bloom's Level: 1. Remember Learning Outcome: 02.04.03 Recognize how monomers are joined to form polymers. Learning Outcome: 02.05.01 Identify the structural components of a carbohydrate. Section: 02.04 Section: 02.05 Topic: Carbohydrates
45. is a polysaccharide that is found in plant cell walls and accounts for their strength. A. Cellulose B. Chitin C. Glycogen D. Starch E. Cholesterol Cellulose is a structural polysaccharide found in plant cell walls. Starch and glycogen are storage forms of glucose found in plants and animals, respectively. Chitin is a polysaccharide found in the cell walls of fungi and the exoskeletons of arthropods. DNA is a nucleic acid, not a polysaccharide, and is found in cell nuclei.
Bloom's Level: 1. Remember Learning Outcome: 02.05.02 List several examples of important monosaccharides and polysaccharides. Section: 02.05 Topic: Carbohydrates
46. Hydrolysis of a fat results in A. glycerol only. B. fatty acids only. C. glucose only. D. two monosaccharides. E. both glycerol and fatty acids. A triglyceride molecule (commonly known as a fat) is composed of three fatty acids covalently bonded to glycerol. No monosaccharides, including glucose, make up a triglyceride.
Bloom's Level: 2. Understand Learning Outcome: 02.06.01 Compare the structures of fats, phospholipids, and steroids. Section: 02.06 Topic: Lipids
47. A long chain of carbon atoms with hydrogen atoms attached, ending in the acidic group COOH would be a(n) A. triglyceride. B. amino acid. C. fatty acid. D. nucleic acid. E. monosaccharide. A fatty acid is a hydrocarbon chain ending with a -COOH group, which is acidic; a triglyceride is composed of three fatty acids bound to glycerol. Amino acids are the monomers of proteins, and monosaccharides are carbohydrates. Nucleic acids are polymers of nucleotides such as DNA and RNA.
Bloom's Level: 1. Remember Learning Outcome: 02.06.01 Compare the structures of fats, phospholipids, and steroids. Section: 02.06 Topic: Lipids
48. Nucleic acids are polymers of A. amino acids. B. nucleotides. C. glycerol. D. monosaccharides. E. fatty acids. Nucleic acids such as DNA and RNA are polymers of nucleotides. Polysaccharides are polymers of monosaccharides, triglycerides are made up of fatty acids bound to glycerol, and proteins are composed of amino acids.
Bloom's Level: 1. Remember Learning Outcome: 02.08.01 Compare the structure and function of DNA and RNA. Section: 02.08 Topic: Nucleic Acids
49. The molecular structure shown here is
A. a glucose molecule. B. a fatty acid molecule. C. a glycerol molecule. D. a protein molecule. E. an amino acid. An amino acid has a central carbon bonded to a hydrogen atom, an amino (-NH2) group, a carboxyl (-COOH) group, and an R group where R stands for the remainder of the molecule. Amino acids are the monomers of proteins. Glucose is a monomer for certain polysaccharides, and a triglyceride is composed of fatty acids and a glycerol molecule.
Bloom's Level: 1. Remember Figure: 02.24 Learning Outcome: 02.07.03 Compare the four levels of protein structure. Section: 02.07 Topic: Proteins
50. Enzymes are organic compounds classified as A. nucleic acids. B. carbohydrates. C. lipids. D. steroids. E. proteins. Enzymes are proteins that speed chemical reactions in living things. Their function is very dependent upon their structure. Steroids and other lipids do not function as enzymes, nor do carbohydrates. With a very few exceptions, neither do nucleic acids.
Bloom's Level: 1. Remember Learning Outcome: 02.07.01 Describe the functions of proteins in cells. Section: 02.07 Topic: Proteins
51. The structure of a protein consists of the sequence of the amino acids joined together by peptide bonds. A. primary B. secondary C. tertiary D. quaternary E. molecular Amino acids joined together by peptide bonds constitute the primary level of protein structure. In secondary structure, hydrogen bonding between amino acids causes the polypeptide to form an alpha helix or a beta pleated sheet. In tertiary structure, interactions such as covalent bonds between R groups cause the polypeptide to fold and twist. When two or more polypeptides join together, this represents a quaternary level of structure. There is no level of protein structure termed the molecular level.
Bloom's Level: 1. Remember Learning Outcome: 02.07.03 Compare the four levels of protein structure. Section: 02.07 Topic: Proteins
52. Hemoglobin is a protein composed of two pairs of polypeptide chains. What is the highest level of protein structure represented by hemoglobin? A. primary B. secondary C. tertiary D. quaternary E. molecular Amino acids joined together by peptide bonds constitute the primary level of protein structure. In secondary structure, hydrogen bonding between amino acids causes the polypeptide to form an alpha helix or a beta pleated sheet. In tertiary structure, interactions such as covalent bonds between R groups cause the polypeptide to fold and twist. When two or more polypeptides join together, this represents a quaternary level of structure. There is no level of protein structure termed the molecular level.
Bloom's Level: 2. Understand Learning Outcome: 02.07.03 Compare the four levels of protein structure. Section: 02.07 Topic: Proteins
53. The proposed cause of CJD and kuru in humans, mad cow disease, and scrapie in sheep is a change in a brain protein. Disease victims appear to have a protein that should normally contain alpha helices but instead they have changed into a protein made of beta pleated sheets. The disease appears to spread when the abnormal protein comes into contact with the normal protein, causing it to become deformed. Which level of protein structure is associated with these diseases? A. primary B. secondary C. tertiary D. quaternary E. molecular Amino acids joined together by peptide bonds constitute the primary level of protein structure. In secondary structure, hydrogen bonding between amino acids causes the polypeptide to form an alpha helix or a beta pleated sheet—this is the level affected in CJD and similar brain diseases. In tertiary structure, interactions such as covalent bonds between R groups cause the polypeptide to fold and twist. When two or more polypeptides join together, this represents a quaternary level of structure. There is no level of protein structure termed the "molecular level."
Bloom's Level: 3. Apply Learning Outcome: 02.07.01 Describe the functions of proteins in cells. Learning Outcome: 02.07.03 Compare the four levels of protein structure. Section: 02.07 Topic: Proteins
54. Which of these combinations would be found in a nucleotide? A. base-acid-salt B. adenine-thymine-uracil C. base-sugar-phosphate D. DNA-RNA-nucleus E. sugar-protein-fat A nucleotide is composed of a nitrogenous base, a pentose sugar, and a phosphate group. Adenine, thymine, and uracil are bases. Combining an acid and a base yields a salt. Sugars, proteins, and fats are all biological organic molecules. DNA and RNA are nucleic acids composed of nucleotides. Both are made in the nucleus of a cell.
Bloom's Level: 2. Understand Learning Outcome: 02.08.01 Compare the structure and function of DNA and RNA. Section: 02.08 Topic: Nucleic Acids
55. The backbone of a nucleic acid strand is composed of A. glycerol. B. "R" groups. C. nitrogenous bases. D. alternating pentose sugars and phosphate groups. E. alternating adenines and thymines. The backbone of a nucleic acid such as DNA is composed of alternating pentose sugars and phosphate groups. Glycerol is the backbone of a triglyceride. Although nitrogenous bases such as adenine and thymine are part of DNA, they do not make up the backbone. R groups are part of amino acids, not nucleotides.
Bloom's Level: 1. Remember Learning Outcome: 02.08.01 Compare the structure and function of DNA and RNA. Section: 02.08 Topic: Nucleic Acids
56. In the search to discover the agents that cause mad cow disease, scrapie in sheep, and CJD and kuru in humans, diseased brain tissues were passed through a fine filter to remove bacteria. The filtrate was still infectious, indicating that something smaller than bacteria, either viruses or organic molecules, must be the causative agent. If a virus was responsible for these brain diseases, then the infectious agent would contain either RNA or DNA. Other possibilities were that the agent was a carbohydrate, fat, or protein. Tissue filtrates were treated with agents that destroyed just one of these chemicals and then injected into a healthy animal, with the results as follows. What is the infectious agent? • Amylase digests carbohydrates; tissue filtrate still infects healthy test animal. • Lipase digests fats; tissue filtrate still infects healthy test animal. • Formaldehyde and/or heat denatures DNA and RNA; tissue filtrate still infects healthy test
animal. • Trypsin digests protein; tissue filtrate does not infect healthy test animal. A. carbohydrate B. fat C. protein D. DNA or RNA E. Could be carbohydrate, fat, or DNA or RNA but not protein. CJD, kuru, mad cow disease, and scrapie are caused by prions—infectious proteins. This was supported by the finding that only trypsin, which digests proteins, was able to deactivate the infectious agent.
Bloom's Level: 5. Evaluate Learning Outcome: 02.07.01 Describe the functions of proteins in cells. Section: 02.07 Topic: Proteins
57. The final shape of a protein is very important to its function. When proteins undergo an irreversible change in shape called they perform their usual functions. A. naturation/can B. naturation/cannot C. denaturation/can D. denaturation/cannot E. dehydration reaction/cannot Denaturation is when a protein loses its shape and cannot function. Although dehydration reactions do join amino acids together, this represents only the most basic (primary) level of protein structure.
Bloom's Level: 1. Remember Learning Outcome: 02.07.02 Explain how a polypeptide is constructed from amino acids. Section: 02.07 Topic: Proteins
58. The primary function of carbohydrates is A. quick fuel and short-term energy storage. B. structural reinforcement of plant and fungal cell walls. C. encoding the hereditary information. D. to speed chemical reactions in cells. E. to transport molecules across cell membranes. Carbohydrates are primarily fuel and short-term energy storage molecules, although some polysaccharides do reinforce cell walls in certain organisms. DNA, a nucleic acid, encodes hereditary information. Proteins can function as enzymes to speed chemical reactions, or as transporters to move molecules across cell membranes.
Bloom's Level: 1. Remember Learning Outcome: 02.05.02 List several examples of important monosaccharides and polysaccharides. Section: 02.05 Topic: Carbohydrates
59. Which of the following types of lipids is the most abundant constituent of cell membranes? A. cholesterol B. phospholipid C. triglyceride D. neutral fat E. fat Phospholipids are the most abundant type of lipids in cell membranes. Animal cells also have cholesterol in their membranes, but it is less abundant than phospholipids. Triglycerides, also known as neutral fats or simply fats, are energy-storage molecules, not structural molecules.
Bloom's Level: 1. Remember Learning Outcome: 02.06.02 Identify the functions lipids play in our bodies. Section: 02.06 Topic: Lipids
60. Which type of lipid molecule is characterized by a backbone of four fused rings? A. DNA B. phospholipid C. triglyceride D. steroid E. amino acid Only steroids are characterized by their backbone of four fused rings. Phospholipids and triglycerides are lipids, but they do not share the same structure as steroids. Amino acids and DNA are not lipids.
Bloom's Level: 1. Remember Learning Outcome: 02.06.01 Compare the structures of fats, phospholipids, and steroids. Section: 02.06 Topic: Lipids
61. Which statement about the cellular nucleic acids DNA and RNA is incorrect? A. DNA is double-stranded, and RNA is single-stranded. B. The sugar in DNA is deoxyribose, and in RNA the sugar is ribose. C. DNA has a helix shape; RNA does not. D. RNA and DNA have the same four nitrogen-containing bases. E. Both DNA and RNA are polymers of nucleotides. In RNA, the base uracil replaces thymine. All the other answer choices are accurate statements.
Bloom's Level: 2. Understand Learning Outcome: 02.08.01 Compare the structure and function of DNA and RNA. Section: 02.08 Topic: Nucleic Acids
62. Which of the following radioactive isotopes are used to detect whether or not an individual has a healthy thyroid? A. I131 B. C14 C. glucose D. H2 E. All of the above can be used. I131 is a radioactive isotope that is used to detect whether the thyroid is healthy or not. C14 is used to date the age of fossils. Glucose is not taken up by the thyroid. H2 is a gas commonly found in the atmosphere.
Bloom's Level: 4. Analyze Learning Outcome: 02.01.04 Identify the beneficial and harmful uses of radiation. Section: 02.01 Topic: Atomic Structure
63. Which of the following subatomic particles will be found within the nucleus of the atom? A. protons and neutrons B. protons and electrons C. electrons and neutrons D. only neutrons E. only protons The nucleus of an atom will contain the protons and neutrons. Electrons are found in the electron orbitals that circle the nucleus.
Bloom's Level: 2. Understand Learning Outcome: 02.01.01 Describe how protons, neutrons, and electrons relate to atomic structure. Section: 02.01 Topic: Atomic Structure
64. Which of the following radiation uses is the one that is most likely to have both beneficial and harmful consequences? A. using radiation to treat a cancer patient B. using radiation to sterilize mail C. using radiation to sterilize surgical equipment prior to a surgery D. radiating fruits and vegetables prior to storage E. All of the answer choices will have beneficial and harmful consequences. When using radiation on a person for cancer treatment, there is the possibility of destroying healthy cells along with the cancer cells. Sterilizing mail, surgical equipment, and fruits and vegetables tends not to have any type of consequence on people.
Bloom's Level: 4. Analyze Learning Outcome: 02.01.04 Identify the beneficial and harmful uses of radiation. Section: 02.01 Topic: Chemical Reactions
65. Which of the following sequences correctly lists the bonds in order of strongest to weakest? A. double covalent - single covalent - ionic - hydrogen B. single covalent - double covalent - ionic - hydrogen C. ionic- double covalent - single covalent - hydrogen D. hydrogen - double covalent - single covalent - ionic E. double covalent - single covalent - hydrogen - ionic The strongest bond listed here is the double covalent followed by the single covalent, ionic, and then hydrogen bonds.
Bloom's Level: 4. Analyze Learning Outcome: 02.02.03 Compare the relative strengths of ionic, covalent, and hydrogen bonds. Section: 02.02 Topic: Chemical Bonds
66. Which type of bond formation is responsible for the properties of water? A. hydrogen B. polar covalent C. ionic D. nonpolar covalent E. None of these bonds play a role in the properties of water. Hydrogen bonds are the attraction between the hydrogen of one water molecule and the oxygen of a second water molecule. This attraction sets up the properties of water. A polar covalent bond forms between the hydrogen and oxygen of a particular water molecule. Water doesn't form ionic or nonpolar covalent bonds.
Bloom's Level: 2. Understand Learning Outcome: 02.02.03 Compare the relative strengths of ionic, covalent, and hydrogen bonds. Section: 02.02 Topic: Chemical Bonds
67. Which functional groups are associated with a dehydration reaction? A. H and OH B. OH and NHH C. OH and NHH D. SH and OH E. COOH and SH H and OH are associated with the dehydration reactions.
Bloom's Level: 2. Understand Learning Outcome: 02.04.02 Identify the role of a functional group. Section: 02.04 Topic: Chemical Bonds
68. Which of the following functional groups is present in amino acids? A. SH B. NHH C. COOH D. OH E. All of the answer choices are present in amino acids. All of these functional groups are present in amino acids.
Bloom's Level: 2. Understand Learning Outcome: 02.04.02 Identify the role of a functional group. Section: 02.04 Topic: Proteins
69. Removal of the hydroxyl functional group would disrupt the structure of A. sugars and some amino acids. B. sugars. C. fatty acids and amino acids. D. nucleotides and phospholipids. E. nucleotides and fatty acids. Hydroxyl is a functional group found within both sugars and some amino acids. If it was removed it would disrupt their structure.
Bloom's Level: 4. Analyze Learning Outcome: 02.04.02 Identify the role of a functional group. Section: 02.04 Topic: Chemical Reactions
70. Which of the following reactions is most likely to occur if an individual was to ingest a large dose of lemon juice? A. There would be an increase in the amount of carbonic acid within the bloodstream. If the carbonic acid didn't form, then the pH of the individual's blood could shift toward 7.2. B. There would be a decrease in the amount of carbonic acid within the bloodstream. If the carbonic acid didn't form, then the pH of the individual's blood could shift toward 7.2. C. There would be a decrease in the amount of carbonic acid within the bloodstream. If the carbonic acid didn't form, then the pH of the individual's blood could shift toward 7.2. D. There would be a decrease in the amount of carbonic acid within the bloodstream. If the carbonic acid didn't form, then the pH of the individual's blood could shift toward 7.8. Lemon juice is acidic so it contains large amounts H+. When H+ enters the bloodstream it combines with HCO3- to form H2CO3 (carbonic acid). The carbonic acid prevents the blood pH from shifting toward a more acidic or basic range. If it didn't form, in this case the individual's blood would become more acidic and shift toward 7.2.
Bloom's Level: 5. Evaluate Learning Outcome: 02.03.03 Describe how buffers are important to living organisms. Section: 02.03 Topic: Chemical Reactions
Short Answer Questions 71. Briefly describe the major functions of lipids in the human body. Answers will vary but should include the following information. Fats and oils function as energy storage molecules. Phospholipids form the cell membrane and inner compartments of the cell. Steroids include the sex hormones.
Bloom's Level: 6. Create Learning Outcome: 02.06.02 Identify the functions lipids play in our bodies. Section: 02.06 Topic: Lipids
Multiple Choice Questions 72. During the formation of a polymer, two monomers are joined by the removal of A. OH and H. B. OH and SH. C. H and COH. D. COOH and SH. E. NHH and COOH. Polymers are formed when H and OH are removed from the monomers during a dehydration reaction. COH, SH, NHH, and COOH are all functional groups.
Bloom's Level: 2. Understand Learning Outcome: 02.04.03 Recognize how monomers are joined to form polymers. Section: 02.04 Topic: Chemical Reactions
73. Cholesterol is a component of cell membranes and is an example of which type of lipid? A. steroids B. phospholipids C. fatty acids D. triglycerides E. oils Due to the structure of cholesterol it is classified as a steroid within the body.
Bloom's Level: 2. Understand Learning Outcome: 02.06.02 Identify the functions lipids play in our bodies. Section: 02.06 Topic: Lipids
74. What type of reaction is necessary to produce a dipeptide from individual amino acids? A. dehydration reaction B. hydrolysis reaction C. denaturation D. Dipeptides are not formed from amino acids. E. None of the answer choices will form a dipeptide. Two amino acids are joined during a dehydration reaction to form a dipeptide. Hydrolysis reactions will break apart a dipeptide into individual amino acids. Denaturation is the change in shape of a protein. Dipeptides are formed from amino acids.
Bloom's Level: 2. Understand Learning Outcome: 02.07.02 Explain how a polypeptide is constructed from amino acids. Section: 02.07 Topic: Chemical Reactions
75. Determine what would happen to an individual's proteins if they developed a fever of 103o F for several days. A. The proteins would denature due to the increase in body temperature and would become unable to function correctly. B. The proteins would increase in their ability to perform their functions because of the increase in body temperature. C. Nothing would happen to the proteins as a result of the increase in temperature. D. The proteins would denature due to the increase in body temperature and would increase in their ability to function correctly. E. One protein would alter in shape which would then cause the next protein to alter in shape which would cause a third protein to alter in shape and so forth until all of the proteins were altered in shape. An increase in body temperature for several days would cause the proteins to denature. Once they denatured they would not be able to function correctly. Proteins would not increase in their ability to function due to an increase in body temperature.
Bloom's Level: 5. Evaluate Learning Outcome: 02.07.01 Describe the functions of proteins in cells. Section: 02.07 Topic: Proteins
76. Which group of lipids will contain a barrier formed by hydrophilic heads that face outwards and hydrophobic tails that face inwards? A. phospholipids B. steroids C. triglycerides D. saturated acids E. trans-fatty acids Phospholipids have hydrophilic heads and hydrophobic tails that will form a barrier. Steroids have a backbone of four fused carbon rings. Triglycerides is formed from three fatty acids and a glycerol. Saturated acids and trans-fatty acids are both made from hydrocarbon chains with an acidic group on the end.
Bloom's Level: 2. Understand Learning Outcome: 02.06.01 Compare the structures of fats, phospholipids, and steroids. Section: 02.06 Topic: Lipids
77. Which nutrient source is the easiest one for humans to break down and form ATP? A. glucose B. protein C. cellulose D. phospholipids E. chitin Glucose is the easiest substance to break down into ATP. Proteins, cellulose, phospholipids, and chitin are not easily broken down into ATP.
Bloom's Level: 2. Understand Learning Outcome: 02.08.02 Explain the role of ATP in the cell. Section: 02.08 Topic: Nucleic Acids
78. Which of the following is not a function of ATP within the cell? A. All of the answer choices are functions of ATP within the cell. B. conduction of nerve impulses C. contraction of muscle cells D. synthesis of macromolecules E. energy currency of the cell All of the answer choices are functions of ATP within the cell.
Bloom's Level: 2. Understand Learning Outcome: 02.08.02 Explain the role of ATP in the cell. Section: 02.08 Topic: Nucleic Acids
Short Answer Questions 79. Briefly describe how ATP is broken down and turned into ADP. Answers will vary but should include the following: The last two phosphate bonds in ATP are unstable and easily broken. The terminal phosphate bond is hydrolyzed, leaving ADP and an inorganic phosphate. Energy is released when the phosphate is broken off.
Bloom's Level: 6. Create Learning Outcome: 02.08.02 Explain the role of ATP in the cell. Section: 02.08 Topic: Nucleic Acids
Multiple Choice Questions 80. What type of bond will connect the amino acids in a protein? A. peptide B. triple covalent C. polar covalent D. ionic E. double covalent A peptide bond will connect the amino acids together to form a protein. The atoms associate with a peptide bond unevenly because the oxygen is more electronegative than nitrogen.
Bloom's Level: 1. Remember Learning Outcome: 02.07.02 Explain how a polypeptide is constructed from amino acids. Section: 02.07 Topic: Chemical Bonds
81. Determine which of the following components are required for a molecule to be classified as organic. A. presence of carbon, hydrogen, four electrons in the outer orbital, and the attachment of a functional group B. presence of sodium, chlorine, four electrons in the outer orbital, and the attachment of a functional group C. presence of carbon, hydrogen, three electrons in the outer orbital, and the attachment of a functional group D. presence of oxygen, hydrogen, four electrons in the outer orbital, and the attachment of a functional group E. presence of carbon, hydrogen, four protons in the outer orbital, and the attachment of a functional group Organic molecules will have carbon, hydrogen, four electrons in the outer orbital, and a functional group attached to it. Sodium, chloride, and oxygen are not necessary for organic molecules. Protons are not found in the outer orbitals.
Bloom's Level: 5. Evaluate Learning Outcome: 02.04.01 Compare inorganic molecules to organic molecules. Section: 02.04 Topic: Chemical Bonds
Chapter 03 - Cell Structure and Function
Chapter 03 Cell Structure and Function
Multiple Choice Questions 1. Which of the following is not true concerning the cell theory? A. Virchow asserted that all cells come from preexisting cells. B. Schleiden determined that all plants are made of cells. C. Schwann researched animal tissues and discovered that they were all made of cells. D. Hooke observed the cells of cork. E. Leeuwenhoek first used the term "cells." All of the statements are true except Hooke, not Leeuwenhoek, was the first to use the term "cells" because the tiny chambers he observed in cork reminded him of the rooms, or cells, in a monastery.
Bloom's Level: 1. Remember Learning Outcome: 03.01.02 List the basic principles of the cell theory. Section: 03.01 Topic: Cell Theory
2. According to cell theory A. all organisms are composed of tissues. B. the smallest unit of life is a nucleus. C. animals but not plants are composed of cells. D. a multicellular organism is composed of many cells. E. new cells arise only from preexisting cells. Cell theory states that all organisms are made up of basic living units called cells, and that all cells come only from previously existing cells. While multicellular organisms are composed of many cells, this is not part of cell theory. Some organisms are single celled, the smallest unit of life is a cell, and both animals and plants are composed of cells.
Bloom's Level: 1. Remember Learning Outcome: 03.01.02 List the basic principles of the cell theory. Section: 03.01 Topic: Cell Theory
3-1 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 03 - Cell Structure and Function
3. The body of a slime mold that flows over a rotten log appears to lack any partitioning into distinct cells; however it does become cellular when it changes form and produces spores. The surfaces of parasitic flatworms and some insect tissues are "syncytial" layers of living material that developed from a single cell but now contains many nuclei but lack partitioning by cell membranes. These tissues actively consume food and produce wastes. Considering that the cell theory states that "all living things are composed of cells," then A. these tissues are not living because they violate the cell theory. B. this proves some vital force is involved beyond normal cell structures in order to give life to living organisms. C. these tissues are obviously a bridge between nonliving and primitive living cells. D. the general concept of life-is-cellular still holds since sometime in their life these organisms still utilize cells, but this shows cell membranes can be abandoned. E. this demonstrates that cell membranes are a necessity to be classified as a living cell. These organisms are living and were once composed of cells, even if the internal membranes between cells have been done away with. Since these organisms exhibit other characteristics of living cells, they are not nonliving, nor do they do not form a bridge between nonliving and living. Since they were once cells, no other vital force needs to be implicated. Since these cells do not have complete cell membranes, cell membranes are not a necessity to be classified as living.
Bloom's Level: 5. Evaluate Learning Outcome: 03.01.01 Explain why cells are the basic unit of life. Section: 03.01 Topic: Cell Theory
3-2 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 03 - Cell Structure and Function
4. The cell theory states A. all organisms are composed of only one cell. B. organelles are the basic living unit of structure and function of organisms. C. all cells come only from other cells. D. all organisms are composed of only one cell AND organelles are the basic living unit of structure and function of organisms. E. all multicellular organisms are unicellular at some point in their life cycle. Cell theory states that all organisms are made up of basic living units called cells, and that all cells come only from previously existing cells. Many living organisms are multicellular and cells, not organelles, are the basic living unit. Whereas all multicellular organisms may begin as a unicellular organism at some point in their life cycle, this is not part of cell theory.
Bloom's Level: 1. Remember Learning Outcome: 03.01.02 List the basic principles of the cell theory. Section: 03.01 Topic: Cell Theory
5. Surface-area-to-volume ratios indicate A. cells must exceed a certain minimum size. B. as cells get larger, their surface area actually decreases. C. that the largest cells have a less proportionate need for food intake and waste removal. D. a chicken egg is one cell thus demonstrating the upper limit for metabolizing cell size. E. as cells get larger, their surface area gets larger but at a slower rate than the volume increases. As a cell increases in size, the volume increases by the cube of the sides, while the surface area increases by the square of the sides. Therefore, the surface area increases but at a slower rate than the volume increases. There is no minimum cell size and large cells still need surface area for food intake and waste removal. A chicken egg is one cell, but it is not actively metabolizing. It divides into many smaller cells before it begins to metabolize.
Bloom's Level: 2. Understand Learning Outcome: 03.01.03 Explain the difference between the surface-area-to-volume ratios for large and small cells. Section: 03.01 Topic: Cell Theory
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Chapter 03 - Cell Structure and Function
6. Compared with a eukaryotic cell, a prokaryotic cell A. lacks organelles beyond ribosomes. B. is larger. C. does not require energy. D. is not living. E. has no method of movement. A prokaryotic cell lacks membrane bounded organelles. It does have ribosomes. Prokaryotic cells are smaller than eukaryotic cells and do require energy. Prokaryotic cells are living and many have methods of movement.
Bloom's Level: 1. Remember Learning Outcome: 03.02.01 Describe the fundamental components of a bacterial cell. Section: 03.02 Topic: Prokaryote Structure
7. Circulating red blood cells in your body do not contain a nucleus and other organelles. Are these cells living? A. Yes, because they are actively metabolizing and once contained organelles. B. Yes, because they are capable of moving throughout the body in the circulation. C. No, because they do not contain a nucleus, they cannot be living. D. No, because red blood cells do not actively metabolize. E. No, because they are now only part of a once living cell. Red blood cells are living because they actively metabolize. The organelles were lost in the maturation process and thus, red blood cells are fairly short lived. Red blood cells do move but that is not why they are considered living.
Bloom's Level: 3. Apply Learning Outcome: 03.03.02 Identify the cellular structures unique to both plant and animal cells. Section: 03.03 Topic: Eukaryote Structure
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Chapter 03 - Cell Structure and Function
8. An agent would make a good antibiotic if it affected a structure or process only found in bacterial cells and not in our (eukaryotic) cells. Which of the following actions would theoretically make a good antibiotic? A. prevents repair of the peptidoglycan cell wall B. damages the nuclear membrane C. damages DNA D. prevents ribosomes from producing proteins E. stops cellular respiration Only prokaryotic cells have a peptidoglycan cell wall. Both prokaryotic and eukaryotic cells have a nuclear membrane, DNA, ribosomes, and cellular respiration.
Bloom's Level: 3. Apply Learning Outcome: 03.02.01 Describe the fundamental components of a bacterial cell. Section: 03.02 Topic: Prokaryote Structure
9. Which of the following is a prokaryotic cell? A. plant cell B. liver cell C. muscle cell D. paramecium E. bacterium Only a bacterium is a prokaryotic cell. Plants, animals (liver and muscle), and paramecia are all eukaryotes.
Bloom's Level: 2. Understand Learning Outcome: 03.02.02 Identify the key differences between the archaea and bacteria. Section: 03.02 Topic: Prokaryote Structure
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Chapter 03 - Cell Structure and Function
10. Which structure regulates passage of molecules into and out of the bacterial cell? A. plasma membrane B. nucleus C. mitochondria D. chloroplast E. flagella The plasma membrane is the boundary that separates the living contents of the cell from the nonliving surrounding environment. Nuclei, mitochondria, and chloroplasts are organelles within the eukaryotic cell. While flagella are on the surface of the cell, they do not regulate the passage of molecules into and out of the cell.
Bloom's Level: 1. Remember Learning Outcome: 03.02.01 Describe the fundamental components of a bacterial cell. Section: 03.02 Topic: Prokaryote Structure
11. In moving from the outside environment to the inside of a bacterium, the first layer encountered would be A. the nucleus. B. the capsule. C. the cell wall. D. the cell membrane. E. the DNA. Bacteria are often surrounded by a capsule or slime layer. This covers the cell wall which is outside of the cell membrane. The DNA is internal to the cell. Prokaryotes do not have a nucleus.
Bloom's Level: 2. Understand Learning Outcome: 03.02.01 Describe the fundamental components of a bacterial cell. Section: 03.02 Topic: Prokaryote Structure
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Chapter 03 - Cell Structure and Function
12. Considering that minute differences in organelle structure are enough to provide selective toxicity, which of the following cell organelles would be a good target for an antibiotic so it doesn't attack your cells? A. plasma membrane B. DNA C. ribosomes D. cytoplasm E. endoplasmic reticulum Prokaryotic ribosomes are smaller and structurally different from eukaryotic ribosomes so they make a good target for antibiotics. The plasma membrane, DNA, and cytoplasm of a bacterial cell are very similar to that of a eukaryotic cell. Prokaryotic cells do not have an endoplasmic reticulum.
Bloom's Level: 3. Apply Learning Outcome: 03.02.01 Describe the fundamental components of a bacterial cell. Section: 03.02 Topic: Prokaryote Structure
13. A researcher samples the waters of the boiling hot springs in Yellowstone National Park. Under the microscope, she observes what appear to be living cells in the water. If these indeed are living cells, what type of cell would they most likely be? A. eukaryotic, animal B. eukaryotic, plant C. eukaryotic, fungi D. prokaryotic, archaea E. prokaryotic, bacteria Archaea are able to live in water temperatures above boiling. None of the other cell types are capable of living in such extreme conditions.
Bloom's Level: 2. Understand Learning Outcome: 03.02.02 Identify the key differences between the archaea and bacteria. Section: 03.02 Topic: Prokaryote Structure
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Chapter 03 - Cell Structure and Function
14. Within eukaryotic cells, the is surrounded by a double membrane and carries the coding that determines protein synthesis. A. smooth endoplasmic reticulum B. chloroplast C. nucleolus D. nucleus E. rough endoplasmic reticulum The nucleus is surrounded by a double membrane and contains the DNA with the code for protein synthesis. None of the others either contain DNA or are surrounded by a double membrane.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Eukaryote Structure
15. Which nuclear structure(s) contain(s) the hereditary material? A. nuclear envelope B. chloroplasts C. chromatin D. nucleoplasm E. mitochondria Chromatin, housed within the nucleus, is composed of DNA and proteins and thus contains the hereditary structure. While the nuclear envelope and nucleoplasm are part of the nucleus, they do not contain DNA. The chloroplasts and mitochondria are not located within the nucleus.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Eukaryote Structure
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Chapter 03 - Cell Structure and Function
16. The nucleolus A. contains RNA and is found in the nucleus. B. contains DNA and is found in the nucleus. C. contains RNA and is found in the cytoplasm. D. contains DNA and is found in the cytoplasm. E. contains ribosome subunits and is found in the cytoplasm. The nucleolus contains ribosomal RNA and is found inside the nucleus.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
17. Which of the following structures are NOT found in the cytoplasm? A. mitochondria B. chloroplast C. chromosomes D. nucleus E. lysosome Chromosomes are found within the nucleus. Each of the others is found within the cytoplasm.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
18. Which eukaryotic organelle is the largest? A. ribosome B. nucleus C. cytoskeleton D. mitochondria E. Golgi apparatus The nucleus is the largest of the organelles listed here.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
19. What evidence would not suggest that the fluid in the nucleus is different from the cytoplasm? A. There is a pH difference between the two fluids. B. The nuclear membrane encloses the nucleoplasm. C. Nuclear pores only permit passage of certain molecules into and out of the nucleus. D. Cytoplasm and nucleoplasm both flow freely into and out of the nucleus. E. Electron microscopy reveals nonmembranous components associated with the nuclear pores. If the cytoplasm and nucleoplasm freely mixed, there would not be differences between them. The other choices suggest a compartmentalization of the nucleoplasm from the cytoplasm which allows the two to be different.
Bloom's Level: 4. Analyze Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Eukaryote Structure
20. Ribosomal RNA is produced in the A. nucleolus. B. Golgi apparatus. C. ribosomes. D. rough endoplasmic reticulum. E. smooth endoplasmic reticulum. Ribosomal RNA is made within the nucleolus. Ribosomes contain rRNA and the rough endoplasmic reticulum contains ribosomes but the rRNA is made in those structures, not in the Golgi apparatus or the smooth ER.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
21. Ribosomes are composed of A. DNA and RNA. B. DNA and protein. C. only protein. D. RNA and protein. E. only DNA. Ribosomes are composed of both ribosomal RNA and proteins.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
22. If a cell lacked ribosomes, it would NOT be able to A. form a spindle apparatus. B. synthesize proteins. C. respire. D. hydrolyze fat. E. form a Golgi vesicle. Ribosomes are responsible for the synthesis of proteins using messenger RNA as a template. Other organelles are associated with the remaining functions listed.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
23. Which cellular structure is responsible for packaging materials with the cell? A. mitochondria B. chloroplasts C. Golgi apparatus D. ribosomes E. lysosomes The Golgi apparatus collects, sorts, packages, and distributes materials such as proteins and lipids. The other organelles are involved in different functions.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
24. Proteins are processed and modified in the interior of the A. mitochondria. B. nucleus. C. chloroplasts. D. rough endoplasmic reticulum. E. smooth endoplasmic reticulum. The processing and modification of proteins occurs in the interior of the rough endoplasmic reticulum (RER). The other organelles are associated with different functions.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
25. What evidence suggests that proteins are synthesized and modified in the rough ER as opposed to the smooth ER? A. Ribosomes are associated with the surface of the rough ER. B. Ribosomes are associated with the surface of the smooth ER. C. The smooth ER functions in the synthesis of phospholipids. D. The smooth ER functions in the synthesis of phospholipids. E. Smooth ER is continuous with rough ER. Since ribosomes are associated with protein synthesis and the rough ER is studded with ribosomes, this suggests a link between these. The surface of the smooth ER is not studded with ribosomes. The other statements are true but are not evidence to support the role of the rough ER in protein synthesis and modification.
Bloom's Level: 4. Analyze Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
26. Which of the following would be a way of finishing this hypothesis about the function of the Golgi apparatus? If the Golgi apparatus is involved in packaging products for secretion, then A. vesicles must travel from the Golgi to the cell surface. B. vesicles must travel from the rough ER and smooth ER to the Golgi apparatus. C. the Golgi apparatus must be part of the endomembrane system. D. the Golgi apparatus must consist of 3 to 20 slightly curved sacs. E. the Golgi apparatus would contain proteins. In order to be secretory, vesicles must travel from the Golgi to the cell surface. Although vesicles do travel from the rough ER and smooth ER to the Golgi apparatus, this does not apply to secretion, only to intracellular transport. The Golgi apparatus is part of the endomembrane system, does consist of 3-20 slightly curved sacs, and does contain proteins, but none of these apply to a secretory function.
Bloom's Level: 5. Evaluate Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
27. If an active cell produces an important protein for secretion, what is the correct sequence of organelles that are involved in the production of the protein? A. endoplasmic reticulum-to-ribosomes-to-Golgi apparatus B. ribosomes-to-endoplasmic reticulum-to-Golgi apparatus C. endoplasmic reticulum-to-Golgi apparatus-to-ribosomes D. Golgi apparatus-to-endoplasmic reticulum-to-ribosomes E. ribosomes-to-Golgi apparatus-to-endoplasmic reticulum The protein must be made on ribosomes, processed in the endoplasmic reticulum, and then packaged into vesicles sent to the Golgi apparatus.
Bloom's Level: 4. Analyze Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
28. Tissues that produce large amounts of secretions, such as the gastric glands of the stomach, contain cells with large numbers of A. lysosomes. B. nucleoli. C. centrioles. D. mitochondria. E. Golgi apparati. Golgi apparati are involved in the transportation of proteins/lipids for secretion. The other organelles are involved with different functions.
Bloom's Level: 3. Apply Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
29. The Golgi apparatus directs its protein products to the correct location in the cell based on A. the specific molecule that is added and the molecule determines the destination in or out of the cell. B. the DNA extending its long molecular helix to direct the protein through the Golgi apparatus and on to the final destination. C. vesicles that constantly shuttle back and forth from the various cell destinations and these vesicles determine which protein in the Golgi apparatus to pick up and deliver. D. proteins drifting away in all directions and are only used at the cell sites that need them. E. how long it takes the vesicle to travel from the endoplasmic reticulum to the Golgi. Proteins made in the rough ER have specific molecular tags that serve as "zip codes" to tell the Golgi apparatus where to send them. DNA does not leave the nucleus. Vesicles do not make the determination, either by picking up particular proteins, or by how long they take to travel from the ER to the Golgi. Proteins are specifically carried by vesicles to their destination.
Bloom's Level: 2. Understand Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
30. The will store pigments and toxins in plant cells. A. chloroplast B. lysosomes C. peroxisomes D. vacuole E. mitochondria Vacuoles store water, sugars, salts, pigments, and toxic molecules. Both lysosomes and peroxisomes enclose enzymes. Mitochondria and chloroplasts are involved in energy production.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
31. When a lysosome fuses with a vacuole, A. they both disappear. B. they both return to the Golgi apparatus. C. protein synthesis begins. D. division begins. E. its contents are digested. Lysosomes are the "garbage disposals" of the cell which digest the contents of the vacuole.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
32. Each time water in a cell freezes slowly, long sharp crystals spear through the membrane structures of the cell. The most reasonable explanation for the bad taste of meat that has "freezer burn" from repeated freezing is the destruction of A. the nuclear membrane, causing mixing of nucleoplasm and cytoplasm. B. ribosomes, causing them to break into subunits. C. The breakage of the lysosomes would result in the release of digestive enzymes which would self-digest and, therefore, alter the taste of the meat. None of the other organelles would destroy the cells of alter the taste. D. rough endoplasmic reticulum causing the release of ribosomes. E. the Golgi apparatus and vesicles. The breakage of the lysosomes would result in the release of digestive enzymes which would self-digest and, therefore, alter the taste of the meat. None of the other organelles would destroy the cells of alter the taste.
Bloom's Level: 3. Apply Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
33. Tay-Sachs is a serious childhood metabolic disorder where lipids. A. lysosomes B. nucleoli C. mitochondria D. Golgi apparati E. centrioles
fail to digest certain
Tay-Sachs disease results from the inability to make a lysosomal storage enzyme. The other organelles are not involved in this disease.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
34. Which organelle primarily functions to package or regulate the production of H2O2? A. lysosome B. vacuole C. Golgi apparatus D. endoplasmic reticulum E. peroxisome The peroxisome contains enzymes that break down hydrogen peroxide, a toxic molecule. The other organelles are not involved in this process.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
35. Which of the following would not be a function of the peroxisomes? A. break down fats in the liver B. produce bile salts in the liver C. help germinating seeds convert fatty acids into molecules that can be converted to sugars D. store toxic materials in the cell E. metabolism that uses oxygen and releases carbon dioxide in leaves The peroxisomes have all of the functions listed except they do not store toxic materials.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
36. The failure of which organelle during development can result in a child being born with webbed toes? A. lysosomes B. peroxisomes C. vacuoles D. rough ER E. Golgi apparati One function of the lysosomes is to autodigest the toe webbing found in the human embryo.
Bloom's Level: 3. Apply Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
37. Cellular respiration is best associated with the A. Golgi apparatus. B. ribosome. C. mitochondrion. D. chloroplast. E. microtubule. Cellular respiration occurs within mitochondria. The other organelles have different functions.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
38. Both plant and animal cells have mitochondria because they both A. carry on photosynthesis. B. have a large central vacuole. C. have endoplasmic reticulum. D. need ATP for energy. E. have lysosomal hydrolytic enzymes. Mitochondria provide ATP for energy from glucose; therefore, both plant and animal cells need them. Mitochondria do not carry out the other functions listed.
Bloom's Level: 2. Understand Learning Outcome: 03.03.02 Identify the cellular structures unique to both plant and animal cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
39. The "powerhouse" of the cell is the A. chloroplast. B. mitochondria. C. Golgi apparatus. D. ribosome. E. centriole. Mitochondria provide energy for the cell in the form of ATP so they are considered the "powerhouse" of the cell. The other organelles carry out other functions.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
True / False Questions 40. Since all organisms carry on cellular respiration, all living cells must contain mitochondria. FALSE Prokaryotic cells carry out cellular respiration but they do not contain mitochondria. The elements of the respiratory pathway are present in the plasma membrane.
Bloom's Level: 2. Understand Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
Multiple Choice Questions
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Chapter 03 - Cell Structure and Function
41. Which organelle will use up oxygen and give off carbon dioxide and water? A. lysosomes B. Golgi apparatus C. mitochondria D. chloroplasts E. endoplasmic reticulum Oxygen is used up and carbon dioxide and water are given off during photosynthesis. Chloroplasts carry out photosynthesis. The other organelles have other functions.
Bloom's Level: 2. Understand Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
42. In plants, chloroplasts are necessary for A. respiration. B. secretion. C. photosynthesis. D. storage. E. cell movement. Chloroplasts carry out photosynthesis for plant cells.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
43. Which of these is present in plant, but not animal, cells? A. nucleus B. centrioles C. cell membrane D. Golgi apparatus E. chloroplasts Plants have chloroplasts but animal cells do not. Both plant and animal cells have a cell membrane, Golgi apparatus, and nucleus. Only animal cells have centrioles.
Bloom's Level: 2. Understand Learning Outcome: 03.03.02 Identify the cellular structures unique to both plant and animal cells. Section: 03.03 Topic: Organelles
44. During photosynthesis, solar energy is converted to chemical energy by the chloroplasts. The process is represented by A. carbohydrate + oxygen → carbon dioxide + water + energy B. carbohydrate + carbon dioxide → oxygen + water + energy C. solar energy + oxygen + water → carbohydrate + oxygen D. solar energy + carbon dioxide + water → carbohydrate + oxygen E. solar energy → carbohydrate + oxygen The formula for photosysnthesis is solar energy + carbon dioxide + water goes to carbohydrate + oxygen. This occurs in the chloroplast.
Bloom's Level: 1. Remember Learning Outcome: 03.03.02 Identify the cellular structures unique to both plant and animal cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
45. Thylakoids and cristae are, respectively, structures of A. lysosomes and the endoplasmic reticulum. B. the nucleus and nucleolus. C. chloroplasts and mitochondria. D. mitochondria and chloroplasts. E. the cell wall and the plasma membrane. Thylakoids are found in chloroplasts, while cristae are found in mitochondria.
Bloom's Level: 1. Remember Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Learning Outcome: 03.03.02 Identify the cellular structures unique to both plant and animal cells. Section: 03.03 Topic: Organelles
46. Which organelle will give off oxygen and use up carbon dioxide in plants? A. Golgi apparatus B. rough endoplasmic reticulum C. lysosome D. mitochondrium E. chloroplast Chloroplasts give off oxygen and use up carbon dioxide during photosynthesis. Mitochondria will give off carbon dioxide and use up oxygen during respiration. The other organelles have different functions.
Bloom's Level: 2. Understand Learning Outcome: 03.03.02 Identify the cellular structures unique to both plant and animal cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
47. Mitochondria are bounded by a double membrane. The inner-filled space is called the matrix, which contains _ that break down carbohydrate products, releasing to be used for ATP production. A. DNA/energy B. ribosomes/carbon dioxide C. RNA/energy D. enzymes/energy The matrix of mitochondria contains enzymes that break down carbohydrates, releasing energy that is used to synthesize ATP. Although mitochondria contain DNA, RNA, and ribosomes, these do not break down carbohydrates.
Bloom's Level: 2. Understand Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
48. Which of the following would be considered evidence that chloroplasts are involved in photosynthesis? A. Isolated chloroplasts give off oxygen when exposed to sunlight. B. Plants that are green contain chloroplasts. C. Both plants and algae have chloroplasts. D. Plants have both chloroplasts and mitochondria. E. Chloroplasts contain thylakoids which are stacked in structures called grana. While all of the statements are true, only the statement that isolated chloroplasts give off oxygen when exposed to sunlight applies to photosynthesis. In photosynthesis, solar energy results in photosynthesis, which gives off oxygen.
Bloom's Level: 4. Analyze Learning Outcome: 03.03.02 Identify the cellular structures unique to both plant and animal cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
True / False Questions 49. Since cyanobacteria do not contain chloroplasts and yet they photosynthesize, chloroplasts are not required for photosynthesis. TRUE Although they do not have chloroplasts, cyanobacteria do have thylakoids. Therefore, thylakoids are required for photosynthesis.
Bloom's Level: 2. Understand Learning Outcome: 03.02.01 Describe the fundamental components of a bacterial cell. Section: 03.02 Topic: Prokaryote Structure
Multiple Choice Questions 50. In a person with a mitochondrial disease in which the mitochondria do not produce enough energy for the cell, which organs would most likely be affected? A. muscles B. liver C. skin D. stomach E. ovaries/testes High energy demand tissues include muscles, the central nervous system, and the eye. Although the other organs would be affected, the high energy demand tissues would be most noticeably affected earlier.
Bloom's Level: 3. Apply Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
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Chapter 03 - Cell Structure and Function
Yes / No Questions 51. Brock's father has inherited a mitochondrial disease. Is Brock in danger of also having this disease? NO Mitochondria are inherited from mother to child. Therefore, Brock is not in danger of inheriting this disease since his father, and not his mother, carries the disease.
Bloom's Level: 2. Understand Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
Multiple Choice Questions 52. Which is NOT a function of the cytoskeleton? A. maintains a cell's shape B. anchors organelles in place within a cell C. allows cell and its organelles to move D. secretes the calcium for bone tissue E. allows the formation of pseudopodia All of the above are functions of the cytoskeleton except that it does not secrete the calcium for bone tissue.
Bloom's Level: 1. Remember Learning Outcome: 03.04.01 Compare and contrast the structural differences between actin filaments, intermediate filaments, and microtubules. Section: 03.04 Topic: Cytoskeleton
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Chapter 03 - Cell Structure and Function
53. When viewed through a microscope, one characteristic of living cells is that their internal structures move. What organelles are most directly responsible for this motion we see? A. cell membrane and nucleus B. ribosomes and endoplasmic reticulum C. centrioles and cell wall D. mitochondria and cytoskeleton E. chloroplasts and lysosomes Mitochondria provide the energy for organelle movement, while the cytoskeleton provides the structures needed for movement. The other organelles are involved in different functions.
Bloom's Level: 2. Understand Learning Outcome: 03.04.01 Compare and contrast the structural differences between actin filaments, intermediate filaments, and microtubules. Section: 03.04 Topic: Cytoskeleton
54. If cells were moved to a low-gravity environment, such as in space exploration, a change would most likely be expected in the A. nucleus and the genetic process. B. chloroplast and energy capture. C. cytoskeleton. D. vacuoles and lysosomes. E. mitochondrion and energy release. Since the cytoskeleton is responsible for cellular movement, it would be most affected by a low-gravity environment.
Bloom's Level: 3. Apply Learning Outcome: 03.04.02 Identify the cellular structures that are composed of the different cytoskeleton components. Section: 03.04 Topic: Cytoskeleton
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Chapter 03 - Cell Structure and Function
55. Microtubules A. are composed mostly of lipid. B. form a spindle that distributes chromosomes in an orderly manner. C. may be associated with metabolism. D. are found in the nucleus. E. are composed of actin. The function of microtubules is to form a spindle that distributes chromosomes in an orderly manner. They are composed of the protein tubulin and are found throughout the cytoplasm.
Bloom's Level: 1. Remember Learning Outcome: 03.04.01 Compare and contrast the structural differences between actin filaments, intermediate filaments, and microtubules. Section: 03.04 Topic: Cytoskeleton
56. Which organelle has 13 rows of tubulin dimers around an empty central core? A. flagella B. microtubules C. actin filaments D. intermediate filaments E. cilia Microtubules are composed of 13 rows of tubulin dimers, surrounding what appears in electron micrographs to be an empty central core.
Bloom's Level: 1. Remember Learning Outcome: 03.04.01 Compare and contrast the structural differences between actin filaments, intermediate filaments, and microtubules. Section: 03.04 Topic: Cytoskeleton
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Chapter 03 - Cell Structure and Function
57. In a certain group of people, males are more subject to respiratory infections and are sterile. The most likely explanation for this disease is malfunctioning A. lysosomes. B. intermediate filaments. C. actin filaments. D. rough endoplasmic reticulum. E. cilia and flagella. The respiratory system uses cilia to keep the passageways clear while sperm use flagella to swim. There is most likely a problem with microtubule formation.
Bloom's Level: 3. Apply Learning Outcome: 03.04.02 Identify the cellular structures that are composed of the different cytoskeleton components. Section: 03.04 Topic: Cytoskeleton
58. Centrioles A. consist of a 9 + 2 pattern. B. are composed of microtubule triplets. C. carry on cellular respiration. D. are found in plant cells. E. are composed of actin. Centrioles are composed of a 9 + 0 pattern of microtubule triplets. They are not found in plant cells and are not composed of actin. They are part of the cytoskeleton.
Bloom's Level: 1. Remember Learning Outcome: 03.04.02 Identify the cellular structures that are composed of the different cytoskeleton components. Section: 03.04 Topic: Cytoskeleton
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Chapter 03 - Cell Structure and Function
59. Which organelle has a 9 + 0 pattern of microtubules? A. cilia B. flagella C. centriole D. kinesin E. mitochondria Centrioles have a 9 + 0 pattern of microtubule triplets. Cilia and flagella have a 9 + 2 pattern. Kinesin is a motor molecule.
Bloom's Level: 1. Remember Learning Outcome: 03.04.02 Identify the cellular structures that are composed of the different cytoskeleton components. Section: 03.04 Topic: Cytoskeleton
60. Kinesin and dynein are associated with A. microtubules. B. actin filaments. C. Golgi apparatus. D. intermediate filaments. E. endoplasmic reticulum. Kinesin and dynein are motor molecules associated with microtubules.
Bloom's Level: 1. Remember Learning Outcome: 03.04.02 Identify the cellular structures that are composed of the different cytoskeleton components. Section: 03.04 Topic: Cytoskeleton
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Chapter 03 - Cell Structure and Function
61. Which of these is mismatched? A. cilia-microtubules B. centrioles-spindle C. mitochondria-stroma D. lysosomes-hydrolytic enzymes E. ribosomes-RNA Mitochondria contain a matrix, not a stroma. A stroma is found within chloroplasts.
Bloom's Level: 4. Analyze Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
62. Which is NOT a correct association of cell organelles and function? A. lysosome-intracellular digestion B. mitochondrion-cell respiration C. ribosome-production of proteins D. cell wall-regulate molecule passage in and out of animal cells E. vacuole-storage of chemicals The cell membrane, not the cell wall, regulates the passage of molecules in and out of animal cells.
Bloom's Level: 4. Analyze Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Eukaryote Structure
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Chapter 03 - Cell Structure and Function
63. Which is NOT a correct association of cell organelles and their structure? A. cell wall-cellulose fibers and lignin B. centriole-microtubules C. endoplasmic reticulum-concentrated chromatin, RNA, and nucleoli D. plasma membrane-two layers of phospholipids with embedded proteins E. ribosome-RNA and protein in two subunits The endoplasmic reticulum synthesizes and processes proteins. Concentrated chromatin, RNA, and nucleoli are associated with the nucleus.
Bloom's Level: 4. Analyze Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
64. The motor molecules associated with actin filaments are called A. centrosomes. B. centrioles. C. dynein. D. myosin. E. kinesin. Myosin is the motor molecule associated with actin. Kinesin and dynein are motor molecules associated with microtubules. Centrioles and centrosomes are not motor molecules.
Bloom's Level: 1. Remember Learning Outcome: 03.04.02 Identify the cellular structures that are composed of the different cytoskeleton components. Section: 03.04 Topic: Cytoskeleton
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Chapter 03 - Cell Structure and Function
True / False Questions 65. Since plant cells do not have centrioles, they are not required for all organisms to conduct mitosis and meiosis. TRUE Centrioles may be involved in microtubule assembly and disassembly in animal cells, but they are not required for cell division (mitosis and meiosis) in plant cells.
Bloom's Level: 2. Understand Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Organelles
66. Actin filaments, intermediate filaments, and microtubules are all similar in that each is composed of a single type of protein. FALSE Actin filaments are composed of actin and microtubules are composed of tubulin, but intermediate filaments are composed of a variety of different proteins depending on the tissue.
Bloom's Level: 2. Understand Learning Outcome: 03.04.01 Compare and contrast the structural differences between actin filaments, intermediate filaments, and microtubules. Section: 03.04 Topic: Cytoskeleton
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Chapter 03 - Cell Structure and Function
Multiple Choice Questions 67. The endosymbiotic theory argues that prokaryotes became some of the organelles of early eukaryotic cells. All of the following support this hypothesis EXCEPT A. mitochondria and chloroplasts have their own DNA. B. mitochondria and chloroplasts have their own DNA. C. mitochondria and chloroplasts are nearly identical to some free-living prokaryotes. D. the mitochondria and chloroplasts have their own DNA and are nearly identical to some free-living prokaryotes. E. mitochondria and chloroplasts divide by splitting. Whereas vacuoles can come and go across a cell membrane, this fact does particularly apply to the endosymbiotic theory.
Bloom's Level: 2. Understand Learning Outcome: 03.05.02 Describe how the endosymbiotic theory explains eukaryotic cell division. Section: 03.05 Topic: Origin of Cells
68. All of the following are possible origins of cell organelles in eukaryotes EXCEPT A. invagination of the plasma membrane to form endoplasmic reticulum. B. incorporation of engulfed heterotrophic bacteria to form mitochondria. C. incorporation of engulfed autotrophic cyanobacteria to form chloroplasts. D. a symbiotic relationship between a host cell and a prokaryote that was taken up but not destroyed. E. groups of prokaryotic cells begin to live in a small group, sharing products of metabolism. Groups of prokaryotic cells do live together in groups and share products of metabolism but this is not considered part of how the first eukaryotic cells came to be.
Bloom's Level: 2. Understand Learning Outcome: 03.05.01 Define endosymbiosis. Section: 03.05 Topic: Organelles
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Chapter 03 - Cell Structure and Function
69. If the mitochondrion is the result of endosymbiosis, what membrane of the mitochondrion represents the engulfing vesicle? A. the outer mitochondrial membrane B. vacuole C. crista D. thylakoid membrane E. granum The outer mitochondrial membrane would come from the engulfing vesicle. The crista is the inner mitochondrial membrane and would have represented the original prokaryotic membrane. The thylakoid membrane and grana are part of chloroplasts. A vacuole membrane is not found in mitochondria.
Bloom's Level: 2. Understand Learning Outcome: 03.05.02 Describe how the endosymbiotic theory explains eukaryotic cell division. Section: 03.04 Section: 03.05 Topic: Origin of Cells
70. If the chloroplast is the result of endosymbiosis, what membrane of the chloroplast represents the plasma membrane of the original prokaryote? A. the outer chloroplast membrane B. the inner chloroplast membrane C. crista D. thylakoid membrane E. granum The outer chloroplast membrane would represent the membrane of the engulfing vesicle. The inner chloroplast membrane would represent the membrane of the original prokaryotic cell. The thylakoid membrane and grana are interior to the chloroplast. Cristae are found in mitochondria.
Bloom's Level: 2. Understand Learning Outcome: 03.05.02 Describe how the endosymbiotic theory explains eukaryotic cell division. Section: 03.05 Topic: Eukaryote Structure
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Chapter 03 - Cell Structure and Function
True / False Questions 71. Because plants evolved from bacterial ancestors, the primary plant cell wall and the bacterial cell wall are both composed of peptidoglycan. FALSE The primary plant cell wall is not composed of peptidoglycan.
Bloom's Level: 3. Apply Learning Outcome: 03.03.02 Identify the cellular structures unique to both plant and animal cells. Section: 03.03 Topic: Eukaryote Structure
72. According to the endosymbiotic theory, eukaryotes became the mitochondria and chloroplasts found in prokaryotic cells. FALSE Prokaryotic cells do not contain mitochondria and chloroplasts. The endosymbiotic theory proposes that prokaryotic cells became the mitochondria and chloroplasts found in eukaryotic cells.
Bloom's Level: 2. Understand Learning Outcome: 03.05.02 Describe how the endosymbiotic theory explains eukaryotic cell division. Section: 03.05 Topic: Origin of Cells
73. Because mitochondria were once independently living prokaryotic cells, if removed from a eukaryotic cell they can survive independently. FALSE Mitochondria have now evolved to be totally dependent on the nucleus of the eukaryotic cell for many of the proteins and genes necessary for mitochondrial structure and function.
Bloom's Level: 3. Apply Learning Outcome: 03.03.01 Recognize the structure and function of the organelles within eukaryotic cells. Section: 03.03 Topic: Origin of Cells
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Chapter 03 - Cell Structure and Function
Multiple Choice Questions 74. Which type of cells will have modifications that increase the surface-area-to-volume ratio? A. cells that specialize in absorption B. cells that specialize in gas exchange C. cells that secrete proteins D. cells that secrete lipids E. cells that form a waterproof barrier Cells that specialize in absorption will have modifications that increase the surface-area-tovolume ratio. These modifications increase the surface area that allows them to absorb more materials.
Bloom's Level: 2. Understand Learning Outcome: 03.01.02 List the basic principles of the cell theory. Section: 03.01 Topic: Cell Theory
Short Answer Questions 75. List the key differences between archaea and bacteria. Answers may vary but should include the following: Archaea cell wall is present without peptidoglycan. Bacterial cell is present with peptidoglycan. Archaea live in harsh environments. Bacteria tend to live in more hospital environments. Archaea are rarely associated with disease. Bacteria are more often associated with diseases.
Bloom's Level: 6. Create Learning Outcome: 03.02.02 Identify the key differences between the archaea and bacteria. Section: 03.02 Topic: Origin of Cells
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Chapter 03 - Cell Structure and Function
Multiple Choice Questions 76. Which of the following events would increase the cell's ability to exchange materials across the membrane? A. multiple cellular divisions that restore the amount of surface area B. the growth of microvilli on the cell membrane C. specialization in absorption D. All of the answer choices would increase the cell's ability to exchange materials across the membrane. E. None of the answer choices would increase the cell's ability to exchange materials across the membrane. All of these events would increase the cell's ability to exchange materials across the membrane.
Bloom's Level: 5. Evaluate Learning Outcome: 03.01.03 Explain the difference between the surface-area-to-volume ratios for large and small cells. Section: 03.01 Topic: Cell Theory
77. Which of the following cells were most likely the first type of cells on Earth? A. bacteria B. fungi C. archaea D. plants E. animals Bacteria were the first cell types to appear on Earth. Archaea are more closely related to eukaryotes, so it would indicate that they evolved after the bacteria. The rest of the groups are eukaryotic types of cells.
Bloom's Level: 1. Remember Learning Outcome: 03.05.01 Define endosymbiosis. Section: 03.05 Topic: Origin of Cells
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Chapter 03 - Cell Structure and Function
Short Answer Questions 78. Explain how the nuclear envelope of eukaryotic cells could have arisen. The cell membrane of a prokaryotic cell invaginated. As it invaginated it would have pinched away from the rest of the cell membrane. The portion that pinched inward would have been retained within the cell, potentially forming the nuclear envelope.
Bloom's Level: 6. Create Learning Outcome: 03.05.01 Define endosymbiosis. Section: 03.05 Topic: Origin of Cells
79. Compare the structural differences between actin filaments, intermediate filaments, and microtubules. Answers may vary but should include the following features: Actin filaments are long, extremely thin flexible fibers. They occur in bundles or meshlike networks. Intermediate filaments are intermediate in size. They are a ropelike assembly of fibrous polypeptides. Microtubules are made up of tubulin and have 13 rows of tubulin dimers that surround an empty central core.
Bloom's Level: 6. Create Learning Outcome: 03.04.01 Compare and contrast the structural differences between actin filaments, intermediate filaments, and microtubules. Section: 03.04 Topic: Cytoskeleton
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Chapter 04 - Membrane Structure and Function
Chapter 04 Membrane Structure and Function
Multiple Choice Questions 1. The plasma membrane is composed of A. proteins and microtubules. B. lipids and actin filaments. C. lipids and microtubules. D. lipids and proteins. E. proteins and actin filaments. The plasma membrane is a phospholipid bilayer in which protein molecules are embedded. Actin filaments and microtubules make up the cytoskeleton.
Bloom's Level: 1. Remember Learning Outcome: 04.01.01 Describe the fluid-mosaic model of membrane structure. Section: 04.01 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
2. Which of the following statements is true regarding the inside and outside of the plasma membrane? A. Both the inside and outside layers make up the phospholipid bilayer and both layers contain identical proteins. B. Both the inside and outside layers make up the phospholipid bilayer but cytoskeletal filaments are present on the outside layer and carbohydrate chains of glycolipids and proteins are present on the inside layer. C. Both the inside and outside layers make up the phospholipid bilayer but cytoskeletal filaments are present on the inside layer and carbohydrate chains of glycolipids and proteins are present on the outside layer. D. The inside layer consists of a phospholipid bilayer while the outside layer consists of carbohydrate chains of glycolipids and proteins. E. The outside layer consists of a phospholipid bilayer while the inside layer consists of carbohydrate chains of glycolipids and proteins. The phospholipid bilayer is identical but the proteins on each side of the membrane are different. The cytoskeleton is attached to the interior face while glycolipids and carbohydrates are attached to the outside face of the membrane.
Bloom's Level: 3. Apply Learning Outcome: 04.01.01 Describe the fluid-mosaic model of membrane structure. Section: 04.01 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
3. Which function does the lipid bilayer component of the plasma membrane NOT provide for the cell? A. defines a permeable boundary between the organized interior and the chaotic external environment B. provides an impermeable, self-sealing membrane that keeps all internal living processes sealed inside C. controls the exchange of molecules between one cell and adjacent cells or the environment D. provides a fluid and flexible boundary that insulates the interior from the variations in humidity, food, and other external conditions E. provides a matrix for the placement of proteins that regulate the exchange of molecules between the inside and outside of the cell The membrane is not impermeable. Many things cross the membrane in both directions.
Bloom's Level: 2. Understand Learning Outcome: 04.01.01 Describe the fluid-mosaic model of membrane structure. Section: 04.01 Topic: Plasma Membrane
4. According to the fluid-mosaic model of membrane structure, A. proteins make up the bulk of the membrane. B. only lipids are found in the membrane. C. cholesterol is the main constituent of the membrane. D. glycolipids form a mosaic pattern inside the cell. E. proteins float inside or within the phospholipid bilayer. The fluid mosaic model states that proteins form a mosaic pattern just outside or within the phospholipid layer of the membrane. Cholesterol is found in membranes but it is not the main constituent. Glycolipids are on the outside of the cell.
Bloom's Level: 2. Understand Learning Outcome: 04.01.01 Describe the fluid-mosaic model of membrane structure. Section: 04.01 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
5. In the new procedures developed to clone a mammalian cell, the nucleus is removed from an egg cell and the nucleus from the adult to be cloned is injected inside this cell. If we did this process with a basketball, we would permanently damage the shell. What property of the cell repairs this damage? A. diffusion of cell membrane proteins B. carrier proteins restore the membrane lipids C. isotonic solutions D. the outer and inner sides of the plasma membrane have identical carbohydrate chains E. the fluid-mosaic nature of the membrane Because of the fluid-mosaic nature of the membrane it is self-sealing. Cell membrane proteins do diffuse in the membrane but they are not responsible for the self-sealing property. Carrier proteins do not restore lipids and the outer and inner sides of the plasma membrane do not have identical carbohydrate chains. Isotonic solutions do not damage the cell membrane but they do not seal it either.
Bloom's Level: 4. Analyze Learning Outcome: 04.01.01 Describe the fluid-mosaic model of membrane structure. Section: 04.01 Topic: Plasma Membrane
6. Which of the following is not a function of the cell membrane? A. It provides mechanical strength to the cell. B. It gives shape to the cell. C. It regulates passage of molecules into and out of the cell. D. It is largely responsible for cellular homeostasis. E. It serves as a site for protein synthesis. The endoplasmic reticulum is the site for protein synthesis.
Bloom's Level: 1. Remember Learning Outcome: 04.01.01 Describe the fluid-mosaic model of membrane structure. Section: 04.01 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
7. What property of phospholipids makes them suitable for the formation of the double layer found in membranes? A. They are uncharged. B. They can interact with proteins. C. They are composed of fatty acids. D. They have both a hydrophobic and a hydrophilic end. E. They serve as an attachment site for carbohydrates. Because phospholipids have both a hydrophobic and a hydrophilic end, the hydrophobic ends interact to form a bilayer with the hydrophilic ends facing the exterior. Phospholipids are charged. The remaining statements are true but do not explain why phospholipids can form a bilayer.
Bloom's Level: 3. Apply Learning Outcome: 04.01.01 Describe the fluid-mosaic model of membrane structure. Section: 04.01 Topic: Plasma Membrane
8. Which of the following molecules add stiffness and strength to the plasma membrane? A. glycoproteins B. cholesterol C. phospholipids D. enzymatic proteins E. phosphate groups Cholesterol stiffens and strengthens the membrane. The other things are present in membranes but do not fulfill that function.
Bloom's Level: 1. Remember Learning Outcome: 04.01.01 Describe the fluid-mosaic model of membrane structure. Section: 04.01 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
9. Radioactive isotopes are used to tag proteins in the cell membranes of mouse cells. These cells are then fused with human cells in cultures. What is the likely outcome of the tagged mouse proteins? A. The tagged proteins remain in the mouse cells and keep their position on the membrane. B. The tagged proteins remain in mouse cells but move anywhere across the mouse cell membrane. C. The tagged proteins drift across cell membranes and are soon found dispersed across both human and mouse cell membranes. D. None of the answer choices will occur since mouse cell membranes are unlike human cell membranes. E. None of the answer choices will occur since radioactive cells will soon die. Once the membranes fuse, proteins are free to move around the membrane within the phospholipid bilayer. This is a consequence of the fluid mosaic nature of the membrane.
Bloom's Level: 4. Analyze Learning Outcome: 04.01.02 Describe the diverse roles of proteins in membranes. Section: 04.01 Topic: Plasma Membrane
10. Which is a correct association of protein types? A. carrier proteins-provide a channel for substances to move across a membrane B. cell recognition proteins-allow a specific molecule to bind to it and may cause a cell response C. channel proteins-combine with a substance to help it move across a membrane D. receptor proteins-variations in these proteins cause tissue transplant rejection E. enzymatic proteins-carry out metabolic reactions directly Enzymatic proteins carry out metabolic reactions. Carrier proteins combine with a substance to help it move across a membrane. Variations in cell recognition proteins cause tissue transplant rejection. Channel proteins provide a pathway for substances to move across the membrane. Receptor proteins allow a specific molecule to bind to it and may cause a cell response.
Bloom's Level: 4. Analyze Learning Outcome: 04.01.02 Describe the diverse roles of proteins in membranes. Section: 04.01 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
11. Which components of a cell membrane are used for cell-to-cell recognition? A. phospholipids B. channel proteins C. glycolipids D. glycoproteins E. cholesterol molecules Glycoproteins in a cell membrane are used for cell recognition. The other proteins are present in the membrane but do not function in this way.
Bloom's Level: 1. Remember Learning Outcome: 04.01.02 Describe the diverse roles of proteins in membranes. Section: 04.01 Topic: Plasma Membrane
12. Which type of membrane protein will be used to bind to hormones? A. channel proteins B. carrier proteins C. cell recognition proteins D. receptor proteins E. enzymatic proteins Hormones are signal molecules that bind to receptor proteins. The others are types of membrane proteins but they do not function in this way.
Bloom's Level: 1. Remember Learning Outcome: 04.01.02 Describe the diverse roles of proteins in membranes. Section: 04.01 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
13. Our body recognizes the foreign nature of a parasite or disease agent and we soon build up antibodies. A few parasites can change their identity to evade our immune system. How could cells do this on a regular basis? A. Replace the cholesterol within the membrane with host cholesterol. B. Wait for the evolution of new protein membrane molecules. C. Because the membrane is "set," a cell must reproduce and then the cell with the old membrane must die. D. A whole new phospholipid bilayer is generated to replace the old layer. E. New glycoproteins are produced in the cell and moved into the plasma membrane. Cell recognition membrane markers are usually proteins, so new proteins, similar enough to the host proteins to not be recognized by antibodies, need to be made and presented on the membrane. This procedure must occur fairly quickly or the immune system will recognize and destroy the invader.
Bloom's Level: 3. Apply Learning Outcome: 04.01.02 Describe the diverse roles of proteins in membranes. Section: 04.01 Topic: Plasma Membrane
14. Red blood cells come in many blood types including A, B, AB, and type O (lacking proteins A and B), Rh positive, Rh negative (lacking Rh+), and many others. If blood is transfused, the recipient detects any new or foreign proteins. The blood type proteins are A. in the red blood cell nucleus. B. inside the red blood cell cytoplasm. C. on the outer surface of the red blood cell membrane. D. in the nuclear membrane. E. in the endoplasmic reticulum of the red blood cells. In order to be used as cell recognition proteins, the red blood cell surface markers must be present on the outer surface of the red blood cell membrane.
Bloom's Level: 3. Apply Learning Outcome: 04.01.02 Describe the diverse roles of proteins in membranes. Section: 04.01 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
True / False Questions 15. The same signaling molecule that binds to the receptor on the outside of the cell is used to alter gene expression in the nucleus. FALSE The signaling molecule initiates a transduction pathway that results in altered gene expression, but it is not the same molecule that causes the end result.
Bloom's Level: 3. Apply Learning Outcome: 04.01.02 Describe the diverse roles of proteins in membranes. Section: 04.01 Topic: Plasma Membrane
Multiple Choice Questions 16. Certain individuals with a Y chromosome develop into females because testosterone is not able to induce the formation of male sexual characteristics. This is most likely due to a defect in A. the signaling molecule. B. the receptor. C. the transduction pathway. D. the target protein. E. the Y chromosome. These individuals lack receptors for testosterone. Although the testosterone is produced and is functional, it cannot bind to the cells and cause altered gene expression resulting in male sexual characteristics. These individuals develop as females.
Bloom's Level: 3. Apply Learning Outcome: 04.01.02 Describe the diverse roles of proteins in membranes. Section: 04.01 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
17. In certain cancerous cells, the cell divides continuously even in the absence of a growth factor (signaling molecule) that indicates the cell should divide. Which of the following could NOT explain this? A. a receptor molecule that is always turned off B. a receptor molecule that is always activated C. a transduction pathway that is always turned on D. a target protein that is always activated E. a gene for cell division that is expressed when it should not be The cell is behaving as if a signal has bound to a receptor even when one has not. This could be due to any of the products of a signaling pathway being activated when it should not be. The only thing it could not be is a receptor that is inactive or turned off.
Bloom's Level: 4. Analyze Learning Outcome: 04.01.02 Describe the diverse roles of proteins in membranes. Section: 04.01 Topic: Plasma Membrane
18. Which statement is false concerning movement of molecules across the cell membrane? A. Water and gas molecules have no difficulty. B. Small uncharged molecules pass through easily. C. Large molecules do not pass through easily. D. Charged molecules do not pass through easily. E. Lipid molecules do not pass through easily. The membrane is differentially permeable. Lipids, small uncharged molecules, water, and gases have no difficulty passing through the membrane. Charged molecules do not pass through easily.
Bloom's Level: 2. Understand Learning Outcome: 04.02.01 Explain why a membrane is selectively permeable. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
19. Permeability refers to A. the movement of molecules from an area of greater concentration to an area of lesser concentration. B. the extent to which a membrane allows a substance to pass through. C. the amount of solute in a solution. D. the state of being permanent. E. the ability to establish a permanent solute level in a solution. The ability of a membrane to allow substances to pass through is permeability. The movement of molecules from greater to lesser concentration is diffusion. The amount of solute in a solution is concentration. The state of being permanent is permanence. It is not possible to establish a permanent solute level in a solution.
Bloom's Level: 1. Remember Learning Outcome: 04.02.01 Explain why a membrane is selectively permeable. Section: 04.02 Topic: Passive Transport
20. Proteins do NOT pass through cell membranes because A. the membrane is made of protein. B. they contain nitrogen. C. they are very large molecules. D. they cause emulsification. E. they cause digestion of the cell. Proteins are too large to pass through the membrane. Proteins do contain nitrogen and may digest the cell but that does not explain why they cannot pass through the membrane. The membrane is made of proteins and phospholipids but again, that does not explain why proteins cannot pass. Proteins do not cause emulsification.
Bloom's Level: 2. Understand Learning Outcome: 04.02.01 Explain why a membrane is selectively permeable. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
21. A student sitting on the back row opened a bottle of foul-smelling perfume and dabbed it on her wrists. One by one (beginning from the back of the room) the students began to cough due to the foul smell. This phenomena was due to A. osmosis. B. molecules moving from an area of low concentration to high concentration. C. an allergic reaction. D. diffusion. E. active transport. Diffusion is the movement of molecules from a higher concentration (the back of the room) to a lower concentration (the other parts of the room). Osmosis and active transport require a membrane. Smelling something is not the same as an allergic reaction.
Bloom's Level: 3. Apply Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
22. Of the following conditions, which is absolutely necessary for diffusion to take place? A. a differentially permeable membrane B. a true solution C. a concentration difference D. a non-permeable membrane E. a living cell Since diffusion is just the movement of molecules from a higher to a lower concentration, only a concentration difference is required.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
4-12 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 04 - Membrane Structure and Function
23. What will happen to dye crystals if placed in the bottom of a beaker of water over a long period of time? A. They will undergo osmosis to a different location. B. They will undergo active transport to a different location. C. They will all diffuse to the top of the beaker. D. They will diffuse equally throughout the beaker. E. They will stay at the bottom of the beaker. Since diffusion is the movement of molecules from an area of higher to lower concentration, the dye molecules will move around the beaker until they are evenly distributed. Active transport and osmosis both require a membrane.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
24. Oxygen diffuses into the capillaries of the lungs because there is A. a higher concentration of O2 in the alveoli than the capillaries. B. a higher concentration of O2 in the capillaries than the alveoli. C. a higher concentration of CO2 in the alveoli than the capillaries. D. a higher concentration of CO2 in capillaries than the alveoli. E. an equal concentration of O2 in the alveoli and the capillaries. For diffusion to take place there must be a concentration difference. Oxygen is present at a higher level in the alveoli than in the capillaries so oxygen diffuses from the lungs into the blood.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
25. When you add sugar to your coffee or tea, the sugar is the A. permeable. B. solvent. C. gradient. D. solution. E. solute. The substance dissolved in a solution is the solute.
Bloom's Level: 1. Remember Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
26. Once a solute and a solvent are evenly distributed in a solution, they will A. stop moving about. B. move back toward a concentration of the solvent. C. continue to move about but with no net movement to higher concentration. D. be totally out of equilibrium. E. move from a liquid to gaseous solution. The molecules will continue to move about but there will be no net gain in any direction. The solution will retain an equal distribution.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
27. When blood supplies oxygen and food to a cell surface and these molecules diffuse across, we might expect some "resting time." However, if we inspected most cell membranes in a living human body, we would find a nearly continuous flow of substances back-and-forth across the plasma membrane. Which of the following is not a reason why? A. Cell metabolism is constantly using up oxygen and raw materials; this lowers their concentration so more will diffuse into the cell. B. Cell metabolism is constantly producing waste molecules; this increases their concentration so more will diffuse out of the cell. C. Cells produce useful secretions; this increases their concentration and more will diffuse out of the cell. D. Cell metabolism constantly produces carbon dioxide in respiration; this increases its concentration so more will diffuse out of the cell. E. Cells are living entities and in order to live can never “rest.” The concentrations of various molecules inside and outside of a cell are constantly changing and therefore, there is a continuous exchange of molecules across the cell membrane. Living entities are able to "rest."
Bloom's Level: 3. Apply Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
28. A semipermeable membrane sac filled with water and large molecules of starch is suspended in a beaker of distilled water. What will happen? A. The starch will leave and the water will enter until both sides reach equal concentrations. B. Water will enter the sac and it will swell. C. Because the starch cannot leave, the water cannot enter. D. We cannot determine the outcome unless we know the tonicity of the solutions. E. The starch will leave the sac and it will shrink. Since the starch is a large molecule, it cannot cross the membrane. However, the water will enter the sac due to osmosis (moving from a higher concentration outside of the sac to a lower concentration inside the sac). This will cause the sac to swell.
Bloom's Level: 3. Apply Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
29. If 0.9% NaCl were isotonic to a cell, then A. 0.9% would also be hypotonic. B. 0.9% would also be hypertonic. C. 1.0% would be hypertonic. D. 1.0% would be hypotonic. E. 0.1% would be hypertonic. Since the 1.0% solution contains more NaCl than the 0.9% solution, it is hypertonic. Hypertonic solutions contain a higher concentration of solute than the cell.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
30. Osmosis can occur when a membrane is A. impermeable. B. semipermeable. C. permeable. D. plasmolyzed. E. absent. Osmosis is the diffusion of water across a semi- or differentially permeable membrane. Water can pass through the membrane but other things cannot.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
31. The definition of osmosis is the A. the movement of solute molecules from a higher concentration to a lower concentration. B. diffusion of water across a selectively permeable membrane. C. likelihood that water will diffuse in a particular direction. D. lower concentration of a solute in a solution. E. movement of water toward a higher water concentration. Osmosis is the diffusion of water across a selectively permeable membrane. Diffusion is the movement of solute molecules from a higher to a lower concentration.
Bloom's Level: 1. Remember Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Plasma Membrane
32. The pressure that is generated when water flows through a differentially permeable membrane is termed A. osmotic pressure. B. turgor pressure. C. plasmolysis. D. crenation. E. tonicity. Osmotic pressure is the pressure that develops in a system due to osmosis. Turgor pressure results from the swelling of a plant cell in a hypotonic solution. Plasmolysis is the shrinking of the cytoplasm due to osmosis. Crenation refers to the shrinking of an animal cell in a hypertonic environment. Tonicity refers to the osmotic pressure or tension of the solution.
Bloom's Level: 1. Remember Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
33. Kidney dialysis relies on diffusion of various dissolved waste molecules (solutes) across a non-living semipermeable membrane. If the concentration of solutes in the blood increases over time before dialysis is administered, what will happen to the rate of dialysis when it occurs? A. It will remain the same since there are no carrier molecules in a non-living membrane. B. The rate will slow down since the solute is more viscous. C. The rate will speed up since the concentration gradient is higher. D. It will remain the same since there is no active transport in a non-living membrane. E. The rate will depend on the amount of time given for diffusion to take place. If the concentration difference is greater, the rate of transfer will also be greater. Carrier molecules and active transport are not needed for osmosis. The solution may be more viscous, but that will not slow down the rate. Rate is amount transferred per time, so it is not affected by increasing times.
Bloom's Level: 3. Apply Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
34. An isotonic solution means that the solute concentration outside the cell A. is greater than inside the cell. B. is less than inside the cell. C. is the same as inside the cell. D. has no effect on the cell. E. is greater than outside the cell. Iso- means "the same as," so in an isotonic solution the solute and water concentration both inside and outside the cell are equal.
Bloom's Level: 1. Remember Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
35. When you cut into most active plant tissue, water appears almost immediately because plant cells contain A. a hypertonic solution that produces turgor pressure. B. a hypotonic solution that produces turgor pressure. C. an isotonic condition and you cut the cell open. D. a hypertonic condition and cutting it reversed this to hypotonic. E. a hypotonic condition and cutting it reversed this to hypertonic. Plant cells are normally in a hypotonic solution and turgor pressure is the swelling of a plant cell in a hypotonic solution. When the cell is cut open, the pressure forces the water out.
Bloom's Level: 3. Apply Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
36. A cell placed in a hypotonic solution will A. lose water. B. gain water. C. neither gain nor lose water. D. lose water initially and then gain water. E. gain water initially and then lose water. A cell in a hypotonic solution will gain water because there is a lower concentration of solution (higher concentration of water) outside than inside the cell.
Bloom's Level: 1. Remember Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
37. The term hypertonic means A. to lose water. B. to gain water. C. a higher solute concentration. D. a lower solute concentration. E. an equal solute concentration. Hyper- means "more than" and refers to a solution with a higher percentage of solute (lower concentration of water) than the cell.
Bloom's Level: 1. Remember Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
38. If you have a 10% sugar solution and a 35% sugar solution, how does the 10% solution compare to the 35% solution? A. isotonic B. sweeter C. osmotic D. hypotonic E. hypertonic The 10% solution has less solute than the 35% solution, so it would be considered hypotonic to the 35% solution. It would not be sweeter than the 35% solution because it contains less sugar.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
39. Which is a correct example of tonicity? A. Water is hypertonic to red blood cells. B. Turgor pressure is created when a plant cell swells in a hypotonic solution. C. Plasmolysis results from plant cells in hypotonic solutions. D. Crenated red blood cells result when they are placed in a hypotonic solution. E. If a cell is placed in a hypertonic solution, water enters the cell. The definition of turgor pressure is correct. Water is hypotonic to red blood cells. Plasmolysis results from plant cells in hypertonic solutions. Crenated red blood cells result when placed in a hypertonic solution. If cells are placed in a hypertonic solution, water leaves the cell.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
40. When water enters a plant cell A. it bursts. B. the vacuole becomes enlarged. C. the vacuole shrinks. D. it undergoes plasmolysis. E. it undergoes crenation. The water will enter the plant vacuole since it is a hypotonic solution. This causes the vacuole to enlarge until it pushes against the cell wall, creating turgor pressure. The cell wall prevents the cell from bursting. Plasmolysis (plants) and crenation (animals) occur in hypertonic solutions.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
41. When a plant cell is placed in a hypotonic solution, the cell wall prevents A. diffusion. B. active transport. C. the cell from bursting. D. the cell from shrinking. E. water from entering the cell. The cell will swell until the vacuole pushes against the cell wall, creating turgor pressure. This prevents the cell from bursting. The water enters the cell until this point is reached. Active transport is not involved in osmosis. Diffusion does not require a membrane.
Bloom's Level: 3. Apply Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
42. A differentially permeable membrane separates a molasses solution from distilled water. Over time, the A. molasses will be found in the water. B. molasses will become more dilute. C. molasses solution will become more concentrated. D. molasses will be found in water and it will be more concentrated. E. solutions will remain the same. The molasses cannot cross the membrane, but the water will enter the membrane, diluting the molasses solution.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
43. Dead plants seen alongside a salted roadway died because the salt solution caused the cells to A. undergo crenation. B. undergo lysis. C. undergo hemolysis. D. undergo plasmolysis. E. the saline solution did not have an effect on the plants. The salt solution is hypertonic to the plant cells and causes them to lose water and undergo plasmolysis. Crenation occurs in animal cells. Lysis occurs when a cell takes in water and bursts. Hemolysis occurs when red blood cells lyse.
Bloom's Level: 3. Apply Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
44. In the emergency room, saline solutions are often run into a person's vein. The saline solution must be A. 0.1% NaCl. B. 0.5% NaCl. C. 0.75% NaCl. D. 0.9% NaCl. E. 9.0% NaCl. The IV saline solution must be isotonic to red blood cells, which is 0.9% NaCl.
Bloom's Level: 1. Remember Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
45. How can marine animals such as sharks live in salt water? A. They increase or decrease the amount of urea in their blood until the blood is isotonic with the environment. B. They increase the amount of urea in their blood until the blood is hypertonic with the environment. C. They decrease the amount of urea in their blood until the blood is hypotonic with the environment. D. They contain impermeable membranes which do not allow the salt water to cross into their cells. E. They block the movement of salt into their cells and maintain a constant blood salt concentration. Sharks increase or decrease the urea in their blood until it is isotonic with the salt water. Their cell membranes are permeable to the water and only by making their blood isotonic can they prevent their cells from shrinking.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
46. What will happen to the protein solution of side A in Figure 4.1? A. It will become less concentrated since protein will move from A to B. B. It will become more concentrated since water passes from B to A. C. It will become more concentrated since water passes from A to B. D. It will become less concentrated since water passes from B to A. E. It will become more concentrated since protein will move from B to A. The protein cannot cross the membrane, so only the water moves. The water will move from the higher concentration (side B) to the lower concentration (side A). This will dilute the protein on side A.
Bloom's Level: 4. Analyze Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
47. What will happen to the water level in Figure 4.1? A. It will rise in side A. B. It will rise in side B. C. It will remain the same on both sides. D. It will rise on both sides of the membrane. E. It will decrease on both sides of the membrane. As the water passes from side B to side A, the level on side A will rise.
Bloom's Level: 4. Analyze Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
48. Molecules such as glucose and amino acids are NOT lipid soluble and therefore they A. easily pass across the cell membrane. B. require active transport to cross the cell membrane. C. must be converted to lipids before they can enter a cell. D. combine with carrier proteins and pass across by facilitated transport. E. must be engulfed by a cell using endocytosis. Substances like glucose and amino acids cross the membrane with the help of carrier proteins via facilitated transport. They move from a region of higher to lower concentration. Active transport would be required if they were moving against the concentration gradient. Glucose and amino acids are not converted to lipids before they enter and are not engulfed via endocytosis.
Bloom's Level: 2. Understand Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
49. Which is true of facilitated transport by carrier proteins? A. Facilitated transport only applies to small and lipid soluble molecules. B. It is represented by the glucose carrier that can transport hundreds of molecules a second. C. After a carrier has transported a molecule, it is unable to transport any more. D. Facilitated transport requires expenditure of chemical energy and is therefore active transport. E. One carrier protein can carry a variety of different molecules. The glucose carrier is an example of facilitated transport, which does not require an expenditure of chemical energy. Small, lipid soluble molecules can cross a membrane by diffusion. The carrier protein is specific for the molecule that it transports and can be used over and over again.
Bloom's Level: 1. Remember Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Passive Transport
4-26 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 04 - Membrane Structure and Function
50. Carrier molecules are required for A. osmosis. B. both osmosis AND diffusion. C. facilitated diffusion. D. active transport. E. both facilitated diffusion AND active transport. A carrier molecule is required for both facilitated diffusion and active transport. Osmosis and diffusion do not require any carrier molecules.
Bloom's Level: 2. Understand Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
51. Which membrane transport process can continue whether the cell is alive or dead? A. sodium/potassium pump B. pinocytosis C. phagocytosis D. exocytosis E. diffusion Diffusion is the only process that does not require energy, so it would continue in a nonliving cell.
Bloom's Level: 2. Understand Learning Outcome: 04.02.02 Predict the movement of molecules in diffusion and osmosis. Section: 04.02 Topic: Passive Transport
4-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 04 - Membrane Structure and Function
52. The number of mitochondria in a cell would be a general indicator of the extent of A. diffusion. B. facilitated transport. C. active transport. D. osmosis. E. both osmosis and diffusion. Of the choices, only active transport requires energy. A high number of energy-producing mitochondria would suggest a cell that requires a lot of energy.
Bloom's Level: 2. Understand Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
53. Which of the following is not true regarding active transport? A. moves molecules or ions against their concentration gradient B. involves protein pumps C. is associated with large numbers of mitochondria D. is associated with nerve and muscle cells E. cannot be done by animal cells All of the statements are true except "cannot be done by animal cells." Animal cells are capable of active transport.
Bloom's Level: 2. Understand Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
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Chapter 04 - Membrane Structure and Function
54. If a cell lacks ATP, which of the following processes would cease to operate immediately? A. diffusion B. sodium/potassium pump C. facilitated diffusion D. osmosis E. tonicity The sodium/potassium pump is a form of active transport which requires energy. Without ATP, it would cease to function. All of the other processes do not require energy.
Bloom's Level: 2. Understand Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
55. When a substance moves from an area of low concentration to an area of high concentration while using energy, the process is termed A. diffusion. B. osmosis. C. facilitated diffusion. D. pinocytosis. E. active transport. Active transport requires energy to move substances against their concentration gradient, from an area of low concentration to high concentration.
Bloom's Level: 1. Remember Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
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Chapter 04 - Membrane Structure and Function
56. In the Malpighian tubules of an insect (blind, threadlike excretory tubule attached to the gut of an insect), salt molecules are actively transferred from body fluids to the inside of the tubule. Which of the following is NOT true? A. Water is also "pulled" by osmosis into the Malpighian tubule. B. The concentration of salt is normally higher in the body fluids than it is inside the tubule. C. The process decreases the salt concentration of the body fluids. D. Insects must excrete excess salt through the Malpighian tubules. E. The tubules have the same tonicity as the body fluids. The body fluid would contain more salt than the Malpighian tubules so it would not have the same tonicity as the tubules. The remaining statements would be true.
Bloom's Level: 5. Evaluate Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
57. A cell is placed in a solution of large nutrient molecules tagged with a red dye. Soon the cell is dark red, showing a concentration of the nutrient much higher than the external solution. We add a reagent that blocks the use of ATP. What result would you expect from this experiment? A. The nutrient would continue to rapidly enter the cell by diffusion because as a nutrient it is constantly being used in cell metabolism, so the cell will get redder. B. The color will remain the same since all future transfer will stop. C. The color will fade as the import of the nutrient stops and diffustion evens the concentrations as it moves the nutrient molecules out of the cell. D. The cell will continue to get darker since the import of the nutrient does not involve ATP. E. The cell will die without access to ATP. Because of its large size, the red dye is being actively transported across the membrane into the cell. If ATP usage is blocked, active transport will stop, and because the membrane is impermeable to the dye, the color will remain the same.
Bloom's Level: 4. Analyze Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
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Chapter 04 - Membrane Structure and Function
58. Macrophages, a type of white blood cells, are able to remove bacteria from our bloodstream and tissues by A. passive transport. B. facilitated diffusion. C. osmosis. D. pinocytosis. E. phagocytosis. Phagocytosis is the process whereby a large substance, such as another cell, is taken into a cell by endocytosis. None of the other processes can move such a large substance.
Bloom's Level: 1. Remember Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
59. When an intestinal cell ingests substances inside very small vesicles that can only be seen with an electron microscope, this is A. pinocytosis. B. phagocytosis. C. exocytosis. D. active transport. E. diffusion. Pinocytosis occurs when vesicles form around a liquid or a very small particle. Phagocytosis involves very large particles. Exocytosis is secreting substances. Active transport and diffusion cannot be seen with a microscope.
Bloom's Level: 1. Remember Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
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Chapter 04 - Membrane Structure and Function
60. Which of the following refers to materials only leaving the cell? A. diffusion B. exocytosis C. endocytosis D. pinocytosis E. phagocytosis Exocytosis is the only process listed in which substances exit the cell.
Bloom's Level: 1. Remember Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
61. Receptor-mediated endocytosis involves all of the following EXCEPT A. receptor proteins to bind to specific molecules. B. a coated pit due to a layer of fibrous protein on the cytoplasmic side of the membrane. C. various outcomes from destruction of the vesicle to restoration of the surface configuration. D. the mechanism for regulating exchange between a mother and fetus. E. secretion of materials from a cell. Endocytosis involves bringing substances inside the cell, not secretion.
Bloom's Level: 2. Understand Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
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Chapter 04 - Membrane Structure and Function
True / False Questions 62. Exocytosis is carried out by proteins in the cell membrane. FALSE Exocytosis and endocytosis involve a vesicle fusing with or being released from the plasma membrane, so phospholipids are also involved in these two processes.
Bloom's Level: 1. Remember Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
63. If a molecule crosses a plasma membrane faster than it diffuses in water, then the process is likely to involve active transport. TRUE The only way a molecule can cross a membrane faster than diffusion is if it is actively being transported into the cell.
Bloom's Level: 1. Remember Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
64. Oxygen leaves the alveoli in the lungs and enters the capillaries by endocytosis. FALSE Oxygen is capable of diffusing from the alveoli into the blood so it does not require endocytosis.
Bloom's Level: 2. Understand Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Passive Transport
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Chapter 04 - Membrane Structure and Function
65. One way to determine whether something is being actively transported across a membrane is to compare its rate of transport with and without a chemical that blocks ATP production. TRUE Because active transport requires energy, blocking ATP production would stop active transport. If the rate of transport did not change, then active transport is not involved.
Bloom's Level: 2. Understand Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
66. If a particular molecule is brought into the cell by receptor-medicated endocytosis, then changing the receptors will change the molecule that is being transported. TRUE Receptor-medicated endocytosis is specific for the molecules that bind to the receptor, so changing the receptor will change the molecule being transported.
Bloom's Level: 3. Apply Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
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Chapter 04 - Membrane Structure and Function
Multiple Choice Questions 67. The extracellular matrix is responsible for which of the following features? A. All of the answer choices are performed by the extracellular matrix. B. The extracellular matrix helps the cell resist stretching. C. The extracellular matrix enables the cell to adhere to neighboring cells. D. The extracellular matrix plays a role in cell signaling. E. The extracellular matrix acts as a structural feature. The extracellular matrix does all of these features.
Bloom's Level: 2. Understand Learning Outcome: 04.03.01 Explain the role of the extracellular matrix in animal cells. Section: 04.03 Topic: Plasma Membrane
68. Which component of the extracellualr matrix is responsible for forming proteoglycans? A. amino sugars B. fibronectin C. collagen D. elastin E. integrin The amino sugars form multiple polysaccharides that attach to proteins forming proteoglycans. Collagen and elastin are structural proteins, fibronectin is an adhesive protein, and integrin is a protein within the plasma membrane, none of which form proteoglycans.
Bloom's Level: 1. Remember Learning Outcome: 04.03.01 Explain the role of the extracellular matrix in animal cells. Section: 04.03 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
69. What two components are commonly found in the extracellular matrix that help resist stretching and provide resilience? A. proteins and polysaccharides B. proteins and phospholipids C. polysaccharides and phospholipids D. amino acids and phospholipids E. nucleic acids and proteins Both proteins and polysaccharides are commonly found in the extracellular matrix that help resist stretching and provide resilience. Phospholipids are associated with the actual cell membrane. Amino acids are the building blocks of the proteins. Nucleic acids are the building blocks of DNA and RNA.
Bloom's Level: 1. Remember Learning Outcome: 04.03.01 Explain the role of the extracellular matrix in animal cells. Section: 04.03 Topic: Plasma Membrane
70. Which cell junctions are commonly found in areas that are subject to stretching, such as the skin? A. adhesion junctions B. gap junctions C. tight junctions D. plasmodesmata E. None of the answer choices is found between skin cells. Adhesion junctions are the most common type of intercellular junction between skin cells. Gap junctions are common among heart cells and smooth muscle. Tight junctions are common in the intestines. Plasmodesmata are found connecting plant cells.
Bloom's Level: 1. Remember Learning Outcome: 04.03.02 Compare the structure and function of adhesion, tight, and gap junctions. Section: 04.03 Topic: Cell Junctions
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Chapter 04 - Membrane Structure and Function
Short Answer Questions 71. List the three types of intercelluar junctions found in animal cells and indicate their function. Adhesion junctions: form a sturdy but flexible sheet of cells. Tight junctions: compress the cells together to form a barrier. Gap junctions: form a channel between two cells, allowing small molecules and ions to pass freely between them.
Bloom's Level: 6. Create Learning Outcome: 04.03.02 Compare the structure and function of adhesion, tight, and gap junctions. Section: 04.03 Topic: Cell Junctions
Multiple Choice Questions 72. Which intercellular junction allows for the rapid movement of small molecules or ions to flow from one animal cell to the next? A. gap junctions B. tight junctions C. adhesion junctions D. desmosomes E. plasmodesmata Gap junctions allow for the rapid movement of small molecules or ions to flow from one animal cell to the next. Tight junctions form a solid barrier. Adhesion junctions form a porous membrane that filters substances. Desmosomes are a type of adhesion junctions. Plasmodesmatas are found in plant cells.
Bloom's Level: 1. Remember Learning Outcome: 04.03.02 Compare the structure and function of adhesion, tight, and gap junctions. Section: 04.03 Topic: Cell Junctions
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Chapter 04 - Membrane Structure and Function
Short Answer Questions 73. Compare and contrast exocytosis and endocytosis. Both processes require the input of energy and are associated with the movement of macromolecules that are too large to fit through the cell membrane by diffusion or forms of active transport. Vesicle formation contains the macromolecules that are moving across the cell membrane. During exocytosis macromolecules are being pushed out of the cell while in endocytosis macromolecules are being moved into the cell.
Bloom's Level: 6. Create Learning Outcome: 04.02.03 Describe the role of proteins in the movement of molecules across a membrane. Section: 04.02 Topic: Active Transport
74. Explain the difference between peripheral proteins and integral proteins as they relate to the cell membrane. Peripheral proteins are associated with only one side of the cell membrane. The integral proteins span the cell membrane and can protrude from one or both sides of the membrane. Integral proteins can move laterally within the membrane.
Bloom's Level: 6. Create Learning Outcome: 04.01.02 Describe the diverse roles of proteins in membranes. Section: 04.01 Topic: Plasma Membrane
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Chapter 04 - Membrane Structure and Function
Multiple Choice Questions 75. If the proteins forming the cell junctions within the stomach were denatured, what is a likely consequence? A. The desmosome junctions would not hold together, causing the cells in the stomach to pull apart. B. The tight junctions would not hold together, causing the cells in the stomach to pull apart. C. The desmosome junctions would be bonded together more tightly, causing the cells in the stomach to form a solid barrier. D. The gap junctions would not hold together, causing the cells in the stomach to be unable to pass ions back and forth. E. The desmosome junctions would not hold together, causing the cells in the kidney to pull apart. Desmosomes are protein junctions found in the stomach forming a porous barrier that can stretch. If the proteins denatured the cells would pull apart and would not form solid barriers. Tight junctions and gap junctions are not found in the stomach. Tight junctions are found in the kidneys.
Bloom's Level: 5. Evaluate Learning Outcome: 04.03.02 Compare the structure and function of adhesion, tight, and gap junctions. Section: 04.03 Topic: Cell Junctions
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Chapter 04 - Membrane Structure and Function
76. Which of the following factors could cause the cell membrane to become less permeable? A. if the channel proteins were to denature and become inactive B. f the molecules trying to enter the cell were decreased in size C. if the hydrophilic head of the phospholipids were to become hydrophobic while the hydrophobic tails become hydrophilic D. if the size of the protein channel was to increase E. All of the answer choices would cause the cell membrane to become less permeable. If the channel proteins were to denature and become inactive this would prevent the movement of small molecules through the membrane decreasing its permeability. Decreasing the size of the molecules entering the cell as well as the size of the protein channels would increase the permeability. If the phospholipid heads and tails were to change position, the membrane would still retain the same permeability.
Bloom's Level: 5. Evaluate Learning Outcome: 04.02.01 Explain why a membrane is selectively permeable. Section: 04.02 Topic: Plasma Membrane
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Chapter 05 - Cell Division
Chapter 05 Cell Division
Multiple Choice Questions 1. The mitotic stage of cell division consists of A. DNA synthesis and DNA degradation. B. mitosis and cytokinesis. C. duplication and division. D. cell growth and cell death. E. meiosis and mitosis. Both mitosis and cytokinesis occur during the mitotic stage of cell division. Duplication occurs during the S phase. Cell death and meiosis are not parts of mitosis.
Bloom's Level: 1. Remember Learning Outcome: 05.01.02 Describe the stages of the cell cycle and what occurs in each stage. Section: 05.01 Topic: Cell Cycle
2. The cell cycle consists of A. mitosis and interphase. B. meiosis and interphase. C. prophase and interphase. D. metaphase and interphase. E. changes from a haploid to a diploid state. The cell cycle includes mitosis and interphase. Prophase and metaphase are parts of mitosis. The cell stays in the diploid state. Meiosis is not part of the cell cycle.
Bloom's Level: 1. Remember Learning Outcome: 05.01.02 Describe the stages of the cell cycle and what occurs in each stage. Section: 05.01 Topic: Cell Cycle
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Chapter 05 - Cell Division
3. Following a cut or scrape, which process repairs your skin? A. meiosis B. meiosis and mitosis C. mitosis D. mitosis and apoptosis E. apoptosis Mitosis is responsible for repair of injury. Meiosis occurs in the reproductive cells and apoptosis results in cell death.
Bloom's Level: 1. Remember Learning Outcome: 05.01.01 Distinguish between the two processes that change the number of cells in the body. Section: 05.01 Topic: Mitosis
4. Which is NOT a correct association? A. S stage-DNA synthesis B. M stage-mitosis and cytokinesis C. interphase-shortest stage of the cell cycle D. G1 stage-cell grows in size and cell organelles increase in number E. G2 stage-metabolic preparation for mitosis Interphase is the longest part of the cell cycle.
Bloom's Level: 1. Remember Learning Outcome: 05.01.02 Describe the stages of the cell cycle and what occurs in each stage. Section: 05.01 Topic: Cell Cycle
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Chapter 05 - Cell Division
5. When does apoptosis occur? A. during development B. during interphase C. during mitosis D. during meiosis E. during injury repair Apoptosis occurs normally during devleopment to remove unwanted tissue.
Bloom's Level: 1. Remember Learning Outcome: 05.01.01 Distinguish between the two processes that change the number of cells in the body. Section: 05.01 Topic: Apoptosis
True / False Questions 6. The only function of apoptosis is to destroy healthy cells when they become damaged or infected. FALSE Apoptosis is a normal part of the development process.
Bloom's Level: 2. Understand Learning Outcome: 05.01.01 Distinguish between the two processes that change the number of cells in the body. Section: 05.01 Topic: Apoptosis
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Chapter 05 - Cell Division
Multiple Choice Questions 7. Which of the following is the correct sequence for the cell cycle? A. S-M-G1-G2 B. S-M-G2-G1 C. S-G1-G1-M D. S-G2-M-G1 E. S-G1-M-G2 The order of the cell cycle is G1, S, G2, and M. Since it is a cycle, you can start at any point, but the order of the stages will be the same.
Bloom's Level: 2. Understand Learning Outcome: 05.01.02 Describe the stages of the cell cycle and what occurs in each stage. Section: 05.01 Topic: Cell Cycle
8. During interphase, A. the cell begins to die. B. the cell rounds up and detaches from its neighbors. C. two nuclei split. D. the cell is at rest. E. hereditary material duplicates itself. During the S phase of interphase, DNA duplicates itself. Interphase is a very active time. The nucleus splits during mitosis. A cell rounds up and begins to die in apoptosis.
Bloom's Level: 2. Understand Learning Outcome: 05.01.02 Describe the stages of the cell cycle and what occurs in each stage. Section: 05.01 Topic: Cell Cycle
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Chapter 05 - Cell Division
True / False Questions 9. Technically, mitosis refers to nuclear division and NOT cytoplasmic division. TRUE Mitosis is nuclear division while cytokinesis is cytoplasmic division.
Bloom's Level: 2. Understand Learning Outcome: 05.01.02 Describe the stages of the cell cycle and what occurs in each stage. Section: 05.01 Topic: Mitosis
Multiple Choice Questions 10. Upon examination, a cell is found to have twice as much DNA as the normal diploid state but is no longer in the process of replicating the DNA. All of the DNA is found within a single nucleus. Which stage of the cell cycle is this cell in? A. M phase B. S phase C. G1 phase D. G2 phase E. cytokinesis The cell has finished DNA replication (S phase) but has not started mitosis (M). Therefore, it must be in G2.
Bloom's Level: 3. Apply Learning Outcome: 05.01.02 Describe the stages of the cell cycle and what occurs in each stage. Section: 05.01 Topic: Cell Cycle
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Chapter 05 - Cell Division
11. Which of the following descriptions does not occur during apoptosis? A. cell rounds up B. cells lose contact with neighbors C. nucleus fragments D. DNA is duplicated E. plasma membrane develops blisters DNA is not duplicated during apoptosis. All of the other stages occur.
Bloom's Level: 2. Understand Learning Outcome: 05.01.01 Distinguish between the two processes that change the number of cells in the body. Section: 05.01 Topic: Apoptosis
12. During which stage of the cell cycle do the chromosomes duplicate? A. prophase B. anaphase C. interphase D. telophase E. mitosis DNA synthesis and the duplication of the chromosomes occurs during the S phase of interphase.
Bloom's Level: 2. Understand Learning Outcome: 05.01.02 Describe the stages of the cell cycle and what occurs in each stage. Section: 05.01 Topic: Cell Cycle
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Chapter 05 - Cell Division
13. Which of the following best characterizes apoptosis? A. Apoptosis is bad for an organism. B. The stages of apoptosis are different in each cell type. C. Apoptosis happens accidentally to healthy cells. D. Apoptosis plays a normal role in development and cancer prevention. E. Apoptosis results in our going from a fertilized egg to trillions of cells. Apoptosis plays a normal role in the body, so it is not always bad. It is highly regulated and does not happen accidentally. The stages of apoptosis are always the same. Cell division results in our going from a fertilized egg to trillions of cells.
Bloom's Level: 5. Evaluate Learning Outcome: 05.01.01 Distinguish between the two processes that change the number of cells in the body. Section: 05.01 Topic: Apoptosis
14. There are two sets of caspases operating during apoptosis. Which of the following statements does not correctly characterize the role of these caspases? A. One set of caspases are called initators because they receive a signal to activate the second set. B. One set of capsases are called executioners because they dismantle the cell. C. Caspases can be unleashed by either internal or external signals. D. Caspases are normally active during the entire cell cycle. E. Caspases are responsible for the typical stages a cell goes through during apoptosis. Caspases are normally held inactive by inhibitors inside the cell.
Bloom's Level: 4. Analyze Learning Outcome: 05.01.01 Distinguish between the two processes that change the number of cells in the body. Section: 05.01 Topic: Apoptosis
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Chapter 05 - Cell Division
15. Which of the following is not true concerning the checkpoints in the cell cycle? A. Mitosis stops if chromosomes are not properly aligned. B. Mitosis will not occur if DNA is damaged or not replicated. C. If DNA is damaged, apoptosis may occur. D. The first checkpoint is located in the S phase of interphase. E. Cyclins must be present at certain stages to advance to the next stage. The first checkpoint is located in the G1 phase of interphase.
Bloom's Level: 2. Understand Learning Outcome: 05.02.02 Describe the checkpoints for the cell cycle. Section: 05.02 Topic: Cell Cycle
16. Which of the following is not involved in the regulation of the cell cycle? A. cyclins B. internal and external signals C. growth factors D. checkpoints E. caspases Caspases are involved in apoptosis.
Bloom's Level: 2. Understand Learning Outcome: 05.02.01 Distinguish between internal and external controls of the cell cycle. Section: 05.02 Topic: Cell Cycle
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Chapter 05 - Cell Division
17. Growth factors that are transported through the blood play what role in the cell cycle? A. an internal signal that tells the cell to stop dividing and repair its DNA B. an internal signal that functions as an executioner C. an internal signal to tell the cell which stage of the cell cycle to enter next D. an external signal to tell the cell how long it should spend in each phase of the cell cycle E. an external signal to tell the cell whether or not to divide A growth factor would be an external signal. External signals tell a cell whether or not to divide.
Bloom's Level: 3. Apply Learning Outcome: 05.02.01 Distinguish between internal and external controls of the cell cycle. Section: 05.02 Topic: Mitosis
18. If a cell stops at the G1 checkpoint, this is most likely due to what problem? A. The DNA has not finished replicating. B. The chromosomes are not aligned properly. C. There is DNA damage. D. The cell is cancerous. E. There is no problem. The cell normally stops at the G1 checkpoint. The cell stops at the first checkpoint during G1 if there is DNA damage or if there is not enough building blocks available to proceed.
Bloom's Level: 2. Understand Learning Outcome: 05.02.02 Describe the checkpoints for the cell cycle. Section: 05.02 Topic: Cell Cycle
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Chapter 05 - Cell Division
19. Which of the following proteins would you expect to be a coded for by a protooncogene? A. a growth factor receptor B. a DNA repair protein C. a protein involved with the G1 checkpoint D. a caspase E. a DNA replication protein Proto-oncogenes code for proteins that stimulate the cell cycle. Therefore, a growth factor receptor would be a proto-oncogene. The other proteins are involved in apoptosis (caspase), stopping the cell cycle (G1 checkpoint protein), or replicating the DNA after the signal has been given (DNA repair and replication proteins).
Bloom's Level: 3. Apply Learning Outcome: 05.02.03 Differentiate between the role of proto-oncogenes and tumor suppressor genes in regulating the cell cycle. Section: 05.02 Topic: Cell Cycle
20. Which type of genes, when abnormally activated so that protein is always present and active, can result in cancer? A. tumor suppressor genes B. initator caspase genes C. proto-oncogenes D. DNA repair protein genes E. executioner caspase genes When proto-oncogenes are abnormally activated, they become oncogenes which cause cancer. If the other genes are abnormally activated, they would prevent cancer.
Bloom's Level: 3. Apply Learning Outcome: 05.02.03 Differentiate between the role of proto-oncogenes and tumor suppressor genes in regulating the cell cycle. Section: 05.02 Topic: Cancer
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Chapter 05 - Cell Division
21. The genes that code for the proteins of the signaling pathway could be classified as A. proto-oncogenes. B. caspase genes. C. DNA repair genes. D. tumor suppressor genes. E. cyclin genes. The signaling pathway promotes cell division. Therefore these genes would be considered proto-oncogenes.
Bloom's Level: 2. Understand Learning Outcome: 05.02.03 Differentiate between the role of proto-oncogenes and tumor suppressor genes in regulating the cell cycle. Section: 05.02 Topic: Cancer
22. Eukaryotic chromosomes are composed of A. DNA. B. protein. C. histones. D. chromatin. E. RNA. Chromosomes are composed of chromatin, a combination of both DNA and protein.
Bloom's Level: 1. Remember Learning Outcome: 05.03.01 Explain the role of mitosis and how it maintains the chromosome number of a cell. Section: 05.03 Topic: Chromosome Structure
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Chapter 05 - Cell Division
23. Which is NOT a correct association? A. cytokinesis-division of the cytoplasm B. centromere-point where sister chromatids remain attached C. haploid-one of each chromosome D. sister chromatids-two identical chromosome strands still attached at the centromere E. mitosis-when a cell duplicates and then divides twice to reduce chromosome number by half Mitosis is when a cell duplicates and then divides once, leaving the chromosome number the same.
Bloom's Level: 2. Understand Learning Outcome: 05.03.01 Explain the role of mitosis and how it maintains the chromosome number of a cell. Section: 05.03 Topic: Mitosis
24. Which of the following is not true of the chromosomes in a cell undergoing mitosis? A. They are highly compacted. B. They are available for RNA synthesis. C. They have histones bound to them. D. They are found within the cell nucleus. E. They are present in the diploid number. During mitosis, the chromosomes are highly compacted and are not available for RNA synthesis.
Bloom's Level: 2. Understand Learning Outcome: 05.03.01 Explain the role of mitosis and how it maintains the chromosome number of a cell. Section: 05.03 Topic: Chromosome Structure
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Chapter 05 - Cell Division
25. In humans, the diploid number of chromosomes is A. 20. B. 46. C. 23. D. variable depending on whether male or female. E. 92. Humans have 46 chromosomes. 2n = 46.
Bloom's Level: 1. Remember Learning Outcome: 05.03.01 Explain the role of mitosis and how it maintains the chromosome number of a cell. Section: 05.03 Topic: Chromosome Structure
True / False Questions 26. The length of DNA in each chromosome is longer than the diameter of the animal cell. TRUE A human cell contains at least 2 m of DNA, yet the nucleus is only about 5 um in diameter.
Bloom's Level: 2. Understand Learning Outcome: 05.03.01 Explain the role of mitosis and how it maintains the chromosome number of a cell. Section: 05.03 Topic: Chromosome Structure
27. Microtubules found in spindle fibers are capable of assembling and disassembling. TRUE Microtubules assemble and disassemble to move the chromosomes.
Bloom's Level: 2. Understand Learning Outcome: 05.03.01 Explain the role of mitosis and how it maintains the chromosome number of a cell. Section: 05.03 Topic: Mitosis
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Chapter 05 - Cell Division
28. The chromatids are held together at a region called the centromere. TRUE The two sister chromatids are held together by the centromere.
Bloom's Level: 1. Remember Learning Outcome: 05.03.02 Summarize the major events that occur during mitosis and cytokinesis. Section: 05.03 Topic: Chromosome Structure
29. The overall cell cycle is dramatically different for animals than for plant cells. FALSE There are only minor differences in the cell cycle between animals and plant cells.
Bloom's Level: 2. Understand Learning Outcome: 05.03.03 Compare and contrast mitosis and cytokinesis in plant and animal cells. Section: 05.03 Topic: Mitosis
30. Centrioles are necessary to the process of mitosis in all organisms. FALSE Because plants do not have centrioles, they cannot be absolutely necessary to the process of mitosis.
Bloom's Level: 2. Understand Learning Outcome: 05.03.03 Compare and contrast mitosis and cytokinesis in plant and animal cells. Section: 05.03 Topic: Mitosis
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Chapter 05 - Cell Division
31. During mitosis, separation of the sister chromatids ensures that each daughter cell will receive two copies of each type of chromosome. FALSE Each daughter cell only receives one copy of each type of chromosome.
Bloom's Level: 2. Understand Learning Outcome: 05.03.01 Explain the role of mitosis and how it maintains the chromosome number of a cell. Section: 05.03 Topic: Chromosome Structure
Multiple Choice Questions 32. Which of the following best describes the role of histones? A. Histones play a role in coding for hereditary features. B. Histones make RNA synthesis possible. C. Histones determine the structure of the chromosome. D. Histones are not thought to play any role in the cell. E. Histones determine the diploid number of chromosomes. Histones play a structural role in the chromosome.
Bloom's Level: 2. Understand Learning Outcome: 05.03.01 Explain the role of mitosis and how it maintains the chromosome number of a cell. Section: 05.03 Topic: Chromosome Structure
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Chapter 05 - Cell Division
True / False Questions 33. Chromosomes are attached to the spindle fibers in order to move to and from the equator. TRUE Spindle fibers made up of microtubules are responsible for the movement of the chromosomes during mitosis.
Bloom's Level: 2. Understand Learning Outcome: 05.03.02 Summarize the major events that occur during mitosis and cytokinesis. Section: 05.03 Topic: Chromosome Structure
Multiple Choice Questions 34. You are looking at chromosome 1 in a human being. Assuming there is no crossing over, what is the source of all the genes on this chromosome? A. mother B. father C. either the mother or father D. a combination of both mother and father E. it is impossible to tell given this information If there was no crossing over, all of the genes on a particular chromosome would originate with either the mother or the father.
Bloom's Level: 4. Analyze Learning Outcome: 05.04.01 Summarize the purpose of meiosis. Section: 05.04 Topic: Meiosis
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Chapter 05 - Cell Division
35. If the total number of chromosomes in a cell is six, then after mitosis there will be A. six chromosomes in each daughter cell. B. three chromosomes in each daughter cell. C. 12 chromosomes in each daughter cell. D. two chromosomes in each daughter cell. E. three chromosomes in one daughter cell and six chromosomes in the other cell. Mitosis results in daughter cells with the exact same number of chromosomes as in the mother cell.
Bloom's Level: 3. Apply Learning Outcome: 05.03.01 Explain the role of mitosis and how it maintains the chromosome number of a cell. Section: 05.03 Topic: Mitosis
36. If a crayfish has 200 total chromosomes in its body cells (not ovaries or testes), A. any 100 could have been from its father and any 100 from its mother. B. they would consist of 100 pairs with one of each pair from the father, one of each pair from the mother. C. as many as none to 200 came from the father and conversely, from 200 to none would have come from the mother. D. 50 pairs or 100 total would come from the father and 50 pairs from the mother. E. all 200 come from the mother in a female crayfish, all 200 from the father in a male crayfish. In the diploid organism, there are two of each type of chromosome (a pair), one from the mother and one from the father.
Bloom's Level: 3. Apply Learning Outcome: 05.04.01 Summarize the purpose of meiosis. Section: 05.04 Topic: Meiosis
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Chapter 05 - Cell Division
37. Which of the following are genetically identical? A. both members of a pair of chromosomes B. sister chromatids C. the mother's and the father's chromosomes D. the mother's and her child's chromosomes E. the father's and his child's chromosomes Sister chromatids are genetically identical. Both members of a pair of chromosomes and the mother's and father's chromosomes have the same genes in the same order, but they are not genetically identical. A child's chromosomes are not identical to either parent's chromosomes.
Bloom's Level: 3. Apply Learning Outcome: 05.04.01 Summarize the purpose of meiosis. Section: 05.04 Topic: Chromosome Structure
38. During what stage of mitosis does the nuclear envelope disappear and the chromosomes become distinct? A. interphase B. prophase C. metaphase D. anaphase E. telophase During prophase, the nuclear membrane breaks down and the chromosomes appear.
Bloom's Level: 1. Remember Learning Outcome: 05.03.02 Summarize the major events that occur during mitosis and cytokinesis. Section: 05.03 Topic: Mitosis
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Chapter 05 - Cell Division
39. Which sequence of stages in mitosis is correct? A. prophase, metaphase, interphase, telophase B. metaphase, anaphase, prophase, telophase C. anaphase, interphase, telophase, prophase D. prophase, metaphase, anaphase, telophase E. interphase, metaphase, prophase, anaphase The stages of mitosis, in the correct order, are prophase, metaphase, anaphase, and telophase.
Bloom's Level: 1. Remember Learning Outcome: 05.03.02 Summarize the major events that occur during mitosis and cytokinesis. Section: 05.03 Topic: Mitosis
40. In which stage of mitosis do the chromosomes line up in the center of the cell? A. interphase. B. prophase. C. metaphase. D. anaphase. E. telophase. The chromosomes line up on the metaphase plate (generally in the middle of the cell) during metaphase.
Bloom's Level: 1. Remember Learning Outcome: 05.03.02 Summarize the major events that occur during mitosis and cytokinesis. Section: 05.03 Topic: Mitosis
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Chapter 05 - Cell Division
41. During which phases of the cell cycle do the chromosomes look like Xs? A. prophase, anaphase B. anaphase, metaphase C. metaphase, telophase D. prophase, metaphase E. anaphase, telophase The X shape is due to the presence of both sister chromatids. Both are present during prophase and metaphase, but separate during anaphase and telophase.
Bloom's Level: 2. Understand Learning Outcome: 05.03.02 Summarize the major events that occur during mitosis and cytokinesis. Section: 05.03 Topic: Mitosis
42. The nuclear membrane reappears in mitosis during A. interphase. B. prophase. C. metaphase. D. anaphase. E. telophase. The nuclear membrane reappears during telophase.
Bloom's Level: 1. Remember Learning Outcome: 05.03.02 Summarize the major events that occur during mitosis and cytokinesis. Section: 05.03 Topic: Mitosis
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Chapter 05 - Cell Division
43. In plant cells, what is responsible for organizing the spindle? A. centrosomes B. centrioles C. microtubules D. asters E. centromeres The centrosomes organize the spindle in both animals and plants. Centrioles and asters are not present in plant cells.
Bloom's Level: 2. Understand Learning Outcome: 05.03.03 Compare and contrast mitosis and cytokinesis in plant and animal cells. Section: 05.03 Topic: Mitosis
44. How does cell division differ between animal and plant cells? A. Animal cells lack centrioles and no spindle forms during cell division. B. Plant cells form a cleavage furrow or indentation of membrane between new daughter cells. C. Plant cells resort to binary fission. D. The cell plate is the final partitioning of plant cells. E. There is no difference. Plant cells and animal cells undergo the same cellular processes during mitosis. Plants form a cell plate instead of a cleavage furrow during telophase. Plant cells lack centrioles. Bacteria undergo binary fission.
Bloom's Level: 4. Analyze Learning Outcome: 05.03.03 Compare and contrast mitosis and cytokinesis in plant and animal cells. Section: 05.03 Topic: Cell Cycle
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Chapter 05 - Cell Division
45. You are examining a cell under the microscope that is undergoing cell division and cannot find any asters. What else would not be present in this cell? A. a cell plate B. centrosomes C. a spindle D. a cleavage furrow E. chromosomes This must be a plant cell, so it would also not have a cleavage furrow.
Bloom's Level: 2. Understand Learning Outcome: 05.03.03 Compare and contrast mitosis and cytokinesis in plant and animal cells. Section: 05.03 Topic: Mitosis
46. What forms the contractile ring in animal cells? A. membrane vesicles B. actin filaments C. cellulose fibrils D. cell wall E. histones The contractile ring is formed of actin filaments.
Bloom's Level: 1. Remember Learning Outcome: 05.03.03 Compare and contrast mitosis and cytokinesis in plant and animal cells. Section: 05.03 Topic: Mitosis
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Chapter 05 - Cell Division
47. Meiosis results in a change in chromosome number indicated by A. 2n to 2n. B. 2n to n. C. n to 2n. D. n to n. E. 2n to 2n in diploid organisms, n to n in haploid. Meiosis reduces the chromosome number by half, 2n to n.
Bloom's Level: 2. Understand Learning Outcome: 05.04.01 Summarize the purpose of meiosis. Section: 05.04 Topic: Meiosis
48. If the diploid chromosome number is 16, the chromosome number of each gamete will be A. 4. B. 8. C. 12. D. 16. E. 32. A gamete results from meiosis, so the chromosome number would be divided in half.
Bloom's Level: 2. Understand Learning Outcome: 05.04.01 Summarize the purpose of meiosis. Section: 05.04 Topic: Meiosis
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Chapter 05 - Cell Division
49. An organism has 8 pairs of chromosomes. If the sperm and egg did not undergo meiosis, how many chromosomes would be present in a cell that resulted from the fusion of a sperm and egg? A. 8 B. 16 C. 24 D. 32 E. 64 Eight pairs of chromosomes are 16 chromosomes. Fusion without meiosis would result in the next generation having 32 chromosomes.
Bloom's Level: 2. Understand Learning Outcome: 05.04.01 Summarize the purpose of meiosis. Section: 05.04 Topic: Meiosis
50. We presume that meiosis evolved later than mitosis. What process would not have to evolve in cells undergoing meiosis in order to make this transition? A. Homologous chromosomes would have to pair up. B. An additional division would have to occur, but without a second DNA replication. C. Sister chromosomes do not separate at first anaphase; centromeres hold until second division. D. a method of lining up single chromosomes along the metaphase plate E. fertilization processes to restore the diploid number in an organism The chromosomes line up along the metaphase plate in both meiosis and mitosis. The other steps are unique to meiosis.
Bloom's Level: 4. Analyze Learning Outcome: 05.04.01 Summarize the purpose of meiosis. Section: 05.04 Topic: Meiosis
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Chapter 05 - Cell Division
51. Synapsis occurs during what stage of meiosis? A. anaphase I B. telophase II C. metaphase II D. prophase I E. anaphase II Synapsis occurs during prophase I.
Bloom's Level: 1. Remember Learning Outcome: 05.04.03 Describe the events of meiosis. Section: 05.04 Topic: Meiosis
52. Without considering the variation provided by crossing-over, how much will two siblings from one set of parents vary from each other? A. All siblings from the same parents are identical to each other and demonstrate traits that are exactly halfway between their parents' traits. B. Siblings receive 100% of one parent's genes or the other parent's genes so they will either be identical if they receive the same parent's genes, or nothing alike if they receive different parent's genes. C. Siblings inherit essentially 50% of their genes from each parent, but two sibling offspring may share with each other from zero to 23 chromosomes in common from each parent and therefore vary widely from each other. D. Siblings inherit copies of the same 23 pair of chromosomes from each parent and only crossing-over provides any differences between siblings. E. There is one chance in 23 of getting identical sets of chromosomes from one parent, times two because there are two parents; therefore two siblings out of 46 will be identical except for the extent of crossing-over. Siblings each receive half of their chromosomes from each parent. Which half they receive is random, so siblings can share from zero to 23 chromsomes in common from each parent.
Bloom's Level: 4. Analyze Learning Outcome: 05.04.01 Summarize the purpose of meiosis. Section: 05.04 Topic: Meiosis
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Chapter 05 - Cell Division
True / False Questions 53. Homologous chromosomes are genetically identical to each other. FALSE Homologous chromosomes have the same genes in the same order, but they are not genetically identical.
Bloom's Level: 2. Understand Learning Outcome: 05.04.02 Explain what is meant by the term homologous chromosomes. Section: 05.04 Topic: Homologous Chromosomes
Multiple Choice Questions 54. During crossing-over, A. chromosomes switch poles. B. mitosis becomes meiosis. C. chromatin becomes chromosomes. D. chromosomes become chromatin. E. chromatids exchange segments of genetic material. Crossing-over results in the exchange of genetic material between chromatids.
Bloom's Level: 1. Remember Learning Outcome: 05.04.03 Describe the events of meiosis. Section: 05.04 Topic: Meiosis
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Chapter 05 - Cell Division
55. Would it be possible to observe crossing-over between sister chromatids? Why or why not? A. No. Crossing-over is not observable. B. No. Sister chromatids are not present within the same nucleus. C. No, since crossing-over does not occur between sister chromatids. D. Yes. Sister chromatids look different from each other. E. Yes. One sister chromatid comes from mother and one comes from father. Since sister chromatids are genetically identical, you cannot observe crossing-over between them unless you label each chromatid in some manner.
Bloom's Level: 3. Apply Learning Outcome: 05.04.03 Describe the events of meiosis. Section: 05.04 Topic: Meiosis
56. During what stage do homologous chromosomes separate from each other? A. prophase B. anaphase I C. anaphase II D. metaphase I E. metaphase II Homologous chromsomes separate during anaphase I.
Bloom's Level: 1. Remember Learning Outcome: 05.04.02 Explain what is meant by the term homologous chromosomes. Section: 05.04 Topic: Homologous Chromosomes
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Chapter 05 - Cell Division
57. During what stage of meiosis do sister chromatids separate from each other? A. prophase B. anaphase I C. anaphase II D. metaphase I E. metaphase II Sister chromatids separate from each other in anaphase II.
Bloom's Level: 1. Remember Learning Outcome: 05.04.03 Describe the events of meiosis. Section: 05.04 Topic: Meiosis
58. You are examining a cell under the microscope and see what appear to be four sister chromatids bound together and remnants of the nuclear envelope. What stage of meiosis are you looking at? A. prophase I B. prophase II C. metaphase I D. metaphase II E. telophase I The two homologous chromosomes, composed of four sister chromatids, are only together during prophase I. The remnants of the nuclear envelope are also seen during prophase I.
Bloom's Level: 3. Apply Learning Outcome: 05.04.03 Describe the events of meiosis. Section: 05.04 Topic: Meiosis
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Chapter 05 - Cell Division
59. You are looking at a cell under the microscope, and see what appears to be several "X" structures being pulled to each side of the cell. What stage of meiosis are you looking at? A. prophase I B. prophase II C. metaphase I D. anaphase I E. anaphase II The X structure is composed of two sister chromatids. Therefore, you are looking at homologous chromosomes separating. Therefore, this is anaphase I.
Bloom's Level: 1. Remember Learning Outcome: 05.04.03 Describe the events of meiosis. Section: 05.04 Topic: Meiosis
True / False Questions 60. Replication of the DNA for meiosis occurs between telophase I and prophase II. FALSE Replication of the DNA for meiosis occurs prior to prophase I.
Bloom's Level: 1. Remember Learning Outcome: 05.04.03 Describe the events of meiosis. Section: 05.04 Topic: Meiosis
61. Crossing-over is the only factor that is responsible for the variation that exists between offspring of one set of parents. FALSE Both crossing-over and independent assortment are both responsible for the variation between offspring.
Bloom's Level: 2. Understand Learning Outcome: 05.04.01 Summarize the purpose of meiosis. Section: 05.04 Topic: Meiosis
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Chapter 05 - Cell Division
Multiple Choice Questions 62. The second meiotic division is essentially a mitotic division except for the fact that the cells produced are A. haploid. B. diploid. C. polyploid. D. autosomal. E. somatic. The cells are haploid following meiosis.
Bloom's Level: 1. Remember Learning Outcome: 05.05.02 Identify the differences in the behavior of homologous chromosomes in meiosis and mitosis. Section: 05.05 Topic: Meiosis Versus Mitosis
True / False Questions 63. Mitotic division is the normal process of cell reproduction to build and maintain the body of an organism, while meiosis takes place only in the formation of gametes for reproduction. TRUE This correctly describes the role of meiosis and mitosis.
Bloom's Level: 2. Understand Learning Outcome: 05.05.01 Compare and contrast the processes of meiosis and mitosis. Section: 05.05 Topic: Meiosis Versus Mitosis
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Chapter 05 - Cell Division
Multiple Choice Questions 64. What lines up at the metaphase plate during both metaphase I of meiosis and metaphase of mitosis? A. each chromosome is composed of two sister chromatids at both B. homologous chromosomes at both C. each chromosome composed of two sister chromatids for meiosis, each chromosome is composed of one sister chromatid for mitosis D. homologous chromosomes for meiosis, each chromosome is composed of two sister chromatids for mitosis E. each chromosome is composed of two sister chromatids for meiosis, homologous chromosomes for mitosis Homologous chromsomes line up at the metaphase plate during metaphase I of meiosis. Each chromosome is composed of two sister chromatids that line up at the metaphase plate of metaphase during mitosis.
Bloom's Level: 2. Understand Learning Outcome: 05.05.01 Compare and contrast the processes of meiosis and mitosis. Section: 05.05 Topic: Meiosis Versus Mitosis
65. What lines up at the metaphase plate during metaphase II of meiosis and metaphase of mitosis? A. each chromosome composed of two sister chromatids at both B. homologous chromosomes at both C. each chromosome composed of two sister chromatids for meiosis, each chromosome composed of one sister chromatid for mitosis D. homologous chromosomes for meiosis, each chromosome composed of two sister chromatids for mitosis E. each chromosome composed of two sister chromatids for meiosis, homologous chromosomes for mitosis Each chromosome composed of two sister chromatids lines up at the metaphase plate during both metaphase II of meiosis and metaphase of mitosis.
Bloom's Level: 2. Understand Learning Outcome: 05.05.02 Identify the differences in the behavior of homologous chromosomes in meiosis and mitosis. Section: 05.05 Topic: Meiosis Versus Mitosis
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Chapter 05 - Cell Division
66. The formation of sperm gametes is termed A. oogenesis. B. homologous formation. C. spermatogenesis. D. meiosis. E. mitosis. Spermatogenesis is the formation of sperm in human males.
Bloom's Level: 1. Remember Learning Outcome: 05.06.02 Explain the process of gamete production in both males and females. Section: 05.06 Topic: Gametogenesis
67. A basic difference between spermatogenesis and oogenesis is that in oogenesis A. four functional eggs are produced. B. mitosis instead of meiosis occurs. C. both sperm and egg are produced. D. one functional egg is produced. E. the chromosome number is not divided in half. Only one functional, haploid egg is produced via meiosis in oogenesis.
Bloom's Level: 3. Apply Learning Outcome: 05.06.02 Explain the process of gamete production in both males and females. Section: 05.06 Topic: Gametogenesis
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Chapter 05 - Cell Division
68. Which of the following statements is NOT true about oogenesis in humans? A. It occurs in the ovary. B. The egg will contain 23 chromosomes. C. Four equal size daughter cells will form. D. At least two nonfunctional polar bodies will form. E. Once started, it will not necessarily go to completion. One egg and two or three much smaller polar bodies are formed during oogenesis. The resulting cells are not all of the same size.
Bloom's Level: 2. Understand Learning Outcome: 05.06.02 Explain the process of gamete production in both males and females. Section: 05.06 Topic: Gametogenesis
69. Which of the following is not a reason why two children born to the same parents (not identical twins) are unlikely to have the same genetic makeup? A. Crossing-over recombines portions of sister chromatids. B. Fertilization provides a new pairing of chromosomes. C. Each sperm or egg has a random assortment of chromosomes, half from each pair. D. Chromosomes align themselves independently during meiosis I. E. Since they are produced at different times, all sperm will be genetically different from all other sperm. Sperm are not all genetically identical due to independent assortment and crossing-over during meiosis, not because they are produced at different times.
Bloom's Level: 5. Evaluate Learning Outcome: 05.06.02 Explain the process of gamete production in both males and females. Section: 05.06 Topic: Gametogenesis
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Chapter 05 - Cell Division
True / False Questions 70. It is the secondary oocyte and not the egg that the sperm fertilizes to form a diploid organism. TRUE If the secondary oocyte is fertilized, it completes meiosis II to form an egg and a second polar body. The sperm combines with the egg and forms a diploid organism.
Bloom's Level: 4. Analyze Learning Outcome: 05.06.01 Describe the human life cycle in terms of haploid and diploid cells. Section: 05.06 Topic: Gametogenesis
Multiple Choice Questions 71. What is the difference between the chromosomes in a secondary spermatocyte and those in a spermatid? A. the chromosomes in secondary spermatocytes consist of two chromatids, the ones in the spermatid consist of only one chromatid B. the chromosomes in a secondary spermatocyte consist of homologous chromosomes, the ones in the spermatid consist of two chromatids C. the chromosomes in the secondary spermatocyte consist of one chromatid, the ones in the spermatid consist of two chromatids D. the secondary spermatocyte is diploid, the spermatid is haploid E. the secondary spermatocyte is haploid, the spermatid is diploid Both cells are haploid, but the secondary spermatocyte has not completed meiosis II and so the chromosomes consist of two chromatids. The spermatid has completed meiosis and so the chromosomes consist of only one chromatid.
Bloom's Level: 3. Apply Learning Outcome: 05.06.02 Explain the process of gamete production in both males and females. Section: 05.06 Topic: Gametogenesis
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Chapter 05 - Cell Division
True / False Questions 72. The only haploid stage in the animal life cycle is the gamete. TRUE Only the sperm and egg are haploid in animals.
Bloom's Level: 1. Remember Learning Outcome: 05.06.01 Describe the human life cycle in terms of haploid and diploid cells. Section: 05.06 Topic: Gametogenesis
Short Answer Questions 73. Indicate the portions of the cell cycle that contain checkpoints and what the checkpoint controls. G1 contains the first checkpoint. This one determines if conditions are favorable to begin the cell cycle. G2 also contains a checkpoint that determines if the DNA replication is proceeding correctly. Metaphase contains a checkpoint to determine if the chromosomes are aligned correctly.
Bloom's Level: 6. Create Learning Outcome: 05.02.02 Describe the checkpoints for the cell cycle. Section: 05.02 Topic: Cell Cycle
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Chapter 05 - Cell Division
Multiple Choice Questions 74. Which of the following events would be controlled by an external signal? A. cell division B. Prophase proceeds into metaphase. C. The chromosomes align at the equatorial plane before the cell begins anaphase. D. All of the DNA is copied before the cell enters into the G2 phase. E. All of the answer choices are controlled by an external signal. External signals tell the cell whether or not to divide. Cell division is the only one that would be regulated by external signals. Internal signals ensure that the stages follow one another in the normal sequence and that each stage is properly completed before the next stage begins.
Bloom's Level: 2. Understand Learning Outcome: 05.02.01 Distinguish between internal and external controls of the cell cycle. Section: 05.02 Topic: Cell Cycle
Short Answer Questions 75. Describe the checkpoints for the cell cycle. The first checkpoint during G1 allows the cell to determine if the conditions are favorable to begin the cell cycle. The second checkpoint is in G2 in which the cell evaluates the quality of the DNA. If it is damaged the cell will initiate repair or apoptosis. The third checkpoint occurs during metaphase during mitosis. The cell evaluates if the chromosomes are lined up evenly across the equatorial plane.
Bloom's Level: 6. Create Learning Outcome: 05.02.02 Describe the checkpoints for the cell cycle. Section: 05.02 Topic: Cell Cycle
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Chapter 05 - Cell Division
76. List in order the stages of the cell cycle. Interphase - prophase - metaphase - anaphase - telophase
Bloom's Level: 6. Create Learning Outcome: 05.01.02 Describe the stages of the cell cycle and what occurs in each stage. Section: 05.01 Topic: Cell Cycle
Multiple Choice Questions 77. What is a potential outcome if one of your father's gametes did not go through meiosis correctly and one of the chromosomes did not separate correctly? A. You inherited a normal amount of genetic material from your mother but inherited an extra chromosome from your father. One of your pairs of chromosomes will not be homologous. B. You inherited an abnormal amount of genetic material from both your mother and father. One of your pairs of chromosomes will not be homologous. C. You inherited a normal amount of genetic material from your mother but inherited an extra chromosome from your father. Two of your pairs of chromosomes will not be homologous. D. You inherited a normal amount of genetic material from both your mother and father resulting in the inheritance of a normal amount of genetic material. E. You inherited multiple genetic abnormalities, due to the faulty chromosomes from your father, resulting in a significant number of health issues. If one of your father's gametes did not go through meiosis correctly and one of the chromosomes did not separate correctly, one of his sperm would carry an extra chromosome. If this sperm fertilized your mother's egg, you would have inherited an extra chromosome resulting in one of your pairs of chromosomes not being homologous.
Bloom's Level: 5. Evaluate Learning Outcome: 05.04.02 Explain what is meant by the term homologous chromosomes. Section: 05.04 Topic: Homologous Chromosomes
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Chapter 05 - Cell Division
78. During which phase will homologous chromosomes separate? A. anaphase I B. anaphase II C. prophase D. anaphase E. metaphase I Homologous chromosomes will separate during anaphase I of meiosis I. The sister chromatids will separate during anaphase II and anaphase. Chromatin condenses into chromosomes during prophase. Homologous chromosomes align during metaphase I.
Bloom's Level: 2. Understand Learning Outcome: 05.05.02 Identify the differences in the behavior of homologous chromosomes in meiosis and mitosis. Section: 05.05 Topic: Homologous Chromosomes
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Chapter 06 - Metabolism: Energy and Enzymes
Chapter 06 Metabolism: Energy and Enzymes
Multiple Choice Questions 1. Which energy association is correct? A. Kinetic energy is energy of motion. B. Water stored behind a dam is an example of kinetic energy. C. Energy in the chemical bonds of a molecule is kinetic energy. D. Potential energy must be used immediately or it is lost. E. Light energy is a form of chemical energy. Kinetic energy is the energy of motion. Water stored behind a dam is potential energy which does not have to be used immediately. Energy in chemical bonds is chemical energy. Light energy is a form of solar energy.
Bloom's Level: 2. Understand Learning Outcome: 06.01.01 Describe the different forms of energy. Section: 06.01 Topic: Energy
2. Which of the following statements most accurately describes the first law of thermodynamics? A. One usable form of energy cannot be completely converted into another usable form. B. One usable form of energy can be completely converted into another usable form. C. Energy cannot be created or destroyed. D. Energy can be neither created nor destroyed but it can be changed from one form to another. E. Energy cannot be transformed. The first law is the law of conservation of energy. It states that energy cannot be created or destroyed, but it can be changed from one form to another.
Bloom's Level: 2. Understand Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
3. Which statement most accurately describes the second law of thermodynamics? A. Energy cannot be changed from one form to another without a loss of usable energy. B. One usable form of energy can be completely converted into another usable form. C. Energy cannot be created nor destroyed but it can be changed from one form to another. D. Energy cannot be created or destroyed. E. Energy cannot be transformed. The second law of thermodynamics states that energy cannot be changed from one form to another without a loss of usable energy.
Bloom's Level: 2. Understand Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
True / False Questions 4. Ultimately, humans get their energy from the sun. TRUE Humans get their energy from their food, which comes from organisms that either photosynthesize or eat those organisms that photosynthesize. Photosynthesizers get their energy from the sun.
Bloom's Level: 2. Understand Learning Outcome: 06.01.01 Describe the different forms of energy. Section: 06.01 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
5. All of the energy that a plant stores in the bonds of glucose are available to an animal to power its muscles. FALSE No, some of the energy is lost as heat, so not all of the energy is available.
Bloom's Level: 2. Understand Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
6. Because the inside of a cell is more organized than the outside, the inside of the cell has increased entropy. FALSE Increased organization means decreased entropy.
Bloom's Level: 2. Understand Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
Multiple Choice Questions 7. What is a metabolic reaction called that involves one reaction releasing energy to provide energy for a second reaction that requires energy? A. an endergonic reaction B. an exergonic reaction C. a coupled reaction D. a reversible reaction E. anabolism A coupled reaction is one in which one reaction provides energy for another. Exergonic reactions release energy while endergonic reactions require energy. Reversible reactions can go either direction. Anabolism refers to the building up of molecules.
Bloom's Level: 2. Understand Learning Outcome: 06.02.02 Summarize the ATP cycle and the role of ATP in the cell. Section: 06.02 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
8. ATP can be broken down into A. ADP. B. ADP plus phosphate. C. ADP plus phosphate plus energy. D. pyruvate. E. lactate. ATP is broken down into ADP plus phosphate plus energy.
Bloom's Level: 2. Understand Learning Outcome: 06.02.02 Summarize the ATP cycle and the role of ATP in the cell. Section: 06.02 Topic: ATP
9. Most cells use a tremendous amount of ATP, A. so our bodies therefore produce and maintain a huge amount of ATP in storage. B. but only a little ATP is needed because it is an enzyme and does not participate in a reaction. C. but the amount of stored ATP is not great because our body constantly generates ATP from ADP + P. D. mainly because ATP is not very efficient as an energy currency. E. but only for chemical work; transport and mechanical work must use transport and mechanical molecules. The amount of ATP on hand at any one moment is minimal because ATP is constantly being generated from ADP and inorganic phosphate. ATP is not an enzyme, is very efficient, and is used for all sorts of work within the cell.
Bloom's Level: 2. Understand Learning Outcome: 06.02.02 Summarize the ATP cycle and the role of ATP in the cell. Section: 06.02 Topic: ATP
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Chapter 06 - Metabolism: Energy and Enzymes
Yes / No Questions 10. Will the reaction A + B turn into C and spontaneously "move" forward if the free energy of A + B is 25 units and the free energy of C is 30 units? NO No. In order for a reaction to go forward, the substrates must have more free energy than the products.
Bloom's Level: 2. Understand Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Metabolic Pathways
Multiple Choice Questions 11. Which is NOT a correct attribute of a metabolic pathway? A. Reactants are the input molecules. B. Reactants act as substrates for specific enzymes. C. The product of one reaction can become the reactant for the next. D. If several metabolic pathways have a molecule in common, one pathway can lead to several others. E. A constant supply of new enzymes must be produced to keep the metabolic pathway active. Enzymes can be used over and over again, so they do not need to be constantly resupplied.
Bloom's Level: 2. Understand Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Metabolic Pathways
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Chapter 06 - Metabolism: Energy and Enzymes
True / False Questions 12. A reactant will always produce the same type of product, regardless of the enzymes present. FALSE A particular reactant can be a substrate for various enzymes, each resulting in a different product.
Bloom's Level: 2. Understand Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Metabolic Pathways
Multiple Choice Questions 13. During an enzymatic reaction, what happens to the enzyme? A. It becomes the product. B. It becomes the substrate. C. It is used up. D. The enzyme and the substrate form a permanent complex. E. The enzyme and the substrate form a temporary complex. The enzyme and the substrate form a temporary enzyme-substrate complex, which is broken down to release the enzyme unchanged and the product. The enzyme is not used up by the reaction.
Bloom's Level: 2. Understand Learning Outcome: 06.03.02 Recognize how enzymes influence the activation energy rates of a chemical reaction. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
14. Which of the following statements is true about enzymes? A. Their 3D shape can vary and still be active. B. Boiling temperatures do not affect their activity. C. They catalyze only one specific type of reaction. D. They can associate with a wide variety of substrates. E. They are unaffected by changes in pH. Enzymes are specific for their substrate and only perform one type of reaction. They are affected by temperature and pH. An enzyme with an altered 3D shape will not bind its substrate.
Bloom's Level: 2. Understand Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
15. Which of the following molecules is NOT an enzyme? A. lipase B. maltase C. urease D. lactose E. ribonuclease Enzymes normally end with the suffix -ase.
Bloom's Level: 2. Understand Learning Outcome: 06.03.02 Recognize how enzymes influence the activation energy rates of a chemical reaction. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
16. Enzymes are specific. This means that they A. have a preferred pH. B. have a preferred temperature. C. have a particular substrate. D. are only in certain cells. E. require ATP and cofactors in order to work properly. Although enzymes do have preferred temperatures and pHs, require ATP and cofactors, and are present only in certain cells, the term "specific" refers to their particular substrate. An enzyme is specific because it has only one substrate.
Bloom's Level: 2. Understand Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Enzymes
17. Each enzyme has a particular substrate because enzymes A. increase the energy of activation. B. decrease the productivity of the cell. C. always require coenzymes. D. have active sites complementary in shape to their substrates. E. are named for their substrate. The specificity of an enzyme is due to the complementarity of the active site and the shape of the substrate. Although enzymes are named for their substrates, this does not account for their specificity.
Bloom's Level: 2. Understand Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
True / False Questions 18. The addition of an enzyme will change the end result of a reaction. In other words, a particular reaction will result in a different product if an enzyme is used. FALSE Enzymes only lower the energy of activation of a reaction. They do not change the end result.
Bloom's Level: 2. Understand Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
Multiple Choice Questions 19. The reason an increase in substrate concentration will not increase the reaction rate indefinitely is because A. eventually the enzyme will no longer recognize the substrate. B. the active site will become saturated with substrate and the enzyme will not be able to go any faster. C. another enzyme will take advantage of the excess substrate and use it for a different reaction. D. enzymes cannot be used over and over in a reaction. E. the enzyme will become denatured. Once the active site is saturated with substrate, the enzyme has achieved its maximum rate and cannot increase any more.
Bloom's Level: 2. Understand Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
20. Which is true about energy of activation? A. Energy of activation is measured as the energy that is released after a reaction occurs. B. Adding the correct enzyme can lower the energy of activation. C. In any one metabolic pathway, all steps will have the same energy of activation. D. Energy of activation is the difference between the energy of the reactant and the energy of the product. E. The energy of activation is always lower than the energy of the reactant. The energy of activation is the energy that must be overcome for a reaction to proceed. Enzymes lower this energy of activation. Free energy is the difference between the energy of the reactant and the energy of the product.
Bloom's Level: 2. Understand Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Enzymes
21. If an enzymatic reaction was controlled by feedback inhibition, we would expect it to stop A. only if the substrate was exhausted. B. when the cofactors are exhausted. C. when the product changed the pH. D. as soon as a critical level of end product builds up. E. only if a metabolic poison is added. Feedback inhibition involves the binding of the end product of an active pathway to a site on the enzyme causing it to change shape.
Bloom's Level: 2. Understand Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Metabolic Pathways
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Chapter 06 - Metabolism: Energy and Enzymes
22. If you wished to increase enzyme activity, you would do all of the following except A. increase the temperature moderately. B. increase the concentration of the enzyme. C. increase the amount of substrate. D. change to optimum pH for the reaction. E. decrease the temperature. All of the above will increase the rate of the reaction except for decreasing the temperature.
Bloom's Level: 2. Understand Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
23. The metabolic pathways of photosynthesis and cellular respiration A. involve the same substrates and therefore the same enzymes. B. are exactly the same. C. involve oxidation reactions during photosynthesis and reduction reactions during respiration. D. involve oxidation reactions during respiration and reduction reactions during photosynthesis. E. both involve oxidation and reduction reactions. Although the equation for respiration and photosynthesis are the same, the reactions are not the same, nor do they have the same substrates and enzymes. Both reactions involve both oxidation and reduction.
Bloom's Level: 2. Understand Learning Outcome: 06.04.01 Explain how equations for photosynthesis and cellular respiration represent oxidation-reduction reactions. Section: 06.04 Topic: Metabolic Pathways
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Chapter 06 - Metabolism: Energy and Enzymes
True / False Questions 24. A reaction is only considered an oxidation reaction when oxygen is involved. FALSE An oxidation reaction involves the loss of electrons, whether or not oxygen is involved.
Bloom's Level: 2. Understand Learning Outcome: 06.04.01 Explain how equations for photosynthesis and cellular respiration represent oxidation-reduction reactions. Section: 06.04 Topic: Metabolic Pathways
Multiple Choice Questions 25. We often say that we need food for energy. In a biological sense, is this correct? A. Yes, because the smallest units inside the atoms that make up the food are simply pure energy. B. Yes, because the food must move through the digestive system, and motion is kinetic energy. C. Yes, because the food we eat has potential energy in its structure and this chemical energy can be converted into mechanical energy. D. No, because food consists of matter and cannot be transformed into energy. E. No, since all food matter stays matter, and energy remains energy. Food contains energy in the form of chemical bonds that is being stored. Our bodies break these bonds and utilize the energy.
Bloom's Level: 3. Apply Learning Outcome: 06.01.01 Describe the different forms of energy. Section: 06.01 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
True / False Questions 26. When heat dissipates into the environment, it is no longer usable by the individual who produced it. TRUE Once heat dissipates it is not available to do work for the individual who produced it.
Bloom's Level: 1 Remember Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
27. Heat is not a form of energy. FALSE Heat is not a usable form of energy once it dissipates into the environment, but it is a form of energy.
Bloom's Level: 1 Remember Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
Multiple Choice Questions 28. Which hypothesis, based on the two laws of thermodynamics as applied to cells, would determine if something is a living cell or not? A. If something uses energy to make itself more organized, then it is alive. B. If something is organized, then it is alive. C. If something moves, then it is alive. D. If something loses heat, then it is alive. E. If something uses energy to move, then it is alive. Non-living things can be organized, move, lose heat, and use energy to move. Living things use energy to make themselves organized.
Bloom's Level: 4. Analyze Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
29. Occasionally someone claims to have built a machine that can run forever, producing as much energy as it consumes. This has always been disproved because it violates A. the first law of thermodynamics. B. the second law of thermodynamics. C. entropy. D. laws preventing any conversion between types of energy. E. coupled reaction equations. This violates the second law which states that energy cannot be changed from one form to another without a loss of usable energy.
Bloom's Level: 3. Apply Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
30. Which of the following is NOT a correct statement about the second law of thermodynamics and entropy? A. The amount of disorder in the universe is always increasing. B. To maintain organization of a cell, a continual input of energy is required. C. Living cells without energy would become less organized. D. While the total amount of energy in a system is unchanged, energy lost as heat is no longer useful to the cell in doing work. E. Carbon dioxide and water form glucose without the input of energy. Glucose, which is highly organized, tends to break down to carbon dioxide and water which is the direction that entropy prefers it to go.
Bloom's Level: 5. Evaluate Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
31. Which of the following is consistent with the laws of physics governing energy? A. Eventually, sunlight that is absorbed on the earth returns to the atmosphere as dispersed heat. B. You eat a "quarter-pounder" hamburger and add exactly a quarter-pound of additional weight to your body. C. When a liter of gasoline is burned in a car engine, 100% of its energy goes into moving the car along the road. D. A calorie of sunlight becomes a calorie of plant tissue. E. A calorie of plant tissue, which when eaten by you, becomes a calorie of muscle "power." Energy does not cycle through the environment. All of the energy that is absorbed on Earth will eventually be lost as heat back into the atmosphere. No conversion of energy is 100% efficient so none of the other answers are possible.
Bloom's Level: 5. Evaluate Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
True / False Questions 32. Cells are not as efficient in their use of energy as gasoline engines. FALSE Gasoline engines are 20-30% efficient, while cells are ~40% efficient.
Bloom's Level: 1 Remember Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
33. Energy, but NOT chemicals, may cycle through living things. FALSE Energy flows but chemicals cycle through living things.
Bloom's Level: 1 Remember Learning Outcome: 06.01.01 Describe the different forms of energy. Section: 06.01 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
Multiple Choice Questions 34. When a cell uses substantial energy to organize its cell contents to a greater degree than the organization that exists outside the living cell, A. energy has been created. B. entropy has increased. C. entropy has decreased. D. entropy remains the same. E. energy has been destroyed. Increasing organization results in decreasing entropy. Energy can be neither created nor destroyed.
Bloom's Level: 3. Apply Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
35. While science cannot describe "thinking" in physical terms, we are certain that the process involves metabolism inside brain cells. With positron emission tomography (PET scan), it is possible to inject short-lived isotopes and image the regions of the brain that have the most active metabolism during various mental activities. However, A. thought cannot be linked to cell processes because energy is not related to matter. B. since thoughts can occur over and over, the requirement for a continual input of energy to prevent entropy does not apply to this cell activity. C. the cellular energy expended in thinking must be less than the chemical bond energy supplied in food to these brain cells. D. "thinking" is beyond the scope of science to study. E. "thinking" is not a cellular process and therefore does not require energy. Since thinking is a metabolic process, it requires energy in the form of food. Energy is required for all metabolic activities.
Bloom's Level: 3. Apply Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
36. The amount of energy available to do work after a chemical reaction has occurred is called A. entropy. B. metabolic energy. C. potential energy. D. kinetic energy. E. free energy. Free energy is that amount of energy still "free" to do work after a chemical reaction has occurred.
Bloom's Level: 1 Remember Learning Outcome: 06.02.01 Identify how the terms anabolic, catabolic, endergonic, and exergonic relate to metabolic reactions. Section: 06.02 Topic: Energy
37. The change in free energy of a reaction, delta G, is calculated by A. subtracting the free energy content of the products from that of the reactants. B. subtracting the free energy content of the reactants from that of the products. C. adding the free energy content of the products to that of the reactants. D. the amount of free energy content of the reactants. E. the amount of free energy content of the products. The amount of free energy is calculated by subtracting the free energy content of the reactants from that of the products.
Bloom's Level: 1 Remember Learning Outcome: 06.02.01 Identify how the terms anabolic, catabolic, endergonic, and exergonic relate to metabolic reactions. Section: 06.02 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
38. Chemical reactions that require the input of energy are A. exergonic reactions. B. endergonic reactions. C. coupled reactions. D. kinetic reactions. E. catabolic reactions. Endergonic reactions require energy. Exergonic reactions release energy. Catabolic reactions break down molecules. Coupled reactions combine an endergonic with an exergonic reaction. Kinetic reactions involve motion.
Bloom's Level: 1 Remember Learning Outcome: 06.02.01 Identify how the terms anabolic, catabolic, endergonic, and exergonic relate to metabolic reactions. Section: 06.02 Topic: Energy
39. Where within the cell is ATP built up? A. mitochondria B. cytoplasm C. nucleus D. endoplasmic reticulum E. Golgi body Mitochondria are the powerhouses of the cell where ATP is produced.
Bloom's Level: 1 Remember Learning Outcome: 06.02.02 Summarize the ATP cycle and the role of ATP in the cell. Section: 06.02 Topic: ATP
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Chapter 06 - Metabolism: Energy and Enzymes
True / False Questions 40. The unique value of the ATP molecule is that when glucose is broken down during cellular respiration, all of the free energy of glucose is transformed into ATP. FALSE Only 39% of the free energy of glucose is transformed to ATP; the rest is lost as heat.
Bloom's Level: 1 Remember Learning Outcome: 06.02.02 Summarize the ATP cycle and the role of ATP in the cell. Section: 06.02 Topic: ATP
41. The term "metabolism" refers to all of the chemical reactions that occur in a cell. TRUE Cellular metabolism is the sum of all the chemical reactions that occur in a cell.
Bloom's Level: 1 Remember Learning Outcome: 06.02.01 Identify how the terms anabolic, catabolic, endergonic, and exergonic relate to metabolic reactions. Section: 06.02 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
Multiple Choice Questions 42. The various uses of ATP include all of the following except A. being a structural component of the cell membrane. B. chemical work. C. mechanical work. D. transport work. E. moving substances into a cell. ATP can do chemical, mechanical, and transport work. Moving substances into a cell is a type of transport work. ATP is not a structural component of the cell membrane.
Bloom's Level: 3. Apply Learning Outcome: 06.02.02 Summarize the ATP cycle and the role of ATP in the cell. Section: 06.02 Topic: ATP
43. The main reason that ATP is considered the energy currency in cells is because it A. is a small molecule. B. carries a positive charge. C. contains accessible energy in phosphate bonds. D. contains an adenine base. E. contains a sugar ring. While ATP is small and does contain an adenine base and a sugar ring, these are not responsible for its energy-carrying capacity. That is a result of the energy within the phosphate bonds. ATP carries a negative charge.
Bloom's Level: 4. Analyze Learning Outcome: 06.02.02 Summarize the ATP cycle and the role of ATP in the cell. Section: 06.02 Topic: ATP
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Chapter 06 - Metabolism: Energy and Enzymes
44. ATP contains A. an adenine base and one phosphate group. B. an adenine base and two phosphate groups. C. an adenine base and three phosphate groups. D. an adenine base, a ribose sugar, and two phosphate groups. E. an adenine base, a ribose sugar, and three phosphate groups. ATP is adenosine (an adenine base plus a ribose sugar) plus three phosphates (triphosphate).
Bloom's Level: 1 Remember Learning Outcome: 06.02.02 Summarize the ATP cycle and the role of ATP in the cell. Section: 06.02 Topic: ATP
45. If A has a free energy of 38 units, and C has a free energy of 45 units, and the reaction is exergonic, based on the calculation of free energy, which is the substrate and which is the product? A. A is the substrate and C is the product. B. C is the substrate and A is the product. C. It is impossible to tell with this information. D. Both A and C are the products. E. Both A and C are the substrates. An exergonic reaction will release energy, therefore the substrate must have more energy than the products beacuse of the loss of energy to the environment. C is therefore the substrate and A is the product.
Bloom's Level: 4. Analyze Learning Outcome: 06.02.01 Identify how the terms anabolic, catabolic, endergonic, and exergonic relate to metabolic reactions. Section: 06.02 Topic: Metabolic Pathways
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Chapter 06 - Metabolism: Energy and Enzymes
46. The high energy bond in ATP is found in or between A. the adenine base. B. the adenine and the ribose. C. the ribose sugar. D. the adenine and the phosphates. E. the phosphate groups. The phosphate groups contain the high energy bonds.
Bloom's Level: 1 Remember Learning Outcome: 06.02.02 Summarize the ATP cycle and the role of ATP in the cell. Section: 06.02 Topic: ATP
True / False Questions 47. In a coupled reaction, some of the energy is not released to the environment because the energy from one reaction is used to drive the other reaction forward. TRUE Energy released from the exergonic reaction is used to drive the endergonic reaction.
Bloom's Level: 1 Remember Learning Outcome: 06.02.01 Identify how the terms anabolic, catabolic, endergonic, and exergonic relate to metabolic reactions. Section: 06.02 Topic: Energy
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Chapter 06 - Metabolism: Energy and Enzymes
Multiple Choice Questions 48. Which of the following is an example of chemical work within a cell? A. catabolism B. anabolism C. moving glucose across the cell membrane D. the beating of cilia E. the contraction of muscle fibers Anabolism, or synthesis of macromolecules, is an example of chemical work. The others are examples of transport or mechanical work.
Bloom's Level: 1 Remember Learning Outcome: 06.02.01 Identify how the terms anabolic, catabolic, endergonic, and exergonic relate to metabolic reactions. Section: 06.02 Topic: Energy
49. If A → B → C → D → E represents a metabolic pathway, then letter E would be A. a substrate. B. a product. C. energy. D. an enzyme. E. an enzyme-substrate complex. E is the final product of the metabolic pathway. The energy and enzymes are not listed.
Bloom's Level: 1 Remember Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Metabolic Pathways
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Chapter 06 - Metabolism: Energy and Enzymes
True / False Questions 50. Enzymes, coenzymes, and cofactors provide a mechanism for a cell to regulate the metabolism of a large amount of cell products using small quantities of chemicals. TRUE Since enzymes, coenzymes, and cofactors are not used up in metabolic reactions, very small amounts can regulate metabolism.
Bloom's Level: 1 Remember Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Metabolic Pathways
51. Enzymes maintain their same shape when they bind to a substrate. FALSE The active site undergoes a slight modification in shape when the substrate binds.
Bloom's Level: 1 Remember Learning Outcome: 06.03.02 Recognize how enzymes influence the activation energy rates of a chemical reaction. Section: 06.03 Topic: Enzymes
52. Less energy is needed to bring about a reaction when an enzyme is present. TRUE An enzyme reduces the energy of activation for a reaction.
Bloom's Level: 1 Remember Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
Multiple Choice Questions 53. Most enzymes are A. lipid molecules. B. carbohydrate molecules. C. protein molecules. D. DNA molecules. E. ATP molecules. Enzymes are proteins, although some RNA molecules can speed chemical reactions.
Bloom's Level: 1 Remember Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Enzymes
54. The function of an enzyme is to A. provide the energy for metabolic reactions. B. increase the rate of a metabolic reaction. C. change the direction of metabolic reactions. D. act as a buffer in metabolic reactions. E. raise the energy of activation for a reaction. An enzyme lowers the energy of activation for a reaction and allows it to proceed faster.
Bloom's Level: 1 Remember Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
55. While eating a container of yogurt, you have to leave, so you store the yogurt in the refrigerator. A day later you return and find the surface of the yogurt is no longer smooth but has broken into several liquified products. You correctly guess that enzymes from your saliva, via the spoon, have continued digesting the yogurt in your absence. What will happen over time? A. The reaction has probably stopped because the amount of saliva is small and you would have to add more saliva to continue the degradation. B. The reaction will continue indefinitely since the enzyme is not consumed by the reaction. C. The reaction will continue until half is digested and then stop because the reaction between substrate and product will be balanced. D. The reaction has probably stopped because the yogurt has denatured the enzyme. E. As long as the enzyme is not denatured by environmental conditions, the reaction will continue until all of the substrate is used up. Enzymes continue to function until all of the substrate is converted into product. The enzymes you introduced in your saliva would continue to digest the sugars in the yogurt until all the sugars were gone.
Bloom's Level: 4. Analyze Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
56. A forest contains vegetation that is in constant battle with the many insect caterpillars that eat the vegetation. Which of the following strategies would not be a driving force in this evolutionary battle? A. Plants are selected that evolve strong chemicals (plant alkaloids) that are toxic to the caterpillar and for which the caterpillars lack digestive enzymes. B. Caterpillars are selected that evolve new enzymes to digest or detoxify the plant alkaloids. C. Plants are selected that evolve chemicals that slow down or inhibit the digestive enzymes in the caterpillars so that ingested leaf material passes through before much nutrition can be gained. D. The caterpillars are selected that eat the least leaf material. E. Caterpillars are selected that evolve enzymes to break down the chemicals that interrupt their digestive enzymes. Those characteristics that give the plant more chance of survival will be selected for. Likewise, those characteristics that give the caterpillar more chance of survival will be selected for. Therefore, caterpillars that eat the least leaf material would not be selective.
Bloom's Level: 5. Evaluate Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Enzymes
57. If there are 12 different intermediate products produced during production of a molecule in a cell, we can expect that there A. is one enzyme that carries this process through all 12 stages to the end product. B. are about 12 enzymes, at least one responsible for each step in the metabolic pathway. C. is one enzyme for degradation and another enzyme for synthesis. D. there may not be any enzymes involved if this is a natural cell product. E. must be 12 different raw materials combined in the cell by one enzyme. Each step of a metabolic pathway must be carried out by a different enzyme. Therefore, if there are 12 intermediates, there must be 12 enzymes.
Bloom's Level: 3. Apply Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Metabolic Pathways
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Chapter 06 - Metabolism: Energy and Enzymes
58. The specificity of an enzyme to a substrate is currently best explained by A. the "lock and key" model. B. the induced-fit model. C. the allosteric model. D. the receptor model. E. the synthase complex model. The active site of the enzyme undergoes a slight change in shape in order to accommodate the substrate(s). This is called the induced-fit model.
Bloom's Level: 1 Remember Learning Outcome: 06.03.02 Recognize how enzymes influence the activation energy rates of a chemical reaction. Section: 06.03 Topic: Enzymes
True / False Questions 59. Since enzymes are not used up during a reaction, they do not play any role in the reaction. FALSE Enzymes are not used up in the reaction. Instead, they play an active role in increasing the rate at which the reaction occurs. They are also reusable.
Bloom's Level: 1 Remember Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
Multiple Choice Questions 60. The location in which the enzyme and substrate complexes is called the A. active site. B. inhibitor site. C. receptor site. D. enzyme-substrate complex. E. enzyme-inhibitor complex. The enzyme and substrate will complex at the active site.
Bloom's Level: 1 Remember Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Enzymes
61. Which of the following statements is not true concerning enzymatic activity? A. Each enzyme has a preferred pH at which the enzyme reaction rate is highest. B. Above a certain temperature, an enzyme will become denatured. C. As the temperature increases, most enzymatic reactions will still proceed at the same rate. D. Enzymatic reactions proceed quite rapidly. E. Enzyme activity increases as substrate concentration increases until the maximum rate is achieved. As the temperature rises, enzyme activity will increase.
Bloom's Level: 3. Apply Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
62. Lactose is milk sugar that is broken down by the enzyme lactase. The reason that some people are "lactose intolerant" could be because A. the enzyme lactase only works in the laboratory, not in the human body. B. the pH of the digestive system does not allow lactase to function properly. C. the temperature of the digestive system does not allow lactase to function properly. D. they drink too much milk and have inactivated the lactase in their digestive system. E. they are missing the enzyme lactase in their digestive system. Most people that are lactose intolerant are missing the enzyme lactase in their digestive system.
Bloom's Level: 3. Apply Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Enzymes
63. A piece of meat and a piece of fruit are fed to both a lizard and a person. The extent to which these foods are digested by digestive enzymes along these two gastrointestinal tracts varies. Which of the following is not a reason for the variation in digestion? A. pH of the digestive organs B. body temperature C. variation in enzymes produced D. the amount of protein in the meat and fruit E. bacteria present in the digestive tract All of the answers, except for the amount of protein in the meat and fruit, are particular for the species of animal and could be responsible for the variability in digestion.
Bloom's Level: 3. Apply Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
64. In feedback inhibition of a metabolic pathway, where does the inhibitor bind? A. to the substrate of the first reaction B. to the product of the first reaction C. to the enzyme of the first reaction D. to the enzyme of the last reaction E. to a substrate or the product of the last reaction In feedback inhibition, the final product binds to the enzyme, either at the active site or at another site.
Bloom's Level: 1 Remember Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Metabolic Pathways
65. Whether or not an enzyme is active and present within a cell depends on all of the following except A. if the gene for the enzyme has been turned on and the enzyme made. B. if the enzyme has been activated or inactivated by the addition of a phosphate group. C. whether an inhibitor is bound to the enzyme. D. whether the necessary cofactor is present. E. whether the enzyme catalyzes an endergonic or exergonic reaction. Enzyme presence and activity can be affected by everything but whether the enzyme catalyzes an ender- or exergonic reaction. If the substrates are present and the products are needed, the enzyme will be synthesized and active regardless of whether the reaction is endergonic or exergonic.
Bloom's Level: 4. Analyze Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
True / False Questions 66. When a kinase adds phosphates to an enzyme, it will only activate the enzyme. FALSE Sometimes phosphates activate an enzyme, while other times they inactivate the enzyme.
Bloom's Level: 1 Remember Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
Multiple Choice Questions 67. Poisons are often A. enzyme inhibitors. B. cofactors. C. coenzymes. D. vitamins. E. kinases. Poisons are often enzyme inhibitors.
Bloom's Level: 1 Remember Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
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Chapter 06 - Metabolism: Energy and Enzymes
68. The removal of electrons and/or hydrogen atoms from a substrate is called what? A. reduction B. oxidation C. phosphorylation D. metabolism E. an enzyme-substrate complex Oxidation is the loss of electrons or loss of hydrogen atoms.
Bloom's Level: 1 Remember Learning Outcome: 06.04.01 Explain how equations for photosynthesis and cellular respiration represent oxidation-reduction reactions. Section: 06.04 Topic: Metabolic Pathways
69. What is the equation for photosynthesis? A. C6H12O6 + 6O2 → 6CO2 + 6H2O + energy B. energy + C6H12O6 → 6CO2 +6H2O C. energy + C6H12O6 → 6CO2 +6H2O D. energy + 6CO2 + 6H2O → C6H12O6 + 6O2 E. C6H12O6 + 6CO2 → 6O2 + 6H2O + energy Photosynthesis is energy plus 6 carbon dioxide plus 6 water goes to 6 glucose plus 6 oxygen.
Bloom's Level: 1 Remember Learning Outcome: 06.04.01 Explain how equations for photosynthesis and cellular respiration represent oxidation-reduction reactions. Section: 06.04 Topic: Metabolic Pathways
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Chapter 06 - Metabolism: Energy and Enzymes
70. Which of the following is a by-product of photosynthesis? A. carbon dioxide B. water C. oxygen D. glucose E. energy Although all of these are involved in photosynthesis, oxygen is the by-product of the production of glucose from carbon dioxide and water.
Bloom's Level: 1 Remember Learning Outcome: 06.04.02 Summarize the relationship between the metabolic reactions of photosynthesis and cellular respiration. Section: 06.04 Topic: Metabolic Pathways
71. During cellular respiration, what is reduced? A. glucose B. oxygen C. carbon dioxide D. water E. ATP (energy) During respiration, oxygen is reduced and glucose is oxidized.
Bloom's Level: 1 Remember Learning Outcome: 06.04.02 Summarize the relationship between the metabolic reactions of photosynthesis and cellular respiration. Section: 06.04 Topic: Metabolic Pathways
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Chapter 06 - Metabolism: Energy and Enzymes
True / False Questions 72. Less energy is needed to bring about a reaction when an enzyme is present. TRUE An enzyme reduces the energy of activation for a reaction.
Bloom's Level: 1 Remember Learning Outcome: 06.03.01 Explain the purpose of a metabolic pathway and how enzymes help regulate it. Section: 06.03 Topic: Enzymes
Multiple Choice Questions 73. Chloroplasts and mitochondria are involved in a redox cycle because A. carbon dioxide is reduced during photosynthesis and carbohydrates are oxidized during cellular respiration. B. both use carbon dioxide. C. photosynthesis uses energy and cellular respiration releases energy. D. both are required to be in all cells. E. energy cycles between the two organelles. A redox cycle involves both oxidation and reduction. Carbon dioxide is reduced during photosynthesis which occurs in chloroplasts and carbohydrates are oxidized during cellular respiration in mitochondria.
Bloom's Level: 4. Analyze Learning Outcome: 06.04.02 Summarize the relationship between the metabolic reactions of photosynthesis and cellular respiration. Section: 06.04 Topic: Organelles
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Chapter 06 - Metabolism: Energy and Enzymes
True / False Questions 74. Glucose is broken down during cellular respiration so that the released energy can be stored and converted into ATP. TRUE The purpose of cellular respiration is to break glucose down step by step in order to make ATP.
Bloom's Level: 1 Remember Learning Outcome: 06.04.02 Summarize the relationship between the metabolic reactions of photosynthesis and cellular respiration. Section: 06.04 Topic: Metabolic Pathways
Short Answer Questions 75. Define the first law of thermodynamics. The first law of thermodynamics is the law of conservation of energy. It states that energy cannot be created or destroyed, but it can be changed from one form to another.
Bloom's Level: 6. Create Learning Outcome: 06.01.02 Summarize the two laws of thermodynamics and how these laws apply to cells. Section: 06.01 Topic: Energy
76. Explain the factors that will cause a protein to become denatured and what will occur structurally to the protein when it is denatured. The two factors that will denature a protein are changes in pH and/or temperature. When an enzyme is denatured, the shape of the protein is altered. This new shape is unable to bind effectively to the substrate so the reaction will not proceed in the normal fashion.
Bloom's Level: 6. Create Learning Outcome: 06.03.03 Identify how environmental conditions influence the activity of an enzyme. Section: 06.03 Topic: Enzymes
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Chapter 07 - Cellular Respiration
Chapter 07 Cellular Respiration
Multiple Choice Questions 1. If the carbons from glucose were radioactively labeled, what molecule(s) would end up with the label? A. pyruvate B. citric acid C. carbon dioxide D. carbon dioxide and citrate E. pyruvate and carbon dioxide The carbons from the glucose are converted into carbon dioxide in the preparatory reaction. During the citric acid cycle the carbons are incorporated into citrate before being released into the carbon dioxide.
Bloom's Level: 4. Analyze Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Cellular Respiration
2. Cellular respiration involves all of the following except A. the breakdown of molecules. B. the release of energy. C. the synthesis of ATP. D. breathing in and out. E. the release of carbon dioxide. Cellular respiration is the release of energy from molecules such as glucose accompanied by the use of this energy to synthesize ATP molecules. It is an aerobic process that uses oxygen and releases carbon dioxide. Do not confuse cellular respiration with the respiratory system.
Bloom's Level: 2. Understand Learning Outcome: 07.01.01 Describe the overall equation for cellular respiration. Section: 07.01 Topic: Cellular Respiration
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Chapter 07 - Cellular Respiration
3. Which of the following is a substrate of cellular respiration? A. carbon dioxide B. water C. glucose D. ATP E. energy Cellular respiration involves glucose + oxygen + ADP + P (substrates) going to carbon dioxide + water + ATP (products).
Bloom's Level: 2. Understand Learning Outcome: 07.01.01 Describe the overall equation for cellular respiration. Section: 07.01 Topic: Cellular Respiration
4. Cellular respiration is an aerobic process. This means that it A. uses energy. B. produces energy. C. requires carbon dioxide. D. requires water. E. requires oxygen. An aerobic process is one that requires oxygen.
Bloom's Level: 2. Understand Learning Outcome: 07.01.01 Describe the overall equation for cellular respiration. Section: 07.01 Topic: Cellular Respiration
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Chapter 07 - Cellular Respiration
5. How efficient is the breakdown of one glucose via cellular respiration? A. 100% B. 58% C. 39% D. 20% E. less than 10% The breakdown of one glucose molecule results in 36-38 ATP or approximately 39% of the potential energy within the glucose molecule. No reaction is 100% efficient.
Bloom's Level: 1. Remember Learning Outcome: 07.01.01 Describe the overall equation for cellular respiration. Section: 07.01 Topic: Cellular Respiration
6. The equation: C6H12O6 + 6O2 → 38 ATP + 6CO2 + 6H2O represents which cellular process? A. glycolysis B. electron transport system C. citric acid cycle D. photosynthesis E. cellular respiration This is the formula for cellular respiration—the breakdown of glucose plus oxygen to form carbon dioxide, water, and ATP.
Bloom's Level: 1. Remember Learning Outcome: 07.01.01 Describe the overall equation for cellular respiration. Section: 07.01 Topic: Cellular Respiration
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Chapter 07 - Cellular Respiration
7. Which molecules are the products of aerobic respiration? A. glucose and carbon dioxide B. glucose and water C. glucose and oxygen D. lactate and carbon dioxide E. carbon dioxide and water Carbon dioxide and water (plus energy) are the products of aerobic respiration. Glucose plus oxygen are the substrates.
Bloom's Level: 1. Remember Learning Outcome: 07.01.01 Describe the overall equation for cellular respiration. Section: 07.01 Topic: Cellular Respiration
8. Which molecules are the reactants or substrates for aerobic respiration? A. glucose and carbon dioxide B. lactate and oxygen C. carbon dioxide and water D. oxygen and glucose E. glucose and water Glucose and oxygen are the substrates for aerobic respiration. Carbon dioxide and water are the products.
Bloom's Level: 2. Understand Learning Outcome: 07.01.01 Describe the overall equation for cellular respiration. Section: 07.01 Topic: Cellular Respiration
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Chapter 07 - Cellular Respiration
9. Why is ATP used as the energy molecule of the cell? A. ATP is easy to synthesize within the cell. B. ATP contains large amounts of energy so at least some of it can be used by the cell. C. ATP contains just about the amount of energy required for most cellular reactions. D. ATP is stored inside the cell where it is readily available whenever needed. E. ATP can be broken down into carbon dioxide and water. ATP contains just the right amount of energy necessary for most cellular reactions. It is not stored within the cell or broken down into carbon dioxide and water. Although it is relatively easy to synthesize, that does not make it a useful energy source.
Bloom's Level: 3. Apply Learning Outcome: 07.01.01 Describe the overall equation for cellular respiration. Section: 07.01 Topic: ATP Yield
10. What role does NAD+ play in cellular respiration? A. It is an electron carrier. B. It produces the ATP. C. It is an enzyme. D. It provides the oxygen. E. It provides the energy. NAD+ is a coenzyme which functions as an electron carrier.
Bloom's Level: 2. Understand Learning Outcome: 07.01.02 Explain the role of electron carriers in respiration. Section: 07.01 Topic: Cellular Respiration
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Chapter 07 - Cellular Respiration
True / False Questions 11. NADH and FADH2 bring electrons to the electron transport chain during cellular respiration. TRUE Both NADH and FADH2 are electron carriers that bring electrons to the electron transport chain during cellular respiration .
Bloom's Level: 1. Remember Learning Outcome: 07.01.02 Explain the role of electron carriers in respiration. Section: 07.01 Topic: Electron Transport Chain
Multiple Choice Questions 12. Which of the following does not describe the role of NAD+ in cellular respiration? A. It accepts two electrons. B. It is a coenzyme. C. It helps to oxidize the substrate. D. It is reduced. E. It provides the energy for the reaction. NAD+ does not provide energy for cellular respiration.
Bloom's Level: 1. Remember Learning Outcome: 07.01.02 Explain the role of electron carriers in respiration. Section: 07.01 Topic: Cellular Respiration
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Chapter 07 - Cellular Respiration
13. Which of the following reactions is NOT part of the cellular respiration pathway? A. glycolysis B. citric acid cycle C. Calvin cycle D. electron transport chain E. preparatory reaction (prep) Cellular respiration includes glycolysis, the preparatory reaction, the citric acid cycle, and the electron transport chain. The Calvin cycle is part of photosynthesis.
Bloom's Level: 1. Remember Learning Outcome: 07.01.03 Summarize the phases of cellular respiration and indicate where they occur in a cell. Section: 07.01 Topic: Cellular Respiration
14. Why is cellular respiration organized into four phases? A. So that the energy within the glucose molecule can be released in a stepwise fashion. B. So that it can take place within different cells. C. So that most of the energy can be released as body heat. D. So that oxidation can occur without reduction. E. So that the body can make energy from different substrates. The four phases of cellular respiration allow the energy of the glucose molecule to be released in a stepwise fashion and converted into ATP. If the breakdown occurred all at once, most of the energy would be released as heat. The body can make ATP from different substrates but that is not the reason for the phases.
Bloom's Level: 3. Apply Learning Outcome: 07.01.03 Summarize the phases of cellular respiration and indicate where they occur in a cell. Section: 07.01 Topic: Cellular Respiration
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Chapter 07 - Cellular Respiration
15. What is the correct order of phases in cellular respiration? A. citric acid cycle, prep reaction, glycolysis, electron transport chain B. electron transport chain, glycolysis, prep reaction, citric acid cycle C. prep reaction, glycolysis, electron transport chain, citric acid cycle D. glycolysis, prep reaction, citric acid cycle, electron transport chain E. glycolysis, citric acid cycle, prep reaction, electron transport chain The order of phases in cellular respiration is glycolysis, prep reaction, citric acid cycle, and electron transport chain.
Bloom's Level: 1. Remember Learning Outcome: 07.01.03 Summarize the phases of cellular respiration and indicate where they occur in a cell. Section: 07.01 Topic: Cellular Respiration
16. Which of the following statements about glycolysis is true? A. Glycolysis results in the release of carbon dioxide. B. Glycolysis is a cyclical reaction. C. Glycolysis is a reduction reaction where only glucose is reduced. D. Glycolysis occurs twice per glucose molecule. E. Glycolysis breaks glucose down to two pyruvate molecules. Glycolysis is the breakdown of glucose to 2 pyruvate molecules. It is an oxidation reaction that produces 2 ATP.
Bloom's Level: 4. Analyze Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
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Chapter 07 - Cellular Respiration
17. Glycolysis is likely to have evolved before the citric acid cycle and electron transport chain because A. it has an older fossil record. B. it is found in all living organisms. C. it occurs inside the mitochondria. D. the process is found in all primitive bacteria but lacking in many advanced organisms. E. it produces more ATP than do the citric acid cycle and electron transport chain. Since all organisms, even the most simple, use glycolysis, it most likely evolved first. Glycolysis does not require oxygen and the environmental conditions when cells evolved were anaerobic.
Bloom's Level: 5. Evaluate Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
18. Where does glycolysis take place within the cell? A. endoplasmic reticulum B. nucleus C. mitochondrial matrix D. mitochondrial membrane E. cytoplasm Glycolysis takes place within the cytoplasm.
Bloom's Level: 1. Remember Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
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Chapter 07 - Cellular Respiration
True / False Questions 19. The most efficient method to produce ATP in animal cells is by glycolysis. FALSE Glycolysis only results in a net of 2 ATP so it is not as efficient as the electron transport chain.
Bloom's Level: 2. Understand Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
20. Glycolysis is linked to the citric acid cycle when oxygen is not available. FALSE If no oxygen is available, the end product of glycolysis is fermented. Glycolysis is only linked to the citric acid cycle when oxygen is available.
Bloom's Level: 2. Understand Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
21. The final product of glycolysis is oxygen. FALSE The final product of glycolysis is pyruvate.
Bloom's Level: 1. Remember Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
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Chapter 07 - Cellular Respiration
Multiple Choice Questions 22. Why are two ATP needed to begin glycolysis? A. to activate the glucose B. to donate electrons to NAD+ C. to compensate for the lack of oxygen D. to accept electrons from glucose E. to move the reaction into the mitochondria The two ATPs are used to activate the glucose by adding two high energy phosphate bonds.
Bloom's Level: 3. Apply Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
23. You have discovered an organism that lives inside the gastrointestinal tract and does not contain a nucleus. Based on what you know of cellular respiration, how does it obtain its energy? A. photosynthesis B. glycolysis C. glycolysis and the citric acid cycle D. photosynthesis and the citric acid cycle E. glycolysis, the citric acid cycle, and the electron transport chain All living organisms utilize glycolysis. Since it does not have a nucleus, it does not contain mitochondria either, so it cannot use the citric acid cycle. Since it does not have access to sunlight, it cannot photosynthesize.
Bloom's Level: 4. Analyze Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
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Chapter 07 - Cellular Respiration
24. How many molecules are formed from the cleavage of glucose during the first step of glycolysis? How many carbons are found in the resulting molecules? A. 6 molecules, each with 2 carbons B. 4 molecules, each with 3 carbons C. 3 molecules, each with 4 carbons D. 2 molecules, each with 3 carbons E. 3 molecules, each with 2 carbons Glucose is a 6-carbon molecule. It is broken into two molecules of glyceraldehyde 3phosphate, each with three carbons.
Bloom's Level: 2. Understand Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
True / False Questions 25. There is a net gain of four ATPs as a result of glycolysis. FALSE Four ATPs are formed from glycolysis, however two ATPs are used to activate the glucose, resulting in a net gain of only two ATP.
Bloom's Level: 2. Understand Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
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Chapter 07 - Cellular Respiration
Multiple Choice Questions 26. Which of the following best defines substrate-level ATP synthesis? A. An enzyme passes a high-energy phosphate to ADP, resulting in ATP. B. ATP is used to activate glucose for glycolysis. C. An exergonic reaction is used to drive the synthesis of ATP from ADP plus phosphate. D. Oxidation of a substrate results in a high energy bond. E. ATP is synthesized from adenosine plus three phosphates. Substrate-level ATP synthesis results from the passing of a high-energy phosphate bond by an enzyme to ADP. ATP is used to activate glucose and an exergonic reaction is used to drive the synthesis of ATP, but those do not define substrate-level ATP synthesis.
Bloom's Level: 2. Understand Learning Outcome: 07.02.02 Explain why ATP is both an input and output of glycolysis. Section: 07.02 Topic: ATP Yield
27. The final products of glycolysis are A. 2 pyruvate, 2 ATP, and 2 NADH + H+. B. 1 pyruvate and 2 ATP. C. 1 pyruvate and 2 NADH + H+. D. 2 ATP and 2 NADH + H+. E. 2 pyruvate. Glycolysis results in 2 ATP, 2 NADH + H+, and 2 pyruvate.
Bloom's Level: 1. Remember Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
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Chapter 07 - Cellular Respiration
True / False Questions 28. Pyruvate contains less chemical energy than glucose. TRUE Glucose is broken down, releasing energy, during glycolysis. Therefore pyruvate does contain less energy than glucose.
Bloom's Level: 2. Understand Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: ATP Yield
Multiple Choice Questions 29. Pyruvate is converted to a two-carbon acetyl group attached to coenzyme A (CoA), and CO2 is given off. This phase is called A. substrate-level ATP synthesis. B. the preparatory reaction. C. the electron transport chain. D. the citric acid cycle. E. fermentation. This reaction is called the preparatory reaction.
Bloom's Level: 2. Understand Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Preparatory Reaction
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Chapter 07 - Cellular Respiration
30. Which is NOT correct about the preparatory reaction? A. It connects glycolysis to the citric acid cycle. B. CO2 is given off. C. Pyruvate is converted to a two-carbon acetyl group. D. NAD+ goes to NADH + H+ as acetyl-CoA forms. E. The reaction occurs once per glucose molecule. All of the statements are true, except that the reaction occurs twice per glucose molecule, because one glucose is converted into 2 pyruvates.
Bloom's Level: 4. Analyze Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Preparatory Reaction
31. Where does the preparatory reaction take place within the cell? A. mitochondrial matrix B. mitochondrial cristae C. nucleus D. cytoplasm E. endoplasmic reticulum The preparatory reaction takes place within the mitochondrial matrix.
Bloom's Level: 1. Remember Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Preparatory Reaction
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Chapter 07 - Cellular Respiration
32. The molecule that is found at the beginning and end of the citric acid cycle is A. four-carbon molecule. B. pyruvate. C. lactate. D. glucose. E. RuBP. The first and last molecule in the citric acid cycle is a C4 molecule. Pyruvate is the last molecule of glycolysis while glucose is the first molecule of glycolysis. Lactate is a fermentation production and RuBP is in the Calvin cycle.
Bloom's Level: 2. Understand Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Citric Acid Cycle/Krebs Cycle
33. Which pathway in cellular respiration will produce ATP, NADH2, and carbon dioxide? A. glycolysis B. preparatory reaction C. citric acid cycle D. electron transport chain E. photosynthesis Only the citric acid cycle produces all three products.
Bloom's Level: 1. Remember Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Citric Acid Cycle/Krebs Cycle
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Chapter 07 - Cellular Respiration
34. Which stage(s) will produce carbon dioxide in cellular respiration? A. glycolysis B. both the preparatory reaction and the citric acid cycle C. citric acid cycle D. both glycolysis and the electron transport chain E. both the preparatory reaction and the citric acid cycle Both the preparation reaction and the citric acid cycle produce carbon dioxide.
Bloom's Level: 2. Understand Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Cellular Respiration
35. Which of the following statements is NOT true? A. The end product of glycolysis is pyruvate. B. The citric acid cycle begins and ends with pyruvate. C. NADH2 will eventually produce three ATP molecules. D. Aerobic respiration of glucose has four phases. E. Aerobic respiration uses oxygen and releases carbon dioxide. The citric acid cycle begins and ends with a C4 molecule.
Bloom's Level: 2. Understand Learning Outcome: 07.04.03 Identify how each stage of the aerobic pathway contributes to the generation of ATP in a cell. Section: 07.04 Topic: Cellular Respiration
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Chapter 07 - Cellular Respiration
36. For the complete respiration of one molecule of glucose, the citric acid cycle must turn a total of A. one time. B. two times. C. three times. D. four times. E. six times. Because two pyruvate are formed from one glucose molecule, the citric acid cycle must turn two times per glucose molecule.
Bloom's Level: 2. Understand Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Citric Acid Cycle/Krebs Cycle
37. An acetyl group contains how many carbons? A. one B. two C. three D. four E. six An acetyl group contains two carbons. Pyruvate is converted into one acetyl group (two carbons) and one carbon dioxide molecule (one carbon).
Bloom's Level: 1. Remember Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Preparatory Reaction
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Chapter 07 - Cellular Respiration
38. The number of ATPs produced directly as a result of one turn of the citric acid cycle is A. 1. B. 2. C. 12. D. 14. E. 16. Each turn of the citric acid cycle produces one ATP. Per glucose molecule, two ATPs are formed.
Bloom's Level: 1. Remember Learning Outcome: 07.04.03 Identify how each stage of the aerobic pathway contributes to the generation of ATP in a cell. Section: 07.04 Topic: ATP Yield
39. How many NADH + H+ are produced by the citric acid cycle per turn? A. one B. two C. three D. four E. six Three NADH + H+ molecules are produced during one turn of the citric acid cycle.
Bloom's Level: 1. Remember Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Citric Acid Cycle/Krebs Cycle
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Chapter 07 - Cellular Respiration
40. Which is a correct association of mitochondrion structure and phase of cellular respiration? A. matrix - electron transport chain B. cristae - preparatory reaction C. matrix - citric acid cycle D. matrix - glycolysis E. cristae - glycolysis Glycolysis occurs in the cytoplasm. The preparatory reaction and citric acid cycle occur in the matrix, while the electron transport chain occurs in the cristae.
Bloom's Level: 4. Analyze Learning Outcome: 07.04.01 Recognize the role of the mitochondria in cellular respiration. Section: 07.04 Topic: Citric Acid Cycle/Krebs Cycle
41. Some desert organisms can live out their life without ever drinking liquid water. What phase(s) of cellular respiration could provide the organism with water? A. glycolysis B. citric acid cycle C. electron transport chain D. glycolysis and electron transport chain E. citric acid cycle and electron transport chain Both glycolysis and the electron transport chain produce water.
Bloom's Level: 4. Analyze Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Electron Transport Chain
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Chapter 07 - Cellular Respiration
42. What phase(s) of cellular respiration produce(s) NADH + H+? A. glycolysis B. preparatory reaction C. citric acid cycle D. glycolysis and preparatory reaction E. glycolysis, preparatory reaction, and citric acid cycle All three phases of these reactions produce NADH + H+.
Bloom's Level: 2. Understand Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Cellular Respiration
43. The final acceptor for hydrogen ions in aerobic cellular respiration is A. oxygen. B. pyruvate. C. CoA. D. glucose. E. carbon dioxide. Oxygen takes the hydrogen ions and becomes water.
Bloom's Level: 1. Remember Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Electron Transport Chain
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Chapter 07 - Cellular Respiration
44. The production of ATP as a result of an electrochemical gradient is called A. glycolysis. B. substrate-level phosphorylation. C. chemiosmosis. D. deamination. E. oxidative phosphorylation. Chemiosmosis is when ATP production is tied to an electrochemical gradient, as in mitochondria.
Bloom's Level: 1. Remember Learning Outcome: 07.04.01 Recognize the role of the mitochondria in cellular respiration. Section: 07.04 Topic: ATP Yield
45. Which of the following would not slow down the electron transport system's production of ATP? A. Limit the amount of oxygen available. B. Decrease use of ATP and thus reduce available ADP. C. Reduce synthesis of NAD+ and FAD. D. Add an NADH reductase inhibitor. E. Increase the pumping of hydrogen ions into the intermembrane space of the mitochondria. Increasing the pumping of hydrogen ions into the intermembrane space would actually increase the amount of ATP produced.
Bloom's Level: 4. Analyze Learning Outcome: 07.04.03 Identify how each stage of the aerobic pathway contributes to the generation of ATP in a cell. Section: 07.04 Topic: ATP Yield
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Chapter 07 - Cellular Respiration
46. If we developed a molecule that blocked the ATP channel protein, what would be the first immediate effect? A. accumulation of ATP in the mitochondrial matrix B. accumulation of NADH in the mitochondrial matrix C. no energy to pump hydrogen ions into the intermembrane space of a mitochondrion D. reversal of the hydrogen ion gradient E. immediate blockage of the cytochrome molecules The ATP channel protein allows ATP to leave the mitochondrial matrix and enter the intermembrane space. Therefore, if it were blocked, ATP would accumulate in the matrix.
Bloom's Level: 4. Analyze Learning Outcome: 07.04.03 Identify how each stage of the aerobic pathway contributes to the generation of ATP in a cell. Section: 07.04 Topic: ATP Yield
47. The carriers of the electron transport chain are located A. in the matrix of the mitochondria. B. on the outer mitochondrial membrane. C. on the cristae of the mitochondria. D. within the intermembrane space. E. within the cytoplasm of the cell. The electron transport chain carriers are located within the mitochondrial cristae.
Bloom's Level: 1. Remember Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Electron Transport Chain
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Chapter 07 - Cellular Respiration
48. Why does chemiosmosis require a membrane? A. to anchor proteins within the mitochondria B. because the phospholipids are involved in the electron transport chain C. to separate two compartments of the cell to allow for gradient formation D. to generate H+ from water E. to provide a large surface area Chemiosmosis requires two separate compartments to allow for the formation of a gradient. The membrane is similar to using a dam to hold back water in order to generate electricity.
Bloom's Level: 4. Analyze Learning Outcome: 07.04.03 Identify how each stage of the aerobic pathway contributes to the generation of ATP in a cell. Section: 07.04 Topic: ATP Yield
49. The ATP synthase complex is located in the A. cytoplasm outside the mitochondria. B. matrix inside the mitochondria. C. cristae of the mitochondria. D. outer membrane of the mitochondria. E. plasma membrane of the cell. The ATP synthase complex is in the cristae of the mitochondria.
Bloom's Level: 1. Remember Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: ATP Yield
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Chapter 07 - Cellular Respiration
50. The largest number of ATP molecules is produced in which phase of cellular respiration? A. glycolysis B. preparation reaction C. citric acid cycle D. electron transport chain E. Calvin cycle The electron transport chain produces the most ATP: 32-34 ATP.
Bloom's Level: 2. Understand Learning Outcome: 07.04.03 Identify how each stage of the aerobic pathway contributes to the generation of ATP in a cell. Section: 07.04 Topic: Electron Transport Chain
51. Why does FADH2 result in fewer ATP than NADH + H+? A. FADH2 only contains one electron. B. FADH2 only contains one hydrogen ion. C. FADH2 drops its electrons off lower on the electron transport chain. D. FADH2 drops its electrons off higher on the electron transport chain. E. FADH2 drops its electrons off to oxygen. FADH2 drops its electrons off lower on the electron transport chain than does NADH + H+.
Bloom's Level: 3. Apply Learning Outcome: 07.04.03 Identify how each stage of the aerobic pathway contributes to the generation of ATP in a cell. Section: 07.04 Topic: ATP Yield
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Chapter 07 - Cellular Respiration
52. In cellular respiration, what is the relationship between the energy content of the reactants (glucose and O2) and products (CO2 and H2O)? A. The energy content of the reactants (glucose and O2) must equal the energy content of the products (CO2 and H2O). B. The energy content of the reactants (glucose and O2) must be greater than the energy content of the products (CO2 and H2O). C. The energy content of the reactants (glucose and O2) must be less than the energy content of the products (CO2 and H2O). D. There is no relationship between the energy content of the reactants (glucose and O2) and the energy content of the products (CO2 and H2O). E. There is no relationship between O2 and H2O, but the energy content of glucose is less than that of CO2. Glucose is a high-energy molecule while carbon dioxide is low energy. The energy content of the reactants is greater than that of the products because ATP is formed.
Bloom's Level: 4. Analyze Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Cellular Respiration
53. The energy difference between the reactant glucose and oxygen molecules and the product water and carbon dioxide is 686 kilocalories, yet the 36 ATP molecules produced are only storing 263 kilocalories in their outermost phosphate bond. The rest of the energy went A. to the ADP fragment of the ATP molecule. B. into providing the energy for life. C. to power the electron transport chain. D. to provide heat that is soon lost from the organism. E. nowhere, but was destroyed. The remainder of the energy is lost as heat. Energy cannot be created or destroyed.
Bloom's Level: 2. Understand Learning Outcome: 07.04.01 Recognize the role of the mitochondria in cellular respiration. Section: 07.04 Topic: ATP Yield
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Chapter 07 - Cellular Respiration
True / False Questions 54. All of the ATP molecules formed by the complete oxidation of glucose result from the electron transfer chain. FALSE Some of the ATP come from glycolysis and the citric acid cycle.
Bloom's Level: 1. Remember Learning Outcome: 07.04.03 Identify how each stage of the aerobic pathway contributes to the generation of ATP in a cell. Section: 07.04 Topic: ATP Yield
55. The citric acid cycle turns once for each original glucose molecule. FALSE Because each original glucose molecule is broken into two pyruvate molecules, the citric acid cycle turns twice for each original glucose molecule.
Bloom's Level: 1. Remember Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Citric Acid Cycle/Krebs Cycle
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Chapter 07 - Cellular Respiration
Multiple Choice Questions 56. Pyruvate can be converted to lactate instead of going to the preparatory reaction. Why does this occur? A. The cells need lactate to produce ATP. B. There is no sunlight. C. The cells doing the reaction are prokaryotes. D. Oxygen is not available. E. There is a shortage of glucose. Pyruvate is converted into lactate during fermentation, which occurs when oxygen is not available.
Bloom's Level: 2. Understand Learning Outcome: 07.03.01 Explain how ATP can continue to be produced in the absence of oxygen. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
57. Which of the following statements about fermentation is not true? A. Fermentation involves the citric acid cycle. B. Fermentation does not require oxygen. C. Fermentation can produce lactic acid. D. Fermentation produces a net two ATP molecules. E. Fermentation can produce alcohol. Fermentation does not use the citric acid cycle.
Bloom's Level: 2. Understand Learning Outcome: 07.03.01 Explain how ATP can continue to be produced in the absence of oxygen. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
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Chapter 07 - Cellular Respiration
58. Why is fermentation essential to humans? A. It is the best reaction we can utilize to produce ATP. B. It produces lactate, which is essential for humans. C. It produces carbon dioxide, which is necessary for our cells. D. It generates the NAD+ we need for metabolism. E. It is essential when we need a rapid burst of energy. Fermentation provides a rapid burst of ATP and thus muscle cells use fermentation when working vigorously over a short period of time when oxygen is temporarily limited.
Bloom's Level: 3. Apply Learning Outcome: 07.03.02 Describe the advantages and disadvantages of fermentation. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
59. Muscles undergo fermentation when A. no oxygen is available. B. no water is available. C. no carbon dioxide is available. D. no ATP is available. E. no pyruvate is available. Fermentation occurs when oxygen is not available.
Bloom's Level: 2. Understand Learning Outcome: 07.03.02 Describe the advantages and disadvantages of fermentation. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
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Chapter 07 - Cellular Respiration
60. Why do organisms without oxygen need to convert pyruvate to lactate? A. in order to regenerate NAD+ B. because lactate is needed to produce ATP C. because pyruvate is toxic to the cells D. in order to use lactate in the citric acid cycle E. because the conversion provides much more ATP for the cell Pyruvate is reduced to lactate when oxygen is unavailable in order to regenerate NAD+ to be used during glycolysis.
Bloom's Level: 3. Apply Learning Outcome: 07.03.01 Explain how ATP can continue to be produced in the absence of oxygen. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
61. Fermentation is inefficient in using the energy found in food molecules and the resulting alcohol can be toxic to the organism producing it. Which of the following is not a reason why an organism would utilize fermentation? A. If glucose levels are not high, there may be time to disperse the toxic alcohol waste. B. The organism can survive short spells of anaerobic conditions and maintain growth and reproduction. C. Fermentation can provide a rapid burst of ATP since it does not have to go through the full breakdown cycle. D. Fermentation is the preferred process, even when oxygen is available. E. For very small organisms, fermentation can be a simple process and is less complicated than cellular respiration. Fermentation is not preferred but is utilized only when oxygen is unavailable.
Bloom's Level: 4. Analyze Learning Outcome: 07.03.02 Describe the advantages and disadvantages of fermentation. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
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Chapter 07 - Cellular Respiration
62. Which of the following is not a consequence of the lactate that results from fermentation in the muscles? A. It triggers ongoing heavier breathing to provide further oxygen for the muscles. B. Much of the lactate is transported to the liver where it is converted to pyruvate. C. Some lactate converted to pyruvate is converted back to glucose. D. Some lactate converted to pyruvate is then completely broken down. E. It is stored in muscle cells for later ATP production. Lactate causes that muscle "burn" and is not stored in muscle cells.
Bloom's Level: 2. Understand Learning Outcome: 07.03.02 Describe the advantages and disadvantages of fermentation. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
63. Which of the following is not an example of fermentation used by organisms to produce ATP? A. bacterial cells producing lactate B. bacterial cells producing alcohol C. human cells producing lactate D. human cells producing lactate E. yeast cells producing alcohol Humans do not produce alcohol from fermentation.
Bloom's Level: 2. Understand Learning Outcome: 07.03.02 Describe the advantages and disadvantages of fermentation. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
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Chapter 07 - Cellular Respiration
True / False Questions 64. The only end product of alcoholic fermentation is the alcohol. FALSE Carbon dioxide is produced along with the alcohol.
Bloom's Level: 1. Remember Learning Outcome: 07.03.02 Describe the advantages and disadvantages of fermentation. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
65. Fermentation follows glycolysis in some cells if oxygen is not available. TRUE Glycolysis converts glucose to pyruvate. Pyruvate can either be fermented if oxygen is not available or enter the preparatory reaction if oxygen is available.
Bloom's Level: 2. Understand Learning Outcome: 07.03.01 Explain how ATP can continue to be produced in the absence of oxygen. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
66. Fermentation is the process that allows bread to rise. TRUE Fermentation produces alcohol and carbon dioxide. The carbon dioxide produces bubbles that cause bread to rise.
Bloom's Level: 3. Apply Learning Outcome: 07.03.02 Describe the advantages and disadvantages of fermentation. Section: 07.03 Topic: Fermentation and Anaerobic Respiration
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Chapter 07 - Cellular Respiration
Short Answer Questions 67. List, in order, the stages of aerobic cellular respiration and indicate where they occur in the cell. Glycolysis is first and it occurs within the cytoplasm of the cell. The preparatory reaction is second and it occurs within the matrix of the mitochondria. The citric acid cycle is third and it occurs within the matrix of the mitochondria. The electron transport chain is fourth and it occurs within the cristae of the mitochondria.
Bloom's Level: 6. Create Learning Outcome: 07.01.03 Summarize the phases of cellular respiration and indicate where they occur in a cell. Section: 07.01 Topic: Cellular Respiration
Multiple Choice Questions 68. What are the correct outputs, during the citric acid cycle, from one molecule of glucose? A. 6 NADH, 4 CO2, 2 ATP, 2 FADH2 B. 3 NADH, 2 CO2, ATP, FADH2 C. 2 acetyl - CoA & 2 CO2 D. 6 NADH, 2 CO2, 4 ATP, 6 FADH2 E. 2 pyruvates, 2 NADH, 2 ADH, and 4 ATP The correct outputs, during the citric acid cycle from one molecule of glucose is 6 NADH, 4 CO2, 2 ATP, 2 FADH2. The glucose will split into two pyruvates, each of which will be converted into acetyl CoA which then will go through the citric acid cycle. Two acetyl, CoA & 2 CO2, are the end products of the preparatory reaction. Two pyruvates, 2 NADH, 2 ADH, and 4 ATP are the end products of glycolysis.
Bloom's Level: 5. Evaluate Learning Outcome: 07.04.02 Summarize the inputs and outputs of the preparatory reaction, the citric acid cycle, and the electron transport chain. Section: 07.04 Topic: Citric Acid Cycle/Krebs Cycle
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Chapter 07 - Cellular Respiration
69. What is the most likely reason that cellular respiration would shift from an aerobic state into an anaerobic state? A. The individual was exercising at a level beyond their capability and entered into oxygen debt. B. A smoker has developed emphysema and has decreased lung capacity. C. The individual has had a stroke in the left hemisphere of their brain. D. None of the answer choices could cause an individual to shift into an anaerobic state. E. All of the answer choices are reasons that an individual could shift into an anaerobic state. All of the choices are reasons that an individual could shift into an anaerobic state. Any reason that decreases the amount of available oxygen can cause a shift into anaerobic respiration.
Bloom's Level: 5. Evaluate Learning Outcome: 07.03.01 Explain how ATP can continue to be produced in the absence of oxygen. Section: 07.03 Topic: Cellular Respiration
Short Answer Questions 70. List the inputs and outputs of glycolysis. Inputs of glycolysis include: glucose, 2 NAD+, 2 ATP, and 4 ADP + 4 P. Outputs of glycolysis include: 2 pyruvates, 2 NADH, 2 ADP, and 4 ATP. The net gain of glycolysis is 2 ATP.
Bloom's Level: 6. Create Learning Outcome: 07.02.02 Explain why ATP is both an input and output of glycolysis. Section: 07.02 Topic: Glycolysis
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Chapter 07 - Cellular Respiration
Multiple Choice Questions 71. What are the input requirements of glycolysis? A. 2 ATP, 2 NAD+, glucose, and 4 ADP B. 4 ATP, 2 NAD+, glucose, and 4 ADP C. 2 pyruvates, 2 NADH, and 2 ADP D. 2 ATP, 4 NAD+, glucose, and 4 ADP E. 2 ATP, 2 NAD+, glucose, and 2 ADP The inputs of glycolysis are 2 ATP, 2 NAD+, glucose, and 4 ADP. 2 pyruvates, 2 NADH, and 2 ADP are the outputs of glycolysis. 2 ATP, 2 NAD+, glucose, and 2 ADP are the inputs of glycolysis.
Bloom's Level: 2. Understand Learning Outcome: 07.02.01 Describe the location and inputs and outputs of glycolysis. Section: 07.02 Topic: Glycolysis
Short Answer Questions 72. Write out the overall equation for cellular respiration. C6H12O6 + 6O2 = 6CO2 + 6H2O ADP + P = 36 - 38 ATP
Bloom's Level: 6. Create Learning Outcome: 07.01.01 Describe the overall equation for cellular respiration. Section: 07.01 Topic: Cellular Respiration
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Chapter 08 - Photosynthesis
Chapter 08 Photosynthesis
Multiple Choice Questions 1. Heterotrophs A. have the ability to synthesize organic molecules from inorganic molecules. B. are ultimately dependent upon preformed organic molecules made by producers. C. are the origin of all food for the rest of the living world. D. are also called autotrophs. E. None of the answer choices is characteristic of heterotrophs. Heterotrophs are also known as consumers. They must take in preformed organic molecules in order to survive.
Bloom's Level: 2. Understand Learning Outcome: 08.01.01 Compare and contrast autotrophs and heterotrophs. Section: 08.01 Topic: Photosynthesis
2. Which statement about producers and/or consumers is true? A. Consumers not only feed themselves but also feed producers. B. All animals must be either producers or consumers. C. Producers produce more food than they use. D. Consumers are not dependent upon other organisms for food. E. Only one species of producer and consumer can exist in any single community. Producers produce more food than they use, which then provides foods for the consumers.
Bloom's Level: 1. Remember Learning Outcome: 08.01.01 Compare and contrast autotrophs and heterotrophs. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
3. Which of the following is not an autotroph? A. yeast B. oak tree C. cyanobacterium D. algae E. grass All of the organisms are capable of photosynthesizing except the yeast.
Bloom's Level: 1. Remember Learning Outcome: 08.01.01 Compare and contrast autotrophs and heterotrophs. Section: 08.01 Topic: Photosynthesis
True / False Questions 4. At the base of most food chains are autotrophs. TRUE Autotrophs produce their own food and are able to sustain themselves and all other living things on Earth.
Bloom's Level: 1. Remember Learning Outcome: 08.01.01 Compare and contrast autotrophs and heterotrophs. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
Multiple Choice Questions 5. Which of the following allows photosynthetic organisms to capture solar energy? A. pigments B. nucleus C. chloroplasts D. glucose E. NAD+ Pigments, such as chlorophyll, allow organisms to capture solar energy.
Bloom's Level: 2. Understand Learning Outcome: 08.01.01 Compare and contrast autotrophs and heterotrophs. Section: 08.01 Topic: Photosynthesis
True / False Questions 6. The red maple tree cannot photosynthesize because it is not green. FALSE The red maple is capable of photosynthesis because it has other pigments besides chlorophyll that enable it to capture solar energy.
Bloom's Level: 3. Apply Learning Outcome: 08.02.01 Identify the photosynthetic pigments required to absorb the various wavelengths of light necessary for photosynthesis. Section: 08.02 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
Multiple Choice Questions 7. The main function of stomata is to A. expose the chlorophyll to sunlight. B. transport water to the chlorophyll. C. allow passage of CO2 and O2 into the leaf. D. store glucose. E. store pyruvate and provide the site of the Calvin cycle. The stomata are small openings in the leaf that allow the exchange of carbon dioxide and oxygen.
Bloom's Level: 2. Understand Learning Outcome: 08.01.04 Describe the process of photosynthesis in terms of two sets of reactions that take place in a chloroplast. Section: 08.01 Topic: Photosynthesis
8. The substance that initially traps solar energy in photosynthesis is A. chlorophyll. B. RuBP. C. water. D. glucose. E. pyruvate. Pigments, such as chlorophyll, capture solar energy.
Bloom's Level: 1. Remember Learning Outcome: 08.02.01 Identify the photosynthetic pigments required to absorb the various wavelengths of light necessary for photosynthesis. Section: 08.02 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
9. The flattened sacs within the stroma of a chloroplast, which are connected to form a single inner compartment, are called A. stomata. B. thylakoids. C. mesophyll. D. carotenoids. E. CAM units. The thylakoids are the flattened sacs within the chloroplast.
Bloom's Level: 1. Remember Learning Outcome: 08.01.04 Describe the process of photosynthesis in terms of two sets of reactions that take place in a chloroplast. Section: 08.01 Topic: Light Reactions
True / False Questions 10. In the process of photosynthesis, the light reactions follow the Calvin cycle reactions. FALSE The light reactions occur first. This provides the energy for the Calvin cycle reactions.
Bloom's Level: 2. Understand Learning Outcome: 08.01.04 Describe the process of photosynthesis in terms of two sets of reactions that take place in a chloroplast. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
Multiple Choice Questions 11. What does the term stroma refer to? A. the cytochrome system in the membranes of the thylakoids B. a stack of thylakoid membrane structures C. the double membrane of the chloroplast D. a flattened disk or sac in the chloroplast E. the central fluid-filled space of the chloroplast The stroma is the fluid-filled interior of the chloroplast.
Bloom's Level: 1. Remember Learning Outcome: 08.02.03 Describe the organization of the thylakoid and how this organization is critical to the production of ATP during photosynthesis. Section: 08.02 Topic: Calvin Cycle/Carbon Reactions
12. The raw materials or reactants of the photosynthetic process include A. glucose and oxygen. B. carbon dioxide and glucose. C. carbon dioxide and water. D. carbon dioxide and oxygen. E. glucose and water. The chloroplast requires carbon dioxide and water for photosynthesis. Glucose and oxygen are the end products.
Bloom's Level: 1. Remember Learning Outcome: 08.01.03 Recognize the overall chemical equation for photosynthesis. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
13. What are the products of photosynthesis? A. water and carbon dioxide B. water and oxygen C. carbohydrate and water D. oxygen and carbohydrate E. carbon dioxide and carbohydrate The end products of photosynthesis are a carbohydrate (glucose) and oxygen. Carbon dioxide and water are the substrates of the reaction.
Bloom's Level: 1. Remember Learning Outcome: 08.01.03 Recognize the overall chemical equation for photosynthesis. Section: 08.01 Topic: Photosynthesis
14. Which molecule would you need to radioactively label in order to produce radioactive oxygen during photosynthesis? A. carbon dioxide B. water C. cytochrome D. glucose E. G3P Water contains oxygen which is broken down during photosynthesis to produce oxygen gas.
Bloom's Level: 3. Apply Learning Outcome: 08.01.03 Recognize the overall chemical equation for photosynthesis. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
15. At the cellular level, photosynthesis occurs within A. the chloroplast. B. the cristae of the mitochondria. C. both chloroplasts and mitochondria. D. all plant cell organelles. E. the nucleus of plants but not of animals. Photosynthesis occurs within chloroplasts.
Bloom's Level: 1. Remember Learning Outcome: 08.01.04 Describe the process of photosynthesis in terms of two sets of reactions that take place in a chloroplast. Section: 08.01 Topic: Photosynthesis
16. How many membranes does a chloroplast have? A. one B. two C. three D. four E. five A chloroplast has three membranes: an outer and inner membrane and a thylakoid membrane.
Bloom's Level: 4. Analyze Learning Outcome: 08.01.04 Describe the process of photosynthesis in terms of two sets of reactions that take place in a chloroplast. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
17. To what does the term grana refer? A. the cytochrome system in the membranes of the thylakoids B. a stack of thylakoid membrane structures C. the double membrane of the chloroplast D. a flattened disk or sac in the chloroplast E. the central fluid-filled space of the chloroplast Grana are stacks of thylakoid membranes within the chloroplast.
Bloom's Level: 1. Remember Learning Outcome: 08.01.04 Describe the process of photosynthesis in terms of two sets of reactions that take place in a chloroplast. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
18. In which plant structure does photosynthesis primarily occur? A. root B. stem C. leaf D. petiole E. bark The leaves of a plant are specialized to carry out photosynthesis.
Bloom's Level: 2. Understand Learning Outcome: 08.01.03 Recognize the overall chemical equation for photosynthesis. Section: 08.01 Topic: Photosynthesis
19. What are the two sets of reactions for photosynthesis? A. light reactions, Calvin cycle reactions B. glycolysis, citric acid cycle C. light reactions, glycolysis D. Calvin cycle reactions, citric acid cycle E. electron transport chain, light reactions The two sets of reactions in photosynthesis are the light reactions and the Calvin cycle reactions.
Bloom's Level: 1. Remember Learning Outcome: 08.01.04 Describe the process of photosynthesis in terms of two sets of reactions that take place in a chloroplast. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
20. An early experiment by Van Helmont (1600s) describes how he grew a tree in a large pot and found that after five years, the amount of soil in the pot had not changed. He concluded that the increase in weight was due to the addition of water. At that time, the compounds of the air had not yet been identified. Today, it is known that he only discovered half the story. Which of these experiments would NOT provide evidence that materials from a source other than water were involved in photosynthesis? A. Carbon is discovered as a major element in trees and is lacking in water molecules. B. Radioactive carbon in carbon dioxide in the air is identified as part of tree structures. C. A very careful analysis of the water taken in and lost by the tree would have revealed only part of the added weight gained by the tree was from water. D. After adding radioactive water with labeled oxygen to a plant, radioactive oxygen is given off. E. Radioactive carbon is found in stored glucose molecules after supplying a plant with radioactive carbon dioxide. Adding radioactive water with labeled oxygen would not address the other requirement for photosynthesis—carbon.
Bloom's Level: 5. Evaluate Learning Outcome: 08.01.04 Describe the process of photosynthesis in terms of two sets of reactions that take place in a chloroplast. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
21. The heavy oxygen isotope (18O) could be provided to plants either in the CO2 or in H2O or in both. Today, we know from experimental results that O2 released from chloroplasts comes from H2O and not from CO2. Which of the following experimental results would show conclusively that O2 is released from H2O and not CO2? A. When heavy oxygen is part of water given to the plant, the plant produces heavy O2. B. When heavy oxygen is part of both water and CO2 given to the plant, the plant produces heavy O2. C. When no heavy oxygen is part of water given to the plant, the plant produces no heavy O2. D. When no heavy oxygen is part of CO2 given to the plant, the plant produces no heavy O2. E. When heavy oxygen is part of CO2 given to the plant, the plant produces heavy O2. Because the oxygen comes from water, only heavy oxygen as part of water will produce heavy oxygen. In order to differentiate between water and carbon dioxide as the source of the oxygen, you could not give both water and carbon dioxide labeled with heavy oxygen.
Bloom's Level: 4. Analyze Learning Outcome: 08.01.04 Describe the process of photosynthesis in terms of two sets of reactions that take place in a chloroplast. Section: 08.01 Topic: Photosynthesis
22. Which two organelles are most directly involved in the flow of energy from the sun through all living things? A. Golgi apparatus and mitochondria B. lysosomes and chloroplasts C. chloroplasts and mitochondria D. mitochondria and ribosomes E. ribosomes and Golgi apparatus Chloroplasts are involved in the capture of solar energy to form carbohydrates. Mitochondria are involved in the breakdown of carbohydrates to form ATP.
Bloom's Level: 2. Understand Learning Outcome: 08.01.04 Describe the process of photosynthesis in terms of two sets of reactions that take place in a chloroplast. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
23. Which cofactor is involved in the Calvin cycle reactions? A. NAD+ B. FAD C. NADP+ D. both NAD+ and FAD E. both FAD and NADP+ NADP+ is the cofactor involved in the Calvin cycle reactions.
Bloom's Level: 1. Remember Learning Outcome: 08.03.01 Describe the three phases of the Calvin cycle. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
24. Which of the following statements is true concerning sunlight radiation used for photosynthesis? A. All of the sunlight that hits the atmosphere is used for photosynthesis. B. Only the highest energy wavelengths are used for photosynthesis. C. All of the visible light is used for photosynthesis. D. Only the red, blue, and violet wavelengths of visible light are used for photosynthesis. E. Only the green visible light is used for photosynthesis. Only the red, blue, and violet wavelengths of visible light are used for photosynthesis.
Bloom's Level: 1. Remember Learning Outcome: 08.02.01 Identify the photosynthetic pigments required to absorb the various wavelengths of light necessary for photosynthesis. Section: 08.02 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
25. Why are plant leaves green? A. They absorb only green wavelengths of light. B. They absorb only yellow and blue wavelengths of light. C. They reflect nearly all wavelengths of light. D. They reflect green wavelengths of light. E. They reflect yellow and blue wavelengths of light. Leaves reflect green light so they appear green to us.
Bloom's Level: 2. Understand Learning Outcome: 08.02.01 Identify the photosynthetic pigments required to absorb the various wavelengths of light necessary for photosynthesis. Section: 08.02 Topic: Photosynthesis
26. The reason that deciduous leaves turn bright colors of red and yellow in the fall is A. chlorophyll is converted into bright carotenoid pigments. B. when chlorophyll breaks down, the remaining pigments other than green will show through. C. it is a chemical reaction involving the formation of ice crystals. D. bright pigments are shifted from the roots and stem to the leaves while sugars and chlorophyll are pumped down to the roots. E. a shift in the reflectance values of light is caused by the sun being lower in the sky; all leaf pigments actually remain the same. The yellow and orange carotenoids are present but masked by the green chlorophyll. When chlorophyll breaks down in the fall, the other pigments are uncovered.
Bloom's Level: 3. Apply Learning Outcome: 08.02.01 Identify the photosynthetic pigments required to absorb the various wavelengths of light necessary for photosynthesis. Section: 08.02 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
27. We set up an experiment where the same species of plant is grown in boxes covered with plastic that only allows a single color of light through. In which box would we expect the least plant growth and even perhaps the death of the plant to occur? A. white B. blue C. red D. green E. indigo White light is a combination of all the colors so it would allow the plants to grow. Chlorophyll absorbs light in the indigo, blue, violet, and red region so those colors would allow the plant to grow. Chlorophyll does not absorb green light so that plant would not be able to photosynthesize and might die.
Bloom's Level: 3. Apply Learning Outcome: 08.02.01 Identify the photosynthetic pigments required to absorb the various wavelengths of light necessary for photosynthesis. Section: 08.02 Topic: Photosynthesis
28. Photosynthesis shows higher activity for violet/blue and orange/red and a lower absorption for green/yellow. If we could create a different photosynthetic pigment that absorbed absolutely all visible wavelengths of light, the leaves would appear which color? A. white B. black C. red D. green E. orange Since it absorbs all the colors, it does not reflect any, so it would appear black to us.
Bloom's Level: 3. Apply Learning Outcome: 08.02.01 Identify the photosynthetic pigments required to absorb the various wavelengths of light necessary for photosynthesis. Section: 08.02 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
True / False Questions 29. Visible light has more energy than radio waves. TRUE Visible light is in the middle of the spectrum and contains more energy than radio waves.
Bloom's Level: 1. Remember Learning Outcome: 08.02.01 Identify the photosynthetic pigments required to absorb the various wavelengths of light necessary for photosynthesis. Section: 08.02 Topic: Photosynthesis
30. There is an inverse relationship between wavelength of light and amount of energy. TRUE The greater the wavelength, the less the energy. This is an inverse relationship.
Bloom's Level: 1. Remember Learning Outcome: 08.02.01 Identify the photosynthetic pigments required to absorb the various wavelengths of light necessary for photosynthesis. Section: 08.02 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
Multiple Choice Questions 31. The "antenna" system of a plant that gathers the solar energy consists of A. the grana. B. the epidermal cells of the leaf. C. the pigment complex. D. the ATP synthase complex. E. the hydrogen ions (H+) The pigment complex gathers the solar energy.
Bloom's Level: 1. Remember Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Light Reactions
32. Water is split and oxygen is released in A. the electron transport chain. B. the cyclic electron pathway. C. the Calvin cycle reactions. D. the noncyclic electron pathway. E. photosystem I. In the noncyclic electron pathway, water is split in order to replace the electrons that are lost by photosystem II.
Bloom's Level: 2. Understand Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Light Reactions
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Chapter 08 - Photosynthesis
33. Which part of photosynthesis does not occur inside the thylakoid membrane? A. noncyclic electron pathway B. cyclic electron pathway C. electron transport chain D. light reactions E. Calvin cycle reactions The Calvin cycle reactions occur within the stroma of the chloroplast. All of the others occur in the thylakoid membrane.
Bloom's Level: 2. Understand Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Calvin Cycle/Carbon Reactions
34. The cyclic electron pathway produces only A. NADPH. B. ATP. C. NAD+. D. oxygen. E. carbon dioxide. The cyclic pathway produces only ATP. The noncyclic produces NADPH, ATP, H+, and O2.
Bloom's Level: 2. Understand Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Light Reactions
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Chapter 08 - Photosynthesis
35. Which of these is most closely associated with the process of electron transport? A. a sequence of molecules in the membranes of the thylakoids B. a stack of thylakoid membrane structures C. the double membrane of the chloroplast D. a flattened disk or sac in the chloroplast E. the central fluid-filled space of the chloroplast The electron transport chain consists of a series of electron carriers in the thylakoid membrane.
Bloom's Level: 2. Understand Learning Outcome: 08.02.03 Describe the organization of the thylakoid and how this organization is critical to the production of ATP during photosynthesis. Section: 08.02 Topic: Light Reactions
True / False Questions 36. Photosystem I comes before photosystem II in the noncyclic electron pathway. FALSE The photosystems were named for the order in which they were discovered, not the order in which they participate in photosynthesis. Photosystem II comes before photosystem I.
Bloom's Level: 1. Remember Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Light Reactions
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Chapter 08 - Photosynthesis
37. ATP synthase is an enzyme complex located in the thylakoid membranes. TRUE The ATP synthase in the thylakoid membranes participates in the production of ATP for the Calvin cycle.
Bloom's Level: 1. Remember Learning Outcome: 08.02.03 Describe the organization of the thylakoid and how this organization is critical to the production of ATP during photosynthesis. Section: 08.02 Topic: Light Reactions
38. The first step of the cyclic electron pathway involves the removal of the electrons from water. FALSE The electrons for the cyclic electron pathway recycle. Water is used to replace the electrons in photosystem II in the noncyclic electron pathway.
Bloom's Level: 2. Understand Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Light Reactions
Multiple Choice Questions 39. What are the products of the noncyclic electron pathway? A. H+, O2, ATP, and NADPH B. H+ and O2 C. H+, ATP, and NADPH D. ATP and NADPH E. O2 and ATP The noncyclic electron pathway produces H+, O2, ATP, and NADPH.
Bloom's Level: 1. Remember Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Light Reactions
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Chapter 08 - Photosynthesis
40. What pathway moves the electrons from water through PS II to PS I and then on to NADP+? A. noncyclic electron pathway B. cyclic electron pathway C. CO2 fixation stage of Calvin cycle reactions D. citric acid cycle E. CO2 reduction phase of Calvin cycle reactions This occurs in the noncyclic electron pathway of the light reactions.
Bloom's Level: 2. Understand Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Light Reactions
41. What will cycle the energized electrons from the reaction center of PS I back to the reaction center in PS I? A. noncyclic electron pathway B. cyclic electron pathway C. CO2 fixation stage of Calvin cycle reactions D. citric acid cycle E. CO2 reduction phase of Calvin cycle reactions This occurs in photosystem I of the cyclic electron pathway.
Bloom's Level: 2. Understand Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Light Reactions
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Chapter 08 - Photosynthesis
42. Biologists are fairly certain that oxygen was built up in the atmosphere by the development of photosynthesis. The production of oxygen and, therefore, the beginning of extensive aerobic life was made possible by the evolution of A. fermentation. B. photosystem I. C. photosystem II. D. glycolysis. E. the nucleus. Photosystem II splits water to form H+ and O2.
Bloom's Level: 2. Understand Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Light Reactions
43. Why does the chloroplast utilize a cyclic electron pathway as well as the noncyclic electron pathway? A. The noncyclic electron pathway does not provide enough NADPH for the cell. B. The noncyclic electron pathway does not provide enough oxygen for the cell. C. The Calvin cycle utilizes more NADPH than ATP and the cyclic electron pathway produces NADPH. D. The Calvin cycle utilizes more ATP than NADPH and the cyclic electron pathway produces ATP. E. The cyclic electron pathway only runs when oxygen is limiting. The cell utilizes both because the noncyclic pathway provides both NADPH and ATP but the Calvin cycle reactions require more ATP than NADPH. Because the cyclic electron pathway only produces ATP it can provide the extra needed by the Calvin cycle.
Bloom's Level: 4. Analyze Learning Outcome: 08.02.02 Explain the role of the noncyclic electron pathway and the cyclic electron pathway. Section: 08.02 Topic: Light Reactions
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Chapter 08 - Photosynthesis
44. The H+ (protons) accumulate in the thylakoid space during electron transport between photosystems I and II. The excess of protons in the thylakoid space A. enters the photorespiration pathway. B. raises the pH of the space until the processes stop. C. is small enough to diffuse back out through the lipid bilayer. D. converts NADP to NADPH and generates ATP in the process. E. moves from the thylakoid space to the stroma through the ATP synthase complex. These excess H+ move through the ATP synthase complex and synthesize ATP from ADP + P. This is chemiosmosis.
Bloom's Level: 2. Understand Learning Outcome: 08.02.03 Describe the organization of the thylakoid and how this organization is critical to the production of ATP during photosynthesis. Section: 08.02 Topic: Light Reactions
45. Which of the following statements comparing the ATP synthase complex in cellular respiration and photosynthesis is not true? A. Both require the movement of H+ from inside a compartment to outside of the compartment. B. Both produce ATP. C. Both work via chemiosmosis. D. Both require a H+ gradient. E. Both are located in a membrane. In photosynthesis, the H+ move from inside the thylakoid space to outside while in cellular respiration, the H+ move from outside the matrix inside.
Bloom's Level: 5. Evaluate Learning Outcome: 08.02.03 Describe the organization of the thylakoid and how this organization is critical to the production of ATP during photosynthesis. Section: 08.02 Topic: Light Reactions
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Chapter 08 - Photosynthesis
46. The reservoir for hydrogen ions for chemiosmotic ATP synthesis during photosynthesis is the A. stroma. B. thylakoid membrane. C. thylakoid space. D. cytoplasm. E. matrix. The H+ accumulate in the thylakoid space.
Bloom's Level: 1. Remember Learning Outcome: 08.02.03 Describe the organization of the thylakoid and how this organization is critical to the production of ATP during photosynthesis. Section: 08.02 Topic: Light Reactions
47. The formation of carbohydrate occurs within the A. stroma. B. outer chloroplast membrane. C. inner chloroplast membrane. D. thylakoid membranes. E. thylakoid space. The Calvin cycle, which produces carbohydrate operates within the stroma.
Bloom's Level: 2. Understand Learning Outcome: 08.03.01 Describe the three phases of the Calvin cycle. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
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Chapter 08 - Photosynthesis
48. NADPH and ATP are used in the A. noncyclic electron pathway. B. cyclic electron pathway. C. Calvin cycle reactions. D. citric acid cycle. E. light reactions. The NADPH and ATP generated in the light reactions are used by the Calvin cycle reactions to produce carbohydrates.
Bloom's Level: 1. Remember Learning Outcome: 08.03.01 Describe the three phases of the Calvin cycle. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
49. In the Calvin cycle, the three-carbon molecule that will be immediately formed after carbon dioxide fixation occurs is A. PS I. B. glucose. C. 3PG. D. G3P. E. RuBP. The first three-carbon molecule formed is 3PG (3-phosphoglycerate).
Bloom's Level: 1. Remember Learning Outcome: 08.03.01 Describe the three phases of the Calvin cycle. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
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Chapter 08 - Photosynthesis
50. What are the stages of the Calvin cycle? A. carbon dioxide fixation and reduction B. carbon dioxide fixation and regeneration of RuBP C. the noncyclic electron pathway and the cyclic electron pathway D. the light reactions, regeneration of RuBP, and cyclic electron pathway E. carbon dioxide fixation, carbon dioxide reduction, and regeneration of RuBP The Calvin cycle has three stages: carbon dioxide fixation, carbon dioxide reduction, and the regeneration of RuBP.
Bloom's Level: 2. Understand Learning Outcome: 08.03.01 Describe the three phases of the Calvin cycle. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
51. How many total carbons are involved in carbon dioxide fixation with 3 molecules of carbon dioxide and 3 molecules of RuBP? A. 6 B. 12 C. 18 D. 24 E. 30 Carbon dioxide contains 1 carbon each while RuBP contains 5 carbons each, for a total of 18 carbons.
Bloom's Level: 2. Understand Learning Outcome: 08.03.01 Describe the three phases of the Calvin cycle. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
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Chapter 08 - Photosynthesis
52. The Calvin cycle reactions are dependent upon a supply of A. water and carbon dioxide. B. carbon dioxide and NADPH. C. carbon dioxide, NADPH, and ATP. D. oxygen and carbon dioxide. E. glucose and carbon dioxide. The Calvin cycle fixes carbon dioxide using NADPH and ATP so it needs all three molecules.
Bloom's Level: 1. Remember Learning Outcome: 08.03.01 Describe the three phases of the Calvin cycle. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
53. Some herbicides inhibit the electron transport chain in the thylakoid membrane. Without the movement of electrons, hydrogen ions would not be pumped from the stroma to the thylakoid space and the hydrogen ion gradient would not be established. How would this affect the Calvin cycle reactions? A. ATP would not be produced and, as a result, the Calvin cycle reactions would not occur. B. CO2 would not enter the cell as a result, and the Calvin cycle reactions would not occur. C. RuBP carboxylase would not function properly, so CO2 fixation would not occur. D. Sunlight could no longer be used by the chloroplast, but this would have no effect on the Calvin cycle reactions because they do not require light. E. Since the Calvin cycle reactions occur in a different part of the chloroplast, there would be no effect. The electron transport chain produces ATP. Without ATP the Calvin cycle reactions would not run.
Bloom's Level: 5. Evaluate Learning Outcome: 08.03.01 Describe the three phases of the Calvin cycle. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
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Chapter 08 - Photosynthesis
54. What is the function of ribulose bisphosphate (RuBP) during photosynthesis? A. It produces the CO2 needed for photosynthesis. B. It combines with CO2 to produce a 6-carbon compound. C. It combines with ATP to form G3P. D. It splits water to release oxygen. E. It splits carbon dioxide to release oxygen. RuBP joins with carbon dioxide during the first step of the Calvin cycle, carbon dioxide fixation, to form three 6-carbon molecules.
Bloom's Level: 2. Understand Learning Outcome: 08.03.01 Describe the three phases of the Calvin cycle. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
55. How much of the protein content in chloroplasts is made up of the enzyme RuBP carboxylase? Why? A. less than 1%; the enzyme is very fast B. less than 1%; the enzyme also catalyzes several side reactions C. 20-50%; the enzyme also catalyzes several side reactions D. 20-50%; the enzyme is very slow E. almost 100%; the enzyme is very slow RuBP carboxylase makes up 20-50% of the chloroplast proteins because it is unusually slow. It does catalyze a side reaction but that is not the reason it is so abundant.
Bloom's Level: 3. Apply Learning Outcome: 08.03.01 Describe the three phases of the Calvin cycle. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
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Chapter 08 - Photosynthesis
56. During photosynthesis, CO2 is reduced. This means that A. the molecule gains electrons. B. CO2 is broken down into two smaller molecules. C. the molecule loses electrons. D. the molecule loses protons. E. CO2 is released into the environment. Reduction is the gain of electrons. CO2 will bond to hydrogen and oxygen to form glucose.
Bloom's Level: 1. Remember Learning Outcome: 08.03.02 Explain how the products of the Calvin cycle are used to form the other molecules found in plants. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
57. One of the products of the Calvin cycle is A. PS I. B. RuBP carboxylase. C. 3PG. D. G3P. E. RuBP. G3P (glyceraldehyde-3-phosphate) is the end product of the Calvin cycle. Two of these make up a glucose.
Bloom's Level: 1. Remember Learning Outcome: 08.03.02 Explain how the products of the Calvin cycle are used to form the other molecules found in plants. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
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Chapter 08 - Photosynthesis
58. Plants need other molecules besides glucose. Where do these molecules, such as cellulose and fructose, come from? A. Glucose must always be produced first; glucose can then be used as the monomer to form everything else. B. G3P is directly converted to many other organic molecules besides glucose. C. Plants absorb those molecules from the environment through their roots. D. Any molecule beyond glucose must be converted from plant tissues already present. E. Alternative forms of photosynthesis beyond those described must be producing those molecules. G3P can be converted to other molecules a plant needs, such as glucose, sucrose, starch, cellulose, fatty acids, and amino acids.
Bloom's Level: 2. Understand Learning Outcome: 08.03.02 Explain how the products of the Calvin cycle are used to form the other molecules found in plants. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
59. The hydrocarbon skeleton used to form fatty acids, glycerol, and amino acids (when N is added) is derived from A. starch. B. sucrose. C. cellulose. D. G3P. E. glucose phosphate. G3P can be converted to other molecules a plant needs, such as glucose, sucrose, starch, cellulose, fatty acids, and amino acids.
Bloom's Level: 2. Understand Learning Outcome: 08.03.02 Explain how the products of the Calvin cycle are used to form the other molecules found in plants. Section: 08.03 Topic: Calvin Cycle/Carbon Reactions
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Chapter 08 - Photosynthesis
60. Most major food plants such as wheat, oats, and rice are in the group of A. nonphotosynthetic plants. B. C3 plants. C. C4 plants. D. C5 plants. E. CAM plants. Wheat, rice, and oats are examples of C3 plants.
Bloom's Level: 1. Remember Learning Outcome: 08.04.01 Contrast C3, C4, and CAM photosynthesis. Section: 08.04 Topic: C3, C4, and CAM Photosynthesis
61. Which of the following is a CAM plant? A. rice B. cactus C. corn D. wheat E. sugarcane A cactus is an example of a CAM plant. Wheat and rice are C3 plants. Corn and sugarcane are C4 plants.
Bloom's Level: 1. Remember Learning Outcome: 08.04.01 Contrast C3, C4, and CAM photosynthesis. Section: 08.04 Topic: C3, C4, and CAM Photosynthesis
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Chapter 08 - Photosynthesis
62. Which statement is not true about C3 and C4 plants? A. C4 plants are more successful in hot climates than C3 plants. B. C3 plants fix carbon dioxide in the mesophyll cells. C. In C3 plants, O2 competes with CO2 for the active site of RuBP carboxylase. D. C4 plants deliver CO2 to the Calvin cycle using bundle sheath cells sheltered from leaf air spaces. E. CO2 is fixed at night to decrease water loss. CAM plants fix carbon dioxide at night to decrease water loss.
Bloom's Level: 4. Analyze Learning Outcome: 08.04.01 Contrast C3, C4, and CAM photosynthesis. Section: 08.04 Topic: C3, C4, and CAM Photosynthesis
63. Over time, what would we expect in the evolution of C3, C4, and CAM strategies? A. As the most complicated plants, CAM plants are most likely to go extinct. B. As the probable ancestral form of photosynthesis, C3 plants are primitive and most likely to go extinct. C. Because CAM plants make better usage of metabolism during the night, they are superior and will eventually become the dominant plants. D. Because CO2 is delivered by the bundle sheath cells in C4 plants, they are superior and will eventually dominate. E. Each form of photosynthesis has advantages in a heterogeneous world and, as long as environmental conditions vary, all forms will have an adaptive advantage in their unique niche. Because the environmental conditions around Earth vary, each type of carbon dioxide fixation has its advantages.
Bloom's Level: 3. Apply Learning Outcome: 08.04.02 Explain how different photosynthetic modes allow plants to adapt to a particular environment. Section: 08.04 Topic: C3, C4, and CAM Photosynthesis
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Chapter 08 - Photosynthesis
64. CAM plants A. do not perform any part of photosynthesis during the day. B. fix CO2 in mesophyll cells during the day. C. include economically important food crops like corn and sugarcane. D. open their stomata at night to minimize water loss. E. typically grow very quickly because they are found in hot, moist environments. CAM plants open their stomata at night to minimize water loss. They include plants such as cacti and are found in hot, dry environments.
Bloom's Level: 2. Understand Learning Outcome: 08.04.01 Contrast C3, C4, and CAM photosynthesis. Section: 08.04 Topic: C3, C4, and CAM Photosynthesis
65. Which of the following is NOT a characteristic shared by cellular respiration and photosynthesis? A. Both photosynthesis and cellular respiration occur in plant cells. B. Both processes produce ATP by chemiosmosis. C. Both processes produce oxygen. D. Both processes use an electron transport chain located in membranes of organelles. E. One or more electron carriers are used in both processes. Photosynthesis produces oxygen, while cellular respiration produces carbon dioxide.
Bloom's Level: 2. Understand Learning Outcome: 08.05.02 Describe the similarities and differences between cellular respiration and photosynthesis. Section: 08.05 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
66. Which of the following statements is false? A. During cellular respiration, carbohydrate energy is converted into ATP. B. During cellular respiration, mitochondria release carbon dioxide. C. During photosynthesis, energy from the sun is used to form carbohydrates. D. During photosynthesis, carbon dioxide is used. E. During photosynthesis, oxygen is used. During photosynthesis, oxygen is produced.
Bloom's Level: 2. Understand Learning Outcome: 08.05.02 Describe the similarities and differences between cellular respiration and photosynthesis. Section: 08.05 Topic: Photosynthesis
True / False Questions 67. Photosynthesis and cellular respiration utilize the same chemical pathways but in reverse of each other. FALSE The chemical equation for photosynthesis and cellular respiration are the reverse of each other, but the pathways are not the same.
Bloom's Level: 2. Understand Learning Outcome: 08.05.02 Describe the similarities and differences between cellular respiration and photosynthesis. Section: 08.05 Topic: Photosynthesis
68. Only organisms with chloroplasts are capable of photosynthesis. FALSE Cyanobacteria do not have organelles, yet they carry on photosynthesis. They do have thylakoids.
Bloom's Level: 2. Understand Learning Outcome: 08.05.02 Describe the similarities and differences between cellular respiration and photosynthesis. Section: 08.05 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
69. At night, plants cannot run metabolic pathways because there is no sunlight. FALSE Plants carry out cellular respiration at night. They cannot carry out the light reactions of photosynthesis without sunlight, but they carry out other chemical reactions.
Bloom's Level: 3. Apply Learning Outcome: 08.05.02 Describe the similarities and differences between cellular respiration and photosynthesis. Section: 08.05 Topic: Photosynthesis
Multiple Choice Questions 70. In what ways are photosynthesis and cellular respiration not alike? A. Both will use the electron carrier NADH. B. Both require ATP. C. Both require a membrane. D. Both require a cyclical enzymatic pathway. E. Both require an electron transport chain. Cellular respiration requires the electron carrier NADH, while photosynthesis requires NADPH.
Bloom's Level: 4. Analyze Learning Outcome: 08.05.02 Describe the similarities and differences between cellular respiration and photosynthesis. Section: 08.05 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
71. Which statement about photosynthesis and cellular respiration is true? A. Photosynthesis produces water, while cellular respiration uses water. B. Photosynthesis produces oxygen, while cellular respiration uses oxygen. C. Photosynthesis occurs in mitochondria, while cellular respiration occurs in chloroplasts. D. Photosynthesis breaks down carbohydrates, while cellular respiration produces carbohydrates. E. Photosynthesis requires oxygen, while cellular respiration requires sunlight. It is true that photosynthesis produces oxygen, while cellular respiration uses oxygen. Photosynthesis does not produce water, does not occur in the mitochondria, does not break down carbohydrates, and does not require oxygen.
Bloom's Level: 4. Analyze Learning Outcome: 08.05.02 Describe the similarities and differences between cellular respiration and photosynthesis. Section: 08.05 Topic: Photosynthesis
72. What is the value of photosynthesis to humans? A. It results in the production of fatty acids and oxygen. B. It results in the production of proteins and oxygen. C. It will convert carbohydrates into usable forms of energy for humans. D. It will reduce the amount of methane in the atmosphere. E. It will reduce the amount of carbon dioxide in the atmosphere. Photosynthesis will reduce the amount of carbon dioxide in the atmosphere and form sugars and oxygen. The reduction of carbon dioxide in the atmosphere will help reduce climate change.
Bloom's Level: 3. Apply Learning Outcome: 08.01.02 Explain the role of photosynthesis for all organisms on Earth. Section: 08.01 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
73. Which pathway is the foundation for the majority of ecosystems and food chains on Earth? A. photosynthesis B. anaerobic respiration C. aerobic respiration D. chemosynthesis E. decomposition Photosynthesis results in the formation of energy in the form of various carbohydrates. These carbohydrates provide fuel for the majority of food chains. Anaerobic respiration, aerobic respiration, and decomposition all require the input of carbohydrates. Chemosynthesis can form energy but it is not highly used.
Bloom's Level: 2. Understand Learning Outcome: 08.01.02 Explain the role of photosynthesis for all organisms on Earth. Section: 08.01 Topic: Photosynthesis
74. Which photosynthetic mode would be most favorable in a desert environment? A. CAM B. C3 C. C4 D. both CAM and C3 E. CAM, C3, and C4 Plants that are adapted to a desert environment will run the CAM form of photosythesis. C3 plants are more adapted to cool, moist environments since they do not do well when drought conditions occur. C4 plants are better suited to the hot, dry periods during the growing season due to their ability to avoid photorespiration, but they are not able to survive in a desert environment.
Bloom's Level: 2. Understand Learning Outcome: 08.04.02 Explain how different photosynthetic modes allow plants to adapt to a particular environment. Section: 08.04 Topic: C3, C4, and CAM Photosynthesis
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Chapter 08 - Photosynthesis
75. Which of the following organisms do not have a dependence on photosynthesis for their energy? A. white-tailed deer B. wolves C. squirrels D. raccoons E. All of the answer choices will have a dependency on photosynthesis. All of the answer choices will have a dependency on photosynthesis. White-tailed deer, squirrels, and raccoons will directly eat plant products that are produced by photosynthesis. Wolves will eat animals that have eaten plant products that are products of photosynthesis.
Bloom's Level: 1. Remember Learning Outcome: 08.01.02 Explain the role of photosynthesis for all organisms on Earth. Section: 08.01 Topic: Photosynthesis
Short Answer Questions 76. Explain how CAM plants are adapted to a desert type of environment. CAM plants separate components of the photosynthesis pathway by time. CO2 is fixed into a four-carbon molecule at night when it is cooler and they can keep their stomata open. During the day, the stomata will be closed and the four-carbon molecule will release the CO2 to the Calvin cycle.
Bloom's Level: 6. Create Learning Outcome: 08.04.02 Explain how different photosynthetic modes allow plants to adapt to a particular environment. Section: 08.04 Topic: C3, C4, and CAM Photosynthesis
77. Write the overall chemical equation for photosynthesis and cellular respiration. Energy + 6CO2 + 6H2O = C6H12O6 + 6H2O
Bloom's Level: 6. Create Learning Outcome: 08.05.01 Compare the overall chemical equation for photosynthesis and cellular respiration. Section: 08.05 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
Multiple Choice Questions 78. Which statement is accurate about how C4 plants are able to survive hot, dry spells during the growing season? A. When the stomata close during heat stress, RuBP carboxylase is not exposed to the increase in oxygen concentration experienced in the leaf. This allows the plant to avoid photorespiration. B. When the stomata close during heat stress, RuBP carboxylase is exposed to the increase in oxygen concentration experienced in the leaf. This allows the plant to avoid photorespiration. C. When the stomata open during heat stress, RuBP carboxylase is not exposed to the increase in oxygen concentration experienced in the leaf. This allows the plant to avoid photorespiration. D. When the stomata close during heat stress, RuBP carboxylase is exposed to the increase in oxygen concentration experienced in the leaf. This allows the plant to undergo photorespiration. E. When the stomata open during heat stress, RuBP carboxylase is not exposed to the increase in oxygen concentration experienced in the leaf. This allows the plant to undergo photorespiration. C4 plants have an advantage during hot, dry spells because the stomata will close during heat stress. This prevents RuBP carboxylase from being exposed to the increase in oxygen concentration experienced in the leaf, avoiding photorespiration. Photorespiration is an inefficient way to produce carbohydrates.
Bloom's Level: 4. Analyze Learning Outcome: 08.04.02 Explain how different photosynthetic modes allow plants to adapt to a particular environment. Section: 08.04 Topic: C3, C4, and CAM Photosynthesis
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Chapter 08 - Photosynthesis
Short Answer Questions 79. Explain why chlorophyll appears green to our eyes. When the wavelengths of visible light enter the chlorophyll, the pigment molecules will absorb various wavelengths of light while reflecting back others. It is the reflected wavelengths that are interpreted by our brain as various colors. The wavelength spectrum absorbed by chlorophyll includes violet, indigo, blue, and red. The wavelength spectrum that is reflected back includes the greens, which is why we interpret chlorophyll as green.
Bloom's Level: 6. Create Learning Outcome: 08.02.01 Identify the photosynthetic pigments required to absorb the various wavelengths of light necessary for photosynthesis. Section: 08.02 Topic: Photosynthesis
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Chapter 08 - Photosynthesis
Multiple Choice Questions 80. Which of the following is a substrate in the photosynthesis equation? A. CO2 B. O2 C. C6H12O6 D. H2 The substrates of photosynthesis include CO2 and H2O. C6H12O6 and O2 are the end products of the reaction.
Bloom's Level: 1. Remember Learning Outcome: 08.05.01 Compare the overall chemical equation for photosynthesis and cellular respiration. Section: 08.05 Topic: Photosynthesis
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Chapter 09 - Plant Organization and Function
Chapter 09 Plant Organization and Function
Multiple Choice Questions 1. In roots, which structure(s) increases surface area for the absorption of water and minerals from the soil? A. root hairs B. root cap C. Casparian strip D. vascular cylinder E. zone of elongation Root hairs on the plant roots increase the surface area for absorption.
Bloom's Level: 1. Remember Learning Outcome: 09.01.01 Explain the function of meristematic tissue. Section: 09.01 Topic: Plant Tissues
2. You are planning to build a swimming pool in your backyard. How close can you get to the trees in your backyard without disturbing their root structure? A. all the way up to the trunk B. within a foot of the trunk C. up to the crown of the tree D. at least 2–4 times the diameter of the crown of the tree E. at least 10–20 times the diameter of the crown of the tree The roots of a tree extend out much farther than the crown of the tree; on average, 2–4 times the diameter of the crown of the tree.
Bloom's Level: 3. Apply Learning Outcome: 09.02.02 Compare the structure and function of roots, stems, and leaves. Section: 09.02 Topic: Plant Organs
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Chapter 09 - Plant Organization and Function
3. Most of the water obtained by a plant enters through the A. cambium layer. B. palisade cells. C. root hairs. D. stomata. E. bark. The root hairs of the roots absorb the majority of the plant's water.
Bloom's Level: 2. Understand Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
4. Which of the following is not a major function of roots? A. absorb water B. absorb minerals C. produce hormones D. anchor and support the plant E. transport of materials throughout the plant Xylem and phloem, not roots, are responsible for transportation of materials throughout the plant.
Bloom's Level: 2. Understand Learning Outcome: 09.02.02 Compare the structure and function of roots, stems, and leaves. Section: 09.02 Topic: Plant Organs
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Chapter 09 - Plant Organization and Function
5. Which of the following is a reproductive structure of a plant? A. roots B. stems C. flowers D. leaves E. shoot system Flowers are structures involved in plant reproduction.
Bloom's Level: 1. Remember Learning Outcome: 09.02.01 Describe the vegetative and reproductive organs of plants and indicate their major functions. Section: 09.02 Topic: Plant Organs
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Chapter 09 - Plant Organization and Function
True / False Questions 6. As soon as a beautiful plant in your backyard blooms, you remove the flowers. By doing so, you have now inhibited the growth of the plant. FALSE By removing the flowers you have inhibited the reproduction of the plant, but the root and shoot system will still be able to grow.
Bloom's Level: 2. Understand Learning Outcome: 09.02.01 Describe the vegetative and reproductive organs of plants and indicate their major functions. Section: 09.02 Topic: Plant Organs
7. Both plants and animals have organs and tissues. TRUE We are more familiar with organs and tissues in animals, but plants also have these.
Bloom's Level: 1. Remember Learning Outcome: 09.01.01 Explain the function of meristematic tissue. Section: 09.01 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
Multiple Choice Questions 8. The point on a stem to which a leaf or bud is attached is termed the A. bud scale. B. axillary bud. C. node. D. internode. E. lenticel. A node occurs where leaves are attached to the stem.
Bloom's Level: 1. Remember Learning Outcome: 09.02.02 Compare the structure and function of roots, stems, and leaves. Section: 09.02 Topic: Plant Organs
9. The stalk that attaches a leaf blade to a stem is a(n) A. bud scale. B. axillary bud. C. node. D. internode. E. petiole. The petiole attaches the blade of the leaf to the stem.
Bloom's Level: 1. Remember Learning Outcome: 09.02.02 Compare the structure and function of roots, stems, and leaves. Section: 09.02 Topic: Plant Organs
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Chapter 09 - Plant Organization and Function
True / False Questions 10. The cells that make up xylem are no longer living. TRUE Xylem consists of nonliving cells that form a continuous pipeline for water and mineral transport.
Bloom's Level: 1. Remember Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
11. A stem is identified by being aboveground and green. FALSE A stem is identified by the presence of nodes and internodes, even if it is not aboveground or green.
Bloom's Level: 1. Remember Learning Outcome: 09.02.02 Compare the structure and function of roots, stems, and leaves. Section: 09.02 Topic: Plant Organs
12. The tendrils that support vining plants are a type of stem. FALSE Tendrils are most often specialized leaves.
Bloom's Level: 1. Remember Learning Outcome: 09.06.02 Describe some types of leaf modifications. Section: 09.06 Topic: Plant Organs
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Chapter 09 - Plant Organization and Function
Multiple Choice Questions 13. Epidermal tissue, ground tissue, and vascular tissue systems are derived from A. sclerenchyma cells. B. vascular cambium. C. cork cambium. D. apical meristem. E. vascular bundles. Apical meristem continually produces three types of primary meristem, and these develop into the three types of primary tissues of the plant.
Bloom's Level: 1. Remember Learning Outcome: 09.01.01 Explain the function of meristematic tissue. Section: 09.01 Topic: Plant Tissues
14. Which of these associations is incorrect? A. epidermal tissue-epidermal cell B. ground tissue-parenchymal cell C. ground tissue-tracheid D. vascular tissue-sieve-tube member E. vascular tissue-vessel element Ground tissue can be paired with parenchyma, collenchyma, or sclerenchyma. Tracheids are part of vascular tissue.
Bloom's Level: 4. Analyze Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
15. The function of meristematic tissue is A. production of new tissues. B. protection. C. transport. D. support. E. reproduction. Meristematic tissue produces all of the tissue types of the plant.
Bloom's Level: 1. Remember Learning Outcome: 09.01.01 Explain the function of meristematic tissue. Section: 09.01 Topic: Plant Tissues
16. Cells of the ground tissue system are called A. epidermal cells. B. tracheids and vessel elements. C. sieve-tube members and companion cells. D. periderm and cork cells. E. parenchyma, collenchyma, and sclerenchyma cells. Ground tissue is composed of parenchyma, collenchyma, and sclerenchyma cells.
Bloom's Level: 1. Remember Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
17. Plant vascular tissue contains A. collenchyma and lignin. B. xylem and phloem. C. epidermis and cuticle. D. periderm and cork. E. parenchyma and sclerenchyma cells. Vascular tissue is composed of xylem and phloem.
Bloom's Level: 1. Remember Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
True / False Questions 18. Upon visiting a redwood forest, you realize that because of their meristematic tissue these ancient trees are still capable of growth. TRUE Meristematic tissue enables plants to grow their entire lives.
Bloom's Level: 2. Understand Learning Outcome: 09.01.01 Explain the function of meristematic tissue. Section: 09.01 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
Multiple Choice Questions 19. What organic substance makes the walls of sclerenchyma cells tough and hard? A. suberin B. stomata C. trichomes D. cuticle E. lignin Lignin is a highly resistant organic substance that makes walls tough and hard.
Bloom's Level: 2. Understand Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
20. Sugars are conducted through plant tissues by the A. xylem. B. phloem. C. epidermis. D. trichomes. E. procambium. Xylem and phloem are transport tissues. Xylem transports water and minerals, while phloem transports organic molecules such as sugars.
Bloom's Level: 1. Remember Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
21. An aphid is an insect that has mouthparts to sip plant sap. When placed on an herbaceous stem, it soon has its mouthparts embedded and is withdrawing plant sap. Sometimes so much plant sap is taken that it cannot use all of it and the aphid secretes droplets of the sugar to attract ants that in turn protect and care for the aphids. What layer of cells has the aphid's mouthparts tapped? A. xylem B. phloem C. parenchyma D. trichomes E. collenchyma Phloem transports sugars around the cell. Therefore, the aphid has tapped into the phloem in order to harvest sugars.
Bloom's Level: 2. Understand Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
22. The fibers that get stuck between your teeth when you eat celery are A. sclerenchyma. B. meristems. C. trichomes. D. parenchyma. E. collenchyma. The strands in celery stalks are composed mostly of collenchyma cells.
Bloom's Level: 1. Remember Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
23. Which of these associations is incorrect? A. epidermal cell--protection B. stomata--gas exchange C. sclerenchyma cells--cell division D. sieve-tube members--sugar transport E. xylem cells--water transport Sclerenchyma cells support the mature regions of the plant. Meristems are associated with cell division.
Bloom's Level: 2. Understand Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
24. Which cells are relatively unspecialized and correspond to a generalized plant cell? A. epidermal B. tracheid C. parenchyma D. sclerenchyma E. meristem Parenchyma cells are the least specialized of the cell types and correspond best to the typical plant cell.
Bloom's Level: 1. Remember Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
25. Xylem tissue contains A. vessel elements and tracheids. B. sieve-tube members and companion cells. C. apical meristem. D. epidermis and ground tissue. E. parenchyma and sclerenchyma cells. Xylem contains two types of conducting cells: tracheids and vessel elements.
Bloom's Level: 1. Remember Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
26. Which cell found in phloem tissue has a nucleus? A. vessel element B. tracheids C. sieve-tube member D. guard cell E. companion cell The companion cell has a nucleus. The other conducting cells of phloem are sieve-tube members, which contain cytoplasm but no nucleus.
Bloom's Level: 1. Remember Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
27. Which association is incorrect? A. vessel element--xylem B. guard cells--epidermis C. parenchyma--ground tissue D. sieve-tube member--bark E. cuticle--epidermis A sieve-tube member is part of phloem, not bark.
Bloom's Level: 2. Understand Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
28. If you wanted to grow a new plant in the laboratory with the appropriate technology, which tissue type would you not start with? A. companion cells B. parenchyma cells C. sieve-tube members D. epidermis E. meristem All of the answer choices are living and still contain a nucleus, except for the sieve-tube members.
Bloom's Level: 3. Apply Learning Outcome: 09.01.02 Describe the types of tissues found in plants. Section: 09.01 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
True / False Questions 29. Some plant leaves and stems are covered with hairs. TRUE Epidermal cells produce hairs called trichomes in plants.
Bloom's Level: 1. Remember Learning Outcome: 09.06.01 Identify the structures of a leaf. Section: 09.06 Topic: Plant Organs
Multiple Choice Questions 30. Monocots differ from eudicots in that monocots have A. two cotyledons in the seed. B. leaf veins that are parallel. C. flower parts in 4s or 5s. D. stem vascular bundles in a distinct ring. E. phloem in a root located between areas of xylem. Monocots have one cotyledon, root xylem and phloem in a ring, scattered vascular bundles in the stem, leaf veins in a parallel pattern, and flower parts in threes or multiples of three.
Bloom's Level: 2. Understand Learning Outcome: 09.03.01 List and describe the key features of monocots and eudicots. Section: 09.03 Topic: Plants
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Chapter 09 - Plant Organization and Function
31. The iris is a plant with long, sword-like leaves. Its flowers have six petals that form a tongue and "flags." When you cut through the stem to cut flowers for a bouquet arrangement, the many vascular bundles form tough "strings" throughout the stem. From this description, it is obvious that the iris is a A. woody plant. B. combination of monocot and eudicot traits. C. monocotyledon. D. eudicotyledon. E. primitive plant that has not yet evolved to have either monocot or eudicot features. This is a description of a monocotyledon: scattered vascular bundles and flower parts in threes or multiples of three.
Bloom's Level: 4. Analyze Learning Outcome: 09.03.01 List and describe the key features of monocots and eudicots. Section: 09.03 Topic: Plants
32. You need to identify a plant but it is winter and the leaves have fallen to the ground mixed with others. The parts of flowers, fruits, and seed leaves of germinating seeds are not available either. Your best clue to whether it is a monocot or a eudicot is to A. inspect the stem for woody bark since only eudicots have woody tissue. B. determine if there is an apical meristem at the tip of the plant, which indicates it is a monocot. C. determine if the roots lack phloem and xylem, which indicates a monocot. D. cut the stem and if the vascular bundles are in a distinct ring, it is a monocot. E. cut the stem and if the vascular bundles are scattered in the stem, it is a monocot. A monocot has vascular bundles scattered in the stem. If only the leaves and roots are available, you could also look at the root to see if the xylem and phloem are in a ring. Monocots do have both phloem and xylem in the roots and eudicots also have apical meristems. Palm trees are monocots and have wood.
Bloom's Level: 3. Apply Learning Outcome: 09.03.01 List and describe the key features of monocots and eudicots. Section: 09.03 Topic: Plants
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Chapter 09 - Plant Organization and Function
33. Which of the following is a monocot? A. dandelion B. oak tree C. maple tree D. palm tree E. rose A palm tree is a monocot. The others are eudicots.
Bloom's Level: 1. Remember Learning Outcome: 09.03.01 List and describe the key features of monocots and eudicots. Section: 09.03 Topic: Plants
34. The waxy material known as the Casparian strip is found on A. the cuticle. B. xylem vessels. C. the vascular cambium. D. endodermal cells. E. epidermal cells. The Casparian strip in the roots is found on endodermal cells.
Bloom's Level: 2. Understand Learning Outcome: 09.04.02 Describe the anatomy of eudicot and monocot roots. Section: 09.04 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
35. The specialized tissue in a root that functions in food storage is the A. epidermis. B. endodermis. C. cortex. D. vascular cylinder. E. Casparian ring. The cortex is a type of ground tissue that functions in food storage.
Bloom's Level: 2. Understand Learning Outcome: 09.04.03 Identify different types of roots and root specializations. Section: 09.04 Topic: Plant Tissues
36. Your task in laboratory is to find cells that have projections to increase surface area for absorption of water and minerals. Which of the following slides would you use to find these cells? A. eudicot woody stem slide B. leaf slide of moncot or eudicot C. monocot stem slide D. root slide showing root cap E. root slide showing zone of maturation You need to find root hairs, which would be present on a root slide showing the zone of maturation.
Bloom's Level: 2. Understand Learning Outcome: 09.04.01 List the zones in the root involved in primary growth. Section: 09.04 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
37. Entrance of minerals into the vascular cylinder of a root are regulated by the A. cortex tissues. B. epidermis tissues. C. endodermis tissues. D. pericycle tissues. E. root hairs. The endodermis tissues contain the Casparian strip, which regulates what enters the vascular cylinder of the root.
Bloom's Level: 2. Understand Learning Outcome: 09.04.02 Describe the anatomy of eudicot and monocot roots. Section: 09.04 Topic: Plant Tissues
38. A cross section through the zone of maturation of a root would show A. cells of the root cap. B. the apical meristem. C. transport tissues. D. greatly elongated, undifferentiated cells. E. dividing cells. The zone of maturation would contain vascular tissue, endodermis, and root hairs.
Bloom's Level: 3. Apply Learning Outcome: 09.04.02 Describe the anatomy of eudicot and monocot roots. Section: 09.04 Topic: Plant Transport
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Chapter 09 - Plant Organization and Function
39. Which of the following statements is not correct about the arrangement of tissues in the eudicot root? A. The epidermis of the root consists of a single layer of cells. B. The specialized root tissues can be seen at the zone of maturation. C. The pericycle is the outer layer of cells in the vascular cylinder and has the ability to form secondary or branch roots. D. The Casparian strip is a waxy lining on four sides of the cells of the endodermis that regulates the movement of materials into and out of the vascular cylinder. E. The star-shaped phloem is located in the center of the vascular cylinder, with xylem arranged between the arms of the phloem. It is the star-shaped xylem that is located in the center of the vascular cylinder, with phloem arranged between the arms.
Bloom's Level: 4. Analyze Learning Outcome: 09.04.02 Describe the anatomy of eudicot and monocot roots. Section: 09.04 Topic: Plant Tissues
40. Root hairs are found in which area of the root? A. zone of maturation B. zone of elongation C. zone of cell division D. apical meristem E. root cap The zone of maturation contains root hairs.
Bloom's Level: 1. Remember Learning Outcome: 09.04.03 Identify different types of roots and root specializations. Section: 09.04 Topic: Plant Organs
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Chapter 09 - Plant Organization and Function
41. If the pericycle in a plant was missing, what would be affected? A. There would be no branch roots. B. There would be no root hairs. C. There would be no xylem and phloem. D. The meristem would not be present. E. The root would not be able to store starches. The pericycle is responsible for the origination and growth of branch roots.
Bloom's Level: 2. Understand Learning Outcome: 09.04.02 Describe the anatomy of eudicot and monocot roots. Section: 09.04 Topic: Plant Tissues
42. The layer that makes the Casparian strip waterproof is made of A. meristem. B. cuticle. C. suberin and lignin. D. plasmodesmata. E. spongy layer. Suberin and lignin border the four sides of the endodermal cells, making them impermeable.
Bloom's Level: 2. Understand Learning Outcome: 09.04.02 Describe the anatomy of eudicot and monocot roots. Section: 09.04 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
43. Which of the following is the correct sequence of zones in a longitudinal section of a root, starting from the bottom? A. zone of cell division → zone of elongation → zone of maturation B. zone of cell division → zone of maturation → zone of elongation C. zone of maturation → zone of elongation → zone of cell division D. zone of maturation → zone of cell division → zone of elongation E. zone of elongation → zone of maturation → zone of cell division After the root cap comes the zone of cell division, the zone of elongation, and the zone of maturation.
Bloom's Level: 1. Remember Learning Outcome: 09.04.02 Describe the anatomy of eudicot and monocot roots. Section: 09.04 Topic: Plants
44. Parasitic plants have specialized roots for invading a host plant's tissues. What are these roots called? A. adventitious B. taproots C. mycorrhizal D. fibrous E. haustoria Dodders and broomrapes are parasitic on other plants. Their stems form haustoria that grow into the host plant.
Bloom's Level: 1. Remember Learning Outcome: 09.04.03 Identify different types of roots and root specializations. Section: 09.04 Topic: Plant Organs
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Chapter 09 - Plant Organization and Function
45. Beans, peas, and other legumes have root nodules that are caused by _ role of A. bacteria; taking up and reducing atmospheric nitrogen. B. bacteria; improving the extraction of water and minerals from the soil. C. fungi; storing food for the plant in the fungal bodies. D. fungi; improving the extraction of water and minerals from the soil. E. fungi; taking up and reducing atmospheric nitrogen.
and have the
Nitrogen-fixing bacteria live in the roots of legumes. Fungi do form associations with plant roots but do not form nodules.
Bloom's Level: 2. Understand Learning Outcome: 09.04.03 Identify different types of roots and root specializations. Section: 09.04 Topic: Plant Organs
46. The roots of many plants have a relationship with fungi called A. root nodules. B. root hairs. C. trichomes. D. adventitious roots. E. mycorrhizae. Mycorrhizae are fungi associated with plant roots.
Bloom's Level: 1. Remember Learning Outcome: 09.04.03 Identify different types of roots and root specializations. Section: 09.04 Topic: Plant Organs
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Chapter 09 - Plant Organization and Function
47. The role of mycorrhizae is to A. take up and reduce atmospheric nitrogen. B. improve the extraction of water and minerals from the soil. C. store food for the plant in the fungal bodies. D. increase vegetative reproduction. E. acquire oxygen for cellular respiration in root tissues. Mycorrhizae increase the surface area available for water and mineral uptake and break down organic matter.
Bloom's Level: 1. Remember Learning Outcome: 09.04.03 Identify different types of roots and root specializations. Section: 09.04 Topic: Plant Organs
48. If you had a problematic hillside that continually eroded when it rained, plants with which type of roots would best solve your problem? A. adventitious B. taproots C. mycorrhizal D. fibrous E. haustoria Fibrous roots would hold the soil in place.
Bloom's Level: 3. Apply Learning Outcome: 09.04.03 Identify different types of roots and root specializations. Section: 09.04 Topic: Plant Organs
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Chapter 09 - Plant Organization and Function
True / False Questions 49. Only leaves carry out photosynthesis. FALSE Sometimes stems are green and can carry out photosynthesis.
Bloom's Level: 2. Understand Learning Outcome: 09.05.02 Describe the anatomy of eudicot and monocot stems. Section: 09.05 Topic: Plant Organs
50. Primary growth of a stem is due to the activity of the vascular meristem. FALSE The apical meristem is responsible for the primary growth of a stem.
Bloom's Level: 2. Understand Learning Outcome: 09.05.02 Describe the anatomy of eudicot and monocot stems. Section: 09.05 Topic: Plant Tissues
51. You can tell the age of a stem by counting the bud scale scars. TRUE There is one bud scale scar for each year's growth.
Bloom's Level: 1. Remember Learning Outcome: 09.05.01 Identify the structures of a woody twig. Section: 09.05 Topic: Plants
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Chapter 09 - Plant Organization and Function
52. When a plant grows upward, the distance between the internodes increases in length. TRUE As a stem grows, the internodes increase in length.
Bloom's Level: 1. Remember Learning Outcome: 09.05.02 Describe the anatomy of eudicot and monocot stems. Section: 09.05 Topic: Plants
Multiple Choice Questions 53. Wood is composed of A. cork. B. cork cambium. C. secondary phloem. D. secondary xylem. E. sieve-tube members. Wood is composed of secondary xylem that builds up year after year, increasing the girth of trees.
Bloom's Level: 1. Remember Learning Outcome: 09.05.03 Explain the secondary growth of stems. Section: 09.05 Topic: Plants
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Chapter 09 - Plant Organization and Function
54. Which is NOT a correct association? A. strawberry-stolon B. onion-bulb C. iris-rhizome D. gladiolus-bulb E. potato-tuber A gladiolus forms a corm, a thick stem covered by papery leaves.
Bloom's Level: 4. Analyze Learning Outcome: 09.05.04 Identify the adaptations that lead to stem diversity. Section: 09.05 Topic: Plants
55. What specialized types of primary meristems develop from a shoot apical meristem? A. epidermis, pith, cortex B. protoderm, ground meristem, procambium C. primary xylem, primary phloem, cortex D. secondary xylem, secondary phloem, pith E. vascular cambium, ground meristem, and pith Protoderm, ground meristem, and procambium are the three primary meristems produced by the apical meristem.
Bloom's Level: 2. Understand Learning Outcome: 09.05.02 Describe the anatomy of eudicot and monocot stems. Section: 09.05 Topic: Plants
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Chapter 09 - Plant Organization and Function
56. Which of these is paired incorrectly? A. protoderm-epidermis B. protoderm-primary xylem C. ground meristem-pith D. ground meristem-cortex E. procambium-vascular cambium Protoderm gives rise to epidermis. Procambium gives rise to primary xylem, primary phloem, and vascular cambium.
Bloom's Level: 2. Understand Learning Outcome: 09.05.02 Describe the anatomy of eudicot and monocot stems. Section: 09.05 Topic: Plants
57. A 15-foot tree growing in a fence row is used as a fence post to which barbed wire is stapled at a height of 5 feet. Years later, the tree is 30 feet high. The barbed wire of the fence is embedded in the tree bark at A. 5 feet high, since the lengthening of the trunk occurs at the apical meristem. B. 15 feet up, since the tree doubled in height. C. 20 feet up, still 10 feet from the top of the tree. D. 20 feet up, but because it added 15 feet to the original 5-foot height. E. 10 feet up, half of the increase in size; the second half is an increase in the girth of the tree. The growth of the tree in length is considered primary growth and comes from the apical meristem. Therefore, the barbed wire would remain at the same height.
Bloom's Level: 3. Apply Learning Outcome: 09.05.03 Explain the secondary growth of stems. Section: 09.05 Topic: Plants
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Chapter 09 - Plant Organization and Function
58. Evolutionary advantages of being "woody" include A. very little energy is required to produce wood. B. tall growth that allows greater leaf access to sunlight. C. less need for defense mechanisms against herbivores and parasites. D. not being as affected by varying amounts of rainfall. E. reproduction before several seasons pass. The taller plants get great leaf access to sunlight. It takes energy to produce secondary growth and woody plants need more defense mechanisms. They usually do not reproduce until they have grown for several seasons. With adequate rainfall they can outgrow herbaceous plants, but not with limited rainfall.
Bloom's Level: 3. Apply Learning Outcome: 09.05.03 Explain the secondary growth of stems. Section: 09.05 Topic: Plants
59. Which of the following statements is NOT true about wood? A. In the temperate regions, wood forms annual rings. B. The vessels will be larger in size in the spring due to more moisture. C. The wood produced in the summer will have a lower proportion of vessels than wood produced in the spring. D. In large trees, the most recently formed layer of cells transports water. E. Heartwood transports water to the leaves throughout the life of the tree. Heartwood no longer functions in water transport and can rot away.
Bloom's Level: 4. Analyze Learning Outcome: 09.05.03 Explain the secondary growth of stems. Section: 09.05 Topic: Plants
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Chapter 09 - Plant Organization and Function
60. Which of the following is not true of primary and secondary growth in plants? A. occur in woody stems B. increase both the length and girth of plants, respectively C. involve two types of meristem tissues D. occur in both woody and herbaceous stems E. occur in trees and shrubs Herbaceous stems do not undergo secondary growth.
Bloom's Level: 5. Evaluate Learning Outcome: 09.05.03 Explain the secondary growth of stems. Section: 09.05 Topic: Plants
61. The meristematic tissue used for secondary growth that is found between the xylem and phloem of each vascular bundle in eudicot stems is termed the A. apical meristem. B. root meristem. C. cork cambium. D. vascular cambium. E. pericycle. Trees and shrubs undergo secondary growth due to the activities of the vascular cambium.
Bloom's Level: 2. Understand Learning Outcome: 09.05.03 Explain the secondary growth of stems. Section: 09.05 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
62. Considering that leaf stomata have the conflicting functions of allowing gaseous carbon dioxide into the leaf, yet preserving the plant's supply of water, which of the following is not reasonable? A. Tree leaves that orient flat and present the top side to the sun would have most of their stomata on the bottom of the leaf. B. A submerged water plant would have no stomata. C. Corn, oats, rice, and wheat (all monocots) have relatively upright blades with sunlight hitting both sides and would have stomata distributed fairly evenly on both sides. D. Leaves of plants in dry areas will have more stomata than leaves of plants in humid areas. E. Plants in arid environments open their stomata at night when temperatures are cooler and evaporation occurs less. Since stomata would increase water loss, leaves of plants in dry areas would have fewer stomata than leaves of plants in humid areas.
Bloom's Level: 4. Analyze Learning Outcome: 09.06.01 Identify the structures of a leaf. Section: 09.06 Topic: Plants
63. Which statement about leaves is NOT correct? A. The photosynthetic mesophyll is made up of an upper spongy layer and a lower palisade layer. B. The epidermis is covered by a waxy layer of cuticle that reduces water loss. C. Gas exchange occurs through tiny openings in the leaf surface called stomata. D. Vascular tissue is arranged in a net pattern in eudicots and a parallel pattern in monocots. E. Regulation of gas exchange through the leaf is the responsibility of the guard cells. The photosynthetic mesophyll is made up of an upper palisade and a lower spongy mesophyll.
Bloom's Level: 2. Understand Learning Outcome: 09.06.01 Identify the structures of a leaf. Section: 09.06 Topic: Plants
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Chapter 09 - Plant Organization and Function
64. Cabbage and onions come from which part of the plant? A. roots B. stems C. petioles D. leaves E. bark These are all leaf vegetables.
Bloom's Level: 2. Understand Learning Outcome: 09.06.02 Describe some types of leaf modifications. Section: 09.06 Topic: Plant Organs
65. Stomata A. are usually found in the upper epidermis of a leaf. B. are surrounded by four guard cells. C. function for primary growth. D. function in secondary growth. E. allow gas exchange. Stomata allow for gas exchange and are usually found in the lower epidermis of the leaf. They are surrounded by two guard cells.
Bloom's Level: 2. Understand Learning Outcome: 09.06.01 Identify the structures of a leaf. Section: 09.06 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
66. The vascular tissue in the leaf is referred to as a A. vascular bundle. B. vascular cylinder. C. vein. D. vascular pit. E. vascular cambium. Leaf veins contain the vascular tissue.
Bloom's Level: 1. Remember Learning Outcome: 09.06.01 Identify the structures of a leaf. Section: 09.06 Topic: Plant Tissues
67. The tissue layers of a leaf, from the upper surface to the lower surface, are A. epidermis, cuticle, mesophyll layer, epidermis. B. cuticle, epidermis, spongy mesophyll, palisade mesophyll, epidermis, cuticle. C. epidermis, palisade layer, spongy layer, epidermis. D. cuticle, epidermis, palisade mesophyll, spongy mesophyll, epidermis, cuticle. E. epidermis, cuticle, palisade mesophyll, spongy mesophyll, cuticle, epidermis. The cuticle is found on the top and bottom of the upper and lower epidermis respectively. The palisade mesophyll is above the spongy mesophyll.
Bloom's Level: 2. Understand Learning Outcome: 09.06.01 Identify the structures of a leaf. Section: 09.06 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
68. The function of the spongy layer of a leaf is primarily A. photosynthesis. B. gas exchange. C. vascular transport. D. protein synthesis. E. as a shock absorber. The spongy mesophyll contains irregular cells bounded by air spaces for gas exchange.
Bloom's Level: 1. Remember Learning Outcome: 09.06.01 Identify the structures of a leaf. Section: 09.06 Topic: Plant Tissues
69. Which feature is not found on the epidermis of a plant leaf? A. bundle sheath cells B. stomata C. cuticle D. trichomes E. guard cells Bundle sheath cells are within the leaf, not on its surface.
Bloom's Level: 2. Understand Learning Outcome: 09.06.01 Identify the structures of a leaf. Section: 09.06 Topic: Plants
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Chapter 09 - Plant Organization and Function
70. A plant with small leaves and sunken stomata is most likely to be found in a A. rain forest. B. desert. C. grassland. D. pond. E. vegetable garden. Small leaves and sunken stomata are adaptations to conserve water. Therefore, this plant is probably found in the desert.
Bloom's Level: 3. Apply Learning Outcome: 09.06.02 Describe some types of leaf modifications. Section: 09.06 Topic: Plants
71. During the daytime, stomata in leaves A. protect the plant. B. undergo photosynthesis. C. manufacture carbohydrates. D. take in water. E. give off oxygen and take in carbon dioxide. Stomata function in gas exchange for the plant. They give off oxygen and take in carbon dioxide.
Bloom's Level: 2. Understand Learning Outcome: 09.06.01 Identify the structures of a leaf. Section: 09.06 Topic: Plant Tissues
9-35 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 09 - Plant Organization and Function
72. Trees are often identified by their leaves. What information about the leaves would you need to observe in order to identify the tree? A. function of the leaves B. whether the leaves are lost in the fall C. blades, pattern of venation, and arrangement on stem D. size and color E. whether the veins are parallel or netted Trees can be identified by blade (simple or compound), pattern of venation (pinnate or palmate), and arrangement on stem (alternate, opposite, or whorled).
Bloom's Level: 3. Apply Learning Outcome: 09.06.02 Describe some types of leaf modifications. Section: 09.06 Topic: Plants
73. Which of the following is the pathway for water transport in a plant? A. root hairs → cortex → vascular cylinder → leaves B. root hairs → pith → endodermis → xylem → leaves C. epidermis → cortex → phloem → root D. epidermis → pericycle → cortex → xylem → leaves E. root hairs → endodermis → leaves → xylem → pith Water comes in through the root hairs. It crosses the cortex of the root and enters the vascular cylinder where it is transported to the leaves.
Bloom's Level: 2. Understand Learning Outcome: 09.07.01 Explain the movement of water in a plant according to the cohesion-tension model of xylem transport. Section: 09.07 Topic: Plant Transport
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Chapter 09 - Plant Organization and Function
74. For water transport to occur there must be A. palisade and spongy mesophyll. B. sugar in the phloem. C. atmospheric pressure available to do the work. D. low pressure in the roots and high pressure in the leaves. E. a continuous column of water in xylem and transpiration of water at the leaves. In order for water to be transported by the xylem, according to the cohesion-tension model, there must be a continuous column of water in the xylem and the loss of water from the leaves.
Bloom's Level: 3. Apply Learning Outcome: 09.07.01 Explain the movement of water in a plant according to the cohesion-tension model of xylem transport. Section: 09.07 Topic: Plant Transport
75. Which of the following is NOT true about the cohesion-tension model of water transport? A. Water must completely fill the pipeline. B. Water molecules attract one another. C. Transpiration exerts a tension as water evaporates from leaves. D. Water is transported in the phloem. E. Water must be attracted to the sides of the xylem. Water is transported in the xylem, not the phloem.
Bloom's Level: 4. Analyze Learning Outcome: 09.07.01 Explain the movement of water in a plant according to the cohesion-tension model of xylem transport. Section: 09.07 Topic: Plant Transport
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Chapter 09 - Plant Organization and Function
76. When you have waxed your car, rainfall makes very large droplets that roll around on the hood. This property where the water molecules stick together is known as A. adhesion. B. turgor pressure. C. cohesion. D. transpiration. E. negative pressure potential. Cohesion is when water sticks to itself. Adhesion is when water sticks to the sides of a container.
Bloom's Level: 2. Understand Learning Outcome: 09.07.01 Explain the movement of water in a plant according to the cohesion-tension model of xylem transport. Section: 09.07 Topic: Plant Transport
77. The loss of water by evaporation from a leaf is termed A. transpiration. B. perspiration. C. transduction. D. translocation. E. transcription. Transpiration is the loss of water from a leaf.
Bloom's Level: 1. Remember Learning Outcome: 09.07.01 Explain the movement of water in a plant according to the cohesion-tension model of xylem transport. Section: 09.07 Topic: Plant Transport
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Chapter 09 - Plant Organization and Function
78. Which statement is NOT true regarding phloem transport during the summer in plants? A. The leaves must make sugar. B. A higher pressure exists in the leaves than in the roots. C. Water follows sugar into sieve-tube members in the leaves. D. Sugar is transported by passive diffusion into sieve-tube members. E. There is a fluid flow from the leaves to the roots. Sugar is actively transported into phloem.
Bloom's Level: 4. Analyze Learning Outcome: 09.07.02 Explain the movement of organic nutrients in a plant according to the pressure-flow model of phloem transport. Section: 09.07 Topic: Plant Transport
79. Girdling a tree involves cutting a deep groove all the way around a tree trunk; it would have the effect of A. triggering additional auxin production and increasing leaf growth. B. increasing the input of carbon dioxide and increasing photosynthesis. C. interrupting the flow of minerals and stunting the growth of the tree. D. interrupting the passage of phloem sap and killing the tree. E. interrupting the passage of water and killing the tree. The phloem is removed in girdling while the xylem is left intact. This results in the death of the tree.
Bloom's Level: 3. Apply Learning Outcome: 09.07.02 Explain the movement of organic nutrients in a plant according to the pressure-flow model of phloem transport. Section: 09.07 Topic: Plant Transport
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Chapter 09 - Plant Organization and Function
True / False Questions 80. Transpiration rates will NOT vary among plants. FALSE Transpiration is dependent upon temperature, humidity, and how often and long the stomata open.
Bloom's Level: 2. Understand Learning Outcome: 09.07.01 Explain the movement of water in a plant according to the cohesion-tension model of xylem transport. Section: 09.07 Topic: Plant Transport
Short Answer Questions 81. Explain the structural differences that help a botanist differentiate between a monocot and a eudicot. Monocots have the following structural features: one cotyledon, root xylem and phloem in a ring, vascular bundles are scattered in the stem, leaf veins are parallel, and the flower parts are in multiples of three. Eudicots have the following structural features: two cotyledons, root phloem between the arms of xylem, vascular bundles in a distinct ring, leaf veins form a net pattern, and the flower parts are in multiples of four or five.
Bloom's Level: 6. Create Learning Outcome: 09.03.01 List and describe the key features of monocots and eudicots. Section: 09.03 Topic: Plants
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Chapter 09 - Plant Organization and Function
82. List the three tissue types found in plants and give several examples of cell types in each tissue category. Epidermal tissues: cell types include epidermis and periderm Ground tissue: cell types include parenchyma, colenchyma, and sclerenchyma Vascular tissue: cell types include tracheids and vessel elements
Bloom's Level: 6. Create Learning Outcome: 09.02.02 Compare the structure and function of roots, stems, and leaves. Section: 09.02 Topic: Plant Tissues
Multiple Choice Questions 83. While transplanting flowers you accidentally damaged the roots. What is the most likely scenario that will occur as a result of damage to the roots? A. The slimy sheath on the root cap has been wiped off and the parenchyma cells were damaged, resulting in damage to the zone of cell division. B. The slimy sheath on the root cap has been wiped off and the parenchyma cells were damaged, resulting in damage to the zone of elongation. C. The slimy sheath on the root cap has been wiped off and the parenchyma cells were damaged, resulting in damage to the zone of maturation. D. The slimy sheath on the root cap has been wiped off and the epidermal cells were damaged, resulting in damage to the zone of cell division. E. The slimy sheath on the root cap has been wiped off and the epidermal cells were damaged, resulting in damage to the zone of maturation. The slimy sheath protects the root cap, which will contain parenchyma cells. The root cap protects the zone of cell division.
Bloom's Level: 5. Evaluate Learning Outcome: 09.04.01 List the zones in the root involved in primary growth. Section: 09.04 Topic: Plant Tissues
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Chapter 09 - Plant Organization and Function
84. Which two structures are associated with the terminal bud? A. apical meristem and leaf primordia B. shoot tip and zone of maturation C. apical meristem and zone of cell division D. zone of cell division and leaf primordia E. None of the answer choices is associated with the terminal bud. The apical meristem and leaf primordia are associated with the terminal bud. The zone of cell division and zone of maturation are associated with the root.
Bloom's Level: 2. Understand Learning Outcome: 09.05.01 Identify the structures of a woody twig. Section: 09.05 Topic: Plants
85. Which of the following is not one of the zones found in a growing root? A. epidermis B. elongation C. cell division D. maturation E. None of the answer choices is found in growing roots. The epidermis is found within the growing root but it is not classified as a zone within the growing root.
Bloom's Level: 1. Remember Learning Outcome: 09.04.01 List the zones in the root involved in primary growth. Section: 09.04 Topic: Plants
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Chapter 09 - Plant Organization and Function
86. Which stem modifications are underground, thin, papery leaves and have a thick stem that lie dormant during the winter? A. corms B. rhizomes C. stolons D. runners E. nodes Corms are underground, thin, papery leaves and have a thick stem that lie dormant during the winter. Rhizomes are underground stems but they tend to be long and thin. Stolons, runners, and nodes are all associated with aboveground vertical stems.
Bloom's Level: 1. Remember Learning Outcome: 09.05.04 Identify the adaptations that lead to stem diversity. Section: 09.05 Topic: Plants
87. Which structure is directly associated with the terminal bud? A. shoot tip B. zone of cell division C. zone of maturation D. zone of elongation E. None of the answer choices is associated with the terminal bud. The shoot tip is associated with the terminal bud. The zone of cell division, maturation, and elongation are associated with the growing root.
Bloom's Level: 1. Remember Learning Outcome: 09.05.01 Identify the structures of a woody twig. Section: 09.05 Topic: Plants
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Chapter 09 - Plant Organization and Function
Short Answer Questions 88. Explain the difference between stolons and rhizomes. Both are stem modifications, but stolons are aboveground horizontal stems that produce new plants where the nodes touch the ground. Rhizomes are thin and long stem modifications that are found underground. Some rhizomes have enlarged portions called tubers.
Bloom's Level: 6. Create Learning Outcome: 09.05.04 Identify the adaptations that lead to stem diversity. Section: 09.05 Topic: Plants
Multiple Choice Questions 89. Which events and features are correctly associated with the roots during the pressure-flow model of phloem transport? A. Sugar is stored in the sink. Cells use the sugar for cellular respiration. Water exits by osmosis and returns to the xylem. B. Sugars are stored in the source. Cells use the sugar for cellular respiration. Water exits by osmosis and returns to the xylem. C. Sugar is stored in the sink. Cells use the sugar for photosynthesis. Water exits by osmosis and returns to the xylem. D. Sugar is stored in the sink. Cells use the sugar for cellular respiration. Water exits by diffusion and returns to the phloem. E. Sugar is stored in the source. Cells use the sugar for photosynthesis. Water exits by osmosis and returns to the xylem. Sugar is stored in the sink. Cells use the sugar for cellular respiration. Water exits by osmosis and returns to the xylem. Sugars are not stored in the source and cells do not use it for photosynthesis. Water will not exit by diffusion and will not return to the phloem.
Bloom's Level: 5. Evaluate Learning Outcome: 09.07.02 Explain the movement of organic nutrients in a plant according to the pressure-flow model of phloem transport. Section: 09.07 Topic: Plant Transport
9-44 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 10 - Plant Reproduction and Responses
Chapter 10 Plant Reproduction and Responses
Multiple Choice Questions 1. What accounts for the attractiveness of many flowers that helps enhance their reproductive opportunities? A. petals B. sepals C. carpels D. stamens E. ovaries The petals of a flower are usually the showy, colorful part of the flower. This is what will often attract a pollinator to the flower.
Bloom's Level: 1. Remember Learning Outcome: 10.01.02 Identify the reproductive parts of a flower and describe the function of each part. Section: 10.01 Topic: Flower Structure and Function
2. Which of the following is NOT a component of the carpel of a flower? A. anther B. stigma C. style D. ovary E. ovule The anther is part of the stamen (male) part of the flower.
Bloom's Level: 1. Remember Learning Outcome: 10.01.02 Identify the reproductive parts of a flower and describe the function of each part. Section: 10.01 Topic: Flower Structure and Function
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Chapter 10 - Plant Reproduction and Responses
3. Which of the following is not part of a flower? A. stem B. sepal C. petal D. stamen E. carpel The stem of a plant is a vegetative, not a reproductive, structure.
Bloom's Level: 2. Understand Learning Outcome: 10.01.02 Identify the reproductive parts of a flower and describe the function of each part. Section: 10.01 Topic: Flower Structure and Function
4. The male gametophyte in flowering plants will be found within the A. carpel. B. ovary. C. sepal. D. filament. E. pollen grain. The pollen grain is the male gametophyte which produces the sperm.
Bloom's Level: 2. Understand Learning Outcome: 10.01.02 Identify the reproductive parts of a flower and describe the function of each part. Section: 10.01 Topic: Flower Structure and Function
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Chapter 10 - Plant Reproduction and Responses
True / False Questions 5. Alternation of generations means that a diploid sporophyte produces haploid spores by meiosis and a haploid gametophyte produces gametes by mitosis. TRUE Alternation of generations means that a diploid sporophyte alternates with a haploid gametophyte.
Bloom's Level: 1. Remember Learning Outcome: 10.01.01 Describe the alternation of generations life cycle including the role of the sporophyte and gametophyte generations. Section: 10.01 Topic: Angiosperm Reproduction
Multiple Choice Questions 6. Hemp (Cannabis sativa) is a dioecious plant. This means that the flowers of hemp A. are complete. B. are incomplete. C. are perfect. D. contain both stamens and carpels. E. contain both petals and sepals. A dioecious plant means that the staminate and carpellate flowers are on separate plants. Therefore the flowers are incomplete.
Bloom's Level: 3. Apply Learning Outcome: 10.01.02 Identify the reproductive parts of a flower and describe the function of each part. Section: 10.01 Topic: Flower Structure and Function
10-3 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 10 - Plant Reproduction and Responses
7. The pollen grain contains the A. sperm. B. egg. C. embryo. D. endosperm. E. anther. The pollen grain is the male gametophyte and contains the sperm.
Bloom's Level: 2. Understand Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Angiosperm Reproduction
8. Which statement is NOT true about the life cycle of flowering plants? A. The sporophyte is diploid. B. The sporophyte produces two different types of spores. C. The female gametophyte is the seed. D. The male gametophyte is the pollen grain. E. The female gametophyte is retained within the body of the sporophyte parent generation. The female gametophyte is the embryo sac. The seed comes from the zygote.
Bloom's Level: 4. Analyze Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Angiosperm Reproduction
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Chapter 10 - Plant Reproduction and Responses
9. When you pick a bouquet of anther-less flowers, you have picked A. male gametophytes retained within the body of sporophytes. B. female gametophytes retained within the body of sporophytes. C. sporophytes retained within the body of male gametophytes. D. sporophytes retained within the body of female gametophytes. E. male gametophytes retained within the body of female gametophytes. In flowering plants, the female gametophytes are retained within the body of the sporophyte.
Bloom's Level: 3. Apply Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Flower Structure and Function
10. What flower mechanism would not help prevent pollination by a foreign plant species? A. A pollen tube only grows inside the style of its own species. B. The stigma is sticky mainly to its own species of pollen. C. The timing of flowering keeps pollinators moving among a limited number of species. D. Flowers attract a small number of specialized pollinators, and therefore pollen is not spread equally to all available species of flowers. E. Flowers of different plant species bloom at the same time with flowers of similar shape, size, and color. Flowers that all bloom at the same time with similar flowers would have a difficult time preventing cross-pollination.
Bloom's Level: 4. Analyze Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Angiosperm Reproduction
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Chapter 10 - Plant Reproduction and Responses
11. The megaspore divides mitotically, resulting in a female gametophyte or embryo sac that differentiates into an egg and a central cell with two nuclei called the A. generative cell. B. pollen grain. C. microspores. D. endosperm. E. polar nuclei. The two nuclei within the central cell of the mature female gametophyte are called polar nuclei. This will develop into the endosperm.
Bloom's Level: 2. Understand Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Angiosperm Reproduction
12. Double fertilization in an angiosperm produces A. a diploid zygote and a haploid polar nucleus. B. a diploid zygote and a diploid endosperm. C. a diploid embryo and a triploid zygote. D. a triploid embryo and a diploid endosperm. E. a diploid zygote and a triploid endosperm. Double fertilization results in a diploid zygote (one sperm and one egg) and a triploid endosperm (one sperm and the two polar nuclei).
Bloom's Level: 2. Understand Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Angiosperm Reproduction
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Chapter 10 - Plant Reproduction and Responses
13. Fertilization in flowering plants occurs A. on the stigma. B. in the style. C. in the ovule. D. in the anther. E. in the endosperm. Fertilization occurs when the sperm reaches the ovule and joins with the egg and polar nuclei.
Bloom's Level: 1. Remember Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Angiosperm Reproduction
14. Which of the following statements is NOT true about flowering plants? A. Sporophyte generation produces spores. B. Gametophyte generation produces gametes. C. Ovules contain the microspore mother cell. D. Microspores undergo mitosis to produce pollen grains. E. Megaspores undergo mitosis to produce an embryo sac. The ovules are the female part of the plant and contain the megaspore mother cell.
Bloom's Level: 2. Understand Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Angiosperm Reproduction
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Chapter 10 - Plant Reproduction and Responses
True / False Questions 15. A flower that is missing the stamens would be considered female. TRUE The stamens are the "male" portion of the flower.
Bloom's Level: 1. Remember Learning Outcome: 10.01.02 Identify the reproductive parts of a flower and describe the function of each part. Section: 10.01 Topic: Flower Structure and Function
Multiple Choice Questions 16. Which of the following is not part of the female portion of the flower? A. filament B. carpel C. pistil D. stigma E. style The female part of the flower is the carpel or pistil, composed of the stima, style, and ovary.
Bloom's Level: 1. Remember Learning Outcome: 10.01.02 Identify the reproductive parts of a flower and describe the function of each part. Section: 10.01 Topic: Flower Structure and Function
10-8 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 10 - Plant Reproduction and Responses
17. How many nuclei are present inside a pollen grain? A. none B. one C. two D. three E. four The pollen grain contains three nuclei: the two sperm nuclei and the tube cell nucleus.
Bloom's Level: 1. Remember Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Angiosperm Reproduction
18. You find a plant that appears to contain stamens and carpels but no petals or sepals. What could you conclude about this "flower"? A. The plant is pollinated by the wind. B. The plant is sterile. C. The plant is dioecious. D. The flower is considered complete. E. The plant is really a gametophyte. Without petals and carpels to attract a pollinator, the plant is most likely pollinated by the wind. Since it has both male and female parts it is not sterile or dioecious. Gametophytes do not produce flowers. A flower missing petals and carpels would be incomplete.
Bloom's Level: 4. Analyze Learning Outcome: 10.01.02 Identify the reproductive parts of a flower and describe the function of each part. Section: 10.01 Topic: Angiosperm Reproduction
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Chapter 10 - Plant Reproduction and Responses
True / False Questions 19. Cross-pollination occurs when pollen from an anther of one plant is transferred to the stigma of another plant of the same species. TRUE Pollination is the transfer of pollen from an anther to a stigma. Cross-pollination would involve two different plants.
Bloom's Level: 1. Remember Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Angiosperm Reproduction
Multiple Choice Questions 20. Which of the following statements is NOT correct about flowering plants? A. A sperm nucleus combines with one polar nucleus to form the endosperm. B. A sperm nucleus combines with the egg nucleus to eventually form the embryo. C. A pollen grain may be carried to the stigma of the ovary by the wind. D. A pollen tube grows down the style. E. They exhibit double fertilization. The sperm nucleus combines with the two polar nuclei to form the endosperm.
Bloom's Level: 2. Understand Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Flower Structure and Function
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Chapter 10 - Plant Reproduction and Responses
21. The function of endosperm is to A. form the seedling. B. develop the fruit. C. provide water to the embryo. D. provide nutrients to the embryo. E. provide a protective coating for the embryo. The endosperm is the nutrient source for the developing plant.
Bloom's Level: 1. Remember Learning Outcome: 10.01.03 Diagram and describe the development of male and female gametophytes and the development of the sporophyte of flowering plants. Section: 10.01 Topic: Angiosperm Reproduction
22. A seed does NOT contain the A. embryo. B. endosperm. C. seed coat. D. ovary. E. developing sporophyte. The ovary is contained within the sporophyte of the plant.
Bloom's Level: 2. Understand Learning Outcome: 10.02.03 Label seed structure and describe germination and dispersal. Section: 10.02 Topic: Angiosperm Reproduction
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Chapter 10 - Plant Reproduction and Responses
23. Which of the following is NOT a fruit? A. corn kernel B. peas and beans C. tomato D. cucumber E. onion An onion is a bulb, derived from a stem, not a fruit.
Bloom's Level: 2. Understand Learning Outcome: 10.02.02 Identify different types of fruits. Section: 10.02 Topic: Angiosperms
24. Compound fruits A. include legumes like peas and beans. B. are derived from a simple ovary of a single carpel. C. develop from several individual ovaries. D. include peaches, tomatoes, and cherries. E. develop from a compound ovary of several fused carpels. A compound fruit develops from several individual ovaries. Examples include raspberries and blackberries.
Bloom's Level: 4. Analyze Learning Outcome: 10.02.02 Identify different types of fruits. Section: 10.02 Topic: Angiosperms
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Chapter 10 - Plant Reproduction and Responses
25. When an ice cap moves south or arid conditions push across a continent, a plant species can "migrate" although not as fast as individual animals usually travel. Which of the following generally allows a plant population to migrate the fastest? A. pollination B. fertilization C. germination D. seed dispersal E. asexual or vegetative propagation Since new sporophytes develop from seeds, the farther the seeds can travel, the farther the plant can migrate.
Bloom's Level: 3. Apply Learning Outcome: 10.02.03 Label seed structure and describe germination and dispersal. Section: 10.02 Topic: Plants
26. Which of the following is not a part of a developing eudicot embryo? A. epicotyl B. hypocotyl C. cotyledon D. radicle E. coleoptile The coleoptile is part of a developing monocot embryo.
Bloom's Level: 2. Understand Learning Outcome: 10.02.01 Recognize the developmental steps of a eudicot embryo and compare the function of its cotyledons to that of a cotyledon in monocots. Section: 10.02 Topic: Angiosperms
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Chapter 10 - Plant Reproduction and Responses
27. Which structure of the developing eudicot embryo within a seed is NOT correctly identified? A. The radicle is the embryonic root. B. A cotyledon is a seed leaf. C. The endosperm is the only nutrient source for the embryo. D. The epicotyl is the shoot above the attachment of the cotyledons. E. The hypocotyl is the embryo plant below the attachment of the cotyledons. The nutrient source for the eudicot embryo is the cotyledons, which replace the endosperm. The endosperm is the food storage part of a monocot seed.
Bloom's Level: 4. Analyze Learning Outcome: 10.02.01 Recognize the developmental steps of a eudicot embryo and compare the function of its cotyledons to that of a cotyledon in monocots. Section: 10.02 Topic: Angiosperms
28. Which of the following parts of a plant embryo grows downward? A. plumule B. epicotyl C. seed coat D. radicle E. shoot The radicle becomes the primary root and grows downward.
Bloom's Level: 1. Remember Learning Outcome: 10.02.03 Label seed structure and describe germination and dispersal. Section: 10.02 Topic: Angiosperms
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Chapter 10 - Plant Reproduction and Responses
29. Stored food in a cotyledon was derived from A. part of the stigma. B. the fruit. C. triploid endosperm. D. the egg nucleus. E. the ovary. The cotyledon takes food from the endosperm and passes it on to the embryo.
Bloom's Level: 2. Understand Learning Outcome: 10.02.03 Label seed structure and describe germination and dispersal. Section: 10.02 Topic: Angiosperms
30. Which of these associations is NOT correct? A. plumule—leaves B. radicle—roots C. seed coat—ovary wall D. cotyledons—stored food E. hypocotyl—stem The ovary wall develops into the fruit, not the seed coat.
Bloom's Level: 4. Analyze Learning Outcome: 10.02.03 Label seed structure and describe germination and dispersal. Section: 10.02 Topic: Angiosperms
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Chapter 10 - Plant Reproduction and Responses
31. The protective sheath that first surrounds the shoot is called the A. cotyledon. B. coleoptile. C. plumule. D. radicle. E. epicotyl. The coleoptile is the protective sheath that protects the plumule in a monocot.
Bloom's Level: 2. Understand Learning Outcome: 10.02.03 Label seed structure and describe germination and dispersal. Section: 10.02 Topic: Angiosperms
32. Which of the following is arranged in proper order? A. fertilization → pollination → seed formation → germination B. seed formation → fertilization → germination → pollination C. pollination → fertilization → seed formation → germination D. germination → seed formation → fertilization → pollination E. pollination → fertilization → germination → seed formation Pollination is required for fertilization, which is required for seed formation. Seeds then germinate.
Bloom's Level: 4. Analyze Learning Outcome: 10.02.01 Recognize the developmental steps of a eudicot embryo and compare the function of its cotyledons to that of a cotyledon in monocots. Section: 10.02 Topic: Angiosperm Reproduction
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Chapter 10 - Plant Reproduction and Responses
33. After a seed has been produced, it often has a time during which no growth occurs called A. pre-germination. B. senescence. C. suspended animation. D. dormancy. E. hibernation. Seeds lie dormant before they germinate. Usually this is when the conditions are not right for germination.
Bloom's Level: 1. Remember Learning Outcome: 10.02.03 Label seed structure and describe germination and dispersal. Section: 10.02 Topic: Angiosperms
34. Which of the following would be considered a dry fruit? A. one that doesn't split open B. one derived from a simple ovary C. one derived from the receptacle and not the ovary D. one derived from several individual ovaries E. one derived from many different carpels A dry fruit is one that doesn't split open. These would include wheat, rice, and corn.
Bloom's Level: 2. Understand Learning Outcome: 10.02.02 Identify different types of fruits. Section: 10.02 Topic: Angiosperms
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Chapter 10 - Plant Reproduction and Responses
True / False Questions 35. A corn kernel is classified as a seed. FALSE A corn kernel is actually a fruit, not a seed; however, the dry pericarp adheres to the seed.
Bloom's Level: 3. Apply Learning Outcome: 10.02.02 Identify different types of fruits. Section: 10.02 Topic: Angiosperms
36. When you eat an apple, the core left over is the ovary. TRUE The apple flesh is actually an accessory fruit. Only the core is derived from the ovary.
Bloom's Level: 3. Apply Learning Outcome: 10.02.02 Identify different types of fruits. Section: 10.02 Topic: Angiosperms
37. A pea is actually a seed. TRUE The pea pod is the fruit. The peas inside the pod are the seeds.
Bloom's Level: 3. Apply Learning Outcome: 10.02.02 Identify different types of fruits. Section: 10.02 Topic: Angiosperms
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Chapter 10 - Plant Reproduction and Responses
Multiple Choice Questions 38. Peas and beans are considered legumes. Why? A. A legume is a fruit that splits along two sides when mature. B. They are both simple fruits. C. The actual seed is inside the pod. D. Both are green when ripe. E. Both develop relationships with nitrogen-fixing bacteria. Although all of the answer choices are true, the reason peas and beans are considered legumes is because a legume is a fruit that splits along two sides when mature.
Bloom's Level: 3. Apply Learning Outcome: 10.02.02 Identify different types of fruits. Section: 10.02 Topic: Angiosperms
39. You are examining a seed with tiny white plumes attached to it. How is this type of seed most likely dispersed? A. by animals who eat it B. by floating on water C. by the wind D. by attaching to the fur of animals E. by bursting forth from the seed pod Tiny white plumes are most likely a type of parachute for being carried on the wind. Dandelions have this type of dispersal mechanism.
Bloom's Level: 3. Apply Learning Outcome: 10.02.03 Label seed structure and describe germination and dispersal. Section: 10.02 Topic: Angiosperms
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Chapter 10 - Plant Reproduction and Responses
40. If digestive enzymes are used to remove the plant cell wall, the naked plant cell that remains is called a A. plastid. B. plantlet. C. callus. D. protoplast. E. phytoplast. Protoplasts are plant cells without cell walls.
Bloom's Level: 1. Remember Learning Outcome: 10.03.02 Describe how plants are propagated in tissue culture. Section: 10.03 Topic: Asexual Reproduction in Plants
41. Which of the following methods of reproduction is NOT asexual? A. runners and stolons B. rhizomes C. cuttings D. grafting and suckers E. flowering Flowers are part of sexual reproduction.
Bloom's Level: 2. Understand Learning Outcome: 10.03.01 Recognize how asexual reproduction in plants differs from sexual reproduction. Section: 10.03 Topic: Asexual Reproduction in Plants
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Chapter 10 - Plant Reproduction and Responses
42. Asexual reproduction in plants does NOT involve A. daughter plants being genetically identical to the parent plant. B. the production of gametes. C. a portion of one plant giving rise to a completely new plant. D. vegetative propagation. E. the use of runners. Gametes (sperm and eggs) are part of sexual reproduction.
Bloom's Level: 1. Remember Learning Outcome: 10.03.01 Recognize how asexual reproduction in plants differs from sexual reproduction. Section: 10.03 Topic: Asexual Reproduction in Plants
43. Which of the following is NOT true concerning plant tissue culture? A. New plants can be cultivated from any existing plant cell. B. Most plant cells are totipotent, meaning an entire plant can be produced from one cell. C. Hormones can be used to cause the production of shoots from tissue. D. Plants can be produced from shoot tips, leaves, and anthers. E. Foreign genes such as those from other plants or even animals can be inserted into plant cells. New plants cannot be produced from plant cells that lose their nuclei or are dead at maturity.
Bloom's Level: 4. Analyze Learning Outcome: 10.03.02 Describe how plants are propagated in tissue culture. Section: 10.03 Topic: Asexual Reproduction in Plants
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Chapter 10 - Plant Reproduction and Responses
44. Which of the following is NOT an advantage of tissue culture? A. Plant tissue can be used to produce medicines such as digitoxin and other products for the treatment of human disease. B. Tissue culture allows the production of healthier plant varieties. C. New plant varieties can be formed by using somatic embryos. D. The use of transgenic plants has led to a serious decline in crop production throughout the world. E. Engineered plants are resistant to herbicides and disease. A serious decline in crop production is not an advantage.
Bloom's Level: 4. Analyze Learning Outcome: 10.03.02 Describe how plants are propagated in tissue culture. Section: 10.03 Topic: Asexual Reproduction in Plants
True / False Questions 45. In asexual reproduction in plants, there is only one parent, instead of two. TRUE Asexual reproduction begins with one plant and is essentially the production of another plant from that one plant. Sexual reproduction requires two parents.
Bloom's Level: 1. Remember Learning Outcome: 10.03.01 Recognize how asexual reproduction in plants differs from sexual reproduction. Section: 10.03 Topic: Asexual Reproduction in Plants
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Chapter 10 - Plant Reproduction and Responses
Multiple Choice Questions 46. Strawberry plants do not produce seeds that are used for next year's crop. How are strawberry plants propagated? A. from flowers B. from pollen C. from the nodes of stolons D. from tissue culture E. from protoplasts Strawberry plants are propogated from runners, the nodes of stolons.
Bloom's Level: 3. Apply Learning Outcome: 10.03.01 Recognize how asexual reproduction in plants differs from sexual reproduction. Section: 10.03 Topic: Asexual Reproduction in Plants
47. How are potato plants propagated? A. by "eyes" (a bud) B. by runners C. by pollen D. by flowers E. by protoplasts Potatoes are actually underground stems and are propogated by "eyes" or buds that produce a new plant.
Bloom's Level: 3. Apply Learning Outcome: 10.03.01 Recognize how asexual reproduction in plants differs from sexual reproduction. Section: 10.03 Topic: Asexual Reproduction in Plants
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Chapter 10 - Plant Reproduction and Responses
True / False Questions 48. Commercial micropropagation methods result in the cloning of plants. TRUE These methods produce thousands to millions of identical, or cloned, seedlings.
Bloom's Level: 2. Understand Learning Outcome: 10.03.02 Describe how plants are propagated in tissue culture. Section: 10.03 Topic: Asexual Reproduction in Plants
Multiple Choice Questions 49. Which of the following is not a characteristic of a somatic embryo? A. These are asexually produced. B. These are produced in large tanks called bioreactors. C. These are genetically quite diverse. D. These are encapsulated in a protective gel for shipping. E. These will generate an entire mature plant. Somatic embryos are clones, so they are genetically identical. However, mutations do arise during the production process. This does not make them quite diverse.
Bloom's Level: 2. Understand Learning Outcome: 10.03.02 Describe how plants are propagated in tissue culture. Section: 10.03 Topic: Asexual Reproduction in Plants
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Chapter 10 - Plant Reproduction and Responses
True / False Questions 50. One of the new technologies developed from tissue culture and genetic engineering is plant hybridization. FALSE Plant hybridization is an ancient technique of producing plants with desirable traits. It does not utilize these new technologies.
Bloom's Level: 2. Understand Learning Outcome: 10.03.03 Explain how genetic engineering can be used to alter plant traits. Section: 10.03 Topic: Asexual Reproduction in Plants
51. In order to produce a new plant via meristem culture, you need to remove the meristem from one plant and place it in a water containing various minerals. The meristem will then develop new shoots and roots. FALSE Hormones must be added to the meristem culture to produce shoots and roots.
Bloom's Level: 2. Understand Learning Outcome: 10.03.02 Describe how plants are propagated in tissue culture. Section: 10.03 Topic: Asexual Reproduction in Plants
52. Cereal grains are one of the easiest plants to genetically modify. FALSE Cereal grains have proved difficult to modify genetically because protoplasts do not regenerate well.
Bloom's Level: 1. Remember Learning Outcome: 10.03.03 Explain how genetic engineering can be used to alter plant traits. Section: 10.03 Topic: Plants
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Chapter 10 - Plant Reproduction and Responses
Multiple Choice Questions 53. Which modern technology is not used in the creation of transgenic plants? A. protoplast regeneration B. high voltage electric pulses C. a gene gun D. plant hybridization E. a bacterial plasmid Plant hybridization is not considered a modern technology. Each of the other techniques can be used to genetically modify plants.
Bloom's Level: 3. Apply Learning Outcome: 10.03.03 Explain how genetic engineering can be used to alter plant traits. Section: 10.03 Topic: Plants
54. Which of the following is not an advantage of genetically engineered plants? A. decreased yield B. herbicide resistance C. salt tolerance D. drought tolerance E. disease protection All of the answer choices have been engineered into transgenic plants except decreased yield. This would not be an advantage.
Bloom's Level: 2. Understand Learning Outcome: 10.03.03 Explain how genetic engineering can be used to alter plant traits. Section: 10.03 Topic: Plants
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Chapter 10 - Plant Reproduction and Responses
55. Which substance will help in seed germination and indicates to the plant that the stem must elongate to reach the light? A. cytokinins B. ethylene C. auxin D. agent orange E. phytochrome In addition to flowering, phytochrome is involved in indicating to seeds of some plants to germinate and that conditions are favorable for germination.
Bloom's Level: 2. Understand Learning Outcome: 10.04.02 Identify the various types of plant hormones and their function. Section: 10.04 Topic: Plant Hormones
56. Which plant hormone will act as an inhibitor of growth and maintains dormancy? A. auxin B. indoleacetic acid C. gibberellin D. cytokinin E. abscisic acid Abscisic acid leads to seed and bud dormancy.
Bloom's Level: 1. Remember Learning Outcome: 10.04.02 Identify the various types of plant hormones and their function. Section: 10.04 Topic: Plant Hormones
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Chapter 10 - Plant Reproduction and Responses
57. Which plant hormone is used by industry to cause fruits to drop from a tree? A. indoleacetic acid B. gibberellin C. auxin D. ethylene E. abscisic acid Ethylene is a gas that causes abscission, or fruit to drop.
Bloom's Level: 3. Apply Learning Outcome: 10.04.02 Identify the various types of plant hormones and their function. Section: 10.04 Topic: Plant Hormones
58. A hormone called cytokinin was discovered in coconut milk and found to stimulate cell division. With this function, where else in a plant would you expect to find cytokinin? A. bark B. pith C. root tips D. spongy mesophyll E. xylem Root tips are actively growing, so you would expect to find a growth hormone present there.
Bloom's Level: 3. Apply Learning Outcome: 10.04.01 Explain the importance of plant hormones. Section: 10.04 Topic: Plant Hormones
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Chapter 10 - Plant Reproduction and Responses
59. Which of the following plant hormones prevents senescence and can initiate leaf growth? A. indoleacetic acid B. gibberellin C. cytokinin D. abscisic acid E. ethylene Cytokinins promote cell division, prevent senescence (aging), and can initiate leaf growth.
Bloom's Level: 1. Remember Learning Outcome: 10.04.01 Explain the importance of plant hormones. Section: 10.04 Topic: Plant Hormones
60. Which of the following plant hormones could be used by industry to increase the size of sugarcane production? A. indoleacetic acid B. gibberellins C. zeatin D. abscisic acid E. ethylene Gibberellins cause stem elongation in plants. Since the harvested part of the sugarcane is the stem, gibberelllins could increase yields.
Bloom's Level: 3. Apply Learning Outcome: 10.04.01 Explain the importance of plant hormones. Section: 10.04 Topic: Plant Hormones
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Chapter 10 - Plant Reproduction and Responses
61. Clover is an example of a what type of flowering plant? A. long-day B. short-day C. day-neutral D. intermediate-day E. nonflowering Clover flowers when the day length is longer than a critical length. Therefore, it is a longday plant.
Bloom's Level: 1. Remember Learning Outcome: 10.04.03 Recognize how plants respond to stimuli. Section: 10.04 Topic: Plant Responses
62. Phototropism and gravitropism occur from the action(s) of A. auxin in both cases. B. gibberellin and auxin, respectively. C. cytokinin and auxin, respectively. D. abscisic acid and ethylene, respectively. E. ethylene and auxin, respectively. Auxin is involved in both phototropism and gravitropism.
Bloom's Level: 2. Understand Learning Outcome: 10.04.03 Recognize how plants respond to stimuli. Section: 10.04 Topic: Plant Hormones
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Chapter 10 - Plant Reproduction and Responses
63. Which action would cause a long-day plant to flower? A. Interrupt the day with a period of darkness. B. Interrupt a long night with a flash of light. C. Give the plant some phytochrome. D. Allow an insect to pollinate the plant. E. Keep the plant in the dark. A long-day plant flowers when the day length is longer than a critical length. Interrupting the darkness with a flash of light causes it to flower.
Bloom's Level: 3. Apply Learning Outcome: 10.04.03 Recognize how plants respond to stimuli. Section: 10.04 Topic: Plant Responses
64. From the discussion of long-day and short-day plants, it is obvious that "day-neutral" plants A. flower when the night is shorter than a critical length. B. flower when the night is longer than a critical length. C. are not dependent on the length of the night or day for flowering. D. have short stems or long stems, depending on the photoperiod. E. have flowers that only open during the daytime. A day-neutral plant does not depend on day length for flowering.
Bloom's Level: 2. Understand Learning Outcome: 10.04.03 Recognize how plants respond to stimuli. Section: 10.04 Topic: Plant Responses
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Chapter 10 - Plant Reproduction and Responses
65. Plants could readily develop seasonal responses based on temperature changes, but have evolved their rhythms based on a comparison of daylight and darkness. Why has the day-night ratio system been selected? A. Temperature has no effect upon plant biochemistry. B. Temperature is a gradation, but day/night is all or nothing. C. A plant's nervous system is tuned to light stimuli and not to temperature stimuli. D. Temperatures fluctuate widely across seasons and from year to year, but photoperiod comparisons are a more reliable indicator of season. E. Having a day-night system makes plants more likely to be an evolutionary success since animals follow day-night cycles as well. The day-night ratio is tied to the movement of Earth and is fairly dependable. Although temperatures affect plants, they are less reliable than the movement of Earth. Plants do not have nervous systems and are not necessarily tied to animal evolution.
Bloom's Level: 5. Evaluate Learning Outcome: 10.04.03 Recognize how plants respond to stimuli. Section: 10.04 Topic: Plant Responses
66. The response of plants to the relative length of light and darkness is termed A. phototropism. B. photosynthesis. C. photoperiodism. D. translocation. E. transpiration. The photoperiod is the ratio of the length of day to the length of night over a 24-hour period.
Bloom's Level: 2. Understand Learning Outcome: 10.04.03 Recognize how plants respond to stimuli. Section: 10.04 Topic: Plant Responses
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Chapter 10 - Plant Reproduction and Responses
67. A friend complains that all of her houseplants have bent stems. You examine the plants and find that their stems all curve toward the light. What would you recommend to your friend to correct the problem? A. Apply an auxin to the shady side of the stems. B. Water the plants more frequently. C. Apply gibberellins to the bases of the stems. D. Turn each plant occasionally. E. Repot the plants. By turning the plants, different sides of the plants will face the light and the stems will grow straighter.
Bloom's Level: 3. Apply Learning Outcome: 10.04.03 Recognize how plants respond to stimuli. Section: 10.04 Topic: Plant Responses
True / False Questions 68. Plant hormones are small organic molecules produced by the plants themselves. TRUE Plants produce their own hormones, which are small, organic molecules.
Bloom's Level: 1. Remember Learning Outcome: 10.04.01 Explain the importance of plant hormones. Section: 10.04 Topic: Plant Hormones
69. Plant hormones bring about a physiological response in the plant after binding to the nucleus of the plant cells. FALSE Plant hormones bind to a specific receptor protein in the plasma membrane of the plant cell.
Bloom's Level: 2. Understand Learning Outcome: 10.04.01 Explain the importance of plant hormones. Section: 10.04 Topic: Plant Hormones
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Chapter 10 - Plant Reproduction and Responses
Multiple Choice Questions 70. What pathway is directly responsible for producing the gametophytes during the alternation of generations in plants? A. mitosis B. meiosis C. binary fission D. both mitosis and meiosis E. None of the answer choices will produce gametophytes. The gametophytes are produced during mitosis. During meiosis, spores are produced. Binary fission is an asexual process found in bacteria.
Bloom's Level: 1. Remember Learning Outcome: 10.01.01 Describe the alternation of generations life cycle including the role of the sporophyte and gametophyte generations. Section: 10.01 Topic: Plants
True / False Questions 71. The two-stage life cycle in plants evolved as an adaptation to life on land. TRUE The two-stage life cycle evolved as an adaptation to life on land for plants. This enabled the plant to produce stages of the life cycle that would be resistant to drying out in the form of spores. It also produces genetic variation through the process of meiosis.
Bloom's Level: 1. Remember Learning Outcome: 10.01.01 Describe the alternation of generations life cycle including the role of the sporophyte and gametophyte generations. Section: 10.01 Topic: Angiosperm Reproduction
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Chapter 10 - Plant Reproduction and Responses
Short Answer Questions 72. List the features that enable us to distinguish the difference between a simple fruit and a compound fruit. A simple fruit is derived from a simple ovary of a single carpel or from a compound ovary of several fused carpels. A compound fruit develops from several individual ovaries.
Bloom's Level: 6. Create Learning Outcome: 10.02.02 Identify different types of fruits. Section: 10.02 Topic: Plants
73. Describe the function of the following plant structures: petals, stamens, and carpel. The petals are the showy part of the plant that are used to attract a pollinator. The stamen is the male portion of the flower. The stamen will contain the anther that produces the pollen and a filament that supports the anther. The carpel is the female portion of the flower. The carpel will contain the stigma, the sticky end that catches the pollen; the style, the stalk that supports the stigma and the ovary; and the base that contains the ovules.
Bloom's Level: 6. Create Learning Outcome: 10.01.02 Identify the reproductive parts of a flower and describe the function of each part. Section: 10.01 Topic: Flower Structure and Function
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Chapter 10 - Plant Reproduction and Responses
Multiple Choice Questions 74. What would be the outcome of applying ethylene to a bushel of oranges? A. Ethylene stimulates cellulase, which will hydrolyze the cellulose in plant cell walls. This will cause the fruit to ripen. B. Ethylene will initiate and maintain seed dormancy as well as bring about the closure of the stomata. C. Ethylene stimulates cellulase, which will hydrolyze the proteins in plant cell walls. This will cause the fruit to ripen. D. Ethylene stimulates cellulase, which will hydrolyze the lipids in plant cell walls. This will cause the fruit to ripen. E. Ethylene stimulates cell division, which will cause the fruit to ripen. Ethylene stimulates cellulase, which will hydrolyze the cellulose in plant cell walls. This will cause the fruit to ripen. Abscisic acid initiates and maintains seed dormancy as well as brings about the closure of the stomata. Cellulase hydrolyzes the cellulase, not the proteins or lipids. Cytokinins are the hormones that promote cell division.
Bloom's Level: 5. Evaluate Learning Outcome: 10.04.02 Identify the various types of plant hormones and their function. Section: 10.04 Topic: Plant Hormones
Short Answer Questions 75. Explain the various uses of gibberellins in the agricultural industry. Gibberellins are growth-producing hormones that bring about elongation of the resulting cells. If gibberellins are applied externally to plants, stem elongation occurs. Dwarf plants will grow to a normal height. Dormancy of seeds can be broken and germination can also be initiated.
Bloom's Level: 6. Create Learning Outcome: 10.04.02 Identify the various types of plant hormones and their function. Section: 10.04 Topic: Plant Hormones
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Chapter 10 - Plant Reproduction and Responses
Multiple Choice Questions 76. Which of the following is the correct sequence for the germination of the eudicot seed? A. The two cotyledons have absorbed the endosperm and supply the nutrients to the embryo. As the embryo emerges, from the soil it consumes the nutrients and develops the shoot. The hypocotyl becomes part of the stem and the radicle develops into the roots. B. The cotyledon was absorbed by the endosperm and supplies the nutrients to the embryo. As the embryo emerges from the soil, it consumes the nutrients and develops the shoot. The hypocotyl becomes part of the stem and the radicle develops into the roots. C. The two cotyledons have absorbed the endosperm and supply the nutrients to the embryo. As the embryo emerges from the soil, it consumes the nutrients and develops the shoot. The radicle becomes part of the stem and the hypocotyl develops into the roots. D. The pericarp has absorbed the endosperm and supplies the nutrients to the embryo. As the embryo emerges from the soil, it consumes the nutrients and develops the shoot. The hypocotyl becomes part of the stem and the radicle develops into the roots. E. The two cotyledons have absorbed the endosperm and supply the nutrients to the embryo. As the embryo emerges from the soil, it consumes the nutrients and develops the shoot. The plumule and the radicle push through the coleorhiza during germination. The two cotyledons have absorbed the endosperm and supply the nutrients to the embryo. As the embryo emerges from the soil, it consumes the nutrients and develops the shoot. The hypocotyl becomes part of the stem and the radicle develops into the roots. Eudicots have two cotyledons, while monocots have one. The plumule and the radicle push through the coleorhiza during germination in the monocots.
Bloom's Level: 5. Evaluate Learning Outcome: 10.02.03 Label seed structure and describe germination and dispersal. Section: 10.02 Topic: Angiosperm Reproduction
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Chapter 11 - Human Organization
Chapter 11 Human Organization
Multiple Choice Questions 1. Which of the following is NOT a basic type of tissue? A. muscular B. nervous C. urinary D. epithelial E. connective Urinary is not a type of tissue.
Bloom's Level: 1. Remember Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
2. Simple squamous epithelium has A. a single layer of cube-shaped cells. B. a single layer of column-shaped cells. C. a single layer of flat cells. D. many layers of flat cells. E. many layers of cube-shaped cells. Simple means that there is a single layer. Squamous means that the cells are flat.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
3. Which of the following tissue types would line the mouth? A. muscular B. connective C. epithelial D. nervous E. glandular Epithelial tissue covers surfaces and lines body cavities.
Bloom's Level: 3. Apply Learning Outcome: 11.01.02 Identify the common locations of each tissue type in the human body. Section: 11.01 Topic: Animal Tissues
4. Which type of tissue contains cells close together and connected to a basement membrane? A. hyaline cartilage B. elastic cartilage C. muscle tissue D. nerve tissue E. epithelial tissue Epithelial tissue consists of tightly packed cells that form a continuous layer. A basement membrane usually joins the epithelium to underlying connective tissue.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
5. Which type of tissue has cube-shaped cells and lines kidney tubules? A. hyaline cartilage B. nerve tissue C. columnar epithelium D. cuboidal epithelium E. squamous epithelium Epithelial tissue covers surface and lines body cavities. Cube-shaped cells are called cuboidal.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
6. What type of epithelium contains thin, flat cells stacked upon each other? A. simple squamous B. stratified squamous C. simple cuboidal D. statified cuboidal E. simple columnar Thin, flat cells are called squamous. Cells that are piled one on top of another are classified as stratified.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
7. Glands that secrete their products into ducts are classified as A. serous glands. B. mucous glands. C. endocrine glands. D. exocrine glands. E. salivary glands. Glands that secrete their product into ducts are called exocrine glands. Those that secrete their product into the blood stream are endocrine glands. The others are more specific types of glands.
Bloom's Level: 2. Understand Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
8. Epithelial tissue can be modified to A. store fat. B. perform absorption. C. receive stimuli. D. move body parts. E. carry impulses. The surface area of epithelial tissue can be modified to perform absorption. The other choices belong to connective, nervous, or muscular tissue.
Bloom's Level: 2. Understand Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
9. Which two types of tissue come together to make a "basement membrane"? A. muscle and connective B. connective and nervous C. nervous and epithelial D. epithelial and connective E. nervous and muscle A basement membrane usually joins an epithelium to underlying connective tissue.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
10. Which statement is NOT true about epithelial tissue? A. Epithelial tissue helps to protect the body from bacterial invasion. B. Epithelial tissue can be glandular. C. Epithelial tissue can be found lining the tubules of the kidney. D. Epithelial tissue may contain column-shaped cells. E. Epithelial cells have spaces between them. Epithelial cells are tightly packed cells that form a continuous layer.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
11. Which of the following tissues may be specialized to secrete mucous? A. epithelial B. connective C. muscle D. nervous E. cartilage Epithelial tissue can be modified to secrete a product such as mucous.
Bloom's Level: 2. Understand Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
12. Which of the following tissues is INCORRECTLY matched with a function? A. epithelial—capillary lining B. connective—storage of fat C. muscle—protection from pathogens D. nervous—integration of data E. connective—production of blood cells Muscle moves body parts. The blood (connective tissue) provides protection from pathogens.
Bloom's Level: 4. Analyze Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
True / False Questions 13. A tight junction forms an impermeable barrier between two cells. TRUE The adjacent plasma membrane proteins actually join, forming an impermeable barrier.
Bloom's Level: 1. Remember Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
Multiple Choice Questions 14. The urine in your bladder is unable to pass out of the bladder through its epithelial lining. Which type of junction exists between these cells? A. tight junction B. There is no special junction between these cells. C. basement membrane D. gap junction E. adhesion junction Tight junctions form a barrier that is impermeable to liquid.
Bloom's Level: 3. Apply Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
15. Which of the following associations is NOT correct? A. collagen—white fibers B. elastin—yellow fibers C. large matrix—epithelial tissue D. muscle—contraction E. nerve—impulse conduction Connective tissue cells are widely separated by a matrix. Epithelial cells form a continuous layer.
Bloom's Level: 4. Analyze Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
16. Which is NOT a correct association? A. adhesion junctions—skin B. adhesion junctions—bladder C. gap junctions—stomach D. tight junctions—intestines E. tight junctions—bladder Gap junctions form when two adjacent plasma membrane channels join. This lends strength to the membrane. Tight junctions are impermeable to liquid while adhesion junctions are found in flexible tissues.
Bloom's Level: 4. Analyze Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
17. Which type of junction permits small molecules to pass between two cells? A. desmosome B. tight junction C. gap junction D. striations E. cross junction A gap junction is permeable to ions, sugars, and small molecules.
Bloom's Level: 1. Remember Learning Outcome: 11.01.02 Identify the common locations of each tissue type in the human body. Section: 11.01 Topic: Animal Tissues
18. Collagenous and elastic fibers are found in what type of tissue? A. epithelial B. muscle C. nerve D. connective E. simple squamous Connective tissue contains a matrix often with collagen, reticular, or elastic fibers.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
19. Muscles are connected to bones by A. nerve tissue. B. epithelial tissue. C. ligaments. D. tendons. E. a basement membrane. Tendons connect muscle to bone. Ligaments connect bone to bone.
Bloom's Level: 1. Remember Learning Outcome: 11.01.02 Identify the common locations of each tissue type in the human body. Section: 11.01 Topic: Animal Tissues
20. Which tissue lacks a direct blood supply and heals very slowly? A. bone B. cartilage C. muscle D. epithelial tissue E. nerve Because cartilage lacks a direct blood supply, it heals very slowly.
Bloom's Level: 1. Remember Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
21. The tissue that has a white, translucent appearance and is found at the ends of long bones is called A. hyaline cartilage. B. fibrocartilage. C. elastic cartilage. D. compact bone. E. dense fibrous connective tissue. Hyaline cartilage has a white, translucent appearance and is found at the ends of long bones and ribs.
Bloom's Level: 2. Understand Learning Outcome: 11.01.02 Identify the common locations of each tissue type in the human body. Section: 11.01 Topic: Animal Tissues
22. Which function is NOT correctly matched to its type of tissue? A. Adipose tissue stores fat. B. Reticular connective tissue produces blood cells. C. Connective tissues bind organs together and provide support and protection. D. Loose, fibrous connective tissue encloses internal organs, including muscles. E. Dense connective tissue is muscle that contracts. Connective tissue is not muscle tissue.
Bloom's Level: 4. Analyze Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
23. Which is NOT a correct description of cartilage? A. Hyaline cartilage is white and translucent and found at the ends of long bones. B. Hyaline cartilage contains only very fine collagen fibers. C. Fibrocartilage contains strong collagen fibers. D. Fibrocartilage forms the nose. E. Elastic cartilage forms the flexible outer ear. Fibrocartilage forms the pads between the vertebrae in the back and the wedges in the knee joint. Hyaline cartilage forms the nose.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
24. What tissue is found in pads between the vertebrae? A. bone B. elastic cartilage C. fibrocartilage D. hyaline cartilage E. stratified squamous epithelial tissue Fibrocartilage forms the pads between the vertebrae in the backbone.
Bloom's Level: 1. Remember Learning Outcome: 11.01.02 Identify the common locations of each tissue type in the human body. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
25. What type of bone contains numerous bony plates separated by irregular spaces? A. compact B. spongy C. marrow D. elastic E. fibrous Spongy bone contains numerous bony bars and plates, separated by irregular spaces.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
26. Bone A. is made of only a hard matrix of inorganic (mainly calcium) salts. B. lacks the collagen fibers found in cartilage. C. is lacking blood vessels and canaliculi that deliver nutrients. D. is built by bone cells called osteons. E. is formed by concentric rings of bone cells in canaliculi. Bone contains cells and collagen fibers as well as matrix and is laden with blood and canaliculi to provide nutrients. An osteon is the cylindrical structural unit of the bone that forms concentric rings. Bone cells are located in spaces called lacunae.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
27. What is the name of the cavity that contains the bone cells themselves? A. osteons B. fibrocartilage C. lacunae D. central canals E. adipose Bone cells are located in spaces called lacunae between the rings of matrix.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
28. Which is NOT a true statement about the blood component? A. Red blood cells provide the major transport for oxygen. B. Blood contains a matrix of collagen fibers made by cells. C. Platelets are not complete cells. D. White blood cells either engulf disease organisms or form antibodies for immunity. E. Plasma makes up slightly over half of human blood volume to provide a liquid in which blood solids can flow. Blood is unlike other types of connective tissue in that the matrix (plasma) is not made by the cells.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
29. Name the tissue in which the cells are separated by a liquid. A. cartilage B. nerve C. blood D. muscle E. bone The matrix in blood is a liquid called plasma.
Bloom's Level: 1. Remember Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
30. Some very important persons such as Japanese Emperor Hirohito and China's Premier Zhou Enlai were kept alive during their last days by many transfusions of blood. Without knowing further details of their ailments, which of the following functions most likely provided the most significant contribution to keeping them alive? A. oxygen and carbon dioxide carrying capacity B. supplement of hormone levels C. provision of organic nutrients such as glucose, amino acids, and fats D. provision of albumin and fibrinogen E. provision of immune antibodies Each of the other components could be provided in a manner other than a transfusion. However, the oxygen and carbon dioxide carrying capacity could only be delivered by a transfusion of red blood cells.
Bloom's Level: 3. Apply Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
31. Which are cell fragments that help initiate blood clotting? A. erythrocytes B. leukocytes C. platelets D. osteocytes E. fibroblasts Platelets are responsible for blood clotting. They are pieces of a larger cell that is broken apart.
Bloom's Level: 2. Understand Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
32. Which type of muscle is involuntary and lacks striations? A. skeletal B. smooth C. cardiac D. triceps E. the heart Both smooth muscle cells and cardiac muscle cells are under involuntary control but only smooth muscle lacks striations.
Bloom's Level: 1. Remember Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
33. The muscle group in your upper arm is composed of what type of muscle tissue? A. skeletal muscle B. smooth muscle C. cardiac muscle D. both cardiac and skeletal E. adipose The muscle group in your upper arm is a type of skeletal muscle. Cardiac muscle is found only in the heart. Adipose is a type of fat tissue.
Bloom's Level: 3. Apply Learning Outcome: 11.01.02 Identify the common locations of each tissue type in the human body. Section: 11.01 Topic: Animal Tissues
34. Which is a correct description of muscle tissues? A. Cardiac muscle is found in the heart, blood vessels, and nerves leading to the heart. B. Smooth muscle is voluntary and represented by most skeletal muscles. C. Skeletal muscle cells are often striated or banded where actin and myosin filaments overlap. D. Smooth muscle cells are voluntary. E. Skeletal muscle cells are branched and tangled and bound end-to-end at intercalated disks. Skeletal muscle cells are striated or banded.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
35. Some anesthetics are also very effective in stopping the food from moving through the intestine. Therefore, after an operation, the surgeon will not permit a patient to eat until the digestive system is moving food again, and a patient has to suck on ice until there is some evidence the digestive tract is functioning. Without knowing more about the anesthetics, what is the likely mode of action of the anesthetic? A. The anesthetic probably slows the cardiac muscle, which in turn slows all other organ systems. B. The anesthetic probably affects nerves that slow all skeletal muscle actions. C. The anesthetic probably affects nerves that slow all smooth muscle actions. D. The anesthetic probably affects nerves that stop all homeostatic actions. E. The anesthetic probably affects nerves that stop all muscle contraction. Since the digestive system is lined with smooth muscle, the anesthetic probably affects nerves that slow all smooth muscle actions.
Bloom's Level: 3. Apply Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
36. Dendrites and an axon are parts of a(n) A. epithelial cell. B. muscle cell. C. reproductive cell. D. neuron. E. cartilage cell. A dendrite is a process that conducts signals toward a neuron and an axon conducts them away.
Bloom's Level: 1. Remember Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
37. In which direction does a nerve impulse usually move? A. cell body → dendrite → axon B. dendrite → axon → cell body C. dendrite → cell body → axon D. axon → cell body → dendrite E. axon → dendrite → cell body A dendrite conducts signals toward the cell body and an axon conducts them away.
Bloom's Level: 4. Analyze Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
38. What is the function of neuroglial cells? A. support and nourish neurons B. conduction of nerve impulses C. production of new neurons D. deposition of inorganic salts E. produce the matrix Neuroglia support and nourish neurons.
Bloom's Level: 1. Remember Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
39. What is the numerical relationship of neuroglial cells compared to neurons? A. Because they are big, one neuroglial cell can serve many neurons, so there are about nine neurons for every neuroglial cell. B. There are about an equal number of both; each neuron has a companion neuroglial cell. C. There are about nine neuroglial cells for every neuron. D. The numbers vary widely from animal to animal and depend on the size of the animal. E. Brain volume is made of approximately two-thirds neurons and one-third neuroglial cells. Neuroglia cells outnumber neurons nine to one and take up more than half the volume of the brain.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
40. Until recently, we only described neurons as cellular units of the nervous system. Which is a probable reason science overlooked the importance of neuroglial cells? A. Neuroglial cells have long processes and look just like neurons. B. Neuroglial cells are rare compared to neurons and make up a very small volume of the brain. C. Neuroglial cells were thought to merely support neurons and their complex chemical role was not yet detected. D. Neuroglial cells are distant from the other neuron tissue. E. Neuroglial cells are developing larger today as people use more memory. As more research has been done, more information about the role of neuroglial cells has come to light. They were initially thought to only support the neurons but are now known to play a complex chemical role.
Bloom's Level: 3. Apply Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
41. Which is a type of neuroglial cell? A. astrocyte B. lacunae C. dendrocytes D. macroglia E. myelin Astrocytes, microglia, oligodendrocytes, and ependymal cells are the four types of neuroglia in the brain.
Bloom's Level: 1. Remember Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
42. Which of these is NOT a part of a neuron? A. cell body B. dendrite C. axon D. neuroglial cell E. nucleus A neuroglial cell is not part of a neuron.
Bloom's Level: 2. Understand Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
43. Which type of tissue will respond to stimuli and transmit impulses? A. epithelial B. connective C. nervous D. muscle E. blood The nervous tissue responds to stimuli and transmits impulses.
Bloom's Level: 1. Remember Learning Outcome: 11.01.03 Explain how the cells that make up various types of tissues are specialized to perform their functions. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
44. Which is NOT a correct description of body cavity partitioning? A. The meninges protect and isolate the brain and spinal cord. B. The diaphragm separates the pericardial cavity from the thoracic cavity. C. The cranial cavity contains the brain. D. The intestines must share space with the liver and stomach in the abdominal cavity. E. Reproductive organs are in the pelvic cavity or an extension of it. The diaphragm separates the thoracic cavity from the abdominal cavity.
Bloom's Level: 2. Understand Learning Outcome: 11.02.01 List the two main cavities of the human body, and the major organs found in each. Section: 11.02 Topic: Animals
45. Which of these organs are found in the coelom? A. brain, spinal cord, kidneys, heart B. salivary glands, vertebrae, heart C. intestines, heart, lungs, liver D. intestines, brain, lungs, liver E. reproductive organs, heart, lungs, brain The coelom is the body cavity that develops into the ventral cavity which divides into the thoracic, abdominal, and pelvic cavities.
Bloom's Level: 5. Evaluate Learning Outcome: 11.02.01 List the two main cavities of the human body, and the major organs found in each. Section: 11.02 Topic: Animals
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Chapter 11 - Human Organization
46. During a medical test for spinal meningitis the physician orders a "spinal tap". What cavity would he be removing the fluid from? A. thoracic B. abdominal C. pericardial D. dorsal E. pleural The spinal cord and brain are part of the dorsal cavity.
Bloom's Level: 3. Apply Learning Outcome: 11.02.01 List the two main cavities of the human body, and the major organs found in each. Section: 11.02 Topic: Animals
47. Usually it is not possible to penetrate inside a body cavity without cutting through protective tissues. However, sperm reach the egg via the vagina, uterus, and fallopian tubes (oviducts). Without crossing any epidermal barrier, the sperm can find itself inside which body cavity? A. thoracic B. pelvic C. pericardial D. dorsal E. pleural The uterus is within the pelvic cavity.
Bloom's Level: 2. Understand Learning Outcome: 11.02.01 List the two main cavities of the human body, and the major organs found in each. Section: 11.02 Topic: Animals
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Chapter 11 - Human Organization
48. The type of membrane that covers the heart is called a A. mucous membrane. B. serous membrane. C. basement membrane. D. pleural membrane. E. plasma membrane. Serous membranes line the thoracic and abdominal cavities and cover the organs they contain.
Bloom's Level: 2. Understand Learning Outcome: 11.02.02 Compare and contrast the location(s) and function(s) of the different types of body membranes. Section: 11.02 Topic: Animals
49. Which of the following is matched to the incorrect body system? A. helps control pH of body fluids—urinary system B. protects against disease and infections—cardiovascular system C. produces body heat—muscular system D. helps regulate metabolism—endocrine system E. produces blood cells—skeletal system The lymphatic system protects the body against disease and infections.
Bloom's Level: 2. Understand Learning Outcome: 11.03.02 Describe the general function(s) of each organ system. Section: 11.03 Topic: Body Systems
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Chapter 11 - Human Organization
50. Which of the following body systems is NOT involved in the control of fluid balance in the body? A. cardiovascular B. muscular C. lymphatic D. urinary E. endocrine The muscular system does not specifically control fluid balance in the body.
Bloom's Level: 2. Understand Learning Outcome: 11.03.02 Describe the general function(s) of each organ system. Section: 11.03 Topic: Body Systems
True / False Questions 51. The endocrine system is the only system that is involved in coordinating and regulating the functioning of all the body's other systems. FALSE The nervous and the endocrine system coordinate and regulate the other body systems.
Bloom's Level: 2. Understand Learning Outcome: 11.03.02 Describe the general function(s) of each organ system. Section: 11.03 Topic: Body Systems
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Chapter 11 - Human Organization
Multiple Choice Questions 52. Which of the following functions would not be lost or seriously compromised if someone suffered a severe burn over most of their body? A. prevention of heat and water loss B. prevention of massive and continuous infection from invasion by external bacteria, viruses, and parasites C. production of sweat that evaporates and cools the body when it is overheating D. protection of muscle and other surface tissues from abrasion and pain E. exchange of gases The skin does not serve as a site of gas exchange in humans.
Bloom's Level: 4. Analyze Learning Outcome: 11.03.02 Describe the general function(s) of each organ system. Section: 11.03 Topic: Integumetary System
53. If a burn victim loses most of his skin, what symptom would NOT be present? A. a decrease in thirst and water intake due to dry surface tissues B. intense pain from exposed nerve endings C. constantly cold due to a lack of body insulation D. serious risk of infection from microbial agents E. vitamin D deficiency A patient may experience an increase in thirst due to dry body surfaces, but not a decrease.
Bloom's Level: 3. Apply Learning Outcome: 11.04.03 List some common disorders of human skin, and how these may be treated. Section: 11.04 Topic: Integumetary System
11-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 11 - Human Organization
54. A person enters a heat chamber carrying a raw egg on a plate. As the temperature is gradually increased, the exposed egg is cooked while the subject continues to have a normal internal temperature by mouth thermometer. What mechanism is lacking in the egg but is present in the human subject that maintains homeostasis? A. Lack of a circulatory system prevents the egg from distributing the heat. B. The egg cannot "breathe away" hot air to "air condition" its body. C. The surface-area-to-volume ratio prevents the smaller body from losing heat as fast. D. Increased sweating causes evaporation that carries away heat. E. The human subject contains more fluid volume to absorb heat. The skin helps to regulate body temperature by sweating (in the case of excess heat).
Bloom's Level: 3. Apply Learning Outcome: 11.05.01 Define homeostasis, and describe why it is essential to living organisms. Section: 11.05 Topic: Homeostasis
55. Which layer of the skin protects the body from bacterial infection and water loss? A. dermis B. epidermis C. connective tissue layer D. hyaline cartilage E. basement membrane The epidermis is the outside layer of the skin that protects from pathogens and prevents water loss.
Bloom's Level: 2. Understand Learning Outcome: 11.04.01 Identify the two main regions of skin, and how these are distinguished from the subcutaneous layer. Section: 11.04 Topic: Integumetary System
11-28 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 11 - Human Organization
56. Which molecule is responsible for the waterproofing of the skin? A. mucus B. keratin C. hyaline cartilage D. myosin E. lanolin Keratin is a waterproof protein.
Bloom's Level: 1. Remember Learning Outcome: 11.04.02 Describe the makeup and function of the accessory structures of human skin. Section: 11.04 Topic: Integumetary System
57. The part of the skin that has adipose tissue to help insulate the body is the A. epidermis. B. dermis. C. subcutaneous layer. D. connective tissue. E. muscle layer. The subcutaneous layer is composed of loose connective tissue and adipose tissue, which stores fat.
Bloom's Level: 2. Understand Learning Outcome: 11.04.01 Identify the two main regions of skin, and how these are distinguished from the subcutaneous layer. Section: 11.04 Topic: Integumetary System
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Chapter 11 - Human Organization
58. Melanin is a substance responsible for A. cell growth. B. muscle contraction. C. nerve impulses. D. skin pigmentation to protect from harmful light. E. increases blood flow to the epidermis. Melanin is the main pigment responsible for skin color. Melanin provides protection from damage by UV radiation.
Bloom's Level: 2. Understand Learning Outcome: 11.04.02 Describe the makeup and function of the accessory structures of human skin. Section: 11.04 Topic: Integumetary System
59. If you wash your skin and hair many times a day, you will soon have dry skin that cracks and bleeds because you have removed the secreted by that is necessary to keep the skin supple. A. salty sweat; sweat glands B. pili; erectors C. lymph; lymphatic ducts D. sebum; sweat glands E. sebum; oil glands Sebum is an oily substance secreted by oil glands that lubricates the hair within the follicle and the skin itself.
Bloom's Level: 3. Apply Learning Outcome: 11.04.02 Describe the makeup and function of the accessory structures of human skin. Section: 11.04 Topic: Integumetary System
11-30 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 11 - Human Organization
True / False Questions 60. "Goose bumps" are caused by the arrector pili muscles making the hair follicles stand up. TRUE When the arrector pili muscles attached to the hair follicles contract, they cause the hairs to become erect. We call this "goose bumps."
Bloom's Level: 1. Remember Learning Outcome: 11.04.02 Describe the makeup and function of the accessory structures of human skin. Section: 11.04 Topic: Integumetary System
Multiple Choice Questions 61. Which of the following statements is FALSE about homeostasis? A. Homeostasis is controlled by hormones. B. Homeostasis is primarily controlled by the circulatory system. C. Homeostasis holds fluctuations near to the body's average levels. D. Homeostasis requires coordination between internal organs. E. Homeostasis is often controlled by negative feedback. All of the body systems participate in homeostasis.
Bloom's Level: 4. Analyze Learning Outcome: 11.05.01 Define homeostasis, and describe why it is essential to living organisms. Section: 11.05 Topic: Homeostasis
11-31 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 11 - Human Organization
62. When discussing homeostasis in body temperature, it is NOT true to state that the body responds to A. heat by stimulating sweat release. B. heat by constricting the blood vessels in the skin. C. cold by contracting the arrector pili muscles. D. cold by contracting skeletal muscles, causing shivering. E. cold by rerouting blood away from the skin. The body responds to heat by dilating the blood vessels in the skin to carry heat from the core to the surface of the body.
Bloom's Level: 3. Apply Learning Outcome: 11.05.01 Define homeostasis, and describe why it is essential to living organisms. Section: 11.05 Topic: Homeostasis
63. Which of the following is an example of homeostasis? A. Lymph nodes regulate homeostasis. B. Human oral body temperature is always close to 37o C (98.6 F). C. There are more red cells than white cells. D. An organism is limited in how high it can jump. E. An organism makes about as many turns to the right as to the left. Homeostasis is the maintenance of a relatively constant internal environment. Body temperature is an example.
Bloom's Level: 3. Apply Learning Outcome: 11.05.01 Define homeostasis, and describe why it is essential to living organisms. Section: 11.05 Topic: Homeostasis
11-32 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 11 - Human Organization
64. The disruption of homeostasis can lead to what type of disease that can be lifethreatening, but usually occurs and is resolved quickly? A. chronic B. systemic C. acute D. localized E. simple An acute disease tends to occur suddenly and generally lasts a short time.
Bloom's Level: 1. Remember Learning Outcome: 11.05.03 Discuss the roles of specific organ systems in homeostasis, and the health consequences if homeostasis is disrupted. Section: 11.05 Topic: Homeostasis
65. Which system is one of the two main body systems that helps maintain homeostasis by regulating other body systems? A. digestive B. endocrine C. respiratory D. lymphatic E. cardiovascular The endocrine and nervous system regulate the other body systems.
Bloom's Level: 2. Understand Learning Outcome: 11.05.03 Discuss the roles of specific organ systems in homeostasis, and the health consequences if homeostasis is disrupted. Section: 11.05 Topic: Homeostasis
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Chapter 11 - Human Organization
True / False Questions 66. When you drink coffee, which is more acidic than your bloodstream, the pH of your blood drops. FALSE Homeostasis maintains the pH of your blood even if you consume acidic or basic foods.
Bloom's Level: 3. Apply Learning Outcome: 11.05.03 Discuss the roles of specific organ systems in homeostasis, and the health consequences if homeostasis is disrupted. Section: 11.05 Topic: Homeostasis
67. A home heating system is an example of a positive feedback mechanism. FALSE A home heating system is an example of a negative feedback mechanism.
Bloom's Level: 1. Remember Learning Outcome: 11.05.02 Differentiate between positive and negative feedback mechanisms, and list specific examples of each. Section: 11.05 Topic: Feedback Mechanisms
68. An example of a negative feedback system is when there is too much of a particular chemical in the body and a sensor detects it and brings about a response that causes the levels to drop. TRUE This is an example of a negative feedback system.
Bloom's Level: 2. Understand Learning Outcome: 11.05.02 Differentiate between positive and negative feedback mechanisms, and list specific examples of each. Section: 11.05 Topic: Feedback Mechanisms
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Chapter 11 - Human Organization
Multiple Choice Questions 69. Why would it make sense to have the birthing process controlled by a positive feedback loop? A. because there is a definite cutoff point to be reached B. because birth is a positive experience C. because there is no sensor involved D. because there is no stimulus involved E. because there is a fluctuation of hormones above and below the level necessary to induce birth Positive feedback brings about an ever greater change in the same direction. This tends to be involved in processes that have a definite cutoff point, in this case, birth of the baby.
Bloom's Level: 3. Apply Learning Outcome: 11.05.02 Differentiate between positive and negative feedback mechanisms, and list specific examples of each. Section: 11.05 Topic: Feedback Mechanisms
70. During a football game an offensive lineman is hit in the legs from the side. Upon examination by the physical trainer it was determined that the player has a dislocated knee. There was a significant amount of damage to the entire knee joint. What is the main muscle, tissues, and membrane that would have been damaged? A. skeletal muscle, connective tissues, and the synovial membrane B. skeletal muscle, connective tissues, and the mucous membrane C. smooth muscle, connective tissues, and the synovial membrane D. skeletal muscle, connective tissues, and the serous membrane E. None of the answer choices is correct. The skeletal muscle, connective tissues, and the synovial membrane are the main tissues and membranes that would have been damaged. Smooth muscle is not a main part of the knee joint. The mucous and serous membranes are not part of the knee joint.
Bloom's Level: 5. Evaluate Learning Outcome: 11.02.02 Compare and contrast the location(s) and function(s) of the different types of body membranes. Section: 11.02 Topic: Animal Tissues
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Chapter 11 - Human Organization
71. Which membrane is composed of loose connective tissue and lines moveable joint cavities? A. synovial B. serous C. mucous D. meninges E. pleurae Synovial membranes are composed of loose connective tissue and line moveable joint cavities. Serous membranes cover organs. Mucous membranes are composed of loose, fibrous connective tissue. Meninges serve as a covering for the brain and spinal cord. Pleurae line the thoracic cavity and cover the lungs.
Bloom's Level: 1. Remember Learning Outcome: 11.02.02 Compare and contrast the location(s) and function(s) of the different types of body membranes. Section: 11.02 Topic: Animal Tissues
72. Which organ is not part of the digestive system? A. kidneys B. mouth C. esophagus D. large intestine E. pancreas The kidneys are part of the urinary system. The digestive system consists of the mouth, esophagus, stomach, small intestine, and large intestine. The accessory organs are the teeth, tongue, salivary glands, liver, gallbladder, and pancreas.
Bloom's Level: 1. Remember Learning Outcome: 11.03.01 List the major organs that make up each organ system. Section: 11.03 Topic: Body Systems
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Chapter 11 - Human Organization
73. Which structure can be classified as an endocrine and reproductive organ? A. testes B. lungs C. lymphocytes D. uterus E. oocytes The testes produce testosterone and the sex cells called sperm. Endocrine organs produce hormones, while reproductive organs are involved in the production of sex cells. The lungs are part of the respiratory system. The lymphocytes are part of the lymphatic system. The uterus is not part of the endocrine system. The oocytes are the egg cells produced by the reproductive system.
Bloom's Level: 2. Understand Learning Outcome: 11.03.01 List the major organs that make up each organ system. Section: 11.03 Topic: Body Systems
74. Which feature(s) helps distinguish the subcutaneous layer from the epidermis and dermis? A. The subcutaneous layer is composed of loose connective tissue and adipose tissue. B. The subcutaneous layer is composed of loose connective tissue and contains various sensory receptors. C. The epidermis is composed of loose connective tissue and adipose tissue. D. The subcutaneous layer is composed of fibrous connective tissue, such as collagen and elastic fibers. E. The subcutaneous layer contains melanocytes, which will give skin its pigmentation. The subcutaneous layer is composed of loose connective tissue and adipose tissue. Sensory receptors, collagen, and elastic fibers are found in the dermis. The melanocytes are found in the epidermis.
Bloom's Level: 5. Evaluate Learning Outcome: 11.04.01 Identify the two main regions of skin, and how these are distinguished from the subcutaneous layer. Section: 11.04 Topic: Integumetary System
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Chapter 11 - Human Organization
75. To which system do the kidneys and bladder belong? A. urinary B. nervous C. integumentary D. endocrine E. musculoskeletal The kidneys and bladder are part of the urinary system.
Bloom's Level: 1. Remember Learning Outcome: 11.03.01 List the major organs that make up each organ system. Section: 11.03 Topic: Body Systems
76. Which skin disorder is the result of a proliferation caused by the human papillomavirus? A. warts B. psoriasis C. acne D. All of the answer choices are caused by the human papillomavirus. E. None of the answer choices is caused by the human papillomavirus. Warts are due to a proliferation caused by the human papillomavirus. Skin cancer is most often caused by exposure to UV radiation. Acne is often caused by the growth of the bacteria Propionibacterium acnes.
Bloom's Level: 1. Remember Learning Outcome: 11.04.03 List some common disorders of human skin, and how these may be treated. Section: 11.04 Topic: Integumetary System
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Chapter 11 - Human Organization
77. What is the main reason for the presence of a pimple? A. All of the answer choices are reasons for the presence of a pimple. B. increased production of sebum C. increased levels of hormones D. the presence of Propionibacterium acnes E. inflammatory response All of the answer choices are reasons for the presence of a pimple.
Bloom's Level: 2. Understand Learning Outcome: 11.04.03 List some common disorders of human skin, and how these may be treated. Section: 11.04 Topic: Integumetary System
Short Answer Questions 78. List the subdivisions of the ventral body cavity and indicate the main organs that are found within each of the subdivisions. The subdivisions of the ventral body cavity include the thoracic, abdominal, and pelvic cavities. The thoracic cavity includes the right and left lungs and the heart. The abdominal cavity includes the stomach, liver, spleen, gallbladder, and small and large intestine. The pelvic cavity includes the rectum, urinary bladder, internal reproductive organs, and a portion of the large intestine.
Bloom's Level: 6. Create Learning Outcome: 11.02.01 List the two main cavities of the human body, and the major organs found in each. Section: 11.02 Topic: Body Systems
79. List the four major tissue types found in the human body. The four major types include: epithelial tissues, connective tissues, muscular tissues, and nervous tissues.
Bloom's Level: 6. Create Learning Outcome: 11.01.01 List the four major types of tissues found in the human body, and describe the structural features of each. Section: 11.01 Topic: Animal Tissues
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Chapter 11 - Human Organization
80. List the three layers of the integumentary system and indicate the main functions of each layer. The epidermis is the most superficial layer. Its main function is to provide protection for the body. The dermis is the middle layer. Its main function is to provide structural support and nourishment for the epidermis. The subcutaneous layer is the deepest layer. Its main function is to insulate the body from gaining or losing heat.
Bloom's Level: 6. Create Learning Outcome: 11.04.01 Identify the two main regions of skin, and how these are distinguished from the subcutaneous layer. Section: 11.04 Topic: Integumetary System
11-40 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 12 - Cardiovascular System
Chapter 12 Cardiovascular System
Multiple Choice Questions 1. Which blood vessels carry blood away from the heart? A. capillaries B. venules C. veins D. arteries E. lymph vessels Arteries carry blood away from the heart while veins carry blood toward the heart.
Bloom's Level: 1. Remember Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
2. Which statement is NOT a correct association? A. arteries—have a thick muscular layer B. capillaries—oxygen and nutrient exchange C. arteries—can control blood pressure D. capillaries—contain most of overall blood volume E. veins—valves to return blood to heart Veins contain most of the overall blood volume, as much as 70%.
Bloom's Level: 2. Understand Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
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Chapter 12 - Cardiovascular System
True / False Questions 3. All arteries contain oxygenated blood, and all veins contain nonoxygenated blood. FALSE Arteries carry blood away from the heart, while veins carry blood toward the heart. Most, but not all, arteries contain oxygenated blood, and most, but not all, veins contain nonoxygenated blood.
Bloom's Level: 2. Understand Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
4. Sphincter muscles regulate the opening and closing of arterioles leading to capillary beds, directing the flow of blood to different areas of the body depending on the body's needs. TRUE Precapillary and postcapillary sphincter muscles regulate the opening and closing of capillary beds.
Bloom's Level: 1. Remember Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
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Chapter 12 - Cardiovascular System
Multiple Choice Questions 5. Which blood vessels will have walls only one cell thick? A. capillaries B. venules C. veins D. arterioles E. arteries Capillaries have thin walls composed only of a single layer of endothelium with a basement membrane.
Bloom's Level: 2. Understand Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
6. How many layers does an arterial wall possess? Which is the thickest? A. two, the inside layer B. three, the inside layer C. two, the outside layer D. three, the outside layer E. three, the middle layer The arterial wall has three layers: endothelium, smooth muscle, and fibrous connective tissue. The smooth muscle (middle layer) is the thickest layer.
Bloom's Level: 4. Analyze Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
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Chapter 12 - Cardiovascular System
True / False Questions 7. When arterioles constrict, blood pressure decreases. FALSE When arterioles constrict, this causes blood pressure to rise.
Bloom's Level: 3. Apply Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
8. You are unable to see arteries and arterioles without a microscope. FALSE Arteries and arterioles are large enough to be seen with the naked eye. The aorta is ~25 mm wide.
Bloom's Level: 2. Understand Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
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Chapter 12 - Cardiovascular System
Multiple Choice Questions 9. Which of the following is true about capillaries? A. Capillaries are extremely narrow. B. Capillaries join arteries to arterioles. C. Capillaries walls have three layers. D. Capillaries are present in only a limited number of internal organs. E. Capillaries work individually to carry out their function. Capillaries are extremely narrow—their walls consist of only a single layer of cells. They join arterioles to venules and are present throughout the body, working in networks.
Bloom's Level: 4. Analyze Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
True / False Questions 10. Capillaries exchange oxygen and nutrients for wastes in the tissues. TRUE Capillaries play an important role in homeostasis because they exchange nutrients and wastes through their thin walls.
Bloom's Level: 2. Understand Learning Outcome: 12.01.02 Explain the process of capillary exchange. Section: 12.01 Topic: Blood Vessels
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Chapter 12 - Cardiovascular System
Multiple Choice Questions 11. Which organ of the body contains very few, if any, capillary beds? A. the heart B. the cornea C. the liver D. the stomach E. the skin The cornea is nearly capillary-free so that light can pass through.
Bloom's Level: 1. Remember Learning Outcome: 12.01.02 Explain the process of capillary exchange. Section: 12.01 Topic: Blood Vessels
True / False Questions 12. In the tissues, oxygen and nutrients enter the capillaries and carbon dioxide and wastes leave the capillaries. FALSE Just the opposite occurs. In the tissues, waste enters the capillaries while nutrients exit.
Bloom's Level: 2. Understand Learning Outcome: 12.01.02 Explain the process of capillary exchange. Section: 12.01 Topic: Blood Vessels
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Chapter 12 - Cardiovascular System
Multiple Choice Questions 13. Which of the following is true concerning a comparison of arteries and veins? A. Both are attached to arterioles. B. Both are attached to venules. C. Both contain three layers in their walls. D. Both contain valves. E. Both allow nutrient exchange across their walls. Both arteries and veins have three layers in their walls: endothelium, smooth muscle, and connective tissue. However, veins have less smooth muscle and connective tissue than arteries.
Bloom's Level: 5. Evaluate Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
14. The outer sac that encloses the heart and provides lubrication is the A. myocardium. B. endocardium. C. ectocardium. D. pericardium. E. coronary capsule. The membrane that surrounds the heart is called the pericardium. The myocardium is the muscle tissue of the heart.
Bloom's Level: 1. Remember Learning Outcome: 12.03.01 Identify the major components of the heart, including the four chambers and four valves. Section: 12.03 Topic: Heart
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Chapter 12 - Cardiovascular System
15. Which of these is NOT true of the heart? A. The heart is a double pump. B. The heart helps maintain blood pressure. C. The heart helps make red blood cells. D. The heart is composed primarily of muscle. E. The heart has an inner lining of endothelium. Red blood cells are made in the bone marrow.
Bloom's Level: 2. Understand Learning Outcome: 12.03.01 Identify the major components of the heart, including the four chambers and four valves. Section: 12.03 Topic: Heart
16. The pulmonary semilunar valve prevents blood from reentering the A. right atrium. B. left atrium. C. right ventricle. D. left ventricle. E. pulmonary trunk. The pulmonary semilunar valve lies between the right ventricle and the pulmonary trunk and prevents blood from reentering the right ventricle.
Bloom's Level: 2. Understand Learning Outcome: 12.03.01 Identify the major components of the heart, including the four chambers and four valves. Section: 12.03 Topic: Heart
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Chapter 12 - Cardiovascular System
17. In the fetal heart, the lungs are deflated and the fetus receives oxygen from its mother across the placenta through the umbilical cord. There is an opening in the wall between the left and right side of the fetal heart. Normally, this closes before birth. What symptoms would be expected if this heart wall opening remained open? A. It would cause hypertension; fetal blood pressure would be abnormally high. B. This would balance the diastolic and systolic blood pressure readings. C. Deoxygenated blood would mix with oxygenated blood and systemic circulation would have too little oxygen to sustain healthy growth. D. This would not actually cause any bad effects at all since the heart pumping would be equally distributed between left and right ventricles. E. This would cause blood to flow back out of the heart through the right atrium and cause drastic effects in the fetus. If the opening between the two sides of the heart does not close, deoxygenated and oxygenated blood are able to mix in the ventricles. This leads to too little oxygenation of the blood.
Bloom's Level: 3. Apply Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Heart
18. A heart murmur that sounded "lub-dup-shush" would indicate which of the following? A. atrioventricular valve is not closing properly B. semilunar valve is not closing properly C. high blood pressure in the atria D. ventricles are not contracting properly E. low blood pressure in the ventricles The lub occurs when the atrioventricular valves close. The dup results when the semilunar valves close. A heart murmur, or a slight whooshing sound, is due to blood flowing back through an ineffectively closed semilunar valve.
Bloom's Level: 5. Evaluate Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Heart
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Chapter 12 - Cardiovascular System
19. There is a condition where a baby is born with an opening (foramen ovale) in the septum or wall between the atria. What are reasonable consequences? A. more efficient pumping of blood through the body with pressure from both ventricles B. an equal reading for blood pressure, that is diastole equals systole C. disability when deoxygenated blood is mixed and sent to the body and partly oxygenated blood is sent to the lungs D. death due to loss of blood from the circulatory system E. no significant differences would occur This would result in a mixing of deoxygenated and oxygenated blood in the ventricles and not enough oxygenation of the blood sent to the rest of the body, resulting in disability.
Bloom's Level: 4. Analyze Learning Outcome: 12.03.01 Identify the major components of the heart, including the four chambers and four valves. Section: 12.03 Topic: Heart
20. What is the correct sequence in which a drop of blood returning from the body encounters the heart chambers? A. left atrium → left ventricle → right atrium → right ventricle B. left ventricle → left atrium → right ventricle → right atrium C. right atrium → left atrium → left ventricle → right ventricle D. right atrium → left ventricle → left atrium → right ventricle E. right atrium → right ventricle → left atrium → left ventricle Blood moves from the vena cavae into the right atrium, then to the right ventricle, through the lungs, to the left atrium, and to the left ventricle.
Bloom's Level: 2. Understand Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Heart
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Chapter 12 - Cardiovascular System
21. What is the correct sequence in which a drop of blood returning from the body encounters the heart valves? A. right atrioventricular valve → aortic semilunar valve → left atrioventricular valve → pulmonary semilunar valve B. right atrioventricular valve → pulmonary semilunar valve → left atrioventricular valve → aortic semilunar valve C. left atrioventricular valve → pulmonary semilunar valve → right atrioventricular valve → aortic semilunar valve D. pulmonary semilunar valve → left atrioventricular valve → aortic semilunar valve → right atrioventricular valve E. left atrioventricular valve → aortic semilunar valve → right atrioventricular valve → pulmonary semilunar valve The blood goes from the right atrium to the right ventricle through the atrioventricular valve. It then goes from the right ventricle to the lungs through the pulmonary semilunar valve. It then goes from the lungs to the left atrium to the left ventricle through the atrioventricular valve. It then goes from the left ventricle to the aorta through the aortic semilunar valve.
Bloom's Level: 2. Understand Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Heart
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Chapter 12 - Cardiovascular System
22. Early doctors did not understand that blood flowed in a circuit but assumed that blood ebbed and flowed like a tide in the body. William Harvey was the first to demonstrate the circulation of blood and convince the medical community. Which of the following observations would be evidence that blood was pumped through the body or flowed in a circuit? A. Dissecting a live fish shows the heart beating and blood moving, gushing one direction. B. A very tight tourniquet causes the arteries to swell on the side toward the heart, but not on the side of the ligature away from the heart. C. Measuring the volume of the heart and counting the beats per minute gives a value of 520 pounds of blood an hour, many times a person's volume. D. In surface veins, blood can be pressed past a valve toward the heart but not away from the heart. E. A cut made anywhere on the body results in blood seeping from the wound. Only a cut made anywhere on the body which results in seeping of blood suggests that the blood is in a circuit. All of the others could not differentiate between a one way and a circular flow.
Bloom's Level: 5. Evaluate Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Circulatory System
23. Contraction of the right ventricle forces blood initially into the A. left atrium. B. right atrium. C. aorta. D. pulmonary vein. E. pulmonary artery. The right ventricle pumps blood into the lungs through the pulmonary artery.
Bloom's Level: 1. Remember Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Heart
12-12 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 12 - Cardiovascular System
24. If you are tracing the path of blood from the right to the left side of the heart, you must mention the A. lungs. B. hepatic portal vein. C. aorta. D. superior vena cava. E. inferior vena cava. The blood from the right ventricle must enter the lungs before it goes to the left atrium.
Bloom's Level: 2. Understand Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Heart
25. In humans, the superior vena cava A. carries blood to the right atrium. B. carries blood away from the right atrium. C. joins with the aorta. D. has a high blood pressure. E. carries O2-rich blood. The superior vena cava drains oxygen-depleted blood from the tissues above the heart and enters the right atrium.
Bloom's Level: 2. Understand Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Heart
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Chapter 12 - Cardiovascular System
26. The pacemaker of the heart is termed the A. AV node. B. cardiac center in the medulla. C. SA node. D. vagus nerve. E. Purkinje fibers. The SA node is called the cardiac pacemaker because it usually keeps the heartbeat regular.
Bloom's Level: 1. Remember Learning Outcome: 12.03.03 Describe the intrinsic and extrinsic control of the heartbeat. Section: 12.03 Topic: Heart
27. The alternating expansion and recoil of an arterial wall is termed a(n) A. ECG. B. EEG. C. pulse. D. hypertension. E. heart murmur. The pulse is a wave effect that passes down the walls of arteries when the aorta expands and then recoils with each ventricular contraction.
Bloom's Level: 1. Remember Learning Outcome: 12.03.03 Describe the intrinsic and extrinsic control of the heartbeat. Section: 12.03 Topic: Circulatory System
12-14 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 12 - Cardiovascular System
28. The "lub" heart sound is due to the A. onward rush of blood to the heart. B. contraction of aortic muscle tissue. C. closing of the semilunar valves. D. closing of the atrioventricular valves. E. the recoil of the aorta following the ventricle contraction. The lub sound is caused when the atrioventricular valves close due to ventricular contraction.
Bloom's Level: 1. Remember Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Heart
29. Systole occurs when A. the heart tissue is contracting. B. the heart tissue is relaxing. C. the semilunar valves open and close. D. the aortic valves open and close. E. there is no heartbeat. The word systole refers to contraction of heart muscle.
Bloom's Level: 1. Remember Learning Outcome: 12.03.03 Describe the intrinsic and extrinsic control of the heartbeat. Section: 12.03 Topic: Heart
True / False Questions 30. The major portion of the heart is the myocardium, made up mainly of cardiac muscle. TRUE The myocardium, consisting of cardiac muscle, makes up most of the heart.
Bloom's Level: 1. Remember Learning Outcome: 12.03.01 Identify the major components of the heart, including the four chambers and four valves. Section: 12.03 Topic: Heart
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Chapter 12 - Cardiovascular System
Multiple Choice Questions 31. Which is NOT a correct association describing aspects of the cardiac cycle? A. Each heartbeat constitutes a cardiac cycle. B. Diastole is contraction of the heart and systole is relaxation. C. A normal heart rate varies from 60 to 80 beats per minute. D. A pulse is the expansion and recoil of artery walls felt near the body surface. E. Valves prevent backflow of blood into the heart when the heart relaxes. Systole refers to contraction of heart muscle, while diastole refers to relaxation of heart muscle.
Bloom's Level: 4. Analyze Learning Outcome: 12.03.03 Describe the intrinsic and extrinsic control of the heartbeat. Section: 12.03 Topic: Heart
32. Intrinsic control of heartbeat rate involves what mechanism(s)? A. The AV node initiates heartbeat, although the SA node can generate a slower beat rate. B. The atrioventricular bundle initiates signals that travel to the Purkinje fibers. C. The Purkinje fibers initiate signals that travel to the atrioventricular bundle. D. The SA node initiates heartbeat and the AV node inhibits the heartbeat. E. The SA node initiates heartbeat, although the AV node can generate a slower beat rate. The SA node initiates the heartbeat, causing the atria to contract. The impulses reach the AV node, which then signals the ventricles to contract by sending the impulse through the atrioventricular bundle to the Purkinje fibers. The AV node can generate a slower beat rate on its own.
Bloom's Level: 5. Evaluate Learning Outcome: 12.03.03 Describe the intrinsic and extrinsic control of the heartbeat. Section: 12.03 Topic: Heart
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Chapter 12 - Cardiovascular System
33. The heart rate speeds up or slows down because the A. AV node detects lower levels of oxygen and sends faster signals to the SA node. B. brain consciously detects the need for more oxygen and intentionally stimulates the heart. C. medulla oblongata, through the autonomic system, stimulates or depresses heart rate based on activity and stress. D. hormone epinephrine inhibits the heartbeat. E. SA node detects lower levels of oxygen and sends faster signals to the AV node. The extrinsic control of the heartbeat occurs in the medulla oblongata, which can alter the heartbeat by way of the autonomic system.
Bloom's Level: 4. Analyze Learning Outcome: 12.03.03 Describe the intrinsic and extrinsic control of the heartbeat. Section: 12.03 Topic: Heart
True / False Questions 34. The sound of a heartbeat results when the ventricles contract and blood is surged out of the heart. FALSE The sound of the heartbeat is due to the sound of the valves within the heart closing.
Bloom's Level: 2. Understand Learning Outcome: 12.04.01 Describe the flow of blood from the heart through all major parts of the body. Section: 12.04 Topic: Heart
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Chapter 12 - Cardiovascular System
Multiple Choice Questions 35. When the paramedic shouts "stand back" and applies the paddles to administer a strong current, this is an attempt to A. restart ventricular contractions that have completely stopped. B. reverse control of heartbeat from AV to SA nodes. C. align the P-QRS waves with the electrical paddle waves. D. stop the ventricles from uncontrolled contractions so the SA node can reestablish a coordinated beat. E. destroy heart tissue that is out of control. Ventricular fibrillation causes uncoordinated contraction of the ventricles. They must be defibrillated by applying a strong electrical current for a short period of time, allowing the SA node to reestablish a coordinated beat.
Bloom's Level: 3. Apply Learning Outcome: 12.03.03 Describe the intrinsic and extrinsic control of the heartbeat. Section: 12.03 Topic: Heart
36. In a hypothetical experiment, the blood flow between two dogs is joined and one dog is stressed so that its heart rate goes up rapidly. What will happen to the heart rate of the second dog? A. It will remain calm because the heart rate is determined intrinsically in each heart. B. Heart rate will go up due to an increase in epinephrine and norepinephrine in the blood. C. Heart rate will go up due to signals from the first dog's SA node. D. It will remain calm because there is no connection of the dogs' sympathetic nervous systems. E. The heart rate of the second dog will decrease to compensate for the speed of the first dog's heart. The heart rate will go up because of the hormones epinephrine and norepinephrine, released by the adrenal medulla of the first dog, are present in the blood and will stimulate the heart of the second dog.
Bloom's Level: 4. Analyze Learning Outcome: 12.03.03 Describe the intrinsic and extrinsic control of the heartbeat. Section: 12.03 Topic: Heart
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Chapter 12 - Cardiovascular System
37. The electrical changes that occur in the heart during contraction are recordings termed A. EEGs. B. ECGs. C. EGGs. D. EMGs. E. MIGs. ECG stands for an electrocardiogram, a recording of the electrical changes that occur in the myocardium during a cardiac cycle.
Bloom's Level: 1. Remember Learning Outcome: 12.03.03 Describe the intrinsic and extrinsic control of the heartbeat. Section: 12.03 Topic: Heart
38. Which phase of the ECG indicates ventricular excitation and contraction? A. P wave B. QRS wave C. T wave D. U wave E. I wave The QRS wave occurs just prior to ventricular contraction. The P wave occurs prior to atrial contraction and the T wave occurs when the ventricles are recovering from contraction.
Bloom's Level: 2. Understand Learning Outcome: 12.03.03 Describe the intrinsic and extrinsic control of the heartbeat. Section: 12.03 Topic: Heart
12-19 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 12 - Cardiovascular System
39. The part of the circulation involved with pumping blood to and from the lungs is the A. systemic circuit. B. internal respiration circuit. C. microcirculation circuit. D. pulmonary circuit. E. ECG circuit. The pulmonary circuit circulates blood through the lungs. The systemic circuit serves the needs of body tissues.
Bloom's Level: 1. Remember Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Cardiovasular System
40. Which blood vessel has the highest concentration of carbon dioxide? A. pulmonary artery B. pulmonary vein C. aorta D. renal vein E. hepatic portal vein The pulmonary artery contains deoxygenated, carbon-dioxide-rich blood on its way to the lungs.
Bloom's Level: 2. Understand Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Cardiovasular System
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Chapter 12 - Cardiovascular System
41. Oxygenated blood enters the heart at the A. left atrium. B. left ventricle. C. right atrium. D. right ventricle. E. pulmonary arteries. The pulmonary vein brings oxygenated blood from the lungs to the left atrium.
Bloom's Level: 1. Remember Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Cardiovasular System
42. The systemic circuit circulates blood from the A. right ventricle to left atrium. B. left ventricle to left atrium. C. right ventricle to right atrium. D. left ventricle to right atrium. E. pulmonary artery to the lungs. Oxygenated blood leaves the left ventricle to go throughout the body. Deoxygenated blood from the body then enters the right atrium.
Bloom's Level: 2. Understand Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Circulatory System
12-21 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 12 - Cardiovascular System
43. Which of the following correctly traces the path of blood from the heart to the kidneys and back to the heart again? A. left ventricle → vena cava → renal vein → kidney → renal artery → aorta → right atrium B. right atrium → aorta → renal artery → kidney → renal vein → hepatic portal vein → vena cava → left atrium C. left ventricle → renal artery → kidney → renal vein → vena cava → right atrium D. left ventricle → aorta → renal artery → kidney → renal vein → vena cava → right atrium E. right ventricle → aorta → renal artery → kidney → renal vein → vena cava → hepatic portal vein → right atrium Blood leaves the heart to go to the rest of the body from the left ventricle into the aotra. From there it goes to the renal artery, the kidney, the renal vein, and into the vena cava. The vena cava empties into the right atrium.
Bloom's Level: 3. Apply Learning Outcome: 12.04.01 Describe the flow of blood from the heart through all major parts of the body. Section: 12.04 Topic: Circulatory System
44. Blood traveling from the intestines to the liver travels through the A. pulmonary vein. B. inferior vena cava. C. superior vena cava. D. coronary vessels. E. hepatic portal vein. The hepatic portal vein connects the digestive tract with the liver.
Bloom's Level: 2. Understand Learning Outcome: 12.04.01 Describe the flow of blood from the heart through all major parts of the body. Section: 12.04 Topic: Circulatory System
12-22 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 12 - Cardiovascular System
45. The blood vessel that provides oxygen to the heart tissue is the A. pulmonary artery. B. coronary artery. C. left carotid artery. D. renal artery. E. left subclavian artery. The heart is nourished by the coronary arteries. It does not draw nourishment from the blood within the heart itself.
Bloom's Level: 2. Understand Learning Outcome: 12.03.01 Identify the major components of the heart, including the four chambers and four valves. Section: 12.03 Topic: Heart
46. The blood that enters the coronary artery comes from the A. pulmonary artery. B. pulmonary vein. C. aorta. D. superior vena cava. E. inferior vena cava. The coronary arteries are the first branches off of the aorta.
Bloom's Level: 2. Understand Learning Outcome: 12.03.01 Identify the major components of the heart, including the four chambers and four valves. Section: 12.03 Topic: Heart
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Chapter 12 - Cardiovascular System
47. Usually there is only one capillary bed between an artery and vein; an exception to this is the A. coronary vessels that serve the heart. B. pulmonary system that serves the lungs and where all factors are reversed. C. hepatic portal system which runs from the capillary bed in the intestine to the capillary bed in the liver via a hepatic portal vein. D. brain circulation which requires special continuous blood flow. E. large bones that lack capillaries but are fed by an artery/vein system. The digestive tract and liver capillary beds are connected by the hepatic portal vein. A portal system connects two beds of capillaries.
Bloom's Level: 1. Remember Learning Outcome: 12.04.01 Describe the flow of blood from the heart through all major parts of the body. Section: 12.04 Topic: Circulatory System
48. Blood flow in the capillaries is A. slower because there are more capillaries and they represent more flow volume. B. faster because they are smaller and blood must speed up to get through. C. the same speed as in the arteries and veins since they are the input and output points. D. highly variable due to different tissues, but must average the same as speeds in arteries and the veins. E. slower than in other blood vessels because of the presence of valves. In capillaries, blood is under minimal pressure and has the least velocity. Blood pressure and velocity drop off because capillaries have a greater total cross-sectional area than arterioles.
Bloom's Level: 2. Understand Learning Outcome: 12.01.02 Explain the process of capillary exchange. Section: 12.01 Topic: Blood Vessels
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Chapter 12 - Cardiovascular System
49. If a person has a blood pressure of 120/80, the "120" refers to the A. diastolic pressure. B. systolic pressure. C. pulse pressure. D. capillary pressure. E. venous pressure. The first number in the blood pressure is the systolic pressure.
Bloom's Level: 1. Remember Learning Outcome: 12.04.02 Explain the factors that affect blood pressure in arteries, capillaries, and veins. Section: 12.04 Topic: Cardiovasular System
50. Which structure(s) would be subject to the lowest blood pressure? A. vena cava B. aorta C. atrioventricular valve D. pulmonary arteries E. semilunar valve The blood pressure results from blood being forced into the arteries during ventricular contraction. The vena cava is the vein furthest from the arteries, so it would feel the effects of this contraction least.
Bloom's Level: 3. Apply Learning Outcome: 12.04.02 Explain the factors that affect blood pressure in arteries, capillaries, and veins. Section: 12.04 Topic: Cardiovasular System
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Chapter 12 - Cardiovascular System
51. Why is it advantageous to have blood move slowly in the capillaries? A. It will not cause damage to the valves. B. It maintains the high pressure in the capillaries. C. It provides time for substances to be exchanged between blood and tissue fluids. D. It creates a greater osmotic pressure for cells. E. It is the major way to control blood pressure. Capillaries are where the exchange of nutrients and wastes occur. A slower velocity allows more time for this exchange to take place.
Bloom's Level: 4. Analyze Learning Outcome: 12.02.03 Define capillary exchange, and describe the two major forces involved. Section: 12.02 Topic: Circulatory System
52. Venous return does not depend on A. respiratory movements. B. skeletal muscle contractions. C. the presence of valves. D. limb movement. E. blood pressure. Since blood pressure is minimal in veins, blood return via veins does not depend on blood pressure. All of the other factors are involved.
Bloom's Level: 2. Understand Learning Outcome: 12.04.02 Explain the factors that affect blood pressure in arteries, capillaries, and veins. Section: 12.04 Topic: Circulatory System
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Chapter 12 - Cardiovascular System
53. A waitress that stands for hours and has a habit of crossing her legs when she sits will increase the chances of having blood accumulate in her A. arteries. B. veins. C. capillaries. D. left ventricle. E. right ventricle. Without muscle contraction and with the cutting off of blood return from the legs to the heart, blood will accumulate in the veins in her legs. This may result in serious swelling.
Bloom's Level: 3. Apply Learning Outcome: 12.04.02 Explain the factors that affect blood pressure in arteries, capillaries, and veins. Section: 12.04 Topic: Circulatory System
True / False Questions 54. A reading of 120/80 indicates blood velocity. FALSE This is an indication of blood pressure, not velocity.
Bloom's Level: 1. Remember Learning Outcome: 12.04.02 Explain the factors that affect blood pressure in arteries, capillaries, and veins. Section: 12.04 Topic: Cardiovasular System
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Chapter 12 - Cardiovascular System
Multiple Choice Questions 55. Which of the following is NOT a correct association? A. arteries—blood away from heart B. ventricles—receive blood from body C. valves—veins D. right heart—lungs E. left atrium—receive oxygenated blood The atria, not the ventricles, receive blood from the body.
Bloom's Level: 4. Analyze Learning Outcome: 12.03.01 Identify the major components of the heart, including the four chambers and four valves. Section: 12.03 Topic: Blood Vessels
56. "Artery" and "vein" are defined by whether they leave or enter the heart. If they were defined by whether they carried oxygenated or deoxygenated blood, we would have to change the name(s) of A. the hepatic portal vein to a hepatic portal artery. B. the hepatic and renal systems (they would have to be switched). C. the pulmonary artery and vein (they would have to be switched). D. the coronary artery and vein (they would have to be switched). E. the aorta and vena cava. The pulmonary artery carries deoxygenated blood from the heart to the lungs, while the pulmonary vein carries oxygenated blood from the lungs to the heart. Those two names would need to be switched.
Bloom's Level: 4. Analyze Learning Outcome: 12.04.01 Describe the flow of blood from the heart through all major parts of the body. Section: 12.04 Topic: Cardiovasular System
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Chapter 12 - Cardiovascular System
57. Which association is NOT correct? A. red blood cells—transport oxygen B. white blood cells—fight infection C. water—maintains blood volume D. oxygen—cellular respiration E. albumin—fights infection Albumin is a blood protein that transports bilirubin.
Bloom's Level: 2. Understand Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
58. All of the following are normally found in the blood, except A. fibrinogen. B. glucose. C. urea. D. oxygen. E. glycogen. All of the answer choices are components of the blood, except for glycogen. Glycogen is the storage form of glucose found within animal cells.
Bloom's Level: 2. Understand Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
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Chapter 12 - Cardiovascular System
59. Researchers have long attempted to find a substitute for red blood cells, specifically the hemoglobin carrying ability. With the low but possible risk of disease transfer from blood transfusion, such a synthetic substance could be guaranteed safe if it was effective. Which of the following properties would be important for "artificial red blood cells"? A. must attach to and never release oxygen B. must release oxygen to any other medium despite concentration gradient C. would bind to iron molecules already bound to other artificial blood units D. must provide the clotting reaction E. would bind to oxygen in a high oxygen concentration solution and release oxygen when in a low oxygen concentration solution Red blood cells bind to oxygen in a high oxygen concentration solution (such as in the lungs) and release oxygen when in a low oxygen concentration solution (as in the tissues). This is an important characteristic of something that must shuttle oxygen within the body.
Bloom's Level: 3. Apply Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
60. Hemoglobin is NOT A. carried in red blood cells. B. an oxygen transporter. C. made of protein and heme. D. normally found free in the plasma. E. a red pigment. Hemoglobin is normally found within red blood cells, not free in the plasma component of the blood.
Bloom's Level: 2. Understand Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
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Chapter 12 - Cardiovascular System
61. Which specific component in hemoglobin combines with oxygen? A. heme B. globin C. iron in the heme D. iron in the globin E. the polypeptide Oxygen binds to the iron in the heme portion of the hemoglobin. The globin part is the polypeptide.
Bloom's Level: 2. Understand Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
62. Which of the following proteins is found in abundance within red blood cells? A. albumin B. hemoglobin C. globulin D. fibrinogen E. myoglobin Red blood cells contain large amounts of hemoglobin.
Bloom's Level: 1. Remember Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
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Chapter 12 - Cardiovascular System
63. The reason red blood cells only live about 120 days is because red blood cells A. lack a nucleus and therefore cannot defend themselves from a pathogen attack. B. lack a nucleus and therefore cannot repair and rebuild cell structures. C. are attacked and eaten by white blood cells. D. are physiologically inferior cells that are not made to last very long. E. are destroyed by the high levels of oxygen they carry. Red blood cells lack a nucleus and are unable to repair/resynthesize new cellular structures. The nucleus is lost to enable them to carry more hemoglobin.
Bloom's Level: 2. Understand Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
64. Aged red blood cells are destroyed in the A. red bone marrow. B. lungs. C. lymph nodes. D. spleen and liver. E. capillaries. The spleen and liver destroy old red blood cells. The iron is mostly salvaged and reused.
Bloom's Level: 1. Remember Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
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Chapter 12 - Cardiovascular System
65. A person with normal blood volume and white cell count but low red cell count or low hemoglobin content may be suffering from A. hemophilia. B. anemia. C. leukemia. D. mononucleosis. E. blood poisoning. Anemia is when an individual has an insufficient number of red blood cells or the red blood cells do not contain enough hemoglobin.
Bloom's Level: 2. Understand Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
66. Neutrophils and lymphocytes are types of A. platelets. B. red blood cells. C. white blood cells. D. plasma cells. E. epithelial cells. Neutrophils and lymphocytes are types of white blood cells.
Bloom's Level: 1. Remember Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
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Chapter 12 - Cardiovascular System
67. Which statement is NOT true about neutrophils? A. They are phagocytic. B. They have a multilobed nucleus. C. They are agranular leukocytes. D. They do not produce antibodies. E. They are the most abundant white blood cell. Neutrophils are granular leukocytes because they are filled with spheres that contain enzymes and proteins.
Bloom's Level: 2. Understand Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
68. The two enzymes in the clotting process are A. fibrin and prothrombin activator. B. prothrombin activator and thrombin. C. platelets and fibrin. D. prothrombin and calcium. E. albumin and hemoglobin. Prothrombin activator is an enzyme that converts prothrombin to thrombin. Thrombin is an enzyme that converts fibrinogen into fibrin.
Bloom's Level: 2. Understand Learning Outcome: 12.02.02 Identify the major molecular and cellular events that result in a blood clot. Section: 12.02 Topic: Blood
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Chapter 12 - Cardiovascular System
69. The element in your blood necessary for the clotting process is A. calcium. B. phosphorus. C. chlorine. D. iron. E. sodium. The conversion of prothrombin to thrombin requires calcium ions.
Bloom's Level: 1. Remember Learning Outcome: 12.02.02 Identify the major molecular and cellular events that result in a blood clot. Section: 12.02 Topic: Blood
70. The long insoluble threads of a blood clot originate from A. fibrinogen. B. thrombin. C. collagen. D. prothrombin. E. platelets. The long insoluble threads are fibrin. These originate from the cleavage of fibrinogen.
Bloom's Level: 2. Understand Learning Outcome: 12.02.02 Identify the major molecular and cellular events that result in a blood clot. Section: 12.02 Topic: Blood
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Chapter 12 - Cardiovascular System
71. Platelets A. phagocytize bacteria. B. are responsible for blood type. C. initiate clotting. D. transport oxygen in the blood. E. are responsible for the buildup of plaques on artery walls. Platelets are involved in the process of blood clotting or coagulation.
Bloom's Level: 1. Remember Learning Outcome: 12.02.02 Identify the major molecular and cellular events that result in a blood clot. Section: 12.02 Topic: Blood
72. Stem cells A. are formed from the formed elements in the blood. B. are found only in bone marrow. C. are found only in embryos. D. can differentiate into many cell types, including neurons. E. will remain as stem cells no matter how many cell divisions take place. A stem cell is a cell that is ever capable of dividing and producing new cells that go on to differentiate.
Bloom's Level: 1. Remember Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
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Chapter 12 - Cardiovascular System
73. Lymph, such as that collected under a blister, contains all the components of blood except A. water. B. salts. C. sugars. D. dissolved gases. E. blood cells, albumin, and other plasma proteins. Lymph does not contain blood cells and other plasma proteins. It is considered tissue fluid.
Bloom's Level: 2. Understand Learning Outcome: 12.02.03 Define capillary exchange, and describe the two major forces involved. Section: 12.02 Topic: Blood
74. Which of the following is a true statement about tissue fluid? A. Tissue fluid exits from white blood cells. B. Tissue fluid exits the capillary at the venous end. C. Tissue fluid is collected by the lymphatic capillaries. D. Tissue fluid contains more protein than blood. E. Tissue fluid forms at the venous end of a capillary. The lymphatic capillaries serve as collectors of excess tissue fluid.
Bloom's Level: 1. Remember Learning Outcome: 12.02.03 Define capillary exchange, and describe the two major forces involved. Section: 12.02 Topic: Blood
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Chapter 12 - Cardiovascular System
True / False Questions 75. Only blood pressure controls movement of fluid through the capillary wall into the tissue. Therefore, people with high blood pressure often have swelling issues. FALSE Both blood pressure and osmotic pressure control the movement of fluid through the capillary wall.
Bloom's Level: 2. Understand Learning Outcome: 12.02.03 Define capillary exchange, and describe the two major forces involved. Section: 12.02 Topic: Blood
76. Hemophilia is an inherited clotting disorder found most often in males. TRUE The most common type of hemophila, a clotting disorder, almost always occurs in males due to the faulty gene being found on the X chromosome.
Bloom's Level: 1. Remember Learning Outcome: 12.02.02 Identify the major molecular and cellular events that result in a blood clot. Section: 12.02 Topic: Blood
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Chapter 12 - Cardiovascular System
Multiple Choice Questions 77. Which of the following variables does not increase an individual's likelihood of developing a cardiovascular disease? A. gender B. family history of heart attack under 55 years of age C. smoking D. hypertension E. type 2 diabetes Cardiovascular diseases are not associated with gender. Each of the other choices increases your risk of developing a cardiovascular disease.
Bloom's Level: 4. Analyze Learning Outcome: 12.05.02 Define hypertension and explain its most common causes. Section: 12.05 Topic: Cardiovasular System
78. Accumulation of cholesterol beneath the inner linings of arteries is called A. hypertension. B. heart attack. C. stroke. D. atherosclerosis. E. thrombus. Atherosclerosis is an accumulation of soft masses of fatty materials beneath the inner linings of arteries.
Bloom's Level: 1. Remember Learning Outcome: 12.05.01 Describe the major categories of cardiovascular disease that occur in the United States. Section: 12.05 Topic: Cardiovasular System
12-39 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 12 - Cardiovascular System
79. Before major surgery, a patient is often asked if they have been taking large amounts of aspirin, and if so, they may have to postpone surgery and go without aspirin for some days. What is the reason for this? A. Aspirin interferes with the action of prothrombin activator. B. Aspirin causes the blood to be thinner or less viscous. C. Aspirin decreases red blood cell production. D. Aspirin reduces the stickiness of platelets and reduces the ability to clot. E. Aspirin eliminates pain and some pain is necessary for healing. Aspirin interferes with blood clotting by reducing the stickiness of platelets. Blood clotting after surgery is important for healing.
Bloom's Level: 3. Apply Learning Outcome: 12.05.03 List three possible treatments for a blocked coronary artery. Section: 12.05 Topic: Blood
80. What is a clot called that remains at its origin? What is a clot called that breaks away and becomes lodged in the lungs? A. thrombus; cerebral embolism B. embolus; cerebral embolism C. thrombus; pulmonary embolism D. embolus; pulmonary embolism E. thrombus; stroke As long as a clot remains stationary, it is called a thrombus. If it dislodges and moves, it is called an embolus. One that moves into the lungs would be called a pulmonary embolism.
Bloom's Level: 3. Apply Learning Outcome: 12.05.01 Describe the major categories of cardiovascular disease that occur in the United States. Section: 12.05 Topic: Blood
12-40 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 12 - Cardiovascular System
True / False Questions 81. In a coronary bypass operation, a stent is placed within the blockage and the clogged area is compressed. FALSE In a coronary bypass surgery, a segment of another blood vessel is used to bypass a blocked area in the coronary artery. An angioplasty involves a stent placed within a blockage.
Bloom's Level: 1. Remember Learning Outcome: 12.05.03 List three possible treatments for a blocked coronary artery. Section: 12.05 Topic: Cardiovasular System
82. It may be possible to "fix" a coronary blockage by gene therapy. FALSE The results of some clinical trials have suggested that stem cell therapy can repair damaged heart muscles and blood vessels in the heart.
Bloom's Level: 2. Understand Learning Outcome: 12.05.03 List three possible treatments for a blocked coronary artery. Section: 12.05 Topic: Cardiovasular System
83. Hyptertension is sometimes called the silent killer because no symptoms may be experienced until a stroke or heart attack occurs. TRUE Hypertension usually exhibits no symptoms.
Bloom's Level: 1. Remember Learning Outcome: 12.05.01 Describe the major categories of cardiovascular disease that occur in the United States. Section: 12.05 Topic: Cardiovasular System
12-41 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 12 - Cardiovascular System
Short Answer Questions 84. List the five different types of white blood cells found in the human body in order of frequency, with the most common one first. Neutrophils are the most common type of white blood cell. Lymphocytes are the second most common type of white blood cell. Monocytes are the third most common type of white blood cell. Eosinophils are the fourth most common type of white blood cell. Basophils are the least common type of white blood cell.
Bloom's Level: 6. Create Learning Outcome: 12.02.01 List the major types of blood cells, and summarize their functions. Section: 12.02 Topic: Blood
Multiple Choice Questions 85. Which blood pressure is considered abnormally high for individuals who are under the age of 45? A. 130/90 mm Hg B. 90/130 mm Hg C. 110/95 mm Hg D. 95/140 mm Hg E. 120/90 mm Hg 130/90 mm Hg is considered abnormally high blood pressure for an individual under the age of 45.
Bloom's Level: 1. Remember Learning Outcome: 12.05.02 Define hypertension and explain its most common causes. Section: 12.05 Topic: Cardiovasular System
12-42 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 12 - Cardiovascular System
Short Answer Questions 86. Compare the differences between the walls of arteries and veins. Both arterial and venous walls are composed of three layers. The middle layer in the arterial wall is much thicker than that in the veins due to the smooth muscle. The thicker wall can contract to help regulate blood flow and blood pressure. The innermost layer of the veins will often have valves that help prevent the back flow of blood.
Bloom's Level: 6. Create Learning Outcome: 12.01.01 Identify the structural features and function of each of the three main types of blood vessels. Section: 12.01 Topic: Blood Vessels
87. List the path a drop of blood would go through, starting with entry into and eventually leaving the heart to go to the entire body. Blood enters the heart through the superior and inferior vena cava - right atrium - right atrioventricular (tricuspid) valve - right ventricle - pulmonary semilunar valve - pulmonary arteries - lungs - pulmonary veins - left atrium - bicuspid (mitral) valve - left ventricle - aortic semilunar valve - aorta - body.
Bloom's Level: 6. Create Learning Outcome: 12.03.02 Trace the path of blood through the heart and lungs. Section: 12.03 Topic: Heart
12-43 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 13 - Lymphatic and Immune Systems
Chapter 13 Lymphatic and Immune Systems
Multiple Choice Questions 1. Which of the following statements is NOT correct about the lymphatic system? A. The lymphatic system protects the body from disease. B. The lymphatic system returns excess tissue fluid to the bloodstream. C. The fluid inside lymphatic vessels is called lymph. D. The lymphatic system absorbs fats at the villi and transports them to the bloodstream. E. The lymphatic system consists of blood vessels and lymphoid organs. Blood vessels are part of the cardiovascular system. The lymph system contains lymph vessels.
Bloom's Level: 2. Understand Learning Outcome: 13.01.01 Describe the major functions of the lymphatic system. Section: 13.01 Topic: Lymphatic System
2. The larger lymphatic vessels are most similar in structure to A. arteries. B. cardiovascular veins. C. capillaries. D. intestinal villi. E. nephron. The structure of the larger lymph vessels resembles that of veins.
Bloom's Level: 1. Remember Learning Outcome: 13.01.01 Describe the major functions of the lymphatic system. Section: 13.01 Topic: Lymphatic System
13-1 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 13 - Lymphatic and Immune Systems
3. Edema can be caused by A. accumulation of tissue fluid. B. abnormally high amount of plasma proteins. C. the kidney excreting too little protein. D. too much tissue fluid drainage. E. drinking more water than normal. Edema is a localized swelling caused by the accumulation of tissue fluid not collected by the lymphatic system.
Bloom's Level: 2. Understand Learning Outcome: 13.01.01 Describe the major functions of the lymphatic system. Section: 13.01 Topic: Lymphatic System
4. Edema cannot be caused by A. increased amounts of tissue fluid. B. not enough tissue fluid being drained away. C. increased blood pressure. D. increased capillary osmotic pressure. E. poor circulation such as constricted veins during pregnancy. Increased capillary osmotic pressure would lead to less tissue fluid, not more.
Bloom's Level: 2. Understand Learning Outcome: 13.01.01 Describe the major functions of the lymphatic system. Section: 13.01 Topic: Lymphatic System
13-2 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 13 - Lymphatic and Immune Systems
5. How is a lymph capillary like a blood capillary? A. Both contain blood. B. Both contain valves. C. Both have thin walls. D. Both are connected to the vena cava. E. Both systems run parallel through all body tissues. Lymphatic and blood capillaries both have very thin walls to assist the transfer of molecules.
Bloom's Level: 4. Analyze Learning Outcome: 13.01.01 Describe the major functions of the lymphatic system. Section: 13.01 Topic: Lymphatic System
6. Lymph is A. excreted by the kidneys. B. returned to the circulatory system via arteries. C. returned to the circulatory system at the subclavian vein. D. absorbed by tissues. E. excreted as sweat. The lymphatic duct returns lymph from the body into the right and left subclavian veins.
Bloom's Level: 2. Understand Learning Outcome: 13.01.01 Describe the major functions of the lymphatic system. Section: 13.01 Topic: Lymphatic System
13-3 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 13 - Lymphatic and Immune Systems
7. Which statement is NOT true about the lymphatic system? A. It takes up excess tissue fluid at the site of capillaries. B. Lymph contains the same proteins as found in the blood. C. It helps defend the body against disease. D. It contains lymph that may have fat in it. E. Lymph is tissue fluid that has entered a lymph vessel. Lymph does not contain plasma proteins.
Bloom's Level: 4. Analyze Learning Outcome: 13.01.01 Describe the major functions of the lymphatic system. Section: 13.01 Topic: Lymphatic System
8. Which lymphatic duct will drain the abdomen? A. lymph capillaries B. right lymphatic duct C. left lymphatic duct D. thoracic duct E. inguinal duct The thoracic duct returns lymph collected from the body below the thorax plus the left arm, the left side of the head, and the neck, into the left subclavian vein.
Bloom's Level: 3. Apply Learning Outcome: 13.01.01 Describe the major functions of the lymphatic system. Section: 13.01 Topic: Lymphatic System
13-4 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 13 - Lymphatic and Immune Systems
9. Elephantiasis is a tropical disease where small nematode larvae block the lymphatic ducts. As a result, the legs become large like those of an elephant and the scrotum and arms may also swell. This problem would be a severe form of A. hemorrhoids. B. phlebitis. C. anemia. D. angina. E. edema. Edema is a swelling caused by the accumulation of tissue fluid that has not been collected by the lymphatic system.
Bloom's Level: 3. Apply Learning Outcome: 13.01.01 Describe the major functions of the lymphatic system. Section: 13.01 Topic: Lymphatic System
10. Which of the following are NOT considered lymphoid organs? A. bone marrow B. thymus C. spleen D. lymph nodes E. kidneys Kidneys are not part of the lymphatic system. They are part of the urinary system.
Bloom's Level: 2. Understand Learning Outcome: 13.01.02 Distinguish between the roles of primary and secondary lymphoid organs, and list examples of each. Section: 13.01 Topic: Lymphatic System
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Chapter 13 - Lymphatic and Immune Systems
11. Earlier in the history of Western medicine, surgeons did not know the immune function of the thymus gland and it was sometimes removed in children. What symptoms would you now predict these patients might display afterward? A. increased inflammatory and allergic responses to foreign material including viruses B. reduced ability to recognize foreign material and distinguish it from body proteins C. serious problems after receiving blood transfusions of any type of blood D. complete rejection of all normal body tissues E. failure to produce any white blood cells Since the thymus is where the T cells mature, this individual would not have T cells and would have a reduced ability to recognize foreign material.
Bloom's Level: 3. Apply Learning Outcome: 13.01.02 Distinguish between the roles of primary and secondary lymphoid organs, and list examples of each. Section: 13.01 Topic: Lymphatic System
12. Which lymphoid organ has sinuses filled with blood instead of lymph? A. thymus B. lymph nodes C. spleen D. bone marrow E. kidneys Most of the spleen is red pulp that filters the blood. All of the others are lymph organs, except the kidneys, that contain lymph but not blood.
Bloom's Level: 2. Understand Learning Outcome: 13.01.02 Distinguish between the roles of primary and secondary lymphoid organs, and list examples of each. Section: 13.01 Topic: Lymphatic System
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Chapter 13 - Lymphatic and Immune Systems
13. The spleen has a relatively thin outer capsule and can actually burst due to a blow to the abdomen. A person whose spleen has burst A. cannot survive. B. cannot synthesize red blood cells. C. cannot synthesize white blood cells. D. may need antibiotic therapy. E. may need blood transfusions to prevent anemia. A person without a spleen is more susceptible to infections and may have to receive antibiotic therapy indefinitely.
Bloom's Level: 2. Understand Learning Outcome: 13.01.02 Distinguish between the roles of primary and secondary lymphoid organs, and list examples of each. Section: 13.01 Topic: Lymphatic System
14. Disease-causing agents like viruses or bacteria are considered a(n) A. antibody. B. vaccine. C. thrombin. D. allergen. E. pathogen. Pathogens are disease-causing agents, such as viruses and bacteria.
Bloom's Level: 1. Remember Learning Outcome: 13.01.02 Distinguish between the roles of primary and secondary lymphoid organs, and list examples of each. Section: 13.01 Topic: Lymphatic System
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Chapter 13 - Lymphatic and Immune Systems
15. Which of the following lymphoid organs is largest during a person's childhood but shrinks in adulthood? A. tonsils B. lymph nodes C. spleen D. bone marrow E. thymus The thymus is largest in children and shrinks as we get older.
Bloom's Level: 2. Understand Learning Outcome: 13.01.02 Distinguish between the roles of primary and secondary lymphoid organs, and list examples of each. Section: 13.01 Topic: Lymphatic System
16. During surgery for breast cancer, surgeons will often remove lymph nodes in the armpit. What is the most likely reason for this? A. If any cancer cells have entered the lymph, they will travel to the closest lymph nodes, which are in the armpit. B. Lymph nodes in the armpit are responsible for breast cancer. C. Lymph nodes that are close to the surgical incision would prevent healing following the surgery. D. Lymph nodes in the armpit become useless because of the surgery and can be removed without problem. E. Lymph nodes do not help a person and are better removed before they become infected. If cancer has metastasized or spread via the lymph, cancer cells may be present in the lymph nodes. Removing and examining the lymph nodes allows physicians to determine how far the cancer may have spread.
Bloom's Level: 3. Apply Learning Outcome: 13.01.02 Distinguish between the roles of primary and secondary lymphoid organs, and list examples of each. Section: 13.01 Topic: Lymphatic System
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Chapter 13 - Lymphatic and Immune Systems
17. An antihistamine counters the effects of a histamine. Which of the following is NOT an effect of a histamine? A. Increased blood flow brings more white blood cells to the site of injury or invasion. B. Capillary dilation increases blood flow and causes surface redness. C. Increased blood flow and permeability increase the fluids in the tissues, causing them to swell. D. Increased blood flow brings more warmth to the area, causing a local "fever" that helps white blood cells and suppresses microbes. E. Decreased blood flow inhibits a pathogen from spreading throughout the body. Histamine dilates blood vessels and causes increased blood flow to the area.
Bloom's Level: 2. Understand Learning Outcome: 13.02.02 Describe the purpose of the inflammatory response. Section: 13.02 Topic: Innate Immunity
18. Which statement is NOT true in regards to the inflammatory reaction? A. The skin becomes reddened and swollen. B. Swelling initiates pain nerve impulses. C. Increased capillary permeability causes more proteins and fluids to escape. D. Neutrophils phagocytize bacteria in the wound. E. Neutrophils change into macrophages that act as scavengers. Monocytes, not neutrophils, change into macrophages that act as scavengers.
Bloom's Level: 4. Analyze Learning Outcome: 13.02.02 Describe the purpose of the inflammatory response. Section: 13.02 Topic: Innate Immunity
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Chapter 13 - Lymphatic and Immune Systems
True / False Questions 19. Innate immunity involves B and T cells since you were born with them. FALSE B and T cells are part of adaptive immunity. Innate immunity is fully functional without previous exposure to antigens.
Bloom's Level: 2. Understand Learning Outcome: 13.02.02 Describe the purpose of the inflammatory response. Section: 13.02 Topic: Innate Immunity
Multiple Choice Questions 20. Inflammation does not involve A. phagocytes. B. neutrophils. C. mast cells. D. macrophages. E. B cells. Inflammation is an innate immune response, while B cells are part of an adaptive immune response.
Bloom's Level: 2. Understand Learning Outcome: 13.02.02 Describe the purpose of the inflammatory response. Section: 13.02 Topic: Innate Immunity
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Chapter 13 - Lymphatic and Immune Systems
21. In order to decrease the effects of inflammation, you would take what type of medication? A. antihistamine B. vaccine C. interferon D. immunoglobulin E. thromboplastin activator Many of the effects of inflammation are due to the release of histamine by mast cells. Therefore, an antihistamine would counter these effects.
Bloom's Level: 2. Understand Learning Outcome: 13.02.02 Describe the purpose of the inflammatory response. Section: 13.02 Topic: Innate Immunity
22. The complement system contains A. either half of an antigen-antibody complex. B. certain plasma proteins. C. a serious infection. D. an autoimmune reaction. E. cells that secrete antibodies. The complement system is composed of a number of blood plasma proteins that are involved in an innate immune response.
Bloom's Level: 2. Understand Learning Outcome: 13.02.04 Describe the three main functions of the complement system. Section: 13.02 Topic: Innate Immunity
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Chapter 13 - Lymphatic and Immune Systems
True / False Questions 23. Inflammation or the inflammatory response is a harmful process that should be avoided. FALSE Inflammation is part of the innate immune system and is very useful when a pathogen has entered the body.
Bloom's Level: 3. Apply Learning Outcome: 13.02.02 Describe the purpose of the inflammatory response. Section: 13.02 Topic: Innate Immunity
Multiple Choice Questions 24. Which of these statements is NOT true about complement? A. Complement is a complex of proteins. B. Complement is required for the formation of antigen-antibody complexes. C. Complement forms pores in bacterial membranes and allows fluid and salts to enter the bacterial cell. D. Complement attracts phagocytes to the bacterial cell and enhances its likelihood of being engulfed. E. Complement can assist in the inflammatory response. Complement is not involved in the formation of antigen-antibody complexes.
Bloom's Level: 2. Understand Learning Outcome: 13.02.04 Describe the three main functions of the complement system. Section: 13.02 Topic: Innate Immunity
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Chapter 13 - Lymphatic and Immune Systems
25. Which protein forms pores in bacterial cell membranes? A. interferon B. cytokines C. histamine D. complement E. thrombin The end of the complement cascade is a membrane attack complex that produces holes in the surface of bacteria.
Bloom's Level: 1. Remember Learning Outcome: 13.02.04 Describe the three main functions of the complement system. Section: 13.02 Topic: Innate Immunity
26. Which of the following is NOT considered to be an innate defense mechanism against infection? A. stomach acids B. skin C. mucus in respiratory tract D. normal bacteria in intestine E. antibodies in mother's milk Antibodies are part of the adaptive immune response.
Bloom's Level: 4. Analyze Learning Outcome: 13.02.01 List examples of physical and chemical barriers to infection. Section: 13.02 Topic: Innate Immunity
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Chapter 13 - Lymphatic and Immune Systems
27. Which of the following is NOT an example of innate immunity? A. natural killer cells B. complement C. inflammation D. histamine E. plasma cells Plasma cells are antibody-secreting cells and are part of the adaptive immune response.
Bloom's Level: 2. Understand Learning Outcome: 13.02.01 List examples of physical and chemical barriers to infection. Section: 13.02 Topic: Innate Immunity
28. Which of the following is specifically responsible for antibody-mediated immunity? A. T cell B. B cell C. platelets D. RBC E. complement B cells mature into plasma cells that produce antibodies.
Bloom's Level: 1. Remember Learning Outcome: 13.03.02 Compare and contrast the activities of B cells versus T cells. Section: 13.03 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
29. Which of the following is NOT true of both T cells and B cells? A. They are both lymphocytes. B. They both pass through the thymus. C. They both function in immunity. D. They both are specific. E. They both originated in bone marrow. T cells, but not B cells, pass through the thymus during their maturation process.
Bloom's Level: 4. Analyze Learning Outcome: 13.03.02 Compare and contrast the activities of B cells versus T cells. Section: 13.03 Topic: Adaptive Immunity
30. Which type of cell will produce antibodies that are released into the bloodstream? A. B cell B. T cell C. plasma cell D. memory cell E. mast cell A plasma cell produces antibodies. An activated B cell differentiates into a plasma cell but the B cell itself does not secrete antibodies. The memory cell is also a form of B cell but it does not secrete antibodies.
Bloom's Level: 2. Understand Learning Outcome: 13.03.02 Compare and contrast the activities of B cells versus T cells. Section: 13.03 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
31. Which of the following is true about the clonal selection theory? A. An antigen selects certain B cells to produce a clone of plasma cells. B. An antibody stimulates the multiplication of B cell antigens against it. C. T cells select those B cells that should produce antibodies, regardless of antigens. D. T cells suppress all those B cells except the ones that should multiply and divide. E. One B cell will give rise to plasma cells that produce antibodies against several antigens. A B cell that binds an antigen is activated and clonally expands to produce plasma cells that secrete antibodies against the original antigen. T cells assist B cells that have been activated by a particular antigen.
Bloom's Level: 2. Understand Learning Outcome: 13.03.02 Compare and contrast the activities of B cells versus T cells. Section: 13.03 Topic: Adaptive Immunity
32. Adaptive immunity depends upon the activity of what type of cells? A. natural killer cells and B cells B. natural killer cells and T cells C. neutrophils and B cells D. neutrophils and T cells E. B cells and T cells Adaptive immunity depends primarily upon the activity of B and T cells.
Bloom's Level: 2. Understand Learning Outcome: 13.03.02 Compare and contrast the activities of B cells versus T cells. Section: 13.03 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
33. Which statement is NOT true of the B cells? A. B cells develop and mature in the bone marrow. B. B cells carry receptor molecules on their cell surfaces. C. B cells, when stimulated and changed into plasma cells, produce antibodies. D. B cells require presentation of antigens by macrophages before they are able to recognize the antigen. E. B cells, when stimulated, divide into memory cells. T cells, not B cells, require presentation of antigens before they are able to recognize the antigen.
Bloom's Level: 4. Analyze Learning Outcome: 13.03.02 Compare and contrast the activities of B cells versus T cells. Section: 13.03 Topic: Adaptive Immunity
34. If we inject three different antigens into a human at the same time, A. one plasma cell will produce three different antibodies. B. plasma cells will produce three times the amount of one antibody. C. at least three different plasma cells will be involved in antibody production since each plasma cell only produces one specific antibody. D. three different immunoglobulins will assist one plasma cell in responding to the antigen. E. three complement plasma proteins will assist one plasma cell in responding to the antigen. Since each antigen is different, it will be recognized and bound by three different B cells, which will differentiate into three different plasma cells, producing three different antibodies. Each antibody will be specific for the original antigen.
Bloom's Level: 3. Apply Learning Outcome: 13.03.01 Describe the major steps in the clonal selection theory. Section: 13.03 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
35. The clonal selection theory states that A. all B cells are exact clones of one another. B. all T cells are exact clones of one another. C. when a B or T cell is activated by binding to/recognizing an antigen, then that cell will be cloned into cells that recognize the same antigen. D. when a B or T cell is stimulated by encountering the correct antibody, it divides to form a clone of cells that will respond to the same antibody. E. T cells are stimulated by B cells to divide and form plasma cells, which secrete antibodies. When a B or T cell is activated by binding to/recognizing an antigen, then that cell will be cloned into cells that recognize the same antigen.
Bloom's Level: 4. Analyze Learning Outcome: 13.03.01 Describe the major steps in the clonal selection theory. Section: 13.03 Topic: Adaptive Immunity
36. A cell that is responsible for the ability of an organism to produce a rapid antibody response against a foreign protein when it is encountered years after the first encounter is a(n) A. memory B cell. B. memory T cell. C. cytotoxic T cell. D. plasma cell. E. antigen-presenting cell. Memory B cells enable the body to respond to an antigen that has been seen previously, through the production of antibodies.
Bloom's Level: 2. Understand Learning Outcome: 13.03.02 Compare and contrast the activities of B cells versus T cells. Section: 13.03 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
37. Which is NOT part of a description of an antibody molecule? A. contains a constant and a variable region on each chain B. forms an X-shaped protein molecule C. contains an antigen-binding site on each arm D. contains two light chains and two heavy chains E. contains hypervariable regions where antigens bind An antibody forms a Y-shaped protein molecule.
Bloom's Level: 2. Understand Learning Outcome: 13.03.03 Describe the basic structure of an antibody molecule, and explain the different functions of IgG, IgA, IgM, and IgE. Section: 13.03 Topic: Adaptive Immunity
38. Most antibodies found in the blood belong to what class of immunoglobulins? A. IgA B. IgD C. IgE D. IgG E. IgM IgG antibodies are the major type found in blood, lymph, and tissue fluid.
Bloom's Level: 1. Remember Learning Outcome: 13.03.03 Describe the basic structure of an antibody molecule, and explain the different functions of IgG, IgA, IgM, and IgE. Section: 13.03 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
39. Which portion of the antibody binds to the antigen? A. variable region B. constant region C. heavy chain D. light chain The variable regions of both the light and heavy chain bind to the antigen. This variability enables antibodies to recognize different antigens.
Bloom's Level: 1. Remember Learning Outcome: 13.03.03 Describe the basic structure of an antibody molecule, and explain the different functions of IgG, IgA, IgM, and IgE. Section: 13.03 Topic: Adaptive Immunity
40. Which of the following is NOT true about T cells? A. They are responsible for cell-mediated immunity. B. They can stimulate B cells. C. They produce antibodies. D. Helper T cells can secrete cytokines. E. They require an antigen-presenting cell. Only plasma cells, produced from B cells, produce antibodies.
Bloom's Level: 4. Analyze Learning Outcome: 13.03.02 Compare and contrast the activities of B cells versus T cells. Section: 13.03 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
41. Which type of T cell improves the immune response? A. helper T cell B. suppressor T cell C. cytotoxic T cell D. memory T cell E. IgT cell Helper T cells "help" or augment the immune response.
Bloom's Level: 2. Understand Learning Outcome: 13.03.02 Compare and contrast the activities of B cells versus T cells. Section: 13.03 Topic: Adaptive Immunity
42. The AIDS virus attacks A. helper T cells. B. suppressor T cells. C. cytotoxic T cells. D. memory T cells. E. B cells. HIV infects helper T cells and thereby suppresses the immune response.
Bloom's Level: 1. Remember Learning Outcome: 13.03.04 Distinguish between the functions of helper T cells and cytotoxic T cells, and explain the role of MHC molecules in these responses. Section: 13.03 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
43. Which type of lymphocyte is sometimes called a "killer cell" that brings about destruction of antigen-bearing cells? A. helper T cell B. suppressor T cell C. cytotoxic T cell D. memory T cell E. macrophage A cytotoxic T cell is sometimes called a "killer" T cell because it kills the cell that presents antigen to it.
Bloom's Level: 2. Understand Learning Outcome: 13.03.04 Distinguish between the functions of helper T cells and cytotoxic T cells, and explain the role of MHC molecules in these responses. Section: 13.03 Topic: Adaptive Immunity
44. Which of the following statements is NOT true in regard to active immunity? A. It forms a specific defense against a particular antigen. B. It will last for some time. C. The particular antibody that is formed will react against any virus or bacterial cell. D. It is the result of B and T lymphocyte activity. E. An antigen-antibody reaction must occur. Antibodies are specific for the original antigen that stimulated the B cell.
Bloom's Level: 4. Analyze Learning Outcome: 13.04.01 Distinguish between active and passive immunity. Section: 13.04 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
45. A vaccine contains A. penicillin. B. horse serum. C. antigens that are no longer virulent. D. antibodies made by plasma cells. E. antibiotics against a particular bacterial strain. A vaccine is a substance that contains an antigen to which the immune system responds. Traditionally these are pathogens or products of pathogens that have been treated so they are no longer able to cause disease.
Bloom's Level: 2. Understand Learning Outcome: 13.04.01 Distinguish between active and passive immunity. Section: 13.04 Topic: Adaptive Immunity
46. Babies generally appear to be immune to diseases for approximately the first six months of life. After this, they develop more diseases. What could be responsible for this immunity? A. active immunity due to the mother B. passive immunity from the mother C. babies not being exposed to diseases until around six months D. babies born with excellent active immunity that wanes the older the child gets E. active immunity due to being exposed to numerous antigens during the birthing process The newborn baby has antibodies from the mother. This is an example of passive immunity since the baby did not make the antibodies.
Bloom's Level: 3. Apply Learning Outcome: 13.04.01 Distinguish between active and passive immunity. Section: 13.04 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
47. Edward Jenner promoted the use of cowpox infection, which is very similar to the more dangerous smallpox, as a vaccination to provide patients immunity from future smallpox epidemics. This was a case of these subjects A. developing passive immunity. B. developing active immunity. C. developing an inflammatory response that increased the IgG production. D. developing high levels of complement. E. developing high levels of interferon, although Jenner did not realize these diseases were viral. The patients actively developed antibodies against the cowpox vaccine. It happens that antibodies against cowpox cross react with small pox and so this immunization provided protection against smallpox. This is evidence of active immunity.
Bloom's Level: 3. Apply Learning Outcome: 13.04.01 Distinguish between active and passive immunity. Section: 13.04 Topic: Adaptive Immunity
True / False Questions 48. When a person is bitten by a venomous snake, a donation of blood plasma from another person who has recently survived the same type of snake bite would provide some passive immunity. TRUE The person who recently survived the same type of snake bite would have antibodies against the venom. You would receive those antibodies passively.
Bloom's Level: 3. Apply Learning Outcome: 13.04.01 Distinguish between active and passive immunity. Section: 13.04 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
Multiple Choice Questions 49. When you have been bitten by a highly venomous snake and received a dose of venom sufficient to kill several people, you do not have a week to build up your own immunity. However, hospitals carry "antivenom" made from injecting horses with small-to-larger dosages of snake venom, and then harvesting the antibodies from the horse plasma. This is a case of A. passive immunity. B. active immunity. C. developing an inflammatory response to increase the IgG production. D. high levels of complement. E. high levels of interferon. This is passive immunity. You are receiving antibodies that you did not make yourself by injection.
Bloom's Level: 3. Apply Learning Outcome: 13.04.01 Distinguish between active and passive immunity. Section: 13.04 Topic: Adaptive Immunity
50. Which of the following is NOT true regarding monoclonal antibodies? A. The antibodies produced are all of the same type. B. They are produced from plasma cells derived from different B cells. C. They are formed when B lymphocytes are fused with malignant plasma cells. D. They are used in the diagnosis of various diseases. E. They are used to treat some cancers. Monoclonal antibodies are identical because they are derived from a single type of plasma cell fused to a myeloma.
Bloom's Level: 4. Analyze Learning Outcome: 13.04.03 Explain the major steps in the production of monoclonal antibodies, and some of their practical uses. Section: 13.04 Topic: Adaptive Immunity
13-25 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 13 - Lymphatic and Immune Systems
True / False Questions 51. Cytokines may be used in cancer and AIDS because they stimulate white blood cell formation and/or function. TRUE Cytokines are chemical messengers produced by T cells, macrophages, and other cells, which regulate white blood cell formation and/or function.
Bloom's Level: 2. Understand Learning Outcome: 13.04.02 Describe some specific clinical applications of cytokine therapies. Section: 13.04 Topic: Immune System
Multiple Choice Questions 52. In the late 1800s, a Frenchman observed that individuals who recovered from a disease were not susceptible to that same disease again. Therefore, he proposed that the agent that caused syphilis should be given to all young, healthy people as a vaccine to prevent them from getting a serious case of the disease. Although the mode of action of vaccines was not yet known, Pasteur argued vehemently against this proposal. Today, we do not follow this procedure for syphilis. Why? A. Vaccines only work against a few pathogens. B. Vaccines do not work for pathogens that are vulnerable to antibiotics. C. Syphilis is an autoimmune disease. D. It might work, but Pasteur was a famous authority and his ideas prevail today. E. Vaccines are made from weakened disease agents and using the syphilis organisms would spread a full-blown serious disease agent. Infecting someone with a virulent agent such as the bacterium that causes syphilis would be unethical and would result in giving the person syphilis. Vaccines are made from avirulent or weakened strains of pathogens.
Bloom's Level: 3. Apply Learning Outcome: 13.04.01 Distinguish between active and passive immunity. Section: 13.04 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
53. In the later 1800s, the brilliant French chemist Louis Pasteur worked with chicken cholera, passing the disease agent from sick to healthy chickens. He suspended his research and went on vacation for several weeks. When he returned, the old cholera culture he gave to healthy chickens failed to kill them. He assumed his culture was bad and injected these same chickens with a fresh culture of chicken cholera. These chickens remained healthy while new chickens injected with the fresh culture of chicken cholera rapidly became sick and died. From our current knowledge of immunity, how would these observations be explained? A. Pasteur had accidentally discovered a strain of chickens that had natural immunity. B. His first disease culture had weakened over time and stimulated the development of active immunity. C. His first disease culture had weakened over time and the chickens developed a form of passive immunity. D. The chickens apparently had better B cells for development of more effective antibodies. E. The first set of chickens served as a control group to the second set that showed the development of antibodies. The weakened culture of cholera initiated an active immune response in the chickens. When they were re-exposed to the same cholera, they were already immune. However, chickens that had not been exposed to the weakened culture died. The injection of a weakened culture is similar to how vaccinations are carried out today.
Bloom's Level: 3. Apply Learning Outcome: 13.04.01 Distinguish between active and passive immunity. Section: 13.04 Topic: Adaptive Immunity
54. Which antibody class is most responsible for the allergic reaction? A. IgA B. IgD C. IgE D. IgG E. IgM Allergic responses are caused by IgE antibodies.
Bloom's Level: 1. Remember Learning Outcome: 13.05.01 Discuss the most common immunological mechanisms responsible for allergies. Section: 13.05 Topic: Immune System
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Chapter 13 - Lymphatic and Immune Systems
55. Anaphylactic shock may occur due to A. exposure to the AIDS virus. B. a bee sting. C. organ donation. D. use of interferon. E. antihistamines. Anaphylactic shock is an immediate allergic response that occurs because the allergen has entered the bloodstream. Bee stings are known to cause this reaction in certain people.
Bloom's Level: 1. Remember Learning Outcome: 13.05.01 Discuss the most common immunological mechanisms responsible for allergies. Section: 13.05 Topic: Immune System
True / False Questions 56. In order to prevent an allergic reaction, injections of the allergen are sometimes given. TRUE Although it seems counter-intuitive, very small amounts of the allergen can be given so that the body will build up high quantities of IgG antibodies and prevent an allergic response to the allergen.
Bloom's Level: 3. Apply Learning Outcome: 13.05.01 Discuss the most common immunological mechanisms responsible for allergies. Section: 13.05 Topic: Immune System
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Chapter 13 - Lymphatic and Immune Systems
57. A response that does not occur immediately cannot be considered an allergic response, even if it appears like one. FALSE Delayed allergic responses, initiated by memory T cells, exist. These sometimes take 1 to 2 days to appear.
Bloom's Level: 2. Understand Learning Outcome: 13.05.01 Discuss the most common immunological mechanisms responsible for allergies. Section: 13.05 Topic: Immune System
Multiple Choice Questions 58. A person who is blood type AB has A. only antigen A on the surface of RBCs. B. only antigen B on the surface of RBCs. C. antigens A and B on the surface of RBCs. D. antibodies A and B on the surface of RBCs. E. only antibody A on the surface of RBCs. A person who is AB has both antigen A and antigen B on the surface of RBCs.
Bloom's Level: 1. Remember Learning Outcome: 13.05.02 Compare the adverse reactions involving the ABO blood system with those involving the Rh system. Section: 13.05 Topic: Immune System
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Chapter 13 - Lymphatic and Immune Systems
59. A person who is B positive will have A. A antigens, anti-B antibody, Rh antigen. B. B antibodies, anti-A antibody, Rh antigen. C. B antigens, anti-A antibody, no Rh antigen. D. B antigens, anti-A antibody, Rh antigen. E. B antigens, anti-B antibody, Rh antigen. A B blood type means the person has the B antigen on the surface of the RBCs. This means that they will have antibodies against A. The Rh positive means that the person has the Rh antigen.
Bloom's Level: 2. Understand Learning Outcome: 13.05.02 Compare the adverse reactions involving the ABO blood system with those involving the Rh system. Section: 13.05 Topic: Immune System
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Chapter 13 - Lymphatic and Immune Systems
60. What blood type does an individual have according to the blood typing results illustrated in this figure?
A. O positive B. A negative C. B positive D. AB negative E. O negative The lack of agglutination in the anti-A and anti-B wells means the person does not have A or B antigens on their RBCs. Therefore, they must be O. The agglutination in the anti-Rh well means they do have the Rh antigen so they must be Rh positive.
Bloom's Level: 3. Apply Learning Outcome: 13.05.02 Compare the adverse reactions involving the ABO blood system with those involving the Rh system. Section: 13.05 Topic: Immune System
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Chapter 13 - Lymphatic and Immune Systems
61. The textbook describes the problem with Rh factor hemolytic disease of the newborn in the case where the mother is Rh- and the father is Rh+. Which of the following statements is true? A. This would also be a problem where the father is Rh- and the mother is Rh+. B. This would be a problem any time the mother is Rh-, regardless of the father's blood type,since the mother has the immune problem. C. This is only a problem in the given case because the Rh- mother lacks the Rh+ protein and only a Rh+ father could produce an Rh+ child. D. This is simply a matter of a mother having a lower threshold for rejecting foreign tissues since all babies have half-foreign tissues. E. The problem is substantially more complicated since the hemolytic disease is also dependent on ABO blood groups. The mother must be Rh- in order to develop antibodies against the blood cells of an Rh+ fetus. The father must be Rh+ in order for the fetus to be Rh+.
Bloom's Level: 4. Analyze Learning Outcome: 13.05.02 Compare the adverse reactions involving the ABO blood system with those involving the Rh system. Section: 13.05 Topic: Immune System
62. Tissue rejection following organ transplantation is due to all of the following except A. MHC incompatibility between donor and recipient. B. transplantation of tissues from one part of the body to another in the same person. C. the destruction of transplanted tissue by the immune system. D. B cells and plasma cells that do not recognize the transplanted tissue as "self." E. T cells that do not recognize the transplanted tissue as "self." If the tissue comes from the same person, it would be recognized as "self" and no rejection would occur.
Bloom's Level: 5. Evaluate Learning Outcome: 13.05.03 Explain the types of precautions that must be taken when transplanting organs. Section: 13.05 Topic: Immune System
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Chapter 13 - Lymphatic and Immune Systems
True / False Questions 63. Xenotransplantation is the use of animal organs instead of human organs in transplant patients. Scientists are considering using dogs as organ donors because they have long been domesticated by humans. FALSE Scientists have chosen to use the pig because pigs are raised as a meat source and are prolific.
Bloom's Level: 1. Remember Learning Outcome: 13.05.03 Explain the types of precautions that must be taken when transplanting organs. Section: 13.05 Topic: Immune System
Multiple Choice Questions 64. Which of the following is NOT an autoimmune disease? A. multiple sclerosis B. systemic lupus C. rheumatoid arthritis D. inflammation E. type I diabetes Inflammation is part of the innate immune system.
Bloom's Level: 1. Remember Learning Outcome: 13.06.01 Define autoimmune disease, and list several specific examples or these diseases. Section: 13.06 Topic: Immune System
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Chapter 13 - Lymphatic and Immune Systems
True / False Questions 65. An autoimmune disease means that the body attacks itself because it does not recognize itself. TRUE When a person has an autoimmune disease, cytotoxic T cells or antibodies mistakenly attack the body's own cells as if they bear foreign antigens.
Bloom's Level: 1. Remember Learning Outcome: 13.06.01 Define autoimmune disease, and list several specific examples or these diseases. Section: 13.06 Topic: Immune System
Yes / No Questions 66. Would untreated SCID be considered a lethal disease? YES In severe combined immunodeficiency disease (SCID), both antibody and cell-mediated immunity are lacking or inadequate. Without treatment, this is fatal.
Bloom's Level: 2. Understand Learning Outcome: 13.06.02 Differentiate between acquired and congenital immunodeficiencies. Section: 13.06 Topic: Immune System
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Chapter 13 - Lymphatic and Immune Systems
Multiple Choice Questions 67. Which of the following would not lead to an immunodeficiency? A. inherited lack of B cells B. blood transfusion C. surgical removal of the thymus D. SCID E. AIDS A blood transfusion may result in a severe immune reaction, but it would not cause an immunodeficiency.
Bloom's Level: 2. Understand Learning Outcome: 13.06.02 Differentiate between acquired and congenital immunodeficiencies. Section: 13.06 Topic: Immune System
68. In the movie Bubble Boy, Jimmy lives inside a plastic bubble where he is protected from all outside agents. Which type of disease would require a child to live inside a plastic bubble? A. an autoimmune disease B. systemic lupus erythematosus C. multiple sclerosis D. AIDS E. a total immune deficiency Only a total immune deficiency requires someone to live inside a plastic bubble protected from all pathogens. Several "real-life" children have had to live in this way.
Bloom's Level: 3. Apply Learning Outcome: 13.06.01 Define autoimmune disease, and list several specific examples or these diseases. Section: 13.06 Topic: Immune System
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Chapter 13 - Lymphatic and Immune Systems
69. Which of the following is an example of a mechanical barrier to infection? A. mucous membrane B. oil gland secretions C. pH of the stomach D. bacteria that live in the intestine E. inflammatory response The mucous membrane is the only mechanical barrier to infection. All of the other answer choices are examples of chemical barriers.
Bloom's Level: 1. Remember Learning Outcome: 13.02.01 List examples of physical and chemical barriers to infection. Section: 13.02 Topic: Innate Immunity
Short Answer Questions 70. Describe the three major functions of the lymphatic system. 1. Lymphatic capillaries absorb excess tissue fluid and return it to the bloodstream. 2. Lymphatic capillaries absorb fats from the digestive tract and transport them to the bloodstream. 3. Lymphoid organs help to defend the body against disease.
Bloom's Level: 6. Create Learning Outcome: 13.01.01 Describe the major functions of the lymphatic system. Section: 13.01 Topic: Lymphatic System
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Chapter 13 - Lymphatic and Immune Systems
Multiple Choice Questions 71. Which type of T cell has a TCR that can recognize an antigen fragment presented by antigen-presenting cells with MHC class II proteins? A. helper T cells B. cytotoxic T cells C. major histocompatibility complex D. B cells E. monocytes Helper T cells have a TCR that can recognize an antigen fragment presented by antigenpresenting cells with MHC class II proteins.
Bloom's Level: 1. Remember Learning Outcome: 13.03.04 Distinguish between the functions of helper T cells and cytotoxic T cells, and explain the role of MHC molecules in these responses. Section: 13.03 Topic: Adaptive Immunity
72. Which blood cells are usually the first to arrive at the site of an infection? A. neutrophils B. natural killer cells C. red blood cells D. interferons E. monocytes Neutrophils are the first white blood cells to arrive at the site of an infection. Natural killer cells and monocytes enter the site of infection at later stages. Red blood cells don't enter the site of infection. Interferons are proteins produced by virus-infected cells.
Bloom's Level: 1. Remember Learning Outcome: 13.02.03 Explain the function of phagocytes and natural killer cells. Section: 13.02 Topic: Innate Immunity
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Chapter 13 - Lymphatic and Immune Systems
73. What are the two most common applications of cytokine therapies? A. stimulation of blood cells and treatment of cancer B. stimulation of blood cells and treatment of HIV C. treatment of cancer and HIV D. production of antibodies E. providing immunization against various diseases The two most common applications of cytokine therapies are the stimulation of blood cells and treatment of cancer.
Bloom's Level: 3. Apply Learning Outcome: 13.04.02 Describe some specific clinical applications of cytokine therapies. Section: 13.04 Topic: Adaptive Immunity
74. What is the major drawback to cytokine therapies? A. Side effects can be life-threatening. B. The therapies are still untested. C. The treatments are very expensive. D. The therapies only work for women. E. The therapies only work for men. The major drawback to cytokine therapies is that the side effects can be life-threatening.
Bloom's Level: 2. Understand Learning Outcome: 13.04.02 Describe some specific clinical applications of cytokine therapies. Section: 13.04 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
75. During organ transplant, which feature is the most important to recognize? A. the difference between the MHC proteins of the donor and the recipient B. the difference between the blood types of the donor and the recipient C. the similarities between the blood types of the donor and the recipient D. the type of organ being transplanted E. the age of the recipient of the transplanted organ The most important feature to recognize during organ transplant is the difference between the MHC proteins of the donor and the recipient. This difference is what leads to tissue rejection in the recipient.
Bloom's Level: 5. Evaluate Learning Outcome: 13.05.03 Explain the types of precautions that must be taken when transplanting organs. Section: 13.05 Topic: Immune System
76. What is the originator cell for monoclonal antibodies? A. B cells B. T cells C. red blood cells D. white blood cells E. macrophages B cells are the originator cell for monoclonal antibodies.
Bloom's Level: 1. Remember Learning Outcome: 13.04.03 Explain the major steps in the production of monoclonal antibodies, and some of their practical uses. Section: 13.04 Topic: Adaptive Immunity
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Chapter 13 - Lymphatic and Immune Systems
77. Which of the following is an example of an acquired immunodeficiency? A. AIDS B. SCID C. flu D. None of the answer choices is an example of an acquired immunodeficiency. E. All of the answer choices are examples of an acquired immunodeficiency. AIDS is an example of an acquired immunodeficiency disease.
Bloom's Level: 1. Remember Learning Outcome: 13.06.02 Differentiate between acquired and congenital immunodeficiencies. Section: 13.06 Topic: Immune System
Short Answer Questions 78. List the function of the following antibodies: IgG, IgM, IgA, IgD, and IgE. IgG: binds to pathogens, activates complement, and enhances phagocytosis IgM: activates complement, clumps cells IgA: prevents pathogens from attaching to epithelial cells in digestive and respiratory tract IgD: presence signifies readiness of B cells to respond to antigens IgE: responsible for immediate allergic response and protection against certain parasitic worms
Bloom's Level: 6. Create Learning Outcome: 13.03.03 Describe the basic structure of an antibody molecule, and explain the different functions of IgG, IgA, IgM, and IgE. Section: 13.03 Topic: Innate Immunity
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Chapter 13 - Lymphatic and Immune Systems
Multiple Choice Questions 79. Which of the following health problems are likely to occur if a person is unable to form the IgE antibodies? A. The person would have problems defending themselves against parasitic worm infections as well as lacking an immediate allergic response. B. The person would be unable to activate the complement. C. Pathogens would be able to attach to their epithelial cells much easier. D. They would be unable to have their cells clump. E. Their B cells would be unable to respond to antigens. The person would have problems defending themselves against parasitic worm infections as well as lacking an immediate allergic response. The IgE antibody is responsible for immediate allergic response and protection against certain parasitic worms. The complement is activated by the IgG antibody. The IgA antibody prevents pathogens from attaching to epithelial cells. The IgM antibody is responsible for clumping cells. The presence of the IgD antibody signifies the readiness of the B cells to respond to antigens.
Bloom's Level: 5. Evaluate Learning Outcome: 13.03.03 Describe the basic structure of an antibody molecule, and explain the different functions of IgG, IgA, IgM, and IgE. Section: 13.03 Topic: Innate Immunity
Short Answer Questions 80. List the steps involved with the production of monoclonal antibodies. Immunized cells are derived from a donor with a specific antigen. These cells are then fused with a myeloma cell. The hybrid cells are then cultured and allowed to produce clones. The cells containing the desired antibodies are selected and expanded.
Bloom's Level: 6. Create Learning Outcome: 13.04.03 Explain the major steps in the production of monoclonal antibodies, and some of their practical uses. Section: 13.04 Topic: Adaptive Immunity
13-41 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 14 - Digestive System and Nutrition
Chapter 14 Digestive System and Nutrition
Multiple Choice Questions 1. In what order does a dentist drill through tooth layers? A. dentin → enamel → pulp B. enamel → dentin → pulp C. enamel → pulp → dentin D. pulp → dentin → enamel E. dentin → pulp → enamel From the outside of the tooth in, the layers are enamel, dentin, and pulp.
Bloom's Level: 1. Remember Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
2. Inflammation of the gums is called A. gingivitis. B. periodontitis. C. dental caries. D. gumma infections. E. gummitis. Gingivitis is an inflammation of the gums.
Bloom's Level: 1. Remember Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
3. Food is prevented from entering the trachea by the A. pharynx. B. larynx. C. epiglottis. D. glottis. E. sphincter muscle. The epiglottis is the flap that covers the trachea when swallowing to prevent food from entering.
Bloom's Level: 1. Remember Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
4. The main function of the esophagus is to A. digest proteins. B. digest carbohydrates. C. transport food from the small intestine to the large intestine. D. transport food from the mouth to the stomach. E. transport food from the stomach to the large intestine. The sole purpose of the esophagus is to conduct the food bolus from the mouth to the stomach.
Bloom's Level: 1. Remember Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
Figure 14.1
5. In Figure 14.1, which structure is the liver? A. 1 B. 2 C. 3 D. 4 E. 5 The largest organ visible in the abdominal cavity is the liver (1).
Bloom's Level: 1. Remember Figure 14.1 Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
6. In Figure 14.1, which structure is the primary site for digestion and absorption of food? A. 1 B. 2 C. 3 D. 4 E. 5 The small intestine (5) fulfills these roles.
Bloom's Level: 2. Understand Figure 14.1 Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
7. Which structure in Figure 14.1 stores food, has limited digestion, and secretes HCl? A. 1 B. 2 C. 3 D. 4 E. 5 The stomach (2) fulfills these roles.
Bloom's Level: 2. Understand Figure 14.1 Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01
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Chapter 14 - Digestive System and Nutrition
8. Muscles that encircle tubes and act as valves are called A. constrictors. B. dilators. C. sphincters. D. lacteals. E. mucosa. A sphincter is a muscle that encircles tubes and acts as a valve. The tubes close when the sphincter contracts and opens when the sphincter relaxes.
Bloom's Level: 1. Remember Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
9. Which layer of the intestinal wall will first absorb nutrients from the lumen? A. serosa B. longitudinal muscle C. circular muscle D. submucosa E. mucosa The mucosa is the layer that lines the lumen of the digestive tract.
Bloom's Level: 1. Remember Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
10. The pH of the stomach is usually about A. 2.0. B. 4.0. C. 7.0. D. 9.0. E. 11.0. The stomach is highly acidic, with a pH of 1–2.
Bloom's Level: 1. Remember Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
11. Which organ of the digestive tract functions to store food, kill bacteria, and partially digest proteins? A. mouth B. esophagus C. stomach D. small intestine E. large intestine These are all roles of the stomach.
Bloom's Level: 1. Remember Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
12. The fingerlike projections along the surface of the small intestine are called A. mucosa. B. lacteals. C. capillaries. D. villi. E. appendices. The wall of the small intestine contains fingerlike projections called villi.
Bloom's Level: 1. Remember Learning Outcome: 14.01.03 Compare the structural features and functions of the small intestine versus the large intestine. Section: 14.01 Topic: Digestive System
13. Water is absorbed primarily by the A. large intestine. B. duodenum. C. stomach. D. anal canal. E. esophagus. One of the roles of the large intestine is to absorb water.
Bloom's Level: 1. Remember Learning Outcome: 14.01.03 Compare the structural features and functions of the small intestine versus the large intestine. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
14. Which of the following is responsible for the storage of extra bile? A. small intestine B. liver C. stomach D. gallbladder E. pancreas The gallbladder is responsible for the storage of the extra bile. The liver produces the bile.
Bloom's Level: 1. Remember Learning Outcome: 14.02.01 Summarize the major functions of the pancreas, the liver, and the gallbladder. Section: 14.02 Topic: Digestive System
15. In the body, glucose is stored in the liver as A. starch. B. protein. C. glycogen. D. fat. E. amino acids. Glycogen is the storage form of glucose in animals.
Bloom's Level: 1. Remember Learning Outcome: 14.02.01 Summarize the major functions of the pancreas, the liver, and the gallbladder. Section: 14.02 Topic: Carbohydrates
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Chapter 14 - Digestive System and Nutrition
16. Which organ in the human body produces maltase and peptidases? A. mouth B. esophagus C. stomach D. small intestine E. large intestine The small intestine produces both of these.
Bloom's Level: 1. Remember Learning Outcome: 14.03.02 Compare the specific types of chemical digestion that take place in the mouth, stomach, and small intestine. Section: 14.03 Topic: Digestive Enzymes
True / False Questions 17. The body is unable to make any amino acids so all 20 of them must be acquired from a balanced diet. FALSE The body is capable of making 12 different types of amino acids. The remaining eight must be obtained from the diet.
Bloom's Level: 1. Remember Learning Outcome: 14.04.01 List the six major classes of nutrients and their major functions in the body. Section: 14.04 Topic: Proteins
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Chapter 14 - Digestive System and Nutrition
Multiple Choice Questions 18. Which of the following types of lipids do not promote cardiovascular disease? A. cholesterol B. trans saturated fatty acids C. saturated fatty acids D. unsaturated fatty acids E. animal fats Unsaturated fatty acids, usually liquid at room temperature, do not promote cardiovascular disease.
Bloom's Level: 1. Remember Learning Outcome: 14.04.02 Describe the relationship between diet and obesity, type 2 diabetes, and cardiovascular disease. Section: 14.04 Topic: Lipids
19. Antioxidants are A. toxic chemicals that must be avoided in the diet. B. the breakdown products of protein digestion. C. another name for free radicals. D. chemicals found in fruits and vegetables that defend against free radicals. E. only found in food supplements in pill form. Antioxidants defend the body against free radicals and are found in deep yellow/orange and leafy green vegetables as well as fruits.
Bloom's Level: 1. Remember Learning Outcome: 14.04.03 Distinguish between vitamins, coenzymes, and minerals. Section: 14.04 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
True / False Questions 20. Sodium is considered one of the major minerals needed in the human diet. TRUE We need under 1500 mg/day of sodium. It is easy, however, in the American diet to consume far more than this.
Bloom's Level: 1. Remember Learning Outcome: 14.04.03 Distinguish between vitamins, coenzymes, and minerals. Section: 14.04 Topic: Human Nutrition
21. People who are obese lack self-control and should just not eat as much. FALSE Obesity is most likely caused by a combination of hormonal, metabolic, and social factors. Eating disorders can be difficult to treat.
Bloom's Level: 1. Remember Learning Outcome: 14.04.02 Describe the relationship between diet and obesity, type 2 diabetes, and cardiovascular disease. Section: 14.04 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
Multiple Choice Questions 22. If the large intestine becomes irritated and peristalsis increases, what disease may result? A. constipation B. diarrhea C. appendicitis D. ulcers E. hernias Diarrhea due to infectious disease agents results in an irritation of the intestine and increased peristalsis.
Bloom's Level: 1. Remember Learning Outcome: 14.06.01 Describe a common disorder that affects each part of the digestive system. Section: 14.06 Topic: Human Digestive System
23. Which eating disorder is characterized by an irrational fear of getting fat, often associated with a perceived body defect? A. anorexia nervosa B. bulimia nervosa C. binge-eating and purging D. binge-eating E. pica Anorexia nervosa is characterized by an irrational fear of getting fat, often associated with a perceived body defect. Bulimia nervosa and binge-eating and purging are often found in individuals who are overly concerned with their body shape and weight. Pica is the repeated digestion of nonfood items.
Bloom's Level: 1. Remember Learning Outcome: 14.05.02 Discuss some of the psychological and social factors that can lead to eating disorders. Section: 14.05 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
24. Which eating disorder is most commonly diagnosed among people who seek treatment? A. EDNOS (eating disorders not otherwise specified) B. anorexia nervosa C. purging episodes D. bulimia nervosa E. binge-eating followed by purging According to the National Institute of Mental Health, EDNOS (eating disorders not otherwise specified) are the most common diagnosis among people who seek treatment. Purging episodes are characteristic of anorexia nervosa. Binge-eating followed by purging is a characteristic of bulimia nervosa.
Bloom's Level: 1. Remember Learning Outcome: 14.05.01 Describe the three main categories of eating disorders. Section: 14.05 Topic: Human Nutrition
25. Which eating disorder is most likely due to a person being depressed? A. binge-eating disorder B. bulimia nervosa C. pica D. anorexia nervosa E. None of the answer choices is associated with depression. Binge-eating disorder is often associated with depression. With bulimia nervosa and anorexia nervosa individuals are overly concerned with their body shape and weight. Pica is often associated with nutritional deficiencies and stress.
Bloom's Level: 1. Remember Learning Outcome: 14.05.02 Discuss some of the psychological and social factors that can lead to eating disorders. Section: 14.05 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
26. Which eating disorder is associated with a binge of high-calorie foods and feelings of guilt followed by a purge cycle. A. bulimia nervosa B. binge-eating disorder C. pica D. anorexia nervosa E. All of the answer choices are associated with a binge of high-calorie foods, and a feeling of guilt followed by a purge cycle. Only bulimia nervosa is associated with a binge of high-calorie foods, and a feeling of guilt followed by a purge cycle. Binge-eating disorder does not have a purge cycle. Pica is the ingestion of nonfood items. Anorexia nervosa is often associated with a starvation diet.
Bloom's Level: 1. Remember Learning Outcome: 14.05.02 Discuss some of the psychological and social factors that can lead to eating disorders. Section: 14.05 Topic: Human Nutrition
27. Which eating disorder is the most common? A. binge-eating disorder B. pica C. anorexia nervosa D. bulimia nervosa E. stomach ulcers Binge-eating disorder is the most common eating disorder.
Bloom's Level: 1. Remember Learning Outcome: 14.05.03 Distinguish between compulsive eating, binge eating, and pica. Section: 14.05 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
28. Which of the following features is associated with pica? A. person is a greater risk for various poisoning B. person has an irrational fear of getting fat C. individual engages in extreme physical activity to avoid weight gain D. person conducts self-induced vomiting and laxative abuse E. individual frequently eats large amount of foods while feeling a loss of control Individuals are at a greater risk for various poisoning if they have pica. People with an irrational fear of getting fat and engaging in extreme physical activity to avoid weight gain are associated with anorexia nervosa. Self-induced vomiting and laxative abuse are associated with bulimia. Frequently eating large amount of foods while feeling a loss of control is associated with binge-eating disorder.
Bloom's Level: 1. Remember Learning Outcome: 14.05.03 Distinguish between compulsive eating, binge eating, and pica. Section: 14.05 Topic: Human Nutrition
29. Which item will most likely be eaten by an individual who suffers from pica? A. charcoal B. cookies C. cakes D. ice cream E. All of the answer choices are associated with pica. Individuals who suffer from pica often ingest nonfood items such as ash, charcoal, cotton, soap, soil, wool, or other nonfood items.
Bloom's Level: 1. Remember Learning Outcome: 14.05.03 Distinguish between compulsive eating, binge eating, and pica. Section: 14.05 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
30. The digestive system does NOT function in the A. chemical breakdown of food. B. mechanical breakdown of food. C. elimination of waste products. D. absorption of food molecules. E. manufacture of food molecules. The digestive system does not manufacture food molecules, but takes in preformed food.
Bloom's Level: 2. Understand Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
31. Which of the following is not involved in digestion? A. transport of nutrients from food via the blood B. chewing of the food C. churning and mixing of the food in the stomach D. chemical digestion E. enzymes breaking down proteins The transport of nutrients from food is done by the cardiovascular system.
Bloom's Level: 2. Understand Learning Outcome: 14.01.02 Distinguish between mechanical and chemical digestion, and note where each step specifically occurs. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
32. Which is NOT true of human salivary glands? A. They produce an enzyme to begin the digestion of starch. B. The set near the ears swell when infected by the virus that causes mumps. C. There are three pairs that open by ducts into the mouth. D. Paired ducts open into the mouth near the ear, the inside back cheek, and under the tongue. E. They can become inflamed and cause tonsillitis. The tonsils become inflamed and cause tonsillitis.
Bloom's Level: 2. Understand Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
33. Which of the following is not true of the tongue? A. The tongue functions to mix food with saliva and assist in swallowing. B. A person's tongue is moved by external muscles. C. The tongue consists of skeletal muscle under voluntary control. D. The tongue is attached to the floor of the mouth. E. The tongue secretes salivary amylase. The salivary glands secrete salivary amylase.
Bloom's Level: 2. Understand Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
34. In humans, digestion of starch begins in the A. mouth. B. esophagus. C. stomach. D. duodenum. E. large intestine. Digestion of starch begins in the mouth with salivary amylase, produced by the salivary glands.
Bloom's Level: 2. Understand Learning Outcome: 14.01.02 Distinguish between mechanical and chemical digestion, and note where each step specifically occurs. Section: 14.01 Topic: Carbohydrates
35. Which of these organs has contact with food for the least amount of time and does not participate in digestion? A. mouth B. esophagus C. stomach D. small intestine E. large intestine The esophagus plays no role in the chemical digestion of food. Its sole purpose is to conduct the food bolus from the mouth to the stomach.
Bloom's Level: 2. Understand Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
36. Starting from the inside of the esophagus and going to the outside, the layers of the digestive tract in sequence are A. serosa → smooth muscle → submucosa → mucosa. B. serosa → submucosa → smooth muscle → mucosa. C. mucosa → smooth muscle → submucosa → serosa. D. mucosa → submucosa → serosa → smooth muscle. E. mucosa → submucosa → smooth muscle → serosa. From the lumen out, the layers are mucosa, submucosa, muscularis (circular and longitudinal smooth muscle), and serosa.
Bloom's Level: 2. Understand Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
37. A correct function of the human stomach is A. to produce highly basic fluids to kill most bacteria. B. to begin the absorption of most small food molecules, especially sugars. C. storage of food and mixing with digestive fluids to allow chemical digestion. D. to pass food from the esophagus to the large intestine. E. absorption of iron to build hemoglobin for red blood cells. The stomach stores food, starts the digestion of food, and passes food into the small intestine.
Bloom's Level: 2. Understand Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
38. The digestive juices found in the stomach include A. trypsin and bile. B. trypsin and pepsin. C. bile and hydrochloric acid. D. amylase and trypsin. E. pepsin and hydrochloric acid. The stomach is highly acidic because the gastric juice contains hydrochloric acid (HCl). In addition, the gastric juice contains pepsinogen, which becomes pepsin when exposed to HCl.
Bloom's Level: 2. Understand Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive Enzymes
39. Name the structure and organ of the digestive tract that contain microvilli and maltase, and absorb nutrients. A. villus, large intestine B. villus, small intestine C. villus, stomach D. folds, stomach E. pyloric sphincter, appendix The villi of the small intestine contain microvilli; contain maltase, an enzyme that digests maltose; and absorbs nutrients.
Bloom's Level: 2. Understand Learning Outcome: 14.01.03 Compare the structural features and functions of the small intestine versus the large intestine. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
40. In humans, fats are absorbed A. in the large intestine. B. in the rugae of the stomach. C. primarily into the blood capillaries. D. primarily into the ducts of the lacteal of the small intestine. E. by the liver. Fats are digested into fatty acids and glycerol that enter the epithelial cells of the villi where they are joined and packaged as lipoprotein droplets that enter the lacteals.
Bloom's Level: 2. Understand Learning Outcome: 14.01.03 Compare the structural features and functions of the small intestine versus the large intestine. Section: 14.01 Topic: Digestive System
41. Which of the following is NOT true of the duodenum? A. It is a duct from the liver that introduces bile. B. It is a pancreatic duct that introduces basic fluids to neutralize the stomach acid. C. Highly acidic stomach contents enter the small intestine here. D. It contains villi for absorption. E. It produces HCl. The stomach produces HCl, not the small intestine.
Bloom's Level: 2. Understand Learning Outcome: 14.01.03 Compare the structural features and functions of the small intestine versus the large intestine. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
42. Secretin functions to stimulate the A. release of bile. B. release of pancreatic juices, particularly sodium bicarbonate. C. release of gastrin. D. release of CCK. E. release of bile from the gallbladder and the release of pancreatic juices from the pancreas. Secretin stimulates the pancreas to secrete its digestive juices and the gallbladder to release bile.
Bloom's Level: 2. Understand Learning Outcome: 14.01.02 Distinguish between mechanical and chemical digestion, and note where each step specifically occurs. Section: 14.01 Topic: Digestive Enzymes
43. The release of secretin is stimulated by the action of A. the alkaline material in the duodenum. B. bile. C. muscular contractions in the stomach. D. nerves to the duodenum. E. acid in the chyme within the duodenum. Acid present in the chyme stimulates the release of secretin from the cells of the duodenal wall.
Bloom's Level: 2. Understand Learning Outcome: 14.01.02 Distinguish between mechanical and chemical digestion, and note where each step specifically occurs. Section: 14.01 Topic: Digestive Enzymes
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Chapter 14 - Digestive System and Nutrition
44. Bacteria living in the large intestine A. are a source of fiber. B. can produce vitamins. C. are found only in the appendix. D. can produce excess acid. E. are found only in the rectum. Bacteria in the digestive tract produce some vitamins that our bodies can absorb and use.
Bloom's Level: 2. Understand Learning Outcome: 14.01.03 Compare the structural features and functions of the small intestine versus the large intestine. Section: 14.01 Topic: Digestive System
45. As food is processed and moved through the digestive tract, what is the correct order that it is seen? A. bolus → feces → chyme B. bolus → chyme → feces C. chyme → feces → bolus D. chyme → bolus → feces E. feces → bolus → chyme The bolus of food passes through the esophagus to the stomach where it is digested into chyme which passes into the intestines. Feces leave the large intestine.
Bloom's Level: 2. Understand Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
46. Which of the following is considered an accessory gland in the digestive system? A. pharynx B. esophagus C. stomach D. cecum E. pancreas The pancreas is an accessory gland. The remaining choices are all part of the digestive tract.
Bloom's Level: 2. Understand Learning Outcome: 14.02.01 Summarize the major functions of the pancreas, the liver, and the gallbladder. Section: 14.02 Topic: Digestive System
47. What is the function of sodium bicarbonate in the digestive tract? A. destruction of bacteria B. digestion of fat in the small intestine C. neutralization of chyme arriving at the small intestine D. breakdown of connective tissue in the stomach E. activation of pepsin in the stomach Sodium bicarbonate helps neutralize the stomach acid in the small intestine.
Bloom's Level: 2. Understand Learning Outcome: 14.01.03 Compare the structural features and functions of the small intestine versus the large intestine. Section: 14.01 Topic: Digestive Enzymes
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Chapter 14 - Digestive System and Nutrition
48. Which of the following is not a function of the liver? A. the production of bile B. the production of digestive enzymes C. the production of blood proteins D. the removal of blood toxins E. the storage of iron and certain vitamins All of the answer choices are functions of the liver, except the production of digestive enzymes.
Bloom's Level: 2. Understand Learning Outcome: 14.02.01 Summarize the major functions of the pancreas, the liver, and the gallbladder. Section: 14.02 Topic: Digestive System
49. The hepatic portal vein is located between A. the gallbladder and the liver. B. the gallbladder and the small intestine. C. the pancreas and the small intestine. D. the small intestine and the liver. E. the large intestine and the small intestine. Since hepatic means liver, the vein brings nutrients to the liver. It does so from the small intestine.
Bloom's Level: 2. Understand Learning Outcome: 14.02.02 Describe the structure and function of the hepatic portal system. Section: 14.02 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
True / False Questions 50. The hepatic portal vein empties into the inferior vena cava. FALSE The hepatic vein, not the hepatic portal vein, empties into the inferior vena cava.
Bloom's Level: 2. Understand Learning Outcome: 14.02.02 Describe the structure and function of the hepatic portal system. Section: 14.02 Topic: Digestive System
Multiple Choice Questions 51. Which of the following human digestive enzymes is incorrectly matched to its substrate? A. pepsin—protein B. trypsin—nucleic acid C. salivary amylase—starch D. lipase—fat E. maltase—maltose Trypsin digests proteins.
Bloom's Level: 2. Understand Learning Outcome: 14.03.01 Describe the overall function of digestive enzymes. Section: 14.03 Topic: Digestive Enzymes
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Chapter 14 - Digestive System and Nutrition
52. Protein digestion takes place in the A. mouth, stomach, and small intestine. B. stomach and small intestine. C. stomach, esophagus, and small intestine. D. mouth and large intestine. E. small intestine, large intestine, and mouth. Protein digestion begins in the stomach with pepsin, which acts on proteins, converting them to peptides. Peptides are broken down to amino acids in the small intestine.
Bloom's Level: 2. Understand Learning Outcome: 14.03.02 Compare the specific types of chemical digestion that take place in the mouth, stomach, and small intestine. Section: 14.03 Topic: Digestive System
53. The importance of dietary fiber includes all of the following EXCEPT A. dilution of bile salts. B. addition of bulk to the digestive tract. C. a decreased risk of constipation. D. a possible decreased risk of polyp formation. E. increased blood cholesterol levels. Soluble fiber actually decreases blood cholesterol levels by binding to cholesterol in the intestine and preventing it from being absorbed.
Bloom's Level: 2. Understand Learning Outcome: 14.04.01 List the six major classes of nutrients and their major functions in the body. Section: 14.04 Topic: Carbohydrates
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Chapter 14 - Digestive System and Nutrition
54. Which of the following statements about proteins in the diet is not true? A. Complete protein sources include meat, milk, and eggs. B. Legumes, seeds, nuts, and grains can be good protein sources, but contain insufficient amounts of at least one essential amino acid. C. The body is harmed if an excess of protein is consumed. D. Amino acids are stored in the body, so we do not need a regular supply of protein in the diet. E. Two servings of meat a day, each approximately the size of a deck of cards, is considered enough protein. Amino acids are not stored in the body. Therefore, a regular supply of protein is required.
Bloom's Level: 2. Understand Learning Outcome: 14.04.01 List the six major classes of nutrients and their major functions in the body. Section: 14.04 Topic: Proteins
55. Which of the following is NOT a mineral required by the body? A. iron B. folic acid (folacin) C. copper D. sodium E. calcium Folic acid is a vitamin, not a mineral.
Bloom's Level: 2. Understand Learning Outcome: 14.04.03 Distinguish between vitamins, coenzymes, and minerals. Section: 14.04 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
56. Which of the following will NOT increase the risk of obesity? A. sedentary activities B. poor eating habits as a family C. exercise and sensible diet D. excessive intake of fatty foods E. body fat content above 20% ideal body weight Exercise and a sensible diet will reverse obesity.
Bloom's Level: 2. Understand Learning Outcome: 14.04.02 Describe the relationship between diet and obesity, type 2 diabetes, and cardiovascular disease. Section: 14.04 Topic: Human Nutrition
57. Obesity, bulimia nervosa, and anorexia nervosa are alike in that A. all are difficult to treat and may be caused by a number of factors. B. all involve eating far less than is required for the body to function. C. all involve vomiting or purging after eating. D. in all cases the person is actually starving to death regardless of their outside appearance. E. in all cases the person is on a very restrictive diet. All three eating disorders are difficult to treat and are caused by a combination of factors.
Bloom's Level: 2. Understand Learning Outcome: 14.05.01 Describe the three main categories of eating disorders. Section: 14.05 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
58. What is the association of Helicobacter pylori in the digestive system? A. This is the major infectious cause of diarrhea. B. Most ulcers are due to stomach infection with this bacteria. C. This is the causative agent of "strep throat." D. This species of bacteria is the main cause of the odor of feces. E. This species of bacteria converts roughage into vitamin D. This bacteria is associated with an infection of the stomach that causes stomach ulcers.
Bloom's Level: 2. Understand Learning Outcome: 14.06.01 Describe a common disorder that affects each part of the digestive system. Section: 14.06 Topic: Human Digestive System
59. Which is NOT a correct association of intestinal disorders? A. diarrhea—rapid movement of feces that may result in dehydration B. constipation—too slow movement of feces, permitting hard and dry feces C. polyps—small growths on the epithelial lining that can become cancerous D. hemorrhoids—a sexually contracted disease E. hepatitis—inflammation of the liver Hemorrhoids are enlarged, inflamed blood vessels of the anus.
Bloom's Level: 2. Understand Learning Outcome: 14.06.02 Classify the digestive disorders by the type of cause (i.e., infectious, cancer, inflammatory, etc.). Section: 14.06 Topic: Human Digestive System
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Chapter 14 - Digestive System and Nutrition
60. Which of the following is an infectious cause of jaundice? A. trauma B. gallstones C. cancer D. alcoholism E. hepatitis B Only hepatitis B is an infectious agent.
Bloom's Level: 2. Understand Learning Outcome: 14.06.02 Classify the digestive disorders by the type of cause (i.e., infectious, cancer, inflammatory, etc.). Section: 14.06 Topic: Human Digestive System
True / False Questions 61. Crohn's disease is considered an autoimmune disease. TRUE Crohn's disease seems to be caused by a misdirected immune response against one's own intestinal tissues, or perhaps against some of the normal bacteria that live in the gut.
Bloom's Level: 2. Understand Learning Outcome: 14.06.01 Describe a common disorder that affects each part of the digestive system. Section: 14.06 Topic: Human Digestive System
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Chapter 14 - Digestive System and Nutrition
Multiple Choice Questions 62. If you vomit and stomach contents move out through the nose, there has been a failure of the A. reflex action of reverse peristalsis. B. trachea to move up under the epiglottis to cover the glottis. C. soft palate to move back to close off the nasopharynx. D. larynx to move up. E. glottis to retain food. It is the responsibility of the soft palate to move back and close off the nasopharynx.
Bloom's Level: 3. Apply Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
63. Heartburn occurs when A. the heart responds painfully to chemical changes due to diet. B. bacteria cause the stomach lining to ulcerate. C. the glottis fails to prevent food from entering the trachea. D. some of the stomach contents escape back into the lower esophagus. E. the bile duct is blocked. Heartburn occurs when some of the stomach contents escape back into the lower esophagus. It is called "heart" burn because people believe this is where their heart is located.
Bloom's Level: 3. Apply Learning Outcome: 14.01.02 Distinguish between mechanical and chemical digestion, and note where each step specifically occurs. Section: 14.01 Topic: Human Digestive System
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Chapter 14 - Digestive System and Nutrition
64. If the hepatic portal vein was blocked, where would the blood that was supposed to flow into the hepatic portal vein back up? A. the inferior vena cava B. the capillaries of the small intestine C. the capillaries of the liver D. the hepatic artery E. the hepatic vein The hepatic portal vein connects the capillaries of the small intestine to the capillaries of the liver. If it were blocked, the contents would back up in the capillaries of the small intestine.
Bloom's Level: 3. Apply Learning Outcome: 14.02.02 Describe the structure and function of the hepatic portal system. Section: 14.02 Topic: Digestive System
65. If we set up an experimental design to explore the full digestion of starch to glucose, we would need to include which enzymes? A. pepsin and hydrochloric acid B. amylase and maltase C. trypsin and maltase D. pepsin and lipase E. trypsin and lipase Salivary amylase converts starch to maltose. Maltase converts maltose to glucose.
Bloom's Level: 3. Apply Learning Outcome: 14.03.01 Describe the overall function of digestive enzymes. Section: 14.03 Topic: Digestive Enzymes
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Chapter 14 - Digestive System and Nutrition
66. Considering the U.S.D.A. recommended diet, which of the following statements is NOT true? A. Protein should come from poultry, fish, and plant sources. B. Whole grain products are a good source of complex carbohydrates and fiber. C. The daily diet should contain large amounts of saturated fats. D. White bread and other refined carbohydrates should be eaten rarely, if at all. E. Salt intake should be low. The diet should be moderate in total fat intake and low in saturated fats.
Bloom's Level: 3. Apply Learning Outcome: 14.04.01 List the six major classes of nutrients and their major functions in the body. Section: 14.04 Topic: Human Nutrition
67. Which of the following does NOT represent a category of the MyPlate food guide symbol? A. skim milk B. orange C. chicken breast D. whole wheat bread E. coffee Only coffee does not represent one of the categories of the USDA food guide pyramid. Chicken breast represents protein, whole wheat bread represents complex carbohydrates, an orange represents fruit, and skim milk represents dairy.
Bloom's Level: 3. Apply Learning Outcome: 14.04.01 List the six major classes of nutrients and their major functions in the body. Section: 14.04 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
True / False Questions 68. A person can fulfill all of their protein requirements without consuming any animal products. TRUE Beans, peas, nuts, and seeds are alternative sources of proteins.
Bloom's Level: 3. Apply Learning Outcome: 14.04.01 List the six major classes of nutrients and their major functions in the body. Section: 14.04 Topic: Proteins
69. Soft drinks are a good source of carbohydrates for the body. FALSE Soft drinks are a good source of simple sugars; however, complex sugars are preferred. Excess amounts of simple sugars may result in various heath problems.
Bloom's Level: 3. Apply Learning Outcome: 14.04.01 List the six major classes of nutrients and their major functions in the body. Section: 14.04 Topic: Carbohydrates
70. You can eliminate all fat from your diet without suffering any harmful effects. FALSE There are essential fatty acids that your body cannot produce that must be obtained from your diet.
Bloom's Level: 3. Apply Learning Outcome: 14.04.02 Describe the relationship between diet and obesity, type 2 diabetes, and cardiovascular disease. Section: 14.04 Topic: Lipids
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Chapter 14 - Digestive System and Nutrition
Multiple Choice Questions 71. Which of the following diseases is not associated with deficient or excess nutrients in the diet? A. scurvy B. anemia C. cardiovascular disease D. hepatitis E. osteoporosis Hepatitis is an infectious disease. It is not caused by too much/little of a nutrient.
Bloom's Level: 3. Apply Learning Outcome: 14.04.02 Describe the relationship between diet and obesity, type 2 diabetes, and cardiovascular disease. Section: 14.04 Topic: Human Nutrition
72. A woman is not getting enough calcium in her diet. One of the first symptoms she may experience would be A. a broken bone. B. anemia C. irregular heart beat D. muscle spasms E. night blindness Calcium is required for strong bones and teeth. A lack of calcium may result in a broken bone due to osteoporosis.
Bloom's Level: 3. Apply Learning Outcome: 14.04.03 Distinguish between vitamins, coenzymes, and minerals. Section: 14.04 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
73. Stacey is 5 foot 8 inches tall and weighs approximately 80 pounds, yet she thinks she is very fat. Which of the following statements is probably not true about Stacey? A. Stacey most likely suffers from anorexia nervosa. B. Stacey is actually starving. C. Stacey has probably stopped menstruating. D. Stacey is in danger of dying. E. Stacey probably binge-eats and then forces herself to throw up. Stacey most likely has anorexia nervosa. Binge-eating and throwing up are symptoms of bulimia nervosa, not anorexia. It would be advisable for Stacey to seek professional medical help immediately.
Bloom's Level: 3. Apply Learning Outcome: 14.05.01 Describe the three main categories of eating disorders. Section: 14.05 Topic: Human Nutrition
74. Public health workers look for ways to prevent diseases. Which of the following jaundice-causing diseases has the potential to be reduced by public health measures? A. obstructive jaundice caused by gallstones B. obstructive jaundice caused by a cancerous tumor C. hepatitis A by sewage control D. cirrhosis caused by excessive alcohol consumption E. hemolytic jaundice by blood screening Public health workers monitor environmental and infectious agents. Hepatitis A, an infectious agent, can be controlled by sewage treatment.
Bloom's Level: 3. Apply Learning Outcome: 14.06.02 Classify the digestive disorders by the type of cause (i.e., infectious, cancer, inflammatory, etc.). Section: 14.06 Topic: Human Digestive System
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Chapter 14 - Digestive System and Nutrition
75. You suddenly develop explosive and frequent diarrhea. Which of the following is most likely not the cause? A. stress B. consumption of spicy food C. infection by a protozoan D. infection by a virus E. infection by a bacteria Consumption of spicy food does not contribute to explosive diarrhea. Sudden explosive diarrhea is most likely due to food poisoning (either a virus, bacteria, or protozoan) or stress.
Bloom's Level: 3. Apply Learning Outcome: 14.06.02 Classify the digestive disorders by the type of cause (i.e., infectious, cancer, inflammatory, etc.). Section: 14.06 Topic: Human Digestive System
76. Which association is NOT correct? A. Periodontitis is a loss of bone and loosening of teeth that occurs in old age. B. Gingivitis is an inflammation of the gums that can spread to the periodontal membrane. C. The loss of 20 deciduous teeth is a natural process that occurs as a child. D. Tooth decay or dental caries is a natural erosion of teeth from wear and tear. E. Tonsillitis is an inflammation of the tonsils. Tooth decay is not a natural process. It is caused by inadequate dental hygiene, not wear and tear.
Bloom's Level: 4. Analyze Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
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Chapter 14 - Digestive System and Nutrition
77. Which is NOT a correct association of structure with function among layers of the digestive tract? A. The outermost serosa lubricates most of the tract among other abdominal organs. B. The inner mucosa contains goblet cells that secrete mucus. C. The submucosa contains blood vessels and Peyer's patches that circulate lymphatic fluid. D. The muscularis consists of two layers of smooth muscles to force food through the tract. E. The mucosa and submucosa are made of epithelial tissue. The mucosa is a layer of epithelial tissue but the submucosa is a broad band of loose connective tissue.
Bloom's Level: 4. Analyze Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
78. In humans, what is the sequence of fat digestion from the lumen of the small intestine to the bloodstream? A. lipids → fatty acid and glycerol (converted in liver) → bloodstream B. lipids → fatty acid and glycerol (converted in lacteal) → bloodstream C. lipids → fatty acid and glycerol (converted in epithelial cell) → bloodstream D. lipids → fatty acid and glycerol (converted in epithelial cell) → lacteal → bloodstream E. lipids → fatty acid and glycerol (converted in liver) → lacteal → bloodstream After lipids are converted into fatty acids and glycerol in the epithelial cells, they are packed and enter the lacteal. From the lacteals within the lymphatic system, they can enter the bloodstream.
Bloom's Level: 4. Analyze Learning Outcome: 14.01.03 Compare the structural features and functions of the small intestine versus the large intestine. Section: 14.01 Topic: Lipids
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Chapter 14 - Digestive System and Nutrition
79. Secretin, gastrin, and CCK in humans are A. all enzymes used to digest food. B. all hormones that control digestive secretions. C. all breakdown products of digestion. D. produced in the gut by an acid pH. E. proteins found in the intestinal submucosa. These three are all hormones. Gastrin is produced by the stomach, while secretin and CCK are produced by the small intestine.
Bloom's Level: 4. Analyze Learning Outcome: 14.01.02 Distinguish between mechanical and chemical digestion, and note where each step specifically occurs. Section: 14.01 Topic: Digestive Enzymes
80. At body temperature, which of these combinations is most likely to result in the digestion of proteins? A. water, pepsin B. HCl, pepsin C. bile, pepsin D. sodium bicarbonate, pepsin E. pepsin, HCl, sodium bicarbonate In this case, protein is the substrate for digestion so that is required. Pepsin digests proteins when exposed to HCl.
Bloom's Level: 4. Analyze Learning Outcome: 14.03.01 Describe the overall function of digestive enzymes. Section: 14.03 Topic: Digestive Enzymes
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Chapter 14 - Digestive System and Nutrition
81. At body temperature, which of these combinations is most likely to result in the digestion of fat? A. bile salts, sodium bicarbonate, lipase B. bile salts, sodium bicarbonate, pepsin C. HCl, maltase D. bile salts, HCl, sodium bicarbonate, and trypsin E. sodium bicarbonate, lipase In this case, fat is the substrate for digestion. Lipase digests fats in the basic conditions of the small intestine caused by sodium bicarbonate after the fat has been emulsified by bile salts.
Bloom's Level: 4. Analyze Learning Outcome: 14.03.01 Describe the overall function of digestive enzymes. Section: 14.03 Topic: Digestive Enzymes
82. Which of the following is not a major class of nutrients required by the human body? A. fiber B. carbohydrates C. proteins D. lipids E. vitamins Although fiber is required for the human diet, it is not considered a nutrient for the body.
Bloom's Level: 4. Analyze Learning Outcome: 14.04.01 List the six major classes of nutrients and their major functions in the body. Section: 14.04 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
83. Which of the following associations is not true? A. Vitamin E is necessary for blood clotting. B. A deficiency in Vitamin C can cause delayed healing of wounds. C. A vital source of Vitamin A is deep yellow/orange vegetables. D. Many B vitamins can be obtained from green leafy vegetables or whole grains. E. Vitamin D is necessary for proper calcium absorption. Vitamin E is an antioxidant. It defends the body against free radicals. Vitamin K is needed for blood clotting.
Bloom's Level: 4. Analyze Learning Outcome: 14.04.03 Distinguish between vitamins, coenzymes, and minerals. Section: 14.04 Topic: Human Nutrition
84. In order for fats to be digested into fatty acids and glycerol efficiently A. the pH must be around 2 and pepsin must be activated. B. pancreatic juice must be secreted and pepsin must be activated. C. bile must be present and the pH must be around 2. D. the pH must be around 2 and pancreatic juice must be secreted. E. pancreatic juice must be secreted and bile must be present. Lipase requires pancreatic juice to neutralize the pH and bile to emulsify the fat.
Bloom's Level: 5. Evaluate Learning Outcome: 14.03.02 Compare the specific types of chemical digestion that take place in the mouth, stomach, and small intestine. Section: 14.03 Topic: Lipids
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Chapter 14 - Digestive System and Nutrition
85. What pH is required in order for protein to be digested into peptides with the use of pepsin? A. 3.0 B. 7.0 C. 7.6 D. 9.3 E. Pepsin will not digest proteins into peptides. Pepsin requires an acidic pH to be activated from the precursor pepsinogen. 3.0 is the only acidic pH listed.
Bloom's Level: 5. Evaluate Learning Outcome: 14.03.01 Describe the overall function of digestive enzymes. Section: 14.03 Topic: Proteins
86. Anemia is one symptom of deficiency of all the following vitamins and minerals except A. iron. B. folic acid (folacin). C. copper. D. vitamin B12. E. calcium. Copper and iron are involved in hemoglobin synthesis so a lack of these would result in anemia. The lack of vitamin B12 results in pernicious anemia, while the lack of folic acid results in megaloblastic anemia. The lack of calcium does not result in anemia.
Bloom's Level: 5. Evaluate Learning Outcome: 14.04.01 List the six major classes of nutrients and their major functions in the body. Section: 14.04 Topic: Human Nutrition
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Chapter 14 - Digestive System and Nutrition
87. In the past, ulcers were treated by a bland diet and drugs to reduce the acidity of the stomach. Today, ulcers are often treated with antibiotics. What evidence caused the change in treatment? A. Experiments showed that antibiotics reduce stomach acidity better than the drugs used previously. B. Experiments showed that Helicobacter pylori infection was the cause of many ulcers. C. Experiments showed that the healing of ulcers required a more acidic, not a less acidic, environment. D. Experiments showed that bland diets actually led to an increase in ulcers. E. Experiments showed that ulcers were due to stress. Antibiotics are used to treat infectious pathogens. When Marshall and Warren discovered that most stomach ulcers were caused by a bacterial infection with Helicobacter pylori, the treatment regimen was changed.
Bloom's Level: 5. Evaluate Learning Outcome: 14.06.02 Classify the digestive disorders by the type of cause (i.e., infectious, cancer, inflammatory, etc.). Section: 14.06 Topic: Human Digestive System
Short Answer Questions 88. List the nine major digestive enzymes and the substrate each is responsible for breaking down. Salivary amylase—starch Pancreatic amylase—starch Maltase—maltose Pepsin—proteins Trypsin—proteins Peptidases—peptides Nuclease—RNA & DNA Nucleosidases—nucleotide Lipase—fat droplets
Bloom's Level: 6. Create Learning Outcome: 14.03.01 Describe the overall function of digestive enzymes. Section: 14.03 Topic: Digestive Enzymes
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Chapter 14 - Digestive System and Nutrition
89. List the seven functions of the liver. 1. Detoxifies the blood. 2. Stores iron and vitamins A, D, E, K, and B12. 3. Makes plasma proteins. 4. Stores glucose as glycogen. 5. Produces urea. 6. Removes bilirubin. 7. Helps regulate blood cholesterol levels.
Bloom's Level: 6. Create Learning Outcome: 14.02.01 Summarize the major functions of the pancreas, the liver, and the gallbladder. Section: 14.02 Topic: Digestive System
90. List the organs that food will pass through, starting with the mouth and ending with the anus. Mouth - pharynx - esophagus - stomach - small intestine - large intestine
Bloom's Level: 6. Create Learning Outcome: 14.01.01 Summarize the function of each organ of the digestive system. Section: 14.01 Topic: Digestive System
14-45 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
Chapter 15 Respiratory System
Multiple Choice Questions 1. The entrance and exit of air into and out of the lungs is called A. ventilation. B. external respiration. C. internal respiration. D. cellular respiration. E. inspiration. Ventilation, another term for breathing, encompasses both inspiration and expiration.
Bloom's Level: 1. Remember Learning Outcome: 15.01.01 Summarize the major functions of the respiratory system. Section: 15.01 Topic: Respiratory System
2. The type of respiration in which gases are exchanged between the blood and tissue fluid is termed A. breathing. B. external respiration. C. internal respiration. D. cellular respiration. E. inspiration. Internal respiration is the exchange of gases between the blood and tissue fluid.
Bloom's Level: 1. Remember Learning Outcome: 15.01.01 Summarize the major functions of the respiratory system. Section: 15.01 Topic: Respiratory System
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Chapter 15 - Respiratory System
3. The vocal cords are found in the A. bronchi. B. nose. C. larynx. D. lungs. E. pharynx. The larynx is called the voice box because it houses the vocal cords.
Bloom's Level: 1. Remember Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
4. The Adam's apple is actually a part of the A. pharynx. B. larynx. C. glottis. D. vocal cords. E. trachea. The Adam's apple is caused by the growth of the larynx at the time of puberty.
Bloom's Level: 1. Remember Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
5. Which portion of the respiratory tract is commonly referred to as the "throat"? A. pharynx B. larynx C. glottis D. trachea E. epiglottis The pharynx is commonly referred to as the throat.
Bloom's Level: 1. Remember Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
6. After the trachea, what is the next structure that air passes through? A. bronchi B. alveolus C. larynx D. pharynx E. bronchioles The trachea divides into the right and left primary bronchi.
Bloom's Level: 1. Remember Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
7. You are looking at a structure in the respiratory tract that is filled with small sacs or air pockets. What structure are you looking at? A. bronchi B. alveoli C. larynx D. pharynx E. nose The alveoli are composed of a multitude of air pockets or sacs.
Bloom's Level: 1. Remember Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
Figure 15.1
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Chapter 15 - Respiratory System
8. Which point in Figure 15.1 produces vocal sounds when air passes through it? A. 1 B. 2 C. 3 D. 4 E. 5 Air produces sound when it passes through the vocal cords, which are #2.
Bloom's Level: 1. Remember Figure 15.01 Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
9. The exchange of oxygen and carbon dioxide occurs at which location in Figure 15.1? A. 1 B. 2 C. 3 D. 4 E. 5 Gases are exchanged in the alveoli, which are #5.
Bloom's Level: 1. Remember Figure 15.01 Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
True / False Questions 10. In an average adult, only about 1/3 of the air inhaled actually reaches the alveoli. FALSE In an average adult, about 70% of the air reaches the alveoli. About 1/3 remains in the airways.
Bloom's Level: 1. Remember Learning Outcome: 15.02.01 Identify the different respiratory volumes that can be measured with a spirometer. Section: 15.02 Topic: Human Respiratory System
Multiple Choice Questions 11. When we are relaxed, only a small amount of air moves in and out with each breath. This is called A. static volume. B. tidal volume. C. inspiratory reserve volume. D. expiratory reserve volume. E. residual volume. The tidal volume is the small amount of air that moves in and out with each breath. It is about 500 ml.
Bloom's Level: 1. Remember Learning Outcome: 15.02.01 Identify the different respiratory volumes that can be measured with a spirometer. Section: 15.02 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
12. The maximum volume of air that can be moved in plus the maximum volume of air that can be moved out during a single breath is the A. tidal volume. B. inspiratory reserve volume. C. expiratory reserve volume. D. vital capacity. E. residual volume. The vital capacity is the maximum volume that can be moved in and out during a single breath.
Bloom's Level: 1. Remember Learning Outcome: 15.02.01 Identify the different respiratory volumes that can be measured with a spirometer. Section: 15.02 Topic: Human Respiratory System
13. The exchange of gases between the lungs and the blood occurs by the process of A. diffusion. B. osmosis. C. filtration. D. active transport. E. ionic bonding. The gases diffuse out of the area of higher concentration into the area of lower concentration.
Bloom's Level: 1. Remember Learning Outcome: 15.03.03 Review how the differences in PO2 and PCO2 in arterial and venous blood determine how gases are exchanged in the lungs versus the tissues. Section: 15.03 Topic: Respiratory System
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Chapter 15 - Respiratory System
14. Carbonic anhydrase A. is a digestive enzyme. B. is dissolved in plasma. C. speeds up the conversion of carbonic acid to carbon dioxide and water. D. speeds up the conversion of carbon dioxide to oxygen. E. speeds up the conversion of oxyhemoglobin. Carbonic anhydrase speeds up the conversion of carbonic acid to water and carbon dioxide.
Bloom's Level: 1. Remember Learning Outcome: 15.03.03 Review how the differences in PO2 and PCO2 in arterial and venous blood determine how gases are exchanged in the lungs versus the tissues. Section: 15.03 Topic: Respiratory System
True / False Questions 15. Blood in the pulmonary capillaries has a higher partial pressure of carbon dioxide than atmospheric air. TRUE This is why carbon dioxide diffuses out of the plasma into the lungs.
Bloom's Level: 1. Remember Learning Outcome: 15.03.03 Review how the differences in PO2 and PCO2 in arterial and venous blood determine how gases are exchanged in the lungs versus the tissues. Section: 15.03 Topic: Respiratory System
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Chapter 15 - Respiratory System
Multiple Choice Questions 16. A lung disease caused by bacteria that become encapsulated is called A. emphysema. B. pneumonia. C. rheumatic fever. D. tuberculosis. E. pulmonary fibrosis. Tuberculosis is caused by the bacterium Mycobacterium tuberculosis. The body walls of the lungs form tubercles to surround and encapsulate the invading bacteria.
Bloom's Level: 1. Remember Learning Outcome: 15.04.02 Classify disorders of the respiratory system according to whether they are caused by allergies, infections, a genetic defect, or exposure to toxins. Section: 15.04 Topic: Human Respiratory System
17. Which of the following is an upper respiratory tract disease? A. pharyngitis B. acute bronchitis C. asthma D. pneumonia E. cystic fibrosis Pharyngitis is the only disorder of the upper respiratory tract. The rest of the disorders are found in the lower respiratory tract.
Bloom's Level: 1. Remember Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
18. Which of the following symptoms would not be typical of otitis media? A. pain B. sense of fullness in the ear C. hearing loss D. headache E. dizziness Otitis media, an inflammation of the middle ear, does not normally cause headaches.
Bloom's Level: 1. Remember Bloom's Level: 3. Apply Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
True / False Questions 19. Most respiratory infections are caused by viruses and can successfully be treated with antibiotics. FALSE Viral infections cannot be treated with antibiotics.
Bloom's Level: 1. Remember Learning Outcome: 15.04.02 Classify disorders of the respiratory system according to whether they are caused by allergies, infections, a genetic defect, or exposure to toxins. Section: 15.04 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
Multiple Choice Questions 20. Which disease causes the lungs to fill with thick fluid? A. pharyngitis B. sinusitis C. emphysema D. tuberculosis E. pneumonia Pneumonia is an infection of the lungs in which the bronchi and alveoli fill with thick fluid.
Bloom's Level: 1. Remember Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
Yes / No Questions 21. If a person has chronic bronchitis, will they cough up mucus? YES In chronic bronchitis, the airways are inflamed and filled with mucus. A cough that brings up mucus is common.
Bloom's Level: 1. Remember Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
True / False Questions 22. Cystic fibrosis has a genetic cause. TRUE Cystic fibrosis is caused by a defective gene that codes for the cystic fibrosis transmembrane regulator.
Bloom's Level: 1. Remember Learning Outcome: 15.04.02 Classify disorders of the respiratory system according to whether they are caused by allergies, infections, a genetic defect, or exposure to toxins. Section: 15.04 Topic: Human Respiratory System
23. The majority of cases of lung cancer are associated with smoking. TRUE About 87% of lung cancers are associated with cigarette smoking.
Bloom's Level: 1. Remember Learning Outcome: 15.04.02 Classify disorders of the respiratory system according to whether they are caused by allergies, infections, a genetic defect, or exposure to toxins. Section: 15.04 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
Multiple Choice Questions 24. Which of the following treatments can cure lung cancer? A. pneumonectomy B. steroids C. antibiotics D. mucus-thinning drugs E. tracheotomy Pneumonectomy, in which a lobe or the whole lung, is removed, is the only possible way to cure lung cancer, assuming it has not spread.
Bloom's Level: 1. Remember Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
25. What happens to the pH of the bloodstream if a person hypoventilates? A. The pH of the blood will become more acidic. B. The pH of the blood will increase. C. The pH of the blood will initially increase and then dramatically decrease. D. The pH of the blood will initially decrease and then dramatically increase. E. Nothing will happen to the pH of the blood. If a person hypoventilates, hydrogen ions (H+) build up in the blood, causing the blood to become more acidic.
Bloom's Level: 1. Remember Learning Outcome: 15.03.02 Interpret the chemical reaction discussed in this section to explain the effect of hyperventilation or hypoventilation on blood pH. Section: 15.03 Topic: Respiratory System
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Chapter 15 - Respiratory System
26. What happens to the blood pH if a person hyperventilates? A. The pH will increase. B. The pH will remain the same. C. The pH will decrease. D. The pH will initially increase but then dramatically decrease. E. The pH will initially decrease but then dramatically increase. When a person hyperventilates, they remove more CO2 from the bloodstream. This will in turn leave the blood with fewer hydrogen (H+) ions, causing the pH to shift towards the basic range (respiratory alkalosis).
Bloom's Level: 1. Remember Learning Outcome: 15.03.02 Interpret the chemical reaction discussed in this section to explain the effect of hyperventilation or hypoventilation on blood pH. Section: 15.03 Topic: Human Respiratory System
27. Compared to outside air, describe the air that reaches the lungs. A. warmer and with more moisture B. warmer and with less moisture C. cooler and with more moisture D. cooler and with less moisture E. exactly the same As the air moves in along the airways, it is cleansed, warmed, and moistened.
Bloom's Level: 2. Understand Learning Outcome: 15.01.01 Summarize the major functions of the respiratory system. Section: 15.01 Topic: Respiratory System
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Chapter 15 - Respiratory System
28. Which is NOT a major function of the respiratory tract? A. warming incoming air before it reaches the lungs B. moistening incoming air C. filtering out dust and water particles D. conducting air with higher O2 to lungs and conducting air with higher CO2 back out E. transferring and extracting nutrients The respiratory track does not transfer and extract nutrients.
Bloom's Level: 2. Understand Learning Outcome: 15.01.01 Summarize the major functions of the respiratory system. Section: 15.01 Topic: Respiratory System
29. Trace the path of an inhaled air molecule. A. nasal cavity → glottis → larynx → trachea → pharynx → bronchi → bronchioles → alveoli B. nasal cavity → pharynx → larynx → glottis → trachea → bronchi → bronchioles → alveoli C. nasal cavity → pharynx → glottis → larynx → trachea → bronchi → bronchioles → alveoli D. nasal cavity → pharynx → glottis → larynx → trachea → bronchioles → bronchi → alveoli E. nasal cavity → pharynx → glottis → trachea → larynx → bronchi → bronchioles → alveoli In the upper respiratory tract, the air moves through the nasal cavities to the pharynx, the glottis, and the larynx. Then it enters the lower respiratory tract, moving through the trachea, bronchi, bronchioles, and alveoli.
Bloom's Level: 2. Understand Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
30. Which respiratory organ normally allows both air and food passage? A. bronchi B. trachea C. larynx D. pharynx E. nasal cavity The pharynx is a funnel-shaped passageway that connects the nasal and oral cavities to the larynx.
Bloom's Level: 2. Understand Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
31. Which of the following statements does not correctly describe air movement along the respiratory tract? A. Cilia and hairs in the nose help to screen out foreign particles in the air. B. Food and air both enter the pharynx, presenting a potential danger to respiration. C. Air reaching the lungs has been warmed to body temperature by passage through the nose and upper respiratory system. D. The glottis is the passageway through the larynx by which air enters the trachea. E. The bronchi serve as passageways into the trachea. The bronchi serve as passageways into the bronchioles and the alveoli.
Bloom's Level: 2. Understand Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
32. A voice changes from high to low pitch in speaking by A. changing the volume of air that flows across the vocal cords. B. controlling the growth of the tissue. C. increasing the amplitude of the vocal cord vibrations. D. changing the tension on the vocal cords. E. changing the tension on the epiglottis. The pitch of the voice is regulated by changing the tension on the vocal cords. The greater the tension, the higher the pitch.
Bloom's Level: 2. Understand Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
33. Food is prevented from entering the trachea by the A. pharynx. B. larynx. C. bronchioles. D. saliva. E. epiglottis. The epiglottis is a flap of tissue that prevents food from passing into the larynx and thus the trachea.
Bloom's Level: 2. Understand Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
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Chapter 15 - Respiratory System
34. Which part of the respiratory system is composed of C-shaped cartilaginous rings and cilia? A. pharynx B. larynx C. trachea D. bronchioles E. alveoli The trachea lies in front of the esophagus and is held open by C-shaped cartilaginous rings. The cilia that project from the epithelial lining keep the lungs clean by sweeping mucus and debris toward the pharynx.
Bloom's Level: 2. Understand Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
35. Which association is NOT correct? A. larynx—voice box B. trachea—windpipe C. alveoli—gas exchange D. pharynx—vocal cords E. hair and mucus—nasal passages The pharynx is called the throat. The larynx contains the vocal cords.
Bloom's Level: 2. Understand Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
15-18 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
36. Which of the following respiratory structures will have the least amount of cartilage and the thinnest walls? A. larynx B. trachea C. bronchi D. bronchioles E. pharynx As the bronchial tubes divide and subdivide, their walls become thinner, and the small rings of cartilage are no longer present.
Bloom's Level: 2. Understand Learning Outcome: 15.01.03 Describe the layers of cells a molecule of oxygen must pass through when diffusing from a lung alveolus into a red blood cell, and from the red blood cell into the tissues. Section: 15.01 Topic: Human Respiratory System
True / False Questions 37. Oxygen in the lungs diffuses from the capillary walls to enter the alveolar walls. FALSE Oxygen in the lungs must diffuse through the alveolar walls in order to reach the capillary walls.
Bloom's Level: 2. Understand Learning Outcome: 15.01.03 Describe the layers of cells a molecule of oxygen must pass through when diffusing from a lung alveolus into a red blood cell, and from the red blood cell into the tissues. Section: 15.01 Topic: Human Respiratory System
15-19 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
Multiple Choice Questions 38. In humans, the lungs inflate when the A. rib muscles contract. B. diaphragm muscle contracts. C. external intercostal rib muscles and diaphragm contract. D. diaphragm relaxes. E. rib muscles and diaphragm relax. The lungs inflate when the diaphragm and rib muscles contract. The ribs move upward and outward and the diaphragm moves down.
Bloom's Level: 2. Understand Learning Outcome: 15.02.02 Contrast the mechanisms of inspiration and expiration. Section: 15.02 Topic: Human Respiratory System
39. Which of the following statements is NOT true regarding the normal mechanism of ventilation? A. There must be a continuous column of air from the pharynx to the alveoli. B. The lungs must lie within a sealed-off thoracic cavity. C. The inner and outer pleural membranes must be separated by a thin film of fluid. D. The intrapleural pressure must be greater than the atmospheric pressure. E. The diaphragm, when contracted, will be lowered. During inspiration, the intrapleural pressure must be less than atmospheric pressure so the air rushes into the lungs.
Bloom's Level: 2. Understand Learning Outcome: 15.02.02 Contrast the mechanisms of inspiration and expiration. Section: 15.02 Topic: Respiratory System
15-20 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
True / False Questions 40. Both inspiration and expiration require active engagement of the diaphragm and the rib muscles. FALSE Expiration is the passive phase of breathing. No effort is required to bring it about.
Bloom's Level: 2. Understand Learning Outcome: 15.02.02 Contrast the mechanisms of inspiration and expiration. Section: 15.02 Topic: Human Respiratory System
Multiple Choice Questions 41. The breathing rate is controlled by chemoreceptors that detect A. levels of oxygen in the alveolar air space. B. levels of oxygen in the blood. C. levels of carbon dioxide in the alveolar air space. D. levels of carbon dioxide in the blood. E. level of stretching of the rib muscles and diaphragm. The chemoreceptors in the blood detect levels of carbon dioxide and hydrogen ions.
Bloom's Level: 2. Understand Learning Outcome: 15.02.03 Indicate how ventilation is controlled by nervous and chemical mechanisms. Section: 15.02 Topic: Human Respiratory System
15-21 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
42. Aortic and carotid bodies are A. the same as the SA and AV nodes. B. types of valves in the heart. C. present in the lungs. D. sensitive to changes in CO2 and H+ in the blood. E. sensitive to oxygen changes. These are chemoreceptors in the carotid artery and aorta that are sensitive to the level of oxygen in the blood.
Bloom's Level: 2. Understand Learning Outcome: 15.02.03 Indicate how ventilation is controlled by nervous and chemical mechanisms. Section: 15.02 Topic: Human Respiratory System
43. Most of the carbon dioxide transported in the plasma is in the form of A. gas bubbles. B. oxyhemoglobin. C. bicarbonate ions. D. carbaminohemoglobin. E. carbon monoxide. Most of the carbon dioxide combines with water to form a bicarbonate ion, which is how carbon dioxide is carried in the blood.
Bloom's Level: 2. Understand Learning Outcome: 15.03.01 Distinguish between external and internal respiration. Section: 15.03 Topic: Respiratory System
15-22 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
44. Oxygen is transported in blood most efficiently by A. being dissolved in the plasma. B. chemical conversion to water. C. combining with hemoglobin. D. combining with carbon dioxide to form bicarbonate ions. E. a free oxygen ion or radical. Oxygen is carried most efficiently by binding to hemoglobin to form oxyhemoglobin.
Bloom's Level: 2. Understand Learning Outcome: 15.03.03 Review how the differences in PO2 and PCO2 in arterial and venous blood determine how gases are exchanged in the lungs versus the tissues. Section: 15.03 Topic: Respiratory System
45. What components does respiration include? A. only the exchange of oxygen in the lungs B. exchange of gases in the lungs and in the tissues C. only the exchange of oxygen in the tissues D. only exchange of carbon dioxide in the lungs E. only the exchange of carbon dioxide in the tissues Respiration includes external respiration (exchange of gases in the lungs) and internal respiration (exchange of gases in the tissues).
Bloom's Level: 2. Understand Learning Outcome: 15.03.01 Distinguish between external and internal respiration. Section: 15.03 Topic: Respiratory System
15-23 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
46. The highest carbon dioxide concentration will be found in (or at) the A. atmosphere. B. trachea. C. alveoli. D. tissue cells. E. blood. Carbon dioxide is the waste that builds up in the tissue cells. The pressure of carbon dioxide in tissue fluid is higher than that of blood.
Bloom's Level: 2. Understand Learning Outcome: 15.03.03 Review how the differences in PO2 and PCO2 in arterial and venous blood determine how gases are exchanged in the lungs versus the tissues. Section: 15.03 Topic: Respiratory System
47. Which of the following diseases is caused by exposure to particles inhaled primarily in the workplace (silica, coal dust, etc.)? A. emphysema B. pneumonia C. rheumatic fever D. tuberculosis E. pulmonary fibrosis Pulmonary fibrosis occurs when fibrous connective tissue builds up in the lungs, causing a lack of elasticity. This is increased by environmental exposure to silica, various types of dust, and asbestos.
Bloom's Level: 2. Understand Learning Outcome: 15.04.02 Classify disorders of the respiratory system according to whether they are caused by allergies, infections, a genetic defect, or exposure to toxins. Section: 15.04 Topic: Human Respiratory System
15-24 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
48. Scientific studies have shown that which of the following is NOT associated with smoking? A. increased risk of lung cancer B. increased severity of allergies C. increased risk of ulcers D. increased risk of stroke E. increased risk of diabetes Smoking increases the risk of all of these except for diabetes.
Bloom's Level: 2. Understand Learning Outcome: 15.04.02 Classify disorders of the respiratory system according to whether they are caused by allergies, infections, a genetic defect, or exposure to toxins. Section: 15.04 Topic: Human Respiratory System
True / False Questions 49. Almost all of the disorders of the respiratory system are due to infectious diseases. Those that are not infectious are most likely inherited diseases. FALSE Disorders of the respiratory system can be caused by allergies and exposure to toxins and pollutants as well as infectious diseases and genetic diseases.
Bloom's Level: 2. Understand Learning Outcome: 15.04.02 Classify disorders of the respiratory system according to whether they are caused by allergies, infections, a genetic defect, or exposure to toxins. Section: 15.04 Topic: Human Respiratory System
15-25 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
50. An infection with Mycobacterium can cause tubercles in the lungs. TRUE When Mycobacterium tuberculosis invades the lung tissue, the body walls them off, forming tubercles.
Bloom's Level: 2. Understand Learning Outcome: 15.04.02 Classify disorders of the respiratory system according to whether they are caused by allergies, infections, a genetic defect, or exposure to toxins. Section: 15.04 Topic: Human Respiratory System
Multiple Choice Questions 51. Which of the following diseases make up chronic obstructive pulmonary disease (COPD)? A. emphysema, chronic bronchitis, and asthma B. emphysema, tuberculosis, and cystic fibrosis C. chronic bronchitis, tuberculosis, and pulmonary fibrosis D. tuberculosis, cystic fibrosis, and asthma E. pulmonary fibrosis, asthma, and cystic fibrosis COPD is made up of emphysema, chronic bronchitis, and asthma, which are often seen in the same patient.
Bloom's Level: 2. Understand Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
15-26 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
52. A short while after a person has died, a "gasp" of air may occur as the lungs collapse. This flow would represent the A. tidal volume. B. expiratory reserve volume. C. total lung volume. D. vital capacity. E. residual volume. The residual volume is the air that always remains in the lungs. When the lungs totally collapse, this air would be expelled.
Bloom's Level: 3. Apply Learning Outcome: 15.02.01 Identify the different respiratory volumes that can be measured with a spirometer. Section: 15.02 Topic: Respiratory System
53. If a lung is punctured in a car accident, that lobe fails to inflate even though there is no obstruction of the air passageway to that lobe. Why? A. The trauma must have damaged the nerve path controlling ribs on that side of the lung. B. Negative feedback prevents the ribs and diaphragm on that side from causing pain. C. When the chest volume expands, air can now rush in through the puncture without filling the alveoli. D. Mucus or blood must be filling the lobe. E. Stimulation of stretch receptors in the alveolar walls initiate inhibitory nerve impulses. Breathing depends on negative pressure inside the lungs. If air can get inside without filling the alveoli, the lung lobe will collapse.
Bloom's Level: 3. Apply Learning Outcome: 15.02.02 Contrast the mechanisms of inspiration and expiration. Section: 15.02 Topic: Human Respiratory System
15-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
54. A pressurized airplane cabin and an oxygen tent over a patient in a hospital both address the need of a person for oxygen. How do these compare? A. The oxygen tent increases the air pressure to increase the PO2. B. Airplane cabin pressure lowers the partial pressures for normal breathing. C. Both use the same mechanism to force hemoglobin to increase affinity for oxygen. D. The oxygen tent increases concentration of O2 while the airplane maintains total air pressure without changing concentrations. E. Both increase the PO2 in order to force the exchange of oxygen for carbon dioxide. The oxygen tent increases the concentration of oxygen available to the patient. The pressurized cabin inside an airplane mimics normal air pressure at ground level but does not change the concentrations of the various gases.
Bloom's Level: 3. Apply Learning Outcome: 15.03.03 Review how the differences in PO2 and PCO2 in arterial and venous blood determine how gases are exchanged in the lungs versus the tissues. Section: 15.03 Topic: Human Respiratory System
55. Many lung ailments are "not curable but treatable." Which of the following conditions is most curable? A. advanced lung cancer with metastases B. standard tuberculosis C. asthma D. emphysema E. COPD Tuberculosis is an infectious disease that can be cured. The others are chronic and incurable.
Bloom's Level: 3. Apply Learning Outcome: 15.04.02 Classify disorders of the respiratory system according to whether they are caused by allergies, infections, a genetic defect, or exposure to toxins. Section: 15.04 Topic: Human Respiratory System
15-28 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
True / False Questions 56. You should treat the common cold with antibiotics to shorten the duration of symptoms. FALSE The common cold is caused by viruses which cannot be treated with antibiotics.
Bloom's Level: 3. Apply Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
57. The symptoms of laryngitis include a severe sore throat, high fever, and white patches on the throat. FALSE These are the symptoms of "strep throat." Laryngitis is an inflammation of the larynx with accompanying hoarseness.
Bloom's Level: 3. Apply Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
15-29 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
Multiple Choice Questions 58. Sarah comes home from school with a horrible pressure behind her eyes. Her nose is running and she has a headache. She recently had a cold. What is Sarah most likely suffering from? A. otitis media B. sinusitis C. bronchitis D. asthma E. pneumonia These are the symptoms of sinusitis, an inflammation of the cranial sinuses.
Bloom's Level: 3. Apply Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
59. Which disease has an allergic cause? A. otitis media B. laryngitis C. asthma D. pneumonia E. emphysema Asthma is a disease in which the airways are unusually sensitive to specific allergens.
Bloom's Level: 3. Apply Learning Outcome: 15.04.02 Classify disorders of the respiratory system according to whether they are caused by allergies, infections, a genetic defect, or exposure to toxins. Section: 15.04 Topic: Human Respiratory System
15-30 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
60. What keeps the alveoli open so that gas exchange can take place? A. rings of cartilage B. surfactant C. smooth muscle D. mucus E. air Surfactant lines the alveoli and lowers the surface tension to prevent them from closing due to the surface tension of water.
Bloom's Level: 4. Analyze Learning Outcome: 15.01.02 Trace the path of air from the human nose to the lungs. Section: 15.01 Topic: Human Respiratory System
61. From the inside of the lungs, what is the first cell type the air must pass through to reach the blood? A. simple squamous epithelium B. cartilage C. muscle D. red blood cells E. pseudostratified ciliated columnar epithelium The walls of the alveoli are made up of simple squamous epithelium that the air must pass through first.
Bloom's Level: 4. Analyze Learning Outcome: 15.01.03 Describe the layers of cells a molecule of oxygen must pass through when diffusing from a lung alveolus into a red blood cell, and from the red blood cell into the tissues. Section: 15.01 Topic: Human Respiratory System
15-31 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
62. Phillip is breathing at approximately 40 ventilations per minute. What is the respiratory center indicating? A. low oxygen concentration in the lungs B. high carbon dioxide concentration in the blood C. low oxygen concentration in the blood D. the firing of the intercostal nerves E. stretch receptors in the alveolar walls This is about twice the normal breathing rate. The respiratory center is indicating that the body needs to breath more due to high carbon dioxide concentrations in the blood.
Bloom's Level: 4. Analyze Learning Outcome: 15.02.03 Indicate how ventilation is controlled by nervous and chemical mechanisms. Section: 15.02 Topic: Human Respiratory System
63. How does expired air compare with inspired air? A. Expired air contains less oxygen but more carbon dioxide. B. Expired air contains less oxygen and less carbon dioxide. C. Expired air contains more oxygen and more carbon dioxide. D. Expired air contains more oxygen but less carbon dioxide. E. Expired and inspired air contain the same amount of oxygen and carbon dioxide. Expired air contains less oxygen because it has lost oxygen to the blood in the lungs. It contains more carbon dioxide because it has picked up carbon dioxide in the lungs.
Bloom's Level: 4. Analyze Learning Outcome: 15.03.01 Distinguish between external and internal respiration. Section: 15.03 Topic: Respiratory System
15-32 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
64. Blood richest in oxygen is found in the A. inferior vena cava. B. superior vena cava. C. pulmonary arteries. D. pulmonary veins. E. systemic capillaries. The pulmonary veins have just picked up oxygen from the lungs so they contain the highest concentration of oxygen.
Bloom's Level: 4. Analyze Learning Outcome: 15.03.03 Review how the differences in PO2 and PCO2 in arterial and venous blood determine how gases are exchanged in the lungs versus the tissues. Section: 15.03 Topic: Respiratory System
65. By what route does CO2 leave the body? A. pulmonary artery → alveolus → bronchus → bronchiole → trachea → pharynx → larynx B. pulmonary vein → bronchiole → alveolus → bronchus → trachea → pharynx → larynx C. pulmonary artery → alveolus → bronchiole → bronchus → trachea → larynx → pharynx D. pulmonary vein → alveolus → bronchus → bronchiole → trachea → larynx → pharynx E. pulmonary artery → alveolus → bronchiole → bronchus → trachea → pharynx → larynx Basically, carbon dioxide leaves the opposite way that oxygen enters. The pulmonary artery transfers carbon dioxide to the alveolus, and then to the bronchiole and bronchus. This then enters the trachea, larynx, and pharynx.
Bloom's Level: 4. Analyze Learning Outcome: 15.03.01 Distinguish between external and internal respiration. Section: 15.03 Topic: Respiratory System
15-33 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
66. Which is NOT a correct association of respiratory problems and symptoms? A. laryngitis—inflammation of the larynx that causes hoarseness or loss of voice B. tonsillitis—inflammation of the uvula often cured by removal of the useless tissue C. otitis media—ear pain due to bacterial infection of the middle ear D. sinusitis—pain and tenderness in lower forehead and over the cheeks E. pharyngitis—infection of the throat Tonsillitis occurs when the tonsils become inflamed and enlarged. The tonsils can be removed surgically but they help initiate immune responses to many pathogens that enter via the pharynx and are an important component of the innate immune system.
Bloom's Level: 4. Analyze Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
67. Which is NOT a correct association of lower respiratory problems and causes or symptoms? A. chronic bronchitis—ongoing inflammation and possible infection of bronchi with loss of cilia B. pulmonary fibrosis—loss of lung elasticity due to breathing silica, coal dust, asbestos, etc. C. emphysema—reduced lung and chest volume due to collapsed alveoli and overventilation D. pulmonary tuberculosis—a bacterial infection that the body tries to isolate and seal off E. Pneumocystis carinii—pneumonia caused by a protozoan in absence of a healthy immune system Emphysema is a chronic disorder in which the alveoli are distended and their walls damaged.
Bloom's Level: 4. Analyze Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
15-34 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
68. Why does the surface of the lungs need to be covered with pleura? A. to allow the exchange of oxygen within the alveoli B. to allow the exchange of carbon dioxide within the alveoli C. to produce the muscular contractions that expand the lungs D. to enable the lungs to slide past the internal chest wall and diaphragm E. to prevent the alveoli from collapsing due to pressure The pleura covers the lungs, internal chest wall, and diaphragm. These membranes produce lubricating fluids that help them slide freely against each other during respiration.
Bloom's Level: 5. Evaluate Learning Outcome: 15.01.03 Describe the layers of cells a molecule of oxygen must pass through when diffusing from a lung alveolus into a red blood cell, and from the red blood cell into the tissues. Section: 15.01 Topic: Human Respiratory System
69. What happens if you hyperventilate? A. Oxygen diffuses out of the blood into the lungs. B. The pH of the blood becomes more basic. C. There are more hydrogen ions in the blood. D. Carbon dioxide diffuses out of the lungs into the blood. E. More bicarbonate ion is formed. More carbon dioxide is lost, driving the bicarbonate ions to convert to carbonic acid and thereby deleting the concentration of hydrogen ions. This causes the pH of the blood to become more basic.
Bloom's Level: 5. Evaluate Learning Outcome: 15.03.02 Interpret the chemical reaction discussed in this section to explain the effect of hyperventilation or hypoventilation on blood pH. Section: 15.03 Topic: Respiratory System
15-35 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
70. Imagine you are trying to fit down an aisle at the local store that is filled with boxes. In order to squeeze between two stacks of boxes you need to "suck it in." Which of the following descriptions best explains why "sucking it in" isn't the best way to fit through a tight space? A. While sucking it in, your diaphragm contracts and moves down, causing the lung volume to increase. This causes the rib cage to move up and out, making your chest expand. B. While sucking it in, your diaphragm contracts and moves upward, causing the lung volume to increase. This causes the rib cage to move up and out, making your chest larger. C. While sucking it in, your diaphragm contracts and moves down, causing the lung volume to decrease. This causes the rib cage to move up and out, making your chest larger. D. While sucking it in, your diaphragm contracts and moves down, causing the lung volume to increase. This causes the rib cage to move down and out, making your chest larger. E. While sucking it in, your diaphragm contracts and moves upward, causing the lung volume to increase. This causes the rib cage to move down and out, making your chest larger. While sucking it in, your diaphragm contracts and moves down, causing the lung volume to increase. This causes the rib cage to move up and out, making your chest larger.
Bloom's Level: 5. Evaluate Learning Outcome: 15.02.02 Contrast the mechanisms of inspiration and expiration. Section: 15.02 Topic: Human Respiratory System
Short Answer Questions 71. Identify the four respiratory volumes that can be measured with a spirometer. 1. Tidal volume (500 ml) 2. Inspiratory reserve volume (3000 ml) 3. Expiratory reserve volume (1500 ml) 4. Residual volume (1000 ml)
Bloom's Level: 6. Create Learning Outcome: 15.02.01 Identify the different respiratory volumes that can be measured with a spirometer. Section: 15.02 Topic: Human Respiratory System
15-36 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 15 - Respiratory System
72. List and describe the three major functions of the respiratory system. 1. External respiration: exchange of gases between the air and the blood. 2. Transport of gases to and from the lungs and the tissues. 3. Internal respiration: exchange of gases between the blood and tissue fluid.
Bloom's Level: 6. Create Learning Outcome: 15.01.01 Summarize the major functions of the respiratory system. Section: 15.01 Topic: Respiratory System
73. Identify the structures and disorders associated with the upper respiratory system. The structures of the upper respiratory system include the nasal cavities, pharynx, and the larynx. Disorders of the upper respiratory system include the common cold, pharyngitis, tonsillitis, laryngitis, sinusitis, and otitis media.
Bloom's Level: 6. Create Learning Outcome: 15.04.01 Summarize the characteristics of the disorders that affect the upper and lower respiratory tracts. Section: 15.04 Topic: Human Respiratory System
15-37 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
Chapter 16 Urinary System and Excretion
Multiple Choice Questions 1. The tube that transports urine from the kidney to the urinary bladder is the A. urethra. B. ureter. C. collecting duct. D. proximal convoluted tubule. E. loop of the nephron (loop of Henle). The ureters conduct urine from the kidneys to the bladder. The urethra conducts urine from the bladder out of the body.
Bloom's Level: 1. Remember Learning Outcome: 16.01.02 Describe the role of the urinary system in homeostasis. Section: 16.01 Topic: Excretory/Urinary System
True / False Questions 2. The wall of the ureter has three layers: an inner mucosa, a smooth muscle layer, and an outer fibrous coat of connective tissue. TRUE These are the three layers of the wall of the ureter.
Bloom's Level: 1. Remember Learning Outcome: 16.01.02 Describe the role of the urinary system in homeostasis. Section: 16.01 Topic: Human Urinary System
16-1 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
3. On average, a healthy adult human produces 10–20 liters of urine per day. FALSE On average, only 1–2 liters of urine is produced per day.
Bloom's Level: 1. Remember Learning Outcome: 16.01.02 Describe the role of the urinary system in homeostasis. Section: 16.01 Topic: Human Urinary System
Multiple Choice Questions 4. How many openings does the bladder have? A. one B. two C. three D. four E. five The bladder has three openings: two for the ureters and one for the urethra.
Bloom's Level: 1. Remember Learning Outcome: 16.01.02 Describe the role of the urinary system in homeostasis. Section: 16.01
True / False Questions 5. Each kidney has about one thousand structural units called nephrons. FALSE Each kidney is composed of over one million nephrons.
Bloom's Level: 1. Remember Learning Outcome: 16.02.01 Identify the structures of a human kidney. Section: 16.02 Topic: Excretory/Urinary System
16-2 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
6. The kidney has three regions: the renal pelvis, the renal atria, and the renal ventricle. FALSE The three regions of the kidney are the renal pelvis, the renal cortex, and the renal medulla.
Bloom's Level: 1. Remember Learning Outcome: 16.02.01 Identify the structures of a human kidney. Section: 16.02 Topic: Excretory/Urinary System
7. The inner layer of the glomerular capsule is composed of renocytes that have long cytoplasmic extensions. FALSE The inner layer of the glomerular capsule is composed of podocytes that have long cytoplasmic extensions.
Bloom's Level: 1. Remember Learning Outcome: 16.02.01 Identify the structures of a human kidney. Section: 16.02 Topic: Excretory/Urinary System
Multiple Choice Questions 8. The portion of the kidney which is continuous with the ureter is the A. renal cortex. B. renal medulla. C. glomerulus. D. renal pelvis. E. nephron. The renal pelvis becomes the ureter.
Bloom's Level: 1. Remember Learning Outcome: 16.02.01 Identify the structures of a human kidney. Section: 16.02 Topic: Human Urinary System
16-3 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
9. What capillaries are enclosed by the glomerular (Bowman's) capsule? A. peritubular B. glomerular C. loop of nephron (loop of Henle) D. collecting tubule E. pulmonary The glomerulus is surrounded by the glomerular capsule.
Bloom's Level: 1. Remember Learning Outcome: 16.02.02 Identify the parts of a nephron and state the function of each. Section: 16.02 Topic: Excretory/Urinary System
10. Which structure releases aldosterone? A. glomerular (Bowman's) capsule B. proximal convoluted tubule cells C. loop of the nephron (loop of Henle) D. collecting tubule E. adrenal cortex The adrenal cortex secretes aldosterone.
Bloom's Level: 1. Remember Learning Outcome: 16.03.02 Explain the functions of ADH, ANH, and aldosterone in homeostasis. Section: 16.03 Topic: Excretory/Urinary System
True / False Questions
16-4 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
11. There is an osmotic gradient within the tissues of the renal medulla: the concentration of salt is greater in the direction of the inner medulla. TRUE This solute gradient is responsible for the hypertonic gradient which draws water out of the descending limb and the collecting duct.
Bloom's Level: 1. Remember Learning Outcome: 16.03.01 Summarize how the kidneys maintain the water-salt balance in the body. Section: 16.03 Topic: Excretory/Urinary System
16-5 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
Multiple Choice Questions 12. The physical principle upon which kidney dialysis is based is A. active transport. B. diffusion. C. heat transfer. D. hydrogen ion transfer. E. pinocytosis. Dialysis is the diffusion of dissolved molecules through a semipermeable natural or synthetic membrane having pore sizes that allow only small molecules to pass through.
Bloom's Level: 1. Remember Learning Outcome: 16.04.02 Summarize the different methods of treating kidney failure. Section: 16.04 Topic: Human Urinary System
13. Urinary tract infections confined to the urethra are called A. urethritis. B. kidneyitis. C. cystitis. D. nephritis. E. prostatitis. Infection of the urethra is called urethritis.
Bloom's Level: 1. Remember Learning Outcome: 16.04.03 Describe three common disorders of the bladder and urethra. Section: 16.04 Topic: Human Urinary System
16-6 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
14. Urinary tract infections that involve the bladder are called A. urethritis. B. kidneyitis. C. cystitis. D. nephritis. E. prostatitis. Infections resulting in inflammation of the bladder are called cystitis.
Bloom's Level: 1. Remember Learning Outcome: 16.04.03 Describe three common disorders of the bladder and urethra. Section: 16.04 Topic: Human Urinary System
True / False Questions 15. Urethra and bladder infections are easily treated with antibiotics, but even simple kidney infections are much more difficult to treat. FALSE All three infections can be treated with antibiotics.
Bloom's Level: 1. Remember Learning Outcome: 16.04.03 Describe three common disorders of the bladder and urethra. Section: 16.04 Topic: Human Urinary System
16. Currently, the one-year survival rate following a kidney transplant is 95–98%. TRUE Patients with renal failure may undergo a kidney transplant, which is very successful, but donor organs are in short supply.
Bloom's Level: 1. Remember Learning Outcome: 16.04.02 Summarize the different methods of treating kidney failure. Section: 16.04 Topic: Human Urinary System
16-7 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
17. Urine excreted from the body should be sterile. TRUE Urine leaving the kidneys is normally free of bacteria.
Bloom's Level: 1. Remember Learning Outcome: 16.04.03 Describe three common disorders of the bladder and urethra. Section: 16.04 Topic: Human Urinary System
18. Although it sounds like science fiction, human bladders can be grown in the laboratory. TRUE Clinical trials of these human bladders grown in the lab may begin soon.
Bloom's Level: 1. Remember Learning Outcome: 16.04.02 Summarize the different methods of treating kidney failure. Section: 16.04 Topic: Human Urinary System
Multiple Choice Questions 19. What is the main reason for hemolytic uremic syndrome? A. the bacteria Escherichia coli B. increased animal protein in the diet C. accumulation of granules within the renal pelvis D. increased usage of diuretics E. All of the answer choices are reasons for hemolytic uremic syndrome. The bacteria Escherichia coli is the main reason for hemolytic uremic syndrome. Increased animal protein and an accumulation of granules in the renal pelvis are components of kidney stones. An increased usage of diuretics could lead to dehydration and an increased heart rate.
Bloom's Level: 1. Remember Learning Outcome: 16.04.01 Summarize how damage to the glomeruli/nephrons can lead to uremia and edema. Section: 16.04 Topic: Human Urinary System
16-8 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
20. Which is NOT a function of the urinary system? A. monitoring and maintaining blood pH at about 7.4 B. regulating blood pressure by regulating salt balance in the blood C. elimination of nitrogenous wastes including urea, uric acid, and creatinine D. hormone secretion to stimulate red blood cell production and regulate sodium ion levels E. production of water from oxygen and bicarbonate ions (HCO3) The urinary system does not produce water from oxygen and bicarbonate ions.
Bloom's Level: 2. Understand Learning Outcome: 16.01.01 Describe four major functions of the urinary system. Section: 16.01 Topic: Excretory/Urinary System
21. Which substance is not removed from the body via the urinary system? A. uric acid B. salt C. feces D. bicarbonate E. urea Feces are eliminated from the digestive system, not the urinary system.
Bloom's Level: 2. Understand Learning Outcome: 16.01.01 Describe four major functions of the urinary system. Section: 16.01 Topic: Excretory/Urinary System
16-9 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
22. Nitrogenous wastes are produced by A. the breakdown of complex sugars. B. salts that are not incorporated in cellular metabolism. C. production of proteins. D. breakdown of proteins. E. production of nucleic acids. Urea, the primary nitrogenous end product of metabolism in human beings, is a by-product of the breakdown of proteins to amino acids followed by amino acid metabolism.
Bloom's Level: 2. Understand Learning Outcome: 16.01.02 Describe the role of the urinary system in homeostasis. Section: 16.01 Topic: Human Urinary System
23. The buildup of the products of nucleic acid metabolism can lead to A. gout. B. creatinine. C. erythropoietin. D. renal failure. E. urethritis. The breakdown of nucleotides, such as those containing adenine and thymine, produces uric acid. Crystals of uric acid sometimes collect in the joints, producing gout.
Bloom's Level: 2. Understand Learning Outcome: 16.01.01 Describe four major functions of the urinary system. Section: 16.01 Topic: Excretory/Urinary System
16-10 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
24. Deamination of amino acids will result initially in the formation of A. ammonia. B. urea. C. glucose. D. creatinine. E. uric acid. The breakdown of amino acids in the liver releases ammonia. This is later converted into urea.
Bloom's Level: 2. Understand Learning Outcome: 16.01.01 Describe four major functions of the urinary system. Section: 16.01 Topic: Excretory/Urinary System
25. Which of the following is NOT considered to be a metabolic waste? A. CO2 formed by cellular respiration B. urea formed during amino acid metabolism C. creatinine formed in the muscles D. fiber that's not absorbed from the digestive tract E. renin from the kidneys Renin is a hormone produced by the kidneys. It is not considered a waste product.
Bloom's Level: 2. Understand Learning Outcome: 16.01.01 Describe four major functions of the urinary system. Section: 16.01 Topic: Excretory/Urinary System
16-11 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
26. Which of the following is NOT a nitrogenous waste product? A. ammonia B. urea C. uric acid D. creatinine E. bile pigments Bile pigments are not a nitrogenous, or nitrogen-containing, waste.
Bloom's Level: 2. Understand Learning Outcome: 16.01.01 Describe four major functions of the urinary system. Section: 16.01 Topic: Excretory/Urinary System
27. It is advantageous to have a bladder because it A. helps with defecation. B. is a place where white cells attack bacteria. C. stores urine to prevent constant urination. D. transports urine to the outside of the body. E. filters wastes and recycles nutrients back into the bloodstream. The bladder stores urine, which can be voided when convenient.
Bloom's Level: 2. Understand Learning Outcome: 16.01.02 Describe the role of the urinary system in homeostasis. Section: 16.01 Topic: Excretory/Urinary System
16-12 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
28. Which of the following is the main difference between the urinary system of males and females? A. Men have two urethras, while women have only one urethra. B. The urethra is shorter in females. C. Men have longer ureters than females. D. Kidneys tend to be larger in females. E. The female urethra passes through the prostate gland. In females, the urethra is shorter. It is about 4 cm long in females and 20 cm long in males.
Bloom's Level: 2. Understand Learning Outcome: 16.01.03 Summarize the functions of each organ in the urinary system. Section: 16.01 Topic: Human Urinary System
29. The urethra serves what other body system in males? A. endocrine B. lymphatic C. reproductive D. digestive E. immune The urethra also conducts sperm out of the body in males. Therefore, it is part of the reproductive system.
Bloom's Level: 2. Understand Learning Outcome: 16.01.03 Summarize the functions of each organ in the urinary system. Section: 16.01 Topic: Human Urinary System
16-13 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
30. Nervous control of urination involves what mechanism(s)? A. The brain determines need for urination from an internal clock. B. Stretch receptors in the urinary bladder directly send impulses to the sphincters to cause urination. C. Stretch receptors in the urinary bladder send impulses to the spine, causing a urinary reflex, but the brain can delay the reflex. D. Stretch receptors in the urinary bladder directly send impulses to the brain that consciously decides to stimulate sphincters to relax. E. Chemical receptors in the bladder detect urine concentration and trigger a reflex action. When the urinary bladder fills, stretch receptors send sensory nerve impulses to the spinal cord. Motor nerve impulses from the spinal cord cause the urinary bladder to contract and the sphincters to relax so that urination is possible. However, the brain is able to control this reflex, delaying urination until a suitable time.
Bloom's Level: 2. Understand Learning Outcome: 16.01.02 Describe the role of the urinary system in homeostasis. Section: 16.01 Topic: Excretory/Urinary System
True / False Questions 31. The bladder is expandable because it contains layers of elastic fibers embedded in thick layers of connective tissue. FALSE The bladder wall is expandable because it contains circular and longitudinal muscle layers.
Bloom's Level: 2. Understand Learning Outcome: 16.01.02 Describe the role of the urinary system in homeostasis. Section: 16.01 Topic: Human Urinary System
16-14 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
Multiple Choice Questions 32. Which of the following structures does not normally contain urine? A. renal artery B. kidney C. ureter D. urethra E. bladder The bladder, urethra, ureters, and kidneys should all contain urine. The renal artery should not.
Bloom's Level: 2. Understand Learning Outcome: 16.01.02 Describe the role of the urinary system in homeostasis. Section: 16.01 Topic: Excretory/Urinary System
33. The microscopic anatomical unit of excretion found in the kidney is the A. nephron. B. glomerular (Bowman's) capsule. C. glomerulus. D. alveoli. E. microvilli. The functional unit of the kidney is called the nephron or renal tubule.
Bloom's Level: 2. Understand Learning Outcome: 16.02.01 Identify the structures of a human kidney. Section: 16.02 Topic: Excretory/Urinary System
16-15 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
34. Which of these is the correct sequence of blood vessels associated with the nephron? A. renal artery → peritubular capillary network → afferent arteriole → efferent arteriole → renal vein B. efferent arteriole → glomerulus → venule → afferent arteriole → collecting duct C. afferent arteriole → glomerulus → efferent arteriole → peritubular capillary network → venule → renal vein D. afferent arteriole → glomerulus → peritubular capillary network → efferent arteriole → venule → renal vein E. renal vein → venule → efferent arteriole → peritubular capillary network → glomerulus → afferent arteriole The renal artery becomes the afferent arteriole, the glomerulus, and the efferent arteriole. This is followed by the peritubular capillary network, the venule, and the renal vein.
Bloom's Level: 2. Understand Learning Outcome: 16.02.02 Identify the parts of a nephron and state the function of each. Section: 16.02 Topic: Excretory/Urinary System
35. What is the correct order an excreted water molecule moves from glomerulus to collecting duct? A. loop of nephron → glomerular capsule → proximal convoluted tubule → distal convoluted tubule B. glomerular capsule → proximal convoluted tubule → loop of nephron → distal convoluted tubule C. glomerular capsule → proximal convoluted tubule → distal convoluted tubule → loop of nephron D. proximal convoluted tubule → glomerular capsule → loop of nephron → distal convoluted tubule E. distal convoluted tubule → glomerular capsule → proximal convoluted tubule → loop of nephron Water moves from the glomerulus to the proximal convoluted tubule, the loop of Henle, and the distal convoluted tubule, and is collected in the collecting duct.
Bloom's Level: 2. Understand Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
16-16 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
36. Which of the following statements about the glomerulus is INCORRECT? A. Blood pressure primarily accounts for filtration. B. Small molecules move from the glomerular (Bowman's) capsule into the glomerulus. C. It is composed of capillaries. D. It is connected to arterioles at both ends. E. It is surrounded by the glomerular (Bowman's) capsule. Small molecules move from the glomerulus into the glomerular capsule.
Bloom's Level: 2. Understand Learning Outcome: 16.02.02 Identify the parts of a nephron and state the function of each. Section: 16.02 Topic: Excretory/Urinary System
37. Which of the following is NOT one of the steps (processes) involved in urine formation? A. glomerular filtration B. tubular reabsorption C. creatinine production D. tubular secretion E. filtration of small molecules into glomerular filtrate Creatinine production is not part of urine production, although creatinine is a product that must be excreted. All of the others are steps in the process.
Bloom's Level: 2. Understand Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Human Urinary System
16-17 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
38. Reabsorption of water, nutrients, and required salts occurs in the what part of the nephron? A. glomerulus B. distal convoluted tubule C. proximal convoluted tubule D. collecting duct E. loop of the nephron (loop of Henle) These substances are reabsorbed by the blood at the proximal convoluted tubule.
Bloom's Level: 2. Understand Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Human Urinary System
39. Tubular secretion allows hydrogen ions, creatinine, and penicillin to be excreted in the distal convoluted tubule. How are these molecules moved across? A. fluid pressure similar to what happens in the glomerulus B. passive diffusion C. osmosis D. active transport E. reverse reabsorption These particular molecules are removed from the blood by active transport.
Bloom's Level: 2. Understand Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
16-18 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
40. Normally, concentrations of metabolically important substances, such as glucose, are A. high in glomerular filtrate, but only a trace in urine. B. low in glomerular filtrate, but high in urine. C. high in both glomerular filtrate and urine. D. low in both glomerular filtrate and urine. E. excluded from entering the excretory system altogether. Filterable blood components are found in high concentration in the glomerular filtrate, but in low concentration in the urine because they are reabsorbed in the proximal convoluted tubule.
Bloom's Level: 2. Understand Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
41. The glomerular (Bowman's) capsule A. returns material to the blood. B. filters material from the blood. C. makes urea. D. contains urine in its final state. E. reabsorbs materials not to be excreted. The glomerular capsule filters material from the blood.
Bloom's Level: 2. Understand Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
16-19 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
42. Which mechanism used to form urine requires a high blood pressure? A. glomerular filtration B. tubular reabsorption C. tubular secretion D. phagocytosis E. countercurrent mechanism Blood pressure is responsible for the filtering of the blood at the glomerular capsule.
Bloom's Level: 2. Understand Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
43. Which of the following is not normally found in the glomerular filtrate? A. water B. red blood cells C. nitrogenous wastes D. glucose and other nutrients E. salts (ions) Red blood cells are nonfilterable blood components and should not be found in the glomerular filtrate.
Bloom's Level: 2. Understand Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
16-20 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
44. Which of the following is a nonfilterable component of the blood? A. urea B. glucose C. amino acids D. albumin E. sodium chloride Plasma proteins, such as albumin, are not filtered into the glomerular capsule.
Bloom's Level: 2. Understand Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
45. The loop of the nephron (loop of Henle) A. is necessary for glucose reabsorption. B. creates a low osmotic pressure in the medulla. C. creates a high osmotic pressure in the medulla. D. creates a low osmotic pressure in the cortex. E. creates a high osmotic pressure in the cortex. The presence of the high osmotic pressure in the medulla allows for the necessary movement of molecules across the loop of Henle.
Bloom's Level: 2. Understand Learning Outcome: 16.03.01 Summarize how the kidneys maintain the water-salt balance in the body. Section: 16.03 Topic: Excretory/Urinary System
16-21 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
46. An increased amount of ADH leads to A. an increased amount of urine. B. a decreased amount of urine. C. no change in the amount of urine. D. kidney failure. E. a decreased amount of filtrate. Antidiuretic hormone (ADH) leads to a decreased amount of urine. Diuresis means an increased amount of urine, so antidiuresis means a decreased amount.
Bloom's Level: 2. Understand Learning Outcome: 16.03.02 Explain the functions of ADH, ANH, and aldosterone in homeostasis. Section: 16.03 Topic: Human Urinary System
47. Which statement is NOT true about ADH? A. It increases the permeability of the collecting duct. B. ADH increases water reabsorption. C. ADH decreases urine volume. D. It is secreted by the adrenal gland. E. Hot, sweaty days will cause more ADH release. The posterior pituitary releases ADH.
Bloom's Level: 2. Understand Learning Outcome: 16.03.02 Explain the functions of ADH, ANH, and aldosterone in homeostasis. Section: 16.03 Topic: Human Urinary System
16-22 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
48. Sodium concentration in the urine will be maintained due to the loop of the nephron (loop of Henle) exhibiting A. an active release of sodium from the descending limb. B. a passive and active release of sodium from the ascending limb. C. a passive release of sodium from the descending limb. D. the release of water from the ascending limb. E. the active transport of water. Sodium passively diffuses out of the lower portion of the ascending limb, but is actively extruded from the upper portion of the ascending limb.
Bloom's Level: 2. Understand Learning Outcome: 16.03.01 Summarize how the kidneys maintain the water-salt balance in the body. Section: 16.03 Topic: Excretory/Urinary System
49. The structure which monitors blood pressure in the kidney is called the A. juxtaglomerular apparatus. B. glomerular (Bowman's) capsule. C. juxtanephritic apparatus. D. ascending tubule. E. renal pelvis. When the blood volume and pressure are not sufficient to promote glomerular filtration, the juxtaglomerular apparatus secretes renin. Renin results in an increase in blood volume and pressure.
Bloom's Level: 2. Understand Learning Outcome: 16.03.01 Summarize how the kidneys maintain the water-salt balance in the body. Section: 16.03 Topic: Excretory/Urinary System
16-23 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
50. The overall effect of renin release will be to A. lower blood pressure. B. lower blood volume. C. convert angiotensin II into angiotensin I. D. inhibit aldosterone secretion. E. increase blood pressure for homeostasis. Renin converts angiotensinogen into angiotensin I, eventually resulting in the reabsorption of water and a blood volume and pressure increase.
Bloom's Level: 2. Understand Learning Outcome: 16.03.01 Summarize how the kidneys maintain the water-salt balance in the body. Section: 16.03 Topic: Human Urinary System
51. Which is NOT a correct step in the hormone regulation of blood pressure and reabsorption? A. When blood volume and pressure are low, the juxtaglomerular apparatus secretes renin. B. When blood pressure is high, the heart atria secrete ANH. C. High blood pressure causes the kidneys to secrete aldosterone. D. Renin leads to production of angiotensin I and II and eventual reabsorption of sodium ions and water. E. If you drink more water, the pituitary reduces ADH and more water and urine are excreted. The adrenal cortex secretes aldosterone when blood volume and pressure are too low.
Bloom's Level: 2. Understand Learning Outcome: 16.03.01 Summarize how the kidneys maintain the water-salt balance in the body. Section: 16.03 Topic: Human Urinary System
16-24 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
52. If the blood is alkaline, A. more hydrogen ions and bicarbonate ions are excreted. B. less hydrogen ions and more bicarbonate ions are reabsorbed. C. less hydrogen ions are excreted and less bicarbonate are reabsorbed. D. the kidneys reabsorb increased amounts of water. E. the diet must be low in necessary acids. If the blood is basic, hydrogen ions are not excreted, thereby lowering the pH, and bicarbonate ions are not reabsorbed, thereby lowering the pH (by not raising it).
Bloom's Level: 2. Understand Learning Outcome: 16.03.03 Describe two ways that the kidneys regulate blood pH. Section: 16.03 Topic: Human Urinary System
53. In the kidney mechanism for buffering the acidity of the blood, when would hydrogen ions be excreted and bicarbonate ions be reabsorbed? A. only when the lung system is not working B. when the blood is acidic C. when the blood is basic D. when the blood is neutral pH, which is not normal E. The kidney never excretes or reabsorbs hydrogen and bicarbonate ions. When the blood is too acidic, the kidneys need to excrete hydrogen ions, thereby increasing the pH, and reabsorb more bicarbonate ions, thereby increasing the pH.
Bloom's Level: 2. Understand Learning Outcome: 16.03.03 Describe two ways that the kidneys regulate blood pH. Section: 16.03 Topic: Human Urinary System
16-25 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
54. Ammonia in urine is NOT evidence of A. tubule cells deaminating amino acids. B. a mechanism to neutralize acidic urine. C. buffering of excess hydrogen ions in urine (NH3 + H+ → NH4+). D. the need to reabsorb more water. E. regulation of pH of body fluids. Ammonia in the urine deals with buffering and removing hydrogen ions. It has nothing to do with water reabsorption.
Bloom's Level: 2. Understand Learning Outcome: 16.03.03 Describe two ways that the kidneys regulate blood pH. Section: 16.03 Topic: Human Urinary System
55. Damage to the nephrons can lead to uremia, which is characterized by A. blood cells in the urine. B. urea in the blood. C. edema. D. ionic imbalance. E. excessive urine. Uremia is the presence of urea and other waste products in the blood.
Bloom's Level: 2. Understand Learning Outcome: 16.04.01 Summarize how damage to the glomeruli/nephrons can lead to uremia and edema. Section: 16.04 Topic: Human Urinary System
16-26 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
56. Which of the following is INCORRECTLY matched? A. prostate enlargement—decreased urethral flow B. cystitis—bladder infection C. pyelonephritis—urethral infection D. uremia—nephron damage E. kidney stones—granules in the renal pelvis Pyelonephritis is an infection of the kidneys.
Bloom's Level: 2. Understand Learning Outcome: 16.04.03 Describe three common disorders of the bladder and urethra. Section: 16.04 Topic: Human Urinary System
57. The kidney was one of the first organs to be successfully transplanted because it is so encapsulated and has limited vessels feeding into or out of it. How many major vessels must be cut in transplant surgery for one kidney? A. two: the renal artery and vein B. three: the renal artery and vein and one urethra C. three: the renal artery and vein and one ureter D. four: the renal artery and vein and the ureter and urethra E. five: the renal artery and vein, two ureters and one urethra Each kidney has one renal artery, one renal vein, and a ureter.
Bloom's Level: 3. Apply Learning Outcome: 16.01.02 Describe the role of the urinary system in homeostasis. Section: 16.01 Topic: Excretory/Urinary System
16-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
58. The prostate gland in older men may block the flow of fluid in the A. ureter. B. urethra. C. collecting tubule. D. proximal convoluted tubule. E. renal vein. The prostate gland surrounds the urethra in males. Inflammation of the prostate gland causes constriction of the urethra.
Bloom's Level: 3. Apply Learning Outcome: 16.01.03 Summarize the functions of each organ in the urinary system. Section: 16.01 Topic: Human Urinary System
59. Patients with diabetes mellitus have frequent urination and increased thirst because A. less water passes from the glomerulus to the glomerular capsule. B. more water is driven from the glomerulus to the glomerular capsule than normal. C. more salt is reabsorbed at the proximal convoluted tubule. D. an increase in sodium ions causes more nerve stimulation and triggers urination. E. increased glucose in the urine increases its osmolarity and less water is reabsorbed by blood. In diabetes mellitus, excess glucose appears in the blood. The kidneys cannot reabsorb all the glucose in the filtrate and this increases the osmolarity of the filtrate. Therefore, less water is reabsorbed into the peritubular capillary network.
Bloom's Level: 3. Apply Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Human Urinary System
16-28 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
60. Glucose is found in glomerular filtrate but NOT in the urine because A. the kidney stores glucose as glycogen. B. kidney cells require glucose because energy is needed for active transport. C. glucose is reabsorbed back into the bloodstream. D. glucose is converted to amino acids in the kidney. E. glucose molecules are too large to pass through the loop of the nephron (loop of Henle). Glucose molecules are filtered into the glomerular cavity but are reabsorbed in the proximal convoluted tubule.
Bloom's Level: 3. Apply Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
61. An antibiotic such as penicillin that is taken orally is soon excreted in the urine. Since infections last for a short period of time, what is the best strategy for overcoming this continual loss? A. Provide another oral medication that stops cellular metabolism. B. Provide a medication that prevents the filtration of all metabolites in the kidney. C. Take a much larger single dose of penicillin. D. Take continual doses of penicillin sufficient to maintain it in the bloodstream. E. Provide a medication that prevents all reabsorption in the kidneys. Penicillin is taken over several days in order to maintain a sufficient level in the bloodstream in spite of it being actively excreted.
Bloom's Level: 3. Apply Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
16-29 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
True / False Questions 62. Urine contains substances that underwent pressure filtration and substances that underwent tubular secretion. TRUE Both items that were filtered under pressure in the glomerular capsule and actively or passively transported back into the tubules make up the urine.
Bloom's Level: 3. Apply Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
Multiple Choice Questions 63. Why do physicians prescribe diuretic drugs to patients with high blood pressure? A. Diuretic drugs decrease urine flow and the more fluid and watery blood is easier to pump. B. Diuretic drugs increase urine flow and decrease the blood volume that must be pumped. C. Diuretic drugs decrease urine flow and the kidney does not have to work as hard. D. Diuretic drugs increase urine flow and the loss of nitrogenous wastes helps the heart. E. Diuretic drugs increase urine flow and salt loss. Salt is bad for the heart. Diuretics increase urine volume, thereby decreasing blood volume and blood pressure.
Bloom's Level: 3. Apply Learning Outcome: 16.03.02 Explain the functions of ADH, ANH, and aldosterone in homeostasis. Section: 16.03 Topic: Human Urinary System
16-30 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
64. "Artificial kidneys" do not function in all respects like a natural kidney. What is one inherent problem and what is its practical solution? A. Artificial systems are less efficient; run the artificial kidney 24 hours a day. B. Artificial kidneys are complex and must be operated at hospitals; train home care nurses. C. Dialysis uses passive diffusion and some kidney functions require selective active transport; restrict diet to limit the variety of waste molecules. D. Artificial kidneys treat the blood directly, requiring IV tubing, etc.; invent an alternative to avoid the bloodstream. E. Blood coagulates any time it leaves the body; dilute the blood. Dialysis uses passive diffusion, but real kidneys use passive diffusion as well as active transport. Molecules that need to be actively transported should be avoided in the diet.
Bloom's Level: 3. Apply Learning Outcome: 16.04.02 Summarize the different methods of treating kidney failure. Section: 16.04 Topic: Human Urinary System
65. Which of the following would be an indication of kidney failure? A. urea in the urine B. salts in the urine C. uric acid in the urine D. large amount of protein in the urine E. dilute urine Protein should not be in the urine, so a large amount of protein in the urine would indicate kidney damage.
Bloom's Level: 3. Apply Learning Outcome: 16.04.01 Summarize how damage to the glomeruli/nephrons can lead to uremia and edema. Section: 16.04 Topic: Human Urinary System
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Chapter 16 - Urinary System and Excretion
66. Jackie's ankles are very swollen at night after work. When she went to the doctor about the problem, the doctor requested a urine specimen. Why? A. Edema could be a sign of kidney failure. A urine sample might contain protein or urea. B. Jackie is probably drinking too much water and the doctor can see this because the urine will be very dilute. C. Jackie probably stands all day long and this prevents the urinary system from working correctly. D. Edema is a sign of kidney stones and the doctor is looking for small stones in the urine. E. Edema could be a sign of an inactive bladder. The doctor wants to see if Jackie is capable of urinating. Edema, the accumulation of fluid in body tissues, could be a sign of kidney damage. Other signs would be the presence of proteins, red or white blood cells, or urea and other waste products in the urine.
Bloom's Level: 3. Apply Learning Outcome: 16.04.03 Describe three common disorders of the bladder and urethra. Section: 16.04 Topic: Human Urinary System
True / False Questions 67. Smoking may increase your risk of lung cancer, but it does not affect the risk of bladder cancer. FALSE It is thought that by-products of tobacco are secreted in the urine, so it is estimated that smoking increases the risk of bladder cancer five-fold.
Bloom's Level: 3. Apply Learning Outcome: 16.04.03 Describe three common disorders of the bladder and urethra. Section: 16.04 Topic: Human Urinary System
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Chapter 16 - Urinary System and Excretion
Multiple Choice Questions 68. One morning when Jon is urinating, he experiences an incredible stabbing pain in his groin. What is most likely causing this pain? A. a bladder infection B. kidney failure C. inflammation of the urethra D. bladder cancer E. a kidney stone, perhaps stuck in the urethra Kidney stones are known to cause incredible pain, especially when they lodge in the urethra.
Bloom's Level: 3. Apply Learning Outcome: 16.04.03 Describe three common disorders of the bladder and urethra. Section: 16.04 Topic: Human Urinary System
69. Which of the following does not contribute to homeostasis? A. excretion of metabolic wastes B. maintenance of water-salt balance C. maintenance of acid-base balance D. excretion of proteins and nucleotides E. secretion of hormones Excretion of proteins and nucleotides would indicate that the body was not in homeostasis. The other choices are all functions of the urinary system that contribute to homeostasis.
Bloom's Level: 4. Analyze Learning Outcome: 16.01.01 Describe four major functions of the urinary system. Section: 16.01 Topic: Excretory/Urinary System
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Chapter 16 - Urinary System and Excretion
True / False Questions 70. The cells of the distal convoluted tubules have numerous mitochondria to support active transport. TRUE Active transport of particular molecules occurs at the distal convoluted tubules. Active transport requires energy provided by mitochondria.
Bloom's Level: 4. Analyze Learning Outcome: 16.02.02 Identify the parts of a nephron and state the function of each. Section: 16.02 Topic: Excretory/Urinary System
Multiple Choice Questions 71. The concentration of protein in the blood is higher in postglomerular blood as compared with that in arterial blood because A. protein is digested in the glomerular (Bowman's) capsule. B. the kidney manufactures protein. C. reabsorption of protein occurs. D. as water passes into the capsule, the concentration of protein in the blood will increase. E. water is a small molecule. Protein is not filtered into the glomerular capsule, but water is. Removing water results in a higher concentration of protein in the remaining blood postglomerulus.
Bloom's Level: 4. Analyze Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
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Chapter 16 - Urinary System and Excretion
72. The liquid that collects in the cavity of the glomerular (Bowman's) capsule is A. concentrated urine. B. blood plasma minus blood proteins. C. old bile ready for excretion. D. glycogen and water. E. albumin. Water, nitrogenous wastes, nutrients, and salts filter from the blood into the glomerular cavity. This is essentially blood plasma without the plasma proteins.
Bloom's Level: 4. Analyze Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
73. Normally, concentrations of wastes, such as urea, are A. high in urine in comparison to plasma. B. high in the filtrate but low in urine. C. in trace amounts in urine but are high in blood. D. not in the filtrate but are present in low concentrations in the urine. E. only found in the excretory organs where they originate from metabolism. Wastes such as urea are filtered from the blood in the glomerular capsule and are not reabsorbed. Therefore, they are in a much higher concentration in urine than in the blood.
Bloom's Level: 4. Analyze Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
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Chapter 16 - Urinary System and Excretion
74. How can urine contain a higher concentration of wastes (be hypertonic) than the blood? A. Urine cannot since the process is passive diffusion. B. Reabsorption of water in the loop of the nephron and collecting ducts leaves behind a more concentrated urine. C. All stages in the kidney involve active transport. D. Urine can be hypertonic for small molecules if it is hypotonic for an equal number of big molecules. E. The whole nephron is impermeable to water. As water is reabsorbed back into the bloodstream, the urine becomes more concentrated. This leaves urine that is hypertonic to blood.
Bloom's Level: 4. Analyze Learning Outcome: 16.03.01 Summarize how the kidneys maintain the water-salt balance in the body. Section: 16.03 Topic: Excretory/Urinary System
75. Which of the following statements about the role of the respiratory system in maintenance of blood pH is not true? A. Increasing the breathing rate rids the body of hydrogen ions because more carbon dioxide is breathed out. B. The respiratory center in the medulla oblongata increases the breathing rate if the hydrogen ion concentration of the blood rises. C. The lungs are able to rid the body of a wide variety of acidic and basic substances, thereby directly adjusting the pH of the blood. D. Breathing readjusts the proportion of carbonic acid and bicarbonate ions in the blood. E. When the buffer system of the blood is overwhelmed by an acidic or basic substance, the respiratory center adjusts the breathing rate to compensate. Only the kidneys are able to rid the body of a wide variety of acidic and basic substances, thereby directly adjusting the pH of the blood.
Bloom's Level: 4. Analyze Learning Outcome: 16.03.03 Describe two ways that the kidneys regulate blood pH. Section: 16.03 Topic: Human Urinary System
16-36 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
76. Urine and feces are both considered excretions or wastes. How do they biologically compare? A. Undigested food passes through as feces; urine is just unused or surplus water. B. Although they are processed by separate systems, feces and urine are bacteria-laden and unsanitary. C. Feces is undigested food plus bacteria that was never inside the tissue; urine carries metabolic wastes from cell metabolism but should be bacteria-free. D. Both feces and urine were molecules that were, for a short time, part of cell metabolism. E. Both are ways of excreting excess water from the body. Although both are considered waste products, their origins are very different. Feces is undigested food plus bacteria that were never part of cell metabolism and do not contain much water. Urine is metabolic waste from cell metabolism. It contains water and wastes but should not contain bacteria.
Bloom's Level: 5. Evaluate Learning Outcome: 16.01.01 Describe four major functions of the urinary system. Section: 16.01 Topic: Excretory/Urinary System
77. Since the proximal convoluted tubule reabsorbs molecules by diffusion, what structure would you expect to find at the microscopic level? A. microvilli to increase surface area B. smooth muscle to stretch as the volume inside increases C. numerous mitochondria to allow active transport D. loose connective tissue so that liquids can pass through the cells E. many neurons to stimulate the cells to reabsorb Cells that are specialized for reabsorption generally have numerous tightly packed microvilli to increase the surface area.
Bloom's Level: 5. Evaluate Learning Outcome: 16.02.02 Identify the parts of a nephron and state the function of each. Section: 16.02 Topic: Excretory/Urinary System
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Chapter 16 - Urinary System and Excretion
78. Where would the blood be most viscous (least fluid) as it passes through the kidney? A. before the glomerulus B. after passing through the glomerulus but before reaching the loop of nephron C. after the loop of nephron D. equally viscous at all points in the human body, including the kidney nephrons E. in the renal vein After passing through the glomerulus, the blood has lost water and is at its most viscous. As it proceeds through the loop of Henle, water is reabsorbed as needed, diluting the blood.
Bloom's Level: 5. Evaluate Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
79. Albumin is the large protein molecule found in egg white and blood plasma. Table salt dissociates into sodium and chloride ions. Urea is assembled in the liver. In a healthy person, what is the fate of the majority of these molecules as they pass through a nephron? A. All three remain in the bloodstream. B. All three pass across to the glomerular capsule and are excreted in urine. C. All three pass across to the glomerular capsule, but the salt ions are mostly reabsorbed. D. Albumin remains in the blood; salt ions and urea are passed across and excreted. E. Albumin remains in the blood, salt ions pass across but are reabsorbed, and the urea is passed across and excreted. Albumin is a plasma protein, so it does not get filtered into the glomerular capsule. Salt ions are filtered, but they are reabsorbed in the proximal convoluted tubule. Urea, a waste, is filtered and excreted.
Bloom's Level: 5. Evaluate Learning Outcome: 16.02.03 Distinguish between glomerular filtration, tubular reabsorption, and tubular secretion. Section: 16.02 Topic: Excretory/Urinary System
16-38 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 16 - Urinary System and Excretion
Short Answer Questions 80. List, in order, the structures of the nephron that will be involved in the formation of urine.
Glomerular capsule (Bowman's capsule) - proximal convoluted tubule - loop of the nephron (loop of Henle) - distal convoluted tubule - collecting ducts
Bloom's Level: 6. Create Learning Outcome: 16.02.02 Identify the parts of a nephron and state the function of each. Section: 16.02 Topic: Excretory/Urinary System
81. Describe the four major functions of the urinary system. 1. Excretion of metabolic wastes. The kidneys excrete metabolic wastes from the body. 2. Osmoregulation. The kidneys maintain the proper balance of water and salt in the blood as well as maintain blood volume. 3. Regulation of acid-base balance. The kidneys monitor the blood and keep the blood pH around 7.4 by excreting hydrogen ions (H+). 4. Secretion of hormones. The kidneys assist the endocrine system in hormone secretion. They release renin, which leads to the secretion of aldosterone.
Bloom's Level: 6. Create Learning Outcome: 16.01.01 Describe four major functions of the urinary system. Section: 16.01 Topic: Excretory/Urinary System
82. Explain the role of ANH (atrial natriuretic hormone) in homeostasis. ANH is secreted by the atria of the heart when cardiac cells are stretched due to an increase in blood volume. ANH inhibits the secretion of renin by the juxtaglomerular apparatus and the secretion of aldosterone by the adrenal cortex. This will in turn promote the excretion of Na+.
Bloom's Level: 6. Create Learning Outcome: 16.03.02 Explain the functions of ADH, ANH, and aldosterone in homeostasis. Section: 16.03 Topic: Excretory/Urinary System
16-39 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
Chapter 17 Nervous System
Multiple Choice Questions 1. Dendrites A. may be several feet in length. B. are always myelinated. C. are found only in the CNS. D. are solely responsible for nervous conduction. E. carry impulses toward a cell body. Dendrites carry impulses toward the cell body of a neuron.
Bloom's Level: 1. Remember Learning Outcome: 17.01.02 Describe the basic structure of a neuron, and compare the functions of the three classes of neurons. Section: 17.01 Topic: Neurons
2. In the axon, the nerve impulses travel A. toward the cell body. B. away from the cell body. C. bidirectionally toward AND away from the cell body. D. away from the synapse. E. toward the dendrites. The axon conducts nerve impulses away from the cell body toward other neurons or effectors.
Bloom's Level: 1. Remember Learning Outcome: 17.01.02 Describe the basic structure of a neuron, and compare the functions of the three classes of neurons. Section: 17.01 Topic: Neurons
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Chapter 17 - Nervous System
3. The sodium-potassium pump is primarily responsible for the A. resting potential. B. action potential. C. excretion of salts. D. contraction of muscle fibers. E. maintenance of isotonic water balance. The sodium-potassium pump actively transports Na+ out of the axon and K+ into the axon. This causes a charge difference across the axonal membrane, which makes the axon inside negative compared to the outside. This is called the resting potential.
Bloom's Level: 1. Remember Learning Outcome: 17.02.01 Summarize the activities that generate and propagate an action potential. Section: 17.02 Topic: Action Potential/Nerve Impulse
4. A nerve impulse is caused by A. the movement of a sodium ion all the way from dendrite to axon tip. B. the movement of a potassium ion all the way from dendrite to axon tip. C. the movement of an electron all the way from dendrite to axon tip. D. a change in the difference in positive and negative ions on the outer and inner surfaces of the neuron membrane, a change that opens adjacent channels and propagates its flow. E. a change in the difference in sugar molecules on the outer and inner surfaces of the neuron membrane that opens adjacent channels. A nerve impulse is an electrical signal resulting from a change in polarity across an axonal membrane. This difference in positive and negative charges on the outer and inner surfaces of the neuron membrane causes the impulse to flow down the neuron.
Bloom's Level: 1. Remember Learning Outcome: 17.02.04 Explain the role of neurotransmitters in the transmission of a nerve impulse across a synapse. Section: 17.02 Topic: Action Potential/Nerve Impulse
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Chapter 17 - Nervous System
5. The summing up of excitatory and inhibitory signals in a neuron is called A. repolarization. B. inhibition. C. synapse. D. synaptic integration. E. sympathetic response. Synaptic integration is the summing up of excitatory and inhibitory signals received by the postsynaptic cell.
Bloom's Level: 1. Remember Learning Outcome: 17.02.03 Define synaptic integration. Section: 17.02 Topic: Synapse and Neurotransmitters
Figure 17.4 Learning Outcome: 17.07.01 Describe two abnormalities seen in the brain of Alzheimer disease patients.
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Chapter 17 - Nervous System
6. In Figure 17.4 of an action potential, number 1 indicates the movement of A. Na+ to the inside. B. Na+ to the outside. C. K+ to the inside. D. K+ to the outside. E. Cl- to the outside. Number 1 indicates depolarization due to Na+ moving to the inside of the axon.
Bloom's Level: 1. Remember Figure: 17.04 Learning Outcome: 17.02.02 Describe saltatory conduction. Section: 17.02 Topic: Action Potential/Nerve Impulse
7. In Figure 17.4, number 3 represents A. depolarization. B. repolarization. C. action potential. D. threshold. E. resting potential. Number 3 represents the action potential as visualized by voltage changes.
Bloom's Level: 1. Remember Figure: 17.04 Learning Outcome: 17.02.02 Describe saltatory conduction. Section: 17.02 Topic: Action Potential/Nerve Impulse
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Chapter 17 - Nervous System
8. The enzyme that breaks down acetylcholine within the synaptic cleft is A. acetylcholinesterase. B. monoamine oxidase. C. GABA. D. lipase. E. maltase. Acetylcholinesterase (AChE) breaks down acetylcholine in the postsynaptic membrane.
Bloom's Level: 1. Remember Learning Outcome: 17.02.04 Explain the role of neurotransmitters in the transmission of a nerve impulse across a synapse. Section: 17.02 Topic: Synapse and Neurotransmitters
9. The membranes that protect the brain and spinal cord are called A. cerebrospinal membranes. B. meninges. C. ventricles. D. epithelium. E. gray matter. The protective membranes of the brain and spinal cord are known as meninges (sing., meninx).
Bloom's Level: 1. Remember Learning Outcome: 17.03.01 Understand the relationship between the CNS and PNS in the nervous system. Section: 17.03 Topic: Central Nervous System
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Chapter 17 - Nervous System
10. What are the interconnecting spaces that produce and serve as a reservoir for cerebrospinal fluid in the brain called? A. meninges B. ventricles C. corpus callosum D. insula E. ganglia The brain has hollow interconnecting cavities termed ventricles that produce and serve as a reservoir for cerebrospinal fluid.
Bloom's Level: 1. Remember Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
11. The primary functions of the spinal cord involve A. intelligence and memory. B. speech, taste, smell, vision, and hearing. C. reflex actions and communication between the brain and peripheral nerves. D. controlling muscle activity and maintaining balance. E. local control and decision making for local anatomy. The spinal cord provides a means of communication between the brain and the peripheral nerves that leave the cord. The spinal cord is also the center for thousands of reflex arcs.
Bloom's Level: 1. Remember Learning Outcome: 17.03.01 Understand the relationship between the CNS and PNS in the nervous system. Section: 17.03 Topic: Central Nervous System
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Chapter 17 - Nervous System
12. Which part of the brain contains centers for the heartbeat and respiration? A. medulla oblongata B. hypothalamus C. cerebellum D. cerebrum E. pons The medulla oblongata contains a number of reflex centers for regulating heartbeat, breathing, and blood pressure.
Bloom's Level: 1. Remember Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
13. Which part of the brain is used to integrate incoming information and send it to the appropriate portion of the cerebrum? A. hypothalamus B. cerebellum C. thalamus D. limbic system E. basal nuclei This is the role of the thalamus.
Bloom's Level: 1. Remember Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
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Chapter 17 - Nervous System
14. Consciousness is best associated with the A. whole brain. B. frontal lobe only. C. cerebrum. D. whole central nervous system. E. cerebellum. The cerebral cortex is the region of the brain that accounts for sensation, voluntary movement, and all the thought processes we associate with consciousness.
Bloom's Level: 1. Remember Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
15. Which lobe of the cerebrum is responsible for vision? A. frontal B. parietal C. temporal D. occipital E. insular The occipital lobe contains the primary visual area and the visual association area.
Bloom's Level: 1. Remember Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
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Chapter 17 - Nervous System
True / False Questions 16. The cerebrum is the largest portion of the brain in humans. TRUE The cerebrum is larger than the other parts of the brain.
Bloom's Level: 1. Remember Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
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Chapter 17 - Nervous System
Multiple Choice Questions 17. What are the two parts of the diencephalon? A. the cerebellum and the cerebrum B. the hypothalamus and the thalamus C. the midbrain and the pons D. the third and fourth ventricles E. the medulla oblongata and the brain stem The hypothalamus and the thalamus compose the diencephalon.
Bloom's Level: 1. Remember Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
18. Language is NOT dependent on A. memory. B. the thalamus. C. Broca's area. D. Wernicke's area. E. the spinal cord. The spinal cord does not play a role in language development.
Bloom's Level: 1. Remember Learning Outcome: 17.04.02 Summarize how the limbic system is involved in memory, language, and speech. Section: 17.04 Topic: Central Nervous System
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Chapter 17 - Nervous System
19. Which part of the brain is a seahorse-shaped structure deep in the temporal lobe? A. ganglia B. medulla C. corpus callosum D. hippocampus E. occipital area The hippocampus is a seahorse-shaped structure deep in the temporal lobe.
Bloom's Level: 1. Remember Learning Outcome: 17.04.01 Identify the structures of the limbic system. Section: 17.04 Topic: Central Nervous System
20. A nerve is A. a neuron. B. composed of sensory axons and motor dendrites. C. composed of the long fibers of long axons. D. a part of the central nervous system. E. any cell located in the brain or spinal region. In the peripheral nervous system, a nerve is composed of bundles of axons.
Bloom's Level: 1. Remember Learning Outcome: 17.05.01 Describe the overall anatomy of the peripheral nervous system, including the cranial nerves and spinal nerves. Section: 17.05 Topic: Peripheral Nervous System
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Chapter 17 - Nervous System
21. A reflex A. is an automatic, involuntary response. B. does not require the central nervous system. C. is normally controlled consciously. D. has no protective value. E. involves only sensory neurons. A reflex is a programmed, built-in circuit that allows for protection and survival. It is involuntary and involves the spinal cord.
Bloom's Level: 1. Remember Learning Outcome: 17.05.01 Describe the overall anatomy of the peripheral nervous system, including the cranial nerves and spinal nerves. Section: 17.05 Topic: Nervous System
22. Which part of a simple reflex takes the message away from the CNS? A. sensory neuron B. receptor C. interneuron D. motor neuron E. effector The motor neuron takes the information from the spinal cord to the effector muscle. The sensory neuron takes the information from the sensory receptor to the spinal cord.
Bloom's Level: 1. Remember Learning Outcome: 17.05.02 Distinguish between the somatic and autonomic divisions of the peripheral nervous system. Section: 17.05 Topic: Peripheral Nervous System
Figure 17.16
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Chapter 17 - Nervous System
23. In Figure 17.16, the structure at "a" is a(n) A. interneuron. B. sensory neuron. C. receptor. D. motor neuron. E. effector. A is the effector muscle.
Bloom's Level: 1. Remember Figure: 17.16 Learning Outcome: 17.05.01 Describe the overall anatomy of the peripheral nervous system, including the cranial nerves and spinal nerves. Learning Outcome: 17.07.01 Describe two abnormalities seen in the brain of Alzheimer disease patients. Section: 17.05 Topic: Peripheral Nervous System
24. In Figure 17.16, the structure at "c" is a(n) A. interneuron. B. sensory neuron. C. receptor. D. motor neuron. E. effector. C is an interneuron that connects the sensory neuron with the motor neuron.
Bloom's Level: 1. Remember Figure: 17.16 Learning Outcome: 17.05.01 Describe the overall anatomy of the peripheral nervous system, including the cranial nerves and spinal nerves. Section: 17.05 Topic: Peripheral Nervous System
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Chapter 17 - Nervous System
25. Heroin binds to receptors on neurons that travel from the spinal cord to the region of the brain that feels pleasure. These receptors are normally meant for A. epinephrine. B. norepinephrine. C. dopamine. D. endorphins. E. acetylcholine. Heroin binds to receptors meant for the endorphins, naturally occurring neurotransmitters.
Bloom's Level: 1. Remember Learning Outcome: 17.06.01 Explain the specific mechanisms whereby the most common illicit drugs affect the brain. Section: 17.06 Topic: Nervous System
True / False Questions 26. A stroke results from a disruption of the blood supply to the brain. TRUE In hemorrhagic stroke, bleeding occurs into the brain due to leakage from small arteries and in an ischemic stroke, there is a sudden loss of blood supply to an area of the brain.
Bloom's Level: 1. Remember Learning Outcome: 17.07.02 Compare the mechanism causing the following diseases: Parkinson disease, MS, stroke, meningitis, and prion disease. Section: 17.07 Topic: Nervous System
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Chapter 17 - Nervous System
27. Both meningitis and kuru involve infectious agents. TRUE Meningitis is the result of a bacterial or viral infection, while kuru is the result of a prion infection.
Bloom's Level: 1. Remember Learning Outcome: 17.07.02 Compare the mechanism causing the following diseases: Parkinson disease, MS, stroke, meningitis, and prion disease. Section: 17.07 Topic: Nervous System
28. Myasthenia gravis can be treated with immunosuppressive drugs, although there is no cure. TRUE MG patients often respond very well to treatment with immunosuppressive drugs.
Bloom's Level: 1. Remember Learning Outcome: 17.07.03 Describe the symptoms seen in several disorders of the brain and spinal cord and discuss why it is difficult to find cures for each. Section: 17.07 Topic: Nervous System
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Chapter 17 - Nervous System
Multiple Choice Questions 29. Which description best fits the central nervous system (CNS)? A. consists of the brain and spinal cord, located in the midline of the body B. consists of the brain and spinal cord, connecting the muscles and the brain C. consists of the spinal cord, located in the midline of the body D. onsists of the brain and spinal cord, located on the edges of the body E. transmits sensory impulses to the PNS and motor impulses to the brain The central nervous system (CNS) consists of the brain and spinal cord, located in the midline of the body.
Bloom's Level: 1. Remember Learning Outcome: 17.01.01 Compare and contrast the location and function of the central nervous system and the peripheral nervous system. Section: 17.01 Topic: Central Nervous System
30. Which disease is associated with mild tremors in one or more limbs, abnormally high levels of glutamate in the cerebrospinal fluid, and onset between the ages of 40 and 60? A. amyotrophic lateral sclerosis B. Guillain-Barre syndrome C. myasthenia gravis D. multiple sclerosis E. Parkinson disease These are all symptoms of amyotrophic lateral sclerosis (ALS).
Bloom's Level: 1. Remember Learning Outcome: 17.07.03 Describe the symptoms seen in several disorders of the brain and spinal cord and discuss why it is difficult to find cures for each. Section: 17.07 Topic: Nervous System
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Chapter 17 - Nervous System
31. Which types of cells are found only in the central nervous system? A. interneurons B. sensory neurons C. motor neurons D. interneurons and sensory neurons E. interneurons and motor neurons Interneurons are found strictly in the central nervous system. Sensory neurons take messages to the CNS, while the motor neurons take messages away from the CNS to an effector. Both are located in the peripheral nervous system.
Bloom's Level: 1. Remember Learning Outcome: 17.01.01 Compare and contrast the location and function of the central nervous system and the peripheral nervous system. Section: 17.01 Topic: Central Nervous System
32. Which description best fits the peripheral nervous system (PNS)? A. nerves that carry sensory messages to the CNS and motor commands to the muscles and glands B. nerves that carries motor commands to the CNS and sensory messages to the muscles and glands C. consists of the brain and spinal cord, located in the midline of the body D. consists of the brain and spinal cord, located on the peripheral aspect of the body E. None of the answers are descriptions of the Peripheral Nervous System (PNS). The peripheral nervous system (PNS) consists of nerves that carry sensory messages to the CNS and motor commands to the muscles and glands.
Bloom's Level: 1. Remember Learning Outcome: 17.01.01 Compare and contrast the location and function of the central nervous system and the peripheral nervous system. Section: 17.01 Topic: Peripheral Nervous System
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Chapter 17 - Nervous System
33. Which structures are associated with the limbic system? A. cerebral lobes, basal nuclei, and diencephalon B. cerebral lobes, basal nuclei, and spinal cord C. cerebral lobes, peripheral nerves, and diencephalon D. cerebellum, basal nuclei, and diencephalon E. spinal cord, basal nuclei, and diencephalon The structures associated with the limbic system include the cerebral lobes, basal nuclei, and diencephalon.
Bloom's Level: 1. Remember Learning Outcome: 17.04.01 Identify the structures of the limbic system. Section: 17.04 Topic: Nervous System
34. Which body part is associated with only the primary somatosensory area? A. genitals B. pharynx C. tongue D. face E. thumb The genitals are associated with only the primary somatosensory area. The rest of the body parts are associated with both the primary motor area and primary somatosensory area.
Bloom's Level: 1. Remember Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
17-18 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
35. Which structure communicates with the prefrontal area of the brain and is involved in learning and memory? A. hippocampus B. amygdala C. pineal gland D. thalamus E. hypothalamus The hippocampus communicates with the prefrontal area of the brain and is involved in learning and memory.
Bloom's Level: 1. Remember Learning Outcome: 17.04.02 Summarize how the limbic system is involved in memory, language, and speech. Section: 17.04 Topic: Central Nervous System
36. What is the definition of learning? A. ability to retain and utilize past memories B. ability to hold a thought in mind or recall events from the past C. ability to forget unfavorable experiences D. ability to store information E. ability to retain and form memories Learning is the ability to retain and utilize past memories. The rest of the choices are types of memories.
Bloom's Level: 1. Remember Learning Outcome: 17.04.02 Summarize how the limbic system is involved in memory, language, and speech. Section: 17.04 Topic: Nervous System
17-19 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
37. Which type of memory is associated with specific facts, persons, and events? A. episodic B. skill C. semantic D. short term E. All of the answer choices are associated with specific facts, persons, and events. Episodic memory is associated with specific facts, persons, and events. Skill memory is involved with performing motor activities. Semantic memory is associated with ideas, concepts, and meanings. Short-term memory is is associated with items that were recently experienced.
Bloom's Level: 1. Remember Learning Outcome: 17.04.03 Distinguish between short-term, long-term, semantic, episodic, and skill memory. Section: 17.04 Topic: Nervous System
38. Due to experimentation, it was discovered that which structure was involved in the conversion of short-term memories to long-term memories? A. hippocampus B. diencephalon C. amygdala D. brain stem E. cerebellum The hippocampus is involved in the conversion of short-term memories to long-term memories.
Bloom's Level: 1. Remember Learning Outcome: 17.04.04 Explain how certain diseases, accidents, and experiments have helped scientists understand some basic components of how memories are made. Section: 17.04 Topic: Central Nervous System
17-20 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
39. Which of the following drugs is a powerful CNS stimulant? A. methamphetamine B. heroin C. marijuana D. alcohol E. All of the answer choices are powerful CNS stimulants. Methamphetamine is a powerful CNS stimulant. Heroin kills pain and produces a feeling of tranquility. Marijuana produces mild euphoria along with alterations in vision and judgment. Alcohol produces a feeling of relaxation.
Bloom's Level: 1. Remember Learning Outcome: 17.06.02 Classify illicit drugs as to whether they have a depressant, stimulant, or psychoactive effect on the body. Section: 17.06 Topic: Nervous System
40. Which is NOT a correct association of structure and function? A. axons—outgoing signals B. sensory neuron—controls sensory organs such as eyes C. cell body—nucleus and organelles D. dendrites—incoming signals E. interneuron—sums up input before sending signals to muscle or gland Sensory neurons take messages to the CNS.
Bloom's Level: 2. Understand Learning Outcome: 17.01.02 Describe the basic structure of a neuron, and compare the functions of the three classes of neurons. Section: 17.01 Topic: Action Potential/Nerve Impulse Topic: Nervous System
17-21 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
41. In dissection, most nerve fibers appear gray to white because A. the neuron is gray to white colored. B. sodium ions are part of the salt compound and they flow along the surface. C. they rapidly die and dead tissues soon become gray or white. D. they are mostly myelin sheath made of lipid or fat molecules. E. they consume large amounts of energy in the form of sugar. The myelin sheath gives nerve fibers their white, glistening appearance.
Bloom's Level: 2. Understand Learning Outcome: 17.01.03 Explain the structure, function, and locations of myelin. Section: 17.01 Topic: Neurons
42. Myelinated axons that run together in bundles in the CNS are A. ventricles. B. meninges. C. cortices. D. tracts. E. ganglia. Myelinated axons that run together in bundles in the CNS are called tracts and make up the white matter.
Bloom's Level: 2. Understand Learning Outcome: 17.01.03 Explain the structure, function, and locations of myelin. Section: 17.01 Topic: Neurons
17-22 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
43. Which of the following does NOT pertain to neurons? A. sodium pumps B. Schwann cells C. myelin D. nodes of Ranvier E. cell walls Neurons do not have cell walls.
Bloom's Level: 2. Understand Learning Outcome: 17.01.02 Describe the basic structure of a neuron, and compare the functions of the three classes of neurons. Section: 17.01 Topic: Neurons
44. Which of the following is NOT true about the myelin sheath? A. It is composed of layers of cellular membrane, containing myelin around nerve fibers. B. It gives nerve fibers their white, glistening appearance. C. It provides a pathway for new fiber growth if the axon is severed. D. It decreases the speed of nerve impulse conduction. E. It is formed from Schwann cells. The myelin sheath increases the speed of nerve impulse conduction.
Bloom's Level: 2. Understand Learning Outcome: 17.01.03 Explain the structure, function, and locations of myelin. Section: 17.01 Topic: Neurons
17-23 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
45. Since the membrane is more permeable to K+ than to Na+, A. there are always more positive ions outside the membrane than inside. B. there are always more negative ions outside the membrane than inside. C. the outside of the membrane is always negative. D. the outside of the membrane is always neutral. E. there really is never a resting potential. The sodium-potassium pumps actively transport Na+ out of and K+ into the axon. Because the membrane is more permeable to K+ than to Na+, there are always more positive ions outside the membrane than inside.
Bloom's Level: 2. Understand Learning Outcome: 17.02.01 Summarize the activities that generate and propagate an action potential. Section: 17.02 Topic: Action Potential/Nerve Impulse
46. The difference between a weak stimulus and an intense stimulus is A. the action potential is gradated and a weak stimulus causes a small change in polarity. B. weak stimuli only open sodium gates, and strong stimuli also open potassium gates. C. the axon fires at a greater frequency for an intense stimulus. D. a strong stimulus does not allow repolarization, but sends a constant flow of ions. E. an intense stimulus causes more sodium ions to diffuse into the cell. An action potential is an all-or-none phenomena. If the stimulus causes the axonal membrane to depolarize to the threshold, an action potential occurs. The strength of an action potential does not change, but an intense stimulus can cause an axon to start an axon potential more often in a given time interval than a weak stimulus.
Bloom's Level: 2. Understand Learning Outcome: 17.02.01 Summarize the activities that generate and propagate an action potential. Section: 17.02 Topic: Action Potential/Nerve Impulse
17-24 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
47. Which statement is NOT true about the development of an action potential? A. There is a rapid change in polarity from about -65 mV to about +40 mV. B. It requires two types of gated channels. C. The action potential ends when the polarity across the membrane reaches +40 mV. D. Depolarization occurs when sodium gates open and allow sodium ions to enter the cell. E. Potassium gates open after the sodium gates and allow potassium ions to leave the cell. The action potential will end when it moves down the axon and reaches the synapse.
Bloom's Level: 2. Understand Learning Outcome: 17.02.01 Summarize the activities that generate and propagate an action potential. Section: 17.02 Topic: Action Potential/Nerve Impulse
48. Saltatory conduction occurs when A. salt (or sodium) levels are highest. B. the sodium (salt) gates open. C. the resting potential fails to be reset. D. there is no synapse and the axon is continuous with the next dendrite. E. the action potential jumps from node to node. Saltatory means "to jump." Saltatory conduction means that the action potential jumps from node to node.
Bloom's Level: 2. Understand Learning Outcome: 17.02.02 Describe saltatory conduction. Section: 17.02 Topic: Action Potential/Nerve Impulse
17-25 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
49. A postsynaptic membrane A. can be a myelin sheath. B. can be a cell body membrane. C. can be an axon membrane. D. can be a dendrite membrane. E. can be all of the answer choices, except being a myelin sheath. The axon terminal connects to either a dendrite or the cell body of another neuron.
Bloom's Level: 2. Understand Learning Outcome: 17.02.03 Define synaptic integration. Section: 17.02 Topic: Synapse and Neurotransmitters
50. In order for transmission across the synapse to occur, A. synaptic vesicles fuse with the postsynaptic membrane. B. neurotransmitters are released from the postsynaptic membrane. C. synaptic vesicles fuse with the presynaptic membrane. D. neurotransmitters are actively transported from the presynaptic to the postsynaptic membrane. E. the postsynaptic membrane must be in the refractory period. The first step of transmission across a synapse is that the synaptic vesicles in presynaptic membrane fuse with the membrane.
Bloom's Level: 2. Understand Learning Outcome: 17.02.03 Define synaptic integration. Section: 17.02 Topic: Synapse and Neurotransmitters
17-26 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
51. Transmission of a nerve impulse from one neuron to another neuron is NOT dependent on A. the presence of calcium ions. B. a neurotransmitter substance. C. a synaptic cleft. D. a presynaptic and postsynaptic membrane. E. a refractory period. The refractory period is the time a nerve is resistant to firing a second time because it has not been "reset."
Bloom's Level: 2. Understand Learning Outcome: 17.02.04 Explain the role of neurotransmitters in the transmission of a nerve impulse across a synapse. Section: 17.02 Topic: Synapse and Neurotransmitters
52. The main function of the cerebrospinal fluid is A. to provide saltatory transmission. B. to provide nutrients for the neurons. C. defense of the nervous system against pathogens. D. to produce gray matter. E. to cushion and protect the central nervous system. Cerebrospinal fluid cushions and protects the CNS.
Bloom's Level: 2. Understand Learning Outcome: 17.03.02 Describe the structure and function of the spinal cord. Section: 17.03 Topic: Central Nervous System
17-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
53. Which is NOT a correct association of brain structures and functions? A. midbrain—reflex centers for visual, auditory, and tactile responses B. cerebellum—coordinates smooth and coordinated motions C. cerebrum—higher thought processes, including learning, memory, and speech D. pons—regulates breathing rate E. thalamus—regulates heartbeat and blood pressure The thalamus integrates sensory input from the visual, auditory, taste, and somatosensory systems and sends it on to the appropriate portions of the cerebrum.
Bloom's Level: 2. Understand Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
54. An impulse traveling up the spinal cord first enters the brain at the A. medulla. B. thalamus. C. hypothalamus. D. cerebellum. E. cerebrum. The spinal cord becomes the medulla in the brain stem.
Bloom's Level: 2. Understand Learning Outcome: 17.03.02 Describe the structure and function of the spinal cord. Section: 17.03 Topic: Central Nervous System
17-28 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
55. Which of the following is NOT true about the cerebrum? A. It is the largest part of the human brain. B. The gray matter is the outermost. C. It is responsible for maintenance of breathing and heart rate. D. The two cerebral hemispheres are connected by the corpus callosum. E. Association areas integrate information from different lobes. The medulla oblongata is responsible for the maintenance of breathing and heart rate.
Bloom's Level: 2. Understand Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
56. Which of the following is not part of the brain? A. the cerebrum B. the diencephalon C. the cerebellum D. the brain stem E. the meninges The meninges are not part of the brain. They are the covering of the brain.
Bloom's Level: 2. Understand Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
17-29 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
57. Which of the following is not a function of the cerebrum? A. maintains posture and balance B. higher thought processes required for learning and memory C. higher thought processes required for language and speech D. receives sensory input E. commands voluntary motor responses The cerebellum maintains posture and balance.
Bloom's Level: 2. Understand Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
58. When individuals experience fear and pleasure, what part of the brain has been stimulated? A. hypothalamus B. pons C. reticular activating system D. limbic system E. thalamus The limbic system blends primitive emotions such as rage, fear, and joy, with higher mental functions such as reason and memory. It accounts for why activities seem pleasurable.
Bloom's Level: 2. Understand Learning Outcome: 17.04.02 Summarize how the limbic system is involved in memory, language, and speech. Section: 17.04 Topic: Central Nervous System
17-30 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
True / False Questions 59. The limbic system is a separate part of the brain located behind the cerebellum on top of the brain stem. FALSE The limbic system is a complex network of tracts and nuclei that incorporates portions of the cerebral lobes, the basal nuclei, and the diencephalon.
Bloom's Level: 2. Understand Learning Outcome: 17.04.01 Identify the structures of the limbic system. Section: 17.04 Topic: Central Nervous System
Multiple Choice Questions 60. People claim that once you have learned to ride a bicycle you can always ride a bicycle, even if you haven't ridden for a long time. What type of memory is associated with this? A. short-term memory B. episodic memory C. semantic memory D. long-term memory E. skill memory Skill memory is involved in performing motor activities. After you have initially learned the activity, the actions become automatic.
Bloom's Level: 2. Understand Learning Outcome: 17.04.03 Distinguish between short-term, long-term, semantic, episodic, and skill memory. Section: 17.04 Topic: Nervous System
17-31 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
61. Which is NOT a correct association of memory type? A. skill memory—ability to perform motor activities B. long-term memory—telephone number you "know by heart" C. short-term memory—brief or temporary recall of recent events D. semantic memory—non-language memory E. episodic memory—persons or events Semantic memory involves ideas, concepts, and meanings.
Bloom's Level: 2. Understand Learning Outcome: 17.04.03 Distinguish between short-term, long-term, semantic, episodic, and skill memory. Section: 17.04 Topic: Nervous System
62. Spinal nerves contain A. only sensory fibers. B. only motor fibers. C. both sensory and motor fibers. D. only parasympathetic fibers. E. only ganglia. Spinal nerves contain a dorsal root with sensory fibers and a ventral root with motor fibers.
Bloom's Level: 2. Understand Learning Outcome: 17.05.01 Describe the overall anatomy of the peripheral nervous system, including the cranial nerves and spinal nerves. Section: 17.05 Topic: Peripheral Nervous System
17-32 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
63. Which of these is NOT true of the autonomic nervous system? A. It controls heartbeat, peristalsis, and secretion of glands. B. It is composed of sympathetic and parasympathetic systems. C. It is composed only of fibers that have an inhibitory function on various organs of the body. D. The impulses require two motor neurons to reach their destination. E. The system coordinates organ responses. The autonomic system contains stimulatory as well as inhibitory fibers.
Bloom's Level: 2. Understand Learning Outcome: 17.05.02 Distinguish between the somatic and autonomic divisions of the peripheral nervous system. Section: 17.05 Topic: Peripheral Nervous System
64. Which of the following is NOT true about the autonomic nervous system? A. It uses two motor neurons and one ganglion for each impulse. B. It uses preganglionic and postganglionic axons. C. It is a part of the peripheral nervous system. D. It functions voluntarily and consciously. E. It innervates all internal body organs. The autonomic nervous system functions automatically and usually in an involuntary manner.
Bloom's Level: 2. Understand Learning Outcome: 17.05.03 Contrast the overall functions of the sympathetic and parasympathetic divisions of the autonomic nervous system. Section: 17.05 Topic: Nervous System
17-33 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
True / False Questions 65. Most illicit drugs affect the action of the brain by interfering with the Na+-K+ pump and nerve impulses. FALSE Most illicit drugs affect the action of a particular neurotransmitter at synapses in the brain.
Bloom's Level: 2. Understand Learning Outcome: 17.06.01 Explain the specific mechanisms whereby the most common illicit drugs affect the brain. Section: 17.06 Topic: Nervous System
66. Nicotine is not considered a drug because it does not affect the body physically. FALSE Nicotine does affect the body physically because it causes a release of epinephrine from the adrenal cortex, causing a feeling of stimulation.
Bloom's Level: 2. Understand Learning Outcome: 17.06.02 Classify illicit drugs as to whether they have a depressant, stimulant, or psychoactive effect on the body. Section: 17.06 Topic: Nervous System
17-34 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
Multiple Choice Questions 67. Which of the following is the best description of the cause of multiple sclerosis (MS)? A. gradual death of brain cells due to calcium influx B. myelin sheath lesions impair normal coordination of impulses C. synaptic uptake of dopamine is inhibited D. degeneration of dopamine-releasing neurons in the brain E. excessive excitatory signals from the motor cortex Multiple sclerosis affects the myelin sheath of the neurons in the white matter of the brain.
Bloom's Level: 2. Understand Learning Outcome: 17.07.02 Compare the mechanism causing the following diseases: Parkinson disease, MS, stroke, meningitis, and prion disease. Section: 17.07 Topic: Nervous System
True / False Questions 68. The brains of Alzheimer disease patients show no abnormalities postmortem. FALSE Brains of patients with AD show plaques and neurofibrillary tangles.
Bloom's Level: 2. Understand Learning Outcome: 17.07.01 Describe two abnormalities seen in the brain of Alzheimer disease patients. Section: 17.07 Topic: Nervous System
17-35 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
Multiple Choice Questions 69. Which limbic structure allows us to respond to and display anger and also prompts the release of adrenaline into the bloodstream? A. amygdala B. hippocampus C. cerebellum D. cerebrum E. pons The amygdala allows us to respond to and display anger and also prompts the release of adrenaline into the bloodstream. The hippocampus is involved in learning and memory. The cerebellum, cerebrum, and pons are not structures associated with the limbic system.
Bloom's Level: 2. Understand Learning Outcome: 17.04.02 Summarize how the limbic system is involved in memory, language, and speech. Section: 17.04 Topic: Nervous System
70. Which brain structure serves as a bridge between the sensory association areas and the prefrontal area? A. hippocampus B. amygdala C. pons D. brain stem E. cerebellum The hippocampus serves as a bridge between the sensory association areas and the prefrontal area.
Bloom's Level: 2. Understand Learning Outcome: 17.04.04 Explain how certain diseases, accidents, and experiments have helped scientists understand some basic components of how memories are made. Section: 17.04 Topic: Central Nervous System
17-36 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
71. Which of the following is a long-term effect of heroin use? A. The body decreases the production of endorphins. B. The body increases the production of endorphins. C. a decrease in dopamine production D. a release of excess dopamine E. All of the answer choices are long-term side effects of heroin use. Heroin binds to the receptors meant for the endorphins, causing the body to decrease the natural production of endorphins. A decrease in dopamine production is associated with the continued use of cocaine. The release of excess dopamine production is associated with the use of nicotine.
Bloom's Level: 2. Understand Learning Outcome: 17.06.03 List the long-term effects of drug use on the body. Section: 17.06 Topic: Nervous System
72. What are the potential effects of heavy marijuana usage? A. hallucinations, anxiety, depression, body image distortion, and other psychotic symptoms B. a decrease in dopamine production C. The body decreases the natural production of endorphins. D. the release of excess dopamine production E. All of the answer choices are potential effects of heavy marijuana usage. Hallucinations, anxiety, depression, body image distortion, and other psychotic symptoms are associated with marijuana use. A decrease in dopamine production is associated with the continued use of cocaine. Heroin binds to the receptors meant for the endorphins, causing the body to decrease the natural production of endorphins. The release of excess dopamine production is associated with the use of nicotine.
Bloom's Level: 2. Understand Learning Outcome: 17.06.03 List the long-term effects of drug use on the body. Section: 17.06 Topic: Nervous System
17-37 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
73. What are the long-term consequences associated with the use of cocaine? A. a decrease in dopamine production B. a decrease in the natural production of endorphins C. the release of excess dopamine D. hallucinations, anxiety, depression, body image distortion, and other psychotic symptoms E. All of the answer choices are consequences associated with the use of cocaine. A decrease in dopamine production is associated with the continued use of cocaine. Heroin binds to the receptors meant for the endorphins, causing the body to decrease the natural production of endorphins. The release of excess dopamine production is associated with the use of nicotine. Hallucinations, anxiety, depression, body image distortion, and other psychotic symptoms are associated with marijuana use.
Bloom's Level: 2. Understand Learning Outcome: 17.06.03 List the long-term effects of drug use on the body. Section: 17.06 Topic: Nervous System
74. Carpal tunnel syndrome results in damage to the median nerve, resulting in lack of control to the wrist and also numbness. This indicates that A. the nerve contained sensory neurons. B. the nerve contained motor neurons. C. the nerve contained both sensory and motor neurons. D. the damage was to a central body in a ganglion. E. the damage was to the spinal cord interneuron. The numbness indicates damage to a sensory neuron, while the lack of control of the wrist indicates damage to a motor neuron.
Bloom's Level: 3. Apply Learning Outcome: 17.01.02 Describe the basic structure of a neuron, and compare the functions of the three classes of neurons. Section: 17.01 Topic: Nervous System
17-38 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
75. How fast a person can type or play the piano is ultimately limited by the number of impulses a person can send to their finger muscles per second. This in turn is limited by A. primarily the type of muscle. B. whether the signal is pain, sound, motor, etc. C. the magnitude or strength of the nerve impulse. D. the number of neurons and synapses involved. E. the speed with which sodium ion can be pumped back outside the neuron membrane. The speed of a nervous impulse is limited by the speed with which sodium ions can be pumped back outside the neuron membrane resetting the membrane for a new nervous impulse.
Bloom's Level: 3. Apply Learning Outcome: 17.02.01 Summarize the activities that generate and propagate an action potential. Section: 17.02 Topic: Action Potential/Nerve Impulse
76. The likely effect on a neuron of 2 excitatory signals and 20 inhibitory signals is A. transmission of a nerve impulse. B. transmission of a nerve impulse, releasing excitatory neurotransmitters at the next synapse. C. transmission of a nerve impulse, releasing inhibitory neurotransmitters at the next synapse. D. prohibiting the axon from firing at all. E. confused integration. Since there are more inhibitory than excitatory signals, the summing up of these signals may prohibit the axon from firing.
Bloom's Level: 3. Apply Learning Outcome: 17.02.03 Define synaptic integration. Section: 17.02 Topic: Synapse and Neurotransmitters
17-39 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
77. You spray an insect with common insecticide that destroys the ability of acetylcholinesterase to recycle acetylcholine. What then happens? A. This kills the neuron directly. B. The lack of recycled acetylcholine brings the cell metabolism to a halt. C. The insect is unable to move anything because nerve impulses quit flowing. D. The insect runs out of acetylcholine and becomes paralyzed. E. The insect loses control of body functions as nerve impulses flow continuously. By not recycling acetylcholine, its continued presence in the synapse causes nerve impulses to flow continuously.
Bloom's Level: 3. Apply Learning Outcome: 17.02.04 Explain the role of neurotransmitters in the transmission of a nerve impulse across a synapse. Section: 17.02 Topic: Synapse and Neurotransmitters
78. When the dorsal root of a spinal nerve is cut, A. the organism is killed. B. incoming sensory nerve impulses are lost. C. outgoing motor nerve impulses are lost. D. incoming motor nerve impulses are lost. E. impulses do not cross over to the other side of the body. The dorsal root of a spinal nerve contains sensory fibers entering the gray matter. Cutting it would mean that incoming sensory information is lost.
Bloom's Level: 3. Apply Learning Outcome: 17.03.01 Understand the relationship between the CNS and PNS in the nervous system. Section: 17.03 Topic: Central Nervous System
17-40 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
True / False Questions 79. When anyone severs his spinal cord, as Christopher Reeve did in a horse riding accident, they suffer a loss of sensation and a loss of voluntary control. TRUE The spinal cord controls our ability to voluntarily move our limbs and the sensations we receive from around the body.
Bloom's Level: 3. Apply Learning Outcome: 17.03.02 Describe the structure and function of the spinal cord. Section: 17.03 Topic: Central Nervous System
80. If you have extreme difficulties learning how to hit a baseball or play the piano, there may be a problem with your cerebellum. TRUE The cerebellum assists the learning of new motor skills, such as playing the piano or hitting a baseball.
Bloom's Level: 3. Apply Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
17-41 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
Multiple Choice Questions 81. Certain people have an inability to speak due to brain damage. The most likely area of the brain that has been damaged is A. Broca's area. B. the meninges. C. the ventricles. D. the amygdala. E. the limbic system. Broca's area is involved in motor speech. Damage to this area results in an inability to speak.
Bloom's Level: 3. Apply Learning Outcome: 17.04.02 Summarize how the limbic system is involved in memory, language, and speech. Section: 17.04 Topic: Central Nervous System
82. Which of the following is not a symptom of addiction to a certain drug? A. The drug continues to cause physical or psychological changes in your body. B. It takes more drug to cause the same changes as before. C. You cannot function without having the drug. D. You spend more and more time trying to get the drug and/or thinking about using the drug. E. You quit eating as long as you can use the drug. Drugs are intended to cause physical or psychological changes in the body. That is not a sign of addiction to the drug.
Bloom's Level: 3. Apply Learning Outcome: 17.06.02 Classify illicit drugs as to whether they have a depressant, stimulant, or psychoactive effect on the body. Section: 17.06 Topic: Neurons
17-42 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
83. Which amino acid, in mice, was discovered to play a role in memory? A. glutamate B. valine C. serine D. histidine E. methionine It was discovered that glutamate functions as a neurotransmitter in the hippocampus, which is associated in the formation of memories.
Bloom's Level: 3. Apply Learning Outcome: 17.04.04 Explain how certain diseases, accidents, and experiments have helped scientists understand some basic components of how memories are made. Section: 17.04 Topic: Nervous System
84. Which is the correct arrangement of tissues from outermost to innermost? A. meninges → vertebrae → spinal cord → central canal B. meninges → spinal cord → vertebrae → central canal C. vertebrae → meninges → spinal cord → central canal D. meninges → vertebrae → central canal → spinal cord E. vertebrae → meninges → central canal → spinal cord From the outside in are the vertebrae, the meninges, the spinal cord, and the central canal.
Bloom's Level: 4. Analyze Learning Outcome: 17.03.01 Understand the relationship between the CNS and PNS in the nervous system. Section: 17.03 Topic: Central Nervous System
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Chapter 17 - Nervous System
85. Jeremy has developed a problem with regulating his waking/sleeping cycle and regulating his body temperature. Doctors suspect he may have a brain lesion. Where might this lesion be located? A. in the cerebellum B. in the prefrontal area of the cerebrum C. in Broca's area of the cerebrum D. in the pons E. in the hypothalamus The hypothalamus is an integrating center that helps maintain homeostasis by regulating hunger, sleep, thirst, body temperature, and water balance.
Bloom's Level: 4. Analyze Learning Outcome: 17.03.03 Identify the major regions of the brain and provide the main functions of each region. Section: 17.03 Topic: Central Nervous System
86. One group of neurotransmitters retards heart rate, promotes digestion, contracts the eye pupil, etc. The other group has the opposite effect. These drugs are affecting the A. motor and sensory nerves. B. cerebrum and cerebellum, respectively. C. central and peripheral nervous systems. D. neurostimulatory and neuroinhibitory synapses. E. sympathetic and parasympathetic nervous systems. The sympathetic division is responsible for the "fight or flight" response, while the parasympathetic division is responsible for the "rest and digest" response.
Bloom's Level: 4. Analyze Learning Outcome: 17.05.03 Contrast the overall functions of the sympathetic and parasympathetic divisions of the autonomic nervous system. Section: 17.05 Topic: Peripheral Nervous System
17-44 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
87. Which is a combination of physiological and psychological dependence? A. tolerance and withdrawal B. increasing a drug to get the same effect and preoccupation with trying to get more drugs C. seizures and craving for cocaine D. the rush of cocaine and the hallucinations of marijuana E. a cocaine binge and a nicotine craving Addiction involves developing a physiological dependence on the drug such that they require larger amounts of the drug to create the same sensation. A psychology dependence involves prioritizing the drug over food, hygiene, etc.
Bloom's Level: 4. Analyze Learning Outcome: 17.06.01 Explain the specific mechanisms whereby the most common illicit drugs affect the brain. Section: 17.06 Topic: Nervous System
88. Which of the following is not a way that a drug affects the brain? A. cause the neurotransmitter to leak out of a synaptic vesicle into the axon terminal B. promote the release of the neurotransmitter into the synaptic cleft C. block the enzyme that causes breakdown of the neurotransmitter D. mimics the action of a neurotransmitter by binding to the receptor E. builds up along the outside of the brain because it cannot pass the blood-brain barrier Drugs must be able to enter the brain in order to function. Therefore, one that cannot pass the blood-brain barrier will not affect the neurotransmitters.
Bloom's Level: 4. Analyze Learning Outcome: 17.06.01 Explain the specific mechanisms whereby the most common illicit drugs affect the brain. Section: 17.06 Topic: Nervous System
17-45 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 17 - Nervous System
89. Which is NOT a correct association of disease with cause or symptoms? A. multiple sclerosis—myelin sheath lesions impair normal conduction of impulses B. Parkinson disease—malfunction of the basal nuclei C. Alzheimer disease—gradual death of brain cells in hippocampus due to calcium ion influx D. fetal alcohol syndrome—increased intelligence and physical performance E. cocaine overdose—seizures and cardiac or respiratory arrest Fetal alcohol syndrome results in decreased intelligence and physical performance.
Bloom's Level: 4. Analyze Learning Outcome: 17.07.02 Compare the mechanism causing the following diseases: Parkinson disease, MS, stroke, meningitis, and prion disease. Section: 17.07 Topic: Nervous System
90. Generally, nerve impulses do not move "backward" along a series of neurons. What is the reason for this? A. There is a short refractory period during which sodium gates cannot open and an action potential cannot move backwards. B. Once the signal has passed, the nerves separate from each other so it cannot go backwards. C. At the synapse, the presynaptic membrane has receptors, ensuring the transmission only goes one way. D. Once the sodium channels open, they cannot be closed. E. The resting potential of the neuron prevents it from conducting an impulse. As soon as an action potential has moved on, the previous section undergoes a refractory period, during which the sodium gates are unable to open. Therefore, that action potential cannot move backward and instead always moves down an axon toward its terminals.
Bloom's Level: 5. Evaluate Learning Outcome: 17.02.01 Summarize the activities that generate and propagate an action potential. Section: 17.02 Topic: Action Potential/Nerve Impulse
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Chapter 17 - Nervous System
91. Neurotransmitters are molecules that cross the synaptic cleft via A. active transport and always inhibit the postsynaptic neuron. B. acitve transport and always excite the postsynaptic neuron. C. passive diffusion and either excite or inhibit the postsynaptic neuron. D. passive diffusion and always inhibit the postsynaptic neuron. E. passive diffusion and always excite the postsynaptic neuron. Neurotransmitters cross the synaptic cleft by passive diffusion, not by active transport. Depending on the type of neurotransmitter and receptor, the response can be excitatory or inhibitory.
Bloom's Level: 5. Evaluate Learning Outcome: 17.02.03 Define synaptic integration. Section: 17.02 Topic: Synapse and Neurotransmitters
92. Why does the primary somatosensory area have a larger section dedicated to the fingers and hand, than to the feet and toes? A. The fingers and hand are how we take in more sensory information from our surroundings rather than the feet and toes. B. The fingers and hand are above the spinal cord termination, while the feet and toes are below it. C. The feet and toes require more motor area so they get less somatosensory area. D. The feet and toes are not innervated. E. The fingers and hand do not have motor area sections, so they are controlled by the somatosensory area. Since much of our sensory information requires fine sensation in our hands and fingers, more of the primary somatosensory area is devoted to those than to the feet and toes.
Bloom's Level: 5. Evaluate Learning Outcome: 17.03.04 Compare the functions of the primary motor and somatosensory areas of the brain versus the association areas and processing centers. Section: 17.03 Topic: Central Nervous System
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Chapter 17 - Nervous System
93. Administration of norepinephrine would A. dilate the bronchi and increase oxygen for blood. B. slow heartbeat. C. stimulate the digestive system to supply more sugar to the blood. D. cause the pupil of the eye to contract. E. promote an overall relaxed state. Norepinephrine is the neurotransmitter of the sympathetic division associated with "fight or flight." An increase in breathing rate and dilation of the bronchi would be associated with this emergency response.
Bloom's Level: 5. Evaluate Learning Outcome: 17.05.02 Distinguish between the somatic and autonomic divisions of the peripheral nervous system. Section: 17.05 Topic: Synapse and Neurotransmitters
94. Which of the following is an accurate comparison of Alzheimer and Parkinson diseases? A. In both cases, the patient slowly loses memory and slowly drifts into inactivity. B. In both cases, the patient becomes increasingly excited and overstimulated. C. Patients with AD produce too much beta amyloid, while those with PD produce too little dopamine. D. Patients with AD produce too little beta amyloid, while those with PD produce too little dopamine. E. Patients with AD produce too much beta amyloid, while those with PD produce too much dopamine. Alzheimer disease results in neurons containing too much beta amyloid. Parkinson disease is characterized by the degeneration of dopamine-releasing neurons in the brain.
Bloom's Level: 5. Evaluate Learning Outcome: 17.07.03 Describe the symptoms seen in several disorders of the brain and spinal cord and discuss why it is difficult to find cures for each. Section: 17.07 Topic: Nervous System
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Chapter 17 - Nervous System
95. Which statement is true regarding the sympathetic and parasympathetic divisions of the autonomic motor pathways? A. Both divisions have two neurons per message. B. The sympathetic division is under voluntary control, while the parasympathetic is under involuntary control. C. The sympathetic divisions uses acetylcholine as the neurotransmitter. D. The parasympathetic division uses norepinephrine as the neurotransmitter. E. Only the sympathetic division controls smooth and cardiac muscle. Both the sympathetic and parasympathetic divisions of the autonomic motor pathways have two neurons per message. Both divisions are under involuntary control. The sympathetic divisions uses norepinephrine, while the parasympathetic uses acetylcholine as the neurotransmitter. Both divisions control smooth and cardiac muscle.
Bloom's Level: 5. Evaluate Learning Outcome: 17.05.03 Contrast the overall functions of the sympathetic and parasympathetic divisions of the autonomic nervous system. Section: 17.05 Topic: Nervous System
Short Answer Questions 96. Explain the transmission across a synapse from one neuron to another. 1. An action potential arrives at an axon terminal; Ca2+ enters and synaptic vesicles fuse with the presynaptic membrane. 2. Neurotransmitter molecules are released and bind to receptors on the postsynaptic membrane. 3. When an excitatory neurotransmitter binds to a receptor, Na+ diffuses into the postsynaptic neuron, and an action potential begins.
Bloom's Level: 6. Create Learning Outcome: 17.02.04 Explain the role of neurotransmitters in the transmission of a nerve impulse across a synapse. Section: 17.02 Topic: Synapse and Neurotransmitters
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Chapter 17 - Nervous System
97. List and compare the three classes of neurons. 1. Sensory neuron: takes messages from the PNS to the CNS. 2. Interneuron: lies within the CNS. Receives input from sensory neurons and communicates the information to the motor neurons. 3. Motor neurons: take messages from the CNS to an effector in order to carry out a response to an environmental change.
Bloom's Level: 6. Create Learning Outcome: 17.01.02 Describe the basic structure of a neuron, and compare the functions of the three classes of neurons. Section: 17.01 Topic: Neurons
98. Explain why a blow to the base of the skull, damaging the medulla oblongata, can result in death. The medulla oblongata regulates vital functions like heartbeat, breathing, and blood pressure. If the medulla oblongata was damaged, it could cause the person to quit breathing or their heart to stop beating correctly, which could result in death.
Bloom's Level: 6. Create Learning Outcome: 17.03.02 Describe the structure and function of the spinal cord. Section: 17.03 Topic: Central Nervous System
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Chapter 18 - Senses
Chapter 18 Senses
Multiple Choice Questions 1. The two major types of sensory receptors, based on the origin of stimuli, are A. chemoreceptors and mechanoreceptors. B. photoreceptors and thermoreceptors. C. chemoreceptors and nonchemoreceptors. D. central nervous system and peripheral nervous system receptors. E. exteroceptors and interoceptors. Based on the source of the stimulus, sensory receptors can be classified as interoceptors (receive stimuli from inside the body) or exteroceptors (detect stimuli from outside the body).
Bloom's Level: 1. Remember Learning Outcome: 18.01.01 Distinguish between interoceptors and exteroceptors. Section: 18.01 Topic: Senses
2. Receptors that are directly involved in maintaining homeostasis are primarily A. chemoreceptors. B. photoreceptors. C. interoceptors. D. proprioceptors. E. exteroceptors. Interoceptors receive stimuli from inside the body and are directly involved in homeostasis.
Bloom's Level: 1. Remember Learning Outcome: 18.01.01 Distinguish between interoceptors and exteroceptors. Section: 18.01 Topic: Senses
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Chapter 18 - Senses
True / False Questions 3. Interoceptors are regulated by a positive feedback mechanism. FALSE Interoceptors are regulated by a negative feedback mechanism. In this way, they can maintain homeostasis within the body.
Bloom's Level: 1. Remember Learning Outcome: 18.01.01 Distinguish between interoceptors and exteroceptors. Section: 18.01 Topic: Senses
Multiple Choice Questions 4. The monitoring of blood oxygen levels within the body is done by what type of receptor? A. thermoreceptor B. chemoreceptor C. mechanoreceptor D. photoreceptor E. proprioceptor Chemoreceptors respond to chemical substances, such as oxygen.
Bloom's Level: 1. Remember Learning Outcome: 18.01.02 Describe the four categories of sensory receptors. Section: 18.01 Topic: Senses
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Chapter 18 - Senses
5. When a stimulus is continuous but we no longer respond to it, the process is called A. integration. B. proprioception. C. somatic sensing. D. countersensation. E. sensory adaptation. Sensory adaptation results in a decrease in the response to a stimulus.
Bloom's Level: 1. Remember Learning Outcome: 18.01.03 Explain the significance of sensory transduction and sensory adaptation. Section: 18.01 Topic: Senses
True / False Questions 6. Senses whose receptors are associated with the skin, muscles, joints, and viscera are called internal senses. FALSE These are called somatic senses.
Bloom's Level: 1. Remember Learning Outcome: 18.02.01 Compare and contrast the functions of proprioceptors, cutaneous receptors, and pain receptors. Section: 18.02 Topic: Senses
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Chapter 18 - Senses
Multiple Choice Questions 7. The receptors that detect position and movement of limbs are called A. mechanoreceptors. B. chemoreceptors. C. proprioceptors. D. photoreceptors. E. nociceptors. Proprioceptors are mechanoreceptors that maintain muscle tone and the body's equilibrium and posture.
Bloom's Level: 1. Remember Learning Outcome: 18.02.01 Compare and contrast the functions of proprioceptors, cutaneous receptors, and pain receptors. Section: 18.02 Topic: Peripheral Nervous System
8. Which of the following is not one of the four types of taste? A. salty B. hot C. sour D. sweet E. bitter Hot is not one of the basic four types of taste.
Bloom's Level: 1. Remember Learning Outcome: 18.03.01 Identify the structures that are involved in taste, including the five types of taste receptors. Section: 18.03 Topic: Senses
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Chapter 18 - Senses
True / False Questions 9. The tongue contains approximately a million taste buds. FALSE The tongue contains about 10,000 taste buds.
Bloom's Level: 1. Remember Learning Outcome: 18.03.01 Identify the structures that are involved in taste, including the five types of taste receptors. Section: 18.03 Topic: Senses
10. There is perhaps a fifth type of taste bud sensitive to umami. TRUE Umami is a meaty or "delicious" taste.
Bloom's Level: 1. Remember Learning Outcome: 18.03.01 Identify the structures that are involved in taste, including the five types of taste receptors. Section: 18.03 Topic: Senses
11. Approximately 80–90% of what we perceive as "taste" is actually due to a sense of smell. TRUE This is why food tastes dull when we have a head cold or a stuffed-up nose.
Bloom's Level: 1. Remember Learning Outcome: 18.03.03 Compare and contrast how the brain receives information about taste versus smell. Section: 18.03 Topic: Senses
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Chapter 18 - Senses
Multiple Choice Questions 12. On the retina there is a specialized region in which vision for color is most acute. This area is called the A. tectorial membrane. B. otolith. C. fovea centralis. D. blind spot. E. ampulla. The retina has a special region called the fovea centralis where cone cells are densely packed and where vision is most acute.
Bloom's Level: 1. Remember Learning Outcome: 18.04.01 Identify the structures of the human eye. Section: 18.04 Topic: Peripheral Nervous System
13. The structure that regulates the size of the opening for light in the eye is the A. pupil. B. retina. C. iris. D. ciliary muscle. E. lens. The iris contains smooth muscle that regulates the size of the pupil through which light enters the eyeball.
Bloom's Level: 1. Remember Learning Outcome: 18.04.01 Identify the structures of the human eye. Section: 18.04 Topic: Nervous System
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Chapter 18 - Senses
14. The sclera is continuous with the A. cornea. B. iris. C. retina. D. choroid coat. E. fovea centralis. The sclera is the outer layer of the eye that is continuous with the cornea. The sclera is white and fibrous, except for at the cornea.
Bloom's Level: 1. Remember Learning Outcome: 18.04.01 Identify the structures of the human eye. Section: 18.04 Topic: Nervous System
15. Which type(s) of receptor is(are) found in the retina? A. rods and cones B. organ of Corti C. muscle spindles D. olfactory cells E. hair cells in ampullae Rods and cones are the photoreceptors in the eye.
Bloom's Level: 1. Remember Learning Outcome: 18.04.03 Discuss the function and distribution of rods and cones. Section: 18.04 Topic: Nervous System
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Chapter 18 - Senses
16. Color vision depends on three types of cones that contain pigments sensitive to either A. red, green, or blue light. B. red, green, or orange light. C. red, yellow, or blue light. D. red, white, or blue light. E. green, blue, or yellow light. Color vision depends on cones, which contain the pigments blue, green, and red.
Bloom's Level: 1. Remember Learning Outcome: 18.04.03 Discuss the function and distribution of rods and cones. Section: 18.04 Topic: Senses
17. Where are the photoreceptors of the eye located? A. retina B. optic nerve C. choroid layer D. sclera E. organ of Corti The rods and cones are located in the retina at the back of the eye.
Bloom's Level: 1. Remember Learning Outcome: 18.04.01 Identify the structures of the human eye. Section: 18.04 Topic: Nervous System
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Chapter 18 - Senses
18. The sense of rotational equilibrium is accomplished by hair cells found in the A. cochlear duct. B. outer ear. C. middle ear. D. ampulla. E. cochlea (organ of Corti). Rotational equilibrium involves the three semicircular canals, the base of which is called an ampulla.
Bloom's Level: 1. Remember Learning Outcome: 18.06.02 Review the structures involved in rotational and gravitational equilibrium. Section: 18.06 Topic: Senses
19. Vertigo would most likely result from A. continuous stimulation of the tympanic membrane. B. bipolar cells stimulating the retina. C. continuous fluid movement in the semicircular canals. D. increased vibrations in the basilar membrane. E. lack of red or green cones. Continuous fluid movement in the semicircular canals would disrupt our rotational sense of equilibrium, resulting in vertigo.
Bloom's Level: 1. Remember Learning Outcome: 18.07.04 Describe some common disorders of hearing and equilibrium. Section: 18.07 Topic: Senses
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Chapter 18 - Senses
20. If light rays are not focused evenly on the retina, a fuzzy image is formed. This condition is called A. nearsightedness. B. farsightedness. C. astigmatism. D. cataracts. E. conduction deafness. When the cornea or lens is uneven, the image is fuzzy. The light rays cannot be evenly focused on the retina. This is known as astigmatism.
Bloom's Level: 1. Remember Learning Outcome: 18.07.02 Explain the anatomical abnormalities that cause color blindness, nearsightedness, farsightedness, and astigmatism. Section: 18.07 Topic: Senses
21. In nearsightedness, light rays are brought into focus A. in front of the retina. B. in back of the retina. C. on the retina. D. on the hair cells of the ampulla. E. unevenly across the retina. In nearsighted individuals, the eyeball is elongated so that when they attempt to look at a distant object, the image focuses in front of the retina.
Bloom's Level: 1. Remember Learning Outcome: 18.07.02 Explain the anatomical abnormalities that cause color blindness, nearsightedness, farsightedness, and astigmatism. Section: 18.07 Topic: Senses
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Chapter 18 - Senses
True / False Questions 22. As people age, they tend to lose their sense of smell and their hearing. TRUE In most people the senses of smell and hearing decrease after age 60.
Bloom's Level: 1. Remember Learning Outcome: 18.07.01 List three common disorders that can affect the senses of taste and smell. Learning Outcome: 18.07.04 Describe some common disorders of hearing and equilibrium. Section: 18.07 Topic: Senses
Multiple Choice Questions 23. The doctor tells you that you have anosmia. What is your problem? A. no sense of smell B. an inability to fall asleep C. an inability to stay asleep D. vertigo E. deafness Anosmia is the complete lack of a sense of smell.
Bloom's Level: 1. Remember Learning Outcome: 18.07.01 List three common disorders that can affect the senses of taste and smell. Section: 18.07 Topic: Senses
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Chapter 18 - Senses
True / False Questions 24. The most frequent cause of blindness in adults are retinal disorders. TRUE Retinal disorders are varied but are the most frequent causes of blindness in adults, ahead of glaucoma and cataracts.
Bloom's Level: 1. Remember Learning Outcome: 18.07.03 Characterize the three most common causes of blindness in the United States. Section: 18.07 Topic: Senses
Multiple Choice Questions 25. Which cutaneous receptors are concentrated in the fingertips and palms? A. Meissner's corpuscles B. Pacinian corpuscles C. Ruffini endings D. root hair plexus E. free nerve endings The Meissner's corpuscles are concentrated in the fingertips and palms. Pacinian corpuscles lie deep within the dermis. Ruffini endings are encapsulated by sheaths of connective tissues. Root hair plexus wind around the base of a hair follicle. Free nerve endings are not concentrated in the fingertips and palms.
Bloom's Level: 1. Remember Learning Outcome: 18.02.02 List specific types of cutaneous receptors that are sensitive to fine touch, pressure, and temperature. Section: 18.02 Topic: Nervous System
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Chapter 18 - Senses
26. What is the location of the sensory receptors associated with the sense of smell? A. roof of the nasal cavity B. inner lining of the nose C. back of the pharynx D. inferior portion of the nasal cavity E. frontal lobe of the cerebral hemisphere The olfactory cells are associated with the sense of taste. These cells are located within the olfactory epithelium in the roof of the nasal cavity.
Bloom's Level: 1. Remember Learning Outcome: 18.03.02 Describe the sensory receptors involved in smell. Section: 18.03 Topic: Nervous System
27. Which health problem is associated with a decrease in the ability to taste certain foods? A. allergies B. upper respiratory tract infection C. brain tumors D. exposure to drugs E. All of the answer choices can decrease a person's ability to taste certain foods. All of the answer choices can decrease a person's ability to taste certain foods.
Bloom's Level: 1. Remember Learning Outcome: 18.07.01 List three common disorders that can affect the senses of taste and smell. Section: 18.07 Topic: Senses
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Chapter 18 - Senses
28. What part of the brain do the olfactory bulbs connect to? A. limbic system B. pons C. medulla D. brain stem E. temporal lobe The olfactory bulbs connect to the limbic system in the brain.
Bloom's Level: 1. Remember Learning Outcome: 18.03.03 Compare and contrast how the brain receives information about taste versus smell. Section: 18.03 Topic: Central Nervous System
29. What are the three areas of the inner ear? A. semicircular canals, vestibule, and cochlea B. semicircular canals, vestibule, and organ of Corti C. malleus, vestibule, and cochlea D. malleus, incus, and stapes E. malleus, incus, and vestibule The semicircular canals, vestibule, and cochlea are the three areas of the inner ear. The malleus, incus, and vestibule are bones of the inner ear.
Bloom's Level: 1. Remember Learning Outcome: 18.05.03 Explain how the inner ear is able to distinguish sounds of different pitches and volumes. Section: 18.05 Topic: Senses
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Chapter 18 - Senses
True / False Questions 30. Taste and smell are examples of exteroceptors. TRUE Exteroceptors receive stimuli from outside the body. Taste and smell, along with vision, hearing, and equilibrium are examples.
Bloom's Level: 2. Understand Learning Outcome: 18.01.01 Distinguish between interoceptors and exteroceptors. Section: 18.01 Topic: Senses
Multiple Choice Questions 31. Mechanoreceptors do not respond to A. touch. B. sound waves. C. motion. D. gravity. E. temperature. Mechanoreceptors are stimulated by mechanical forces, most often pressure of some sort. Temperature requires a thermoreceptor.
Bloom's Level: 2. Understand Learning Outcome: 18.01.02 Describe the four categories of sensory receptors. Section: 18.01 Topic: Senses
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Chapter 18 - Senses
32. Which is NOT a correct association of sensory receptors? A. thermoreceptors—pressure waves sensed as "hearing" B. photoreceptors—sense of "vision" C. chemoreceptors—detection of carbon dioxide level in blood D. proprioceptors—sense of position E. mechanoreceptors—sense of touch Thermoreceptors are stimulated by changes in temperature.
Bloom's Level: 2. Understand Learning Outcome: 18.01.02 Describe the four categories of sensory receptors. Section: 18.01 Topic: Senses
33. Taste buds and olfactory cells are considered A. mechanoreceptors. B. chemoreceptors. C. radioreceptors. D. proprioceptors. E. general receptors. Taste buds respond to chemicals in food. Olfactory cells respond to chemicals in air. Therefore, both are considered chemoreceptors.
Bloom's Level: 2. Understand Learning Outcome: 18.01.02 Describe the four categories of sensory receptors. Section: 18.01 Topic: Senses
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Chapter 18 - Senses
34. Which is not a general receptor found in the skin to detect pressure or fine touch? A. free nerve endings B. Merkel's disks C. Krause end bulbs D. Meissner's corpuscles E. Pacinian corpuscles Free nerve endings in the skin detect pain, heat, and cold.
Bloom's Level: 2. Understand Learning Outcome: 18.02.02 List specific types of cutaneous receptors that are sensitive to fine touch, pressure, and temperature. Section: 18.02 Topic: Peripheral Nervous System
35. When an internal pain is felt in a surface region of the skin, it is called A. referred pain. B. proprioception. C. somatic sensing. D. countersensation. E. sensory adaptation. Sometimes stimulation of internal pain receptors is felt as pain from the skin as well as the internal organs. This is called referred pain.
Bloom's Level: 2. Understand Learning Outcome: 18.02.03 Explain the importance of pain receptors to the survival of an organism. Section: 18.02 Topic: Senses
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Chapter 18 - Senses
True / False Questions 36. There are separate nociceptors for mechanical damage and damage due to toxic substances. FALSE Nociceptors are sensitive to chemicals released by damaged cells, regardless of the cause of the damage.
Bloom's Level: 2. Understand Learning Outcome: 18.02.03 Explain the importance of pain receptors to the survival of an organism. Section: 18.02 Topic: Peripheral Nervous System
Multiple Choice Questions 37. When you perceive a sense of taste, your brain A. responds to the one taste sense (sweet, sour, etc.) that is sending the most signals. B. takes an average of the taste messages being received. C. uses memory to form a taste image after eliminating those signals (sweet, salty, etc.) not received. D. identifies the chemical profile as a mix of sweet, sour, etc. and remembers the taste from previous experiences. E. takes a near-infinite variety of taste signals and generalizes them into one of the four primary tastes. The brain appears to survey the overall pattern of incoming sensory impulses and to take a "weighted average" of their taste messages as the perceived taste.
Bloom's Level: 2. Understand Learning Outcome: 18.03.01 Identify the structures that are involved in taste, including the five types of taste receptors. Section: 18.03 Topic: Senses
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Chapter 18 - Senses
38. Which of the following is not unique to the sense of smell? A. It can often be associated with vivid memories. B. Smell has direct connections to the limbic system and associated emotions. C. The sense of smell uses a variety of olfactory cell types to permit a wide combination of smell perceptions. D. Perception of smell is determined by which the pattern or signature of neurons that are stimulated. E. The sense of smell uses chemoreceptors. The sense of taste also uses chemoreceptors, so this is not unique to the sense of smell alone.
Bloom's Level: 2. Understand Learning Outcome: 18.03.02 Describe the sensory receptors involved in smell. Section: 18.03 Topic: Senses
39. The order in which light reaches the lens of a human eye is A. pupil → cornea → aqueous humor → lens. B. lens → aqueous humor → pupil → cornea. C. cornea → aqueous humor → pupil → lens. D. cornea → pupil → aqueous humor → lens. E. cornea → vitreous humor → pupil → lens. Light from the outside must go through the cornea, the aqueous humor, and the pupil before it reaches the lens.
Bloom's Level: 2. Understand Learning Outcome: 18.04.01 Identify the structures of the human eye. Section: 18.04 Topic: Nervous System
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Chapter 18 - Senses
40. When you look at an image, it is inverted on the back of the retina and yet appears rightside-up when perceived in the brain. What part of the body is responsible for inverting the image? A. the retina B. the rods and cones C. the optic nerve D. the brain E. the lens The brain is responsible for correctly righting the image.
Bloom's Level: 2. Understand Learning Outcome: 18.04.04 Trace the path of a nerve impulse from the retina to the visual cortex. Section: 18.04 Topic: Senses
True / False Questions 41. The brain has taken the visual field apart even though we see it as a unified visual field. TRUE The visual association areas are believed to rebuild the field the brain has taken apart and to give us an understanding of it as well.
Bloom's Level: 2. Understand Learning Outcome: 18.04.04 Trace the path of a nerve impulse from the retina to the visual cortex. Section: 18.04 Topic: Central Nervous System
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Chapter 18 - Senses
Multiple Choice Questions 42. The bones of the middle ear A. respond to a change in the position of the head. B. transmit sound waves. C. are sense receptors connected to the auditory nerve. D. are named the malleus, incus, and otolith. E. are connected to the tectorial membrane and oval window. The bones of the inner ear, the malleus, incus, and stapes, touch the tympanic membrane and oval window. Their job is to transmit sound waves.
Bloom's Level: 2. Understand Learning Outcome: 18.05.01 List the parts of the ear that make up the outer ear, middle ear, and inner ear. Section: 18.05 Topic: Senses
43. Which of the following statements is true? A. Hearing is not dependent on the inner ear. B. All parts of the organ of Corti hear all ranges of sound. C. Loud music does not damage your ears. D. Hearing is dependent on mechanical pressure. E. Sound is similar to light insofar as it is transmitted through the vacuum of space. Hearing is the result of mechanoreceptors, which distinguish mechanical pressure in the form of sound waves.
Bloom's Level: 2. Understand Learning Outcome: 18.05.02 Describe the mechanism by which sound waves in the outer ear are converted into nerve impulses in the inner ear. Section: 18.05 Topic: Senses
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Chapter 18 - Senses
44. The malleus, incus, and stapes are A. located in the inner ear. B. equivalent terms for the saccule, utricle, and the otolith. C. surrounded by fluid in a bony cavity. D. located between the tympanic membrane and the oval window. E. located between the tectorial membrane and the basilar membrane. The three bones of the middle ear connect the tympanic membrane to the oval window.
Bloom's Level: 2. Understand Learning Outcome: 18.05.01 List the parts of the ear that make up the outer ear, middle ear, and inner ear. Section: 18.05 Topic: Senses
45. Which correctly traces the path of a sound vibration until the signal is sent to the brain? A. auditory canal → tympanic membrane → malleus → incus → stapes → oval window → cochlea → cochlear nerve B. auditory canal → malleus → incus → stapes → tympanic membrane → oval window → cochlea → cochlear nerve C. auditory canal → tympanic membrane → malleus → incus → stapes → cochlea → oval window → cochlear nerve D. auditory canal → tympanic membrane → oval window → cochlea → malleus → incus → stapes → cochlear nerve E. auditory canal → malleus → incus → stapes → oval window → tympanic membrane → cochlea → cochlear nerve Sound enters the auditory canal where it strikes the tympanic membrane. The malleus passes the pressure to the incus and then the stapes, which passes it on to the oval window. The oval window causes the fluid within the cochlea to vibrate, which stimulates the cochlear nerve.
Bloom's Level: 2. Understand Learning Outcome: 18.05.02 Describe the mechanism by which sound waves in the outer ear are converted into nerve impulses in the inner ear. Section: 18.05 Topic: Senses
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Chapter 18 - Senses
46. Which pair is mismatched? A. semicircular canals—inner ear B. ampulla—outer ear C. auditory canal—outer ear D. ossicles—middle ear E. stapes—oval window The ampulla is located within the inner ear.
Bloom's Level: 2. Understand Learning Outcome: 18.06.02 Review the structures involved in rotational and gravitational equilibrium. Section: 18.06 Topic: Senses
47. Rotational equilibrium is movement in which plane? What does it depend upon? A. angular; utricle and saccule B. angular; semicircular canals C. vertical; utricle and saccule D. vertical; semicircular canals E. horizontal; utricle and saccule Rotational equilibrium involves the semicircular canals, which recognize movement in the rotational and/or angular movement of the head.
Bloom's Level: 2. Understand Learning Outcome: 18.06.02 Review the structures involved in rotational and gravitational equilibrium. Section: 18.06 Topic: Senses
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Chapter 18 - Senses
48. The direction of movement of the head relative to gravity is determined by A. stereocilia in a gelatinous material called a cupula. B. movement of fluid in the ampulla. C. displacement of otoliths on the otolithic membrane. D. asymmetric nerve impulses from proprioceptors on the left and right side of the head. E. integration of the visual and auditory signals in the equilibria center of the cortex. Gravitational equilibrium depends on the otolithic membrane and calcium carbonate granules called otoliths. which rest on the membrane.
Bloom's Level: 2. Understand Learning Outcome: 18.06.03 Compare the role of the fluid found in the semicircular canals with the otoliths present in the utricle and saccule. Section: 18.06 Topic: Senses
49. Glaucoma is caused by A. hardening of the lens. B. inadequate drainage of the aqueous humor. C. loss of the otoliths. D. loss of taste cells. E. overstimulation of the ampulla. Glaucoma occurs when the drainage system of the eyes fails, so fluid builds up and the resulting increase in pressure destroys the nerve fibers responsible for peripheral vision.
Bloom's Level: 2. Understand Learning Outcome: 18.07.03 Characterize the three most common causes of blindness in the United States. Section: 18.07 Topic: Senses
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Chapter 18 - Senses
50. A person who is farsighted cannot see close objects because A. the lens is too small. B. the optic nerve is damaged. C. the eyeball is too short. D. the eyeball is too long. E. of a lack of rod receptors. Farsighted individuals have a shortened eyeball, and when they try to see close objects, the image is focused behind the retina.
Bloom's Level: 2. Understand Learning Outcome: 18.07.02 Explain the anatomical abnormalities that cause color blindness, nearsightedness, farsightedness, and astigmatism. Section: 18.07 Topic: Senses
51. When a baseball player is hit in the back by a ball, what is detected by the pain receptors? A. Chemicals that are released by the damaged cells. B. The pressure produced by the trauma. C. The body's position during the time of the injury. D. The temperature of the day. E. None of the answer choices would be detected by the pain receptors. The pain receptors would detect the chemicals that are released by the damaged cells. Pacinian corpuscles would detect the pressure produced by the trauma. Proprioceptors would detect the body's position during the time of the injury. Free nerve endings would detect the temperature of the day.
Bloom's Level: 2. Understand Learning Outcome: 18.02.03 Explain the importance of pain receptors to the survival of an organism. Section: 18.02 Topic: Nervous System
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Chapter 18 - Senses
52. What structures are responsible for detecting odor? A. receptor proteins on the olfactory cells B. receptor lipids on the olfactory cells C. receptor carbohydrates on the olfactory cells D. receptor proteins on the sclera E. receptor proteins on the ossicles The receptor proteins on the olfactory cells are responsible for detecting odor. The lipids and carbohydrates on the olfactory cells do not detect odor. The sclera is part of the eye and the ossicles are part of the ear.
Bloom's Level: 2. Understand Learning Outcome: 18.03.03 Compare and contrast how the brain receives information about taste versus smell. Section: 18.03 Topic: Nervous System
53. What two structures are associated with gravitational equilibrium? A. utricle and saccule B. semicircular canal and cupula C. utricle and semicircular canal D. saccule and cupula E. utricle and cupula The utricle and saccule are associated with gravitational equilibrium. The semicircular canal and cupula are associated with rotational equilibrium.
Bloom's Level: 2. Understand Learning Outcome: 18.06.01 Explain how mechanoreceptors are involved in the sense of equilibrium. Section: 18.06 Topic: Senses
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Chapter 18 - Senses
54. It is very hot in the room but Jamie does not seem to notice. Her exteroreceptors have detected the heat and sent the message to her brain, but her brain has not correctly determined the message. What does Jamie have a problem with? A. exteroreceptors B. detection C. sensation D. perception E. thermoreceptors The thermoreceptors (a type of exteroreceptor) have properly detected the environmental change. Sensation occurs when nerve impulses arrive at the brain. Therefore, Jamie has a problem with perception, when the brain interprets the meaning of the stimulus.
Bloom's Level: 3. Apply Learning Outcome: 18.01.03 Explain the significance of sensory transduction and sensory adaptation. Section: 18.01 Topic: Central Nervous System
55. Which of the following is an example of sensory adaptation? A. The act of trying to read a vibrating book while riding in a car causes eyestrain. B. When it is totally dark, the rods and cones of the eyes do not send any visual impulses. C. After the wound healed, the pain also went away. D. You notice the smell of liver and onions cooking when you enter the kitchen, but soon you do not notice the odor although they are still cooking. E. At first you did not like rock music but as you became older, you found you liked it. Since the odor of the liver and onions are still present, but you don't notice them anymore, you have experienced sensory adaptation.
Bloom's Level: 3. Apply Learning Outcome: 18.01.03 Explain the significance of sensory transduction and sensory adaptation. Section: 18.01 Topic: Senses
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Chapter 18 - Senses
56. Jennifer's doctor tells her that she is experiencing a problem with her proprioceptors. What symptoms must Jennifer be experiencing? A. falling down B. memory failure C. intense pain D. a fight-or-flight response E. blindness Proprioceptors are mechanoreceptors that maintain muscle tone and the body's equilibrium and posture. Problems with these would result in falling down.
Bloom's Level: 3. Apply Learning Outcome: 18.02.01 Compare and contrast the functions of proprioceptors, cutaneous receptors, and pain receptors. Section: 18.02 Topic: Peripheral Nervous System
57. Scott has lost his sense of smell due to an accident, although to look at him you would never know anything was wrong. Upon external examination, his doctor cannot see any damage either. What part of Scott's body has been damaged? A. his oral cavity B. his nasal cavity C. his olfactory tract to the brain D. his pharynx E. his nose Since you cannot see any damage, it is unlikely that Scott has damaged his pharynx, nose, nasal cavity, or oral cavity. He has most likely severed his olfactory tract to the brain or some region of his brain that receives the olfactory signals.
Bloom's Level: 3. Apply Learning Outcome: 18.03.02 Describe the sensory receptors involved in smell. Section: 18.03 Topic: Central Nervous System
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Chapter 18 - Senses
58. David has a problem with his vision. He sees very well but everything is in black and white. The problem is most likely with what part of his eye? A. the cornea B. the cones C. the lens D. the fovea centralis E. the retina The cones make color vision possible so David must have some difficulty with the cones of his eye.
Bloom's Level: 3. Apply Learning Outcome: 18.04.03 Discuss the function and distribution of rods and cones. Section: 18.04 Topic: Nervous System
59. Julie can see very well close up, but has more problems with distance vision. She really has trouble when looking from her paper up to the professor at the front of the class and back again. What part of her eye is most likely the problem? A. optic nerve B. fovea C. cornea D. ciliary muscle E. suspensory ligaments The ciliary muscle controls the shape of the lens and accommodates for close-up vs. faraway vision.
Bloom's Level: 3. Apply Learning Outcome: 18.04.02 Explain how the eye focuses on near and far objects. Section: 18.04 Topic: Senses
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Chapter 18 - Senses
60. Puncturing the eardrum would A. make you totally deaf. B. prevent the normal transmission of sound vibration. C. destroy the sense receptors for hearing. D. account for why some people who hear still cannot sing a tune. E. account for dizziness and equilibrium problems. Rupture of the ear drum would prevent the normal transmission of sound vibration. Some sound would still be transmitted but it would not be as loud or as clear.
Bloom's Level: 3. Apply Learning Outcome: 18.05.01 List the parts of the ear that make up the outer ear, middle ear, and inner ear. Section: 18.05 Topic: Senses
61. From your understanding of the anatomy of the ear, what is the most likely explanation for a dog's ability to hear higher frequencies (pitch)? A. more ear bones for greater amplification B. larger outer ears C. a thinner and therefore more sensitive tympanum (eardrum) D. a wider range to the tectorial membrane and hair cells E. a more advanced brain section devoted to hearing The spiral organ, containing the tectorial membrane and hair cells, are responsible for the ability to hear different pitches (frequencies).
Bloom's Level: 3. Apply Learning Outcome: 18.05.03 Explain how the inner ear is able to distinguish sounds of different pitches and volumes. Section: 18.05 Topic: Senses
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Chapter 18 - Senses
62. From what you currently know, in a spacecraft at a point of weightlessness, which effect is most likely? A. Rotational equilibrium would be lacking and you could not detect head movement. B. Gravitational equilibrium would be altered since the otoliths would not be pulled down by gravity. C. Without gravity, there would be no vertigo. D. All balance systems would work similar to being on Earth with normal gravity. E. There would be absolutely no sense of balance signals of either type without gravity. Without gravity, the otoliths would not rest on the otolithic membrane, and gravitational equilibrium would be altered.
Bloom's Level: 3. Apply Learning Outcome: 18.06.01 Explain how mechanoreceptors are involved in the sense of equilibrium. Section: 18.06 Topic: Senses
True / False Questions 63. Deafness due to injury to the auditory center of the brain could be alleviated by the use of a hearing aid. FALSE A hearing aid may amplify sound waves, but it cannot fix the perception of sound in the auditory center of the brain.
Bloom's Level: 3. Apply Learning Outcome: 18.07.04 Describe some common disorders of hearing and equilibrium. Section: 18.07 Topic: Central Nervous System
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Chapter 18 - Senses
Multiple Choice Questions 64. You are a sun worshipper, constantly sunbathing without sunscreen or other protective devices. In addition you are also a smoker. These behaviors have significantly increased your chances of getting what disease? A. glaucoma B. cataracts C. vertigo D. Meniere's disease E. anosmia The risk of cataracts is greatly increased with exposure to ultraviolet radiation and smoking.
Bloom's Level: 3. Apply Learning Outcome: 18.07.03 Characterize the three most common causes of blindness in the United States. Section: 18.07 Topic: Senses
65. You go to the doctor because you have been experiencing dizziness. Where does the doctor look first? A. in your ears B. in your mouth C. in your eyes D. in your nose E. at your skin Dizziness is a sign of vertigo, which is related to ear problems, perhaps from an ear infection. The doctor would look in your ear to see if there was any sign of an infection.
Bloom's Level: 3. Apply Learning Outcome: 18.07.04 Describe some common disorders of hearing and equilibrium. Section: 18.07 Topic: Senses
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Chapter 18 - Senses
66. Impulses from the eye result in sight perception and impulses from the ear are perceived as sound. This is due to A. the neurons of the receptor, which use different ions to send signals to the brain. B. the neurons of the receptor, which send signals at different frequencies. C. the final destination of the signals. The signals from the eye travel to the visual perception region of the brain; auditory signals go to the hearing section. D. the type of signal sent by the neuron, for example, either a visual pattern or a sound pattern as the signal travels through the nerves. E. the different receptors (photoreceptors, mechanoreceptors) that are completely different types of nervous tissue cells. The different receptors send their signals to different destinations. It is the destination that determines the sensation and, therefore, the meaning of the stimulus.
Bloom's Level: 4. Analyze Learning Outcome: 18.01.03 Explain the significance of sensory transduction and sensory adaptation. Section: 18.01 Topic: Senses
67. Generally, the senses of the skin are somewhat specialized with A. Pacinian corpuscles, Ruffini's endings, and Krause's end bulbs detecting pain. B. Meissner's corpuscles and Merkel's disks detecting fine touch. C. one type of nerve cell detecting both hot and cold. D. all types spread uniformly in the dermal tissue throughout the body. E. temperature receptors, free nerve endings in the dermis. Meissner's corpuscles, Krause's end bulbs, Merkel's disks, and root hair plexuses are sensitive to fine touch. These are spread in a mosaic throughout the skin. Temperature receptors are free nerve endings in the epidermis. Some respond to cold, while others respond to warmth.
Bloom's Level: 4. Analyze Learning Outcome: 18.02.02 List specific types of cutaneous receptors that are sensitive to fine touch, pressure, and temperature. Section: 18.02 Topic: Peripheral Nervous System
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Chapter 18 - Senses
68. The correct order in which a light ray enters the eye is A. cornea → vitreous humor → lens → aqueous humor → retina. B. lens → vitreous humor → cornea → aqueous humor → retina. C. lens → cornea → aqueous humor → vitreous humor → retina. D. cornea → aqueous humor → lens → vitreous humor → retina. E. lens → aqueous humor → cornea → vitreous humor → retina. From the outside of the eye, light passes through the cornea, aqueous humor, lens, vitreous humor, and retina.
Bloom's Level: 4. Analyze Learning Outcome: 18.04.01 Identify the structures of the human eye. Section: 18.04 Topic: Nervous System
69. Which of the following is a FALSE statement? A. The iris regulates the amount of light reaching the retina. B. The rods are involved in black-and-white vision. C. There is a blind spot where the optic nerve joins the retina. D. The sclera is the outer coat of the eye. E. The anterior chamber is filled with vitreous humor. The anterior chamber of the eye is filled with aqueous humor. The posterior chamber is filled with vitreous humor.
Bloom's Level: 4. Analyze Learning Outcome: 18.04.01 Identify the structures of the human eye. Section: 18.04 Topic: Nervous System
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Chapter 18 - Senses
70. Which of the following is NOT a true statement about the rod photoreceptors? A. They contain the pigment rhodopsin. B. They are functional only in dim light. C. They detect even the slightest motion. D. They detect fine detail. E. The stimulus formed results from the breakdown of rhodopsin. The cone cells allow us to detect the fine detail and the color of an object.
Bloom's Level: 4. Analyze Learning Outcome: 18.04.03 Discuss the function and distribution of rods and cones. Section: 18.04 Topic: Senses
71. Which of the following is NOT true about cones? A. They detect color of an object. B. They detect detail of an object. C. They require high intensity light. D. Rods are more numerous in the retina than cones. E. They are absent from the fovea centralis. Cones are more concentrated in the fovea centralis where vision is most acute.
Bloom's Level: 4. Analyze Learning Outcome: 18.04.03 Discuss the function and distribution of rods and cones. Section: 18.04 Topic: Senses
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Chapter 18 - Senses
72. There are more rods than cones, but many rods may synapse with one ganglion cell while only one cone does. Therefore, when light intensity drops at nighttime, we can expect a lit city street scene to be A. intense in color and very sharp. B. more black and white, but very sharp. C. intense in color, but grainy. D. more black and white and grainy. E. identical to the image formed of the street in daytime. Since the light levels are dim, the rods are seeing the image. Therefore, we expect it to be more black and white and grainy.
Bloom's Level: 4. Analyze Learning Outcome: 18.04.03 Discuss the function and distribution of rods and cones. Section: 18.04 Topic: Senses
73. Which of the following is NOT necessary to "see" an object? A. retina B. optic nerve C. frontal lobe of the cerebrum D. nerve impulse E. bipolar cells The visual area is in the occipital lobe; therefore, the frontal lobe is not required for vision.
Bloom's Level: 4. Analyze Learning Outcome: 18.04.04 Trace the path of a nerve impulse from the retina to the visual cortex. Section: 18.04 Topic: Nervous System
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Chapter 18 - Senses
74. Which of the following groups contain all membranous structures? A. ossicles, tympanic membrane, oval window B. cochlea, ampulla, otoliths C. tympanic membrane, oval window, round window D. malleus, incus, stapes E. tympanic membrane, oval window, stapes The tympanic membrane, oval window, and round window are all three membranous structures.
Bloom's Level: 4. Analyze Learning Outcome: 18.05.01 List the parts of the ear that make up the outer ear, middle ear, and inner ear. Section: 18.05 Topic: Senses
True / False Questions 75. If you did not have hair within your inner ear, you would not be able to hear. TRUE It is the hair cells within the spiral organ that stimulate the cochlear nerve. Therefore, without these you would be deaf.
Bloom's Level: 4. Analyze Learning Outcome: 18.05.02 Describe the mechanism by which sound waves in the outer ear are converted into nerve impulses in the inner ear. Section: 18.05 Topic: Senses
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Chapter 18 - Senses
76. The fluid within the semicircular canals and the otoliths present in the utricle and saccule play very similar roles in our sense of equilibrium. TRUE The fluid in the semicircular canals bends the stereocilia of the hair cells and sends information to the brain about our rotational equilibrium. Similarly, the otoliths in the utricle and saccule bend the stereocilia of the utricle and saccule and send information to the brain about our gravitational equlibrium.
Bloom's Level: 4. Analyze Learning Outcome: 18.06.03 Compare the role of the fluid found in the semicircular canals with the otoliths present in the utricle and saccule. Section: 18.06 Topic: Senses
77. Stereocilia on hair cells are involved in both our rotational and our gravitational sense of equilibrium. TRUE The fluid in the semicircular canals bends the stereocilia on hair cells for rotational equilibrium, while the otoliths bend the stereocilia on hair cells for our gravitational sense of equilibrium.
Bloom's Level: 4. Analyze Learning Outcome: 18.06.02 Review the structures involved in rotational and gravitational equilibrium. Section: 18.06 Topic: Senses
18-38 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 18 - Senses
Multiple Choice Questions 78. Which is NOT a correct association of sensory problems? A. color blindness—lack of red or green cones B. deafness—infection with certain viruses such as the rubella virus C. cataracts—lens becomes opaque with aging and sun exposure D. nearsighted—unable to see near objects in detail E. otosclerosis—new bones overgrow the stirrup, requiring surgery Nearsighted individuals are unable to see far objects in detail.
Bloom's Level: 4. Analyze Learning Outcome: 18.07.02 Explain the anatomical abnormalities that cause color blindness, nearsightedness, farsightedness, and astigmatism. Learning Outcome: 18.07.04 Describe some common disorders of hearing and equilibrium. Section: 18.07 Topic: Senses
79. What change in shape will happen to the lens when an individual tries to focus on an image that is close? A. The lens will round up in order to bring the image to focus on the retina. B. The lens will flatten out in order to bring the image to focus on the retina. C. The lens will round up in order to bring the image to focus on the sclera. D. The lens will flatten out in order to bring the image to focus on the sclera. E. The lens does not change shape in order to focus on an image that is close up. When an individual focuses on an image that is close up, the lens will round up in order to bring the image to focus on the retina. The lens will flatten out when focusing on an image that is far away. The sclera does not play a role in focusing the image.
Bloom's Level: 4. Analyze Learning Outcome: 18.04.02 Explain how the eye focuses on near and far objects. Section: 18.04 Topic: Senses
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Chapter 18 - Senses
80. What is the correct order of sound transmission in the inner ear in order to determine pitch? A. tympanic membrane - malleus - incus - stapes - oval window B. tympanic membrane - incus - malleus - stapes - oval window C. oval window - tympanic membrane - incus - malleus - stapes D. tympanic membrane - incus - malleus - oval window - stapes E. stapes - tympanic membrane - incus - malleus - oval window The correct order of sound transmission in the inner ear in order to determine pitch is from the tympanic membrane - malleus - incus - stapes - oval window.
Bloom's Level: 4. Analyze Learning Outcome: 18.05.03 Explain how the inner ear is able to distinguish sounds of different pitches and volumes. Section: 18.05 Topic: Senses
81. What substance will displace the cupula during rotational equilibrium? A. fluid within the semicircular canal B. fluid within the utricle C. the otoliths within the semicircular canal D. the otoliths within the vestibule E. fluid within the auditory tube Fluid within the semicircular canal displaces the cupula during rotational equilibrium. The utricle does not contain fluid. The utricle is not found within the semicircular canals. The otoliths within the vestibule detect gravitational equilibrium. The auditory tube equalizes air pressure.
Bloom's Level: 4. Analyze Learning Outcome: 18.06.03 Compare the role of the fluid found in the semicircular canals with the otoliths present in the utricle and saccule. Section: 18.06 Topic: Senses
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Chapter 18 - Senses
82. What is the greatest benefit of having three semicircular canals involved in detecting rotational equilibrium? A. It allows for one to be in each dimension of space. B. It increases the surface area for detection of direction. C. It provides backups in case one gets damaged. D. It can process information faster. E. It produces larger volumes of calcium carbonate. The greatest benefit of having three semicircular canals involved in detecting rotational equilibrium is that it allows for one to be in each dimension of space. This enables the individual to determine which dimension the body is in at all times.
Bloom's Level: 4. Analyze Learning Outcome: 18.06.01 Explain how mechanoreceptors are involved in the sense of equilibrium. Section: 18.06 Topic: Senses
83. Which of these is true of all receptors? A. All receptors are sensitive to all the different types of stimuli. B. They are all stimulated by a change in the external environment. C. They are all found at the ends of neurons and initiate a nerve impulse that is carried to the brain. D. They not only receive stimuli but also interpret them. E. They are all located in the hypothalamus and skin. Each receptor responds to a different stimulus, some internal and some external. They do not interpret stimuli and they are located in different places all over the body. They are all found at the ends of neurons and initiate a nerve impulse that is carried to the brain.
Bloom's Level: 5. Evaluate Learning Outcome: 18.01.02 Describe the four categories of sensory receptors. Section: 18.01 Topic: Nervous System
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Chapter 18 - Senses
84. Both eyes "see" the entire visual field. However, because of the optic chiasma A. only data from the left half of the retina goes to the visual cortex where it is "seen." B. data from the left half of the retina goes to the right visual cortex and data from the right half of the retina goes to the left visual cortex. C. data from the right half of the retina goes to the right visual cortex and data from the left half of the retina goes to the left visual cortex. D. the image on the back of the retina is inverted compared to reality. E. objects in the blind spot will disappear. The optic chiasma has an X shape formed by crossing-over of the optic nerve fibers. Fibers from the right half of each retina converge on the right visual cortex, while data from the left half of each retina converges on the left visual cortex.
Bloom's Level: 5. Evaluate Learning Outcome: 18.04.04 Trace the path of a nerve impulse from the retina to the visual cortex. Section: 18.04 Topic: Senses
85. What causes "loudness"? A. The wave frequency or number of vibrations per second increases. B. The fluid in the cochlea vibrates with more force and this causes the basilar membrane to move more up and down with greater force. C. The distribution of hair cells that are stimulated changes. D. The louder the noise, the more both ears send the same nerve impulses. E. The auditory reception region of the brain simply assigns greater volume to sounds coming from sources that are visually recognized as being louder. Loud noises cause the fluid within the vestibular canal to exert more pressure and the basilar membrane to vibrate to a greater extent. The brain interprets this as volume.
Bloom's Level: 5. Evaluate Learning Outcome: 18.05.02 Describe the mechanism by which sound waves in the outer ear are converted into nerve impulses in the inner ear. Section: 18.05 Topic: Senses
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Chapter 18 - Senses
Short Answer Questions 86. List the function of each of the parts of the eye: Sclera: Ciliary body: Pupil: Rods: Optic nerve:
The functions of the following structures include: Sclera: Protects and supports the eyeball Ciliary body: Holds the lens in place and aids in accommodation Pupil: Admits light Rods: Makes black-and-white vision possible Optic nerve: Transmits impulses to the brain
Bloom's Level: 6. Create Learning Outcome: 18.04.01 Identify the structures of the human eye. Section: 18.04 Topic: Senses
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Chapter 18 - Senses
87. Complete the following table of exteroreceptors. Sensory Receptor Stimulus Category Taste Cells Olfactory cells Rods and cones in retina Hair cells in spiral organ Hair cells in semicircular canals Hair cells in vestibule
Sense
The completed table should look like this: Sensory Receptor Stimulus Category Sense Taste Cells Chemicals Chemoreceptors Taste Olfactory cells Chemicals Chemoreceptors Smell Rods and cones in retina Light Rays Photoreceptors Vision Hair cells in spiral organ Sound Rays Mechanoreceptors Hearing Hair cells in semicircular canals Motion Mechanoreceptors Rotational equilibrium Hair cells in vestibule Motion Mechanoreceptors Gravitational equilibrium
Bloom's Level: 6. Create Learning Outcome: 18.01.03 Explain the significance of sensory transduction and sensory adaptation. Section: 18.01 Topic: Senses
88. List the parts of the ear that make up the outer ear, middle ear, and inner ear. The outer ear consists of the pinna, earlobe, auditory canal, and modified sweat glands. The middle ear consists of the tympanic membrane, malleus, incus, stapes, auditory tube, oval window, and the round window. The inner ear consists of the semicircular canals, vestibule, and cochlea.
Bloom's Level: 6. Create Learning Outcome: 18.05.01 List the parts of the ear that make up the outer ear, middle ear, and inner ear. Section: 18.05 Topic: Senses
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Chapter 19 - Musculoskeletal System
Chapter 19 Musculoskeletal System
Multiple Choice Questions 1. What is the order you would encounter these tissues entering from the outside of the end of a long bone of the arm? A. articular cartilage → spongy bone → marrow → periosteum B. articular cartilage → periosteum → spongy bone → marrow C. periosteum → articular cartilage → spongy bone → marrow D. articular cartilage → marrow → spongy bone → periosteum E. spongy bone → articular cartilage → periosteum → marrow From the end of the bone, you would go through hyaline cartilage (articular cartilage), the periosteum, spong bone, and the bone marrow.
Bloom's Level: 2. Understand Learning Outcome: 19.01.01 Explain the macroscopic and microscopic structure of bone. Section: 19.01 Topic: Bone and Cartilage Structure
2. Which of these would you NOT expect to find in a long bone? A. compact bone B. spongy bone C. sarcomeres D. cartilage E. medullary cavity Sarcomeres are part of muscles, not bones.
Bloom's Level: 2. Understand Learning Outcome: 19.01.01 Explain the macroscopic and microscopic structure of bone. Section: 19.01 Topic: Bone and Cartilage Structure
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Chapter 19 - Musculoskeletal System
3. What structure allows the distribution of nutrients to osteocytes? A. canaliculi B. spongy bone C. osteoblasts D. growth plates E. diaphysis Canaliculi bring nutrients from the blood vessel in the central canal to the cells in the lacunae.
Bloom's Level: 1. Remember Learning Outcome: 19.01.01 Explain the macroscopic and microscopic structure of bone. Section: 19.01 Topic: Bone and Cartilage Structure
4. The type of bone tissue that has lacunae, concentric circles around a central canal, and are found in the middle portion of a long bone is A. spongy bone. B. compact bone. C. red marrow. D. yellow marrow. E. epiphyses. Compact bone is highly organized and composed of tubular units, called osteons, and a central canal. Marrow fills the cavities of the bone, and epiphyses are the ends of a long bone composed largely of spongy bone.
Bloom's Level: 2. Understand Learning Outcome: 19.01.01 Explain the macroscopic and microscopic structure of bone. Section: 19.01 Topic: Bone and Cartilage Structure
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Chapter 19 - Musculoskeletal System
5. A red bone marrow transplant would be used to A. increase fat storage. B. supply bone forming material or osteoblasts. C. supply blood cell forming material. D. help bones grow longer. E. increase the amount of dense fibrous connective tissue. Red bone marrow is a specialized tissue that produces all types of blood cells.
Bloom's Level: 3. Apply Learning Outcome: 19.01.01 Explain the macroscopic and microscopic structure of bone. Section: 19.01 Topic: Bone and Cartilage Structure
6. In modern surgery, it is possible to lengthen a short leg bone because new bone tissue can rapidly grow to fill a cavity. However, it would be important to A. administer a chemical to kill all osteoblasts. B. leave the periosteum intact to provide nutrients and define the boundary for bone growth. C. ensure the growth plates are all ossified. D. remove calcium from the diet. E. remove the red bone marrow from the bone. The periosteum contains blood vessels that enter the bone and give off branches that service the various types of cells found in bone. It would need to be intact in order to provide nutrients to the new bone tissue.
Bloom's Level: 4. Analyze Learning Outcome: 19.01.01 Explain the macroscopic and microscopic structure of bone. Section: 19.01 Topic: Bone and Cartilage Structure
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Chapter 19 - Musculoskeletal System
7. Which of the following is NOT a correct association? A. fibrocartilage—thick collagen fibers B. elastic cartilage—flexible ear flaps C. hyaline cartilage—larynx and trachea D. articular cartilage—a type of elastic cartilage E. hyaline cartilage—appears uniform and glassy Articular cartilage is a type of hyaline cartilage.
Bloom's Level: 2. Understand Learning Outcome: 19.01.02 Distinguish the structure and function of the three types of cartilage. Section: 19.01 Topic: Bone and Cartilage Structure
8. Which of the following attach muscle to bone? A. ligaments B. joints C. tendons D. adipose tissue E. sarcomere Tendons connect muscle to bone. Ligaments bind bone to bone.
Bloom's Level: 1. Remember Learning Outcome: 19.01.01 Explain the macroscopic and microscopic structure of bone. Section: 19.01 Topic: Human Skeletal System
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Chapter 19 - Musculoskeletal System
9. Scientists utilize carpet beetles to reduce an animal carcass to bones. Unlike bacteria that decompose all softer tissues, the beetles leave the bones of the skeleton connected to one another. Therefore, they have avoided consuming the A. ligaments. B. tendons. C. spongy bone. D. marrow. E. periosteum. Ligaments attach bone to bone. They would need to be intact for the skeleton to "stay together."
Bloom's Level: 3. Apply Learning Outcome: 19.01.01 Explain the macroscopic and microscopic structure of bone. Section: 19.01 Topic: Human Skeletal System
10. Which is NOT a correct association of bone structures and functions? A. osteoblasts—cells that build up bone tissue B. osteoclasts—cells that break down bone tissue C. osteons—tubular units of compact bone D. canaliculi—holes through which blood vessels run E. lacunae—tiny chambers in concentric circles around a central canal The canaliculi are tiny channels through which nutrients run. The hole through which blood vessels run is called the central canal.
Bloom's Level: 2. Understand Learning Outcome: 19.01.01 Explain the macroscopic and microscopic structure of bone. Section: 19.01 Topic: Bone and Cartilage Structure
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Chapter 19 - Musculoskeletal System
11. Bone-forming cells are termed A. chondrocytes. B. blood cells. C. osteoclasts. D. osteocytes. E. osteoblasts. Osteoblasts form bone. Osteoclasts break down bone. Osteocytes are the bone cells.
Bloom's Level: 1. Remember Learning Outcome: 19.01.03 Describe how bones change during a person’s lifetime. Section: 19.01 Topic: Bone and Cartilage Structure
12. What begins as stem cells in the red bone marrow and then specializes to break down bone and release the calcium into the blood? A. menisci B. chondrocytes C. osteoclasts D. osteoblasts E. sarcomeres Osteoclasts are large, multinucleated, macrophage-like cells that break down the spongy bone.
Bloom's Level: 1. Remember Learning Outcome: 19.01.03 Describe how bones change during a person’s lifetime. Section: 19.01 Topic: Bone and Cartilage Structure
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Chapter 19 - Musculoskeletal System
13. What cell type takes calcium from the blood? What cell type deposits calcium into the blood? A. osteoclasts; osteoblasts B. osteocytes; osteoclasts C. osteocytes; osteoblasts D. osteoblasts; osteocytes E. osteoblasts; osteoclasts Osteoblasts take calcium from the blood in order to build bone. Osteoclasts break down bone returning calcium to the blood. Osteocytes are bone cells.
Bloom's Level: 2. Understand Learning Outcome: 19.01.03 Describe how bones change during a person’s lifetime. Section: 19.01 Topic: Bone and Cartilage Structure
14. In young persons who are still developing, what is the band of cartilage called that remains between the primary ossification center and each secondary ossification center? A. chondrocytic mediastinum B. osteoclastic border C. osteon promoter D. fontanel E. growth plate A growth plate remains between the primary ossification center and each secondary center. The limbs keep increasing in length as long as the growth plates are present.
Bloom's Level: 1. Remember Learning Outcome: 19.01.03 Describe how bones change during a person’s lifetime. Section: 19.01 Topic: Bone and Cartilage Structure
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Chapter 19 - Musculoskeletal System
True / False Questions 15. All types of cartilage are associated with bones. TRUE All three types of cartilage, hyaline, fibro, and elastic, are associated with bones.
Bloom's Level: 1. Remember Learning Outcome: 19.01.02 Distinguish the structure and function of the three types of cartilage. Section: 19.01 Topic: Bone and Cartilage Structure
16. Jeremy has broken his leg during a soccer match. Although this is bad, from a healing standpoint, it is not as bad as if he had torn his cartilage. TRUE Cartilage has no blood supply and heals very slowly.
Bloom's Level: 3. Apply Learning Outcome: 19.01.02 Distinguish the structure and function of the three types of cartilage. Section: 19.01 Topic: Bone and Cartilage Structure
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Chapter 19 - Musculoskeletal System
Multiple Choice Questions 17. If your bones were "pulled out," you would perhaps resemble a large pancake with eyes. Which of the following would NOT be affected? A. substantial body movement (walking, etc.) B. storage of calcium, phosphorus, and fat C. blood cell production D. protection of brain and other soft organs E. detection of stimuli The skeletal system provides all of the above functions except for the detection of stimuli. That is a function of the nervous system.
Bloom's Level: 4. Analyze Learning Outcome: 19.02.01 Summarize the major functions of the skeleton. Section: 19.02 Topic: Skeletal System
18. Which of the following is NOT a function of bones? A. provide protection B. provide support C. allow flexible and controlled movement D. provide a storehouse of inorganic salts E. provide a nonliving architecture to build the body around Although the skeleton does provide structure for the body, it is a living system.
Bloom's Level: 2. Understand Learning Outcome: 19.02.01 Summarize the major functions of the skeleton. Section: 19.02 Topic: Skeletal System
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Chapter 19 - Musculoskeletal System
19. Which of the following bones is NOT part of the axial skeleton? A. ribs B. sternum C. scapula D. vertebrae E. skull The scapula, or shoulder blade, is part of the appendicular skeleton.
Bloom's Level: 1. Remember Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
20. Which one of the following is NOT a part of the appendicular skeleton? A. ribs B. fibula C. humerus D. coxal E. radius The ribs are part of the axial skeleton.
Bloom's Level: 1. Remember Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
Figure 19.1
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Chapter 19 - Musculoskeletal System
21. The humerus bone in Figure 19.1 is the bone labeled A. a B. b C. c D. d E. e The humerus is the large bone in the top of the arm—a.
Bloom's Level: 1. Remember Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
22. The radius bone in Figure 19.1 is the bone labeled A. a B. b C. c D. d E. e The radius is the bone in the lower arm on the thumb side—d.
Bloom's Level: 1. Remember Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
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Chapter 19 - Musculoskeletal System
23. A structure that reduces the weight of the skull and drains into the middle ear would be A. a fontanel. B. a diaphysis. C. yellow bone marrow. D. red bone marrow. E. a sinus. Some of the bones of the cranium contain sinuses, air spaces lined by mucous membranes. The sinuses reduce the weight of the skull, give a resonant sound to the voice, and, if filled with fluid, drain into the middle ear.
Bloom's Level: 1. Remember Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
24. Which bone is the only bone in the body that does not articulate with another bone? A. zygomatic bone B. mandible C. floating rib D. hyoid bone E. wrist bone The hyoid bone in the throat does not articulate with any other bone.
Bloom's Level: 1. Remember Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
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Chapter 19 - Musculoskeletal System
25. Which is NOT a correct association of facial bones? A. zygomatic bones—bridge of the nose B. mandible—lower jaw C. maxillae—upper jaw and hard palate D. temporal—temples E. frontal—forehead The zygomatic bones form the cheekbone prominences. The bridge of the nose is formed by the nasal bones.
Bloom's Level: 2. Understand Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
26. Which of the following is NOT true about the vertebrae? A. They are part of the appendicular skeleton. B. They form a dorsal backbone. C. They help form four curvatures of the spine. D. They are separated by intervertebral disks. E. They help protect the spinal cord. The vertebral column is part of the axial skeleton.
Bloom's Level: 2. Understand Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
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Chapter 19 - Musculoskeletal System
27. The correct order of regions of vertebrae, from superior to inferior, is A. thoracic → lumbar → sacrum → cervical → coccyx. B. cervical → lumbar → sacrum → thoracic → coccyx. C. cervical → thoracic → lumbar → sacrum → coccyx. D. lumbar → sacrum → cervical → thoracic → coccyx. E. sacrum → cervical → thoracic → lumbar → coccyx. From the head down are the cervical, thoracic, lumbar, sacrum, and coccyx vertebrae.
Bloom's Level: 2. Understand Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
28. What disease is an abnormal lateral curvature of the spine? A. rheumatoid arthritis B. bursitis C. osteoarthritis D. osteoporosis E. scoliosis Scoliosis is an abnormal lateral (sideways) curvature of the spine.
Bloom's Level: 1. Remember Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
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Chapter 19 - Musculoskeletal System
29. Which of the following would NOT be found in the pectoral girdle and associated appendages? A. clavicle B. scapula C. humerus D. tibia E. phalanges The tibia is in the leg. It is found in the pelvic, not the pectoral girdle.
Bloom's Level: 2. Understand Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
30. If you change your hands from a position where they are folded together to where they are cupped to collect a handful of water, you have mainly rotated which arm bone(s)? A. humerus B. scapula C. ulna D. radius E. carpels You have rolled the radius, the bone in the forearm on the thumb side, out.
Bloom's Level: 3. Apply Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
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Chapter 19 - Musculoskeletal System
31. The radius is to the ulna as the A. tibia is to the femur. B. fibula is to the ulna. C. fibula is to the tibia. D. humerus is to the femur. E. humerus is the radius. The radius and ulna are the two bones in the lower part of the arm. The tibia and fibula are the two bones in the lower part of the leg.
Bloom's Level: 4. Analyze Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
32. The patella is primarily held in place by tendons that attach it to the A. femur. B. fibula. C. tibia. D. calcaneus. E. tarsal bones. The patella is held in place by the quadriceps tendon, which attaches to the tibia.
Bloom's Level: 1. Remember Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
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Chapter 19 - Musculoskeletal System
33. Which of these bones is found in the pelvic girdle? A. humerus B. ulna C. femur D. coxal E. tibia The coxal forms the hip bone, which is part of the pelvic girdle.
Bloom's Level: 1. Remember Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
34. In Hong Kong, as in many Asian countries, there is not enough countryside to have everyone buried in cemeteries. Therefore, they require reburial of the human bones only in an ossarium, a wall that consists of many shoe box–sized cubicles, or concrete boxes, that hold all of the bones intact but disarticulated. The size of this box is about 10 inches high, 12 inches wide, and about 2 feet long. What are the largest skeletal structures that will determine the size of the box? A. humerus and skull B. ulna and pelvis C. femur and skull D. femur and pelvis E. tibia and humerus The femur is the longest bone in the skeleton that requires a box of this length. The skull is the widest bone that requires the box of this width.
Bloom's Level: 4. Analyze Learning Outcome: 19.02.02 List the major bones that comprise the axial and appendicular skeletons. Section: 19.02 Topic: Skeletal System
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Chapter 19 - Musculoskeletal System
35. Classification of joints is based on A. their location in the body. B. the presence or absence of tendons. C. their ability to move. D. the length of bones concerned. E. the type of muscles attached. Joints are classified based on their structure and their ability to move: immovable, slightly movable, or freely movable.
Bloom's Level: 2. Understand Learning Outcome: 19.02.03 Compare the structure and function of fibrous joints, cartilaginous joints, and synovial joints. Section: 19.02 Topic: Skeletal System
36. The crescent-shaped pieces of cartilage between the bones in the knee are called A. ligaments. B. tendons. C. patellas. D. bursae. E. menisci. The knee contains menisci, crescent-shaped pieces of hyaline cartilage between the bones. These give added stability and act as shock absorbers.
Bloom's Level: 1. Remember Learning Outcome: 19.02.03 Compare the structure and function of fibrous joints, cartilaginous joints, and synovial joints. Section: 19.02 Topic: Skeletal System
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Chapter 19 - Musculoskeletal System
37. The joints found between the vertebrae are classified as being A. immovable. B. slightly movable. C. freely movable. D. synovial joints. E. discoidal. The joints between the vertebrae are considered slightly movable. They are composed of fibrocartilage.
Bloom's Level: 1. Remember Learning Outcome: 19.02.03 Compare the structure and function of fibrous joints, cartilaginous joints, and synovial joints. Section: 19.02 Topic: Skeletal System
38. The knee is an example of a(n) A. immovable joint. B. slightly movable joint. C. fibrous joint. D. immovable synovial joint. E. movable, synovial joint. The knee and elbow joints are movable, synovial hinge joints.
Bloom's Level: 1. Remember Learning Outcome: 19.02.03 Compare the structure and function of fibrous joints, cartilaginous joints, and synovial joints. Section: 19.02 Topic: Skeletal System
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Chapter 19 - Musculoskeletal System
True / False Questions 39. The origin of a muscle is found on the bone that moves. FALSE The origin is on the stationary bone. The insertion is on the bone that moves.
Bloom's Level: 1. Remember Learning Outcome: 19.03.01 Explain how antagonistic muscle groups can move bones in different directions. Section: 19.03 Topic: Muscular System
40. When the biceps brachii contracts, the lower arm extends. FALSE When the biceps brachii contracts, the lower arm flexes (comes toward the upper arm).
Bloom's Level: 2. Understand Learning Outcome: 19.03.01 Explain how antagonistic muscle groups can move bones in different directions. Section: 19.03 Topic: Muscular System
Multiple Choice Questions 41. Skeletal muscles A. are found in and about internal organs. B. work in antagonistic pairs. C. never get longer when they contract. D. contain calcium throughout the cell. E. have the same origin and insertion point. Skeletal muscles work in antagonistic pairs. When one contracts, the antagonist relaxes.
Bloom's Level: 2. Understand Learning Outcome: 19.03.01 Explain how antagonistic muscle groups can move bones in different directions. Section: 19.03 Topic: Muscular System
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Chapter 19 - Musculoskeletal System
42. In an accident, the extensor muscles across the back of the fingers of one hand are completely severed. What is the result if no surgery is done to rejoin them? A. The fingers will dangle and extend out limp. B. The fingers will curl and remain permanently flexed. C. The fingers will forcibly extend and remain extended. D. Hand movements will continue as before the accident. E. It depends on how the muscles heal. The extensor muscles extend the fingers. Therefore, if they are cut, they fingers will curl and remain permanently flexed.
Bloom's Level: 3. Apply Learning Outcome: 19.03.02 Identify the major muscles in the human body. Section: 19.03 Topic: Muscular System
43. What is the name of the muscle that you sit on? A. deltoid B. quadriceps femoris group C. hamstring group D. gluteus maximus E. pectoralis major Your "bottom" is the gluteus maximus.
Bloom's Level: 1. Remember Learning Outcome: 19.03.02 Identify the major muscles in the human body. Section: 19.03 Topic: Muscular System
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Chapter 19 - Musculoskeletal System
44. Timothy has really "ripped" abdominal muscles. What muscle(s) has he exercised extensively to get that appearance? A. rectus abdominis B. biceps brachii C. latissimus dorsi D. external oblique E. adductor longus Your abdominal muscles are called rectus abdominis.
Bloom's Level: 3. Apply Learning Outcome: 19.03.02 Identify the major muscles in the human body. Section: 19.03 Topic: Muscular System
True / False Questions 45. The antagonistic muscle to the quadriceps femoris group is the gastrocnemius. FALSE The quadriceps femoris group is in the upper leg (thigh muscle), while the gastrocnemius is in the lower leg (calf muscle). Therefore, they cannot be an antagonistic pair.
Bloom's Level: 4. Analyze Learning Outcome: 19.03.02 Identify the major muscles in the human body. Section: 19.03 Topic: Muscular System
46. The deltoid muscle is shaped like a triangle. TRUE The Greek letter delta has the appearance of a triangle. This muscle is named for its shape.
Bloom's Level: 1. Remember Learning Outcome: 19.03.03 Describe six characteristics that are used to name skeletal muscles. Section: 19.03 Topic: Muscular System
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Chapter 19 - Musculoskeletal System
Multiple Choice Questions 47. Which of the following criteria is not used in the naming of muscles? A. size B. shape C. location D. name(s) of the bone(s) it attaches to E. number of attachments Bones are named for all of the above, except for the name of the bone to which it attaches.
Bloom's Level: 2. Understand Learning Outcome: 19.03.03 Describe six characteristics that are used to name skeletal muscles. Section: 19.03 Topic: Muscular System
48. Which of these gives the correct order from large to small? A. muscle → muscle cell → myofibril → sarcomeres → filaments B. muscle → muscle fibers → sarcomeres → filaments → myofibrils C. muscle → sarcomeres → myofibrils → actin filaments → myosin filaments D. muscle cells → myofibrils → filaments → sarcomeres E. muscle → myofibrils → sarcomeres → muscle cells → filaments A muscle is composed of many muscle cells. Within these muscle cells are many myofibrils, composed of sarcomeres. The sarcomeres are composed of filaments.
Bloom's Level: 2. Understand Learning Outcome: 19.04.01 Describe the microscopic structure of a muscle fiber. Section: 19.04 Topic: Skeletal Muscle Structure
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Chapter 19 - Musculoskeletal System
49. Actin and myosin filaments are both present in the dense region called the A. A band. B. Z line. C. H zone. D. I band. E. M band. The actin and myosin filaments overlap within the A band.
Bloom's Level: 2. Understand Learning Outcome: 19.04.02 Review the sliding filament model of muscle contraction. Section: 19.04 Topic: Skeletal Muscle Structure
50. In contraction A. the thin filaments become shorter. B. the thin filaments become longer. C. the thick filaments become shorter. D. the thick filaments become longer. E. neither of the filaments change length. Neither the thin actin or thick myosin filaments change in length. They change in position relative to one another.
Bloom's Level: 2. Understand Learning Outcome: 19.04.02 Review the sliding filament model of muscle contraction. Section: 19.04 Topic: Muscle Physiology
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Chapter 19 - Musculoskeletal System
51. The muscle action potential will cause the sarcoplasmic reticulum to release calcium ions when stimulated directly by the A. sarcolemma. B. neuromuscular junction. C. tropomyosin. D. axon terminal. E. T tubule. Nerve impulses travel down a T tubule and calcium is released from the sarcoplasmic reticulum.
Bloom's Level: 2. Understand Learning Outcome: 19.04.02 Review the sliding filament model of muscle contraction. Section: 19.04 Topic: Muscle Physiology
52. What is the correct order of structures that a motor nerve stimulation encounters when triggering a contraction? A. motor nerve → synaptic cleft → sarcoplasmic reticulum → sarcolemma → troponin B. motor nerve → synaptic cleft → sarcolemma → sarcoplasmic reticulum → troponin C. motor nerve → sarcoplasmic reticulum → synaptic cleft → sarcolemma → troponin D. motor nerve → sarcolemma → sarcoplasmic reticulum → synaptic cleft → troponin E. motor nerve → sarcolemma → synaptic cleft → sarcoplasmic reticulum → troponin The motor nerve causes the release of a neurotransmitter at the synaptic junction. The neurotransmitter binds to receptors in the sarcolemma, which generates an impulse that spreads down T tubules to the sarcoplasmic reticulum, causing the release of calcium ions. The calcium ions bind to troponin, initiating the movement of the filaments past each other.
Bloom's Level: 4. Analyze Learning Outcome: 19.04.02 Review the sliding filament model of muscle contraction. Section: 19.04 Topic: Muscle Physiology
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Chapter 19 - Musculoskeletal System
53. Name the neuromuscular transmitter that is released at the axon terminal. A. sodium ions B. calcium ions C. ATP D. creatine phosphate E. acetylcholine Axon terminals contain synaptic vesicles that are filled with the neurotransmitter acetylcholine (ACh).
Bloom's Level: 1. Remember Learning Outcome: 19.04.03 Summarize how activities within the neuromuscular junction control muscle fiber contraction. Section: 19.04 Topic: Muscle Physiology
54. Which statement is NOT true regarding cross bridges? A. They involve the use of Ca2+. B. They are used to split ATP. C. They form a complex with actin. D. They are extensions of myosin molecules. E. They attach to active sites on myosin. The cross bridges attach to active sites on the actin filament.
Bloom's Level: 4. Analyze Learning Outcome: 19.04.02 Review the sliding filament model of muscle contraction. Section: 19.04 Topic: Muscle Physiology
19-26 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
True / False Questions 55. The plasma membrane of a muscle cell is called a sarcolemma. TRUE The muscle fiber is a cell containing the usual cellular components, but special names have been assigned to some of them. The plasma membrane is called a sarcolemma.
Bloom's Level: 1. Remember Learning Outcome: 19.04.01 Describe the microscopic structure of a muscle fiber. Section: 19.04 Topic: Skeletal Muscle Structure
56. A neuromuscular junction operates much like a synapse between two nerve cells. TRUE They are very similar, except that the postsynaptic cell is a muscle cell instead of another nerve cell.
Bloom's Level: 5. Evaluate Learning Outcome: 19.04.03 Summarize how activities within the neuromuscular junction control muscle fiber contraction. Section: 19.04 Topic: Muscle Physiology
19-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
Multiple Choice Questions 57. The conversion that supplies the required energy for muscle contraction might be illustrated as A. ATP → ADP + (P). B. AMP → ATP. C. glucose → O2 + lactic acid. D. O2 → lactic acid + CO2. E. glucose + O2 → water + lactic acid. ATP being broken down to ADP + (P) is what powers muscle contraction.
Bloom's Level: 1. Remember Learning Outcome: 19.04.04 Indicate three ways that muscle cells can generate ATP. Section: 19.04 Topic: Muscle Physiology
58. Which of the following will increase during strenuous muscular activity? A. oxygen B. carbon monoxide C. lactic acid D. liver glycogen E. myosin Strenuous muscle activity is anaerobic. Therefore, glucose is being broken down to lactic acid.
Bloom's Level: 2. Understand Learning Outcome: 19.04.04 Indicate three ways that muscle cells can generate ATP. Section: 19.04 Topic: Muscle Physiology
19-28 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
59. Which molecule can be used directly to regenerate ATP? A. ADP B. glucose C. lactate D. creatine phosphate E. tropomyosin Creatine phosphate is a high-energy compound built up when the muscle is resting. It can regenerate ATP directly.
Bloom's Level: 2. Understand Learning Outcome: 19.04.04 Indicate three ways that muscle cells can generate ATP. Section: 19.04 Topic: Muscle Physiology
60. Oxygen debt may be associated with A. low oxygen demand by the cells. B. lactic acid buildup. C. too much oxygen in the body. D. too little carbon dioxide in the body. E. failure to convert potential energy into kinetic energy. When glucose is broken down anaerobically to lactic acid, it creates an oxygen debt. Oxygen is required to metabolize the glucose completely. The lactate can be changed back to pyruvate and metabolized in the mitochondria.
Bloom's Level: 2. Understand Learning Outcome: 19.04.04 Indicate three ways that muscle cells can generate ATP. Section: 19.04 Topic: Muscle Physiology
19-29 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
61. Which part of a muscle twitch occurs when the sarcomere distance is decreasing? A. latent period B. contraction period C. relaxation period D. refractory period E. expansion period When the sarcomere distance is decreasing, the muscle cell is contracting. The contraction period is when the muscle shortens.
Bloom's Level: 2. Understand Learning Outcome: 19.05.01 Explain what is happening to a muscle fiber during summation, tetanus, and fatigue. Section: 19.05 Topic: Muscle Physiology
62. If all of the muscle fibers in the neck, trunk, and legs were to totally relax, the body would collapse. What is the state called where some of the fibers in a particular muscle are always contracted? A. twitch B. threshold C. tetanus D. tone E. summation Even when muscles appear to be at rest, they exhibit muscle tone, in which some of their fibers are always contracting.
Bloom's Level: 2. Understand Learning Outcome: 19.05.02 Describe how muscles generate different levels of contractive force. Section: 19.05 Topic: Muscle Physiology
19-30 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
63. A maximal sustained muscle contraction is called A. a twitch. B. a threshold. C. tetanus. D. tone. E. summation. Maximal sustained contraction is called tetanus.
Bloom's Level: 1. Remember Learning Outcome: 19.05.01 Explain what is happening to a muscle fiber during summation, tetanus, and fatigue. Section: 19.05 Topic: Muscle Physiology
64. Judo is a sport that requires an explosion of energy rather than endurance. The better type of muscle would be A. slow-twitch fibers with many mitochondria and high blood supply. B. slow-twitch fibers with fewer mitochondria and lower blood supply. C. fast-twitch fibers with many mitochondria and high blood supply. D. fast-twitch fibers with fewer mitochondria and lower blood supply. E. fast-twitch fibers with a reserve of glycogen and fat. Fast-twitch fibers tend to be anaerobic and provide explosions of energy. They have fewer mitochondria and a lower blood supply.
Bloom's Level: 3. Apply Learning Outcome: 19.05.03 Explain the types of exercise that require more fast-twitch versus slow-twitch muscle fibers. Section: 19.05 Topic: Muscle Physiology
19-31 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
True / False Questions 65. A muscle but not the muscle fiber will obey the all-or-none law. FALSE A muscle fiber obeys the all-or-none law. However, a whole muscle has degrees of contraction.
Bloom's Level: 1. Remember Learning Outcome: 19.05.01 Explain what is happening to a muscle fiber during summation, tetanus, and fatigue. Section: 19.05 Topic: Muscle Physiology
66. Depending on the degree of stimulation, more and more individual muscle fibers contract, resulting in a stronger muscle contraction. TRUE Although each muscle fiber has an all-or-none contraction, a whole muscle can have varying degrees of contraction. Those varying degrees depend on the strength of the stimulation. The stronger stimulation recruits more muscle fibers to contract, resulting in a stronger whole muscle contraction.
Bloom's Level: 2. Understand Learning Outcome: 19.05.01 Explain what is happening to a muscle fiber during summation, tetanus, and fatigue. Section: 19.05 Topic: Muscle Physiology
67. Muscles that are not used or used for very weak contractions experience hypertrophy. FALSE Hypertrophy is an increase in muscle size. Muscles that are not used will atrophy or decrease in size.
Bloom's Level: 1. Remember Learning Outcome: 19.05.02 Describe how muscles generate different levels of contractive force. Section: 19.05 Topic: Muscle Physiology
19-32 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
Multiple Choice Questions 68. Jack broke his right leg in a football accident. He has had a cast on for about 2 months. Now that the cast is off, Jack has found that his right calf is significantly smaller than his left calf. Why is this? A. The unused right calf atrophied. B. When you break a bone, the attached muscle is also torn. Therefore, it gets smaller. C. Calf muscles are composed of fast-twitch fibers and tend to grow larger over time. D. The left calf probably hypertrophied in compensating for the broken leg during walking. E. This is highly unusual and suggests some type of muscle disease. The right calf has been inactive for 2 months and has therefore atrophied.
Bloom's Level: 3. Apply Learning Outcome: 19.05.03 Explain the types of exercise that require more fast-twitch versus slow-twitch muscle fibers. Section: 19.05 Topic: Muscle Physiology
69. In the movies, there are often scenes where someone is holding a person to prevent them from falling. All of a sudden, the person's grip gives way and the other person falls. What is this most likely due to? A. The person changes their mind and lets go. B. Fatigue, following sustained muscle contraction, is apparent when a muscle relaxes eventhough stimulation continues. C. A muscle twitch has a contraction period followed by a relaxation period. D. Some of the muscle fibers contract while others relax at any given time. When more muscle fibers relax than contract, the grip would be broken. E. The slow-twitch muscle fibers take over after the fast-twitch muscle fibers fail. These do not have the same strength and endurance as the fast-twitch fibers and the person lets go. Sustained muscle contraction results in tetanus. Tetanus continues until the muscle becomes fatigued and relaxes even though stimulation continues.
Bloom's Level: 3. Apply Learning Outcome: 19.05.01 Explain what is happening to a muscle fiber during summation, tetanus, and fatigue. Section: 19.05 Topic: Muscle Physiology
19-33 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
70. What is the condition in many older women in which the bones become weak and thin? A. rheumatoid arthritis B. bursitis C. osteoarthritis D. osteoporosis E. meningitis Osteoporosis is a condition in which bone loses mass and mineral content. It is found most often in older women.
Bloom's Level: 1. Remember Learning Outcome: 19.06.01 List three common disorders that affect the bones and joints and three that affect the muscles. Section: 19.06 Topic: Human Skeletal System
71. Which is NOT a correct association of diseases and description or symptoms? A. atrophy—decrease in size of muscles from lack of use or stimulation B. fibromyalgia—chronic pain in the muscles and ligaments C. osteoarthritis—degenerative joint disease D. osteoporosis—weakening or thinning of bone sometimes due to low estrogen levels E. rheumatoid arthritis—simple wear and tear on joints Rheumatoid arthritis is an autoimmune disease that results in debilitating effects on the joints.
Bloom's Level: 2. Understand Learning Outcome: 19.06.01 List three common disorders that affect the bones and joints and three that affect the muscles. Section: 19.06 Topic: Human Skeletal System
19-34 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
72. Which is not true of osteoarthritis? A. typically begins after age 40 B. is more common than rheumatoid arthritis C. may be treated by replacing damaged joints D. may be caused by overuse of a joint E. result of the immune system attacking the joints Rheumatoid arthritis is the result of an immune system attacking the joints.
Bloom's Level: 2. Understand Learning Outcome: 19.06.01 List three common disorders that affect the bones and joints and three that affect the muscles. Section: 19.06 Topic: Human Skeletal System
73. Which of the following can prevent osteoporosis? A. aspirin therapy B. estrogen replacement therapy C. treatment with chondroitin sulfate D. treatment with glucosamine E. treatment with corticosteroids Estrogen replacement therapy has been shown to increase bone mass and reduce fractures.
Bloom's Level: 2. Understand Learning Outcome: 19.06.02 Identify some of the most common steps that can be taken to help prevent osteoporosis. Section: 19.06 Topic: Human Skeletal System
True / False Questions 74. In a complete bone fracture, the break does not pass all the way through the bone. FALSE In a complete bone fracture the break passes all the way through the skin.
Bloom's Level: 1. Remember Learning Outcome: 19.06.01 List three common disorders that affect the bones and joints and three that affect the muscles. Section: 19.06 Topic: Human Skeletal System
19-35 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
75. A muscle cramp happens when the actin and myosin filaments "seize up." FALSE Muscle cramps actually don't originate in the muscle itself. They are caused by abnormal nerve activity.
Bloom's Level: 2. Understand Learning Outcome: 19.06.01 List three common disorders that affect the bones and joints and three that affect the muscles. Section: 19.06 Topic: Muscular System
Short Answer Questions 76. List the five major functions of the skeletal system. 1. The skeleton supports the body. 2. The skeleton protects the soft body parts. 3. The skeleton produces blood cells. 4. The skeleton stores minerals and fat. 5. The skeleton, along with the muscles, permits flexible body movements.
Bloom's Level: 6. Create Learning Outcome: 19.02.01 Summarize the major functions of the skeleton. Section: 19.02 Topic: Skeletal System
77. List the three types of cartilage and give a location of each type. 1. Hyaline cartilage is found in the joints at the ends of the long bones, in the nose, at the ends of the ribs and the larynx and trachea. 2. Fibrocartilage is found in the discs between the vertebrae and in the disks between the bones of the knees. 3. Elastic cartilage is found in the ear flaps and epiglottis.
Bloom's Level: 6. Create Learning Outcome: 19.01.02 Distinguish the structure and function of the three types of cartilage. Section: 19.01 Topic: Skeletal System
19-36 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
Multiple Choice Questions 78. What problems would occur if an individual were to tear their biceps brachii? A. They would be unable to flex the forearm. B. They would be unable to extend the forearm. C. They would be unable to abduct their arm. D. They would be able to curl their fingers. E. All of the answer choices could not be performed. The biceps brachii originates on the head of the humerus and the scapula and inserts on the radius. The action of the biceps brachii is to flex the forearm and rotate the hand outwards. The triceps brachii is responsible for extending the forearm. The deltoids are responsible for abducting the arm. The flexor digitorum is responsible for flexing or curling the fingers.
Bloom's Level: 5. Evaluate Learning Outcome: 19.03.01 Explain how antagonistic muscle groups can move bones in different directions. Section: 19.03 Topic: Muscular System
Short Answer Questions 79. Describe the six characteristics used to name skeletal muscles. 1. The size of the muscle is used to help name it (i.e., maximus means greatest) 2. The shape of the muscle. The deltoid is shaped like a triangle. 3. Location. The frontalis overlies the frontal bone. 4. Direction of the muscle fiber. Orbicularis oculi is a circular muscle around the eye. 5. Number of attachments. The biceps brachii has two attachments, or origins. 6. Action. The extensor muscles will extend the joint.
Bloom's Level: 6. Create Learning Outcome: 19.03.03 Describe six characteristics that are used to name skeletal muscles. Section: 19.03 Topic: Muscular System
19-37 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 19 - Musculoskeletal System
Multiple Choice Questions 80. If an athlete is training for the Olympic weight lifting team, what type of exercising should they be doing? A. Their training routine should include weight lifting, sprinting, or any other type of activity that requires an explosion of energy in order to increase their fast-twitch fibers. B. Their training routine should include weight lifting, sprinting, or any other type of activity that requires an explosion of energy in order to increase their slow-twitch fibers. C. Their training routine should include swimming, biking, jogging, and any other type of activity that requires an explosion of energy in order to increase their fast-twitch fibers. D. Their training routine should include swimming, biking, jogging, and any other type of activity that produces their energy aerobically in order to increase their slow-twitch fibers. Their training routine should include weight lifting, sprinting, or any other type of activity that requires an explosion of energy in order to increase their fast-twitch fibers. Weight lifting requires a large amount of fast-twitch fibers in order to produce an explosion of energy. Slow-twitch fibers are not used during weight lifting. Activities like swimming, biking, and jogging will help increase the number of slow-twitch fibers.
Bloom's Level: 5. Evaluate Learning Outcome: 19.05.03 Explain the types of exercise that require more fast-twitch versus slow-twitch muscle fibers. Section: 19.05 Topic: Muscle Physiology
19-38 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 20 - Endocrine System
Chapter 20 Endocrine System
Multiple Choice Questions 1. Which three endocrine glands are affected by hormones from the anterior pituitary? A. thymus, adrenal cortex, gonads B. thymus, parathyroids, pancreas C. thyroid, adrenal cortex, gonads D. adrenal cortex, gonads, pineal gland E. pineal gland, pancreas, parathyroid The anterior pituitary produces a hormone that stimulates the thyroid, an adrenocorticotropic hormone that stimulates the adrenal cortex, and gonadotropic hormones that stimulate the gonads.
Bloom's Level: 2. Understand Learning Outcome: 20.02.03 Recognize which three pituitary hormones act on other endocrine glands. Section: 20.02 Topic: Endocrine System
2. Which body system coordinates activities of body parts by releasing hormones into the blood? A. nervous system B. digestive system C. respiratory system D. circulatory system E. endocrine system The endocrine system consists of glands and tissue that secrete hormones that regulate the rest of the body.
Bloom's Level: 1. Remember Learning Outcome: 20.01.01 Identify the endocrine glands of the human body. Section: 20.01 Topic: Endocrine System
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Chapter 20 - Endocrine System
3. The production of hormones is usually controlled in what two ways? A. positive and negative feedback mechanisms B. neurosecretions and endocrine gland secretions C. cooperative hormones and antagonist hormones D. negative feedback and other hormone actions E. positive feedback and cooperative hormones The production of hormones is usually controlled in two ways: by negative feedback, and by the action of other hormones.
Bloom's Level: 2. Understand Learning Outcome: 20.01.01 Identify the endocrine glands of the human body. Section: 20.01 Topic: Hormones
True / False Questions 4. Endocrine glands have no ducts. TRUE Endocrine glands usually secrete their hormones into tissue fluid without ducts.
Bloom's Level: 1. Remember Learning Outcome: 20.01.01 Identify the endocrine glands of the human body. Section: 20.01 Topic: Endocrine System
20-2 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 20 - Endocrine System
Multiple Choice Questions 5. Which of the following does not pertain to hormones? A. are all carried in the bloodstream B. affect the behavior by acting on the brain C. influence the metabolism of cells D. essential components of the body's response to stress E. can promote cell division and mitosis Not all hormones are carried in the bloodstream. Some affect neighboring cells.
Bloom's Level: 4. Analyze Learning Outcome: 20.01.02 Compare the mechanism of action of peptide hormones and steroid hormones. Section: 20.01
6. Which of the following is not an endocrine gland? A. hypothalamus B. salivary glands C. pituitary D. thyroid E. adrenal cortex The salivary glands send saliva into the mouth by way of the salivary ducts. Therefore, they are not endocrine glands.
Bloom's Level: 4. Analyze Learning Outcome: 20.01.01 Identify the endocrine glands of the human body. Section: 20.01 Topic: Endocrine System
20-3 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 20 - Endocrine System
7. Jason is having problems with his metabolic rate. He cannot seem to burn calories and gains weight even on a severely reduced caloric intake. What endocrine gland is probably not involved in this problem? A. anterior pituitary B. thyroid C. thymus D. hypothalamus E. pancreas The thymus is involved in the maturation of T lymphocytes and would not be involved. The hypothalamus regulates the anterior pituitary. The anterior pituitary regulates the thyroid, and the thyroid increases metabolic rate. The pancreas regulates the blood sugar levels.
Bloom's Level: 3. Apply Learning Outcome: 20.01.01 Identify the endocrine glands of the human body. Section: 20.01 Topic: Endocrine System
8. When Lauren reached the age of puberty, she did not undergo the expected changes. Today, at 17, she has still not undergone puberty. What endocrine gland could be the issue? A. posterior pituitary B. parathyroid C. thymus D. pancreas E. adrenal cortex The adrenal cortex produces sex hormones that stimulate reproductive organs and bring about sex characteristics.
Bloom's Level: 3. Apply Learning Outcome: 20.01.01 Identify the endocrine glands of the human body. Section: 20.01 Topic: Endocrine System
20-4 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 20 - Endocrine System
9. Which of the following is NOT a correct association? A. thyroid—thyroxine B. anterior pituitary—FSH and LH C. adrenal gland—norepinephrine D. posterior pituitary—growth hormone E. pancreas—insulin The posterior pituitary produces antidiuretic hormone and oxytocin. Growth hormone is produced by the anterior pituitary.
Bloom's Level: 2. Understand Learning Outcome: 20.01.01 Identify the endocrine glands of the human body. Section: 20.01 Topic: Endocrine System
10. Which statement is NOT true about steroid hormones? A. It includes hormones such as estrogen. B. It does not bind to cell surface receptors. C. The hormone-receptor complex can enter the nucleus. D. The hormone-receptor complex can bind to chromatin. E. Steroid hormones act faster than peptide hormones. Because it takes longer to synthesize new proteins than to activate enzymes already present in the cells (the mechanism by which steroid hormones work), steroid hormones generally act more slowly than peptide hormones.
Bloom's Level: 5. Evaluate Learning Outcome: 20.01.02 Compare the mechanism of action of peptide hormones and steroid hormones. Section: 20.01 Topic: Hormones
20-5 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 20 - Endocrine System
11. Which statement is NOT true about peptide hormones? A. They are derived from peptides, proteins, polypeptides, and derivatives of amino acids. B. They bind to receptors on the cell surface. C. They initiate the formation of cyclic AMP inside the cell. D. They create an enzyme cascade effect. E. They enter the cell in order to have an effect. Peptide hormones bind to a receptor on the cell surface and do not need to enter the cell in order to have an effect.
Bloom's Level: 5. Evaluate Learning Outcome: 20.01.02 Compare the mechanism of action of peptide hormones and steroid hormones. Section: 20.01 Topic: Hormones
12. In the following pairs, which lists the first messenger followed by a second messenger? A. insulin; glucagon B. ACTH; cortisol C. TSH; thyroxin D. glucose; insulin E. peptide hormone; cAMP Peptide hormones or steroid hormones are first messengers. cAMP is a second messenger.
Bloom's Level: 2. Understand Learning Outcome: 20.01.03 Differentiate between first messengers and second messengers. Section: 20.01 Topic: Hormones
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Chapter 20 - Endocrine System
13. If a chemical agent acted to make the nuclear envelope pores impermeable to the passage of hormones, what effect would it have? A. Neither steroid nor peptide hormones would be effective. B. Steroid hormones would continue to be effective but peptide hormones would be ineffective. C. Peptide hormones would continue to be effective but steroid hormones would be ineffective. D. Both steroid and peptide hormones would continue to be effective. E. It would depend on the cell being affected. Since steroid hormones must enter the cell through the membrane, steroid hormones would be ineffective. However, since peptide hormones bind to a receptor on the cell surface, they would continue to be effective.
Bloom's Level: 4. Analyze Learning Outcome: 20.01.02 Compare the mechanism of action of peptide hormones and steroid hormones. Section: 20.01 Topic: Hormones
True / False Questions 14. Steroid hormones always have the same complex of four carbon rings. TRUE Steroid hormones always have the same complex of four carbon rings; however, the side chains differ. These side chains account for their different actions.
Bloom's Level: 1. Remember Learning Outcome: 20.01.02 Compare the mechanism of action of peptide hormones and steroid hormones. Section: 20.01 Topic: Hormones
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Chapter 20 - Endocrine System
15. The role of the second messenger is to carry the message of the hormone on to surrounding cells. FALSE The role of the second messenger is to start an enzyme cascade within the cell that was activated by the hormone.
Bloom's Level: 1. Remember Learning Outcome: 20.01.03 Differentiate between first messengers and second messengers. Section: 20.01 Topic: Hormones
Multiple Choice Questions 16. The part of the brain controlling the anterior pituitary gland secretions is the A. medulla. B. thalamus. C. cerebral cortex. D. hypothalamus. E. cerebellum. The hypothalamus regulates the internal environment in two ways: the autonomic system, and by controlling the glandular secretions of the pituitary gland.
Bloom's Level: 2. Understand Learning Outcome: 20.02.01 Describe the relationship between the hypothalamus and the pituitary gland. Section: 20.02 Topic: Nervous System
20-8 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 20 - Endocrine System
17. Melanocyte-stimulating hormone A. causes the mammary glands in the breasts to develop and produce milk. B. is a major regulator of daily cycles and timing of puberty. C. is in relatively high concentration in most humans. D. causes dramatic skin color changes in some fish, amphibians, and reptiles. E. is responsible for sleep and wake cycles. Melanocyte-stimulating hormone causes skin color changes in many fishes, amphibians, and reptiles. Its concentration in humans is very low.
Bloom's Level: 2. Understand Learning Outcome: 20.02.02 List two hormones released by the posterior pituitary and six produced by the anterior pituitary. Section: 20.02 Topic: Hormones
18. In contrast to most other hormones, oxytocin has what unique property? A. Oxytocin only has one effect on one tissue. B. Oxytocin is a sex hormone found in equal amounts in both males and females. C. The action of oxytocin builds in a positive feedback loop, rather than maintaining an equilibrium. D. Oxytocin affects all cells in the human body, not just one target tissue. E. Oxytocin is produced by the embryo and transferred across the placenta to the mother. Oxytocin causes uterine contractions during childbirth. It is regulated by a positive feedback system.
Bloom's Level: 2. Understand Learning Outcome: 20.02.02 List two hormones released by the posterior pituitary and six produced by the anterior pituitary. Section: 20.02 Topic: Hormones
20-9 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 20 - Endocrine System
19. Which of the following endocrine glands does NOT produce its own hormones, but stores hormones produced by the hypothalamus? A. thyroid B. adrenal cortex C. adrenal medulla D. posterior pituitary E. anterior pituitary The posterior pituitary stores antidiuretic hormone and oxytocin made by the hypothalamus.
Bloom's Level: 2. Understand Learning Outcome: 20.02.01 Describe the relationship between the hypothalamus and the pituitary gland. Section: 20.02 Topic: Endocrine System
20. The hypothalamus controls the anterior pituitary via A. nerve stimulation. B. blood osmotic concentrations. C. blood glucose concentrations. D. hormones in a portal blood system. E. ACTH. The hypothalamus controls the anterior pituitary by releasing hormones that it sends through a portal system.
Bloom's Level: 2. Understand Learning Outcome: 20.02.01 Describe the relationship between the hypothalamus and the pituitary gland. Section: 20.02 Topic: Nervous System
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Chapter 20 - Endocrine System
21. Which hormone stimulates the rate at which amino acids enter the cell, increasing protein synthesis? A. oxytocin B. melanocyte-stimulating hormone C. anti-diuretic hormone D. growth hormone E. gonadotropic hormone
Bloom's Level: 1. Remember Learning Outcome: 20.02.02 List two hormones released by the posterior pituitary and six produced by the anterior pituitary. Section: 20.02 Topic: Hormones
22. When Rebekah went in to deliver her first child, her labor did not progress. The doctors gave her a hormone to speed the birthing process. Which hormone did they give her? A. oxytocin B. prolactin C. ADH D. HGH E. epinephrine Oxytocin causes uterine contracts during childbirth and milk let-down when a baby is nursing.
Bloom's Level: 3. Apply Learning Outcome: 20.02.02 List two hormones released by the posterior pituitary and six produced by the anterior pituitary. Section: 20.02 Topic: Hormones
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Chapter 20 - Endocrine System
True / False Questions 23. The sex hormones produced by the gonads will create negative feedback and regulate the production of hormones from the anterior pituitary. TRUE The blood level of the last hormone in the hypothalamus-anterior pituitary-target gland (in this case gonad) control system exerts negative feedback over the secretions of the hypothalamus and the anterior pituitary.
Bloom's Level: 2. Understand Learning Outcome: 20.02.03 Recognize which three pituitary hormones act on other endocrine glands. Section: 20.02 Topic: Hormones
24. The thyroid is in the abdomen, on top of the kidneys. FALSE The thyroid is a large gland located in the neck. The adrenal glands are in the abdomen, on top of the kidneys.
Bloom's Level: 1. Remember Learning Outcome: 20.03.01 Identify the anatomical location of the thyroid and parathyroid glands. Section: 20.03 Topic: Endocrine System
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Chapter 20 - Endocrine System
Multiple Choice Questions 25. In the past, the parathyroid glands were sometimes mistakenly removed during surgery when another part of the body was removed. Which part was intended to be removed? A. thyroid B. tonsils C. adenoids D. appendix E. uterus The parathyroid glands are located on the posterior surface of the thyroid gland and were sometimes removed along with the thyroid because they were so small.
Bloom's Level: 1. Remember Learning Outcome: 20.03.01 Identify the anatomical location of the thyroid and parathyroid glands. Section: 20.03 Topic: Endocrine System
26. Which of the following hormones is produced by the thyroid gland and controls blood calcium levels? A. mineralocorticoids B. parathyroid hormone C. T4 D. calcitonin E. T3 Calcitonin is produced by the thyroid and helps control blood calcium levels. T3 and T4 are also produced by the thyroid but they do not control blood calcium levels. Parathyroid hormone helps control blood calcium levels but it is produced by the parathyroid glands.
Bloom's Level: 2. Understand Learning Outcome: 20.03.02 Distinguish between the functions of T3, T4, calcitonin, and parathyroid hormone. Section: 20.03 Topic: Hormones
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Chapter 20 - Endocrine System
27. Which of the following statements about T3 and T4 is not true? A. T4 has more potent activity than T3. B. They are both produced from tyrosine. C. They are both considered peptide hormones. D. T3 contains 3 iodine atoms while T4 contains 4 iodine atoms. E. Both increase the metabolic rate. T3 is much more potent than T4.
Bloom's Level: 4. Analyze Learning Outcome: 20.03.02 Distinguish between the functions of T3, T4, calcitonin, and parathyroid hormone. Section: 20.03 Topic: Hormones
28. Weakened bones can result from an oversecretion of the A. thyroid gland. B. adrenal gland. C. pancreas. D. parathyroid gland. E. pituitary. The parathyroid glands secrete parathyroid hormone, which promotes the activity of osteoclasts and the release of calcium from the bones.
Bloom's Level: 1. Remember Learning Outcome: 20.03.02 Distinguish between the functions of T3, T4, calcitonin, and parathyroid hormone. Section: 20.03 Topic: Hormones
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Chapter 20 - Endocrine System
29. Which statement is NOT correct about PTH? A. When calcium level rises, PTH secretion is inhibited. B. When calcium level lowers, PTH secretion is stimulated. C. PTH has the opposite effect of calcitonin. D. PTH stimulates calcium absorption from the gut. E. PTH decreases the activity of osteoclasts. PTH increases the activity of osteoclasts and thereby releases calcium from bones.
Bloom's Level: 2. Understand Learning Outcome: 20.03.02 Distinguish between the functions of T3, T4, calcitonin, and parathyroid hormone. Section: 20.03 Topic: Hormones
True / False Questions 30. We purchase and consume iodized salt because the thyroid needs iodine and the lack of iodine interferes with thyroid function. TRUE The primary source for iodine in our diets is iodized salt. The thyroid gland uses this iodine to make T3 and T4.
Bloom's Level: 2. Understand Learning Outcome: 20.03.02 Distinguish between the functions of T3, T4, calcitonin, and parathyroid hormone. Section: 20.03 Topic: Endocrine System
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Chapter 20 - Endocrine System
Multiple Choice Questions 31. Too much urine indicates A. too little ADH. B. too much ADH. C. too little ACTH. D. too much ACTH. E. an imbalance between the ADH and ACTH levels. Aldosterone (ADH) promotes the uptake of water, therefore too little ADH would result in more urine. ACTH does not control the release of the mineralocorticoids such as ADH.
Bloom's Level: 3. Apply Learning Outcome: 20.04.03 Distinguish between mineralocorticoid and glucocorticoid hormones. Section: 20.04 Topic: Hormones
32. Which structure is NOT stimulated by the anterior pituitary? A. thyroid B. adrenal cortex C. adrenal medulla D. testes and ovaries E. mammary glands The adrenal medulla is stimulated by the nervous system, not the anterior pituitary.
Bloom's Level: 2. Understand Learning Outcome: 20.04.01 Describe the location of the adrenal glands in the body. Section: 20.04 Topic: Nervous System
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Chapter 20 - Endocrine System
33. Where are the adrenal glands located? A. near the thyroid gland B. near the kidneys C. near the posterior pituitary D. near the pancreas E. near the gonads The adrenal glands sit atop the kidneys.
Bloom's Level: 1. Remember Learning Outcome: 20.04.01 Describe the location of the adrenal glands in the body. Section: 20.04 Topic: Endocrine System
34. What is the cascade of events that follows a stress or trauma to produce adrenal reaction? A. hypothalamus (ACTH-releasing hormone) → anterior pituitary (ACTH) → adrenal cortex (glucocorticoids) B. hypothalamus (ACTH-releasing hormone) → anterior pituitary (ACTH) → adrenal medulla (epinephrine and norepinephrine) C. anterior pituitary (ACTH) → hypothalamus (ACTH-releasing hormone) → adrenal medulla (epinephrine and norepinephrine) D. adrenal cortex (glucocorticoids) → anterior pituitary (ACTH) → hypothalamus (ACTHreleasing hormone) E. adrenal medulla (epinephrine and norepinephrine) → hypothalamus (ACTH-releasing hormone) → anterior pituitary (ACTH) The hypothalamus produces ACTH-releasing hormone, which causes the anterior pituitary to secrete ATCH. This acts upon the adrenal cortex and causes it to release glucocorticoids.
Bloom's Level: 4. Analyze Learning Outcome: 20.04.02 List the hormones produced by the adrenal medulla and adrenal cortex and provide a function for each. Section: 20.04
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Chapter 20 - Endocrine System
35. What hormones produced by the adrenal cortex regulate salt and water balance? A. adrenal steroids B. adrenalins C. glucocorticoids D. thymosins E. mineralocorticoids Mineralocorticoids regulate sodium ion and water reabsorption by the kidney.
Bloom's Level: 1. Remember Learning Outcome: 20.04.02 List the hormones produced by the adrenal medulla and adrenal cortex and provide a function for each. Section: 20.04 Topic: Hormones
36. Which of the following hormones will allow us to react to emergency situations? A. estrogen B. progesterone C. testosterone D. atrial natriuretic hormone E. norepinephrine Norepinephrine, along with epinephrine, bring about all the body changes associated with the "fight-or flight" reaction.
Bloom's Level: 2. Understand Learning Outcome: 20.04.02 List the hormones produced by the adrenal medulla and adrenal cortex and provide a function for each. Section: 20.04 Topic: Hormones
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Chapter 20 - Endocrine System
37. Cortisone is derived from the hormone cortisol that is released from the A. adrenal medulla. B. adrenal cortex. C. thyroid. D. parathyroid. E. posterior pituitary. The adrenal cortex releases cortisol.
Bloom's Level: 1. Remember Learning Outcome: 20.04.02 List the hormones produced by the adrenal medulla and adrenal cortex and provide a function for each. Section: 20.04 Topic: Hormones
38. Which of the following hormones is considered a glucocorticoid? A. aldosterone B. insulin C. thyroxin D. cortisol E. parathyroid hormone Cortisol is an example of a glucocorticoid. It promotes the breakdown of muscle proteins to amino acids and promotes the metabolism of fatty acids.
Bloom's Level: 2. Understand Learning Outcome: 20.04.03 Distinguish between mineralocorticoid and glucocorticoid hormones. Section: 20.04 Topic: Hormones
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Chapter 20 - Endocrine System
39. Which hormone is mismatched? A. thyroxin—thyroid B. parathyroid—calcium C. insulin—glucose D. aldosterone—iodine E. thymus—thymosin Aldosterone should be paired with Na+ regulation. T3 or T4 should be paired with iodine.
Bloom's Level: 4. Analyze Learning Outcome: 20.04.02 List the hormones produced by the adrenal medulla and adrenal cortex and provide a function for each. Section: 20.04 Topic: Hormones
40. Which of the following glands has both an endocrine and an exocrine function? A. mammary gland B. pancreas C. pituitary D. adrenal gland E. thyroid gland The pancreas has an exocrine function in the production of digestive juices and an endocrine function in the production of glucagon, insulin, and somatostatin.
Bloom's Level: 1. Remember Learning Outcome: 20.05.01 Explain how the pancreas has both endocrine and exocrine functions. Section: 20.05 Topic: Endocrine System
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Chapter 20 - Endocrine System
41. Which hormone will decrease blood glucose levels? A. thyroxin B. aldosterone C. cortisol D. insulin E. glucagons Insulin stimulates the uptake of glucose by cells, thereby decreasing the blood level.
Bloom's Level: 1. Remember Learning Outcome: 20.05.02 Describe the functions of insulin, glucagon, and somatostatin. Section: 20.05 Topic: Hormones
True / False Questions 42. Glucagon stimulates the liver to take up glucose in order to store it as glycogen. FALSE Glucagon stimulates the liver to break down glycogen to produce glucose.
Bloom's Level: 1. Remember Learning Outcome: 20.05.02 Describe the functions of insulin, glucagon, and somatostatin. Section: 20.05 Topic: Hormones
43. Somatostatin is produced by the pancreas, stomach, and small intestine. TRUE Somatostatin, also known as growth hormone inhibiting hormone, is produced by all three of these.
Bloom's Level: 1. Remember Learning Outcome: 20.05.02 Describe the functions of insulin, glucagon, and somatostatin. Section: 20.05 Topic: Hormones
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Chapter 20 - Endocrine System
Multiple Choice Questions 44. Which of the following is not an effect of insulin? A. The small intestine decreases the absorption of nutrients from the digestive tract. B. The liver stores glucose from the blood as glycogen. C. Muscle cells store glycogen and build protein. D. Adipose tissue uses glucose from blood to form fat. E. The glucose level drops. Somatostatin decreases the absorption of nutrients from the digestive tract.
Bloom's Level: 5. Evaluate Learning Outcome: 20.05.02 Describe the functions of insulin, glucagon, and somatostatin. Section: 20.05 Topic: Hormones
True / False Questions 45. The beta cells of the pancreas produce somatostatin, the alpha cells produce glucagon, and the delta cells produce insulin. FALSE The beta cells produce insulin and the delta cells produce somatostatin.
Bloom's Level: 1. Remember Learning Outcome: 20.05.02 Describe the functions of insulin, glucagon, and somatostatin. Section: 20.05 Topic: Hormones
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Chapter 20 - Endocrine System
Multiple Choice Questions 46. One side effect of taking testosterone as a steroid to increase muscle bulk is that the testes often shrink. Since the testes also produce testosterone, what is the likely mechanism for this shrinkage? A. The anterior pituitary assumes overproduction and drops the level of hormones that cause the testes to develop. B. Artificial testosterone causes a negative feedback to the anterior pituitary that in turn tries to stimulate ovary development. C. Testosterone has a negative feedback effect directly on testes. D. Artificial testosterone is toxic to testicular cells. E. Excess testosterone actually backfires and reverses the effects of puberty. The anterior pituitary/hypothalamus measures the testosterone levels and no longer sends out testes-stimulating hormones.
Bloom's Level: 3. Apply Learning Outcome: 20.06.01 Discuss the role of the testes and ovaries in the development of male and female secondary sex characteristics. Section: 20.06 Topic: Hormones
47. Which is NOT an effect of the hormone estrogen? A. accumulation of a fat layer beneath the skin B. pelvic girdle grows wider C. breast development D. egg maturation and menstrual cycle control E. growth of hair and muscular tissue Growth of hair and muscular tissue is an effect of testosterone.
Bloom's Level: 1. Remember Learning Outcome: 20.06.01 Discuss the role of the testes and ovaries in the development of male and female secondary sex characteristics. Section: 20.06 Topic: Hormones
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Chapter 20 - Endocrine System
48. Which gland will produce melatonin? A. pancreas B. pineal gland C. adrenal gland D. thyroid gland E. pituitary The pineal gland produces melatonin.
Bloom's Level: 1. Remember Learning Outcome: 20.06.02 Describe the functions of thymosin, melatonin, leptin, and prostaglandins. Section: 20.06 Topic: Hormones
49. Prostaglandins A. are distributed by blood like other hormones. B. cause muscles to relax. C. alleviate pain. D. are proteins produced in the brain. E. act locally. Prostaglandins are potent chemical signals produced within cells from a fatty acid that are not distributed in the blood. Instead, they act locally. They often cause muscle contraction and pain.
Bloom's Level: 2. Understand Learning Outcome: 20.06.02 Describe the functions of thymosin, melatonin, leptin, and prostaglandins. Section: 20.06 Topic: Hormones
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Chapter 20 - Endocrine System
50. Aspirin interferes with the synthesis of which molecule? A. leptin B. growth factors C. prostaglandins D. melatonin E. thymosins Aspirin controls pain because it inhibits the synthesis of prostaglandins.
Bloom's Level: 1. Remember Learning Outcome: 20.06.02 Describe the functions of thymosin, melatonin, leptin, and prostaglandins. Section: 20.06 Topic: Hormones
51. Which is NOT a correct association? A. leptin—signals satiety B. platelet-derived growth factor—increases the size of platelets C. nerve growth factor—promotes growth of nerves D. epidermal growth factor—stimulates growth of skin E. granulocyte and macrophage colony-stimulating factor—helps stem cells decide on blood cell production Platelet-derived growth factors help in wound healing and increase the number of fibroblasts, smooth muscle cells, and certain nervous system cells.
Bloom's Level: 5. Evaluate Learning Outcome: 20.06.03 Define and list two examples of growth factors. Section: 20.06 Topic: Hormones
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Chapter 20 - Endocrine System
52. A hormone that acts on the hypothalamus to signal satiety is A. melatonin. B. thymosin. C. leptin. D. growth hormone. E. prostaglandin. Leptin, produced by adipose tissue, acts on the hypothalamus to signal satiety.
Bloom's Level: 1. Remember Learning Outcome: 20.06.02 Describe the functions of thymosin, melatonin, leptin, and prostaglandins. Section: 20.06 Topic: Hormones
53. Older people tend to be more susceptible to infections because their immune systems get weaker. Which gland decreases in size as we age and could be responsible for this? A. thymus B. thyroid C. parathyroid D. pancreas E. testes in male; ovaries in females The thymus gland is the location of the maturation of the T lymphocytes of the immune system. With aging, it gets smaller and becomes fatty.
Bloom's Level: 3. Apply Learning Outcome: 20.06.02 Describe the functions of thymosin, melatonin, leptin, and prostaglandins. Section: 20.06 Topic: Endocrine System
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Chapter 20 - Endocrine System
54. College students are notorious for having a reversed day-night schedule. They stay up all night and sleep during the day. This alteration changes what hormone levels? A. melatonin B. leptin C. growth hormone D. testosterone in males; estrogen in females E. prostaglandins Melatonin helps establish our circadian rhythms, such as our wake-sleep cycle.
Bloom's Level: 3. Apply Learning Outcome: 20.06.02 Describe the functions of thymosin, melatonin, leptin, and prostaglandins. Section: 20.06 Topic: Hormones
True / False Questions 55. Leptin, growth factors, and prostaglandins are not produced by organs we call "glands." TRUE Leptin is produced by adipose tissue. Growth factors are produced by different types of organs and cells. Prostaglandins are produced by cells.
Bloom's Level: 2. Understand Learning Outcome: 20.06.02 Describe the functions of thymosin, melatonin, leptin, and prostaglandins. Section: 20.06 Topic: Hormones
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Chapter 20 - Endocrine System
Multiple Choice Questions 56. The condition that results when there is an increased production of human growth hormone in an adult is termed A. Cushing syndrome. B. Addison disease. C. gigantism. D. dwarfism. E. acromegaly. When growth hormone is overproduced in the adult, acromegaly results. When growth hormone is overproduced during childhood, the person may become a giant.
Bloom's Level: 2. Understand Learning Outcome: 20.07.01 Identify the cause of each of the following conditions: diabetes insipidus, pituitary dwarfism, gigantism, acromegaly, Cushing syndrome, and Addison disease. Section: 20.07 Topic: Endocrine System
57. Which is NOT a correct association for hormonal disorders? A. acromegaly—overproduction of GH in adult B. pituitary dwarf—too little GH produced during childhood C. Cushing syndrome—oversecretion of ACTH due to an adrenal cortex tumor D. diabetes—either the pancreas produces too much insulin or liver and muscle cells have too many receptors E. goiter—thyroid is unable to produce thyroxine due to too little iodine in diet In diabetes, either the pancreas produces too little insulin, or the liver and muscle cells have insulin receptor malfunctions.
Bloom's Level: 4. Analyze Learning Outcome: 20.07.01 Identify the cause of each of the following conditions: diabetes insipidus, pituitary dwarfism, gigantism, acromegaly, Cushing syndrome, and Addison disease. Section: 20.07 Topic: Hormones
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Chapter 20 - Endocrine System
58. Oversecretion of thyroid hormone results in A. Cushing syndrome. B. Addison disease. C. Graves disease. D. a simple goiter. E. acromegaly. The overproduction of thyroid hormones causes Graves disease.
Bloom's Level: 1. Remember Learning Outcome: 20.07.02 Discuss the specific causes and likely outcomes of hypothyroidism and hyperthyroidism. Section: 20.07 Topic: Hormones
59. A simple goiter can be prevented by A. surgery to remove the thyroid gland. B. removal of the pituitary. C. administration of ACTH. D. administration of insulin. E. increasing intake of iodine in the diet. If iodine is lacking in the diet, the thyroid gland is unable to produce sufficient amounts of T3 and T4 and the thyroid gland enlarges, resulting in a simple goiter.
Bloom's Level: 2. Understand Learning Outcome: 20.07.02 Discuss the specific causes and likely outcomes of hypothyroidism and hyperthyroidism. Section: 20.07 Topic: Hormones
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Chapter 20 - Endocrine System
60. Exophthalmia can sometimes be cured by A. injections of thyroid hormone. B. adding iodine to the diet. C. a pancreatic implant. D. surgically removing a portion of the pancreas, or destroying part of it with radiation. E. surgically removing a portion of the thyroid, or destroying part of it with radiation. Exophthalmia, excessive protrusion of the eyes, is a result of hyperthyroidism. Destroying or removing part of the thyroid can sometimes be curative.
Bloom's Level: 5. Evaluate Learning Outcome: 20.07.02 Discuss the specific causes and likely outcomes of hypothyroidism and hyperthyroidism. Section: 20.07 Topic: Endocrine System
61. Absence of the parathyroid hormone causes A. lethargy and hair loss. B. tetany. C. sleepiness. D. mental retardation. E. cretinism. When insufficient parathyroid hormone production leads to a dramatic drop in blood calcium levels, tetany results.
Bloom's Level: 2. Understand Learning Outcome: 20.07.02 Discuss the specific causes and likely outcomes of hypothyroidism and hyperthyroidism. Section: 20.07 Topic: Hormones
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Chapter 20 - Endocrine System
62. What is "tetany"? A. continuous muscle contractions due to high blood calcium B. continuous muscle contractions from low blood calcium C. the random contractions of fibers in a muscle that keep a person upright D. inability to function caused by low blood sugar due to diabetes E. ultimate muscle fatigue where the oxygen debt exceeds all ATP available Tetany is when the body shakes from continuous muscle contraction. It can be caused by a dramatic drop in blood calcium levels.
Bloom's Level: 2. Understand Learning Outcome: 20.07.02 Discuss the specific causes and likely outcomes of hypothyroidism and hyperthyroidism. Section: 20.07 Topic: Endocrine System
63. Which is NOT a correct consequence of surgical removal of portions of these glands? A. adrenal cortex—bronzing of skin, no increase in blood glucose levels in response to stress, dehydration and death B. pineal gland—decrease in T cell immune function C. parathyroid glands—drop in blood calcium level and tetany (muscles shake) D. ovaries—alteration in menstrual cycle and change in secondary sex characteristics E. adult thyroid—low pulse rate and body temperature and lethargy The removal of the thymus will result in the lack of T cell maturation. Removal of the pineal gland will result in the loss of circadian rhythms.
Bloom's Level: 4. Analyze Learning Outcome: 20.07.01 Identify the cause of each of the following conditions: diabetes insipidus, pituitary dwarfism, gigantism, acromegaly, Cushing syndrome, and Addison disease. Section: 20.07 Topic: Endocrine System
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Chapter 20 - Endocrine System
64. Which disease results when the adrenal cortex produces too much hormone? A. diabetes insipidus B. diabetes mellitus C. Cushing syndrome D. Addison disease E. myxedema Increased production of cortisol by the adrenal glands is known as Cushing syndrome.
Bloom's Level: 1. Remember Learning Outcome: 20.07.01 Identify the cause of each of the following conditions: diabetes insipidus, pituitary dwarfism, gigantism, acromegaly, Cushing syndrome, and Addison disease. Section: 20.07 Topic: Endocrine System
65. Which of the following is NOT true of type I diabetes? A. It is caused when the pancreas is not producing insulin. B. It is likely due to an environmental agent, perhaps a virus, that causes cytotoxic T cells to destroy islets. C. It is treated with insulin injections and pancreatic transplants. D. It is dangerous due to risk of acidosis, coma, and death. E. It is caused by obesity. Obesity is a risk factor for type II diabetes not type I.
Bloom's Level: 2. Understand Learning Outcome: 20.07.03 Compare and contrast type 1 versus type 2 diabetes mellitus. Section: 20.07 Topic: Endocrine System
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Chapter 20 - Endocrine System
66. Which of the following symptoms is NOT characteristic of diabetes mellitus? A. cells unable to take up glucose B. increased breakdown of fats and protein C. frequent urination D. sugar in the urine E. bronzing of the skin Bronzing of the skin is a characteristic of Addison disease, not diabetes mellitus.
Bloom's Level: 2. Understand Learning Outcome: 20.07.03 Compare and contrast type 1 versus type 2 diabetes mellitus. Section: 20.07 Topic: Endocrine System
67. What are the chemical signals that act between two individuals of the same species? A. pheromones B. first messenger C. second messenger D. growth factors E. None of the answer choices are chemical signals that communicate between two individuals of the same species. Pheromones are the chemical signals that act between two individuals of the same species. First and second messengers are a type of chemical signal that binds to receptor cells and causes the enzyme cascade to occur. Growth factors are types of hormones that promote cell division.
Bloom's Level: 1. Remember Learning Outcome: 20.01.04 Define pheromones, and provide an example of pheromone activity in humans. Section: 20.01 Topic: Hormones
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Chapter 20 - Endocrine System
68. What is the role of the second messenger? A. to set in motion the enzyme cascade B. convey information between two individuals of the same species C. binds to the cell surface D. All of the answer choices are functions of the second messenger. E. None of the answer choices are functions of the second messenger. The second messenger sets in motion the enzyme cascade. Pheromones convey information between two individuals of the same species. The first messenger binds to the cell surface.
Bloom's Level: 2. Understand Learning Outcome: 20.01.03 Differentiate between first messengers and second messengers. Section: 20.01 Topic: Hormones
69. What is the potential benefit for a woman to choose a mate who has a different major histocompatibility complex than themselves? A. increased immune response for their offspring B. increased intellectual ability of the offspring C. greater chance of fertility D. greater chance for attracting a successful mate E. increased chance of producing multiple offspring Having a different major histocompatibility complex can potentially lead to an increased immune response for their offspring. The major histocompatibility complex plays no role in intellectual ability, fertility, chances for attracting successful mates, or producing multiple offspring.
Bloom's Level: 3. Apply Learning Outcome: 20.01.04 Define pheromones, and provide an example of pheromone activity in humans. Section: 20.01 Topic: Hormones
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Chapter 20 - Endocrine System
70. Which region of the human body is least likely to produce pheromones? A. palms B. scalp C. oral cavity D. axilla E. genital area The palms are the area least likely to produce a pheromone. All other body areas produce pheromones.
Bloom's Level: 1. Remember Learning Outcome: 20.01.04 Define pheromones, and provide an example of pheromone activity in humans. Section: 20.01 Topic: Endocrine System
71. Which group of pituitary hormones will act upon other endocrine glands? A. TSH, ACTH, and gonadotropic hormones B. TSH, ADH, and gonadotropic hormones C. oxytocin, ACTH, and gonadotropic hormones D. ADH, oxytocin, and gonadotropic hormones E. PRL, MSH, and gonadotropic hormones TSH, ACTH, and gonadotropic hormones are produced by the anterior pituitary. These will all act upon other endocrine glands. ADH, oxytocin, PRL, and MSH do not act upon endocrine glands. They act upon the kidneys, uterus, mammary glands, and melanocytes in the skin, respectively.
Bloom's Level: 4. Analyze Learning Outcome: 20.02.03 Recognize which three pituitary hormones act on other endocrine glands. Section: 20.02 Topic: Hormones
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Chapter 20 - Endocrine System
72. Which statement is correct about the location of the parathyroid glands? A. The parathyroid glands are located on the posterior surface of the thyroid gland. B. The parathyroid glands are located on the anterior surface of the thyroid gland. C. The parathyroid glands are located on the posterior surface of the thymus gland. D. The parathyroid glands are located on the anterior surface of the thymus gland. E. The parathyroid glands are located on the superior surface of the thyroid gland. The parathyroid glands are located on the posterior surface of the thyroid gland.
Bloom's Level: 1. Remember Learning Outcome: 20.03.01 Identify the anatomical location of the thyroid and parathyroid glands. Section: 20.03 Topic: Endocrine System
73. Which statement is correct about the role of mineralocorticoids? A. Mineralocorticoids are used to regulate salt and water balances in the body, leading to an increase in the blood volume and blood pressure. B. Mineralocorticoids are used to regulate salt and water balances in the body, leading to a decrease in the blood volume and blood pressure. C. Mineralocorticoids are used to regulate carbohydrate, protein, and fat metabolism, leading to an increase in the blood glucose levels. D. Mineralocorticoids are used to regulate carbohydrate, protein, and fat metabolism, leading to an increase in the blood glucose levels. E. Mineralocorticoids are used to regulate carbohydrate, protein, and fat metabolism, leading to a decrease in the blood glucose levels. Mineralocorticoids are used to regulate salt and water balances in the body, leading to an increase in the blood volume and blood pressure. Glucocorticoids are used to regulate carbohydrate, protein, and fat metabolism, leading to an increase in the blood glucose levels.
Bloom's Level: 4. Analyze Learning Outcome: 20.04.03 Distinguish between mineralocorticoid and glucocorticoid hormones. Section: 20.04 Topic: Hormones
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Chapter 20 - Endocrine System
74. Which statement is correct about the location of adrenal glands in the body? A. The adrenal glands sit on the superior surface of the kidneys. B. The adrenal glands sit on the inferior surface of the kidneys. C. The adrenal glands sit on the superior surface of the thyroid. D. The adrenal glands sit on the lateral surface of the thyroid. E. The adrenal glands sit on the lateral surface of the kidneys. The adrenal glands sit on the superior surface of the kidneys.
Bloom's Level: 1. Remember Learning Outcome: 20.04.01 Describe the location of the adrenal glands in the body. Section: 20.04 Topic: Endocrine System
75. What is the exocrine role of the pancreas? A. The pancreas secretes digestive juices into the small intestine. B. The pancreas secretes digestive juices into the stomach. C. The pancreas secretes digestive juices into the large intestine. D. The pancreas produces insulin that is secreted into the bloodstream. E. The pancreas produces glucagon that is taken up by the liver. The pancreas secretes digestive juices into the small intestine, not into the stomach or large intestine. The secretion of insulin and glucagon are endocrine functions.
Bloom's Level: 2. Understand Learning Outcome: 20.05.01 Explain how the pancreas has both endocrine and exocrine functions. Section: 20.05 Topic: Endocrine System
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Chapter 20 - Endocrine System
76. What is an endocrine function of the pancreas? A. The pancreas secretes glucagon that stimulates the liver to break down glycogen to glucose. B. The pancreas secretes glucagon that stimulates the liver to break down glucose into glycogen. C. The pancreas secretes digestive juices into the small intestine. D. The pancreas secretes digestive juices into the stomach. E. The pancreas secretes digestive juices into the large intestine. The secretion of glucagon to stimulate the liver to break down glycogen to glucose is an endocrine function of the pancreas. The secretion of digestive juices into the small intestine is an exocrine function.
Bloom's Level: 2. Understand Learning Outcome: 20.05.01 Explain how the pancreas has both endocrine and exocrine functions. Section: 20.05 Topic: Endocrine System
77. What is the role of the epidermal growth factors? A. stimulate the healing of wounds B. increase the number of fibroblasts C. increase the number of smooth muscle cells D. Bone marrow stem cells form into different types of white blood cells. E. increase the chances of developing a wound The epidermal growth factors stimulate the healing of wounds. Platelet-derived growth factors increase the number of fibroblasts and the number of smooth muscle cells. Granulocyte-macrophage colony-stimulating factors cause the bone marrow stem cells to form into different types of white blood cells.
Bloom's Level: 4. Analyze Learning Outcome: 20.06.03 Define and list two examples of growth factors. Section: 20.06 Topic: Hormones
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Chapter 20 - Endocrine System
78. Which statement is correct about the role of the ovaries? A. Under the influence of gonadotropic hormones, the ovaries produce eggs as well as estrogen and progesterone. B. Under the influence of gonadotropic hormones, the ovaries produce eggs as well as androgens. C. The gonadotropic hormones are secreted by the posterior pituitary that will influence the ovaries to produce eggs as well as estrogen and progesterone. D. Under the influence of gonadotropic hormones, the ovaries produce sperm as well as estrogen and progesterone. E. Under the influence of gonadotropic hormones, the ovaries produce eggs as well as testosterone. Under the influence of gonadotropic hormones, the ovaries produce eggs as well as estrogen and progesterone. The production of androgens is a role of the testes. Gonadotropic hormones are secreted by the anterior pituitary, not the posterior pituitary. The production of sperm is done by the testes. Testosterone is an androgen produced by the testes not the ovaries.
Bloom's Level: 4. Analyze Learning Outcome: 20.06.01 Discuss the role of the testes and ovaries in the development of male and female secondary sex characteristics. Section: 20.06 Topic: Endocrine System
True / False Questions 79. Type 1 diabetes is the most common form of diabetes in the United States. FALSE Only about 10% of diabetics in the United States are type 1.
Bloom's Level: 1. Remember Learning Outcome: 20.07.03 Compare and contrast type 1 versus type 2 diabetes mellitus. Section: 20.07 Topic: Endocrine System
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Chapter 20 - Endocrine System
Short Answer Questions 80. List and explain the role of the main growth factors. Granulocyte-macrophage colony-stimulating factors cause bone marrow stem cells to form various types of white blood cells. Platelet-derived growth factors help in wound healing and cause an increase in the number of fibroblasts, smooth muscle cells, and cells of the nervous system. Epidermal growth factor and nerve growth factors stimulate the skin and nerve cells to heal.
Bloom's Level: 6. Create Learning Outcome: 20.06.03 Define and list two examples of growth factors. Section: 20.06 Topic: Hormones
81. List the six hormones produced by the anterior pituitary and indicate their primary function. 1. Thyroid-stimulating hormone: stimulates the thyroid to produce the thyroid hormones. 2. Adrenocorticotropic hormone: stimulates the adrenal cortex to produce cortisol. 3. Gonadotropic hormones: stimulate the gonads to produce gametes and sex hormones. 4. Prolactin: causes the mammary glands to develop and produce milk and suppress ovulation. 5. Melanocyte-stimulating hormone: causes skin color change in lower vertebrates. 6. Growth hormones: promote skeletal and muscular growth.
Bloom's Level: 6. Create Learning Outcome: 20.02.02 List two hormones released by the posterior pituitary and six produced by the anterior pituitary. Section: 20.02 Topic: Hormones
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Chapter 20 - Endocrine System
Multiple Choice Questions 82. What is the role of the granulocyte-macrophage colony-stimulating factors? A. Granulocyte-macrophage colony-stimulating factors cause the bone marrow stem cells to form into different types of white blood cells. B. Granulocyte-macrophage colony-stimulating factors cause the bone marrow stem cells to form into different types of red blood cells. C. increase the number of fibroblasts D. increase the number of smooth muscle cells E. stimulate the healing of wounds Granulocyte-macrophage colony-stimulating factors cause the bone marrow stem cells to form into different types of white blood cells, not red blood cells. Platelet-derived growth factors increase the number of fibroblasts and the number of smooth muscle cells. Epidermal growth factors stimulate wound healing.
Bloom's Level: 4. Analyze Learning Outcome: 20.06.03 Define and list two examples of growth factors. Section: 20.06 Topic: Hormones
Short Answer Questions 83. List the hormones produced by the adrenal medulla and give a function of each. Epinephrine and norepinephrine are produced by the adrenal medulla. Both bring about rapid changes that occur when an individual reacts to an emergency situation in a fight-or-flight manner. They provide a short-term response to stress.
Bloom's Level: 6. Create Learning Outcome: 20.04.02 List the hormones produced by the adrenal medulla and adrenal cortex and provide a function for each. Section: 20.04 Topic: Hormones
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Chapter 21 - Reproductive System
Chapter 21 Reproductive System
Multiple Choice Questions 1. Sperm mature within the A. prostate gland. B. vas deferens. C. seminal vesicles. D. ejaculatory duct. E. epididymis. While all of these structures are part of the male reproductive system the site of sperm maturation is the epididymis.
Bloom's Level: 1. Remember Learning Outcome: 21.01.01 Describe the path of sperm from the testes to the urethra. Section: 21.01 Topic: Human Male Reproductive System
2. The male reproductive system includes all EXCEPT which of the following? A. testes B. epididymis C. penis D. prostate E. fimbriae The fimbriae functions in the female reproductive system to sweep an oocyte into an oviduct.
Bloom's Level: 2. Understand Learning Outcome: 21.01.01 Describe the path of sperm from the testes to the urethra. Section: 21.01 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
3. Semen production does NOT involve the A. testes. B. prostate gland. C. seminal vesicles. D. bulbourethral glands. E. Graafian follicle. Sperm are produced by the testes and mature within the epididymides. The seminal vesicles, the prostate gland, and the bulbourethral glands add secretions to semen. The Graafian follicle is a vesicular follicle found in egg production in the female.
Bloom's Level: 2. Understand Learning Outcome: 21.01.02 List the organs that produce components of seminal fluid. Section: 21.01 Topic: Human Male Reproductive System
4. Seminal fluid contains A. androgens B. progesterones C. fructose D. prostaglandins E. estrogens
, chemicals that cause the uterus to contract.
Semen contains prostaglandins. These hormones are thought to cause the uterus to contract and help propel the sperm toward the egg. Androgens are male sex hormones that are secreted by cells that lie between the seminiferous tubules. Energy is needed for sperm to swim and semen contains the sugar fructose, which serves as an energy source. Estrogen and progesterone are female sex hormones.
Bloom's Level: 1. Remember Learning Outcome: 21.01.02 List the organs that produce components of seminal fluid. Section: 21.01 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
5. Trace the correct path of the sperm during ejaculation. A. urethra → vas deferens → seminal vesicles → testes B. testes → urethra → vas deferens → penis C. seminiferous tubules → epididymis → vas deferens → urethra D. seminiferous tubules → vas deferens → epididymis → urethra E. vas deferens → seminiferous tubules → epididymis → urethra The testes contain seminiferous tubules, where spermatogenesis occurs. Sperm are produced here and mature within the epididymides. When they leave the epididymides, they enter the vas deferens where they may also be stored for a time. At the time of ejaculation, sperm leave the penis through the urethra.
Bloom's Level: 3. Apply Learning Outcome: 21.01.01 Describe the path of sperm from the testes to the urethra. Section: 21.01 Topic: Human Male Reproductive System
6. The energy source for sperm to swim comes from A. stored fat molecules in the acrosome. B. stored sugar molecules in the sperm head. C. fructose sugar in the surrounding seminal fluids added by the seminal vesicles. D. protein breakdown by mitochondria in the middle of the sperm. E. dissolved lipids in the seminal fluids added by the prostate gland. The energy source for sperm is fructose sugar that is available to the sperm from the surrounding seminal fluid. Proteins and lipids (fats) are not used by the mitochondria of the sperm as an energy source.
Bloom's Level: 3. Apply Learning Outcome: 21.01.01 Describe the path of sperm from the testes to the urethra. Section: 21.01 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
7. In the human male, the tube used to carry both sperm and urine is the A. ureter. B. seminiferous tubule. C. vas deferens. D. urethra. E. fallopian tube. Semen enters the urethra, where rhythmic muscle contractions cause it to be ejaculated from the penis in spurts. During ejaculation, a sphincter normally closes off the urinary bladder so urine cannot enter the urethra, and no semen enters the bladder. (Notice that the urethra carries urine or semen at different times.) The ureter carries only urine, while the vas deferens and seminiferous tubules carry only sperm. The fallopian tube is part of the female reproductive system.
Bloom's Level: 2. Understand Learning Outcome: 21.01.01 Describe the path of sperm from the testes to the urethra. Section: 21.01 Topic: Human Male Reproductive System
8. Which association concerning structures in the human male is incorrect? A. testes—produce sperm B. vas deferens—carries urine C. prostate gland—seminal fluid D. urethra—conducts sperm E. seminal vesicle—seminal fluid When sperm leave an epididymis, they enter a vas deferens, where they may also be stored for a time. Each vas deferens passes into the abdominal cavity, where it curves around the urinary bladder and empties into an ejaculatory duct. The urethra carries urine from the bladder.
Bloom's Level: 2. Understand Learning Outcome: 21.01.01 Describe the path of sperm from the testes to the urethra. Section: 21.01 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
9. The time interval following ejaculation during which stimulation does not cause erection is called A. the luteal phase. B. the proliferative phase. C. the secretory phase. D. spermatogenesis. E. the refractory period. After ejaculation there is loss of sexual arousal and the penis returns to its normal, flaccid state. A male typically experiences a period of time when stimulation does not bring about an erection; this is called the refractory period. The length of the refractory period increases with age.
Bloom's Level: 1. Remember Learning Outcome: 21.01.02 List the organs that produce components of seminal fluid. Section: 21.01 Topic: Human Male Reproductive System
10. A functional advantage of having human testes in the scrotum rather than in the abdomen is A. lack of room in the abdomen. B. a shorter sperm duct. C. a more direct blood supply. D. lower temperature. E. greater protection to the testes. Since the scrotum extends outside the body, it provides lower temperatures than the abdomen, allowing sperm to be produced at a lower temperature. This is a functional advantage, because sperm produced at higher temperatures are not viable and there is a reduction in the production of sperm.
Bloom's Level: 3. Apply Learning Outcome: 21.01.01 Describe the path of sperm from the testes to the urethra. Section: 21.01 Topic: Human Male Reproductive System
21-5 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
11. In the human male, sperm cells are produced in the A. interstitial tissue. B. urethra. C. seminiferous tubules. D. vas deferens. E. seminal vesicles. Interstitial tissues produce male sex hormones called androgens. The vas deferens and seminal vesicles are part of the system that delivers sperm to the urethra. The testes contain seminiferous tubules, where spermatogenesis occurs.
Bloom's Level: 1. Remember Learning Outcome: 21.01.01 Describe the path of sperm from the testes to the urethra. Section: 21.01 Topic: Human Male Reproductive System
12. Which part of the sperm contains ATP-producing mitochondria? A. head B. mid-piece C. tail D. acrosome E. zona pellucida The head of the sperm is enclosed by an acrosome and contains DNA. The tail is a flagellum that is driven by sugars contained in the seminal fluid, which is converted into ATP by mitochondria found in the mid-piece. The zona pellucida is part of the mature egg.
Bloom's Level: 1. Remember Learning Outcome: 21.01.02 List the organs that produce components of seminal fluid. Section: 21.01 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
13. The role of the acrosome during the fertilization process is to A. contain enzymes that help a sperm head digest its way into an egg. B. increase the sperm's motility. C. direct the sperm to the egg. D. transport the chromosomes into the egg. E. store energy for swimming to the egg. The acrosome is a covering on the head of the sperm that contains enzymes that help the sperm penetrate the zona pellucida of the egg.
Bloom's Level: 2. Understand Learning Outcome: 21.01.02 List the organs that produce components of seminal fluid. Section: 21.01 Topic: Human Male Reproductive System
14. If we cut a cross-section through the testes, we would see A. individual sperm follicles with all future immature sperm present. B. large chambers lined with developing sperm. C. small tubules lined with developing sperm. D. flat layers of tissue like pages, lined with developing sperm. E. fluid-filled structures with developing sperm. A cross-section of the testes would show small, tightly coiled tubes called seminiferous tubules, which are the sites of spermatogenesis.
Bloom's Level: 3. Apply Learning Outcome: 21.01.02 List the organs that produce components of seminal fluid. Section: 21.01 Topic: Human Male Reproductive System
21-7 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
15. What is the function of the Sertoli cells in the male reproductive tract? A. produce an alkaline fluid added to the semen B. produce sperm cells through meiosis C. activate the sperm cells so they can swim rapidly D. support and regulate the spermatogenic cells E. produce a fluid added to the semen just before ejaculation Sertoli cells are found in the testes and are present as spermatogonia move through mitosis and meiosis to produce spermatids. Stertoli cells function to nourish and regulate primary spermatocytes until they develop into sperm.
Bloom's Level: 2. Understand Learning Outcome: 21.01.02 List the organs that produce components of seminal fluid. Section: 21.01 Topic: Human Male Reproductive System
16. Ultimate control of testes development is by the A. anterior pituitary; GnRH B. anterior pituitary; testosterone C. anterior pituitary; ICSH D. hypothalamus; GnRH E. hypothalamus; testosterone
, which secretes
.
The hypothalmus secretes the hormone gonadotropin-releasing hormone (GnRH), which stimulates the anterior pituitary to secrete follicle-releasing hormone (FSH) and luteinizing hormone (LH). FSH promotes the production of sperm.
Bloom's Level: 1. Remember Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.01 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
17. The male sex hormones are secreted by which of the following? A. spermatids B. interstitial cells C. vas deferens D. prostate E. seminiferous tubules Interstitial cells secrete androgens, or male sex hormones, that include testosterone. The other choices are all part of the male reproductive system, but do not secrete these hormones.
Bloom's Level: 1. Remember Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.01 Topic: Human Male Reproductive System
18. Which of the following is responsible for controlling the production of male sex hormones? A. testosterone B. progesterone C. GnRH D. FSH E. LH There are two gonadotropic hormones, follicle-stimulating hormone (FSH) and luteinizing hormone (LH). They are found in both males and females. LH controls the production of testosterone by the interstitial cells. Testosterone is the main sex hormone in males and is essential for the normal development and functioning of male reproductive systems.
Bloom's Level: 3. Apply Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.01 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
19. Why do testes shrink when male athletes take synthetic steroid testosterone hormones? A. Testosterone itself has the direct effect of shrinking the testes. B. Synthetic chemicals are not the same in action as natural chemicals. C. The guilt reaction in the brain causes a reverse hormonal action. D. The pituitary detects high levels of testosterone in the bloodstream and reduces FSH and LH. E. Scientists have no explanation for this phenomenon that is opposite of expected. FSH and LH are gonadotropic hormones that control the production of sperm. These hormones are involved in a negative feedback relationship and maintain relatively constant concentrations of sperm and testosterone. If there is an artificial source, the body will decrease the natural source of testosterone causing a decrease in size of the testes.
Bloom's Level: 3. Apply Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.01 Topic: Human Male Reproductive System
20. LH stimulates the of the testis to produce A. seminiferous tubules; sperm B. seminiferous tubules; testosterone C. interstitial cells; sperm D. interstitial cells; testosterone E. interstitial cells; estrogen
.
Luteinizing hormone (LH) controls the production of testosterone by the interstitial cells.
Bloom's Level: 1. Remember Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.01 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
21. LH is an abbreviation for luteinizing hormone, which is present in both males and females. In the male, LH A. has no known physiological function. B. exists in rudimentary levels since LH is made by the anterior pituitary. C. is exactly the opposite chemical from male hormones, in an antibody-antigen fashion. D. controls production of testosterone. E. causes the development of spermatids. Luteinizing hormone (LH), in males, controls the production of testosterone in the interstitial cells.
Bloom's Level: 1. Remember Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.01 Topic: Human Male Reproductive System
22. Which structure in humans produces FSH? A. hypothalamus B. ovary C. uterus D. pituitary E. interstitial cells The hypothalamus secretes the hormone gonadotropin-releasing hormone (GnRH), which stimulates the anterior pituitary to secrete follicle-releasing hormone (FSH) and luteinizing hormone (LH). FSH promotes the production of sperm.
Bloom's Level: 2. Understand Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.01 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
23. Secondary sexual characteristics in the male are directly maintained by the hormone A. testosterone. B. estrogen. C. FSH. D. LH. E. progesterone. Estrogen and progesterone are hormones found in females. FSH promotes the production of sperm and LH controls the production of testosterone, which is the main sex hormone in males and is essential for the normal development and functioning of the male reproductive system. LH leads to the secondary characteristics associated with males.
Bloom's Level: 1. Remember Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.01 Topic: Human Male Reproductive System
24. Castrating prepubescent bulls causes them to grow in female proportions because A. castration damages the brain, which then switches its hormonal control from the pituitary to produce more estrogens. B. small levels of estrogens are produced by the adrenal glands and fat tissue; with the lack of testosterone, a body develops along female patterns. C. castration allows ovaries to develop and secrete estrogens. D. the presence of live sperm is necessary for the growth of horns, shoulder muscles, and other bull features. E. the resulting lack of gonadal hormone causes the adrenal cortex to overproduce sex hormones. Castration is the removal of the testes. When they are removed, the production of testosterone in the interstitial cells of the testes can no longer occur. In this case, the influences of estrogen in the adrenal glands and fat tissue leads to the development of the body according to this female hormone, producing a female pattern.
Bloom's Level: 4. Analyze Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.01 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
25. Another name for the oviduct is the A. uterus. B. urethra. C. cervix. D. fallopian tube. E. ureter. The fallopian tube, also called the oviduct, is the means of transport of an oocyte from the ovary to the uterus.
Bloom's Level: 1. Remember Learning Outcome: 21.02.01 Identify where an oocyte is produced and how it is transported to the uterus. Section: 21.02 Topic: Human Female Reproductive System
26. Fertilization in humans normally occurs in the A. uterus. B. vagina. C. abdominal cavity. D. ovarian follicle. E. oviducts. Oocytes are released from the ovary and then move into the oviduct. If fertilization takes place, the fertilized egg then moves into the uterus, where it becomes implanted in the uterine lining.
Bloom's Level: 1. Remember Learning Outcome: 21.02.01 Identify where an oocyte is produced and how it is transported to the uterus. Section: 21.02 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
27. The egg is propelled down the oviduct by A. cilia on the layer of nurse cells surrounding the egg. B. simply drifting on the prevailing current of body fluids. C. oviduct cilia and tubular muscle contractions. D. suction from the uterine contractions. E. flagella present on the egg. Once in the oviduct, the oocyte is propelled by ciliary movement and tubular muscle contraction toward the uterus. The egg itself does not have cilia or flagella to propell it forward, yet the movement must be controlled so that the reproductive cycle occurs in a regulated time frame.
Bloom's Level: 2. Understand Learning Outcome: 21.02.01 Identify where an oocyte is produced and how it is transported to the uterus. Section: 21.02 Topic: Human Female Reproductive System
28. Which one of these is NOT part of the human female external genitalia? A. mons pubis B. glans clitoris C. labia minora D. labia majora E. uterus All of the answer choices are external features of the female reproductive system except the uterus, which is an internal structure.
Bloom's Level: 2. Understand Learning Outcome: 21.02.02 Describe the major components of the female external genitalia. Section: 21.02 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
29. The glans clitoris in the female is homologous to the A. testes B. vas deferens C. penis D. scrotum E. prostate gland
in the male.
In females, the glans clitoris, which is extremely sensitive, plays a role in the sexual response of females and swells to two or three times its usual size during intercourse. The penis in the male also swells during an erection as its spongy erectile tissue becomes distended with blood.
Bloom's Level: 2. Understand Learning Outcome: 21.02.02 Describe the major components of the female external genitalia. Section: 21.02 Topic: Human Female Reproductive System
30. The order of the structures in the vulva, from the frontal mons pubis to the anus, is A. glans clitoris → vagina → urethra. B. glans clitoris → urethra → vagina. C. urethra → glans clitoris → vagina. D. vagina → urethra → glans clitoris. E. vagina → glans clitoris → urethra. When studying the external features of female reproduction, they are aligned from the mons pubis to the anus in this order: glans clitoris → urethra → vagina.
Bloom's Level: 2. Understand Learning Outcome: 21.02.02 Describe the major components of the female external genitalia. Section: 21.02 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
31. How does orgasm in the human male differ from orgasm in the female? A. The female refractory period is longer than the male refractory period. B. Blood vessel dilation and tissue congestion only occur in the male. C. The intensity of pleasure is far greater for the male. D. The female can experience multiple orgasms during one sexual experience. E. All of the answer choices are accurate distinctions. The female refractory period is shorter than the male refractory period, but the intensity of pleasure experienced varies by individual, not by gender. Both males and females experience the dilation of blood vessels as arousal occurs. Males can experience one orgasm while females can experience multiple orgasms.
Bloom's Level: 3. Apply Learning Outcome: 21.02.03 Describe the events that occur during a female orgasm. Section: 21.02 Topic: Human Female Reproductive System
32. Which of the following statements is incorrect? A. The follicles in the ovary produce estrogen. B. The corpus luteum produces progesterone. C. The placenta can produce both estrogen and progesterone. D. LH stimulates the formation of the corpus luteum. E. A surge of FSH is believed to promote ovulation. FSH stimulates a follicle within the ovary to produce estrogen, causing the endometrium to thicken. An estrogen spike causes a sudden secretion of a large amount of GnRH from the hypothalamus, leading to a surge of luteinizing hormone (LH) from the anterior pituitary. This, in turn, causes ovulation at about the 14th day of a 28-day cycle.
Bloom's Level: 2. Understand Learning Outcome: 21.03.01 List the stages of the ovarian cycle and explain what is occurring in each stage. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
33. The structure from which an egg is released during ovulation is called a(n) A. ovarian cyst. B. primary follicle. C. corpus luteum. D. secondary follicle. E. vesicular (Graafian) follicle. Primary and secondary follicles are precursor stages to produce the oocyte within the vesicular (Graafian) follicle. The corpus luteum is the structure that remains in the ovary once the secondary oocyte has been released. Ovarian cysts are fluid-filled sacs that develop on the ovaries.
Bloom's Level: 1. Remember Learning Outcome: 21.03.01 List the stages of the ovarian cycle and explain what is occurring in each stage. Section: 21.03 Topic: Human Female Reproductive System
34. LH stimulates the A. seminiferous tubules of the testis. B. pituitary to produce ACTH. C. follicle to produce estrogen. D. corpus luteum to produce progesterone. E. interstitial cells to produce estrogen. After ovulation and during the luteal phase of the ovarian cycle, LH promotes the development of the corpus luteum. The corpus luteum, in turn, produces increased levels of progesterone, which causes the endometrium to double or triple in thickness.
Bloom's Level: 3. Apply Learning Outcome: 21.03.01 List the stages of the ovarian cycle and explain what is occurring in each stage. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
35. Which of the following hormones is best associated with maturation of the egg? A. follicle-stimulating hormone B. luteinizing hormone C. testosterone D. human chorionic gonadotropin hormone E. progesterone Follicle-stimulating hormone (FSH) is produced by the anterior pituitary and during the follicular phase it promotes the development of a follicle in the ovary.
Bloom's Level: 3. Apply Learning Outcome: 21.03.02 Describe the process of oogenesis. Section: 21.03 Topic: Gametogenesis
36. Which of the following is NOT true about estrogen? A. Estrogen causes the endometrium to thicken. B. Estrogen causes the endometrium to become vascular and glandular. C. Estrogen causes a positive feedback on the hypothalamus to secrete GnRH. D. Estrogen causes a surge in LH production. E. Estrogen stimulates the release of FSH. FSH stimulates the follicle to produce estrogen and, as the estrogen level in the blood rises, it exerts negative feedback control over the anterior pituitary's secretion of FSH so that the follicular phase comes to an end.
Bloom's Level: 2. Understand Learning Outcome: 21.03.03 Summarize how estrogen and progesterone affect the uterus. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
37. In the human female, estrogen and progesterone exert feedback control over the A. pituitary. B. ovary. C. uterus. D. testes. E. prostate gland. Estrogen and progesterone exert feedback control over the pituitary gland leading to changes in the production of FSH so that the follicular phase comes to and end.
Bloom's Level: 2. Understand Learning Outcome: 21.03.03 Summarize how estrogen and progesterone affect the uterus. Section: 21.03 Topic: Human Female Reproductive System
38. In the human female, the uterine cycle is an average A. 20 days. B. 24 days. C. 28 days. D. 32 days. E. 36 days. The average uterine cycle of a human female is based on a 28-day cycle. This is further divided into predictable events that occur during those 28 days.
Bloom's Level: 1. Remember Learning Outcome: 21.03.01 List the stages of the ovarian cycle and explain what is occurring in each stage. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
39. During a 28-day uterine cycle, ovulation usually occurs on the A. 5th day. B. 9th day. C. 14th day. D. 20th day. E. 25th day. The average uterine cycle of a human female is based on a 28-day cycle. This is further divided into predictable events that occur during those 28 days, including ovulation, which usually occurs on the 14th day of the cycle.
Bloom's Level: 1. Remember Learning Outcome: 21.03.02 Describe the process of oogenesis. Section: 21.03 Topic: Human Female Reproductive System
40. After ovulation, the ruptured follicle A. disappears and all of its cells disintegrate. B. passes on as waste material down the oviduct with the egg. C. mends itself and begins the maturation of another egg. D. becomes a part of the epithelial covering of the ovary. E. differentiates into the corpus luteum. After ovulation, the ruptured follicle becomes the corpus luteum, which produces progesterone. If fertilization does not occur, the corpus luteum degenerates and the low level of sex hormones in the female body leads to the breakdown of the endometrium during menstruation.
Bloom's Level: 1. Remember Learning Outcome: 21.03.01 List the stages of the ovarian cycle and explain what is occurring in each stage. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
41. Menstruation begins because A. the pituitary triggers it. B. LH activity is at its peak. C. estrogen activity is at a peak. D. progesterone production is highest. E. progesterone and estrogen levels are both declining. Once the follicle has ruptured, the remains of the follicle become the corpus luteum and produce progesterone and some estrogen. These hormones, in turn, lead to the thickening of the endometrium as it is prepared it to receive the fertilized egg. If fertilization does not occur, the corpus luteum degenerates and levels of progesterone and estrogen drop, leading to menstruation.
Bloom's Level: 3. Apply Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Human Female Reproductive System
42. The uterine cycle at day 12 would primarily be under the influence of A. estrogen. B. progesterone. C. epinephrine. D. thyroxin. E. testosterone. On day 12 there is increased production of estrogen arising from a new follicle in the ovary, causing the endometrium to thicken.
Bloom's Level: 2. Understand Learning Outcome: 21.03.03 Summarize how estrogen and progesterone affect the uterus. Section: 21.03 Topic: Human Female Reproductive System
21-21 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
43. The lining of the uterus that is discharged during the menstrual phase is the A. endocardium. B. oviduct. C. endometrium. D. hymen. E. myocardium. The endometrium, or uterine lining, thickens during the uterine cycle and is discharged at the end of the cycle if pregnancy does not occur.
Bloom's Level: 2. Understand Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Human Female Reproductive System
44. Days 6–13 of a 28-day uterine cycle is called the A. menstrual period. B. proliferative phase. C. ovulation period. D. secretory phase. E. fertilization period. The proliferation phase occurs when an increased production of estrogen by the new ovarian follicle causes the endometrium to proliferate; it thickens and becomes vascular and glandular. This happens during days 6 to 13 of the 28-day uterine cycle.
Bloom's Level: 1. Remember Learning Outcome: 21.03.01 List the stages of the ovarian cycle and explain what is occurring in each stage. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
45. In the human female, the luteal phase is part of the A. ovarian cycle. B. uterine cycle. C. sexual cycle. D. pituitary cycle. E. testicular cycle. The luteal phase is when ovulation occurs and a secondary oocyte is expelled from the ovary, making it part of the ovarian cycle.
Bloom's Level: 1. Remember Learning Outcome: 21.03.01 List the stages of the ovarian cycle and explain what is occurring in each stage. Section: 21.03 Topic: Human Female Reproductive System
46. Which of these descriptions could be associated with the secretory phase of the uterine cycle? A. low FSH → high estrogen → developing follicle → increase in uterine lining B. high LH → high estrogen → developing follicle → uterine lining breakdown C. decrease in LH → increase in progesterone → corpus luteum present → secretory uterine lining D. decrease in LH → decrease in progesterone → corpus luteum present → secretory uterine lining E. increase in LH → increase in progesterone → corpus luteum present → endometrium released While all of the answers give elements of the secretory phase of the uterine cycle, they are not listed in the correct order after ovulation and, during the luteal phase of the ovarian cycle, LH promotes the development of the corpus luteum. This structure produces increasing levels of progesterone, which causes the endometrium to become secretory.
Bloom's Level: 3. Apply Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
47. Although estrogen is primarily produced by ovaries, it is also produced by the adrenal glands and in fat tissue. Young female athletes who train strenuously often find that their menstrual cycle shuts down. What would be a reasonable direct cause of this shutdown? A. The increased need for ATP for the muscle cells deprives ATP for ovaries. B. The increased use of ATP for muscles deprives ATP needed by the brain where the hypothalamus and pituitary control the menstrual cycle. C. The cycle requires minimal threshold of estrogen and although the ovaries produce their regular levels, the loss of body fat would lower the estrogen level to below the threshold, thus shutting down the menstrual cycle. D. Physical trauma to the ovaries themselves would shut down their estrogen production. E. Less food would result in less hormone production. Estrogen functions to affect the endometrium, causing the uterus to undergo a cyclical series of events called the uterine cycle. Increased production of estrogen causes the endometrium to thicken and become vascular and glandular. Later in the cycle when there are low levels of estrogen, the endometrium disintegrates and its blood vessels rupture, leading to menstruation when there is a flow of blood and tissues out of the vagina. Without the threshold level of estrogen, the endometrium would not thicken so there would be no menstruation. The estrogen produced by the adrenal glands and in fat tissue is necessary to enter the proliferative phase and an athlete who has too low a level of body fat does not have a sufficient level to enter the uterine cycle.
Bloom's Level: 4. Analyze Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Human Female Reproductive System
21-24 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
48. Although estrogen is primarily produced by ovaries, it is also produced by the adrenal glands and in fat tissue. Loss of the normal amount of fat lowers the level of estrogen so that the menstrual cycle and ovulation stops. Female Olympic athletes who train severely often shut off their cycles. What would be a reasonable evolutionary cause of this shutdown? A. This mechanism shifts use of energy from reproduction to life activities for staying alive. B. Since an unborn infant gets priority on blood nutrient pools, pregnancy during famine will likely be fatal for both mother and offspring; this mechanism reduces this risk and increases survival. C. The adrenal glands have evolved to produce fight-or-flight responses and natural selection has shifted them to this function and away from estrogen. D. Less food would result in less hormone production. E. Evolution always selects for simpler systems. From an evolutionary standpoint, mothers who continued to become pregnant during times of famine would not survive as frequently to pass on their genetic material to the next generation as those who did not become pregnant during this time and, therefore, had a better chance to survive and reproduce when conditions were better.
Bloom's Level: 5. Evaluate Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
49. Which is a correct comparison of the role(s) of hormones in male and female cycles and sexual development? A. The same hormones are used by both systems; the names are simply different. B. The hypothalamus-pituitary uses a completely separate set of hormones to communicate with ovaries in contrast to testes. C. In males, the hypothalamus-pituitary and testes communicate; in females the uterus is an additional actor along with the ovaries and the hypothalamus-pituitary. D. In both males and females, there is a monthly cycle of changing hormone levels, although in males, this only results in sperm production. E. The hypothalamus controls the male sexual development and keeps sperm production level; the pituitary constantly cycles the female production of eggs. The hypothalamus and pituitary are the master controller of both systems but the hormones produced act on different structures in the two genders.
Bloom's Level: 4. Analyze Learning Outcome: 21.03.02 Describe the process of oogenesis. Section: 21.03 Topic: Gametogenesis
50. Pregnancy occurs when A. the sperm and the egg combine. B. the fertilized egg implants in the ovary. C. the developing embryo implants itself in the endometrium. D. sperm are in the vagina. E. sperm are in the uterus. Until the embryo is implanted in the uterus, it is not capable of developing into a fetus because it does not have a supply of nutrients or a way to exchange gases or eliminate waste.
Bloom's Level: 2. Understand Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Animal Reproduction and Development
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Chapter 21 - Reproductive System
51. Pregnancy tests detect the presence of A. HCG B. progesterone C. estrogen D. FSH E. LH
_ in the urine.
Progesterone, estrogen, FSH, and LH are all present at times when pregnancy has not occurred. The placenta, which sustains the developing embryo and later fetus, originates from both maternal and fetal tissues. It produces human chorionic gonadotropin (HCG), which maintains the corpus luteum in the ovary until the placenta begins its own production of progesterone and estrogen. Therefore, HCG is only present when pregnancy occurs.
Bloom's Level: 1. Remember Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Animal Reproduction and Development
52. The organ that exchanges molecules between fetal and maternal blood is the A. vagina. B. oviduct. C. uterus. D. placenta. E. mammary gland. The vagina, oviduct, and uterus are parts of the reproductive system of the mother. The mammary glands are located in the breast and produce milk for the child after it is born. The placenta provides nutrients and removes wastes for the developing embryo and later for the fetus. It arises from both maternal and fetal tissues and allows for the exchange of materials between fetal and maternal blood, although the two rarely mix.
Bloom's Level: 1. Remember Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Human Female Reproductive System
21-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
53. Trace the path of the sperm through the female reproductive tract. A. vagina → uterus → oviduct → cervix B. urethra → vagina → oviduct → cervix C. vagina → cervix → uterus → oviduct D. urethra → uterus → cervix → fallopian tube E. cervix → fallopian tube → vagina → uterus Sperm is ejaculated into the vagina of the female; to reach the egg it must then travel past the cervix and into the uterus and from there into the oviduct where the unfertilized egg is located.
Bloom's Level: 3. Apply Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Human Female Reproductive System
54. What two hormones influence the development of the secondary sexual characteristics of the female? A. testosterone and estrogen B. progesterone and estrogen C. androgen and estrogen D. testosterone and progesterone E. FSH and cortisol Testosterone is a type of androgen, or male sex hormone. Cortisol is a hormone associated with stress. Estrogen and progesterone affect not only the uterus but other parts of the body as well. Estrogen is largely responsible for the secondary sex characteristics in females, including body hair and fat distribution. Both estrogen and progesterone are also required for breast development.
Bloom's Level: 1. Remember Learning Outcome: 21.03.03 Summarize how estrogen and progesterone affect the uterus. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
55. Menopause refers to A. the menstrual phase. B. premenstrual phase. C. a time when men pause. D. cessation of menstruation. E. onset of menstruation. In menopause the ovaries stop responding to FSH and LH and they no longer produce progesterone and estrogen. Without these hormones the ovarian cycle and the uterine cycle cease; this usually occurs between ages 45 and 55.
Bloom's Level: 1. Remember Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Human Female Reproductive System
56. Symptoms of menopause include A. end of uterine cycles but not ovarian cycles. B. absence of the hot flashes, dizziness, insomnia, and depression so common during regular menstrual cycles. C. decreased sex drive probably due to androgen production by the adrenal cortex. D. cessation of menstruation. E. increased fat deposition and breast development. In menopause the ovaries stop responding to FSH and LH and they no longer produce progesterone and estrogen. Without these hormones the ovarian cycle and the uterine cycle cease. Estrogen functions to affect the endometrium, causing the uterus to undergo a cyclical series of events called the uterine cycle. Increased production of estrogen causes the endometrium to thicken and become vascular and glandular. Later in the cycle when there are low levels of estrogen, the endometrium disintegrates and its blood vessels rupture, leading to menstruation when there is a flow of blood and tissues out of the vagina. Without the threshold level of estrogen, the endometrium would not thicken so there would not be menstruation.
Bloom's Level: 2. Understand Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
57. Normal human sperm and eggs are similar in which one of the following respects? A. They have the same number of chromosomes in their nuclei. B. They share a common form of locomotion. C. They have the same relative amount of cytoplasm surrounding their nuclei. D. Approximately the same number of gametes are produced from both gonads. E. Both use meiosis to produce four sperm or four eggs from the parent cell. Both sperm and eggs are haploid, meaning that they have one set of DNA; one-half of the amount that the fertilized egg will contain. In the testes four sperm are produced for each spermatogonium that begins spermatogenesis, while in ovaries only one egg is produced for each follicle that begins oogenesis. Sperm have a flagellum which moves it through the uterus and into the oviduct where the egg is located. The egg is moved to this location by cilia lining the oviduct along with muscular contractions. Sperm contribute only their DNA to the embryo, while the egg contributes both DNA and the cytoplasmic components necessary for the cell to function.
Bloom's Level: 3. Apply Learning Outcome: 21.03.02 Describe the process of oogenesis. Section: 21.03 Topic: Gametogenesis
58. Female birth control pills act by inhibiting the production of A. ACTH and LH. B. TSH and FSH. C. ACTH and LTH. D. FSH and LH. E. estrogen and progesterone. FSH and LH are produced by the pituitary, which regulates the ovarian and uterine cycles. Birth control pills typically mimic the levels of hormones present when a female is pregnant. This stops the production of ova and thereby prevents pregnancy.
Bloom's Level: 3. Apply Learning Outcome: 21.04.01 Distinguish several birth control methods that are designed to prevent ovulation from those that prevent fertilization. Section: 21.04 Topic: Animal Reproduction and Development
21-30 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
59. Of the following, which is the most effective method of birth control? A. male condom B. abstinence C. female condom D. diaphragm E. IUD Apart from abstinence, there is a chance that pregnancy will occur; no other method is 100% certain.
Bloom's Level: 1. Remember Learning Outcome: 21.04.01 Distinguish several birth control methods that are designed to prevent ovulation from those that prevent fertilization. Section: 21.04 Topic: Animal Reproduction and Development
60. Which is NOT a physical barrier method of birth control? A. condom B. cervical cap C. diaphragm D. contraceptive implant E. IUD A contraceptive implant is a method of delivery of progesterone. The others listed all physically block sperm from contacting the egg.
Bloom's Level: 2. Understand Learning Outcome: 21.04.01 Distinguish several birth control methods that are designed to prevent ovulation from those that prevent fertilization. Section: 21.04 Topic: Animal Reproduction and Development
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Chapter 21 - Reproductive System
61. Which of the following is true regarding Preven? A. kills sperm as they enter the female reproductive tract B. is a synthetic progesterone pill that disrupts the uterine cycle C. is another name for mifepristone or RU-486 D. destroys the embryo which in turn causes a spontaneous miscarriage E. is a synthetic progesterone that is injected Preven is intended to be used to prevent unintended pregnancies after fertilization of the egg has occurred. It disrupts the normal uterine cycle and decreases the likelihood that an embryo will become implanted in the endometrium.
Bloom's Level: 1. Remember Learning Outcome: 21.04.03 Describe two medications used as emergency contraceptives. Section: 21.04 Topic: Animal Reproduction and Development
62. HIV attacks A. helper T lymphocytes. B. Sertoli cells. C. B lymphocytes. D. interstitial cells. E. follicle cells. Helper T lymphocytes are part of the normal immune system of humans. HIV damages the immune response by infecting these cells.
Bloom's Level: 1. Remember Learning Outcome: 21.05.02 Describe how HIV specifically affects the immune system and how this explains the three categories of HIV infection. Section: 21.05 Topic: Sexually Transmitted Diseases
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Chapter 21 - Reproductive System
63. The cure for HIV is A. vaccination. B. changing the diet. C. antibiotic treatment. D. antiviral drugs. E. not in existence yet. Although there is no cure for HIV, there are drugs that can inhibit the replication of the virus and increase the amount of time that an infected individual has before AIDS develops and therefore extend the person's life expectancy.
Bloom's Level: 1. Remember Learning Outcome: 21.05.02 Describe how HIV specifically affects the immune system and how this explains the three categories of HIV infection. Section: 21.05 Topic: Sexually Transmitted Diseases
64. Opportunistic infections resulting from AIDS do NOT include A. bacterial pneumonia. B. mycobacterial infections. C. toxoplasmosis. D. Kaposi's sarcoma. E. strokes. Strokes are caused by blockage of blood and therefore oxygen to the brain. They are not caused by an infection and so are not controlled by the immune system.
Bloom's Level: 2. Understand Learning Outcome: 21.05.02 Describe how HIV specifically affects the immune system and how this explains the three categories of HIV infection. Section: 21.05 Topic: Sexually Transmitted Diseases
21-33 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
65. Which of the following STDs is not the result of a virus? A. AIDS B. herpes C. chlamydia D. genital warts E. hepatitis B Hepatitis B, genital warts, herpes, and HIV are all caused by viruses; chlamydia is caused by a bacterium.
Bloom's Level: 2. Understand Learning Outcome: 21.05.01 List the specific cause and common symptoms of three STDs that are caused by viruses and three caused by bacteria. Section: 21.05 Topic: Sexually Transmitted Diseases
66. HAART would most likely be used to treat which of the following? A. herpes B. syphilis C. HIV infection D. hepatitis E. none of these Highly active antiretroviral therapy (HAART) is the use of three or four of the available retroviral drugs used to treat HIV. It is unlikely that the virus will be resistant to all of the drugs in use.
Bloom's Level: 2. Understand Learning Outcome: 21.05.03 Discuss four types of antiretroviral drugs. Section: 21.05 Topic: Sexually Transmitted Diseases
21-34 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
67. The cure for herpes is A. vaccination. B. changing the diet. C. antibiotic treatment. D. antiviral drugs. E. not possible due to it being a virus. Herpes is a viral infection and no cure currently exists. In part this is because the virus is able to "hide" by becoming latent and residing in the ganglia of the sensory nerves.
Bloom's Level: 1. Remember Learning Outcome: 21.05.01 List the specific cause and common symptoms of three STDs that are caused by viruses and three caused by bacteria. Section: 21.05 Topic: Sexually Transmitted Diseases
68. A liver-damaging disease that is more likely to be contracted from sexual contact than HIV is A. AIDS. B. HBV. C. chlamydia. D. herpes. E. gonorrhea. HBV, hepatitis B, is more contagious than HIV; it attacks the liver and symptoms may be either chronic or acute. Chlamydia and gonorrhea do not attack the liver.
Bloom's Level: 2. Understand Learning Outcome: 21.05.01 List the specific cause and common symptoms of three STDs that are caused by viruses and three caused by bacteria. Section: 21.05 Topic: Sexually Transmitted Diseases
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Chapter 21 - Reproductive System
69. The cure for gonorrhea is A. vaccination. B. changing the diet. C. antibiotic treatment. D. antiviral drugs. E. no cure exists for gonorrhea. Since gonorrhea is caused by a bacterium it is possible to cure it with appropriate antibiotics. While antibiotics can be used to treat gonorrhea, there is concern because antibiotic resistant strains are becoming more common.
Bloom's Level: 1. Remember Learning Outcome: 21.05.01 List the specific cause and common symptoms of three STDs that are caused by viruses and three caused by bacteria. Section: 21.05 Topic: Sexually Transmitted Diseases
70. Blindness in newborn infants occurs when the mother A. is overweight. B. has gonorrhea or chlamydia. C. has a yeast infection. D. has syphilis. E. uses birth control pills without a prescription. Newborns can contract gonorrhea or chlamydia as they pass through the birth canal, resulting in eye infections that can lead to blindness. To combat the likelihood that blindness will result from an infection, all infants delivered vaginally receive eye drops containing antibacterial agents.
Bloom's Level: 3. Apply Learning Outcome: 21.05.01 List the specific cause and common symptoms of three STDs that are caused by viruses and three caused by bacteria. Section: 21.05 Topic: Sexually Transmitted Diseases
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Chapter 21 - Reproductive System
71. Syphilis is caused by A. Candida albicans. B. Trichomonas vaginalis. C. a virus. D. Treponema pallidum. E. genital warts. Treponema pallidum is the bacteria that causes syphilis. Candida albicans is a fungus that can cause a "yeast infection." Trichomonas vaginaliis a parasite that can infect the vagina. Genital warts are caused by a virus (HPV).
Bloom's Level: 1. Remember Learning Outcome: 21.05.01 List the specific cause and common symptoms of three STDs that are caused by viruses and three caused by bacteria. Section: 21.05 Topic: Sexually Transmitted Diseases
72. If a person had recently undergone major doses of antibiotics accompanying surgery, which STD(s) would likely be collaterally treated? A. AIDS B. syphilis C. genital warts D. HBV E. HIV Antibiotics can be used to treat bacterial infections. Syphilis is the only choice listed that is not caused by a virus. Viruses would not be affected by an antibiotic.
Bloom's Level: 3. Apply Learning Outcome: 21.05.04 Explain why antibiotics should not be used to treat viral infections. Section: 21.05 Topic: Sexually Transmitted Diseases
21-37 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
73. A male accessory organ that often becomes enlarged in older men is the A. testis. B. prostate gland. C. epididymis. D. bulbourethral gland. E. vas deferens. The prostate gland can become enlarged as an enzyme that converts testosterone to a substance that promotes prostate growth continues to function after the developmental time (puberty) when it is needed. This leads to enlargement of the gland in older men.
Bloom's Level: 1. Remember Learning Outcome: 21.06.01 Describe some common disorders of the male reproductive system. Section: 21.06 Topic: Human Male Reproductive System
74. A more precise term for impotency is A. sterility. B. frigidity. C. sexual psychosis. D. refractory period. E. erectile dysfunction. Impotency is the inability to produce or maintain an erection sufficient to perform intercourse; the more correct term is erectile dysfunction (ED).
Bloom's Level: 1. Remember Learning Outcome: 21.06.01 Describe some common disorders of the male reproductive system. Section: 21.06 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
75. The impotency drug Viagra acts by A. increasing whole body blood pressure. B. stimulating erotic mental images. C. increasing blood flow to the penis. D. decreasing blood flow to the penis. E. reducing the amount of seminal fluid. In a normal erection the spongy tissue of the penis that contains distensible blood spaces and the veins that normally carry blood away from the penis are compressed to maintain the erection. Viagra and similar medications increase blood flow to the penis to enable an erection to be maintained.
Bloom's Level: 1. Remember Learning Outcome: 21.06.01 Describe some common disorders of the male reproductive system. Section: 21.06 Topic: Human Male Reproductive System
76. A PAP smear is used to detect A. cervical cancer. B. ovarian cancer. C. infertility problems. D. sperm count. E. AIDS. A PAP smear can detect cervical cancer but does not usually detect ovarian cancer or AIDS. Since this is performed on women, it would not reveal a sperm count nor would it reveal infertility since it does not test for the presence of the hormones that regulate the ovarian cycle.
Bloom's Level: 2. Understand Learning Outcome: 21.06.02 Describe some common disorders of the female reproductive system. Section: 21.06 Topic: Human Female Reproductive System
21-39 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
77. Which of the following is not a factor that may contribute to infertility? A. low sperm count B. body weight of the female C. endometriosis D. environmental factors E. release of fibrinolsin Fibrinolsin is an enzyme released by dying cells that controls clotting of blood. The other answers all affect fertility.
Bloom's Level: 2. Understand Learning Outcome: 21.06.02 Describe some common disorders of the female reproductive system. Section: 21.06 Topic: Animal Reproduction and Development
78. A sophisticated procedure where one sperm is injected into an egg is used in cases where a man has severe infertility problems and is called A. artificial insemination. B. vasectomy. C. cloning. D. surrogate mothering. E. intracytoplasmic sperm injection. The medical procedure where sperm are placed into a woman’s vagina is called artificial insemination. A vasectomy is a surgical procedure to prevent a man from fathering a child. Cloning involves taking a cell that is not a gamete and using it to create an embryo. In some instances, women are contracted and paid to have babies. These women are called surrogate mothers.
Bloom's Level: 1. Remember Learning Outcome: 21.06.03 Discuss four assisted reproductive technologies that can help infertile couples conceive a child. Section: 21.06 Topic: Animal Reproduction and Development
21-40 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 21 - Reproductive System
79. Luteinizing hormone will stimulate the initial production of the first haploid cells during spermatogenesis. Which of the following are the first haploid cells produced? A. primary spermatocytes B. spermatogonium C. spermatids D. sertoli cells E. follicles Primary spermatocytes are the first haploid set of haploid cells. Spermatogonium are diploid cells. Spermatids are not the first haploid cells to develop. Sertoli cells are diploid cells that "nurse" the developing sperm. Follicles are found in the ovary.
Bloom's Level: 2. Understand Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.01 Topic: Gametogenesis
80. A mutation, involving luteinizing hormone, that causes a change in the enzymes contained in the acrosome, would affect A. the production of sperm. B. the ability of the sperm to travel to the uterus. C. the ability of the sperm to fertilize the egg. D. the implantation of the egg. E. None of the answer choices is correct. The enzymes in the acrosome dissolve the zona pellucida, allowing the nucleus of the sperm to enter and fertilize the egg.
Bloom's Level: 5. Evaluate Learning Outcome: 21.01.03 Analyze the role of gonadotropin-releasing hormone, luteinizing hormone, follicle-stimulating hormone, and testosterone in male sexual reproduction. Section: 21.03 Topic: Gametogenesis
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Chapter 21 - Reproductive System
81. The reason that morning-after pills can be taken over the period of one to three days after unprotected sex is A. the sperm does not reach the egg until the third day after intercourse. B. the release of FSH and LH is at a relatively stable concentration until the third day after intercourse. C. implantation of the embryo does not take place until around the third day after fertilization. D. the placenta will begin production of NCG on the third day after fertilization. E. the corpus luteum does not develop until the third day. After fertilization occurs in the oviduct, the fertilized egg must travel down the oviduct to the uterus where it will become implanted in the endometrium. This takes around three days from the time of fertilization.
Bloom's Level: 3. Apply Learning Outcome: 21.04.03 Describe two medications used as emergency contraceptives. Section: 21.04 Topic: Animal Reproduction and Development
82. What is the function of reverse transcriptase in the reproduction cycle of HIV? A. to transcribe viral DNA into RNA for the production of viral proteins B. to convert human DNA into viral DNA C. to produce single-stranded DNA from single-stranded viral RNA D. to produce double-stranded DNA from single-stranded viral RNA E. to produce mRNA from viral RNA Reverse transcriptase is an enzyme that uses single-stranded viral RNA to produce two strands of DNA which are then able to join into the double-stranded DNA of the host cell. This allows the virus to use the machinery of the cell to make copies of its genetic material.
Bloom's Level: 3. Apply Learning Outcome: 21.05.02 Describe how HIV specifically affects the immune system and how this explains the three categories of HIV infection. Section: 21.05 Topic: Sexually Transmitted Diseases
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Chapter 21 - Reproductive System
83. HIV has a high rate of mutation because A. HIV reverse transcriptase is likely to make errors when transcribing viral RNA into DNA. B. HIV has RNA instead of DNA. C. HIV infects T cells in the human immune system, allowing the virus to respond directly to changes in the immune system. D. HIV can only replicate in a host cell. E. HIV is a bacteria and they have a higher expected rate of mutation. Reverse transcriptase does not check for mistakes as it transcribes viral RNA into DNA so that errors go uncorrected, causing the virus to constantly undergo changes in the proteins it produces. This allows the virus to constantly change.
Bloom's Level: 2. Understand Learning Outcome: 21.05.02 Describe how HIV specifically affects the immune system and how this explains the three categories of HIV infection. Section: 21.05 Topic: Sexually Transmitted Diseases
84. The inactivation of the host cell protein p53 benefits human papillomavirus (HPV) A. by removing all of the components in the cell that can attack the infection. B. by causing the cell to lyse and spill the viral load, allowing the virus to move to other cells. C. because p53 is a signal to produce T cells to fight the infection. D. by allowing the infected cell to go through repeated cell division, thus replicating the virus. E. All of the answer choices are benefits to HPV. The protein p53 in the host cell serves as a "brake" to cell division; by inactivating it, the HPV causes the cell to divide repeatedly and therefore produce more virus along with more cells.
Bloom's Level: 4. Analyze Learning Outcome: 21.05.01 List the specific cause and common symptoms of three STDs that are caused by viruses and three caused by bacteria. Section: 21.05 Topic: Sexually Transmitted Diseases
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Chapter 21 - Reproductive System
85. Changes in which organ system can lead to erectile dysfunction (ED)? A. The cardiovascular system because it can lead to an inadequate blood supply to the penis. B. The endocrine system because this can lower the sperm count. C. The digestive system because with poor digestion the body does not extract sufficient energy from food. D. The skeleto-muscular system because a reduction in muscle fibers will make an erection impossible. E. The endocrine system because this will limit the production of testosterone. The cardiovascular system provides the blood that is used to engorge the spongy tissues of the penis allowing an erection.
Bloom's Level: 4. Analyze Learning Outcome: 21.06.01 Describe some common disorders of the male reproductive system. Section: 21.06 Topic: Human Male Reproductive System
86. In the uterine cycle, as the estrogen level in the blood rises A. it leads to negative feedback control over the anterior pituitary secretions of FSH. B. the proliferation phase ends. C. it leads to a large amount of GnRH from secretions of the hypothalamus. D. the proliferation phase ends and it leads to negative feedback control over the anterior pituitary secretions of FSH. E. the proliferation phase ends, which leads to negative feedback control over the anterior pituitary secretions of FSH as well as the production of a large amount of GnRH from secretions of the hypothalamus. Estrogen levels control the uterine cycle and bring the proliferation phase to an end. Estrogen levels also affect the anterior pituitary and the hypothalmus, causing the first to lower FSH production and the latter to increase production of LH.
Bloom's Level: 4. Analyze Learning Outcome: 21.03.04 Describe changes that occur in the uterus during menstruation, pregnancy, and menopause. Section: 21.03 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
87. Which of the following statement is not true? A. Viagra acts to increase blood flow to the penis during sexual arousal. B. BPH, a benign prostatic hyperplasia is suspected when men have a more frequent urge to urinate and produce a weaker stream of urine. C. Prostate enlargement is often due to the presence of a normal hormone (5-alpha reductase) present at an inappropriate time in development. D. Enlargement of the prostate puts pressure on the vas deferens, leading to difficulty urinating. E. All of the answer choices are true. Enlargement of the prostate puts pressure on the urethra, leading to difficulty urinating. The vas deferens carries sperm, not urine.
Bloom's Level: 3. Apply Learning Outcome: 21.06.01 Describe some common disorders of the male reproductive system. Section: 21.06 Topic: Human Male Reproductive System
88. Why are men who have a heart condition warned not to take Viagra? A. Viagra can cause the buildup of plaque in the arteries, leading to another heart attack. B. Viagra closes down the arteries to the heart to shunt the blood supply to the penis. C. Viagra increases blood flow to the penis through vasodilation which, combined with some cardiac medications, may dangerously lower blood pressure. D. Viagra shuts down the immune system in order to trick the body into sending more blood to the penis and this can open up the heart to increased infection rates. E. Viagra slows the beating of the heart and this could result in a heart attack. The lessening of blood available to be pumped by the heart as it is diverted to the penis can lower blood pressure, and this, combined with a heart condition, could be fatal.
Bloom's Level: 4. Analyze Learning Outcome: 21.06.01 Describe some common disorders of the male reproductive system. Section: 21.06 Topic: Human Male Reproductive System
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Chapter 21 - Reproductive System
True / False Questions 89. During arousal, the vagina will secrete a lubricating fluid. TRUE Blood vessels in the vaginal wall will release small droplets of fluid that seep into the vagina and lubricate it.
Bloom's Level: 1. Remember Learning Outcome: 21.02.03 Describe the events that occur during a female orgasm. Section: 21.02 Topic: Human Female Reproductive System
90. The refractory period of female orgasm is significantly longer than that of a male. FALSE The refractory period of female orgasm is very short or none at all, enabling females to have multiple orgasms in a single sexual experience.
Bloom's Level: 1. Remember Learning Outcome: 21.02.03 Describe the events that occur during a female orgasm. Section: 21.02 Topic: Human Female Reproductive System
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Chapter 21 - Reproductive System
Multiple Choice Questions 91. What occurs when a woman takes the Preven Emergency Contraceptive Kit? A. The progesterone pills upset the uterine cycle and prevent the embryo from implanting. B. The estrogen pills upset the uterine cycle and prevent the embryo from implanting. C. The progesterone pills upset the uterine cycle and prevent the sperm from fertilizing the egg. D. The progesterone pills upset the uterine cycle and prevent the egg from being ovulated from the ovary. E. The polyurethane lining prevents the sperm from reaching the egg and fertilizing it. The Preven Emergency Contraceptive Kit contains four synthetic progesterone pills that will upset the uterine cycle and prevent the embryo from implanting. The kit doesn't contain estrogen pills. Preventing the sperm from fertilizing the egg or the egg from being ovulated from the ovary are forms of birth control methods such as condoms and hormonal birth control methods. A polyurethane lining is a form of birth control method known as the diaphragm or cervical cap.
Bloom's Level: 5. Evaluate Learning Outcome: 21.04.03 Describe two medications used as emergency contraceptives. Section: 21.04 Topic: Animal Reproduction and Development
True / False Questions 92. The only 100% method of preventing pregnancy and the transmission of STDs is the use of a male condom along with a spermicidal jelly. FALSE The only 100% method of preventing pregnancy and the transmission of STDs is to practice abstinence. Male condoms are about 85% effective in preventing pregnancy. The addition of spermicidal jelly further decrease the possibility of pregnancy and the transmission of STDs but doesn't make it 100%.
Bloom's Level: 3. Apply Learning Outcome: 21.04.02 Summarize the birth control methods that are effective at preventing the spread of sexually transmitted diseases. Section: 21.04 Topic: Animal Reproduction and Development
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Chapter 21 - Reproductive System
Multiple Choice Questions 93. Which of the following birth control methods is effecting in preventing pregnancy as well as the spread of many STDs? A. male condom B. coitus interruptus C. spermicidal jellies D. vaginal film E. intrauterine device The male condom is effective in in preventing pregnancy as well as the spread of STDs. All of the other birth control methods have various levels of effectiveness in preventing pregnancy, but they do not protect against the spread of STDs.
Bloom's Level: 2. Understand Learning Outcome: 21.04.02 Summarize the birth control methods that are effective at preventing the spread of sexually transmitted diseases. Section: 21.04 Topic: Animal Reproduction and Development
True / False Questions 94. The female condom is effective in preventing pregnancy as well as the transmission of many sexually transmitted diseases. TRUE The female condom is a polyurethane liner that is fitted inside the vagina that blocks the entrance of sperm to the uterus and prevents STDs.
Bloom's Level: 1. Remember Learning Outcome: 21.04.02 Summarize the birth control methods that are effective at preventing the spread of sexually transmitted diseases. Section: 21.04 Topic: Animal Reproduction and Development
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Chapter 21 - Reproductive System
95. Chlamydia is a common STD that can be treated by appropriate antibiotics. TRUE Chlamydia is a common STD that can be treated by appropriate antibiotics.
Bloom's Level: 1. Remember Learning Outcome: 21.05.04 Explain why antibiotics should not be used to treat viral infections. Section: 21.05 Topic: Animal Reproduction and Development
Multiple Choice Questions 96. Which of the following is not a class of antiretroviral drugs that is currently available? A. reverse transcriptase inhibitors B. protease inhibitors C. fusion/entry inhibitors D. integrase inhibitors E. All of the answer choices are classes of antiretroviral drugs that are currently available. All of the answer choices are classes of antiretroviral drugs that are currently available.
Bloom's Level: 2. Understand Learning Outcome: 21.05.03 Discuss four types of antiretroviral drugs. Section: 21.05 Topic: Sexually Transmitted Diseases
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Chapter 21 - Reproductive System
Short Answer Questions 97. List four structures involved in the production of seminal fluid and the component of seminal fluid that they produce. Testes produce the sperm that are found in the semen. Seminal vesicles produce nutrients and fluid to the semen. Prostate gland produces a basic fluid to the semen. Bulbourethral gland produces a viscous fluid to the semen.
Bloom's Level: 6. Create Learning Outcome: 21.01.02 List the organs that produce components of seminal fluid. Section: 21.01 Topic: Human Male Reproductive System
Multiple Choice Questions 98. Which reproductive technology involves having a physician place the man's sperm into the woman's vagina in an attempt to fertilize her egg? A. artificial insemination B. in vitro fertilization C. gamete intrafallopian transfer D. surrogate mother E. intracytoplasmic sperm injection During artificial insemination a physician will place the man's sperm into the woman's vagina in an attempt to fertilize her egg. In vitro fertilization involves fertilizing the egg within laboratory glassware and then inserting the embryo into the uterus. Gamete intrafallopian transfer involves fertilizing the egg within laboratory glassware and then immediately inserting it back into the uterus. A surrogate mother involves having a different woman carry and deliver the baby. Intracytoplasmic sperm injection involves injecting a single sperm into the egg and then inserting it back into the mother.
Bloom's Level: 1. Remember Learning Outcome: 21.06.03 Discuss four assisted reproductive technologies that can help infertile couples conceive a child. Section: 21.06 Topic: Animal Reproduction and Development
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Chapter 21 - Reproductive System
Short Answer Questions 99. List and describe the function of five of the female reproductive organs. Ovaries: produce the oocyte and sex hormones Oviducts: conduct the oocyte, site of fertilization, transportation of zygote Uterus: houses developing fetus Cervix: contains opening to the uterus Vagina: receives penis during intercourse, birth canal and exit for menstrual flow
Bloom's Level: 6. Create Learning Outcome: 21.02.01 Identify where an oocyte is produced and how it is transported to the uterus. Section: 21.02 Topic: Human Female Reproductive System
100. List and briefly describe several assisted reproductive technologies that can help infertile couples have children. Artificial insemination by donor: sperm are placed directly in the woman's vagina by a physician. In vitro fertilization: immature oocytes are harvested from the woman and the man's sperm is collected. The oocyte is fertilized in the laboratory glassware and allowed to develop into embryos. These are then transferred back into the mother's uterus. Gamete intrafallopian transfer: immature oocytes are harvested from the woman and the man's sperm is collected. The oocyte is fertilized in the laboratory glassware and then transferred back into the mother's uterus. Surrogate mother: women are contracted to have a baby for another individual. Intracytoplasmic sperm injection: a single sperm is injected into an oocyte.
Bloom's Level: 6. Create Learning Outcome: 21.06.03 Discuss four assisted reproductive technologies that can help infertile couples conceive a child. Section: 21.06 Topic: Animal Reproduction and Development
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Chapter 21 - Reproductive System
Multiple Choice Questions 101. If a person believes that they have contracted a STD, taking antibiotics may or may not cure them. Antibiotics would be useless against which of the following STDs? A. genital herpes B. chlamydia C. gonorrhea D. syphilis E. All of the answer choices can be cured with antibiotics. Genital herpes is a viral STD, so it would be useless to take antibiotics in an attempt to cure it. Gonorrhea, chlamydia, and syphilis are all bacterial infections. The appropriate antibiotics can be effective in curing these STDs.
Bloom's Level: 1. Remember Learning Outcome: 21.05.04 Explain why antibiotics should not be used to treat viral infections. Section: 21.05 Topic: Sexually Transmitted Diseases
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Chapter 22 - Development and Aging
Chapter 22 Development and Aging
Multiple Choice Questions 1. How does a sperm penetrate an egg? A. Enzymes in the acrosome dissolve the zona pellucida. B. The sperm melts its way inside, using the nucleic acids in its DNA. C. The forward pressure of the flagellum forces it through the plasma membrane. D. Spermatids are wedge-shaped and are chemically attracted to the inside cytoplasm of the egg cell. E. The haploid sets of chromosomes match across the membrane and this triggers the membrane to dissolve. In order for the DNA contained in the sperm to reach the interior of the egg and merge with the DNA from the egg (fertilization), the coverings of the egg must be penetrated. The sperm is able to pass through the zona pellucida using enzymes contained in the acrosome, fuses with the membrane of the egg, and then releases its nucleus into the egg.
Bloom’s Level: 2. Understand Learning Outcome: 22.01.01 Identify the major structures of a sperm cell and an oocyte, and provide a function for each. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
2. Generally only one sperm fertilizes an egg because A. sperm compete against one another before entering and only the most fit one is accepted. B. sperm are so few that two are unlikely to arrive at the same time. C. one small entry hole called the blastopore allows only one sperm to fit through and then seals. D. many sperm enter but only one set of chromosomes can fuse with the egg nucleus; the other sperm are absorbed. E. when the first sperm fuses with the egg membrane, it causes the membrane to depolarize and also separates the fertilization membrane and this gap forms a barrier to any other sperm. Prevention of the entry of multiple sperm depends on changes in the oocyte’s plasma membrane and in the zona pellucida. As soon as a sperm touches an oocyte, the oocyte’s plasma membrane depolarizes (from –65 m V to 10 mV), and this prevents the binding of any other sperm.
Bloom’s Level: 3. Apply Learning Outcome: 22.01.01 Identify the major structures of a sperm cell and an oocyte, and provide a function for each. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
3. What of the following would NOT be a method that would prevent one species' sperm from fertilizing another species' egg? A. The chemical and other behavioral cues for mating are unlikely to lead to insemination by a male of another species. B. The reactions between the sperm acrosome and the surface of the egg are generally species-specific. C. Any successful fertilization would require a successful match of chromosomes and genes to produce a viable zygote. D. The chemical elements of the layers of the egg match the chemical elements of the sperm acrosome. E. The joining of two differing sets of chromosomes does not provide the proper genetic information for the development of an embryo. There are many mechanisms that prevent the fertilization of the egg of one species by the sperm of another species. Among these are those that prevent the males and females of two different species from mating, such as behavioral cues that are species specific. Other mechanisms act at the molecular level, these include chemical signals for mating, the inability of the acrosomes of one species to penetrate the egg of another. If the sperm is able to enter the egg, there can be a lack of developmental control by the "mismatched" DNA. If the sperm acrosome matched the chemical of the layers of the egg, that would have the opposite effect and allow fertilization to occur.
Bloom’s Level: 4. Analyze Learning Outcome: 22.01.02 Describe the steps involved in fertilization. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
4. A fertilized egg undergoes cell division without further growth in the process called A. cleavage. B. gastrulation. C. differentiation. D. morphogenesis. E. embryology. Gastrulation is the process that occurs when the fertilized egg develops into three layers of cells that will, in turn, develop into adult organs. Cellular differentiation occurs when cells become specialized in structure and function. Morphogenesis is the process where the overall body pattern is formed. Embryology is the study of how embryos develop into adult organisms.
Bloom’s Level: 1. Remember Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Animal Reproduction and Development
5. During the formation of an embryo, a hollow sphere of cells with a fluid-filled cavity inside is termed a A. morula. B. blastula. C. blastocoel. D. gastrula. E. neurula. A morula is a ball of cells that occurs early in development. A blastocoel is a fluid-filled cavity in a ball of cells called a blastula. The gastrula is formed when the cells begin to different into layers of cells. At the neurula stage, each of the germ layers can be associated with the later development of particular parts including the neural tube that becomes the brain and spinal cord.
Bloom’s Level: 1. Remember Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
6. The blastocoel is formed during the A. morula B. blastula C. gastrula D. zygote E. neurula
stage of development.
The blastocoel is the fluid-filled cavity of the blastula.
Bloom’s Level: 1. Remember Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Animal Reproduction and Development
7. Which one of the following is NOT derived from endoderm? A. gut lining B. hormonal glands C. lungs D. bones E. bladder lining The ectoderm (outer layer) becomes the nervous system, the epidermis of skin, and the epithelial lining of oral cavity and rectum. The mesoderm (middle layer) becomes the musculoskeletal system. The endoderm (inner layer) becomes the epithelial lining of digestive tract and respiratory tract, associated glands of these systems, and epithelial lining of urinary bladder.
Bloom’s Level: 2. Understand Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
8. The last tissue layer to be formed during gastrulation is the A. endoderm. B. ectoderm. C. mesoderm. D. gastroderm. E. epiderm. An early gastrula has two layers of cells. The outer layer of cells is called the ectoderm, and the inner layer is called the endoderm. In addition to ectoderm and endoderm, the late gastrula has a middle layer of cells called the mesoderm.
Bloom’s Level: 2. Understand Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Animal Reproduction and Development
9. During gastrulation in vertebrates, the pore created by invagination will become the A. brain. B. heart. C. mouth. D. anus. E. ear. The hole created by invagination is called the blastopore and becomes the anus. The brain will develop from the neural tube, and the heart will arise as a tube from the mesoderm.
Bloom’s Level: 1. Remember Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
10. Which of the following develop from the germ layer called the ectoderm? A. skin and nervous system B. muscles and blood C. gut and pancreas D. nervous system and muscles E. reproductive and urinary The ectoderm (outer layer) becomes the nervous system, the epidermis of skin, and the epithelial lining of oral cavity and rectum. The mesoderm (middle layer) becomes the musculoskeletal system. The endoderm (inner layer) becomes the epithelial lining of digestive tract and respiratory tract, associated glands of these systems, and epithelial lining of urinary bladder.
Bloom’s Level: 1. Remember Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Animal Reproduction and Development
11. The cardiovascular and urinary systems form from the A. endoderm B. ectoderm C. mesoderm D. blastoderm E. gastroderm
layer.
The ectoderm (outer layer) becomes the nervous system, the epidermis of skin, and the epithelial lining of oral cavity and rectum. The mesoderm (middle layer) becomes the musculoskeletal system. The endoderm (inner layer) becomes the epithelial lining of digestive tract and respiratory tract, associated glands of these systems, and epithelial lining of urinary bladder.
Bloom’s Level: 1. Remember Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
12. A failure of the cells in the ectoderm to differentiate correctly would most affect which of the following body systems? A. nervous system B. cardiovascular system C. urinary system D. digestive system E. urinary system Since the nervous system is the only one of those listed that arise from the ectoderm, it is the only one that would be affected.
Bloom’s Level: 2. Understand Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Human Embryonic Development
13. The is a dorsal supporting rod present in the early embryonic development of the vertebrates. A. neural plate B. neural tube C. notochord D. neurula E. vertebral column The notochord is present in early embryos and lies just below the neural tube. In vertebrates it will become the vertebral column. Both the neural plate and neural tube are early developmental forms of the spinal cord. The neurula is a stage of development where each of the germ layers can be associated with the later development of particular parts.
Bloom’s Level: 1. Remember Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Human Embryonic Development
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Chapter 22 - Development and Aging
14. The neural tube of vertebrates develops during morphogenesis of the nervous system by the A. folding of ectoderm tissue. B. migration of mesoderm cells. C. fusion of ectoderm and mesoderm. D. contraction of the endoderm away from the mesoderm. E. extension of endoderm into a thin spinal column. The nervous system develops from midline ectoderm located just above the notochord. At first, a thickening of cells, called the neural plate, is seen along the dorsal surface of the embryo. Then, neural folds develop on either side of a neural groove, which becomes the neural tube when these folds fuse.
Bloom’s Level: 3. Apply Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Human Embryonic Development
15. Which system below develops first in the embryo? A. reproductive B. circulatory C. excretory D. respiratory E. nervous The nervous system is part of embryonic development (1–2 months) while the major organs form during fetal development (3–9 months).
Bloom’s Level: 2. Understand Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Human Embryonic Development
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Chapter 22 - Development and Aging
16. The egg from a A. fish B. frog C. reptile D. chicken E. human
has been observed to form a gray crescent after fertilization.
A frog’s egg has anterior/posterior and dorsal/ventral axes that correlate with the position of the gray crescent.
Bloom’s Level: 1. Remember Learning Outcome: 22.02.01 Explain how cellular differentiation and morphogenesis play a role in development. Section: 22.02 Topic: Animal Reproduction and Development
17. Which of the following describes the process in which cells become specialized in structure and function? A. development B. pattern formation C. cellular differentiation D. morphogenesis E. embryology Development is the general term used to describe the steps a fertilized egg goes through to reach its final adult form. Pattern formation, or morphogenesis, is the determination of the overall pattern of how the body is laid out. Embryology is the study of how a fertilized egg develops into an adult organism. Cellular differentiation occurs when cells become specialized in structure and function.
Bloom’s Level: 1. Remember Learning Outcome: 22.02.01 Explain how cellular differentiation and morphogenesis play a role in development. Section: 22.02 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
18. Another term for programmed cell death is A. morphogenesis. B. apoptosis. C. pattern formation. D. cellular differentiation. E. mitosis. Morphogenesis is the establishment of the overall body plan. Pattern formation is the ultimate in morphogenesis. Cellular differentiation is when cells become specialized and occurs when cells become specialized in structure and function. Mitosis is the process where cells replicate their DNA and divide to form identical cells.
Bloom’s Level: 1. Remember Learning Outcome: 22.02.03 Describe how morphogen genes, homeotic genes, and apoptosis help determine pattern formation during development. Section: 22.02 Topic: Animal Reproduction and Development
19. Which of these is not associated with development? A. Cells divide and get larger. B. Cells become specialized in structure and function. C. Body parts are shaped and patterned into a specific form. D. One cell eventually becomes four sperm. E. Tissues and organs become arranged in the body. One cell dividing to become four sperm would be part of spermatogenesis, which must happen before the egg is fertilized.
Bloom’s Level: 2. Understand Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
20. Morphogenesis is A. the total life cycle of any animal. B. a process by which tissues are shaped into organs. C. the mechanism by which gametes are formed. D. a study of the beginning and ending of life. E. the study of aging. Morphogenesis occurs when shapes emerge during the formation of tissues, organs, or an embryo during development.
Bloom’s Level: 1. Remember Learning Outcome: 22.02.01 Explain how cellular differentiation and morphogenesis play a role in development. Section: 22.02 Topic: Animal Reproduction and Development
21. is the parceling out of maternal determinants as mitosis occurs in embryonic cells. A. Homeostasis B. Induction C. Morphogenesis D. Cytoplasmic segregation E. Aging Homeostasis refers to the steady-state condition that the body typically maintains. Induction is the influencing of cells during development by cells located nearby; this is typically done through cell signaling. Morphogenesis is the formation of the overall body pattern and organization of organ systems. Aging is the process where an organism undergoes changes due to the length of time that the organism has been in existence.
Bloom’s Level: 2. Understand Learning Outcome: 22.02.02 Explain how cytoplasmic segregation and induction are able to influence cellular differentiation. Section: 22.02 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
22. Which of the following is NOT correct? A. Embryonic cells are totipotent. B. Differentiation is due to parceling out of genes into various embryonic cells. C. The contents of the cytoplasm of an egg are not distributed uniformly. D. If an egg from a frog divides without receiving the gray crescent, it will not become a tadpole. E. Differentiation is influenced by chemical signals given off by neighboring cells. Differentiation occurs in cells which have identical genomes. The genes are expressed differently because some of the genes are turned off, depending on the cells' type.
Bloom’s Level: 4. Analyze Learning Outcome: 22.02.03 Describe how morphogen genes, homeotic genes, and apoptosis help determine pattern formation during development. Section: 22.02 Topic: Animal Reproduction and Development
23. The process in which neighboring cells influence the development of each other, either by direct contact or by production of chemical signals, is A. neurulation. B. gastrulation. C. induction. D. cytoplasmic segregation. E. homeotic pattern formation. Neurulation is the process of forming the neural tube. Gastrulation is the formation of different layers of cells, which give rise to different tissues in the embryo. Cytoplasmic segregation is the parceling out of maternal determinants as mitosis occurs. Homeotic pattern formation refers to the establishment of specific sections of cells that give rise to specific structures.
Bloom’s Level: 1. Remember Learning Outcome: 22.02.02 Explain how cytoplasmic segregation and induction are able to influence cellular differentiation. Section: 22.02 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
24. The formation of the lens of the vertebrate eye from a skin layer in contact with an outgrowth of the embryonic brain is an example of A. cellular aggregation. B. induction. C. morphogenesis. D. gastrulation. E. neurulation. Cells close to the region where the lens will form send a molecular signal to nearby cells to induce them to express the genes that will cause the cells to become lens cells. This process is called induction.
Bloom’s Level: 3. Apply Learning Outcome: 22.02.02 Explain how cytoplasmic segregation and induction are able to influence cellular differentiation. Section: 22.02 Topic: Animal Reproduction and Development
25. In experiments with the roundworm Caenorhabditis elegans, researchers were able to trace the development of the generalized zygote into the specialized cells of the adult roundworm, a schematic that formed a(n) A. homeobox. B. pattern formation. C. cytoplasmic segregation. D. induction. E. fate map. Morphogenesis produces the shape and form of the body. A homeobox refers to a particular sequence of amino acids called a homeodomain that helps determine which genes are turned on. Pattern formation refers to the establishment of the axes of the body. Cytoplasmic segregation is the parceling out of maternal determinants as mitosis occurs. A fate map shows that as cells arise by cell division, they are destined to become particular structures.
Bloom’s Level: 3. Apply Learning Outcome: 22.02.02 Explain how cytoplasmic segregation and induction are able to influence cellular differentiation. Section: 22.02 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
26. Researchers have discovered individual parts. A. embryonic signposts B. morphogen genes C. cytoplasmic segregation routes D. induction genes E. fate maps
_ that determine the pattern of an animal and its
Certain genes called morphogen genes determine the relationship of individual parts of an organism such as the anterior and posterior ends.
Bloom’s Level: 2. Understand Learning Outcome: 22.02.03 Describe how morphogen genes, homeotic genes, and apoptosis help determine pattern formation during development. Section: 22.02 Topic: Animal Reproduction and Development
27. Which statement is NOT true about homeotic genes? A. Each homeotic gene has a variable region and a sequence called the homeobox that is very similar in all such genes. B. They have been found in almost all eukaryotic organisms. C. They are master genes that control all other genes in the organism. D. They are sequentially arranged on the chromosome in the same sequence they are activated during development of the embryo in Drosophila. E. All homeotic genes are thought to be derived from an original DNA sequence that has been highly conserved in evolution because of its importance to animal evolution. Homeotic genes encode master regulatory proteins that act by controlling the identity of each segment. They do not control all genes.
Bloom’s Level: 4. Analyze Learning Outcome: 22.02.03 Describe how morphogen genes, homeotic genes, and apoptosis help determine pattern formation during development. Section: 22.02 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
28. The presence of the same homeotic genes in a wide range of organisms for producing set patterns is an indication that A. evolution solved certain developmental problems just once. B. all organisms are composed of exactly the same proteins. C. the same homeotic mutations are likely to appear in all organisms. D. essentially we can complete the genome mapping of all animals in a short time. E. structures remain fairly constant, but regulation of animal development varies widely across the kingdom. Because the homeotic genes of so many different organisms contain the same homeodomain, we know that this nucleotide sequence arose early in the history of life and that it has been largely conserved as evolution occurred.
Bloom’s Level: 5. Evaluate Learning Outcome: 22.02.03 Describe how morphogen genes, homeotic genes, and apoptosis help determine pattern formation during development. Section: 22.02 Topic: Animal Reproduction and Development
29. A homeotic gene codes for a(n) protein, a sequence of 60 amino acids, sets of which determine the genes to be turned on during development. A. induction B. morphogenesis C. growth D. regulatory E. homeodomain A homeodomain consists of a particular sequence of 60 amino acids; this protein binds to DNA and acts to determine which genes are turned on.
Bloom’s Level: 2. Understand Learning Outcome: 22.02.03 Describe how morphogen genes, homeotic genes, and apoptosis help determine pattern formation during development. Section: 22.02 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
30. Genes that determine how segments develop during morphogenesis are termed genes. A. induction B. cleavage C. differentiation D. homeotic E. home alone Homeotic genes have been found in many organisms. They encode master regulatory proteins that act by controlling the identity of each segment.
Bloom’s Level: 1. Remember Learning Outcome: 22.02.03 Describe how morphogen genes, homeotic genes, and apoptosis help determine pattern formation during development. Section: 22.02 Topic: Animal Reproduction and Development
31. The is responsible for gas exchange in the egg of a chicken. A. yolk sac B. amnion C. embryo D. umbilical cord E. chorion The yolk sac surrounds the remaining yolk, which provides nourishment to the developing chick. The amnion is the layer that contains the protective amniotic fluid, which bathes the developing embryo. Embryo refers to the early developmental stages. The umbilical cord connects the developing embryo to the placenta. The chorion lies next to the shell and carries on gas exchange.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
32. The provides nourishment for the embryo during chick development. A. umbilical cord B. allantois C. yolk sac D. embryo E. chorion The umbilical cord connects the developing embryo to the placenta. The allantois is the membrane that collects nitrogenous wastes. The yolk sac surrounds the remaining yolk, which provides nourishment to the developing chick. Embryo refers to the early developmental stages. The chorion lies next to the shell and carries on gas exchange.
Bloom’s Level: 1. Remember Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Animal Reproduction and Development
33. The allantoic blood vessels A. are analogous to the umbilical blood vessels in humans. B. are a structure composed of two germ layers. C. lie next to the shell in chicks. D. have the same function in chicks as humans. E. become the chorionic villi. Both chicks and humans have the same extra embryonic membranes. In chicks, the chorion carries on gas exchange and the yolk sac supplies nutrients. In humans, the placenta, which carries on gas exchange and supplies nutrients to the fetus, arises from the chorion. The yolk sac serves as the site of blood cell formation. This shows that humans have modified the extraembryonic membranes to perform with internal development.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Animal Reproduction and Development
22-18 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
34. The develops into the fetal portion of the placenta. A. amnion B. archenteron C. chorion D. yolk sac E. allantois In humans, the placenta, which carries on gas exchange and supplies nutrients to the fetus, arises from the chorion.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.02 Summarize, in chronological order, the major events that occur during fetal development from three to nine months. Section: 22.03 Topic: Human Fetal Development
35. Which part of the blastocyst will develop into the human embryo? A. archenteron B. blastopore C. chorion D. trophoblast E. inner cell mass The archenteron is a cavity in the developing gastrula that will eventually become the gut. The blastopore is a pore created by invagination during gastrulation that will eventually become the anus. The chorion is a membrane that develops into the fetal half of the placenta. The trophoblast is the single layer of cells around the blastocyst; later, it gives rise to the chorion. The inner cell mass of the blastocyst eventually becomes the embryo, which develops into a fetus. Embryonic stem cells are derived from this inner cell mass.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Human Embryonic Development
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Chapter 22 - Development and Aging
36. A pregnancy test is based on the production of A. FSH; anterior pituitary B. estrogen; ovary C. progesterone; corpus luteum D. hCG; trophoblast E. GnRH; hypothalamus
by the
.
FSH, estrogen, progesterone, and GnRH are all present whether or not there is a pregnancy. hCG is produced only during pregnancy. The developing trophoblast secretes the hormone human chorionic gonadotropin (hCG), which helps to maintain the corpus luteum past the time it normally disintegrates.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Animal Reproduction and Development
37. The human blastocyst A. gives rise to umbilical blood vessels. B. contains a fluid-filled central cavity. C. is the expanded end of the gastrula. D. has three germ layers. E. is the fertilized egg. The blastocyst has a fluid-filled cavity. The single outer layer of cells is called the trophoblast. Also, there is a group of cells inside the blastocyst called the inner cell mass which gives rise to the embryo.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Human Embryonic Development
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Chapter 22 - Development and Aging
38. Implantation occurs during the A. fertilization. B. first week. C. second week. D. second month. E. fourth month. During the first week the developing embryo travels down the oviduct. It reaches the uterus around the end of the first week and the trophoblast develops and serves as the implantation medium and eventually develops into the chorion.
Bloom’s Level: 1. Remember Learning Outcome: 22.03.02 Summarize, in chronological order, the major events that occur during fetal development from three to nine months. Section: 22.03 Topic: Animal Reproduction and Development
39. In human development, the A. umbilical cord B. placenta C. amniotic fluid D. yolk sac E. allantois
is the site of early blood cell formation for the fetus.
In humans, the yolk sac serves as the site of blood cell formation. The placenta is the site of gas exchange and nutrient transfer between the mother and the fetus. Amniotic fluid serves to cushion and protect the fetus. The yolk sac serves as the site of blood cell formation. The allantois is one of the extraembryonic membranes that serves to form the umbilical blood vessels.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Human Fetal Development
22-21 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
40. The sac that surrounds the fetus and usually ruptures just before childbirth is the A. gastrula. B. allantois. C. yolk sac. D. amnion. E. archenteron. The gastrula is an early stage of development where distinct layers of cells have formed that will later give rise to specific tissues. The allantois is one of the extraembryonic membranes that serves to form the umbilical blood vessels. The yolk sac is the site of blood cell formation. The amnion contains amniotic fluid, which protects and cushions the embryo; this membrane ruptures shortly before childbirth, releasing the fluid. The common name for this process is that the mother's "water broke." The archenteron is a cavity in the developing gastrula that will eventually become the gut.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Human Fetal Development
41. Which association is incorrect? A. chorion—gas exchange B. amnion—blood vessels C. allantois—waste storage D. yolk sac—food storage E. yolk—nourishment The amnion contains amniotic fluid, which protects and cushions the embryo, while the site of blood vessel formation is the yolk sac.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Human Fetal Development
22-22 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
42. Which of the following serves as an insulator for the embryo? A. amniotic fluid B. allantois C. chorion D. primitive streak E. yolk sac The amnion is a membrane that contains the amniotic fluid, which protects and cushions the fetus while it is in the womb.
Bloom’s Level: 1. Remember Learning Outcome: 22.03.01 Identify the extraembryonic membranes and provide a function for each. Section: 22.03 Topic: Human Embryonic Development
43. Which statement is NOT correct about the placenta? A. Oxygenated blood flows to the fetus via the umbilical vein. B. Fetal blood flows to the placenta via the umbilical artery. C. Oxygen and nutrients go to the fetus from the maternal circulatory system. D. Maternal and fetal blood mix with each other. E. The placenta produces progesterone and estrogen. The circulatory systems of the fetus and mother are completely separate. Gas exchange, nutrient uptake and the removal of wastes takes place through the placenta. The choronic villi are the specific sites of exchange and are linked into the umbilical veins and arteries.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.03 Describe the flow of blood in a fetus and explain the role of the placenta. Section: 22.03 Topic: Human Fetal Development
22-23 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
44. The sex of the fetus can be first determined in the A. first B. second C. third or fourth D. six or seventh E. eighth or ninth
month of development.
Sometime during the third or fourth month the testes descend into the scrotal sacs and can be visualized so gender may be determined. It is possible that the testes will fail to descend during this time and an operation will be required after birth to place them in their proper location.
Bloom’s Level: 1. Remember Learning Outcome: 22.03.02 Summarize, in chronological order, the major events that occur during fetal development from three to nine months. Section: 22.03 Topic: Human Fetal Development
45. The in fetal circulation allows blood from the right atrium to directly enter the left atrium. A. umbilical cord B. arterial duct C. foramen ovale D. venous duct E. right ventricle Since the fetus does not use its lungs, it is necessary for circulation to differ from the final adult process. In order to accommodate this, most of the blood entering the right atrium is shunted to the left atrium through a structure called the foramen ovale, which is an opening between the two atria. This opening normally closes after birth.
Bloom’s Level: 1. Remember Learning Outcome: 22.03.03 Describe the flow of blood in a fetus and explain the role of the placenta. Section: 22.03 Topic: Human Fetal Development
22-24 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
46. The placenta develops from A. fetal membranes only. B. maternal tissue only. C. both fetal and maternal tissue. D. separate polar bodies that develop just the placental tissues. E. from the yolk sac. The chorion, which is an extraembryonic membrane, and the mother’s uterine tissue are both part of the placenta.
Bloom’s Level: 3. Apply Learning Outcome: 22.03.02 Summarize, in chronological order, the major events that occur during fetal development from three to nine months. Section: 22.03 Topic: Human Fetal Development
47. During the month, the fetal heartbeat is first loud enough to be heard by a physician with a stethoscope. A. first B. second C. fourth D. sixth E. eighth If a physician uses a stethoscope during the fourth month they can hear the heartbeat of the fetus since by that time the development of the heart is sufficient to beat that strongly.
Bloom’s Level: 1. Remember Learning Outcome: 22.03.02 Summarize, in chronological order, the major events that occur during fetal development from three to nine months. Section: 22.03 Topic: Human Fetal Development
22-25 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
48. In order to provide for easier birth, a newborn baby has membranous areas of skull called A. oval windows. B. bursa. C. sutures. D. septa. E. fontanels. Fontanels are large membranous areas of the skull which allow for flexibility during birth. They will close as the skull bones fuse when the child is about 16 months of age.
Bloom’s Level: 3. Apply Learning Outcome: 22.03.02 Summarize, in chronological order, the major events that occur during fetal development from three to nine months. Section: 22.03 Topic: Human Fetal Development
49. From months 5 to 7, a fetus is covered in a fine down called A. chorion. B. pubescence. C. lanugo. D. vernosa. E. afterbirth. The skin of the fetus is covered by lanugo, a fine down, which is in turn covered by vernix caseosa, a white, greasy cheese-like substance. These structures appear to protect the very delicate skin against the amniotic fluid.
Bloom’s Level: 1. Remember Learning Outcome: 22.03.02 Summarize, in chronological order, the major events that occur during fetal development from three to nine months. Section: 22.03 Topic: Human Fetal Development
22-26 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
50. Which of the following ducts connect the pulmonary trunk to the aorta in the fetus? A. oval opening B. venous duct C. arterial duct D. umbilical artery E. umbilical vein Since the fetus does not use its lungs, it is necessary for circulation to differ from the final adult process. As part of this change in flow, the ductus arteriosus is used to shunt blood that enters the right ventricle and passes into the pulmonary trunk into the aorta. This blood then moves into the various branches that connect to the umbilical arteries, allowing the blood to move to the placenta and undergo exchange of materials with maternal blood.
Bloom’s Level: 2. Understand Learning Outcome: 22.03.03 Describe the flow of blood in a fetus and explain the role of the placenta. Section: 22.03 Topic: Human Fetal Development
51. When a baby is born, what change(s) do NOT need to occur in fetal circulation for it to survive? A. The oval opening (foramen ovale) must close. B. The arterial duct must close. C. The umbilical arteries and veins close. D. Blood must be able to pass from the right side to the left side. E. Fetal blood must be shunted toward the lungs after tying off the umbilical cord. Fetal blood has been moved from the right atrium to the left atrium of the fetal heart but it has not passed through the lungs of the fetus since no gas exchange took place there. The foramen ovale and the ductus arteriosus have been used to circumvent the normal adult flow of blood through the lungs. These two structures must close for normal circulation to occur and the blood must flow through the lungs for gas exchange to take place.
Bloom’s Level: 4. Analyze Learning Outcome: 22.03.03 Describe the flow of blood in a fetus and explain the role of the placenta. Section: 22.03 Topic: Human Fetal Development
22-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
52. The afterbirth refers to the A. expulsion of the placenta and membranes. B. expulsion of the uterine lining. C. expulsion of the amniotic membrane. D. expulsion of the fetus. E. period of time from birth to nursing. After the birth of the child, contractions continue and, approximately 15 minutes later, the placenta is expelled through the vagina.
Bloom’s Level: 1. Remember Learning Outcome: 22.04.01 Describe changes that occur in female physiology during pregnancy. Section: 22.04 Topic: Animal Reproduction and Development
53. During which stage of labor will the cervix be dilated? A. stage 1 B. stage 2 C. stage 3 D. stage 4 E. The cervix does not need to dilate for labor. Dilation of the cervix begins during stage 1 of birth as the cervical canal slowly disappears due to uterine contractions. The baby’s head then serves to act as a wedge, leading to further cervical dilation. Stage 1 ends when the cervix is completely dilated.
Bloom’s Level: 1. Remember Learning Outcome: 22.04.02 Outline the stages of birth. Section: 22.03 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
54. Which of the following hormones causes the uterus to contract? A. oxytocin B. estrogen C. progesterone D. testosterone E. prolactin Uterine contractions are begun by stretching of the cervix and also leads to the release of oxytocin by the posterior pituitary.
Bloom’s Level: 1. Remember Learning Outcome: 22.04.01 Describe changes that occur in female physiology during pregnancy. Section: 22.04 Topic: Animal Reproduction and Development
55. The hormone from the causing milk to flow. A. prolactin; hypothalamus B. prolactin; anterior pituitary C. oxytocin; hypothalamus D. oxytocin; anterior pituitary E. estrogen; ovaries
causes mammary lobules in the breast to contract,
The hypothalamus signals the pituitary gland to produce oxytocin in response to the stimulus of suckling on the breast. The arrival of oxytocin in the breast results in contractions of the lobules that cause milk to flow into the ducts and allows the milk to be drawn out.
Bloom’s Level: 2. Understand Learning Outcome: 22.04.01 Describe changes that occur in female physiology during pregnancy. Section: 22.04 Topic: Animal Reproduction and Development
22-29 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
56. Milk production is triggered in breasts following several days of . A. prolactin; hypothalamus B. prolactin; anterior pituitary C. oxytocin; hypothalamus D. oxytocin; anterior pituitary E. estrogen; ovaries
production by the
The production of milk is begun in response to the hormone prolactin. Prolactin production is suppressed during pregnancy due to the increased levels of estrogen and progesterone. After the birth of the baby this suppression of the production of prolactin is no longer in place and the anterior pituitary begins to secrete the hormone.
Bloom’s Level: 2. Understand Learning Outcome: 22.04.03 Summarize the advantages of breast-feeding. Section: 22.04 Topic: Animal Reproduction and Development
57. After milk production begins, it continues as long as the baby suckles, but ends otherwise. This maintenance is provided by nervous stimulation sent to the , which eventually causes production of . A. ovaries; prolactin B. pituitary; prolactin C. hypothalamus; oxytocin D. ovaries; oxytocin E. uterus; estrogen The hypothalamus signals the pituitary gland to produce oxytocin in response to the stimulus of suckling on the breast. The arrival of oxytocin in the breast results in contractions of the lobules that cause milk to flow into the ducts and allows the milk to be drawn out. For milk production to continue, it is necessary for the stimulus of suckling to continue to stimulate oxytocin production.
Bloom’s Level: 3. Apply Learning Outcome: 22.04.03 Summarize the advantages of breast-feeding. Section: 22.04 Topic: Animal Reproduction and Development
22-30 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
58. Which statement is NOT true regarding aging? A. A genetic factor is indicated when children of long-lived parents live a long time. B. Different body processes wear out. C. Lack of calcium in a person's diet may lead to bone loss. D. The immune system increases in activity. E. Some proteins become crosslinked. The immune system no longer functions as it did in the past. Cancer and autoimmune diseases often increase in incidence as we age.
Bloom’s Level: 3. Apply Learning Outcome: 22.05.01 Describe the factors that contribute to aging. Section: 22.05 Topic: Human Development and Aging
59. Which of the following would NOT provide evidence for a genetic basis for aging? A. Children of short-lived parents live shorter lives than children of long-lived parents. B. Cells only divide for a set number of divisions. C. Cell lines become nonfunctional before their maximum number of divisions is reached. D. Smoking and excessive alcohol intake are linked to shorter life spans. E. Certain genes that code for antioxidant enzymes to detoxify free radicals are present in animals that live longer. Since smoking and alcohol consumption are not genetic factors, they would not provide a genetic basis for aging. They can, however, act to increase the effects of aging.
Bloom’s Level: 5. Evaluate Learning Outcome: 22.05.02 Discuss the effects of aging on various body systems. Section: 22.05 Topic: Human Development and Aging
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Chapter 22 - Development and Aging
60. As we become older, the A. skin becomes more elastic. B. heart decreases in size. C. blood vessels become harder. D. production of gametes continues in both sexes. E. liver receives increased blood flow. With aging, the skin becomes less elastic as cross-connections form between collagen fibers. Typically the heart shrinks in size due to a loss of cardiac muscle, and blood flow to the liver is reduced affecting its ability to metabolize drugs. Gamete production in both sexes is reduced, with females typically ceasing around 50 years of age when menopause occurs. Blood vessels do become harder as the fibers which make the elastic become increasingly cross-linked.
Bloom’s Level: 3. Apply Learning Outcome: 22.05.01 Describe the factors that contribute to aging. Section: 22.05 Topic: Human Development and Aging
22-32 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
61. Researchers have observed that in villages in Russian Georgia where people have a lowcalorie diet but are not malnourished, the average life expectancy is longer. Less food intake would result in fewer free radicals from metabolism, compared to societies where individuals ate more than they required. The text explains that genes for antioxidant enzymes that destroy free radicals would be inherited. How could we test, in a practical manner, whether this longevity was genetic or a matter of chosen diet? A. Continue keeping records of villagers versus a control population with a normal diet until the numbers are statistically significant. B. Track the fate of villagers who move away to more affluent communities where people eat a better diet, or supplement the diet of a portion of the villagers over time. C. Conduct test-crosses to determine the life span of F1 and F2 generations. D. Examine a cell sample for developmental or homeotic genes. E. Take DNA samples from people and test for similarity of genes. To track villagers with their diet and non-villagers with the normal diet for their population would not address the issue. The diets of the two populations would need to be reversed. It would be unethical to conduct test-crosses with humans, so testing F1 and F2 generations is not possible. Since homeotic genes are highly conserved across species, it is reasonable to assume that genetic differences would not exist in-between two human populations. While DNA samples might reveal genetic differences between the two populations, it would not explain whether or not those genetic differences would be difficult to directly relate to the aging process without a great deal of further research. The option of looking at villagers who change their diets would address the question directly and provide much clearer results.
Bloom’s Level: 5. Evaluate Learning Outcome: 22.05.02 Discuss the effects of aging on various body systems. Section: 22.05 Topic: Human Development and Aging
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Chapter 22 - Development and Aging
62. Free radicals, unpaired electrons, can cause aging because A. they collect in the organs and skin, forming large molecules that interfere with organ function. B. they bond with molecules that are nearby, like DNA and proteins, causing damage to the molecule. C. unpaired electrons pair up together and attack DNA. D. free radicals seek to free other electrons so that all electrons are stripped from DNA. E. they speed up metabolic processes, causing the organs to wear out faster. Free radicals (unpaired electrons) are attracted to nearby molecules so that they can bond with these molecules. The addition of an electron would change the charge and shape of the molecule, possibly changing the functioning of the molecule. This is especially true of DNA since it contains the directions to direct the cell's functioning. Free radicals do not play a role in the speed of metabolic processes.
Bloom’s Level: 4. Analyze Learning Outcome: 22.05.02 Discuss the effects of aging on various body systems. Section: 22.05 Topic: Human Development and Aging
63. As aging occurs, changes are seen in the function of body systems. Among these changes are all but which of the following? A. A decline in the production of hormones leads to menopause and other changes. B. A decline in the function of the immune system leads to increases in certain diseases, like cancer. C. An increase in cross-links between proteins leads to decreased flexibility. D. An increase in the production of hormones leads to menopause and other changes. E. An increase in the size of the heart muscle occurs. A decline in the production of estrogen leads to menopause. When the production of other hormones declines, it also leads to the decrease in functionality of organs. The heart decreases in strength so it can increase in size in order to compensate.
Bloom’s Level: 3. Apply Learning Outcome: 22.05.01 Describe the factors that contribute to aging. Section: 22.05 Topic: Human Development and Aging
22-34 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 22 - Development and Aging
64. Some of the changes that are attributed to aging are actually brought about by poor health habits. One of these is osteoporosis, which could be prevented by A. adequate calcium intake and eating a high-protein diet. B. not smoking cigarettes, adequate calcium intake, and exercise. C. exercise and a high-protein diet. D. drinking adequate amounts of water, and exercising. E. a low-fat and protein diet. Osteoporosis is a condition where bone density decreases so that bones are more likely to fracture with only minor trauma. The condition can be lessened or eliminated by stopping smoking and heavy alcohol intake as well as positive lifestyle changes such as an increase in the consumption of calcium and an exercise program. The levels of protein in the diet and water consumed have not been shown to have an effect on the condition.
Bloom’s Level: 2. Understand Learning Outcome: 22.05.01 Describe the factors that contribute to aging. Section: 22.05 Topic: Human Development and Aging
65. Which of these is not a change in the skin that occurs with aging? A. loss of elasticity B. loss of adipose tissue C. loss of sweat glands D. loss of sebaceous (oil) glands E. All of the answer choices are changes seen in the skin with aging. All of the answer choices occur to the skin with aging. The skin becomes less elastic with the cross-linking of collagen fibers that make up the skin, in addition to a loss of those elastic fibers and adipose tissue in the subcutaneous tissue. In addition, sweat glands in the skin are reduced, as are the number of sebaceous glands. The loss of sweat glands makes it harder for the individual to adjust to temperature changes and the loss of sebaceous glands leads to drying and cracking of the skin.
Bloom’s Level: 2. Understand Learning Outcome: 22.05.01 Describe the factors that contribute to aging. Section: 22.05 Topic: Human Development and Aging
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Chapter 22 - Development and Aging
66. Women are less likely than men to suffer from a heart attack. This is thought to be true because A. women have less stressful lives than men. B. men have testosterone and women do not. C. women have estrogen and men do not. D. women tend to take better care of their health. E. women have a genetic resistance that prevents them from having heart attacks. Estrogen is thought to affect the cardiovascular system so that heart attacks are less likely. As women enter menopause and estrogen levels decrease, they have an increased incidence and eventually surpass men in the incidence of heart attacks. There is no gene in women that would prevent them from having heart attacks.
Bloom’s Level: 3. Apply Learning Outcome: 22.05.01 Describe the factors that contribute to aging. Section: 22.05 Topic: Human Development and Aging
67. Which of the following is not true about age-related issues? A. Neurons are extremely sensitive to oxygen deficiency, so loss of brain function may occur due to narrowing of the blood vessels rather than aging. B. With age, some sensory receptors require more stimulation to function at the same level as before. C. Strength in the muscles that control breathing may be lost, leading to breathing difficulties. D. Osteoporosis, the loss of bone density, cannot be prevented and all women will suffer from it to some degree. E. All of the answer choices are true. Osteoporosis is a condition where bone density decreases so that bones are more likely to fracture with only minor trauma. The condition can be lessened or eliminated by stopping smoking and heavy alcohol intake as well as positive lifestyle changes such as an increase in the consumption of calcium and an exercise program.
Bloom’s Level: 2. Understand Learning Outcome: 22.05.01 Describe the factors that contribute to aging. Section: 22.05 Topic: Human Development and Aging
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Chapter 22 - Development and Aging
68. What is the difference between a skin cell and a stem cell? A. A skin cell gives rise only to other skin cells and a stem cell gives rise only to stems. B. A skin cell has a full complement of DNA, but a skin cell does not. C. Specific genes are turned on in skin cells, while stem cells still have all of their genes turned on. D. Stem cells are not found in humans, while skin cells are. E. Stem cells arise from skin cells when select genes are turned on. Stem cells are undifferentiated cells that can possibly become any type of tissue. Stem cells can give rise to any of a number of types of cells. Both skin and stem cells have the same complement of DNA, but in skin cells not all of the DNA is active. Stem cells are not derived from skin cells; it is the other way around.
Bloom’s Level: 4. Analyze Learning Outcome: 22.02.01 Explain how cellular differentiation and morphogenesis play a role in development. Section: 22.02 Topic: Animal Reproduction and Development
69. Which of the following does not occur during the process of fertilization? A. The sperm cell releases acrosomal enzymes. B. The sperm cell enters the egg and the nucleus is released. C. The nucleus of the sperm and egg cell fuse. D. The plasma membrane of the sperm head and egg fuse. E. The tail bores a hole in the zona pellucida, which the head then penetrates. It is the enzymes in the acrosome that bore the hole in the zona pellucida.
Bloom’s Level: 2. Understand Learning Outcome: 22.01.02 Describe the steps involved in fertilization. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
70. The statement "a child is more closely related to its mother than to its father" is A. not true because there is an equal contribution from both parents. B. true because the egg contains more than just DNA and the sperm only contributes DNA. C. not true because the egg only contributes DNA, while the sperm contributes enzymes from the acrosome and mitochondria. D. true because the mother's blood flows into the umbilical cord and, therefore, the baby shares the mother's blood. E. true because it is possible for a woman to become pregnant without the sperm. Since the egg contains the machinery that is necessary for the development of the fetus and since that cellular machinery is inherited from the mother, the statement is true. Notably, the mitochondria contain DNA that is not part of the nuclear DNA. This mitochondrial DNA is therefore inherited completely from the mother. If the mother has a disorder carried in the mitochondrial DNA, then the child must inherit that disorder. While in some animal species the female can give birth without sperm fertilizing the egg, it is not possible in humans.
Bloom’s Level: 5. Evaluate Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.02 Topic: Animal Reproduction and Development
Short Answer Questions 71. List and describe the major events during the three stages of birth. Stage 1: The cervical canal disappears as the lower part of the uterus is pulled towards the baby's head. The cervix dilates. The amniotic membrane ruptures, releasing the amniotic fluid. Stage 2: Uterine contractions occur every 1 to 2 minutes. The baby's head gradually descends into the vagina and begins to exit the vagina. The mother must push to help propel the head and shoulders outward. The rest of the baby emerges from the vagina. Once the child is breathing normally, the umbilical cord is cut and tied. Stage 3: The placenta or afterbirth is delivered as the uterine contractions shrink the uterus.
Bloom’s Level: 6. Create Learning Outcome: 22.04.02 Outline the stages of birth. Section: 22.04 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
Multiple Choice Questions 72. What role does the acrosome play in fertilization? A. It contains enzymes that digest a pathway for the sperm through the zona pellucida. B. It contains lipids that digest a pathway for the sperm through the zona pellucida. C. It contains enzymes that digest a pathway for the sperm through the corona radiata. D. It contains lipids that digest a pathway for the sperm through the corona radiata. E. It causes the zona pellucida to become the fertilization membrane. The acrosome contains enzymes that digest a pathway for the sperm through the zona pellucida. There are no lipids in the acrosome. The sperm will wiggle its way through the corona radiata; it doesn't require the acrosome for this. The cortical granules release enzymes that cause the zona pellucida to become the fertilization membrane.
Bloom’s Level: 2. Understand Learning Outcome: 22.01.01 Identify the major structures of a sperm cell and an oocyte, and provide a function for each. Section: 22.01 Topic: Animal Reproduction and Development
73. Which of the following is an advantage of breast-feeding? A. All of the answer choices are advantages of breast-feeding. B. Colostrum is rich in proteins and antibodies. C. physiological benefits for the mother D. Suckling causes uterine contractions that help reduce the uterus to its normal size. E. Breast-feeding uses up calories that can help a woman return to her normal weight. All of the answer choices are advantages of breast-feeding.
Bloom’s Level: 2. Understand Learning Outcome: 22.04.03 Summarize the advantages of breast-feeding. Section: 22.04 Topic: Human Development and Aging
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Chapter 22 - Development and Aging
74. During which stage of birth is the placenta delivered? A. third stage B. first stage C. second stage D. It is done prior to stage one. E. It is done after the third stage. The placenta, or afterbirth, is delivered during the third stage of birth. Once it is delivered, the third stage of birth is complete.
Bloom’s Level: 1. Remember Learning Outcome: 22.04.02 Outline the stages of birth. Section: 22.04 Topic: Animal Reproduction and Development
Short Answer Questions 75. Describe the six steps involved in fertilization of an egg. Step 1: The sperm makes its way through the corona radiata. Step 2: Acrosomal enzymes digest a portion of zona pellucida. Step 3: Sperm binds to and fuses with oocyte plasma membrane. Step 4: The sperm nucleus enters cytoplasm of oocyte. Step 5: Cortical granules release enzymes; zona pellucida becomes fertilization membrane. Step 6: Sperm and egg pronuclei are enclosed in a nuclear envelope.
Bloom’s Level: 6. Create Learning Outcome: 22.01.02 Describe the steps involved in fertilization. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 22 - Development and Aging
76. List the three embryonic germ layers and indicate a structure that forms from each. Ectoderm layer: nervous system, epidermis of skin, epithelial lining of oral cavity and rectum Mesoderm layer: musculoskeletal system, dermis of skin, cardiovascular system, urinary system, reproductive system, outer layers of respiratory and digestive systems Endoderm layer: epithelial lining of digestive tract and respiratory tract, associated glands of digestive and respiratory tracts, epithelial lining of urinary bladder
Bloom’s Level: 6. Create Learning Outcome: 22.01.03 Describe the cellular, tissue, and organ stages of development. Section: 22.01 Topic: Animal Reproduction and Development
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Chapter 23 - Patterns of Gene Inheritance
Chapter 23 Patterns of Gene Inheritance
Multiple Choice Questions 1. Alternative forms of a gene that influence the same trait and are found at the same location in homologous chromosomes are called A. alleles. B. phenotypes. C. genotypes. D. codominant. E. incomplete dominance. Genotype describes the genetic makeup of an individual, while phenotype describes the appearance of the individual based on which allele is expressed. Codominant and incomplete dominance describe instances where one allele is not completely dominant over the other.
Bloom’s Level: 1. Remember Learning Outcome: 23.01.02 Define the term allele, and explain what it means for an allele to be dominant or recessive. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
2. What are alleles? A. genes for different traits, such as hair color and eye color B. alternative forms of a gene for a single trait, such as blue eyes or brown eyes C. the locations of genes on a chromosome D. recessive forms of a kind of characteristic carried by genes E. dominant forms of a kind of characteristic carried by genes Genes may have more than one form such as hair color, eye color, or length of fingers; these alternate forms are called alleles. The gene is located at a specific point on a particular chromosome and the alternate forms are located on homologous chromosomes. One form may be expressed in preference to the other; this is termed the dominant trait.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.02 Define the term allele, and explain what it means for an allele to be dominant or recessive. Section: 23.01 Topic: Mendelian Genetics
3. Mendel carried out most of his research with A. livestock. B. pea plants. C. guinea pigs. D. fruit flies. E. bacteria. Mendel used pea plants for his research, since they are easily manipulated and are found in several varieties. Plants are generally suitable for the experiments because of their ability to self-pollinate.
Bloom’s Level: 1. Remember Learning Outcome: 23.01.01 Describe how Mendel’s law of segregation and law of independent assortment are related to the movements of chromosomes during gamete formation and fertilization. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
4. Mendel's law of segregation implies that the two members of an allele pair A. are distributed to separate gametes. B. are distributed to the same gamete. C. are assorted dependently. D. are segregated pairwise. E. must always code for the identical trait or feature. By examining the F2, generation Mendel was able to see that a heterozygous plant had two alternate forms for the trait that he was studying. Either of these two traits could be passed on to offspring. This demonstrated that the two forms separated during the formation of gametes.
Bloom’s Level: 3. Apply Learning Outcome: 23.01.01 Describe how Mendel’s law of segregation and law of independent assortment are related to the movements of chromosomes during gamete formation and fertilization. Section: 23.01 Topic: Mendelian Genetics
5. For Mendel's law of segregation to occur, the alleles must be A. at different loci on the same chromosome. B. at the same loci on the same chromosome. C. on a nonhomologous pair of chromosomes. D. on a homologous pair of chromosome at the same loci. E. on different loci on different chromosomes. Alleles are located on homologous chromosomes. These two chromosomes are separated during the formation of gametes. The homologous chromosomes carry the same genes; they code for the same traits, but may have different forms of the trait. Alleles are not only located on homologous chromosomes; they are also located at the same locus on the two chromosomes.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.01 Describe how Mendel’s law of segregation and law of independent assortment are related to the movements of chromosomes during gamete formation and fertilization. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
6. Which is NOT true according to Mendel's law of segregation? A. Each individual contains two alleles for each trait. B. An individual can have either both dominant alleles, both recessive alleles, or a dominant and recessive allele. C. Alleles separate from each other during gamete formation. D. Each gamete contains one copy of each allele. E. Fertilization restores the presence of two alleles. While one allele may be dominant over another, it is not necessary for the individual to contain one of each choice. The individual could have two dominant alleles (homozygous dominant) or two recessive alleles (homozygous recessive).
Bloom’s Level: 2. Understand Learning Outcome: 23.01.01 Describe how Mendel’s law of segregation and law of independent assortment are related to the movements of chromosomes during gamete formation and fertilization. Section: 23.01 Topic: Mendelian Genetics
7. Which of the following represents the physical characteristics of the individual? A. phenotype B. genotype C. alleles D. homozygous E. dominance The phenotype of an individual refers to the visible characteristics of the organism; this could be physical appearance or behavior. Genotype refers to the genetic makeup of the individual. Remember that some traits can be dominant and some recessive. Recessive traits do not show in the individual if a dominant allele is present.
Bloom’s Level: 1. Remember Learning Outcome: 23.01.03 Contrast genotype and phenotype, and use appropriate terms for describing genotypes. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
8. The indicates the gene combination of an individual. A. phenotype B. loci C. genotype D. dominance E. homozygous Genotype refers to the genetic makeup of the individual. Remember that some traits can be dominant and some recessive; recessive traits do not show in the individual if a dominant allele is present, but the recessive allele can still be passed on to offspring. The phenotype of an individual refers to the visible characteristics of the organism; this could be physical appearance or behavior.
Bloom’s Level: 1. Remember Learning Outcome: 23.01.03 Contrast genotype and phenotype, and use appropriate terms for describing genotypes. Section: 23.01 Topic: Mendelian Genetics
9. If an organism shows a recessive phenotype, such as short pea plants, its genotype can be A. either TT or Tt. B. either Tt or tt. C. only TT. D. only tt. E. TT, Tt, or tt. Since the recessive trait shows only if the dominant trait is not present, then an individual showing the recessive trait must be homozygous recessive. In this instance tt is the only possible genotype.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.03 Contrast genotype and phenotype, and use appropriate terms for describing genotypes. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
10. If you had a black male guinea pig and brown female guinea pig, but you didn't know which was the dominant characteristic, how would you find out the dominant color, assuming the dominant characteristic is homozygous,? A. Mate them and see what color the offspring are—that will be the dominant color. B. Mate them and see what color the offspring are—the other will be the dominant color. C. Mate them, and then mate their offspring to see what color the next generation is—that will be the dominant color. D. Mate them together, and then mate their offspring to see what color the next generation is—the other color will be the dominant color. E. The dominant color could not be determined by mating the two, and then mating their offspring; further generations would have to be produced to determine the dominant color. If there are two forms of a trait and one is dominant over the other, then an organism that is heterozygous will show the dominant trait. The recessive trait is present, but is not seen in the organism. The F2 generation should show both parental phenotypes, but in different percentages. This will demonstrate which trait is dominant (it will be the highest percentage), but the F1 will show the dominant trait directly.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
11. If you were to breed a black male guinea pig to a brown female guinea pig in an attempt to determine which one was homozygous dominant and which one was homozygous recessive, how could you be certain that the guinea pigs are truly homozygous? A. The guinea pigs would be homozygous for black (or brown) coat color if each strain could be bred for many generations and only black- (or brown-) colored offspring were produced. B. If the immediate parents of the black (or brown) guinea pigs were both of that color, it proves they are homozygous. C. If a cross between the black and brown guinea pig produced three black and one brown offspring, the black guinea pig would have to be homozygous for black coat color. D. If a cross between the black and brown guinea pig produced four all black offspring, this would prove the black guinea pig was recessive. E. Only microscopic examination of the guinea pig's genes could absolutely confirm homozygosity. Homozygous organisms are also called true-breeding because they will only show a single form of a trait no matter how many generations are examined as long as there is no outcrossing. Many generations need to be examined, since a heterozygous organism would show only one allele (the dominant one), but the recessive allele could also be present.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
12. In humans, brown eyes (B) is a simple dominant trait over blue eyes (b). What is the genotype of a brown-eyed woman who has a blue-eyed child? A. bb. B. Bb. C. BB. D. BBB. E. BbBb. In order for a gamete to be created that has the recessive allele, it must be present in the parent. Since the trait is recessive, it could be hidden by the dominant version of the allele. In this case, the mother would need to be Bb. B would give her brown eyes, and b would allow her to pass the blue-eye allele on to her child.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
13. The ability to roll the edges of the tongue upward in a U-shape has been considered to be an inherited ability. The standard assumption is that tongue-rolling is a dominant allele at a single gene locus. Which of the following would cast doubt on this assumption? A. Behaviors are not inherited, only structures are inherited. B. A student who cannot roll his tongue has a mother and father, both of whom can. C. A student who can roll his tongue has a mother and father, both of whom cannot. D. With very little effort the non-tongue-rollers can learn to also roll their tongues. E. A student who can roll his tongue has a mother and father, both of whom cannot roll their tongues AND with very little effort the non-tongue-rollers can learn to roll their tongues. An organism's behavior is part of the phenotype of the individual and is at least partly inherited. If the child cannot roll his tongue but both parents can, this could simply mean that they carry the recessive trait but show the dominant one. While having a student who can roll their tongue when neither parent could would give some indication that the trait is learned rather than inherited. Having the additional information that non-tongue-rollers can learn to roll their tongues would give even stronger support to the hypothesis that the ability is learned and not inherited.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
14. In humans, widow's peak (W) is dominant over straight hairline (w). If a heterozygous male marries a female with a straight hairline, what percent of their children can be expected to have widow's peak? A. 0% B. 25% C. 50% D. 100% E. can't be determined The parents' genotypes are Ww (male) and ww (female). The father could give either a W or a w to any gamete created, while the mother could only give a w. When looking at the offspring, you would find 50% would be heterozygous (Ww) and have a widow's peak and 50% would be homozygous recessive (ww) and have a straight hairline.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
15. What will the genotypic ratio be of a monohybrid cross of two individuals who are both heterozygous for a trait? A. 100% homozygous dominant B. 100% homozygous recessive C. 50% homozygous dominant, 50% homozygous recessive D. 25% homozygous dominant, 50% heterozygous, 25% homozygous recessive E. 50% homozygous dominant, 50% heterozygous Two heterozygous individuals would have the genotype of Aa and they could donate either of these allelic choices to a gamete that was created. This would lead to 25% as homozygous dominant (AA), 50% heterozygous (Aa), and 25% homozygous recessive (aa). The phenotypic ratio would be 75% showing the dominant trait and 25% showing the recessive.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
16. Tay-Sachs is a recessive genetic disease in humans. If one parent is homozygous normal and the other is a carrier (heterozygous), what are the chances that their child will have this disease? A. 0% B. 25% C. 50% D. 75% E. 100% The parents' genotypes are TT and Tt; both of these individuals would function normally. The homozygous dominant parent could only give a dominant allele for the normal gene, while the heterozygote could give either T or t. Since any child would have received T from the homogygous dominant parent, all children would have a normal phenotype even though 25% could be carriers with a Tt genotype.
Bloom’s Level: 2. Understand Learning Outcome: 23.02.02 Analyze pedigrees to determine the probability of a genetic disorder. Section: 23.02 Topic: Pedigrees
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Chapter 23 - Patterns of Gene Inheritance
17. In which kind of cross would you expect to find a phenotypic ratio of 3:1 among the F2 offspring? A. monohybrid cross B. dihybrid cross C. test cross D. homozygous recessive homozygous dominant E. homozygous dominant heterozygous A monohybrid cross would give the genotypes of 25% homozygous dominant, 50% heterozygous, and 25% homozygous recessive. The first two categories would have the dominant phenotype and only the homozygous recessive offspring would show the recessive trait. A dihybrid cross would show a 9:3:3:1 phenotypic ratio. A homozygous recessive homozygous dominant cross (AA aa) would show 100% the dominant trait. A homozygous dominant heterozygous (AA Aa) would show 100% the dominant trait.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
18. In a Mendelian monohybrid cross involving two homozygous genotypes, the generation is always completely heterozygous. A. F1 B. F2 C. F3 D. P E. P2 If the parents are homozygous (AA and aa), then any offspring could only have one of each of the two allele choices (Aa) and be heterozygous. The F2 being a cross between two heterozygotes would have a mixture of all three possible genotypes (AA, Aa, aa). An F3 generation would also show a mixture of the three possible genotypes.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
19. In a testcross, an organism with a dominant phenotype, but unknown genotype is crossed with which of the following to establish its genotype? A. homozygous dominant B. heterozygous C. homozygous recessive D. heterozygous dominant E. monohybrid dominant In a testcross, a known genotype must be used to cross with the unknown genotype. The only sure genotype would be an organism that showed the recessive trait, which would have to be homozygous recessive. Depending on the genotype, the organism with the dominant characteristic would be either (AA aa) 100% dominant phenotype or (Aa aa) 50% dominant phenotype and 50% recessive phenotype.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
20. In peas, yellow is dominant over green in seeds. With which of these is it best to cross a yellow-seeded pea plant to determine whether it is homozygous or heterozygous? A. a green-seeded plant B. a heterozygous yellow-seeded plant C. a pure yellow-seeded plant D. a heterozygous yellow-seeded plant or a pure yellow-seeded plant E. There is no way to determine if the plant is pure or hybrid. Where Y = yellow and y = green, in a testcross a known genotype must be used to cross with the unknown genotype. In this case it would be yy for green seeds since this is the recessive trait. Depending on the genotype, the organism with the dominant characteristic would be either (YY yy) 100% dominant phenotype or (Yy yy) 50% dominant phenotype and 50% recessive phenotype.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
21. When crossing individuals with differing genotypes, in which kind of cross would you expect to have your actual results match the predicted results 100% of the time the cross is made? A. monohybrid cross B. dihybrid cross C. test cross D. independent assortment cross E. crosses of differing genotypes result in differing answers—no cross will produce expected ratios every time Living systems seldom behave exactly as expected. While the ratios will be very close to what is predicted, that will not necessarily be exactly what is expected.
Bloom’s Level: 3. Apply Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
22. A woman who can roll her tongue (presumably dominant) is married to a man who cannot. Two of their four children can roll their tongues and two cannot. If A = roll tongue and a = cannot roll tongue, then what is the genotype of the parents? A. woman Aa; man Aa B. woman AA; man aa C. woman Aa; man AA D. woman Aa; man aa E. woman aa; man aa Since "cannot roll tongue" is recessive, the father must be aa. If some of the children can roll their tongue and some others cannot, then the mother must have had both alleles (heterozygous), Aa.
Bloom’s Level: 3. Apply Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
23. A pheasant breeder starts with two birds in the P generation, one of which is AA and the other is aa. If he takes two of the birds from the F1 generation and breeds them together, what can he expect in his F2 offspring? A. AA and Aa. B. Aa and aa. C. AA, Aa, and aa. D. AA only. E. Aa only. If the parents (P generation) are AA and aa, then all of the F1 offspring must be Aa because they will get one allele from each parent, and each parent only has one allele choice to give. When the F1 Aa individuals are crossed, the F2 generation will have the genotypes AA, Aa, and aa.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
24. Some plants fail to produce chlorophyll, due to a recessive trait. If we locate a pea plant that is heterozygous for this trait, self-pollinate it, and harvest the seeds, what are the likely phenotypes of the resulting offspring? A. All will be green with chlorophyll since that is the dominant trait. B. All will be white and lack chlorophyll since this is self-pollinated. C. About one-half will be green and one-half white since that is the distribution of the genes in the parents. D. About one-fourth will be white and three-fourths green since it is similar to a monohybrid cross. E. About one-fourth will be green and three-fourths white since it is similar to a monohybrid cross. Where g = no chlorophyll and G = chlorophyll, a heterozygote would have the genotype Gg. If the plant self-pollinates, then the genotype for ova production is Gg and the genotype for pollen production would be Gg. This monohybrid cross would yield the 1:2:1 ratio (25%, 50%, 25%), typical of such a cross.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
25. Since each child of two heterozygous parents has a 1/4 chance of receiving a recessive trait from each parent, A. if the first child is phenotypically recessive, then the next child must be phenotypically dominant. B. if the first child is phenotypically recessive, then the next child has a 3/4 chance of being phenotypically recessive. C. if the first child is phenotypically recessive, then the next child has a 1/2 chance of being phenotypically recessive. D. if the first child is phenotypically recessive, then the next child has a 1/4 chance of being phenotypically dominant. E. no matter what the first child's phenotype, the next child will have a 1/4 chance of being phenotypically recessive. The chance of any one child possessing a particular phenotype does not depend on the genotype of any sibling. The parents still have the same genotype and the eggs and sperm are created from germ cells, which contain both allele choices that the parents have. Thus, the probability is the same for each child to have any one particular genotype.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
26. Computer simulations are sometimes used to demonstrate the outcome of monohybrid fruit fly crosses, where a student can run generation after generation of fruit flies with 100 offspring produced each generation, half male and half female, and a 3-to-1 phenotype ratio (or 75 to 25) in the F1 generation. Compared to real genetics results, which of the following is NOT true? A. Rarely would exactly 100 fly offspring be produced or survive. B. An exact balance between males and females would be rare. C. A precise 3-to-1 ratio would be uncommon. D. It is usually possible to have exactly 100 offspring, half male and half female, and a 3-to-1 phenotypic outcome in real life. E. Ratios and outcomes will not be exact but will be very close to the expected numbers. In real life it is very uncommon for exact numbers or ratios to be generated in an experiment. With a computer simulation it is possible to obtain exactly the expected ratios or numbers since the computer model is designed by humans who program in these expected values.
Bloom’s Level: 3. Apply Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
27. If an albino (recessive) woman is married to a man with normal coloring and they have an albino child, what was the genotype of the man? A. homozygous dominant B. heterozygous C. homozygous recessive D. incomplete dominance E. sex-linked recessive Given that A = normal coloring and a = albino coloring, an albino recessive mother is aa and with normal coloring the father can either be AA or Aa. If the father is AA, then all of the children would have normal coloring. If the father is Aa, then there is a 50% probability of a child with normal coloring and a 50% probability of an albino child. Therefore, the father must be heterozygous for the trait.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
28. If an individual with a dominant phenotype is crossed with an individual with a recessive phenotype, 4 of their 9 offspring show the recessive phenotype. What is the genotype of the dominant parent? A. AA B. Aa C. aa D. either AA or Aa E. either Aa or aa If one of the parents has a recessive phenotype then the other parent must have a dominant allele in order to have a child with the dominant phenotype.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
29. In 1940, two researchers named Weiner and Landsteiner discovered that about 85 percent of the human population sampled possessed a blood cell protein called the Rh factor. The presence of Rh factor was labeled Rh positive (Rh+) and was found to be dominant over the absence of the blood factor (Rh-). Under normal Mendelian inheritance, which of the following statements is/are FALSE? A. Two Rh+ parents could have an Rh- child. B. Two Rh- parents could have an Rh+ child. C. An Rh- child would require that both parents be carriers of at least one Rh- allele. D. It is possible with just one pair of parents to have children where some siblings are Rhand some are Rh+. E. Two Rh+ parents can have Rh- or Rh+ children. Rh+Rh- would still show an Rh+ phenotype, so if both parents were heterozygous, they could have an Rh- child. Both parents would need to carry the Rh- allele. Since the parents also have the Rh+ allele, they could also have children of this phenotype. If both parents have the Rhphenotype, then their genotype must be Rh- Rh- and it would not be possible for them to have a child with the Rh+ phenotype.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
30. In cattle, consider that the lack of horns is dominant to the presence of horns. A bull without horns is crossed with a horned cow. Of four offspring, one is horned and the other three lack horns. The bull is A. homozygous dominant. B. homozygous recessive. C. heterozygous. D. sex-lined recessive. E. heterozygous recessive. Given that H = lack of horns and h = horns, if the mother is horned and the father is hornless and calves with both phenotypes are produced by the cross, then the female who has has horns must be (hh) and the bull must be Hh. Since the mother only has one allelic choice to give (h), then the father must have both alleles.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
31. In humans, assume red hair is recessive to dark hair. What are the chances of a darkhaired couple having a red-haired child if each had a red-haired parent? A. 0% B. 25% C. 50% D. 75% E. 100% Given that D = dark hair and d = red hair, if both of the parents, who have dark hair, each had a parent with red hair, which is recessive, then both must be heterozygous for hair color (Dd). This would be a monohybrid cross and the chance of a homozygous recessive from this match would be 25%.
Bloom’s Level: 3. Apply Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
32. Both parents are heterozygous for Tay-Sachs disease (an autosomal recessive disorder). Three children in a row were born with Tay-Sachs disease. What is the chance that a fourth child will have Tay-Sachs disease? A. 0% B. 25% C. 50% D. 75% E. 100% The chance of any one child possessing a particular phenotype does not depend on the genotype of any sibling. The parents still have the same genotype and the eggs and sperm are created from germ cell,s which contain both allele choices that the parents have. Therefore, the probability is the same for each child to have any one particular genotype. With two heterozygotes, this would mean a 75% chance of a normal phenotype and a 25% chance of a recessive phenotype.
Bloom’s Level: 4. Analyze Learning Outcome: 23.02.01 Recognize autosomal dominant and autosomal recessive patterns of inheritance when examining a pedigree. Section: 23.02 Topic: Pedigrees
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Chapter 23 - Patterns of Gene Inheritance
33. Unattached earlobes (EE or Ee) are dominant over attached earlobes (ee). A couple both have unattached earlobes. Both notice that one of their parents on both sides have attached earlobes (ee). Therefore, they correctly assume that they are carriers for attached earlobes (Ee). If the couple proceeds to have four children, then A. they can be certain that three will be heterozygous and one homozygous recessive. B. if the first three are heterozygous, the fourth must be homozygous recessive. C. the children must repeat the grandparents' genotype (Ee). D. all children must have unattached earlobes since both parents possess the dominant gene for it. E. two heterozygous, one homozygous dominant, and one homozygous recessive is a likely outcome, but all genotype ratios are possible. The chance of any one child possessing a particular phenotype does not depend on the genotype of any sibling. The parents still have the same genotype and the eggs and sperm are created from germ cells, which contain both allele choices that the parents have. Therefore, the probability is the same for each child to have any one particular genotype. In this case, the probabilities would be 75% chance of unattached earlobes and a 25% chance of attached earlobes.
Bloom’s Level: 3. Apply Learning Outcome: 23.02.01 Recognize autosomal dominant and autosomal recessive patterns of inheritance when examining a pedigree. Section: 23.02 Topic: Pedigrees
34. If the probability of event A is 3/4 and the probability of event B is 1/4, then the probability of both A and B occurring at the same time is A. 3/4. B. 1/4. C. 1 or absolute certainty. D. 1/2. E. 3/16. The multiplication rule applies to probability; that is, we multiply the probability of one event by the probability of the other event. In this case A B = 3/4 1/4 = 3/16.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
35. Which of the following is NOT correct concerning the law of independent assortment? A. It is based upon the process of meiosis. B. Each pair of factors separates independently. C. All possible combinations of factors can occur in the gametes. D. It was the second law of heredity established by Mendel. E. It follows the observation that all maternal chromosomes end up in the egg. Independent assortment occurs because homologous chromosomes are separated during meiosis I. This separated the alleles that are contained on these chromosomes. Because the probability of either allele ending up in the gamete is equal, then all possible combinations of alleles from the parents is equally likely. Mendel discovered this law through his experiments using monohybrid crosses. Only half of the chromosomes from the mother end up in the egg.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.01 Describe how Mendel’s law of segregation and law of independent assortment are related to the movements of chromosomes during gamete formation and fertilization. Section: 23.01 Topic: Mendelian Genetics
36. A cross that involves two traits is called a A. monohybrid B. dihybrid C. Punnett D. test E. homozygous
cross.
A monohybrid cross follows one trait, while a dihybrid cross follows two traits that assort independently of one another. A Punnett square may be used to determine the outcome of either of these types of crosses.
Bloom’s Level: 1. Remember Learning Outcome: 23.01.01 Describe how Mendel’s law of segregation and law of independent assortment are related to the movements of chromosomes during gamete formation and fertilization. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
37. If pairs of factors separate independently of other pairs of factors, you are dealing with A. the law of segregation. B. the law of independent assortment. C. the law of dominance. D. multiple alleles. E. codominance. The law of independent assortment addresses the fact that unlinked genes will be inherited independent of how the other is inherited. The law of segregation addresses the fact that two alleles for the same gene are separated and only one is inherited from each parent. The law of dominance says that given two alleles for the same trait, one can be dominant over the other; if both are present in the organism, then only one will be shown and it will always be the same one. With codominance, both of the alleles are expressed equally.
Bloom’s Level: 1. Remember Learning Outcome: 23.01.01 Describe how Mendel’s law of segregation and law of independent assortment are related to the movements of chromosomes during gamete formation and fertilization. Section: 23.01 Topic: Mendelian Genetics
38. If the genotype of an organism is YySsTt, then yST would represent A. the genotype of the offspring. B. a possible phenotype of the offspring. C. a gamete of the parent. D. a possible zygote. E. the P generation. Given three genes, Y, S, and T, then YySsTt could represent the genotype of an individual who is heterozygous for all three traits and yST could represent a possible combination of the alleles that could be produced in a gamete. If the genotype of either the offspring or a parent were to be listed, then there would need to be two alleles for each gene since a zygote or an organism would have the full complement of chromosomes.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.03 Contrast genotype and phenotype, and use appropriate terms for describing genotypes. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
39. Which of the following crosses will yield only homozygous recessive offspring? A. RRYY RRYY B. RRYY rryy C. RrYy rryy D. rryy rryy E. RrYy RrYy To produce only homozygous recessive offspring, only homozygous recessive parents could be mated. If any dominant alleles were present in the parents, there would be a probability that they would be present in the offspring.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
40. Which of the following would be considered a heterozygous dihybrid genotype? A. Ww B. sS C. WWSs D. WWss E. WwSs A dihybrid would mean that there were two different genes involved. WwSs would be heterozygous for both traits. WWss or WWSs would not be heterozygous and sS or Ww would not show the information for two traits.
Bloom’s Level: 1. Remember Learning Outcome: 23.01.03 Contrast genotype and phenotype, and use appropriate terms for describing genotypes. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
41. Which of the following crosses will yield four phenotypes in a ratio of 9:3:3:1? A. RRYY RRYY B. RRYY rryy C. RrYy RrYy D. RrYy rryy E. rryy rryy For a dihybrid cross, the parents would need to be heterozygous for both of two traits. RrYy RrYy would produce RRYY, RRYy, RRyy, RrYY, Rryy, rrYY, rrYy, or rryy as genotypes. The phenotypes of these would be 9 dominant for both traits to 3 recessive for the first trait and dominant for the second trait to 3 dominant for the first trait and recessive for the second and 1 recessive for both traits.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
42. In which kind of cross would you expect to find a ratio of 9:3:3:1 among the F2 offspring? A. monohybrid cross B. dihybrid cross C. test cross D. homozygous recessive homozygous dominant E. homozygous dominant heterozygous For a dihybrid cross, the parents would need to be heterozygous for both of two traits. Example: RrYy RrYy would produce RRYY, RRYy, RRyy, RrYY, Rryy, rrYY, rrYy, or rryy as genotypes. The phenotypes of these would be 9 dominant for both traits to 3 recessive for the first trait and dominant for the second trait to 3 dominant for the first trait and recessive for the second and 1 recessive for both traits.
Bloom’s Level: 1. Remember Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
43. Which of the following crosses will yield four phenotypes in the 1:1:1:1 ratio? A. RRYY RRYY B. RRYY rryy C. RrYy RrYy D. RrYy rryy E. rryy rryy A cross between an individual that is heterozygous for two traits and one that is homozygous recessive for those two traits would yield a ratio of 1 dominant for both traits to 1 dominant for the first trait and recessive for the second to 1 recessive for the first trait and dominant for the second to 1 recessive for both traits.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
44. Two parents with widow's peak and short fingers have a child with a continuous hairline and long fingers. What are the genotypes of the two parents? (W = widow's peak; S = short fingers) A. WWSS wwss B. wwss WWSS C. WWss WWss D. WwSs WwSs E. WwSs wwss If the child's phenotype is different for both dominant traits from that of the parents, then the parents must be heterozygous for both traits. This would mean that they possess the recessive alleles to pass on to their offspring.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
45. A man with a widow's peak (dominant) and can curl his tongue (dominant) has a child who has a continuous hairline and cannot curl the tongue. What is the genotype of the father? A. WWTT B. wwtt C. WwTt D. Wwtt E. WT Given that W = widow's peak and w = continuous hair line and T = roll tongue while t = cannot roll tongue, then if the child's phenotype is different for both dominant traits from that of the parent then the parent must be heterozygous for both traits. This would mean that he possess the recessive alleles to pass on to his offspring. His genotype would be WwTt.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
46. A blue-eyed, left-handed woman marries a brown-eyed, right-handed man who is heterozygous for both of his traits. If blue eyes and left-handedness are recessive, how many different phenotypes are possible in their children? A. one B. two C. three D. four E. five If B = brown eyes and b = blue eyes and R = right handed while r = left handed, then the mother's genotype is bbrr and the father is BbRr. A cross between an individual that is homozygous recessive for two traits and one that is heterozygous for those two traits then there would be a ratio of 1 dominant for both traits to 1 dominant for the first trait and recessive for the second to 1 recessive for the first trait and dominant for the second to 1 recessive for both traits, yielding four different phenotypes.
Bloom’s Level: 5. Evaluate Learning Outcome: 23.02.01 Recognize autosomal dominant and autosomal recessive patterns of inheritance when examining a pedigree. Section: 23.02 Topic: Pedigrees
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Chapter 23 - Patterns of Gene Inheritance
47. In guinea pigs, B = black, b = brown, S = short hair, s = long hair. Two heterozygous individuals reproduce. The expected results are A. 9 black long hair, 3 black short hair, 3 brown long hair, 1 brown short hair. B. 9 black short hair, 6 black long hair, 3 brown long hair, 1 black short hair. C. 9 black short hair, 3 black long hair, 3 brown short hair, 1 brown long hair. D. 9 brown short hair, 3 black long hair, 3 brown long hair, 1 black short hair. E. 1 black short hair, 1 black long hair, 1 brown short hair, 1 brown long hair. For a dihybrid cross, the parents would need to be heterozygous for both of two traits. BbSs BbSs would produce BBSS, BBSs, BBss, BbSS, Bbss, bbSS, bbSs, or bbss as genotypes. The phenotypes of these would be 9 dominant for both traits (black and short hair) to 3 recessive for the first trait and dominant for the second trait (brown with short hair) to 3 dominant for the first trait and recessive for the second (black with long hair) and 1 recessive for both traits (brown with long hair).
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
48. In guinea pigs, B = black, b = brown, S = short hair, s = long hair. A heterozygous black, short-haired animal reproduces with a brown, long-haired animal. What is the expected phenotypic ratio of the offspring? A. 1 black short hair, 1 black long hair, 1 brown short hair, 1 brown long hair B. 3 black short hair, 1 brown long hair, 3 black long hair, 1 brown short hair C. 9 black short hair, 3 black long hair, 3 brown long hair, 1 brown short hair D. 9 black short hair, 3 black long hair, 3 brown short hair, 1 brown long hair E. 9 black short hair, 6 black long hair, 3 brown long hair, 1 brown short hair The cross of BbSs bbss would yield BbSs, bbSs, Bbss, and bbss as possible genotypes. The ratio of phenotypes would be 1 black short hair to 1 black long hair to 1 brown short hair to 1 brown long hair.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
49. In pea plants, the gene for round seed (R) is dominant, and wrinkled seeds (r) are recessive. The endosperm of the pea is also either starchy, a dominant gene (S), or waxy (s). What CANNOT be said of a cross between two parents that are heterozygous for both traits? A. It is possible to secure offspring that are homozygous for both dominant genes. B. It is possible to secure offspring that are homozygous for both recessive genes. C. It is possible to secure offspring that are homozygous for one dominant gene, such as round seed, and homozygous recessive for the other recessive waxy gene. D. It is possible to secure offspring that are heterozygous for both genes. E. It is impossible to secure offspring that are homozygous for both dominant genes. A dihybrid cross would be RrSs RrSs, and this would produce RRSS, RRSs, RRss, RrSS, Rrss, rrSS, rrSs, or rrss as genotypes. The phenotypes of these would be 9 dominant for both traits ( round and starchy seeds) to 3 recessive for the first trait and dominant for the second trait (wrinkled and starchy seeds) to 3 dominant for the first trait and recessive for the second (round and waxy seeds) and 1 recessive for both traits (wrinkled and waxy seeds).
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
50. A cross is made between two parents with genotypes AaBB and aabb. If there are 32 offspring, how many of them would be expected to exhibit both dominant characteristics? A. 32 B. 24 C. 16 D. 8 E. 0 A cross between AaBB aabb would produce AaBb and aaBb as possible genotypes. This means that there is a 50% chance that the offspring would be dominant for both traits. If 32 offspring are born, then half of those, 16, would be expected to have this phenotype.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
51. In which kind of cross could you expect to find ratios of 1:1:1:1 among the offspring? A. monohybrid cross B. dihybrid cross C. one-trait test cross D. two-trait test cross E. no cross will yield offspring in 1:1:1:1 ratios The 1:1:1:1 ratio is expected when an organism heterozygous for two traits is crossed with an organism that is homozygous recessive for both traits. This is also a test cross that is frequently used to determine the genotype of organisms that show the dominant traits.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
52. In horses, black coat (B) is dominant over brown coat (b), and being a trotter (T) is dominant over being a pacer (t). A black horse who is a pacer is crossed with a brown horse who is a trotter. The offspring is a brown pacer. What is the genotype of the brown trotter parent? A. BBTT B. BbTt C. Bbtt D. bbTt E. bbtt The cross given is B tt bbT with the blanks showing that the genotype for the second allele is not known because the dominant phenotype is seen. To produce a brown pacer (bbtt), the parents would need to have the recessive allele for both traits, so their genotypes would be Bbtt and bbTt.
Bloom’s Level: 3. Apply Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
53. In humans, aniridia, a type of blindness, is due to a dominant allele A. Migraine headaches are due to another dominant allele M. If a man who suffers from both conditions (AaMm) marries a woman who suffers from both (AaMm), what are the chances of an offspring expressing both traits. A. 9/16 B. 3/16 C. 1/2 D. 1/16 E. 4/16 The man is AaMm and he marries a woman who is AaMm. This is a dihybrid cross and would produce AAMM, AAMm, AAmm, AaMM, Aamm, aaMM, aaMm, or aamm as genotypes. The phenotypes of these would be 9 dominant for both traits to 3 recessive for the first trait and dominant for the second trait to 3 dominant for the first trait and recessive for the second and 1 recessive for both traits.
Bloom’s Level: 4. Analyze Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
54. The four o'clock flower is an example of incomplete dominance: R = red, r = white, and Rr = pink. If two hybrids are crossed, what are the chances that an offspring will have pink flowers? A. 0% B. 25% C. 50% D. 75% E. 100% Rr Rr will produce RR, Rr, and rr flowers. Since the parents have both alleles, they have the ability to pass both on to their offspring. This is the cross that Mendel made when he crossed the F2 with itself. We can calculate that the offspring's genotypes will be 25% RR, 50% Rr, and 25% rr.
Bloom’s Level: 2. Understand Learning Outcome: 23.03.01 Explain how the phenotypes for incompletely dominant and codominant traits differ from dominant/recessive trait phenotypes. Section: 23.03 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
55. What are the chances that two individuals with wavy hair (an incomplete trait) will have a curly-haired child? Curly hair and straight hair exhibit incomplete dominance. A. none B. 25% C. 50% D. 75% E. 100% Given that WW = curly hair, ww = straight hair, and Ww = wavy hair, if both parents have wavy hair then their genotype is Ww. A cross of Ww Ww would yield a 25% chance for WW (curly,) a 50% chance for Ww (wavy), and a 25% chance for ww (straight), The ratio would be 1:2:1.
Bloom’s Level: 2. Understand Learning Outcome: 23.03.01 Explain how the phenotypes for incompletely dominant and codominant traits differ from dominant/recessive trait phenotypes. Section: 23.03 Topic: Non-Mendelian Inheritance
56. Which of the following is an example of the blending of phenotypes? A. codominance B. sex-linked inheritance C. polygenic inheritance D. simple Mendelian inheritance E. incomplete dominance Blending is when the offspring of homozygous parents do not show the phenotype of either of the parents, but rather a unique third phenotype.
Bloom’s Level: 1. Remember Learning Outcome: 23.03.01 Explain how the phenotypes for incompletely dominant and codominant traits differ from dominant/recessive trait phenotypes. Section: 23.03 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
57. Mr. X has been accused by Miss Z of fathering her child. The child's blood type is AB and Miss Z's blood type is A. Mr. X could NOT be the father if his blood type is A. only A. B. only B. C. AB. D. only O. E. either A or O. The child's blood type is AB and Miss Z's blood type is A. Miss Z could be either AA or AO. If the child is AB, then she must have passed the information for blood type A to her child. The father must have a blood type that has the information for B-type blood to pass on the the child. The father could be AB, BB, or BO and be the father of the child; he could not be AA, AO, or OO and be the father of the child.
Bloom’s Level: 3. Apply Learning Outcome: 23.03.02 Explain how ABO blood types are an example of multiple allele inheritance. Section: 23.03 Topic: Non-Mendelian Inheritance
58. The inheritance of blood types in humans can be explained by A. simple Mendelian genetics (one gene pair). B. linked genes. C. simple dominance. D. multiple alleles. E. polygenic inheritance. Blood types in humans involve multiple alleles that are codominant. Simple Mendelian genetics would indicate a simple dominant/recessive relationship and polygenetic inheritance would mean that there were multiple genes on different chromosomes. Linked genes are passed as a group, whereas blood types are a single gene with multiple alleles.
Bloom’s Level: 1. Remember Learning Outcome: 23.03.02 Explain how ABO blood types are an example of multiple allele inheritance. Section: 23.03 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
59. If one parent has type B blood and the other has type AB, what type might the child have? A. A or O only B. B or O only C. A or B only D. A, B, AB only E. A, B, O, or AB The parent with the AB blood type has the genotype AB. The parent with the blood type B could have the genotype of BB or BO. If the genotype of the second parent is BB, then the possibilities for offspring are AB or BB. If the genotype of the second parent is BO, then the possibilities for offspring are AB or A. The only phenotype that can be definitely eliminated is O.
Bloom’s Level: 4. Analyze Learning Outcome: 23.03.02 Explain how ABO blood types are an example of multiple allele inheritance. Section: 23.03 Topic: Non-Mendelian Inheritance
60. A man with blood type AB could NOT be the father of a child with A. blood type A. B. blood type B. C. blood type AB. D. blood type O. E. a child of any blood type. If the father has blood type AB he could father a child with blood types AB, A, or B; the only blood type that could be eliminated would be O.
Bloom’s Level: 3. Apply Learning Outcome: 23.03.02 Explain how ABO blood types are an example of multiple allele inheritance. Section: 23.03 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
61. Traits that are controlled by several sets or pairs of alleles, such as skin color and height in humans, are the result of what form of inheritance? A. polygenic B. incomplete dominance C. multiple allele systems D. simple Mendelian inheritance E. codominance Polygenic inheritance would mean that there were multiple genes at different locations and possibly on different chromosomes all controlling one trait. Simple Mendelian genetics would indicate a simple dominant/recessive relationship, whereas codominance would be a system where alleles both show in the offspring and incomplete domanance would be a case where heterozygotes would show a different phenotype than either of the homozygotes.
Bloom’s Level: 1. Remember Learning Outcome: 23.03.03 Describe polygenic inheritance. Section: 23.03 Topic: Non-Mendelian Inheritance
62. Consider skin color is regulated by three genes: A, B, and C. All recessive alleles would be 0 or white on the color scale. All dominant alleles would be 6 or black on the color scale. What color is a person with AAbbCc genotype? A. 6 or black B. 5 or dark C. 3 or medium brown D. 2 or light E. 0 or white Since the darkness of the skin depends on how many dominant alleles are inherited, someone with AABBCC would be the darkest skin possible and aabbcc would be the lightest skin possible. AAbbCc would be a color in the middle of the range of colors or a brown color.
Bloom’s Level: 5. Evaluate Learning Outcome: 23.03.03 Describe polygenic inheritance. Section: 23.03 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
63. When two or more sets of alleles affect the same trait in an additive fashion, it is termed A. a single-trait cross. B. a double-trait cross. C. codominant. D. incomplete dominance. E. polygenic inheritance. Polygenic inheritance involves more than one gene and, therefore. more than one allele pair. These genes are located in different places on the genome and possibly on different chromosomes.
Bloom’s Level: 1. Remember Learning Outcome: 23.03.03 Describe polygenic inheritance. Section: 23.03 Topic: Non-Mendelian Inheritance
64. Cold weather can change the A. genotype B. phenotype C. alleles D. sex E. rate of genetic mutations
of a Himalayan rabbit.
Environmental factors can affect the way that the genotype is expressed; it does not change the genotype itself. This means that the alleles are still the same ones that were inherited but that the phenotype is altered. The sex of the organism is also an inherited character and it is not affected by environmental factors. The rate of genetic mutations is not controlled by temperature.
Bloom’s Level: 1. Remember Learning Outcome: 23.04.01 Describe ways in which the environment can influence genetic traits. Section: 23.04 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
65. Humans can have any of several eye colors, This is because we possess a variety of A. particulates. B. genotypes. C. loci. D. alleles. E. homologous. Alleles are defined as alternate forms of a gene; they are located on homologous chromosomes at the same loci on those two chromosomes.
Bloom’s Level: 2. Understand Learning Outcome: 23.01.01 Describe how Mendel’s law of segregation and law of independent assortment are related to the movements of chromosomes during gamete formation and fertilization. Section: 23.01 Topic: Mendelian Genetics
66. Huntington disease is an autosomal dominant disorder that leads to progressive degeneration of brain cells. There is no effective treatment for the disorder and individuals usually die in their 40s or 50s, 10 to 15 years after the onset of symptoms. Which of these statements about Huntington disease is true? A. As an autosomal dominant disease, Huntington disease is expressed only if both parents carry the trait. B. Huntington disease usually skips a generation. C. Huntington disease can only be inherited when a parent is showing the trait. D. Huntington disease usually appears during middle age. E. Huntington disease is caused by a defect in the elastic connective tissue protein. Huntington disease is inherited if one dominant allele is inherited but the disease frequently does not appear until the individual has reached middle age and already has children. There is a test to detect the disorder, but since there is no cure most individuals choose not to be tested.
Bloom’s Level: 3. Apply Learning Outcome: 23.02.02 Analyze pedigrees to determine the probability of a genetic disorder. Section: 23.02 Topic: Pedigrees
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Chapter 23 - Patterns of Gene Inheritance
67. You have two true-breeding rose bushes, one with red flowers and one with white flowers. A cross between these two roses yields a bush with pink flowers. What condition does this demonstrate? A. codominance B. incomplete dominance C. environmental effects D. polygenetic inheritance E. monohybrid inheritance Incomplete dominance occurs when the heterozygote does not show the phenotype of either parent. The combination of red and white flowers leading to pink flowers would be an example of this. Incomplete dominance is when the organism can express three different variations for a particular trait. Environmental effects is when the environment influences traits. Polygenic inheritance is when two or more genes regulate a trait. Monohybrid inheritance is when a trait has a dominant and recessive variation.
Bloom’s Level: 2. Understand Learning Outcome: 23.03.01 Explain how the phenotypes for incompletely dominant and codominant traits differ from dominant/recessive trait phenotypes. Section: 23.03 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
68. You have two true-breeding rose bushes, one with red flowers and one with white flowers. A cross between these two roses yields a bush with white flowers that have red splotches. What condition does this demonstrate? A. codominance B. incomplete dominance C. environmental effects D. polygenetic inheritance E. monohybrid inheritance Codominance occurs when the heterozygote shows both of the parental phenotypes equally. Incomplete dominance is when the organism can express three different variations for a particular trait. Environmental effects is when the environment influences traits. Polygenic inheritance is when two or more genes regulate a trait. Monohybrid inheritance is when a trait has a dominant and recessive variation.
Bloom’s Level: 2. Understand Learning Outcome: 23.03.01 Explain how the phenotypes for incompletely dominant and codominant traits differ from dominant/recessive trait phenotypes. Section: 23.03 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
69. Hydrangeas are a flowering plant with large showy blooms. When a plant is grown in aluminum-rich soil it has blue flowers. If the same plant is transplanted into soil that is lacking aluminum, the flowers produced will be pink. This is an example of A. codominance. B. incomplete dominance. C. environmental effects. D. polygenetic inheritance. E. monohybrid inheritance. Codominance occurs when the heterozygote shows both of the parental phenotypes equally. Incomplete dominance is when the organism can express three different variations for a particular trait. Since the phenotype of the flowers depends on an external condition, this is an example of environmental effects on gene expression. Polygenic inheritance is when two or more genes regulate a trait. Monohybrid inheritance is when a trait has a dominant and recessive variation.
Bloom’s Level: 2. Understand Learning Outcome: 23.04.01 Describe ways in which the environment can influence genetic traits. Section: 23.04 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
70. The leaves of a maple tree vary in size depending on the location of the leaves on the tree. Those growing at the outer region of the tree tend to be smaller than those growing closer to the trunk of the tree. This allows leaves that are growing in the shade to have a larger surface area. This show that genes have the ability to A. respond to environmental factors. B. show two different genetic variations. C. be turned off if they are not needed. D. choose whether or not they are expressed. E. mutated by exposure to sunlight. Phenotypic variations allow organisms to respond to conditions in the environment. Leaves that receive less sunlight require a larger surface area in order to carry out photosynthesis efficiently. Those that are exposed to full sunlight do not need that same increased surface area. The plant is not showing two different genetic variations of leaves. The genes are not being turned on or off due to their position on the tree and they are not choosing to be expressed or not. Sunlight is not causing them to mutate.
Bloom’s Level: 2. Understand Learning Outcome: 23.04.01 Describe ways in which the environment can influence genetic traits. Section: 23.04 Topic: Non-Mendelian Inheritance
71. To try and discover if the environment has an effect on a particular characteristic, scientists often examine A. siblings that share only one parent and were separated at birth. B. fraternal human twins that were separated at birth. C. identical human twins that were separated at birth. D. siblings that share both parents and were separated at birth. E. unrelated children who have been adopted by the same family and grown up together. Identical twins have the same genetic makeup, so studying two individuals with the same genotype that have been exposed to two different sets of environmental conditions would allow any differences in phenotype to be analyzed.
Bloom’s Level: 3. Apply Learning Outcome: 23.04.02 Understand how scientists determine the effect of nature versus nurture. Section: 23.04 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
72. All but one of the following experiments would allow a scientist to discover if the phenotype of an organism would be affected by environmental changes. Choose the experiment that would NOT show environmental effects. A. Cuttings are taken from an English Ivy plant and are then grown in soils with different pH levels. B. A Siamese cat has patches of fur removed from its hips and an ice pack is secured over the area. C. A mouse with the genetic disorder phenylketonuria (PKU) is fed a diet low in phenylalanine for a year and then changed to a diet rich in the same molecule. D. A group of mice that show the same dominant trait are fed a variety of identical diets to determine if they show different body types. E. A hydrangea is grown for one year in aluminum-poor soil and then grown for one year in aluminum-rich soil. It is necessary to determine that the organisms involved have the same genotype, so exposure to different test conditions in the experiment is the one variable that is different. Cuttings from one plant or altering the conditions for a single organism so it experiences two different conditions would meet these conditions. Even if the mice all show the same trait since it is dominant, the genotype is unknown.
Bloom’s Level: 5. Evaluate Learning Outcome: 23.04.02 Understand how scientists determine the effect of nature versus nurture. Section: 23.04 Topic: Non-Mendelian Inheritance
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Chapter 23 - Patterns of Gene Inheritance
73. The skin pigment melanin is expressed differently according to exposure to sunlight, with a greater production of the molecule with greater exposure. However, the degree of production can never exceed the inherited genetic level. What evolutionary advantage would be gained from having this phenotypic range for skin pigmentation? A. Skin color is an example of polygenic inheritance. B. It allows someone to get a tan when they want to. C. Production of the molecule would only occur when necessary. D. Changes in skin color allow humans to look healthier. E. Skin color is an example of a codominant gene. The production of molecules requires large amounts of cellular energy; therefore, only producing the molecule when it is necessary because of exposure to increased sunlight would allow for a more efficient use of cellular energy.
Bloom’s Level: 3. Apply Learning Outcome: 23.04.02 Understand how scientists determine the effect of nature versus nurture. Section: 23.04 Topic: Non-Mendelian Inheritance
74. Which allele combination represents a recessive monohybrid trait? A. aa B. Aa C. AA D. ab E. Ab aa represents a recessive monohybrid trait. Aa and AA represent the dominant variation. ab and Ab do not represent a trait. Both alleles need to be represented by the same letter.
Bloom’s Level: 1. Remember Learning Outcome: 23.01.02 Define the term allele, and explain what it means for an allele to be dominant or recessive. Section: 23.01 Topic: Mendelian Genetics
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Chapter 23 - Patterns of Gene Inheritance
75. How many alleles are required to express a monohybrid trait? A. 2 B. 1 C. 3 D. 5 E. 4 Monohybrid traits are made up of two alleles.
Bloom’s Level: 1. Remember Learning Outcome: 23.01.02 Define the term allele, and explain what it means for an allele to be dominant or recessive. Section: 23.01 Topic: Mendelian Genetics
76. Which genetic disorder is associated with a defect in the elastic connective tissue protein called fibrillin? A. Marfan syndrome B. Huntington disease C. sickle cell disease D. phenylketonuria (PKU) E. cystic fibrosis (CF) Marfan syndrome is associated with a defect in the elastic connective tissue protein called fibrillin. Huntington disease is caused by a mutation to the protein huntingtin. Sickle cell disease is an irregular shape of the red blood cells. Phenylketonuria (PKU) is associated with the lack of the enzyme necessary for the metabolism of phenylalanine. Cystic fibrosis (CF) is associated with a faulty protein channel in the cell membrane.
Bloom’s Level: 1. Remember Learning Outcome: 23.02.02 Analyze pedigrees to determine the probability of a genetic disorder. Section: 23.02 Topic: Pedigrees
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Chapter 23 - Patterns of Gene Inheritance
77. Which genetic disorder is associated with an irregular shape of the red blood cells? A. sickle cell disease B. Marfan syndrome C. Huntington disease D. phenylketonuria (PKU) E. cystic fibrosis (CF) Sickle cell disease is an irregular shape of the red blood cells. Marfan syndrome is associated with a defect in the elastic connective tissue protein called fibrillin. Huntington disease is caused by a mutation to the protein huntingtin. Phenylketonuria (PKU) is associated with the lack of the enzyme necessary for the metabolism of phenylalanine. Cystic fibrosis (CF) is associated with a faulty protein channel in the cell membrane.
Bloom’s Level: 1. Remember Learning Outcome: 23.02.02 Analyze pedigrees to determine the probability of a genetic disorder. Section: 23.02 Topic: Pedigrees
78. What genetic disorder is associated with the lack of an enzyme necessary for the normal metabolism of the amino acid phenylalanine? A. phenylketonuria (PKU) B. Marfan syndrome C. Huntington disease D. sickle cell disease E. cystic fibrosis (CF) Phenylketonuria (PKU) is associated with the lack of the enzyme necessary for the metabolism of phenylalanine. Marfan syndrome is associated with a defect in the elastic connective tissue protein called fibrillin. Huntington disease is caused by a mutation to the protein huntingtin. Sickle cell disease is an irregular shape of the red blood cells. Cystic fibrosis (CF) is associated with a faulty protein channel in the cell membrane.
Bloom’s Level: 1. Remember Learning Outcome: 23.02.02 Analyze pedigrees to determine the probability of a genetic disorder. Section: 23.02 Topic: Pedigrees
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Chapter 23 - Patterns of Gene Inheritance
79. Which of the following is an autosomal dominant disorder? A. Marfan syndrome B. Tay-Sachs C. sickle cell disease D. cystic fibrosis (CF) E. phenylketonuria (PKU) Marfan syndrome is an autosomal dominant disorder. The rest of the disorders are autosomal recessive.
Bloom’s Level: 1. Remember Learning Outcome: 23.02.01 Recognize autosomal dominant and autosomal recessive patterns of inheritance when examining a pedigree. Section: 23.02 Topic: Pedigrees
Short Answer Questions 80. Indicate the genotypic and phenotypic ratios that you could expect if an individual who is heterozygous for Huntington disease has children with a normal individual. The individual who is heterozygous for Huntington disease would be Hh, since it is a dominant disorder. The normal individual would be hh. Hh hh would produce a genotypic ratio of 50% Hh and 50 hh. The phenotypic ratio would be 50% Huntington disease and 50% normal.
Bloom's Level: 6. Create Learning Outcome: 23.02.02 Analyze pedigrees to determine the probability of a genetic disorder. Section: 23.02 Topic: Pedigrees
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Chapter 23 - Patterns of Gene Inheritance
81. In the human population, short fingers are dominant over long fingers. Indicate the genotypic and phenotypic ratios if a heterozygous short-fingered female reproduces with a long-fingered male. The female's genotype would be Ss and the male's genotype would be ss. This would produce a genotypic ratio of 50% Ss and 50% ss. The phenotypic ratio would be 50% short and 50% long fingers.
Bloom's Level: 6. Create Learning Outcome: 23.01.04 Predict outcome ratios and probabilities for problems based on Mendelian patterns of inheritance. Section: 23.01 Topic: Mendelian Genetics
82. Indicate the phenotypic and genotypic ratios expected for the offspring from a straighthaired man and a wavy-haired woman. Straight and curly hair are incompletely dominant variations. The man's genotype would be H1H1 and the woman's genotype would be H1H2. This would produce a phenotypic ratio of 50% straight hair and 50% wavy hair. The genotypic ratio would be 50% H1H1 and 50% H1H2.
Bloom's Level: 6. Create Learning Outcome: 23.03.01 Explain how the phenotypes for incompletely dominant and codominant traits differ from dominant/recessive trait phenotypes. Section: 23.03 Topic: Non-Mendelian Inheritance
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
Chapter 24 Chromosomal Inheritance and Genetic Disorders
Multiple Choice Questions 1. Which of the following would not be the same in a male and female? A. the total number of autosomes B. the loci for the majority of their genes C. the types of sex chromosomes D. the inheritance of one X chromosome from their mother E. the need for two sex chromosomes Both males and females have the same number of autosomal chromosomes with the same genes found in the same loci on the same chromosomes for all of the traits located on these chromosomes. Both males and females also inherit one X chromosome from their mother. The mother only has an X to give and the sex chromosome from the father determines the gender of the child. Males have one X and one Y chromosome, while females have two X chromosomes.
Bloom’s Level: 1. Remember Learning Outcome: 24.02.01 Explain X-linked genetic inheritance. Section: 24.02 Topic: Sex-Linked Inheritance
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
2. All of the genes on a single chromosome are A. a locus. B. homologous. C. a linkage map. D. a linkage group. E. an allele. The genes on a single chromosome are often inherited as a group, with those closer together more likely to be passed as a unit or a linkage group.
Bloom’s Level: 1. Remember Learning Outcome: 24.01.01 Describe what is meant by the term linkage group. Section: 24.01 Topic: Non-Mendelian Inheritance
3. The location of a gene on a chromosome is called A. a locus. B. homologous. C. a linkage map. D. a linkage group. E. an allele. A gene is found at a particular location (locus) on a chromosome. The two alleles for that gene are found at the same loci on homologous chromosomes.
Bloom’s Level: 1. Remember Learning Outcome: 24.01.01 Describe what is meant by the term linkage group. Section: 24.01 Topic: Mendelian Genetics
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
4. If there is complete linkage with genes Ss (for smooth or wrinkled seeds) and Gg (for green or yellow seeds), what phenotypes would you expect in the offspring of a cross with one heterozygous parent (Ss, Gg), which would all produce all smooth and green seeds, and a homozygous recessive parent (ss, gg), which would produce all wrinkled yellow seeds? A. smooth green seeds or wrinkled yellow seeds only B. smooth green seeds only C. smooth green seeds, wrinkled yellow seeds, smooth yellow seeds, and wrinkled green seeds D. wrinkled yellow seeds only Since the traits are linked, the SG or sg will pass from the one parent and combine with the sg from the other parent. This would result in SsGs (smooth green) or ssgg (wrinkled yellow) offspring.
Bloom’s Level: 4. Analyze Learning Outcome: 24.01.02 Predict which alleles in a linkage group will cross over more frequently. Section: 24.01 Topic: Mendelian Genetics
5. What is the relationship between linked genes and independent assortment? A. Linked genes show independent assortment. B. Linked genes do not show independent assortment. C. Whether or not genes are linked does not influence whether or not they show independent assortment. D. There is no relationship between linked genes and independent assortment. E. One of the linked genes will be inherited from the mother and the other one from the father. Linked genes do not completely show independent assortment. They only assort independently if crossing over occurs and this is based on the distance between the linked genes.
Bloom’s Level: 2. Understand Learning Outcome: 24.01.01 Describe what is meant by the term linkage group. Section: 24.01 Topic: Mendelian Genetics
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
6. What determines whether or not genes located on the same chromosome can assort independently? A. Genes located on the same chromosome can never assort independently. B. Genes located on the same chromosome always assort independently. C. Whether or not independent assortment occurs is due to random chance. D. The distance between genes located on the same chromosome determines whether or not they can show any level of independent assortment. E. The phenotype decides what that the gene will express. The farther apart the genes are the more likely that crossing over will occur and the more they behave as if they were not located on the same chromosome.
Bloom’s Level: 2. Understand Learning Outcome: 24.01.02 Predict which alleles in a linkage group will cross over more frequently. Section: 24.01 Topic: Mendelian Genetics
7. Color-blindness is inherited as an X-linked recessive trait. A male who is color-blind marries a heterozygous woman. What percent of their total children will be color-blind? A. 0% B. 25% C. 50% D. 75% E. 100% Given that XB= normal color vision and Xb = color blind (recessive), a heterozygous female would be XBXb and a color-blind man would be XbY. Daughters would have a 50% chance of inheriting normal color vision, but they will be heterozygous for the trait, the other 50% will be color-blind. Sons have the same 50/50 chance of normal versus color blind vision. Notice that this is based on the probability of the X chromosome that they will inherit from their mother since they will not receive an X from their father. For daughters it is still the mother's X chromosome that makes the difference because the X chromosome they would inherit from their father must carry the color-blind allele.
Bloom’s Level: 4. Analyze Learning Outcome: 24.02.02 Solve crosses involving sex-linked traits. Section: 24.02 Topic: Sex-Linked Inheritance
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
8. A normal male marries a color-blind woman. What percent of their female children will be color-blind? A. 0% B. 25% C. 50% D. 75% E. 100% Given that XB = normal color vision and Xb = color blind (recessive), a color-blind female would be XbXb and a normal man would be XBY. Daughters would have normal color vision, but they will be heterozygous for the trait. Sons have a 100% probability of color-blind vision. Notice that this is based on the probability of the X chromosome that they will inherit from their mother since they will not receive an X from their father. For daughters it is the father's X chromosome that makes the difference because the X chromosome they would inherit from their mother must carry the color-blind allele and the one from their father must carry the normal color vision allele.
Bloom’s Level: 4. Analyze Learning Outcome: 24.02.02 Solve crosses involving sex-linked traits. Section: 24.02 Topic: Sex-Linked Inheritance
9. When homologous chromosomes fail to separate during meiosis, this is termed A. linked genes. B. disjunction. C. nondisjunction. D. crossover. E. monosomy. Linked genes are ones located on the same chromosome sometimes. Disjunction refers to the separation of homologous chromosomes during meiosis I or sister chromatids during meiosis II. This is the normal process. If it fails to occur, then it is termed nondisjunction and extra genetic information is present. Crossing over is the equal exchange of genetic information between two nonsister chromotids. It results in a reshuffling of parental genotypes. Monosomy refers to the loss of an entire chromosome through nondisjunction during meiosis.
Bloom’s Level: 1. Remember Learning Outcome: 24.03.01 Describe how chromosome number disorders arise. Section: 24.03 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
10. Which refers to the loss of a complete chromosome? A. inversion B. translocation C. deletion D. duplication E. monosomy An inversion refers to the changing of direction of a segment of DNA while it stays in the same location on the same chromosome. Translocations occur when a segment of DNA moves from one location in the genome to another. Deletion is loss of a portion of a chromosome. Duplications occur when the genetic information from one sister chromatid is placed onto the other chromatid, along with the original copy of the genetic material, resulting in an extra or duplicate section of DNA. Monosomy refers to the loss of an entire chromosome through nondisjunction during meiosis.
Bloom’s Level: 1. Remember Learning Outcome: 24.03.01 Describe how chromosome number disorders arise. Section: 24.03 Topic: Mutations
11. A person who has an extra copy of a chromosome is said to have A. monosomy. B. bisomy. C. trisomy. D. nondisjunction. E. duplication. Duplication occurs when the genetic information from one sister chromatid is placed onto the other chromatid, along with the original copy of the genetic material, resulting in an extra or duplicate section of DNA. Monosomy refers to the loss of an entire chromosome through nondisjunction during meiosis, while a trisomy refers to the gain of an entire chromosome also through nondisjunction during meiosis. Nondisjunction occurs when homologous chromosomes fail to separate during meiosis I or sister chromatids fail to separate during meiosis II.
Bloom’s Level: 1. Remember Learning Outcome: 24.03.01 Describe how chromosome number disorders arise. Section: 24.03 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
12. A female with two Barr bodies in her cells would have what combination of sex chromosomes? A. XXY B. XXXY C. XXX D. XX E. XYY Barr bodies are inactivated X chromosomes. Cells that contain more than one X have one copy randomly deactivated during fetal development. Males do not typically have Barr bodies since they have only one X chromosome. A female with two Barr bodies would have a total of three X chromosomes.
Bloom’s Level: 2. Understand Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
13. Which statement is true regarding the Barr body? A. It is found in the nuclei of females. B. It is found in the nuclei of males. C. It is found in the cytoplasm of males. D. It is found in the cytoplasm of females. E. It is a condensed, inactive Y chromosome. Barr bodies are inactivated X chromosomes. Cells that contain more than one X have one copy randomly deactivated during fetal development. Males do not typically have Barr bodies since they have only one X chromosome, while a normal female would have one in every cell. Since these are chromosomes, they are located in the nucleus. They must be copied when cells undergo division and a copy passed on to each daughter cell. Once division is completed, they are deactivated again.
Bloom’s Level: 2. Understand Learning Outcome: 24.03.01 Describe how chromosome number disorders arise. Section: 24.03 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
14. The most common autosomal abnormality present in people is A. XXX. B. XXY. C. XO. D. a deletion in chromosome 5. E. an extra chromosome 21. Since the question refers to autosomal abnormalities, any changes in the number of X or Y chromosomes are not relevant. The deletion of a segment of chromosome 5 leads to the genetic condition called cri du chat syndrome. This syndrome is usually fatal within a short time of the child's birth. Trisomy 21 occurs when nondisjunction of the 21st chromosome leads to an extra copy of the chromosome and is commonly called Down syndrome. It is frequently seen when older women give birth.
Bloom’s Level: 2. Understand Learning Outcome: 24.03.01 Describe how chromosome number disorders arise. Section: 24.03 Topic: Mutations
15. Karyotyping can be used to diagnose which of the following genetic disorders? A. Down syndrome B. phenylketonuria C. neurofibromatosis D. cystic fibrosis E. hemophilia Karyotyping is used to study the physical, structural appearance; it does not reveal the exact alleles that are contained in the chromosome. Down syndrome is the only one listed that is based on a physical characteristic of the genome; an extra chromosome 21. The others are all caused by the inheritance of particular alleles which are not revealed by this test.
Bloom’s Level: 3. Apply Learning Outcome: 24.03.01 Describe how chromosome number disorders arise. Section: 24.03 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
16. If a person has short stature, a fold in the eyelids, fissured tongue, and mental retardation, they have A. cri du chat syndrome. B. Down syndrome. C. fragile X syndrome. D. Turner syndrome. E. Klinefelter syndrome. Cri du chat syndrome is a deletion of a portion of chromosome 5. Down syndrome, or trisomy 21, is nondisjunction of the 21st chromosome. This leads to an extra copy of the chromosome. Fragile X syndrome results from a repeated segment of nucleotides on the X chromosome. Turner syndrome is the inheritance of a single X chromosome (XO). Klinefelter syndrome is the inheritance of two X chromosomes and one Y chromosome.
Bloom’s Level: 1. Remember Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
17. Down syndrome A. is due to disjunction of chromosomes. B. individuals have two number 21 chromosomes. C. may occur at a lower rate in women over 40. D. can occur if the sperm has an extra number 21 chromosome. E. persons have normal-appearing eyelids. Down syndrome, or trisomy 21, occurs when nondisjunction of the 21st chromosome leads to an extra copy of the chromosome and it is frequently seen when older women give birth, although the extra chromosome 21 can come from the father as well.
Bloom’s Level: 2. Understand Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
18. Which statement is NOT true about Down syndrome? A. It is caused by a third copy of chromosome 21. B. Greatly increased incidence occurs with fathers over age 40. C. It is usually associated with chromosomal nondisjunction in meiosis. D. Characteristics include mental retardation and extra eyelid folds. E. Affected individuals display mental retardation. Down syndrome, or trisomy 21, occurs when nondisjunction of the 21st chromosome leads to an extra copy of the chromosome and it is frequently seen when older women give birth, although the extra chromosome 21 can come from the father as well. There is less evidence that suggests that male's age has an effect on the incidences of nondisjunction in males.
Bloom’s Level: 2. Understand Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
19. Which genetic trait is NOT associated with the chromosome involved in Down syndrome? A. increased incidence of leukemia B. cataracts C. mental retardation D. accelerated aging E. cystic fibrosis Down syndrome, or trisomy 21, occurs when nondisjunction of the 21st chromosome leads to an extra copy of the chromosome and it is frequently seen when older women give birth. The symptoms of the syndrome include short stature, a flat face with eyelid folds, stubby fingers, a round head, and mental impairment. The syndrome has also been linked to an increased tendency of leukemia, cataracts, and an increased rate of aging.
Bloom’s Level: 2. Understand Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
20. The
gene causes mental retardation by increasing the levels of purines in the blood.
A. Gart B. Dart C. RFLP D. SRY E. Barr An extra copy of the Gart gene is found in individuals with Down syndrome who have three copies of the 21st chromosome. The extra gene leads to an increased level of purines in the blood which in turn leads to mental impairment.
Bloom’s Level: 1. Remember Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
21. Genes on the A. X B. Y C. 21st D. 5th E. 19th
chromosome determine if the sex of a child will be male or female.
All humans are female by default, but if the Y chromosome is present and produces the hormones associated with maleness, then the individual will be male.
Bloom’s Level: 1. Remember Learning Outcome: 24.02.01 Explain X-linked genetic inheritance. Section: 24.02 Topic: Sex-Linked Inheritance
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
22. If a person inherits two X chromosomes, this individual will be A. female. B. male. C. color-blind. D. sterile. E. a poly-X female. A female is typically XX. This would be the normal condition for females. As long as a Y chromosome is not present, the individual will be female.
Bloom’s Level: 1. Remember Learning Outcome: 24.02.01 Explain X-linked genetic inheritance. Section: 24.02 Topic: Sex-Linked Inheritance
23. An individual who has an XXY combination of sex chromosomes is said to have syndrome. A. Klinefelter B. Turner C. Down D. fragile X E. cri du chat Klinefelter syndrome is the inheritance of two X chromosomes and one Y chromosome. Turner syndrome is the inheritance of a single X chromosome (XO). Down syndrome, or trisomy 21, is nondisjunction of the 21st chromosome. This leads to an extra copy of the chromosome. Fragile X syndrome results from a repeated segment of nucleotides on the X chromosome. Cri du chat syndrome is a deletion of a portion of chromosome 5.
Bloom’s Level: 1. Remember Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Sex-Linked Inheritance
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
24. Considering that males can have Klinefelter (XXY) syndrome, XYY, and normal XY chromosomal combinations, and females can have Turner (XO) syndrome, poly-X (XXX, XXXX), and normal XX combinations, it is obvious that A. maleness results from the presence of only one X chromosome. B. maleness results from the absence of two or more X chromosomes. C. maleness results from the minimal presence of one Y chromosome. D. femaleness results from the presence of two or more X chromosomes. E. sex determination is a delicate balance between X and Y chromosomes. If the Y chromosome is present and produces the hormones associated with maleness, then the individual will be male. Males can have one or two Y chromosomes.
Bloom’s Level: 3. Apply Learning Outcome: 24.02.01 Explain X-linked genetic inheritance. Section: 24.02 Topic: Sex-Linked Inheritance
25. The reason that missing an X chromosome and having an extra X chromosome do NOT cause more harm than they do is best explained by the A. inactivation of any X beyond the first as a Barr body. B. presence of genes on the Y chromosome that determine maleness. C. presence of genes on the X chromosomes that determine femaleness. D. loss of a sex chromosome in normal cells as embryo development occurs. E. a higher level of gene-repair enzyme activity on sex chromosomes. As long as one X chromosome is present, the individual can develop in a relatively normal fashion, although there can be problems with reproduction. Additional X chromosomes will simply be deactivated, forming a Barr body for each one.
Bloom’s Level: 3. Apply Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
26. If an individual has a XYY genotype, they are classified as having A. Down syndrome. B. cri du chat syndrome. C. Turner syndrome. D. Klinefelter syndrome. E. Jacobs syndrome. Down syndrome, or trisomy 21, is nondisjunction of the 21st chromosome; this leads to an extra copy of the chromosome. Cri du chat syndrome is a deletion of a portion of chromosome 5. Turner syndrome is the inheritance of a single X chromosome (XO). Klinefelter syndrome is the inheritance of two X chromosomes and one Y chromosome. Jacobs syndrome results when an individual inherits an extra Y chromosome (XYY).
Bloom’s Level: 2. Understand Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
27. A person with an XO genotype is classified as having A. Down syndrome. B. cri du chat syndrome. C. Turner syndrome. D. Klinefelter syndrome. E. a poly-X state. Down syndrome, or trisomy 21, is nondisjunction of the 21st chromosome; this leads to an extra copy of the chromosome. Cri du chat syndrome is a deletion of a portion of chromosome 5. Turner syndrome is the inheritance of a single X chromosome (XO). Klinefelter syndrome is the inheritance of two X chromosomes and one Y chromosome. Poly-X females have more than two X chromosomes and no Y chromosome.
Bloom’s Level: 1. Remember Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
28. A female that does not undergo puberty or menstruate or lacks breast development may have A. Klinefelter syndrome. B. Down syndrome. C. cri du chat syndrome. D. Turner syndrome. E. poly-X female state. Klinefelter syndrome is the inheritance of two X chromosomes and one Y chromosome. Down syndrome, or trisomy 21, is nondisjunction of the 21st chromosome; this leads to an extra copy of the chromosome. Cri du chat syndrome is a deletion of a portion of chromosome 5. Turner syndrome is the inheritance of a single X chromosome (XO). Poly-X females have more than two X chromosomes and no Y chromosome.
Bloom’s Level: 4. Analyze Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
29. An XXX female would most likely result from nondisjunction in A. the mother during egg formation. B. the father during sperm formation. C. in both the mother and father during gamete formation. D. the fertilized egg. E. neither the mother nor the father. Since the father usually has only one X chromosome to give to a gamete, extra copies of X chromosomes must arise from nondisjunction in the mother.
Bloom’s Level: 3. Apply Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
30. Generally, it is not possible to determine whether nondisjunction failed to occur in oogenesis or spermatogenesis. However, it is possible to assert that resulted in nondisjunction in . A. XXY; oogenesis B. XYY; spermatogenesis C. XXX; oogenesis D. XXY; spermatogenesis E. XO; oogenesis Extra Y chromosomes must come from the father during spermatogenesis since the mother would not have any Y chromosomes; if she did she would be male not female.
Bloom’s Level: 5. Evaluate Learning Outcome: 24.03.01 Describe how chromosome number disorders arise. Section: 24.03 Topic: Mutations
31. If a chromosomal segment is turned around 180, the chromosomal mutation is termed a(n) A. translocation. B. duplication. C. deletion. D. inversion. E. monosomy. Translocations occur when a segment of DNA moves from one location in the genome to another. Deletions are loss of a portion of a chromosome. Duplications occur when the genetic information from one sister chromatid is placed onto the other chromatid, along with the original copy of the genetic material resulting in an extra or duplicate section of DNA. An inversion refers to the changing of direction of a segment of DNA while it stays in the same location on the same chromosome. Monosomy refers to the loss of an entire chromosome through nondisjunction during meiosis.
Bloom’s Level: 2. Understand Learning Outcome: 24.04.01 Describe a chromosomal deletion, duplication, translocation, and inversion. Section: 24.04 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
32. Which refers to the movement of a piece of one chromosome to another nonhomologous chromosome? A. inversion B. translocation C. deletion D. duplication E. monosomy Translocations occur when a segment of DNA moves from one location in the genome to another. Deletions are loss of a portion of a chromosome. Duplications occur when the genetic information from one sister chromatid is placed onto the other chromatid, along with the original copy of the genetic material, resulting in an extra or duplicate section of DNA. An inversion refers to the changing of direction of a segment of DNA while it stays in the same location on the same chromosome. Monosomy refers to the loss of an entire chromosome through nondisjunction during meiosis.
Bloom’s Level: 2. Understand Learning Outcome: 24.04.01 Describe a chromosomal deletion, duplication, translocation, and inversion. Section: 24.04 Topic: Mutations
33. Which refers to the loss of a portion of a chromosome? A. inversion B. translocation C. deletion D. duplication E. monosomy Translocations occur when a segment of DNA moves from one location in the genome to another. Deletions are loss of a portion of a chromosome. Duplications occur when the genetic information from one sister chromatid is placed onto the other chromatid, along with the original copy of the genetic material, resulting in an extra or duplicate section of DNA. An inversion refers to the changing of direction of a segment of DNA while it stays in the same location on the same chromosome. Monosomy refers to the loss of an entire chromosome through nondisjunction during meiosis.
Bloom’s Level: 2. Understand Learning Outcome: 24.04.01 Describe a chromosomal deletion, duplication, translocation, and inversion. Section: 24.04 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
34. Which refers to the addition of a repeat segment of a chromosome? A. inversion B. translocation C. deletion D. duplication E. monosomy An inversion refers to the changing of direction of a segment of DNA while it stays in the same location on the same chromosome. Translocations occur when a segment of DNA moves from one location in the genome to another. Deletions are a loss of part of a chromosome. Duplications occur when the genetic information from one sister chromatid is placed onto the other chromatid, along with the original copy of the genetic material, resulting in an extra or duplicate section of DNA. Monosomy refers to the loss of an entire chromosome through nondisjunction during meiosis.
Bloom’s Level: 2. Understand Learning Outcome: 24.04.01 Describe a chromosomal deletion, duplication, translocation, and inversion. Section: 24.04 Topic: Mutations
35. If a person has a portion of number 5 chromosome missing, they may have A. Down syndrome. B. cri du chat syndrome. C. Turner syndrome. D. Klinefelter syndrome. E. Jacobs syndrome. Down syndrome, or trisomy 21, is nondisjunction of the 21st chromosome; this leads to an extra copy of the chromosome. Cri du chat syndrome is a deletion of a portion of chromosome 5. Turner syndrome is the inheritance of a single X chromosome (XO). Klinefelter syndrome is the inheritance of two X chromosomes and one Y chromosome. Jacobs syndrome results when an individual inherits an extra Y chromosome (XYY).
Bloom’s Level: 2. Understand Learning Outcome: 24.04.02 Summarize the disorders caused by changes in chromosome structure. Section: 24.04 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
36. If a chromosomal segment appears more than once in the same chromosome, it is termed a(n) A. translocation. B. duplication. C. deletion. D. inversion. E. polyploid. Translocations occur when a segment of DNA moves from one location in the genome to another. Deletions are losses of portions of a chromosome and duplications occur when the genetic information from one sister chromatid is placed onto the other chromatid, along with the original copy of the genetic material, resulting in an extra or duplicate section of DNA. An inversion refers to the changing of direction of a segment of DNA while it stays in the same location on the same chromosome. Polyploid refers to the inheritance of additional sets of chromosomes; such individuals would be 3n, 4n, etc. depending on how many additional sets they have.
Bloom’s Level: 2. Understand Learning Outcome: 24.04.01 Describe a chromosomal deletion, duplication, translocation, and inversion. Section: 24.04 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
37. The cri du chat syndrome represents a chromosomal mutation type termed A. translocation. B. duplication. C. deletion. D. inversion. E. polyploidy. Cri du chat syndrome results from the loss of a portion of chromosome 5. Deletions are losses of portions of a chromosome. Translocations occur when a segment of DNA moves from one location in the genome to another. Duplications occur when the genetic information from one sister chromatid is placed onto the other chromatid, along with the original copy of the genetic material, resulting in an extra or duplicate section of DNA. An inversion refers to the changing of direction of a segment of DNA while it stays in the same location on the same chromosome. Polyploid refers to the inheritance of additional sets of chromosomes; such individuals would be 3n, 4n, etc. depending on how many additional sets they have.
Bloom’s Level: 1. Remember Learning Outcome: 24.04.02 Summarize the disorders caused by changes in chromosome structure. Section: 24.04 Topic: Mutations
Figure 24.1
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
38. The trait diagrammed in Figure 24.1 is a(n) A. dominant X-linked trait. B. recessive X-linked trait. C. recessive Y-linked trait. D. autosomal dominant trait. E. dominant Y-linked trait. The trait is seen at a higher frequency among males than among females, indicating that the trait is X-linked. Also, a male in the second generation who does not show the trait and a woman who does not show the trait can have an affected son. This is also an indicator of an X-linked trait. The trait must be recessive because the female in the second generation who is a daughter of an affected father and a mother who does not show the trait also does not show the trait but is able to pass it along to her son.
Bloom’s Level: 5. Evaluate Learning Outcome: 24.02.01 Explain X-linked genetic inheritance. Section: 24.02 Topic: Pedigrees
39. Which of the following conditions is NOT due to a sex-linked gene? A. color-blindness B. hemophilia C. muscular dystrophy D. Klinefelter syndrome E. inability to see red or green Color-blindness is an X-linked recessive disorder; individuals with this condition see one color as another one. Hemophilia is an X-linked recessive disorder. Hemophilia is due to the absence or minimal presence of some clotting factor. Duchenne muscular dystrophy is an Xlinked recessive disorder characterized by wasting away of the muscles. Klinefelter syndrome occurs when an individual inherits two complete X chromosomes and one Y chromosome; the individual is male. Because this syndrome deals with an entire chromosome, it is not based on a sex-linked gene.
Bloom’s Level: 2. Understand Learning Outcome: 24.02.01 Explain X-linked genetic inheritance. Section: 24.02 Topic: Sex-Linked Inheritance
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
40. If a woman is a carrier for the color-blind recessive allele and her husband has normal vision, what are their chances that a son will be color-blind? A. None, because the father is normal. B. 50%, since the mother is only a carrier. C. 100% because the mother has the gene. D. 25% because the mother is a hybrid. E. None since the son will also be just a carrier. The mother is XBXb and the father is XBY. With this combination of genotypes, any son will have a 50% chance of inheriting the color-blind allele because the father can only give the Y chromosome and the mother can give either the normal allele or the color-blind allele.
Bloom’s Level: 3. Apply Learning Outcome: 24.02.02 Solve crosses involving sex-linked traits. Section: 24.02 Topic: Sex-Linked Inheritance
41. In fruit flies, bar eye is inherited by an X-linked allele (B for bar). If a heterozygous bareyed female is mated to a non-bar-eyed male, what will be the expected ratio of phenotype given four offspring? A. two bar-eyed females, two bar-eyed males B. one bar-eyed and one non-bar-eyed female, one bar-eyed and one non-bar-eyed male C. three bar-eyed females and one non-bar-eyed male D. bar-eyed females only E. non-bar-eyed males only The mother is XBXb and the father is XbY. Given these genotypes, females can be either homozygous recessive or heterozygous; the heterozygote will have the bar-eye phenotype, while the homozygous recessive will be the non-bar-eye phenotype. Males of the same mating will have a 50% chance of inheriting either of the two alleles from the mother, so the expected ratio would be 1:1 for males.
Bloom’s Level: 3. Apply Learning Outcome: 24.02.02 Solve crosses involving sex-linked traits. Section: 24.02 Topic: Sex-Linked Inheritance
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
42. Which disorder is characterized by a lack of the protein dystrophin? A. hemophilia B. color-blindness C. muscular dystrophy D. Down syndrome E. cystic fibrosis Duchenne muscular dystrophy is an X-linked recessive disorder characterized by wasting away of the muscles due to the protein dystrophin. It has been determined that the lack of dystrophin leads to calcium leaking into the cell. This promotes the action of an enzyme that dissolves muscle fibers.
Bloom’s Level: 2. Understand Learning Outcome: 24.02.03 Describe four X-linked recessive disorders. Section: 24.02 Topic: Mutations
43. Which of the following is NOT true about a karyotype? A. Homologous chromosomes are arranged in pairs. B. Sex chromosomes are identified separately from autosomes. C. All chromosome pairs are numbered differently for males and females. D. Chromosome pairs are assorted by both size and shape. E. Banding patterns are used in pairing chromosomes. A karyotype is a picture of the chromosomes of an individual. They are arranged by homologous pairs and in descending order of size, with the sex chromosomes placed together, not as part of the homologues. In addition, the chromosomes are treated with stains to show banding patterns. This allows comparison of the chromosomes and can determine if changes in chromosome structure or number are present.
Bloom’s Level: 2. Understand Learning Outcome: 24.03.02 Give several examples of chromosome number disorders. Section: 24.03 Topic: Mendelian Genetics
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
44. A color-blind (recessive trait) woman will pass the allele to A. her sons only. B. all her children. C. her daughters only. D. none of her children. E. her husband. If color-blindness is recessive, then an affected woman must carry a copy of the allele on both of her X chromosomes. This is the only allele she has to pass to any of her offspring. Daughters may or may not be color-blind, depending on the X donated by the father. Since she is the one who gives the X chromosome to her sons (the father gives the Y that determines maleness), all of her sons must be color-blind.
Bloom’s Level: 2. Understand Learning Outcome: 24.02.01 Explain X-linked genetic inheritance. Section: 24.02 Topic: Sex-Linked Inheritance
45. Hemophilia (h) is a sex-linked recessive trait. If a hemophiliac male marries a carrier female, A. 50% of their daughters will be hemophiliac. B. 75% of their daughters will be hemophiliac. C. 25% of their daughters will be hemophiliac. D. 25% of their sons will be hemophiliac. E. 75% of their sons will be hemophiliac. Father XhY and mother XHXh. The possible offspring genotypes are XhY and XHY for sons and XHXh or XhXh for daughters. So, in this instance, sons and daughters both have a 50% chance of being affected by hemophilia.
Bloom’s Level: 3. Apply Learning Outcome: 24.02.02 Solve crosses involving sex-linked traits. Section: 24.02 Topic: Sex-Linked Inheritance
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
46. A karyotype will NOT reveal which of the following genetic diseases? A. Turner syndrome B. Klinefelter syndrome C. Down syndrome D. cri du chat syndrome E. hemophilia Turner syndrome has the genotype X0, Klinefelter syndrome XXY, Down syndrome has three 21 chromosomes, and cri du chat is missing part of the 5th chromosome. All of these can be seen in a karyotype. Hemophilia is an allele for a gene; it cannot be seen in a karyotype.
Bloom’s Level: 3. Apply Learning Outcome: 24.03.01 Describe how chromosome number disorders arise. Section: 24.03 Topic: Mutations
47. Which one of the following disorders listed below is NOT caused by a change in chromosome structure? A. fragile X syndrome B. Klinefelter syndrome C. Williams syndrome D. cri du chat syndrome E. All of the answer choices are caused by a change in chromosome structure. All of the syndromes listed are caused by changes in the structure of the chromosome, except Klinefelter ,which results from the addition of one or more X chromosomes. In fragile X syndrome, there are repeated sequences of CGG. In Williams syndrome, part of the number 7 chromosome is missing. In cri du chat, chromosome 5 is missing a piece at the end of the chromosome.
Bloom’s Level: 2. Understand Learning Outcome: 24.04.02 Summarize the disorders caused by changes in chromosome structure. Section: 24.04 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
48. A mother is a carrier for blue eyes (autosomal recessive) and for hemophilia (X-linked recessive). Which of these is a correct statement? A. All sons will have blue eyes and be hemophiliacs. B. It depends on the father's genotype as to whether the sons will have blue eyes and/or be hemophiliacs. C. The father's genotype can determine whether a son will have blue eyes but cannot determine whether a son will be a hemophiliac. D. Regardless of the father, no sons will have blue eyes or be hemophiliacs. E. There is a linkage between eye color and hemophilia. Mother's genotype is Bb XHXh. Because the mother has both of the possible alleles for eye color and both alleles for normal blood clotting, or hemophilia, then the genotype of the father will determine if the recessive blue allele is expressed (he must be able to pass on the b allele for a son to possibly have blue eyes), but the mother alone will determine if a son has hemophilia because she will pass on the only X chromosome that a son will have.
Bloom’s Level: 4. Analyze Learning Outcome: 24.02.02 Solve crosses involving sex-linked traits. Section: 24.02 Topic: Sex-Linked Inheritance
49. Which of the following sex-linked diseases is characterized by the absence of a clotting factor? A. hemophilia B. fragile X syndrome C. color-blindness D. Duchenne muscular dystrophy E. None of the answer choices is true. Hemophilia is characterized by the absence of a clotting factor.
Bloom’s Level: 1. Remember Learning Outcome: 24.02.03 Describe four X-linked recessive disorders. Section: 24.02 Topic: Sex-Linked Inheritance
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
50. Which of the following pairs of alleles would be most likely to cross over? A. A. ........ B B. A...B...... C. A......B.......C D. AB............. E. ........... AB The greater the distance between the alleles on a linkage group, the more likely they are to cross over.
Bloom’s Level: 5. Evaluate Learning Outcome: 24.01.02 Predict which alleles in a linkage group will cross over more frequently. Section: 24.01 Topic: Non-Mendelian Inheritance
Short Answer Questions 51. List the possible zygote combinations that could occur if a man has nondisjunction occur during meiosis II, but the woman's eggs are normal. Two of the zygotes would be normal (2n) because the eggs (n) are fertilized by a sperm (n) containing a normal amount of DNA. One of the zygotes would have an additional component of DNA (2n+1) because the sperm contains an extra piece of DNA (2n). The other zygote would be short a piece of DNA (2n-1) because the sperm is lacking a piece of DNA.
Bloom's Level: 6. Create Learning Outcome: 24.03.01 Describe how chromosome number disorders arise. Section: 24.03 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
52. List the four X-linked recessive disorders. 1. Color-blindness 2. Duchenne muscular dystrophy 3. Fragile X syndrome 4. Hemophilia
Bloom's Level: 6. Create Learning Outcome: 24.02.03 Describe four X-linked recessive disorders. Section: 24.02 Topic: Sex-Linked Inheritance
Multiple Choice Questions 53. Which of the following sex-linked disorders is characterized by an abnormal number of repeat sequences in the genome? A. fragile X syndrome B. hemophilia C. color-blindness D. Duchenne muscular dystrophy E. None of the answer choices is true. Fragile X syndrome is characterized by an abnormal number of repeat sequences in the genome.
Bloom’s Level: 1. Remember Learning Outcome: 24.02.03 Describe four X-linked recessive disorders. Section: 24.02 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
Short Answer Questions 54. Describe a genetic disorder associated with a deletion in chromosomal number. Williams syndrome occurs when chromosome 7 loses a tiny end piece. Children who have this syndrome look like pixies because they have turned-up noses, wide mouths, small chins, and large ears. The gene for the protein elastin is missing, causing problems with the cardiovascular system and causing their skin to age prematurely. Cri du chat syndrome occurs when chromosome 5 is missing an end piece. The individual has a small head and is mentally impaired. Abnormal development of the glottis and larynx results in an infant's cry that resembles a cat.
Bloom's Level: 6. Create Learning Outcome: 24.04.02 Summarize the disorders caused by changes in chromosome structure. Section: 24.04 Topic: Mutations
Multiple Choice Questions 55. Which genetic disorder is the result of a deletion of a section of an individual's chromosome? A. Williams syndrome B. Klinefelter syndrome C. Down syndrome D. Turner syndrome E. None of the answer choices is the result of a deletion of a section of an individual's chromosome. Williams syndrome is the result of a deletion of a section of an individual's chromosome. Klinefelter syndrome is the result of XXY genome. Down syndrome can be caused by a translocation of a chromosomal section. Turner syndrome is the result of an X genome due to missing a chromosome.
Bloom’s Level: 1. Remember Learning Outcome: 24.04.02 Summarize the disorders caused by changes in chromosome structure. Section: 24.04 Topic: Mutations
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Chapter 24 - Chromosomal Inheritance and Genetic Disorders
56. Which trait is not part of the linkage group found on chromosome 19? A. production of elastin B. muscular dystrophy C. hair color (brown) D. eye color (green/blue) E. colorectal cancer The production of elastin is found on chromosome 7.
Bloom’s Level: 1. Remember Learning Outcome: 24.01.01 Describe what is meant by the term linkage group. Section: 24.01 Topic: Non-Mendelian Inheritance
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Chapter 25 - DNA Structure and Control of Gene Expression
Chapter 25 DNA Structure and Control of Gene Expression
Multiple Choice Questions 1. The hereditary material found in all cells is A. DNA. B. rRNA. C. mRNA. D. tRNA. E. ATP. While mRNA, tRNA, and rRNA are all necessary to express the genetic code of an organism, it is DNA that carries the information to make the other molecules and to make the proteins that carry out the necessary functions of the cell.
Bloom’s Level: 1. Remember Learning Outcome: 25.01.01 Summarize the experiments that enabled scientists to determine that DNA made up the genetic material. Section: 25.01 Topic: DNA Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
2. If the protein coat of a virus that infects a bacterium is labeled with radioactive sulfur and the DNA of the virus is labeled with radioactive phosphorus, over time A. both the sulfur and the phosphorus will be found within the bacterium. B. only the sulfur will be found inside the bacterium. C. only the phosphorus will be found inside the bacterium. D. both the sulfur and the phosphorus will be found outside the bacterium. E. the radioactivity of the sulfur and phosphorus will decay very quickly and not be detectable. The protein coat of the virus is not taken inside the cell; a bacteriophage instead injects its DNA into the bacterium. This would mean that the radioactive labeling for the protein coat would remain outside the cell while the labeling on the viral DNA would enter the cell.
Bloom’s Level: 3. Apply Learning Outcome: 25.01.01 Summarize the experiments that enabled scientists to determine that DNA made up the genetic material. Section: 25.01 Topic: DNA Structure
3. Hershey and Chase experimented with radioactively labeled phosphorus and sulfur to determine that DNA and not protein is the genetic material. Which of the following would NOT be an essential part of this experiment? A. Sulfur is present in amino acids in the protein coat of bacteria. B. Phosphorus is present in high amounts in DNA. C. Sulfur is not present in DNA. D. Phosphorus is not present in amino acids in the protein coat of bacteria. E. DNA contains nitrogen. Since nitrogen is not used in the experiment, its presence would not affect the experiment. It is important to ensure that the two different radioactively labeled molecules are located only in one part of the virus or the other.
Bloom’s Level: 3. Apply Learning Outcome: 25.01.01 Summarize the experiments that enabled scientists to determine that DNA made up the genetic material. Section: 25.01 Topic: DNA Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
4. Which statement is NOT correct about the results of the Hershey-Chase experiment with the T2 virus? A. Radioactively labeled protein was found in the bacteria, not in the virus coats. B. Radioactively labeled DNA was found in the bacteria, not in the virus coats. C. DNA was labeled with radioactive phosphorus. D. Protein was labeled with radioactive sulfur. E. Two separate experiments were actually run, one with radioactive phosphorus and one with radioactive sulfur. There were two experiments performed by Hershey and Chase; in the first one radioactive 32P was used to label viral DNA and the viruses were allowed to inject their DNA into a bacterium. After this, most of the labeled DNA was found in the bacterium. In the second experiment, the capsid proteins were labeled with 32S and again the virus was allowed to inject its DNA into a bacterium. This time the radioactive substance was found outside the bacterium.
Bloom’s Level: 3. Apply Learning Outcome: 25.01.01 Summarize the experiments that enabled scientists to determine that DNA made up the genetic material. Section: 25.01 Topic: DNA Structure
5. Information from X-ray crystallographic data collected by and Crick in their development of the model of DNA. A. Chargaff B. Griffith C. McClintock D. Franklin E. Hershey and Chase
was used by Watson
Rosalind Franklin was working at King’s College in London with Maurice Wilkins. She prepared X-ray diffraction photographs of DNA. It was this data that allowed Watson and Crick to determine that DNA was a helix-shaped molecule. She died before the Nobel Prize was awarded for the discovery of the structure of DNA and was not mentioned in the award since Nobel Prizes cannot be awarded posthumously.
Bloom’s Level: 1. Remember Learning Outcome: 25.01.01 Summarize the experiments that enabled scientists to determine that DNA made up the genetic material. Section: 25.01 Topic: DNA Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
6. The X-ray diffraction photography of Rosalind Franklin and Maurice Wilkins was critical evidence A. indicating that DNA has a double helix structure. B. showing the bases of DNA were held together by hydrogen bonds. C. showing equal numbers of purines and pyrimidines. D. revealing the structure of the deoxyribose sugar. E. of the location of each adenine, guanine, cytosine, and thymine. The X-shaped pattern in the center of the X-ray diffraction photograph told Watson and Crick that DNA is a helix. The dark portions at the top and the bottom of the photograph told them that some feature is repeated over and over and Watson and Crick were able to determine that the repeated portion were hydrogen bonds between the nitrogenous bases.
Bloom’s Level: 2. Understand Learning Outcome: 25.01.01 Summarize the experiments that enabled scientists to determine that DNA made up the genetic material. Section: 25.01 Topic: DNA Structure
7. The individual(s) credited with the discovery of the structure of DNA is (are) A. Alfred Hershey and Martha Chase. B. James Watson and Francis Crick. C. Erwin Chargaff. D. Rosalind Franklin and Maurice Wilkins. E. Charles Darwin and Gregor Mendel. James Watson and Francis Crick used data collected by Erwin Chargaff, a chemist, and Rosalind Franklin, who performed X-ray diffraction photography of DNA, to determine the chemical structure of DNA.
Bloom’s Level: 1. Remember Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
8. The fact that for a given species the amount of purines in the DNA always matches the number of pyrimidines was first determined by A. Watson and Crick. B. Franklin and Wilkins. C. Hershey and Chase. D. Chargaff. E. Mendel. Erwin Chargaff determined that the number of purines in DNA always equals the number of pyrimidines. He also determined that amount of adenine equals the amount of thymine (A = T), and the amount of guanine equals the amount of cytosine (G = C). These findings came to be known as Chargaff’s rules.
Bloom’s Level: 1. Remember Learning Outcome: 25.01.01 Summarize the experiments that enabled scientists to determine that DNA made up the genetic material. Section: 25.01 Topic: DNA Structure
9. Chargaff's rules of DNA structure states that A. A, T = G, C. B. A, G = T, C. C. A = C, T = G. D. A = G, T = C. E. the number of purines in DNA never equals the number of pyrimidines. Erwin Chargaff determined that the amount of adenine equals the amount of thymine (A = T), and the amount of guanine equals the amount of cytosine (G = C). These findings came to be known as Chargaff’s rules.
Bloom’s Level: 2. Understand Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
10. If a species contains 23% A in its DNA, what is the percentage of guanine it would contain? A. 23% B. 46% C. 25% D. 44% E. 27% Following Chargaff’s rules, A = T and G = C. So, if the DNA is 23% A it is also 23% for a total of 46%. This means that the other 54% is composed of G and C in equal amounts. This means that 27% is G and 27% is C.
Bloom’s Level: 3. Apply Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
11. The amount of adenine is always equal to the amount of A. cytosine B. uracil C. guanine D. thymine E. ATP
in DNA.
According to Chargaff’s rules, the amount of adenine equals the amount of thymine (A = T), and the amount of guanine equals the amount of cytosine (G = C).
Bloom’s Level: 1. Remember Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
12. Which of the following have nitrogenous bases correctly paired in DNA? A. adenine—guanine; thymine—cytosine B. adenine—uracil; guanine—cytosine C. adenine—thymine; guanine—cytosine D. adenine—adenine; guanine—guanine E. adenine—cytosine; guanine—thymine Since the amount of adenine (A) equals the amount of thymine (T) and the amount of guanine (G) equals the amount of cytosine (C), this means that A must pair with T and C with G.
Bloom’s Level: 1. Remember Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
13. A nucleotide contains A. DNA and RNA. B. a sugar and a phosphate. C. complementary purines and pyrimidines. D. RNA, protein, and lipids. E. a sugar, a phosphate, and a nitrogen base. A nucleotide is composed of a phosphate group, a pentose sugar (deoxyribose for DNA and ribose for RNA), and one of four possible nitrogenous bases. The bases are either a purine (G and A) or a pyrimidine (C and T).
Bloom’s Level: 1. Remember Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
14. In the Watson-Crick model of DNA, the "sides" of the ladder are composed of A. sugars. B. bases. C. hydrogen bonds. D. sugar-phosphate molecules. E. phosphate groups. The sides of the DNA "ladder" are the portions of the nucleotides that are identical, regardless of the type of nucleotide. The portion that is always the same is the pentose sugar and the phosphate group. This "backbone" part of the molecule is held together by covalent bonds between the phosphate group of one nucleotide and the pentose sugar of the next; forming a long strand of nucleotides that is held to another such strand by hydrogen bonds between the nitrogenous bases.
Bloom’s Level: 2. Understand Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
15. Which of the following is NOT true about DNA? A. has a double helix B. bases held together by hydrogen bonds C. bases are complementary to each other D. has a deoxyribose sugar E. contains adenine, guanine, cytosine, and uracil DNA contains the bases adenine, guanine, cytosine, and thymine; uracil is only found in RNA. A nucleotide is composed of a phosphate group, a pentose sugar (deoxyribose for DNA and ribose for RNA), and one of four possible nitrogenous bases.
Bloom’s Level: 2. Understand Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
16. Which of the following statements about DNA replication is NOT correct? A. Unwinding of the DNA molecule occurs as hydrogen bonds break. B. Replication occurs as each base is paired with another exactly like it. C. The process is known as semi-conservative replication because one old strand is conserved in the new molecule. D. The enzyme that catalyzes DNA replication is DNA polymerase. E. Complementary base pairs are held together with hydrogen bonds. In DNA replication a molecule unwinds the helix, breaking the hydrogen bonds between the two DNA strands, and then a complementary nucleotide is joined to each of the template (or parent) strands. Using the parent strand as a guide, the new strand of DNA is made by DNA polymerase to exactly mirror the original strand. The new strand of DNA and the parent strand are held together by hydrogen bonds that form between the nitrogenous bases.
Bloom’s Level: 4. Analyze Learning Outcome: 25.02.01 Explain why DNA replication is semiconservative. Section: 25.02 Topic: DNA Replication
17. The enzyme that is used to join complementary DNA nucleotides together is A. DNA polymerase. B. RNA polymerase. C. helicase. D. ribozyme. E. lipase. In DNA replication helicase unwinds the helix, breaking the hydrogen bonds between the two DNA strands, and then a complementary nucleotide is joined to each of the template (or parent) strands by DNA polymerase. Using the parent strand as a guide, the new strand of DNA is made to exactly mirror the original strand.
Bloom’s Level: 1. Remember Learning Outcome: 25.02.02 Summarize the events that occur during the process of DNA replication. Section: 25.02 Topic: DNA Replication
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Chapter 25 - DNA Structure and Control of Gene Expression
18. Because one original strand of the double-stranded helix is found in each daughter cell, the replication process is called A. proofreading. B. semi-conservative. C. redundant. D. freeing of DNA. E. mutation positive. The term semi-conservative replication is used to describe the replication of DNA since each newly made DNA molecule is composed of one strand that serves as the template and one strand that has been newly formed by DNA polymerase.
Bloom’s Level: 2. Understand Learning Outcome: 25.02.01 Explain why DNA replication is semiconservative. Section: 25.02 Topic: DNA Replication
19. Which does NOT describe a function of the DNA polymerase molecule? A. recognize the free nucleotide that pairs with the base on the old strand of DNA B. read the strand of old DNA and recognize the base there C. proofread to ensure that the proper base has been incorporated D. make the proper nucleotide to match with the base read on the old strand E. nucleotides are joined by hydrogen bonds DNA polymerase is able to join new nucleotides to the 3’ end of a growing strand of new DNA. It does this by matching the new nucleotide to its complement on the parent strand. Since the DNA polymerase can determine which new nucleotide belongs in the sequence through this matching process, it is also able to proofread the strand it is composing to make sure that the correct nucleotide has been used. DNA polymerase does not make the nucleotides it uses; they are already present in the nucleus.
Bloom’s Level: 3. Apply Learning Outcome: 25.02.02 Summarize the events that occur during the process of DNA replication. Section: 25.02 Topic: DNA Replication
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Chapter 25 - DNA Structure and Control of Gene Expression
20. In semiconservative DNA replication, each new double helix formed will have A. two new strands and two old strands. B. one new and one old strand in each helix. C. three new strands in one helix and three old strands in the second helix. D. two new and one old strand in one helix and two old and one new strand in second helix. E. two new strands in one helix and two old strands in the other helix. The term semiconservative replication is used to describe the replication of DNA, since each newly made DNA molecule is composed of one strand that serves as the template and one strand that has been newly formed by DNA polymerase. Thus, every molecule of DNA is one-half parent DNA and one-half newly made strand.
Bloom’s Level: 2. Understand Learning Outcome: 25.02.01 Explain why DNA replication is semiconservative. Section: 25.02 Topic: DNA Replication
21. The enzyme is responsible for unwinding the double-helix structure of DNA during replication. A. helicase B. polymerase C. ligase D. extendase E. windase Helicase unwinds the DNA strand so that DNA polymerase can bind to the opened DNA and replicate it. To complete replication, the enzyme DNA ligase seals any breaks in the sugarphosphate backbone.
Bloom’s Level: 1. Remember Learning Outcome: 25.02.02 Summarize the events that occur during the process of DNA replication. Section: 25.02 Topic: DNA Replication
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Chapter 25 - DNA Structure and Control of Gene Expression
Figure 25.4
22. In Figure 25.4, if adenine is located on strand 4, then on strand 2 at the same location must be present. A. adenine B. thymine C. cytosine D. guanine E. uracil Strands 1 and 4 are the parent strands of DNA and as such would be complementary to one another. This means that strands 2 and 4 are identical, so they would contain the same bases in the same order.
Bloom’s Level: 3. Apply Learning Outcome: 25.02.01 Explain why DNA replication is semiconservative. Section: 25.02 Topic: DNA Replication
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Chapter 25 - DNA Structure and Control of Gene Expression
23. In Figure 25.4, when replication is finished, strand 1 will have the same base composition as strand , while strand 2 will have the same base composition as strand . A. 2; 3 B. 1; 3 C. 3; 4 D. 1; 4 E. 2; 4 Strands 1 and 2 are complementary and strands 3 and 4 are complementary. So, once replication is complete, strands 1 and 2 form one molecule of DNA and strands 3 and 4 form an identical molecule of DNA.
Bloom’s Level: 3. Apply Learning Outcome: 25.02.01 Explain why DNA replication is semiconservative. Section: 25.02 Topic: DNA Replication
24. A gene is physically located on a A. chromosome B. messenger RNA C. enzymes D. proteins E. nucleus
in the nucleus of a cell.
Genes are portions of DNA that give the information to make a particular molecule. This can be a protein or an RNA molecule.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.01 Describe the roles of RNA molecules in gene expression. Section: 25.03 Topic: Chromosome Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
25. Which of the following is NOT true about RNA? A. RNA transfers messages from DNA to ribosomes. B. RNA contains the sugar ribose. C. RNA contains adenine, guanine, uracil, and cytosine. D. RNA is single stranded. E. RNA forms a helix. As a single-stranded molecule, RNA does not make a helix shape. The DNA helix is formed because of the hydrogen bonds between the bases of the two antiparallel strands.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.01 Describe the roles of RNA molecules in gene expression. Section: 25.03 Topic: RNA Structure
26. Which of the following nucleotide bases is found only in RNA, not in DNA? A. guanine B. adenine C. thymine D. uracil E. cytosine DNA contains the bases cytosine, guanine, adenine, and thymine while RNA contains the bases cytosine, guanine, adenine, and uracil which replaces thymine.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.01 Describe the roles of RNA molecules in gene expression. Section: 25.03 Topic: RNA Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
27. Which of the following nucleotide bases is found only in DNA, not in RNA? A. guanine B. adenine C. thymine D. uracil E. cytosine DNA contains the bases cytosine, guanine, adenine, and thymine, while RNA contains the bases cytosine, guanine, adenine, and uracil which replaces thymine.
Bloom’s Level: 1. Remember Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
28. The function of transfer RNA is to A. carry amino acids to ribosomes. B. transfer nucleotides to the nucleus. C. transmit coded information to the cytoplasm. D. turn DNA on and off. E. act as the site for protein synthesis. Transfer RNA (tRNA) serves to bind and carry amino acids to the site of protein synthesis. The tRNA then binds its anticodon to the condon of the mRNA to ensure that the appropriate amino acid is joined into the growing polypeptide.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.01 Describe the roles of RNA molecules in gene expression. Section: 25.03 Topic: RNA Types
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Chapter 25 - DNA Structure and Control of Gene Expression
29. Which of the following is NOT part of gene expression? A. transcription B. replication C. translation D. amino acids E. protein synthesis Replication is the copying of DNA that takes place before cell division can occur. This does not express genes, but does serve to preserve them so that they are passed on to any daughter cells. The process of gene expression is 1) transcription, the manufacture of mRNA from DNA; and 2) translation, the manufacture of a polypeptide using mRNA, tRNA, and ribosomes. Translation is the synthesis of a protein by joining the correct amino acids together in the correct order.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Gene Regulation: Eukaryotes
30. Which of the classes of RNA molecules carries the genetic information from the nucleus as it is needed for the construction of a protein? A. ribosomal RNA B. transfer RNA C. messenger RNA D. primary mRNA transcript E. mature RNA Ribsomal RNA is the site of complementary base pairing between tRNA anticodons and mRNA codons. Transfer RNA brings the amino acids to the ribosomes. Messenger RNA carries the genetic information as it is needed for the construction of a protein. The primary mRNA transcript is the initial form of mRNA found within the nucleus. The mature RNA is the RNA code that has had the introns cut out of it.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.01 Describe the roles of RNA molecules in gene expression. Section: 25.03 Topic: RNA Types
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Chapter 25 - DNA Structure and Control of Gene Expression
31. Which of the classes of RNA molecules carries the amino acids that are added to the growing polypeptide chain? A. ribosomal RNA B. transfer RNA C. messenger RNA D. primary mRNA transcript E. ribozyme Transfer RNA (tRNA) serves to bind and carry amino acids to the site of protein synthesis. The tRNA then binds its anticodon to the condon of the mRNA to ensure that the appropriate amino acid is joined into the growing polypeptide.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.01 Describe the roles of RNA molecules in gene expression. Section: 25.03 Topic: RNA Types
32. Which of the classes of RNA molecules is linked with proteins in forming the large and small subunits of a cytoplasmic structure? A. ribosomal RNA B. transfer RNA C. messenger RNA D. primary mRNA transcript E. ribozymes Translation is the process that makes a polypeptide. Once mRNA is transcribed from DNA, it is processed, and to yield its final functional form called mature mRNA and exits the nucleus. Outside the nucleus it binds to the small subunit of a ribosome and a tRNA-carrying methionine, which is the amino acid that matches the start codon. Next, the large subunit of the ribosome joins and translation of the mRNA into a polypeptide begins.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.01 Describe the roles of RNA molecules in gene expression. Section: 25.03 Topic: RNA Types
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Chapter 25 - DNA Structure and Control of Gene Expression
33. Prior to protein synthesis, the DNA A. attracts tRNAs with appropriate amino acids. B. serves as a template for the production of mRNA. C. adheres to ribosomes for protein synthesis. D. contains anticodons that become codons. E. must first undergo replication. Transcription is the manufacture of a complementary copy of the DNA as mRNA. One side of the DNA molecule serves as the template for mRNA transcription. The DNA is opened temporarily to allow RNA polymerase access to the template strand; once the polymerase has transcribed, a section immediately reforms hydrogen bonds with the nontemplate DNA strand.
Bloom’s Level: 3. Apply Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Gene Regulation: Eukaryotes
34. Artificial RNA with a known base sequence was added to a medium containing bacterial ribosomes and a mixture of amino acids. By chemical identification of the resulting polypeptide, researchers could discover the A. transcription factors involved. B. mutation rate. C. codon. D. Chargaff rule. E. difference between DNA and RNA. By allowing a known nucleotide sequence to be translated into a protein, it would then be possible to determine the amino acid that matches each of the predetermined codons.
Bloom’s Level: 4. Analyze Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Gene Regulation: Eukaryotes
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Chapter 25 - DNA Structure and Control of Gene Expression
35. In modern biochemical genetics, the flow of inherited information is from A. protein → RNA → DNA. B. DNA → RNA → protein. C. DNA → protein → RNA. D. RNA → DNA → protein. E. RNA → protein → DNA. Inherited information is stored in DNA, so that would be the starting point for the information passed down to generations. That information is then transcribed into mRNA, which is used to produce a protein with the aid of ribosomes and tRNA.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Gene Regulation: Eukaryotes
36. Transcription of part of a DNA molecule with a nucleotide sequence of AAACAACTT results in a mRNA molecule with the complementary sequence of A. GGGAGAACC. B. UUUGUUGAA. C. TTTGAAGCC. D. CCCACCTCC. E. AAACAACTT. When DNA is transcribed into mRNA it is done by matching complementary RNA nucleotides to the template strand of the DNA. The base-pairing rules are A binds to U and C to G. This is the same as it would be with the replication of DNA, except that uracil replaces thymine.
Bloom’s Level: 4. Analyze Learning Outcome: 25.03.03 Determine the sequence of amino acids in a peptide, given the messenger RNA sequence. Section: 25.03 Topic: Transcription
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Chapter 25 - DNA Structure and Control of Gene Expression
37. If one strand of DNA has the base sequence AAGCAA, the complementary strand has which of the following sequences? A. UUCGUU B. TTCGTT C. AAGCAA D. UTCGTU E. TTCGTG When DNA is replicated, it is done by matching complementary DNA nucleotides to the parent strand of the DNA. The base-pairing rules are A binds to T and C to G.
Bloom’s Level: 3. Apply Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
38. During the process of transcription, the information in A. a protein is converted into RNA information. B. RNA is converted into protein information. C. RNA is converted into DNA information. D. DNA is converted into RNA information. E. DNA is converted into protein information. Transcription is the manufacture of a complementary copy of the DNA as mRNA. One side of the DNA molecule serves as the template for mRNA transcription. The DNA is opened temporarily to allow RNA polymerase access to the template strand; once the polymerase has transcribed, a section immediately reforms hydrogen bonds with the nontemplate DNA. strand.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Transcription
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Chapter 25 - DNA Structure and Control of Gene Expression
39. An intervening sequence of DNA that is not expressed is called a(n) A. exon. B. intron. C. replicon. D. promoter. E. gene. During mRNA processing, segments of the mRNA that do not code for an amino acid necessary for the protein are removed. These noncoding sequences that interrupt the "message" of the mRNA are called introns.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.04 Explain the purpose of mRNA processing. Section: 25.03 Topic: RNA Processing
40. The A. introns B. exons C. promoter D. ribosomes E. nucleoli
contain(s) the information for the structure of the protein.
During mRNA processing, segments of the mRNA that do not code for an amino acid necessary for the protein are removed. These noncoding sequences that interrupt the "message" of the mRNA are called introns. The segments that are necessary for the formation of the correct protein are called exons.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.04 Explain the purpose of mRNA processing. Section: 25.03 Topic: RNA Processing
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Chapter 25 - DNA Structure and Control of Gene Expression
41. The addition of a poly-A tail to mRNA occurs after which of the following? A. replication B. transcription C. translation D. translocation E. termination Processing of the primary mRNA transcript includes the addition of nucleotides to each end of the transcript. A modified guanine is added to the 5’ end of the mRNA and a poly-A tail is added to the 3’ end. These additions serve to help orient the mRNA correctly on the ribosome and to prevent degradation of the mRNA so it can continue to function and produce multiple copies of the protein. Once these modifications have been made, the mRNA is mature and can leave the nucleus.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.04 Explain the purpose of mRNA processing. Section: 25.03 Topic: RNA Processing
42. Which is the process by which a protein is constructed in the cytoplasm of eukaryotic cells? A. translation B. transcription C. transposition D. transformation E. translocation Translation is the process that makes a polypeptide. Once mRNA is transcribed from DNA, it is processed to yield its final functional form called mature mRNA and exits the nucleus. Outside the nucleus, it binds to the small subunit of a ribosome and a tRNA carrying methionine, which is the amino acid that matches the start codon. Next, the large subunit of the ribosome joins and translation of the mRNA into a polypeptide begins. Once all of these components are in place, translation of the protein continues until the polypeptide is complete.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Translation
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Chapter 25 - DNA Structure and Control of Gene Expression
43. For translation to take place, which of the following would NOT be required to be present? A. DNA B. mRNA C. tRNA amino acid complex D. rRNA E. ribosome Once mRNA is transcribed from DNA, it is processed to yield its final functional form called mature mRNA and exits the nucleus. Outside the nucleus, it binds to a ribosome and a tRNA carrying methionine, which is the amino acid that matches the start codon. tRNA continue to bring amino acids to join into the growing polypeptide until the mRNA has been completely translated. The DNA is not used in this process because it never leaves the nucleus.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Translation
44. The genetic code consists of A. 1 B. 2 C. 3 D. 4 E. 20
bases that stand for one codon.
Codons consist of three nucleotides; they are sometimes called triplets. The order of the three nucleotides determines which amino acid appears in the polypeptide sequence.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.03 Determine the sequence of amino acids in a peptide, given the messenger RNA sequence. Section: 25.03 Topic: RNA Processing
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Chapter 25 - DNA Structure and Control of Gene Expression
45. Which is the process that synthesizes mRNA? A. translation B. transcription C. transposition D. transformation E. translocation Transcription is the manufacture of a complementary copy of the DNA as mRNA. One side of the DNA molecule serves as the template for mRNA transcription.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Transcription
46. Messenger RNA is produced in the A. cytoplasm. B. ribosomes. C. nucleus. D. endoplasmic reticulum. E. lysosome. Transcription is the manufacture of a complementary copy of the DNA as mRNA. One side of the DNA molecule serves as the template for mRNA transcription. This occurs in the nucleus of the cell, allowing DNA to be kept in conditions that are less likely to cause damage to the message it contains.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.04 Explain the purpose of mRNA processing. Section: 25.03 Topic: RNA Types
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Chapter 25 - DNA Structure and Control of Gene Expression
47. Which is most directly responsible for the sequence of amino acids in a protein? A. the sequence of the anticodons B. the number of codons in mRNA C. the enzyme that attaches the amino acid to tRNA D. the proteins associated with rRNA E. the sequence of codons in mRNA The codons of mRNA determine the amino acids that are joined into the polypeptide. If a change were to occur to the information of the mRNA so that one or more nucleotides were changed, then it would change the polypeptide.
Bloom’s Level: 5. Evaluate Learning Outcome: 25.03.03 Determine the sequence of amino acids in a peptide, given the messenger RNA sequence. Section: 25.03 Topic: RNA Processing
48. Which of the following processes does NOT take place during translation? A. attachment of a ribosome to mRNA B. growth of a polypeptide chain C. binding of two tRNA molecules/ribosome D. liberation of polypeptide from ribosome E. production of mRNA Once mRNA is transcribed from DNA, it exits the nucleus. Outside the nucleus, the process of translation begins when mRNA binds to the small subunit of a ribosome and a tRNA carrying methionine, which is the amino acid that matches the start codon. Next, the large subunit of the ribosome joins and translation of the mRNA into a polypeptide begins. Once all of these components are in place, translation of the protein continues until the polypeptide is complete.
Bloom’s Level: 4. Analyze Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Translation
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Chapter 25 - DNA Structure and Control of Gene Expression
49. Which is NOT true about the genetic code? A. It is exactly the same in all organisms. B. It is composed of a triplet code of three bases per codon. C. It produces 64 different possibilities of codon sequences. D. Some amino acids are coded for by more than one codon. E. It contains start and stop codons as instructions. While the genetic code of all living organisms is based on the same DNA structure and uses the same chemical code, the exact sequence of nucleotides varies from one organism to the next. Even organisms of the same species do not carry exactly the same sequence.
Bloom’s Level: 3. Apply Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Gene Regulation: Eukaryotes
50. Segments of DNA that are NOT part of the gene are called A. exons. B. introns. C. transposons. D. inducers. E. promoters. During mRNA processing, segments of the mRNA that do not code for an amino acid necessary for the protein are removed. These noncoding sequences that interrupt the "message" of the mRNA are called introns. The segments that are necessary for the formation of the correct protein are called exons.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.04 Explain the purpose of mRNA processing. Section: 25.03 Topic: RNA Processing
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Chapter 25 - DNA Structure and Control of Gene Expression
51. The of a tRNA molecule will base-pair with the codon of a mRNA molecule. A. DNA sequence B. anticodon C. amino acid binding site D. RNA polymerase E. promoter site tRNA has one end that binds a specific amino acid, while the other end contains an anticodon, the opposite of the codon that is located on the mRNA. The anitcodon binds to the codon on the mRNA, bringing the attached amino acid into proximity of the growing polypeptide chain and the new amino acid is joined into the amino acid sequence.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.01 Describe the roles of RNA molecules in gene expression. Section: 25.03 Topic: RNA Types
52. Initiation, elongation, and termination are three stages in A. DNA replication. B. error correction by proofreading enzymes. C. mRNA processing. D. translation. E. intron removal. Initiation occurs when the mRNA, ribosome, and tRNA all come together. Elongation is the adding in of amino acids until the stop codon is reached. Once the stop codon is reached and a termination factor binds to the mRNA, the entire apparatus disassembles and the polypeptide is released. These are all steps in translation.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Translation
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Chapter 25 - DNA Structure and Control of Gene Expression
53. The correct sequence of events in the production of a polypeptide is A. initiation → termination → elongation. B. elongation → termination → initiation. C. termination → elongation → initiation. D. elongation → initiation → termination. E. initiation → elongation → termination. Initiation occurs when the mRNA, ribosome, and tRNA all come together. Elongation is the adding in of amino acids until the stop codon is reached. Once the stop codon is reached and a termination factor binds to the mRNA, the entire apparatus disassembles and the polypeptide is released.
Bloom’s Level: 2. Understand Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Gene Regulation: Eukaryotes
54. The site of translation is the A. nucleus. B. nucleolus. C. ribosome. D. ribozyme. E. mitochondria. Translation is the manufacture of a polypeptide using mRNA, tRNA, and ribosomes. mRNA and tRNA bind to the ribosome, which is the site of protein synthesis.
Bloom’s Level: 1. Remember Learning Outcome: 25.03.02 Summarize the sequence of events that occurs during gene expression. Section: 25.03 Topic: Translation
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Chapter 25 - DNA Structure and Control of Gene Expression
55. Gene mutations are A. always deleterious. B. always beneficial. C. radiation-induced changes only. D. alterations in the normal sequence of bases within a gene. E. alterations in the normal sequence of bases outside a gene. Mutations are defined as changes in the DNA. These changes can be neutral if they do not change the resulting polypeptide, harmful if they result in a nonfunctional protein, or beneficial if they lead to a molecule that increases the fitness of the organism. Most mutations fall into the category of neutral changes.
Bloom’s Level: 1. Remember Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Mutations
56. If the normal nucleotide sequence was TACGGCATG, what type of gene mutation is present if the resulting sequence becomes TAGGCATG? A. chromosomal mutation B. germinal mutation C. addition mutation D. deletion mutation E. substitution mutation Since the third nucleotide in the sequence is missing from the altered sequence, this is a deletion. This would result in a frameshift mutation and all of the codons after this point would be changed.
Bloom’s Level: 5. Evaluate Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Mutations
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Chapter 25 - DNA Structure and Control of Gene Expression
57. Which of the following is NOT a frameshift mutation of the nucleotide sequence CATUAUCCC? A. CATTUAUCCC B. ATUAUCCC C. CTUAUCCC D. CATUAUCGC E. CCATUAUCCC In the sequence CATUAUCGC, the difference between the original CATUAUCCC is a substitution of G for C in the last triplet. This would not result in a frameshift since no new nucleotide was inserted or deleted.
Bloom’s Level: 5. Evaluate Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Mutations
58. Specific DNA sequences that have the ability to move within and out of chromosomes are called A. transposons. B. introns. C. exons. D. operators. E. ribozymes. Transposons are specific DNA sequences that have the remarkable ability to move within and between chromosomes. Their movement to a new location sometimes alters neighboring genes, particularly by increasing or decreasing their expression.
Bloom’s Level: 1. Remember Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Mutations
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Chapter 25 - DNA Structure and Control of Gene Expression
59. Which of these is NOT a form of a point mutation? A. nonsense mutation B. missense mutation C. silent mutation D. frameshift mutation E. UAC becomes UAU Point mutations involve a change in a single DNA nucleotide and, therefore, a possible change in a specific amino acid. They can result in missense, nonsense, or silent mutations. A frameshift occurs when a single nucleotide is added or removed from a nucleotide sequence, resulting in a change in the way that nucleotides are grouped into triples and read as codons.
Bloom’s Level: 2. Understand Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Mutations
60. Which of the following does NOT occur when lactose is present in the environment of E. coli? A. The lac repressor binds to the operator. B. The RNA polymerase binds to the promoter. C. Three different genes are transcribed. D. The repressor binds to lactose. E. The enzymes that break down lactose are synthesized. In the case of the lac operon, the structural genes code for enzymes that metabolize lactose. When lactose is absent, the repressor binds to the operator and shuts off the operon. When lactose is present, it binds to the repressor so that it can no longer bind to the operator, and this allows RNA polymerase to transcribe the structural genes.
Bloom’s Level: 3. Apply Learning Outcome: 25.04.01 Describe the relationship between gene regulation and gene activity in a cell. Section: 25.04 Topic: Gene Regulation: Prokaryotes
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Chapter 25 - DNA Structure and Control of Gene Expression
61. Which part of an operon is incorrectly matched with its function? A. promoter—where RNA polymerase first binds to DNA B. regulator—binds to the repressor protein C. structural—makes mRNA by transcription D. operator—if unbound, allows RNA polymerase to bind to DNA E. All of the answer choices are incorrectly matched with its function. A protein called a repressor binds to the operator. When the repressor is bound to the operator, RNA polymerase cannot transcribe the three structural genes of the operon. The repressor is encoded by a regulator gene located outside of the operon.
Bloom’s Level: 3. Apply Learning Outcome: 25.04.02 Compare regulation in prokaryotes to regulation in eukaryotes. Section: 25.04 Topic: Gene Regulation: Prokaryotes
62. Two strains of E. coli—one of which can turn on its lactase production and one that cannot—are grown together with lactose in the medium. If all other conditions are equal, what will occur? A. Both strains will continue to be present in equal amounts. B. The two strains will kill each other off. C. The strain that can turn on lactase production will overgrow the culture and be the only one left. D. The strain that cannot turn on lactase production will overgrow the culture and be the only one left. E. The strain that can turn on lactase production will make its own lactose once the lactose present in the medium is completely used. The metabolism of lactose in E. coli is dependent on the ability of the cell to produce the enzyme lactase. Since one bacterium will be able to use this resource by turning on the gene for lactase and one will not be able to do so, the bacterium with lactase will outcompete the other and will be able to continue growing beyond any other food supplied in the medium.
Bloom’s Level: 5. Evaluate Learning Outcome: 25.04.02 Compare regulation in prokaryotes to regulation in eukaryotes. Section: 25.04 Topic: Gene Regulation: Prokaryotes
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Chapter 25 - DNA Structure and Control of Gene Expression
63. The function of the regulator gene is to A. code for enzymes in a metabolic pathway. B. serve as an on/off switch for transcription. C. code for a repressor molecule that can bind to the operator. D. bind to the RNA polymerase molecule. E. bind to the promoter. Regulator genes are located outside the operon and code for a protein that will bind to the operator close to the promoter region where RNA polymerase must bind to initiate transcription of the gene. This protein is a repressor that keeps the RNA polymerase from binding and transcribing the gene.
Bloom’s Level: 3. Apply Learning Outcome: 25.04.01 Describe the relationship between gene regulation and gene activity in a cell. Section: 25.04 Topic: Gene Regulation: Prokaryotes
64. Active genes are found in the A. heterochromatin B. euchromatin C. Barr bodies D. promoters E. operons
of eukaryotic chromosomes.
Active genes in eukaryotic cells are associated with more loosely packed chromatin, called euchromatin. However, even euchromatin must be “unpacked” before it can be transcribed. The presence of nucleosomes limits access to the DNA by the transcription machinery. A chromatin remodeling complex pushes the nucleosomes aside to open up sections of DNA for expression.
Bloom’s Level: 2. Understand Learning Outcome: 25.04.02 Compare regulation in prokaryotes to regulation in eukaryotes. Section: 25.04 Topic: Gene Regulation: Eukaryotes
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Chapter 25 - DNA Structure and Control of Gene Expression
65. regulation occurs within a cell when the newly formed protein is modified. A. Transcriptional B. Posttranscriptional C. Translational D. Posttranslational E. Operon Posttranslational modifications of proteins allow final changes to be made to the gene product. These modifications include additions of phosphates, lipids, and carbohydrates; removal of some of the amino acids that were originally part of the polypeptide; and the joining of two or more polypeptides to give the protein quaternary structure.
Bloom’s Level: 2. Understand Learning Outcome: 25.04.03 Discuss the many ways genes are regulated in eukaryotes. Section: 25.04 Topic: Gene Regulation: Eukaryotes
66. In eukaryotes, control of gene expression is NOT by A. translational control in cytoplasm after mRNA leaves the nucleus and before protein is produced. B. transcriptional control in the nucleus based on which genes are transcribed and how fast they are transcribed. C. posttranscriptional control in the nucleus after DNA is transcribed, including the speed with which mRNA leaves the nucleus. D. posttranslational control in the cytoplasm that occurs after protein synthesis. E. the use of repressor proteins and operators. In eukaryotes genes are not arranged into operons so operators and repressor proteins would not allow for efficient control of genes. Eukaryotes do use a system of controls that determine what genes are turned on or whether or not they are available to be expressed.
Bloom’s Level: 2. Understand Learning Outcome: 25.04.03 Discuss the many ways genes are regulated in eukaryotes. Section: 25.04 Topic: Gene Regulation: Eukaryotes
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Chapter 25 - DNA Structure and Control of Gene Expression
67. Which of the following levels of control involves the use of a poly-A tail? A. protein activity B. translation of mRNA C. mRNA processing D. transcriptional control E. chromatin structure The poly-A tail is added to mRNA before it is allowed to leave the nucleus. This tail allows the mRNA transcript to be used repeatedly before it begins to degrade. Without the poly-A tail, the gene would need to be transcribed more often or the amount of protein made would be reduced.
Bloom’s Level: 4. Analyze Learning Outcome: 25.04.03 Discuss the many ways genes are regulated in eukaryotes. Section: 25.04 Topic: Gene Regulation: Eukaryotes
68. Which level of primary control in eukaryotic gene activity involves the life span of the mRNA molecule and the ability of the mRNA to bind to ribosomes? A. feedback control B. translational control C. transcriptional control D. posttranscriptional control E. posttranslational control Translational control is based on the length of time that a transcript can be used to take part in translation. The longer an mRNA remains in the cytoplasm before it is broken down, the more gene product can be translated. Differences in the poly-A tail or the guanine cap can determine how long a particular transcript remains active before it is destroyed by a ribonuclease associated with ribosomes.
Bloom’s Level: 4. Analyze Learning Outcome: 25.04.03 Discuss the many ways genes are regulated in eukaryotes. Section: 25.04 Topic: Gene Regulation: Eukaryotes
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Chapter 25 - DNA Structure and Control of Gene Expression
69. Which level of primary control in eukaryotic gene activity involves processing early RNA transcripts to mRNA and control of the rate at which mRNA leaves the nucleus? A. feedback control B. translational control C. transcriptional control D. posttranscriptional control E. posttranslational control The primary mRNA is converted to the mature mRNA by the addition of a poly-A tail and a guanine cap, and by the removal of the introns and splicing back together of the exons. The same mRNA can be spliced in different ways to make slightly different products in different tissues. The speed of transport of mRNA from the nucleus into the cytoplasm can ultimately affect the amount of gene product following transcription. There is a difference in the length of time it takes various mRNA molecules to pass through a nuclear pore.
Bloom’s Level: 4. Analyze Learning Outcome: 25.04.03 Discuss the many ways genes are regulated in eukaryotes. Section: 25.04 Topic: Gene Regulation: Eukaryotes
70. Which level of primary control in eukaryotic gene activity involves changes in the polypeptide chain before it becomes functional? A. feedback control B. translational control C. transcriptional control D. posttranscriptional control E. posttranslational control Some proteins are not active immediately after synthesis. Posttranslational modifications of proteins allow final changes to be made to the gene product. These modifications include additions of phosphates, lipids. and carbohydrates; removal of some of the amino acids that were originally part of the polypeptide; and the joining of two or more polypeptides to give the protein quaternary structure.
Bloom’s Level: 2. Understand Learning Outcome: 25.04.03 Discuss the many ways genes are regulated in eukaryotes. Section: 25.04 Topic: Gene Regulation: Eukaryotes
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Chapter 25 - DNA Structure and Control of Gene Expression
71. If the gene product needed is either transfer RNA or ribosomal RNA, the most efficient way to produce this product is A. activating chromatin. B. increasing transcription factors. C. increasing the rate mRNA matures. D. increasing translational control. E. increasing posttranscriptional control. Eukaryotic genes exhibit control of gene expression at five different levels: 1) pretranscriptional control, 2) transcriptional control, 3) posttranscriptional control, 4) translational control, and 5) posttranslational control. In pretranscriptional control, eukaryotes use DNA methylation and chromatin packing as a way to keep genes turned off. Genes within highly condensed portions of chromatin are called heterochromatin, and are inactive. Active genes in eukaryotic cells are associated with more loosely packed chromatin, called euchromatin. However, even euchromatin must be “unpacked” before it can be transcribed. Since tRNA and rRNA do not produce mRNA or a protein, posttranscriptional control, translational control, and posttranslational control would not be possible.
Bloom’s Level: 5. Evaluate Learning Outcome: 25.04.03 Discuss the many ways genes are regulated in eukaryotes. Section: 25.04 Topic: Gene Regulation: Eukaryotes
72. Eukaryotic transcription factors assist the A. repressor; operator B. RNA polymerase; promoter C. DNA polymerase; promoter D. helicases; operator E. operator; repressor
in binding to the
.
Eukaryotic transcriptional control is dependent on the interaction of proteins with particular DNA sequences. The proteins are called transcription factors and activators, while the DNA sequences are called enhancers and promoters. In eukaryotes, each gene has its own promoter. Transcription factors are proteins that help RNA polymerase bind to a promoter.
Bloom’s Level: 2. Understand Learning Outcome: 25.04.03 Discuss the many ways genes are regulated in eukaryotes. Section: 25.04 Topic: Gene Regulation: Eukaryotes
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Chapter 25 - DNA Structure and Control of Gene Expression
73. bind to enhancers in eukaryotic cells to increase the rate of transcription. A. Repressors B. RNA polymerases C. Ligases D. Transcription activators E. DNA polymerases Eukaryotic transcriptional control is dependent on the interaction of proteins with particular DNA sequences. The proteins are called transcription factors and activators, while the DNA sequences are called enhancers and promoters; each gene has its own promoter. Transcription factors are proteins that help RNA polymerase bind to a promoter. Transcription activators are proteins that speed transcription dramatically by binding to a DNA region called an enhancer and stimulating transcription. In general, transcription activators are themselves activated in response to a signal received and transmitted by a signal transduction pathway.
Bloom’s Level: 2. Understand Learning Outcome: 25.04.03 Discuss the many ways genes are regulated in eukaryotes. Section: 25.04 Topic: Gene Regulation: Eukaryotes
74. Which of the following is NOT true about cancer cells? A. They exhibit uncontrolled growth. B. They can form benign or malignant tumors. C. They exhibit disorganized growth. D. They exhibit contact inhibition. E. They may undergo metastasis. Normal cells exhibit contact inhibition, which limits cell division when they are crowded together. This may be due to a lack of growth factors that signal a cell to begin dividing. One theory is that cancer cells manufacture their own growth factor, thus constantly signaling to themselves to continue dividing.
Bloom’s Level: 3. Apply Learning Outcome: 25.05.03 Describe the characteristics of cancer cells. Section: 25.05 Topic: Mutations
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Chapter 25 - DNA Structure and Control of Gene Expression
75. Which is NOT associated with cancer? A. It can establish new tumors distant from the site of the primary tumor. B. It forms new blood vessels and brings nutrients and oxygen to the tumor. C. Cancer cells can undergo cell division repeatedly and indefinitely. D. Tumors invade surrounding tissues and are filtered by lymph nodes. E. A malignant mass of cells is encapsulated and does not invade adjacent tissue. Many of the changes that must occur for cancer cells to move throughout the body and form new tumors are not understood. It seems that cancer cells are able to detach from their normal location and travel through the blood and lymphatic systems to new areas of the body which they then invade and form tumors.
Bloom’s Level: 3. Apply Learning Outcome: 25.05.03 Describe the characteristics of cancer cells. Section: 25.05 Topic: Mutations
76. Apoptosis refers to cell death and A. is always biologically detrimental to an organism. B. can be programmed and is essential to normal development. C. is the accumulation of genetic errors. D. is a failure in the translation or transcription mechanism. E. is any failure of the genetic machinery to work correctly. A cell that has genetic damage or problems with the cell cycle will initiate apoptosis, or programmed cell death. Also, cells from the immune system, when they detect an abnormal cell, will send signals to that cell, inducing apoptosis.
Bloom’s Level: 2. Understand Learning Outcome: 25.05.02 Describe why cancer is a failure of genetic control. Section: 25.05 Topic: Mutations
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Chapter 25 - DNA Structure and Control of Gene Expression
77. DNA segments called are at the ends of chromosomes and as they shorten they eventually signal the cell to enter apoptosis. A. transposons B. mutagens C. activated chromatin D. operons E. telomeres Most normal cells have a built-in limit to the number of times they can divide before they die. One of the reasons normal cells stop entering the cell cycle is that the telomeres become shortened. Telomeres are sequences at the ends of the chromosomes that protect the genetic information by allowing cells to go through repeated divisions without loss of genes. With each cell division, the telomeres shorten, eventually becoming short enough to signal apoptosis.
Bloom’s Level: 1. Remember Learning Outcome: 25.05.03 Describe the characteristics of cancer cells. Section: 25.05 Topic: Chromosome Structure
78. When the sheep Dolly was successfully cloned, it was produced by growing an in vitro fertilized egg where the normal egg nucleus had been removed and replaced by a nucleus from an adult. Since this nucleus is from an old mature animal, we would expect it to . Interestingly, tests show that this did not happen, a fact that currently puzzles researchers. A. have additional Barr bodies B. be mutated C. have shorter telomeres D. have longer telomeres E. express transcription and translation more rapidly In an older cell the telomeres should be shorter since the cell has gone through repeated cell divisions. Gametes typically have long telomeres on their DNA because germ cells have not gone through repeated cell divisions. This is important since the zygote will need to undergo repeated cell division to develop into a fetus and then into an adult.
Bloom’s Level: 5. Evaluate Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Chromosome Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
79. Which of the following is NOT true about oncogenes? A. They are normal genes that have undergone a mutation. B. They cause growth factors or growth factor receptors on a cell to malfunction. C. When mutated, oncogenes turn into proto-oncogenes. D. They cause a cell to divide repeatedly. E. They are not alien to the cell. Proto-oncogenes stimulate cell division but are usually turned off in fully differentiated nondividing cells. When proto-oncogenes mutate, they become oncogenes that are active all the time. Proto-oncogenes code for transcription factors or proteins that control transcription factors.
Bloom’s Level: 3. Apply Learning Outcome: 25.05.02 Describe why cancer is a failure of genetic control. Section: 25.05 Topic: Mutations
80. Which of the following serve to directly accelerate the cell cycle? A. tumor suppressor genes B. carcinogens C. repressors D. oncogene proteins E. p53 gene Tumor suppressor genes help block the growth of tumors. Carcinogens are substances that can cause normal cells to behave differently. Repressors regulate the expression of genes. Proto-oncogenes stimulate cell division but are usually turned off in fully differentiated nondividing cells. The p53 gene acts as a transcription factor that turns on the expression of genes.
Bloom’s Level: 2. Understand Learning Outcome: 25.05.03 Describe the characteristics of cancer cells. Section: 25.05 Topic: Chromosome Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
81. Tumor suppressor genes are A. oncogenes that code for proteins that inhibit cell division. B. anti-oncogenes that code for proteins that inhibit cell division. C. oncogenes that code for proteins that stimulate cell division. D. anti-oncogenes that code for proteins that stimulate cell division. E. transposons that can be used as cancer therapy. Tumor supressor genes can be thought of as anti-oncogenes since they put the brakes on cell division, especially if it is abnormal growth; they code for transcription factors or proteins that control transcription factors. Oncogenes, on the other hand, continually stimulate cell growth.
Bloom’s Level: 3. Apply Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Mutations
82. An oncogene is A. a viral gene with no relation to the host cells genes. B. a mutated form of a proto-oncogene. C. a bacterial gene that causes cancer in the host. D. always seen in human cancer cells. E. a gene that turns off cellular reproduction. Proto-oncogenes stimulate cell division but are usually turned off in fully differentiated nondividing cells. When proto-oncogenes mutate, they become oncogenes that are active all the time constantly stimulating cell division.
Bloom’s Level: 2. Understand Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Mutations
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Chapter 25 - DNA Structure and Control of Gene Expression
83. Which of the following is NOT associated with a higher cancer risk? A. hormone therapy B. safe sex C. smoking D. use of alcohol E. exposure to certain hazardous chemicals in the workplace Hormone therapy, smoking, excessive alcohol use, exposure to hazardous chemicals, as well as unprotected sex can expose a person to agents that can cause mutations in DNA that will lead to cancerous growths. Practicing safe sex would prevent exposure to human papillomaviruses (HPVs).
Bloom’s Level: 2. Understand Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Mutations
84. Which of the following behaviors will NOT help prevent cancer? A. avoid carcinogenic chemicals B. avoid excessive sunlight C. lower total fat intake D. decrease intake of high-fiber foods E. increase consumption of broccoli High-fiber foods would be part of a healthy diet that would assist in preventing the mutations that lead to cancer.
Bloom’s Level: 2. Understand Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Mutations
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Chapter 25 - DNA Structure and Control of Gene Expression
85. Which of the following will NOT assist you in protecting yourself from cancer? A. use of alcohol B. eating vitamin-rich foods C. reducing the amount of salt-cured and nitrate-cured foods D. reducing your weight E. regular exercise Alcohol should be consumed in moderation. Cancers of the mouth, throat, esophagus, larynx, and liver occur more frequently among heavy drinkers, especially when accompanied by smoking cigarettes or chewing tobacco.
Bloom’s Level: 2. Understand Learning Outcome: 25.05.01 Summarize the causes of gene mutations. Section: 25.05 Topic: Mutations
86. A cluster of prokaryotic genes under the control of one regulatory sequence is called a(n) A. repressor. B. promoter. C. operon. D. operator. E. regulator gene. In prokaryotes, an operon is a cluster of genes usually coding for proteins, related to a particular metabolic pathway, along with the short DNA sequences that coordinately control their transcription. The control sequences consist of a promoter, a sequence of DNA where RNA polymerase first attaches to begin transcription, and an operator, a sequence of DNA where a repressor protein binds.
Bloom’s Level: 1. Remember Learning Outcome: 25.04.02 Compare regulation in prokaryotes to regulation in eukaryotes. Section: 25.04 Topic: Gene Regulation: Prokaryotes
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Chapter 25 - DNA Structure and Control of Gene Expression
87. Which of the following is NOT true of ribosomes? A. They are produced in the nucleus of a eukaryotic cell. B. They are made of RNA and proteins. C. They are composed of two subunits that only come together when protein synthesis is about to occur. D. They bind only one type of amino acid. E. DNA serves as a template for rRNA. Transfer RNA binds only one specific type of amino acid. Ribosomes do not actually bind amino acids, rather they bind to mRNA which in turn forms temporary bonds with tRNAcarrying amino acids. The amino acids form covalent bonds with one another as they are brought into proximity to one another.
Bloom’s Level: 3. Apply Learning Outcome: 25.03.01 Describe the roles of RNA molecules in gene expression. Section: 25.03 Topic: RNA Types
88. Of the following types of RNA, which is one of the determining factors in cell differentiation? A. tRNA B. rRNA C. mRNA D. No type of RNA is a determing factor in cell differentiation. E. cRNA All cell must contain rRNA and all forms of tRNA for protein synthesis to occur. The transcription of mRNA from DNA is a means of controlling which genes are expressed in each cell. Therefore, the mRNA produced by different cell types is different and determines the functions carried out by the cell.
Bloom’s Level: 3. Apply Learning Outcome: 25.03.01 Describe the roles of RNA molecules in gene expression. Section: 25.03 Topic: RNA Types
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Chapter 25 - DNA Structure and Control of Gene Expression
89. A cell of the pancreas would specialize in producing insulin, among other functions, while a muscle cell would specialize in the production of the protein myosin. This means that A. pancreas and muscle cells have the genes for both insulin and myosin active. B. the active gene in pancreas cells is the one for insulin, while the muscle cells have genes for insulin and myosin. C. pPancreas cells activate the gene that would produce insulin, while muscle cells activate the gene that would produce myosin. D. neither pancreas cells nor muscle cells are specialized. E. only muscle cells can produce proteins. While cells do have molecules that they specialize in producing, they do still need to make other molecules. Pancreas cells specialize in producing insulin so that gene is active in these cells and muscle cell specialize in the production of myosin so the gene for myosin would be active in them.
Bloom’s Level: 4. Analyze Learning Outcome: 25.04.01 Describe the relationship between gene regulation and gene activity in a cell. Section: 25.04 Topic: Gene Regulation: Eukaryotes
Short Answer Questions 90. Indicate the complementary nucleotide sequence for the following section of DNA: AATGCGTATACG DNA forms a complementary sequence, which means for each A (adenine) it will pair bond to T (thymine). For each G (guanine) it will pair bond to C (cytosine). The complementary nucleotide sequence would be TTACGCATATGC.
Bloom's Level: 6. Create Learning Outcome: 25.01.02 Describe the structure of a DNA molecule. Section: 25.01 Topic: DNA Structure
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Chapter 25 - DNA Structure and Control of Gene Expression
Multiple Choice Questions 91. Which factor is likely to lead to a change in the genetic control of the individual and possibly cancer? A. X-rays B. ultraviolet radiation C. cigarette smoke D. pesticides E. All of the answer choices can lead to cancer. All of the answer choices can cause mutations to occur in the cell, which can lead to cancer.
Bloom’s Level: 1. Remember Learning Outcome: 25.05.02 Describe why cancer is a failure of genetic control. Section: 25.05 Topic: Mutations
Short Answer Questions 92. Identify the amino acids that would be assembled based upon the following codons: CGG AUA GAC GUA. CGG would code for arginine. AUA would code for isoleucine. GAC would code for asparatate. GUA would code for valine.
Bloom's Level: 6. Create Learning Outcome: 25.03.03 Determine the sequence of amino acids in a peptide, given the messenger RNA sequence. Section: 25.03 Topic: RNA Processing
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Chapter 25 - DNA Structure and Control of Gene Expression
93. Summarize the process of DNA replication. 1. The enzyme DNA helicase unwinds and unzips the DNA strand. It breaks the hydrogen bonds between the bases. 2. New complementary DNA nucleotides are attached to both sides of the strand by the DNA polymerase. 3. The DNA polymerase will add the new nucleotides in the 5' to 3' directions. This produces a leading strand and a lagging strand that is composed of short segments of the DNA. 4. DNA ligase connects the short strands and seals any breaks in the sugar phosphate backbones. 5. The two double helix molecules are identical to each other.
Bloom's Level: 6. Create Learning Outcome: 25.02.02 Summarize the events that occur during the process of DNA replication. Section: 25.02 Topic: DNA Replication
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Chapter 26 - Biotechnology and Genomics
Chapter 26 Biotechnology and Genomics
Multiple Choice Questions 1. When human DNA is inserted into a bacterial plasmid, the resulting bacterium will then A. produce the products of the gene that has been inserted into the bacteria. B. be placed into a vaccine and used to fight off viruses that attack humans. C. be used to study the evolutionary relationship between humans and bacteria. D. produce restriction enzymes that will help fight off viruses that attack humans. E. be used to create DNA to insert the human genome. Transgenic bacteria are used to produce molecules that can be used in pharmaceuticals or in manufacture. Cellular machinery will produce the molecules that are encoded in the inserted DNA.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.01 Describe the steps in forming recombinant DNA. Section: 26.01 Topic: DNA Technology
2. The molecule DNA ligase that is used to seal the piece of foreign DNA into the vector is A. created for each specific plasmid. B. created for each specific foreign DNA to be inserted. C. naturally occurring in humans and is removed from humans and placed in the bacteria to create rDNA. D. naturally occurring in both humans and bacteria and functions in DNA replication. E. genetically engineered for each experiment. Ligase is a molecule that occurs naturally in all cells and functions in the normal replication of DNA.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.01 Describe the steps in forming recombinant DNA. Section: 26.01 Topic: DNA Technology
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Chapter 26 - Biotechnology and Genomics
3. Why must an engineered DNA polymerase be used in PCR and not the one that occurs naturally? A. It needs to work at very cold temperatures. B. It needs to attach to DNA without a primer. C. It must work in the 3' to 5' direction, adding nucleotides to the 5' end. D. It must work on single-stranded DNA at high temperatures and attach to a primer. E. It must attach without a primer, work in the 3' to 5' direction and function at very cold temperatures. Naturally occurring DNA polymerase attaches to an RNA primer on the DNA and functions at body temperature.
Bloom’s Level: 5. Evaluate Learning Outcome: 26.01.02 Discuss how the polymerase chain reaction works. Section: 26.01 Topic: DNA Technology
4. The purpose of PCR is to create A. fragments of DNA that are different lengths. B. recombinant DNA. C. billions of copies of a segment of DNA. D. plasmids to be used as vectors in cloning. E. transgenic crop plants. Polymerase chain reaction is used to take very small amounts of DNA and to replicate it to produce a sufficient amount of DNA to use for testing. This is especially useful in forensics when very small quantities may be all that can be found at a crime scene.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.02 Discuss how the polymerase chain reaction works. Section: 26.01 Topic: DNA Technology
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Chapter 26 - Biotechnology and Genomics
5. PCR is often used in forensic analysis since A. it requires no specialized equipment. B. untrained personnel can carry out the procedure. C. very small samples are frequently all that are available. D. juries always trust the method. E. it has been proven to determine the criminal. Polymerase chain reaction is used to take very small amounts of DNA and to replicate it to produce a sufficient amount of DNA to use for testing. This is especially useful in forensics when very small quantities may be all that can be found at a crime scene.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.02 Discuss how the polymerase chain reaction works. Section: 26.01 Topic: DNA Technology
6. Which of the following is NOT a use of DNA fingerprinting? A. paternity tests B. identifying Superbowl footballs C. forensic analysis D. determining a predisposition to cancer E. All of the answer choices are uses of DNA fingerprinting. DNA fingerprinting can be used to compare DNA from any two or more samples to determine if they are from the same source or, in the case of paternity tests, if the sources of the two samples are related.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.03 Explain what is meant by a DNA “fingerprint.” Section: 26.01 Topic: DNA Technology
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Chapter 26 - Biotechnology and Genomics
7. In STR, short tandem repeat, profiling, both are used to identify an individual. A. length or the repeated sequence; number of fragments B. rDNA; tRNA C. specific genetic sequences; types of mRNA D. mitochondrial DNA; number of fragments E. mitochondrial DNA; specific genetic sequences
and
STR is a technique where short repetitive sequences of DNA are replicated through PCR, and then labeled with fluorescence that can be detected by an automated system. The system looks for the length of the repeated sequence, which differs because each person has their own unique number of repeats. It also looks for whether there are one or two lengths involved. If there is only one length, then the person is homozygous for the trait or two different lengths if the person is heterozygous for the trait.
Bloom’s Level: 3. Apply Learning Outcome: 26.01.03 Explain what is meant by a DNA “fingerprint.” Section: 26.01 Topic: DNA Technology
8. In producing transgenic plants, cells called protoplasts are often used. Protoplasts are plant cells from which the cell wall has been removed. Why would the cell wall need to be removed? A. The foreign DNA enters the cell through self-healing holes in the membrane but it could not pass through the cell wall. B. The cell wall prevents the transgenic cell from growing into a mature plant. C. The cell wall makes it difficult for the biotechnician to determine if the foreign DNA has been incorporated into the cell’s genome. D. The cell wall prevents the microinjection of foreign DNA into the cell because the needle cannot pass through it. E. The foreign DNA passes through the plasmodesmata of the plant cell, which is blocked by the cell wall. The protoplast is treated with an electric current and causes holes to develop in the membrane. These holes are self-healing and allow the uptake of foreign DNA that is mixed into the solution containing the protoplasts.
Bloom’s Level: 5. Evaluate Learning Outcome: 26.02.01 Identify ways in which human society benefits from genetically modified bacteria, plants, and animals. Section: 26.02 Topic: Genetically Modified Organisms
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Chapter 26 - Biotechnology and Genomics
9. The two techniques typically used to create transgenic animals are A. microinjection of eggs and vortex mixing of eggs. B. microinjection of eggs and electroshock of eggs. C. electroshock of protoplasts and electroshock of eggs. D. microinjection of protoplasts and electroshock of eggs. E. microinjection of eggs and vortex mixing of protoplasts. The eggs of animals can be directly microinjected with foreign DNA by hand or the eggs can be agitated by a vortex mixer in a solution containing foreign DNA. The eggs are pricked by needles that disrupt the membrane and allow the DNA to enter.
Bloom’s Level: 1. Remember Learning Outcome: 26.02.02 Describe the steps involved in the production of a transgenic animal. Section: 26.02 Topic: Genetically Modified Organisms
10. When creating transgenic organisms, one major difference between plants and animals is that A. only embryos can be used to create transgenic animals, while any plant cell can be used for transgenic plants. B. only eggs can be used to create transgenic animals, while embryos can be used for transgenic plants. C. only eggs can be used to create transgenic animals, while any plant cell can be used for transgenic plants. D. only embryos can be used to create transgenic animals, while eggs must be used to create transgenic plants. E. any cell may be used to create transgenic animals, while embryos can be used for transgenic plants. The creation of transgenic animals can only be accomplished by using eggs and then placing them in a host to carry them to term. Transgenic plants and bacteria can be grown from any cell and do not require a host to allow the cell to develop into a viable organism.
Bloom’s Level: 4. Analyze Learning Outcome: 26.02.02 Describe the steps involved in the production of a transgenic animal. Section: 26.02 Topic: Genetically Modified Organisms
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Chapter 26 - Biotechnology and Genomics
11. When creating transgenic bacteria, plants, and animals A. any cell may be used as long as is does not have a cell wall. B. only eggs may be used for animals, while any cell may be used for plants and bacteria. C. only eggs may be used for plants and animals, while any cell may be used for bacteria. D. only eggs may be used to create any transgenic organism. E. transgenic bacteria are created first and then used to create transgenic plants and animals. The creation of transgenic animals can only be accomplished by using eggs and then placing them in a host to carry them to term. Transgenic plants and bacteria can be grown from any cell and do not require a host to allow the cell to develop into a viable organism.
Bloom’s Level: 4. Analyze Learning Outcome: 26.02.02 Describe the steps involved in the production of a transgenic animal. Section: 26.02 Topic: Genetically Modified Organisms
12. Transgenic crop plants have been created that do all of the following except A. grow two kinds of crops, like the pomato which produces both tomatoes and potatoes. B. are resistant to insect damage. C. are resistant to herbicides. D. produce human hormones or antibodies. E. transgenic plants have been produced that can do all of the things listed. Transgenic plants have been created that can carry the DNA for a wide variety of traits including all of the above and many others. Crop plants have been the subject of intensive experimentation in transgenic techniques, many of which are then patented.
Bloom’s Level: 2. Understand Learning Outcome: 26.02.01 Identify ways in which human society benefits from genetically modified bacteria, plants, and animals. Section: 26.02 Topic: Genetically Modified Organisms
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Chapter 26 - Biotechnology and Genomics
13. Gene pharming is A. the use of bacterial DNA to impart drug resistance to plants. B. the use of transgenic farm animals to produce pharmaceuticals. C. the creation of new genes to be implanted in farm animals. D. the creation of crop plants to replace the use of farm animals as a source of protein. E. the application of genetic principles to breeding farm animals. The addition of foreign DNA to the egg of an animal that is then used to produce molecules to be used in the pharmaceutical industry is gene pharming.
Bloom’s Level: 1. Remember Learning Outcome: 26.02.01 Identify ways in which human society benefits from genetically modified bacteria, plants, and animals. Section: 26.02 Topic: Genetically Modified Organisms
14. Goats and cattle are used in gene pharming because A. they are the easiest farm animals to clone. B. therapeutic and diagnostic proteins are produced in the milk of these animals and can be harvested for use. C. the diagnostic and therapeutic proteins produced can be passed to humans who consume their meat. D. there are more of these two types of farm animals than other types. E. it is easier to create recombinant DNA in these two farm animals than it is in other types. Since they produce milk, cattle and goats are desirable agents for gene pharming. The transgenic animals can then produce a supply of the desired molecule for their entire life span.
Bloom’s Level: 3. Apply Learning Outcome: 26.02.01 Identify ways in which human society benefits from genetically modified bacteria, plants, and animals. Section: 26.02 Topic: Genetically Modified Organisms
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Chapter 26 - Biotechnology and Genomics
15. Gene therapy may be used in the future to fight cancer by inserting genes that A. fight off mutations of the patient’s DNA. B. produce radioactive isotopes. C. cause cell death. D. produce anticancer drugs. E. All of the answer choices are correct. Cancer is caused by uncontrolled cell growth. Cells should have a tumor suppressor protein called p53. If it is missing, the cell will grow beyond the normal life span of a cell. Gene therapy could serve to insert the gene for this protein into a cancerous cell and, thereby, cause cell death.
Bloom’s Level: 3. Apply Learning Outcome: 26.03.01 Compare and contrast in vivo and ex vivo gene therapy. Section: 26.03 Topic: Gene Therapy
16. One of the problems that limits the use of gene therapy is A. the finding of suitable donors. B. some patients are allergic to the bacterial DNA used. C. mutation of the patient’s genes caused by the therapy. D. detrimental side effects from the inserted genes. E. All of the answer choices are problems with gene therapy. Gene therapy has detrimental side effects for many patients, such as causing leukemia.
Bloom’s Level: 2. Understand Learning Outcome: 26.03.01 Compare and contrast in vivo and ex vivo gene therapy. Section: 26.03 Topic: Gene Therapy
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Chapter 26 - Biotechnology and Genomics
17. The human genome project discovered many small regions of DNA that vary among individuals called single nucleotide polymorphisms (SNPs). Certain SNP patterns change an individual’s susceptibility to disease and after response to medical treatments. This knowledge has led to the idea A. of designer drugs that are made to match one specific individual’s DNA profile. B. of designer babies that have known genetic traits. C. that everyone will respond to a drug in the same way. D. that not all humans are as well suited for survival. E. that some DNA does not code for proteins. If the defective proteins can be determined, then it would be possible to determine the exact drugs to treat an illness rather than using the broad spectrum approach that treats all of the symptoms or the perceived causes of the disease.
Bloom’s Level: 4. Analyze Learning Outcome: 26.04.01 Discuss the implications of knowing the human genome sequence. Section: 26.04 Topic: Genomics
18. The human genome project led to the discovery that much of the genome consists of repeated sequences of nucleotides. These are thought to A. have no function. B. function as protection on the ends of DNA and as an attachment site on the centromere. C. function in DNA replication. D. produce rRNA. E. function as a protection against cancer. Repeated sequences of nucleotides are found at the centromere and may play a role in the adhesion of sister chromatids and occur at the ends of chromosomes, where they are called telomeres, and function to protect the ends of the chromosome from degradation during repeated replications.
Bloom’s Level: 2. Understand Learning Outcome: 26.04.02 Describe a major insight from comparative genomics. Section: 26.04 Topic: Genomics
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Chapter 26 - Biotechnology and Genomics
19. Transposable elements make up 44% of the human genome. What is thought to be their function? A. They fight cancer. B. They code for rRNA. C. They determine gender. D. They help in the replication of DNA. E. We are not sure of their function. While it is speculated that transposable elements, or transposons, are agents of DNA variability and may affect the expression of genes when they jump to a location near the gene their function is not known.
Bloom’s Level: 2. Understand Learning Outcome: 26.04.02 Describe a major insight from comparative genomics. Section: 26.04 Topic: Genomics
20. Transposons are transposable elements in the DNA because A. they transport genetic information. B. they transfer organelles inside a cell. C. they transmit signals in the cell indicating the genes to be copied. D. they move from one location to another in the DNA. E. they cause the DNA to take on a specific shape or pose. Transposons are called transposable elements because they are able to transpose or relocate within the DNA.
Bloom’s Level: 2. Understand Learning Outcome: 26.04.02 Describe a major insight from comparative genomics. Section: 26.04 Topic: Genomics
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Chapter 26 - Biotechnology and Genomics
21. How can comparative genomics assist in the treatment of HIV? A. The genome of HIV can now be compared to the human genome to find similarities between the two. B. Knowing the genome of HIV allows for the manufacture of designer drugs to treat specific strains. C. Understanding the genetic code of HIV allows scientists to replicate it in the laboratory. D. Understanding the evolution of HIV in an individual will help scientists understand how the virus responds to different drug regimes and will lead to better treatments. E. Knowing the proteins produced by HIV will allow the development of drugs to mimic those proteins. Because comparative genomics allows changes in the genome over time to be analyzed, it can show researchers how HIV responds to specific drug therapies. Watching this evolutionary response can lead to the anticipation of these changes and permit drugs to be developed to counteract those changes.
Bloom’s Level: 4. Analyze Learning Outcome: 26.04.02 Describe a major insight from comparative genomics. Section: 26.04 Topic: Genomics
22. The study of proteomics is more complex than the study of genomics because A. proteins are more complex than DNA. B. proteins are harder to sequence than DNA. C. there are 20 amino acids that comprise proteins and only 4 nucleotides that compose DNA. D. scientists understand less about amino acids than they do about nucleotides. E. each cell in an organism has exactly the same DNA but different cell types each produce different types of proteins. While the entire human genome is the same for every cell in a human and is similar for all humans, the same gene is located in the same location on the same chromosome for all humans. Different cell types express that genome according to the function that they have. This means that each different cell type contains different proteins.
Bloom’s Level: 3. Apply Learning Outcome: 26.04.03 Compare and contrast functional genomics and proteomics. Section: 26.04 Topic: Proteomics
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Chapter 26 - Biotechnology and Genomics
23. Which of the following terms applies to the study of all of the proteins produced by an organism? A. genomics B. bioinformatics C. proteomics D. genetic profiling E. genetics Genomics refers to the study of the structure of nucleotide sequences in DNA. Frequently, bioinformatics, a type of computer technology, is used to study the sequences. Proteomics is the study of the structure, function, and interactions of proteins. Genetic profiling is an individual's complete genome, including any mutations. Genetics is the study of inherited information that is passed from parent to offspring.
Bloom’s Level: 1. Remember Learning Outcome: 26.04.03 Compare and contrast functional genomics and proteomics. Section: 26.04 Topic: Proteomics
24. is the application of computer technologies to the study of the genome. A. Proteomics B. Genomics C. Bioinformatics D. DNA fingerprinting E. Genetic profiling Proteomics is the study of the structure, function, and interactions of proteins. Genomics refers to the study of the structure of nucleotide sequences in DNA. Frequently, bioinformatics, a type of computer technology, is used to study the sequences. DNA fingerprinting is a kind of genetic testing that produces fragments of DNA from a person's genome; those fragments are then separated according to size. Genetic profiling is an individual's complete genome, including any mutations.
Bloom’s Level: 1. Remember Learning Outcome: 26.04.03 Compare and contrast functional genomics and proteomics. Section: 26.04 Topic: Genomics
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Chapter 26 - Biotechnology and Genomics
25. The treatment of a disorder by inserting genetic material into an organism is called A. karyotyping. B. amniocentesis. C. genetic profiling. D. chorionic villi sampling. E. gene therapy. Karyotyping, amniocentesis, genetic profiling, and chorionic villi sampling are all forms of genetic testing. Gene therapy is used to treat disorders that are genetic in origin by either introducing agents into the cells that allow the correct genetic expression to occur or by removing, directly treating, and then reinserting cells.
Bloom’s Level: 1. Remember Learning Outcome: 26.03.01 Compare and contrast in vivo and ex vivo gene therapy. Section: 26.03 Topic: Gene Therapy
26. Which of the following is NOT used in the production of recombinant DNA (rDNA)? A. vectors B. restriction enzymes C. DNA ligase D. RNA E. plasmids Recombinant DNA (rDNA) is composed from DNA from at least two different organisms. A vector is the manipulated DNA and is commonly a plasmid or small circular accessory piece of DNA from a bacterium. The vector is then treated with a restriction enzyme that cuts the DNA in specific locations and DNA ligase, which inserts the bonds into the opening created by the restriction enzyme. RNA is not used in this process.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.01 Describe the steps in forming recombinant DNA. Section: 26.01 Topic: DNA Technology
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Chapter 26 - Biotechnology and Genomics
27. A piece of DNA that contains sequences from two or more different organisms is called DNA. A. inverse B. species-specific C. recombinant D. transgenic E. semiconservative Recombinant DNA has a new combination of genes, in DNA technology this refers to DNA that has been created with the DNA of at least two separate organisms.
Bloom’s Level: 1. Remember Learning Outcome: 26.01.01 Describe the steps in forming recombinant DNA. Section: 26.01 Topic: DNA Technology
28. Enzymes from the thermophile organism Thermus aquaticus have allowed the development of which of the following procedures? A. genetic engineering B. recombinant DNA C. DNA cloning D. polymerase chain reaction E. protein synthesis The process that allows DNA to be copied in the laboratory to allow for large samples to be obtained from minute amount of DNA is the polymerase chain reaction (PCR). This process requires that DNA be subject to temperatures that are higher than the enzyme, which copies DNA (DNA polymerase), can usually function. The organism Thermus aquaticus, which normally exists in hot springs, has a DNA polymerase that can function at these temperatures.
Bloom’s Level: 1. Remember Learning Outcome: 26.01.02 Discuss how the polymerase chain reaction works. Section: 26.01 Topic: DNA Technology
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Chapter 26 - Biotechnology and Genomics
29. Which of the following is NOT correct regarding restriction enzymes? A. They exist in bacteria to restrict the growth of viruses. B. They cut double-stranded DNA at specific sites. C. They are used during the manufacture of proteins. D. They produce "sticky ends" that can bind foreign DNA. E. They facilitate inserting foreign DNA into vector DNA. Restriction enzymes occur naturally in bacteria where they act as a primitive immune system by cutting up any viral DNA that enters the cell. They can also be used as molecular scissors to cut double-stranded DNA at a specific site. One side of the double strand will have several base pairs that do not have hydrogen bonds to another base; this makes the ends "sticky," or willing to bind to nucleotides, which allows DNA from a different organism to be joined into these regions. Restriction enzymes function by manipulating DNA that do not affect protein synthesis.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.01 Describe the steps in forming recombinant DNA. Section: 26.01 Topic: DNA Technology
30. uses an electrical field to separate DNA fragments based upon their length. A. DNA cloning B. Polymerase chain reaction C. Transgenics D. Gel electrophoresis E. DNA digestion Restriction enzymes cut the DNA at unique sites dependent on the genetic makeup of an individual. This will yield fragments of varying lengths which are then separated through gel electrophoresis according to their length, producing a distinctive pattern as fragments of a specific length separate from those of different lengths.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.03 Explain what is meant by a DNA “fingerprint.” Section: 26.01 Topic: DNA Technology
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Chapter 26 - Biotechnology and Genomics
31. Which of the following procedures would most likely be used to identify a specific individual of a population? A. DNA cloning B. transgenic DNA C. polymerase chain reaction D. DNA fingerprinting E. genetic engineering DNA fingerprinting refers to the process where DNA is cut into fragments with unique lengths. Restriction enzymes cut the DNA at sites that are dependent on the genetic makeup of an individual. This will yield fragments of varying lengths which are then separated through gel electrophoresis according to their length, producing a distinctive pattern as fragments of a specific length separate from those of different lengths. Each individual will produce a different pattern of fragments because their DNA is unique.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.03 Explain what is meant by a DNA “fingerprint.” Section: 26.01 Topic: DNA Technology
32. Which of the following is NOT a concern over the use of genetically engineered corn? A. resistance among populations of certain pests B. resistance to herbicides and pesticides C. genetic exchange between engineered plants and related species D. impact of engineered plants on nontarget plant species E. presence of allergens Concerns regarding the creation of genetically modified organisms (GMOs), such as corn, generally pertain to the "escape" of the genetic modification into the general population of corn plants. Pollen can easily be transferred over long distances and could fertilize nonengineered corn in adjacent fields. Some of the genetic modifications are made to create plants that are resistant to herbicides as a beneficial trait.
Bloom’s Level: 2. Understand Learning Outcome: 26.02.01 Identify ways in which human society benefits from genetically modified bacteria, plants, and animals. Section: 26.02 Topic: Genetically Modified Organisms
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Chapter 26 - Biotechnology and Genomics
33. Which of the following is not part of the process for PCR (polymerase chain reaction)? A. the use of an engineered DNA polymerase B. the denaturing of the double-stranded DNA to yield single-strand DNA C. multiple cycles of the chain reaction to yield a greater number of DNA copies D. final product is single-stranded DNA E. All of the answer choices are parts of the process for PCR. The product of PCR is a double-stranded DNA. The single strand is only necessary while the complementary strand is formed against the template strand.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.02 Discuss how the polymerase chain reaction works. Section: 26.01 Topic: DNA Technology
34. A method used to determine if the DNA found at a crime scene belongs to a particular individual is short tandem repeat (STR). This method employs A. repeated sequences found at multiple sites in the DNA. B. one particular gene found in the individual in question. C. fluorescent labeling of DNA found in the mitochondria. D. repeated sequences found at one specific site in the DNA. E. the repeated segments at the end of chromosomes called telomeres. STR is a procedure where multiple repeating segments of DNA are amplified through PCR, fluorescently labeled and analyzed to determine their length. The repeated sequences will vary in length from one individual to the next since each individual has their own number of times that the sequence is repeated.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.03 Explain what is meant by a DNA “fingerprint.” Section: 26.01 Topic: DNA Technology
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Chapter 26 - Biotechnology and Genomics
35. In DNA fingerprinting the products of PCR are cut by restriction enzymes into unique lengths; these fragments are then A. subjected to fluorescent dyes and visualized under an electron microscope. B. separated according to their relative positive and negative charges. C. separated according to the lengths of the fragments. D. arranged into a karyotype. E. used to produce the genes that give the individual a unique set of fingerprints. Restriction enzymes cut the DNA at unique sites, dependent on the genetic makeup of an individual. This will yield fragments of varying lengths which are then separated through gel electrophoresis according to their length, producing a distinctive pattern as fragments of a specific length separate from those of different lengths.
Bloom’s Level: 2. Understand Learning Outcome: 26.01.03 Explain what is meant by a DNA “fingerprint.” Section: 26.01 Topic: DNA Technology
36. A major difference in the production of transgenic bacteria, plants, and animals is that A. transgenic bacteria do not pass on their genetic modification to offspring, while plants and animals do. B. transgenic animals require the use of host animals to carry the genetically modified embryo, while bacteria and plants do not require this. C. transgenic bacteria and plants receive genes from other species of organisms, while animals are only able to incorporate genes from their own species. D. transgenic animals are the most useful of the GMOs, while transgenic plants and bacteria are only created to perfect the techniques. E. transgenic bacteria, animals, and plants can only incorporate genes from their own species, so the process for each species differs from the process for any other species. Neither plants nor bacteria need a host to carry an embryo, while many species of animals do. Even if the embryo of a goat is created in the laboratory a surrogate mother must provide the correct environment for development; in contrast, plants and bacteria can develop in a nutrient media.
Bloom’s Level: 4. Analyze Learning Outcome: 26.02.02 Describe the steps involved in the production of a transgenic animal. Section: 26.02 Topic: Genetically Modified Organisms
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Chapter 26 - Biotechnology and Genomics
37. When creating transgenic animals, as contrasted with the creation of transgenic bacteria, A. it is necessary to carry out the procedure completely by hand. B. animals need donor eggs to contain the genetic material, while bacteria can have the new genes inserted into any bacterial cell. C. transgenic animals can be created in large numbers, while bacteria must be created one at a time to control contamination. D. the membranes of the bacteria must be disrupted to allow the genes to be taken up into the cell and animals do not require the host cell membrane to be disrupted. E. the membranes of the animal cell must be disrupted to allow the genes to be taken up into the cell and bacteria do not require the host cell membrane to be disrupted. Both animals and bacteria can be modified through mechanized means that disrupt the membrane and allow the new genetic material to be taken into the host cell. In bacteria any cell can be used as the host, while in animals only an egg can be used for this purpose.
Bloom’s Level: 4. Analyze Learning Outcome: 26.02.02 Describe the steps involved in the production of a transgenic animal. Section: 26.02 Topic: Genetically Modified Organisms
38. Bacteria, plants, and animals have all been used to create transgenic forms that are useful in A. producing high-yield tomato plants. B. creating pest-resistant versions of corn. C. cleaning up oil spills. D. generating pharmaceuticals. E. producing new types of high-yield beef. Bacteria, plants, and animals have all had genetic materials inserted in their genomes that allow them to produce products that can then be used by the pharmaceutical industry. Bacteria and plants generate such products as human hormones, insulin, clotting factors, and antibodies. Animals such as goats and cattle have been used to produce therapeutic and diagnostic proteins in their milk.
Bloom’s Level: 2. Understand Learning Outcome: 26.02.01 Identify ways in which human society benefits from genetically modified bacteria, plants, and animals. Section: 26.02 Topic: Genetically Modified Organisms
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Chapter 26 - Biotechnology and Genomics
39. All of the answer choices listed below are uses for transgenic bacteria, except A. bioremediation, when bacteria are used to clean up oil spills or other toxic substances. B. the production of organic chemicals, such as phenylalanine, used in the production of aspartame. C. the production of chemicals toxic to insects that can be used to protect plants from insects. D. the production of plants like the pomato. It has a starchy root like the potato and also produces tomatoes on the stem portion of the plant. E. the production of human growth hormones. Bacteria are not used in the production of the pomato, which was developed from the addition of the genes of one plant to another plant.
Bloom’s Level: 3. Apply Learning Outcome: 26.02.01 Identify ways in which human society benefits from genetically modified bacteria, plants, and animals. Section: 26.02 Topic: Genetically Modified Organisms
40. Why are genetically modified crop plants (GMOs) are able to express new traits? A. They now contain genes from another organism. B. They grow bacteria on the surface of leaves or roots. C. There is no natural method for processes to be degraded in the organism. D. Their normally occurring processes have been modified to serve different functions. E. They have not been genetically modified enough to change the plant's genetics. GMOs are able to manufacture molecules that they do not normally contain because the genes from another organism have been added to their genome. These genes code for the production of the new molecule and will be passed to any offspring.
Bloom’s Level: 5. Evaluate Learning Outcome: 26.02.01 Identify ways in which human society benefits from genetically modified bacteria, plants, and animals. Section: 26.02 Topic: Genetically Modified Organisms
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Chapter 26 - Biotechnology and Genomics
41. Ex vivo gene therapy A. requires the removal of cells from the individual undergoing the treatment and then the reintroduction of these cells once they have been genetically modified. B. requires the use of an agent to introduce genetic material directly into the body using a vector such as an adenovirus vector. C. has few or no side effects on the individual receiving the treatment. D. allows for a wider range of disorders to be treated than in vivo treatments. E. is considered to be more dangerous for the patient than in vivo treatment. Ex vivo means “outside the body” so the treatment involves the removal of cells from the patient so that they can be modified genetically and reinserted. The cells then express the gene that was inserted and this in some way treats the disorder. There are side effects associated with the treatment and they are often severe. Both ex vivo and in vivo treatments are still relatively new procedures and have a limited number of conditions for which uses have been developed.
Bloom’s Level: 2. Understand Learning Outcome: 26.03.01 Compare and contrast in vivo and ex vivo gene therapy. Section: 26.03 Topic: Gene Therapy
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Chapter 26 - Biotechnology and Genomics
42. All of the following are benefits of sequencing the human genome except A. knowing the exact sequence of DNA in an individual could allow pharmaceutical companies to design drugs for their exact genotype. B. the discovery of the chemical makeup of DNA. C. determining individuals who have a predisposing factor that could lead to future illness could allow for earlier treatment. D. the improvement of sequencing technology that will allow the genomes of other species to be more readily sequenced. E. enabling scientists to sequence the genome of various other organisms. The chemical makeup of DNA was discovered in the 1950s by Francis Crick and James Watson. Forty years later in 1990 the human genome project began. This project has illuminated the individual differences that are present in any human’s DNA, leading each individual to react slightly differently to medications and showing when there might be a genetic predisposition to illnesses. In addition, the technological advances made through this project have greatly improved knowledge in this area. Once the technique was figured out, it was applied to a variety of other organisms as well.
Bloom’s Level: 2. Understand Learning Outcome: 26.04.01 Discuss the implications of knowing the human genome sequence. Section: 26.04 Topic: Genomics
43. Which of the following statements was discovered by information gained from the human genome project? A. Large sections of the human genome do not code for a polypeptide. B. Nucleotides are composed of a pentose sugar, a phosphate group, and one of four nitrogenous bases. C. DNA carries the information to make a polypeptide. D. Regulation of gene expression determines the specialization of cells in humans. E. DNA carries the information to make mRNA. While all of these statements is true, only the information that large sections of the genome do not give the information on how to make a polypeptide was discovered during this project.
Bloom’s Level: 2. Understand Learning Outcome: 26.04.01 Discuss the implications of knowing the human genome sequence. Section: 26.04 Topic: Genomics
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Chapter 26 - Biotechnology and Genomics
44. The study of comparative genomics has given researchers the ability to look at multiple genomes from different species, which has led to new ideas about the evolutionary history of organisms. The conclusion is that A. modern vertebrate species all evolved from very different ancestral species, so there is little similarity between their genomes. B. modern vertebrate species all evolved from the same ancestral species, but they have changes so that their genomes are very different in modern times. C. ancestral species contained very similar genomes, but modern vertebrate species have significant differences in their genomes. D. modern vertebrate species evolved from the same ancestors and a large portion of their genome is conserved. E. neither the ancestral species nor the modern one resemble each other genetically. Research in comparative genomics has found that the genome of humans and chimps has 98% similarity, which was not a surprise, but they have also found that humans and mice share 85% of their genome. It was surprising that two organisms with such different phenotypes share so much of their genetic material.
Bloom’s Level: 4. Analyze Learning Outcome: 26.04.02 Describe a major insight from comparative genomics. Section: 26.04 Topic: Genomics
45. What structure is used to seal the DNA into an opening created by the restriction enzyme during recombinant DNA technology? A. DNA ligase B. restriction enzymes C. plamids D. vectors E. DNA helicase DNA ligase is used to seal the DNA into an opening created by the restriction enzyme. Restriction enzymes cleave the vector DNA during recombinant DNA technology. Plasmids are small accessory rings of DNA from bacteria that are capable of self-replicating. Vectors are pieces of DNA that can be manipulated. DNA helicase is used to break apart the H+ bonds in the DNA strand.
Bloom’s Level: 1. Remember Learning Outcome: 26.01.01 Describe the steps in forming recombinant DNA. Section: 26.01 Topic: DNA Technology
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Chapter 26 - Biotechnology and Genomics
46. What is the substance required to cleave the vector DNA during recombinant DNA technology? A. restriction enzymes B. DNA ligase C. plasmids D. vectors E. DNA helicase Restriction enzymes cleave the vector DNA during recombinant DNA technology. DNA ligase is used to seal the DNA into an opening created by the restriction enzyme. Plasmids are small accessory rings of DNA from bacteria that are capable of self-replicating. Vectors are pieces of DNA that can be manipulated. DNA helicase is used to break apart the H+ bonds in the DNA strand.
Bloom’s Level: 1. Remember Learning Outcome: 26.01.01 Describe the steps in forming recombinant DNA. Section: 26.01 Topic: DNA Technology
47. Which technique is preferred when investigators are trying to DNA fingerprint an individual? A. short tandem repeat (STR) B. polymerase chain reaction (PCR) C. recombinant DNA (rDNA) D. gene cloning E. None of the answer choices is correct. Short tandem repeat (STR) is the method of choice when investigators are trying to DNA fingerprint an individual because it doesn't require the use of restriction enzymes. Polymerase chain reaction (PCR) is used to create billions of copies of segments of DNA. Recombinant DNA (rDNA) is the blending together of DNA from different organisms. Gene cloning is the production of identical copies of a single gene.
Bloom’s Level: 1. Remember Learning Outcome: 26.01.03 Explain what is meant by a DNA “fingerprint.” Section: 26.01 Topic: DNA Technology
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Chapter 26 - Biotechnology and Genomics
48. Which technique is used to create billions of copies of DNA in a short amount of time? A. polymerase chain reaction B. recombinant DNA technology C. short tandem repeat (STR) profiling D. gene therapy E. None of the answer choices is correct. The polymerase chain reaction (PCR) can create billions of copies of DNA in a short amount of time. Recombinant DNA technology is used to blend DNA from different organisms together. Short tandem repeat (STR) profiling is used for DNA analysis. Gene therapy is the insertion of genetic material into new cells.
Bloom’s Level: 1. Remember Learning Outcome: 26.01.02 Discuss how the polymerase chain reaction works. Section: 26.01 Topic: DNA Technology
Short Answer Questions 49. List the steps required for insertion of foreign genes into animal eggs by using vortex mixing. 1. The eggs are placed in an agitator with DNA and silicon-carbide needles. 2. The needles make tiny holes in the eggs through which the DNA can pass. 3. The eggs containing the foreign genes are inserted into a host mother for development. 4. The offspring produced carry the foreign genes.
Bloom's Level: 6. Create Learning Outcome: 26.02.02 Describe the steps involved in the production of a transgenic animal. Section: 26.02 Topic: Genetically Modified Organisms
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Chapter 26 - Biotechnology and Genomics
Multiple Choice Questions 50. Which animals have had the bovine growth hormone injected into them in order to produce larger individuals? A. All of the answer choices are correct. B. cows C. fish D. pigs E. sheep All of these animals have had the bovine growth hormone injected into them in order to produce larger individuals.
Bloom’s Level: 1. Remember Learning Outcome: 26.02.02 Describe the steps involved in the production of a transgenic animal. Section: 26.02 Topic: Genetically Modified Organisms
Short Answer Questions 51. List three sites of ex vivo gene therapy and indicate the type of therapy being conducted. 1. Skin: gene transfer by modifying blood cells (skin cancer) 2. Liver: gene transfer by modified implants (familial hypercholesterolemia) 3. Endothelium, blood vessel lining: gene transfer by implantation of modified implants (hemophilia and diabetes mellitus) 4. Bone marrow: gene transfer by implantation of modified stem cells (SCID and sickle-cell disease)
Bloom's Level: 6. Create Learning Outcome: 26.03.01 Compare and contrast in vivo and ex vivo gene therapy. Section: 26.03 Topic: Gene Therapy
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Chapter 26 - Biotechnology and Genomics
Multiple Choice Questions 52. Which of the following is(are) sites of in vivo gene therapy? A. muscle cells B. bone marrow C. endothelium D. liver E. skin Muscle cells are a site of in vivo gene therapy. Researchers are trying to cure Duchenne muscular dystrophy by gene transfer injections directly into the muscle cells. The rest of the sites are examples of ex vivo gene therapy.
Bloom’s Level: 1. Remember Learning Outcome: 26.03.01 Compare and contrast in vivo and ex vivo gene therapy. Section: 26.03 Topic: Gene Therapy
53. Which species has an estimated 6,300 genes in its genome? A. yeast B. roundworms C. human beings D. fruit flies E. mice The genome of yeast is estimated to be 6,300 genes.
Bloom’s Level: 1. Remember Learning Outcome: 26.04.02 Describe a major insight from comparative genomics. Section: 26.04 Topic: Genomics
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Chapter 26 - Biotechnology and Genomics
54. Which of the following is a site of ex vivo gene therapy? A. bone marrow B. muscle C. blood D. lungs E. brain Bone marrow is a site of ex vivo gene therapy. All of the others are sites of in vivo gene therapy.
Bloom’s Level: 1. Remember Learning Outcome: 26.03.01 Compare and contrast in vivo and ex vivo gene therapy. Section: 26.03 Topic: Gene Therapy
Short Answer Questions 55. List the number of chromosomes and estimated number of genes for each of the following organisms: 1. Homo sapiens 2. Mus musculus 3. Drosophilia melanogaster 4. Caenorhabditis elegans 5. Saccharomyces cerevisiae Species # of Chromosomes 1. Homo sapiens 46 2. Mus musculus 40 3. Drosophilia melanogaster 8 4. Caenorhabditis elegans 12 5. Saccharomyces cerevisiae 32
# of Genes 25,000 25,000 13,6000 19,100 6,300
Bloom's Level: 6. Create Learning Outcome: 26.04.02 Describe a major insight from comparative genomics. Section: 26.04 Topic: Genomics
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Chapter 26 - Biotechnology and Genomics
True / False Questions 56. The aim of functional genomics is to understand the function of the various genes discovered within each genomic sequence and how these genes interact. TRUE It is true that the aim of functional genomics is to understand the function of the various genes discovered within each genomic sequence and how these genes interact.
Bloom’s Level: 1. Remember Learning Outcome: 26.04.03 Compare and contrast functional genomics and proteomics. Section: 26.04 Topic: Genomics
57. The genome of Homo sapiens and Mus musculus are approximately the same size and composed of the same number of genes. FALSE The genome of Homo sapiens and Mus musculus is composed of the same number of genes (25,000) but they are not of the same size. Homo sapiens have approximately 3.200 million base pairs while Mus musculus has approximately 2.500 million base pairs.
Bloom’s Level: 1. Remember Learning Outcome: 26.04.02 Describe a major insight from comparative genomics. Section: 26.04 Topic: Genomics
58. Ex vivo gene therapy is being used to treat patients with Huntington disease, Alzheimer disease, Parkinson disease, and brain tumors. FALSE In vivo gene therapy is being used to treat patients with Huntington disease, Alzheimer disease, Parkinson disease, and brain tumors.
Bloom’s Level: 1. Remember Learning Outcome: 26.03.01 Compare and contrast in vivo and ex vivo gene therapy. Section: 26.03 Topic: Gene Therapy
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Chapter 27 - Evolution of Life
Chapter 27 Evolution of Life
Multiple Choice Questions 1. Which evidence for evolution uses impressions of plants and animals pressed into sedimentary rock? A. fossil record B. biogeography C. comparative anatomy D. comparative embryology E. comparative biochemistry The fossil record consists of the remains of ancient organisms that have been replaced with minerals or the impressions of ancient organisms such as footprints, soft body tissues, or feathers. These fossils provide evidence for evolution by showing the ancestral species from which existing species are descended. There are also transitional species evident in the record that show the development of distinctive features such as feathers.
Bloom’s Level: 1. Remember Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
2. The earliest forms of life on Earth are found in rocks that are estimated to be A. 1 million years old. B. 4 million years old. C. 1 billion years old. D. 3.5 billion years old. E. 4.6 billion years old. Earth is 4.6 billion years old, and the earliest fossils of prokaryotes are 3.5 billion years old.
Bloom’s Level: 1. Remember Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
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Chapter 27 - Evolution of Life
3. Prokaryotic cells are represented by fossils that are dated back as far as ago. A. 1.5 B. 2.5 C. 3.5 D. 4.5 E. 10
billion years
The earliest evidence of life on Earth are prokaryotic cells that date from 3.5 billion years old.
Bloom’s Level: 1. Remember Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Origin of Cells
4. The age of the earth is estimated to be about A. 1 million years old. B. 4 million years old. C. 1 billion years old. D. 2.5 billion years old. E. 4.6 billion years old. The planet Earth existed for a long time before the first cell arose and is 4.6 billion years old.
Bloom’s Level: 1. Remember Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Geologic Timescale
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Chapter 27 - Evolution of Life
5. Which pairing of occurrence and date is NOT correct? A. origin of invertebrates—630 million years ago B. origin of life—4.6 billion years ago C. origin of eukaryotic cells—2.1 billion years ago D. origin of prokaryotic cells—3.5 billion years ago E. chordate evolution—545 million years ago Earth is 4.6 billion years old, and the earliest fossils of prokaryotes are 3.5 billion years old. The eukaryotic cell arose about 2.1 billion years ago; chordates first evolved about 545 million years ago.
Bloom’s Level: 2. Understand Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Geologic Timescale
6. Which fossil evidence is considered an intermediate between reptiles and birds? A. Archaeopteryx B. Eusthenopteron C. Seymouria D. synapsids E. therapods A fossil of Archaeopteryx, now considered the first bird, has features of both birds and dinosaurs. Fossils indicate it had feathers and wing claws. Most likely, it was a poor flier. Perhaps it ran over the ground on strong legs and climbed into trees with the assistance of these claws.
Bloom’s Level: 1. Remember Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
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Chapter 27 - Evolution of Life
7. Transitional forms are seen in the fossil record that link all of these groups EXCEPT A. reptiles and birds. B. amphibians and reptiles. C. fish and amphibians. D. birds and mammals. E. reptiles and mammals. A fossil of Archaeopteryx, now considered the first bird, has features of both birds and dinosaurs. Other transitional links among fossil vertebrates suggest that fishes evolved before amphibians, which evolved before reptiles, which evolved before mammals in the history of life.
Bloom’s Level: 2. Understand Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
8. The geological time scale indicates the order of eras is Cenozoic as the most recent; the middle era is the Mesozoic, with dinosaurs as the major group; and the earliest era is the Paleozoic. Consider a hillside of rock that surfaces in your backyard that is identified as dating back to the Mesozoic era. What is possible? A. Fossils from all other eras can still be found underneath if you dig far enough. B. All periods of fossils were once present in the backyard but older ones were eroded away. C. Fossils were mainly deposited when your backyard was sediment, so many layers might be missing, but any layer from Cenozoic to Paleozoic might be found. D. Fossils were mainly deposited when your backyard was sediment, so many layers might be missing, but any layer from earlier Mesozoic through Paleozoic might be found. E. Only one layer of fossils can be found at any site. According to the geologic time scale the most recent era is the Cenozoic era, the middle Mesozoic era comes next, proceeded by the early Paleozoic era. If the topmost layer is Mesozoic, then only that era and those proceeding it could be present.
Bloom’s Level: 4. Analyze Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Geologic Timescale
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Chapter 27 - Evolution of Life
9. The position of mammals in evolution is best described as A. they dominate life in the sea. B. they were among the first animals to live on land and their variety outstrips all living things known. C. they arose from mammal-like reptiles in the Triassic but remained small and insignificant while dinosaurs dominated the land. D. mammals gave rise to birds. E. mammals have been present since plants and other organisms first appeared on land. Reptiles arose during the Carboniferous period of the Mesozoic era 359.2 to 299 million years ago. Mammals are seen first in the Triassic period 251 to 199.6 million years ago as are the first dinosaurs. During the Jurassic period, 199.6 to 145.5 million years ago, dinosaurs flourished while mammals remained small in both physical size and population size. It is not until the Paleocene period, 65.5 to 55.8 million years ago, after the mass extinction of dinosaurs that mammals increase in both size and numbers.
Bloom’s Level: 3. Apply Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Geologic Timescale
10. Consider that the rock outcrops at the surface of the land in your neighborhood are Permian. What would be true if you dig downward? A. You may find dinosaur fossils below. B. You may find fossils of first flowers. C. The Mesozoic layers will be present underneath. D. A road cut in your area would reveal exactly the same species of fossils as a road cut through a Permian layer in Asia. E. You may find fossils of the first reptiles or jawed fishes below. The Permian era occurred 299 to 251 million years ago, which is before dinosaurs arose in the Triassic period 251 to 199.6 million years ago. Flowers are seen even later in the geologic time scale during the Jurassic period 199.6 to 145.5 million years ago. The first reptiles arose during the Carboniferous period of the Mesozoic era 359.2 to 299 million years ago so they could be present in layers beneath the surface, while jawed fishes arose even earlier in the Ordovician period of the Paleozoic era 488.3 to 443.7 million years ago.
Bloom’s Level: 3. Apply Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
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Chapter 27 - Evolution of Life
11. Which of the following is NOT a method by which the general dates of the geological time scale can be determined? A. Lower rock layers are older than surface layers. B. Common assemblages of indicator fossils occur in layers with primitive layers lower. C. Radioisotopes decay at constant rates and the ratio of isotopes-to-stable elements provides a relative age. D. Carbon 14 should double in a sample every 5,730 years and so can be used to date organic matter. E. Absolute and relative dating methods can both be used. There are two ways to age fossils. The relative dating method determines the relative order of fossils and strata depending on the layer of rock in which they were found, but it does not determine the actual date they were formed. The absolute dating method relies on radioactive dating techniques to assign an actual date to a fossil. Assuming a fossil contains organic matter, half of the 14C will have changed to nitrogen 14 (14N) in 5,730 years. So the carbon 14 in a sample decreases instead of increasing over time.
Bloom’s Level: 4. Analyze Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
12. The extinction of the dinosaurs has been associated with A. a virus leading to disease and massive loss of life. B. dinosaurs preying upon each other until they became extinct. C. a meteor impact at the end of the Cretaceous era that caused major cooling. D. new chemicals and ions in the environment which caused dinosaurs to become sterile. E. mammals who were more adapted and more able to acquire similar resources. It has been proposed that the Cretaceous period extinction was due to an asteroid that exploded, producing meteorites that fell to Earth. A large meteorite striking Earth could have produced a cloud of dust that mushroomed into the atmosphere, which would block out the sun and cause plants to freeze and die.
Bloom’s Level: 2. Understand Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Extinction
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Chapter 27 - Evolution of Life
13. Mass extinctions A. are the result of humans damaging the natural environment. B. are the result of humans exploiting wildlife and using pesticides. C. occurred in cycles when organisms fail to evolve. D. have occurred about 4 or 5 times in fossil record, due in some cases to catastrophic changes. E. require life to evolve again from protocells. Mass extinctions are seen in the fossil record, long before humans evolved, and thus far five have been designated. During mass extinctions, a large percentage of species become extinct within a relatively short period of time. There have been five major mass extinctions; these occurred at the ends of the Ordovician, Devonian, Permian, Triassic, and Cretaceous periods, and a sixth is likely occurring now, probably as a result of human activities.
Bloom’s Level: 2. Understand Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Extinction
14. What evidence would NOT be studied by a biogeographer? A. continental drift or the movement of tectonic plates over time B. amount of genetic similarity among current populations C. ocean currents and wind patterns D. ranges of animals and ability to migrate E. the genetic makeup of organisms that evolved in separate but similar conditions Biogeography is the study of the range and distribution of plants and animals in different places throughout the world. It would be necessary to look at continental drift to see how the movements of continents has changed climate or when species that evolved together were separated by this movement and whether or not the organism migrates so that populations come together again. They would also study the movement of ocean currents and winds to determine if organisms could move over long distances through these mediums. The one area that biogeographers would not study would be the genetics of organisms that evolved in separate but similar conditions.
Bloom’s Level: 2. Understand Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
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Chapter 27 - Evolution of Life
15. Africa, Asia, South America, and Antarctica share some patterns of primitive (fossil) plants and early reptiles, but do not have similar mammal populations. This therefore A. casts serious doubts upon the theory of continental drift and fused land masses. B. is an unsolved puzzle probably due to the random nature of biological evolution. C. suggests that a pattern of land bridges existed at different times in geological history. D. suggests that the earlier plants and reptiles evolved while continents were joined but mammals radiated into diverse groups after separation. E. suggests that the mammals evolved earlier while continents were joined but plants and reptiles radiated into diverse groups after separation. For more primitive organisms, such as flowering plants and early reptiles, to be found in Africa, Asia, South America, and Antarctica these now-separate continents would have to have been joined when these groups arose. The fact that mammals on these continents are very dissimilar indicates that they evolved separately.
Bloom’s Level: 5. Evaluate Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
16. Homologous structures such as the bones in wings, flippers, and arms are most closely concerned with A. comparative anatomy. B. biogeography. C. the fossil record. D. comparative embryology. E. comparative biochemistry. Comparative anatomy would look at structures that are anatomically similar because they are inherited from a common ancestor; these are called homologous structures.
Bloom’s Level: 1. Remember Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
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Chapter 27 - Evolution of Life
17. Anatomical features that are fully developed and functional in one group of organisms but reduced and functionless in a similar group are termed A. vestigial. B. homologous. C. analogous. D. polygenic. E. sympatric. Structures that are anatomically similar because they are inherited from a common ancestor are called homologous structures.
Bloom’s Level: 1. Remember Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
18. Almost all living organisms use DNA, ATP, and the same DNA triplet code. This is an example of evolution based on A. the fossil record. B. biogeography. C. comparative anatomy. D. comparative embryology. E. biochemistry. Almost all living organisms use the same basic biochemical molecules, including DNA, ATP, and many identical or nearly identical enzymes. Further, organisms use the same DNA triplet code for the same 20 amino acids in their proteins. All of these examples are based on the makeup of molecules found in various organisms; this is the study of the biochemical makeup of these molecules.
Bloom’s Level: 1. Remember Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
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Chapter 27 - Evolution of Life
19. A line of evidence NOT considered by Darwin in his development of the theory of natural selection is A. comparative anatomy. B. biogeography. C. the fossil record. D. geography. E. biochemistry. The study of comparative biochemistry would not have been possible for Darwin to use since it had not been developed as yet.
Bloom’s Level: 3. Apply Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
20. Which of the following can be used to investigate the accumulation of small changes in the gene pool of a population over time? A. structural proteins. B. basic biochemical molecules that are universal. C. blood type molecules. D. similarities only based on available dietary proteins. E. There are no restrictions, any amino acids for any structure would provide a legitimate comparison among organisms. In order to compare molecules across diverse organisms it would be necessary to choose molecules that are present in all of the organisms. These would be basic to the biochemical makeup of all of the organisms involved.
Bloom’s Level: 4. Analyze Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
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Chapter 27 - Evolution of Life
21. In comparing the number of differences in amino acids for cytochrome c: between a moth and a rattlesnake there are 31 different amino acids in the sequence, between a moth and a dog there are only 25, and there are 6 differences between a dog and a horse. Therefore, A. this is sufficient evidence to prove the moth is related more to the dog than to the horse. B. dogs had to evolve earlier than horses, since moths are more primitive. C. evolution used to be considered a unifying theory until biochemical data showed only differences. D. added to anatomical similarities and fossil records, this provides additional evidence for constructing a phylogeny or family tree. E. biochemical information is not relevant to the evolution of organisms. Similarities and differences in amino acid sequences provide additional lines of evidence for the formation of a family tree.
Bloom’s Level: 5. Evaluate Learning Outcome: 27.02.02 Describe, with examples, four lines of evidence for evolution. Section: 27.02 Topic: Evidence for Evolution
22. All the members of a single species that occupy a particular area and are able to interbreed are a A. subspecies. B. gene pool. C. population. D. domain. E. community. A population is a group of organisms of the same species that live in close enough contact to be able to exchange genetic information through breeding. This will keep the group of organisms genetically similar.
Bloom’s Level: 1. Remember Learning Outcome: 27.03.01 List the five conditions necessary for the allele frequencies of a population to be in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
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Chapter 27 - Evolution of Life
23. The is the total number of alleles of all the gene loci in all the members of a population. A. genetic drift B. gene flow C. gene pool D. adaptive radiation E. community The gene pool consists of all of the available genetic information that is available to a population of organisms. In theory, this information can be passed to future generations of offspring of that population. It is customary to describe this gene pool in terms of allele frequencies for the various genes.
Bloom’s Level: 1. Remember Learning Outcome: 27.03.01 List the five conditions necessary for the allele frequencies of a population to be in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
24. What is the term used to describe the accumulation of small changes in the gene pool of a species over time? A. genetic drift B. founder effect C. microevolution D. directional selection E. mutation rate Microevolution is defined as a change in the frequency at which an allele is seen in a population. As changes in these frequencies occur, the phenotype of the population changes and over generations the population evolves.
Bloom’s Level: 1. Remember Learning Outcome: 27.03.01 List the five conditions necessary for the allele frequencies of a population to be in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
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Chapter 27 - Evolution of Life
25. If the Hardy-Weinberg equilibrium is met, what is the net effect? A. evolution leading to a population better adapted to an unchanging environment B. evolution leading to a population better adapted to a changing environment C. very slow and continuous evolution with no increased adaptation D. no evolution because the alleles in the population remain the same E. no evolution because the alleles in the population will change The Hardy-Weinberg principle states that allele frequencies in a gene pool will remain at equilibrium, and so are constant after one generation of random mating in a large, sexually reproducing population as long as certain conditions are met. A population in HardyWeinberg equilibrium is not experiencing changes in allele frequencies and is therefore not evolving.
Bloom’s Level: 3. Apply Learning Outcome: 27.03.01 List the five conditions necessary for the allele frequencies of a population to be in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
26. If the frequency of the dominant gene is 0.7, what is the frequency of the recessive gene? A. also 0.7 B. 0.49 or (0.07)2 C. approximately 0.27 (square root of 0.7) D. 0.3 (or 1.00 - 0.7) E. 0.14 The Hardy and Weinberg equation is p2 + 2pq + q2 = 1, with p representing the frequency of the dominant allele and q representing the frequency of the recessive allele. The frequency of the occurrence of alleles cannot be greater than 100% or 1, therefore p + q = 1. In this case p = 0.7 so the equation becomes 0.7 + q = 1. Solving for q gives a frequency of 0.3 for the recessive allele.
Bloom’s Level: 4. Analyze Learning Outcome: 27.03.02 Calculate genotype frequencies of a population in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
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Chapter 27 - Evolution of Life
27. The equation, p2 + 2 pq + q2 = 1.0, describes A. the process of evolution. B. the size of a population. C. the rate of speciation of species p and q. D. genotype frequencies of a nonevolving population. E. evolution of a population. The Hardy-Weinberg equation is p2 + 2pq + q2 = 1, with p representing the frequency of the dominant allele and q representing the frequency of the recessive allele. The frequency of the occurrence of alleles cannot be greater than 100% or 1, therefore p + q = 1.
Bloom’s Level: 2. Understand Learning Outcome: 27.03.02 Calculate genotype frequencies of a population in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
28. According to the Hardy-Weinberg equilibrium, in a population in which 1% does NOT have freckles (recessive), what percentage of the next population is likely to NOT have freckles? A. None, since it is a recessive condition. B. 1% C. 2% D. 50% E. 99% Since the Hardy-Weinberg equilibrium states that allele frequencies are not changing from one generation to the next if the original frequency is 0.1 or 1%, then it must remain at that frequency.
Bloom’s Level: 3. Apply Learning Outcome: 27.03.03 Identify an evolving population from a change in its allele frequencies over generations. Section: 27.03 Topic: Microevolution
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Chapter 27 - Evolution of Life
29. A student proposes that left-handedness is a recessive trait that is therefore hidden in much of the human population. A survey of a class of 36 students finds that 27 (0.75) are right-handed and 9 (0.25) are left-handed. Using the Hardy-Weinberg formula, what would the expected genotype and allele frequencies be in this theoretical population? A. 0.75 right-handed homozygous dominant and 0.25 recessive homozygous for 3-to-1 rightto left-handed alleles in the population B. 0.25 right-handed homozygous, 0.50 heterozygous, and 0.25 recessive homozygous for a 3-to-1 right- to left-handed alleles in the population C. 0.25 right-handed homozygous, 0.50 heterozygous, and 0.25 recessive homozygous for a 0.5 allele frequency for each allele D. 0.50 right-handed homozygous, 0.25 heterozygous, and 0.25 recessive homozygous for a 0.5 allele frequency for each allele E. They cannot be estimated using these limited data. The Hardy-Weinberg equation is p2 + 2pq + q2 = 1, with p representing the frequency of the dominant allele and q representing the frequency of the recessive allele. The frequency of the occurrence of alleles cannot be greater than 100% or 1, therefore p + q = 1. Since 9 students are recessive (rr) then the frequency of q2 would be .25, the q = .5. The frequency of the dominant allele would be p = .5. The frequency of the homozygous dominant genotype would be p2 (.5 x .5 = .25). The frequency of the heterozygous genotype is 2pq (2 x .5 x .5 = .5).
Bloom’s Level: 4. Analyze Learning Outcome: 27.03.02 Calculate genotype frequencies of a population in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
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Chapter 27 - Evolution of Life
30. Which of these conditions is NOT among the requirements of the Hardy-Weinberg equilibrium? A. no net mutations B. no net migration of alleles into or out of the population C. small population with genetic drift D. no selection of one genotype over another E. sexually reproducing and random mating population The five conditions that must be met for Hardy-Weinberg Equilibrium to be met are; 1) no mutation, 2) random mating, 3) an infinitely large population, 4) a population that is reproductively isolated from other populations of the same species, and 5) all phenotypes must be equally able to survive and reproduce.
Bloom’s Level: 1. Remember Learning Outcome: 27.03.01 List the five conditions necessary for the allele frequencies of a population to be in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
31. A variety of genotypes and phenotypes in a population is useful because it A. makes life more interesting. B. allows the species to survive if the environment changes. C. means that the gene pool is constant and unchanging. D. makes genetic drift an unlikely occurrence. E. will lead to nonrandom mating. Natural selection can only act on the phenotypes that are present in the population, the more diverse the population (and its genetic makeup) the more likely it is that one or more of the phenotypes will be able to survive as environmental conditions change.
Bloom’s Level: 3. Apply Learning Outcome: 27.03.01 List the five conditions necessary for the allele frequencies of a population to be in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
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Chapter 27 - Evolution of Life
32. The raw material for evolutionary change is A. gene flow. B. genetic drift. C. mutation. D. nonrandom reproduction. E. natural selection. While all of the processes listed take part in how populations change over time, or evolve, the only one that contributes new alleles to the gene pool is mutation. The others only act on already existing traits.
Bloom’s Level: 2. Understand Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Mutation
33. The Amish of Lancaster, PA. have a higher frequency of an unusual form of dwarfism than the population at large. This is probably due to A. natural selection. B. a bottleneck effect. C. a founder effect. D. vestigial structures. E. industrial melanism. The founder effect is seen when a small number of individuals establish a new population. While the larger original population may contain a variety of different alleles for each trait, only those carried by the individuals making up the new population will be seen in the new gene pool. With this more limited gene pool, any random shifts in allele frequencies can change the frequency of one or more alleles.
Bloom’s Level: 2. Understand Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Macroevolution
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Chapter 27 - Evolution of Life
34. When only a few individuals survive unfavorable times, thereby losing the majority of genotypes in the next generation, it is called A. natural selection. B. a bottleneck effect. C. a founder effect. D. vestigial structures. E. industrial melanism. A bottleneck occurs when a random occurrence reduces a large population to a small one. While the larger original population may contain a variety of different alleles for each trait, only those carried by the individuals surviving to make up the new population will be seen in the new gene pool. With this more limited gene pool, any random shifts in allele frequencies can change the frequency of one or more alleles.
Bloom’s Level: 1. Remember Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Microevolution
35. If there is a change in the allele frequency of a gene pool due to chance, primarily in a small population, it is termed A. mutations. B. genetic drift. C. gene flow. D. nonrandom mating. E. natural selection. Genetic drift is random changes in allele frequencies by chance. If a population is very large, changes in allele frequencies due to chance alone are insignificant. In small populations these same changes can have profound effects on the overall genetic makeup of the population. It can even lead to the complete loss of one or more alleles.
Bloom’s Level: 1. Remember Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Microevolution
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Chapter 27 - Evolution of Life
36. A random alteration in the sequence of DNA nucleotides that provides a new variant of the gene is A. gene mutation. B. polymorphism. C. gene frequency. D. disruption. E. allele frequency. A mutation occurs when nucleotides in the organisms DNA are in some way changed and passed on to offspring. This can lead to a new allelic choice in the population and therefore changes the frequency of alleles.
Bloom’s Level: 2. Understand Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Microevolution
37. Which statement is NOT true about the founder effect? A. It is a form of genetic drift. B. It produces a high frequency of some rare alleles in a small isolated population. C. Founding members contain a tiny fraction of the alleles found in the original population. D. The founder effect occurs when a population is subjected to near extinction and then recovers, so that only a few alleles are left in survivors. E. Chance determines which alleles are carried by the original founders. A bottleneck occurs when a random occurrence reduces a large population to a small one. The founder effect is seen when a small number of individuals establish a new population. Both of these are forms of genetic drift, random changes in allele frequencies by chance. If a population is very large, changes in allele frequencies due to chance alone are insignificant. In small populations these same changes can have profound effects on the overall genetic makeup of the population. It can even lead to the complete loss of one or more alleles or to a more frequent occurrence of rare alleles.
Bloom’s Level: 2. Understand Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Microevolution
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Chapter 27 - Evolution of Life
38. Which of the following does NOT reflect(s) the likely presence of a gene mutation(s)? A. Fruit flies subjected to intense radiation breed a wider array of variable offspring. B. A chemical leaking from the surface of an old abandoned coal mine alters a regulatory gene so that a cricket nymph develops an extra set of eyes. C. The bacteria that cause gonorrhea, a common sexually transmitted disease, have previously been killed by penicillin; however, after continuous usage of the antibiotic, penicillin-resistant strains are now becoming prevalent. D. Radiation causes an alteration of a DNA nucleotide sequence which is discovered when mapped, but which appears to be neither increasing nor decreasing in successive generations. E. Offspring always have the same coloration as the parents. All of these show the effects of some type of mutagen, leading to changes in DNA that are passed on to offspring except for offspring and parent types, maintaining the same color patterns.
Bloom’s Level: 5. Evaluate Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Microevolution
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Chapter 27 - Evolution of Life
39. Social research indicates that a person is most likely to marry someone from the same village or city, or a high school or college classmate. Therefore, the Hardy-Weinberg equilibrium does not apply well to human populations because A. allelic changes in one direction are balanced by changes in the opposite direction. B. there is no directional trend in selection of mates since most individuals marry someone. C. individuals are not pairing up by chance across the whole population, and genetic drift is more likely to change gene frequencies. D. this increases gene flow. E. we accumulate adaptive traits that improve the population. One of the requirements of the Hardy-Weinberg equilibrium is that mating be random. Any selection of mate will lead to a change in allele frequencies based on the character that led to the choice of that mate. Nonrandom mating occurs when individuals pair up, not by chance, but according to their genotypes or phenotypes. Inbreeding, or mating between relatives to a greater extent than by chance, is an example of nonrandom mating. Inbreeding decreases the proportion of heterozygotes and increases the proportions of homozygotes at all gene loci. In a human population, inbreeding increases the frequency of recessive abnormalities.
Bloom’s Level: 3. Apply Learning Outcome: 27.03.01 List the five conditions necessary for the allele frequencies of a population to be in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
40. Which of the following is true about genetic drift? A. It is more likely to occur in a large population than in a small population. B. It may lead to an allele becoming fixed in a population when its alternative allele is lost from the population. C. It increases the number of heterozygotes in a population. D. It increases the level of rare alleles in a population. E. It reduces the chances of mutation in a population. Genetic drift is a random change in allele frequencies in a population. If a population is very large, changes in allele frequencies due to chance alone are insignificant. In small populations, these same changes can have profound effects on the overall genetic makeup of the population. It can even lead to the complete loss of one or more alleles or to a more frequent occurrence of rare alleles.
Bloom’s Level: 2. Understand Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Microevolution
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Chapter 27 - Evolution of Life
41. The observation by that human reproductive potential would rapidly outstrip available food and living space was an important concept adopted by Darwin in formulating his theory of natural selection. A. Stanley Miller B. Graham Cairns-Smith C. Thomas Malthus D. Aleksandr Oparin E. Sidney Fox Before formulating the theory, Darwin read an essay on human population growth written by Thomas Malthus. Malthus observed that although humans have a great reproductive potential, many environmental variables, such as availability of food and living space, tend to keep the human population in check with factors such as disease and famine.
Bloom’s Level: 1. Remember Learning Outcome: 27.01.03 Summarize the four main observations that make up Darwin’s theory of natural selection. Section: 27.01 Topic: History of Evolutionary Theory
42. Which of the following conditions does NOT contribute to evolution? A. mutations B. gene flow C. genetic drift D. natural selection E. unchanging environmental conditions In unchanging environmental conditions there would be no selection pressure so the allele frequencies of the population would not change. Mutation would change allele frequencies because new alleles would be added. Natural selection would lead to changes in allele frequencies as some members of the population were able to survive and reproduce at a higher rate than others with different alleles. Gene flow and genetic drift also lead to different frequencies of existing alleles and therefore contribute to the processes of evolution.
Bloom’s Level: 2. Understand Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Microevolution
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Chapter 27 - Evolution of Life
43. Which of the following mechanisms does NOT result in variation in the gene frequencies from the ancestral population? A. founder effect B. differential reproduction C. genetic equilibrium D. genetic drift E. natural selection Differential reproduction occurs as natural selection leads to certain phenotypes with an advantage in both survival and reproduction, changing allele frequencies. Genetic drift and the founder effect also lead to different frequencies of existing alleles and, therefore, contribute to the processes of evolution. Genetic equilibrium is the change in the gene frequency of a population.
Bloom’s Level: 2. Understand Learning Outcome: 27.03.03 Identify an evolving population from a change in its allele frequencies over generations. Section: 27.03 Topic: Microevolution
44. Lamarck's proposal of the inheritance of acquired characteristics included the idea that A. the continual stretching of giraffe's necks to reach leaves led to longer necks. B. local catastrophes cause mass extinctions of species. C. species are only produced through special creation. D. species are fixed and unchanging over time. E. organisms are acted on by the environment. Lamarck proposed the theory of inheritance of acquired characteristics, which states that the environment can bring about inherited change. One example he gave is that the long neck of a giraffe developed over time because animals stretched their necks to reach food high in trees and then passed gradually longer necks to their offspring.
Bloom’s Level: 2. Understand Learning Outcome: 27.01.02 Identify how Darwin’s and Lamarck’s theories of evolution are different. Section: 27.01 Topic: History of Evolutionary Theory
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Chapter 27 - Evolution of Life
45. Lamarck's ideas on evolution were adopted by some Russian scientists, including Michurin and Lysenko in Stalinist Russia. Their textbooks printed Lysenko's assertions that a wheat plant could be made cold-hardy by conditioning in cold storage, or that workmen who develop strong muscles due to working in the mines would produce children who would be born stronger. Which of the major elements of evolution by natural selection is violated by these examples? A. The organisms vary in traits. B. The acquired characteristics are inherited. C. More young are born than can survive. D. Some individuals are better adapted to the environment. E. The environment selects for phenotype. Lamarck’s idea that changes that are acquired by a parent due to the environment can be passed on to offspring. Only genetically controlled traits can be passed along. There must be a molecular mechanism to allow traits to be passed on to future generations.
Bloom’s Level: 3. Apply Learning Outcome: 27.01.02 Identify how Darwin’s and Lamarck’s theories of evolution are different. Section: 27.01 Topic: History of Evolutionary Theory
46. The organisms examined by Darwin on the Galápagos Islands that were most important in his development of the theory of natural selection were A. rabbits and hares. B. plants. C. finches. D. fish. E. monkeys and armadillos. When Darwin arrived at the Galápagos Islands, he began to study the diversity of finches; each of these finches was adapted to gathering and eating a different type of food. These adaptations could best be explained by assuming they had diverged from a common ancestor. He found such a hypothetical ancestor on the mainland of South America, supporting his theory. With this type of evidence, Darwin concluded that species evolve (change) with time.
Bloom’s Level: 1. Remember Learning Outcome: 27.01.03 Summarize the four main observations that make up Darwin’s theory of natural selection. Section: 27.01 Topic: History of Evolutionary Theory
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Chapter 27 - Evolution of Life
47. When he arrived at the Galápagos Islands, Darwin did not observe the amazing tool-using "woodpecker finch" that can modify twigs to pry out grubs. Because there are no true woodpeckers on the Galápagos Islands, this behavior allows this finch to exploit an untapped food source. However, not all members of this species exhibit this behavior, which is probably learned from watching other finches. Therefore, which of the following is NOT true? A. Young isolated at hatching will not know how to do this. B. It is probably not "hardwired" in the brain as a behavior passed on genetically. C. There must be a great advantage to reaching this food source for this learned behavior to be repeated by most descendants each generation. D. As an acquired characteristic, in a strict sense this is not part of the adaptive radiation of finches on the Galápagos. E. This "learned" behavior will not lead to evolutionary change in the woodpecker population. In order for a trait to be passed genetically, and therefore lead to a change in allele frequencies, it must be determined by the sequence of nucleotides contained in the DNA. A trait that is learned is not one that can be passed from parent to offspring through the DNA inherited by the offspring.
Bloom’s Level: 4. Analyze Learning Outcome: 27.01.01 List two observations Darwin made on his voyage that influenced the development of his theory. Section: 27.01 Topic: History of Evolutionary Theory
48. Which of the following is true about natural selection? A. It acts on genotypes rather than phenotypes. B. It assures the survival of each fit individual. C. On average, it favors the survival of young with adaptive characteristics. D. It always selects for more complex forms. E. It always selects for forms that are a mutated variation. Natural selection acts on the phenotypes of organisms. The organism with a phenotype that gives it an advantage in survival and reproduction will lead to the phenotype being seen in the population more and more frequently.
Bloom’s Level: 2. Understand Learning Outcome: 27.04.02 List the four requirements of evolution by natural selection. Section: 27.04 Topic: Natural Selection
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Chapter 27 - Evolution of Life
49. Movement of alleles between populations such as by the migration of breeding individuals is called A. mutations. B. genetic drift. C. gene flow. D. nonrandom mating. E. natural selection. Gene flow is defined as the movement of alleles among populations; it occurs when individuals migrate from one population to another and breed in that new population. Gene flow among populations keeps their gene pools similar. It also prevents close adaptation to a local environment.
Bloom’s Level: 1. Remember Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Microevolution
50. If two adjacent populations of the same species show gene flow, then the two populations will A. become more similar in their gene pools. B. become isolated from each other. C. develop into different species. D. adapt to different conditions and become separate. E. become the same single population. Gene flow is defined as the movement of alleles among populations; it occurs when individuals migrate from one population to another and breed in that new population. Gene flow among populations keeps their gene pools similar. It also prevents close adaptation to a local environment.
Bloom’s Level: 2. Understand Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Microevolution
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Chapter 27 - Evolution of Life
51. Inbreeding within a population is an example of A. mutations. B. genetic drift. C. gene flow. D. nonrandom mating. E. natural selection. Inbreeding, or mating between relatives to a greater extent than by chance, is an example of nonrandom mating. Inbreeding decreases the proportion of heterozygotes and increases the proportions of homozygotes at all gene loci. In a human population, inbreeding increases the frequency of recessive abnormalities.
Bloom’s Level: 1. Remember Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Microevolution
52. Which type of natural selection increases the frequency of one extreme phenotype? A. directional B. diversifying C. stabilizing D. disruptive E. nonrandom Directional selection occurs when one or the other of the phenotypes at the extremity of the phenotypic range is favored and the distribution curve shifts in that direction. This changes the average phenotype in a population. Such a shift can occur when a population is adapting to a changing environment.
Bloom’s Level: 1. Remember Learning Outcome: 27.04.03 Identify examples of the three types of natural selection. Section: 27.04 Topic: Natural Selection
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Chapter 27 - Evolution of Life
53. Which statement is NOT true about nonrandom mating? A. Inbreeding is mating between relatives more often than by chance. B. Inbreeding is a change in allele frequencies that increases the proportion of heterozygotes in the population. C. An example is when a tall man marries a tall woman. D. Nonrandom mating tends to cause subdivision into two phenotypic classes with reduced gene flow between them. E. An example is when the strongest male wins the opportunity to mate with his choice of females. Inbreeding, or mating between relatives to a greater extent than by chance, is an example of nonrandom mating. Inbreeding decreases the proportion of heterozygotes and increases the proportions of homozygotes at all gene loci. In a human population, inbreeding increases the frequency of recessive abnormalities.
Bloom’s Level: 3. Apply Learning Outcome: 27.04.01 Define the five agents of evolutionary change. Section: 27.04 Topic: Natural Selection
54. Which type of natural selection occurs when an intermediate phenotype is favored? A. disruptive selection B. directional selection C. stabilizing selection D. genetic drift selection E. adaptive radiation With stabilizing selection, extreme phenotypes are selected against and individuals near the average are favored. Stabilizing selection can improve adaptation of the population to those aspects of the environment that remain constant.
Bloom’s Level: 1. Remember Learning Outcome: 27.04.03 Identify examples of the three types of natural selection. Section: 27.04 Topic: Natural Selection
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Chapter 27 - Evolution of Life
55. Which of the following agents of evolutionary change results in adaptation? A. mutation B. genetic drift C. gene flow D. nonrandom mating E. natural selection An adaptation is a feature of an organism that is suitable to the environment and increases the organism’s reproductive success. Natural selection is the process that leads to organisms with more adaptive traits surviving and reproducing more frequently.
Bloom’s Level: 2. Understand Learning Outcome: 27.04.02 List the four requirements of evolution by natural selection. Section: 27.04 Topic: Natural Selection
56. Which is NOT one of the preconditions in a population where natural selection is at work? A. There is variation that can be inherited in a population. B. The population always becomes adapted to its environment. C. Many more individuals are produced by a population than can survive and reproduce. D. Adaptive characteristics in some individuals make them more likely to survive and reproduce. E. Heritable variations must be passed on the subsequent generations. Natural selection does not require that the organisms adapt to the environment; if the phenotype does not exist in the population that will allow the population to adapt to the environmental conditions, the population may become extinct.
Bloom’s Level: 3. Apply Learning Outcome: 27.04.02 List the four requirements of evolution by natural selection. Section: 27.04 Topic: Natural Selection
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Chapter 27 - Evolution of Life
57. Which statement is NOT true about the maintenance of variation in a population? A. Selection for adaptation to a particular environment ensures that the population will become stronger and more viable under any and all conditions. B. Only phenotypes are acted on by selection, so heterozygotes serve as a reservoir of recessive alleles that may be adaptive in a different environment. C. Heterozygote superiority may lead to selection for the heterozygote above either homozygote. D. Variation is maintained through mutation, recombination, gene flow, and changed conditions. E. Variations allow for the presence of conditions such as sickle-cell anemia. Adaptation to a particular environment does not guarantee that the organism will be equally well adapted to all environments. Natural selection can select for a different phenotype if the environment changes in such a way that the original trait is no longer adaptive.
Bloom’s Level: 2. Understand Learning Outcome: 27.04.03 Identify examples of the three types of natural selection. Section: 27.04 Topic: Natural Selection
58. Disruptive selection is described in the text with the case of British land snails. In the grassy fields, the light-banded snails escape bird predators. In the darker forest, the dark snails survive and the light-banded snails are eaten. As long as the snails continue to cruise across the British landscape mating at the same season and having access to each other, why doesn't this "disruptive selection" eventually lead to two separate species? A. There is no reproductive isolation to prevent gene flow. B. They are already two separate species and the intermediate forms are hybrids. C. The color forms are probably not genetically determined. D. There must be some unknown factor producing an equal stabilizing selection "to hold the species together." E. This will result in the formation of two species if given long enough time. Gene flow functions to keep the variation in the population and can prevent natural selection from completely removing an allele from the population. Heterozygotes who carry variations that may, or may not, be seen in the population still maintain the allele. As long as reproductive isolation does not occur, the snails will continue to be one species.
Bloom’s Level: 5. Evaluate Learning Outcome: 27.04.03 Identify examples of the three types of natural selection. Section: 27.04 Topic: Natural Selection
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Chapter 27 - Evolution of Life
59. What is the correct order of the steps regarding speciation? A. geographic isolation → one species → reproductive isolation → two species B. two species → geographic isolation → one species → reproductive isolation C. one species → geographic isolation → reproductive isolation → two species D. reproductive isolation → one species → reproductive isolation → two species E. reproductive isolation → two species → reproductive isolation → one species If geographic isolation occurs with a species that is divided into two populations, then the two populations can become reproductively isolated from one another. Once reproductive isolation has occurred, then they become two species, not one.
Bloom’s Level: 2. Understand Learning Outcome: 27.05.02 Explain how adaptive radiation can lead to speciation. Section: 27.05 Topic: Speciation
60. According to the evolutionary theory, which statement is NOT true? A. Evolution explains the unity of life. B. Evolution explains the diversity of life. C. All living things are NOT descended from a common ancestor. D. Diversity occurs because various living things are adapted to different ways of life. E. All living things share the same fundamental characteristics. Evolutionary theory states that once a true cell formed, biological evolution began. Biological evolution is all the changes that have occurred in living things since the beginning of life due to differential reproductive success. This has led to a unity of all life since the basic processes still work mainly in the same way in all life. It has also led to the diversity of life as different species of organisms have adapted to survive in different environments.
Bloom’s Level: 3. Apply Learning Outcome: 27.02.01 Define biological evolution. Section: 27.02 Topic: History of Evolutionary Theory
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Chapter 27 - Evolution of Life
61. During allopatric speciation A. gene flow continues between subpopulations. B. reproduction between all subpopulations is impossible. C. a geographic separation occurs between subpopulations. D. wide phenotype differences appear between subpopulations. E. subpopulations are still able to interbreed. During allopatric speciation a single population becomes physically separated by a geographic barrier and divided into two populations. Once this has occurred and gene flow between the two has ceased, the accumulation of differences will lead to two populations who can no longer interbreed; at this point they have become separate species.
Bloom’s Level: 1. Remember Learning Outcome: 27.05.01 Distinguish between prezygotic and postzygotic isolation mechanisms. Section: 27.05 Topic: Speciation
62. During sympatric speciation A. evolution ceases for a time. B. wide phenotype differences disappear between subpopulations. C. reproductive isolation between certain subpopulations occurs. D. a geographic separation occurs between certain subpopulations. E. mutations begin to appear, making the subpopulations distinctly different. During sympatric speciation, gene flow between two populations ceases and the accumulation of differences will lead to two populations who can no longer interbreed; at this point they have become separate species.
Bloom’s Level: 1. Remember Learning Outcome: 27.05.01 Distinguish between prezygotic and postzygotic isolation mechanisms. Section: 27.05 Topic: Speciation
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Chapter 27 - Evolution of Life
63. Gulls recognize their own species by the type of ring around the eye. This is an example of A. geographical isolation. B. ecological isolation. C. behavioral isolation. D. physiological isolation. E. temporal isolation. Any form of behavior, such as courtship dances, different songs, calls, pheromones, or other signals can lead to reproductive isolation. Identifying a potential mate through visual cues like the eye ring would be based on the organisms following the appropriate behavior.
Bloom’s Level: 2. Understand Learning Outcome: 27.05.01 Distinguish between prezygotic and postzygotic isolation mechanisms. Section: 27.05 Topic: Speciation
64. The model of speciation that requires some time with geographic barriers between two populations, allowing evolution of reproductive isolation, is A. allopatric speciation. B. phyletic gradualism. C. sympatric speciation. D. punctuated equilibrium. E. prezygotic isolation. One type of speciation, called allopatric speciation, usually occurs when populations become separated by a geographic barrier and gene flow is no longer possible. It is also possible that a single population could suddenly divide into two reproductively isolated groups without being geographically isolated; this type of speciation is called sympatric speciation. Traditionally, evolutionists have supported a model of speciation called phyletic gradualism, which states that change is very slow but steady within a lineage before and after a divergence. However another model for speciation is punctuated equilibrium which states that long periods of stasis, or no visible change, are followed by rapid periods of speciation.
Bloom’s Level: 1. Remember Learning Outcome: 27.05.01 Distinguish between prezygotic and postzygotic isolation mechanisms. Section: 27.05 Topic: Speciation
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Chapter 27 - Evolution of Life
65. Which of the following would NOT result in reproductive isolation? A. Snakes found in one state are genetically identical to snakes found in a neighboring state with no geographic barriers separating them. B. Fruit flies on one Hawaiian island live for hundreds of generations and do not come in contact with fruit flies on another island except when blown there by rare tropical storms. C. One brood of the 17-year cicada emerged in 1987 (and will do so every 17 years) and lives a few months as adults; another brood emerged in 1992 (and will do so every 17 years); the larvae of both feed side by side on tree roots. D. A lion and a tiger mate in the artificial confines of a zoo but the offspring is infertile. E. Two populations of crickets are indistinguishable in physical features but the females in each group only come to the different songs of their males. Reproductive isolation would not result from having genetically identical populations spread out over a wide range. As long as the populations are experiencing gene flow reproductive isolation would not be possible.
Bloom’s Level: 4. Analyze Learning Outcome: 27.05.01 Distinguish between prezygotic and postzygotic isolation mechanisms. Section: 27.05 Topic: Speciation
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Chapter 27 - Evolution of Life
66. An insect population lives along the edge of a north-south mountain range. The populations from the east and west slope eventually join in a low northern pass and interbreed, producing fertile offspring; but they do not circle around the southern edge because of a desert barrier. When glaciers move southward, the populations are pushed south of the northern pass and are isolated. While isolated, the two populations develop enough differences over time that when the glaciers retreat north and the insects again share the same pass, they no longer mate at the same time, nor can they produce fertile offspring. These insects A. began as one species and therefore remain one species. B. were originally two species and remain two species. C. were originally two species but are now one species. D. were originally one species but are now two species. E. The number of species cannot be determined from the information given. If geographic isolation occurs with a species that is divided into two populations, then the two populations can become reproductively isolated from one another. Once reproductive isolation has occurred, then they become two species, not one.
Bloom’s Level: 3. Apply Learning Outcome: 27.05.01 Distinguish between prezygotic and postzygotic isolation mechanisms. Section: 27.05 Topic: Speciation
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Chapter 27 - Evolution of Life
67. In the case of Darwin's finches, an ancestral finch species from the mainland arrived on the Galápagos Islands and soon developed into many new species via adaptive radiation. The finches did NOT undergo adaptive radiation back on the mainland. What is the most plausible biological explanation? A. Directional selection works better on islands. B. Competition from many other bird species on the mainland provided stabilizing selection that was absent on the islands. C. The environment on the mainland was completely uniform. D. The founder effect greatly expanded the variation in alleles in the Galápagos finch gene pool. E. The ancestral mainland finch was reproductively isolated. When the founding population of finches reached the Galápagos Islands, they would have encountered a variety of habitats and food sources but there would not have been any other species of birds competing for these resources. This would have resulted in multiple selective forces acting on the population and produces multiple species. On the mainland, where there are many of competing species of birds, the finches would have been adapted to a narrow set of environmental conditions.
Bloom’s Level: 5. Evaluate Learning Outcome: 27.05.02 Explain how adaptive radiation can lead to speciation. Section: 27.05 Topic: Speciation
68. What information is contained in the branches of a cladogram? A. a current species and its descendents B. a recent common ancestor and two outgroups C. a recent common ancestor and all of its descendents D. all of the potential ancestors of a group E. all species classified in a particular genus Branches in cladograms are called clades. A clade contains a most recent ancestor and all of its descendants.
Bloom’s Level: 2. Understand Learning Outcome: 27.06.01 Recognize how phylogenetics, the theory of evolution, and classification are interrelated. Section: 27.06 Topic: Systematics
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Chapter 27 - Evolution of Life
69. The new model of evolution that proposes that organisms spend most of their time in fairly unchanging form and undergo occasional bursts of speciation is called A. genetic drift. B. phyletic gradualism. C. sympatric speciation. D. punctuated equilibrium. E. the biological species concept. One model for speciation is punctuated equilibrium, which states that long periods of stasis, or no visible change, are followed by rapid periods of speciation. The more traditional model of speciation, called phyletic gradualism, states that change is very slow but steady within a lineage before and after a divergence.
Bloom’s Level: 1. Remember Learning Outcome: 27.05.03 Compare and contrast gradualism with punctuated equilibrium. Section: 27.05 Topic: Speciation
70. The three-domain system is primarily based on A. molecular data showing eukaryotes are much more closely related than previously thought. B. structural data showing eukaryotes are far more diverse than previously thought. C. molecular data showing prokaryotes are far more diverse than previously thought. D. new biogeographical data showing prokaryotes are far more diverse than previously thought. E. new classification concepts that have shown everything except molecular data to be irrelevant. The three-domain system is based on data from the sequencing of genes and cellular data. These data provided new information, suggesting that there are two groups of prokaryotes, named Bacteria and Archaea. These two groups are so fundamentally different from each other they have been assigned to separate domains. The Bacteria arose first, followed by the Archaea, and then the Eukarya. The Archaea and Eukarya are more closely related to each other than either is to the Bacteria.
Bloom’s Level: 2. Understand Learning Outcome: 27.06.01 Recognize how phylogenetics, the theory of evolution, and classification are interrelated. Section: 27.06 Topic: Phylogeny
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Chapter 27 - Evolution of Life
71. According to the proposed three-domain system, A. archaea and eukarya are more closely related than either is to the bacteria. B. archaea and bacteria are more closely related than either is to the eukarya. C. bacteria and eukarya are more closely related than either is to the archaea. D. all three groups arose separately. E. All three groups are equally related. The data provide new information suggesting that there are two groups of prokaryotes, named Bacteria and Archaea. These two groups are so fundamentally different from each other they have been assigned to separate domains. The Bacteria arose first, followed by Archaea, and then Eukarya. Archaea and Eukarya are more closely related to each other than either is to the Bacteria.
Bloom’s Level: 1. Remember Learning Outcome: 27.06.02 Define the goal of systematic biology and its two branches of study: taxonomy and phylogenetics. Section: 27.06 Topic: Systematics
72. Phylogeny refers to the A. development of an organism. B. evolutionary history of a species. C. taxonomic hierarchies of cladistics. D. reproductive isolation of species. E. naming of organisms. A phylogeny relies on a combination of data from the fossil record and comparative anatomy and development, with an emphasis today on molecular data, to reconstruct the evolutionary history of a group of organisms.
Bloom’s Level: 1. Remember Learning Outcome: 27.06.01 Recognize how phylogenetics, the theory of evolution, and classification are interrelated. Section: 27.06 Topic: Phylogeny
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Chapter 27 - Evolution of Life
73. Once a true cell was formed, biological evolution began. Biological evolution is all the changes that have occurred in living things since the beginning of life due to differential reproductive success. This means that A. living things have been evolving toward the perfect form. B. living things have begun reproducing. C. living things have all evolved from a common ancestor. D. life has recently begun to change through evolutionary processes. E. life has evolved to become increasingly complex. If life began evolving from the beginning through differences in survival and reproduction, then natural selection has led to many different forms but only those forms that are best suited to the current environment are selected to be passed to future generations; there is no perfect form, only one that is best suited. If a less complex form was best suited to the environment, then it would be the one selected for. Since biological evolution began with the first cell, then all cells must be derived from that first one.
Bloom’s Level: 3. Apply Learning Outcome: 27.02.01 Define biological evolution. Section: 27.02 Topic: Evidence for Evolution
74. All living organisms share characteristics such as DNA, the molecule that passes information between generations; metabolic processes like glycolysis, and the ability to utilize specific energy sources. This is thought to be due to A. biological evolution. B. limited environmental stimuli. C. the presence of transitional forms. D. postzygotic isolation. E. prezygotic isolation. It is thought that all living things share characteristics because they all evolved from an ancestral form. The structures that evolved early on would have been transmitted with modifications from one generation to the next.
Bloom’s Level: 3. Apply Learning Outcome: 27.02.01 Define biological evolution. Section: 27.02 Topic: Evidence for Evolution
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Chapter 27 - Evolution of Life
75. There are two species of larkspur that have two different size flowers; Delphinium decorum has large, fully opened flowers and Delphinium nudicaule has small bud, like flowers. These two species are both fertilized by insects, but they cannot be fertilized by the same type of insect because the smaller flower of D. nudicaule will not permit larger insects to reach the nectar of the flower and the larger flower of the D. decorum is typically fed on by larger insects. Which type of isolating mechanism would this be? A. postzygotic isolation B. temporal isolation C. behavioral isolation D. habitat isolation E. mechanical isolation Since it is not possible for fertilization to occur, because of an inability of insects to transfer pollen between the two species of larkspur, this is a form of mechanical isolation, Postzygotic isolation would occur after fertilization happens. Temporal isolation is based on differences in timing of reproduction. Habitat isolation refers to species that live in different areas of the ecosystem and behavioral isolation is based on signals that pass between the male and female to determine whether or not they will mate.
Bloom’s Level: 3. Apply Learning Outcome: 27.05.01 Distinguish between prezygotic and postzygotic isolation mechanisms. Section: 27.05 Topic: Speciation
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Chapter 27 - Evolution of Life
76. Different species of female fireflies flash different patterns during mating season. This is an example of _ isolation. A. postzygotic isolation B. temporal isolation C. behavioral isolation D. habitat isolation E. mechanical isolation This is behavioral isolation since the different species of female fireflies signal their species to male fireflies. Mechanical isolation describes when fertilization cannot occur because of an inability for gametes to reach each other. Postzygotic isolation would occur if the zygote could not survive after fertilization happens. Temporal isolation is based on differences in timing of reproduction. Habitat isolation refers to species that live in different areas of the ecosystem.
Bloom’s Level: 3. Apply Learning Outcome: 27.05.01 Distinguish between prezygotic and postzygotic isolation mechanisms. Section: 27.05 Topic: Speciation
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Chapter 27 - Evolution of Life
77. You are traveling across the Pacific Ocean when you come across a small island, island #1. There is lush vegetation and several species of beetles that vary in the color of their carapace. The colors you find are green, blue, purple, and black. A second island, 200 miles closer to the mainland (island #2), is mostly sand with some palm trees and has a single species of beetle that has a brown carapace. You examine the colored beetles and discover that they are all related and that they are related to the brown species on the second island. Which of the following is the most likely scenario? A. The brown beetle migrated to island #1 and gave rise to the colored species on island #2 through adaptive radiation. B. The colored species of beetle on island #1 migrated to island #2 and gave rise to the brown beetle through extinction. C. The brown beetle evolved from the colored beetles on island #1 and then migrated to island #2. D. The brown beetle gave rise to all of the colored beetles on island #2 which then migrated to island #1. E. A beetle of some undetermined color gave rise to the beetles that are brown, green, purple, and black and they then migrated to different islands. The brown beetles of island #2 have a color that is adaptive for that environment, while the beetles of island #1 are adaptive for multiple colors of vegetation. Adaptive radiation could explain the multiple species found on island #1 arising from the ancestral species found on island #2.
Bloom’s Level: 4. Analyze Learning Outcome: 27.05.02 Explain how adaptive radiation can lead to speciation. Section: 27.05 Topic: Speciation
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Chapter 27 - Evolution of Life
78. The difference between phyletic gradualism and punctuated equilibrium is that A. phyletic gradualism can only happen with eukaryotes and punctuated equilibrium happens with prokaryotes. B. punctuated equilibrium shows speciation in a relatively short time, while phyletic gradualism shows speciation as changes accumulate over vast periods of time. C. phyletic gradualism shows speciation in a relatively short time, while punctuated equilibrium shows speciation as changes accumulate over vast periods of time. D. phyletic gradualism shows in the fossil record and punctuated equilibrium does not. E. There is no difference between punctuated equilibrium and phyletic gradualism. Both punctuated equilibrium and phyletic gradualism describe processes of speciation. Phyletic gradualism is the term used to describe the process when small changes accumulate in two or more subsets of a population until they are no longer the same species. Punctuated equilibrium refers to the model where a species undergoes rapid change and speciation after a period of no change.
Bloom’s Level: 2. Understand Learning Outcome: 27.05.03 Compare and contrast gradualism with punctuated equilibrium. Section: 27.05 Topic: Speciation
79. Phylogenetics uses evolutionary trees called cladograms to arrange species into clades, or related groups, showing all of the descendents of a common ancestor. The branches of the tree are based on all but which of the following? A. shared derived characteristics B. convergent evolution C. homologous structures D. morphological data E. DNA sequences Phylogenetics uses information on DNA, structures that may look different but are derived from the structure, characteristics that are derived from a common ancestor, and body type. Convergent evolution refers to structures that look similar but are derived from different sources. These would not demonstrate relatedness but rather evolution in similar environments.
Bloom’s Level: 3. Apply Learning Outcome: 27.06.03 Interpret the information provided in both a cladogram and a phylogenetic tree. Section: 27.06 Topic: Phylogeny
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Chapter 27 - Evolution of Life
80. In a cladogram, clades are groups of related organisms. In the past, phylogeneticists built clades using the idea of parsimony, that the pattern that uses the fewest evolutionary changes is the most likely to be correct. Today, with the use of computers and DNA sequences, the explanation that is thought to be the best is A. one that shows the most changes in nucleotide sequences. B. one that eliminates any similar DNA sequences. C. one that shows the DNA unique to that species. D. one that shows the fewest changes in the nucleotide sequences. E. still one that uses the same system of parsimony. More modern approaches use complicated computer simulations to determine the most likely phylogenetic reconstruction based on the data that they have accumulated. Today, these data are most commonly DNA or rRNA sequences. As a result, modern phylogenetic studies are based on the assumption that the more closely species are related, the fewer changes will be found in molecular nucleotide sequences.
Bloom’s Level: 5. Evaluate Learning Outcome: 27.06.03 Interpret the information provided in both a cladogram and a phylogenetic tree. Section: 27.06 Topic: Phylogeny
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Chapter 27 - Evolution of Life
81. Which piece of evidence did Darwin observe during his 5-year journey aboard the HMS Beagle? A. A South American species of finch is most likely the ancestor of the Galápagos Island finches. B. Species do not change over time. C. The environment can bring about inherited change in an individual. D. All species share the same basic genetic and molecular makeup. E. The earth is approximately 4.6 billion years old. During Darwin's voyage on the HMS Beagle he speculated that the finches on the Galápagos Islands were descended from a species of finches found on he mainland of South America. This helped him develop his theory of natural selection. Species do not change over time was a concept Darwin had at the beginning of his journey. The more evidence he saw the more he dismissed the idea. The idea that the environment can bring about inherited change in an individual was proposed by Lamarck. All species share the same basic genetic and molecular makeup was not determined until centuries after Darwin's journey. The age of the earth was not determined until centuries after Darwin's journey.
Bloom’s Level: 1. Remember Learning Outcome: 27.01.01 List two observations Darwin made on his voyage that influenced the development of his theory. Section: 27.01 Topic: History of Evolutionary Theory
Short Answer Questions 82. Summarize the four main observations that make up Darwin's theory of natural selection.
1. Individual organisms within a species exhibit variation that can be passed from one generation to the next. 2. Organisms compete for available resources. 3. Individual organisms within a population differ in their reproductive success. 4. Organisms become adapted to conditions as the environment changes.
Bloom's Level: 6. Create Learning Outcome: 27.01.03 Summarize the four main observations that make up Darwin’s theory of natural selection. Section: 27.01 Topic: Natural Selection
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Chapter 27 - Evolution of Life
Multiple Choice Questions 83. According to Lamarck, which variable would have the greatest influence on the evolution of an organism? A. the environment B. the genetics of the individual C. Both the environment and genetics are equally responsible for the evolution of an organism. D. Neither the environment nor the genetics of an organism play a role in its evolution. E. A creator would determine the evolution of an organism. Lamarck proposed that the environment can bring about inherited change in an organism. Darwin proposed that the genetics of an individual must be passed on to the next generation. Many people supported the idea that a creator fixed organisms as they were and that they did not evolve.
Bloom’s Level: 1. Remember Learning Outcome: 27.01.02 Identify how Darwin’s and Lamarck’s theories of evolution are different. Section: 27.01 Topic: History of Evolutionary Theory
84. Which of the following is not a feature of punctuated equilibrium? A. slow and steady change within a lineage before and after a divergence B. long periods of stasis C. no visible change in a species for long periods D. rapid periods of speciation E. relatively few transitional species Slow and steady change within a lineage before and after a divergence is a feature of phyletic gradualism. Long periods of stasis, no visible change in a species, rapid periods of speciation, and relatively few transitional species are all features of punctuated equilibrium.
Bloom’s Level: 2. Understand Learning Outcome: 27.05.03 Compare and contrast gradualism with punctuated equilibrium. Section: 27.05 Topic: Macroevolution
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Chapter 27 - Evolution of Life
True / False Questions 85. If a population goes from an allele frequency of 0.12 A & 0.88 a to 0.23 A & 0.77 a in the course of seven generations, we can conclude that it is evolving. TRUE One of the measures of evolution is to determine if the allele frequency has changed over the course of several generations. If this populations allele frequencies changed from 0.12 A & 0.88 a to 0.23 A & 0.77 a in the course of seven generations, then we can conclude that it has evolved.
Bloom’s Level: 2. Understand Learning Outcome: 27.03.03 Identify an evolving population from a change in its allele frequencies over generations. Section: 27.03 Topic: Evidence for Evolution
86. Within the field of systematic biology scientists use the characteristics of living and fossil organisms to infer the evolutionary relationships among organisms. TRUE Systematic biology is a quantitative science that uses the characteristics of living and fossil organisms to infer the evolutionary relationships among organisms.
Bloom’s Level: 1. Remember Learning Outcome: 27.06.02 Define the goal of systematic biology and its two branches of study: taxonomy and phylogenetics. Section: 27.06 Topic: Systematics
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Chapter 27 - Evolution of Life
Short Answer Questions 87. Name the isolating mechanisms that exist within nature and indicate which ones are prezygotic and which ones are postzygotic. Prezygotic: habitat isolation, temporal isolation, behavioral isolation, mechanical isolation, and gamete isolation Postzygotic: zygote mortality, hybrid sterility, and F2 fitness
Bloom's Level: 6. Create Learning Outcome: 27.05.01 Distinguish between prezygotic and postzygotic isolation mechanisms. Section: 27.05 Topic: Microevolution
Multiple Choice Questions 88. If you were assigned the task of constructing a cladogram representing the evolutionary relationship between humans, gorillas, tigers, lizards, and sharks, what feature could you use as the shared derived trait? A. presence of a jaw B. presence of hair C. give birth to live young D. lay eggs E. production of milk to nourish their young When constructing a cladogram a shared derived trait is used. This is a trait shared by all of the species being considered. The presence of jaws is the only trait that would be present in humans, gorillas, tigers, lizards, and sharks. Hair is lacking in the lizard and sharks. Giving birth to live young is lacking in the lizards and some sharks. Laying eggs is lacking in the human, gorilla, tigers, and some sharks. Production of milk to nourish their young is lacking in the lizards and sharks.
Bloom’s Level: 3. Apply Learning Outcome: 27.06.03 Interpret the information provided in both a cladogram and a phylogenetic tree. Section: 27.06 Topic: Phylogeny
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Chapter 27 - Evolution of Life
True / False Questions 89. Taxonomy is the branch of systemic biology that studies the evolutionary relatedness among groups of organisms. FALSE Taxonomy is the branch of systemic biology concerned with identifying, naming, and classifying organisms. Phylogenetics is the branch of systemic biology that studies the evolutionary relatedness among groups of organisms.
Bloom’s Level: 1. Remember Learning Outcome: 27.06.02 Define the goal of systematic biology and its two branches of study: taxonomy and phylogenetics. Section: 27.06 Topic: Taxonomy
Short Answer Questions 90. List the five conditions required for the Hardy-Weinberg equilibrium to be present in a population. 1. No mutations. 2. No genetic drift. 3. No gene flow. 4. Random mating. 5. No selection.
Bloom's Level: 6. Create Learning Outcome: 27.03.01 List the five conditions necessary for the allele frequencies of a population to be in Hardy-Weinberg equilibrium. Section: 27.03 Topic: Microevolution
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Chapter 28 - Viruses, Bacteria, and Archaea
Chapter 28 Viruses, Bacteria, and Archaea
Multiple Choice Questions 1. Which type of prokaryotes are able to live in the most extreme environments? A. cyanobacteria B. anaerobic bacteria C. archaea D. heterotrophic E. aerobic bacteria Archaea are sometimes called extremophiles since they can live in conditions that no other organisms can tolerate. They can live in extremely hot, salty, acidic, and anaerobic environments.
Bloom’s Level: 2. Understand Learning Outcome: 28.03.03 Distinguish between halophiles, thermoacidophiles, and methanogens. Section: 28.03 Topic: Archaea
2. The archaea include all of the following EXCEPT the A. methanogens. B. lichens. C. halophiles. D. thermoacidophiles. E. prokaryotes. All of the terms listed describe one or more types of archaea except for lichens, which are organisms that are a combination of algae and fungi.
Bloom’s Level: 3. Apply Learning Outcome: 28.03.03 Distinguish between halophiles, thermoacidophiles, and methanogens. Section: 28.03 Topic: Archaea
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Chapter 28 - Viruses, Bacteria, and Archaea
3. Which relationship is NOT correct? A. halobacteria—high salt concentration B. thermoacidophiles—cold Antarctic ice C. methanogens—swamps and marshes D. cyanobacteria—photosynthesis E. methanogens—biogas and greenhouse warming All of the relationships listed are correct except for thermoacidophile and the Antarctic. Thermoacidophiles would live in extremely hot and acidic environments such as the hot springs at Yellowstone National Park.
Bloom’s Level: 2. Understand Learning Outcome: 28.03.03 Distinguish between halophiles, thermoacidophiles, and methanogens. Section: 28.03 Topic: Archaea
4. Which statement is NOT true about bacteria? A. They lack mitochondria. B. They lack a nucleus but contain DNA. C. They reproduce sexually. D. They occur in three basic shapes. E. They have a single circular chromosome. Bacteria are single-celled organisms; they do not have any membrane-bound organelles including a nucleus. Their DNA is a single circular chromosome that is located in the cytoplasm of the cell. Bacteria are found in one of three shapes; bacilli (rod), cocci (spheres), and spirillium (curved). They reproduce asexually through binary fission.
Bloom’s Level: 2. Understand Learning Outcome: 28.04.01 Identify the major structural features of bacteria. Section: 28.04 Topic: Prokaryote Structure
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Chapter 28 - Viruses, Bacteria, and Archaea
5. The three shapes of bacteria are A. haplontic, diplontic, and alternations. B. anaerobic, aerobic, and facultative. C. heterotrophic, autotrophic, and chemosynthetic. D. bacillus, coccus, and spirillum. E. helical, spiral, and streptococci. Bacteria are found in one of three shapes: bacilli (rod), cocci (spheres), and spirillum (curved).
Bloom’s Level: 1. Remember Learning Outcome: 28.04.01 Identify the major structural features of bacteria. Section: 28.04 Topic: Prokaryote Structure
6. The response of bacteria that allows them to be identified as Gram-positive or Gramnegative is due to A. the type of DNA or RNA they contain. B. their ability or lack of ability to fix nitrogen. C. whether they are aerobic or anaerobic. D. whether their cell walls have a thin or thick layer of peptidoglycan. E. the rRNA sequences they demonstrate. Cell walls that have a thick layer of peptidogylcan outside the plasma membrane stain purple with the Gram stain and are called Gram-positive bacteria. If the peptidogylcan layer is either thin or missing, then the bacterial cells stain pink and are called Gram-negative.
Bloom’s Level: 2. Understand Learning Outcome: 28.04.01 Identify the major structural features of bacteria. Section: 28.04 Topic: Bacteria Topic: Prokaryote Structure
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Chapter 28 - Viruses, Bacteria, and Archaea
7. Which type of organism will form endospores when faced with unfavorable environmental conditions? A. viruses B. bacteria C. algae D. protozoans E. fungi In unfavorable conditions bacteria will form thick-walled dehydrated structures called endospores that are capable of surviving harsh conditions. These structures are not for reproduction but serve to preserve the bacteria until conditions improve.
Bloom’s Level: 1. Remember Learning Outcome: 28.04.02 Describe bacterial reproduction, including three mechanisms by which bacteria increase variation. Section: 28.04 Topic: Bacteria
8. Which statement is true about bacteria? A. They contain a nucleus. B. They lack ribosomes. C. They usually lack a cell wall. D. They divide sexually. E. They contain a single, circular DNA molecule as the genetic material. Bacteria are single-celled organisms; they do not have any membrane-bound organelles, but they do contain ribosomes which are not membrane bound. Their DNA is a single circular chromosome that is located in the cytoplasm of the cell, not in a nucleus. Bacteria are found in one of three shapes: bacilli (rod), cocci (spheres), and spirillum (curved). They reproduce asexually through binary fission.
Bloom’s Level: 2. Understand Learning Outcome: 28.04.01 Identify the major structural features of bacteria. Section: 28.04 Topic: Prokaryote Structure
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Chapter 28 - Viruses, Bacteria, and Archaea
9. Which of these is a correct description of a form of horizontal gene transfer in bacteria? A. Crossing-over occurs between paired chromosomes in meiosis. B. Conjugation occurs when a cell passes DNA to another cell by means of a sex pilus. C. Transformation occurs when a bacteriophage carries a bit of DNA from a previous host cell to a new host cell. D. Transduction occurs when a live bacterium picks up DNA from dead bacteria that have shed it into the environment of the living cell. E. Binary fission occurs when two bacteria cells exchange genetic information. While bacteria do not use sexual reproduction, they do have three known methods of combining genetic material. First, there is conjugation, where genetic material is transferred from one bacterium to another via a sex pilus. Next is the process of transformation where the DNA of a dead bacterium is taken up from the surrounding medium by a live bacterium. And finally there is transduction. In this method a virus carries portions of bacterial DNA between two bacteria.
Bloom’s Level: 3. Apply Learning Outcome: 28.04.02 Describe bacterial reproduction, including three mechanisms by which bacteria increase variation. Section: 28.04 Topic: Prokaryote Reproduction
10. Bacterial cells pick up free pieces of DNA from the medium pieces that were released from dead bacteria in a process called A. transformation. B. transduction. C. conjugation. D. infection. E. replication. While bacteria do not use sexual reproduction, they do have three known methods of combining genetic material. First, there is conjugation, where genetic material is transferred from one bacterium to another via a sex pilus. Next is the process of transformation, where the DNA of a dead bacterium is taken up from the surrounding medium by a live bacterium. And finally there is transduction. In this method, a virus carries portions of bacterial DNA between two bacteria.
Bloom’s Level: 1. Remember Learning Outcome: 28.04.02 Describe bacterial reproduction, including three mechanisms by which bacteria increase variation. Section: 28.04 Topic: Prokaryote Reproduction
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Chapter 28 - Viruses, Bacteria, and Archaea
11. One bacterial cell passes DNA to a second cell through a sex pilus in the process of A. transformation. B. transduction. C. conjugation. D. infection. E. replication. Of the three possible methods of DNA transfer for bacteria, the only one that involves a movement of DNA directly from one bacterium to another is conjugation.
Bloom’s Level: 1. Remember Learning Outcome: 28.04.02 Describe bacterial reproduction, including three mechanisms by which bacteria increase variation. Section: 28.04 Topic: Prokaryote Reproduction
12. Which type of sexual exchange occurs among bacteria in which DNA is carried into a bacterial cell by means of a virus? A. binary fission B. conjugation C. transformation D. transduction E. budding Although bacteria do not use sexual reproduction, they do have three known methods of combining genetic material. First, there is conjugation, where genetic material is transferred from one bacterium to another via a sex pilus. Next is the process of transformation, where the DNA of a dead bacterium is taken up from the surrounding medium by a live bacterium. And finally there is transduction. In this method, a virus carries portions of bacterial DNA between two bacteria.
Bloom’s Level: 2. Understand Learning Outcome: 28.04.02 Describe bacterial reproduction, including three mechanisms by which bacteria increase variation. Section: 28.04 Topic: Prokaryote Reproduction
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Chapter 28 - Viruses, Bacteria, and Archaea
13. To be an effective antibiotic or chemotherapeutic agent, it is necessary A. not to kill the human host or cause allergic reactions. B. to be able to kill a bacterial cell only. C. not to kill the human host or cause allergic reactions AND to be able to kill a bacterial cell only. D. to change the genome of the bacterial disease agent. E. to stop any universal living cell process. Most antibiotics kill or inhibit bacteria by interfering with the unique metabolic pathways that the bacteria possess; therefore, they are not expected to interfere with the metabolism of the host (human). There are sometimes side effects from antibiotic therapy, including allergic reactions and the deaths of beneficial bacteria.
Bloom’s Level: 3. Apply Learning Outcome: 28.04.04 List several major bacterial diseases of humans and describe how they are treated. Section: 28.04 Topic: Bacteria
14. A new classification by domains separates prokaryotes into A. Archaea and Cyanobacteria. B. Bacteria and Cyanobacteria. C. photosynthetic bacteria and chemosynthetic bacteria. D. Archaea and Bacteria. E. autotrophs and heterotrophs. Traditionally, all prokaryotic cells were grouped together into one kingdom called Bacteria. With recent advances in biochemical analysis, the groups have been divided into two new groups called domains. It has also been determined that these two groups are less closely related than was previously thought.
Bloom’s Level: 2. Understand Learning Outcome: 28.03.01 List some of the major criteria that are used to distinguish the domains Archaea, Bacteria, and Eukarya. Section: 28.03 Topic: Archaea
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Chapter 28 - Viruses, Bacteria, and Archaea
15. Metabolically, the archaea are A. photosynthetic. B. not metabolically active. C. heterotrophic and some are autotrophic as well. D. parasitic. E. all autotrophs. Archaea can be either heterotrophs or autotrophs; they perform a unique form of photosynthesis using a pigment similar to retinal, a pigment found in human eyes. They can also be chemoautotrophs, using carbon dioxide and hydrogen as energy sources.
Bloom’s Level: 3. Apply Learning Outcome: 28.03.01 List some of the major criteria that are used to distinguish the domains Archaea, Bacteria, and Eukarya. Section: 28.03 Topic: Archaea
16. How are the archaea different from bacteria? A. Archaea show the earliest form of photosynthetic ability. B. Bacteria lack the peptidoglycan cell wall found in the archaea. C. Archaea have a monolayer of lipids with branched side chains. D. Archaea have evolved far more advanced forms of metabolism than bacteria. E. Archaea have very delicate plasma membranes compared to more durable bacteria. While archaea do not have membrane-bound organelles or a nucleus-like bacteria, they also do not have peptidoglycans in their cell wall like bacteria. Also, unlike bacteria, they have branched hydrocarbons in their cell membranes; bacteria have only unbranched hydrocarbons.
Bloom’s Level: 3. Apply Learning Outcome: 28.03.01 List some of the major criteria that are used to distinguish the domains Archaea, Bacteria, and Eukarya. Section: 28.03 Topic: Archaea
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Chapter 28 - Viruses, Bacteria, and Archaea
17. What is the evolutionary relationship among archaea, bacteria, and eukarya? A. Since archaea are the most primitive, archaean ancestors gave rise to bacteria that in turn gave rise to eukaryotes. B. All three domains are equally distant from the most primitive common ancestor, a protocell. C. Archaea and eukarya share nucleic acid similarities, so eukarya split off from archaea. D. Bacterial ancestors gave rise to both archaea and to eukaryotes as two separate side branches. E. Archaea ancestors gave rise to both bacteria and to eukaryotes as two separate side branches. Based on biochemical evidence, it is thought that domain Bacteria is not closely related to either domain Archaea or domain Eukarya. Archaea and Eukarya share many biochemical similarities and some molecular similarities.
Bloom’s Level: 5. Evaluate Learning Outcome: 28.03.01 List some of the major criteria that are used to distinguish the domains Archaea, Bacteria, and Eukarya. Section: 28.03 Topic: Archaea
18. Which organisms form lichens? A. archaea and algae B. archaea and fungi C. cyanobacteria and algae D. cyanobacteria and fungi E. cyanobacteria and archaea Lichens are a combination of algae or cyanobacteria and fungi where these organisms live in a symbiotic relationship with the algae or cyanobacteria, supplying food through photosynthesis and the fungus supplying a protective environment.
Bloom’s Level: 1. Remember Learning Outcome: 28.04.03 Explain why bacteria are considered to be metabolically diverse. Section: 28.04 Topic: Bacteria
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Chapter 28 - Viruses, Bacteria, and Archaea
19. Which statement is NOT true about cyanobacteria? A. They were formerly called blue-green algae. B. Some release sulfur during photosynthesis. C. They can combine with fungi to form lichens. D. They are eukaryotic like fungi. E. They can colonize rocks. Cyanobacteria are a type of bacteria and are, therefore, prokaryotes, not eukaryotes.
Bloom’s Level: 2. Understand Learning Outcome: 28.04.01 Identify the major structural features of bacteria. Section: 28.04 Topic: Bacteria
20. Which of the following associations is correct? A. herpesvirus—chicken pox B. Epstein-Barr virus—herpes C. viroid—shingles D. prion—influenza E. HIV—chicken pox Chicken pox is caused by the herpesvirus (varicella-zoster virus). The Epstein-Barr virus is associated with mononucleosis, or "mono." Viroids lead to diseases in plants, while prions cause diseases of the nervous system in humans and other animals. HIV leads to AIDS.
Bloom’s Level: 2. Understand Learning Outcome: 28.05.03 Describe several viral diseases of humans, and explain why it is difficult to produce vaccines against some viruses. Section: 28.05 Topic: Viruses
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Chapter 28 - Viruses, Bacteria, and Archaea
21. Which of the following is considered to be acellular? A. bacteria B. fungi C. algae D. protozoans E. viruses Viruses do not have the structures necessary to be considered cellular; nor are they considered living. All living things must be composed of at least one cell and all of the other organisms listed are composed of at least one cell and meet the other criteria for life.
Bloom’s Level: 1. Remember Learning Outcome: 28.05.01 Identify the major structural features of viruses. Section: 28.05 Topic: Viruses
22. Which of these is the most accurate description of a virus? A. a noncellular living organism B. one of the smallest bacteria known C. a member of the kingdom Virusae D. a cell at the boundary between living and nonliving things E. chemical complexes of RNA or DNA protected by protein Viruses are segments of either DNA or RNA enclosed in a capsid composed of protein and possibly a membrane called an envelope. They are generally smaller than the smallest known bacteria and are obligate parasites, meaning that they cannot exist without a host, whose host supplies the molecular machinery to produce more viral genetic material, proteins for the capsid, and the envelope.
Bloom’s Level: 3. Apply Learning Outcome: 28.05.01 Identify the major structural features of viruses. Section: 28.05 Topic: Viruses
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Chapter 28 - Viruses, Bacteria, and Archaea
23. The innermost core of a virus's structure is made up of A. a membranous envelope. B. both DNA and RNA. C. either DNA or RNA. D. a protein capsid. E. a protein spore coat. All viruses are composed of a protein capsid that surrounds the genetic material of the virus; this is either DNA, which can be single stranded or double stranded, or RNA, which can be single stranded or double stranded.
Bloom’s Level: 1. Remember Learning Outcome: 28.05.01 Identify the major structural features of viruses. Section: 28.05 Topic: Viruses
24. Which of the following is NOT true about viruses? A. The genome may be DNA or RNA. B. The nucleic acid may be single or double stranded. C. They contain nucleic acid, protein, and mitochondria. D. They exhibit host specificity. E. They are obligate intracellular parasites. Viruses are segments of either DNA or RNA enclosed in a capsid composed of protein and possibly a membrane called an envelope. The genetic material of the virus may be either DNA, which can be single stranded or double stranded, or RNA, which can be single stranded or double stranded. They cannot exist without a host who supplies the molecular machinery to produce more viral genetic material, proteins for the capsid, and the envelope. Viruses do not have the structures necessary to be considered cellular; nor are they considered living.
Bloom’s Level: 3. Apply Learning Outcome: 28.05.01 Identify the major structural features of viruses. Section: 28.05 Topic: Viruses
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Chapter 28 - Viruses, Bacteria, and Archaea
25. The capsid of a virus is composed of A. RNA. B. protein. C. DNA. D. cellulose. E. lipid. The capsid of a virus may take one of several shapes and is composed of protein subunits.
Bloom’s Level: 1. Remember Learning Outcome: 28.05.01 Identify the major structural features of viruses. Section: 28.05 Topic: Viruses
26. Some, but not all, viruses contain , located on their outer surface. A. a membranous envelope composed of the host's plasma membrane B. both DNA and RNA from the host C. either DNA or RNA D. a protein capsid E. a protein spore coat Some viruses, especially those that infect animals, can have a membrane called the envelope that is formed from the membrane of the host cell. Since the membrane contains the molecules that are normally found in the host’s own cells, this helps the virus fool the host into not attacking the invading virus.
Bloom’s Level: 2. Understand Learning Outcome: 28.05.01 Identify the major structural features of viruses. Section: 28.05 Topic: Viruses
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Chapter 28 - Viruses, Bacteria, and Archaea
27. In order to infect a cell, a virus must A. inject its protein into the cell while the nucleic acid remains attached to the host cell surface. B. have a special protein spike on its surface capsid that can interact with a receptor protein on the surface of the host cell. C. actively burrow through the cell wall or cell membrane of the host cell to reach the cell's nucleus. D. produce a special extension of its cytoplasm when it comes into contact with the appropriate host cell. E. have enzymes to break down the cell wall or cell membrane of the host cell. During attachment, the virus binds to a specific receptor protein on the surface of the cell. Although the receptor has some function that it normally performs for the cell, the virus is able to use the receptor for its own purposes. It then fuses its envelope with the membrane of the cell, allowing the capsid and its contents to enter the cell.
Bloom’s Level: 4. Analyze Learning Outcome: 28.05.02 Describe the steps in a typical viral reproductive cycle, and explain how some viruses become latent. Section: 28.05 Topic: Viral Reproduction
28. Which stage of viral reproduction takes place when the spikes of the virus bind to a specific receptor molecule on the surface of a host cell? A. attachment stage B. penetration stage C. biosynthesis stage D. maturation stage E. release stage During attachment, the virus binds to a specific receptor protein on the surface of the cell. Although the receptor has some function that it normally performs for the cell, the virus is able to use the receptor for its own purposes. It then fuses its envelope with the membrane of the cell, allowing the capsid and its contents to enter the cell.
Bloom’s Level: 1. Remember Learning Outcome: 28.05.02 Describe the steps in a typical viral reproductive cycle, and explain how some viruses become latent. Section: 28.05 Topic: Viral Reproduction
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Chapter 28 - Viruses, Bacteria, and Archaea
29. Most viruses are A. much smaller than bacteria. B. usually more than 1 micron long. C. larger than protists. D. immeasurable. E. single celled. Generally viruses are smaller than the smallest bacteria that are known. They vary in size from 0.03 microns to 0.2 microns, although one virus has been found that is 0.4 microns.
Bloom’s Level: 1. Remember Learning Outcome: 28.05.01 Identify the major structural features of viruses. Section: 28.05 Topic: Viruses
30. If a virus is latent, A. it cannot be a retrovirus. B. the viral genome is reproduced along with the host cell. C. it must be a retrovirus. D. it is gaining a new envelope via "budding." E. it is easy to develop immunity against. Latency describes the reproductive stage of viruses where they are located inside a host cell and the genetic material is reproduced every time the host cell divides, leading to multiple cells that are infected. However, during this time, no viruses are exiting the cells to infect new cells. This is called the lysogenic stage.
Bloom’s Level: 2. Understand Learning Outcome: 28.05.02 Describe the steps in a typical viral reproductive cycle, and explain how some viruses become latent. Section: 28.05 Topic: Viral Reproduction
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Chapter 28 - Viruses, Bacteria, and Archaea
31. The life cycle stage of an animal virus during which a mature capsid forms around copies of the viral RNA genome is A. budding. B. attachment. C. biosynthesis. D. uncoating. E. assembly. Attachment is the first stage where the virus attaches to a receptor protein on the host cell. Entry is the next stage and denotes when the virus actually enters the host cell. Once the virus's genetic material is inside the host cell, replication of the viral genome occurs. In biosynthesis the capsid and other viral components are replicated. Next is the assembly stage, where the capsid forms around a copy of the genetic material and any viral enzymes. Finally, budding occurs so that the new virus may exit the host cell to infect a new cell.
Bloom’s Level: 2. Understand Learning Outcome: 28.05.02 Describe the steps in a typical viral reproductive cycle, and explain how some viruses become latent. Section: 28.05 Topic: Viral Reproduction
32. Which of the following is NOT a characteristic of a virus? A. acellular B. obligate parasite C. can be synthesized from chemicals in the laboratory D. diseases caused by viruses can be treated with antibiotics E. contain DNA or RNA Viruses are segments of either DNA or RNA enclosed in a capsid composed of protein and possibly a membrane called an envelope. The genetic material of the virus may be either DNA, which can be single stranded or double stranded, or RNA, which can be single stranded or double stranded. They cannot exist without a host who supplies the molecular machinery to produce more viral genetic material, proteins for the capsid, and the envelope. Viruses do not have the structures necessary to be considered cellular; nor are they considered living.
Bloom’s Level: 2. Understand Learning Outcome: 28.05.01 Identify the major structural features of viruses. Section: 28.05 Topic: Viruses
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Chapter 28 - Viruses, Bacteria, and Archaea
33. Which statement is NOT true about a retrovirus? A. It contains a DNA genome before transcription. B. It contains the enzyme reverse transcriptase. C. It is able to carry out transcription of RNA to DNA. D. Viral DNA remains in the host genome. E. Can be hidden from the host immune system. Retroviruses are RNA viruses. They contain, in addition to the RNA that serves as their genetic material, reverse transcriptase, an enzyme that can transcribe RNA into DNA. The viral DNA (cDNA) can then be taken up by the host DNA and replicated, along with the cell’s DNA. Because it is now part of the DNA of the host cell, it becomes resistant to treatment by antiviral medications and is not detected by the immune system of the host.
Bloom’s Level: 2. Understand Learning Outcome: 28.05.01 Identify the major structural features of viruses. Section: 28.05 Topic: Viruses
34. When an enveloped animal virus enters a cell during the entry stage, A. the next thing it does is assemble a new virus. B. the envelope is removed after the virus is inside the cell's nucleus. C. the protein capsid is removed through uncoating to expose the viral genome. D. it immediately integrates its nucleic acid genome into the host chromosomes. E. its cycle always results in the death of the animal cell. Attachment is the first stage where the virus attaches to a receptor protein on the host cell. Entry is the next stage and denotes when the virus actually enters the host cell. Once the virus has entered the cell, uncoating of the virus proceeds and the viral genetic material as well as the capsid and other viral components are deposited in the cell.
Bloom’s Level: 2. Understand Learning Outcome: 28.05.02 Describe the steps in a typical viral reproductive cycle, and explain how some viruses become latent. Section: 28.05 Topic: Viral Reproduction
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Chapter 28 - Viruses, Bacteria, and Archaea
35. Influenza strains that sweep around the world often carry names such as Shanghai H1N1 or Mexico City H2N2. The viruses vary in H and N surface spikes because A. the viruses reproduce on their own and attack people in cities more often. B. these viruses emerged as stray DNA from the genomes of people in these cities. C. this is where the antibodies of immune people began to break down and the old virus was again virulent. D. infected people develop immunity to the original strain, and strains that mutate sufficiently to be outside the range of immunity are soon spread in highly populated areas. E. the H and N surface spikes are obtained from the human hosts. The surface spikes of viruses are used to identify different strains of influenza. There are multiple types of each of these two molecules and they can be mixed and matched as the virus undergoes antigenic drift as the virus changes over time or antigenic shift as two unrelated viruses exchange materials as they infect the same host cell.
Bloom’s Level: 5. Evaluate Learning Outcome: 28.05.03 Describe several viral diseases of humans, and explain why it is difficult to produce vaccines against some viruses. Section: 28.05 Topic: Viruses
36. Naked strands of RNA not covered by a capsid are A. archaea. B. retroviruses. C. viroids. D. prions. E. viruses. In addition to viruses which do have a capsid covering them, and possibly have an envelope, there are also naked strands of RNA that are known to attack plants; these circular segments of RNA are called viroids.
Bloom’s Level: 1. Remember Learning Outcome: 28.05.04 Distinguish between viruses, viroids, and prions. Section: 28.05 Topic: Viroids
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Chapter 28 - Viruses, Bacteria, and Archaea
37. Some human diseases appear to be due to protein agents that may convert other normal proteins in the cell to also become these agents. This new disease protein agent is called a(n) A. archeon. B. prion. C. cyanobacterium. D. phage. E. retrovirus. In animals there are proteins that cause disease; in addition to causing problems by their presence, they also are infectious and convert other proteins to change shape. These infectious proteins are called prions and affect normal proteins, leading to degenerative disease of the nervous system.
Bloom’s Level: 2. Understand Learning Outcome: 28.05.04 Distinguish between viruses, viroids, and prions. Section: 28.05 Topic: Viruses
38. Bovine spongiform encephalopathy (BSE), or "mad cow disease," is caused by a(n) A. archeon. B. bacterium. C. cyanobacterium. D. prion. E. virus. One form of infectious protein or prion is bovine spongiform encephalopathy (BSE), or "mad cow disease."
Bloom’s Level: 1. Remember Learning Outcome: 28.05.04 Distinguish between viruses, viroids, and prions. Section: 28.05 Topic: Viruses
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Chapter 28 - Viruses, Bacteria, and Archaea
39. Which group does NOT have members that cause human disease? A. archaea B. bacteria C. fungi D. prions E. viruses Bacteria cause a number of different diseases in humans, including strep throat and tuberculosis. Fungi can cause diseases as mild as athlete’s foot or as serious as “sick building syndrome,” which is caused by black mold. Prions cause Creutzfeldt-Jakob disease and kuru, which occur only infrequently in humans but there is no known cure for these or any other prion disease. Viruses cause a wide variety of diseases in humans, including influenza and measles. The organisms in domain Archaea are not known to cause any human disease, although they can be found in the human gut.
Bloom’s Level: 3. Apply Learning Outcome: 28.05.03 Describe several viral diseases of humans, and explain why it is difficult to produce vaccines against some viruses. Section: 28.05 Topic: Archaea
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Chapter 28 - Viruses, Bacteria, and Archaea
40. Which statement is NOT true about prion diseases? A. Prions are believed to cause Creutzfeldt-Jakob disease (CJD). B. Prions lack a nucleus but must form DNA/RNA at some time in order to replicate. C. Prions are misshapen proteins that cause other proteins to change shape. D. Mad cow disease (BSE) is spread by consumption of cattle feed contaminated with prions. E. The buildup of these abnormal proteins cause a loss of neurons. Prions are infectious proteins that have changed shape into a nonfunctional form and aggregated together. They also recruit other proteins to change to the nonfunctional shape. They result in degenerative diseases of the nervous system. Part of the reason that prions are so startling is that they are single molecules and not living organisms as most disease-causing agents are. Even the nonliving viruses that cause many diseases do contain some kind of nucleic acid (DNA or RNA), while prions are simply proteins. One of the diseases caused by prions is mad cow disease (BSE), which can be transmitted to humans through the consumption of the meat of infected cattle. No amount of cooking or any other treatment of the meat of the infected cow will destroy the prions. In humans, this disease is also known as Creutzfeldt-Jakob disease.
Bloom’s Level: 3. Apply Learning Outcome: 28.05.04 Distinguish between viruses, viroids, and prions. Section: 28.05 Topic: Viruses
41. Antonie van Leeuwenhoek was a Dutch tradesman and scientist who is famous for all but one of the following. Which is the exception? A. He thought that animals were able to arise spontaneously from inorganic sources. B. He was skilled at grinding glass and made improvements in the microscope. C. He was the first to see living organisms in a drop of water. D. He invented the microscope. E. He called the organisms that he saw in a drop of water "wee animals." While Antonie van Leeuwenhoek did contribute to improving the microscope, he worked with an already existing version.
Bloom’s Level: 2. Understand Learning Outcome: 28.01.01 Describe the contributions of Leeuwenhoek and Pasteur to the science of microbiology. Section: 28.01 Topic: Bacteria
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Chapter 28 - Viruses, Bacteria, and Archaea
42. Louis Pasteur conducted an experiment to answer the question of whether or not living things can arise spontaneously from nonliving materials. In this experiment he used two flasks of sterile broth. How could this demonstrate the invalidity of spontaneous generation? A. In the flask in which no outside materials could enter, no growth occurred while growth did occur in the flask that allowed outside air to enter, demonstrating that the living things came from outside the sterile broth. B. No growth occurred in either flask, demonstrating that sterile broth could not promote growth. C. Both flasks showed growth, demonstrating that spontaneous generation does occur. D. While both flasks showed growth, the flask that did not allow air to enter showed a higher growth, demonstrating that while spontaneous generation does not occur that it does not take air to permit growth. E. No growth occurred in either flask, demonstrating that sterilizing the broth removed its nutrients so growth could not occur regardless of whether or not life was present. Louis Pasteur devised an experiment in which he used two flasks of sterile broth. When flasks containing sterilized broth were exposed to either outdoor or indoor air, they often became contaminated with microbial growth. However, if the neck of the flask was curved so that microbes could not enter the broth from the air, no growth occurred. Thus, the bacteria were not arising spontaneously in the broth, but rather were coming from the air.
Bloom’s Level: 2. Understand Learning Outcome: 28.01.01 Describe the contributions of Leeuwenhoek and Pasteur to the science of microbiology. Section: 28.01 Topic: Bacteria
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Chapter 28 - Viruses, Bacteria, and Archaea
43. In addition to his work with bacteria, Louis Pasteur was also the one to name viruses. His work with rabies produced what breakthrough? A. He was the first to discover that viruses could be based on RNA rather than DNA. B. He developed a vaccine that he used on a young rabies patient. C. He was the first to see a virus under a microscope. D. He was the first to determine the life cycle of a virus. E. He invented the microscope. Microscopes were invented several hundred years before Pasteur lived so he could not have invented it. Viruses are too small to be seen under a light microscope and electron microscopes were not yet invented so he could not have seen a virus nor have determined the life cycle of one. Also, DNA and RNA were not known during Pasteur’s lifetime.
Bloom’s Level: 1. Remember Learning Outcome: 28.01.01 Describe the contributions of Leeuwenhoek and Pasteur to the science of microbiology. Section: 28.01 Topic: Viruses
44. Microbiology is the study of microbes. These include all but which of the following? A. archae B. protists C. viroids D. plants E. fungi All of the listed organisms, except plants, fall into the category of microbes. Most of these are small enough that a microscope to see them, although that is not true in all cases.
Bloom’s Level: 1. Remember Learning Outcome: 28.01.02 Explain what is meant by the term microbiota. Section: 28.01 Topic: Bacteria
28-23 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 28 - Viruses, Bacteria, and Archaea
45. Although microbes are often equated with pathogens, not all are pathogenic and some help humans maintain their health. Which of the following are synthesized by bacteria that live in the human intestine? A. folic acid and vitamin B12 B. insulin and vitamin K C. lipids and folic acid D. vitamin K and lipids E. vitamins K and B12 Intestinal bacteria aid in digestion and produce vitamins K and B12. Insulin can be produced by bacteria that have their genetic code altered but these are not usually implanted in humans. Normal insulin production in humans takes place in the pancreas.
Bloom’s Level: 1. Remember Learning Outcome: 28.01.03 Provide several specific examples of the beneficial effects of microbes. Section: 28.01 Topic: Bacteria
46. Bacteria can be beneficial in all but which of the following ways? A. They can be used to clean up an oil spill in a process called bioremediation. B. They can be used in industry to generate products, particularly in food processing. C. They are present on the skin and help crowd out harmful bacteria. D. They are the primary producers of most ecosystems. E. They function in waste water purification to remove both organic and inorganic substances. While some bacteria can carry out photosynthesis or chemosynthesis, they are not the primary producers for most ecosystems. Bacteria can be used to clean up oil and other toxic spills without causing further harm to the environment. Bacteria are used in the production of yogurt, cheese, and wine. Bacteria can also function as decomposers, breaking down organic molecules such as nutrients found in waste water. They are also present in large numbers on the skin of humans where they prevent harmful bacteria from establishing colonies.
Bloom’s Level: 2. Understand Learning Outcome: 28.01.03 Provide several specific examples of the beneficial effects of microbes. Section: 28.01 Topic: Bacteria
28-24 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 28 - Viruses, Bacteria, and Archaea
47. Bacteria cause diseases in humans when they A. rearrange the genes of the host to code for specific virulence factors that determine the type and extent of the illness. B. carry the same genes as the host organism, causing them to be indistinguishable from the host cell. C. carry the genes that code for virulent factors, causing them to be more harmful to humans. D. carry genes that allow them to become symbiotes. E. carry the genes of organisms like birds (bird flu) or pigs (swine flu). Bacteria that might otherwise be harmless can become pathogens if they carry certain virulence factors that cause them to produce substances that are toxic to humans.
Bloom’s Level: 4. Analyze Learning Outcome: 28.04.04 List several major bacterial diseases of humans and describe how they are treated. Section: 28.04 Topic: Bacteria
48. Which genus of bacteria causes more diseases in humans than any other? A. E. coli B. Staphylococcus C. Streptococcus D. Mycobacterium E. Clostridium While all of the bacteria listed cause disease in humans, the one that causes more is Streptococcus. The diseases range from strep throat to necrotizing fasciitis (flesh-eating bacteria) to pneumonia, depending on the species of streptococcus.
Bloom’s Level: 1. Remember Learning Outcome: 28.04.04 List several major bacterial diseases of humans and describe how they are treated. Section: 28.04 Topic: Bacteria
28-25 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 28 - Viruses, Bacteria, and Archaea
49. Pathogenic bacteria are becoming increasingly deadly as A. humans refuse to wash their hands frequently enough. B. antibiotic resistance builds when more and more humans take unnecessary antibiotics. C. they pick up genes from humans, allowing them to fool human body cells into thinking that they are harmless. D. humans lose their natural immunity to many pathogens through lack of exposure. E. there are an increasing number of side effects from antibiotics. As more and more antibiotics are prescribed, natural selection acts on the bacterial populations to select for those that have a resistance. Typically antibiotics act on some molecule in the bacterium, such as an enzyme that plays an important role in the bacterium’s functioning. Bacteria that have a mutation in this molecule are not susceptible to the antibiotic and are the ones that survive and pass on this mutation to future generations. Over time, more and more of the population have the mutation and eventually the antibiotic will no longer affect the bacterial population.
Bloom’s Level: 3. Apply Learning Outcome: 28.04.04 List several major bacterial diseases of humans and describe how they are treated. Section: 28.04 Topic: Bacteria
50. Considering the various theories, the energy used in forming organic molecules in the primitive atmosphere could have come from all EXCEPT A. lightning. B. ultraviolet radiation. C. heat from volcanoes. D. electric spark. E. sound. While all of the choices are forms of energy, it would require an intense form to supply the energy needed for the formation of organic molecules. Sound waves would not contain sufficient energy to stimulate the process.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.03 Explain how the Miller-Urey experiment (among others) provided support for the “primordial soup” hypothesis. Section: 28.02 Topic: Origin of Life
28-26 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 28 - Viruses, Bacteria, and Archaea
51. Which of the following forms of energy is NOT one of those thought to have been involved in the production of large organic molecules in the primitive reducing atmosphere? A. radioactivity B. electrical energy C. heat D. radiation from the sun E. sound While all of the choices are forms of energy, it would require an intense form to supply the energy needed for the formation of organic molecules. Sound waves would not contain sufficient energy to stimulate the process.
Bloom’s Level: 1. Remember Learning Outcome: 28.02.03 Explain how the Miller-Urey experiment (among others) provided support for the “primordial soup” hypothesis. Section: 28.02 Topic: Origin of Life
52. Which of the following kinds of molecules is thought to have been absent from the primitive reducing atmosphere? A. water vapor (H2O) B. methane (CH4) C. hydrogen (H2) D. oxygen (O2) E. nitrogen (N2) In the early earth, volcanoes erupted constantly, and the first atmospheric gases would have consequently contained methane (CH4), ammonia (NH3), hydrogen (H2), and water vapor. Oxygen (O2) would not have been present until photosynthetic organisms had evolved and produced oxygen that collected in the atmosphere.
Bloom’s Level: 1. Remember Learning Outcome: 28.02.03 Explain how the Miller-Urey experiment (among others) provided support for the “primordial soup” hypothesis. Section: 28.02 Topic: Origin of Life
28-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 28 - Viruses, Bacteria, and Archaea
53. Miller's laboratory experiments showed that A. the first cell was probably made of nucleic acids. B. it is possible to form protocells. C. the primitive gases can react together to produce small organic molecules. D. atmospheric pressure is required for life to begin. E. the earth is approximately 4.5 billion years old. To simulate the earth’s early environment, Miller placed the inorganic chemicals mentioned in a closed system, heated the mixture, and circulated it past an electric spark (simulating lightning). After a week, the solution contained a variety of amino acids and organic acids.
Bloom’s Level: 1. Remember Learning Outcome: 28.02.03 Explain how the Miller-Urey experiment (among others) provided support for the “primordial soup” hypothesis. Section: 28.02 Topic: Origin of Life
54. The organic soup in which the first cell is believed to have arisen probably contained A. isolated atoms. B. bubbling hot gases only. C. algae and other microscopic organisms. D. amino acids and other organic acids. E. nucleic acids and proteins. A true cell is a membrane-bound structure that can carry on protein synthesis to produce the enzymes that allow DNA to replicate. The molecules that are needed to construct the first cell membrane would have to be present, along with some form of nucleic acid and amino acids.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.03 Explain how the Miller-Urey experiment (among others) provided support for the “primordial soup” hypothesis. Section: 28.02 Topic: Origin of Life
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Chapter 28 - Viruses, Bacteria, and Archaea
55. Which of the following is the correct order, from simple to complex, showing the origin of life? A. proteins → CH4 and NH3 → amino acids → C, H, O, N B. C, H, O, N → CH4 and NH3 → amino acid → proteins C. C, H, O, N → amino acids → proteins → CH4 and NH3 D. amino acids → proteins → C, H, O, N → CH4 and NH3 E. CH4 and NH3 → proteins → C, H, O, N → amino acids The least complex of the listed components are the component atoms (C, H, O, and N). These came together to form the inorganic molecules CH4 and NH3. Both of these were present in the atmosphere of the early earth and, together with an energy source, they formed amino acids. These amino acids then linked together by covalent bonds to form chains.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.01 Distinguish between chemical and biological evolution. Section: 28.02 Topic: Origin of Life
56. showed that amino acids can polymerize abiotically when exposed to dry heat. A. A. I. Oparin B. Stanley Miller C. Sidney Fox D. Graham Cairns-Smith E. Charles Darwin Sidney Fox has shown that amino acids polymerize abiotically (without life) when exposed to dry heat. He suggests that amino acids collected in shallow puddles along the rocky shore, and the heat of the sun caused them to form proteinoids, small polypeptides that have some catalytic properties. When proteinoids are returned to water, they form microspheres, structures composed only of protein that have many of the properties of a cell. Some of these proteins could have had enzymatic properties. This hypothesis assumes that DNA genes came after protein enzymes arose.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.04 Compare the implications of the protein-first, RNA-first, and membrane-first hypotheses for the evolution of the first living cell. Section: 28.02 Topic: Origin of Life
28-29 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 28 - Viruses, Bacteria, and Archaea
57. The hypothesis that suggests that RNA and polypeptides might have evolved simultaneously was proposed by A. Stanley Miller. B. Alexander Cairns-Smith. C. Charles Darwin. D. Gunter Wachtershauser E. Sidney Fox. A hypothesis put forward by Alexander Cairns-Smith states that clay was especially helpful in causing the polymerization of both polypeptides and RNA at the same time. Stanley Miller proposed the primordial soup hypothesis. Charles Darwin proposed evolution by natural selection. Gunter Wachtershauser proposed the iron-sulfur world hypothesis. Sidney Fox proposed the protein-first hypothesis.
Bloom’s Level: 1. Remember Learning Outcome: 28.02.03 Explain how the Miller-Urey experiment (among others) provided support for the “primordial soup” hypothesis. Section: 28.02 Topic: Origin of Life
58. Which statement is NOT true about the RNA-first hypothesis? A. It is supported by the discovery that RNA can act as a catalyst. B. It is supported by the fact that RNA can act as a substrate and as an enzyme. C. It suggests that there was an "RNA world" about 4 billion years ago. D. It says that first RNA, then DNA, then proteins would have been necessary to interact in the chemical evolution that would have led to the development of the first cells. E. It suggests that only RNA was needed to progress to the formation of the first cells. Proteins would have developed before RNA since they are composed of amino acids that arose from the original molecules found on the earth. Ribozymes, catalytic segments of RNA, are able to perform the same as an enzyme and can act on RNA to modify the nucleotide sequence. If RNA was the first genetic material to evolve, then the first true cell would have had RNA genes rather than the DNA genes that are found in cells today.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.04 Compare the implications of the protein-first, RNA-first, and membrane-first hypotheses for the evolution of the first living cell. Section: 28.02 Topic: Origin of Life
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Chapter 28 - Viruses, Bacteria, and Archaea
59. If the hypothesis that protocells were based on an "RNA world" is correct, which would NOT be necessary to shift to a "DNA world"? A. an enzyme or reaction capable of converting ribose to deoxyribose in nucleotides B. enzymes for reverse transcription of RNA into DNA C. new enzymes to replicate the DNA D. new enzymes for transcribing DNA back to RNA E. the presence of oxygen If RNA came before DNA, then it would be necessary to evolve a method to convert RNA into DNA. This would mean that ribose would need to become deoxyribose and that reverse transcription would have to occur to produce DNA. Additionally, it would be necessary to copy the DNA so that new cells might have an identical copy of the information contained in it. Finally, in order to make proteins, the cell would need to transcribe the DNA into RNA. None of these processes depend on the presence of oxygen.
Bloom’s Level: 4. Analyze Learning Outcome: 28.02.04 Compare the implications of the protein-first, RNA-first, and membrane-first hypotheses for the evolution of the first living cell. Section: 28.02 Topic: Origin of Life
60. The fact that a nucleic acid is a very complicated molecule suggests that A. the RNA-first hypothesis is impossible. B. the protein-first hypothesis is therefore the only plausible hypothesis. C. sophisticated enzymes were not present or available to synthesize it. D. no natural system could ever generate them. E. RNA could not have arisen on its own by chance, but required enzymes to guide the synthesis of nucleotides and nucleic acids. Nucleotides are complex molecules that require a complex system of other molecules to maintain and replicate them and to produce proteins. They, therefore, could not have arisen without first the machinery to produce and maintain them.
Bloom’s Level: 5. Evaluate Learning Outcome: 28.02.04 Compare the implications of the protein-first, RNA-first, and membrane-first hypotheses for the evolution of the first living cell. Section: 28.02 Topic: Origin of Life
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Chapter 28 - Viruses, Bacteria, and Archaea
61. The synthesis of DNA or RNA from the organic soup would have been guided by the actions of enzymes in the A. protein-first hypothesis. B. RNA-first hypothesis. C. clay-catalyst hypothesis. D. chloroplast. E. DNA-first hypothesis. The protein-first hypothesis suggests that proteins, or at least polypeptides, were the first of the three molecules to arise. Only after the protocell developed sophisticated enzymes did it have the ability to synthesize DNA and RNA from small molecules provided by the ocean. Researchers point out that because nucleic acids are very complicated molecules, the likelihood that RNA arose on its own is minimal.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.04 Compare the implications of the protein-first, RNA-first, and membrane-first hypotheses for the evolution of the first living cell. Section: 28.02 Topic: Origin of Life
62. Which theory would suggest that both polypeptides and RNA arose at the same time? A. Aleksandr Oparin's coacervate theory B. Stanley Miller's chemical soup theory C. Sidney Fox's protein-first hypothesis D. Alexander Cairns-Smith's simultaneous evolution of polypeptides and RNA E. Darwin's original theory of evolution, as written in Origin of Species A combination of these RNA-first and the protein-first hypotheses was put forth by Alexander Cairns-Smith. He thinks that clay was especially helpful in causing the polymerization of both polypeptides and RNA at the same time. Clay attracts small organic molecules and contains iron and zinc, which may have served as inorganic catalysts for polypeptide formation. In addition, clay tends to collect energy from radioactive decay and then discharges it when the temperature or humidity changes, possibly providing a source of energy for polymerization. Cairns-Smith suggests that RNA nucleotides and amino acids became associated in such a way that polypeptides were ordered by, and helped synthesize, RNA.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.04 Compare the implications of the protein-first, RNA-first, and membrane-first hypotheses for the evolution of the first living cell. Section: 28.02 Topic: Origin of Life
28-32 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 28 - Viruses, Bacteria, and Archaea
63. According to Fox's laboratory experiments, the protocell could have been composed A. only of carbohydrates. B. mostly of polypeptides. C. of phospholipids. D. of chitinous threads. E. of protein and RNA. After macromolecules formed, something similar to a modern plasma membrane was needed to separate them from the environment. Thus, before the first true cell arose, there would likely have been a protocell, a structure that had a lipid-protein membrane and carried on energy metabolism. Fox has shown that if lipids are made available to microspheres, the two tend to become associated and produce a lipid-protein membrane.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.03 Explain how the Miller-Urey experiment (among others) provided support for the “primordial soup” hypothesis. Section: 28.02 Topic: Origin of Life
64. A complex protein-rich cell shelters and supports the DNA genetic code, and the code is the blueprint for the cell proteins; thus is set the "chicken-or-the-egg" dilemma of which came first. This is probably solved by the discovery that A. polypeptides and RNA could have evolved simultaneously. B. reverse transcriptase can convert RNA back into DNA. C. all DNA and protein can be distributed evenly inside and outside the nucleus. D. some proteins are simply assemblages of guanine, cytosine, thymine, and adenine. E. viruses can give rise to all of the cell molecules. The first true cell would have contained RNA genes that could have replicated because of the presence of proteins. This eliminates the baffling chicken-and-egg paradox: which came first, proteins or RNA? It does mean, however, that two unlikely events would have had to happen at the same time.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.04 Compare the implications of the protein-first, RNA-first, and membrane-first hypotheses for the evolution of the first living cell. Section: 28.02 Topic: Origin of Life
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Chapter 28 - Viruses, Bacteria, and Archaea
65. In a liquid environment, liposomes. A. amino acid B. protein C. glucose D. phospholipid E. nucleic acid
molecules will automatically form droplets called
A liposome is a microscopic sac formed by fatty acids, in this case phospholipids. In an aqueous environment they will arrange themselves into a spherical shape to keep the hydrophobic portion of the molecules out of the liquid.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.04 Compare the implications of the protein-first, RNA-first, and membrane-first hypotheses for the evolution of the first living cell. Section: 28.02 Topic: Origin of Life
66. Which statement is NOT true about a protocell? A. It would have contained DNA as its genetic material. B. It would have been present before the development of a true cell. C. It would have had a lipid and protein membrane surrounding it. D. It would have contained a biochemical pathway for energy metabolism. E. The hydrophilic heads of the membrane would have pointed outwards while the hydrophilic tails would have pointed inwards. Before the first true cell arose, there would likely have been a protocell, a structure that had a lipid-protein membrane and carried on energy metabolism. Fox has shown that if lipids are made available to microspheres, the two tend to become associated, producing a lipid-protein membrane. The hydrophobic heads of the membrane would have pointed outwards while the hydrophilic tails would have pointed inwards.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.02 List and briefly describe the four stages that are thought to have led to the formation of the first cells. Section: 28.02 Topic: Origin of Life
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Chapter 28 - Viruses, Bacteria, and Archaea
67. Which of these would have been most likely to develop only after the formation of a true cell? A. autotroph B. heterotroph C. protocell D. microsphere E. proteinoid The ocean at this time contained a ready supply of preformed food so it is likely that heterotrophs were the first to evolve. Autotrophs would have evolved later.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.02 List and briefly describe the four stages that are thought to have led to the formation of the first cells. Section: 28.02 Topic: Origin of Life
68. Which term does NOT apply to the first protocell? A. fermentation B. autotrophic C. membrane-bound D. single-celled E. self-replicating The first protocells would have a ready supply of food molecules available in the ocean and were therefore, likely to be heterotrophs. With no oxygen present in the atmosphere fermentation would have been the only option for energy production. A cell must, by definition, have a membrane to separate it from the surrounding environment and the first cells would have been single cells with multicellular forms evolving much later.
Bloom’s Level: 2. Understand Learning Outcome: 28.02.02 List and briefly describe the four stages that are thought to have led to the formation of the first cells. Section: 28.02 Topic: Origin of Life
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Chapter 28 - Viruses, Bacteria, and Archaea
69. Some 20th-century authors wrote that the earliest protocells would have to be autotrophs. This concept appears to be A. correct, since heterotrophs would depend upon eating autotrophs. B. correct, since life is not possible without ATP which is only available in living systems. C. correct, since glycolysis and fermentation only occur after oxygen is present from photosynthesis. D. incorrect, since the primordial soup could contain many preformed food molecules. E. incorrect, since glycolysis and fermentation require complex enzymes for catalytic reactions. It has been suggested that the protocell likely was a heterotroph, an organism that takes in preformed food. During the early evolution of life, the ocean contained abundant nutrition in the form of small organic molecules. This suggests that heterotrophs preceded autotrophs, organisms that make their own food.
Bloom’s Level: 4. Analyze Learning Outcome: 28.02.01 Distinguish between chemical and biological evolution. Section: 28.02 Topic: Origin of Life
70. Which type of nutrition is believed to have come first? A. autotrophic B. heterotrophic C. aerobic D. photosynthetic E. catalytic The first protocells would have a ready supply of food molecules available in the ocean and were therefore, likely to be heterotrophs.
Bloom’s Level: 1. Remember Learning Outcome: 28.02.02 List and briefly describe the four stages that are thought to have led to the formation of the first cells. Section: 28.02 Topic: Origin of Life
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Chapter 28 - Viruses, Bacteria, and Archaea
71. The central dogma of genetics states that DNA directs protein synthesis and that information flows from A. protein → RNA → DNA B. protein → DNA → RNA C. RNA → protein → DNA D. RNA → DNA → protein E. DNA → RNA → protein Today’s cell is able to carry on protein synthesis in order to produce the enzymes that allow DNA to replicate. The of genetics states that DNA directs protein synthesis and that information flows from DNA to RNA to protein. It is possible that this sequence developed in stages.
Bloom’s Level: 4. Analyze Learning Outcome: 28.02.01 Distinguish between chemical and biological evolution. Section: 28.02 Topic: Origin of Life
72. Biological evolution (the evolution of living cells) differs from chemical evolution (the evolution of the molecules that make living cells) in that biological evolution would have been possible only after the development of A. a membrane. B. nucleic acids. C. enzymes. D. a metabolic pathway. E. RNA and enzymes. Biological evolution could not have begun until the formation of a plasma membrane that would enclose the cellular components.
Bloom’s Level: 1. Remember Learning Outcome: 28.02.01 Distinguish between chemical and biological evolution. Section: 28.02 Topic: Origin of Life
28-37 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 28 - Viruses, Bacteria, and Archaea
73. Which of the following organisms is considered a microbe? A. bacteria B. snail C. eagle D. oak tree E. roundworm Bacteria is considered a microbe.
Bloom’s Level: 1. Remember Learning Outcome: 28.01.02 Explain what is meant by the term microbiota. Section: 28.01 Topic: Bacteria
74. What general feature is necessary to consider an organism a microbe? A. small enough that a microscope is required to see them B. multicellular C. heterotrophic D. sexual reproduction E. use aerobic respiration for metabolism Microbes are generally small enough that a microscope is required to see them. All of the other features are not necessarily found in microbes.
Bloom’s Level: 2. Understand Learning Outcome: 28.01.02 Explain what is meant by the term microbiota. Section: 28.01 Topic: Prokaryote Structure
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Chapter 28 - Viruses, Bacteria, and Archaea
Short Answer Questions 75. Compare the domains of Archaea and Eukarya in regards to their features. Feature Archaea Eukarya Nucleus No Yes Organelles No Yes Introns Sometimes Yes Histones Yes Yes RNA polymerase Several types Several types Methionine is at start of protein synthesis Yes Yes
Bloom's Level: 6. Create Learning Outcome: 28.03.01 List some of the major criteria that are used to distinguish the domains Archaea, Bacteria, and Eukarya. Section: 28.03 Topic: Prokaryote Structure
Multiple Choice Questions 76. Which of the following is a benefit of microbes? A. break down organic molecules B. Various microbes help protect us from harmful microbes. C. aid in digestion D. synthesize vitamins K and B12 E. All of the answer choices are benefits of microbes. All of the answer choices are benefits of microbes.
Bloom’s Level: 1. Remember Learning Outcome: 28.01.03 Provide several specific examples of the beneficial effects of microbes. Section: 28.01 Topic: Bacteria
28-39 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 28 - Viruses, Bacteria, and Archaea
77. Which feature is lacking in the cell wall of archaea that will distinguish them from the bacteria? A. peptidoglycan B. phospholipids C. proteins D. polysaccharides E. None of the answer choices are lacking in the cell walls of the archaea. Archaea lack peptidoglycan in their cell walls, which will distinguish them from the bacteria. Phospholipids, proteins, and polysaccharides are found in the cell walls of both archaea and bacteria.
Bloom’s Level: 1. Remember Learning Outcome: 28.03.02 Review the structural features of the plasma membranes and cell walls of archaea. Section: 28.03 Topic: Archaea
78. What feature enables the archaea to survive in harsh environments? A. their plasma membranes and cell walls B. their metabolism C. the ability to form multicellular organisms D. their reproductive capabilities E. their nucleus The cell wall is the main feature that enables the archaea to survive in harsh environments. They do not possess a nucleus. Their metabolism and reproductive capabilities are not the main feature that enables them to survive in harsh environments. Archaea do not form multicellular organisms.
Bloom’s Level: 1. Remember Learning Outcome: 28.03.02 Review the structural features of the plasma membranes and cell walls of archaea. Section: 28.03 Topic: Archaea
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Chapter 28 - Viruses, Bacteria, and Archaea
True / False Questions 79. The plasma membrane of archaea is composed of a phospholipid bilayer. FALSE The plasma membrane of archaea is composed of a monolayer of lipids with branched side chains.
Bloom’s Level: 1. Remember Learning Outcome: 28.03.02 Review the structural features of the plasma membranes and cell walls of archaea. Section: 28.03 Topic: Archaea
Multiple Choice Questions 80. Which description best fits a chemoautotroph bacteria? A. They reduce carbon dioxide to an organic compound by using energetic electrons derived from chemicals. B. They reduce carbon dioxide to an inorganic compound by using energetic electrons derived from chemicals. C. They run photosynthesis that uses solar energy to produce their own food. D. They transfer electrons to sulfate, producing hydrogen sulfate. E. All of the answer choices are descriptions of a chemoautotroph bacteria. Chemoautotroph bacteria reduce carbon dioxide to an organic compound by using energetic electrons derived from chemicals. When reduced, carbon dioxide is an organic compound. Running photosynthesis that uses solar energy to produce their own food is the description of a producer. Transferring electrons to sulfate, producing hydrogen sulfate, is a feature of heterotrophic bacteria.
Bloom’s Level: 2. Understand Learning Outcome: 28.04.03 Explain why bacteria are considered to be metabolically diverse. Section: 28.04 Topic: Bacteria
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Chapter 28 - Viruses, Bacteria, and Archaea
True / False Questions 81. Bacterial metabolism is quite diverse and can range from heterotrophic individuals that are anaerobic to ones that use mitochondria to run cellular respiration. FALSE Bacterial metabolism is quite diverse and can range from heterotrophic individuals that are anaerobic but prokaryotes (bacteria) lack mitochondria.
Bloom’s Level: 1. Remember Learning Outcome: 28.04.03 Explain why bacteria are considered to be metabolically diverse. Section: 28.04 Topic: Bacteria
Short Answer Questions 82. Describe the steps in viral replication. 1. Attachment: the spikes of the virus bind to a specific receptor molecule on the host cell. 2. Entry: the viral envelope fuses with the host cell membrane and allows the viral genetic material to enter the host cell. 3. Replication: viral enzymes make copies of the genome. 4. Biosynthesis: some of the RNA molecules serve as mRNA for production of capsids and spike proteins. 5. Assembly: mature caspid forms around a copu of the viral genome. 6. Budding: new viruses are released from the cell surface.
Bloom's Level: 6. Create Learning Outcome: 28.05.02 Describe the steps in a typical viral reproductive cycle, and explain how some viruses become latent. Section: 28.05 Topic: Viral Reproduction
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Chapter 28 - Viruses, Bacteria, and Archaea
83. List the four stages that are thought to have led to the formation of the first cells. Stage 1: Organic monomers evolved from inorganic compounds. Stage 2: Organic polymers formed by the joining of the organic monomers. Stage 3: Protobionts formed as the organic polymers became enclosed in a membrane. Stage 4: Living cells formed as protobionts acquired the ability to self-replicate.
Bloom's Level: 6. Create Learning Outcome: 28.02.02 List and briefly describe the four stages that are thought to have led to the formation of the first cells. Section: 28.02 Topic: Origin of Life
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Chapter 29 - Protistans
Chapter 29 Protistans
Multiple Choice Questions 1. Why are green algae placed in the Protista while plants are placed in a separate kingdom? A. The green algae use chlorophyll a, while plants use chlorophyll b. B. Green algae have a cell wall of chitin, while plants have cell walls of cellulose. C. Plants enclose and protect the embryo within the seed, while green algae do not. D. All plants are vascular, while all green algae are nonvascular. E. Algae reproduce only asexually, while plants reproduce sexually. While algae and plants do share a number of similarities, there are also some very important differences between the two. Among these differences are the protections and nourishment of the developing embryo, the formation of vascular tissue, and the development of seeds as a means of reproduction.
Bloom’s Level: 3. Apply Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
2. Single-celled eukaryotes belong to the kingdom A. Prokaryotae. B. Protista. C. Archaea. D. Plantae. E. Animalia. For the most part, protists are unicellular and microscopic eukaryotes. Other than these characteristics, they are a very diverse group.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.02 Describe the general characteristics of a protist. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
3. The idea that mitochondria may have resulted when a nucleated cell engulfed aerobic bacteria is known as the A. chemiosmotic theory. B. endosymbiotic hypothesis. C. Lyon hypothesis. D. theory of evolution by means of acquired characteristics. E. theory of acquisition of organelles. The endosymbiotic theory states that mitochondria originated when a nucleated cell engulfed aerobic bacteria and chloroplasts originated when a cell with a mitochondria engulfed a cyanobacteria.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.01 Summarize what is known regarding the evolution of the first single-celled protist. Section: 29.01 Topic: Endosymbiosis
4. Protist reproduction A. is asexual only. B. is sexual only. C. is neither sexual nor asexual. D. is mostly sexual, with asexual as a second option. E. is mostly asexual, with sexual as a second option. Most protists carry out asexual reproduction, with sexual reproduction as an option, particularly when conditions are unfavorable. Sexual reproduction often results in a spore or cyst that can survive until environmental conditions improve.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.02 Describe the general characteristics of a protist. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
5. The kingdom Protista includes all of the following kinds of organisms EXCEPT A. protozoa. B. algae. C. water molds. D. yeasts. E. slime molds. Yeasts are a unicellular fungus, not a protist. Protozoa are animal like protists and algae are photosynthetic protists. Water molds and slime molds are both types of protists that function as decomposers.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
6. A member of the green algae is A. Amoeba proteus. B. Plasmodium vivax. C. Chlamydomonas. D. Penicillium. E. Paramecium. Chlamydomonas is a unicellular green algae with a cup-shaped chloroplast and two whiplike flagella.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
7. An example of a colonial green alga is A. Ulva. B. Chlamydomonas. C. Volvox. D. Spirogyra. E. Fucus. Volvox is a colonial green alga with thousands of flagellated cells that are arranged in a single layer around a watery interior.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
8. Members of which genus of algae are found in ponds and have chloroplasts that are arranged in a spiral? A. Ulva B. Spirogyra C. Euglena D. Chlamydomonas E. Volvox Spirogyra is a filamentous green algae in which each cell has a ribbon-like chloroplast.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
9. Spirogyra practice a type of sexual exchange called A. budding. B. aggregation. C. fruiting body. D. conjugation. E. gametogenesis. Spirogyra are able to reproduce by conjugation. In this process physical contact between cells allows for the transmission of DNA between two filaments.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
10. The algae are thought to be closely related to the first plants because they share most of the characteristics of plants. A. brown B. red C. green D. white E. pink Green algae are thought to be the most closely related to plants because they share these characteristics: a cell wall composed of cellulose; chlorophylls a and b; and the storage of starch as a reserve energy source.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.02 Describe the general characteristics of a protist. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
11. An example of a filamentous green alga is A. Ulva. B. Chlamydomonas. C. Volvox. D. Spirogyra. E. Fucus. Spirogyra is a filamentous green alga that has a spiral ribbon of chloroplast.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
12. Diatoms have elaborate shells made of A. carbon. B. magnesium. C. plastic. D. silica. E. cellulose. Diatoms are unicellular algae that have silica shells in a wide variety of shapes. These algae are important primary producers in both fresh- and saltwater ecosystems.
Bloom’s Level: 3. Apply Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
13. Which is grouped with the algae? A. sporozoa B. zooflagellates C. slime molds D. diatoms E. ciliates The algae include green algae, diatoms, dinoflagellates, red algae, and brown algae. Sporozoa, zooflagellates, slime molds, and ciliates are protists but fall under other classifications than algae.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
14. Which type of organism causes a "red tide"? A. euglenoids B. water molds C. brown algae D. green algae E. dinaflagellates Red tides happen when dinaflagellates increase in population size very rapidly. Some species of dinoflagellates also release toxins and can be harmful if humans eat shellfish that have an accumulation of the toxin.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.04 Identify several human diseases caused by protists, and explain why many are difficult to treat. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
15. A rapid increase in the population of "red tide." A. diatoms B. dinaflagellates C. red algae D. brown algae E. multicellular green algae
can often lead to an event known as a
Red tides happen when dinaflagellates increase in population size very rapidly. Some species of dinoflagellates also release toxins and can be harmful if humans eat shellfish that have an accumulation of the toxin.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.04 Identify several human diseases caused by protists, and explain why many are difficult to treat. Section: 29.01 Topic: Protists
16. The kind of algae that help to build a coral reef are A. Porphyra. B. Laminaria. C. coralline algae. D. Paramecium. E. Chlamydomonas. Porphyra is a red algae, but it is used for sushi wraps. Laminaria is a brown algae also known as kelp. Coralline algae is a type of red algae that contains calcium carbonate in their cell walls and can contribute to coral reef formation. Paramecium are ciliates that a a type of protozoan. Chlamydomonas is a single-celled green algae.
Bloom’s Level: 4. Analyze Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
17. A unicellular alga that has both plant-like photosynthesis and animal-like motility is a(n) A. diatom. B. dinoflagellate. C. euglenoid. D. zooflagellate. E. red algae. Dinoflagellates have some plant-like and some animal-like characteristics. Some species are photosynthetic, while others are heterotrophs. Many species have silica-encrusted plates as a covering and they have two flagella. The arrangement of the flagella cause the organism to swim with a whirling motion.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
18. A pellicle, either rigid or flexible, could be found in A. diatoms. B. brown and red algae. C. dinoflagellates. D. green algae. E. euglenoids. A pellicle is the outer membrane composed of protein strips on euglenoids. Diatoms and dinoflagellates both have outer shells that cover and protect that have silica as a component.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
19. A simplified list of characteristics of "algae" includes the following: Green Algae - mostly freshwater, cell wall, chlorophylls a and b Euglenoids - mostly freshwater, no cell wall, chlorophylls a and b Brown Algae - mostly marine, cell wall, chlorophylls a and c Diatoms - marine and freshwater, cell wall, chlorophylls a and c Dinoflagellates - mostly marine, cell wall, chlorophylls a and c Red Algae - mostly marine, chlorophyll a and phycobilin From examination of this restricted set of data, and referring back to discussions of systematics and taxonomy, it is most likely that the most primitive algae A. were marine. B. were freshwater. C. contained cell walls. D. contained chlorophyll a. E. contained chlorophyll c. Examining the list of characteristics that are given above, the one characteristic that is seen in all of the algae is the presence of chlorophyll a. Given its presence in a number of diverse algae it is likely that it was found in more primitive forms of algae.
Bloom's Level: 5. Evaluate Learning Outcome: 29.01.02 Describe the general characteristics of a protist. Section: 29.01 Topic: Protists
20. Which is NOT part of the correct description of a protozoan? A. heterotrophic B. multicellular C. usually motile D. often use an eyespot to detect light E. can be endosymbionts of animals Protozoans are eukaryotic, unicellular protists. Some forms are motile with cilia, flagella, or pseudopods, while others are nonmotile. Protozoans can also be either autotrophic, heterotrophic, or parasitic.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.02 Describe the general characteristics of a protist. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
21. Which of the following organisms move about by means of pseudopodia? A. Plasmodium vivax B. trypanosome C. Amoeba proteus D. Chlamydomonas E. Paramecium caudatum Pseudopods are projections of the plasma membrane formed through cytoplasmic streaming. The process allows some protists to move and to engulf food. The movement is sometimes called amoeboid movement. Organisms that use this form of movement include Amoeba proteus.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
22. The deposits of chalky fossils that build the White Cliffs of Dover were produced by A. radiolaria. B. foraminiferans. C. ciliates. D. diatoms (as diatomaceous earth). E. dinoflagellates. Foraminiferans are a type of amoeba that have an external skeleton called a test. Over millions of years, deposits of the tests of these organisms, followed by an up thrust of the sediments, produced the White Cliffs of Dover.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
23. Which of the following is NOT a fungus? A. mushroom B. yeast C. ringworm D. mold E. amoeba While mushrooms, yeast, ringworm, and molds are all fungi, amoeba are classified as protists.
Bloom’s Level: 1. Remember Learning Outcome: 29.02.01 Explain how most fungi obtain their nutrients. Section: 29.02 Topic: Fungi
24. Which protist is NOT correctly linked to the type of movement it shows? A. amoeboids—pseudopodia B. ciliates—cilia C. zooflagellates—flagella D. sporozoa—flexing the pellicle E. Paramecium—cilia All phases of sporozoa are generally nonmotile and are either internal or external parasites.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
25. Which of the following is an incorrect statement? A. Bacteria, fungi, and protozoans sometimes cause disease. B. Dinoflagellates, diatoms, and trypanosomes are algae. C. Cyanobacteria, diatoms, and euglenoids carry on photosynthesis. D. Protozoans include amoebas, paramecia, and flagellates. E. Amoeboids, zooflagellates, and sporozoans can cause human diseases. While dinoflagellates and diatoms are both unicellular algae, trypanosomes are flagellates that often cause disease in humans.
Bloom’s Level: 4. Analyze Learning Outcome: 29.01.02 Describe the general characteristics of a protist. Section: 29.01 Topic: Protists
26. Which includes the parasites that cause African sleeping sickness? A. sporozoa B. flagellates C. slime molds D. diatoms E. ciliates Unicellular flagellates include Trypanosoma brucei, which causes African sleeping sickness.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.04 Identify several human diseases caused by protists, and explain why many are difficult to treat. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
27. A member of the sporozoan group of protists that can cause diseases is A. Amoeba proteus. B. Plasmodium vivax. C. Chlamydomonas. D. Penicillium. E. Paramecium. The sporozoan group consists of generally nonmotile organisms that form spores and that are internal or external parasites. The parasite Plasmodium vivax causes malaria and is one of these.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.04 Identify several human diseases caused by protists, and explain why many are difficult to treat. Section: 29.01 Topic: Protists
28. Which form of protist has a complicated parasitic life cycle that nearly always involves the production of infective spores? A. sporozoans B. zooflagellates C. slime molds D. diatoms E. ciliates All phases of sporozoa are generally nonmotile and are either internal or external parasites. They generally have a complex life cycle involving the production of spores.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.04 Identify several human diseases caused by protists, and explain why many are difficult to treat. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
29. The most important human parasite among the sporozoa is malaria. A. Plasmodium vivax B. trypanosome C. Amoeba proteus D. Chlamydomonas E. Paramecium caudatum
, the causative agent of
The sporozoan group consists of generally nonmotile organisms that form spores and that are internal or external parasites. The parasite Plasmodium vivax causes malaria and is one of these.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.04 Identify several human diseases caused by protists, and explain why many are difficult to treat. Section: 29.01 Topic: Protists
30. Which organisms exist in a plasmodial form that is a diploid cytoplasmic mass? A. sporozoa B. zooflagellates C. slime molds D. diatoms E. ciliates Plasmodial slime molds usually exist as a diploid multinucleated cytoplasmic mass that is covered by a slime sheath. Sporozoa, zooflagellates, diatoms, and ciliates all exist as a form with a single nucleus.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
31. Water molds are not placed with the fungi because A. unlike fungi, water molds are saprotrophic. B. water molds have cell walls of cellulose, while fungi have cell walls of chitin. C. water molds grow only in water. D. adult water molds are diploid, while adult fungi are haploid. E. unlike fungi, water molds do not have a filamentous body. Water molds usually live in water and decay organic remains. They have a filamentous body with cell walls composed of cellulose. They can reproduce either sexually or asexually. Fungi also decay organic remains and have a filamentous body, but their cell walls are made up of chitin rather than cellulose.
Bloom’s Level: 3. Apply Learning Outcome: 29.01.02 Describe the general characteristics of a protist. Section: 29.01 Topic: Protists
32. Which is the largest group of protozoans? A. amoeboids B. zooflagellates C. ciliates D. sporozoans E. slime molds The protozoans include the groups flagellates, ciliates, amoeboids, and sporozoans. The largest of these are the ciliates with cilia that function in movement, capturing prey, and moving food into their mouthparts.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.02 Describe the general characteristics of a protist. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
33. Molds (except slime and water molds) and mushrooms belong to the kingdom A. Archaea. B. Protists. C. Fungi. D. Plantae. E. Animalia. Molds and mushrooms are both part of the kingdom Fungi. Archaea is composed of prokaryotic organisms that are closely related to eukaryotes. Protists are typically unicellular eukaryotes and do contain water and slime molds, which are not classified as fungi because they have cellulose in their cell walls and engulf their food, unlike fungi which digest outside their bodies. Unlike plants, molds and mushrooms do not contain chlorophyll.
Bloom’s Level: 1. Remember Learning Outcome: 29.02.01 Explain how most fungi obtain their nutrients. Section: 29.02 Topic: Fungi
34. are mostly saprotrophic decomposers that assist in recycling of nutrients in ecosystems. A. Algae B. Viruses C. Fungi D. Protozoans E. Ciliates Fungi are strict heterotrophs that release their digestive enzymes into the environment and digest their food outside of their bodies. This makes any unused nutrients available to other organisms in the ecosystem.
Bloom’s Level: 1. Remember Learning Outcome: 29.02.02 Describe the structural components of a fungal body. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
35. The kingdom absorb nutrients. A. Archaea B. Protista C. Fungi D. Animalia E. Plantae
are characterized by having filaments called hyphae that are used to
Fungi have a filamentous body; the individual filaments are called hyphae, while the collective mass of filaments is called a mycelium.
Bloom’s Level: 2. Understand Learning Outcome: 29.02.01 Explain how most fungi obtain their nutrients. Section: 29.02 Topic: Fungi
36. are organisms that break down dead organic matter in order to absorb the nutrient molecules. A. Prokaryotes B. Eukaryotes C. Parasites D. Saprotrophs E. Heterotrophs Organisms can be classified by the way that they obtain food; heterotrophs are organisms that ingest matter as a food source, parasites spend all or part of their lives feeding off of a host and saprotrophs digest food outside their bodies, allowing any unused nutrients to be used by autotrophs, who convert inorganic energy into an organic molecule.
Bloom’s Level: 2. Understand Learning Outcome: 29.02.01 Explain how most fungi obtain their nutrients. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
37. An organism that will feed on dead plants, animals, and microbes are called A. chemosynthetic. B. autotrophic. C. heterotrophic. D. saprotrophs. E. parasitic. Organisms can be classified by the way that they obtain food; heterotrophs are organisms that ingest matter as a food source, parasites spend all or part of their lives feeding off of a host, and saprotrophs digest food outside their bodies, allowing any unused nutrients to be used by autotrophs, who convert inorganic energy into an organic molecule. Autotrophs can be further broken down into photoautotrophs that use the energy of the sun and chemoautotrophs that use chemical energy.
Bloom’s Level: 1. Remember Learning Outcome: 29.02.01 Explain how most fungi obtain their nutrients. Section: 29.02 Topic: Fungi
38. In what way are fungi like heterotrophic bacteria? A. They both produce gametes. B. They are both heterotrophic and play an important role in ecosystems. C. They both have cell walls of the same material. D. They are both photosynthetic and thus are producers. E. They are both eukaryotic. Fungi are strict heterotrophs that release their digestive enzymes into the environment and digest their food outside of their bodies. This makes any unused nutrients available to other organisms in the ecosystem. Heterotrophic bacteria obtain nutrients in the same way.
Bloom’s Level: 3. Apply Learning Outcome: 29.02.01 Explain how most fungi obtain their nutrients. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
39. Fungi are NOT photosynthetic because they lack A. sporangium. B. xylem. C. cell walls. D. chloroplasts. E. cell membrane. Fungi do not contain chloroplasts which contain the photosynthetic pigment chlorophyll. This molecule allows autotrophs to convert the energy of the sun into chemical potential energy in the form of organic molecules.
Bloom’s Level: 2. Understand Learning Outcome: 29.02.01 Explain how most fungi obtain their nutrients. Section: 29.02 Topic: Fungi
40. The mycelium is a mesh of filaments, each of which is called A. a super-mycelium. B. a conidium. C. an ascospore. D. a basidiospore. E. a hypha. Fungi have a filamentous body; an individual filament is called a hypha, while the collective mass of filaments is called a mycelium.
Bloom’s Level: 1. Remember Learning Outcome: 29.02.02 Describe the structural components of a fungal body. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
41. At one time, biologists thought that fungi were merely forms of plants that had lost their chlorophyll and had returned to saprotrophy to gain food. Why is this no longer considered a solid theory? A. Fungal cell walls contain chitin rather than cellulose. B. Fungi attack and engulf food for internal digestion. C. Fungi have flagella at some stage, providing mobility that plants never have. D. Plants are multicellular and fungi are unicellular or multinucleated noncellular plasmodia. E. Plants store glycogen, while fungi store starch. Since plants and fungi have two different molecules that serve as the major structural component of their cells walls as well as two different methods of obtaining food, it is unlikely that one arose from the other.
Bloom’s Level: 4. Analyze Learning Outcome: 29.02.02 Describe the structural components of a fungal body. Section: 29.02 Topic: Fungi
42. When two mating strains of hyphae merge, the fungal cells A. are just another form of haploid cells. B. are just another name for diploid cells. C. contain two haploid nuclei. D. are nonseptate plasmodial masses with many nuclei and no cell partitioning. E. are potential symbionts that can join with algae to form lichens. The two types of mating types are designated (+) and (-). When these two mating types merge through conjugation, the two haploid nuclei from the hyphae eventually merge to form a zygote that is diploid. The zygote then goes through meiosis, followed by the formation of spores.
Bloom’s Level: 2. Understand Learning Outcome: 29.02.03 Compare the reproductive strategies of the major groups of fungi. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
43. Which of the following are NOT grouped with the Protista? A. protozoans B. algae C. slime molds D. water molds E. black bread mold Black bread mold is classified as a fungus rather than a protist. Unlike protists, it digests organic molecules outside of the body and contains chitin in its cell wall rather than the cellulose of algae.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.02 Describe the general characteristics of a protist. Section: 29.01 Topic: Protists
44. A fungal structure that results when haploid nuclei fuse during sexual reproduction and can survive unfavorable growing conditions is the A. mycelium. B. sporangium. C. ascocarp. D. basidium. E. zygospore. A thick-walled zygospore results from the fusion of the ends of hyphae of opposite mating types. After a period of dormancy, the zygospore undergoes meiosis and then germination. Sporangia form at the tips of aerial hyphae and produce spores.
Bloom’s Level: 2. Understand Learning Outcome: 29.02.03 Compare the reproductive strategies of the major groups of fungi. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
45. In time, bread becomes stale and grows mold on the exposed surfaces. Why does mold not begin growing inside the bread loaf? A. Mold can grow only on dry surfaces. B. Only molds at the surface of the bread can get oxygen to respire. C. Mold actually begins from anywhere inside but produces colored spores only at the surface. D. There is an antibiotic antagonism between the bread mold and the internal yeasts that made the bread rise. E. The bread was sterilized when it was baked and thereafter mold spores settle only on the outside of the cooled bread. Heat from the baking process kills any living organisms that are in the ingredients used to make the bread. Only spores that a carried to the surface of the bread by air currents after baking are present on the bread.
Bloom's Level: 5. Evaluate Learning Outcome: 29.02.04 Identify several fungal diseases of humans, and explain why many can be difficult to treat. Section: 29.02 Topic: Fungi
46. Rhizopus stolonifer (black bread mold) is more likely to grow on home-baked bread than on store bought bread. The most likely reason for this is A. the store bought bread has been refrigerated, while the home-baked bread has not. B. home-baked bread is more likely to contain chemical preservatives than store bought bread. C. store bought bread is more likely to contain chemical preservatives than home-baked bread. D. home ingredients are more likely to contain fungal spores. E. home-baked bread will contain more nutrients than store bought breads. Store bought bread contains chemical preservatives that are used to retard the growth of organisms like fungi. Usually home-baked bread does not contain these same chemicals.
Bloom’s Level: 3. Apply Learning Outcome: 29.02.01 Explain how most fungi obtain their nutrients. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
47. Sac fungi take their name from the shape A. of their spores. B. of the sexual reproductive structure that produces the spores. C. of their fruiting bodies. D. of their hyphae. E. made by clusters of sac fungi together. Sac fungi, phylum Ascomycota, are named for their cup-like reproductive structure called the ascocarp.
Bloom’s Level: 2. Understand Learning Outcome: 29.02.02 Describe the structural components of a fungal body. Section: 29.02 Topic: Fungi
48. A chain of asexual spores produced by a sac fungus is called A. a mycelium. B. a conidia. C. an ascospore. D. a basidiospore. E. a hypha. Sac fungi may reproduce asexually by producing chains of asexual spores called conidia. Conidia are haploid and are produced at the ends of mycelia that arise from the germination of ascospores.
Bloom’s Level: 2. Understand Learning Outcome: 29.02.03 Compare the reproductive strategies of the major groups of fungi. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
49. The above-ground portion of a mushroom is termed basidiocarp, which specializes in A. securing food for the organism. B. reproduction. C. running photosynthesis. D. disintegration. E. parasitizing other organisms. The basidiocarp of a mushroom is a reproductive structure. The basidium forms when hyphae of two different mating types join. Contained within the basidiocarp is the basidium where the + and – nuclei fuse and then undergo meiosis to produce haploid spores.
Bloom’s Level: 1. Remember Learning Outcome: 29.02.03 Compare the reproductive strategies of the major groups of fungi. Section: 29.02 Topic: Fungi
50. A mushroom will produce a sexual spore at the tip of a club-shaped A. mycelia. B. conidia. C. ascus. D. basidia. E. hypha. The cap of a mushroom contains reproductive structures and is called a basidiocarp. The basidiocarp contain gills that radiate out from the stalk and are lined with the dikaryotic cells called basidia that give rise to haploid spores once the nuclei fuse and then undergo meiosis.
Bloom’s Level: 2. Understand Learning Outcome: 29.02.03 Compare the reproductive strategies of the major groups of fungi. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
51. When the cap of an average gilled mushroom is cut off and shaken, a cloud of powdery material is released. The material released are A. mycelia. B. sporangia. C. hyphae. D. spores. E. seeds. A mushroom cap is filled with gills, structures that radiate out from the stalk in a spoke-like pattern. These gills are lined with basidia, which contain haploid spores. They would appear as a powdery material if the spores were collected after they are ejected from the basidium.
Bloom’s Level: 3. Apply Learning Outcome: 29.02.02 Describe the structural components of a fungal body. Section: 29.02 Topic: Fungi
52. A member of the sac fungi that contains an ascocarp is A. Amoeba proteus. B. Plasmodium vivax. C. Chlamydomonas. D. Penicillium species. E. Paramecium species. Amoeba proteus is a member of the amoeboid protists. Plasmodium vivax is a member of sporozoan protists. Chlamydomonas is a type of green algae also classified as a protist. Paramecium sp. would be a member of the ciliate protists. Penicillium sp. would be a member of phyllum Ascomycota, a sac fungi.
Bloom’s Level: 1. Remember Learning Outcome: 29.02.02 Describe the structural components of a fungal body. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
53. Which pair is mismatched? A. sporangium—bread mold B. ascocarp—shelf fungi C. basidium—club fungi D. penicillin—sac fungi E. yeasts—sac fungi Fungi are often recognized by and classified by their method of reproduction. Black bread mold is classed in phylum Zygomycota, which has a reproductive structure called a sporangium that produces spores. Club fungi are from phylum Basidiomycota and have reproductive structures called basidia (basidium singular). Penicillin is classed as Ascomycota (sac fungi) because of DNA sequence data. Yeasts are a member of the Ascomycota as well although they can reproduce by budding. Ascocarps, however, have a cup-shaped reproductive structure called an ascocarp, while mushrooms have basidia and are members of Basidiomycota.
Bloom’s Level: 4. Analyze Learning Outcome: 29.02.03 Compare the reproductive strategies of the major groups of fungi. Section: 29.02 Topic: Fungi
54. Crustose, fruticose, and foliose obtain their nutrients from photosynthetic cells contained within their middle layer. These are varieties of A. rusts and smuts. B. mushrooms. C. yeasts. D. lichens. E. mycorrhizae. Lichens may have one of three structures depending on the type of fungus that is in the association: Crustose, which looks like a crust that has formed on bare rocks; fruticose, which looks like a very small plant and has an upright form; and foliose, which looks like leaves growing out of a hard surface.
Bloom’s Level: 1. Remember Learning Outcome: 29.02.01 Explain how most fungi obtain their nutrients. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
55. The relationship between the fungi and algae in lichens is best described as A. a mutually beneficial relationship between normally free-living strains. B. a complete mixing of the genomes of the two groups at the cellular level. C. an endosymbiotic transfer of the chloroplasts of algae to the hyphae cells. D. a possible controlled parasitism wherein the fungi withdraw food from the algae and the algae can grow better alone. E. a possible controlled parasitism wherein the algae are moistened and protected by the fungi but the fungi can grow better alone. The algae and fungi in lichen both derive benefits from the association and both are exploited by the association. The algae is protected from predators and from drying out by the hyphae of the fungus and also receives nutrients. In turn the fungus is able to take food from the photosynthetic algae. It is thought that the original relationship between the two began as a parasite/host interaction but that over time it became mutualistic.
Bloom's Level: 5. Evaluate Learning Outcome: 29.02.03 Compare the reproductive strategies of the major groups of fungi. Section: 29.02 Topic: Fungi
56. During fermentation, A. ethanol B. carbon dioxide C. oxygen D. spores E. sugar
is produced by yeast cells, causing the dough to rise.
Yeasts produce energy for themselves through fermentation. Fermentation releases carbon dioxide as a waste product of the process. The carbon dioxide bubbles supply lift to the bread and give bread its characteristic texture.
Bloom’s Level: 3. Apply Learning Outcome: 29.02.01 Explain how most fungi obtain their nutrients. Section: 29.02 Topic: Fungi
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Chapter 29 - Protistans
True / False Questions 57. Protists offer a glimpse into the past because they are most likely related to the first prokaryotic cell to have evolved. FALSE Protists are eukaryotic cells due to the presence of the nucleus. They are most likely related to the first eukaryotic cell to have evolved.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.01 Summarize what is known regarding the evolution of the first single-celled protist. Section: 29.01 Topic: Protists
Multiple Choice Questions 58. Which features set most of the members of protista apart from the rest of the kingdoms? A. unicellular and microscopic B. multicellular and microscopic C. photosynthetic and unicellular D. heterotrophic and multicellular E. None of the answer choices is correct. Protists are often unicellular and microscopic. While some of them are photosynthetic, heterotrophic, and even multicellular these are not traits unique to the protists. Animals, plants, and fungi all share some of these traits as well.
Bloom’s Level: 2. Understand Learning Outcome: 29.01.02 Describe the general characteristics of a protist. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
Short Answer Questions 59. List the six eukaryotic supergroups and indicate the types of organisms that belong in each. Archaeplastids: land plants as well as green and red algae Chromalveolates: brown algae, diatoms, ciliates, sporozoans, and water molds Excavates: euglenoids and certain other flagellates Amoebzoans: amoeboids, and plasmodial and cellular slime molds Rhizarians: foraminiferans and radiolarians Opisthokonts: animals, fungi and certain flagellates
Bloom’s Level: 6. Create Learning Outcome: 29.01.03 Summarize the unique and specific structural features of each group of protists, including how they reproduce and obtain food. Section: 29.01 Topic: Eukarya
Multiple Choice Questions 60. Which group of organisms can trace their ancestry to the protists? A. All of the answer choices are correct. B. None of the answer choices is correct. C. fungi D. plants E. animals Protists are believed to be one of the earliest groups of eukaryotic organisms. All other eukaryotes would be related to the protists.
Bloom’s Level: 1. Remember Learning Outcome: 29.01.01 Summarize what is known regarding the evolution of the first single-celled protist. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
Short Answer Questions 61. Explain why chytrid fungi (phylum Chytridomycota) are believed to have been the first fungi to have evolved. Chytrid fungi are unique among fungi because they are aquatic and they produce flagellated reproductive cells. Most reproduce asexually through the production of zoospores, which grow into new chytrids.
Bloom’s Level: 6. Create Learning Outcome: 29.02.03 Compare the reproductive strategies of the major groups of fungi. Section: 29.02 Topic: Fungi Evolution
62. List the six steps in the life cycle of Plasmodium vivax, the protist that causes malaria. 1. Gametes fuse in the gut of the female mosquito, Anopheles. The zygote will undergo divisions to produce sporozoites, which migrate to the salivary gland. 2. When a mosquito bites a person, the sporozoites are passed into the host's bloodstream, which migrate to the host's liver. 3. Asexual spores (merozoites) produced in the liver enter the host's bloodstream and enter the red blood cells. 4. Merozoites reproduce asexually inside the red blood cells. 5. Merozoites and toxins pour into the bloodstream, causing chills and fever. 6. Some merozoites become gametocytes, which enter the bloodstream. These are taken up by a new mosquito when she bites an infected person.
Bloom’s Level: 6. Create Learning Outcome: 29.01.04 Identify several human diseases caused by protists, and explain why many are difficult to treat. Section: 29.01 Topic: Protists
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Chapter 29 - Protistans
Multiple Choice Questions 63. Which of the following organisms does not lead to a fungal disease? A. Varicella B. Dermatophytes C. Systemic mycoses D. Coccidioides immitis E. Trypanosoma brucei Candida albicans can cause yeast infections. Dermatophytes causes infections of the skin known as tineas. Systemic mycoses refers to fungal infections that affect internal organs, such as the lungs. Coccidioides immitis leads to a fungal disease called valley fever, which can cause benign to severe infections. Trypanosoma brucei is a type of flagellate that causes African sleeping sickness.
Bloom’s Level: 4. Analyze Learning Outcome: 29.02.04 Identify several fungal diseases of humans, and explain why many can be difficult to treat. Section: 29.02 Topic: Fungi
True / False Questions 64. Fungal infections are difficult to treat because of the similarity in their nucleus, organelles, and ribosomes to those in human cells. TRUE It is a challenge for doctors to treat fungal infections because of the similarity between fungal cells and those of the patient. It is difficult to develop an antimicrobial that will destroy a fungal cell without destroying the host cells as well. Researchers are trying to exploit any biochemical differences that exist between fungi and host cells. This will enable them to develop antimicrobial agents that can target specific membrane sterols of the fungi.
Bloom’s Level: 2. Understand Learning Outcome: 29.02.04 Identify several fungal diseases of humans, and explain why many can be difficult to treat. Section: 29.02 Topic: Fungi
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Chapter 30 - Plants
Chapter 30 Plants
Multiple Choice Questions 1. Why are green algae placed in the Protista while plants are given their own kingdom? A. The green algae use chlorophyll a, while plants use chlorophyll b. B. Green algae have a cell wall of chitin, while plants have cell walls of cellulose. C. Plants enclose and protect the embryo within the female plant, while green algae do not. D. All plants are vascular, while all green algae are nonvascular. E. Algae are sporophytes that reproduce only asexually, while plants are gametophytes and reproduce sexually. While algae and plants do share a number of similarities, there are also some very important differences between the two. Among these differences are the protections and nourishment of the developing embryo, the formation of vascular tissue, and the development of seeds as a means of reproduction.
Bloom's Level: 4. Analyze Learning Outcome: 30.01.01 Identify the features present in the ancestor of land plants that helped give rise to the land plants. Section: 30.01 Topic: Plants
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Chapter 30 - Plants
2. Which of the following is NOT a significant event in the evolution of plants? A. protection of the embryo B. evolution of vascular tissue C. evolution of the seed D. evolution of the flower E. evolution of the rhizoid Rhizoids are root-like structures that are found in filamentous algae and in nonvascular land plants. They are an ancestral structure rather than a derived structure so they are not significant to the evolution of plants.
Bloom's Level: 2. Understand Learning Outcome: 30.01.01 Identify the features present in the ancestor of land plants that helped give rise to the land plants. Section: 30.01 Topic: Plant Evolution
3. Structures in plants that produce the gametes are called A. sporangia. B. fruit. C. gametophyte. D. homosporophyte. E. sporophyte. Plants undergo a cycle called alternation of generations that alternate between multicellular haploid and diploid organisms. The haploid generation is called the gametophyte and produces gametes; the gametes fuse and give rise to a diploid generation called the sporophyte. The sporophyte has structures that undergo meiosis and produces spores that are haploid and give rise to the sporophyte generation.
Bloom's Level: 1. Remember Learning Outcome: 30.01.03 Describe alternation of generations in plants. Section: 30.01 Topic: Plants
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Chapter 30 - Plants
4. Which of the following statements is NOT true about alternation of generations life cycle? A. Meiosis produces spores that are haploid. B. Spores develop into the gametophyte generation. C. The gametophyte produces eggs and sperm. D. When the sperm fertilizes the egg, a zygote results that develops into the sporophyte. E. The sporophyte generation produces spores by mitosis in structures called sporangia. Since the sporophyte generation is diploid it must undergo meiosis to produce haploid spores. Mitosis would produce cells that have the same genetic component as the original.
Bloom's Level: 4. Analyze Learning Outcome: 30.01.03 Describe alternation of generations in plants. Section: 30.01 Topic: Plants
5. Which sequence is correct in the life cycle of alternation of generations? A. gametophyte → spores → sporophyte B. gametophyte → gametes → spores → sporophyte C. gametophyte → gametes → zygote → spores → sporophyte D. gametophyte → spores → gametes → zygote → sporophyte E. gametophyte → gametes → zygote → sporophyte → spores Plants undergo a cycle called alternation of generations that alternate between multicellular haploid and diploid organisms. The haploid generation is called the gametophyte and produces gametes; the gametes fuse and give rise to a diploid generation called the sporophyte. The sporophyte has structures that undergo meiosis and produces spores that are haploid and give rise to the sporophyte generation.
Bloom's Level: 2. Understand Learning Outcome: 30.01.03 Describe alternation of generations in plants. Section: 30.01 Topic: Plants
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Chapter 30 - Plants
6. A major evolutionary trend among plants is A. a reduction in the size of the gametophyte and an increase in the size of the sporophyte. B. a reduction in the size of the sporophyte and an increase in the size of the gametophyte. C. an increase in the size of both the sporophyte and the gametophyte. D. a decrease in the size of both the sporophyte and the gametophyte. E. the sporophyte nor the gametophyte have evolved in plants and are basically the same size for all plant types. An evolutionary trend seen in plants is that the haploid gametophyte becomes reduced in size and the diploid sporophyte has increased in size. Since the sporophyte has become reduced in size, it can be enclosed and packaged as a seed which can then be disbursed.
Bloom's Level: 3. Apply Learning Outcome: 30.01.02 List the major events in the evolutionary history of plants. Section: 30.01 Topic: Plant Evolution
7. The structure found in nonvascular plants that anchors them to the soil and may serve in asexual reproduction is called the A. antheridia B. archegonia C. rhizoid D. protonema E. sori Root-like structures called rhizoids are found in nonvascular plants, where they anchor the plant in the soil and may assist in the asexual reproduction by producing new plants along the length of the filament.
Bloom's Level: 1. Remember Learning Outcome: 30.02.01 Identify the traits of nonvascular plants that enable them to survive on land. Section: 30.02 Topic: Plant Tissues
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Chapter 30 - Plants
8. The term for the body of a bryophyte such as Marchantia is A. antheridia. B. rhizoid. C. archegonium. D. protonema. E. thallus. Bryophytes are nonvascular plants that have root-like rhizoids, a leaf-like thallus which produces gemmae that function in asexual reproduction, and disk-headed stalks that bear either antheridia or archegonia and function in sexual reproduction.
Bloom's Level: 1. Remember Learning Outcome: 30.02.02 Compare and contrast the features of liverworts and moses. Section: 30.02 Topic: Plant Tissues
9. A leafy plant with no true roots, no stem, and rhizoids that grow into the soil from the leaflike structure is a A. rose. B. liverwort. C. conifer. D. ginkgo. E. fern. Roses, conifers, ginkgo, and ferns are all vascular plants and so have true roots, leaves, and stems. Liverworts are a division of the bryophytes, which are nonvascular plants; they possess a leaf-like thallus and root-like rhizoids.
Bloom's Level: 1. Remember Learning Outcome: 30.02.02 Compare and contrast the features of liverworts and moses. Section: 30.02 Topic: Bryophytes
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Chapter 30 - Plants
10. The gametophyte of mosses has two stages. During the first stage, there is a branching of filament cells called the A. antheridium. B. archegonium. C. rhizoid. D. protonema. E. sori. Mosses are a division of bryophyte or nonvascular plant. The haploid gametophyte arises from a spore and germinates into the filament of cells that first appears; this is either a male or a female protonema, which grows into the recognizable mature gametophyte. Antheridium is the structure that produces the flagellated sperm. Archegonia are the structures that produce the eggs. Rhizoids are the root-like structures that anchor the plant to the soil. Sori are the structures that house the spores of ferns.
Bloom's Level: 2. Understand Learning Outcome: 30.02.02 Compare and contrast the features of liverworts and moses. Section: 30.02 Topic: Bryophytes
11. In the moss life cycle, the A. haploid gametophyte B. diploid gametophyte C. haploid sporophyte D. diploid sporophyte E. diploid antheridium
generation is dominant.
With alternation of generations, plants cycle between generations that are haploid and give rise to gametes. This is called the gametophyte generation. A generation that is diploid and gives rise to spores after meiosis is called the sporophyte generation. Both generations are multicellular, but in mosses the gametophyte is larger. As plants became adapted to land, the size of the gametophyte decreased and the size of the sporophyte increased until the gametophyte was small enough to be enclosed in a seed.
Bloom's Level: 1. Remember Learning Outcome: 30.02.02 Compare and contrast the features of liverworts and moses. Section: 30.02 Topic: Bryophytes
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Chapter 30 - Plants
12. When a moss spore lands on an appropriate site, it germinates into the first stage of the gametophyte, called a(n) A. antheridium. B. rhizoid. C. archegonium. D. protonema. E. thallus. Mosses are a division of bryophytes, or nonvascular plants. The haploid gametophyte arises from a spore and germinates into the filament of cells that first appears. This is either a male or a female protonema, which grows into the recognizable mature gametophyte.
Bloom's Level: 2. Understand Learning Outcome: 30.02.02 Compare and contrast the features of liverworts and moses. Section: 30.02 Topic: Bryophytes
13. Haploid spores are produced in the A. sori B. antheridia C. archegonia D. sporangium E. anther
of a moss.
The sporangium of a moss is the site of meiosis and the production of haploid spores. The antheridia and archegonia are sites of the production of sperm and egg and so serve in sexual reproduction. Sori are the structures found on ferns that cover the sporangia. Anthers are the structures in flowering plants that bear pollen.
Bloom's Level: 2. Understand Learning Outcome: 30.02.02 Compare and contrast the features of liverworts and moses. Section: 30.02 Topic: Bryophytes
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Chapter 30 - Plants
14. Fertilization in the mosses is dependent upon A. insects. B. mammals. C. water. D. wind. E. light. The flagellated sperm produced on the antheridia must swim to the archegonium which contains the egg. If no water is present, then the sperm has no way to travel to the egg.
Bloom's Level: 1. Remember Learning Outcome: 30.02.02 Compare and contrast the features of liverworts and moses. Section: 30.02 Topic: Bryophytes
15. Which life cycle phase is diploid in moss? A. sporophyte generation B. gametophyte generation C. gametes D. spores E. antheridium With alternation of generations, plants cycle between generations that are haploid and give rise to gametes. This is called the gametophyte generation. A generation that is diploid and gives rise to spores after meiosis is called the sporophyte generation. This cycle is found in all plants and only varies as to which generation is larger.
Bloom's Level: 2. Understand Learning Outcome: 30.02.02 Compare and contrast the features of liverworts and moses. Section: 30.02 Topic: Bryophytes
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Chapter 30 - Plants
16. Swimming sperm are produced in the A. antheridia B. archegonia C. rhizoids D. protonema E. sori
of the moss gametophyte stage.
In moss the gametophyte generation produces antheridia, the sites of the production of sperm, and archegonia, the sites of the production of eggs. Rhizoids are root-like structures that serve to attach the plant. Sori are the structures found on ferns that cover the sporangia. Anthers are the structures in flowering plants that bear pollen. Protonema are the single-celled filaments that arise from spores in moss.
Bloom's Level: 1. Remember Learning Outcome: 30.02.02 Compare and contrast the features of liverworts and moses. Section: 30.02 Topic: Bryophytes
17. Which of the following is a nonvascular plant that has the longest evolutionary history on land? A. ferns B. bristlecone pine tree C. moss D. redwood tree E. grass Ferns, pine trees, redwood trees, and grass are all vascular plants. Mosses are nonvascular plants; they do not have true leaves, true roots, or stems.
Bloom's Level: 3. Apply Learning Outcome: 30.02.01 Identify the traits of nonvascular plants that enable them to survive on land. Section: 30.02 Topic: Plant Evolution
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Chapter 30 - Plants
18. Which of the following is NOT correct? A. xylem—conducts water up from the soil B. phloem—transports organic nutrients C. vascular tissue—consists of xylem and phloem D. vascular plants—sporophyte is dominant E. sporophyte—haploid Sporophytes are diploid, not haploid. The diploid sporophyte produces haploid spores after sporangia undergo meiosis.
Bloom's Level: 2. Understand Learning Outcome: 30.02.01 Identify the traits of nonvascular plants that enable them to survive on land. Section: 30.02 Topic: Plants
19. The vascular tissue by which nutrients are transported throughout ferns is called A. xylem. B. stigma. C. phloem. D. absent in roots. E. produced by stomata. Vascular tissue is classed on the material that they conduct. Xylem conducts water and minerals up from the roots in the soil and phloem which moves organic molecules from one part of the plant into another.
Bloom's Level: 1. Remember Learning Outcome: 30.03.01 Explain the importance of vascular tissue to plants. Section: 30.03 Topic: Pteridophytes
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Chapter 30 - Plants
20. Psilotum are most closely related to A. angiosperms. B. gymnosperms. C. conifers. D. green algae. E. ferns. Psilotum have a life cycle that closely resembles that of the fern and suggests that it is a vestigial fern that has no leaves or roots.
Bloom's Level: 1. Remember Learning Outcome: 30.03.01 Explain the importance of vascular tissue to plants. Section: 30.03 Topic: Pteridophytes
21. In the life cycle of the fern, meiosis takes place A. during development of seeds. B. upon development of the gametophyte. C. during the production of egg and sperm. D. during the production of spores. E. during the development of vascular tissue. The diploid sporophyte of the fern produces sporangia, which are sites of meiosis that produce haploid spores.
Bloom's Level: 2. Understand Learning Outcome: 30.03.02 Distinguish between a lycophyte and pteridophyte. Section: 30.03 Topic: Pteridophytes
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Chapter 30 - Plants
22. Which of the following is NOT characteristic of a fern? A. Eggs are produced in antheridia. B. Fern sporophyte is the dominant generation in the fern life cycle. C. Fern gametophyte is nonvascular. D. Fern sporophyte leaves first appear in a curled-up form termed a fiddlehead. E. Fern sporophyte is vascular. In ferns, the dominant generation is the diploid sporophyte that produces sporangia; sites of meiosis that produce haploid spores. The spores give rise to a haploid, leaf-looking structure that does not contain vascular tissue called the thallus. On the thallus are antheridia, which produce sperm, and the archegonium, which produce eggs. Sperm swim to the archegonium and fuse to produce a diploid zygote that remains in the archegonium. The young sporophyte grows from the gametophyte and eventually gives rise to the mature sporophyte.
Bloom's Level: 4. Analyze Learning Outcome: 30.03.01 Explain the importance of vascular tissue to plants. Section: 30.03 Topic: Pteridophytes
23. Compared to the moss, the fern is better adapted to life on land because the fern A. produces seeds, but the moss does not. B. does not have a swimming sperm, but the moss does. C. does not have a gametophyte generation, but the moss does. D. has vascular tissue, but the moss does not. E. can be found in dry areas, but moss cannot. Vascular tissue allows plants to grow in areas that do not have a readily available water source. It also allows plants to grow away from the ground and be taller and larger.
Bloom's Level: 3. Apply Learning Outcome: 30.03.01 Explain the importance of vascular tissue to plants. Section: 30.03 Topic: Pteridophytes
30-12 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 30 - Plants
24. In the life cycle of a fern, the zygote A. becomes the sporophyte generation. B. becomes the gametophyte generation. C. does not occur. D. undergoes meiosis. E. is haploid. On the thallus of the gametophyte are antheridia, which produce sperm, and the archegonium, which produce eggs. Sperm swim to the archegonium and fuse to produce a diploid zygote, which remains in the archegonium. The young sporophyte (embryo) grows from the gametophyte and eventually gives rise to the mature sporophyte.
Bloom's Level: 2. Understand Learning Outcome: 30.03.02 Distinguish between a lycophyte and pteridophyte. Section: 30.03 Topic: Pteridophytes
25. Which statement is NOT true about ferns? A. The gametophyte stage lacks vascular tissue. B. The sporophyte stage lacks vascular tissue. C. Antheridia and archegonia develop on the prothallus of the gametophyte stage. D. Fertilization requires moisture for the sperm to swim to the egg within the archegonia. E. The sporophyte stage is the dominant generation. In ferns, the dominant generation is the diploid sporophyte that produces sporangia, sites of meiosis that produce haploid spores. The spores give rise to a haploid gametophyte, a leaflooking structure that does not contain vascular tissue, called the thallus. On the thallus are antheridia, which produce sperm, and the archegonium, which produce eggs; the sperm require water to swim to the archegonium and fuse to produce a diploid zygote, which remains in the archegonium. The young sporophyte grows from the gametophyte and eventually gives rise to the mature sporophyte. The mature sporophyte is a vascular plant with true roots, leaves, and stems.
Bloom's Level: 4. Analyze Learning Outcome: 30.03.02 Distinguish between a lycophyte and pteridophyte. Section: 30.03 Topic: Pteridophytes
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Chapter 30 - Plants
26. Which statement is NOT true about fiddleheads? A. They are young fronds curled up. B. They are found in ferns. C. They are part of the sporophyte generation. D. They are haploid. E. They are part of a vascular plant. Fiddleheads are the earliest stages of the diploid sporophyte generation of ferns.
Bloom's Level: 2. Understand Learning Outcome: 30.03.01 Explain the importance of vascular tissue to plants. Section: 30.03 Topic: Pteridophytes
27. The fiddlehead is a that unfolds into a A. diploid gametophyte; fern frond B. diploid sporophyte; fern frond C. haploid sporophyte; fern frond D. haploid gametophyte; fern frond E. diploid gametophyte; moss plant
.
Fiddleheads are the earliest stages of the diploid sporophyte generation of ferns. They will eventually mature into the recognizable fern frond.
Bloom's Level: 2. Understand Learning Outcome: 30.03.02 Distinguish between a lycophyte and pteridophyte. Section: 30.03 Topic: Pteridophytes
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Chapter 30 - Plants
28. In the fern life cycle, the A. haploid gametophyte B. diploid gametophyte C. haploid sporophyte D. diploid sporophyte E. monoploid sporophyte
generation is dominant.
In ferns, the dominant generation is the diploid sporophyte that produces sporangia. The mature sporophyte is a vascular plant with true roots, leaves, and stems.
Bloom's Level: 2. Understand Learning Outcome: 30.03.02 Distinguish between a lycophyte and pteridophyte. Section: 30.03 Topic: Pteridophytes
29. Which of these statements is NOT true about fertilization in a fern plant? A. An egg is produced in an archegonium. B. A sperm is produced in an antheridium. C. The sperm is carried by the wind to the egg. D. The eggs and sperm are produced on the same plant. E. Eggs and sperm are produced by the gametophyte generation. Spores give rise to a haploid gametophyte; a leaf-looking structure that does not contain vascular tissue, called the thallus. On the thallus are antheridia, which produce sperm, and the archegonium, which produce eggs; the sperm require water to swim to the archegonium and fuse to produce a diploid zygote, which remains in the archegonium.
Bloom's Level: 2. Understand Learning Outcome: 30.03.02 Distinguish between a lycophyte and pteridophyte. Section: 30.03 Topic: Pteridophytes
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Chapter 30 - Plants
30. It is possible to take the mature frond of many ferns and rub the brown growths on the lower surface over a wet plug of peat moss. Soon, small heart-shaped green structures grow on the nutritious peat moss plugs. What has happened? A. You have broken off male and female structures from the bottom of the fern frond and they united on the wet plug to form the sporophyte generation called a prothallus. B. You have broken the sori on the bottom of the fern frond and spores fell on the plug to germinate and form the gametophyte called a prothallus. C. You have broken the sori on the bottom of the fern frond and spores fell on the plug to germinate and form rhizomes that develop an indusium. D. You have broken the indusium on the bottom of the fern frond and zygotes germinated on the plug to form the sporophyte made of an archegonium and an antheridium. E. You have broken the sori on the bottom of the fern frond and zygotes germinated on the plug to form the fiddleheads that will develop into more fronds. The sporangia of the fern sporophyte is located on the bottom of the fronds and covered by structures called sori that look like brown spots on the bottom of the leaf. These sporangia are the sites of meiosis and contain spores. The spores develop into heart-shaped, leaf-like prothallus when placed in a favorable environment.
Bloom's Level: 5. Evaluate Learning Outcome: 30.03.02 Distinguish between a lycophyte and pteridophyte. Section: 30.03 Topic: Pteridophytes
31. The small, heart-shaped gametophyte called a prothallus is part of the life cycle of the A. mosses. B. ferns. C. liverworts. D. ginkgoes. E. gnetophyte. The spores of a fern sporophyte develop into heart-shaped, leaf-like prothallus when placed in a favorable environment.
Bloom's Level: 1. Remember Learning Outcome: 30.03.02 Distinguish between a lycophyte and pteridophyte. Section: 30.03 Topic: Pteridophytes
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Chapter 30 - Plants
32. Spores are produced in the A. sporangium B. antheridia C. archegonia D. ovule E. ovary
of a fern.
In ferns, the dominant generation is the diploid sporophyte that produces sporangia, sites of meiosis that produce haploid spores.
Bloom's Level: 1. Remember Learning Outcome: 30.03.02 Distinguish between a lycophyte and pteridophyte. Section: 30.03 Topic: Pteridophytes
33. Which of the following is from a division within the gymnosperm group? A. horsetail B. cycad C. moss D. fern E. monocot Gymnosperms are cone-bearing vascular plants. These cones contain ovules that later become seeds. The ovules are called naked because they are not completely enclosed in diploid tissue. Cycads are an ancient form of plant that bears cones and were plentiful in the Mesozoic era. Horsetails and mosses are nonvascular plants. Ferns are vascular plants but do not produce cones. Monocots are seed-bearing plants that are part of the flowering plants.
Bloom's Level: 1. Remember Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Gymnosperms
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Chapter 30 - Plants
34. Which vascular plants have seeds? A. ferns B. club mosses C. whisk ferns D. gymnosperms E. horsetails Ferns, club mosses, whisk ferns, and horsetails all produce either spores or a zygote that is retained in the archegonium. Gymnosperms are cone-bearing vascular plants. These cones contain ovules that later become seeds. The ovules are called naked because they are not completely enclosed in diploid tissue.
Bloom's Level: 1. Remember Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Gymnosperms
35. Plants that flourished in the Mesozoic era along with dinosaurs were the A. ferns. B. bryophytes. C. cycads. D. flowering plants. E. conifers. Cycads are gymnosperms that flourished during the Mesozoic era and were a food source for dinosaurs.
Bloom's Level: 1. Remember Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Gymnosperms
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Chapter 30 - Plants
36. Which of the following types of plants will produce a naked seed? A. moss B. ferns C. algae D. gymnosperms E. angiosperms Gymnosperms are cone-bearing vascular plants. These cones contain ovules that later become seeds. The ovules are called naked because they are not completely enclosed in diploid tissue. The only other plant listed that produces seeds are the angiosperms, which develop from an ovule within an ovary that becomes a fruit. Angiosperms therefore produce covered seeds, unlike gymnosperms.
Bloom's Level: 1. Remember Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Gymnosperms
37. How is a seed like a spore? A. They both are haploid. B. They both contain the sporophyte generation. C. They both contain the gametophyte generation. D. They both are flagellated. E. They both disperse the species. Seeds are the gametophyte generations for gymnosperms and angiosperms, while spores give rise to the sporophyte generations for non-seed-bearing plants. Neither seeds nor spores are flagellated. They do, however, help to disperse the organisms. Spores are frequently blown on the wind for dispersal and seeds are carried in a variety of ways including wind, water, and animals.
Bloom's Level: 3. Apply Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Plants
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Chapter 30 - Plants
38. The megaspore in seed plants will develop into the A. seeds. B. spores. C. pollen grains. D. mature female gametophyte. E. mature male gametophyte. The megaspore undergoes mitosis and develops into a mature female gametophyte with two to six archegonia, each of which contains a single egg near the ovule opening.
Bloom's Level: 2. Understand Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Plants
39. Which statement is NOT correct about the sporophyte generation in gymnosperms? A. It is the diploid generation. B. It contains vascular tissue. C. It is the dominant generation. D. They produce winged naked seeds. E. They produce homospores. The sporophyte generation in gymnosperms produces microspores found in pollen, and megaspores containing an ovule.
Bloom's Level: 2. Understand Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Gymnosperms
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Chapter 30 - Plants
40. Which of these may occur in both angiosperms and gymnosperms? A. Seeds develop within a cone. B. Seeds develop within a flower. C. Seeds are surrounded by a fruit at maturity. D. Pollen is carried by the wind for pollination. E. Pollen is carried by honeybees as pollinators. In gymnosperms, seeds develop within cones and, in angiosperms, seeds develop within a flower. The seeds from a flowering plant are surrounded by a fruit. The pollen of flowering plants is often carried by honeybees but not for gymnosperms. The pollen of both gymnosperms and angiosperms is frequently carried on the wind.
Bloom's Level: 4. Analyze Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Plants
41. Which statement is NOT true about seed plants? A. They generally do not require external water for reproduction. B. Seed plants produce homospores. C. The pollen grain later becomes the male gametophyte. D. The female gametophyte is retained within the sporophyte. E. The seed contains the embryonic sporophyte generation. Seed plants are heterosporous; this means that they produce two different types of spores: megaspores that contain the ovule and microspores that contain pollen. These produce two different types of gametophytes: male and female.
Bloom's Level: 2. Understand Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Plants
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Chapter 30 - Plants
42. Which statement is true regarding both gymnosperms and angiosperms? A. They both have swimming sperm. B. They both lack the gametophyte generation. C. They both have a separate male and female gametophyte generation. D. They both produce cones and flowers. E. They both rely on pollinators. Seed plants are heterosporous; this means that they produce two different types of spores: megaspores that contain the ovule and microspores that contain pollen. These produce two different types of gametophytes: male and female.
Bloom's Level: 2. Understand Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Plants
43. Pine trees have a land-based adaptation. When microsporocytes undergo meiosis to produce microspores, they are then liberated as the A. embryo sacs. B. ovules. C. pollen grains. D. zygotes. E. egg. Microsporocytes, once they have gone through meiosis, produce four haploid microspores that contain the pollen grains.
Bloom's Level: 4. Analyze Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Gymnosperms
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Chapter 30 - Plants
44. In the pine life cycle, a microspore develops into A. a seed. B. a sporophyte. C. a pollen grain. D. a megaspore. E. an ovule. Haploid microspores contain the pollen grains and four cells are produced from each microsporocyte.
Bloom's Level: 1. Remember Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Gymnosperms
45. Which of the following is NOT a reason why the ginkgo is unique among gymnosperm? A. The ginkgo grows well in polluted areas. B. Ginkgo biloba is the only surviving species in its division. C. The tree is propagated vegetatively to provide male trees that lack smelly seeds. D. Ginkgos have naked seeds. E. Ginkgos are popular for big cities along streets and parks. Since all gymnosperms have naked seeds, this would not make ginkgos unique.
Bloom's Level: 2. Understand Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Gymnosperms
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Chapter 30 - Plants
46. Monocots have their floral parts in multiples of those numbers. A. threes; fours or fives B. threes or fours; fives C. fours; threes or fives D. fours or fives; threes E. fives; threes or fours
, while eudicots have their parts in
, or
Monocots are so named because they have only one cotyledon in their seeds and eudicots have two cotyledons. The cotyledons are the seed leaves containing nutrients to nourish the developing embryo.
Bloom's Level: 1. Remember Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
47. Characteristics of eudicots include all EXCEPT A. two cotyledons. B. leaves with netted veins. C. woody or herbaceous plants. D. flower parts in fours and fives. E. vascular bundles arranged irregularly in the stem. Eudicots have two cotyledons, or seed leaves, and have a network of veins through the leaves. They are either woody or herbaceous (non-woody stems) plants and have flower parts that are either arranged in fours or fives or in multiples of these. The vascular tissue is arranged in a ring in the stem.
Bloom's Level: 2. Understand Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
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Chapter 30 - Plants
48. A lily (a monocot) would have A. rhizoids rather than true roots. B. flower parts in threes. C. seeds in a cone. D. netted leaf veins. E. two cotyledons. Monocots have one cotyledon, or seed leaf, and they have a parallel arrangement of veins in the leaves. They are usually herbaceous, have non-woody stems, and have flower parts that are either arranged in multiples of three. The vascular tissue is scattered in bundles in the stem.
Bloom's Level: 1. Remember Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
49. Gladiolus produces leaves with parallel veins and floral parts in multiples of three. This plant is a(n) A. bryophyte. B. eudicot. C. monocot. D. cycad. E. fern. Plants that have parallel venation and have flower parts arranged in multiples of three are classified as monocots.
Bloom's Level: 1. Remember Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
30-25 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 30 - Plants
50. The structure that forms most of the flesh of an apple is the A. carpel. B. sepal. C. ovary. D. petal. E. stamen. A typical fruit consists of a mature ovary and sometimes other parts of the flower.
Bloom's Level: 2. Understand Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
51. Pollen would land on the A. ovary B. stigma C. anther D. filament E. style
during pollination.
The sticky stigma would trap pollen that landed there and a pollen tube would grow from the pollen grain down to the egg in the ovule, fertilizing the egg.
Bloom's Level: 1. Remember Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
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Chapter 30 - Plants
52. All of the flower mechanisms would help prevent pollination by a foreign species except A. A pollen tube only grows in the style of its own species. B. The stigma is sticky mainly to its own species of pollen. C. The timing of flowering varies by species and keeps pollinators moving among a limited number of species. D. Flowers attract a small number of specialized pollinators, and therefore pollen is not spread equally to all available species of flowers. E. Flowers of different species are the same color and attract similar pollinators. All of the choices listed except that flowers of different species are the same color and attract similar pollinators would prevent pollination by a different species of plant. If flowers of different species were the same color and attracted the same pollinators, then the possibility that some pollen from a different species could fertilize the flower.
Bloom's Level: 5. Evaluate Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Angiosperms
53. The reproductive parts of an angiosperm are the A. petals and sepals. B. petals and stamen. C. sepals and stamen. D. sepals and carpel. E. carpel and stamen. The parts of a flower are the sepals that make up the calyx and protect the flower bud. The petals often attract a specific pollinator based on color or shape. In addition to these parts, which do not take direct part in reproduction, are the male and female flower parts. The male parts are the anther, where pollen is produced, and filament, and the female parts are the stigma with a sticky end, the style that connects the stigma and ovule, and the ovule that contains the ovary.
Bloom's Level: 1. Remember Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Angiosperms
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Chapter 30 - Plants
54. During pollination in angiosperms, pollen grains will stick to the enlarged knob in the center of the flower called the A. stigma. B. style. C. ovary. D. ovule. E. anther. The sticky stigma would trap pollen that landed there and a pollen tube would grow from the pollen grain down to the egg in the ovule, fertilizing the egg.
Bloom's Level: 2. Understand Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Angiosperms
55. The stigma, style, and ovary are located in a A. carpel. B. sepal. C. receptacle. D. petal. E. stamen. The carpel, or pistil, of a flower is made up of the stigma, style, and the ovary that contains the ovule.
Bloom's Level: 1. Remember Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
30-28 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 30 - Plants
56. The anther and filament are parts of the A. carpel. B. sepal. C. receptacle. D. petal. E. stamen. The male parts of the flower, called the stamen, are the anther that produces pollen and the filament that supports the anther.
Bloom's Level: 1. Remember Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
57. Ovules are contained within the A. ovary. B. stigma. C. anther. D. filament. E. style. The carpel, or pistil, of a flower is the female part of the flower and is made up of the stigma, style, and the ovary that contains the ovule.
Bloom's Level: 1. Remember Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
30-29 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 30 - Plants
58. The site of formation of pollen grains is in the A. carpel. B. sepal. C. receptacle. D. petal. E. stamen. The anther at the end of the filament is where pollen is produced.
Bloom's Level: 1. Remember Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
59. Botanically, a seed is a structure developed from a(n) A. style. B. ovule. C. stigma. D. anther. E. sepal. A seed is a mature ovule containing a sporophyte. The stigma is the sticky end of a carpel where pollen is trapped and the style is the tube that connects the stigma to the ovule.
Bloom's Level: 2. Understand Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Angiosperms
30-30 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 30 - Plants
60. Which part of a seed provides nourishment for the embryo, enabling seed plants to survive on land? A. ovary B. seed coat C. egg sac D. endosperm E. ectosperm The developing embryo is nourished by the endosperm, a 3n structure that arises from the polar nuclei and one of the two sperm in the ovule.
Bloom's Level: 2. Understand Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Angiosperms
61. Which part of an angiosperm seed is triploid (3n)? A. embryonic sporophyte generation B. embryonic gametophyte generation C. protective seed coat D. fruit around the seed E. endosperm The developing embryo is nourished by the endosperm, a 3n structure that arises from the polar nuclei and one of the two sperm in the ovule.
Bloom's Level: 2. Understand Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Angiosperms
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Chapter 30 - Plants
62. The function of endosperm is to A. form the seedling. B. develop into the fruit. C. provide water to the embryo. D. provide nutrients to the embryo. E. provide a protective coating for the embryo. The developing embryo is nourished by the endosperm, a 3n structure that arises from the polar nuclei and one of the two sperm in the ovule.
Bloom's Level: 1. Remember Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
63. Which statement best describes double fertilization in an angiosperm? A. Two egg cells are fertilized within each ovule. B. A sperm nucleus fuses with an egg cell and a sperm nucleus fuses with polar nuclei. C. Two sperm are required for fertilization of one egg cell. D. A flower can engage in both self-pollination and cross-pollination. E. One sperm fertilizes the egg while two polar nuclei fuse with a second egg. Two sperm move down the pollen tube and enter the ovule where one of the sperm fertilizes the egg and the other fuses with the two polar nuclei to form the 3n endosperm.
Bloom's Level: 2. Understand Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Angiosperms
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Chapter 30 - Plants
64. Double fertilization in an angiosperm produces A. a diploid zygote and a haploid polar nucleus. B. a diploid zygote and a diploid endosperm. C. a diploid embryo and a triploid zygote. D. a triploid embryo and a diploid endosperm. E. a diploid zygote and a triploid endosperm. Two sperm move down the pollen tube and enter the ovule where one of the sperm fertilizes the egg and the other fuses with the two polar nuclei to form the 3n endosperm.
Bloom's Level: 2. Understand Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Angiosperms
65. Seeds contain all of the following EXCEPT A. a seed coat. B. an endosperm. C. an embryo. D. a pollen grain. E. triploid tissue. Seed is composed of several layers; listed from the outside in they are the seed coat derived from the integument of the ovule, the triploid endosperm that is derived from the polar nuclei of the ovule and the embryo.
Bloom's Level: 2. Understand Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Angiosperms
30-33 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 30 - Plants
66. Which feature is common across all seed plants? A. flowers B. fruits C. pollen tubes D. flagellated sperm E. sperm carried by pollinators Gymnosperms are seed-bearing plants that do not have flowers or fruits. Neither gymnosperms nor angiosperms have flagellated sperm nor do all have sperm carried by windborne pollen. Both gymnosperms and angiosperms do use pollen tubes to allow sperm to enter the ovule and fertilize the egg.
Bloom's Level: 3. Apply Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Angiosperms
67. Which statement is NOT true about the life cycle of gymnosperms? A. The sporophyte is diploid. B. The sporophyte produces two distinct types of spores. C. The megasporocyte develops into the pollen grain. D. The microsporocyte develops into the pollen grain. E. Two nonflagellated sperm are released by the male gametophyte. The sporophyte is the diploid portion of the life cycle. The sporophyte produces microspores and megaspores. The megaspore develops into the seed, NOT the pollen grain. The microspore does develop into the pollen grain. Once the pollen grain reaches the seed cone it will mature into the male gametophyte and then release two nonflagellated sperm.
Bloom's Level: 4. Analyze Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Gymnosperms
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Chapter 30 - Plants
68. If a tree has flower petals that are small and the same color as its leaves, it is likely to be pollinated by A. bats. B. bees. C. the wind. D. butterflies. E. moths. Since flower petals are generally of a color and shape to attract specific pollinators, it is likely that flowers that are not showy are not pollinated by either insects or bats but rather by the wind. If the more elaborate colors or shapes are not needed to attract the pollinator, then the plant would be using energy unnecessarily to make such flowers.
Bloom's Level: 4. Analyze Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
69. Which of the following statements is NOT true about angiosperms? A. The reproductive structures are located in flowers. B. The flower protects the pollen grain. C. They are divided into monocots and eudicots. D. The seeds are enclosed by fruit. E. They exhibit double fertilization. Angiosperms are also called flowering plants. They are divided into the monocots and eudicots, based on the number of seed leaves and other characteristics. In angiosperms the ovary of the flower develops into a fruit. A second sperm that passes down the pollen tube fuses with one to two nuclei that are present in the ovule to form the endosperm. This is termed double fertilization.
Bloom's Level: 2. Understand Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Angiosperms
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Chapter 30 - Plants
70. Which of the following is NOT considered to be a fruit according to a biologist? A. apples B. grapes C. tomatoes D. peas in a pod E. celery stalks Celery is the vegetative stalk of a plant, whereas all of the other foods listed are mature ovules and, therefore, fruit.
Bloom's Level: 3. Apply Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Angiosperms
71. A megasporocyte would be found in the A. ovary. B. stigma. C. anther. D. filament. E. style. A megasporocyte is found in the ovary, which is inside the carpel of the flower.
Bloom's Level: 2. Understand Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Angiosperms
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Chapter 30 - Plants
72. A fruit is a A. stigmatized pollen grain. B. mature ovary. C. microspore. D. megaspore. E. seed. A fruit develops from the ovary of the flower.
Bloom's Level: 1. Remember Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Angiosperms
73. Showy flowers are most likely pollinated by A. wind only. B. insects only. C. birds only. D. water only. E. insects or birds. Since flower petals are generally of a color and shape to attract specific pollinators, it is likely that flowers that are showy would be needed to attract birds or insects.
Bloom's Level: 3. Apply Learning Outcome: 30.04.02 Compare and contrast gymnosperms with angiosperms, explaining why angiosperms are more species-rich today. Section: 30.04 Topic: Angiosperms
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Chapter 30 - Plants
74. What is thought to be the first major derived feature that separates land plants from charophytes? A. Both charophytes and land plants have chlorophyll b, but only land plants have chlorophyll a. B. Both charophytes and land plants store excess carbohydrates, but only land plants use glucose. C. Both charophytes and land plants protect their zygotes, but only land plants also protect the resulting embryo. D. Both charophytes and land plants are made up of cells, but only land plants have cell walls. E. Both charophytes and land plants have cell walls, but only land plants use cellulose in their cell walls. In order for land plants to be successful, they need to be able to reproduce effectively. One of the difficulties with this is that the zygote and embryo could dry out and die. Charophytes and land plants both protect the zygote, but land plants protect the embryo as well. This characteristic made possible a successful colonization of land by the ancestral forms that gave rise to land plants.
Bloom's Level: 5. Evaluate Learning Outcome: 30.04.01 List the features of seed plants that make them fully adapted to life on land. Section: 30.04 Topic: Plant Evolution
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Chapter 30 - Plants
75. Both charophytes and land plants share which of these characteristics? A. Both have true roots and leaves. B. Both have chlorophyll a and store carbohydrates as starch. C. Both have vascular tissue and protect the zygote. D. Both protect the embryo and have cell walls with cellulose. E. Both cell walls that contain cellulose and produce seeds. Charophytes and land plants share three characteristics: 1) they both contain chlorophyll a and b; 2) they both use starch as a storage medium and have cellulose in their cell walls, meaning that both make glucose the molecule that is used to produce both of these; and finally, 3) they both protect the zygote once it is formed. In addition, land plants protect the embryo that develops from the zygote. Only some land plants have vascular tissue and, therefore, true roots, stems, and leaves (mosses do not) and only some land plants form seeds (neither mosses nor ferns produce seeds).
Bloom's Level: 4. Analyze Learning Outcome: 30.01.01 Identify the features present in the ancestor of land plants that helped give rise to the land plants. Section: 30.01 Topic: Plant Evolution
76. For the ancestors of land plants, supplying water to all cells of the organism would have been possible because A. algae have vascular tissue that transports water and nutrients to all parts of the organism. B. water would simply diffuse into the cells from the surrounding environment. C. the circulatory system of algae evolved early in the history of life. D. algae take water in through root-like cilia that move water into the cells. E. algae have specialized structures called desmosomes that move water into the cells. Since algae are surrounded by water it would only be necessary for the water to diffuse into the cells from the surrounding medium; this is also how nutrients are moved into algal cells.
Bloom's Level: 4. Analyze Learning Outcome: 30.01.01 Identify the features present in the ancestor of land plants that helped give rise to the land plants. Section: 30.01 Topic: Plant Evolution
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Chapter 30 - Plants
77. Both land plants and algae use starch as a storage medium and cellulose in their cell walls. The best conclusion that can be drawn from this observation is that A. they both produce glucose since this is the molecule that both starch and cellulose are made from. B. they both use chlorophyll as a photosynthetic pigment. C. they both use tubers to store starch and they both use the same enzymes to make cellulose. D. they both produce galactose, which gives rise to maltose the molecule that both starch and cellulose are made from. E. if algae are transported to land they will be able to survive the same as a land plant. Both starch and cellulose are made from molecules of glucose; the difference between the two is that starch is made from -glucose and cellulose is made from -glucose. The chemical difference between the two forms of glucose causes the connecting bonds between the molecules of glucose to change, thus changing the overall structure of the resulting starch or cellulose.
Bloom's Level: 3. Apply Learning Outcome: 30.01.01 Identify the features present in the ancestor of land plants that helped give rise to the land plants. Section: 30.01 Topic: Plant Evolution
78. Which feature possessed by most land plants is missing in mosses, which grow low to the ground in damp, shaded areas? A. cell walls, so they cannot stand upright B. a protected zygote, so they must grow in damp areas to supply the zygote with moisture C. vascular tissue, so they must grow along the ground in damp areas to supply water to the cells D. chlorophyll b, so they can only use filtered light for photosynthesis E. starch production, so they only produce small amounts of glucose, since any beyond what they use immediately cannot be stored Mosses lack vascular tissue and therefore need to grow in places where they are constantly bathed in water. This limits mosses to damp, shaded areas and prevents the plant from growing upward away from the water used to bathe the cells of the plant.
Bloom's Level: 2. Understand Learning Outcome: 30.01.02 List the major events in the evolutionary history of plants. Section: 30.01 Topic: Plant Evolution
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Chapter 30 - Plants
79. One adaptation that was necessary for charophytes to move onto land was the protection of the embryo from desiccation. In mosses the embryo, called a developing sporophyte, is A. protected within the archegonium and nourished by the "foot" growing down into the gametophyte tissue. B. protected by the antheridia and nourished by the rhizoids. C. protected by the protonema and nourished by the "foot" growing down into the gametophyte. D. protected within the antheridia and nourished by the "foot" growing down into the sporophyte tissue. E. not protected since mosses are not vascular plants. Mosses show alternation of generations. The haploid gametophyte produces antheridia and archegonia that contain the structures to produce sperm and egg. The sperm swim to the archegonia and fertilize the egg, which is still inside the archegonia. The resulting zygote and later the embryo remain inside the archegonia, where it is protected. The developing sporophyte (embryo) consists of a foot, stalk, and upper capsule. The developing gametophyte is nourished by the sporophyte to which it is attached by the foot.
Bloom's Level: 4. Analyze Learning Outcome: 30.02.01 Identify the traits of nonvascular plants that enable them to survive on land. Section: 30.02 Topic: Plant Evolution
80. Which of the following is not true of ferns? A. Ferns are seedless vascular plants. B. Ferns can grow in dry, cold conditions. C. The gametophyte of ferns is nonvascular. D. Ferns can reproduce asexually through the growth of rhizomes in the soil. E. The gametophyte is dependent on the sporophyte for nourishment. The gametophyte of ferns is an independent structure. It grows from a spore in damp soil and gives rise to both antheridia and archegonia. Once fertilization has occurred, the young sporophyte grows from the archegonia of the gametophyte.
Bloom's Level: 2. Understand Learning Outcome: 30.03.01 Explain the importance of vascular tissue to plants. Section: 30.03 Topic: Pteridophytes
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Chapter 30 - Plants
Short Answer Questions 81. Identify the traits of nonvascular plants that enabled them to adapt to life on land. Nonvascular plants have rootlike, stemlike, and leaflike structures. The gametophyte is the dominant generation. Flagellated sperm swim in a continuous film of water to reach the egg. The sporophyte is attached to the gametophyte and receives its nourishment from it. They also have a form of protection for the embryo to keep it from drying out.
Bloom's Level: 6. Create Learning Outcome: 30.02.01 Identify the traits of nonvascular plants that enable them to survive on land. Section: 30.02 Topic: Bryophytes
82. List the five major events in the evolutionary history of plants. 1. Some form of protection developed for the embryo to prevent drying out. 2. Vascular tissue developed to transport water and nutrients throughout the plant body. 3. Megaphylls, large leaves, evolved to increase the surface area for photosynthesis. 4. Seeds developed to contain the embryo and supportive nutrients within a protective coat. 5. Flowers evolved to attract pollinators.
Bloom's Level: 6. Create Learning Outcome: 30.01.02 List the major events in the evolutionary history of plants. Section: 30.01 Topic: Plant Evolution
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Chapter 30 - Plants
83. List the key features used to identify if a flowering plant is a monocot or a eudicot. Monocot features: 1. One cotyledon 2. Flower parts in threes or multiples of threes 3. Usually herbaceous 4. Usually parallel venation 5. Scattered bundles in stem 6. Fibrous root system Eudicot features: 1. Two cotyledons 2. Flower parts in fours or fives or multiples of four or five 3. Woody herbaceous 4. Usually net venation 5. Vascular bundles in a ring 6. Taproot system
Bloom's Level: 6. Create Learning Outcome: 30.04.03 List the key features of monocots and eudicots. Section: 30.04 Topic: Angiosperms
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Chapter 31 - Animals: Part I
Chapter 31 Animals: Part I
Multiple Choice Questions 1. Which of the following is generally NOT a characteristic of all animals? A. They ingest food that is digested in a central cavity. B. Usually undergo sexual reproduction. C. They produce an embryo that undergoes development in stages. D. Animals range from unspecialized single-celled to specialized multicellular forms. E. The adult form is diploid. Animals are multicellular and most have cells that are specialized to form tissues and organs.
Bloom’s Level: 2. Understand Learning Outcome: 31.01.01 List the general characteristics of animals. Section: 31.01 Topic: Animals
2. Which statement is NOT true about invertebrate animals? A. The vast majority of animal species are invertebrates. B. Many invertebrate species live in a marine environment. C. Most animal phyla are invertebrates. D. All major invertebrate groups arose from protistan ancestors. E. Invertebrates lack a nervous system. Invertebrates such as sponges do lack nerve tissue, while others have a nervous system that includes cephalization, eyes, and a brain.
Bloom’s Level: 2. Understand Learning Outcome: 31.01.01 List the general characteristics of animals. Section: 31.01 Topic: Animals
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Chapter 31 - Animals: Part I
3. Animals that have no particular symmetry exhibit A. asymmetry. B. radial symmetry. C. bilateral symmetry. D. trilateral symmetry. E. spherical symmetry. Animals that have no discernable symmetry are asymmetric and include sponges. Radial symmetry is found in jellyfish and shows a circular organization radiating out from a central point. Bilateral symmetry is found in humans and shows organization into two halves (right and left) and tend to have cephalization.
Bloom’s Level: 1. Remember Learning Outcome: 31.01.02 Compare and contrast the types of symmetry found in animals. Section: 31.01 Topic: Animals
4. Which of the following is NOT true about protostomes? A. They are types of advanced invertebrates. B. They form a true coelom. C. The first opening during embryonic development becomes the anus. D. They will become mollusks, annelids, and arthropods. E. They are more advanced than acoelomates, but not as advanced as deuterostomes. Protostomes include the mollusks, annelids, and arthropods. In protostomes the mouth develops from the blastopore. In protostomes the mesoderm arises from cells located near the embryonic blastopore and splits to form the coelom.
Bloom’s Level: 2. Understand Learning Outcome: 31.01.03 Describe the differences between protostomes and deuterostomes. Section: 31.01 Topic: Protostomes
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Chapter 31 - Animals: Part I
5. Which difference distinguishes protostomes from deuterostomes? A. One has a nervous system, and one doesn't. B. Their embryonic development is different. C. One has a circulatory system and one doesn't. D. One has cephalization and one doesn't. E. One has a notochord and one doesn't. There are two developmental plans: protostomes or deuterostomes. During the embryonic development of protostomes the mouth develops from the blastopore and in deuterostomes the anus develops from the blastopore.
Bloom’s Level: 2. Understand Learning Outcome: 31.01.03 Describe the differences between protostomes and deuterostomes. Section: 31.01 Topic: Protostomes
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Chapter 31 - Animals: Part I
6. Which of the following characteristics is NOT true about deuterostomes? A. The second opening during embryonic development becomes the mouth. B. They include the echinoderms and humans. C. They have a coelom. D. They include the simplest of invertebrates. E. They are typically more advanced than protostomes. Deuterostomes are comprised of echinoderms and chordates. Chordates all have notochords and includes the vertebrates whose notochord is replaced by a vertebral column.
Bloom’s Level: 2. Understand Learning Outcome: 31.01.03 Describe the differences between protostomes and deuterostomes. Section: 31.01 Topic: Deuterostomes
7. Which of the following is correct in matching the common name with a phylum name? A. planarian—Nematoda B. fluke—Platyhelminthes C. coral—Porifera D. roundworm—Cnidaria E. sponge—Arthropoda Flukes are a type of flatworm and therefore belong in the phylum Platyhelminthes. Planarians are also flatworms and therefore in the phylum Platyhelminthes. Coral are in the phylum Cnidaria. Roundworms are in the phylum Nematoda and sponges are in the phylum Porifera
Bloom’s Level: 2. Understand Learning Outcome: 31.01.01 List the general characteristics of animals. Section: 31.01 Topic: Animals
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Chapter 31 - Animals: Part I
8. Which of the following organisms are considered advanced invertebrates with a true coelom? A. sponges B. hydras C. flatworms D. roundworms E. earthworms Sponges lack true tissue. Hydras have two layers of tissue in the embryo; the outer ectoderm gives rise to a protective epidermis, while the inner tissue derived from the endoderm is the gastrodermis. Flatworms are acoelomates with a digestive cavity. Roundworms are pseudocoelomates.
Bloom’s Level: 4. Analyze Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
9. Sponges are filter feeders and are usually A. active predators. B. sessile or relatively inactive. C. always herbivorous. D. squids. E. snails. Filter feeders are usually sessile, allowing the food carried in the water to flow through a straining device.
Bloom’s Level: 3. Apply Learning Outcome: 31.02.02 Contrast the way of life of a sponge with that of a cnidarian. Section: 31.02 Topic: Porifera
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Chapter 31 - Animals: Part I
10. A recent study of the marine hawksbill turtle's stomach contents revealed that sponges constitute a major portion of their diet. Sponge beds are generally protected from predators by the sponge's calcium and silica crystals, but these were found in quantity in the turtles' digestive systems. These structures that protect the sponges from most predators, but not turtles, are A. amoebocytes. B. the osculum. C. choanocytes. D. spicules. E. spongin. Sponges have an internal skeleton made up of small, needle-like structures that have one to six rays made of calcium or silica crystals; these structures are called spicules.
Bloom’s Level: 3. Apply Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Porifera
11. A sponge actually feeds by A. acting as a net in a current that sweeps food particles through it. B. squeezing the spongocoel cavity and sucking debris in and out through the osculum. C. the beating of the flagella of collar cells, forming a current; food is absorbed by collar cells. D. the beating collar cells form a current from osculum to pores; the food is engulfed by amoebocytes in the central cavity of the sponge. E. water the moves gently through the sponge and the amoebocytes that engulf passing food. Sponges are filter feeders; water carrying food particles passes through pores and are engulfed by the collar cells and digested by them in food vacuoles or are passed to the amoeboid cells for digestion.
Bloom’s Level: 2. Understand Learning Outcome: 31.02.02 Contrast the way of life of a sponge with that of a cnidarian. Section: 31.02 Topic: Porifera
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Chapter 31 - Animals: Part I
12. If a sponge is pressed through a coarse cloth bag and whole cells are extruded through the holes. The cells will then A. reproduce sexually. B. die from being separated. C. form spicules in the pattern of the cloth. D. all go back to primitive cells and soon differentiate into another sponge. E. reassemble into a new organized sponge with each cell resuming its original job. Sponges do not possess true tissues and, if the cells of a sponge are separated, they can reassemble and form a functioning sponge.
Bloom’s Level: 3. Apply Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Porifera
13. Why are real bath sponges so soft? A. The silica is washed away and the calcium carbonate remains. B. The spongin spicules are washed away and the silky silica remains. C. The choanocytes are softer and the amoebocytes are washed away. D. The harder silica or calcium spicules are absent and the softer spongin remains. E. The silica and spicules are washed away and the spongin treated with softening chemicals. A genuine bath sponge has all living tissue removed and only the soft protein spongin remaining.
Bloom’s Level: 3. Apply Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Porifera
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Chapter 31 - Animals: Part I
14. Which of these is NOT a characteristic of sponges? A. sessile filter feeders B. body wall has two cell layers C. Flagellated collar cells move water. D. Water enters through a single cavity, the osculum. E. Amoeboid cells transport food and make skeletal fibers and gametes. Water enters through pores in the outer layer of cells and moves through both cell layers and into the osculum where it then flows out of the sponge.
Bloom’s Level: 2. Understand Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Porifera
15. Which statement about sponges is NOT correct? A. The larval form is flagellated and able to swim to a suitable location. B. Sponges are classified on the basis of their type of skeletal material. C. Sponges resemble a colony of flagellated cells more than a multicellular animal. D. Sponges are thought to be at the base of the evolutionary tree, leading to more complex animals such as corals and worms. E. Sponges can reproduce sexually or asexually. Corals and tube worms are in the phylum Cnidaria (radial symmetry), which is thought to have originated from the same ancestor that gave rise to organisms with bilateral symmetry.
Bloom’s Level: 4. Analyze Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Porifera
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Chapter 31 - Animals: Part I
16. Which of the following types of skeletal material is NOT found in at least some sponges? A. calcium carbonate B. silica C. spicules D. spongin E. muscle Spongin is the protein that makes up the soft body of the sponge. This is reinforced by spicules of calcium carbonate or silica, depending on the type of sponge.
Bloom’s Level: 2. Understand Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Porifera
17. Which of the following statements about sponges is NOT correct? A. Sponges have no nerve fibers. B. Sponges have no fully developed muscle fibers. C. Amoeboid cells capture food particles from the water. D. Sponges may reproduce asexually by budding or by regeneration from a small piece. E. Cells of a single sponge will recognize others of the same kind and reaggregate if the cells are separated and allowed to reassociate. Amoeboid cells in sponges are sites of digestion as well as a means for the transport of nutrients from one cell to another and produce the gametes and spicules.
Bloom’s Level: 4. Analyze Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Porifera
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Chapter 31 - Animals: Part I
18. Which of the following organisms exhibit radial symmetry as adults? A. flatworms B. cnidarians C. roundworms D. clams E. fish Cndiarans are the only group to exhibit radial symmetry and two layers of tissue. The others have three layers of tissue and show bilateral symmetry.
Bloom’s Level: 1. Remember Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Cnidarians
19. The stinging cells or nematocysts are characteristic of A. sponges. B. clams. C. cnidarians. D. flatworms. E. roundworms. Nematocysts are found in cnidarians and are toxin-filled capsules with thread-like hollow structures.
Bloom’s Level: 1. Remember Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Cnidarians
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Chapter 31 - Animals: Part I
20. Which of the following are NOT cnidaria? A. corals B. planaria C. sea anemones D. hydrozoa E. Portuguese man-of-war Planarians are in the phylum Platyhelminthes, not Cnidaria.
Bloom’s Level: 1. Remember Learning Outcome: 31.02.02 Contrast the way of life of a sponge with that of a cnidarian. Section: 31.02 Topic: Cnidarians
21. Which of the following statements is NOT correct about cnidaria? A. The adult body is bilaterally symmetrical. B. Most are found in shallow coastal marine environments. C. There are two tissue layers: an outer epidermis and inner gastrodermis. D. Stinging cells contain a threadlike nematocyst released to capture prey. E. A jellylike mesoglea contains cells that form a nerve net throughout the body. Cnidaria all have radial symmetry.
Bloom’s Level: 2. Understand Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Cnidarians
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Chapter 31 - Animals: Part I
22. Which body structure describes the freshwater hydra? A. a swimming medusa with mouth pointing downward B. a swimming medusa with mouth pointing upward C. a sessile medusa with mouth pointing upward D. a sessile polyp form with mouth pointing downward E. a freely moving polyp with mouth pointing upward The hydra’s body is a small tubular polyp with two layers of tissue with a mouth found in a raised area surrounded by four to six tentacles.
Bloom’s Level: 2. Understand Learning Outcome: 31.02.03 Identify the basic morphology of a hydra. Section: 31.02 Topic: Cnidarians
23. Both corals and calcareous sponges have calcium carbonate structures. What is the relationship between these two groups? A. The calcareous sponges were ancestors of the corals. B. The corals were ancestors of the calcareous sponges. C. Both use the calcium carbonate to form a rock-like or crusty basal skeletal structure. D. The corals are actually colonial forms of the independent calcareous sponges. E. The sponges use calcium carbonate to form sharp spicules, while the corals independently use the same material to form a skeleton. While both sponges and corals use calcium carbonate structures, neither is derived from the other but rather share a common ancestor that was a multicelled protozoan.
Bloom’s Level: 3. Apply Learning Outcome: 31.02.02 Contrast the way of life of a sponge with that of a cnidarian. Section: 31.02 Topic: Cnidarians
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Chapter 31 - Animals: Part I
24. Which statement about cnidaria is NOT true? A. Reproduction is both sexual and asexual. B. Some forms are sessile and others are motile. C. They live in either marine or freshwater environments. D. To capture food, the animal wraps itself around its prey. E. The body plan is tube-within-a-tube, with both mouth and anus. Cnidarians have true tissues and develop two types of tissues from embryonic germ layers of endoderm and ectoderm. The outer layer of tissue in an adult is a protective covering derived from the ectoderm and the inner layer makes up the gastrovascular cavity where digestion takes place.
Bloom’s Level: 2. Understand Learning Outcome: 31.02.02 Contrast the way of life of a sponge with that of a cnidarian. Section: 31.02 Topic: Cnidarians
25. A floating Portuguese man-of-war is A. a cubozoan. B. merely an inflated, noncircular jellyfish. C. a free-swimming polyp in the hydra group. D. the first animal with complete organ systems. E. a colony of polyps composed of different types of individuals. A Portuguese man-of-war is made up of modified polyps and medusa that have formed into a colony. The original polyp becomes gas filled to provide buoyancy and other polyps bud from the original and specialize in feeding or reproduction.
Bloom’s Level: 2. Understand Learning Outcome: 31.02.03 Identify the basic morphology of a hydra. Section: 31.02 Topic: Cnidarians
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Chapter 31 - Animals: Part I
26. If you play with a living anemone with a probe, it will withdraw even if you vibrate the probe like prey would vibrate; it also withdraws if you just add sugar. However, it reaches out and grasps an object that both vibrates and tastes like food. The use of such a double system prevents unnecessary firing of expensive nematocysts. Why are nematocysts "expensive"? A. The organism dies without nematocysts. B. The research equipment to detect them is very expensive. C. Nematocysts are used in medicine and are very expensive to harvest. D. The nematocyst is very complex and most likely costs the anemone much energy to build. E. The nematocyst can only be used for food and doesn't "fire" for non-food objects. Nematocysts are not reused after they have been discharged and it requires that the cnidarians use energy to produce the cells. This means that it is to the advantage of the organism to use them only when it is certain that food will be obtained in return for their use.
Bloom’s Level: 3. Apply Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Cnidarians
27. A hydra moves via A. nematocysts. B. alternation of generations. C. contraction of the nerve net. D. muscle and nerve fibers. E. muscular fibers in the mesoglea. Mesoglea is a jellylike substance found between the epidermal and gastrovascular layers of tissue in cnidarians. It contains a network of fibers in the gel and serves as a means of support and movement.
Bloom’s Level: 1. Remember Learning Outcome: 31.02.03 Identify the basic morphology of a hydra. Section: 31.02 Topic: Cnidarians
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Chapter 31 - Animals: Part I
28. What system do the cnidaria have that sponges lack? A. respiratory B. nervous C. circulatory D. excretory E. reproductive Nerves occur in their simplest form in cnidarians. Sponges do not have true tissues and so could not have any specialized tissues.
Bloom’s Level: 3. Apply Learning Outcome: 31.02.01 Describe the anatomy and life cycle of sponges and cnidarians. Section: 31.02 Topic: Cnidarians
29. In flame cells A. light is detected. B. ciliated sperm are stored. C. undigested food is expelled. D. slow fires actually burn food for energy. E. cilia drive fluids through tubules for excretion of excess water. Flame cells are ciliated cells found in flatworms and function to pull fluid through pores to the outside, helping to maintain osmotic balance with the outside environment.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.01 Describe the basic features of the lophotrochozoa. Section: 31.03 Topic: Trochozoans
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Chapter 31 - Animals: Part I
30. Planaria feed by A. attaching with a sucker(s) and extracting blood. B. tearing and sucking food particles via a muscular pharynx. C. slurping food particles through a mouth and excreting waste via an anus. D. soaking all nutrients from the environment through its thin epidermal surface. E. grabbing food with tentacles and pulling them into a ventrally located mouth. Planaria extend a muscular pharynx to a prey item and tear and suck small particles off of the item; the resulting small pieces of food are sucked into the pharynx and then into the gastrovascular cavity where digestion is completed.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.01 Describe the basic features of the lophotrochozoa. Section: 31.03 Topic: Trochozoans
31. Which of the following statements is NOT correct about the flatworms? A. There are three complete tissue layers. B. Flatworms may be either free living or parasitic. C. Flatworms have a sac body plan, with only a mouth and no anus. D. Parasitic flatworms have a well-developed head with eyespots and nerves concentrated into a brain. E. Most planaria are found in marine environments, but some dwell in freshwater and moist terrestrial environments. Parasitic flatworms often exhibit some reduced systems due to being parasitic; the systems needed to seek and capture prey are not needed and so a well-developed head with eyespots becomes a waste of the organism’s energy. Instead, they have specialized structures like the hooks and suckers found on tapeworms.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
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Chapter 31 - Animals: Part I
32. Which of the following statements about planaria is NOT true? A. There is a rudimentary circulatory system with a small central heart. B. Planaria cut in half can regenerate to form two complete worms. C. Active movement occurs through ciliary cells and muscle movement. D. Tissue levels include endoderm (inner), mesoderm (middle), and ectoderm (outer). E. Cephalization is shown by the formation of a brain and sense organs in a head region. Planaria do not have a true circulatory system. They have an excretory system made up of a network of interconnected canals and water is driven through this system by the beating of cilia in flame cells. The fine branches of the gastrovascular cavity distribute food directly to the cells of the organism.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
33. Which of these statements is NOT correct about the flatworms? A. Tapeworms do not have a larval stage. B. Planaria ingest food through a mouth located in the head region. C. Flukes have two suckers by which they attach to and feed from host tissues. D. Tapeworms have a scolex with hooks and suckers by which they attach to the host's intestinal tissues. E. Tapeworms enter the human body through undercooked meat, especially pork, which contains encysted larval forms. Planaria have a pharynx that extends through the mouth that is located in the body region of the organism.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
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Chapter 31 - Animals: Part I
34. Which of the following statements is NOT true about the flatworms? A. Tapeworms have a ladder-type nervous system similar to other flatworms. B. Planaria contain pigmented, photosensitive eyespots. C. Liver flukes and blood flukes are parasites in humans only. D. Schistosomiasis is a common human blood disease caused by flukes in tropical areas. E. Tapeworms are hermaphroditic, having both male and female reproductive structures in each proglottid. Liver flukes and blood flukes can use multiple kinds of vertebrates, such as cats, dogs, and pigs, in addition to humans.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
35. Planaria move by A. flame cells. B. small pseudopodia. C. lateral undulation of muscles. D. expulsion of fluids as with squid. E. secreting a film of mucus and pushing against it with epidermal cilia. Although planarians have muscles that allow them to move in a variety of ways, planarians propel themselves through the environment by the beating cilia on their epidermis, allowing them to glide through a film of mucus that they secrete.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
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Chapter 31 - Animals: Part I
36. The of the tapeworm contains both male and female sex organs and becomes filled with developing embryos. A. proglottids B. scolex C. cyst D. gastrovascular cavity E. nematocyst A tapeworm contains long series of units called proglottids which contain both male and female sex organs. They fertilize their own eggs. These fertilized units break off and are excreted in the host’s feces.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
37. Tapeworms are specialized for parasitism by an excessive development of which system? A. muscular B. reproductive C. excretory D. circulatory E. digestive Tapeworms are reproductive machines; most of their body is given over to the production of thousands of offspring in each unit.
Bloom’s Level: 3. Apply Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
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Chapter 31 - Animals: Part I
38. Which of the following is characteristic of flukes? A. Flukes are ectoparasites of invertebrates. B. Most flukes have separate sexes. C. Flukes are covered by a ciliated integument. D. Flukes possess an oral sucker that is surrounded by a nonsensory papilla. E. Flukes have a well-developed alimentary canal. Flukes are endoparasites that have a tough nonciliated integument. The anterior end has an oral sucker that is surrounded by sensory papilla. Most flukes are hermaphroditic, containing both male and female reproductive structures. While the digestive system is less developed than free-living flatworms, the alimentary canal is well developed.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
39. Ascaris is considered to be a(n) A. flatworm. B. earthworm. C. roundworm. D. cnidarian. E. arthropod. Roundworms, phylum Nematoda, can be either free-living or parasitic. Ascaris is parasitic and infects humans and pigs.
Bloom’s Level: 1. Remember Learning Outcome: 31.04.01 Identify the characteristics unique to ecdysozoans. Section: 31.04 Topic: Ecdysozoans
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Chapter 31 - Animals: Part I
40. Which of these is NOT associated with a roundworm infection? A. hookworm B. trichinosis C. elephantiasis D. tapeworm E. pinworm Tapeworms are not nematodes (roundworms), but rather a type of flatworm, or Platyhelminthes.
Bloom’s Level: 2. Understand Learning Outcome: 31.04.01 Identify the characteristics unique to ecdysozoans. Section: 31.04 Topic: Nematodes
41. How could you possibly become infected with adult tapeworms? A. eating insufficiently cooked pork B. being bitten by a mosquito C. ingesting eggs in contaminated water D. eating freshly peeled fruit E. wading in contaminated water Larvae in the secondary host, pig, burrow through the intestinal wall and travel in the bloodstream to finally encyst and lodge in the muscle. These cysts can be passed to a new host if the meat is not heated sufficiently to kill the larvae.
Bloom’s Level: 3. Apply Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Platyhelminthes
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Chapter 31 - Animals: Part I
42. An adaptation exhibited by the tapeworm for survival in a host's intestine is A. proglottids coated with calcium carbonate. B. alkaline secretions from the scolex. C. a protective lining of chitin. D. eggs before the proglottids are digested. E. a tough integument resistant to host digestive fluids. Since the adult tapeworm lives in the digestive tract of the host it is important that the digestive juices of the host not kill the parasite.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Platyhelminthes
43. The head region of a tapeworm is called the A. proglottid. B. cercaria. C. sporocyst. D. scolex. E. glycocalyx. The head region of a tapeworm is called the scolex and contains hooks to attach to the intestinal wall and suckers for feeding.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Platyhelminthes
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Chapter 31 - Animals: Part I
44. The segments of a tapeworm consisting primarily of eggs, which hang in bags as a long chain behind the head, are called A. proglottids. B. cercariae. C. sporocysts. D. scolexes. E. glycocalyxes. The proglottids are segments that contain the male and female reproductive structures of the tapeworm. They are located in a long chain behind the scolex.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Platyhelminthes
45. Which statement is NOT correct about the roundworms? A. There are three layers of tissues in the body. B. Roundworms have a tube-within-a-tube body plan. C. There is an internal body cavity called a pseudocoelom. D. Roundworms have a smooth, unsegmented outside body wall. E. Roundworms are essentially no more than parasitic flatworms that have lost their need to be flat. Roundworms, or nematodes, are ecdysozoa which molt their outer covering periodically. They are classified with arthropods who also molt.
Bloom’s Level: 3. Apply Learning Outcome: 31.04.01 Identify the characteristics unique to ecdysozoans. Section: 31.04 Topic: Nematodes
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Chapter 31 - Animals: Part I
46. Which of these parasitic worms is NOT a nematode? A. Ascaris B. Trichinella C. the worm that causes elephantiasis D. hookworm E. the worm that causes schistosomiasis Schistosomiasis mansoni is a type of blood fluke which is classed in the phylum Platyhelminthes not in Nematoda.
Bloom’s Level: 2. Understand Learning Outcome: 31.04.01 Identify the characteristics unique to ecdysozoans. Section: 31.04 Topic: Nematodes
47. Larval Trichinella worms live in the A. liver. B. lungs. C. lymph glands. D. skeletal muscle of pork. E. water. Trichinosis is caused by a roundworm and infects humans who eat pork containing the larval form encysted in the muscles of pigs that is not cooked sufficiently.
Bloom’s Level: 1. Remember Learning Outcome: 31.04.01 Identify the characteristics unique to ecdysozoans. Section: 31.04 Topic: Nematodes
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Chapter 31 - Animals: Part I
48. The molluscs look so different, and yet we can tell they are related because they all A. are segmented. B. are predators. C. have an external skeleton. D. have a mantle, visceral mass, and a foot. E. have shells. All molluscs have three characteristics: 1) a visceral mass that is the soft-bodied portion that contains the organs; 2) a strong, muscular foot used for locomotion; and 3) the mantle, a covering that partially covers the visceral mass and secretes a shell if one is present.
Bloom’s Level: 3. Apply Learning Outcome: 31.03.02 Identify the unique features of the molluscs. Section: 31.03 Topic: Mollusks
49. Clams, snails, sea slugs, and octopuses belong to the phylum A. Annelida. B. Mollusca. C. Porifera. D. Platyhelminthes. E. Arthropoda. Annelida include earthworms and are segmented, with a hydrostatic skeleton and a closed circulatory system. Porifera includes sponges and have no true tissues. Platyhelminthes are flatworms like tapeworms and flukes. Arthropoda includes crustaceans, insects, and spiders and are characterized by an exoskeleton, an open circulatory system, and well-developed sensory organs.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.02 Identify the unique features of the molluscs. Section: 31.03 Topic: Mollusks
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Chapter 31 - Animals: Part I
50. Which are NOT among the phylum Mollusca? A. clams B. squids C. lobsters D. nautiluses E. snails Lobsters are members of phylum Arthropoda and have the characteristic hard exoskeleton found in that phylum.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.02 Identify the unique features of the molluscs. Section: 31.03 Topic: Mollusks
51. The complex eye that can distinguish shapes is found A. only in advanced vertebrates. B. in molluscs and indicates that vertebrate eyes originated with ancestors common with this group. C. in molluscs but the eye structure evolved differently. D. in arthropods and indicates that vertebrate eyes originated with ancestors common with this group. E. in echinoderms and indicates that vertebrate eyes originated with ancestors common with this group. While there is a similarity in the appearance of the eyes in cephalopods and vertebrates, the eye construction is actually different and the two eyes evolved independently in the two groups of animals.
Bloom’s Level: 3. Apply Learning Outcome: 31.03.02 Identify the unique features of the molluscs. Section: 31.03 Topic: Mollusks
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Chapter 31 - Animals: Part I
52. The cephalopods resemble other molluscs because they have A. focusing camera-type eyes. B. a siphon for jet propulsion. C. a parrot-like beak for tearing prey. D. well-developed brains with high learning capacity. E. most of the internal organs located in a visceral mass. Cephalopods have a mantle covering the visceral mass, but the foot has evolved into a funnel or siphon about the head.
Bloom’s Level: 3. Apply Learning Outcome: 31.03.02 Identify the unique features of the molluscs. Section: 31.03 Topic: Mollusks
53. Which of these is a bivalve? A. slug B. octopus C. earthworm D. snail E. oyster Bivalves have shells that come in two parts that is connected by a hinge; the one organism listed that has two shells is the oyster.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.02 Identify the unique features of the molluscs. Section: 31.03 Topic: Mollusks
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Chapter 31 - Animals: Part I
54. If you live on a river delta or coastline, you may not be able to eat the local shellfish because of high levels of toxins present in the water. Why? A. Clams seasonally produce toxic substances. B. Shellfish regularly consume poisonous red tide organisms. C. Most harmless native clams have been replaced by bad species. D. Clams filter-feed and concentrate the dilute pollutants from the water, thus acting as early warning indicators of severe pollution. E. The toxins contained in the shellfish can't be broken down by your stomach. Shellfish are filter feeders that collect food that is suspended in the water. Food particles become trapped in the mucus on the gills and are then swept toward the mouth by the action of cilia. Toxins found in the detritus upon which they feed would be concentrated in the tissues of the shellfish.
Bloom’s Level: 4. Analyze Learning Outcome: 31.03.02 Identify the unique features of the molluscs. Section: 31.03 Topic: Mollusks
55. A closed circulatory system is found in A. insects. B. snails. C. clams. D. earthworms. E. scorpions. Arthropods, insects, and scorpions, have an open circulatory system as do most mollusks, snails, and clams. Earthworms are annelids and have a closed circulatory system.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Arthropods
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Chapter 31 - Animals: Part I
56. Which of the following is NOT characteristic of earthworms? A. segmentation B. expanded dorsal surface of the intestine called a typhlosole C. hermaphroditic D. paired nephridia in each segment E. dorsal solid nerve cord Annelids are segmented and have a well-developed coelom, a specialized digestive system that includes a typhlosole, a closed circulatory system, a nervous system with a ventral nerve cord, and an excretory system that contains nephridia.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
57. Earthworms differ from most of the marine annelids in that earthworms A. lack setae that marine annelids have. B. lack paddle-like parapodia that marine annelids have. C. have segments, while marine annelids do not. D. develop from a larval form, while marine annelids do not. E. have well-developed brains and sensory organs in the head region, while marine annelids do not. Marine annelids, polychaetes, are predatory and have a well-developed head and each segment usually has parapoda and many setae. Earthworms, oligochaetes, are not predatory and lack the well-developed head and parapoda of the polychaetes.
Bloom’s Level: 3. Apply Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
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Chapter 31 - Animals: Part I
58. The chitinous bristles that earthworms use to anchor themselves in the earth and pull themselves along are A. aortic arches. B. lateral ventricles. C. setae. D. parapodia. E. nephridia. Each segment of oligochaetes has four pairs of setae or bristles that help them to move through the soil.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
59. Fertilization in the earthworm is accomplished by the use of the A. nephridia. B. clitellum. C. flame cells. D. trachea. E. typhlosole. The clitellum of an oligochaete secretes mucus that prevents sperm from drying out as they pass between two mating worms. After the worms separate, the clitellum produces a slime tube which slides along the worm and picks up the sperm and eggs and deposits them in the soil.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
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Chapter 31 - Animals: Part I
60. Earthworms are A. annelids. B. molluscs. C. echinoderms. D. platyhelminthes. E. arthropods. Annelida include earthworms and are segmented with a hydrostatic skeleton and a closed circulatory system. Molluscs include clams, snails, and squids and all have a foot, mantle, and visceral mass. Echinoderms include star fish and are characterized by a water vascular system and a thin epidermis covering an endoskeleton of calcareous plates. Platyhelminthes are flatworms like tapeworms and flukes. Arthropoda includes crustaceans, insects, and spiders and are characterized by an exoskeleton, an open circulatory system, and well-developed sensory organs.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
61. Earthworms possess both male and female organs on separate segments; thus, mating consists of "parallel parking" in opposite directions and cross-fertilizing. Given this, which of the following is(are) true? A. This is evidence of a primitive state that is generally less efficient than separate sexes. B. This is primarily to allow them to self-fertilize if no other worm is in their sector of the soil. C. This is an advantage since they can mate with any other earthworm of the same species that they encounter. D. Maintaining any system not clearly male or female is disadvantageous; the earthworm is trapped in an evolutionary dead end. E. This demonstrates the how ineffective self-fertilization is as a means of reproduction. Unlike organisms that have individuals that are one sex, male or female, any other earthworm is a potential mate. It is not possible to meet another earthworm of the same species with whom they cannot mate.
Bloom’s Level: 4. Analyze Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
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Chapter 31 - Animals: Part I
62. When an earthworm secretes a slime layer to receive eggs and sperm and form a ''cocoon,'' this secretion is produced by the A. annulations. B. peritoneum. C. clitellum. D. nephridium. E. prostomium. The clitellum of an oligochaete secretes mucus that prevents sperm from drying out as they pass between two mating worms. After the worms separate, the clitellum produces a slime tube which slides along the worm and picks up the sperm and eggs and deposits them in the soil.
Bloom’s Level: 1. Remember Learning Outcome: 31.03.03 Contrast the differences between the flatworms and annelids. Section: 31.03 Topic: Trochozoans
63. A major characteristic of the arthropods is the presence of A. flame cells. B. radial symmetry. C. a soft exoskeleton. D. a closed circulatory system. E. jointed appendages. Arthropods have a rigid exoskeleton that completely covers the organism. They also have segments that each have a pair of jointed appendages attached.
Bloom’s Level: 1. Remember Learning Outcome: 31.04.02 Describe the characteristics contributing to the success and diversity of arthropods. Section: 31.04 Topic: Arthropods
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Chapter 31 - Animals: Part I
64. The largest animal group, both in number of species and number of individuals, is the A. annelids. B. crustacea. C. insects. D. reptiles. E. fishes. Insects likely have over one million species and are extremely diverse. They are more species-rich and diverse than all other types of life.
Bloom’s Level: 1. Remember Learning Outcome: 31.04.02 Describe the characteristics contributing to the success and diversity of arthropods. Section: 31.04 Topic: Arthropods
65. Which type of organism is characterized by having jointed appendages, ability to molt, and three sets of fused segments? A. molluscs B. echinoderms C. arthropods D. annelids E. chordates Arthropods are covered in a rigid exoskeleton and must molt to increase in size. They have fused segments with a pair of jointed legs for every segment.
Bloom’s Level: 2. Understand Learning Outcome: 31.04.02 Describe the characteristics contributing to the success and diversity of arthropods. Section: 31.04 Topic: Arthropods
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Chapter 31 - Animals: Part I
66. It is NOT correct to say that arthropods have A. a solid ventral nerve cord. B. a variety of respiratory organs. C. a well-developed nervous system. D. jointed appendages and a segmented body. E. a tough exoskeleton that grows by expansion. Once the exoskeleton has been formed it is hard and will not expand; in order to grow it is necessary for arthropods to shed the existing exoskeleton and produce a new, larger one.
Bloom’s Level: 2. Understand Learning Outcome: 31.04.02 Describe the characteristics contributing to the success and diversity of arthropods. Section: 31.04 Topic: Arthropods
67. An arachnid differs from a crustacean because A. arachnids have book gills. B. arachnids are mostly aquatic and crustaceans are mostly terrestrial. C. only arachnids shed their exoskeletons. D. arachnids have two major body segments, the cephalothorax and abdomen, while crustaceans have three body segments. E. arachnids have four pairs of walking legs. Crustaceans are named for their hard shells; their exoskeleton is calcified to a greater extent than other arthropods. In crustaceans, such as the crayfish, the thorax has five sets of walking leg and the first set is a pinching claw. Arachnids have six sets of appendages, four sets of walking legs, and the chelicerae and pedipalps.
Bloom’s Level: 2. Understand Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
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Chapter 31 - Animals: Part I
68. are animals that have three pairs of legs, a tympanum for reception of sound waves, and Malpighian tubules for excretion. A. Starfish B. Lancelets C. Crayfish D. Grasshoppers E. Earthworms The excretory system of grasshoppers consists of Malpighian tubules that extend into the hemocoel and collect nitrogenous wastes that are then excreted into the digestive tract. In addition grasshoppers have a tympanum for the reception of sound waves; grasshoppers use courtship sounds by rubbing body parts (such as legs) together. The tympanum allows the reception of these signals.
Bloom’s Level: 2. Understand Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
69. The major insect body parts are A. head, cephalothorax, and abdomen. B. cephalothorax, midsection, and abdomen. C. head, thorax, and abdomen. D. head, pyothorax, and metathorax. E. head, cephalothorax, and genitalia. Insects have bodies that are divided into three segments: the head, thorax, and abdomen. The head usually bears sensory antennae, eyes, and mouth parts. The thorax bears three pairs of legs and wings, if present. The abdomen contains the internal organs.
Bloom’s Level: 1. Remember Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
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Chapter 31 - Animals: Part I
70. An insect circulatory system is best described as A. totally closed like ours, with arteries and veins. B. similar to ours, with general closed vessels but no distinct arteries and veins. C. an open system that mostly washes hemolymph "blood" through the hemocoel. D. totally absent; each body cell manages on its own to gain food and gets rid of wastes. E. containing five pairs of "hearts." An insect’s circulatory system is an open system. The heart lies against the dorsal wall of the abdomen and pumps hemolymph, the fluid used for circulation, into a short artery where is then sent into the hemocoel, or sinuses surrounding the tissues and organs. It then collects and returns to the heart where the process begins again.
Bloom’s Level: 2. Understand Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
71. The insect organ most equivalent to the human lung in function is the A. tracheae. B. midgut. C. crop. D. gizzard. E. Malpighian tubule. The respiratory system of insects consists of openings in the exoskeleton called spiracles. Air is actively pumped into the spiracles by the alternate contraction and expansion of the body wall. Once inside, the air moves into the trachea, small multibranched tubes that are the actual sites of gas exchange. The trachea is most closely equivalent to the human lung since these are also sites of gas exchange.
Bloom’s Level: 3. Apply Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
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Chapter 31 - Animals: Part I
72. Which of the following organisms exhibit complete metamorphosis? A. crayfish B. starfish C. grasshoppers D. butterfly E. lobsters Many insects undergo a change in form and physiology as they develop into an adult. In incomplete metamorphosis, the young nymphs resemble the adult but are smaller and do not have wings. In complete metamorphosis, the development involves drastic changes and one stage does not resemble the next. The stages are typically larva, pupa, and adult. The butterfly goes through all of these stages.
Bloom’s Level: 4. Analyze Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
73. Which statement about arachnids is NOT true? A. Ticks and mites are often parasitic. B. Spiders and scorpions are predators. C. Spiders have trachea for respiration. D. Spiders have a pair of fangs which release poison. E. The body parts include a cephalothorax and an abdomen. Arachnids have book lungs that function to carry out gas exchange. The folded lamella that make up the "pages" are invaginations of the body wall.
Bloom’s Level: 4. Analyze Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
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Chapter 31 - Animals: Part I
74. Which pair exhibits the closest relationship? A. spider—tick B. housefly—mite C. earthworm—grasshopper D. crayfish—planarian E. scorpion—clam Spiders and ticks are both arachnids and share many of the same traits. Houseflies are insects, while mites are arachnids. Earthworms and grasshoppers are in different phyla; Annelida and Arthropoda, respectively. Crayfish and Planaria are in different phyla; Arthropoda and Platyhelminthes, respectively. Scorpions and clams are in different phyla; Arthropoda and Mollusca, respectively.
Bloom’s Level: 5. Evaluate Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
75. Respiratory structures in insects are A. commonly termed book lungs. B. spiracles and tracheae. C. hemolymphic. D. lamellae. E. lungs. The respiratory system of insects consists of openings in the exoskeleton called spiracles. Air is actively pumped into the spiracles by the alternate contraction and expansion of the body wall. Once inside, the air moves into the trachea, small multibranched tubes that are the actual sites of gas exchange.
Bloom’s Level: 1. Remember Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
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Chapter 31 - Animals: Part I
76. Which of the following would not show the five characteristics typically found in animals? A. sponges, which have no true tissues but have two layers of cells that perform different functions B. planaria, who have an incomplete digestive tract but do have muscles and excretory and reproductive systems C. trypanosome, a single-celled organism that causes sleeping sickness in humans D. leeches, segmented worms that are predatory on other animals E. tapeworms, flatworms that infest the intestines of humans One of the characteristics that define animals is that they are multicellular. As a single-celled organism, a trypanosome is not an animal.
Bloom’s Level: 5. Evaluate Learning Outcome: 31.01.01 List the general characteristics of animals. Section: 31.01 Topic: Animals
77. The life cycle of animals A. demonstrates alternation of generations, where the zygote is haploid and the adult is diploid. B. does not demonstrate alternation of generations. Both the zygote and the adult are haploid. C. demonstrates alternation of generations, where the zygote is diploid and the adult is haploid. D. does not demonstrate alternation of generations because only the spores are haploid. E. does not demonstrate alternation of generations; gametes are haploid and the organism is diploid. Alternation of generations is seen in plants and describes a life cycle where both the haploid and diploid phases are multicellular. In animals, only the diploid organism is multicellular.
Bloom’s Level: 4. Analyze Learning Outcome: 31.01.01 List the general characteristics of animals. Section: 31.01 Topic: Animals
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Chapter 31 - Animals: Part I
78. Which of the following organisms show the typical method of acquiring food that is found in animals? A. euglena, who are autotrophic and heterotrophic B. basidiomycetes, club fungi, who decompose wood and other plant material C. earthworms, annelids, who feed on any available organic matter that they can take into their mouths D. volvox, a colonial alga that is autotrophic E. archaea, who are chemotrophic Animals are all heterotrophs that ingest food. Euglenas are autotrophic in addition to heterotrophic, so they do not always feed as in the typical animal fashion. Fungi are decomposers; they do not ingest food but rather digest it outside the body and only take in the needed nutrients. Volvox is autotrophic not heterotrophic. Chemotropic organisms would use chemical energy in place of solar energy but would still be autotrophic. Earthworms ingest their food and so feed in the typical animal manner.
Bloom’s Level: 4. Analyze Learning Outcome: 31.01.01 List the general characteristics of animals. Section: 31.01 Topic: Animals
79. Which of the following statements about hydra morphology is correct? A. The hydra has a long single tentacle containing numerous nematocysts. B. The hydra has a single polyp that is filled with gas and keeps the organism afloat. C. The hydra is bell shaped and has multiple tentacles. D. The hydra is a tube-shaped polyp that has two layers of tissue. E. The hydra is bilaterally symmetrical. Hydras are tubular with two layers of cells that carry out specific functions. Between these two tissue layers is a gel-like substance that contains muscle fibers that allow the organism to move directionally. Around the mouth are four to six tentacles used to capture prey items.
Bloom’s Level: 2. Understand Learning Outcome: 31.02.03 Identify the basic morphology of a hydra. Section: 31.02 Topic: Cnidarians
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Chapter 31 - Animals: Part I
80. The two layers of tissue in a hydra are A. the epidermis, which covers and protects, and the inner layer that contains flame cells. B. the epidermis that covers and protects, and the gastrodermis, where cells that complete digestion are found. C. the epidermis that contains flame cells and the gastrodermis that contains the nephridia. D. the epidermis that covers and protects and the gastrodermis that contains the nephridia. E. the epidermis that covers and protects and the gastrodermis that contains a ladder-like nervous system. Hydras are tubular with two layers of cells that carry out specific functions. The epidermis develops from the ectoderm and is the outer protective covering that contains nematocysts. The inner gastrodermis lines the gastrovascular cavity, which is the site of primary digestion. Between these two tissue layers is a gel-like substance that contains muscle fibers that allow the organism to move directionally.
Bloom’s Level: 2. Understand Learning Outcome: 31.02.03 Identify the basic morphology of a hydra. Section: 31.02 Topic: Cnidarians
81. Because they have nerve fibers and muscle fibers, hydra are capable of directional movement. These nerve and muscle fibers are found A. in the gastrodermis. B. in the epidermis. C. in the mesoglea. D. in the gastrodermis; muscle fibers are in the ectodermis. E. just under the epidermis; muscle fibers are found in the gastrodermis. The outer tissue layer is a protective epidermis derived from ectoderm. The inner tissue layer, derived from endoderm, is called a gastrodermis. The two tissue layers are separated by mesoglea. There are both circular and longitudinal muscle fibers that are located in the gastrodermis. Nerve cells located below the epidermis, near the mesoglea, interconnect and form a nerve net that communicates with sensory cells throughout the body.
Bloom’s Level: 2. Understand Learning Outcome: 31.02.03 Identify the basic morphology of a hydra. Section: 31.02 Topic: Cnidarians
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Chapter 31 - Animals: Part I
82. Which of the following is a way that crayfish and grasshoppers are similar? A. Both have tympanum for use in mating rituals. B. Both have malpighian tubules that function in the extraction of nitrogenous waste. C. Both have spicules and a tracheal system for respiration. D. Both have bodies that are composed of a cephalothorax and abdomen. E. Both have antennae that have a sensory function. Only grasshoppers have tympanum that they use in their courting rituals and malpighian tubules that are used to remove nitrogenous waste. In crayfish the green gland serves to excrete metabolic waste and they have their bodies fused into two body sections, the cephalothorax and abdomen, while the grasshopper is divided into three sections: the head, thorax, and abdomen. Crayfish use gills to exchange gasses, while the grasshopper uses spicules and tracheal tubes.
Bloom’s Level: 4. Analyze Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
83. Which of the following is a way that crayfish and grasshoppers are different? A. Crayfish have an exoskeleton, while grasshoppers do not. B. Grasshoppers have compound eyes, while crayfish have a camera type of eye. C. Grasshoppers have well-developed nervous systems, while crayfish have a ladder network of nerves. D. Grasshoppers have spicules and tracheal tubes that serve as the means of gas exchange, while crayfish have gills. E. Crayfish have an open circulatory system, while grasshoppers have a closed circulatory system. All arthropods have an external exoskeleton, a body with fused segments, although the numbers of segments varies. All arthropods also have jointed appendages, some of which are specialized in obtaining food and in bearing sensory systems, including compound eyes. In addition, they have well-developed nervous systems and digestive systems as well as reproductive systems that use sexual reproduction. The circulatory system for all arthropods is an open system where a fluid bathes the tissues and internal structures.
Bloom’s Level: 4. Analyze Learning Outcome: 31.04.03: List the major features that distinguish the crustaceans, insects, and arachnids. Section: 31.04 Topic: Arthropods
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Chapter 31 - Animals: Part I
84. Echinoderms and cnidarians A. are both radially symmetrical. B. both have three germ layers. C. both have the sac body plan. D. are both deuterostomes. E. are both found in fresh water. While echinoderms are closely related to chordates, they often have radial symmetry in their adult form.
Bloom’s Level: 2. Understand Learning Outcome: 31.05.01: Recognize the basic morphological characteristics of echinoderms. Section: 31.05 Topic: Cnidarians
85. "Derm" means "skin," so the name Echinodermata literally means A. soft skin. B. circulation of water. C. spiny skin. D. skinless. E. ability to store water. Echinoderm comes from the Greek "echin," meaning "spiny," and "derm," meaning "skin"; so the name literally means spiny skin.
Bloom’s Level: 1. Remember Learning Outcome: 31.05.01: Recognize the basic morphological characteristics of echinoderms. Section: 31.05 Topic: Echinoderms
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Chapter 31 - Animals: Part I
86. If echinoderms are radially symmetrical, why are they not in the same phylum as the cnidarians? A. Echinoderms branched from cnidarians, but are considered more advanced because they also gave rise to chordates. B. The echinoderm larvae are originally bilaterally symmetrical but metamorphose into a radially symmetrical adult. C. Radial symmetry is maintained as a primitive state by all animals in the embryonic stage and can appear anywhere. D. Echinoderms are the most primitive of the deuterostomes and evolution begins in each group with radial symmetry. E. Echinoderms have a five-sided bilateral symmetry that just appears to be radial. Echinoderms and chordates share some developmental similarities; both are deuterostomes. The larval forms are free-swimming filter feeders and have bilateral symmetry. After metamorphosis, the larval forms develop into radially symmetrical adults.
Bloom’s Level: 5. Evaluate Learning Outcome: 31.05.01: Recognize the basic morphological characteristics of echinoderms. Section: 31.05 Topic: Echinoderms
87. The term deuterostome refers to A. having a spiny skin. B. having three germ layers. C. possessing a notochord. D. the second embryonic opening becoming the mouth. E. the coelom forming by outpocketing of the primitive gut. The term deuterostome is from the Greek, "deuteros," meaning "second" and "stoma," meaning "mouth"; so the term refers to the formation of the mouth from the second opening. A deuterostome is a developmental stage where the second embryonic opening becomes the mouth and an outpocketing by the primitive gut becomes the coelom.
Bloom’s Level: 1. Remember Learning Outcome: 31.01.03 Describe the differences between protostomes and deuterostomes. Section: 31.01 Topic: Deuterostomes
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Chapter 31 - Animals: Part I
88. Which statement is NOT true about echinoderms? A. The exoskeleton is made up of bony tissue. B. Gas exchange occurs through gills on the skin. C. Larvae are free-swimming and bilaterally symmetrical. D. A water vascular system allows the tube feet to produce suction. E. Symmetry is radial in the adult, usually with parts in fives or multiples of five. The endoskeleton of echinoderms is made of spine-bearing calcium-rich plates, not bony tissue.
Bloom’s Level: 2. Understand Learning Outcome: 31.05.01: Recognize the basic morphological characteristics of echinoderms. Section: 31.05 Topic: Echinoderms
89. Which of the following is NOT an echinoderm? A. sea star B. sea urchin C. sea cucumber D. sand dollar E. crayfish Crayfish are arthropods, as shown by their jointed appendages and exoskeleton. Echinoderms have no jointed appendages and have an endoskeleton.
Bloom’s Level: 1. Remember Learning Outcome: 31.05.01: Recognize the basic morphological characteristics of echinoderms. Section: 31.05 Topic: Echinoderms
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Chapter 31 - Animals: Part I
90. Which of the following is NOT a characteristic of echinoderms? A. adult is radially symmetrical B. locomotion by muscles C. endoskeleton of spiny, calcium-rich plates D. larva is bilaterally symmetrical E. both sexual and asexual reproduction Echinoderms do not use muscles to move; most are slow moving or sessile. The echinoderms, such as sea stars and sea urchins, which do move, do so by means of tube feet. Tube feet work by means of forcing water into the ampulla and a podium (foot portion). The ampulla squeezes water into the podium, causing it to expand. When the foot comes into contact with a surface, the center withdraws, giving the foot suction and causing it to adhere to the surface. The organism can move along slowly by expanding and contracting the tube feet.
Bloom’s Level: 2. Understand Learning Outcome: 31.05.01: Recognize the basic morphological characteristics of echinoderms. Section: 31.05 Topic: Echinoderms
91. Sea stars and sea urchins operate their tube feet by A. a hydraulic system that regulates water pressure. B. cilia transporting hemolymph from chamber to chamber. C. sticky threads leading from the mouth. D. actin and myosin fibers that slide and give muscle actions to the five arms or "rays." E. an extensive net of nerve and muscle cells. Tube feet work by means of forcing water into the ampulla and a podium (foot portion). The ampulla squeezes water into the podium, causing it to expand. When the foot comes into contact with a surface, the center withdraws giving the foot suction and causes it to adhere to the surface. The organism can move along slowly by expanding and contracting the tube feet. The water enters the system by madreporite, or sieve plate, and passes into the ring canal and then into the radial canal from which the tube feet branch out.
Bloom’s Level: 2. Understand Learning Outcome: 31.05.02: Describe how sea stars move, feed, and reproduce. Section: 31.05 Topic: Echinoderms
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Chapter 31 - Animals: Part I
92. If the water vascular system of the echinoderm was eliminated, which echinoderm function would not exist anymore? A. locomotion B. respiration C. circulation D. digestion E. sensing the environment The water vascular system is used to operate the tube feet of the organism, allowing it to move slowly along. Echinoderms don’t have a respiratory, excretory, or circulatory system. Fluids within the coelomic cavity and the water vascular system carry out many of these functions.
Bloom’s Level: 2. Understand Learning Outcome: 31.05.02: Describe how sea stars move, feed, and reproduce. Section: 31.05 Topic: Echinoderms
93. The madreporite helps the sea star A. reproduce. B. locate food. C. digest food and distribute it to the arms. D. absorb water to regulate its vascular system. E. keep its surface clean of algae and barnacles and debris. The madreporite, or sieve plate, allows water to flow into or out of the water vascular system of a sea star.
Bloom’s Level: 1. Remember Learning Outcome: 31.05.02: Describe how sea stars move, feed, and reproduce. Section: 31.05 Topic: Echinoderms
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Chapter 31 - Animals: Part I
94. A sea star's nervous system includes A. no actual nerve cells. B. a complex of nerve ganglia in each ray. C. a central nervous system, including a brain. D. a central nerve ring that gives off radial nerves in each arm. E. a dorsal hollow nerve cord. The nervous system of a sea star consists of a central nerve ring found in the central disk and radial nerves that extend into each arm. At the end of each arm there is a light-sensitive eyespot.
Bloom’s Level: 2. Understand Learning Outcome: 31.05.02: Describe how sea stars move, feed, and reproduce. Section: 31.05 Topic: Echinoderms
95. Sea stars reproduce A. by budding. B. by asexual processes alone. C. asexually by fragmentation. D. by sexual processes only. E. by sexual means and also by asexual fragmentation. The reproductive system of sea stars consists of gonads and produces either eggs or sperm. They can also reproduce asexually by fragmentation; any fragment can produce a new sea star as long as it contains part of the central disk.
Bloom’s Level: 2. Understand Learning Outcome: 31.05.02: Describe how sea stars move, feed, and reproduce. Section: 31.05 Topic: Echinoderms
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Chapter 31 - Animals: Part I
96. The most likely ancestors of the chordates are considered to be echinoderms because A. tunicates are echinoderms. B. there is a direct lineage from sea stars to tunicates. C. embryos of echinoderms and chordate embryos are deuterostomes. D. adult echinoderms display similar characteristics as adult chordates. E. the nervous system of echinoderms is similar to that of chordates. A deuterostome is a developmental stage where the second embryonic opening becomes the mouth and an outpocketing by the primitive gut becomes the coelom. The fact that this is a developmental process shared by these two phyla and not shared by others demonstrates relatedness between them.
Bloom’s Level: 4. Analyze Learning Outcome: 31.05.01: Recognize the basic morphological characteristics of echinoderms. Section: 31.05 Topic: Echinoderms
97. Which type of animal symmetry produces mirror images of each other no matter how the animal is sliced longitudinally? A. radial symmetry B. asymmetrical symmetry C. bilateral symmetry D. asymmetrical symmetry and bilateral symmetry E. None of the answer choices is correct. Radial symmetry means that the animal is organized circularly, so that no matter how the animal is sliced longitudinally it produces mirror images. Asymmetrical symmetry means that they have no particular symmetry. Bilateral symmetry means that the animal has definite right and left halves. The only cut that produces a mirror image is down the center of the animal.
Bloom’s Level: 2. Understand Learning Outcome: 31.01.02 Compare and contrast the types of symmetry found in animals. Section: 31.01 Topic: Animals
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Chapter 31 - Animals: Part I
Short Answer Questions 98. List the five general characteristics shared by animals. 1. Animals typically have the power of locomotion, by means of muscle fibers, at some stage of their life. 2. Animals are multicellular. 3. They have a life cycle in which the adults are typically diploid. 4. Usually undergo sexual reproduction and produce embryos that go through a series of developmental stages. 5. Heterotrophic; acquire food by ingestion followed by digestion.
Bloom's Level: 6. Create Learning Outcome: 31.01.01 List the general characteristics of animals. Section: 31.01 Topic: Animals
99. Identify the features unique to the molluscs. 1. Molluscs contain a visceral mass that is a soft-bodied portion that contains the internal organs. 2. The foot is a strong, muscular portion used for locomotion. 3. The mantle is a membranous/muscular covering over the visceral mass. The space between the folds of the mantle is called the mantle cavity.
Bloom's Level: 6. Create Learning Outcome: 31.03.02 Identify the unique features of the molluscs. Section: 31.03 Topic: Mollusks
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Chapter 31 - Animals: Part I
True / False Questions 100. Sponges exhibit radial symmetry. FALSE Radial symmetry means that the animal is organized circularly, similar to a wheel, so that no matter how the animal is sliced longitudinally, mirror images are obtained. Sponges have no particular symmetry so they would be classified as being asymmetrical.
Bloom’s Level: 1. Remember Learning Outcome: 31.01.02 Compare and contrast the types of symmetry found in animals. Section: 31.01 Topic: Porifera
101. Lophotrochozoa are animals that are organized circularly, so no matter how they are sliced in half longitudinally, mirror images are obtained. FALSE Lophotrochozoa have bilateral symmetry, which means that they have a definite right and left half. The only cut that would produce mirror images is a longitudinal cut down the center of the animal.
Bloom’s Level: 4. Analyze Learning Outcome: 31.03.01 Describe the basic features of the lophotrochozoa. Section: 31.03 Topic: Lophotrochozoans
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Chapter 31 - Animals: Part I
Multiple Choice Questions 102. Which feature is not found in the lophotrochozoa? A. As embryos, they have two germ layers that develop into their body systems. B. aquatic lifestyle C. bilaterally symmetrical during some stage of their development D. use a lophophore as a feeding apparatus E. They are protostomes. As embryos, lophotrochozoa will have three germ layers that develop into their body systems.
Bloom’s Level: 2. Understand Learning Outcome: 31.03.01 Describe the basic features of the lophotrochozoa. Section: 31.03 Topic: Lophotrochozoans
Short Answer Questions 103. List the five features that have contributed to the success of the arthropods. 1. A rigid but jointed exoskeleton. 2. Segmentation within the body plan. 3. Well-developed nervous system. 4. Variety of respiratory organs. 5. Reduced competition through metamorphosis.
Bloom's Level: 6. Create Learning Outcome: 31.04.02 Describe the characteristics contributing to the success and diversity of arthropods. Section: 31.04 Topic: Arthropods
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Chapter 31 - Animals: Part I
Multiple Choice Questions 104. Which description best fits the lophotrochozoa? A. As embryos, they have three germ layers that will develop into the organ level of organization as adults. B. As embryos, they have two germ layers that will develop into the tissue level of organization as adults. C. Lophotrochozoa are deuterostomes. D. Lophotrochozoa are primarily a terrestrial group. E. The lophophore, a flagellated appendage, is used to define members of this group. As embryos, they have three germ layers that will develop into the organ level of organization as adults. Lophotrochozoa are protostomes. Lophotrochozoa are primarily an aquatic group. The lophophore is a ciliated tentacle-like feeding apparatus that is used to define members of this group.
Bloom’s Level: 4. Analyze Learning Outcome: 31.03.01 Describe the basic features of the lophotrochozoa. Section: 31.03 Topic: Lophotrochozoans
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Chapter 32 - Animals Part II
Chapter 32 Animals: Part II
Multiple Choice Questions 1. Extraembryonic membranes are NOT found in the development of A. amphibians. B. reptiles. C. birds. D. mammals. E. chordates. Since amphibians lay their eggs in water and the resulting larvae live in the water until they undergo the necessary developmental changes to become an adult, they do not require extraembryonic membranes.
Bloom’s Level: 3. Apply Learning Outcome: 32.02.03 Identify the features of the three groups of living amphibians. Section: 32.02 Topic: Amphibians
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Chapter 32 - Animals Part II
2. Which of the following is NOT a unique characteristic of chordates at some time in their life history? A. a dorsal notochord B. a ventral hollow nerve cord C. pharyngeal pouches D. gill clefts E. a postanal tail There are four characteristics that an organism must have at some point in this life history to be classified as a chordate. These are 1) a notochord or dorsally located support rod; 2) a dorsally located nerve cord that is filled with fluid; 3) pharyngeal pouches, these can give rise to gills or can be adapted for other uses; and 4) a postanal tail.
Bloom’s Level: 2. Understand Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
3. In vertebrates, the embryonic A. notochord B. nerve cord C. pharyngeal pouches D. gill clefts E. deuterostomes
is replaced by tissues that form a vertebral column.
Vertebrates have a dorsal support rod or notochord that is present during development of the fetus. Later the flexible notochord is replaced by a vertebral column that is made up of individual vertebra.
Bloom’s Level: 1. Remember Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
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Chapter 32 - Animals Part II
4. In vertebrates, the is/are usually replaced by A. pharyngeal pouches; gill arches B. dorsal hollow nerve cord; a ventral solid nerve cord C. gill arches; pharyngeal pouches D. notochord; a dorsal hollow nerve cord E. notochord; a vertebral column
.
Vertebrates have a dorsal support rod or notochord that is present during development of the fetus. Later the flexible notochord is replaced by a vertebral column that is made up of individual vertebra.
Bloom’s Level: 2. Understand Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
5. Defining characteristics of chordates are A. segmentation, dorsal hollow nerve cord, gill arches or pharyngeal pouches. B. dorsal hollow nerve cord, notochord, bilateral symmetry, and postanal tail. C. bilateral symmetry, segmentation, and well-developed coelom. D. well-developed coelom, gill arches or pharyngeal pouches, and notochord. E. gill arches or pharyngeal pouches, dorsal tubular nerve cord, notochord, and postanal tail. There are four characteristics that an organism must have at some point in this life history to be classified as a chordate. These are 1) a notochord or dorsally located support rod; 2) a dorsally located nerve cord that is filled with fluid; 3) pharyngeal pouches, these can give rise to gills or can be adapted for other uses; and 4) a postanal tail.
Bloom’s Level: 2. Understand Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
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Chapter 32 - Animals Part II
6. Which statement about the chordates is NOT correct? A. Chordates have a notochord. B. Chordates have pharyngeal pouches. C. Invertebrate chordates are marine and filter feeding. D. Adult lancelets have all the characteristics that distinguish chordates from members of other phyla. E. Adult tunicates lack all the characteristics that distinguish chordates from members of other phyla. Tunicates are nonvertebrate chordates whose larvae have the four characteristics that define chordates. As the larvae develop, they lose all but one of these characteristics; they still have a pharynx and gill slits that develop from the pharyngeal pouches.
Bloom’s Level: 2. Understand Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
7. Which of the following sequences correctly represents vertebrate evolution? A. jawless fish → lobe-finned fish → amphibians → reptiles → mammals B. jawless fish → amphibians → reptiles → lobe-finned fish → mammals C. mammals → lobe-finned fish → reptiles → jawless fish → amphibians D. lobe-finned fish → mammals → reptiles → jawless fish → amphibians E. lobe-finned fish → reptiles → mammals → jawless fish → amphibians Jawless fish and all of the other classes have vertebra, the first derived trait. Lobe-finned fish, amphibians, reptiles, and mammals all have lungs. Amphibians, reptiles, and mammals all are tetrapods; they have true limbs. Reptiles and mammals (as well as birds, or feathered reptiles) have amniotic eggs; in conjunction with internal fertilization, the amniotic egg allows these classes to reproduce without the necessity of water to protect and support the egg. Mammals further have the characteristic of producing milk from mammary glands to nourish offspring once they are born.
Bloom’s Level: 4. Analyze Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
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Chapter 32 - Animals Part II
8. Which represents the correct order of evolution of chordate features? A. amnion → limbs → lungs → jaws → vertebrae B. limbs → lungs → jaws → vertebrae → amnion C. lungs → jaws → vertebrae → amnion → limbs D. vertebrae → jaws → lungs → limbs → amnion E. vertebrae → amnion → limbs → lungs → jaws The first defining feature of chordates is the formation of a vertebral column found in vertebrates. Jaws were the next significant feature to evolve; jaws allow animals to feed on other organisms as predators. The evolution of lungs was necessary for animals to move onto land; they allow for gas exchange with the atmosphere. Also necessary on land, animals needed a means of moving around and limbs gave them a way to do this. Finally, in order to reproduce on land, animals needed a way for the sperm to reach the egg (internal fertilization) and a way to protect the embryo from desiccation and to cushion it; the amniotic egg does both.
Bloom’s Level: 4. Analyze Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
9. Which of these is NOT a true statement? A. All chordates are vertebrates. B. All vertebrates are chordates. C. Tunicates and lancelets are invertebrate chordates. D. Echinoderms and chordates are deuterostomes. E. Sea squirts are invertebrate chordates. Lancelets and tunicates are nonvertebrate chordates, so all chordates are not vertebrates. Instead, they retain the notochord as adults.
Bloom’s Level: 3. Apply Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
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Chapter 32 - Animals Part II
10. The lancelets are found A. in tropical Amazon streams. B. as parasites inside the tracts of fish. C. very rarely in mountain streams, where they are almost extinct. D. commonly in the sandy bottoms of shallow coastal waters around the world. E. on the bottom of the ocean. Lancelets are marine chordates that are only a few centimeters long. They typically live along the shorelines on most coasts and bury themselves in the mud or sand.
Bloom’s Level: 2. Understand Learning Outcome: 32.01.02 Describe the features of the two groups of nonvertebrate chordates. Section: 32.01 Topic: Invertebrate Chordates
11. Although adult tunicates are stuck to the ocean bottom, they are considered early relatives of vertebrates because A. tunicate gill slits are like ribs. B. tunicate adults have a notochord. C. tunicate larvae have a notochord. D. all vertebrate embryos go through a stage that resembles adult tunicates. E. tunicate adults have all four chordate characteristics. Tunicates are nonvertebrate chordates whose larvae have the four characteristics that define chordates. These are 1) a notochord, 2) a dorsally located nerve cord, 3) pharyngeal pouches and 4) a postanal tail. As the larvae develop they lose all but one of these characteristics; they still have a pharynx and gill slits that develop from the pharyngeal pouches.
Bloom’s Level: 3. Apply Learning Outcome: 32.01.02 Describe the features of the two groups of nonvertebrate chordates. Section: 32.01 Topic: Invertebrate Chordates
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Chapter 32 - Animals Part II
12. Jaws are believed to have evolved from the A. cranium. B. first ribs. C. first pair of gill arches. D. tooth buds. E. suckers of lampreys. The first pair of gill arches (skeletal rods that support the gill slits) are thought to have become modified so that they became a hinged jaw. The second pair of gill arches also became modified and evolved into support structures for the jaw.
Bloom’s Level: 2. Understand Learning Outcome: 32.02.01 Compare and contrast the characteristics of a jawless fish to those of a jawed fish. Section: 32.02 Topic: Fishes
13. The jawless fish that is parasitic, attaching itself to and sucking fluids from the tissues of a living fish, is the A. hagfish. B. lamprey. C. coelacanth. D. skate. E. ray-finned fish. Lampreys are jawless fish that feed parasitically on larger fish. The mouth is modified so that they clamp on the flank of another fish and a rasping tongue scrapes a hole in the living fish so that the lamprey can ingest the fish’s blood.
Bloom’s Level: 1. Remember Learning Outcome: 32.02.01 Compare and contrast the characteristics of a jawless fish to those of a jawed fish. Section: 32.02 Topic: Fishes
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Chapter 32 - Animals Part II
14. When researchers videotaped the body of a dead whale on the bottom of the ocean, it was rippling with the activity of many scavengers inside the carcass. These scavengers would have been A. hagfish. B. lampreys. C. coelacanths. D. skates. E. ray-finned fishes. Hagfish are jawless scavengers of dead fish. Lampreys only feed on living fish, while coelacanths, skates, and ray-finned fish would not have been inside the carcass but rather outside feeding on the carcass.
Bloom’s Level: 2. Understand Learning Outcome: 32.02.01 Compare and contrast the characteristics of a jawless fish to those of a jawed fish. Section: 32.02 Topic: Fishes
15. The cartilaginous fishes include all EXCEPT A. lampreys. B. dogfish shark. C. skates. D. rays. E. whale shark. Cartilaginous fish, class Chondrichthyes, do not have bony skeletons; of the fish listed, sharks, rays, and skates all belong to this class. Lampreys do not have a skeleton so they are not part of this class but, because they are jawless, they belong to the class Petromyzontida.
Bloom’s Level: 2. Understand Learning Outcome: 32.02.01 Compare and contrast the characteristics of a jawless fish to those of a jawed fish. Section: 32.02 Topic: Fishes
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Chapter 32 - Animals Part II
16. Sharks A. have bone skeletons. B. lack jaws. C. have a keen sense of smell. D. lack the lateral line system of bony fishes. E. do not have paired fins. Sharks have acute senses that go with their predatory behavior. The portion of the brain that goes along with the olfactory system is enlarged relative to other parts. Sharks can detect one drop of blood in 25 gallons of water.
Bloom’s Level: 3. Apply Learning Outcome: 32.02.01 Compare and contrast the characteristics of a jawless fish to those of a jawed fish. Section: 32.02 Topic: Fishes
17. Which of these is NOT a characteristic of a shark? A. ability to sense electric currents in the water B. skeleton of cartilage C. a sucker used to attach to prey D. ability to sense pressure of other swimming objects E. ability to detect blood in minute amounts Sharks are members of class Chondrichthyes, the cartilaginous fishes, and have very acute senses. They are able to detect small quantities of blood, pressure changes in the water, and electric currents. Sharks are also jawed fishes and do not have suckers, but rather a large number of teeth used to catch and devour prey.
Bloom’s Level: 3. Apply Learning Outcome: 32.02.01 Compare and contrast the characteristics of a jawless fish to those of a jawed fish. Section: 32.02 Topic: Fishes
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Chapter 32 - Animals Part II
18. Amphibians most likely evolved from ancestors most similar to the A. hagfish. B. lamprey. C. coelacanth. D. skate. E. ray-finned fish. Coelacanths are lobed fishes; they most likely gave rise to the ancestral species that moved from the water and swimming onto land and walking. The lobed fishes have bone structures in their fins that are homologous to the bones in the appendages of amphibians.
Bloom’s Level: 2. Understand Learning Outcome: 32.02.02 Describe the major evolutionary innovations that distinguish the fishes from the amphibians. Section: 32.02 Topic: Amphibians
19. The living coelacanths discovered off the coast of Africa in the Indian Ocean in 1938, and near Indonesia in 1998, proved that it A. was a modern fish unrelated to fossil coelacanths. B. was the "missing link" between fish and amphibians. C. was the link between cartilage and bony fishes. D. had changed very little and had not gone extinct. E. had gone extinct and then evolved from other fish still present. The coelacanths found in modern times matches the fossilized forms that have been found.
Bloom’s Level: 3. Apply Learning Outcome: 32.02.02 Describe the major evolutionary innovations that distinguish the fishes from the amphibians. Section: 32.02 Topic: Fishes
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Chapter 32 - Animals Part II
20. Amphibians are NOT considered to be fully adapted to life on land because A. they are dependent upon water for external fertilization. B. their skin is used for respiration. C. they have hopping legs. D. they have gills used for respiration. E. most amphibians live their entire lives in water. Amphibians do use their skin to supplement their respiration and require that the skin stay moist for gas exchange to take place but they also have lungs for respiration as well. Hopping legs would not be an advantage in the water but would on land, while webbed feet would be an advantage in the water but would not prevent them from being adapted to life on land. Larval stages usually live in the water and adult stages live on land, but they do not spend all of their life in the water. Water is required for fertilization to take place and this is a disadvantage to living on land and requires that amphibians stay close to a body of water to reproduce.
Bloom’s Level: 2. Understand Learning Outcome: 32.02.02 Describe the major evolutionary innovations that distinguish the fishes from the amphibians. Section: 32.02 Topic: Amphibians
21. The amphibian heart has A. two separate atria and two separate ventricles that partition oxygen-rich and oxygen-poor blood. B. two separate atria and one shared ventricle that allow oxygen-rich and oxygen-poor blood to mix. C. two separate ventricles and one shared atrium that allow oxygen-rich and oxygen-poor blood to mix. D. one atrium and one ventricle similar to the fish. E. one atrium and one ventricle but reroutes the blood through the heart after passing through the gills. The heart of amphibians has a divided atrium and a single ventricle. Deoxygenated blood enters the right atrium from the body and oxygenated blood enters the left atrium from the lungs. These two types of blood mix in the single ventricle and are then sent to the skin where the blood is oxygenated further.
Bloom’s Level: 2. Understand Learning Outcome: 32.02.03 Identify the features of the three groups of living amphibians. Section: 32.02 Topic: Amphibians
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Chapter 32 - Animals Part II
22. Which of these characteristics is NOT found in the amphibians but is developed in reptiles? A. thin, moist skin B. internal fertilization C. small, inefficient lungs D. aquatic larvae E. eggs with gelatinous covering Reptiles, which live in drier environments, are able to carry out reproduction without an external water source. The male is able to deliver the sperm directly into the body of the female so that fertilization takes place internally. The resulting eggs are typically covered in a leathery shell and may either be laid or carried internally until the offspring hatch and are then born.
Bloom’s Level: 2. Understand Learning Outcome: 32.03.01 Compare the major evolutionary innovations that distinguish the reptiles and mammals. Section: 32.03 Topic: Reptiles
23. Which of the following features was NOT necessary for adaptation by reptiles to life on land? A. hard, horny scales B. internal fertilization by means of a penis C. shelled egg D. being cold-blooded E. well-developed lungs It is not necessary for life on land for reptiles to be cold-blooded, ectothermic. Endothermic mammals and birds are quite successful and can actually live in areas where reptiles cannot survive.
Bloom’s Level: 4. Analyze Learning Outcome: 32.03.01 Compare the major evolutionary innovations that distinguish the reptiles and mammals. Section: 32.03 Topic: Reptiles
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Chapter 32 - Animals Part II
24. Which types of embryos have extraembryonic membranes? A. fish B. frog C. salamander D. shark E. mammals Organisms that have developed amniotic eggs must have extraembryonic membranes to carry out gas exchange, the removal of wastes, and the providing of nutrients to the developing embryo. The only organisms listed that are fully adapted to land are mammals; the others all still reproduce in the water and use external fertilization.
Bloom’s Level: 2. Understand Learning Outcome: 32.03.01 Compare the major evolutionary innovations that distinguish the reptiles and mammals. Section: 32.03 Topic: Mammals
25. Which of these characteristics first developed in reptiles? A. amniotic egg B. scales on skin C. four-legged body D. skull and vertebral column E. animals with lungs Scales and the skull and vertebral column are seen in fish and a four-legged body and lungs are seen in amphibians. Both of these phyla developed before reptiles. What is not seen in these earlier forms are amniotic eggs.
Bloom’s Level: 4. Analyze Learning Outcome: 32.03.01 Compare the major evolutionary innovations that distinguish the reptiles and mammals. Section: 32.03 Topic: Reptiles
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Chapter 32 - Animals Part II
26. Animals with a three-chambered heart would include A. amphibians. B. mammals. C. fish. D. sharks. E. birds. Amphibians have a three-chambered heart, while mammals and birds have a four-chambered heart and fish have a two-chambered heart.
Bloom’s Level: 2. Understand Learning Outcome: 32.02.03 Identify the features of the three groups of living amphibians. Section: 32.02 Topic: Amphibians
27. A heart with two separate atria and two separate ventricles is seen among A. fishes and amphibians. B. amphibians and reptiles. C. birds and amphibians. D. mammals and birds. E. fish and amphibians. The hearts of fishes are two-chambered, while amphibians and most reptiles have a threechambered heart. Birds and mammals have four-chambered hearts that prevent the mixing of oxygenated and deoxygenated blood and is therefore more efficient.
Bloom’s Level: 3. Apply Learning Outcome: 32.03.01 Compare the major evolutionary innovations that distinguish the reptiles and mammals. Section: 32.03 Topic: Mammals
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Chapter 32 - Animals Part II
28. Which of these characteristics is NOT associated with flight requirements in birds? A. feet with claws B. light, hollow bones C. a light, horny beak with no teeth D. respiratory air sacs and lungs with one-way air flow E. an enlarged breastbone with strong muscles attached While feet with claws are beneficial for landing on branches and other surfaces, they are not required for flying. Light, hollow bones and a light beak along with the increased efficiency of the avian lungs and the strong muscular muscles attached to the breastbone are necessary characteristics for flight.
Bloom’s Level: 4. Analyze Learning Outcome: 32.03.02 List the features that distinguish birds from the rest of the reptile lineage. Section: 32.03 Topic: Reptiles
29. Because most birds travel through the air at high speeds, they must have A. a crop and a gizzard. B. highly sensitive sensory systems. C. complex songs. D. eggs and no placenta. E. a three-chambered heart. Birds need highly sensitive sensory systems to respond to stimuli in a very short period of time. As birds move very rapidly through the air, they must be able to turn to avoid obstacles or to catch prey.
Bloom’s Level: 3. Apply Learning Outcome: 32.03.02 List the features that distinguish birds from the rest of the reptile lineage. Section: 32.03 Topic: Reptiles
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Chapter 32 - Animals Part II
30. Endothermic animals with four-chambered hearts include the A. amphibians. B. reptiles. C. fish. D. arthropods. E. birds. The hearts of fish are two-chambered, while amphibians and most reptiles have a threechambered heart. In addition fish, amphibians, and reptiles are ectotherms. Birds and mammals have four-chambered hearts that prevent the mixing of oxygenated and deoxygenated blood and is therefore more efficient and they are endotherms.
Bloom’s Level: 2. Understand Learning Outcome: 32.03.02 List the features that distinguish birds from the rest of the reptile lineage. Section: 32.03 Topic: Reptiles
31. Which of the following features is NOT a characteristic used to distinguish birds? A. feathers B. air sacs C. variable body temperature D. hard-shelled egg E. four-chambered heart Birds are endotherms so they do not have a variable body temperature but rather maintain a specific temperature through metabolic processes.
Bloom’s Level: 2. Understand Learning Outcome: 32.03.02 List the features that distinguish birds from the rest of the reptile lineage. Section: 32.03 Topic: Reptiles
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Chapter 32 - Animals Part II
32. Which of these is a proper contrast between birds and mammals? A. Birds are cold-blooded; mammals are warm-blooded. B. Birds are egg laying; no mammals lay eggs. C. Birds have air sacs in addition to lungs; mammals have no air sacs. D. Birds lack a septum between the ventricles; mammals have a septum between the ventricles. E. Birds, unlike mammals, require no parental care. Both birds and mammals are endotherms, warm-blooded, as well as possessing a fourchambered heart and so have a septum between the ventricles. Birds are egg laying and, while most mammals do not lay eggs, there are monotremes that are egg-laying mammals. Birds do care for their offspring for a period of time, although not as long as some mammals care for theirs. Birds do have air sacs that are light and very efficient, while humans do not have them.
Bloom’s Level: 4. Analyze Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Reptiles
33. Vertebrates possessing a four-chambered heart are A. bipedal. B. all birds. C. warm-blooded (maintains a constant body temperature). D. cold-blooded (body temperature varies with external temperature). E. reptiles. All vertebrates, both birds and mammals, with a four-chambered heart are endotherms, warm-blooded. Reptiles have three-chambered hearts and are ectotherms. Both bipedal and quadruped mammals have four-chambered hearts and are endotherms.
Bloom’s Level: 3. Apply Learning Outcome: 32.03.01 Compare the major evolutionary innovations that distinguish the reptiles and mammals. Section: 32.03 Topic: Mammals
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Chapter 32 - Animals Part II
34. The respiratory advantage provided by a four-chambered heart over a three-chambered heart in vertebrates is A. combination of oxygen-rich blood with oxygen-poor blood. B. reduction of hemoglobin requirement. C. elimination of alveolar dependency. D. separation of oxygen-rich blood and oxygen-poor blood. E. more blood gets sent to the lungs. Deoxygenated blood enters the right atrium from the body and passes into the right ventricle where it is pumped into the pulmonary arteries and thence into the lungs; here CO2 is removed and O2 is added. The oxygenated blood then returns to the heart to the left atrium and then into the left ventricle where it is pumped into the aorta; the blood then flows into the body where it gives up O2 and picks up CO2. It then returns to the heart where the cycle begins again. This system separates oxygen and deoxygenated blood leading to a greater efficiency in oxygenation of cells and removal of waste.
Bloom’s Level: 5. Evaluate Learning Outcome: 32.03.01 Compare the major evolutionary innovations that distinguish the reptiles and mammals. Section: 32.03 Topic: Mammals
35. Mammals are different from birds in all these characteristics EXCEPT A. having hair. B. having mammary glands. C. maintaining a constant body temperature. D. having young born alive. E. having a bladder. Both birds and mammals maintain a constant body temperature through metabolic processes.
Bloom’s Level: 4. Analyze Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Mammals
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Chapter 32 - Animals Part II
36. The chief distinguishing characteristic of all mammals is the presence of A. constant internal temperature. B. four limbs. C. a four-chambered heart. D. hair and mammary glands. E. a placenta. Mammals and birds both have a four-chambered heart and maintain a constant body temperature. Mammals, amphibians, birds, and some reptiles all have four limbs. Some but not all mammals have a placenta that supplies the fetus with nutrients and removes wastes.
Bloom’s Level: 2. Understand Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Mammals
37. An animal that is characterized by vertebrae, dorsal hollow nerve cord, hair, and mammary glands must be a(n) A. fish. B. amphibian. C. reptile. D. bird. E. mammal. These are all characteristics of a mammal. Birds, reptiles, amphibians, and fish do not have hair or mammary glands.
Bloom’s Level: 3. Apply Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Mammals
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Chapter 32 - Animals Part II
38. Which type of mammal lays eggs? A. monotremes B. marsupials C. placentals D. snakes E. birds Neither birds nor snakes are mammals. Placentals are mammals that use internal development of the fetus and use the placenta to supply nutrients and remove wastes from the fetus. Marsupials are a type of mammal that give birth to immature young who finish development in a pouch.
Bloom’s Level: 1. Remember Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Mammals
39. The duckbill platypus is an exception to many mammals because it A. lacks hair. B. is ectothermic, or "cold-blooded." C. doesn't secrete milk. D. doesn't give live birth but lays eggs. E. is found in Australia. The duckbilled platypus differs from many other mammals because it lays eggs rather than giving birth to live offspring.
Bloom’s Level: 1. Remember Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Mammals
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Chapter 32 - Animals Part II
40. The organ of nutrition used in fetal development in the most successful mammals is the A. mammary gland. B. egg. C. pouch. D. placenta. E. amniote egg. During fetal development the placenta is used to supply nutrients to the fetus and to remove waste.
Bloom’s Level: 2. Understand Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Mammals
41. The most primitive primates are considered to be the A. prosimians. B. apes. C. New World monkeys. D. Old World monkeys. E. humans. Prosimians, lemurs, tarsiers, and lorises, are thought to be the most primitive of the primates, since they most closely resemble the fossil forms of primates; additionally, they have the smallest brains. Brain size has increased in primates from the most primitive forms to the most recent ones.
Bloom’s Level: 2. Understand Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Primate Evolution
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Chapter 32 - Animals Part II
42. All of the following are characteristics of primates EXCEPT the A. nails on hands and feet. B. stereoscopic vision. C. well-developed brain. D. opposable thumb. E. rapid growth after birth. Primates usually give extended care to offspring so more time and energy is devoted to each offspring. The offspring tend to grow slowly, requiring extended parental care for survival.
Bloom’s Level: 2. Understand Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Primate Evolution
43. Primates are associated with all of the following except A. nails rather than claws. B. a larger brain (cerebral) volume. C. opposable thumb and sometimes large toe. D. eyes on the sides of the head. E. good hand-eye coordination. Primates have shortened muzzles that allowed the eyes to move closer to the center of the face, giving stereoscopic vision. This has led to greater depth perception and better hand-eye coordination.
Bloom’s Level: 2. Understand Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Primate Evolution
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Chapter 32 - Animals Part II
44. Which of the following characteristics is NOT a distinguishing feature of all primates? A. opposable thumb B. fingernails C. usually single birth D. extended period of parental care E. expanded hindbrain The cerebral cortex is expanded in primates. This is a trend among primates where more primitive forms have smaller brains with a less expanded cerebral cortex to the most recent forms that have the largest brains with the most expanded cerebral cortex. The hindbrain houses the functions for homeostasis and helps coordinate large-scale movement. The cerebral cortex is associated with learning, memory, thought, and awareness.
Bloom’s Level: 2. Understand Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Primate Evolution
45. The fossil called Lucy is a member of the species years ago. A. Australopithecus afarensis B. Australopithecus africanus C. Proconsul D. Homo habilis E. Homo erectus
, which lived about 3.18 million
Lucy is the name given to fossilized remains of an Australopithecus afarensis female who has provided evidence on upright, bipedal movement among this group.
Bloom’s Level: 1. Remember Learning Outcome: 32.04.02 Summarize the importance of both ardipithecines and australopithecines in hominin evolution. Section: 32.04 Topic: Human Evolution
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Chapter 32 - Animals Part II
46. Of the following fossils, which is arranged in the correct order from oldest to the most recent? A. Homo erectus, Homo sapiens, Homo habilis B. Homo sapiens, Homo erectus, Homo habilis C. Homo habilis, Homo erectus, Homo sapiens D. Homo erectus, Homo habilis, Homo sapiens E. Homo sapiens, Homo habilis, Homo erectus Homo habilis lived between 2.5 and 1.5 million years ago (MYA). Homo erectus lived between 1.9 and 0.3 MYA. Homo sapiens lived from 0.3 MYA to the present.
Bloom’s Level: 2. Understand Learning Outcome: 32.04.03 Arrange the early species of Homo in evolutionary order. Section: 32.04 Topic: Human Evolution
47. Based on molecular data, the split between apes and humans is thought to have occurred about million years ago. A. one B. two C. three D. five E. six Molecular data show the level of relatedness between groups of organisms by showing the accumulation of differences in their molecules. The more similar the molecules, the more closely they are related. Molecular data comparing the relatedness of humans and apes shows that the two lineages split around 6 to 8 million years ago.
Bloom’s Level: 1. Remember Learning Outcome: 32.04.01 Explain the significance of bipedalism in hominin evolution. Section: 32.04 Topic: Human Evolution
32-24 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 32 - Animals Part II
48. Sometimes biologists are alleged to have said that "man comes from monkeys." The correct way for a modern biologist to explain the apparent sequence of human evolution is A. monkeys have evolved less than humans. B. humans and monkeys share a common ancestor. C. humans and monkeys are biologically identical. D. evolution leads toward more perfect forms and humans therefore came after modern monkeys. E. humans continue to evolve while monkeys do not. Humans and monkeys are descended from the same ancestral form. Humans and monkeys are only distantly related, while humans are more closely related to apes. The common ancestor of humans/apes (hominoids) and monkeys existed more than 40 million years ago (MYA). The lineage of humans and apes diverged 7 MYA.
Bloom’s Level: 4. Analyze Learning Outcome: 32.04.01 Explain the significance of bipedalism in hominin evolution. Section: 32.04 Topic: Human Evolution
49. Which hominid is considered to be the handyman by being the first to use tools? A. gorilla B. Lucy C. Homo habilis D. Homo erectus E. chimpanzee The name Homo habilis means "handyman" and they are considered by some to be the first to use tools.
Bloom’s Level: 1. Remember Learning Outcome: 32.04.03 Arrange the early species of Homo in evolutionary order. Section: 32.04 Topic: Human Evolution
32-25 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 32 - Animals Part II
50. The main features that place Homo habilis in the genus Homo rather than Australopithecus are A. human facial appearance and lack of hair. B. advanced use of language and total carnivorous diet. C. brain size, posture, and dentition. D. care of young and altruistic behavior. E. hand grip, extensive use of tools, and house-building. Homo habilis has a brain size of 600 cc or greater, whereas Australopithecus has a brain size of 370 cc to 515 cc. Australopithecus walks upright but has limbs that are proportioned for apes, whereas Homo habilis walks upright and has slender legs with hips which allow them to walk on two feet over long distances. The dentition of Australopithecus has massive jaws with chewing muscles attached to a prominent bony crest at the top of the skull. Homo habilis tends to have smaller cheek teeth, giving evidence that they were omnivores.
Bloom’s Level: 4. Analyze Learning Outcome: 32.04.03 Arrange the early species of Homo in evolutionary order. Section: 32.04 Topic: Human Evolution
51. The stone tools and the dentition of Homo habilis indicate that early humans probably A. were herbivores. B. were carnivores. C. were omnivores. D. were predators. E. cooked with fire. The dentition of Homo habilis shows smaller cheek teeth and less massive musculature associated with the jaws than Australopithecus, indicating that H. habilis had a diet that included a variety of food stuffs.
Bloom’s Level: 2. Understand Learning Outcome: 32.04.03 Arrange the early species of Homo in evolutionary order. Section: 32.04 Topic: Human Evolution
32-26 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 32 - Animals Part II
52. The first member of genus Homo to use fire is A. Australopithecus robustus. B. Australopithecus africanus. C. Homo habilis. D. Homo erectus. E. Homo sapiens. Homo erectus is thought to have used fire to prepare foods to allow them to eat foods that would otherwise be too tough for their dentition.
Bloom’s Level: 1. Remember Learning Outcome: 32.04.03 Arrange the early species of Homo in evolutionary order. Section: 32.04 Topic: Human Evolution
53. The are believed to be the first group of hominids that buried their dead. A. Homo habilis B. Homo erectus C. Peking man D. Homo neanderthalensis E. Cro-Magnon Homo neanderthalensis remains have been found in burial sites accompanied by tools and other artifacts suggesting that they had some sort of cultural structure and perhaps a form of religion.
Bloom’s Level: 1. Remember Learning Outcome: 32.05.01 Explain how the replacement model relates to evolution of later species of the genus Homo. Section: 32.05 Topic: Human Evolution
32-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 32 - Animals Part II
54. All of these humans are believed to have evolved in Africa EXCEPT A. Homo erectus. B. Homo habilis. C. Homo rudolfensis. D. Neandertals. E. Homo ergaster. Neandertals are thought to have evolved in Europe from H. ergaster.
Bloom’s Level: 2. Understand Learning Outcome: 32.05.01 Explain how the replacement model relates to evolution of later species of the genus Homo. Section: 32.05 Topic: Human Evolution
55. Which of these made advanced tools and produced artwork? A. Homo erectus B. Homo habilis C. Cro-Magnon man D. Neandertal man E. Homo ergaster Cro-Magnons hunted cooperatively, were perhaps the first to have language, and their culture included artwork such as sculpted figurines and drawings.
Bloom’s Level: 1. Remember Learning Outcome: 32.05.02 Discuss the significance of increased tool use, language, and agriculture in Cro-Magnons. Section: 32.05 Topic: Human Evolution
32-28 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 32 - Animals Part II
56. Both echinoderms and chordates are deuterostomes. This means that A. they have a similar development in their embryonic stages. B. they both have eyes that have retinas. C. they both start with radial symmetry in their early embryonic stages. D. they both use calcium-based endoskeletons. E. they are both invertebrates. A deuterostome is a developmental stage where the second embryonic opening becomes the mouth and an outpocketing by the primitive gut becomes the coelom. Since both echinoderms and chordates are deuterostomes, this means that they have similar developmental processes.
Bloom’s Level: 3. Apply Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
57. Nonvertebrate chordates, tunicates, and lancelets, differ from other chordates in that A. they do not have a notochord. B. they have gills, while other chordates have lungs. C. they do not have jaws, while all other chordates do have them. D. the notochord of nonvertebrate chordates does not become a vertebral column. E. nonvertebrate chordates do not have a internal bony skeleton, while all other chordates do. Some portion of the life cycle does have a notochord but it does not develop into a vertebral column. In tunicates the notochord is only present in the larval form, not in the adult.
Bloom’s Level: 3. Apply Learning Outcome: 32.01.02 Describe the features of the two groups of nonvertebrate chordates. Section: 32.01 Topic: Invertebrate Chordates
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Chapter 32 - Animals Part II
58. Larval tunicates differ from adults in all but one of the following ways. Which of the following is not true? A. The adults are sessile and the larvae are free-swimming. B. The adults do not have a notochord, while the larvae do. C. The adults do not have pharynx and gill slits, while the larvae do. D. The adults do not have a dorsal, hollow nerve chord, while the larvae do. E. The adults do not have a postanal tail, while the larvae do. The larvae of tunicates have all four of the characteristics of chordates. They undergo metamorphosis and the resulting adult form only has one characteristic: a pharynx and gill slits. In addition, the larval form is free-swimming and searches out a location to settle and assume the adult form. Adults do not move from this location.
Bloom’s Level: 4. Analyze Learning Outcome: 32.01.02 Describe the features of the two groups of nonvertebrate chordates. Section: 32.01 Topic: Invertebrate Chordates
59. The out-of-Africa hypothesis (replacement model) of human evolution proposes that Homo sapiens evolved from A. several populations of Homo ergaster that existed in Asia, Africa, and Europe. B. one population of Homo ergaster and the other populations were supplanted by H. sapiens. C. Old World monkeys. D. New World monkeys. E. chimpanzees. The out-of-Africa hypothesis states that one population of H. ergaster, the one located in Africa, evolved into H. sapiens and then moved into Asia and Europe, supplanting the populations of H. ergaster living in these regions.
Bloom’s Level: 3. Apply Learning Outcome: 32.05.01 Explain how the replacement model relates to evolution of later species of the genus Homo. Section: 32.05 Topic: Human Evolution
32-30 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 32 - Animals Part II
60. Which feature is used to distinguish modern humans? A. bipedalism B. notochord C. use of tools D. omnivore diet E. forward-facing eyes The key feature that is used to distinguish modern humans is the presence of bipedalism. The presence of a notochord is a feature shared by all chordates. The use of tools is not present in all modern humans. The omnivore diet is found in many different species. Forward-facing eyes is a feature shared by many different species.
Bloom’s Level: 1. Remember Learning Outcome: 32.04.01 Explain the significance of bipedalism in hominin evolution. Section: 32.04 Topic: Human Evolution
Short Answer Questions 61. Identify the four morphological traits unique to the chordates. 1. A dorsal supporting rod called a notochord. 2. A dorsal tubular nerve chord. 3. Pharyngeal pouches. 4. Postanal tail.
Bloom's Level: 6. Create Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
32-31 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 32 - Animals Part II
Multiple Choice Questions 62. Which species is believed to be the first creature to walk erect and possibly be the first hominin? A. Ardipithecus kadabba B. Australopithecus anamensis C. Ardipithecus ramidus D. Australopithecus afarensis E. Homo habilis Ardipithecus kadabba is dated around 5.5 MYA and may have been the first creature to walk erect. Australopithecus anamensis is dated around 4.4 MYA. Ardipithecus ramidus is dated 5.5 MYA to 4.5 MYA and is an ancestral line to Ardipithecus kadabba. Australopithecus afarensis is dated between 4 MYA to 3 MYA. Homo habilis is dated 2 MYA to 1.5 MYA.
Bloom’s Level: 1. Remember Learning Outcome: 32.04.02 Summarize the importance of both ardipithecines and australopithecines in hominin evolution. Section: 32.04 Topic: Human Evolution
True / False Questions 63. The australopithecines evolved and diversified in Africa from 4 MYA until 1.5 MYA and are the possible direct hominin ancestors for humans. TRUE It is true that the australopithecines evolved and diversified in Africa from 4 MYA until 1.5 MYA and are the possible direct hominin ancestors for humans.
Bloom’s Level: 1. Remember Learning Outcome: 32.04.02 Summarize the importance of both ardipithecines and australopithecines in hominin evolution. Section: 32.04 Topic: Human Evolution
32-32 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 32 - Animals Part II
Multiple Choice Questions 64. Which technology was first established by Cro-Magnon? A. throwing spears B. scrapers C. controlled fire D. group life E. All of the answer choices are correct. It is believed that Cro-Magon were the first to develop throwing spears, giving them the ability to kill animals at a distance. Neandertals were known to use scrapers. Homo ergaster was the first species to have knowledge of fire. Homo habilis was known to be socially organized.
Bloom’s Level: 1. Remember Learning Outcome: 32.05.02 Discuss the significance of increased tool use, language, and agriculture in Cro-Magnons. Section: 32.05 Topic: Human Evolution
True / False Questions 65. The ability to throw spears and kill animals from a distance gave Cro-Magnons an advantage in their environment. TRUE It is true that the ability to throw spears and kill animals from a distance gave Cro-Magnons an advantage in their environment. It is hypothesized that they may have been responsible for the Pleistocene overkill.
Bloom’s Level: 1. Remember Learning Outcome: 32.05.02 Discuss the significance of increased tool use, language, and agriculture in Cro-Magnons. Section: 32.05 Topic: Human Evolution
32-33 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 32 - Animals Part II
Multiple Choice Questions 66. What is the most probable evolutionary path for genus Homo? A. Sahelanthropus tchadensis - Ardipithecus kadabba - Australopithecus anamensis Australopithecus afarensis - Homo habilis B. Ardipithecus kadabba - Australopithecus anamensis - Sahelanthropus tchadensis Australopithecus afarensis - Homo habilis C. Australopithecus afarensis - Ardipithecus kadabba - Australopithecus anamensis Sahelanthropus tchadensis - Homo erectus D. Ardipithecus kadabba - Australopithecus afarensis - Australopithecus anamensis Sahelanthropus tchadensis - Homo habilis E. Sahelanthropus tchadensis - Ardipithecus kadabba - Australopithecus anamensis Australopithecus afarensis - Homo ergaster Sahelanthropus tchadensis - Ardipithecus kadabba - Australopithecus anamensis Australopithecus afarensis - Homo habilis is the most probable evolutionary path for genus Homo. This starts with the oldest-known hominid fossil and progressing towards the most recent fossil.
Bloom’s Level: 5. Evaluate Learning Outcome: 32.04.03 Arrange the early species of Homo in evolutionary order. Section: 32.04 Topic: Human Evolution
67. Which feature distinguishes birds from the rest of the reptiles? A. endothermic B. scales C. egg laying D. lungs E. four appendages Birds are endothermic animals, while the reptiles are ectothermic. Both birds and reptiles share all of the other features.
Bloom’s Level: 1. Remember Learning Outcome: 32.03.02 List the features that distinguish birds from the rest of the reptile lineage. Section: 32.03 Topic: Reptiles
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Chapter 32 - Animals Part II
Short Answer Questions 68. Identify the three living lineages of mammals and describe the feature that makes each one unique. 1. Monotremes: presence of a cloaca and lay hard-shelled amniotic eggs. 2. Marsupials: begin development inside of the mother's body but finish inside a pouch on the abdomen. 3. Placental animals: nutrients are exchanged between mother and offspring through a placenta.
Bloom's Level: 6. Create Learning Outcome: 32.03.03 Identify the unique features that define the three living lineages of mammals. Section: 32.03 Topic: Mammals
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Chapter 32 - Animals Part II
Multiple Choice Questions 69. What is the correct evolutionary pathway for the key features of chordates? A. notochord - vertebrae - jaws - bony skeleton - lungs - 4 limbs - amniotic egg - mammary glands B. vertebrae - notochord- jaws - bony skeleton - lungs - 4 limbs - amniotic egg - mammary glands C. notochord - vertebrae - jaws - bony skeleton - lungs - amniotic egg - 4 limbs - mammary glands D. notochord - vertebrae - jaws - bony skeleton - mammary glands - lungs - 4 limbs amniotic egg E. mammary glands - notochord - vertebrae - jaws - lungs - 4 limbs - amniotic egg - bony skeleton The first feature to evolve was the presence of a notochord. The notochord is then replaced by the vertebrae. The evolution of jaws enabled chordates to feed on different food choices. A bony skeleton increased the efficiency of movement. Lungs increased respiration. Four limbs would provide greater support on land. An amniotic egg protected the embryo from drying out on land. Mammary glands increase the nutrition available for the offspring.
Bloom’s Level: 5. Evaluate Learning Outcome: 32.01.01 Identify the four morphological characteristics unique to the chordates. Section: 32.01 Topic: Chordates
True / False Questions 70. The ray-finned fishes are believed to have evolved into the amphibians. FALSE The lobe-finned fishes have fleshy appendages that could be adapted to land locomotion. Most also had a lung that was used for respiration. It is believed that the lobe-finned fishes evolved into the amphibians.
Bloom’s Level: 1. Remember Learning Outcome: 32.02.02 Describe the major evolutionary innovations that distinguish the fishes from the amphibians. Section: 32.02 Topic: Fishes
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Chapter 32 - Animals Part II
Short Answer Questions 71. List the features that make amphibians different than the fishes. 1. jointed appendages 2. four limbs 3. eyelids for keeping their eyes moist 4. ears for picking up sound waves 5. a voice-producing larynx
Bloom's Level: 6. Create Learning Outcome: 32.02.02 Describe the major evolutionary innovations that distinguish the fishes from the amphibians. Section: 32.02 Topic: Amphibians
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Chapter 33 - Animal Behavior
Chapter 33 Animal Behavior
Multiple Choice Questions 1. Scientists can state that all behavior has a genetic basis because A. they have proved that all aspects of behavior are coded in the animal's genome. B. as with other life characteristics, there is no other source beyond what can be coded in DNA. C. scientists do not believe in anything beyond natural phenomena arising from the structure of matter. D. an animal inherits its anatomy and physiology with which it performs behaviors. E. animal behaviorists such as B. F. Skinner have reduced all behavior to biology. While all aspects of behavior cannot be directly traced to genetics, the physical basis for expressing those behaviors can be. The anatomy and physiology necessary for the production of molecules related to behavior (hormones, adrenaline, etc.) are directly created from the DNA.
Bloom’s Level: 3. Apply Learning Outcome: 33.01.02 Describe an experiment that shows behavior can be genetically based. Section: 33.01 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
2. Which of the following supports a genetic basis of behavior? A. two brothers raised in the same household who like the same type of ice cream B. identical twins that are raised in two separate families and grow up to smoke the same brand of cigarettes C. a sister and a brother who are raised in the same household and say things in the same way D. two unrelated individuals who are raised in the same household and have the same taste in clothing E. cousins who grow up in the same family prefer the same foods If the individuals are raised in the same household, then learned behaviors could not be discounted. Identical twins would have the same genetics, but if they are raised apart from one another then it would be possible to discount learned behaviors and relate their behavior to a genetic cause.
Bloom’s Level: 5. Evaluate Learning Outcome: 33.01.02 Describe an experiment that shows behavior can be genetically based. Section: 33.01 Topic: Behavioral Ecology
3. The main result of the experiments with two groups of California garter snakes that varied in habitat and food was A. the behavior was determined by the environment (terrestrial versus aquatic). B. the behavior was learned; one group learned to like to eat slugs, the other, fish. C. the behavior was general; the snakes ate anything that moved. D. the behavior was genetic, and hybrids had an intermediate response to the two food extracts. E. the behavior was imprinted by whatever they ate when they were first born. Aquatic garter snakes normally eat frogs and fish, while a terrestrial garter snake feeds on slugs and aquatic snakes refused slugs when they were offered. When these two species were crossed the resulting offspring showed some response to the offering of slugs but not at the same rate as the terrestrial garter snakes.
Bloom’s Level: 3. Apply Learning Outcome: 33.01.02 Describe an experiment that shows behavior can be genetically based. Section: 33.01 Topic: Behavioral Ecology
33-2 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 33 - Animal Behavior
4. The discovery of a gene in Aplysia slugs that results in protein products including egglaying hormone is evidence that A. the egg-laying behavior is innate or hardwired into the slugs' brain. B. the egg-laying behavior has absolutely no nervous system involvement. C. genes can control endocrine gland secretions that can control behavior. D. this slug could avoid being eaten by garter snakes. E. genes only influence physiology and structure but not behavior. The hormone ELH (egg-laying hormone) in Aplysia slugs can trigger egg-laying behavior in the slug even if mating has not occurred. The hormone directly causes contractions of the smooth muscle in the reproductive tract, leading to the expulsion of the eggs whether or not they are fertilized.
Bloom’s Level: 3. Apply Learning Outcome: 33.01.02 Describe an experiment that shows behavior can be genetically based. Section: 33.01 Topic: Behavioral Ecology
5. Which experiment most closely confirms the linkage between DNA → a gene → a gene product → the behavior that results from that product? A. experiments with the inland and coastal garter snakes and their food preferences B. experiments with the Fischer lovebirds and their nest-building behavior C. experiments with laughing gull chicks and their pecking behavior to solicit food from the mother gull D. experiments with white-crowned sparrows and the critical period for learning bird songs E. experiments with Aplysia snails and their egg-laying behavior The experiments with the Aplysia slugs have allowed researchers to find a gene and then to trace its physiological effects on the organism, leading to a specific behavior. The other experiments do not have the full sequence of DNA leading to a gene product which leads to a behavior, although they do provide information on linking genetics and behavior.
Bloom’s Level: 5. Evaluate Learning Outcome: 33.01.02 Describe an experiment that shows behavior can be genetically based. Section: 33.01 Topic: Behavioral Ecology
33-3 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 33 - Animal Behavior
6. Which two organ systems are most involved in producing physiological changes leading to appropriate behavior patterns? A. nervous and digestive B. digestive and urinary C. nervous and reproductive D. nervous and endocrine E. reproductive and digestive The endocrine system produces hormones, which are secreted into the bloodstream and travel to other parts of the body, to carry the chemical signal that triggers a behavior. The nervous system determines how external stimuli are perceived. These perceptions can trigger behaviors.
Bloom’s Level: 1. Remember Learning Outcome: 33.01.02 Describe an experiment that shows behavior can be genetically based. Section: 33.01 Topic: Behavioral Ecology
7. A digger wasp continues to close its nest even though the researcher has removed the nest's contents. This shows that nest-closing behavior in the digger wasp is A. a learned behavior. B. an imprinted behavior. C. a fixed action pattern. D. an insight learning behavior. E. a classical conditioned behavior. Since this is a behavior that must be completed once it is begun and the behavior is brought about by a specific stimulus, then it meets the definition of an FAP.
Bloom’s Level: 2. Understand Learning Outcome: 33.02.01: Identify an experiment that shows how behavior can be environmentally influenced. Section: 33.02 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
8. Which of the following statements is NOT true about fixed action patterns (FAPs)? A. All the members of a species perform the exact same sequence of behaviors. B. The behavior pattern is stereotyped. C. Many behaviors thought to be fixed action patterns have been shown to develop after practice. D. The fixed action pattern was considered to be initiated by a sign stimulus. E. FAPs are strictly the result of the expression of particular genes. Fixed action patterns (FAPs) are behaviors that are directly linked to a simple stimulus. These patterns are unchangeable and usually must be carried out completely once they have begun. While FAPs do exist in many species, some behaviors that were thought to be strictly FAPs have since been discovered to improve upon practice. At this point no direct link to a specific gene has been found for a specific FAP.
Bloom’s Level: 2. Understand Learning Outcome: 33.02.01: Identify an experiment that shows how behavior can be environmentally influenced. Section: 33.02 Topic: Behavioral Ecology
9. An increase in fitness—the survival of more offspring—that occurs due to competition among males and mate choice by females is called A. altruistic behavior. B. selfish behavior. C. natural selection. D. sociobiology. E. sexual selection. Altruistic behavior is a social interaction that reduces the reproductive success of an individual but increasing the fitness of another individual. Selfish behavior is behavior that promotes the fitness of one individual while reducing the fitness of others. Natural selection is the process where the differential survival and reproduction of organisms with greater fitness changes the allele frequencies in the population. Sociobiology is the study of social interaction and behavior among animals.
Bloom’s Level: 1. Remember Learning Outcome: 33.04.02: Explain why females should be choosier than males in selecting a mate. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
10. The pecking behavior of laughing gull chicks becomes more accurate with time. This is most likely an example of A. a fixed action pattern. B. innate behavior. C. learning. D. imprinting. E. altruistic behavior. While the initial reaction to the stimulus of the parent’s beak is an FAP, the further action of improving in aim and efficiency in the behavior is learned with practice.
Bloom’s Level: 3. Apply Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
11. That behavior often has a learned component was shown by studying the A. migration of birds. B. altruism of human beings. C. feeding behavior of baby gull chicks. D. singing patterns of song birds. E. feeding behavior of baby gull chicks AND singing patterns of song birds. The feeding behavior of gull chicks shows that learning occurs as the chicks practice and is able to improve their aim as their motor skills improve. Additionally, the singing of birds shows learned behavior; they practice the song as they hear it and then sing that song regardless of whether or not it is the dialect or even species that they should sing.
Bloom’s Level: 2. Understand Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
12. Learned behavior that causes a family of baby ducks to follow their mother is called A. imprinting. B. operant conditioning. C. insight learning. D. extinction. E. motivation. Imprinting occurs when a behavior has a strong genetic component. Imprinting has a critical period when certain behaviors can be learned and this distinguishes them from other forms of learning. If this period of time is missed, then the behavior will not be learned.
Bloom’s Level: 1. Remember Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
13. Fifty days after hatching, a white-crowned sparrow will never be able to sing a normal song if reared in isolation. This is an example of A. a sensitive period. B. a sign stimulus. C. an innate behavior. D. fixed action pattern. E. habituation. Since the behavior must be learned in a specific period of time, this is an example of a sensitive period; once the period is passed, the behavior cannot be learned.
Bloom’s Level: 2. Understand Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
14. Imprinting and songbird singing are both examples of A. altruism. B. innate but not learned behavior. C. a type of courtship ritual. D. innate behavior that also requires some learning. E. kin selection. Both of these behaviors have a component that is innate, given a stimulus the organism will respond, but practice or learning are also involved. The chick will imprint on the first object that it sees moving during the sensitive period and the song bird will learn any song that it hears during this time, even if it is from a different species.
Bloom’s Level: 2. Understand Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
15. Male birds court females of the correct species because they look like their mother. This is due to A. an innate behavior. B. a learned behavior. C. an altruistic act. D. imprinting. E. a fixed action pattern. Male birds learn the appropriate appearance of the parents from imprinting on the parents as chicks. Chicks that imprint on something other than a member of their own species will not then recognize a suitable mate.
Bloom’s Level: 2. Understand Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
16. The sensitive period is most closely associated with A. imprinting. B. operant conditioning. C. insight learning. D. extinction. E. motivation. Imprinting must occur during a specific period of time or sensitive period.
Bloom’s Level: 1. Remember Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
17. Imprinting helps birds A. perform the proper courtship ritual. B. defend their territory. C. recognize a member of their own species. D. form a fixed action pattern. E. build a better nest. By imprinting on the parents, the chicks learn the appropriate appearance of others in their species. If they imprint on something other than their species, they will not recognize appropriate mates later when they are adults.
Bloom’s Level: 2. Understand Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
18. You are passing through one of the science buildings on a university campus when you see a researcher come out of his laboratory and go into another room. You hear and then see a newly hatched duckling come out of the lab and run to follow the researcher into the other room. The duck's behavior can best be described as A. a fixed action pattern. B. classical conditioning. C. imprinting. D. operant conditioning. E. altruistic. The duck has imprinted on the researcher instead of its parent and will follow him anytime it is not prevented from doing so.
Bloom’s Level: 3. Apply Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
19. Salmon fish hatcheries add a harmless perfume (morpholine) to a stream when they release young fish. When the fish mature, they attract them back to the home stream by using this chemical (in nature the fish smell the unique odors of soil and trees along the home stream). In effect, the hatchery biologist is using A. the fish's inborn instincts to relocate its own stream. B. a fish's higher order reasoning. C. an experiment based on the ability of fish to learn by trial and error. D. chemotropisms. E. imprinting. This is a form of imprinting since the stimulus must be presented within a window of time or another stimulus (scent) will be used to mark the appropriate stream for the salmon.
Bloom’s Level: 3. Apply Learning Outcome: 33.02.01: Identify an experiment that shows how behavior can be environmentally influenced. Section: 33.02 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
20. Isolated songbirds will sing a more correct song if they A. are never removed from isolation. B. listen to their own species' song within a particular time period. C. listen to other species' songs. D. are punished for not singing a complicated song. E. are given an adult tutor. Birds that are isolated from other birds but who heard songs of their species played for them will be able to sing if the song is played within the sensitive period.
Bloom’s Level: 2. Understand Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
21. What is a change in behavior that involves associating two separate events with each other? A. a fixed action pattern B. innate behavior C. associative learning D. imprinting E. altruistic behavior In associative learning an organism that experiences one stimulus associates some other event with that stimulus. The link exists for that organism but not necessarily for others of its species.
Bloom’s Level: 1. Remember Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
22. Learning that is reinforced by rewards, a gradual strengthening of stimulus-response connections, is called A. imprinting. B. operant conditioning. C. insight learning. D. extinction. E. motivation. In operant conditioning a response to a stimulus is made stronger by a reward for the behavior associated with the stimulus. This is used to train dogs or other animals to perform tricks; when they perform correctly they receive a reward. Later, when the stimulus is given the behavior will be performed, whether or not the reward is offered.
Bloom’s Level: 1. Remember Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
23. Which statement about behavior is NOT true? A. Inborn behavior is usually inherited. B. Innate behavior is triggered by a stimulus and does not vary. C. Fixed action patterns are sometimes subject to modification by learning. D. Animals with simple nervous systems tend to respond to a stimulus with an inherited behavior. E. All behaviors can be explained as fixed action patterns. Many behaviors are much more complex than a FAP. Only the simplest behaviors are based on one stimulus and always result in the same response. It is more usual that a stimulus will result in a range of responses.
Bloom’s Level: 2. Understand Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
24. To train an animal, you try to reward each instance of desired behavior. In this way, you can eventually get pigeons to play ping-pong. This system of reward, where animals are given a reward each time they responded correctly, usually in a boxlike environment, is called A. imprinting. B. operant conditioning. C. insight learning. D. extinction. E. motivation. B. F. Skinner studied the idea of learning by reward, operant conditioning. In his experiments he taught a number of behaviors to a number of species based on this system.
Bloom’s Level: 1. Remember Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
25. is a system in which animals arrange themselves in a pecking order. The animal above takes precedence over the one below. A. Imprinting B. Inclusive fitness C. Dominance hierarchy D. Altruism E. Circadian rhythm Imprinting is learning one type of behavior for a specific stimulus in a specific window of time. Inclusive fitness refers to the fitness not only of an individual but also the close relatives of that individual; this results in the genetic material of the family group being passed on to the next generation. Altruism is a social interaction that reduces the reproductive success of an individual but increasing the fitness of another individual. Circadian rhythms are regular behaviors or physiological changes that are based on a 24-hour cycle.
Bloom’s Level: 1. Remember Learning Outcome: 33.04.03: Summarize the costs and benefits of living in a society. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
26. An example of territoriality is A. barnacles attached to a boat bottom. B. a red-winged blackbird singing to ward off other males and attract a female to the nest in his fence row. C. a cloud of moths attracted to a light. D. a swarm of mosquitoes hovering around a sweaty person. E. bees attracted to a flower garden. Territoriality refers to the defense of a certain resource. Animals are generally territorial and defend resources that will assist them in obtaining a mate or in raising offspring. When the male red-winged black bird sings, he announces that the resources in the area are his and any other red-winged black bird who wants them must battle him to obtain them.
Bloom’s Level: 3. Apply Learning Outcome: 33.04.01: Discuss why territoriality evolved in certain animal groups. Section: 33.04 Topic: Behavioral Ecology
27. The first female aphids to hatch in the spring settle on large leaves and defend them against other females. This is an example of A. territoriality. B. reproductive behavior. C. feeding behavior. D. learned behavior. E. dominance hierarchy. The female aphid is defending the resources of the leaf. Territoriality refers to the defense of a certain resource. Animals are generally territorial and defend resources that will assist them in obtaining a mate or in raising offspring.
Bloom’s Level: 3. Apply Learning Outcome: 33.04.01: Discuss why territoriality evolved in certain animal groups. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
28. The competitive behavior among male elk competing for access to females is advantageous because A. the survivor of the competition is the elk that lives the longest, longer than the females. B. the survivor is the most altruistic and caring. C. the increased number of offspring produced will offset the shorter life of the male elk. D. the best DNA genome is directing the best animal to be superior in combat. E. once the top male elk wins, females ensure that he will have mating opportunities for his entire life. Competition and fighting among elk would increase the likelihood that injury and death would be the result of repeated conflict. However, if the number of offspring resulting from the mating opportunities is larger than it would have been without, then the genetic heritage of the male elk is the same or better than if it had lived longer.
Bloom’s Level: 4. Analyze Learning Outcome: 33.04.03: Summarize the costs and benefits of living in a society. Section: 33.04 Topic: Behavioral Ecology
29. Which of the following is NOT a benefit to living in a group? A. protection against predators B. easier to find food C. easier to find a mate D. amount of work can be divided up among members E. increased chance of disease An increased chance of disease is a result of living in a group but would not be a benefit but rather a detriment.
Bloom’s Level: 2. Understand Learning Outcome: 33.04.03: Summarize the costs and benefits of living in a society. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
30. Kin selection is a form of A. altruism. B. territoriality. C. imprinting. D. confrontation. E. sexual selection. Kin selection refers to the idea that the passing of the genetic material by your relatives allows the genes of the family group to be passed to the next generation even if they are not specifically your genes. This is an altruistic viewpoint.
Bloom’s Level: 1. Remember Learning Outcome: 33.04.03: Summarize the costs and benefits of living in a society. Section: 33.04 Topic: Behavioral Ecology
31. Altruism is described as A. displaced aggressive behavior. B. a means of ranking animals in a dominance hierarchy. C. actions of an animal that immediately benefit others rather than itself, but may harm itself. D. belligerent behavior that helps an animal compete in its society. E. an individual's reproductive success. Altruistic behavior is a social interaction that reduces the reproductive success of an individual but increasing the fitness of another individual.
Bloom’s Level: 1. Remember Learning Outcome: 33.04.03: Summarize the costs and benefits of living in a society. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
32. Unselfish acts that help others in a group so that they increase their chances of surviving and reproducing is called A. altruistic behavior. B. selfish behavior. C. natural selection. D. inclusive fitness. E. sexual selection. Altruistic behavior is a social interaction that reduces the reproductive success of an individual but increasing the fitness of another individual.
Bloom’s Level: 1. Remember Learning Outcome: 33.04.03: Summarize the costs and benefits of living in a society. Section: 33.04 Topic: Behavioral Ecology
33. The ability of an individual and the individual's kin to survive and reproduce is called A. altruistic behavior. B. selfish behavior. C. natural selection. D. fitness. E. inclusive fitness. Inclusive fitness refers to the fitness not only of an individual but also the close relatives of that individual; this results in the genetic material of the family group being passed on to the next generation.
Bloom’s Level: 1. Remember Learning Outcome: 33.04.03: Summarize the costs and benefits of living in a society. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
34. Altruistic behaviors between closely related animals are selected for because they A. reduce fighting between species. B. ensure the survival of the altruistic individual. C. increase the frequency of the altruistic individual's genes in the next generation. D. force individuals to cooperate with one another and thereby increase mating and population growth. E. stimulate new learning behaviors. Since the genes in closely related animals would be very similar, the genes governing altruistic behavior would be passed on to the next generation.
Bloom’s Level: 3. Apply Learning Outcome: 33.04.03: Summarize the costs and benefits of living in a society. Section: 33.04 Topic: Behavioral Ecology
35. Many sterile worker bees give their lives to save a beehive from an attack by bears, etc. In such cases, we now know A. these female worker bees are more related to hive offspring than they would be if they were fertile and mated; it is "calculated selfishness." B. this is really a case of bees consciously understanding the need to preserve the nest for their individual benefit and they might survive if all work together. C. each female worker is waiting to get her chance to reproduce. D. this is a case of sexual selection. E. the system is not yet understood. The queen bee is the mother of all of the workers and of the fertile males and the new queens that will arise from the hive. Altruism among the worker bees to defend the queen and larvae in the hive benefits them by allowing the same genes that they have are passed on by their mother, the queen.
Bloom’s Level: 4. Analyze Learning Outcome: 33.04.03: Summarize the costs and benefits of living in a society. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
36. A very rapid kind of communication that works even at a distance and in the dark is A. chemical communication. B. auditory communication. C. tactile communication. D. visual communication. E. pheromones. Auditory communication is rapid and does not require that the communicating parties have physical or visual contact as tactile or visual communication would. With pheromones or chemical communication, weather conditions can affect the message by diluting the scent or removing it all together.
Bloom’s Level: 2. Understand Learning Outcome: 33.03.02: Describe the advantages and disadvantages of chemical, auditory, visual, and tactile communication. Section: 33.03 Topic: Animal Communication
37. is defined as the transmission of a signal from one animal that influences the behavior of a receiving animal. A. Imprinting B. Dominance hierarchy C. Communication D. Circadian rhythm E. Operant conditioning Imprinting is the learning one type of behavior for a specific stimulus in a specific window of time. Dominance among a group of animals is generally determined by a confrontation between members of the group with the winner of the confrontation having access to resources that the loser does not have access to. Circadian rhythms are regular behaviors or physiological changes that are based on a 24-hour cycle. In operant conditioning, a response to a stimulus is made stronger by a reward for the behavior associated with the stimulus.
Bloom’s Level: 1. Remember Learning Outcome: 33.03.01: Summarize the various ways in which animals communicate. Section: 33.03 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
38. When an ant finds a source of food, such as an open sugar bin in a kitchen, it returns to the nest, leaving a trail that other ants can follow to the sugar. This is a case of A. innate knowledge of where food is located. B. a learned response that is an advantage of social insects. C. a pheromone pathway, a type of chemical communication. D. marking territory. E. higher reasoning, just as a hiker leaving signs along a trail. Ants leave a scent trail so that other ants may find their way to and from a food source. In this way the foraging members of the colony do not become lost and food is efficiently gathered and returned to the developing larva.
Bloom’s Level: 3. Apply Learning Outcome: 33.03.01: Summarize the various ways in which animals communicate. Section: 33.03 Topic: Animal Communication
39. Pheromones are A. visual signals. B. auditory signals. C. chemical attractants. D. tactile responses. E. hormones. Pheromones are chemicals produced by animals to send a signal to another individual. In animals, these chemicals act like hormones in altering the physiology and behavior or the receiving individual.
Bloom’s Level: 1. Remember Learning Outcome: 33.03.01: Summarize the various ways in which animals communicate. Section: 33.03 Topic: Animal Communication
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Chapter 33 - Animal Behavior
40. Which type of communication occurs when one animal touches another? A. chemical communication B. visual communication C. tactile communication D. auditory communication E. pheromone communication Auditory communication uses sound waves to allow distant organisms to exchange information. With pheromones or chemical communication, a signal is placed in a location and organisms that come into proximity with the message are able to pick up the message; the signaling and receiving animals do not have to come into contact for the message to pass between them. Visual communication involves the changing of body posture, which is the message and elicits a response from the animal viewing the signal. Tactile communication occurs when two or more animals have physical contact.
Bloom’s Level: 1. Remember Learning Outcome: 33.03.01: Summarize the various ways in which animals communicate. Section: 33.03 Topic: Animal Communication
41. Primate grooming and honeybee waggle dancing share which of these forms of communication? A. chemical communication B. auditory communication C. tactile communication D. visual communication E. physical communication Tactile communication occurs when two or more animals have physical contact. Primate grooming establishes and reinforces the dominance hierarchy between the individuals. The waggle dance of bees uses changes in position and direction to convey information from one individual to a group of individuals. The positional and directional information is passed as the group touches the "dancing bee" to feel where it has been.
Bloom’s Level: 2. Understand Learning Outcome: 33.03.01: Summarize the various ways in which animals communicate. Section: 33.03 Topic: Animal Communication
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Chapter 33 - Animal Behavior
42. What mechanism allows bees to compensate for the movement of the sun when they return to the hive and do a waggle dance? A. the ability to detect the earth's magnetic field B. always doing the waggle dance outside where the sun is visible C. a biological clock D. imprinting E. a mental map of the region encoded by learning when young Bees have a biological clock that allows them to compensate for the change of the sun in the sky as the day progresses. They use this in conjunction with the magnetic field of the earth to convey the distance and direction of a food source.
Bloom’s Level: 2. Understand Learning Outcome: 33.03.02: Describe the advantages and disadvantages of chemical, auditory, visual, and tactile communication. Section: 33.03 Topic: Animal Communication
43. The waggle dance helps bees A. feed their offspring when they are hungry. B. learn to hum the right tune. C. perfect the avoidance response. D. communicate the location of food. E. sound the alarm to protect the hive when it is invaded. Bees use this series of movements to convey the distance and direction of a food source. Bees from the hive crowd around and touch the dancing bee with their antennae and so receive the signal.
Bloom’s Level: 3. Apply Learning Outcome: 33.03.02: Describe the advantages and disadvantages of chemical, auditory, visual, and tactile communication. Section: 33.03 Topic: Animal Communication
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Chapter 33 - Animal Behavior
44. Which of the following is NOT a tactile communication method? A. baby gulls pecking at parent's beak B. grooming in primates C. the waggle dance of bees D. pheromone release from female moths E. fish using their lateral lines to feel the presence of other fish swimming alongside Pheromones would not require any physical contact with the communicating animals. One organism can leave the chemical signal and, weather permitting, it can remain in that location for days or weeks afterwards.
Bloom’s Level: 3. Apply Learning Outcome: 33.03.01: Summarize the various ways in which animals communicate. Section: 33.03 Topic: Animal Communication
45. You are lying on your favorite beach in Florida and, while looking down the beach, you observe periodic white flashes in the sand. You get up and walk down the beach to see what is causing the flashes. You observe Fiddler crabs waving their enlarged claw in the air producing the flashes. This behavior is an example of A. auditory communication. B. chemical communication. C. visual communication. D. tactile communication. E. territoriality. Visual communication is any form of signaling that involves changes in the appearance or posture of an animal and causes a change in the physiology or behavior one or more other animals. Fiddler crabs wave their claws to attract a female to his burrow. The visual signal in this case is that he is a fit and desirable mate.
Bloom’s Level: 3. Apply Learning Outcome: 33.03.02: Describe the advantages and disadvantages of chemical, auditory, visual, and tactile communication. Section: 33.03 Topic: Animal Communication
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Chapter 33 - Animal Behavior
46. While walking along a trail in the woods you notice that your dog walks over to a tree and urinates on the base of the largest tree on the trail. Your dog's behavior is an example of A. visual communication. B. auditory communication. C. tactile communication. D. chemical communication. E. learning. Dogs use chemical communication to signal to other dogs. They mark their territory by spraying urine in specific locations; the scent of this urine is unique to that individual. The chemical signal tells the next dog who comes by the sex, reproductive condition, and health of the first dog. With chemical signaling the animal leaving the signal does not have to be present to deliver its message and the message lasts as long as weather conditions allow it to.
Bloom’s Level: 3. Apply Learning Outcome: 33.03.01: Summarize the various ways in which animals communicate. Section: 33.03 Topic: Animal Communication
47. When training a dog to perform on a television show the trainer first watches the dog’s behavior, chooses one that approximates what the director has asked to have done and begins rewarding the dog every time he performs this natural behavior. Gradually the trainer changes when the reward is given, altering the behavior toward the desired performance until this is what the dog does each time, in place of the natural behavior. The trainer has used to train the dog. A. tactile communication B. imprinting C. a fixed action pattern D. operant conditioning E. chemical communication In operant conditioning, a response to a stimulus is made stronger by a reward for the behavior associated with the stimulus.
Bloom’s Level: 5. Evaluate Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
48. A parrot fluffs out her feathers and screeches when another female bird approaches. She is A. communicating using only visual signals. B. sick. C. communicating using only auditory signals. D. communicating using visual signals and auditory signals. E. communicating using tactile cues. Any signal from one animal that influences the physiology or behavior of another is a form of communication. The communication, in this case, is both visual and auditory since the fluffing of feathers would be a visual signal, while the screech would be an auditory signal.
Bloom’s Level: 3. Apply Learning Outcome: 33.03.02: Describe the advantages and disadvantages of chemical, auditory, visual, and tactile communication. Section: 33.03 Topic: Animal Communication
49. A male and female crane perform a series of moves in concert that lead up to mating. This is a form of A. operant conditioning. B. altruism. C. imprinting. D. communication. E. learning. The two cranes are performing a courtship ritual that communicates their readiness to mate and leads to physiological changes that prepare them for mating.
Bloom’s Level: 3. Apply Learning Outcome: 33.03.01: Summarize the various ways in which animals communicate. Section: 33.03 Topic: Animal Communication
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Chapter 33 - Animal Behavior
50. Some fish give off a chemical when injured that signals to other fish that a predator is nearby. The chemical causes the uninjured fish to group together in a tight formation, move to the bottom of the water, and cease movement to make it less likely that they will be preyed upon. A. This is communication using a chemical signal. B. This is a futile attempt by the fish to survive. C. This is an unnecessary action since the predator already has killed its prey. D. This is a visual signal to the predator that they are prey items. E. This behavior will only work one time. The chemical signal alerts the other fish that there is a predator so that they can escape. The behavior that they show is an instinctive avoidance so that the predator will not notice them.
Bloom’s Level: 5. Evaluate Learning Outcome: 33.03.02: Describe the advantages and disadvantages of chemical, auditory, visual, and tactile communication. Section: 33.03 Topic: Animal Communication
51. Some animals define a territory and defend it. Maintaining a territory costs in terms of the energy expended in defense and puts the defender and the aggressor at risk of an injury that could cost their life. What benefits do organisms derive from keeping a territory? A. obtaining a mate and eliminates predators B. keeping a constant supply of food and encouraging other species to settle in the area C. encouraging other species to settle in the area and obtaining a mate D. obtaining a mate and keeping a constant supply of food E. defeating predators and encouraging other species to settle in the area The benefits of expending injury in defense of territory can be thought of as increasing the likelihood that the individuals will be able to reproduce and raise young. The territory must attract a mate, feed the defender, his mate(s) and offspring, and perhaps offer a safe place to raise the young. Females often choose their mate based on the ability of the male to provide food.
Bloom’s Level: 5. Evaluate Learning Outcome: 33.04.01: Discuss why territoriality evolved in certain animal groups. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
52. Why do different species of animals have territories that are different sizes? A. They must be able to travel the entire territory in one day to defend it. B. They only need to maintain a territory that is large enough for the male and one mate. C. Different species need different-sized territories depending on their food source. D. They must be able to sing loudly enough to be heard throughout the territory. E. They must be able to mark the entire territory every day so others of their species will not invade it. Territory is maintained in order to provide food and shelter for the male, his mate(s), and offspring. If the species is predatory, it requires more area to provide enough food. Small herbivores need less territory, while large herbivores need larger territories.
Bloom’s Level: 4. Analyze Learning Outcome: 33.04.01: Discuss why territoriality evolved in certain animal groups. Section: 33.04 Topic: Behavioral Ecology
53. In most species the _ invests most of the energy in raising the young and the chooses the mate. A. male; female B. female; male C. male; male D. female; female E. They both invest energy equally in raising the young and mate choice is random. The female generally invests more energy in bearing and raising the offspring so she usually is the one who chooses the male with whom she will mate.
Bloom’s Level: 2. Understand Learning Outcome: 33.04.02: Explain why females should be choosier than males in selecting a mate. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
54. Why is it an advantage for a female cardinal to choose a brightly colored male as her mate? A. A brightly colored male is less attractive to predators. B. Brightly colored males are able to camouflage the nest with their red feathers. C. Brightly colored males are healthier and more likely to produce healthy offspring. D. A brightly colored male sings the loudest. E. A brightly colored male will help her raise their offspring. Generally brightly colored feathers in a male signals health and vitality and by choosing a mate with these characteristics she is more likely to produce offspring that will be healthy.
Bloom’s Level: 4. Analyze Learning Outcome: 33.04.02: Explain why females should be choosier than males in selecting a mate. Section: 33.04 Topic: Behavioral Ecology
55. Why do females have a greater investment in offspring? A. Choosing a mate takes a great deal of the energy from the female. B. Females are the only ones who invest any energy in raising offspring. C. Females must fight off any predators. D. Females invest energy as the embryo develops. E. Females invest energy in fighting off other females to obtain a mate. The body of the female produces eggs and, in some species, supplies energy to the developing embryo and, in mammals, even supplies milk to the offspring once it is born.
Bloom’s Level: 4. Analyze Learning Outcome: 33.04.02: Explain why females should be choosier than males in selecting a mate. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
56. Which body system plays an important role in influencing behavior? A. endocrine B. cardiovascular C. respiratory D. digestive E. integumentary The endocrine system determines the production of hormones in the body. Hormones can then influence the behavior of the individual. All of the other systems play a role in the overall health of the individual but have not been determined to have an influence on the behavior of the individual.
Bloom’s Level: 1. Remember Learning Outcome: 33.01.01 Understand what is meant by the term nurture versus nature. Section: 33.01 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
57. Which of the following experiments would help support the idea that behavior can be influenced by the environment? A. Robins who were raised in the nest of sparrows sing the sparrow song when they are older. B. Laughing gull chicks increase their pecking efficiency when presented with a model of the parents bill. C. A baby goose hatches out and the first object it sees is a red ball. The goose will follow the ball around as it grows up. D. Young blue jays eat a monarch for the first time and become ill. The birds stop preying upon monarchs after the experience. E. All of the answer choices are correct. Robins who were raised in the nest of sparrows sing the sparrow song when they are older is an example of social interaction and learning. Laughing gull chicks increase their pecking efficiency when presented with a model of the parents bill is an example of learning. A baby goose hatches out and the first object it sees is a red ball. The goose will follow the ball around as it grows up because this is an example of imprinting. Young blue jays eat a monarch for the first time and become ill. The birds stop preying upon monarchs after the experience because this is associative learning. All of these types of learning are examples of how the environment can influence behavior.
Bloom’s Level: 4. Analyze Learning Outcome: 33.02.01: Identify an experiment that shows how behavior can be environmentally influenced. Section: 33.02 Topic: Behavioral Ecology
Short Answer Questions 58. List the four main methods of communication among animals. 1. chemical communication 2. auditory communication 3. visual communication 4. tactile communication
Bloom's Level: 6. Create Learning Outcome: 33.03.01: Summarize the various ways in which animals communicate. Section: 33.03 Topic: Animal Communication
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Chapter 33 - Animal Behavior
59. List and describe the two types of associative learning. 1. Classical conditioning is the presentation of two different types of stimuli at the same time that causes an animal to form an association between them. 2. Operant conditioning is when there is a stimulus-response connection that is strengthened through repeated exposure.
Bloom's Level: 6. Create Learning Outcome: 33.02.02: Compare three different types of learned behavior. Section: 33.02 Topic: Behavioral Ecology
Multiple Choice Questions 60. In gibbon society. which parent invests more energy into the success of the young? A. Both parents are equally involved in the success of the young. B. The female invests more energy into the success of the young. C. The male invests more energy into the success of the young. D. Neither parent invests in the success of the young. E. Initially, the female invests in the success of the young but the male takes over and ensures the success of the young. Gibbons are monogamous, meaning that they pair bond. Both parents will invest in the success of the young.
Bloom’s Level: 2. Understand Learning Outcome: 33.04.02: Explain why females should be choosier than males in selecting a mate. Section: 33.04 Topic: Behavioral Ecology
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Chapter 33 - Animal Behavior
61. Which of the following is a negative consequence of having a territory? A. energy expended to defend the territory B. access to food sources C. nesting sites for raising offspring D. protection from predators E. attract potential mates An increased energy expenditure to defend the territory is a disadvantage. All of the others are advantages of having a territory.
Bloom’s Level: 1. Remember Learning Outcome: 33.04.01: Discuss why territoriality evolved in certain animal groups. Section: 33.04 Topic: Behavioral Ecology
62. Which communication method can be effective both during the day and at night? A. chemical B. tactile C. visual D. auditory E. All of the answer choices are correct. All of these can be effective both during the day and at night. As long as the two individuals are in proximity to each other, tactile communication can work. Visual works when the individual has some form of light-producing mechanism (i.e., fireflies). Auditory works as long as the sound is species specific (i.e., mating song of frogs). Chemical works when the scent is left in a location that can be found by other individuals of the same species.
Bloom’s Level: 2. Understand Learning Outcome: 33.03.02: Describe the advantages and disadvantages of chemical, auditory, visual, and tactile communication. Section: 33.03 Topic: Animal Communication
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Chapter 33 - Animal Behavior
True / False Questions 63. In reciprocal altruism the helper "gains" a fitness benefit. TRUE It is true that in reciprocal altruism the helper "gains" a fitness benefit. Often the helper is increasing the success of their relatives; thus, some component of their genetics are being passed on. They may also inherit the territory of the parent in the future or receive help themselves.
Bloom’s Level: 1. Remember Learning Outcome: 33.04.03: Summarize the costs and benefits of living in a society. Section: 33.04 Topic: Behavioral Ecology
Short Answer Questions 64. Describe the classical conditioning experiment of Ivan Pavlov. Pavlov would ring a bell each time the dog was fed. Salivation would be recorded. Eventually the dog would salivate every time the bell was rung, even though no food was presented. Food is an unconditioned stimulus. The sound of the bell is a conditioned stimulus that brings about the response: salivation.
Bloom's Level: 6. Create Learning Outcome: 33.02.01: Identify an experiment that shows how behavior can be environmentally influenced. Section: 33.02 Topic: Behavioral Ecology
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Chapter 34 - Population Ecology
Chapter 34 Population Ecology
Short Answer Questions 1. List the features used to identify a species as an opportunistic species (r-strategist). small individuals short life span fast to mature many offspring little or no parental care of offspring many offspring die before reaching reproductive age early reproductive age
Bloom's Level: 6. Create Learning Outcome: 34.03.01 Identify the difference between density dependence and density independence. Section: 34.03 Topic: Community Interactions
2. Explain the correct sequence of events in secondary succession of a forest. During secondary succession there is already soil present. Assumably there will be seeds within the seed bank of the soil and grasses will grow. Next will come the low shrubs followed by the high shrubs. These will give way to the shrub/tree mix. The trees will crowd out the shrubs and there will be low trees. Over time these will mature into high trees and eventually a climax community.
Bloom's Level: 6. Create Learning Outcome: 34.04.02: Choose the correct sequence of events that occur during ecological succession. Section: 34.04 Topic: Ecological Succession
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Chapter 34 - Population Ecology
3. Explain the species interactions in each of the following symbiotic relationships: parasitism, commensalism, and mutualism. Parasitism: one species benefits while another is harmed. Commensalism: one species benefits while there is no effect upon the other. Mutualism: both species benefit from this relationship.
Bloom's Level: 6. Create Learning Outcome: 34.03.03 List the three types of symbiosis, and discuss the characteristics of each. Section: 34.03 Topic: Community Interactions
Multiple Choice Questions 4. Ecology is best defined as the study of A. populations. B. the rate of population changes. C. communities. D. how populations are restricted by environmental resistance. E. organisms as they interact with other organisms and with their physical surroundings. A population is a group of organisms of the same species living in the same place at the same time. A community is all of the species living in one place at the same time. Logistic growth is the term to describe how a population is restricted by environmental resistance.
Bloom’s Level: 1. Remember Learning Outcome: 34.01.01 Identify what aspects of biology the study of ecology encompasses. Section: 34.01 Topic: Ecosystem Ecology
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Chapter 34 - Population Ecology
5. An ecosystem contains A. only the biotic (living) components of the environment. B. only the abiotic (nonliving) components of the environment. C. only the energy flow components of an environment. D. both the living organisms and the abiotic components of the environment. E. only the food relationships found in an environment. An ecosystem describes all of the factors that affect an organism. This must include all of the living things that surround the organisms as well as the nonliving things that constitute the resources needed by the organisms.
Bloom’s Level: 1. Remember Learning Outcome: 34.01.01 Identify what aspects of biology the study of ecology encompasses. Section: 34.01 Topic: Ecosystem Ecology
6. All of the organisms living in your front yard represent a(n) represent a(n) . A. population; community B. population; ecosystem C. ecosystem; population D. community; population E. ecosystem; community
, while the blue jays
A community is all of the species living in one place at the same time, so all of the organisms living in your front yard would be a community. A population is a group of organisms of the same species living in the same place at the same time. All of the leafhoppers would be looking at a single species out of that group and that would be a population.
Bloom’s Level: 3. Apply Learning Outcome: 34.01.02 Recognize the different levels of ecological study. Section: 34.01 Topic: Ecosystem Ecology
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Chapter 34 - Population Ecology
7. The relatively thin layer on the earth's surface where life is possible is known as the A. ecosystem. B. biosphere. C. biome. D. biopreserve. E. lithosphere. The biosphere includes the soil of the earth, the waters, the air, and the surface; organisms are able to live in all of these places. An ecosystem refers only to one specific area of the biosphere. A biome is the term used to refer to a certain type of ecosystem; it has a particular set of organisms, climatic conditions, and soil conditions. The lithosphere is the outer portion of the surface of the earth; it includes the upper portion of the mantle and all of the earth’s crust.
Bloom’s Level: 1. Remember Learning Outcome: 34.01.01 Identify what aspects of biology the study of ecology encompasses. Section: 34.01 Topic: Ecosystem Ecology
8. The study of the interrelationships of plants and animals with each other and with their environment is known as A. a trophic level. B. a food web. C. a habitat. D. a biosphere. E. ecology. Trophic levels refer to the feeding relationships of organisms; they are based on how far they are from the original energy source, usually the sun. A food web describes the complex feeding relationships found in an ecosystem. A habitat describes the conditions under which an organism can survive and reproduce. The biosphere describes the entire area of the earth where organisms can exist; it includes the soil, the air, the surface of the earth, and the waters of the earth.
Bloom’s Level: 1. Remember Learning Outcome: 34.01.01 Identify what aspects of biology the study of ecology encompasses. Section: 34.01 Topic: Ecosystem Ecology
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Chapter 34 - Population Ecology
9. All the members of the same species that inhabit a particular area are termed a(n) A. ecosystem. B. biosphere. C. ecological niche. D. population. E. community. An ecosystem is all of the living organisms as well as the physical condition found in a particular area of the earth. The biosphere describes the entire area of the earth where organisms can exist; it includes the soil, the air, the surface of the earth, and the waters of the earth. An ecological niche describes the interactions of an organism with other organisms and with the physical environment. A population is a group of organisms of the same species living in the same place at the same time. A community is all of the species living in one place at the same time, so all of the organisms living in your front yard would be a community.
Bloom’s Level: 1. Remember Learning Outcome: 34.01.02 Recognize the different levels of ecological study. Section: 34.01 Topic: Population Ecology
10. A number of populations of different species interacting with one another is called A. competition. B. a community. C. an ecosystem. D. predation. E. symbiosis. Competition is a type of species interaction where both organisms are harmed because both need the same resource. A community is all of the species living in one place at the same time, so all of the organisms living in your front yard would be a community. An ecosystem is all of the living organisms as well as the physical condition found in a particular area of the earth. Predation is a type of species interaction where one organism is harmed and the other is benefited because one organism feeds on the other. Symbiosis is a type of species interaction where both organisms live in a close association with one another.
Bloom’s Level: 1. Remember Learning Outcome: 34.01.02 Recognize the different levels of ecological study. Section: 34.01 Topic: Community Ecology
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Chapter 34 - Population Ecology
11. A number of populations of different species interacting with one another in a natural environment, such as a rotten log, is called A. competition. B. a community. C. a biome. D. predation. E. symbiosis. Competition is a type of species interaction where both organisms are harmed because both need the same resource. A community is all of the species living in one place at the same time, so all of the organisms living in your front yard would be a community. A biome is the term used to refer to a certain type of ecosystem; it has a particular set of organisms, climatic conditions, and soil conditions. Predation is a type of species interaction where one organism is harmed and the other is benefited because one organism feeds on the other. Symbiosis is a type of species interaction where both organisms live in a close association with one another.
Bloom’s Level: 3. Apply Learning Outcome: 34.01.02 Recognize the different levels of ecological study. Section: 34.01 Topic: Community Ecology
12. Consider that a species of salmon lays 20,000 eggs per pair when it spawns and dies. At the end of five years, an average of one pair of mature salmon from this group of hatched eggs returns again to spawn in the parent stream (19,998 have died). What is the per capita rate of increase? A. 10,000, because there were that many eggs produced per parent fish (r) B. 2,000, because this must be divided by five years C. zero, because there is exact replacement of the previous generation D. -2,000, because there was this much average die-off per year E. -19,998, because there was this much total loss If only two mature salmon survive to reproduce per pair of parents then they have only replaced the original parents. It would require that the average survival of the offspring be greater than two for any increase in the population size to occur.
Bloom’s Level: 4. Analyze Learning Outcome: 34.02.01: Recognize how the rate of natural increase in a population is calculated. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
13. Population growth rate would be negative when A. birthrate is greater than death rate. B. death rate is greater than birthrate. C. all couples are married but average less than two children apiece. D. a country becomes poorer, because it is related to economic growth. E. better health care reduces the death rate and increases survivorship of newborns. In order for the population size to decline it would be necessary for there to be a decrease in the number of surviving individuals; in other words, the death rate must be higher than the birthrate. If the two are equal, then the population growth rate would be zero. If the birthrate was higher than the death rate, then population growth rate would be positive and the population would be increasing in size.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.01: Recognize how the rate of natural increase in a population is calculated. Section: 34.02 Topic: Population Growth Models
14. The maximum per capita rate of increase for a population that can occur under ideal conditions is the A. population growth. B. biotic potential. C. environmental resistance. D. carrying capacity. E. doubling time. Population growth describes the increase or decrease in the number of individuals in a population. The biotic potential of an organism is the maximum number of offspring that can be produced in ideal conditions; it depends on the number of offspring produced at one time and the number of times that a female can produce offspring during her lifetime. Environmental resistance accounts for the limitations placed on the growth of a population because of limited resources. The carrying capacity of an ecosystem determines the maximum number of organisms that can exist there for an indefinite period of time and is based on the availability of resources that are needed by that population. Doubling time refers to the amount of time it takes for a population to double in size.
Bloom’s Level: 1. Remember Learning Outcome: 34.02.01: Recognize how the rate of natural increase in a population is calculated. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
15. When the number of births exceeds the number of deaths, this results in A. population growth. B. biotic potential. C. environmental resistance. D. carrying capacity. E. doubling time. Population growth describes the increase or decrease in the number of individuals in a population. The biotic potential of an organism is the maximum number of offspring that can be produced in ideal conditions; it depends on the number of offspring produced at one time and the number of times that a female can produce offspring during her lifetime. Environmental resistance accounts for the limitations placed on the growth of a population because of limited resources. The carrying capacity of an ecosystem determines the maximum number of organisms that can exist there for an indefinite period of time and is based on the availability of resources that are needed by that population. Doubling time refers to the amount of time it takes for a population to double in size.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.01: Recognize how the rate of natural increase in a population is calculated. Section: 34.02 Topic: Population Growth Models
16. Which of the following would allow optimal population growth? A. accumulation of waste products B. predation C. competition D. limited access to food and water E. unlimited resources In order for the biotic potential to be reached it would be necessary for organisms to have all of the necessary resources in an unlimited supply. The other choices all limit access to some resource in some way.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
17. Which of the following does not impact an organism's biotic potential? A. number of offspring per reproduction B. chances of survival until age of reproduction C. age at which reproduction begins D. how often each individual reproduces E. the rate of immigration into an area The rate of immigration into an area would affect the overall size of the population but would not change the physical or physiological factors that determine the biotic potential of an organism.
Bloom’s Level: 5. Evaluate Learning Outcome: 34.02.01: Recognize how the rate of natural increase in a population is calculated. Section: 34.02 Topic: Population Growth Models
18. Which of the following is NOT a form of environmental resistance? A. limited food supply B. accumulation of waste products C. number of offspring produced per litter D. predation E. limited living space Environmental resistance accounts for the limitations placed on the growth of a population because of limited resources. The number of offspring per litter describes a physical characteristic of the species, not an aspect of the environment.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
19. A scientist observes a population of grasshoppers in a farmer's field several times over the summer. She notices that after each application of an insecticide the grasshopper population increases to its pre-pesticide levels very rapidly and that the population never goes near zero at any time. The scientist’s results indicate the grasshopper population is probably A. in the lag phase. B. in the exponential growth phase. C. in the deceleration phase. D. in the stable equilibrium phase. E. at carrying capacity. Exponential growth describes the growth pattern of a population that is increasing geometrically in an ideal environment with unlimited resources. With this growth pattern the population would recover very rapidly from any loss of members.
Bloom’s Level: 3. Apply Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
20. Biotic potential depends on all of the following EXCEPT A. the number of offspring per reproduction. B. how often each individual reproduces. C. how many different mates each individual has. D. chances of survival until the age of reproduction. E. the age at which reproduction begins. The biotic potential of an organism is the maximum number of offspring that can be produced in ideal conditions; it depends on the number of offspring produced at one time and the number of times that a female can produce offspring during her lifetime. The number of mates that an organism has would not affect any of the physical or physiological characteristics of the organism and would not affect its biotic potential.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.01: Recognize how the rate of natural increase in a population is calculated. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
21. When growth proceeds at a rate similar to 2 → 4 → 8 → 16 → 32 → 64...etc., it is called A. arithmetic growth. B. additive growth. C. exponential growth. D. zero population growth. E. carrying capacity. Exponential growth describes the growth pattern of a population that is increasing geometrically in an ideal environment with unlimited resources. The growth pattern 2 → 4 → 8 → 16 → 32 → 64...etc. would be a numeric description of the geometric pattern.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
22. A population with rapidly expanding exponential growth would be best represented by a graph with a(n) -shaped curve. A. bell B. urn C. pyramid D. S E. J A J-shaped curve would fit the geometric pattern seen in exponential growth. The x-axis represents time and the y-axis represents the number of individuals; as the population size increases more and more rapidly during each time period, the line of population growth would become steeper and form the upright portion of the J.
Bloom’s Level: 1. Remember Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
23. The human growth curve is best described as being A. flat. B. J-shaped. C. V-shaped. D. S-shaped. E. W-shaped. The human population is undergoing exponential growth so the graph is a J-shaped curve.
Bloom’s Level: 3. Apply Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Human Population Growth
24. The carrying capacity of the environment for a species is determined by A. the reproductive rate of the organism. B. the number of organisms of that species. C. the state and national wildlife laws pertaining to that species. D. a complex "balance of nature" that still has not been explained in terms that scientists can calculate. E. the limited productivity of the environment and the environmental resistance to the biotic potential of the organism. The carrying capacity of an ecosystem determines the maximum number of organisms that can exist there for an indefinite period of time and is based on the availability of resources that are needed by that population. The productivity of the environment would determine the food available to the organisms and would therefore affect the carrying capacity. Environmental resistance refers to resources and would account for any non-food resources that were needed by the population.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
25. Population size is believed to level off at the A. population growth B. biotic potential C. environmental resistance D. carrying capacity E. doubling time
of an environment.
As a population responds to environmental resistance, the population usually stabilizes around the carrying capacity of that population. Population growth describes the increase or decrease in the number of individuals in a population. The biotic potential of an organism is the maximum number of offspring that can be produced in ideal conditions; it depends on the number of offspring produced at one time and the number of times that a female can produce offspring during her lifetime. Environmental resistance accounts for the limitations placed on the growth of a population because of limited resources. The carrying capacity of an ecosystem determines the maximum number of organisms that can exist there for an indefinite period of time and is based on the availability of resources that are needed by that population. Doubling time refers to the amount of time it takes for a population to double in size.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
26. The maximum population that the environment can support for an indefinite period of time is called the A. biotic potential. B. environmental resistance. C. carrying capacity. D. replacement reproduction. E. demographic transition. The biotic potential of an organism is the maximum number of offspring that can be produced in ideal conditions; it depends on the number of offspring produced at one time and the number of times that a female can produce offspring during her lifetime. Environmental resistance accounts for the limitations placed on the growth of a population because of limited resources. The carrying capacity of an ecosystem determines the maximum number of organisms that can exist there for an indefinite period of time and is based on the availability of resources that are needed by that population. Replacement reproduction refers to a population where females have only two children. This does not mean the same thing as zero population growth since a larger number of reproductive females entering the reproductive base than older females leaving would lead to a positive growth rate. The demographic transition refers to the change seen in the growth rate of human populations as the socioeconomic status of the population changes.
Bloom’s Level: 1. Remember Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
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Chapter 34 - Population Ecology
27. In Figure 34.03, the carrying capacity of the environment occurs at number A. 1. B. 2. C. 3. D. 4. E. 5. F. 6. Carrying capacity = 1; environmental resistance = 2; stable equilibrium = 3; deceleration = 4; exponential growth = 5; and lag = 6.
Bloom’s Level: 3. Apply Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
28. At what point of Figure 34.03, does the population exhibit biotic potential? A. 1 B. 2 C. 3 D. 4 E. 5 F. 6 Carrying capacity = 1; environmental resistance = 2; stable equilibrium = 3; deceleration = 4; exponential growth = 5; and lag = 6. A population experiencing biotic potential would be in exponential growth.
Bloom’s Level: 3. Apply Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
29. In Figure 34.03, where does the effect of environmental resistance show on the graph? A. 1 B. 2 C. 3 D. 4 E. 5 F. 6 Carrying capacity = 1; environmental resistance = 2; stable equilibrium = 3; deceleration = 4; exponential growth = 5; and lag = 6.
Bloom’s Level: 3. Apply Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
30. Limiting factors of food and space will first be observed in Figure 34.03 at number A. 1. B. 2. C. 3. D. 4. E. 5. F. 6. Carrying capacity = 1; environmental resistance = 2; stable equilibrium = 3; deceleration = 4; exponential growth = 5; and lag = 6. The effects of limited resources such as food and space would affect the population's growth rate and would be an expression of environmental resistance.
Bloom’s Level: 3. Apply Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
31. Which of the following describes a type I survivorship curve (drawn by plotting the number of individuals in a given population alive at the beginning of each age interval)? A. Most individuals die of old age. B. Many individuals die early in life. C. Individuals die at a constant rate throughout time. D. Most individuals die during their reproductive years. E. Most individuals die before old age, but after their reproductive years. In a type I survivorship curve, most individuals exist through the expected life span and then die during old age; so, individuals usually die during old age. Examples are humans and elephants.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
32. Which of the following describes a type II survivorship curve (drawn by plotting the number of individuals in a given population alive at the beginning of each age interval)? A. Most individuals die of old age. B. Many individuals die early in life. C. Individuals die at a constant rate throughout time. D. Most individuals die during their reproductive years. E. Most individuals die before old age, but after their reproductive years. In a type II survivorship curve, individuals are lost throughout all ages and death is expected to occur at the same frequency at any age. Examples are songbirds and squirrels.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
33. Which of the following describes a type III survivorship curve (drawn by plotting the number of individuals in a given population alive at the beginning of each age interval)? A. Most individuals die of old age. B. Many individuals die early in life. C. Individuals die at a constant rate throughout time. D. Most individuals die during their reproductive years. E. Most individuals die before old age, but after their reproductive years. In a type III survivorship curve, most individuals die at an early age but those who do reach adulthood live a normal life span. Examples are oysters and frogs.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
34. Consider the life of the praying mantis. The large predatory female lays several hundred eggs in a foam mass in the fall. The young are most vulnerable when they emerge in the spring, but the few that survive spread out over the countryside and, if they find a mate, lay eggs the following fall. Which type of survivorship curve does this represent? A. type I B. type II C. type III D. exponential growth followed by a decline from resource depletion E. maximal exponential growth and minimal use of carrying capacity In a type III survivorship curve, most individuals die at an early age, but those who do reach adulthood live a normal life span. In a type I survivorship curve, most individuals exist through the expected life span and then die during old age; so, individuals usually die during old age. In a type II survivorship curve, individuals are lost throughout all ages and death is expected to occur at the same frequency at any age.
Bloom’s Level: 3. Apply Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
35. If the growth rate of a population increases, the doubling time A. increases. B. decreases. C. remains the same. D. depends on the amount of natural resources. E. is not related to growth rate increase. As the growth rate of a population increases, the doubling time decreases. An increase in the amount of time it took the population to double would indicate that the population was growing more and more slowly.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.02: Compare exponential and logistic population growth curves. Section: 34.02 Topic: Population Growth Models
36. The demographic transition, where population growth slows in developed countries, is due to a(n) A. decreased death rate followed by a decreased birthrate. B. decreased death rate followed by increased birthrate. C. increased death rate followed by a decreased birthrate. D. increased death rate followed by increased birthrate. E. loss of a large number of members of the child-bearing age population. The demographic transition refers to the change seen in the growth rate of human populations as the socioeconomic status of the population changes. In less developed countries the economic base tends to be agrarian, where a large family is beneficial and the death rate is high, leading to a steady population size. As the country becomes developed the population becomes more urban; at first the birthrate stays high based on cultural norms but the death rate falls as conditions and medical care improves. The growth rate of the population during this time increases rapidly. Eventually, the pressure to have a smaller family increases as the benefits of the larger family are removed and the benefits of a small family increase. The population has increased significantly in size by this time and it stabilizes at this larger size.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
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Chapter 34 - Population Ecology
37. The demographic transition, where population growth slows in developed countries, may be attributed at least in part to A. socialization. B. sensitivity. C. an increase in family size. D. the rise of medicine. E. education. The demographic transition refers to the change seen in the growth rate of human populations as the socioeconomic status of the population changes. In less developed countries the economic base tends to be agrarian, where a large family is beneficial. As the country becomes developed, the population becomes more urban and having a large family costs the family more than it returns. Eventually, the pressure to have a smaller family increases as the benefits of the larger family are removed and the benefits of a small family increase. Parents choose to have fewer children so they can give the ones they do have more material advantages.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
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Chapter 34 - Population Ecology
38. If a population has many more young women entering the reproductive years than older women leaving the reproductive years, which of the following statements is NOT true? A. The growth rate will most likely increase. B. The replacement reproduction will give a positive growth rate. C. It is likely to be a MDC. D. It is likely to be a LDC. E. The population will continue to grow as more young women enter the reproductive phase. Since there are more young women of reproductive age it is likely that they will reproduce and increase the population size. Under these circumstances the population will continue to grow even if every woman only has two children, since the birthrate is positive. This age distribution is most commonly seen in less developed countries. In more developed countries it is more normal for there to be fewer young women coming of reproductive age since their mothers and grandmothers chose to have fewer children and the growth rate is close to zero.
Bloom’s Level: 3. Apply Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
39. Which of the following is NOT true? A. Many MDCs have a stabilized age structure diagram. B. Most LDCs have a youthful age structure diagram. C. The quicker the replacement reproduction is achieved, the sooner the zero population growth results. D. The human population growth curve is exponential. E. As the human population increases in size, less energy but more material will be consumed. As the human population increases in size, it is likely that more energy and more materials will be used as all individuals seek to attain the lifestyle of individuals in more developed countries.
Bloom’s Level: 3. Apply Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
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Chapter 34 - Population Ecology
40. Which of the following is a true statement? A. The growth rate can be positive and a population can still be the same size year after year. B. The growth rate has to be zero before a population can be the same size year after year. C. Replacement reproduction automatically gives a growth rate of zero. D. A growth rate of zero means the population is dying. E. The growth rate can be negative and a population can still be the same size year after year. If the growth rate is positive, then the population is increasing in size. Replacement reproduction refers to a population where females have only two children. This does not mean the same thing as zero population growth because a larger number of reproductive females entering the reproductive base than older females leaving would lead to a positive growth rate. A zero growth rate means that the population is staying the same size from year to year. If the growth rate, is negative then the population is decreasing in size.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
41. Age structure diagrams with a broad base suggest that A. the population will continue to grow for some time. B. environmental resistance is not at work. C. the biotic potential must be larger than usual. D. the individuals must be larger than usual. E. there are many older individuals in a population. An age structure diagram with a large base means that there are a large number of individuals entering reproductive age and therefore there is likely to be an increase in the size of the population.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
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Chapter 34 - Population Ecology
42. Which of the following statements accurately describes the less developed countries? A. high birthrate, large pre-reproductive portion of the population B. high birthrate, different age groups of equal size C. high growth rate, reproductive portion of the population is the largest D. low growth rate, different age groups of equal size E. high growth rate, different age groups of equal size In most less developed countries, the reproductive base is large and there is a high birthrate since many of those individuals contribute to the population.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
43. Replacement reproduction occurs when A. the number of births equals the number of deaths. B. each couple has an average of two children, which also results in zero population growth. C. each couple has an average of two children, but population growth may still occur. D. the population reaches carrying capacity. E. all of the adult population is married. Replacement reproduction refers to a population where females have only two children. This does not mean the same thing as zero population growth because a larger number of reproductive females entering the reproductive base than older females leaving would lead to a positive growth rate.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
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Chapter 34 - Population Ecology
44. Which of these is most likely true? A. The less developed countries will always have a high growth rate with no accompanying problems. B. The growth rate will decline only when the less developed countries are industrialized as much as the more developed countries. C. A good family planning program can help the growth rate decline even if the country is not yet industrialized. D. The less developed countries already have a low growth rate. E. The growth rate of more developed countries contributes more to world population growth. In many less developed countries the population growth rate will slow if women are more educated and are given a method to help them plan when they choose to have children. Typically it is lack of choice that leads to a woman having a large number of children.
Bloom’s Level: 4. Analyze Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
45. Replacement reproduction cannot bring about zero population growth unless A. all diseases such as cholera, typhus, and diphtheria are brought under control. B. there are many fatalities due to a world war. C. there are as many older women leaving the reproductive years as there are younger women entering them. D. there are more older women leaving the reproductive years than there are younger women entering them. E. there are fewer older women leaving the reproductive years than there are younger women entering them. In order to reach zero population growth, with each woman of reproductive age having an average of two children, it is necessary for as many women to be leaving their reproductive years as there are entering.
Bloom’s Level: 4. Analyze Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
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Chapter 34 - Population Ecology
46. A population age structure diagram in which the birthrate is high and the population is mainly young would be best represented by a graph with a(n) A. bell shape. B. urn shape. C. pyramid shape. D. S-shaped curve. E. J-shaped curve. A pyramid shape would show the large reproductive base with a young population. The broader base shows the relative numbers at young ages as compared to the more narrow top that has a smaller number of individuals.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
47. A population age structure diagram in which the birthrate is less than the death rate and the post-reproductive group is the largest would be best represented by a graph with a(n) A. bell shape. B. urn shape. C. pyramid shape. D. S-shaped curve. E. J-shaped curve. The urn shape with relatively straight sides would best depict a population where there is a stable number of individuals at each age group until the very oldest age groups which would show a decrease.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
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Chapter 34 - Population Ecology
48. Which of the following human populations would be expected to have the greatest increase in growth rate? A. A population with the majority of their population over 50 years of age. B. A population with its population distributed equally across all age ranges. C. A population with most of its population under the age of 18 years of age. D. A population with most of the population in middle age. E. A population with most of its population over the age of 18 years of age. A population with most of its members under the age of 18 would have more individuals entering reproductive age and would grow at a higher growth rate as those individuals reproduced even if they only had two children each.
Bloom’s Level: 2. Understand Learning Outcome: 34.02.03: Recognize how the proportion of individuals at varying reproductive stages determines a population's age distribution. Section: 34.02 Topic: Human Population Growth
49. Farmers sprayed leechi trees to suppress populations of scale insects. This also killed the populations of a predatory lacewing that controlled the numbers of scales. Soon the spraying did no good, and the damage to the leechi from the scales was greater than before spraying had occurred. Few predatory lacewings existed in the trees now because A. they, too, are insects so they are opportunistic species. B. both insects were equilibrium species and resistance was a matter of chance. C. the predator was an opportunistic species and was therefore more susceptible to the spray. D. the scale insect was an opportunistic species and the high number of young included a resistant strain; the predator was an equilibrium species and will take longer to produce a resistant variety. E. they had reached their carrying capacity and thus their numbers decreased due to environmental resistance. Opportunistic species produce large numbers of offspring that quickly mature to produce the next generation of offspring. They tend to survive disturbances to the ecosystem and repopulate the area. Equilibrium species tend to have more trouble in surviving when the ecosystem is disturbed since they have fewer young and invest more energy in their, and their offspring’s, survival.
Bloom’s Level: 5. Evaluate Learning Outcome: 34.03.01 Identify the difference between density dependence and density independence. Section: 34.03 Topic: Community Ecology
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Chapter 34 - Population Ecology
50. In nature, animals tend to be either an opportunistic or equilibrium species or somewhere along an intermediate point between the two. Why aren't there animals in nature that are both big, fierce, and highly reproductive so they leave many offspring? A. Most animals are both opportunistic and equilibrium species. B. There is no limitation; such an animal could evolve. C. It is structurally impossible for a large animal to have many young. D. You can "spend" your energy to be a large equilibrium species, to produce many offspring as an opportunistic species, or to be somewhere in between, but you can't spend the energy to be both large and highly reproductive. E. Large animals are incapable of producing large numbers of offspring. Equilibrium species invest most of their energy in ensuring the survival of a few large offspring. Opportunistic species spend most of their energy in producing large numbers of small offspring.
Bloom’s Level: 5. Evaluate Learning Outcome: 34.03.01 Identify the difference between density dependence and density independence. Section: 34.03 Topic: Community Ecology
51. Which is NOT a density-dependent factor? A. food supply B. weather C. shelter or nest space D. disease E. predators Density-dependent factors have a greater effect on the population when the number of individuals in the same area are greater. Density-independent factors have the same effect, regardless of the number of individuals making up the population in that area. Weather would act the same on a population, regardless of the number of individuals.
Bloom’s Level: 2. Understand Learning Outcome: 34.03.01 Identify the difference between density dependence and density independence. Section: 34.03 Topic: Population Ecology
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Chapter 34 - Population Ecology
52. The outbreak of a disease, such as the bubonic plague, that kills large numbers of people in a city is an example of A. a density-independent factor. B. a density-dependent factor. C. a commensalistic factor. D. a symbiotic relationship. E. predation. This is an example of a density-dependent factor, since the effects of a disease increase as individuals have greater contact with others to pass the disease.
Bloom’s Level: 2. Understand Learning Outcome: 34.03.01 Identify the difference between density dependence and density independence. Section: 34.03 Topic: Population Ecology
53. The role a species plays in the community is called its A. habitat. B. ecological niche. C. biotic role. D. abiotic role. E. ecology. A habitat describes the conditions under which an organism can survive and reproduce. An ecological niche describes the interactions of an organism with other organisms and with the physical environment. Ecology is the study of the interactions of organisms with one another and with their environment.
Bloom’s Level: 1. Remember Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Population Ecology
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Chapter 34 - Population Ecology
54. In an Asian rice paddy, carp eat decaying material from around the base of rice plants while a snail scrapes algae from the leaves, stems, and roots of the same plant. They can survive at the same time in the same rice paddy because they A. occupy the same habitat but use the resources differently. B. occupy the same habitat and use the same resources. C. use the same resources but occupy different habitats. D. occupy different habitats and use the resources differently. E. work according to the competitive exclusion principle. A habitat describes the conditions under which an organism can survive and reproduce. Resource partitioning refers to the division of resources so that organisms do not use them in the same way and therefore do not compete for the same resources. While these organisms are occupying the same habitat, they partition resources so they do not compete for food.
Bloom’s Level: 3. Apply Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Ecology
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Chapter 34 - Population Ecology
55. Several species of grain beetles can live on dry meal, obtaining water mostly as metabolic water. Many of these beetle species are grain pests that do considerable damage to stored grain. You set up a dozen jars of dry meal and introduce 50 individuals of each species to each jar, being careful to have half of each species from each sex. The food supply is sufficient to last for a year and the size is adequate so that wastes do not become toxic. Most likely, examination of the jars in six months will find A. a totally random variation in numbers of both beetles. B. only one species per jar, similar to the classic competitive exclusion principle experiment with paramecia. C. the same ratio of beetles as when you started, about half from each species. D. only dead beetles in all jars due to intense competition for the niche. E. half the jars with more of one species and the other half of the jars will have more of a second species of beetle. Since all species would be competing directly for the same food resource, the species that could make the most efficient use of the resource would likely out-compete the others and the other species would die out. This is the same as the results that were seen in the experiment with a paramecium.
Bloom’s Level: 4. Analyze Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
56. There are two organisms with overlapping ranges and filling a similar niche. We find that where their niches overlap, there is twice the competition for resources. This leads to A. speciation. B. hypervolume. C. competitive exclusion. D. total extinction of one species. E. a switch in habitat for one of the organisms. Because the two species compete directly for the same resource, it is very likely that one species would be excluded from the resource by the species that is more efficient at using the resource.
Bloom’s Level: 2. Understand Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
57. Which term describes the concept that no two species can have the same "job" in the community at the same time? A. competitive exclusion B. habitat C. density-dependent factor D. mimicry E. symbiosis Competitive exclusion happens when one species directly competes with another for a particular resource. If both species exist in the same habitat, the less efficient species will be excluded from the resource. This means that two species that need the same resource cannot exist in the same ecosystem.
Bloom’s Level: 1. Remember Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
58. Interaction between two species as both attempt to use the same environmental resources is A. competition. B. a community. C. an ecosystem. D. predation. E. symbiosis. Competition is a type of species interaction where both organisms are harmed because both need the same resource. A community is all of the species living in one place at the same time, so all of the organisms living in your front yard would be a community. An ecosystem is all of the living organisms as well as the physical condition found in a particular area of the earth. Predation is a type of species interaction where one organism is harmed and the other is benefited because one organism feeds on the other. Symbiosis is a type of species interaction where both organisms live in a close association with one another.
Bloom’s Level: 1. Remember Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
59. In large natural ecosystems, competition between two species over time will usually result in A. death of all the members of one species within a short time. B. equal numbers of each species persisting for a long time. C. each species occupying a slightly different niche. D. hybridization between the two species, resulting in a third species. E. alternating high numbers of each species as they constantly evolve mechanisms to outcompete each other. Since two species cannot occupy the same habitat and use the same resources equally for long periods of time, over time, the two species usually evolve to use the resource differently.
Bloom’s Level: 2. Understand Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
60. Similar species occupying slightly different niches in the same community is shown by A. the presence of carp in polluted waters in the United States. B. fluctuation of the snowshoe hare and lynx populations in Canada. C. competition between two species of barnacles in Scotland, resulting in one living in the intertidal zone and the other living below this species. D. removal of a starfish species from intertidal communities in Washington. E. killing dingoes in Australia or coyotes in the American West. The two different species of barnacles live in slightly different locations, thereby partitioning the resource of an attachment substrate and allowing both species to exist in the same habitat.
Bloom’s Level: 3. Apply Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
61. Interaction between two species in which one feeds on the other is A. competition. B. a community. C. an ecosystem. D. predation. E. symbiosis. Competition is a type of species interaction where both organisms are harmed because both need the same resource. A community is all of the species living in one place at the same time, so all of the organisms living in your front yard would be a community. An ecosystem is all of the living organisms as well as the physical condition found in a particular area of the earth. Predation is a type of species interaction where one organism is harmed and the other is benefited because one organism feeds on the other. Symbiosis is a type of species interaction where both organisms live in close association with one another.
Bloom’s Level: 1. Remember Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
62. Which of the following is correct when describing a predator-prey cycle? A. A decline in the number of predators causes a decline in the number of prey. B. A decline in the number of prey causes a decline in the number of predators. C. An increase in the number of predators triggers an increase in the number of prey. D. An up-and-down cycle will be seen for the prey animal. E. A seasonal die-off will occur without the other species present. Since the predators are dependent on the number of prey available for food, there is a decrease in the number of surviving predators when the number of prey declines.
Bloom’s Level: 2. Understand Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
63. Overall, the most scientifically correct viewpoint toward predators is A. predators help keep prey populations from overexploiting limited food resources. B. generally, ecosystems support more and healthier populations when the large carnivores are eliminated from the system. C. there is a high level of cruelty and indiscriminate killing among larger predators. D. when we eliminate predators that could harm us and our activities, we also improve conditions for other animal populations. E. predators are a neutral influence on prey populations and our activities merely substitute us for the prey we eliminate. Ecosystems have evolved as complex interactions between the organisms in the community. Predators and prey have one such complex relationship. The predators in a community act to limit the size of the prey organisms and the prey in turn limit the size of the predator population since the prey require them as a food source. If either species is removed, the ecosystem will suffer from the disruption.
Bloom’s Level: 3. Apply Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
64. Which of the following is NOT an anti-predator defense mechanism? A. Plant chemicals, including coffee and tea caffeine, make caterpillars sick or jittery. B. The large eyespots on a moth's wing are exposed abruptly to startle a hungry bird. C. Many trees, vines, and shrubs have stems lines with long thorns. D. Prairie dog towns always have several prairie "watchdogs" to alert them of approaching hawks and snakes. E. Many plants have brightly colored fruit and flowers. Brightly colored flowers and fruit serve to attract other organisms who feed on the fruit or flower and help with reproduction or distribution of offspring.
Bloom’s Level: 2. Understand Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
65. Consider the consequences of a harmless prey that evolves toward mimicry of a model that has an anti-predator defense-Batesian mimicry. What happens to the effectiveness of this defense as more and more prey in the population are harmless mimics? Then consider what happens when a species with bad taste evolves to resemble a model that stings (Müllerian mimicry). A. The number of Müllerian mimics is limited because they dilute the protection afforded by the warning color pattern, but the addition of more Batesian mimics only strengthens the protective mimicry complex. B. The number of Batesian mimics is limited because they dilute the protection afforded by the warning color pattern, but the addition of more Müllerian mimics only strengthens the protective mimicry complex. C. The number of both Batesian and Müllerian mimics is limited because both dilute the protection afforded by the warning color pattern. D. The number of both Batesian and Müllerian mimics is unlimited because they both reinforce the protection afforded by the warning color pattern. E. The number of Batesian mimics is unlimited, but the addition of more Müllerian mimics dilutes the protective mimicry complex. In the case of the harmless prey that evolves toward mimicry of a model that has antipredator defenses, or Batesian mimicry, as the number of prey items that do not have the defense increases the number of predators who encounter a prey item that cannot defend itself. With the increased frequency of a reward (food) for eating the mimic, the more frequently the predator will try the model and mimic for food. With Müllerian mimicry, both the model and the mimic have a defense mechanism and the predator is "punished" every time it attempts to prey on either so they will cease to attempt to prey on either.
Bloom’s Level: 5. Evaluate Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
66. The harmless orange-and-black viceroy butterfly closely resembling the toxic orangeand-black monarch butterfly is an example of A. Müllerian mimicry. B. Batesian mimicry. C. both Müllerian and Batesian mimicry. D. a case of resource partitioning. E. a case of competitive exclusion. This is an example of a harmless organism mimicking one that has an anti-predator defense, or Batesian mimicry.
Bloom’s Level: 2. Understand Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
67. Black-and-yellow stripes are a common warning coloration. Among a set of insects that possess black-and-yellow patterns are: -female wasps that sting -male wasps that do not sting -a tasty, harmless, and nutritious leaf beetle -an acrid blister beetle that makes you sick if you eat it -a harmless hover fly that sounds and looks like a bee This assortment represents A. all organisms that share black-and-yellow because it evolved in a common ancestor. B. all cases of Batesian mimicry. C. all cases of Müllerian mimicry. D. a combination of cases including both Müllerian and Batesian mimicry. E. warning coloration only in the cases of the female wasp and blister beetle; the other cases are not mimicry. Since there are both harmless insects and ones with anti-predator defenses then both Batesian mimicry and Müllerian mimicry are shown. Batesian mimicry is seen with the male and female wasps, one of which stings and the other of which does not. Müllerian mimicry is seen with the female wasp that stings and the acrid blister beetle that is poisonous.
Bloom’s Level: 3. Apply Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
68. An intimate relationship between two different species is called A. competition. B. a community. C. an ecosystem. D. predation. E. symbiosis. Competition is a type of species interaction where both organisms are harmed because both need the same resource. A community is all of the species living in one place at the same time, so all of the organisms living in your front yard would be a community. An ecosystem is all of the living organisms as well as the physical condition found in a particular area of the earth. Predation is a type of species interaction where one organism is harmed and the other is benefited because one organism feeds on the other. Symbiosis is a type of species interaction where both organisms live in close association with one another.
Bloom’s Level: 1. Remember Learning Outcome: 34.03.03 List the three types of symbiosis, and discuss the characteristics of each. Section: 34.03 Topic: Community Interactions
69. Which statement about parasitism is true? A. The host is generally smaller than the parasite. B. Parasites are always severe and kill the host. C. All parasites require a single host for their life cycle. D. There are examples of parasites in every kingdom of life. E. All parasites will remain attached to their host for the entire life span of the host. Parasites tend to be much smaller than the host organism. They seldom kill their hosts because this would limit the time that they could use a host. It is more beneficial for the parasite if the host can be used for longer periods of time. Some parasites will use a primary host and a secondary host in order to complete their life cycle. Other parasites will use the host only long enough for them to obtain nutrients and reproduce. They don't all stay on the host for the entire life of the host.
Bloom’s Level: 2. Understand Learning Outcome: 34.03.03 List the three types of symbiosis, and discuss the characteristics of each. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
70. Parasites are found A. only as viruses, bacteria, and protists. B. only as viruses, bacteria, protists, and animals; no plants are parasites because they cannot move. C. only on host animals. D. only on one host where they can complete their life cycle and disperse to others of that same species. E. in all kingdoms, and sometimes one parasite uses more than one host to disperse. Parasites can be found in all kingdoms: Bacteria have strep infections; protists have malaria; fungi have athlete's foot; plants have mistletoe; and animals have tapeworms. Many parasites use more than one host. Frequently, one is the primary host where the parasite lives most of its life and the other is the secondary host that serves to transmit the parasite to a new primary host.
Bloom’s Level: 2. Understand Learning Outcome: 34.03.03 List the three types of symbiosis, and discuss the characteristics of each. Section: 34.03 Topic: Community Interactions
71. Which is NOT a parasitic relationship? A. The Dutch elm disease fungus infects and kills elm trees. B. The yellow fever virus causes fever and sometimes kills humans. C. A female tick fastens to the skin of a human host and draws blood for nutrition to make her eggs. D. The trichina worm burrows into muscle tissue, forms a cyst, and waits to continue its cycle until a predator consumes the host. E. Unsightly dandelions sprout in a yard and compete with the fescue grass for sunlight. In a parasitic relationship one organism benefits and the other is harmed by the interaction and the parasite lives either in or on the host. When dandelions grow among fescue grass and compete with the grass for resources both organisms are harmed by the competition and neither organism lives in or on the other.
Bloom’s Level: 4. Analyze Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
72. A form of symbiosis in which one participant benefits and the other apparently is not benefited nor harmed is A. commensalism. B. parasitism. C. mutualism. D. coevolution. E. symbiosis. In commensalism, organisms interact so that one organism benefits from the interaction while the interaction is neutral for the other organism. In a parasitic relationship, one organism benefits and the other is harmed by the interaction and the parasite lives either in or on the host. In a mutualistic relationship, both organisms benefit from a mutual exploitation of the other. Coevolution refers to a close interaction between different species so that the two evolve jointly and one influences the other in evolutionary terms. Symbiosis is a type of species interaction where both organisms live in close association with one another.
Bloom’s Level: 1. Remember Learning Outcome: 34.03.03 List the three types of symbiosis, and discuss the characteristics of each. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
73. A small marine amphipod has recently been discovered that carries another organism on its back. The "backpack" organism tastes bad. If a fish ingests the pair, it immediately spits them back out. If the amphipod is alone, however, it is readily eaten and swallowed. There is no apparent benefit in this relationship for the "backpack" organism, which appears to be neither helped nor harmed. This is therefore a case of A. mutualism. B. parasitism. C. commensalism. D. competitive exclusion. E. predation. In a mutualistic relationship, both organisms benefit from a mutual exploitation of the other. In a parasitic relationship, one organism benefits and the other is harmed by the interaction and the parasite lives either in or on the host. In commensalism, organisms interact so that one organism benefits from the interaction while the interaction is neutral for the other organism. Competitive exclusion happens when one species directly competes with another for a particular resource; if both species exist in the same habitat, the less efficient species will be excluded from the resource. Predation is a type of species interaction where one organism is harmed and the other is benefited because one organism feeds on the other.
Bloom’s Level: 3. Apply Learning Outcome: 34.03.03 List the three types of symbiosis, and discuss the characteristics of each. Section: 34.03 Topic: Community Interactions
74. An example of commensalism is A. mycorrhizal fungal roots on the roots of plants. B. ants living on the bullhorn acacia tree. C. termites with protozoa in their digestive tracts. D. flowering plants and their pollinators. E. clown fishes in sea anemones. In commensalism, organisms interact so that one organism benefits from the interaction while the interaction is neutral for the other organism. Clown fish benefit because they are able to escape from predators while hiding among the tentacles of the sea anemones. The sea anemone does not seem to benefit or be harmed by the presence of the clown fish.
Bloom’s Level: 3. Apply Learning Outcome: 34.03.03 List the three types of symbiosis, and discuss the characteristics of each. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
75. A form of symbiosis in which both participants benefit is A. commensalism. B. parasitism. C. mutualism. D. predation. E. competition. In commensalism, organisms interact so that one organism benefits from the interaction while the interaction is neutral for the other organism. In a parasitic relationship, one organism benefits and the other is harmed by the interaction and the parasite lives either in or on the host. In a mutualistic relationship, both organisms benefit; this is a mutual exploitation of each other. Predation is a type of species interaction where one organism is harmed and the other is benefited because one organism feeds on the other. Competition is a type of species interaction where both organisms are harmed because both need the same resource.
Bloom’s Level: 1. Remember Learning Outcome: 34.03.02 Describe how competition and predation can impact population growth. Section: 34.03 Topic: Community Interactions
76. You are on vacation in the Canadian Rocky Mountains and you observe that a massive forest fire has burned all the trees off the side of the mountain and sterilized the soil, leaving no topsoil. The type of succession that occurs following this type of fire is A. secondary succession. B. tertiary succession. C. primary succession. D. primary soil leaching. E. facilitated succession. Primary succession is when a community of plants is established in a new environment where there is not soil. A receding glacier would expose bedrock and this would lead to primary succession. Secondary succession is when a community of plants reestablishes in an area where the ecosystem has been disturbed. A plowed field or a burned area would undergo secondary succession. A facilitated model of succession occurs when each successive community improves conditions in the area for the community that follows. A primary species is one that is able to move and colonize an area that has been disturbed.
Bloom’s Level: 4. Analyze Learning Outcome: 34.04.01: Compare the various types of succession. Section: 34.04 Topic: Ecological Succession
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Chapter 34 - Population Ecology
77. Which would be the climax community in a biome that receives a large amount of rainfall? A. plants with high tolerance to full sun, limited water, limited nutrients B. the pioneer community C. small native plants of the area such as grasses and herbs D. large plants such as shrubs and trees E. lichens, mosses, and ferns In an area with sufficient moisture, the climax community is typically characterized by large complex plants such as trees and shrubs. It is only in areas where there is insufficient water to support these plant forms that smaller plants or pioneer species would be the climax vegetation.
Bloom’s Level: 3. Apply Learning Outcome: 34.04.01: Compare the various types of succession. Section: 34.04 Topic: Ecological Succession
78. The idea that plants cannot grow on a particular area until the soil has been developed enough by an earlier community is the A. climax-pattern model. B. facilitation model. C. inhibition model. D. tolerance model. E. soil development model. A facilitated model of succession occurs when each successive community improves conditions in the area for the community that follows. The improvement usually takes the form of increased nutrients in the soil.
Bloom’s Level: 2. Understand Learning Outcome: 34.04.01: Compare the various types of succession. Section: 34.04 Topic: Ecological Succession
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Chapter 34 - Population Ecology
79. Primary succession takes much longer than secondary succession because it involves A. colonization by more equilibrium species. B. time for weathering of rock and development of the soil horizons. C. more time for development of a seed bank. D. colonization by organisms that are farther away. E. redevelopment of the atmospheric gases. In primary succession, the community starts on ground with no soil present. Since soil must develop for species with more extensive root systems to establish themselves in the area, it takes a longer time to reach the stage where there are shrubs and trees present.
Bloom’s Level: 3. Apply Learning Outcome: 34.04.02: Choose the correct sequence of events that occur during ecological succession. Section: 34.04 Topic: Ecological Succession
80. In a population of 150 organisms over a six-month period there are 75 births and 30 deaths. Calculate the growth rate for those six months. A. 0.7 B. 0.5 C. 0.2 D. 0.3 E. 0.75 Number of births – number of deaths/total number in population = growth rate (75 – 30)/150 = 0.3.
Bloom’s Level: 4. Analyze Learning Outcome: 34.02.01: Recognize how the rate of natural increase in a population is calculated. Section: 34.02 Topic: Population Growth Models
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Chapter 34 - Population Ecology
81. In a population of 10,000 organisms over a one-year period there are 750 births and 130 deaths. Calculate the growth rate for that year. A. 0.062 B. 0.055 C. 0.088 D. 0.013 E. 0.075 Number of births – number of deaths/total number in population = growth rate (750 – 130)/10,000 = 0.062
Bloom’s Level: 4. Analyze Learning Outcome: 34.02.01: Recognize how the rate of natural increase in a population is calculated. Section: 34.02 Topic: Population Growth Models
82. A population of 100 butterflies living on an acre of land loses three-quarters of its members when a sudden freeze in the spring occurs just after they emerge as caterpillars. This population has undergone a reduction in population size due to A. an intrinsic factor. B. a density-independent factor. C. a density-dependent factor. D. natural selection. E. an opportunistic control. A sudden freeze would act the same on the population whether there were 100 organisms per acre or 50, so this is a density-independent control.
Bloom’s Level: 3. Apply Learning Outcome: 34.03.01 Identify the difference between density dependence and density independence. Section: 34.03 Topic: Population Ecology
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Chapter 34 - Population Ecology
83. A population of 1,000 mice living in a barn loses one-half of its members when a sudden virus spreads through the population. This population has undergone a reduction in population size due to A. an intrinsic factor. B. a density-independent factor. C. a density-dependent factor. D. natural selection. E. an opportunistic control. Disease is a density-dependent factor and it acts more heavily on a population that is more crowded over one that is less crowded. If the population had contained 500 individuals rather than 1,000 it might not have had losses that were as severe.
Bloom’s Level: 3. Apply Learning Outcome: 34.03.01 Identify the difference between density dependence and density independence. Section: 34.03 Topic: Population Ecology
84. In a pine forest, few other species of plants grow under the pines because the needles of pine trees are acidic and make the soil more acidic. This is an example of A. the facilitation model. B. the tolerance model. C. the inhibition model. D. an opportunistic model. E. the climax model. Since the pine trees are changing the soil to be less hospitable for new successive species, this is an example of the inhibition model.
Bloom’s Level: 3. Apply Learning Outcome: 34.04.01: Compare the various types of succession. Section: 34.04 Topic: Ecological Succession
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Chapter 34 - Population Ecology
85. One model proposed for succession predicts that different types of plants can all colonize an area at the same time and it is chance alone that determines which one arrives first and begins the process of succession. This model is called A. the facilitation model. B. the tolerance model. C. the inhibition model. D. an opportunistic model. E. the climax model. A facilitated model of succession occurs when each successive community improves conditions in the area for the community that follows. An inhibition model of succession occurs when a species growing in the area makes it more difficult for later species to follow. The tolerance model of succession predicts that different types of plants can all colonize an area at the same time and it is chance alone that determines which one arrives first and begins the process of succession. The climax model states that the same area will always lead to a specific type of community if it is given a sufficient amount of time.
Bloom’s Level: 2. Understand Learning Outcome: 34.04.01: Compare the various types of succession. Section: 34.04 Topic: Ecological Succession
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Chapter 34 - Population Ecology
86. There are two plowed fields that are each one acre and one mile apart with a wood surrounding one and the other surrounded by a pasture; they have the same soil composition and climate. In the first field, as succession begins, the pioneer species are pine trees and shrubs. In the second field, the pioneer species are weeds and shrubs. This provides evidence for which model of succession? A. the facilitation model B. the tolerance model C. the inhibition model D. an opportunistic model E. the climax model It seems that the surrounding vegetation determines the species that reach the area and this supports the tolerance model. A facilitated model of succession occurs when each successive community improves conditions in the area for the community that follows. An inhibition model of succession occurs when a species growing in the area makes it more difficult for later species to follow. The tolerance model of succession predicts that different types of plants can all colonize an area at the same time and it is chance alone that determines which one arrives first and begins the process of succession. The climax model states that the same area will always lead to a specific type of community if it is given a sufficient amount of time.
Bloom’s Level: 5. Evaluate Learning Outcome: 34.04.02: Choose the correct sequence of events that occur during ecological succession. Section: 34.04 Topic: Ecological Succession
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Chapter 34 - Population Ecology
87. The climax pattern model states that the same area will always lead to a specific type of community if it is given a sufficient amount of time. Which of the following statements is true of this model? A. In recent time this has been modified to recognize that while the area may return to the same biome it may not have exactly the same species diversity. B. An area that has been disturbed must always return to the same climax community if given enough time. C. It is necessary for an area to return to the climax community before it is disturbed again or the area will undergo primary succession on a second disturbance. D. The area must undergo primary and then secondary succession before it can reach the climax community. E. In recent time this model has been changed to state that the climax community cycles through a series of different possibilities. The climax pattern model gives the expected community for the area based on the type of biome that was originally present; however, there can be differences in the exact species found in the new ecosystem.
Bloom’s Level: 5. Evaluate Learning Outcome: 34.04.02: Choose the correct sequence of events that occur during ecological succession. Section: 34.04 Topic: Ecological Succession
True / False Questions 88. During mutualism, one species benefits while the other is harmed. FALSE During parasitism one species benefits while the other is harmed. During mutualism both species benefit.
Bloom’s Level: 1. Remember Learning Outcome: 34.03.03 List the three types of symbiosis, and discuss the characteristics of each. Section: 34.03 Topic: Community Interactions
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Chapter 34 - Population Ecology
89. Secondary succession will start with bare ground that is lacking any type of soil. FALSE Primary succession starts with bare ground that is lacking any type of soil. Secondary succession starts with soil.
Bloom’s Level: 1. Remember Learning Outcome: 34.04.02: Choose the correct sequence of events that occur during ecological succession. Section: 34.04 Topic: Ecological Succession
90. At the basic level an ecologist will study how an organism is adapted to its environment. TRUE At the most basic level, ecologists study how organisms are adapted to their environment.
Bloom’s Level: 1. Remember Learning Outcome: 34.01.01 Identify what aspects of biology the study of ecology encompasses. Section: 34.01 Topic: Population Ecology
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Chapter 35 - Nature of Ecosystems
Chapter 35 Nature of Ecosystems
Multiple Choice Questions 1. The is made of a community of organisms, plus its physical environment. A. food chain B. habitat C. niche D. ecosystem E. trophic level A food chain refers to a sequence of feeding levels; it is a one-time snapshot of feeding and does not express the true complexity of trophic levels as does a food web. A habitat describes the conditions under which an organism can survive and reproduce. The term ecosystem refers to both the living organisms and the nonliving physical components of a specified area. An ecological niche describes the interactions of an organism with other organisms and with the physical environment. The term trophic level refers to the feeding levels in an ecosystem. The levels are based on the distance from the original energy input.
Bloom’s Level: 1. Remember Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
2. A disruption in which biogeochemical cycle will lead to cultural eutrophication? A. phosphorus B. carbon C. nitrogen D. hydrological E. All of the answer choices are correct. Cultural eutrophication is the over-enrichment of a body of water that can lead to algal blooms. This process is often due to a large increase in the nutrient runoff that enters the body of water. This is often due to a disruption in the phosphorus cycle.
Bloom’s Level: 1. Remember Learning Outcome: 35.03.03: Identify how human activities can alter each of the biogeochemical cycles. Section: 35.03 Topic: Energy and Chemical Cycling
3. In an experiment carried out to study the energy flow through an ecosystem, scientists measured the total solar energy captured by plants to be 10,000 kilocalories. The portion of this available to the next trophic level will be A. 10,000 kilocalories. B. 1,000 kilocalories. C. 100 kilocalories. D. 10 kilocalories. E. 1 kilocalorie. Since the plants will use all of the energy except approximately 10%, the amount of energy available to the next level will be 1,000 kilocalories.
Bloom’s Level: 4. Analyze Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
4. An ecosystem consists of what two components? A. grazing and detritus food chains B. habitat and niches C. biotic and abiotic factors D. trophic and eutrophic factors E. water and temperature The term ecosystem refers to both the living organisms (biotic) and the nonliving (abiotic) physical components of a specified area.
Bloom’s Level: 1. Remember Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
5. Which of the following is a biotic factor in an ecosystem? A. the food chain B. inorganic nutrients C. water availability D. temperature E. sunlight A biotic factor is a living organism. The only living component listed above is the food chain, which refers to a sequence of feeding levels; it is a one-time snapshot of feeding and does not express the true complexity of trophic levels as does a food web.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
6. Deposits of coal are important in the carbon cycle because they A. keep humans from burning more trees by serving as a fuel source. B. serve as a reservoir of carbon that will eventually be released into the ecosystem. C. lock up carbon so that it can never be returned to the atmosphere to make global warming increase. D. make carbon readily available to plants through their roots growing down into the coal deposit. E. serve as short-term storage for carbons. Coal deposits serve as a long-term storage for carbon that will eventually release the carbon when weathering has released the deposit.
Bloom’s Level: 3. Apply Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.02 Topic: Energy and Chemical Cycling
7. The primary producers in terrestrial ecosystems are the saltwater ecosystems it is . A. autotrophs; heterotrophs B. herbivore; carnivore C. consumers; producers D. algae; green plants E. green plants; algae
, whereas in freshwater and
Primary producers are organisms that use inorganic components to manufacture organic molecules and in a terrestrial ecosystem are green plants. Their counterparts in aquatic ecosystems are photosynthetic algae.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
8. The primary consumers that feed directly on green plants are called A. omnivores. B. decomposers. C. herbivores. D. carnivores. E. producers. Omnivores are animals that are able to feed at multiple trophic levels and consume both plants and animals equally. Decomposers are organisms that feed on organic matter by digesting it outside their bodies and taking in the needed nutrients and energy. They serve to recycle nutrients in an ecosystem because any nutrients not taken in by the decomposer are available to plants in the ecosystem. Herbivores are animals that feed primarily on plant matter. Their digestive systems are adapted to digesting this type of food. Carnivores are animals that feed primarily on other animals (meat). Their digestive systems are adapted to digesting this type of food. Producers are organisms that use inorganic components to manufacture organic molecules and therefore produce the food available in the ecosystem.
Bloom’s Level: 1. Remember Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
9. The three locations where a nutrient may be found in an ecosystem are in order of A. nutrient/producer, reservoir/long-term storage, and organism/short-term storage. B. exchange pool/long-term storage, reservoir/nutrients for producers, and organisms/shortterm storage. C. exchange pool/short-term storage, reservoir/nutrients for producers, and organisms/longterm storage. D. exchange pool/short-term storage, reservoir/long-term storage, and organisms/nutrients available to producers. E. exchange pool/long-term storage, reservoir/short-term storage, and organisms/nutrients available to producers. The exchange pool keeps the necessary elements available to producers while storing them either long term or short term, depending on the needs of the producers. Reservoirs serve as long-term storage but typically not in a form that is usable by producers. Living organisms also serve to store nutrients, typically short term, while the organism is living; upon the death of the organism, the nutrient is released back into the environment through the action of decomposers.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.01: Define what is meant by a biogeochemical cycle. Section: 35.03 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
10. Which kind of organism would be most likely to perform photosynthesis? A. omnivore B. herbivore C. decomposer D. autotroph E. carnivore Omnivores are animals that are able to feed at multiple trophic levels and consume both plants and animals equally. Herbivores are animals that feed primarily on plant matter. Their digestive systems are adapted to digesting this type of food. Decomposers are organisms that feed on organic matter by digesting it outside their bodies and taking in the needed nutrients and energy. They serve to recycle nutrients in an ecosystem because any nutrients not taken in by the decomposer are available to plants in the ecosystem. Autotrophs are producers: they use inorganic materials (carbon dioxide, water, and solar energy) to manufacture organic molecules and are the basis for the food webs. Carnivores are animals that feed primarily on other animals (meat). Their digestive systems are adapted to digesting this type of food.
Bloom’s Level: 1. Remember Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
11. A carbon atom will cycle between an organism and the atmosphere when it is taken up by a producer and is then eaten by an herbivore and eventually passed back into the atmosphere through respiration. The carbon atom has passed through which portions of the nutrient cycle? A. exchange pool to reservoir to exchange pool to reservoir B. living organism to long-term storage to living organism to long-term storage C. living organism to reservoir to living organism to exchange pool D. living organism to exchange pool to living organism to reservoir E. living organism to exchange pool to living organism to exchange pool The carbon atom released through cellular respiration moves from a living organism to the exchange pool of the atmosphere. It is then taken in by another living organism, the producer, where it is then respired back into the exchange pool of the atmosphere.
Bloom’s Level: 4. Analyze Learning Outcome: 35.03.01: Define what is meant by a biogeochemical cycle. Section: 35.03 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
12. Which type of organism produces food in an ecosystem? A. heterotroph B. autotroph C. consumer D. carnivore E. herbivore Heterotrophs are consumers; they must take in organic molecules to provide energy and nutrients for their needs. They can be herbivores (eat plant matter), carnivores (eat other animals), or omnivores (eat both plants and animals). Autotrophs are producers: they use inorganic materials (carbon dioxide, water, and solar energy) to manufacture organic molecules and are the basis for the food webs. Consumers are organisms that take in nutrients and energy by eating organic molecules. Carnivores are animals that feed primarily on other animals (meat). Their digestive systems are adapted to digesting this type of food. Herbivores are animals that feed primarily on plant matter. Their digestive systems are adapted to digesting this type of food.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
13. The one nutrient that cannot be categorized as either a sedimentary or a gaseous cycle is A. nitrogen, because it requires bacteria to convert it into a usable form. B. carbon, because it can dissolve into water. C. water, because it can exist as a solid, a liquid, or a gas. D. carbon, because human activities disturb the natural cycle. E. nitrogen, because it must be an ion to be available to producers. While all of these statements are true, the one that addresses whether or not a nutrient moves through a gaseous or a sedimentary cycle is the fact that water exists in multiple forms so that it cannot be classified the same as most other nutrients.
Bloom’s Level: 3. Apply Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
14. Which of the following organisms would be considered a decomposer? A. green algae B. green plant C. caterpillar D. bird E. mushroom Decomposers are organisms that feed on organic matter by digesting it outside their bodies and taking in the needed nutrients and energy. They serve to recycle nutrients in an ecosystem because any nutrients not taken in by the decomposer are available to plants in the ecosystem. Green algae and green plants are producers. Caterpillars and birds are consumers. Mushrooms are decomposers since they digest organic matter by excreting digestive enzymes and then absorbing the needed nutrients.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
15. Given that most nutrients can dissolve into water, which of the following statements is false? A. Runoff water can pollute a pond, stream, or lake if excess nutrients are carried in the water. B. Water can carry nutrients into the soil, where they are available to plants through their roots. C. Water can leach nutrients out of the soil and carry it away from an ecosystem. D. N2 dissolves into water droplets and is carried to the earth to be used by producers. E. Carbon dioxide dissolves into water and is then available to aquatic producers. Atmospheric nitrogen does not dissolve into water because it is a non-polar molecule. Even if it did it would not be useful to plants since they cannot use N2; they require that the nitrogen be fixed by bacteria into an ionic form.
Bloom’s Level: 5. Evaluate Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
16. How are omnivores, carnivores, and herbivores similar? A. They are all producers. B. They are all consumers and produce oxygen. C. They all produce their own food. D. They all produce oxygen. E. They are all consumers and produce carbon dioxide. Omnivores are animals that are able to feed at multiple trophic levels and consume both plants and animals equally. Carnivores are animals that feed primarily on other animals (meat). Their digestive systems are adapted to digesting this type of food. Herbivores are animals that feed primarily on plant matter. Their digestive systems are adapted to digesting this type of food. All of these descriptions are of some type of consumer and, through the process of cellular respiration, release carbon dioxide as a waste product.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.02 Interpret the energy flow and biogeochemical cycling within and among ecosystems. Section: 35.01 Topic: Trophic Levels
17. Once a nutrient enters an ecosystem, it A. must remain in that ecosystem. B. can cycle endlessly in that ecosystem. C. must leave that ecosystem to enter into an organism. D. cannot leave that ecosystem. E. can only leave that ecosystem if it is carried by a living organism. Nutrients are able to cycle into and out of living organisms and into and out of physical components of the environment. A nutrient can cycle endlessly back and forth from biotic to abiotic components, but it does not have to do this. The nutrient can also leave or enter an ecosystem through runoff or other transport.
Bloom’s Level: 3. Apply Learning Outcome: 35.03.01: Define what is meant by a biogeochemical cycle. Section: 35.03 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
18. Animals that feed on both plants and other animals are called A. herbivores. B. carnivores. C. omnivores. D. decomposers. E. detritivores. Herbivores are animals that feed primarily on plant matter. Their digestive systems are adapted to digesting this type of food. Carnivores are animals that feed primarily on other animals (meat). Their digestive systems are adapted to digesting this type of food. Omnivores are animals that are able to feed at multiple trophic levels and consume both plants and animals equally. Decomposers are organisms that feed on organic matter by digesting it outside their bodies and taking in the needed nutrients and energy. They serve to recycle nutrients in an ecosystem because any nutrients not taken in by the decomposer are available to plants in the ecosystem. Detritivores are animals that consume detritus, bits of decaying organic matter.
Bloom’s Level: 1. Remember Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
19. Humans are biologically adapted to be A. omnivores. B. herbivores. C. decomposers. D. autotrophs. E. carnivores. Humans have a digestive system that can extract the needed nutrients and energy from either plant matter or animal matter; this makes them omnivores.
Bloom’s Level: 1. Remember Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
20. Fungi and bacteria are detritus-feeders, also known as A. omnivores. B. herbivores. C. decomposers. D. autotrophs. E. carnivores. Fungi and bacteria feed on detritus by excreting digestive molecules and then taking in the needed molecules for nutrients and energy. Because they digest externally they are categorized as decomposers.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
21. The only heterotrophs required for an ecosystem to function are A. omnivores. B. herbivores. C. decomposers. D. autotrophs. E. carnivores. Since decomposers make nutrients available to plants that would otherwise be locked up in forms that plants cannot access, they are necessary to recycle these nutrients. Other types of consumers are not essential to the cycling of nutrients.
Bloom’s Level: 4. Analyze Learning Outcome: 35.01.02 Interpret the energy flow and biogeochemical cycling within and among ecosystems. Section: 35.01 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
22. Which is NOT a consumer? A. omnivore B. herbivore C. detritivores D. autotroph E. carnivore Consumers are organisms that take in nutrients and energy by eating organic molecules. All of the classes of feeding methods listed consume organic molecules for energy except autotrophs; they manufacture organic molecules from inorganic materials.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
23. Which eats only plant foods? A. omnivore B. herbivore C. decomposer D. autotroph E. carnivore Omnivores are animals that are able to feed at multiple trophic levels and consume both plants and animals equally. Herbivores are animals that feed primarily on plant matter. Their digestive systems are adapted to digesting this type of food. Decomposers are organisms that feed on organic matter by digesting it outside their bodies and taking in the needed nutrients and energy. Autotrophs are producers: they use inorganic materials (carbon dioxide, water, and solar energy) to manufacture organic molecules and are the basis for the food webs. Carnivores are animals that feed primarily on other animals (meat). Their digestive systems are adapted to digesting this type of food.
Bloom’s Level: 1. Remember Learning Outcome: 35.01.01 Identify the ways that autotrophs and heterotrophs obtain nutrients. Section: 35.01 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
24. Energy flow in an ecosystem begins with A. omnivores. B. herbivores. C. decomposers. D. autotrophs. E. carnivores. Autotrophs are producers: they use inorganic materials (carbon dioxide, water, and solar energy) to manufacture organic molecules and are the basis for the food webs. Because they capture energy in a form that cannot be used by other organisms as an energy source and convert it to a usable form, they are the starting point for the flow of energy in an ecosystem.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.02 Interpret the energy flow and biogeochemical cycling within and among ecosystems. Section: 35.01 Topic: Trophic Levels
25. When a plant produces food by photosynthesis, what is the fate of food stored as sugars, starches, etc.? A. All of the food stored by plants is eventually consumed by animal consumers. B. All of the food stored by plants is eventually consumed by either animals or decomposers. C. Most of the plant is consumed by animals and very little is consumed by decomposers. D. A portion of the food is used by the plant itself in cellular respiration and the rest is consumed by animals or by decomposers. E. A majority of the food is used by the plant itself in cellular respiration and very little is consumed by animals or by decomposers. Most of the molecules manufactured by plants are directly used by the plant to fuel the cellular processes necessary to keep the plant alive. A much smaller portion is stored by the plant or used to build more tissues; these molecules are the only ones available to an organism that feeds on the plant.
Bloom’s Level: 2. Understand Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
26. Energy flow in an ecosystem is NOT cyclic because energy is A. increased as you go up the energy pyramid. B. evenly spread out over many organisms. C. converted to many kinds of useful energy. D. totally destroyed as it is used. E. no longer useful when it is converted to heat. As stated in the second law of thermodynamics, when energy is converted from one form to another part of that energy is converted to a less useful form (heat) for living organisms. When converting the chemical potential energy in a food molecule to energy that the cell can use, only approximately 45% of the energy is converted into a useful form and any other conversion as cellular work is performed further reduces the amount of useful energy. So any energy conversion results in a "loss" of energy to the organism or ecosystem (even though the energy does still exist).
Bloom’s Level: 2. Understand Learning Outcome: 35.01.02 Interpret the energy flow and biogeochemical cycling within and among ecosystems. Section: 35.01 Topic: Energy and Chemical Cycling
27. The ultimate source of energy for the majority of organisms on Earth, including humans, is A. heat from burning fuels. B. heat from muscle contraction. C. sunlight. D. electric light. E. ultraviolet light. Since solar energy drives production of organic molecules by plants, photosynthesis, the sun is the ultimate source of energy for almost all organisms on Earth.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.02 Interpret the energy flow and biogeochemical cycling within and among ecosystems. Section: 35.01 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
28. Ecologists have found that A. life as we know it does not match the energy laws of physics. B. ecology cannot be explained using principles from chemistry and physics. C. it is possible to capture all the photosynthetic energy absorbed in a molecule of glucose. D. energy flows one way through an ecosystem and requires an external input. E. inorganic elements flow through an ecosystem. As predicted from the second law of thermodynamics, energy conversions in an ecosystem result in a reduction of useful energy after each conversion. This means that energy cannot be recycled in an ecosystem but rather flows through the system from the original energy source, the sun, through organisms in the system and then out of the system.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.02 Interpret the energy flow and biogeochemical cycling within and among ecosystems. Section: 35.01 Topic: Energy and Chemical Cycling
29. Which of the following is NOT necessary to keep a balanced (stable) ecosystem going? A. Matter (i.e., elements) has to cycle. B. There must be an outside energy source. C. Various populations of producers and consumers must be proportional. D. Bacteria and fungi should be eliminated. E. Energy must be allowed to flow through the ecosystem. Bacteria and fungi are essential to the cycling of nutrients in an ecosystem; without them the nutrients in an organism when it dies or in detritus would not be returned to the producers of the ecosystem and eventually there would be insufficient amounts of available nutrients.
Bloom’s Level: 3. Apply Learning Outcome: 35.01.02 Interpret the energy flow and biogeochemical cycling within and among ecosystems. Section: 35.01 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
30. In the biosphere, which of the following is NOT constantly recycled? A. carbon B. nitrogen C. water D. energy E. phosphorus As predicted from the second law of thermodynamics, energy conversions in an ecosystem result in a reduction of useful energy after each conversion. This means that energy cannot be recycled in an ecosystem but rather flows through the system from the original energy source, the sun, through organisms in the system and then out of the system.
Bloom’s Level: 2. Understand Learning Outcome: 35.01.02 Interpret the energy flow and biogeochemical cycling within and among ecosystems. Section: 35.01 Topic: Energy and Chemical Cycling
31. Which may be a secondary or tertiary consumer? A. producer B. herbivore C. photoautotrophs D. autotroph E. carnivore Autotrophs and producers are not consumers. Photoautotrophs use solar energy to produce organic molecules. Herbivores feed on plants so are always primary consumers. Carnivores can feed at either the secondary or tertiary level; they may feed on an herbivore or on animals that feed on an herbivore.
Bloom’s Level: 4. Analyze Learning Outcome: 35.02.01 Recognize the difference between a food chain and a food web. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
32. When settlers move into a wooded area, they slash and burn the woods, increasing grass production. Initially, they are likely to see A. foxes increase. B. squirrels increase. C. grasshoppers increase. D. hawks decrease. E. grasshoppers decrease. As the change in the food web from a forested area to an area of grass the consumers that feed on grass would increase. Grasshoppers are the organisms listed that would feed mostly on grass.
Bloom’s Level: 3. Apply Learning Outcome: 35.02.01 Recognize the difference between a food chain and a food web. Section: 35.02 Topic: Trophic Levels
33. Settlers move into an area and harvest most of the foxes for fur. They are likely to see A. hawks die out. B. squirrel populations increase. C. grasshopper populations increase. D. hawk populations immediately increase. E. grasshoppers decrease. Since foxes are likely to prey on small mammals, the squirrel population will probably increase as their predators decrease.
Bloom’s Level: 3. Apply Learning Outcome: 35.02.01 Recognize the difference between a food chain and a food web. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
34. Which is NOT true about a complex food web? A. It remains stable. B. Populations tend to remain about the same size. C. Energy levels remain about the same for all trophic levels. D. Inputs are constant and outputs are minimal except for heat. E. Most of the energy entering the system maintains the whole community. As organisms feed farther and farther from the original source of energy input, there is less and less energy available due to losses in the form of heat.
Bloom’s Level: 3. Apply Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
35. Compared to the numbers of grasshoppers or deer, there are few mountain lions. The reason that large and fierce animals, such as the mountain lion, are rare in an ecosystem is that A. such animals do not tolerate other members like them. B. through evolution, most fierce animals have been killed off by others. C. food chains are only 10–20% efficient at each step and top carnivores require large amounts of calories for their survival. D. only a few such animals are permitted by succession in a biome and there are only a few biomes. E. more large predators bring more disease to an ecosystem. Large predators feed at the top of the food web and, as organisms feed farther and farther from the original source of energy input, there is less and less energy available due to losses in the form of heat. This limits the number of top carnivores that can be maintained in an ecosystem.
Bloom’s Level: 3. Apply Learning Outcome: 35.02.01 Recognize the difference between a food chain and a food web. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
36. Most natural food chains are only four or five links long, rarely more. The number of trophic levels is limited because A. there are more predators than herbivores. B. the efficiency in the conversion of food eaten at each level is very low, about 10 percent. C. winter kills off most insects and stops the food chains. D. This is not correct; most food chains are much longer than four or five links. E. nutrients cannot be passed after four or five levels. As predicted from the second law of thermodynamics, energy conversions in an ecosystem result in a reduction of useful energy after each conversion. This means that the energy available to be passed to the next level is less and less at each trophic level and this limits the number of trophic levels in an ecosystem.
Bloom’s Level: 2. Understand Learning Outcome: 35.02.01 Recognize the difference between a food chain and a food web. Section: 35.02 Topic: Trophic Levels
37. Which sequence correctly illustrates the most plausible food chain? A. algae, insect larvae, fish, humans B. algae, fish, insect larvae, humans C. insect larvae, algae, fish, humans D. fish, insect larvae, algae, humans E. humans, fish, insect larvae, algae A producer must be the first link in a food chain so algae, as the only producer listed, must come first. They can be fed on by either insect larvae or fish but it is unlikely that fish would be fed on by insect larvae so insect larvae must come next, with fish feeding on the insect larvae and humans feeding on fish.
Bloom’s Level: 5. Evaluate Learning Outcome: 35.02.01 Recognize the difference between a food chain and a food web. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
38. In a grazing food chain, A. all the consumers are primary consumers. B. primary consumers eat photosynthetic organisms. C. primary consumers eat detritus. D. secondary consumers eat photosynthetic organisms. E. secondary consumers eat detritus. Primary consumers are herbivores and therefore feed on producers. In a grazing food web the producers would be photosynthetic plants. There would be successive levels of consumes in the food web that would be carnivores so not all are primary consumers. Secondary consumers, by definition, must feed on an animal that has fed on a plant.
Bloom’s Level: 2. Understand Learning Outcome: 35.02.01 Recognize the difference between a food chain and a food web. Section: 35.02 Topic: Trophic Levels
39. The base of an ecological pyramid represents the A. producer trophic level. B. primary consumer trophic level. C. secondary consumer trophic level. D. top predator trophic level. E. decomposer trophic level. The base of a pyramid represents the largest portion of the model so, in this case, it represents the producers who are the largest group in the food web.
Bloom’s Level: 1. Remember Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
40. A detritus food chain begins A. in the ocean. B. with a producer. C. with decaying matter. D. with air pollution. E. with fungi. Since detritivores feed on decaying organic matter, detritus, that can be from any source, it is said that a detrital food web begins with these molecules, although they did come from some other undetermined source.
Bloom’s Level: 2. Understand Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
41. In a certain ecosystem, caterpillars eat grass. Field mice eat the caterpillars and seeds from the grass. Snakes eat the mice. Hawks eat both the snakes and mice. Such an arrangement would be considered a(n) A. food web. B. food chain. C. ecosystem. D. first trophic level. E. second trophic level. This describes a food web since it is more than a onetime snapshot that tells was the food source for that sequence but rather a statement of the possibilities of which organism could feed on another.
Bloom’s Level: 3. Apply Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
42. In the following food chain, which organism is at the third trophic level? grass → rabbits → snakes → hawks → foxes A. grass B. rabbits C. snakes D. hawks E. foxes Snakes feed at the third step away from the original energy source, the sun, which provided the grass with the energy to make organic molecules.
Bloom’s Level: 2. Understand Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
43. How is a carnivore like a herbivore? A. Both pass on all energy received to the next trophic level. B. Both tend to produce nutrients for plants. C. Both pass on much less energy to the next trophic level than they received. D. Both are able to convert organic compounds to ATP without loss of energy. E. Both use nutrients released from decomposers. At every trophic level, regardless of the original energy source (food), there is much less energy passed to the next organism so all organisms are alike in this way.
Bloom’s Level: 2. Understand Learning Outcome: 35.02.01 Recognize the difference between a food chain and a food web. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
44. Which of the following is NOT true about a food web? A. It contains many food chains. B. It contains several trophic levels. C. It is usually stable. D. It interacts with the physical environment. E. It uses only the carbon cycle. The food web passes not only energy but also nutrients to the organisms in the web; the nutrients are cycled through the ecosystem while the energy flows through it. All nutrients, not just carbon, are obtained in this way and are used by the organisms to build the molecules that they need.
Bloom’s Level: 2. Understand Learning Outcome: 35.02.01 Recognize the difference between a food chain and a food web. Section: 35.02 Topic: Trophic Levels
45. What happens to the amount of available energy as you progress up the food pyramid? A. The amount of energy decreases. B. The amount of energy increases. C. The amount of energy stays the same. D. The amount of energy first decreases and then finally increases. E. Energy does not move from one trophic level to the next. As the shape of the pyramid implies, the amount of energy from one trophic level to the next decreases as energy obtained from organisms that are fed upon is converted and some of it lost as heat and some is used by the organisms doing the feeding.
Bloom’s Level: 2. Understand Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
46. What eventually happens to the energy that is passed from one trophic level to the next in a food chain? A. It recycles back to the producers. B. It results in a much larger decomposer population. C. It is dissipated into the atmosphere in the form of heat. D. It is recaptured by another food chain. E. It is eventually sealed in fossil fuels such as coal and oil. All of the energy that enters a food web is eventually converted into heat and dissipated into the atmosphere.
Bloom’s Level: 4. Analyze Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
47. Which pyramid illustrates that at each successive trophic level there is a decreasing amount of energy available? A. a pyramid representing the number of organisms at each level B. biomass pyramid of organisms at each level C. ecological pyramid of organisms at each level D. food chain E. food web Different types of pyramids are used to express the movement of matter and energy in an ecosystem. A pyramid of biomass would express the amount of matter found at each trophic level. A pyramid expressing the numbers of organisms at each level would reflect on the number of each type of organism that feed at each trophic level. An ecological pyramid is constructed to show the passage of energy to the next trophic level. This allows the loss of energy to be visualized as each step in the pyramid is smaller than the one beneath. A food chain refers to a sequence of feeding levels; it is a one-time snapshot of feeding and does not express the true complexity of trophic levels as does a food web. A food web describes the complex feeding relationships found in an ecosystem.
Bloom’s Level: 4. Analyze Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
48. Is the annual productivity of a typical city backyard large enough to support a house cat or will the cat require food supplements in order for it to survive? A. The producer productivity is large enough for the cat to survive. B. The cat must be fed additional food or it will starve. C. The cat will be able to survive if there is at least one fruit tree in the backyard as part of the producer productivity. D. The cat will be able to survive if the backyard consists only of grass. E. The producer productivity can support the cat along with several offspring. The food web that could exist in a backyard in a city would not be extensive enough to produce the energy needed by a top carnivore that was restricted to that area. If an outside food/energy source was not available, the cat would starve.
Bloom’s Level: 5. Evaluate Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
49. Approximately what percentage of the energy in one trophic level becomes incorporated into the next level? A. 1% B. 10% C. 30% D. 60% E. 100% At each trophic level the organism converts some of the energy into ATP to use for cellular work. This process is not efficient and a large portion is lost to the organism (and any subsequent trophic levels) as heat. ATP is then used to perform the work that is necessary to keep the organism alive and this energy is typically lost to subsequent trophic levels. The only energy available to the next trophic level is that stored in the molecules or the organism. This is usually about 10% of the energy taken in by the organism.
Bloom’s Level: 1. Remember Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
50. Less energy is available at the next trophic level because A. all the food is captured and eaten before it can reach the next trophic level. B. all of the food eaten is digested and absorbed. C. only a portion of the food that is digested can be used to make ATP. D. all of the absorbed food molecules are used as an energy source for ATP buildup in mitochondria. E. all of the food is absorbed, but not all can be used to make ATP. At each trophic level the organism converts some of the energy into ATP to use for cellular work. This process is not efficient and a large portion is lost to the organism (and any subsequent trophic levels) as heat. ATP is then used to perform the work that is necessary to keep the organism alive and this energy is typically lost to subsequent trophic levels. The only energy available to the next trophic level is that stored in the molecules or the organism.
Bloom’s Level: 2. Understand Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
51. Consider the components of a four-step food chain: producers, herbivores, carnivores, top predators. If everything is eaten at each level, how much of the energy stored by a producer is passed to the top predator? A. 100% B. 10% C. 1% D. 0.1% E. 0.01% The producer would have access to 100% of the energy it captures and would use much of that energy so that it is lost to the next trophic level. Approximately 10% would pass along to the next level: the herbivore. The herbivore in turn would use all but 10% of the energy it obtained but this is 10% of 10%, or 1% of the original energy captured by the producer. At the next level, the carnivore would use all but 10% of the energy that it obtained; this would then be 10% of the 1% of the original energy, or 0.1% of the original energy captured by the producer.
Bloom’s Level: 4. Analyze Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
52. Inverted ecological pyramids can occur in what habitat? A. forest canopies B. deserts during rare rainfalls C. ocean shorelines with algae D. any ecosystem disturbed by human activity E. If studied closely, inverted pyramids never occur naturally. In aquatic ecosystems, a biomass pyramid may take the form of an inverted pyramid as the algae have much less biomass than the herbivores that feed upon them. The reason that this is possible is that the algae reproduce rapidly and are rapidly fed upon.
Bloom’s Level: 2. Understand Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
53. An inverted energy pyramid for an ecosystem, one where there are more consumers than producers at a given point in time, is possible if which of the following is true? A. Producers are reproducing more slowly than consumers. B. There is intense sunlight for long periods of time, as in the tropics. C. Producers are reproducing much more rapidly than consumers. D. Decomposers are reproducing much more rapidly than consumers. E. Consumers are reproducing more rapidly than producers. In aquatic ecosystems, a biomass pyramid may take the form of an inverted pyramid as the algae have much less biomass than the herbivores that feed upon them. The reason that this is possible is that the algae reproduce rapidly and are rapidly fed upon. The consumers on the other hand, do not reproduce as rapidly.
Bloom’s Level: 3. Apply Learning Outcome: 35.02.02 Explain the energy flow among populations through food webs and ecological pyramids. Section: 35.02 Topic: Trophic Levels
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Chapter 35 - Nature of Ecosystems
54. Which of the following is NOT required in robust chemical cycling of inorganic nutrients in a natural ecosystem? A. a reservoir for the element in the earth B. the biotic community and its food chains C. an exchange pool from which producers draw nutrients D. weather E. plants and animals In order for nutrients to be cycled, the two groups of organisms that are required are autotrophs and decomposers. The method of capturing energy and the energy source can be either chemical, chemosynthesis, or solar, photosynthesis, so the producers can be plants, algae, or bacteria. Decomposers are also necessary to return organic molecules to a form that is usable for the producers. Any other organism is not absolutely required for nutrient cycling.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.01: Define what is meant by a biogeochemical cycle. Section: 35.03 Topic: Energy and Chemical Cycling
55. The two general categories of biogeochemical cycles are A. energy and matter. B. terrestrial and aquatic. C. gaseous and sedimentary. D. chemical and physical. E. organic and inorganic. When looking at cycles other than the hydrologic cycle, it is possible to classify them into one of two categories: either gaseous or sedimentary, depending on how the nutrient is made available to the producers. In a gaseous cycle, such as the carbon cycle, the primary source of carbon available to producers is in a gaseous form. In a sedimentary cycle, such as the phosphorous cycle, the primary source of phosphorous is from the soil.
Bloom’s Level: 1. Remember Learning Outcome: 35.03.01: Define what is meant by a biogeochemical cycle. Section: 35.03 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
56. Fresh water makes up about A. 3 percent B. 10 percent C. 30 percent D. half E. 90 percent
of the world's supply of water.
Only about 3 percent of the world's supply of water is fresh water; the remaining 97 percent is salt water. Of the 3 percent of water that is fresh water, 2 percent of that is in a frozen form in glaciers and polar ice.
Bloom’s Level: 1. Remember Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
57. Which statement is TRUE about the water (hydrologic) cycle? A. Because this is a true cycle, it is impossible to run out of fresh water for human use. B. Some water evaporates from land and from plants. C. All water molecules that evaporate from the ocean precipitate on land and move by gravity through groundwater to the ocean again. D. Once water sinks into the ground, it is safe from human exploitation or pollution until it has rejoined the ocean. E. The main source of water in the atmosphere is from evaporation over landmasses. The evaporation of water occurs both from bodies of water and through transpiration, the process where water evaporates from the pores of leaves. Most of the water that evaporates does so over water and it then moves over land in the form of clouds. Although water is considered a renewable resource, the distribution of the resource across the globe can happen in a way that is uneven and some places, such as deserts, do not receive enough for plants and animals to exist in large numbers. Some water infiltrates the ground remains there, where it is frequently removed for human use through wells; this water can become polluted if water infiltrates the ground, carrying dissolved pollutants.
Bloom’s Level: 3. Apply Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
58. When deep wells drain aquifers, it is called A. transpiration. B. negative hydrology. C. establishing a reservoir. D. increasing the transfer rate. E. groundwater mining. The removal of waters stored deep underground in aquifers can result in a lowering of the level of water in the aquifer if removal happens at a higher rate than recharge. Recharge is usually an extremely slow process so it is not uncommon for withdrawal to exceed recharge.
Bloom’s Level: 1. Remember Learning Outcome: 35.03.03: Identify how human activities can alter each of the biogeochemical cycles. Section: 35.03 Topic: Energy and Chemical Cycling
59. Someone asks if some of the carbon atoms in their body may have once made up the living tissues of another person or animal in earlier times. This is A. possible only if you are a cannibal. B. possible since carbon cycles through the ecosystem. C. impossible since carbon can't enter a living system more than once. D. highly unlikely since carbon is always tied up in molecules in organisms or minerals. E. impossible since carbon molecules are specific for different organisms. The saying that "you are what you eat" is an accurate statement since the atoms that are used to make up your body are obtained from your food. These atoms have cycled from reservoirs and into living organisms and then back into reservoirs. This happens again and again as living things die and decompose and their atoms are released to be reused by producers.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
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Chapter 35 - Nature of Ecosystems
60. Which of the following does NOT add carbon to the atmosphere? A. photosynthesis B. burning fossil fuels C. bicarbonate from aquatic systems D. carbon dioxide produced by cellular respiration of plants E. carbon dioxide produced by cellular respiration of animals Photosynthesis is the one process that removes carbon dioxide from the atmosphere rather than introducing it.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
61. In the carbon cycle, carbon is returned to the atmosphere by A. respiration of animals and plants. B. photosynthesis. C. evaporation of water. D. glycolysis. E. denitrification. Cellular respiration takes organic molecules and releases the energy in the chemical bonds by breaking those bonds and rearranging the atoms into carbon dioxide and water. This returns carbon dioxide to the atmosphere. Neither the evaporation of water nor the denitrification by bacteria affects the carbon cycle. Photosynthesis is the one process that removes carbon dioxide from the atmosphere rather than introducing it. Glycolysis does break down organic molecules into other forms but it does not release carbon dioxide.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
35-32 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 35 - Nature of Ecosystems
62. Which of the following is NOT true of the phosphorus cycle? A. Phosphorus never enters the atmosphere. B. Phosphate only becomes available when a geological upheaval exposes sedimentary rocks. C. Phosphate is often a limiting nutrient in most ecosystems. D. Phosphate detergents are a major source of eutrophication. E. The phosphorous cycle is not affected by human activities. The phosphorous cycle is affected by humans when excess phosphorous is released into ecosystems by fertilizers, detergents, and animal waste.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
63. Concerning the nitrogen cycle, which of these is NOT true? A. It requires a number of different types of bacteria. B. Plants can take in and use nitrogen from the air by using their leaves. C. Ammonium in the soil is converted to nitrite by nitrite-producing bacteria and then nitrite is converted to nitrate by nitrate-producing bacteria. D. Eventually, organic nitrogen becomes inorganic nitrogen again. E. Human activities increase the nitrogen fixation rate. Although the atmosphere is 78 percent nitrogen, it is in the form of nitrogen gas, or N2. This gas cannot be used by plants since they cannot break the bond between the two nitrogen atoms. Bacteria convert it into an ionic form that the plants can use.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
35-33 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 35 - Nature of Ecosystems
64. Nitrification refers to the production of A. nitrogen gas. B. nitrate. C. nitrite. D. ammonia. E. ammonium. Nitrification occurs when soil bacteria convert NH4+ into NO3-; or ammonia into nitrate.
Bloom’s Level: 1. Remember Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
65. The conversion of nitrate to nitrogen gas is achieved by A. denitrifying B. leguminous C. nitrogen-fixing D. nitrate-producing E. nitrite-producing
bacteria.
The reverse process of producing N2, which can return to the atmosphere from nitrate (NO3-), is also carried out by bacteria and is called denitrification.
Bloom’s Level: 1. Remember Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
35-34 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 35 - Nature of Ecosystems
66. The first step in the use of carbon by living organisms occurs in A. humans. B. vertebrates. C. invertebrates. D. green plants. E. soil bacteria. Green plants remove carbon dioxide from the atmosphere and incorporate it into organic molecules that can then be consumed by other organisms.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
67. Which of the following statements is false regarding the carbon cycle? A. Terrestrial plants take up carbon dioxide from the air. B. Aquatic plants exchange carbon dioxide directly with the atmosphere. C. Algae produce food for themselves and for heterotrophs. D. When organisms respire, a portion of carbon is returned to the atmosphere as carbon dioxide. E. The amount of bicarbonate in the water is in equilibrium with the amount of carbon dioxide in the air. Aquatic plants remove carbon from the water surrounding them. Carbon dioxide from the atmosphere dissolves into water to form bicarbonate (HCO3-). Aquatic plants use this form of carbon for photosynthesis.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
35-35 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 35 - Nature of Ecosystems
68. Which of the following is NOT a carbon reservoir? A. living organisms B. dead organisms C. coal D. ammonium and nitrates E. inorganic carbonate in limestone/carbonaceous shells Both living and dead organisms contain molecules made in part from carbon. Coal is a fossil fuel that is formed from the carbon containing remains of organisms found during the Carboniferous period.
Bloom’s Level: 1. Remember Learning Outcome: 35.03.01: Define what is meant by a biogeochemical cycle. Section: 35.03 Topic: Energy and Chemical Cycling
69. Atmospheric CO2 combines with water to produce a bicarbonate which is A. a cause of the greenhouse effect. B. ozone. C. a source of carbon for algae. D. photochemical smog. E. a thermal inversion. Aquatic plants remove carbon from the water surrounding them. Carbon dioxide from the atmosphere dissolves into water to form bicarbonate (HCO3-). Aquatic plants use this form of carbon for photosynthesis.
Bloom’s Level: 3. Apply Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
35-36 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 35 - Nature of Ecosystems
70. The greenhouse effect is A. due to increased sunlight in recent years. B. caused by increased plant life and photosynthesis. C. a natural phenomenon due to absorption of sunlight and re-radiating of the heat. D. a manmade hazard due to decreased levels of carbon dioxide and ozone in the atmosphere in recent years. E. not affected by human activities. The greenhouse effect is a natural process that has allowed life on Earth to exist at relatively warm temperatures, rather than the freezing temperatures that would exist without this process. Concern in recent years has focused on an intensifying of this effect as more greenhouse gases collect in the atmosphere and the earth becomes warmer.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
71. Which of the following is usually a limiting factor (nutrient) for plants? A. nitrogen B. nitrites C. carbon dioxide D. phosphates E. hydrogen Plants do not use nitrogen (N2), nitrates (NO2-), or atoms of hydrogen. Carbon dioxide is in plentiful supply in the atmosphere, so it does not serve as a limiting factor. Phosphates are necessary for plants to manufacture ATP, phospholipids, and nucleotides for DNA and RNA. If phosphorous is in short supply, as it usually is in a natural ecosystem, then it will limit plant growth.
Bloom’s Level: 2. Understand Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
35-37 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 35 - Nature of Ecosystems
72. You buy a house on a small inland lake and your front lawn ends at the water's edge. A lawn care company representative suggests you fertilize the lawn to keep it lush and green. You agree to the treatment and then discover there is a major weed problem in the lake. The most likely cause of the weed problem is A. nitrogen. B. nitrites. C. sulfates. D. phosphates. E. carbon. The runoff of fertilizers carrying phosphorous into aquatic systems is a serious problem. The overabundance of a nutrient that usually limits the ability of plants to reproduce allows the algae to grow uncontrolled.
Bloom’s Level: 3. Apply Learning Outcome: 35.03.03: Identify how human activities can alter each of the biogeochemical cycles. Section: 35.03 Topic: Energy and Chemical Cycling
73. Atmospheric carbon serves as A. an exchange pool for carbon that is available to plants for photosynthesis. B. a long-term storage reservoir that becomes available to plants through the action of bacteria. C. an exchange pool for carbon that is made available to plants through the action of bacteria. D. a reservoir of carbon that will eventually be released into the ecosystem through the process of weathering. E. a storage reservoir of carbon that is part of the sedimentary cycle of carbon. Atmospheric carbon serves as a ready supply of carbon for plants; it is, therefore, an exchange pool.
Bloom’s Level: 3. Apply Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
35-38 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 35 - Nature of Ecosystems
Short Answer Questions 74. When an herbivore eats, only about 10 percent of the energy it consumes is available for the next trophic level. Discuss where the remaining 90 percent of the consumed energy goes. When an herbivore eats approximately 90 percent of the energy consumed is used in the following fashions: 1. cellular respiration, which releases energy into the environment as heat 2. when the organism dies, much of the body will be decomposed, which doesn't make it available for the next trophic level 3. when the organism urinates, the excretion is not consumed by the next trophic level 4. when the organism defecates, the feces are not consumed by the next trophic level
Bloom's Level: 6. Create Learning Outcome: 35.01.02 Interpret the energy flow and biogeochemical cycling within and among ecosystems. Section: 35.01 Topic: Energy and Chemical Cycling
75. List the four biogeochemical cycles that need to be in balance in order to have stable ecosystems. 1. Hydrological cycle 2. Nitrogen cycle 3. Phosphorus cycle 4. Carbon cycle
Bloom's Level: 6. Create Learning Outcome: 35.03.01: Define what is meant by a biogeochemical cycle. Section: 35.03 Topic: Energy and Chemical Cycling
35-39 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 35 - Nature of Ecosystems
Multiple Choice Questions 76. Which biogeochemical cycle is directly associated with fossil fuels? A. carbon B. nitrogen C. hydrological D. phosphorus E. All of the answer choices are correct. Living and dead organisms contain organic carbon. When these organisms die some of them become fossilized, preserving the carbon in their remains.
Bloom’s Level: 1. Remember Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
77. In a grassland biome, which organism would most likely be at the highest trophic level? A. hawk B. big bluestem grass C. sparrow D. caterpillar E. mouse In the grassland biome, the hawk would most likely occupy the highest trophic level. The big bluestem grass is the producer base. The mouse, caterpillar, and sparrow all eat the grass. The hawk would eat the mouse and sparrow.
Bloom’s Level: 1. Remember Learning Outcome: 35.02.01 Recognize the difference between a food chain and a food web. Section: 35.02 Topic: Trophic Levels
35-40 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 35 - Nature of Ecosystems
Short Answer Questions 78. Explain the steps in the hydrological cycle. 1. Evaporation occurs, causing liquid water to change to a gaseous state. 2. Condensation occurs when the gas is changed back into a liquid. The amount of evaporation in the ocean exceeds the amount of condensation over the ocean. The excess water in the gaseous state then moves over land and falls as precipitation on land. Due to gravity, the precipitation will eventually makes its way towards the ocean. Some of it may be captured in freshwater bodies (lakes, rivers, streams, and aquifers) as it makes its journey towards the ocean.
Bloom's Level: 6. Create Learning Outcome: 35.03.02: Identify the steps of the water cycle, the phosphorus cycle, the nitrogen cycle, and the carbon cycle. Section: 35.03 Topic: Energy and Chemical Cycling
35-41 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
Chapter 36 The Biosphere
Multiple Choice Questions 1. What drives the movement of winds across the face of the Earth? A. the unequal heating of the surface of the Earth by the sun B. the currents of the oceans C. the distribution of mountains and valleys D. the rotation of the Earth E. the greenhouse effect The unequal heating of the surface of the Earth leads to warmer air around the equator, rising up into the atmosphere where it cools and then sinks back to the surface of the Earth at around 30 north or 30 south of the equator, forming a large convection current. Some of the descending air flows toward the poles and rises again around 60. Some of this air descends around 30 and some of the air that has reached the upper atmosphere flow farther toward the poles and descends at the pole.
Bloom’s Level: 3. Apply Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
36-1 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
2. The phenomenon called a rain shadow requires what topographical elements? A. For a rain shadow to occur, a mountain range must be located to the leeward of the ocean. B. For a rain shadow to occur, a mountain range must be located in the middle of a desert. C. For a rain shadow to occur, a mountain range must be located next to a large lake. D. For a rain shadow to occur, a mountain range must be located to the windward side of the ocean. E. For a rain shadow to occur, a mountain range must be located in a tropical rainforest. A rain shadows occurs when warm, moisture-laden air blows in from the ocean and encounters a mountain range. In order to cross the mountains, the air rises higher into the atmosphere and as it cools. It can no longer hold the same amount of moisture, which is lost as rainfall on the windward side of the mountains. The air then crosses the mountain and sinks on the leeward side where it warms as it descends and picks up moisture from the surrounding environment, making it very arid.
Bloom’s Level: 3. Apply Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
3. Tundra is divided into alpine tundra at the top of mountains and arctic tundra found close to the North Pole. These two biomes have similar plants and animals even though they can be located in vastly different regions of the world. What causes these two biomes to be similar? A. Both are found close to coniferous forests. B. One is cold and dry because it is at a high latitude and the other is cold and dry because it is at a high altitude. C. Both are located in areas of intense sunshine. D. One grows mostly grasses because the soil is poor and the other grows mostly grasses because the soil is frozen. E. Both have large grazing herds that keep the grasses clipped short. The two factors that determine the type of biome are temperature and precipitation; both of these biomes have extremely cold and dry climates. Alpine tundra is a cold biome because it is located high above sea level, while arctic tundra is cold because it is close to the North Pole.
Bloom’s Level: 3. Apply Learning Outcome: 36.02.01 Identify the geographical distribution of the major terrestrial ecosystems. Section: 36.02 Topic: Terrestrial Ecosystems
36-2 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
4. Why is a rocky seashore inhabited by more types of organisms than a sandy seashore? A. The rocks trap and hold more nutrients. B. The sand drains of water quickly at low tide. C. The sand does not hold nutrients. D. Predators are more likely to hunt along sandy shores. E. It is easier for organisms to attach to the rocks. Organisms have a variety of ways to attach to a rocky substrate to live on a rocky shore. Organisms living on a sandy beach can only remain if they are able to live burrowed into the sand.
Bloom’s Level: 4. Analyze Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
5. If the Earth was standing still (not rotating), wind directions would A. cease. B. all run north and south. C. all run east and west. D. remain the same since Earth's rotation does not affect wind directions. E. all run towards the equator. The wind curves around the Earth due to its rotation; this is called the Coriolis effect. Without the rotation, the air rising from the surface of the Earth would sink back to the Earth at 30 north or south of the equator but would not curve.
Bloom’s Level: 3. Apply Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
36-3 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
6. Which is NOT a correct characterization of tundra? A. All precipitation in the tundra remains permanently frozen, hence the name "permafrost." B. The landscape of pools and mires in the tundra is due to melted snow that does not evaporate. C. The long periods of darkness, sometimes even six months of night, prevents trees and other photosynthetic producers from growing. D. In the summer, the tundra is alive with insects and birds. E. Most animals found in the tundra live there year round. During the short growing season, many organisms migrate to the tundra to take advantage of the extended daylight hours but few have the adaptations necessary to survive the extreme cold, extended darkness and lack of liquid water that characterize winter on the tundra.
Bloom’s Level: 3. Apply Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
7. Which of the following communities will have the highest net primary productivity? A. prairie grassland B. temperate deciduous forest C. tropical rain forest D. tundra E. tropical rain forest and estuaries Tropical rain forests and estuaries have the highest net primary productivity among biomes. They both have high levels of diversity and abundant resources, which contribute to their ability to capture the energy from the sun and convert it into biomass.
Bloom’s Level: 3. Apply Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
36-4 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
8. The poor soil of tropical rain forests is due to A. poor quality substrate, such as bedrock that lacks iron and phosphorus. B. intense sunlight bleaching the substrate. C. lack of litter; the leaves never fall to the ground. D. lack of freezing cold to break apart the substrate bedrock. E. high temperatures that allow rapid decomposition and rapid recycling of nutrients. The nutrients of a rainforest are located in the living organisms and do not stay in the soil for long. Decomposition is extremely rapid and if a leaf falls to the forest floor it is recycled into a living organism very quickly. This prevents nutrients from entering the soil and so leads to poor soils.
Bloom’s Level: 3. Apply Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
9. In the desert, plants and animals are adapted to withstand A. heat. B. dryness. C. cold. D. earthquakes. E. heat and dryness. Deserts are typically characterized by rainfall that is less than 25 cm/year and have hot days and cold nights. The organisms in this biome must be adapted to the low amount of water that is available and to the extreme heat during the day.
Bloom’s Level: 3. Apply Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
36-5 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
10. A foul-smelling lake that is rich in algae and rooted plants but lacks oxygen is A. oligotrophic. B. eutrophic. C. benthic. D. pelagic. E. limnetic. Oligotrophic lakes are nutrient poor; they have clear water and low biological productivity. Eutrophic lakes are rich in organic material and so are nutrient rich; they have large populations of algae, rooted plants, and decomposers that reduce the oxygen content in the water. The benthic region of the lake or the ocean is the bottom, reaching from the shoreline to the deepest regions of the water. The pelagic region of the ocean refers to the open water. The limnetic region of a lake refers to the open water.
Bloom’s Level: 3. Apply Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
11. Which definition is NOT correct? A. littoral lake zone—closest to shore B. benthic zone—lake or ocean floor C. limnetic lake zone—on surface away from shore D. profundal zone—division of the open ocean E. epipelagic—upper part of open waters of ocean The profundal zone is the deepest waters of a lake, deep enough to not receive light. The open ocean is the pelagic zone.
Bloom’s Level: 3. Apply Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
36-6 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
12. Which are NOT strictly a freshwater community? A. lakes B. rivers C. estuaries D. streams E. ponds Estuaries are where rivers flow into the ocean and so are areas where salt water and fresh water mix.
Bloom’s Level: 3. Apply Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
13. The coral reef community would be best described as A. a hard, stony desert that is nevertheless beautiful. B. a common view of the ocean bottom but only seen in shallow water near the shoreline. C. a densely populated community that only occurs in shallow water near the shoreline. D. a community based on animals that migrate from one area of the ocean to another. E. undersea flowers. Coral reefs are the most diverse aquatic ecosystem. In order for coral to grow and survive they need to be in shallow, nutrient-poor waters.
Bloom’s Level: 3. Apply Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
36-7 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
14. The Earth is actually slightly farther from the sun during the period when the Northern Hemisphere is having summer. How can this be explained? A. The sun grows more powerful on a half-year cycle. B. There is a six-month lag phase between sunlight being absorbed and lost. C. The tilt of the Earth presents the Northern Hemisphere in a more perpendicular angle and this increased sunlight per surface area more than offsets the decrease from added distance. D. There must be some mistake in measuring the Earth's orbit because if the Earth was farther away, we would be cooler. E. During the summer in the northern hemisphere the Earth is actually closer to the sun even though the Northern Hemisphere is tilted away from the sun. The Earth’s axis has about a 23 angle. This causes one-half of the Earth to be presented to the sun while the other is tilted farther away. This causes seasonal differences between the North and South Poles.
Bloom’s Level: 4. Analyze Learning Outcome: 36.01.01 Describe how solar radiation produces variations in the Earth's climate. Section: 36.01 Topic: Climate and the Biosphere
15. How does temperature change in relation to the latitude? A. As latitude increases, moving away from the North Pole to the South Pole, the temperature increases. B. As latitude increases, moving away from the equator at the center of the Earth toward either the North Pole or the South Pole, the temperature decreases. C. As latitude increases, moving away from the equator at the center of the Earth toward either the North Pole or the South Pole, the temperature increases. D. As latitude increases, moving away from the equator at the center of the Earth toward either the North Pole or the South Pole, the temperature first increases until 60 and then decreases at the pole. E. There is no relationship between temperature and latitude. Since there is more intense sunshine at the equator, the temperature is greater at this latitude. The sunlight is less intense toward the North and South Poles, so temperatures decrease as latitude increases.
Bloom’s Level: 5. Evaluate Learning Outcome: 36.01.01 Describe how solar radiation produces variations in the Earth's climate. Section: 36.01 Topic: Climate and the Biosphere
36-8 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
16. Why is the energy of the sun less intense at the poles as compared to the equator? A. The tilt of the Earth moves the poles farther away from the most intense sunlight. B. Because the Earth is spherical and the sun shines toward the equator, the sunlight is more spread out at the poles and therefore less intense. C. The movement of wind across the face of the Earth pulls cooler air from over the oceans and blows it toward the poles. This cools the poles while leaving the region around the equator hot. D. The movement of cool waters in ocean currents flows away from the equator and cools the air farther from the equator, causing the polar regions to be cooler than the equatorial region. E. The Earth shifts in its orbit periodically and at times it moves farther from the sun. This movement coincides with times that the Earth is tilted toward the sun so the poles never receive as much sunlight as the equator. The spherical shape of the Earth causes the beams of solar energy to be more diffuse as it covers a larger area than beams that reach the equator.
Bloom’s Level: 2. Understand Learning Outcome: 36.01.01 Describe how solar radiation produces variations in the Earth's climate. Section: 36.01 Topic: Climate and the Biosphere
17. What causes an ocean breeze to blow? A. the presence of mountain ranges along the shoreline B. the fact that water heats rapidly C. the differential heating of land and water D. the fact that lands heats much more slowly than water E. the heating of the air around the equator Land heats much more rapidly than water, so during the day cooler breezes blow from over the water to over the land. At night the land gives up heat more quickly than the water so the wind shifts to blow from the land to the ocean.
Bloom’s Level: 2. Understand Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
36-9 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
18. Which of the following factors will have the greatest impact in determining the type of biome in a location? A. temperature B. soil conditions C. rainfall D. wind patterns E. the presence of mountains The amount of rainfall in a location will determine the type of biome that is present over any of the other factors. If a region is hot and dry it will be a desert; if hot and wet it will be a tropical rainforest.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Climate and the Biosphere
19. What environmental factor makes it very difficult for organisms that live in estuaries? A. the movement of water as the river flows into the ocean B. the constant fluctuation in the salinity of the water as the salt water and fresh water mix C. the fluctuating nutrient levels as the river brings different levels of nutrients to the estuary D. the change in water levels as the tide ebbs and flows E. the change in the temperature of the water since the river water is colder than the ocean The constant change in the salinity of the water is difficult for organisms to live in because they must regulate the water that is taken into their bodies. Marine fish do not survive in fresh water and freshwater fish do not survive well in a marine environment. The same is true of plants.
Bloom’s Level: 5. Evaluate Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
36-10 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
20. Climate is due to temperature and rainfall that in turn are NOT determined by A. variations in the solar radiation received on a spherical Earth. B. the tilt of the Earth's axis as it rotates about the sun. C. the distribution of landmasses and oceans. D. the topography or landscape, such as mountains. E. biomes present. The biomes present would be dependent on the climate, not the other way around.
Bloom’s Level: 2. Understand Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
21. In the water cycle, evaporated equatorial air masses that rise lose most of their water. The air immediately then flows A. all the way to the poles where it causes snow. B. all the way to the poles if it does not encounter mountain ranges, which it usually does. C. to about 60 north or south and then cool and condense, supporting a band of moist vegetation. D. to about 30 north or south and then descends, creating a band of deserts. E. to the west. The air cools as it rises and moves away from the equator so when it falls back toward the Earth around 30 north and south of the equator it contains very little moisture. These areas receive very little rainfall because the air is relatively dry as it returns to the Earth.
Bloom’s Level: 2. Understand Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
36-11 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
22. Zones of low rainfall and, therefore, most deserts are at A. the equator since the light intensity is greatest there. B. the centers of all continents because they are farthest from the oceans that are the source of all water evaporation. C. about 30 north and south latitude. D. about 60 north and south latitude. E. random points wherever wind patterns fail to bring moist air. The air cools as it rises and moves away from the equator so when it falls back toward the Earth around 30 north and south of the equator it contains very little moisture. These are desert areas due to this lack of rainfall.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.01 Identify the geographical distribution of the major terrestrial ecosystems. Section: 36.02 Topic: Terrestrial Ecosystems
23. The spinning of the Earth A. angles winds away from directly north-south toward east or west. B. causes mountain ranges to spin off moisture. C. is the initial driving force for generating winds. D. causes ocean currents to be directed "out" to the equator. E. causes the tidal ebb and flow. As the Earth rotates, the air masses that rise away from the surface seem to curve or angle as the Earth’s surface moves under them; this is called the Coriolis effect.
Bloom’s Level: 2. Understand Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
36-12 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
24. When moist air from over an ocean moves inland and up a mountain slope, it causes heavy precipitation on the windward side. On the leeward side of the mountain, the region receives far less rain. This is referred to as A. lake effect. B. eutrophication. C. a montane successional gradient. D. a monsoon climate. E. rain shadow. Lake effect refers to the increased likelihood of snowfall as warm, moist air moves from over a large body of water, like the Great Lakes, in the winter and loses its moisture as it rises; the precipitation comes in the form of snow because the air cools rapidly as it rises. Eutrophication is the enrichment of water by inorganic nutrients that are used by producers in the lake. The montane successional gradient refers to the change of biome as altitude increases in the mountains. A monsoon climate is one that alternates between dry and wet seasons, each one lasting about half of the year; in the wet season, rainfall is frequent and heavy, resulting in flooding. A rain shadow is caused when the leeward side of a mountain receives less rain than the windward side because the air cools as it rises over the mountain, causing it to drop its moisture.
Bloom’s Level: 1. Remember Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
36-13 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
25. Arctic winds blowing over inland reservoirs become saturated and then lose their moisture as they move across land. This is referred to as A. lake effect. B. eutrophication. C. a montane successional gradient. D. a monsoon climate. E. rain shadow. Lake effect refers to the increased likelihood of snowfall as warm moist air moves from over a large body of water, like the Great Lakes, in the winter and loses its moisture as it rises; the precipitation comes in the form of snow since the air cools rapidly as it rises. Eutrophication is the enrichment of water by inorganic nutrients that are used by producers in the lake. The montane successional gradient refers to the change of biome as altitude increases in the mountains. A monsoon climate is one that alternates between dry and wet seasons each one lasting about half of the year; in the wet season rainfall is frequent and heavy resulting flooding. A rain shadow is caused when the leeward side of a mountain receives less rain than the windward side because the air cools as it rises over the mountain causing it to drop its moisture.
Bloom’s Level: 1. Remember Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
36-14 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
26. When wet ocean winds blow onshore for half the year and slowly rise and cool, causing long periods of rainfall over a large area, this is referred to as A. lake effect. B. eutrophication. C. a montane successional gradient. D. a monsoon climate. E. rain shadow. Lake effect refers to the increased likelihood of snowfall as warm, moist air moves from over a large body of water, like the Great Lakes, in the winter and loses its moisture as it rises; the precipitation comes in the form of snow because the air cools rapidly as it rises. Eutrophication is the enrichment of water by inorganic nutrients that are used by producers in the lake. The montane successional gradient refers to the change of biome as altitude increases in the mountains. A monsoon climate is one that alternates between dry and wet seasons, each one lasting about half of the year; in the wet season rainfall is frequent and heavy, resulting in flooding. A rain shadow is caused when the leeward side of a mountain receives less rain than the windward side because the air cools as it rises over the mountain, causing it to drop its moisture.
Bloom’s Level: 1. Remember Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
36-15 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
27. The largest communities on land are called A. biospheres. B. estuaries. C. tundras. D. taigas. E. biomes. The biosphere includes the soil of the Earth, the waters, the air, and the surface; organisms are able to live in all of these places. Estuaries are places where a river meets the ocean; these are very rich and diverse areas where salt water and fresh water meet. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the treeline region. Taiga is a biome characterized by a coniferous forest and found around the world in a wide belt in the Northern Hemisphere. A biome is the term used to refer to a certain type of ecosystem; it has a particular set of organisms, climatic conditions and soil conditions, and is found in certain predictable locations.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.01 Identify the geographical distribution of the major terrestrial ecosystems. Section: 36.02 Topic: Terrestrial Ecosystems
28. Which of the following statements is NOT true about biomes? A. They vary according to climate. B. They are communities of similar populations. C. They vary according to latitude and altitude. D. They are the smallest communities on land. E. They are characterized by particular animals and plants. Biomes are broad areas characterized by similar climates, soil, topography, plants, and animals.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
29. Which of the following do NOT help determine a biome? A. amount of rainfall B. solar energy C. soil D. atmospheric content E. mountain ranges Any physical aspect such as rainfall, amount of incoming solar radiation, soil composition, or topography would determine the type of biome present.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
30. Which would be characterized by little rainfall, permafrost, and reindeer? A. grasslands B. temperate deciduous forest C. desert D. taiga E. tundra Grasslands are characterized by limited rainfall that is not sufficient to support trees and large grazing animals. A temperate deciduous forest is characterized by relatively high rainfall, well-defined seasons, and trees that lose their leaves seasonally. Deserts are usually found 30 north or south of the equator and are characterized by very low rainfall and organisms that are well adapted to dry conditions. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the tree-line region. Taiga is a biome characterized by a coniferous forest and found around the world in a wide belt in the Northern Hemisphere.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
31. Trees do NOT grow in the tundra because the A. atmosphere is too rarefied. B. land is too swampy. C. land shifts from time to time. D. permafrost is always frozen. E. shrubs are too plentiful. Since the permafrost is always frozen, the roots of trees cannot find a sufficient hold or obtain the necessary nutrients.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
32. As the latitude increases, which biome sequence would be present? A. tundra, coniferous forest, deciduous forest, tropical forest B. tundra, deciduous forest, coniferous forest, tropical forest C. tropical forest, deciduous forest, coniferous forest, tundra D. tropical forest, coniferous forest, deciduous forest, tundra E. tropical forest, deciduous forest, tundra, coniferous forest Latitude refers to the distance from the equator so the biome found closest to the equator would be a tropical rainforest. Deciduous forests are found south of the taiga and north of tropical forests. Coniferous forests also called taiga are found north of deciduous forests and south of tundra and tundra are found closest to the North Pole.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.01 Identify the geographical distribution of the major terrestrial ecosystems. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
33. Which pair is mismatched? A. tundra—short grasses B. savanna—acacia trees C. prairie—epiphytes D. coniferous forest—evergreen trees E. deserts—cacti Prairies are characterized by grasses, not epiphytes. Epiphytes are plants that obtain their nutrients and moisture from the air and typically grow on another plant and are usually found in tropical areas.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
34. A treeless region with little rainfall in the far north is the A. grassland. B. tundra. C. taiga. D. chaparral. E. desert. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the tree-line region, that is characterized by little rainfall and organisms that migrate in and out of the area.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
36-19 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
35. Treeless biomes are treeless primarily because A. of limited sunlight. B. there is a lack of water or the water is frozen. C. there is no source of seed dispersal in the region. D. trees are dependent upon the animals for support. E. trees cannot withstand hot temperatures or below-freezing cold. Trees require a sufficient supply of water to be successful in an ecosystem. Biomes characterized by little rainfall or an area where the precipitation is frozen and therefore unavailable to the trees do not have trees.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
36. A coniferous forest that runs along the west coast of Canada and the United States, due to the moist winds off the Pacific, is called the A. tropical deciduous forest. B. tundra. C. temperate rain forest. D. taiga. E. temperate deciduous forest. Tropical deciduous forests are seasonal tropical forests that have a wet and a dry season and characterized by trees that shed their leaves; organisms include tigers and elephants. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the tree-line region. Taiga is a biome characterized by a coniferous forest and found around the world in a wide belt in the Northern Hemisphere. A temperate deciduous forest is characterized by relatively high rainfall, well-defined seasons, and trees that lose their leaves seasonally.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
36-20 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
37. A forest found in a broad belt in northern Eurasia and North America, with pine trees, mosses, and lichens, is called the A. tropical rain forest. B. tundra. C. temperate deciduous forest. D. chaparral. E. coniferous forest. A tropical rain forest has high rainfall and warm temperatures, leading to great diversity and an abundance of organisms. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the tree-line region. A temperate deciduous forest is characterized by relatively high rainfall, well-defined seasons, and trees that lose their leaves seasonally. Chaparral is a type of shrubland with dry summers and limited rainfall in winter that occurs in California.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.01 Identify the geographical distribution of the major terrestrial ecosystems. Section: 36.02 Topic: Terrestrial Ecosystems
38. Which of the following biomes contains trees? A. chaparral B. tundra C. desert D. taiga E. savanna Taiga are coniferous forests and so are not treeless. Chaparral is a type of shrubland with dry summers and limited rainfall in winter that occurs in California. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the treeline region. Deserts are usually found 30 north or south of the equator and are characterized by very low rainfall and organisms that are well adapted to dry conditions. Savannas are a type of grassland that have a cool, dry season followed by a hot, rainy season and containing a few hardy trees and abundant species of animals.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
36-21 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
39. The taiga is a A. tropical rain forest. B. temperate deciduous forest. C. chaparral. D. grassland. E. coniferous forest. Taiga is a biome characterized by a coniferous forest and found around the world in a wide belt in the Northern Hemisphere.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
40. Temperate deciduous forests are defined by A. a humid climate with seasonal droughts. B. high variation in day and night length. C. a high amount of organisms living in tree canopies. D. a growing season between 140 and 300 days. E. a rich topsoil horizon. A temperate deciduous forest is characterized by relatively high rainfall, well-defined seasons, and trees that lose their leaves seasonally. The trees have broad leaves and there is a well-developed understory beneath them. Squirrels, rabbits, skunks, and other small mammals live in the woods in addition to a number of species of birds.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
41. Trees such as oak and maple and animals such as foxes and deer are found in the A. tropical rain forest. B. tundra. C. taiga. D. chaparral. E. temperate deciduous forest. A tropical rain forest has high rainfall and warm temperatures, leading to great diversity and an abundance of organisms. Many of the animals are found living in the canopy of the forest. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the tree-line region. The animals are generally small mammals that have an adaptation to allow them to survive the winter or animals that migrate. Taiga is a biome characterized by a coniferous forest and found around the world in a wide belt in the Northern Hemisphere. They have evergreen trees with needle-like leaves and a reduced understory under the shade of the trees. These trees are located in a very cold, dry climate where most of the precipitation is in the form of snow. Chaparral is a type of shrubland with dry summers and limited rainfall in winter that occurs in California. The animals tend to be jack rabbits, coyotes, and rodents.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
42. Which of the following statements is NOT true about deciduous trees? A. They include oak, beech, and maple trees. B. They lose their leaves in the fall. C. They grow their leaves in the spring. D. They are found in a moderate climate with a relatively high rainfall. E. They are found north of the taiga in eastern North America. Deciduous forests are found south of the taiga and north of tropical forests across North America, Europe, and Asia.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
43. Which biome has the greatest diversity of species and total amount of living matter? A. epipelagic region of the ocean B. temperate deciduous forest C. tropical rain forest D. taiga E. African savanna Tropical rain forests have a high level of diversity and abundant resources, including a long growing season which contributes to its ability to capture the energy from the sun and convert it into biomass, giving it a high primary productivity.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
44. Epiphytes are most closely associated with A. tropical rain forests. B. tundra. C. deserts. D. deciduous forests. E. temperate deciduous forests. Epiphytes are plants that live in tropical rainforests. They obtain moisture and nutrients from the air and typically grow on another plant.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
45. Which biome has both a high temperature and plentiful water? A. desert B. tundra C. savanna D. tropical forest E. taiga Tropical forests are characterized by the greatest amount of sunlight, leading to a warm temperature and plentiful rainfall. These abundant resources lead to a high level of diversity and productivity.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
46. The following number of ant species were counted in these biomes. Tundra (5) Taiga (18) Conifer Mixed Forest (41) Temperature Mixed Forest (103) The most likely reason(s) for the different number of ant species in this list is (are) A. greater sunlight. B. variations in soil type. C. different species of predators. D. air pressure, humidity, and resulting rainfall. E. the amount of photosynthetic production, length of the warm season, and diversity of plants. Ants are insects that do not have an adaptation for cold climates, so the number of species would be expected to increase as the temperature of the biome increases. In addition, the biomes closer to the equator have greater diversity and greater primary productivity, allowing the ants to feed on a variety of plant life.
Bloom’s Level: 4. Analyze Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
47. Grasslands do NOT have trees because A. they are next to deserts. B. the soil is poor. C. of the limited rain. D. there is too much snow in winter. E. there is too much sun in summer. Trees require a sufficient supply of water to be successful in an ecosystem. Biomes characterized by little rainfall or an area where the precipitation is frozen and therefore unavailable to the trees do not have trees.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
48. Which biome supports the greatest variety and number of large herbivores? A. desert B. savanna C. prairie D. tundra E. chaparral Deserts are usually found 30 north or south of the equator and are characterized by very low rainfall and organisms that are well adapted to dry conditions, like reptiles and rodents. Savannas are a type of grassland that have a cool, dry season followed by a hot, rainy season and contain a few hardy trees and abundant species of animals. These include antelope, zebras, wildebeest, and water buffalo. Prairies are grasslands found in North America; the grasses can be either short grasses or tall grasses and the animals tend to be small mammals like mice, prairie dogs, and rabbits and the predators that feed on them. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the tree-line region. The animals are generally small mammals that have an adaptation to allow them to survive the winter or animals that migrate. Chaparral is a type of shrubland with dry summers and limited rainfall in winter that occurs in California. The animals here tend to be jack rabbits, coyotes, and rodents.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
49. The flat-topped acacia tree will be found in the A. desert B. tundra C. savanna grassland D. prairie grassland E. chaparral
biome.
Deserts are usually found 30 north or south of the equator and are characterized by very low rainfall and organisms that are well adapted to dry conditions, like reptiles and rodents. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the tree-line region. The animals are generally small mammals that have an adaptation to allow them to survive the winter or animals that migrate. Savannas are a type of grassland that have a cool, dry season followed by a hot, rainy season and containing a few hardy trees like acacia and abundant species of animals. These include antelope, zebras, wildebeest, and water buffalo. Prairies are grasslands found in North America; the grasses can be either short grasses or tall grasses and the animals tend to be small mammals like mice, prairie dogs, and rabbits and the predators that feed on them. Chaparral is a type of shrubland with dry summers and limited rainfall in winter that occurs in California. The animals here tend to be jack rabbits, coyotes, and rodents.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
36-27 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
50. Which would be characterized by rich soils, insufficient rainfall for trees, and many herbivores? A. grasslands B. chaparral C. desert D. taiga E. tundra Grasslands are characterized by either short grasses or tall grasses and the animals that graze on these grasses. The soils are rich but the rainfall is limited, which prevents trees from growing. Chaparral is a type of shrubland with dry summers and limited rainfall in winter that occurs in California. The animals here tend to be jack rabbits, coyotes, and rodents. Deserts are usually found 30 north or south of the equator and are characterized by very low rainfall and organisms that are well adapted to dry conditions, like reptiles and rodents. Taiga is a biome characterized by a coniferous forest and found around the world in a wide belt in the Northern Hemisphere. They have evergreen trees with needle-like leaves and a reduced understory under the shade of the trees. These trees are located in a very cold, dry climate where most of the precipitation is in the form of snow. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the tree-line region. The animals are generally small mammals that have an adaptation to allow them to survive the winter or animals that migrate.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
51. An African tropical grassland that varies cool, dry seasons with hot, rainy seasons is the A. savanna. B. chaparral. C. prairie. D. taiga. E. tundra. Savannas are a type of grassland that have a cool, dry season followed by a hot, rainy season and contain a few hardy trees like acacia and abundant species of animals. These include antelope, zebras, wildebeest, and water buffalo. Chaparral is a shrubland found in California with dry summers and limited rainfall in winter. Prairies are grasslands found in North America characterized by limited rainfall that is insufficient to support trees. Taiga is a biome characterized by a coniferous forest and found around the world in a wide belt in the Northern Hemisphere. These trees are located in a very cold, dry climate where most of the precipitation is in the form of snow. Tundra is a biome with permanently frozen subsoil found close the North Pole, below the ice cap and above the tree-line.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
52. Which would be characterized by rainfall under 25 cm/year, hot days, cold nights, and spiny-leafed plants? A. grasslands B. temperate deciduous forest C. desert D. taiga E. tundra Grasslands are typically characterized by rainfall that is more than 25 cm/year but less than 80 cm/year and they are typically hot in the summer and cold in the winter. Temperate deciduous forests are typically characterized by rainfall that is between 75 and 150 cm/year and there are four distinct seasons. Deserts are typically characterized by rainfall that is less than 25 cm/year and have hot days and cold nights. Taigas are typically characterized by rainfall that is between 30 and 85 cm/year and they are located in a very cold, dry climate where most of the precipitation is in the form of snow. Tundra is typically characterized by rainfall that is less than 25 cm/year and the average annual temperature is between -12 to 6C.
Bloom’s Level: 1. Remember Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
53. Which of the following does not affect the distribution of world biomes? A. altitude B. latitude C. rainfall patterns D. distribution of animals E. sunlight The distribution of animals is based on the type of biome present. Altitude affects the climate of a region because higher altitudes tend to be cooler than lower altitudes. Latitude affects the climate of the region because the farther it is from the equator, the cooler it is and certain latitudes have higher rainfall than others. Rainfall determines the type of plants that can be supported in the region as does the amount of sunlight available. Areas closer to the equator have the greatest rainfall, temperature, and sunlight.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
54. Which of the following biomes has the overall harshest conditions for organism survival? A. tropical rain forest B. eastern deciduous forest C. arctic tundra D. grasslands E. savanna The arctic tundra is cold and dark most of the year. Rainfall amounts to about 20 cm per year.
Bloom’s Level: 5. Evaluate Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
55. The biome is characterized by hot days, cold nights, cacti, and the mesquite tree. A. desert B. tundra C. savanna grassland D. prairie grassland E. chaparral Deserts are typically characterized by rainfall that is less than 25 cm/year and have hot days and cold nights. They are usually found 30 north or south of the equator and have organisms that are well adapted to dry conditions, like reptiles and rodents. Tundra is typically characterized by rainfall that is less than 25 cm/year and the average annual temperature is between -12 to -6C. The animals are generally small mammals that have an adaptation to allow them to survive the winter or animals that migrate. Grasslands are typically characterized by rainfall that is more than 25 cm/year but less than 80 cm/year and they are typically hot in summer and cold in winter. Savannas are a type of grassland that have a cool, dry season followed by a hot, rainy season and contain a few hardy trees like acacia and abundant species of animals. These include antelope, zebras, wildebeest, and water buffalo. Prairies are grasslands found in North America. The grasses can be either short grasses or tall grasses and the animals tend to be small mammals like mice, prairie dogs, and rabbits and the predators that feed on them. Chaparral is a type of shrubland with dry summers and limited rainfall in winter that occurs in California. The animals here tend to be jack rabbits, coyotes, and rodents. Chaparral is a shrubland found in California, with dry summers and limited rainfall in winter.
Bloom’s Level: 2. Understand Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
56. In the summer, a deep lake contains three layers: an upper . A. epilimnion; thermocline; hypolimnion B. thermocline; hypolimnion; epilimnion C. hypolimnion; epilimnion; thermocline D. epilimnion; hypolimnion; thermocline E. thermocline; epilimnion; hypolimnion
, a middle
, and a lower
The waters of a lake are divided into three layers. On top is the epilimnion , which is warmed by the sun. The next layer is the thermocline; this layer abruptly changes in temperature as you go deeper. The bottom layer is cold and is called the hypolimnion.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
57. The mixing of oxygenated water with nutrient laden water that occurs in lakes is called A. intertidal zonation. B. upwelling. C. pelagic division. D. fall overturn. E. eutrophication. In the fall, as air temperatures cool, the epilimnion also cools and eventually these waters become cooler than the waters of the hypolimnion. At this time, the upper waters sink to the bottom and the waters of the hypolimnion rise. This mixes the waters until the temperature is uniform throughout the lake. This also mixes oxygen from the upper waters and nutrients from the lower waters throughout the lake.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
36-33 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
58. In a lake, cooler water on top sinks below warmer water on the bottom during A. the autumnal equinox. B. spring overturn. C. summer stratification. D. eutrophication. E. biological magnification. In the fall, as air temperatures cool, the epilimnion also cools and eventually these waters become cooler than the waters of the hypolimnion. At this time, the upper waters sink to the bottom and the waters of the hypolimnion rise. This mixes the waters until the temperature is uniform throughout the lake. This also mixes oxygen from the upper waters and nutrients from the lower waters throughout the lake.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
59. Freshwater and marine microscopic organisms that freely drift in fresh or salt water are called A. plankton. B. benthic. C. abyssal. D. neritic. E. pelagic. Plankton is the term used for microscopic organisms that drift or swim weakly at the surface of ponds, lakes, or oceans. The benthic region of the lake or the ocean is the bottom, reaching from the shoreline to the deepest regions of the water. The abyssal zone is in the ocean and is the bottom of the deepest part of the ocean. The neritic zone is the shallow region of the ocean at the continental shelf. The pelagic region of the ocean refers to the open water.
Bloom’s Level: 1. Remember Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
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Chapter 36 - The Biosphere
60. The open sunlit layer of the body of a lake is the A. littoral zone. B. limnetic zone. C. profundal zone. D. benthic zone. E. abyssal zone. The littoral zone is the area of a lake where the water meets the shore and plants can emerge from the water. The limnetic region of a lake refers to the open water that is exposed to the sunlight. The profundal zone is the deepest waters of a lake deep enough to not receive light. The benthic region of the lake or the ocean is the bottom, reaching from the shoreline to the deepest regions of the water. The abyssal zone is in the ocean and is the bottom of the deepest part of the ocean.
Bloom’s Level: 1. Remember Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
61. The bottom of a lake or stream harbors worms, clams, and other organisms in the A. littoral zone. B. limnetic zone. C. profundal zone. D. benthic zone. E. abyssal zone. The littoral zone is the area of a lake where the water meets the shore and plants can emerge from the water. The limnetic region of a lake refers to the open water that is exposed to the sunlight. The profundal zone is the deepest waters of a lake deep enough to not receive light. The benthic region of the lake or the ocean is the bottom, reaching from the shoreline to the deepest regions of the water. The abyssal zone is in the ocean and is the bottom of the deepest part of the ocean.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
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Chapter 36 - The Biosphere
62. Aquatic plants are rooted in a shallow lake region near the shore called the A. littoral zone. B. benthic zone. C. pelagic zone. D. profundal zone. E. limnetic zone. The littoral zone is the region where the water meets the shore of a lake or pond; this water is shallow enough that plants can root on the bottom and still emerge from the water. The benthic region of the lake or the ocean is the bottom, reaching from the shoreline to the deepest regions of the water. The pelagic region of the ocean refers to the open water. The profundal zone is the deepest waters of a lake deep enough to not receive light. The limnetic region of a lake refers to the open water that is exposed to the sunlight.
Bloom’s Level: 1. Remember Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
63. The part of a lake where light does not penetrate is the A. littoral zone. B. benthic zone. C. pelagic zone. D. profundal zone. E. limnetic zone. The littoral zone is the region where the water meets the shore of a lake or pond sunlight can reach to the bottom of this zone. The benthic region of the lake or the ocean is the bottom, reaching from the shoreline to the deepest regions of the water; depending on the region of the benthic zone, sunlight may or may not be able to reach. The pelagic region of the ocean refers to the open water and its upper reaches are exposed to sunlight. The profundal zone is the deepest waters of a lake deep enough to not receive light. The limnetic region of a lake refers to the open water whose upper reaches are exposed to sunlight.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
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Chapter 36 - The Biosphere
64. Of all the aquatic communities, A. oceans B. lakes C. sandy beaches D. estuaries E. rivers
act as a nutrient trap.
Estuaries are where rivers flow into the ocean; the flowing waters of the river carry nutrients and soils and then deposit them into the area of interface when the flow of the river is slowed by meeting the ocean. This deposits the nutrients and soil in the area of the estuary, making them nutrient rich.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
65. Which of the following is called the nursery of the sea? A. benthic zone B. pelagic region C. rocky seashore D. estuary E. abyssal zone The benthic region of the lake or the ocean is the bottom, reaching from the shoreline to the deepest regions of the water; organisms that live in this region are called the benthos and generally feed on detritus. The pelagic region of the ocean refers to the open water and its upper reaches are exposed to sunlight where phytoplankton carryout photosynthesis. Estuaries are located where rivers flow into the ocean; the flowing waters of the river carry nutrients and soils and then deposit them into the area of interface when the flow of the river is slowed by meeting the ocean. The nutrient-rich waters attract many photosynthetic organisms and, because of this, many marine invertebrates and fish use them as breeding grounds. The abyssal zone is in the ocean and is the bottom of the deepest part of the ocean and the organisms that live here include echinoderms, worms, and chemosynthetic bacteria.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
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Chapter 36 - The Biosphere
66. Which would NOT be a characteristic of an estuary? A. low percentage of immature animals B. high nutrient concentration C. brackish water D. shallow E. good sunlight penetration Estuaries are located where rivers flow into the ocean; the flowing waters of the river carry nutrients and soils and then deposit them into the area of interface when the flow of the river is slowed by meeting the ocean. This deposits the nutrients and soil in the area of the estuary, making them nutrient rich. The nutrient-rich waters attract many photosynthetic organisms and, because of this, many marine invertebrates and fish use them as breeding grounds. This causes there to be a higher percentage of immature animals in this biome.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
67. The largest biome on Earth is the A. tundra. B. taiga. C. temperate deciduous forest. D. desert. E. ocean. The ocean covers approximately 75 percent of the Earth. It is the largest of all of the biomes.
Bloom’s Level: 1. Remember Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
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Chapter 36 - The Biosphere
68. Ocean currents A. are generated by friction between winds and ocean surfaces. B. always move clockwise in an ocean basin. C. are important in shipping but have little effect on living environments. D. are driven by the Earth's core, rotating faster than the ocean water. E. have no effect on inland weather. The flow of prevailing winds across the surface of the ocean generates friction with the surface of the water and sets up a similar movement of water matching the movement of the air; much like blowing on the surface of a cup of coffee and setting up ripples.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
69. The ocean A. lacks life below the first several hundred feet. B. has zooplankton but not phytoplankton. C. covers little of the Earth's surface. D. is not affected by freshwater pollution. E. covers three-fourths of the planet. The ocean covers approximately 75 percent of the Earth. It is the largest of all of the biomes.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
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Chapter 36 - The Biosphere
70. When an ocean current brings rich nutrients to the surface, it is called a(n) A. eutrophication. B. lake effect. C. upwelling. D. monsoon. E. profundal zone. Eutrophication is the enrichment of water by inorganic nutrients that are used by producers in the lake. Lake effect refers to the increased likelihood of snowfall as warm, moist air moves from over a large body of water, like the Great Lakes, in the winter and loses its moisture as it rises; the precipitation comes in the form of snow because the air cools rapidly as it rises. An upwelling occurs when cold waters and nutrients from the bottom of a body of water are pulled to the surface. A monsoon climate is one that alternates between dry and wet seasons, each one lasting about half of the year; in the wet season, rainfall is frequent and heavy, resulting in floods.
Bloom’s Level: 1. Remember Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
71. Which region of the open ocean is brightly lit? A. epipelagic B. mesopelagic C. bathypelagic D. abyssal E. benthic The epipelagic zone of the ocean is the upper surface of the open ocean that is exposed to sunlight. It is the region where photosynthesis occurs and where the organisms that feed on the phytoplankton live.
Bloom’s Level: 1. Remember Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
36-40 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 36 - The Biosphere
72. The ocean region that contains the neritic province and the oceanic province is the A. littoral zone. B. benthic zone. C. pelagic zone. D. profundal zone. E. limnetic zone. The neritic zone is the region of water that covers the continental shelf; it is part of the waters of the ocean found in the pelagic zone.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
73. Which organisms provide most of the nutrient supply in the ocean? A. small fishes B. large fishes C. seaweeds D. phytoplankton E. zooplankton Phytoplankton are the photosynthetic organisms that serve as the basis of marine food webs.
Bloom’s Level: 5. Evaluate Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
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Chapter 36 - The Biosphere
74. Which of the following is the correct sequence of the different zones in the open ocean (top to bottom)? A. abyssal, bathypelagic, mesopelagic, epipelagic B. abyssal, mesopelagic, bathypelagic, epipelagic C. epipelagic, bathypelagic, mesopelagic, abyssal D. epipelagic, abyssal, bathypelagic, mesopelagic E. epipelagic, mesopelagic, bathypelagic, abyssal The waters of the ocean are divided into four layers. On top is the epipelagic zone, which is warmed by the sun and is the region where photosynthesis is carried out. The next layer is the mesopelagic zone; this layer is home to many species of fish and to mollusks like squid. Next is the bathypelagic zone, where sunlight cannot reach; the organisms that live here are adapted to this lightless region. The bottom layer is completely dark and is called the abyssal zone; the organisms that live here are detritivores that feed on matter that falls from the upper regions or members of a chemosynthetic ecosystem around a thermal vent.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
75. In which of the following ocean zone(s) is there no light? A. epipelagic B. mesopelagic C. bathypelagic D. abyssal E. bathypelagic and abyssal Neither the bathypelagic nor the abyssal zones can be reached by sunlight. The epipelagic zone receives the most sunlight but the mesopelagic does receive some wavelengths of light.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
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Chapter 36 - The Biosphere
76. Organisms living in the brittle stars, and sea urchins. A. epipelagic B. mesopelagic C. bathypelagic D. benthic E. littoral
zone of the open ocean include the tube worms, clams,
Tube worms, clams, brittle stars, and sea urchins are all organisms that are classed as benthos, meaning that they live at the interface of water and sediment on the bottom of the ocean.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
77. There are tubeworms and clams that live at the bottom of the ocean near thermal vents. They derive their energy ultimately from A. the sun, by way of burning fossil fuels like we do with coal and oil. B. the sun, but they are deep in the dark ocean and live off dead detritus that sinks to them. C. bacteria that break apart hydrogen sulfide molecules and are therefore not dependent ultimately on sunlight. D. sources not yet known. E. photosynthesis. Tube worms and some clams are part of an ecosystem at the bottom of the ocean where light cannot reach, that is based on chemosynthesis. In this ecosystem, bacteria use chemical energy to drive the synthesis of organic molecules which then serve as the food source for all of the organisms present.
Bloom’s Level: 2. Understand Learning Outcome: 36.03.02 Recognize the differences between the pelagic and benthic divisions of the ocean. Section: 36.03 Topic: Aquatic Ecosystems
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Chapter 36 - The Biosphere
Short Answer Questions 78. Describe the differences in temperature of deep lakes in a temperate environment. Deep lakes are stratified during the summer and winter. In summer the lake will have three different temperature zones. The epilimnion is the surface layer that is warmed by the sun. The thermocline is the middle layer that decreases 1oC per meter of depth. The lowest layer is the hypolimnion and is cold water. The warmer, less dense water floats on top of the colder, more dense water.
Bloom's Level: 6. Create Learning Outcome: 36.03.01 Compare the characteristics of freshwater and saltwater ecosystems. Section: 36.03 Topic: Aquatic Ecosystems
79. Compare the mean annual temperature and precipitation ranges required for the following biomes: temperate deciduous forest, tropical rainforest, and tundra. The temperate deciduous forest has a mean annual precipitation range of 150-260 cm. The mean annual temperature range is 3oC to 17oC. The tropical rain forest has a mean annual precipitation range of 260-450 cm. The mean annual temperature range is 18oC to 29oC. The tundra has a mean annual precipitation range of 10-20 cm. The mean annual temperature range is -15oC to 4oC.
Bloom's Level: 6. Create Learning Outcome: 36.02.02 Summarize the key characteristics of the major terrestrial biomes. Section: 36.02 Topic: Terrestrial Ecosystems
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Chapter 36 - The Biosphere
80. Explain how a mountain range, located near the ocean, can lead to the formation of a desert on the leeward side of the mountain. When cool, moist air carrying condensation comes off the ocean, it will move over land. When it reaches the mountain, it will blow up and over the top. As it rises, the air cools, causing condensation to occur on the windward side of the mountain range. The cool, dry air rises over the mountain and comes down on the leeward side. As the dry air descends, it often produces a dry, arid type of environment that can result in a desert type of biome.
Bloom's Level: 6. Create Learning Outcome: 36.01.02 Explain how global circulation patterns and physical geographic features are associated with the Earth's temperature and rainfall. Section: 36.01 Topic: Climate and the Biosphere
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Chapter 37 - Environmental Concerns
Chapter 37 Environmental Concerns
Multiple Choice Questions 1. Resources that are NOT in limited supply are A. renewable resources. B. nonrenewable resources. C. squandered resources. D. fossil fuels. E. minerals. Some resources are renewed by the natural processes of the Earth and can be harnessed by humans; renewable energy sources include solar and wind energy.
Bloom's Level: 1. Remember Learning Outcome: 37.05.01: Compare renewable versus nonrenewable resources. Section: 37.05 Topic: Sustainability
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Chapter 37 - Environmental Concerns
2. Which of the following statements about conservation biology is NOT true? A. Conservation biology relies on many subdisciplines of biology including genetics, physiology, and systematics. B. Conservation biology relies on both scientists and government officials for conservation management. C. Conservation biology depends on the education of the public to lead to better consumer choices that facilitate conservation practices. D. Conservation biology relies on both biologists and social scientists to determine courses of action. E. Conservation biology proposes that ecosystems take precedence over activities that sustain human populations. Conservation biology is an interdisciplinary science that draws from almost all branches of biology and many other sciences to study conservation issues and propose courses of action. One important aspect of conservation biology is that the issues occur when there is a human ecosystem interface; this interaction usually leads to the problem. It is therefore necessary to use organizations and institutions to help with planning and instituting policies; these include social scientists, government officials, and the public itself. Conservation biology cannot ignore the needs of humans in its plans for preserving and restoring ecosystems.
Bloom's Level: 5. Evaluate Learning Outcome: 37.01.01: Summarize the goals of conservation biology. Section: 37.01 Topic: Conservation Biology
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Chapter 37 - Environmental Concerns
3. The main goal of conservation biology is to A. Promote the needs of certain ecosystems that are determined to be in danger. B. Apply scientific principles to preserving and restoring ecosystems. C. Prevent the hunting and fishing of all species. D. Prevent the building of large structures like shopping malls and businesses. E. Promote the welfare of endangered species. Conservation biology seeks to understand and develop solutions for environmental issues by the application of scientific principles. While some species are protected from hunting and fishing and some development is advised against because it is determined to be damaging to the environment, human needs must be considered. All ecosystems are considered equally important although some are in greater danger at present than others.
Bloom's Level: 1. Remember Learning Outcome: 37.01.01: Summarize the goals of conservation biology. Section: 37.01 Topic: Conservation Biology
4. Ecosystems are based on complex interactions between communities. What are possible consequences from a disruption of these interactions? A. increases in pollution B. increases in biodiversity C. increases in extinctions D. decreases in pollution E. decreases in extinctions Disrupting the complex interactions in a community can lead to loss of species. If one species is lost it can, in turn, lead to the loss of more species.
Bloom's Level: 5. Evaluate Learning Outcome: 37.01.02: Recognize the spectrum of biodiversity. Section: 37.01 Topic: Biodiversity
37-3 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 37 - Environmental Concerns
5. The most familiar level of biodiversity is A. species diversity B. genetic diversity C. ecosystem diversity D. landscape diversity E. conservation diversity Species diversity refers to the number of species present and is the type of diversity that is typically equated with biodiversity. Genetic diversity refers to the number of alleles in the gene pool of a species. Ecosystem diversity refers to the diversity of complex interactions in an ecosystem and the interdependence of all of the components. Landscape diversity refers to the interactions that occur between several ecosystems that are in proximity to one another.
Bloom's Level: 1. Remember Learning Outcome: 37.01.02: Recognize the spectrum of biodiversity. Section: 37.01 Topic: Biodiversity
6. You are studying the effects of the construction of an interstate highway that divides a valley and separates the existing ecosystem into two. Which kind of diversity would be affected by this construction? A. species diversity B. genetic diversity C. ecosystem diversity D. landscape diversity E. conservation diversity Ecosystem diversity would look at the disruption of the complex interactions in an ecosystem and the resulting changes.
Bloom's Level: 3. Apply Learning Outcome: 37.01.02: Recognize the spectrum of biodiversity. Section: 37.01 Topic: Biodiversity
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Chapter 37 - Environmental Concerns
7. Pandas live in the midst of a bamboo forest located in a mountainous area. The study of effects on pandas when farmlands prevent pandas from moving from one bamboo forest on one mountain to another forest on another mountain would utilize the study of which kind of biodiversity? A. species diversity B. genetic diversity C. ecosystem diversity D. landscape diversity E. conservation diversity When looking at the interaction of organisms within or across two or more ecosystems, the type of diversity involved is landscape diversity.
Bloom's Level: 3. Apply Learning Outcome: 37.01.02: Recognize the spectrum of biodiversity. Section: 37.01 Topic: Biodiversity
8. Which of the following is NOT a reason that genetic diversity is important? A. Evolution can only act on phenotypes that are present in the population. B. The greater the genetic diversity of a population, the more likely that one organism will have an adaptive trait when the environment changes. C. Lower genetic diversity can lead to the extinction of species. D. Higher genetic diversity leads to more species in the ecosystem. E. Low genetic diversity can lead to greater susceptibly to pathogens. Genetic diversity refers to the number of alleles in the gene pool of a species. A gene pool that contains more alleles is more likely to contain alleles that will allow the population to respond to changes in the environment. Natural selection only operates on those traits that already exist in the population, so low genetic diversity increases the likelihood that a species will become extinct.
Bloom's Level: 4. Analyze Learning Outcome: 37.01.02: Recognize the spectrum of biodiversity. Section: 37.01 Topic: Biodiversity
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Chapter 37 - Environmental Concerns
9. A partnership between less developed countries and resorts can lead to an increased income for the country. We call this indirect value in biodiversity A. direct value. B. ecotourism. C. exotic species. D. colonization. E. overexploitation. Ecotourism is a partnership that benefits LDCs and preserves the ecosystems in the country. Ecotours offer tourists the opportunity to see environments and species that they do not exist in their own country.
Bloom's Level: 2. Understand Learning Outcome: 37.02.01: Compare the direct and indirect values of biodiversity. Section: 37.02 Topic: Biodiversity
10. What is the relationship between diversity and latitude? A. As latitude increases, moving from the North Pole to the South Pole, the diversity increases. B. As latitude increases, moving away from the equator at the center of the Earth toward either the North Pole or the South Pole, biodiversity decreases. C. As latitude increases, moving away from the equator at the center of the Earth toward either the North Pole or the South Pole, the biodiversity increases. D. As latitude increases, moving away from the equator at the center of the Earth toward either the North Pole or the South Pole, biodiversity first increases until 60 and then decreases toward the pole. E. There is no relationship between diversity and latitude. The equator has the highest diversity and diversity generally decreases toward the poles, with little diversity in the arctic tundra.
Bloom's Level: 2. Understand Learning Outcome: 37.04.01: Describe the value of preserving biodiversity hotspots. Section: 37.04 Topic: Biodiversity
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Chapter 37 - Environmental Concerns
11. Because certain areas have greater diversity, conservation biology would consider them A. prioritized for preservation. B. typically overcrowded. C. sustainable ecosystems. D. subjected to edge effects. E. less stable. Areas of high diversity are a high priority for preservation because preserving the ecosystem would preserve many species.
Bloom's Level: 2. Understand Learning Outcome: 37.04.01: Describe the value of preserving biodiversity hotspots. Section: 37.04 Topic: Biodiversity
12. Each biosphere reserve area is divided into three areas. Which of these do not describe one of those three areas? A. the inner core area that is the reserve B. the inner core that allows only research, light tourism, and limited sustainable cultural use C. the buffer zone that surrounds the core D. the buffer zone that allows sustainable human development E. the transition zone that surrounds the buffer zone The buffer zone only allows low-impact human activities, whereas the transition zone, which surrounds it, allows sustainable human development.
Bloom's Level: 2. Understand Learning Outcome: 37.04.03: Summarize the goals of habitat restoration. Section: 37.04 Topic: Conservation Biology
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Chapter 37 - Environmental Concerns
13. Which of these activities are permitted in core of the biosphere reserve? A. research and hunting B. light ecotourism and hunting C. hunting and fishing D. light ecotourism and research E. None of the answer choices is correct. The core of the biosphere reserve can be used for activities that will have a very minimum impact on the ecosystem. The two listed that are possibilities are research and light ecotourism. Hunting and fishing would not be permitted because they could interfere with the balance of the ecosystem.
Bloom's Level: 1. Remember Learning Outcome: 37.04.03: Summarize the goals of habitat restoration. Section: 37.04 Topic: Conservation Biology
14. When designing a reserve area, there should be a limit to A. the number of species included so it does not become overcrowded. B. how large the area is so that it can be patrolled and monitored. C. how small the area is so that edge effects will be minimized. D. how close the area is to a national park so that wildlife will not be tempted to move from one area to the other. E. the number of conservation corridors that connect to it. Edge effects limit the amount of usable habitat. The edge has slightly different conditions than the interior and species often cannot use the edge in the same way as the interior.
Bloom's Level: 2. Understand Learning Outcome: 37.04.03: Summarize the goals of habitat restoration. Section: 37.04 Topic: Conservation Biology
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Chapter 37 - Environmental Concerns
15. Restoration ecology A. seeks to design reserves that allow minimum impact on the species in the reserve. B. seeks to return damaged ecosystems to their state prior to the damage. C. seeks to use breeding programs in zoos to return extinct species to the wild. D. seeks to import exotic species into specific areas. E. seeks to encourage sustainable human development. Restoration ecology returns damaged ecosystems to their original condition, using fragments from the original habitat to supply wildlife and seeds to restock the ecosystem. It seeks to create ecosystems that are able to sustain themselves once the restoration process is complete.
Bloom's Level: 1. Remember Learning Outcome: 37.04.03: Summarize the goals of habitat restoration. Section: 37.04 Topic: Conservation Biology
16. Which of the following is NOT a goal when restoring an ecosystem? A. to create a sustainable ecosystem that will continue to function after restoration is complete B. to use the wildlife and seeds from fragments of the original ecosystem to restore the original ecosystem C. to begin restoration as soon as possible to salvage existing fragments of the habitat D. to create a sustainable ecosystem that will provide services to humans E. to use the wildlife and plants from endangered ecosystem to establish a new habitat Restoration ecology tries to return damaged ecosystems to their original state using existing plants and wildlife from adjoining areas. These habitats should continue to function without human intervention and should provide humans with the indirect benefits of a functioning ecosystem.
Bloom's Level: 2. Understand Learning Outcome: 37.04.03: Summarize the goals of habitat restoration. Section: 37.04 Topic: Conservation Biology
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Chapter 37 - Environmental Concerns
17. Which of the following is not a method intended to extend nonrenewable resources? A. efficient use B. recycling C. over demand by MDC (more developed countries) D. better extraction methods E. substitution When a resource is nonrenewable, it is necessary to use methods to conserve that resource. These methods include recycling of resources so that materials are used more than once before they are taken out of the manufacturing process. Sometimes substitution allows a different, more abundant, resource to be used in place of a more limited one. Efficient use of a resource saves portions of the resource that might otherwise be squandered. Better extraction methods allow for more efficient gathering of the resource so that portions are not lost during this process.
Bloom's Level: 2. Understand Learning Outcome: 37.05.02: List two ways to transition to renewable energy resources. Section: 37.05 Topic: Sustainability
18. Which of the following is NOT true regarding pollution? A. Pollution is an undesirable alteration of the environment. B. Pollution increases with an increasing world population. C. The impact of pollution by humans is proportional to the world population. D. Pollution does not have an effect on the living environment. E. It is a side effect of resource consumption. Pollution impacts the living environment in many ways; it leads to decreased biodiversity and an increasing portion of the environment that is not habitable by or useful to humans or other organisms.
Bloom's Level: 2. Understand Learning Outcome: 37.03.01: Classify the five causes of extinction. Section: 37.03 Topic: Extinction
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Chapter 37 - Environmental Concerns
19. Coastal degradation, such as the loss of mangrove swamps and coral reefs, is primarily caused by A. erosion of seashores by storms. B. slash and burn agriculture to raise crops. C. increased ranching for beef export. D. the large concentration of people living along coasts. E. increased high-tech industry. Erosion, pollution, introduced species, and habitat destruction are all products of human habitation along coasts. These all lead to loss of coastal or offshore ecosystems.
Bloom's Level: 1. Remember Learning Outcome: 37.03.02: Compare natural and human influenced causes of extinction. Section: 37.03 Topic: Extinction
20. Allowing animals to overgraze land could lead to all of the following consequences EXCEPT A. desertification. B. famine. C. less land suitable for human habitation. D. more rainwater runoff. E. improved nutrition for local human occupants. Overgrazing can remove all, or almost all, vegetation which can lead to land that cannot be used for human habitation or agriculture. Once this has begun it can lead to desertification, the transforming of non-desert areas into desert. This desert land with no vegetation is more prone to soil loss through erosion.
Bloom's Level: 4. Analyze Learning Outcome: 37.03.02: Compare natural and human influenced causes of extinction. Section: 37.03 Topic: Extinction
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Chapter 37 - Environmental Concerns
21. Which of the following statements is NOT true about tropical rain forests? A. Tropical rain forest land contains rich soil that is desirable as farmland. B. Ranchers only temporarily benefit from deforestation. C. People are settling in tropical rain forests due to road construction. D. Humans remove vegetation for fuel or to feed animals. E. Destruction of rain forests can lead to desertification. The soils of tropical rainforests are nutrient poor since all of the nutrients are locked up in the living organisms. As soon as a leaf falls or an insect dies and falls to the forest floor, decomposers immediately begin the process of recycling the nutrients and producers immediately take up those nutrients.
Bloom's Level: 2. Understand Learning Outcome: 37.05.03: Discuss how modern agriculture can be changed to minimize environmental impacts. Section: 37.05 Topic: Sustainability
22. Oftentimes around the world, famine causes severe loss of life in less developed countries. In many cases, what is the initial step that would eventually lead to a famine situation? A. deforestation B. desertification C. coastal pollution D. no access to drinking water E. increase in human habitation If lands that are not suitable for agriculture are converted from their natural state and the indigenous species removed it can lead to the loss of the land as a resource. Deforestation often starts an area into a pattern of degradation; first the trees and nutrients are lost and then the soil is lost through erosion.
Bloom's Level: 5. Evaluate Learning Outcome: 37.05.03: Discuss how modern agriculture can be changed to minimize environmental impacts. Section: 37.05 Topic: Sustainability
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Chapter 37 - Environmental Concerns
23. What is the main use of most of the world's freshwater supply? A. drinking water B. irrigation of crops C. cooking D. bathing and flushing toilets E. household uses Worldwide, around 70 percent of the fresh water withdrawn is used for agricultural purposes. All domestic combined use comprises around 6 percent.
Bloom's Level: 2. Understand Learning Outcome: 37.05.03: Discuss how modern agriculture can be changed to minimize environmental impacts. Section: 37.05 Topic: Sustainability
24. Dams provide water for 40 percent of irrigated land, yet there are disadvantages to dams. Which of the following is NOT a disadvantage of dam use? A. loss of water due to evaporation B. provide electricity for many countries C. increase in salinity due to evaporation and runoff D. increase in sediment buildup leads to less available water E. depletion of river water Providing electricity for human use would be an advantage because this is a renewable energy source and produces little or no pollution.
Bloom's Level: 2. Understand Learning Outcome: 37.05.01: Compare renewable versus nonrenewable resources. Section: 37.05 Topic: Sustainability
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Chapter 37 - Environmental Concerns
25. During summer months many highly populated communities are placed under water restrictions during which people are not allowed to wash their cars or water their lawns. The reason for this would probably be A. low levels of water in underground aquifers. B. saltwater intrusion. C. high costs of pumping water from other cities. D. excessive rainfall leading to nearby rivers being at flood stage. E. restrictions of pollutants draining into local water supplies. Aquifers hold about 1,000 times the amount of fresh water that falls as rainwater on the land but humans remove the water faster than nature can recharge the aquifer. Depletion of the waters of an aquifer cause problems like subsidence, when the surface of the land settles where the water has been removed, or saltwater intrusion when the aquifer is close to the ocean. Once saltwater has entered the aquifer, it is no longer usable for irrigation of crops, human consumption, or for watering livestock.
Bloom's Level: 3. Apply Learning Outcome: 37.05.02: List two ways to transition to renewable energy resources. Section: 37.05 Topic: Sustainability
26. Excessive use of water from aquifers does NOT lead to which of the following? A. subsidence B. sink holes C. flood situations D. saltwater intrusion E. water shortages Excessive use of water from aquifers would not lead to flooding because the water is sent to specific locations for a specific use. Depletion of the waters of an aquifer cause problems like subsidence, when the surface of the land settles where the water has been removed, or saltwater intrusion when the aquifer is close to the ocean. Once saltwater has entered the aquifer, it is no longer usable for irrigation of crops, human consumption, or for watering livestock.
Bloom's Level: 2. Understand Learning Outcome: 37.05.01: Compare renewable versus nonrenewable resources. Section: 37.05 Topic: Sustainability
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Chapter 37 - Environmental Concerns
27. Modern farming methods have resulted in all of the following EXCEPT A. an increase in the overall food supply. B. less efficient irrigation methods. C. heavy use of fertilizers, pesticides, and herbicides. D. excessive fuel consumption. E. planting of fewer genetic varieties of plants. In more modern irrigation methods a more efficient use is made of the water since fields are no longer flooded using a canal or ditch to divert the water to the field. The increased exposure of water to the atmosphere results in increased evaporation. The use of drip irrigation, while it is the most efficient use of water, is only used on about 1 percent of the irrigated lands in the United States.
Bloom's Level: 4. Analyze Learning Outcome: 37.05.03: Discuss how modern agriculture can be changed to minimize environmental impacts. Section: 37.05 Topic: Sustainability
28. Marginal rangeland becomes desert and farmland loses its productivity as a result of lost topsoil. Which of the following is a main cause of topsoil loss? A. Strong winds and rain on bare soil cause erosion, along with the use of large farming machines. B. Farmers plant too many crops on smaller pieces of land. C. Excessive animal grazing on land results in loss of vegetation and, therefore, topsoil. D. The use of fertilizers, along with the use of large farming equipment, leads to topsoil losses. E. The accumulation of mineral salts and the evaporation of excess irrigation water are leading causes of the loss of topsoil. The loss of topsoil results in land that is not sufficient in nutrients for growing crops. Many traditional agricultural practices involve digging into the soil and turning the bottom layers to that they are on top. This practice also exposes the bare ground to the elements of wind and water, which remove layers of the soil and transport it to another location. The dust bowl of the Great Plains of the United States is an extreme example of this phenomenon. Soil from the Dakotas traveled all the way to Asia on the winds.
Bloom's Level: 2. Understand Learning Outcome: 37.05.03: Discuss how modern agriculture can be changed to minimize environmental impacts. Section: 37.05 Topic: Sustainability
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Chapter 37 - Environmental Concerns
29. Which of the following does NOT result in the loss of topsoil? A. planting crops in straight rows B. use of large farming machines C. salinization D. overuse of fertilizer E. wind and rain The planting of crops in straight rows increases the topsoil lost to erosion as water washes down the row. Large farm machinery also exposes soil to both water and wind and leads to erosion. Soil can be "lost" as a resource for growing crops, even if it is not removed from the field through an increase in the salts in the soil, called salinization. The increase of salts in the soil disrupts the water uptake of water into the roots of the plant and kills the plant.
Bloom's Level: 2. Understand Learning Outcome: 37.05.03: Discuss how modern agriculture can be changed to minimize environmental impacts. Section: 37.05 Topic: Sustainability
30. The depletion of the world's fisheries is most likely due to which of the following? A. sport fishing B. commercial large-scale fishing C. discharge of sewage and garbage from cruise ships D. natural climate shifts E. natural extinction of fish population Commercial large-scale fishing is becoming more and more efficient and typically removes the large, mature, reproducing adults and returns the small, young, non-reproducing fish. As the reproductive base is depleted, the population recovers more and more slowly and eventually will not be able to recover.
Bloom's Level: 2. Understand Learning Outcome: 37.03.02: Compare natural and human influenced causes of extinction. Section: 37.03 Topic: Extinction
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Chapter 37 - Environmental Concerns
31. The situation with major ocean fishing areas is A. good since the oceans are so large; they are the one remaining natural ecosystem. B. generally positive, with only 1 or 2 of the 15 major oceanic fishing areas overexploited. C. generally negative, due to increased number and efficiency of fishing boats and the negative impact on biodiversity. D. positive for the bottom dwellers that are protected, but negative for the schools of swimming fish. E. negative for the bottom dwellers that are over-fished by trawling, but positive for the schools of swimming fish that can easily evade capture. Many countries use the resources produced by the open waters of the ocean to supply them with food and with an income. With a larger and larger human population, the pressure on the ocean’s fish populations increases every year. Commercial large-scale fishing is becoming more and more efficient and typically removes the large, mature, reproducing adults and returns the small, young, non-reproducing fish. As the reproductive base is depleted, the population recovers more and more slowly and eventually will not be able to recover.
Bloom's Level: 4. Analyze Learning Outcome: 37.03.02: Compare natural and human influenced causes of extinction. Section: 37.03 Topic: Extinction
32. Which of the following does NOT pertain to fossil fuels? A. nonrenewable resources B. contribute to greenhouse gases C. oil and coal D. formed during the last 1,000 years E. The United States uses more than half of the world's supply of fossil fuels. Fossil fuels are deposits of carbon from organisms that lived during the Carboniferous period around 300 to 350 million years ago. Since they were formed from massive amounts of dead organisms so long ago, they are a nonrenewable resource with a finite amount available. The typical forms are coal and oil and emit pollutants, including greenhouse gases, when burned to produce energy.
Bloom's Level: 4. Analyze Learning Outcome: 37.05.01: Compare renewable versus nonrenewable resources. Section: 37.05 Topic: Sustainability
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Chapter 37 - Environmental Concerns
33. Which of the following is NOT a disadvantage to the use of fossil fuels? A. contribution to the greenhouse gases B. excessive pollution C. nonrenewable resource D. transference of fossil fuel energy into food energy E. increased methane production Fossil fuels are deposits of carbon from organisms that lived during the Carboniferous period around 300 to 350 million years ago. Since they were formed from massive amounts of dead organisms so long ago, they are a nonrenewable resource with a finite amount available. The typical forms are coal and oil and emit pollutants, including greenhouse gases and methane, when burned to produce energy.
Bloom's Level: 2. Understand Learning Outcome: 37.05.01: Compare renewable versus nonrenewable resources. Section: 37.05 Topic: Sustainability
34. Which of the following is a renewable energy sources? A. coal B. oil C. hydropower D. natural gas E. fossil fuels Renewable energy sources are ones that are constantly produced by natural processes and are available in unlimited supply. Any energy sources that are available in limited amounts and were created sometime in the past are nonrenewable resources. We draw from existing reservoirs of these nonrenewable resources. Hydropower is created using the renewable resource of water movement. As water flows downhill (usually over a dam), it is used to turn a turbine and generate electricity.
Bloom's Level: 1. Remember Learning Outcome: 37.05.01: Compare renewable versus nonrenewable resources. Section: 37.05 Topic: Sustainability
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Chapter 37 - Environmental Concerns
35. Which of the following countries is the world's leading producer of hydropower? A. United States B. Brazil C. Canada D. Switzerland E. Germany In Canada, 97 percent of the renewable energy produced is from hydropower. Brazil and Switzerland produce at least 75 percent of their renewable energy through hydroelectric power, whereas the United States produces 67 percent of their renewable energy through hydroelectric power.
Bloom's Level: 1. Remember Learning Outcome: 37.05.02: List two ways to transition to renewable energy resources. Section: 37.05 Topic: Sustainability
36. is when underground water is heated by radioactive decay and the steam it produces is then used for home heating or hot water use. A. Hydropower B. Wind power C. Fossil fuel burning D. Geothermal energy E. Solar energy Geothermal energy is produced by the Earth’s internal heat. When water comes into contact with the heat source, hot water and steam are produced. This water and steam can be captured and piped to the surface to be used either directly as a heat source or indirectly to produce electricity.
Bloom's Level: 1. Remember Learning Outcome: 37.05.01: Compare renewable versus nonrenewable resources. Section: 37.05 Topic: Sustainability
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Chapter 37 - Environmental Concerns
37. Conservation biology supports all of the following ethical principles EXCEPT A. biodiversity is desirable for the biosphere and humans. B. all animals have the same rights or legal status as humans. C. biodiversity has value in and of itself, regardless of any practical human benefits. D. extinctions due to human actions are undesirable. E. the complex interactions of ecosystems support biodiversity and are desirable. While conservation biology proposes that all animals and plants should be valued for their own sake and for the benefits humans gain from them, it does not propose legal right for animals.
Bloom's Level: 2. Understand Learning Outcome: 37.01.01: Summarize the goals of conservation biology. Section: 37.01 Topic: Conservation Biology
38. Which of the following statements is NOT true about biodiversity? A. The really important species of animals and plants are those that we rely on for food. If we can protect them, we can avoid any harmful effects even if other species go extinct. B. The size of tropical rain forests have been reduced from 14 percent to 6 percent. C. Exotic species play a role in 50 percent of extinctions. D. Coral reefs may become extinct in the next 40 years. E. Overexploitation by human activities can lead to a loss of biodiversity. All species are important in maintaining a functioning ecosystem and a functioning biosphere. The species that we rely on for food are only a small part of the complex, interconnected system.
Bloom's Level: 2. Understand Learning Outcome: 37.01.02: Recognize the spectrum of biodiversity. Section: 37.01 Topic: Biodiversity
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Chapter 37 - Environmental Concerns
39. In spite of hundreds of millions of dollars spent on insecticides targeted at pest species, such as mosquitoes and crop pests, we have been unable to drive any to extinction. On the other hand, bird, fish, plant, and insect species that we barely recognize have gone extinct. This has mainly been due to A. hunting. B. farming. C. insecticides such as DDT. D. habitat destruction. E. scientific collecting. While hunting, fishing, and insecticide used to play a role in the extinction of species, by far the most serious threat to the loss of biodiversity is habitat loss. In areas such as a tropical rainforest, the species can be lost when farmers use a slash-and-burn method of agriculture and their existence is never known before they are lost.
Bloom's Level: 3. Apply Learning Outcome: 37.03.01: Classify the five causes of extinction. Section: 37.03 Topic: Extinction
40. Which of the following is true about extinction? A. Overexploitation by hunting and fishing is the foremost threat to tropical animals and plants. B. Without human activity, all species would survive and there would be no extinctions. C. There are many causes of extinctions and an animal or plant may go extinct for any of a combination of these factors. D. Commercial hunting and fishing are careful not to deplete animal reserves. E. Introduction of new alien species helps preserve biodiversity and slow extinction. The five main factors that lead to species extinction are habitat loss, introduction of exotic species, pollution, overexploitation, and disease. Species that are subject to one of these factors are also subject to one or more others and the combination of pressures pushes them closer to extinction.
Bloom's Level: 2. Understand Learning Outcome: 37.03.01: Classify the five causes of extinction. Section: 37.03 Topic: Extinction
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Chapter 37 - Environmental Concerns
41. Which one of the following is NOT a way that invasive (alien) species are introduced into a new habitat? A. intentionally introduced by humans B. unintentionally introduced during transport of crops C. unintentionally introduced during discharge of ship ballast tanks D. natural introductions on the feet, wings, or fur of migrating organisms E. unintentionally introduced in packaged, processed foods Packaged, processed foods would not contain exotic species because the processing of the food is designed to kill any organisms that might be present.
Bloom's Level: 4. Analyze Learning Outcome: 37.03.02: Compare natural and human influenced causes of extinction. Section: 37.03 Topic: Extinction
42. Introduction of alien or exotic species into new ecosystems does NOT occur by A. colonization when new settlers arrive in an area. B. accidental transport by ship or plane without anyone's knowledge. C. introduction as new agriculture by settlers. D. horticulture use. E. expansion and exploration of the new species itself. The natural process of expansion and exploration by a species would allow for natural evolutionary interactions to happen between the species and so would not lead to the disruption caused in an ecosystem by species that have had no prior evolutionary interaction.
Bloom's Level: 3. Apply Learning Outcome: 37.03.02: Compare natural and human influenced causes of extinction. Section: 37.03 Topic: Extinction
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Chapter 37 - Environmental Concerns
43. Which is NOT an example of an alien or exotic species? A. zebra mussels introduced to the Great Lakes B. dandelions planted by pioneers as salad greens C. the mongoose brought to Hawaii to control rats D. the brown tree snake in Guam E. buffalo in the United States Buffalo are indigenous to the United States and so are not an introduced species.
Bloom's Level: 1. Remember Learning Outcome: 37.03.02: Compare natural and human influenced causes of extinction. Section: 37.03 Topic: Extinction
44. Which of the following statements is NOT a correct association about pollution? A. eutrophication—a lake dies due to lack of nutrients B. acid deposition—water vapor and exhausts combine to form wet or dry acid deposits C. climate change—changes in the Earth's climate are causing the sea levels to rise D. organic chemicals—synthetic chemicals mimic hormones and cause abnormal growth and behavior in wild animals E. ozone depletion—CFCs like Freon have chlorine atoms that break down the ozone shield Eutrophication refers to the increased nutrient load in a lake, leading to a rapid increase in the plant and algae populations in the lake. Once these populations reach their peak level and die off, they cause an increase in the number of decomposers in the water and a corresponding loss of oxygen in the water.
Bloom's Level: 2. Understand Learning Outcome: 37.03.01: Classify the five causes of extinction. Section: 37.03 Topic: Extinction
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Chapter 37 - Environmental Concerns
45. Rosy periwinkle, the nine-banded armadillo, and some species of fungus and bacteria are examples of value drawn from biodiversity. A. agricultural B. medicinal C. consumptive use D. indirect E. no Most drugs that are in use today were first derived from a natural source. The rosy periwinkle, nine-banded armadillo, and penicillin are all examples of organisms that have been used to develop medications.
Bloom's Level: 1. Remember Learning Outcome: 37.02.01: Compare the direct and indirect values of biodiversity. Section: 37.02 Topic: Biodiversity
46. The regulation of climate, cleansing of fresh water, and support of ecotourism are examples of value drawn from biodiversity. A. agricultural B. medicinal C. consumptive use D. indirect E. no The natural processes that ecosystems use when functioning correctly benefit humans because they perform much-needed services like cleaning water and air. These services are considered indirect benefits since they are found worldwide and most humans are not even aware that they are occurring.
Bloom's Level: 1. Remember Learning Outcome: 37.02.01: Compare the direct and indirect values of biodiversity. Section: 37.02 Topic: Biodiversity
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Chapter 37 - Environmental Concerns
47. The use of uncultivated fruit, skins, fiber, beeswax, etc. are examples of drawn from biodiversity. A. agricultural B. medicinal C. consumptive use D. indirect E. no
value
Uncultivated products that are produced by nature and consumed by humans are said to have consumptive value because they are used/consumed by humans who have not invested in the production of the product.
Bloom's Level: 1. Remember Learning Outcome: 37.02.01: Compare the direct and indirect values of biodiversity. Section: 37.02 Topic: Biodiversity
48. The use of wheat, corn, rice, and other crops modified from wild ancestors are an example of value drawn from biodiversity. A. agricultural B. medicinal C. consumptive use D. indirect E. no Ancient man began as nomadic hunter gathers, meaning that they hunted for the available animals and gathered the available plant matter but did not grow these resources. After humans settled, they began to cultivate plants for their use and also began to breed them for a better production of the usable part of the plant. Wheat, corn, and rice are among some of the earliest cultivated plants. They resemble little of their wild ancestors that had a much lower yield of gain per plant than the modern versions.
Bloom's Level: 1. Remember Learning Outcome: 37.02.01: Compare the direct and indirect values of biodiversity. Section: 37.02 Topic: Biodiversity
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Chapter 37 - Environmental Concerns
49. Which of the following is NOT correct concerning the indirect values of biodiversity? A. Trees affect the climate and provide shade and cooling during the summer. B. Trees can maintain rainfall to help prevent dry periods. C. If biodiversity cannot maintain the biogeochemical cycles, then technology will have to artificially create those cycles. D. Diverse organisms are critical in breaking down wastes and immobilizing pollutants. E. Erosion can lead to critical imbalances in an ecosystem. Technological means of maintaining an ecosystem are extremely expensive if they do exist. Water purification plants and the cleanup of spills of pollutants are examples of processes that technology can do. Most of the services provided by a functioning ecosystem cannot be replicated by technology. The climate and biogeochemical cycles are examples of services that cannot be mimicked by technological means.
Bloom's Level: 4. Analyze Learning Outcome: 37.02.01: Compare the direct and indirect values of biodiversity. Section: 37.02 Topic: Biodiversity
50. An example of ecotourism would include A. charging an airport tax to compensate for the pollution from jetliners. B. vacationing to hike or watch birds in a natural outdoor environment. C. volunteering to conduct ecological research at a university. D. keeping an organic garden in your backyard. E. planting trees on a vacant city lot. Ecotourism includes activities in which individuals travel to a specific location in order to experience the natural ecosystem of that area.
Bloom's Level: 2. Understand Learning Outcome: 37.02.01: Compare the direct and indirect values of biodiversity. Section: 37.02 Topic: Biodiversity
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Chapter 37 - Environmental Concerns
51. Which of the following is an example of the importance of biodiversity in a natural ecosystem? A. pollination of flowers by bees B. production of antibiotics from fungi C. increasing the survival rate of childhood leukemia from 10 percent to 90 percent D. increasing the availability of food for the human population E. supplies building materials for new construction Pollination of flowers by bees is the only example of the value of biodiversity in a natural ecosystem. All of the others are examples of the value of biodiversity to the human population, which is not a natural ecosystem.
Bloom's Level: 1. Remember Learning Outcome: 37.02.02: Describe the role that biodiversity plays in a natural ecosystem. Section: 37.02 Topic: Biodiversity
Short Answer Questions 52. Compare the direct and indirect values of biodiversity by listing three direct values and four indirect values of biodiversity. Direct values of biodiversity include medicinal, agricultural, and consumptive values. Indirect values of biodiversity include regulating biogeochemical cycles, waste disposal, provision of fresh water, flood prevention, prevention of soil erosion, regulation of climate, and ecotourism.
Bloom's Level: 6. Create Learning Outcome: 37.02.01: Compare the direct and indirect values of biodiversity. Section: 37.02 Topic: Biodiversity
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Chapter 37 - Environmental Concerns
Multiple Choice Questions 53. Which service provided by the natural biodiversity of an ecosystem is unable to be duplicated by human technology? A. biogeochemical cycles B. waste disposal C. provision of fresh water D. prevention of soil erosion E. flood prevention Human technology is unable to artificially replicate natural biogeochemical cycles. Technology has developed methods to dispose of waste, clean, and purify fresh water, prevent soil erosion, and prevent floods. While technology has been able to replicate many of the natural functions of biodiversity, it is not as efficient as the natural biodiversity.
Bloom's Level: 4. Analyze Learning Outcome: 37.02.02: Describe the role that biodiversity plays in a natural ecosystem. Section: 37.02 Topic: Biogeochemical Cycles
True / False Questions 54. Natural ecosystems are able to break down dead organic material and recycle their nutrients. TRUE It is true that natural ecosystems are able to break down dead organic material and recycle their nutrients.
Bloom's Level: 1. Remember Learning Outcome: 37.02.02: Describe the role that biodiversity plays in a natural ecosystem. Section: 37.02 Topic: Biodiversity
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Chapter 37 - Environmental Concerns
Multiple Choice Questions 55. Which of the following species would be considered a flagship species? A. dolphins B. grasshopper C. ants D. bats E. roses Flagship species are those that evoke an emotional response from humans. They are considered charismatic and valued for their beauty, regal nature, or similarity to people's pets. Dolphins are the only species considered a flagship species.
Bloom's Level: 1. Remember Learning Outcome: 37.04.02: Distinguish between keystone species and flagship species. Section: 37.04 Topic: Conservation Biology
56. The loss of a keystone species would lead to A. a great number of secondary extinctions. B. an increase in species diversity within the community. C. an increase in the population of other keystone species. D. immediate collapse of the community. E. no noticeable change in the biodiversity of the community. The loss of a keystone species would lead to a great number of secondary extinctions. Species diversity would decrease within the community due to the loss of the keystone species. There is only one keystone species in a community, so there would not be an increase in the populations of other keystone species. It will take some time for the loss of the keystone species to have an impact on the community, so there would not be an immediate collapse of the community. There will be a noticeable change in the biodiversity of the community as the food chains and food webs become disrupted.
Bloom's Level: 1. Remember Learning Outcome: 37.04.02: Distinguish between keystone species and flagship species. Section: 37.04 Topic: Conservation Biology
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Chapter 37 - Environmental Concerns
57. Which species can be used to help motivate the public to conserve biodiversity? A. giant pandas B. lions C. tigers D. dolphins E. All of the answer choices are correct. All of these species would be considered flagship species. They are species that are considered charismatic and valued for their beauty, regal nature, or similarity to people's pets. Flagship species can motivate the public to conserve biodiversity.
Bloom's Level: 1. Remember Learning Outcome: 37.04.02: Distinguish between keystone species and flagship species. Section: 37.04 Topic: Conservation Biology
True / False Questions 58. Biodiversity hotspots are valuable because they contain large numbers of exotic species. FALSE Biodiversity hotspots are valuable because they contain large numbers of endemic species. Exotic species often disrupt and create problems within ecosystems.
Bloom's Level: 1. Remember Learning Outcome: 37.04.01: Describe the value of preserving biodiversity hotspots. Section: 37.04 Topic: Conservation Biology
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Chapter 37 - Environmental Concerns
Short Answer Questions 59. List and explain the five causes of extinction. 1. Habitat loss: habitat loss and habitat disruption of natural habitats are the most influential factors in biodiversity loss. 2. Exotic species: introduction of non-native species into a community can disrupt the balance of the community. 3. Pollution: environmental change that adversely affects the lives and health of living organisms can lead to species extinction. 4. Overexploitation: the number of individuals taken form the wild population exceeds the ability of the population to replace those individuals. 5. Disease: wildlife is subject to new pathogens that they do not have resistance to.
Bloom's Level: 6. Create Learning Outcome: 37.03.01: Classify the five causes of extinction. Section: 37.03 Topic: Extinction
60. Explain the four ethical principles embodied by conservation biology. 1. Biodiversity is desirable for the biosphere and therefore humans. 2. Human-induced extinctions are therefore undesirable. 3. Complex interactions within ecosystems and communities support biodiversity and the maintenance of such interactions is therefore desirable. 4. Biodiversity generated by evolutionary change has intrinsic value, regardless of any practical benefit.
Bloom's Level: 6. Create Learning Outcome: 37.01.01: Summarize the goals of conservation biology. Section: 37.01 Topic: Conservation Biology
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