Test Bank for Biology How Life Works 3rd Edition by Ace Test Banks

Test Bank for Biology How Life Works 3rd Edition

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Chapter 1 Multiple Choice 1. Nucleic acids are important information storage molecules present in virtually every cell. Which of the processes is carried out by a cell when it accesses that information in the DNA of the genes? a. transcription and translation b. DNA replication and transcription c. DNA replication and translation ANSWER: a 2. Observations are used by scientists to draw tentative explanations called hypotheses. a. true b. false ANSWER: a 3. When small molecules are linked together to form larger molecules, the increase in entropy typically comes from: a. gas produced. b. light captured (for example, in photosynthesis). c. work done (for example, creation of new bonds in larger molecules). d. heat loss. e. enzymes utilized. ANSWER: d 4. Which process is an example of the first law of thermodynamics in action? a. As monomers combine into polymers, the disorder inside the cell decreases. b. Light energy is transformed into chemical energy during photosynthesis. c. Energy is created by cells during ATP synthesis. d. Some energy is released as heat during metabolic processes. ANSWER: b 5. Consider the image. If the ostrich egg shown in the photo is not fertilized, it is composed of approximately how many cells?

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a. 1 b. 100 c. 10,000 d. 1,000,000 e. 100,000,000 ANSWER: a 6. Imagine that you are standing in a field and you see a group of butterflies. You notice an individual butterfly that looks significantly different from the others in the population. Its difference allows the butterfly to escape predation more efficiently than the other butterflies. How might this trait have arisen in the individual? a. There were more predators in the surrounding area, so the butterfly needed the trait in order to escape predation. b. There was a mutation in a gene that led to differences in the ability to attract mates. c. There was a random mutation in a gene that led to differences in the ability to escape predation. d. There were more predators in the surrounding area, so the butterflies allowed themselves to be caught to save the faster butterflies in the population. ANSWER: c 7. All of the classifications are considered domains of life except: a. Bacteria. b. Eukarya. c. Archaea. d. Protists. e. Protists, Eukarya, Archaea, and Bacteria are all domains of life. ANSWER: d 8. Consider the phylogenetic tree. According to this phylogenetic tree, the primate most closely related to humans is the:

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a. lemur. b. chimpanzee. c. gorilla. d. orangutan e. gibbon. ANSWER: b 9. Trees in the desert and trees in the rainforest experience vast differences in the amount of water available for uptake. Water can be lost from the leaf surface very easily in dry and hot regions. What types of differences might you expect between tree species in a rainforest compared with those in a desert? a. Rainforest trees have fewer adaptations for conserving water than desert trees. b. Rainforest trees have adaptations for requiring less water than trees in the desert. c. Rainforest trees do not have any adaptations related to water conservation or loss. d. Rainforest trees have more adaptations for water conservation and loss than trees in the desert. ANSWER: a 10. Ecological relationships reflect the _____ traits of organisms in nature. a. biomechanical b. behavioral and physiological c. behavioral d. physiological e. behavioral, physiological, and biomechanical ANSWER: e 11. The figure illustrates the projected changes in distributions of beech trees and chinquapin oak trees in Japan if human activities continue to cause global temperatures to rise.

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Which of the statements accurately reflects these predictions? a. Beech trees will become extinct b. The distribution of beech and chinquapin oak in areas where they are found together will stay the same. c. Beech distribution will increase to the south of its present-day distribution. d. Chinquapin oak distribution will increase with rising temperatures. ANSWER: d 12. Which one of the statements about the human impact on Earth's ecology is correct? a. All of these statements about the human impact on Earth's ecology are correct. b. Agriculture has increased abundance and distribution of some species, while decreasing abundance and distribution of other species. c. Humans commandeer as much as 25% of all photosynthetic production on land. d. More atmospheric nitrogen is converted to ammonia by humans than by the rest of nature. e. Human activities produce more carbon dioxide than do volcanoes. ANSWER: a 13. What is a hypothesis? a. the same thing as an unproven theory b. a tentative explanation that can be tested by experiments c. a verifiable observation d. an experiment that leads to a prediction e. None of the other answer options is correct. ANSWER: b 14. When carrying out a controlled experiment, it is important to: a. subject different groups to different conditions. b. change multiple variables at once to see the full effect of the variables. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 c. change only one variable at a time. d. All of these choices are correct. ANSWER: c 15. A hypothesis is considered a theory when the: a. results of several experiments do not support the hypothesis. b. results of a single experiment support the hypothesis. c. hypothesis has been revised many times. d. results of several experiments support the hypothesis. ANSWER: d 16. Which one of the steps would occur last as part of the scientific inquiry? a. prediction b. hypothesis c. experimentation d. observation ANSWER: c 17. Which one of the statements explains a characteristic of both all living organisms and all nonliving material? a. They both conform to the basic laws of chemistry and physics. b. They both have the capacity to evolve. c. They both have the ability to reproduce. d. They both are complex, with spatial organization at several levels. e. They both have the ability to change in response to the environment. ANSWER: a 18. The first law of thermodynamics states that the degree of disorder in the universe tends to increase. a. true b. false ANSWER: b 19. In which of the examples is the entropy definitely increasing? a. photosynthesis b. making a house of playing cards c. melting ice in a glass of soda d. placing marbles in a row ANSWER: c 20. Which of the examples are cellular life forms: virus, yeast, bacteria, plant, animal? a. All are cellular except viruses. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 b. All are cellular except bacteria. c. All are cellular except yeast. d. Only plants and animals are cellular e. All of these choices are correct. ANSWER: a 21. A retrovirus hijacks a cell's machinery to turn its own RNA into DNA, which is then inserted into the cell's genome so that viral proteins can be made. Does this follow the central dogma? a. yes b. no ANSWER: b 22. Imagine you are standing in a field and you see a group of butterflies. You notice an individual that looks significantly different from the other butterflies in the population. It has much larger wings and can fly faster than the other butterflies. This difference allows the butterfly to escape predation more efficiently than other butterflies in the population. The difference came about because this butterfly spent more time developing in an area of the habitat that is particularly warm and moist and had more food available than other areas. Do you expect the following generations to be composed of butterflies with larger wings? a. No, this represents nonheritable variation caused by the environment. b. No, this represents heritable variation but this trait does not increase ability to mate so will not be passed to the next generation. c. Yes, this represents a genetic mutation that will be passed onto offspring. d. Yes, this trait is advantageous so more butterflies will choose to develop in warm moist areas of the habitat. ANSWER: a 23. Most of life's diversity is: a. aquatic. b. acellular. c. aerobic. d. terrestrial. e. microbial. ANSWER: e 24. Which one of the types of organisms most closely resembles the first cells on Earth? a. bacteria b. algae c. fungi d. plant cells e. animal cells ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 25. Which one of the statements about viruses is correct? a. Some types of viruses integrate their genomes into the genome of the host organism. b. Viruses use the cellular machinery of other organisms to replicate the virus's genetic material. c. The genomes of viruses are enclosed in a protein coat and occasionally a lipid bilayer envelope. d. Viruses are valuable tools in biological research. e. All of these choices are correct. ANSWER: e 26. Some plants are dependent on animals for: a. seed dispersion only. b. pollination, seed dispersion, and respiration. c. pollination only. d. pollination and seed dispersion. e. respiration only. ANSWER: d 27. If the duration of life on Earth was on a scale of 20,000 days, for how many of those days would the human species be present? a. 1 b. 100 c. 2000 d. 10,000 e. 20,000 ANSWER: a 28. Which one of the organisms has been most damaged by the influence of humans? a. rats b. corn c. cockroaches d. apples e. white rhinos ANSWER: e 29. Humans have affected different organismal populations in which ways? a. All of these choices are correct. b. habitat destruction c. agriculture d. overfishing ANSWER: a 30. Hypotheses can only be tested by doing experiments. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 a. true b. false ANSWER: b 31. Let's say you feel very strongly that cigarette smoke does not increase the probability of getting cancer, and you base your view on something you read on the Internet. This is a good example of a(n): a. experiment. b. observation. c. hypothesis. d. theory. e. None of the other answer options is correct. ANSWER: e 32. Many salmon return to the place where they were born to spawn (reproduce). You hypothesize that they use visual cues to find their way back. To test your hypothesis, you blind salmon and then examine whether or not they are able to return to their birthplace. You find that they are able to find their way back. From this experiment, you: a. have rejected your hypothesis. b. have proven your hypothesis. c. supported your hypothesis. d. can't determine whether your hypothesis is supported or not. e. developed a theory about the role of vision in salmon navigation. ANSWER: a 33. Which order accurately reflects the process of science as described in your textbook? a. Observation → question → hypothesis formulation → experiment → support or refute hypothesis b. Observation → hypothesis formulation → question → experiment → prediction c. Observation → question → hypothesis formulation → experiment → prove or disprove hypothesis d. Observation → question → experiment → hypothesis formulation → prove or disprove hypothesis e. Observation → question → experiment → hypothesis formulation → support or refute hypothesis ANSWER: a 34. An explanation supported by a large body of observations and experimentation is referred to as a(n): a. hypothesis. b. prediction. c. theory. d. supposition. e. investigation. ANSWER: c 35. In the 1600s, Francesco Redi demonstrated that living organisms come from other living organisms. However, it would be inaccurate to say that Redi supported his hypothesis because: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 a. his experiment didn't have the proper controls. b. his experiment was done so long ago. c. his experiment was based on observations. d. his experiment only investigated a single kind of meat. e. his experiment only investigated a single organism. ANSWER: e 36. When you eat a hamburger, some of the energy in the food is converted to ATP that your cells can use to do all kinds of work, some of the energy is stored for later use, and some of the energy is dissipated as heat. The amount of energy before and after eating the hamburger is the same. This illustrates the: a. the theory of evolution. b. second law of thermodynamics. c. cell theory. d. first law of thermodynamics. e. central dogma. ANSWER: d 37. When you eat a hamburger, some of the energy in the food is converted to ATP that your cells can use to do all kinds of work, some of the energy is stored for later use, and some of the energy is dissipated as heat. In other words, you can only make use of a portion of the energy available in the hamburger because some is always lost as heat. This is a consequence of the: a. second law of thermodynamics. b. first law of thermodynamics. c. cell theory. d. theory of evolution. e. central dogma. ANSWER: a 38. Which one of the elements makes up more than 40% of both living organisms and the Earth's crust? a. hydrogen b. oxygen c. silicon d. carbon e. nitrogen ANSWER: b 39. When we say that the cell is the fundamental unit of life, we mean that: a. life doesn't exist in the absence of cells. b. all living things are made up of one or more cells. c. the smallest entity that can be considered living is a cell. d. a single cell can carry out all life processes. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 e. all of these choices are correct. ANSWER: e 40. The metabolic pathway that harvests energy molecules from glucose is highly conserved across many different organisms. This statement means that in each of these organisms the metabolic pathway: a. is subject to the first law of thermodynamics. b. is the same or very similar. c. is subject to the second law of thermodynamics. d. is very different from each other. e. obeys the law of the conservation of energy. ANSWER: b 41. The three main groups, or domains, of organisms are: a. Bacteria, Archaea, and Eukarya. b. animals, plants, and Bacteria. c. animals, plants, and fungi. d. Bacteria, Archaea, and prokaryotes. e. animals, plants, and protists. ANSWER: a 42. The first cells were most similar to: a. prokaryotes. b. eukaryotes. c. multicellular forms. d. viruses. e. None of the other answer options is correct. ANSWER: a 43. A mutation in _____ results in a change in _____ that sometimes produces a(n) _____ with altered structure and function. a. protein; RNA; DNA b. RNA; DNA; protein c. protein; DNA; RNA d. DNA; RNA; protein e. RNA; protein; DNA ANSWER: d 44. Which of the statements is the best description of mutations in DNA? a. They do not affect an organism. b. They arise in order to harm an organism. c. They occur randomly. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 d. They arise in order to benefit an organism. e. All of these choices are correct. ANSWER: c 45. Santiago Elena and Richard Lenski performed long-term artificial selection experiments with bacteria. Over time, the bacteria evolved an ability to use succinate as a food source. Which of the statements is a conclusion of these experiments? a. All of these choices are correct. b. Evolution can occur in the laboratory. c. Bacteria can evolve over time. d. Bacteria can evolve an improved ability to use succinate. e. Natural selection can occur in the laboratory. ANSWER: a 46. Consider the phylogenetic tree. The phylogenetic tree shown represents a phylogeny of different species of butterflies. What is represented by the circled area on the phylogeny?

a. the most recent speciation event b. the appearance of a new mutation c. the appearance of a new genetic variant d. a common ancestor e. a species that must be extinct ANSWER: d 47. The metabolic pathway that harvests energy molecules from glucose is highly conserved across many different organisms. From this observation, scientists conclude that the metabolic pathway: a. is nonessential. b. arose late in the evolution of life. c. was conserved simply by chance. d. arose early in the evolution of life. e. None of the other answer options is correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 ANSWER: d 48. Transcription is the process by which: a. DNA is synthesized from protein. b. proteins are synthesized from RNA molecules. c. proteins are synthesized from DNA molecules. d. RNA is synthesized from protein. e. RNA is synthesized from DNA. ANSWER: e 49. Mutations always result in the death of the organism that acquires them. a. true b. false ANSWER: b 50. Imagine walking through a tropical rainforest. You notice that there are different types of trees, birds, insects, and a plethora of other living things. A few weeks later you are taking a walk through the desert and notice that the trees, birds, insects, and many other living things are different than those you saw in the rainforest. Which of the statements best explains the differences between each of these ecological systems? a. The manner in which organisms interact with each other and their physical environment shapes the diversity found in an ecological system. b. Organisms that evolved in the rainforest found it easier to live in that ecological system, so they have not spread out to evolve adaptations necessary to live in the desert. c. Organisms in each ecological system haven't had enough time to evolve the adaptations necessary to live in the other ecosystem; with enough time, organisms in each ecological system will evolve adaptations for the other ecological system. d. Organisms in each ecological system are there by chance, and their presence in different ecological systems does not have a biological explanation. e. All of these choices are correct. ANSWER: a 51. Interactions between organisms lead to the evolution of particular traits in populations of those organisms over time. a. true b. false ANSWER: a 52. Variation among individuals in a species is usually caused by: a. both environmental and genetic variation. b. environmental, genetic, and infectious variation. c. environmental variation only. d. genetic variation only. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 e. infectious variation only. ANSWER: a 53. The chemical reactions required to sustain life are collectively referred to as a cell's: a. physiology. b. metabolism. c. genetics. d. anatomy. e. All of the choices are correct. ANSWER: b 54. Ultraviolet (UV) light can penetrate the skin and damage DNA, and it can also destroy the B vitamin folate needed for bone-marrow maturation and the development of red blood cells. On the other hand, exposure to ultraviolet light is beneficial in the synthesis of vitamin D3, which is important for growth, calcium absorption, and bone development. The amount of ultraviolet light that penetrates the skin depends on the skin's pigmentation: more melanin (skin pigment) means less penetration. Which of the statements do you think best describes natural selection as it applies to human skin pigmentation? a. Natural selection favors skin with more pigment. b. Natural selection favors skin with less pigment. c. Natural selection favors darker or lighter skin, depending on the intensity of UV in a geographical region. ANSWER: c 55. Imagine you are standing in a field and you see a group of butterflies. You notice that the butterflies are not identical to each other even though they are all from the same species and the same population. Indicate whether the differences could explain the variation you see in the butterflies. Variations in the genetic material among the butterflies: a. could explain the variation seen in the butterflies. b. could not explain the variation seen in the butterflies. ANSWER: a 56. Imagine you are standing in a field and you see a group of butterflies. You notice that the butterflies are not identical to each other even though they are all from the same species and the same population. Indicate whether the differences could explain the variation you see in the butterflies. Differences within the population that result from variation in the environment: a. could explain the variation seen in the butterflies. b. could not explain the variation seen in the butterflies. ANSWER: a 57. Imagine you are standing in a field and you see a group of butterflies. You notice that the butterflies are not identical to each other even though they are all from the same species and the same population. Indicate whether the differences could explain the variation you see in the butterflies. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 Differences in the type of food consumed by each butterfly in the population: a. could explain the variation seen in the butterflies. b. could not explain the variation seen in the butterflies. ANSWER: a 58. Imagine you are standing in a field and you see a group of butterflies. You notice that the butterflies are not identical to each other even though they are all from the same species and the same population. Indicate whether the differences could explain the variation you see in the butterflies. Differences in the temperature in which each butterfly developed: a. could explain the variation seen in the butterflies. b. could not explain the variation seen in the butterflies. ANSWER: a 59. The effect of increasing human populations can be seen in many different ways. Which of the choices describe(s) how human actions have caused evolutionary changes in different organisms? Indicate whether each of the human actions caused evolutionary changes in the organism described. Increased use of antibiotics has caused many bacterial strains to become resistant. a. Human behavior caused this evolutionary change to occur. b. Human behavior did not cause this evolutionary change. ANSWER: a 60. The effect of increasing human populations can be seen in many different ways. Which of the choices describe(s) how human actions have caused evolutionary changes in different organisms? Indicate whether each of the human actions caused evolutionary changes in the organism described. Planting different commercial crops has increased their worldwide distribution. a. Human behavior caused this evolutionary change to occur. b. Human behavior did not cause this evolutionary change. ANSWER: a 61. The effect of increasing human populations can be seen in many different ways. Which of the choices describe(s) how human actions have caused evolutionary changes in different organisms? Indicate whether each of the human actions caused evolutionary changes in the organism described. Worldwide, more forests have been declining due to human interest in acquiring lumber for construction. a. Human behavior caused this evolutionary change to occur. b. Human behavior did not cause this evolutionary change. ANSWER: a 62. Read the scenario: You get in your car to drive to class. You turn the key, and the engine starts making a clicking sound, but does not start (1). You think to yourself, "The battery must be dead" (2). So, you borrow the battery from your neighbor's car (with permission, of course) and exchange it for the one in your car (3). You Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 figure that if the battery in your car is dead and you replace it then the car will start (4). You get in the car again, turn the key, and the car starts right up (5), and you make it to class on time. Notice that there are numbers at the end or parts of some of the sentences in the scenario. Refer to these numbers when answering the question. Which sentence, or part of a sentence, in the story is a hypothesis? a. 2 b. 1 c. 3 d. 4 e. 5 ANSWER: a 63. Read the scenario: You get in your car to drive to class. You turn the key, and the engine starts making a clicking sound, but does not start (1). You think to yourself, "The battery must be dead" (2). So, you borrow the battery from your neighbor's car (with permission, of course) and exchange it for the one in your car (3). You figure that if the battery in your car is dead and you replace it then the car will start (4). You get in the car again, turn the key, and the car starts right up (5), and you make it to class on time. Notice that there are numbers at the end or parts of some of the sentences in the scenario. Refer to these numbers when answering the question. Which sentence, or part of a sentence, in the story provides support that the idea about the battery being dead is correct? a. 5 b. 1 c. 2 d. 3 e. 4 ANSWER: a 64. Read the scenario: You get in your car to drive to class. You turn the key, and the engine starts making a clicking sound, but does not start (1). You think to yourself, "The battery must be dead" (2). So, you borrow the battery from your neighbor's car (with permission, of course) and exchange it for the one in your car (3). You figure that if the battery in your car is dead and you replace it then the car will start (4). You get in the car again, turn the key, and the car starts right up (5), and you make it to class on time. Notice that there are numbers at the end or parts of some of the sentences in the scenario. Refer to these numbers when answering the question. Which sentence, or part of a sentence, in the story is an observation? a. 1 b. 2 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 1 c. 3 d. 4 e. 5 ANSWER: a 65. Read the scenario: You get in your car to drive to class. You turn the key, and the engine starts making a clicking sound, but does not start (1). You think to yourself, "The battery must be dead" (2). So, you borrow the battery from your neighbor's car (with permission, of course) and exchange it for the one in your car (3). You figure that if the battery in your car is dead and you replace it then the car will start (4). You get in the car again, turn the key, and the car starts right up (5), and you make it to class on time. Notice that there are numbers at the end or parts of some of the sentences in the scenario. Refer to these numbers when answering the question. Which sentence, or part of a sentence, in the story is an experiment? a. 3 b. 2 c. 4 d. 1 e. 5 ANSWER: a

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Chapter 2 Multiple Choice 1. 14C is an isotope of carbon that possesses: a. 6 protons, 8 neutrons, and 6 electrons. b. 6 protons, 6 neutrons, and 2 electrons. c. 8 protons, 6 neutrons, and 2 electrons. d. 6 protons, 2 neutrons, and 6 electrons. e. 6 protons, 8 neutrons, and 2 electrons. ANSWER: a 2. Refer to the periodic table.

Using the periodic table provided, select the element that would be found in least abundance in a living cell. a. sodium (Na) b. silicon (Si) c. phosphorous (P) d. zinc (Zn) e. hydrogen (H) ANSWER: b 3. How many electron orbitals does a carbon atom possess? a. 6 b. 4 c. 2 d. 5 e. 12 ANSWER: d 4. Consult the periodic table provided if necessary.

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Chapter 2

Which of the elements would most likely have bonding properties similar to nitrogen (N)? a. carbon (C) b. silicon (Si) c. phosphorus (P) d. sulfur (S) e. oxygen (O) ANSWER: c 5. Which property of water causes ice to float? a. Hydrogen bonds form between the water molecules. b. The highly ordered crystal packing is less dense. c. Its ability to dissolve many other substances. d. Its ability to adhere to polar compounds. e. Its partial positive and partial negative charges on each molecule. ANSWER: b 6. Which statement best describes an effect of the low density of frozen water in a lake? a. When water in a lake freezes, it floats, providing insulation for organisms below the ice. b. Water in a lake freezes from the bottom up, killing most aquatic organisms. c. When water freezes, it contracts, decreasing the water level in the lake. d. Water removes thermal energy from the land around a lake, causing the lake to freeze. ANSWER: a 7. Which number represents the pH of a solution with the highest concentration of hydrogen ions? a. 11.5 b. 4.5 c. 7.0 d. 9.1 e. 1.0 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 ANSWER: e 8. In a solution that has pH = 7.0, the ratio of protons (H+) to hydroxide ions (OH-) equals a. 1 b. 7 c. 70 d. 1/7 e. 1/70 ANSWER: a 9. Refer to the periodic table provided.

Which group ranks the elements carbon, sodium, calcium, and iodine in order of decreasing number of protons? a. I→Ca→Na→C b. C→Na→Ca→I c. I→C→Ca→Na d. C→Ca→Na→I ANSWER: a 10. Which statement about a carbon-carbon double bond is correct? a. Each of the two carbons is capable of bonding to three other atoms. b. The double bond allows free rotation of the molecule at the bond position. c. The double bond is longer than a corresponding carbon-carbon single bond. d. Double bonds can be found in both chain and ring structures. ANSWER: d 11. How many hydrogen atoms are present in a hydrocarbon chain of five carbons joined to each other by single covalent bonds? a. 8 b. 12 c. 10 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 d. 6 ANSWER: b 12. How many hydrogen atoms are present in a hydrocarbon chain of 8 carbon atoms with 3 double bonds and the rest single bonds? a. 12 b. 10 c. 14 d. 16 ANSWER: a 13. Which choice is a pyrimidine found in DNA? a. None of the other answer options is correct. b. guanine c. uracil d. adenine e. thymine ANSWER: e 14. _____ are the subunits of nucleic acids, and _____ are the subunits of proteins. a. Polypeptides; sugars b. Bases; polypeptides c. Nucleotides; amino acids d. Amino acids; nucleic bases e. Nucleoli; amino acids ANSWER: c 15. Carbohydrates and proteins are two types of macromolecules. Which functional characteristic of proteins distinguishes them from carbohydrates? a. large amount of stored information b. efficient storage of usable chemical energy c. tendency to make cell membranes hydrophobic d. ability to catalyze biochemical reactions e. None of the other answer options is correct. ANSWER: d 16. Which choice most accurately describes the ratio of oxygen to carbon to hydrogen in a simple 6-carbon sugar such as glucose? a. 1:1:2 b. 1:2:1 c. 2:1:1 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 d. 1:2:3 e. 1:3:2 ANSWER: a 17. Sucrose is composed of: a. glycerol and three fatty acids. b. a six-carbon sugar and a five-carbon sugar. c. a simple sugar and a nucleotide. d. two ketose sugars. e. an aldose and a ketose. ANSWER: e 18. Which biomolecule is defined by a physical property instead of a chemical structure? a. lipids b. proteins c. monosaccharides d. nucleic acids e. polysaccharides ANSWER: a 19. Which type of fatty acid would be likely to have the lowest melting temperature? a. long tails and low saturation b. long tails and high saturation c. short tails and low saturation d. short tails and high saturation e. All fatty acids have the same melting temperature, regardless of tail length or level of saturation. ANSWER: c 20. Which type of fatty acids would be likely to participate in the greatest amount of van der Waals forces with other fatty acids? Fatty acids with: a. short tails and high saturation. b. long tails and low saturation. c. short tails and low saturation. d. long tails and high saturation. ANSWER: d 21. Miller and Urey's initial simulation resulted in the formation of which one of the molecules? a. DNA b. glucose c. RNA d. phospholipids Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 e. amino acids ANSWER: e 22. Sutherland and colleagues demonstrated the synthesis of which molecules under conditions thought to resemble those of early Earth? a. nucleotides b. amino acids c. phospholipids d. nucleic acid chains e. polypeptides ANSWER: a 23. Refer to the periodic table.

Which of these choices ranks the elements carbon, sodium, calcium, and iodine in order of decreasing number of protons? a. I, Ca, Na, C b. I, Na, Ca, C c. C, Na, Ca, I d. Ca, I, C, Na e. Na, Ca, C, I ANSWER: a 24. Refer to the periodic table.

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Chapter 2

Which of these choices ranks the elements carbon, phosphorus, calcium, and iodine in order of decreasing number of protons. a. C, P, Ca, I b. I, P, Ca, C c. I, Ca, P, C d. Ca, I, C, P e. P, Ca, C, I ANSWER: c 25. Complete the description: ____________ can occur when a molecule has a hydrogen atom covalently linked to an electronegative atom, whereas _______________ occurs between oppositely charged ions and ______________ occurs when atoms share electrons. a. A hydrogen bond; a covalent bond; an ionic bond b. An ionic bond; a hydrogen bond; a covalent bond c. A covalent bond, an ionic bond, a hydrogen bond d. A covalent bond; a hydrogen bond; an ionic bond e. A hydrogen bond; an ionic bond; a covalent bond ANSWER: e 26. What differentiates isotopes of the same element? a. protons b. neutrons c. electrons d. charge ANSWER: b 27. You discover an isotope of an element that has 6 electrons in its second and outermost shell, 8 protons, and 6 neutrons. What element is it? a. nitrogen (N) Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 b. carbon (C) c. oxygen (O) d. fluorine (F) ANSWER: c 28. Which statement accurately describes a polar covalent bond? a. the unequal sharing of electrons between an atom with a partial positive charge and an atom with a partial negative charge b. the interaction of an atom with very high electronegativity and an atom with very low electronegativity c. the interaction of a hydrogen atom connected to an atom with a high electronegativity and an electronegative atom of another molecule d. the equal sharing of electrons between atoms of identical or similar electronegativities e. None of the other answer options is correct. ANSWER: a 29. The oxygen and hydrogens of a water molecule contains what type of bond? a. polar covalent b. ionic c. hydrogen d. van der Waals interactions ANSWER: a 30. Which choice correctly lists the five most abundant elements found in living organisms? a. carbon, hydrogen, oxygen, nitrogen, iron b. sodium, carbon, oxygen, nitrogen, phosphorus c. magnesium, carbon, hydrogen, oxygen, nitrogen d. carbon, hydrogen, oxygen, nitrogen, phosphorus ANSWER: d 31. Which choice ranks the elements carbon, sodium, calcium, and iodine in order of decreasing number of valence electrons? a. I→C→Ca→Na b. I→Ca→Na→C c. C→Na→Ca→I d. C→Ca→Na→I ANSWER: a 32. Which choice ranks the elements carbon, sodium, calcium, and iodine in order of decreasing number of energy shells/levels? a. C→Na→Ca→I b. I→Ca→Na→C Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 c. I→C→Ca→Na d. C→Ca→N→I ANSWER: b 33. Rank the elements carbon, phosphorus, calcium, and iodine in order of decreasing number of valence electrons. a. I→P→C→Ca b. I→Ca→P→C c. C→P→Ca→I d. P→C→Ca→I ANSWER: a 34. Rank the elements carbon, phosphorus, calcium, and iodine in order of decreasing number of energy shells/levels. a. I→P→C→Ca b. C→P→Ca→I c. I→Ca→P→C d. P→C→Ca→I ANSWER: c 35. Helicase is an enzyme that separates the double helix of the DNA into two separate strands. How do you think helicase does this? a. by breaking ionic bonds b. by breaking phosphodiester bonds c. by breaking peptide bonds d. by breaking hydrogen bonds ANSWER: d 36. Two of the main ingredients in plant fertilizer are phosphorus and nitrogen. These elements are found in which classes of biomolecules? a. carbohydrates and DNA b. proteins and carbohydrates c. DNA and proteins d. lipids and carbohydrates ANSWER: c 37. Although monosaccharides can exist in linear form, virtually all cellular monosaccharides are found in circular form. a. true b. false ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 38. Imagine you were there when Stanley Miller performed his experiment to reproduce the building blocks of life. If Miller originally identified 5 different amino acids, how many polypeptides that are 10 amino acids long could be made from just these 5 amino acids? a. The answer cannot be determined from the information provided. b. 105= 100,000 c. 5 × 10 = 50 d. 510 = 9,765,625 ANSWER: d 39. Studies of the origin of life on Earth help us to consider what would be required for life elsewhere in the universe. What is most likely to be true of extraterrestrial life if it exists? a. Carbon will act as the backbone for organic molecules. b. Light from a nearby star will make photosynthesis possible. c. Oxygen will be used to convert energy in cells. d. Water will not be required to sustain life. ANSWER: a 40. Current theory about molecular systems in the very earliest evolution of living cells suggests that which of these types of biomolecules probably provided the stored genetic information? a. DNA b. RNA c. proteins d. carbohydrates e. biomolecules unlike those in modern cells ANSWER: b 41. Which gas was not a component of the simulated atmosphere in the Miller-Urey experiment? a. ammonia b. water vapor c. hydrogen d. methane e. oxygen ANSWER: e 42. Rank the elements carbon, sodium, calcium, and iodine in order of decreasing number of valence electrons. a. C, I, Ca, Na b. I, C, Ca, Na c. Na, Ca, C, I d. Ca, Na, I, C e. I, C, Na, Ca ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 43. Rank the elements carbon, sodium, calcium, and iodine in order of decreasing number of energy shells/levels. a. I, Ca, Na, C b. C, I, Ca, Na c. Na, Ca, C, I d. Ca, Na, I, C e. I, C, Ca, Na ANSWER: a 44. Rank the elements carbon, phosphorus, calcium, and iodine in order of decreasing number of valence electrons. a. Ca, P, I, C b. C, I, Ca, P c. P, Ca, C, I d. I, P, C, Ca e. I, C, Ca, P ANSWER: d 45. Rank the elements carbon, phosphorus, calcium, and iodine in order of decreasing number of energy shells/levels. a. P, Ca, C, I b. C, I, Ca, P c. I, Ca, P, C d. Ca, P, I, C e. I, Ca, C, P ANSWER: c 46. An atom with three electrons has: a. three occupied orbitals, each of which contains one electron. b. one occupied orbital with three electrons. c. two occupied orbitals, one of which has two electrons and the other has one. d. three energy shells, each of which contains one electron. ANSWER: c 47. For an atom that is not an ion, which statement must be true? a. The number of electrons equals the number of protons. b. The number of electrons equals the number of neutrons. c. The number of protons equals the number of neutrons. d. The number of neutrons must be less than the number of electrons. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 48. Which choice correctly pairs the particles of an atom with their physical properties? a. proton - positively charged; neutron - uncharged; electron - negatively charged b. proton - negatively charged; neutron - uncharged; electron - positively charged c. proton- positively charged; neutron - negatively charged; electron - uncharged d. proton - uncharged; neutron - negatively charged; electron - positively charged ANSWER: a 49. The most common isotope of oxygen has 8 protons and an atomic mass of 16. How many neutrons are present in the oxygen nucleus? a. 8 b. 4 c. 6 d. 2 e. 10 ANSWER: a 50. The most common isotope of oxygen has 8 protons and an atomic mass of 16. How many electrons are present in the orbitals around an atom of oxygen? a. 2 b. 4 c. 6 d. 8 e. 10 ANSWER: d 51. The most common isotope of oxygen has an atomic mass of 16 (16O). An isotope with an atomic mass of 18 (18O) is also stable. How many valence electrons are present in 18O? a. the same as in 16O b. more than in 16O c. fewer than 16O d. None of the other answer options is correct. ANSWER: a 52. The description "two of the outermost atomic orbitals of two atoms, each containing one electron, merge into a single orbital containing a full complement of two electrons" refers to which type of bonds? Select all that apply. a. hydrogen bonds b. ionic bonds c. covalent bonds ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 53. Which bonds occur between two atoms in each of which the number of protons does not equal the number of electrons? Select all that apply. a. ionic bonds b. covalent bonds c. hydrogen bonds ANSWER: a 54. Refer to the periodic table.

Decide which molecule is held together by ionic bonds. a. CO2 b. KCl c. NH3 d. NO e. CH4 ANSWER: b 55. Refer to the periodic table.

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Chapter 2 Decide which molecule is held together by polar covalent bonds. a. NH3 b. CO2 c. KCl d. NO e. CH4 ANSWER: a 56. Refer to the periodic table.

Decide which molecule is held together by nonpolar covalent bonds. a. H2O b. NH3 c. KCl d. CO2 ANSWER: d 57. Of the given types of bonds between atoms, which is the strongest? a. van der Waals forces b. hydrogen bond c. ionic bond d. covalent bond ANSWER: d 58. A pair of atoms joined by a polar covalent bond: a. is unlikely to form hydrogen bonds with water. b. has the charge spread evenly across both atoms. c. has a slight positive charge on one atom and a slight negative charge on the other. d. mixes well with nonpolar solvents. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 ANSWER: c 59. The association of individual water molecules with other water molecules is called _____ and occurs through _____ bonds between water molecules. a. adhesion; hydrogen b. cohesion; polar covalent c. adhesion; polar covalent d. cohesion; hydrogen ANSWER: d 60. The unique properties of water are due to the _____ of water molecules and the ability of water to form _____ with other water molecules and with other polar molecules. a. polarity; polar covalent bonds b. polarity; hydrogen bonds c. electronegativity; polar covalent bonds d. hydrophobicity; hydrogen bonds ANSWER: b 61. You have an aqueous solution with a pH of exactly 7.0. What would you add to make the solution more acidic? a. hydrogen chloride (HCl) b. sodium hydroxide (NaOH) c. sodium chloride (NaCl) d. deionized water (H2O) ANSWER: a 62. You have an aqueous solution with a pH of 6.0. What would you add to make the solution more basic? a. hydrogen chloride (HCI) b. sodium hydroxide (NaOH) c. sodium chloride (NaCl) d. deionized water (H2O) ANSWER: b 63. You have an aqueous solution with a pH of 8.0. You add sodium chloride to a concentration of 1 gram per 100 milliliters. What happens to the pH? a. It stays the same. b. It goes down. c. It goes up. d. It depends on the temperature. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 64. A _______ substance could contain __________ bonds which allow it to interact with water. a. hydrophobic; non-polar covalent b. hydrophilic; non-polar covalent c. hydrophobic; hydrogen d. hydrophilic; hydrogen e. hydrophilic; polar covalent ANSWER: e 65. A _______ substance could contain __________ bonds which cause it not to interact with water. a. hydrophobic; nonpolar covalent b. hydrophilic; nonpolar covalent c. hydrophobic; hydrogen d. hydrophilic; hydrogen e. hydrophilic; polar covalent ANSWER: a 66. Which choice ranks the elements carbon, sodium, calcium, and iodine in order of decreasing number of electrons? a. C→Ca→Na→I b. C→Na→Ca→I c. I→C→Ca→Na d. I→Ca→Na→C ANSWER: d 67. Which of these choices ranks the elements hydrogen, carbon, nitrogen, and oxygen in order of decreasing dry mass in living organisms? a. O→C→H→N b. C→O→H→N c. O→H→C→N d. C→N→O→H ANSWER: b 68. Consider the two pie-graph representations of the composition of the human body and human cells presented in the figure shown.

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Chapter 2

Why is oxygen so prevalent in live cells of the human body compared to oxygen in the dry mass of human cells? a. The "dry biomass" does not include the 60-80% of water, mostly oxygen mass, that is present in a typical, live human. b. The human body is composed of much more than just the cells that make up the tissues and organs. c. The human body stores a large amount of oxygen within its tissues and organs for use when oxygen is low. d. These represent two different experimental analyses, so there is natural variation. ANSWER: a 69. Rank the elements carbon, nitrogen, phosphorus, and oxygen in order of decreasing proportion of human cell dry mass. a. C→O→N→P b. O→C→N→P c. C→N→P→O d. P→N→O→C e. N→O→P→C ANSWER: a 70. Rank the elements carbon, phosphorus, calcium, and iodine in order of greatest dry mass in human cells. a. I→P→C→Ca b. I→Ca→P→C c. C→P→Ca→I d. P→C→Ca→I ANSWER: c 71. Single covalent bonds between carbon atoms: a. allow free rotation of the carbon atoms around the bond. b. are strong enough to support long chains of carbon atoms. c. allow a molecule to twist and turn into many different arrangements. d. All of these choices are correct. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 72. Three carbon atoms are linked by single covalent bond such that those carbon atoms and bonds together form the shape of a V. All of the unshared electrons form covalent bonds with hydrogen. How many hydrogen atoms does this molecule contain? a. 10 b. 4 c. 6 d. 2 e. 8 ANSWER: e 73. Because of hydrogen bonding, water is uniquely suited for its central role in life. Many hydrophilic molecules interact freely with water, but a number of hydrophobic molecules are important for life, too. How does the interaction between water and hydrophobic molecules help to organize biological systems? a. Because water molecules preferentially associate with each other, they force hydrophobic molecules to associate with each other and not with water molecules. b. The ionic bonds between water molecules cause hydrophobic molecules to associate with each other and not with water molecules. c. Because cells are not pure water—in that they have many substances dissolved within them—the hydrophilic/hydrophobic effect has a limited role in biological organization. d. None of the other answer options is correct. ANSWER: a 74. Which bonds are covalent bonds? a. peptide bonds b. glycosidic bonds c. phosphodiester bonds d. All of these choices are correct. ANSWER: d 75. In general, colder temperatures would tend to reduce the fluidity of membranes. In response, cells can adjust the composition of their membranes to maintain the proper degree of fluidity. How would the membrane change in response to colder temperatures? a. The amount of saturated triacylglycerols would increase. b. The amount of unsaturated fatty acids would increase. c. The length of the fatty acid chains in the phospholipids would increase. d. The amount of unsaturated fatty acids would decrease. e. The amount of saturated triacylglycerols would decrease. ANSWER: b 76. Consider the structure and function of DNA. Which statement is true? a. Because DNA contains carbohydrates, it provides structural support to the cells. b. The phosphodiester bonds that stabilize the association of the two strands are easily broken and Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 reformed. c. If the sequence of one DNA strand is known, then the sequence of the other strand can be determined. d. Because DNA is made of phosphate groups that are ionized, it could easily pass through a cell membrane. ANSWER: c 77. In DNA molecules, complementary base pairs always include one purine nucleotide and one pyrimidine nucleotide. Suppose you analyzed the DNA from some bacterial cells and you found that 16% of the nucleotides are adenine nucleotides. What are the percentages of the other nucleotides in the bacterial DNA? a. 16% thymine, 34% guanine, 34% cytosine b. 34% uracil, 16% guanine, 16% cytosine c. 34% thymine, 34% guanine, 16% cytosine d. 34% thymine, 16% guanine, 34% cytosine e. None of the other answer options is correct. ANSWER: a 78. Samples of three different triacylglycerol types were tested to determine the melting point of each one. The results of the tests are shown in the graph.

The length of the fatty acids A, B, and C is the same. Which of the three triacylglycerols is likely to have the most double bonds in the fatty acids? a. type 1 b. type 2 c. type 3 d. There is no way of knowing which of the three triacylglycerols would likely have the most double bonds based on the information available. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 ANSWER: a 79. Samples of three different triacylglycerols were tested to determine the melting point of each one. The results of the tests are shown in the graph.

The length of the fatty acids in A, B, and C is the same. Which of the three triacylglycerols is likely to have the fewest number of double bonds in the fatty acids? a. type 2 b. type 1 c. type 3 d. There is no way of knowing which of the three triacylglycerols would likely have the fewest double bonds based on the information available. ANSWER: a 80. Samples of three different triacylglycerols were tested to determine the melting point of each one. The results of the tests are shown in the graph.

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Chapter 2

The length of the fatty acids in A, B, and C is the same. Which of the three fatty acids is likely to have the most saturated fatty acids? a. type 2 b. type 1 c. type 3 d. There is no way of knowing which of the three triacylglycerols would likely have the most saturated fatty acids based on the information available. ANSWER: a 81. Samples of three different triacylglycerols were tested to determine the melting point of each one. The results of the tests are shown in the graph.

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Chapter 2

The length of the fatty acids in A, B, and C is the same. Which of the three fatty acids is likely to have the most unsaturated fatty acids? a. type 3 b. type 2 c. type 1 d. There is no way of knowing which of the three triacylglycerols would likely have the most unsaturated fatty acids based on the information available. ANSWER: c 82. Samples of three different triacylglycerols were tested to determine the melting point of each one. The results of the tests are shown in the graph.

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Chapter 2

The number of double bonds in each of the fatty acids A, B, and C is the same. Which of the three triacylglycerols is likely to have the fatty acids with the longest hydrocarbon chains? a. type 1 b. type 2 c. type 3 d. There is no way of knowing which of the three triacylglycerols would likely have the fatty acids with the longest hydrocarbon chains based on the information available. ANSWER: b 83. Samples of three different triacylglycerols were tested to determine the melting point of each one. The results of the tests are shown in the graph.

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Chapter 2

The number of double bonds in each of the fatty acids in A, B, and C is the same. Which of the three triacylglycerols is likely to have the fatty acids with the shortest hydrocarbon chains? a. There is no way of knowing which of the three triacylglycerols would likely have the fatty acids with the shortest hydrocarbon chains based on the information available. b. type 2 c. type 3 d. type 1 ANSWER: d 84. How many hydrogen atoms are present in a hydrocarbon chain of five carbon atoms with one double bond and the rest single bonds? a. 8 b. 10 c. 6 d. 12 ANSWER: b 85. How many hydrogen atoms are present in a hydrocarbon chain of five carbon atoms with two double bonds and two single bonds? a. 10 b. 6 c. 8 d. 12 ANSWER: c 86. How many hydrogen atoms are present in a five-carbon hydrocarbon molecule with four of the carbons Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 linked in a chain by single covalent bonds and with the fifth carbon atom attached by a single bond as a branch to the second carbon in the chain? a. 6 b. 8 c. 10 d. 12 ANSWER: d 87. How many hydrogen atoms are present in a ring of six carbon atoms held together by alternating single and double bonds? a. 6 b. 8 c. 10 d. 12 ANSWER: a 88. What important feature of noncovalent molecular interactions makes them so important to life? a. They can only occur in cells. b. They are strong in a cellular environment that holds atoms together tightly. c. They are weak in a cellular environment, so they can be made, broken, and reformed easily. d. None of the other answer options is correct. ANSWER: c 89. Peptide bonds are characteristic of: a. nucleic acids. b. carbohydrates. c. lipids. d. fatty acids. e. proteins. ANSWER: e 90. Pyrimidine and purine bases are found in: a. carbohydrates. b. nucleic acids. c. lipids. d. fatty acids. e. proteins. ANSWER: b 91. Aldoses and ketoses are examples of: a. lipids. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 b. nucleic acids. c. proteins. d. carbohydrates. e. fatty acids. ANSWER: d 92. An unsaturated fatty acid contains: a. one or more double bonds between carbon atoms. b. only single covalent bonds between carbon atoms. c. only carbon and hydrogen. d. one or more double bonds between hydrogen atoms. ANSWER: a 93. A phosphodiester bond in nucleic acid polymers is formed between: a. two amino acids. b. a 5' phosphate and a 3' hydroxyl group. c. a base and a sugar. d. a 3' phosphate and a 5' hydroxyl group. e. a fatty acid and a glycerol molecule. ANSWER: b 94. Which component of an amino acid differs from one amino acid to another? a. the hydrogen atom opposite the R group b. the carboxyl group c. the α-carbon atom d. the amino group e. the side chain ANSWER: e 95. If you isolate a single nucleotide from a nucleic acid chain and determine that the nitrogenous ring structure is cytosine, you could say with certainty that the nucleotide may have come from: a. either DNA or RNA. b. RNA but not DNA. c. DNA but not RNA. d. neither DNA nor RNA. ANSWER: a 96. If an atom has three electrons, it will have ____ occupied orbitals _________ of which is/are full. a. three; one b. three; all c. two; both Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 d. three; two e. two; one ANSWER: e 97. An atom that is not an ion is electrically neutral because the number of ___________ charged _____________ always equals the number of _____________ charged ___________. a. negatively; protons; positively; electrons b. negatively; electrons; positively; neutrons c. un-; neutrons; positively; protons d. positively; protons; negatively; electrons e. negatively; neutrons; positively; protons ANSWER: d 98. Which choice correctly pairs the particles in an atom with its electrical charge? a. proton–positively charged; neutron–uncharged; electron–negatively charged b. proton–uncharged; neutron–uncharged; electron–negatively charged c. proton–positively charged; neutron–negatively charged; electron–negatively charged d. proton–negatively charged; neutron–uncharged; electron–negatively charged ANSWER: a 99. Refer to the image.

The molecule shown here is: a. a protein. b. an amino acid. c. a nucleotide. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 d. a monosaccharide. e. a triacylglycerol. ANSWER: b 100. Refer to the image.

In the molecule shown here, label C points to the __________ and label B points to the __________. a. amino group; carboxyl group b. carboxyl group; R group c. R group; carboxyl group d. glycerol; fatty acid e. alpha carbon; R group ANSWER: e 101. Refer to the image.

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Chapter 2

In the molecule shown here, label A points to the __________ and label D points to the __________. a. R group; carboxyl group b. carboxyl group; R group c. carboxyl group; amino group d. glycerol; fatty acid e. alpha carbon; R group ANSWER: c 102. Refer to the image.

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Chapter 2

In the molecule shown here, which labeled parts would form peptide bonds? a. B and D b. A and B c. B and C d. A and D e. C and D ANSWER: d 103. Refer to the image.

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Chapter 2

A polymer consisting of many subunits similar to the molecule shown would be what kind of biomolecule? a. a protein b. an amino acid c. a nucleic acid d. a lipid e. a carbohydrate ANSWER: a 104. Refer to the image.

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Chapter 2

In the molecule shown, which labeled parts would be ionized in the environment of the cell? a. C and D b. A and B c. B and C d. B and D e. A and D ANSWER: e 105. Refer to the image.

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Chapter 2

In the molecule shown, which labeled part or parts would be different in each of the other 19 amino acids? a. B b. D c. B and D d. A and D e. C and D ANSWER: a Multiple Response 106. The structural diversity of carbon-based molecules is determined by which properties? Select all the apply. a. the ability of carbon to form four covalent bonds b. the ability of carbon's covalent bonds to rotate freely c. the orientation of carbon's bonds in the form of a tetrahedron d. carbon's strong electronegativity results in polar covalent bonds e. the ability of carbon to ionize and interact with other ions ANSWER: a, b, c 107. Certain meteorites have been examined and found to carry samples of which molecules? Select all the apply. a. amino acids b. lipids c. polypeptides Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 d. nucleotides e. monosaccharides ANSWER: a, b 108. Which qualities are maintained during all chemical reactions? Select all that apply. a. the identity of the atoms present in the reactants b. the number of atoms present in the reactants c. the arrangement of chemical bonds present in the reactants d. the number of reactant molecules ANSWER: a, b 109. As part of their normal function, many proteins bind to DNA briefly and then release it again. Which types of interactions might be involved in these transient protein-DNA interactions? Select all that apply. a. hydrogen bonds b. ionic bonds c. covalent bonds d. van der Waals forces ANSWER: a, b, d 110. Sometimes, atoms gain or lose particles. The loss of which particles results in a change of atomic mass? Select all that apply. a. a neutron b. a proton c. an electron ANSWER: a, b 111. Sometimes, atoms gain or lose particles. The loss of which particles would result in a change of overall electrical charge? Select all that apply. a. a neutron b. an electron c. a proton ANSWER: b, c 112. Which bonds rely on the attraction of positive and negative charges? Select all that apply. a. ionic bonds b. hydrogen bonds c. covalent bonds ANSWER: a, b 113. Several chemical properties make water uniquely suited for its role as a central "molecule of life". Which statements are true? a. Hydrogen bonding leads to high cohesiveness between water molecules. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 2 b. Water resists temperature changes. c. Water molecules are always polar. d. Water is a good solvent of polar molecules and ions. e. The structure of a water molecule is stabilized by hydrogen bonds. ANSWER: a, b, c, d 114. Which components of an amino acid participate in the formation of the peptide bonds that join amino acids into a chain to form proteins? Select all that apply. a. the amino group b. the carboxyl group c. the α-carbon atom d. the side chain e. the hydrogen atom opposite the R group ANSWER: a, b 115. Which choices can combine to form a triacylglycerol molecule? Select all that apply. a. glycerol b. unsaturated fatty acid c. saturated fatty acid d. cholesterol e. phosphate ANSWER: a, b, c

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Chapter 3 Multiple Choice 1. When a mixture of debris from killed virulent bacteria and live nonvirulent bacteria are injected into mice: a. the mice are killed because the virulent bacteria are revived. b. the mice are killed because the nonvirulent bacteria are transformed into virulent bacteria. c. the mice survive because none of the living bacteria are virulent. d. the mice survive for a period of a few weeks and then die from an unrelated illness. e. half the mice are killed, and the other half survive. ANSWER: b 2. In the DNA sequence, the top strand is the template strand. If the base pair G-C (in bold) is changed to T-A, what would the resulting nucleotide be in the mRNA? 5'-GTAGCCGATAAT-3' 3'-CATCGGCTATTA-5' a. A b. C c. T d. G e. U ANSWER: a 3. In the DNA sequence, the bottom strand is a template strand. If the base pair G-C (in bold) is replaced to AT, what would the resulting nucleotide be on the mRNA? 5'-GTAGCCGATAAT-3' 3'-CATCGGCTATTA-5' a. T b. C c. U d. A e. G ANSWER: c 4. Occasionally, a double-stranded DNA molecule contains a uracil base (U) instead of a (T). If a U were present in the template strand of DNA, what base do you think would be incorporated into the RNA transcript at that position? a. U b. C c. T d. A e. G ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 5. Which statement about RNA is correct? a. The nucleotide at the 5¢ end of an RNA molecule is typically a nucleoside triphosphate. b. RNA is usually found in double-stranded form, just like DNA. c. RNA molecules are typically longer than DNA molecules. d. RNA molecules are synthesized in 3¢ to 5¢ direction. e. RNA molecules are incapable of evolving an enzymatic activity over time. ANSWER: a 6. Which choice is part of a nucleotide? a. a nitrogen-containing base b. one or more phosphate groups c. a five-carbon sugar d. All of these choices are correct. ANSWER: d 7. In a nucleotide, the phosphate is attached to the sugar at the: a. 1' carbon. b. 3' carbon. c. 3' and 5' carbons. d. 5' carbon. e. to the nitrogen of the bigger ring of each base. ANSWER: d 8. In a nucleotide, the base is attached to the sugar at the: a. 1' carbon. b. 5' carbon. c. 3' carbon. d. 3' and 5' carbons. e. to the phosphate group. ANSWER: a 9. Which choice represents an actual Watson–Crick base pair with the largest number of hydrogen bonds? a. thymine and guanine b. adenine and cytosine c. cytosine and guanine d. cytosine and thymine e. adenine and thymine ANSWER: c 10. Which is a function of the RNA-polymerase complex? a. All of these choices are correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 b. to stabilize RNA-DNA duplex c. to perform polymerization reaction d. to disrupt the DNA double helix ANSWER: a 11. What is the name of the enzyme complex that forms at the start of transcription? a. DNA polymerase b. RNA polymerase c. DNA gyrase d. RNA helicase ANSWER: b 12. Which molecule is made during the process of transcription? a. All of these choices are correct. b. mRNA c. tRNA d. rRNA ANSWER: a 13. If you made a change in the promoter sequence in the DNA that inactivates the promoter, what would happen at the RNA level? a. Nothing, because the RNA would be made as usual. b. The RNA polymerase would not be able to recognize and bind the DNA, so no RNA would be made. c. The mutation of the DNA would be carried through to the RNA sequence. d. The DNA helicase would not be able to recognize and bind the DNA, so the RNA would not be made. ANSWER: b 14. Whichever DNA strand is transcribed, the RNA polymerase reads the template strand from 3' to 5'. a. true b. false ANSWER: a 15. Whichever DNA strand is transcribed, the RNA polymerase reads the template strand from 5' to 3'. a. true b. false ANSWER: b 16. In an RNA world: a. None of the other answer options is correct. b. RNA functions in information storage, and proteins perform catalysis. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 c. DNA functions in information storage, and RNA alone catalyzes reactions. d. RNA functions in information storage, and there is no catalysis. e. RNA functions in information storage and also performs catalysis. ANSWER: e 17. Which component is essential, even in an RNA world? a. polymerase b. introns c. exons d. spliceosomes ANSWER: a 18. Catalytic activity of RNA can be positively selected in a laboratory. a. true b. false ANSWER: a 19. RNA catalysis means that RNA can: a. function as an information-storage molecule. b. function like an enzyme. c. function as a signaling molecule. d. form membrane-like structures. e. All of these choices are correct. ANSWER: b 20. In E. coli, the molecule(s) responsible for promoter recognition is/are referred to as: a. the sigma factor. b. the TATA box. c. the mediator complex. d. an enhancer. e. transcription factors. ANSWER: a 21. In eukaryotes, where do activator proteins bind? a. RNA polymerases b. terminators c. DNA polymerases d. promoters e. enhancers ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 22. In eukaryotes, where do general transcription initiation factors bind? a. 5' UTR b. enhancers c. terminators d. promoters e. RNA polymerases ANSWER: d 23. In E. coli, which number is closest to the number of bases of RNA that at any one time are hydrogen bonded to DNA within an RNA polymerase? a. 10 b. 1 c. 100 d. 1000 e. 10,000 ANSWER: a 24. Refer to the image.

Where does the energy come from to add a uracil to the 3' end of a transcript? a. the cell's supply of ATP b. the energy released by allowing the uracil to complementary base pair with an adjacent thymine c. the hydrolysis of a terminal phosphate from the incoming UTP molecule d. the hydrolysis of two phosphate groups from the incoming UTP molecule e. the hydrolysis of pyrophosphate from the incoming UTP molecule Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 ANSWER: d 25. Which statement about RNA is correct? a. RNA uses the same purine bases as DNA. b. RNA is a more stable molecule than DNA. c. RNA has the same 5-carbon sugars as DNA. d. RNA uses the same pyrimidine bases as DNA. e. All of these choices are correct. ANSWER: a 26. Which RNA type is the most abundant in mammalian cells? a. mRNA b. rRNA c. tRNA d. miRNA e. snRNA ANSWER: b 27. The poly(A) sequence that is added to RNA during processing: a. is needed for ribosomes to attach to messenger RNA. b. helps prevent formation of complex three-dimensional structures in the messenger RNA. c. helps prevent rapid breakdown of the messenger RNA. d. aids in the accuracy of translation of the messenger RNA into protein. ANSWER: c 28. A type of noncoding RNA called small interfering RNA: a. destroys RNA transcripts. b. inhibits transcription. c. inhibits RNA processing. d. activates translation. e. None of the other answer options is correct. ANSWER: a 29. In prokaryotes, the messenger RNA consists of the: a. primary RNA transcript with cap added and introns removed. b. primary RNA transcript with cap added. c. primary RNA transcript. d. primary RNA transcript with cap added, introns removed, and poly(A) sequence added. ANSWER: c 30. RNA processing occurs in: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 a. eukaryotes only. b. prokaryotes only. c. viruses only. d. both prokaryotes and eukaryotes. e. both prokaryotes and viruses. ANSWER: a 31. RNA processing occurs in the: a. plasma membrane. b. cytoplasm. c. nucleus. d. nucleus and cytoplasm. ANSWER: c 32. RNA processing consists of the: a. All of these choices are correct. b. addition of a cap. c. removal of introns. d. addition of a poly(A) sequence. ANSWER: a 33. In a messenger RNA, the cap is present at the: a. 3' end. b. 5' end. c. 5' end of the poly(A) sequence. d. 3' end of the poly(A) sequence. ANSWER: b 34. In a messenger RNA, the cap is needed to: a. initiate transcription. b. initiate replication. c. initiate translation. d. initiate processing. ANSWER: c 35. In messenger RNA, the protein-coding sequence is present in: a. exons and the poly(A) sequence. b. introns. c. introns and the poly(A) sequence. d. exons. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 36. In eukaryotes, the messenger RNA consists of: a. an RNA transcript with cap added, introns removed, and poly(A) sequence added. b. primary RNA transcript. c. primary RNA transcript with cap added. d. primary RNA transcript with cap added and introns removed. ANSWER: a 37. What is the central dogma? a. RNA is transcribed into protein, which is translated into DNA. b. RNA is transcribed into DNA, which is translated into protein. c. DNA is transcribed into protein, which is translated into RNA. d. DNA is transcribed into RNA, which is translated into protein. ANSWER: d 38. What is the order of processes that support the central dogma? a. DNA replication, translation, protein synthesis b. DNA replication, transcription, translation c. transcription, DNA replication, translation d. DNA replication, translation, transcription ANSWER: b 39. Why aren't nucleosides incorporated into DNA? a. The sugar is not in the right form. b. The bases are not fully assembled. c. There are no phosphates to make the phosphodiester bonds. d. The peptide bonds don't form. ANSWER: c 40. Sometimes the DNA sequence gets mutated and an adenine is paired to a cytosine. Why is this interaction unstable? a. the chemical groups that form hydrogen bonds are in the wrong positions b. the phosphate groups of the two bases repel each other c. an ionic bond cannot be formed between the two d. the charges on the bases repel one another ANSWER: a 41. In the DNA of certain bacterial cells, 16% of the nucleotides are adenine. What are the percentages of the other nucleotides in the bacterial DNA? a. 16% thymidine, 34% guanine, 34% cytosine b. 34% uracil, 16% guanine, 16% cytosine Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 c. 34% thymidine, 34% guanine, 16% cytosine d. 34% thymidine, 16% guanine, 34% cytosine ANSWER: a 42. Imagine that you are an interior architecture/biology double major. While at home during break, one of your friends asks you to design and help build a spiral staircase that is modeled after a DNA molecule. The staircase must have 12 steps, and each step needs to be 6 feet wide. You say, "OK, as long as you provide the materials." So, your friend asks you for a list of materials needed to build the staircase. Which choice includes the items necessary to make a staircase that is based on the molecular structure of a DNA molecule? a. 1 railing and 24 boards (12 that are 2 feet long, and 12 that are 4 feet long) b. 1 railing and 12 boards (each of which is 6 feet wide) c. 2 railings and 12 boards (each of which is 6 feet wide) d. 2 railings and 24 boards (each of which is 3 feet long) e. 2 railings and 24 boards (12 that are 2 feet long, and 12 that are 4 feet long) ANSWER: e 43. In the DNA sequence 5'-TGAC -3', the phosphodiester linkage between the guanine and the adenine connects: a. the 5' end of the guanine to the 1' end of the adenine. b. the 2' end of the adenine to the 3' end of the guanine. c. the 3' end of the guanine to the 5' end of the adenine. d. the 3' end of the adenine to the 5' end of the guanine. e. the 5' end of the guanine to the 2' end of the adenine. ANSWER: c 44. If a DNA molecule is visualized as a wood screw from a hardware store, the threads in the screw would be spaced identical distances from each other. a. true b. false ANSWER: b 45. What would happen if an enhancer sequence were mutated so that its binding partner was always bound and recruiting the RNA polymerase complex? a. Transcription wouldn't change. b. No transcription would occur, as the site is now blocked to other proteins. c. Transcription would occur continuously. d. No transcription would occur because the mediator complex could not form. ANSWER: c 46. During transcription of a given protein-coding gene, both strands are used as template. a. true b. false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 ANSWER: b 47. In a long double-stranded DNA molecule containing the genetic information for many genes, the template strand for one gene may be the nontemplate strand for another gene at a different region of the DNA. a. true b. false ANSWER: a 48. A template strand of DNA and a newly transcribed RNA molecule are antiparallel. a. true b. false ANSWER: a 49. Transcription continues until: a. a transcription factor signals the end of the gene. b. all bases in the DNA are copied. c. a stop codon is encountered. d. a ribosome pulls RNA polymerase off the DNA. e. a terminator sequence is encountered. ANSWER: e 50. In eukaryotes, among the possible errors listed, which would make an mRNA less stable? a. Either the 5' cap was not added or the poly(A) tail was not formed. b. The 5' cap was not added. c. The poly(A) tail was not formed. d. None of the other answer options is correct. ANSWER: a 51. The discovery that DNA from killed virulent bacteria can transform live harmless bacteria into virulent forms means that DNA: a. codes for proteins. b. is double stranded. c. is transcribed into RNA. d. contains information that controls an organism's traits. ANSWER: d 52. If the Avery–MacLeod–McCarty transformation experiment were carried out in an RNA world, the transforming activity would be destroyed by: a. protease. b. DNase. c. RNase. d. lipase. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 ANSWER: c 53. Which choice correctly describes the complementary base pairing of adenine in both DNA and RNA? a. Adenine pairs with thymine in DNA and with uracil in RNA. b. Adenine pairs with thymine in both DNA and RNA. c. Adenine pairs with uracil in DNA and with thymine in RNA. d. Adenine pairs with cytosine in DNA and with guanine in RNA. e. Adenine pairs with guanine in DNA and with cytosine in RNA. ANSWER: a 54. Imagine you have discovered a new species of bacteria. To begin your investigation of this organism, you run an assay on the total nucleotide content of the bacterial DNA. If the cytosine content of DNA from the bacterial cells is 40%, what is the adenine content? a. 10% b. 20% c. 40% d. 60% e. It is not possible to calculate this number for prokaryotes. ANSWER: a 55. DNA is often described as having a shape like a spiral staircase.

Which of the spiral staircases in the photos is the better analogous representation of a DNA molecule? a. A b. B ANSWER: a 56. In the spiral staircase analogy of DNA structure, each railing and each step represent respectively: a. a sugar-phosphate backbone and a base. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 b. a sugar-phosphate backbone and a pair of bases. c. a base and the sugar-phosphate backbone of the DNA molecule. ANSWER: b 57. Which statement is true of DNA? a. It is used by ribosomes for translation. b. Successive nucleotides in a strand are connected by hydrogen bonds. c. A phosphate group in a nucleotide is attached to the 3' carbon in ribose. d. The percentage of the purine A always equals the percentage of the purine G. e. A purine always forms a complementary base pair with a pyrimidine. ANSWER: e 58. The central dogma of molecular biology refers to: a. replication only. b. transcription and translation. c. transcription only. d. translation only. e. replication and transcription. ANSWER: b 59. In a double-stranded DNA molecule, the strands are said to be antiparallel because: a. they form an uneven pair of grooves on the outside of the molecule. b. only one of them is a template strand. c. each purine of one strand pairs with a pyrimidine of another. d. one strand runs in 5' to 3' direction and the other in 3' to 5'. ANSWER: d 60. DNA is isolated from a virus and found to contain 27% A, 27% T, 23% C, and 23% G. The DNA is likely to be: a. circular. b. contaminated with RNA. c. double stranded. d. single stranded. ANSWER: c 61. In double-stranded DNA, the amount of A equals that of T, and the amount of C equals that of G because: a. the strands have complementary sequences of bases. b. the strands wind around one another. c. pyrimidines always pair with each other, as do purines. d. strands are antiparallel. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 62. In a DNA strand, successive nucleotides are linked by: a. hydrogen bonds. b. 3'-5' phosphodiester bonds. c. base stacking. d. 2'-5' phosphodiester bonds. e. peptide bonds. ANSWER: b 63. In double-stranded DNA, the sugar-phosphate backbones are: a. on the outside, separated by grooves of unequal size. b. on the outside, separated by grooves of equal size. c. on the inside, separated by grooves of equal size. d. on the inside, separated by grooves of unequal size. ANSWER: a 64. Which nucleotide base pairs with adenine in an RNA molecule? a. uracil b. cytosine c. thymine d. adenine e. guanine ANSWER: a 65. Which example correctly lists the components necessary for eukaryotic transcription? a. ribosomes, general transcription factors, DNA, and RNA nucleotides b. RNA polymerase, general transcription factors, DNA, and DNA nucleotides c. RNA polymerase, general transcription factors, DNA, and RNA nucleotides d. ribosomes, general transcription factors, DNA, and DNA nucleotides ANSWER: a 66. RNA polymerase complex can do which actions? a. restore the original DNA strands b. allow RNA-DNA hybrids to form c. release a finished RNA transcript from a DNA template d. separate DNA strands e. All of these choices are correct. ANSWER: e 67. In which cellular process is RNA involved? a. All of these choices are correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 b. transcription c. translation d. splicing e. transcription and translation ANSWER: a 68. What RNA function has been evolved by selection in the experiments of Jack W. Szostak and his collaborators? a. transcription b. catalysis c. translation d. DNA replication ANSWER: b 69. Transcription is sometimes described as a process in which RNA is "copied" from the template strand of DNA. This statement is potentially misleading because: a. All of these choices are correct. b. RNA nucleotides contain ribose and so cannot be an exact copy of DNA. c. the RNA transcript and the DNA template strand are antiparallel. d. the RNA transcript has a complementary sequence of bases to the template strand. e. RNA molecules contain uracil instead of thymine. ANSWER: a 70. A template DNA strand contains the sequence 5'-ATGCTGAC-3'. The corresponding sequence in the RNA transcript is: a. 5'-GUCAGCAU-3'. b. 5'-TACGACTG-3'. c. 5'-GTCAGCAT-3'. d. 5'-UACGACUG-3'. ANSWER: a 71. A template DNA strand contains the sequence 3'-ATGCTGAC-5'. The corresponding sequence in the RNA transcript is: a. 5'-GTCAGCAT-3'. b. 5'-TACGACTG-3'. c. 5'-UACGACUG-3'. d. 5'-GUCAGCAU-3'. ANSWER: c 72. A template DNA strand contains the sequence 3'-ATGCTGAC-5'. This strand is transcribed: a. from left to right. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 b. from right to left. ANSWER: a 73. In the process of transcription, the RNA transcript is synthesized: a. from the 5' end to the 3' end. b. from the 3' end to the 5' end. ANSWER: a 74. A template DNA strand contains 30% A, 20% T, 27% G, and 23% C. The RNA transcript contains: a. 30% A, 20% U, 27% G, and 23% C. b. 30% A, 20% T, 27% G, and 23% C. c. 30% T, 20% A, 27% C, and 23% G. d. 0% U, 20% A, 27% C, and 23% G. ANSWER: d 75. Transcription of RNA from DNA in eukaryotes requires: a. transcription factors. b. activator proteins. c. a promoter sequence. d. RNA polymerase. e. All of these choices are correct. ANSWER: a 76. In transcription, each added ribonucleotide comes into the RNA polymerase complex as a: a. nucleoside triphosphate. b. nucleoside. c. nucleoside monophosphate. d. nucleoside diphosphate. ANSWER: a 77. In transcription, the energy to attach each successive ribonucleotide to the growing RNA chain comes from: a. the RNA polymerase itself. b. cleavage of the high-energy phosphate bonds of the incoming nucleotide. c. cleavage of the high-energy phosphate bonds of the growing transcript. d. cleavage of the 2' hydroxyl group on the ribose of the incoming nucleotide. ANSWER: b 78. In transcription, the energy to attach each successive ribonucleotide to the growing RNA chain comes from: a. creation of a phosphodiester bond. b. hybridization of a free nucleotide to a template. c. breaking of the bond between the 2' carbon and the hydroxyl group. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 d. hydrolysis of a pyrophosphate group. e. the entry of a free nucleotide into an RNA polymerase. ANSWER: d 79. An intron is: a. an RNA sequence that is removed during the processing of an RNA molecule in the nucleus. b. a polypeptide that is excised out of a larger protein post-translationally. c. part of an intact, mature mRNA that leaves the nucleus. d. part of an RNA transcript that is not present in the DNA template. e. a type of transfer RNA. ANSWER: a 80. An exon is: a. a transfer RNA that binds to the codon. b. a protein that is modified post-translationally. c. RNA that is removed during the processing of an RNA molecule and remains inside the nucleus. d. part of an intact, mature mRNA that leaves the nucleus. e. a series of amino acids at the end of a new polypeptide that directs transcription to the ER. ANSWER: d 81. In order to test the effects of a new drug, you isolate the messenger RNA molecules from both treated and untreated eukaryotic cells and separate them according to size, using gel electrophoresis. In each lane of the gel, the shorter RNA molecules migrate more quickly through the gel and end up near the bottom of the gel, whereas the longer RNAs migrate more slowly and remain near the top.

The samples that were loaded into each of the four lanes are as follows: Lane 1: the primary RNA transcripts isolated from the nucleus of untreated cells Copyright Macmillan Learning. Powered by Cognero.

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Chapter 3 Lane 2: the primary RNA transcripts isolated from the nucleus of cells treated with the drug being tested Lane 3: RNA isolated from the cytosol of untreated cells Lane 4: RNA isolated from the cytosol of cells treated with the drug being tested Which conclusion is most likely to be correct? a. These results suggest that the drug affects RNA processing. b. These results suggest that the drug inhibits post-translational processing of this gene. c. These results suggest that the drug contains a protease that targets this gene product. d. These results suggest that the drug digests DNA. e. These results suggest that the drug inhibits DNA replication. ANSWER: a 82. Which type of post-transcriptional modification is common in eukaryotes? a. All of these choices are correct. b. polyadenylation c. intron removal d. 5¢ cap addition e. polyadenylation and intron removal ANSWER: a 83. Alternative splicing allows for: a. increased stability of a mature mRNA. b. different polypeptides to be made from a single gene. c. enhanced recognition of an mRNA by a ribosome. d. multiple genes to be used to code for a single polypeptide chain. ANSWER: b 84. T2 and T4 are two related but different types of bacteriophages (viruses that infect bacterial cells). Scientists have discovered how to make a hybrid phage that has the protein coat of T2 and the DNA of the T4 phage. If this hybrid phage infects bacteria, the new phage produced in the host cell will have: a. the protein and DNA of T2. b. the protein of T2 and the DNA of T4. c. the protein of T4 and the DNA of T2. d. a mixture of the DNA and proteins of both phage types. e. the protein and DNA of T4. ANSWER: e 85. You obtain three samples of nucleic acid and would like to determine each sample’s chemical identity (single-stranded or double-stranded DNA or RNA). You analyze the base composition of each sample and get these results, with the numbers indicating percentage of each base in the strand: A

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Chapter 3 Sample 1 Sample 2 Sample 3

12 31 23

13 0 15

37 31 30

38 15 23

0 23 0

What conclusion can you come to about the nature of these samples? a. Sample 1 is double-stranded DNA, sample 2 is RNA, and sample 3 is single-stranded DNA. b. Sample 1 is single-stranded DNA, sample 2 is RNA, and sample 3 is double-stranded DNA. c. Sample 1 is double-stranded DNA, sample 2 is single-stranded DNA, and sample 3 is RNA. d. Sample 1 is RNA, sample 2 is double-stranded DNA, and sample 3 is single-stranded DNA. e. Both samples 1 and 3 are single-stranded DNA, and sample 2 is RNA. ANSWER: a 86. Transcription and translation differ in prokaryotes versus eukaryotes in which of these ways? a. Prokaryotic ribosomes are located inside their nuclei, and eukaryotic ribosomes are located in the cytoplasm. b. Transcription of mRNA requires RNA polymerase in eukaryotes, but RNA polymerase is not required in prokaryotes. c. Translation of prokaryotic mRNA can occur as the mRNA is being transcribed, which is not possible in eukaryotes. d. None of these choices are correct. ANSWER: c 87. The number of proteins produced by protein coding genes in the human genome is less than the number of protein coding genes. a. true b. false ANSWER: b

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Chapter 4 Multiple Choice 1. How many water molecules would be produced in making a polypeptide that is 14 amino acids long? a. 27 b. 13 c. 14 d. 28 ANSWER: b 2. The individual polypeptide chains in a multi-subunit protein each have their own primary, secondary, and tertiary structure. a. true b. false ANSWER: a 3. Which force can contribute to a protein's tertiary structure? a. ionic bonding b. van der Waal's forces c. covalent bonding d. hydrogen bonding e. of these choices are correct. ANSWER: e 4. The interactions between amino acids are major factors in determining the shape of a protein. These interactions can be affected by the environment surrounding a protein. Which factor would have an effect on the shape of a protein? a. the temperature of the environment. b. the pH of the environment c. whether the other molecules in the environment are predominantly hydrophilic or hydrophobic d. the concentrations of ions present in the environment e. All of these choices are correct. ANSWER: e 5. Amino acids with hydrophobic R groups are most often found in the interior of folded proteins. a. true b. false ANSWER: a 6. Because there are three different possible reading frames in a messenger RNA (mRNA) molecule, most mRNAs can be translated in a cell into three different proteins. a. true b. false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 ANSWER: b 7. Peptide bond formation is catalyzed by the enzymatic activity of an RNA molecule. a. true b. false ANSWER: a 8. A polycistronic mRNA with six protein-coding genes has: a. 6 start codons and 1 stop codon. b. 1 start codon and 6 stop codons. c. 6 start codons and 6 stop codons. d. 1 start codon and 1 stop codon. e. 3 start codons and 3 stop codons. ANSWER: c 9. Referring to the mRNA sequence, 5'—AUGAGACUUACCGAA—3', what would the anticodon look like if the second nucleotide of the fourth codon was mutated to U? a. 3′--AUG--5′ b. 3′--TAG--5′ c. 3′--AAC--5′ d. 3′--UAG--5′ ANSWER: d 10. You are translating an mRNA in the lab and notice that the protein produced is only two amino acids long when you were expecting a length of forty amino acids. You have checked the sequence of the mRNA and everything appears to be correct. What could be happening in your reaction? a. The acceptor site of the ribosome is not functioning. b. The E (exit) site of the ribosome is not functioning. c. The DNA is mutated. d. The peptide bonds are not forming properly. ANSWER: b 11. You want to translate a polycistronic bacterial mRNA in eukaryotic cells. You remove all stop codons except the last stop codon in the bacterial DNA before inserting it into the genome of a eukaryotic cell. Can you get a functional protein using this method? a. yes b. no ANSWER: b 12. A folding domain of a polypeptide chain has a primary structure containing 5 phenylalanine residues, where F represents the side chain of phenylalanine. Consider the possible folding orientations. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4

If this domain folds in one of the two orientations shown, which is more likely, the one on the left or the one on the right?

a. the one on the left b. the one on the right ANSWER: a 13. How many water molecules would be produced in making a polypeptide that is 20 amino acids long? a. 18 b. 19 c. 21 d. 40 ANSWER: b 14. Which of the environmental factors affect the interactions between amino acids that in turn determine the shape of a protein. a. temperature b. pH c. ion concentration d. hydrophilic or hydrophobic properties of the environment e. temperature and ion concentration f. All of these choices are correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 ANSWER: a 15. The histone proteins are among the most slowly evolving proteins. For example, a histone protein called histone H4 in cows and peas differs by only 2 amino acids among a total of about 100. A reason why histone proteins evolve so slowly is their function: These proteins bind with the negatively charged phosphate groups in double-stranded DNA, so the proteins must be rich in the basic amino acids lysine and arginine. Therefore, these amino acids cannot be replaced without disrupting the histone's structure and function. a. true b. false ANSWER: a 16. Which amino acid has its R group covalently linked to the amino group? a. cysteine b. serine c. proline d. glutamic acid e. glycine ANSWER: c 17. Which amino acid is most likely to participate in hydrogen bonding with water? a. alanine b. asparagine c. leucine d. phenylalanine e. valine ANSWER: b 18. At physiological pH, the ionized state of the carboxyl (COOH) group in the R group of aspartic acid is: a. COO-. b. COOH+. c. COOH. d. CO-O-. ANSWER: a 19. At physiological pH, the ionized state of the amino (NH2) group in the R group of lysine is: a. NH4++. b. NH2. c. NH3+. d. NH-. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 20. X-rays are waves of electromagnetic radiation with an extremely short wavelength (0.01 to 10 nanometers). They can be used in X-ray crystallography to infer the arrangement of atoms in a crystal because: a. some X-rays pass through the crystal and scatter in different directions. b. they can be used on live cells without isolating the protein in question. c. their wavelength allows them to determine the amino acid sequence of the protein. d. all X-rays bounce off the surface of the crystal. e. all X-rays pass through the crystal without changing direction. ANSWER: a 21. How many residues separate amino acids that are stabilized by hydrogen bonds in α helices? a. 2 b. 4 c. 10 d. 20 e. 100 ANSWER: b 22. Proteins that do not refold properly after being heated generally require the assistance of which type of molecules in cells? a. polymerases b. ribosomes c. enhancers d. chaperones e. transferases ANSWER: d 23. Which factors are used during translation? a. elongation factors b. release factors c. initiation factors d. All of these choices are correct. ANSWER: d 24. Ribosomes in prokaryotes and eukaryotes are: a. similar in structure and translate using different genetic code. b. identical in structure but translate using different genetic codes. c. identical in structure and translate using the same genetic code. d. similar in structure and translate using the same genetic codes. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 25. Which location is a critical region of a tRNA molecule? a. amino acid attachment site and start codon b. anticodon loop and ribosome binding site c. stop codon and Shine-Dalgarno sequence d. amino acid attachment site and anticodon loop e. ribosome binding site and 5'cap ANSWER: d 26. Which of the codons is capable of terminating translation? a. UAG b. UAA c. UGA d. All of these choices are correct. ANSWER: d 27. Which one is responsible for the addition of amino acids after translation initiation has begun? a. elongation factors b. tRNA c. small ribosomal subunits d. mRNA ANSWER: a 28. In eukaryotes, the AUG codon that starts translation is: a. the AUG nearest the 5' cap on the mRNA. b. adjacent to a Shine-Dalgarno sequence. c. adjacent to the 5' cap on the mRNA. d. at the 5' end of the mRNA. e. None of the other answer options is correct. ANSWER: b 29. In which of the ribosomal sites does the anticodon of a tRNA pair completely with the mRNA codon? a. the A site only b. the P site only c. the A site and the P site d. the P site and the E site e. the A site, the P site, and the E site ANSWER: c 30. Which statements are the steps in the elongation phase of translation? a. introduction of a charged tRNA into an unoccupied A site of the ribosome b. ejection of an uncharged tRNA from the E site of the ribosome Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 c. peptide bond formation d. movement of the ribosome three bases closer to the 3' end of the mRNA e. All of these choices are correct. ANSWER: e 31. Imagine you are following a particular tRNA, called tRNAQ, through the process of translation in a eukaryote. Consider the steps of tRNAQ translation. In what order does tRNAQ go through the steps listed? Note that some steps may be used more than once. 1. The polypeptide is transferred to tRNAQ. 2. tRNAQ binds the A site of the ribosome. 3. tRNAQ binds the P site of the ribosome. 4. The ribosome shifts, with tRNAQ still bound. 5. tRNAQ binds the E site of the ribosome. a. 2, 1, 4, 3, 4, 5 b. 2, 4, 1, 4, 2, 4, 5 c. 3, 1, 4, 2, 4, 5 d. 3, 4, 1, 4, 2, 4, 5 e. 5, 1, 4, 3, 4, 5 ANSWER: a 32. Which step occurs in the A site of the ribosome during translation? a. An uncharged tRNA is ejected from this site as the ribosome slides to the next codon. b. The tRNA carrying the growing polypeptide moves to this site as the ribosome slides to the next codon. c. An incoming charged tRNA binds to this site. d. None of the other answer options is correct. ANSWER: c 33. Which step occurs in the P site of the ribosome during translation? a. An incoming charged tRNA binds to this site. b. The tRNA carrying the growing polypeptide moves to this site as the ribosome slides to the next codon. c. An uncharged tRNA is ejected from this site as the ribosome slides to the next codon. d. None of the other answer options is correct. ANSWER: b 34. Which step occurs in the E site of the ribosome during translation? a. An incoming charged tRNA binds to this site. b. The tRNA carrying the growing polypeptide moves to this site as the ribosome slides to the next codon. c. An uncharged tRNA is ejected from this site as the ribosome slides to the next codon. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 d. None of the other answer options is correct. ANSWER: c 35. After translation has been initiated, how long will elongation continue? a. until the ribosome reaches a UAA, UAG, or UGA codon b. until the ribosome reaches an AUG codon c. until the ribosome reaches the end of the mRNA d. until the ribosome reaches the poly(A) tail of the mRNA e. until the cell runs out of tRNAs ANSWER: a 36. The growing polypeptide remains in the P site throughout translation. a. true b. false ANSWER: b 37. When we say that tRNAs move from the A to P to E sites of a ribosome, what is moving? a. the tRNA b. the mRNA c. the ribosome d. Nothing is moving. ANSWER: c 38. A double-stranded DNA molecule, only part of which is shown, is being transcribed. If the molecule is transcribed from left to right, one of the nucleotides shown in bold would be the first transcribed in this small molecule. 5'—ATGATCGGATCGATCCAT—3' 3'—TACTAGCCTAGCTAGGTA—5' Which sequence is the correct mRNA produced from the transcription of this DNA molecule? a. 3'-AUGAUCGGAUCGAUCCAU-5' b. 5'-AUGAUCGGAUCGAUCCAU-3' c. 5'-UACUAGCCUAGCUAGGUA-3' d. 3'-UACUAGCCUAGCUAGGUA-5' ANSWER: b 39. If the RNA transcript 5'-AUGAUCGGAUCGAUCCAU-3' resulting from the DNA sequence is present in the mRNA and translated codon by codon from one end to the other, which of the polypeptides would correspond to this part of the mRNA? Use the table shown to answer this question.

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a. NH2- Met-Ile-Gly-Ser-Ile-His -COOH b. NH2- Tyr-Leu-Ala-Arg-Lue-Val -COOH c. COOH- Met-Ile-Gly-Ser-Ile-His -NH2 d. COOH- Tyr-Leu-Ala-Arg-Lue-Val -NH2 ANSWER: a 40. The molecule currently being investigated by researchers as a possible means of synthesis of the first proteins is: a. mRNA. b. DNA. c. rRNA. d. tRNA. e. aminoacyl tRNA synthetases. ANSWER: d 41. Which of the cellular processes occurs in the cytoplasm of a eukaryote? a. RNA processing b. translation c. transcription Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 d. transcription and RNA processing e. All of these choices are correct. ANSWER: b 42. Proteins can be used for which cellular roles? a. cell signaling. b. structural support c. biological catalysis d. cell communication e. All of these choices are correct. ANSWER: e 43. One hypothesis for the original function of precursors of tRNA is that: a. they carried nucleotides to a self-replicating RNA. b. they carried amino acids to a catalytic RNA that made peptide bonds. c. they were self-replicating. d. they protected RNA from undergoing mutations. ANSWER: a 44. One hypothesis for the original function of precursors of charged tRNA is that: a. the amino acids catalyzed self-replication. b. the amino acids increased the accuracy of RNA self-replication. c. the amino acids created the proper pH for RNA self-replicating. d. None of the other answer options is correct. ANSWER: b 45. Transcripts of rRNA genes are concentrated in the: a. nuclear envelope. b. nucleolus. c. cytoplasm. d. mitochondria. e. endoplasmic reticulum. ANSWER: b 46. Multiple ribosomes can be simultaneously translating protein from a single mRNA in a eukaryote, but the mRNAs being translated are monocistronic. a. true b. false ANSWER: a 47. In a protein-coding region of DNA, any mutation that replaces a single nucleotide for another will replace Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 any amino acid with any other amino acid. a. true b. false ANSWER: b 48. In a protein-coding region, a mutation that replaces a single nucleotide for another may or may not change the resulting amino acid. a. true b. false ANSWER: a 49. In a protein-coding region of DNA, a mutation that replaces a single nucleotide, but does not change the resulting amino acid, is likely to be: a. at the 5' position in a codon of the transcribed mRNA. b. at the middle position in a codon of the transcribed mRNA. c. at the 3' position in a codon of the transcribed mRNA. d. at any position in a codon of the transcribed mRNA. e. None of the other answer options is correct. ANSWER: c 50. Refer to the table. Which of the amino acid(s) are encoded by just one codon?

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Chapter 4

a. methionine and asparagine b. asparagine c. methionine and tryptophan d. serine and leucine ANSWER: c 51. Refer to the table. Which amino acids are encoded by the greatest number of codons?

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Chapter 4

a. serine and leucine b. tryptophan and asparagine c. methionine d. methionine and tryptophan ANSWER: a 52. Pseudogenes are relics of former genes that no longer possess biological functions. What would you expect to be the rate at which mutations accumulate in a pseudogene over time compared to the original functional gene? a. Accumulation of mutations in a pseudogene cannot occur. b. Mutations will accumulate faster in a pseudogene than in the functional gene. c. Accumulation of mutations in a pseudogene and in the functional gene will be approximately the same. d. The mutation will accumulate slower in a pseudogene than in the functional gene. e. It is impossible to measure mutation rates in a pseudogene. ANSWER: b 53. Folding domains in proteins encoded in pseudogenes will gradually lose their identity as amino acid replacements take place in the molecule. a. true Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 b. false ANSWER: a 54. Similarity between proteins is often judged by means of a distance matrix, such as the one shown, in which each square in the matrix enumerates the number of differences between any pair of sequences. In the distance matrix shown, 6 proteins, each 100 amino acids in length, are compared. For example, the number of amino acid differences between proteins 1 and 5 is 36 as recorded in the corresponding square.

Based on the distance matrix shown, the most closely related pairs of proteins are: a. (1, 4), (2, 6), and (3, 5). b. (1, 2), (3, 4), and (5, 6). c. (1, 3), (2, 5), and (3, 6). d. (1, 4), (2, 5), and (3, 6). e. (1, 5), (2, 6), and (3, 4). ANSWER: a 55. Similarity between proteins is often judged by means of a distance matrix, such as the one shown, in which each square in the matrix enumerates the number of differences between any pair of sequences. In the distance matrix shown, 6 proteins, each 100 amino acids in length, are compared. For example, the number of amino acid differences between proteins 1 and 5 is 36 as recorded in the corresponding square.

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Chapter 4

Based on the distance matrix shown, the most distantly related pairs of proteins are: a. (1, 4), (2, 6), and (3, 5). b. (2, 5), (4, 6), and (3, 5). c. (1, 4), (2, 6), and (5, 6). d. (2, 4), (2, 5), and (3, 6). e. (1, 3), (1, 5), and (3, 4). ANSWER: e 56. Based on the standard genetic code, what polypeptide sequence, or sequences, would you expect to be translated from a messenger RNA with the sequence 5'–AUAUAUAUAU… –3' if translation were occurring in a laboratory setting in which translation can start anywhere along the mRNA molecule. a. Ile-Tyr-Ile-Tyr b. Ile-Tyr-Ile-Tyr… and Tyr-Ile-Tyr-Ile-Tyr… c. Tyr-Ile-Tyr-Ile-Tyr d. Met-Cys-Tyr-Tyr-Tyr ANSWER: b 57. The function of a protein is dependent upon the shape into which the chain of amino acids folds. Many noncovalent interactions are responsible for maintaining the protein's shape. Assume you have isolated a protein from an organism in its proper shape, and you have treated it with an enzyme that selectively targets and breaks only the peptide bonds in the proteins. Would the protein retain its shape under these conditions? a. No; while the noncovalent bonds determine the shape of a protein, the peptide bonds are required to hold the amino acids together. b. Yes; because the noncovalent interactions that determine the shape of a protein are stronger than the peptide bonds. c. Yes; once noncovalent bonding determines the shape of a protein the peptide bonds are no longer necessary. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 d. Yes; but the shape would be affected to a greater extent by changes in temperature. ANSWER: a 58. Two major types of protein secondary structures are referred to as: a. amino and carboxyl. b. helix and sheet. c. residual and permanent. d. elementary and primary. ANSWER: b 59. Secondary structure is characterized by which type of interactions? a. ionic bonding between an acidic R group and a basic R group b. a covalent bond between two cysteines c. hydrogen bonding within the peptide backbone d. hydrogen bonding between R groups of amino acids e. the covalent bond between an amino nitrogen and a carboxyl carbon ANSWER: c 60. Consider the diagram. What is wrong with this diagram of secondary structure in a polypeptide chain? Hint: In the representation of β sheets as paired arrows, the convention is that the amino end is at the base of the arrow and the carboxyl end is at the tip.

a. β sheets cannot form when the arrows are pointed in the same direction. b. The β sheet has an amino end and a carboxyl end. c. The directionality of the polypeptide chain reverses orientation between the arrows representing a β Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 sheet. d. Nothing is wrong with the diagram. ANSWER: c 61. Consider the diagram. What is wrong with this diagram of secondary structure in a polypeptide chain?

a. β sheets cannot form when the arrows are pointed in opposite directions. b. The directionality of the polypeptide chain reverses orientation between the arrows representing a β sheet. c. The β sheet has two amino ends. d. Nothing is wrong with the diagram. ANSWER: c 62. Imagine that you and your colleagues are working in a lab to develop a protein synthesis system for a new type of synthetic cell. During your brainstorming sessions, you propose that polycistronic mRNA inserted directly into the synthetic cell would be much more useful than mRNA that is only translated into one protein. This would allow for multiple proteins necessary for a particular function to be translated together. One of your colleagues says that is a good idea, but if you decide to go with polycistronic mRNA, you had better make sure to use a prokaryotic translation system. Why might it be problematic to use a eukaryotic translation system with polycistronic mRNA? a. Eukaryotic ribosomes initiate translation by binding to the 5' cap, which is only found in eukaryotic mRNAs. b. Introns always prevent translation of the mRNA. c. Prokaryotic ribosomes do not detach from mRNA when they reach a stop codon. d. Eukaryotic start and stop codons are different from prokaryotic start and stop codons. ANSWER: a 63. You are investigating a mutant eukaryotic cell line that makes all of its mRNAs much longer than the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 mRNAs from normal cells. Intrigued, you examine the proteins in these cells and note that many of them are either much longer or much shorter than the normal proteins from nonmutant cells. Assuming there is just one mutant defect in these cells, which of the possibilities is most likely? a. stop codon is mutated. b. RNA polymerase does not recognize the terminator sequence. c. The release factor is defective. d. The spliceosome is nonfunctional. e. The enzyme that adds the poly-A tail is defective. ANSWER: d 64. The genetic code is: a. comprised of codons which usually differ at the first position. b. ambiguous. c. different for bacteria than for eukaryotes. d. a quadruplet code. e. redundant. ANSWER: e 65. When a peptide bond is created between two amino acids: a. the carboxyl group of the first amino acid is joined to the carboxyl group of the second. b. the amino group of the first amino acid is joined to the carboxyl group of the second. c. the carboxyl group of the first amino acid is joined to the amino group of the second. d. the amino group of the first amino acid is joined to the amino group of the second. ANSWER: c 66. Assuming A-U and G-C pairing between the anticodon and the codon, what anticodon in tRNAMet would pair with the codon 5'-AUG-3'? a. 5'-CAU-3' b. 5'-GUA-3' c. 5'-UAC-3' d. 5'-ATG-3' e. 5'-AUG-3' ANSWER: a 67. An mRNA molecule has a sequence 5'—CAGAUCUAAUGCUUAUCGGAU—3'. When translated in a laboratory setting where translation can be initiated anywhere along the molecule, how many reading frames are possible? a. 1 b. 2 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 c. 3 d. 4 e. 6 ANSWER: c 68. A synthesized mRNA molecule has a random sequence consisting of 25% A, 25% U, and 50% C. Among the amino acids in the polypeptide chains resulting from translation in a laboratory setting in which translation can be initiated anywhere along the mRNA molecule, what is the expected frequency of phenylalanine codons? a. 1/16 b. 1/32 c. 1/64 d. 3/32 e. 3/64 ANSWER: e 69. Assuming that transcription and translation both proceed from left to right, which is the correct orientation of the DNA template, the RNA transcript, and the protein product? a. DNA template 3'– ……… –5' RNA transcript 5'– ……… –3' Protein product H2N– ……… –COOH b. DNA template 3'– ……… –5' RNA transcript 5'– ……… –3' Protein product HOOC–……… –NH2 c. DNA template 5'– ……… –3' RNA transcript 3'– ……… –5' Protein product HOOC–……… –NH2 d. DNA template 5'– ……… –3' RNA transcript 5'– ……… –3' Protein product H2N–……… –COOH ANSWER: a 70. Refer to the table. Which of the mRNA sequences would be translated in a test tube as the three different polypeptide chains polyserine, polyleucine, and polyphenylalanine, assuming that laboratory conditions were such that translation could start anywhere along the mRNA molecule?

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Chapter 4

a. None of the other answer options is correct. b. 5'—UCCUCCUCCUCC...—3' c. 5'—GUCGUCGUCGUS...—3' d. 5'—ACUGACUGACUG...—3' e. 5'—CUUCUUCUUCUU...—3' ANSWER: e 71. An mRNA sequence consisting of an alternating sequence of two nucleotides (e.g., 5'—AUAUAU…—3') codes in a test tube for a polypeptide consisting of two alternating amino acids. This means that, in the table organizing the codons in the standard genetic code, the positions of the amino acids must be in: a. the same rows but different columns. b. different rows but the same column. c. different rows and different columns. d. None of the other answer options is correct. ANSWER: c 72. In what order does a charged tRNA move through the sites of a ribosome? a. The tRNA binds with the A site, then is moved to the E site, then the P site as the ribosome shifts. b. The tRNA binds with the A site, then is moved to the P site, then the E site as the ribosome shifts. c. The tRNA binds with the P site, then is moved to the A site, then the E site as the ribosome shifts. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 d. The tRNA binds with the P site, then is moved to the E site, then the A site as the ribosome shifts. e. The tRNA binds with the E site, then is moved to the P site, then the A site as the ribosome shifts. ANSWER: b 73. When a charged tRNA is about to bind to the vacant A site of a ribosome, where is the growing polypeptide? a. in the E site b. in the A site c. in the P site d. The polypeptide is equally likely to be in any of the three sites. ANSWER: c 74. A peptide bond between two amino acids is catalyzed by: a. the rRNA component of the small subunit of a ribosome. b. the rRNA component of the large subunit of a ribosome. c. the rRNA components from both subunits of a ribosome. d. proteins in a large subunit of a ribosome. e. proteins in a small subunit of a ribosome. ANSWER: b 75. Examine the figure shown, which depicts one stage in the process of translation in a eukaryote.

Where will the tRNA be when the large subunit of the ribosome joins the complex? a. in the E site of the ribosome b. in the A site of the ribosome c. in the P site of the ribosome ANSWER: c 76. Examine the figure shown, which depicts a stage in the process of translation in a eukaryote.

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Chapter 4

What happens next? a. A polypeptide bond is formed between the arginine and the valine, transferring the polypeptide to the A site. b. A polypeptide bond is formed between the arginine and the valine, transferring the arginine to the P site. c. A polypeptide bond is formed between the arginine and the methionine, transferring the arginine to the P site. d. A polypeptide bond is formed between the arginine and the methionine, transferring the polypeptide to the A site. ANSWER: a 77. Consider the diagrams. Which of the diagrams shows what one would see in an electron microscope during the transcription of a polycistronic bacterial operon that codes for two proteins?

a. A b. B c. C d. D Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 ANSWER: a 78. A mutation takes place in a protein-coding gene that changes one codon resulting in a single amino acid replacement. Which type of replacement is most likely to result in a disruption of protein structure and function? a. a basic amino acid to a polar amino acid b. a hydrophilic amino acid to a polar amino acid c. a polar amino acid to an acidic amino acid d. a polar amino acid to a basic amino acid e. a hydrophobic amino acid to a hydrophilic amino acid ANSWER: e 79. A mutation takes place in a protein-coding gene that changes one codon resulting in a single amino acid replacement. Which type of replacement is least likely to result in a disruption of protein structure and function? a. a polar amino acid to a hydrophobic amino acid b. a polar amino acid to a polar amino acid c. a hydrophobic amino acid to a hydrophilic amino acid d. a hydrophilic amino acid to a hydrophobic amino acid e. an acidic amino acid to a basic amino acid ANSWER: b 80. 109 bacterial cells are spread evenly on an agar surface of medium containing an antibiotic. After a few days, a few of the cells grow into small clumps of cells. Isolation of individual bacterial cells and further testing reveals that the clumps consist of mutant bacteria that are resistant to the antibiotic. Which of the statements is true? a. The mutations occur because they are beneficial in the presence of the antibiotic. b. The mutations were not in the original population of bacteria but were caused by the presence of the antibiotic. c. The mutations that are resistant to the antibiotic preexisted in the original population of bacteria and were selected by the antibiotic. d. All of these choices are correct. ANSWER: c 81. Proteins with similar primary structure and function can usually be grouped into families based on the degree of similarity in their amino acid sequence. The list shows a small region of 6 aligned amino acids in protein from 5 related species. (1) QMADGCK (2) QIVDLYR (3) QIADGCK (4) QMAEGCK (5) QFVDLYR Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 Similarity is judged by a distance matrix in which each cell (square) enumerates the number, or percent, of differences between any pair of sequences. Which of the distance matrices shown reflects the differences among the five sequences? a.

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Chapter 4 c.

ANSWER: b 82. Proteins with similar primary structure and function can usually be grouped into families based on the degree of similarity in their amino acid sequence. The list shows a small region of 6 aligned amino acids in protein from 5 related species. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 (1) QMADGCK (2) QIVDLYR (3) QIADGCK (4) QMAEGCK (5) QFVDLYR If you were to group the proteins into two families based on the distance matrix, which proteins would you group together? a. Proteins (1), (3), and (4) would be grouped into one family and (2) and (5) into another family. b. Proteins (2), (3), and (4) would be grouped into one family and (1) and (5) into another family. c. Proteins (3), (4), and (5) would be grouped into one family and (1) and (2) into another family. d. These proteins cannot be grouped into families ANSWER: a 83. Proteins that evolve extremely rapidly are often difficult to group into families according to their level of sequence similarity. a. true b. false ANSWER: a 84. Suppose that codons consisted of 4 nucleotides instead of 3 and that there were only 2 different bases. How many amino acids could be encoded by this variant of the genetic code? a. 4 b. 8 c. 16 d. 20 ANSWER: c 85. How many different mRNA sequences could encode the amino acid sequence Met–Leu–Val–His? a. 12 b. 24 c. 48 d. 84 ANSWER: c 86. How many different mRNA sequences could encode the amino acid sequence Ser–Leu–Ser–Arg? a. 216 b. 108 c. 64 d. 20 ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 87. An RNA molecule has a sequence 5'– GCGAUCUAAUGCUUUUCGGAA–3'. How many reading frames are possible if this molecule is translated in a laboratory setting in which translation can be initiated anywhere along the molecule? How many reading frames are possible when this molecule is translated in a cellular environment? a. 3, 3 b. 1, 3 c. 3, 1 d. 6, 1 ANSWER: c 88. Translation of an RNA molecule in a test tube does not require an AUG codon to start but can start anywhere. Once translation begins, however, codons are read in successive triplets just like translation in a cell. Consult the standard genetic code, and identify from the choices listed what polypeptide sequence you would expect to be translated from an mRNA molecule with the sequence 5'–CAGCAGCAGCAG … –3' if translation occurred in a test tube. a. Gln-Gln-Gln-Gln … b. Lys-Lys-Lys-Lys … c. Gln-Ser-Ala-Gln … d. Ser-Ala-Ser-Ala … ANSWER: a 89. In Kohrana's experiments, a mRNA with the sequence 5'–UUUUUU… –3' coded for a polypeptide with the sequence Phe–Phe–Phe–Phe–…. (polyphenylalanine). Imagine that he added a single nucleotide bearing the base G to one end of the mRNA that resulted in a polyphenylalanine with a single Leu at the carboxyl (–COOH) end. This would indicate that Kohrana added the G base to the 5' end of the mRNA. a. true b. false ANSWER: b 90. Which mRNA sequence would be translated in a test tube as a polypeptide consisting of alternating His and Thr. a. 5'—CAUCAUCAU…—3' b. 5'—CCACCACCA…—3' c. 5'—ACCACCACC…—3' d. 5'—CACACACA…—3' ANSWER: d 91. A messenger RNA 5'–GUAGUAGUAGUA… –3' is translated in a test tube where translation does not need to start at an AUG. How many different polypeptides would you expect to be translated from this mRNA? a. 3 b. 2 c. 1 d. 0 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 ANSWER: b 92. An mRNA with the sequence 5'–UGUGUGUG…–3' codes in a test tube for a polypeptide with the alternating amino acid sequence Cys–Val–Cys–Val–…. You purify charged tRNAs carrying cysteine, whose anticodons bind with the 5'–UGU–3' codon, and you chemically change the cysteine into alanine. If the charged tRNA with the altered amino acid is used in the mixture for a test tube translation, under conditions in which translation does not need to start at an AUG, the resulting polypeptide translating 5'–UGUGUGUG…–3' would be Ala–Val–Ala–Val–…. a. true b. false ANSWER: a 93. The three-dimensional shape of a protein is determined by the primary, secondary, tertiary, and in many cases, the quaternary structure of the protein, all of which are listed in the answer options. The sentence is taken from a scientific article on protein structure. Select the level(s) of protein structure from the list of options. Hydrogen bonds between peptide backbone components form a distinct helical structure. a. primary b. secondary c. tertiary d. quaternary ANSWER: b 94. The three-dimensional shape of a protein is determined by the primary, secondary, tertiary, and in many cases, the quaternary structure of the protein, all of which are listed in the answer options. The sentence is taken from a scientific article on protein structure. Select the level(s) of protein structure from the list of options. Hydrogen bonds between peptide 'backbone' components on one polypeptide and R groups on another polypeptide contribute to the overall function. a. primary b. secondary c. tertiary d. quaternary ANSWER: d 95. The three-dimensional shape of a protein is determined by the primary, secondary, tertiary, and in many cases, the quaternary structure of the protein, all of which are listed in the answer options. The sentence is taken from a scientific article on protein structure. Select the level(s) of protein structure from the list of options. There are extensive ionic interactions between positively charged R groups and negatively charged R groups on the polypeptide. a. primary b. secondary c. tertiary Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 d. quaternary ANSWER: c 96. The three-dimensional shape of a protein is determined by the primary, secondary, tertiary, and in many cases, the quaternary structure of the protein, all of which are listed in the answer options. The sentence is taken from a scientific article on protein structure. Select the level(s) of protein structure from the list of options. There are hydrogen bonds between peptide backbone components that form neither an α helix nor a β sheet. a. primary b. secondary c. tertiary d. quaternary ANSWER: c 97. The three-dimensional shape of a protein is determined by the primary, secondary, tertiary, and in many cases, the quaternary structure of the protein, all of which are listed in the answer options. The sentence is taken from a scientific article on protein structure. Select the level(s) of protein structure from the list of options. Peptide bonds form between the monomers. a. primary b. secondary c. tertiary d. quaternary ANSWER: a 98. The codons in mRNA specify the amino acids that are used to make a protein. Is the statement true or false? Sixty-one of the 64 possible codons are specified amino acids, whereas the other 3 are stop codons. Each of the 61 codons specifies just one amino acid. a. true b. false ANSWER: a 99. The codons in mRNA specify the amino acids that are used to make a protein. Is the statement true or false? Some amino acids are specified by just one codon, whereas others are specified by multiple codons. a. true b. false ANSWER: a 100. The codons in mRNA specify the amino acids that are used to make a protein. Is the statement true or false? The limitations of the genetic code are such that a single amino acid may be specified by no more than four Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 codons. a. true b. false ANSWER: b 101. The codons in mRNA specify the amino acids that are used to make a protein. Is the statement true or false? Because there are 4 RNA nucleotides and each codon is 3 nucleotides long, there are 64 possible codons, each of which directs the ribosome to incorporate a different amino acid into a growing polypeptide chain. a. true b. false ANSWER: b 102. Suppose that codons consisted of two nucleotides instead of three, but there were still 20 amino acids to be incorporated into proteins during translation. If this were the case, indicate whether the statement would be true or false. There would still be enough codons to specify 20 amino acids and at least one stop codon. a. true b. false ANSWER: b 103. Suppose that codons consisted of two nucleotides instead of three, but there were still 20 amino acids to be incorporated into proteins during translation. If this were the case, indicate whether the statement would be true or false. Redundancy would be eliminated; in fact, there would not be enough two-base codons to specify all 20 acids or act as stop codons. a. true b. false ANSWER: a 104. Suppose that codons consisted of two nucleotides instead of three, but there were still 20 amino acids to be incorporated into proteins during translation. If this were the case, indicate whether the statement would be true or false. Most codons would probably specify a single amino acid. However, some codons would have to specify more than one amino acid or an amino acid and a stop codon. a. true b. false ANSWER: a Multiple Response Copyright Macmillan Learning. Powered by Cognero.

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Chapter 4 105. Which of the actions would affect the secondary structure of a protein? Select all that apply. a. breaking the hydrogen bonds between amino acids b. breaking the ionic bonds between amino acids c. changing the sequence of amino acids d. disruption of the interactions between two different polypeptide chains ANSWER: a, c 106. Examine the figure shown, which depicts one stage in the process of translation in a eukaryote.

What will happen when the ribosome shifts one codon further on the mRNA, assuming the next codon is not a stop codon? Select all that apply. a. The tRNA carrying the polypeptide will be in the P site. b. The tRNA that is no longer carrying the polypeptide will be ejected from the ribosome. c. The sites of the ribosomes will be relabeled (from left to right) P, A, E. d. A new tRNA will bind to the ribosome. e. The tRNA that is in the A site in the figure shown will be ejected from the ribosome. ANSWER: a, b, d 107. The three-dimensional shape of a protein is determined by the primary, secondary, tertiary, and in many cases, the quaternary structure of the protein, all of which are listed in the answer options. The sentence is taken from a scientific article on protein structure. Select the level(s) of protein structure from the list of options. Disulfide bonds formed between cysteines stabilize the overall structure of this protein isolated from the bacterium. Select all that apply. a. primary b. secondary c. tertiary d. quaternary ANSWER: c, d

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Chapter 5 Multiple Choice 1. The plasma membrane is composed of a phospholipid bilayer and associated proteins. What else is commonly found in the plasma membranes of animal cells? a. nucleic acids b. cholesterol c. amino acids d. ethanol ANSWER: b 2. In response to seasonal changes in temperature, many organisms must alter the composition of their plasma membranes to maintain the proper degree of fluidity. Which change in the fatty acids of phospholipids would be most effective in maintaining membrane fluidity in a colder environment? a. a decrease in phospholipid fatty acid side chain length and a decrease in side chain saturation b. a decrease in phospholipid fatty acid side chain length and an increase in side chain saturation c. an increase in phospholipid fatty acid side chain length and an increase in side chain saturation d. an increase in phospholipid fatty acid side chain length and a decrease in side chain saturation ANSWER: a 3. In order for phospholipids to form bilayers spontaneously in an aqueous environment, the pH should be: a. very acidic. b. moderately acidic. c. approximately neutral. d. moderately basic. e. very basic. ANSWER: c 4. The first cells were surrounded by a membrane composed largely of lipids synthesized by intracellular proteins. As life on earth was starting, from what were the first membranes most likely formed? a. nucleic acids undergoing self-replication b. fatty acids produced by cellular enzymes in the endoplasmic reticulum c. amino acids joining together by peptide bonds d. lipids mixing in the watery environment ANSWER: d 5. Some transmembrane proteins are embedded in patches of plasma membrane with a specialized, non-random assortment of phospholipids, sphingolipids and cholesterol. These patches are referred to as: a. lipid rafts. b. sphingopatches. c. biomembrane aggregation regions (BARs). Copyright Macmillan Learning. Powered by Cognero.

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Chapter 5 d. plaques. e. cholesterol plugs. ANSWER: a 6. Which choice describes the hydrophilic component of cholesterol? a. the carbon tail b. the hydroxyl head group c. the rigid group of planar rings d. the methyl groups attached to the rings ANSWER: b 7. Which choice is considered an integral membrane protein? a. a protein with its amino-terminus in the cytoplasm and its carboxy-terminus in the extracellular space b. a protein attached to a transmembrane protein via hydrogen bonding c. a protein attached to a phospholipid via ionic bonding with the head group of the lipid molecule d. a protein capable of diffusing throughout the cytoplasm of a cell ANSWER: a 8. A beaker contains two solutions of salt dissolved in water. The two solutions have different concentrations of salt (measured by molarity, M) and are separated by a membrane that is permeable to both salt and water. The salt and water will move through the membrane by diffusion.

Which statement is true about the diffusion of these solutions? a. There will be a net movement of salt from side B to side A and net movement of water from side A to side B. b. There will be a net movement of salt from side A to side B and no movement of water. c. There will be a net movement of both salt and water from side B to side A. d. There will be a net movement of water from side A to side B and no movement of salt. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 5 9. In intestinal epithelial cells, a transport protein moves the bulky, polar glucose molecules through the membrane into the cytoplasm, while at the same time it transports Na+ through the membrane into the cell down its electrochemical gradient. Which example correctly describes this cotransport of glucose and sodium? a. secondary active transport by an antiporter b. secondary active transport by a symporter c. primary active transport by a symporter d. primary active transport by an antiporter ANSWER: b 10. The sodium-potassium pump is an example of: a. a symporter. b. passive transport. c. an antiporter. d. channel-mediated diffusion. e. None of the answer options is correct. ANSWER: c 11. Which component is not considered part of the cytoplasm? a. None of the answer options is correct. b. the endoplasmic reticulum c. the Golgi apparatus d. the cytoskeleton e. the nucleus ANSWER: e 12. During translation, the new polypeptides are often directed to specific parts of the cell by the presence or absence of short sequences of amino acids called signal sequences. Which sequence would you expect to find in the signal sequences that will eventually become enzymes important in photosynthesis? a. an amino-terminal signal sequence b. no signal sequence c. an internal signal sequence d. a signal anchor sequence ANSWER: a 13. If a mutation blocked the function of the signal recognition particle, making it unable to bind signal sequence, what would result? a. All proteins would end up in the nucleus. b. All proteins would be translated on free ribosomes in the cytoplasm. c. All proteins would be secreted from the cell. d. All proteins would end up in the mitochondria or chloroplast. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 5 14. Which proteins would be synthesized on the rough endoplasmic reticulum and processed in Golgi apparatus? a. cytoskeletal proteins actin and tubulin, which are found in the cytoplasm b. DNA polymerase that functions during DNA replication c. transcription factors that bind to DNA sequences in the nucleus d. lysosomal enzymes that break down proteins in the lumen of the lysosome ANSWER: d 15. RNA molecules are transported from the nucleus to the cytoplasm in eukaryotes through: a. aquaporins. b. nuclear pores. c. passive diffusion. d. budding off of the nuclear envelope. e. sodium-potassium pumps. ANSWER: b 16. Which particles are pumped across the lysosomal membrane by a transport protein to create a special internal environment in the lysosome? a. protons b. broken-down macromolecules c. enzymes capable of breaking down macromolecules being delivered from the Golgi apparatus d. phospholipids and cholesterol ANSWER: a 17. A protein with an internal signal sequence is most likely to be located in: a. the cytoplasm. b. a mitochondrion. c. the nucleus. d. a chloroplast. e. the extracellular space. ANSWER: c 18. During the translation of mRNA molecules, the new polypeptides are often directed to specific parts of the cell by the presence or absence of short sequences of amino acids called signal peptides. Which peptide would you expect to find in the polypeptides that will eventually fold to become the cytoskeletal proteins tubulin and actin, which are found in the cytoplasm? a. a signal anchor peptide b. an amino-terminal signal peptide c. an internal signal peptide d. no signal peptide ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 5 19. During the translation of mRNA molecules, the new polypeptides are often directed to specific parts of the cell by the presence or absence of short sequences of amino acids called signal peptides. Which peptide would you expect to find in the polypeptides that will eventually fold to become enzymes important in the citric acid cycle, a critical metabolic pathway in the mitochondria? a. a signal anchor peptide b. an amino-terminal signal peptide c. an internal signal peptide d. no signal peptide ANSWER: b 20. Which structures in the plasma membrane allow proteins involved in the same biochemical pathway to associate with each other? a. formation of lipid rafts b. flip-flop of individual phospholipids c. amphipathic nature of the phospholipids d. decreased fluidity of the membrane due to carbon-carbon double bonds ANSWER: a 21. Purified phospholipids mixed in water will only form membranes if the appropriate enzyme is present. a. true b. false ANSWER: b 22. Which component of a phospholipid is found in the interior of a lipid bilayer? a. fatty acids b. glycerol c. phosphate group ANSWER: a 23. Phospholipids spontaneously form a variety of structures in aqueous solution. Which choice best describes this property? a. the ability of vesicles to bud off from the plasma membrane (endocytosis) b. the first cells c. formation of lipid bilayers d. the ability of vesicles to fuse with the plasma membrane (exocytosis) e. All of these choices are correct. ANSWER: e 24. Paramecium is a unicellular organism that lives in fresh water. Suppose you are studying two populations of this organism. One population lives in a pond in a cold environment, and the other lives in a pond in a warm environment. If you examined the fatty acids in plasma membranes of both of these populations, what difference would you expect to find? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 5 a. The fatty acid tails from the cold environment would generally be shorter and less saturated compared to fatty acids from the warm environment. b. The fatty acid tails from the cold environment would generally be longer and more saturated compared to fatty acids from the warm environment. c. The fatty acid tails from the cold environment would generally be longer but less saturated compared to fatty acids from the warm environment. d. The fatty acid tails from the cold environment would generally be shorter and more saturated compared to fatty acids from the warm environment. ANSWER: a 25. Which molecule would most likely require a transport protein to cross the plasma membrane of a red blood cell? a. C6H12O6 b. H2O c. CO2 d. O2 ANSWER: a 26. Some diseases result from defective transport across the membrane. For example, cystic fibrosis results when a chloride ion transporter does not function. What is affected when this transporter does not function? a. the chloride concentration gradient b. the electrical gradient of the cell c. both the chloride concentration gradient and the electrical gradient of the cell d. neither the chloride gradient nor the electrical gradient of the cell ANSWER: c 27. The protein-assisted movement of a polar molecule across the membrane that does not require ATP or the input of energy occurs through: a. The substance is impermeable to the cell membrane. b. active transport. c. endocytosis. d. simple diffusion. e. facilitated diffusion. ANSWER: e 28. The beaker in the illustration contains two solutions of salt with different concentrations (measured by molarity, M). The two solutions are separated by a membrane that is permeable to both salt and water. What will occur in this container?

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Chapter 5

a. diffusion of salt across the membrane but not of water b. net diffusion of water from B to A and of salt from A to B c. net diffusion of water across the membrane but not of salt d. net diffusion of water from A to B and of salt from B to A e. net diffusion of salt from B to A but no net diffusion of water ANSWER: d 29. The beaker in the illustration contains two solutions of salt with different concentrations (measured by molarity, M). The two solutions are separated by a membrane that is permeable to water but not to salt.

What will occur in this container? a. diffusion of salt from B to A, but not of water b. diffusion of water from B to A and of salt from A to B c. diffusion of water from A to B but no diffusion of salt d. diffusion of both water and salt from B to A ANSWER: c 30. Some drugs are made of fairly large polar molecules which cannot easily enter cells. To correct this problem, drugs are now being delivered into cells by containing them inside liposomes, in which the drug is surrounded by a phospholipid bilayer. How do liposomes help the drugs enter into the cell? a. The liposome fuses with the phospholipid bilayer of the cell and delivers its internal contents to the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 5 cytoplasm. b. The liposome enters the cell through endocytosis and remains in a vesicle. c. The liposome ejects the drugs by exocytosis, which then diffuses into the cell. d. The liposomes enter the cell through protein channels on the cell surface. ANSWER: a 31. The plasma membranes of some plant cells use transport proteins to move protons out of the cell against their concentration gradient. This is an example of: a. simple diffusion. b. passive transport. c. endocytosis. d. facilitated diffusion. e. active transport. ANSWER: e 32. Which substance could most easily cross a synthetic membrane composed of phospholipids but not proteins? a. oxygen (O2) b. water (H2O) c. sodium ions (Na+) d. glucose (C6H12O6) ANSWER: a 33. G protein-coupled receptors are transmembrane receptors involved in cell signaling. The amino acid sequence of these proteins reveals that the polypeptide chain of the protein has seven hydrophobic regions. Based on this information, what can you conclude about how the protein is associated with the cell membrane? a. The receptor is likely to be a peripheral membrane protein. b. The polypeptide chain is likely to span the membrane once. c. The polypeptide chain is likely to span the membrane seven times. d. The polypeptide chain is not likely to span the membrane. ANSWER: c 34. Some plant cells create a high concentration of protons outside the cell to move solutes, such as sucrose, across the plasma membrane into the cell where the sucrose concentration is already relatively high. This type of transport is an example of: a. osmosis. b. secondary active transport. c. facilitated diffusion. d. passive transport. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 5 35. What is responsible, in part, for keeping the cytoplasm of red blood cells isotonic with the blood plasma and thereby preventing either lysis or shrinking of the cells? a. facilitated diffusion b. sodium channels c. the sodium-potassium pump d. osmosis ANSWER: c 36. The active maintenance of a constant internal environment is referred to as: a. homeostasis. b. equilibrium. c. stability. ANSWER: a 37. Which statement is true about the presence or absence of plasma membranes? a. Only animal cells have a plasma membrane. Plant cells and bacterial cells have a cell wall. b. Plant cells and bacterial cells have a plasma membrane, but animal cells do not. c. Plant cells and animal cells have a plasma membrane, but bacterial cells do not. d. All cells have a plasma membrane. ANSWER: d 38. Consider the events that describe the progress of a protein that will be secreted from the cell. 1. SRP binds to the growing polypeptide chain and to the ribosome. 2. Translation resumes. 3. SRP binds to its receptor. 4. The signal sequence is cleaved. 5. Protein synthesis begins in the cytosol. 6. Translation pauses. Which sequence of events correctly describes the progress of a protein that will be secreted from the cell? a. 5 → 1 → 6 → 3 → 2 → 4 b. 5 → 6 → 1 → 3 → 2 → 4 c. 3 → 6 → 1 → 2 → 4 → 5 ANSWER: a 39. If a mutation rendered the signal recognition particle nonfunctional, what would be the most obvious effect on the cell? a. Translation would not be completed for most proteins. b. All proteins normally secreted by the cell would remain partially formed and attached to the endoplasmic reticulum. c. Proteins destined for the nucleus would remain in the cytosol. d. No proteins would arrive at their proper destinations within the cell. e. All proteins normally secreted by the cell would remain in the cytosol. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 5 ANSWER: e 40. Which sequence accurately describes the path traveled by a new protein from when it first starts to be translated to its release from the cell? a. cytosol → ER → Golgi → vesicle → plasma membrane → external environment b. plasma membrane → ER → vesicle → Golgi → cytosol → external environment c. nuclear envelope → ER → vesicle → Golgi → plasma membrane → external environment d. cytosol → Golgi → ER → vesicle → plasma membrane → external environment e. nucleus → ER → Golgi → vesicle → plasma membrane → external environment ANSWER: a 41. Some diseases, such as Tay-Sachs, are caused by defects in the breakdown of cellular components. Which organelle could be defective with such a disease? a. endoplasmic reticulum b. ribosome c. lysosome d. Golgi apparatus e. plasma membrane ANSWER: c 42. Disorders of which organelle are often associated with defects in transport from compartment to compartment, resulting in poor sorting of protein components within the cell? a. plasmids b. the Golgi apparatus c. endoplasmic reticulum d. plasma membrane e. nucleus ANSWER: b 43. Insulin is a protein hormone that helps to control the level of glucose in the blood. It is secreted from specialized cells in the pancreas. Based on this information, which path does insulin take out of the cell? a. ER → Golgi apparatus → vesicle → exterior of cell b. vesicle → ER → Golgi apparatus → exterior of cell c. nucleus → ER → Golgi apparatus → vesicle → exterior of cell d. ER membrane → vesicle membrane → plasma membrane → exterior of cell e. ER → nucleus → cytosol → exterior of cell ANSWER: a 44. When physicians perform organ transplants, they make sure that there is a match between the donor (the person donating the organ) and the recipient (the person receiving the organ). What feature of the cell is being matched? a. phospholipids in the plasma membrane Copyright Macmillan Learning. Powered by Cognero.

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Chapter 5 b. cholesterol in the plasma membrane c. protein channels in the plasma membrane d. protein pumps in the plasma membrane e. glycoproteins in the plasma membrane ANSWER: e 45. During the translation of mRNA molecules, the new polypeptides are often directed to specific parts of the cell by the presence or absence of short sequences of amino acids called signal peptides. What would you expect to find in the polypeptide that will eventually fold to become a histone protein? a. a signal anchor peptide b. an amino-terminal signal peptide c. no signal peptide d. a nuclear localization signal ANSWER: d 46. Which cellular compartment does not have a double membrane structure separating it from the rest of the cell? a. lysosomes b. the nucleus c. mitochondria d. ribosome ANSWER: a 47. During the translation of mRNA molecules, the new polypeptides are often directed to specific parts of the cell by the presence or absence of short sequences of amino acids called signal peptides. What would you expect to find in the polypeptide that will eventually fold to become RNA polymerase? a. a nuclear localization signal b. an amino-terminal signal peptide c. no signal peptide d. a signal anchor peptide ANSWER: a 48. Individuals with a condition known as exercise intolerance lack sufficient ATP and suffer extreme fatigue from minimal exertion. Defects in which structures are most likely responsible for this condition? a. Golgi apparatus b. ribosomes c. endoplasmic reticulum d. mitochondria e. lysosome ANSWER: d 49. Molecular oxygen (O2), which is required for the production of ATP by mitochondria, must pass through at Copyright Macmillan Learning. Powered by Cognero.

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Chapter 5 least three membranes to get to the enzymes where it is used. O2 is also produced in chloroplasts and must pass through at least four membranes to be released from the plant. How does O2 move across biological membranes? a. O2 is nonpolar and very small, so it can move by simple diffusion across the membrane. b. There are numerous protein transporters that allow O2 to travel across membranes. c. O2 undergoes a chemical reaction that binds it to other substances that are transported across the membranes. d. O2 is hydrophilic and travels with water across the membranes. ANSWER: a 50. What would be found in the cells of a blade of grass but not in the cells of an insect feeding on that grass? a. nuclei b. mitochondria c. endoplasmic reticulum d. cytoskeletal elements e. chloroplasts ANSWER: e Multiple Response 51. During translation, the new polypeptides are often directed to specific parts of the cell by the presence or absence of short sequences of amino acids called signal sequences. Which peptides would you expect to find in the polypeptide that will eventually fold to become an ion channel protein? Select all that apply. a. a signal-anchor peptide b. an amino-terminal signal peptide c. an internal signal peptide d. no signal peptide ANSWER: a, b

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Chapter 6 Multiple Choice 1. Animals such as cats, worms, and butterflies are classified as: a. photoautotrophs. b. photoheterotrophs. c. chemoautotrophs. d. chemoheterotrophs. ANSWER: d 2. To which of the substances is ATP most closely chemically related? a. testosterone b. guanine c. phospholipid d. glucose e. tryptophan ANSWER: b 3. What is an example of potential energy? a. heat b. light c. an electrochemical gradient d. a moving muscle e. wind ANSWER: c 4. An anabolic reaction decreases entropy within the system because the reaction results in a more ordered macromolecule. a. true b. false ANSWER: a 5. Reactions in which there is a negative change in free energy (-ΔG) are: a. spontaneous and exergonic. b. spontaneous and endergonic. c. nonspontaneous and endergonic. d. nonspontaneous and exergonic. ANSWER: a 6. Which of the statements describes catabolic reactions? a. They are exergonic and have a positive charge in free energy. b. They are endergonic and have a negative change in free energy. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 c. They are exergonic and have a negative change in free energy. d. They are endergonic and have a positive change in free energy. ANSWER: c 7. ATP is a good energy currency for cells because it has a(n) _____ amount of Gibbs free energy. a. very low b. intermediate c. very high ANSWER: b 8. A cellular reaction with a ΔG of 8.5 kcal/mol could be effectively coupled to the hydrolysis of a single molecule of ATP. a. true b. false ANSWER: b 9. A biologist working in a lab adds a compound to a solution that contains an enzyme and substrate. This particular compound binds reversibly to the enzyme at the active site. Once the compound is bound to the enzyme, the catalysis of substrate to product stops. Which of the statements is true of the compound? a. The compound is an allosteric activator. b. The compound is an allosteric inhibitor. c. The compound is a catalyst for the reaction. d. The effect of the compound can be overcome by adding more substrate. e. The effect of the compound cannot be overcome by adding more substrate. ANSWER: d 10. Suppose that three critical amino acids in the active site of a specific enzyme are arginine, lysine, and histidine. Which of the characteristics would you predict the substrate to possess, in order to bind the active site of this enzyme? Consult the figure to help you make your predictions.

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Chapter 6

a. relatively hydrophobic b. hydrophilic, with a negative charge c. hydrophilic, with a positive charge Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 d. hydrophilic, but without a charge ANSWER: b 11. Which of the statements describes the relationship between β-galactosidase and β-thiogalactoside? a. They are synonyms for the same molecule. b. β-galactosidase cleaves β-thiogalactoside. c. β- galactosidase is a precursor for synthesizing β-thiogalactoside. d. β-thiogalactoside is a precursor for synthesizing β-galactosidase. e. β-galactosidase binds to β-thiogalactoside but is unable to cleave it. ANSWER: e 12. Lactose is composed of _____ joined by a _____ linkage. a. glucose and fructose; glycosidic b. two glucoses; peptide c. glucose and galactose; glycosidic d. two amino acids; peptide e. two galactoses; glycosidic ANSWER: c 13. Activators and inhibitors that bind to enzymes at positions other than the active site of the enzyme bind to _____ site on the enzyme. a. an allosteric b. an inhibitory c. the active d. a non-specific ANSWER: a 14. Consider an enzyme that has these three amino acids crucial to substrate binding at the active site: leucine, tryptophan, and alanine. What characteristics would you predict for this enzyme's substrate? Consult the figure to help you make your predictions.

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Chapter 6

a. relatively hydrophobic b. hydrophilic, with a negative charge c. hydrophilic, with a positive charge Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 d. hydrophilic, but without a charge ANSWER: a 15. Plants power their cellular processes by breaking down the sugar they make. a. true b. false ANSWER: a 16. Which of the processes requires energy input in the form of ATP? a. anabolism b. catabolism ANSWER: a 17. Of the choices, which can be a product of a catabolic reaction? a. a motor protein like myosin b. a nucleic acid like RNA c. an amino acid like tryptophan d. a lipid like cholesterol e. a complex carbohydrate like cellulose ANSWER: c 18. The reactions in the pathways of glycolysis and the citric acid cycle break down glucose into smaller molecules. Therefore, these pathways: a. involve the reduction of glucose and its metabolic intermediates. b. are anabolic pathways. c. take place in animal cells, but not in plant cells. d. are catabolic pathways. ANSWER: d 19. An example of potential energy is a ball sitting _____ of the stairs. a. All of these choices are correct. b. at the bottom c. in the middle d. at the top ANSWER: a 20. Why does ADP have less potential energy than ATP? a. Because ATP has ribose as a sugar. b. Because ADP has only one phosphate group. c. Because ADP has only two phosphate groups. d. Because ATP has adenine in it. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 ANSWER: c 21. Entropy is often increased by: a. order. b. heat. c. potential energy. d. kinetic energy. ANSWER: b 22. Which of the statements is a consequence of the first law of thermodynamics? a. The energy at the start of a reaction equals the sum of the energy in the product plus energy released as heat and disorder. b. The amount of usable energy resulting from a reaction is always less than the total energy available at the start of a reaction. c. The entropy of the products of a reaction is always greater than the entropy at the start of the reaction. d. The entropy at the start of a reaction is always greater than the entropy of the products. ANSWER: a 23. Which statement is true about exergonic reactions? a. The products of exergonic reactions have more free energy than the reactants. b. Energy is released from the reactants. c. There is a positive ΔG. ANSWER: b 24. Which of the reactions is most likely to be exergonic? a. the digestion of protein from food into amino acids b. the replication of DNA from free nucleotides c. the formation of cellulose from individual glucose molecules d. the synthesis of a phospholipid from glycerol and fatty acids ANSWER: a 25. You notice that a chemical reaction in your system is happening at a slow rate. You want to speed up the reaction. What do you do? a. add more products b. change the ΔG for the reaction c. add an enzyme that catalyzes the reaction d. increase the activation energy ANSWER: c 26. How might an enzyme inhibitor slow down the action of an enzyme without binding to the active site? a. by binding the substrate Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 b. by binding to another site on the enzyme and changing its shape c. by lowering the activation energy d. by changing the shape of the substrate ANSWER: b 27. Which of the statements is true of allosteric inhibitors of an enzyme? a. Allosteric inhibitors decrease enzyme activity. b. Allosteric inhibitors bind to the active site of the enzyme. c. Allosteric inhibitors are structurally similar to the normal substrate of an enzyme. d. Allosteric inhibitors increase the rate of enzyme activity. ANSWER: a 28. A biologist working in a lab adds a compound to a solution that contains an enzyme and its substrate. This compound binds to the enzyme and decreases the rate at which the enzyme converts substrate to product. However, this decrease can be overcome by increasing the concentration of substrate in the reaction mix. Therefore, which of the statements is true of the compound? a. The compound binds to the active site of the enzyme. b. The compound is an allosteric inhibitor. c. The compound is an allosteric activator. d. The compound is a catalyst for the reaction. ANSWER: a 29. Suppose that three critical amino acids in the active site of a specific enzyme are lysine, histidine, and arginine. Which of the characteristics would you predict the substrate to possess in order to bind the active site of this enzyme? Refer to the figure shown.

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Chapter 6

a. hydrophilic, with a positive charge b. hydrophilic, but uncharged c. relatively hydrophobic Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 d. hydrophilic, with a negative charge ANSWER: d 30. The amino acids that actively contribute to catalysis in the active site of an enzyme do not have to be located close to each other in the primary sequence (the linear sequence of amino acids) of the protein. a. true b. false ANSWER: a 31. Suppose that three critical amino acids in the active site of a specific enzyme are phenylalanine, methionine, and valine. Which of the characteristics would you predict the substrate to possess in order to bind the active site of this enzyme? Refer to the figure shown.

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Chapter 6

a. relatively hydrophobic b. hydrophilic, with a negative charge c. hydrophilic, with a positive charge Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 d. hydrophilic, but without a charge ANSWER: a 32. Anabolic pathways of metabolism are pathways that: a. release stored chemical energy. b. build complex molecules from simple ones. c. take place primarily in skeletal muscle. d. make large quantities of ATP. ANSWER: b 33. Autotrophs typically obtain their carbon from: a. CO2. b. ATP. c. C6H12O6. d. CH4. e. CH3OH. ANSWER: a 34. Based on what you know of ATP's chemistry, which of the substances is most likely to have similar functions in energy-management processes? a. phospholipids b. guanosine nucleotide triphosphate c. lysine (amino acid) d. steroids e. potassium ions (K+) ANSWER: b 35. It is often stated that the phosphate-phosphate bonds in ATP are "high energy," but in fact, they are not notably high in energy. Rather, they are easy to break, and the ΔG of hydrolysis is a "useful" quantity of energy. What makes the phosphate bonds easy to break? a. High alkalinity attacks bonds between phosphate groups. b. Positive charges on amino groups repel each other. c. They are close to the destabilizing nitrogenous base adenosine. d. High acidity attacks bonds between amino acids. e. Negative charges on phosphate groups repel each other. ANSWER: e 36. The potential energy in a molecule of ATP that is harnessed to do the work of the cell is held in part in the: a. the carbon-nitrogen bonds of the base. b. carbon-carbon bonds of the sugar. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 c. repulsion of the phosphate groups from each other. d. the carbon-oxygen bonds of the sugar. e. repulsion of the sugar and the base. ANSWER: c 37. Suppose you ignite a sheet of paper from your notebook using a match and allow the fire to burn until it dies out. If you could measure all the energy in the resulting combustion products and all the energy in the heat released (including whatever increase in disorder has occurred), would you predict this amount to be more than, less than, or the same as the amount of potential energy in the sheet of paper you started with? (You should ignore the activation energy provided by the match to light the paper.) a. the same energy as the paper b. less energy than the paper c. more energy than the paper ANSWER: a 38. Which of the statements is a consequence of the second law of thermodynamics? a. The amount of usable energy resulting from a reaction will always be less than the total energy available in the reactant. b. The energy of the reactant will equal the sum of energies of the products plus energy released as heat and disorder. c. The entropy of the products of a reaction will always be greater than the entropy of the reactant. d. The entropy of the reactant will always be greater than the entropy of the products. ANSWER: a 39. G = H -TS is the equation that describes Gibbs free energy. Rearranged, the equation can be written as H = G + TS. In either case, G stands for the _____, H stands for the _____, and S stands for the _____. a. usable energy available to the cell; free energy; unusable energy b. usable energy available to the cell; total energy; unusable energy c. total energy; free energy; usable energy available to the cell d. total energy; usable energy available to the cell; unusable energy ANSWER: b 40. Which of the reactions would you predict could be coupled to ATP → ADP + Pi? a. glutamic acid + NH3 → glutamine, ΔG +3.4 kcal/mol b. phosphoenolpyruvate + H2O → pyruvate + Pi, ΔG -14.8 kcal/mol c. glucose 6-phosphate + H2O → glucose + Pi, ΔG -3.3 kcal/mol d. glucose 1-phosphate + H2O → glucose + Pi, ΔG -5.0 kcal/mol e. creatine phosphate + H2O → creatine + Pi, ΔG -10.3 kcal/mol ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 41. In the metabolic pathway illustrated, the reactant (substrate A) is converted equally to one of two end products, E or G. Use the diagram of the pathway to answer the questions Letters indicate the substrates and products, and numbers indicate the enzymes. In this pathway, the intermediate, C, is a substrate for both enzyme 3 and enzyme 5 and is converted with equal efficiency to D and F.

If end product E inhibits enzyme 1, which outcome would you expect to observe? a. an increase in the production of A b. a decrease in the production of E, but no change in the production of G c. an increase in the production of G d. a decrease in the production of E and G ANSWER: d 42. In the metabolic pathway illustrated, the reactant (substrate A) is converted equally to one of two end products, E or G. Use the diagram of the pathway to answer the question. Letters indicate the substrates and products, and numbers indicate the enzymes. In this pathway, the intermediate, C, is a substrate for both enzyme 3 and enzyme 5 and is converted with equal efficiency to D and F.

If end product G inhibits enzyme 5, what would you expect to observe as the amount of G increases in the cell? a. an increase in the production of C b. an increase in the production of B c. an increase in the production of E d. a decrease in the production of D e. an increase in the production of G ANSWER: c 43. In the metabolic pathway illustrated, the reactant (substrate A) is converted equally to one of two end products, E or G. Use the diagram of the pathway to answer the question. Letters indicate the substrates and Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 products, and numbers indicate the enzymes. In this pathway, the intermediate, C, is a substrate for both enzyme 3 and enzyme 5 and is converted with equal efficiency to D and F.

If end product G inhibits enzyme 3, what would you expect to observe as the amount of G increases in the cell? a. an increase in the production of C b. an increase in the production of B c. an increase in the production of E d. an increase in the production of D e. an increase in the production of G ANSWER: e 44. Some ions that facilitate enzyme-catalyzed reactions are capable of catalyzing reactions independently of an enzyme. a. true b. false ANSWER: a 45. The highest free energy is found in the _____(s) of a reaction. a. transition state b. starting substrate c. end product ANSWER: a 46. An uncatalyzed reaction has a higher ΔG than the same reaction when catalyzed by an enzyme. a. true b. false ANSWER: b 47. When muscles contract, the _____ energy of ATP is used to drive the _____ energy of contraction. a. potential; potential b. kinetic; potential c. potential; kinetic d. kinetic; kinetic ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 48. Suppose you are sampling the microbes from anaerobic sediments in a swamp and you encounter an unfamiliar bacterium. You determine that this bacterium can use the energy from chemical compounds to convert inorganic carbon, such as CO2, into organic matter for growth. These bacteria would be classified as: a. chemoheterotrophs. b. chemoautotrophs. c. photoheterotrophs. d. photoautotrophs. ANSWER: b 49. Suppose you ignited a sheet of paper composed of pure cellulose (a polymer of C6H12O6, or glucose) and let it burn completely. If you could capture and weigh all of the gases, smoke, and ash resulting from the combustion, they would weigh _____ the starting paper and there would be _____ of carbon atoms. a. less than; slightly larger number b. the same as; half the number c. much less than; half the number d. much more than; twice the number e. the same as; exactly the same number ANSWER: e 50. Consider the graph. In the graph shown, the arrow labeled _____ represents the ΔG for the uncatalyzed reaction, and the arrow labeled _____ represents the ΔG for the catalyzed reaction.

a. B; B b. B; A c. E; D d. C; C e. D; E ANSWER: d 51. Consider the graph. In the graph shown, the arrow labeled _____ represents the activation energy (EA) for Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 the uncatalyzed reaction, and the arrow labeled _____ represents the activation energy for the catalyzed reaction.

a. B; B b. B; A c. E; D d. E; C e. D; E ANSWER: b 52. In cellular chemical pathways, the product(s) of any particular reaction are often quickly consumed by the next reaction in the pathway. This would tend to keep the product concentration _____ and drive the reaction _____. a. high; in the forward direction b. low; in the reverse direction c. low; in the forward direction d. high; in the reverse direction ANSWER: c 53. Suppose that a cellular process resulted only in an increase in the entropy (S) of a system, but no change in enthalpy (H). What would the effect of this be on ΔG? a. ΔG would be negative. b. ΔG would be positive. c. A change in S alone would not affect ΔG. ANSWER: a 54. Consider a substance dissolved in the cytoplasm of a cell and in the extracellular fluid surrounding that cell. If transport processes by the cell acted to increase the concentration of that substance inside the cell, the result would be _____ in entropy and _____ ΔG. a. a decrease; a positive Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 b. an increase; a positive c. a decrease; a negative d. an increase; a negative ANSWER: a 55. In a metabolic pathway, a series of enzymatic reactions catalyzes the conversion of molecule A to molecule E. Several intermediate steps are involved in which the product of one reaction becomes the substrate for the next. The graph illustrates the changes of free energy that occur at each step in the pathway.

Overall, this _____, based on the changes in free energy that take place as A is converted to E. a. is an anabolic pathway b. is a catabolic pathway c. pathway has four exergonic reactions d. pathway has four endergonic reactions ANSWER: a 56. This pathway is a series of _____ reaction(s). a. one b. two c. three d. four e. five ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 57. The reaction that converts C to D is: a. an exergonic reaction. b. an anabolic reaction. ANSWER: a 58. The greatest change in free energy occurs when: a. A is converted to B. b. B is converted to C. c. C is converted to D. d. D is converted to E. ANSWER: b 59. Which of the reactions in the pathway is energetically different from the others? a. the reaction in which A is converted to B b. the reaction in which B is converted to C c. the reaction in which C is converted to D d. the reaction in which D is converted to E ANSWER: c Multiple Response 60. Which statements are true of an inhibitor that binds the active site of an enzyme? Select all that apply. a. These inhibitors compete with the substrate for the active site of the enzyme. b. These inhibitors are structurally similar to the normal substrate of an enzyme. c. Adding more substrate can reduce the effect of these inhibitors. d. These inhibitors increase the rate of enzyme activity. e. These inhibitors are a kind of allosteric regulator that decreases enzyme activity. ANSWER: a, b, c 61. Which of the statements represent the laws of thermodynamics? Select all that apply. a. The amount of energy in the universe is constant. b. All cells arise from pre-existing cells. c. The energy available to do work decreases as energy is transformed from one form to another. d. Matter cycles in the biosphere, but energy flows. e. Enzymes lower the activation energy requirement of reactions. ANSWER: a, c 62. Which of the elements are known to aid in the functioning of cellular enzymes? Select all that apply. a. copper b. iron c. zinc Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 d. magnesium e. mercury ANSWER: a, b, c, d 63. What are the functions of an enzyme? Select all that apply. a. to lower the activation energy for the reaction b. to increase the rate of a specific reaction c. to allow a reaction to be reversible d. to alter the equilibrium of a specific reaction e. to alter the ΔG of a specific reaction ANSWER: a, b 64. The assembly of glucose into polysaccharides: Select all that apply. a. is a catabolic process. b. takes place in some plant cells and in some animal cells. c. takes place in some plant cells but not in animal cells. d. is an anabolic process. e. takes place in some animal cells but not in plant cells. ANSWER: b, d 65. A biologist working in a lab adds a compound to a solution that contains an enzyme and substrate. This particular compound binds reversibly to the enzyme at the active site. Once the compound is bound to the enzyme, the rate of catalysis of substrate to product is greatly reduced. Which of the statements are true of the compound? a. The compound is an enzyme cofactor. b. The compound is an allosteric inhibitor. c. The compound is an allosteric activator. d. The compound competes with the substrate for the active site of the enzyme. e. The effect of the compound can be overcome by adding more substrate. ANSWER: d, e 66. Which of the choices are components of an ATP molecule? Select all that apply. a. three phosphate groups b. deoxyribose c. adenine d. ribose e. three amino groups ANSWER: a, c, d 67. Which of the statements are true? Select all that apply. a. The entropy at the start of a reaction is always greater than the entropy of the products. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 b. The amount of usable energy resulting from a reaction is always less than the total energy available at the start of the reaction. c. The entropy of the products of a reaction is always greater than the entropy at the start of the reaction. d. The energy for a reaction equals the sum of the energy in the product plus energy released as heat and disorder. ANSWER: b, d 68. Which areas can directly participate in creating the binding specificity of an enzyme active site? Select all that apply. a. hydrophilic amino acid side chains b. negatively charged amino acid side chains c. positively charged amino acid side chains d. hydrophobic amino acid side chains e. beta-sheet regions of the polypeptide f. alpha-helical regions of the polypeptide ANSWER: a, b, c, d 69. Which of the statements about the formation of a peptide bond are true? Select all that apply. a. The reaction is anabolic. b. The reaction is spontaneous. c. The reaction has a positive ΔG. d. The reaction is endergonic. ANSWER: a, c, d 70. Which of the examples are considered catabolism? Select all that apply. a. synthesis of new DNA copies prior to cell division b. use of fat (triglyceride) stores as a cellular energy source c. a person losing weight on a calorie restriction diet d. hydrolysis of glycogen (a glucose polymer) during physical activity e. fat cells growing bigger during times of ample nutrition ANSWER: b, c, d 71. Which of the examples are considered anabolism? Select all that apply. a. a person losing weight on a calorie restriction diet b. use of fat (triglyceride) stores as a cellular energy source c. hydrolysis of glycogen (a glucose polymer) during physical activity d. synthesis of new DNA copies prior to cell division e. fat cells growing bigger during times of ample nutrition ANSWER: d, e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 6 72. Which of the reactions would you predict could be coupled to ATP synthesis from ADP + Pi? Select all that apply. a. creatine phosphate + H2O → creatine + Pi, ΔG – 10.3 kcal/mol b. phosphoenolpyruvate + H2O → pyruvate + Pi, ΔG – 14.8 kcal/mol c. glucose 6-phosphate + H2O → glucose + Pi, ΔG – 3.3 kcal/mol d. glucose 1-phosphate + H2O → glucose + Pi, ΔG – 5.0 kcal/mol e. glutamic acid + NH3 → glutamine, ΔG + 3.4 kcal/mol ANSWER: a, b 73. Which of the choices are examples of living cells or organisms exhibiting kinetic energy? Select all that apply. a. single-celled algae swimming towards a light source b. an earthworm burrowing through decaying leaf litter c. muscles in a bird's wing contracting d. fats (triacylglycerides) being mobilized to provide cellular energy e. energy from soft drink sugars being stored as triacylglycerides in fat cells ANSWER: a, b, c 74. Which of the changes could cause a cellular chemical reaction to proceed in the reverse direction? Select all that apply. a. decreasing the substrate concentration b. increasing the product concentration c. decreasing the product concentration d. increasing the substrate concentration ANSWER: a, b

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Chapter 7 Multiple Choice 1. Production of NADH by glycolysis requires an input of the oxidized molecule NAD+. Where does this supply of NAD+ come from in the absence of oxygen? a. It is regenerated by donating electrons to the electron transport chain. b. It is regenerated by reducing pyruvate to ethanol or lactic acid. c. It is transferred from the citric acid cycle, which is not functioning due to lack of oxygen. d. There is a ready supply of NAD+ from the cytoplasm. ANSWER: b 2. What happens to pyruvate during fermentation? a. It is reduced to ethanol or lactic acid. b. It is oxidized to ethanol or lactic acid. c. It is oxidized to CO2. d. It is oxidized to acetyl-CoA. e. It is reduced to acetyl-CoA. ANSWER: a 3. Why is fermentation a required pathway for providing cellular energy when oxygen is unavailable or in insufficient supply? a. NAD+ is not regenerated by the electron transport chain. b. ATP production requires oxygen. c. NADH cannot be reduced to NAD+. d. Cells need either lactic acid or ethanol when oxygen is low. e. Lactic acid or ethanol can be used to generate oxygen. ANSWER: a 4. Not all cell or tissue types in the human body can undergo fermentation. Such cells can be grown in the lab in tissue culture as long as oxygen is supplied, but they will not survive under anoxic (no oxygen) conditions. Which of the answer choices would be an experiment that would likely confirm the limiting substance during anoxic conditions for a lab culture of such a tissue/cell line? a. Add more glucose to the culture medium and monitor cell survival without oxygen. b. Microinject the cells with NADH and monitor cell survival without oxygen. c. Microinject the cells with oxygen and monitor cell survival. d. Add NADH to the culture medium and monitor cell survival without oxygen. e. Microinject the cells with NAD+ and monitor cell survival without oxygen. ANSWER: e 5. The lactate dehydrogenase gene codes for the enzyme that converts pyruvate to lactic acid. If cells in a lab culture are unable to undergo fermentation because of a mutation that blocks activity of the lactate dehydrogenase gene, and those cells are placed in anoxic conditions, which treatment should allow the cells to Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 survive without oxygen? a. adding more glucose to the culture medium b. microinjection of the cells with NADH c. microinjection of the cells with acetyl-CoA d. microinjection of the cells with NAD+ ANSWER: d 6. During the metabolism of glucose (the cellular fuel molecule) by the process of aerobic cellular respiration, glucose is _____ to CO2 via _____ reactions. a. oxidized; redox b. reduced; endergonic c. oxidized; hydrolysis d. reduced; dehydration e. reduced; oxidation ANSWER: a 7. During glycolysis, _____ phosphorylation adds phosphate groups to ADP by _____. a. substrate-level; ATP synthase b. substrate-level; enzymatic transfer c. oxidative; ATP synthase d. oxidative; enzymatic transfer ANSWER: b 8. If you could attach fluorescent marker tags to the enzyme that transfers the acetyl group from acetyl CoA to oxaloacetate, and then detect that fluorescence with a powerful microscope, in which part of a cell would you predict to observe the fluorescence? a. in the cytoplasm b. in the mitochondrial intermembrane space c. embedded in the inner mitochondrial membrane d. embedded in the outer mitochondrial membrane e. in the mitochondrial matrix ANSWER: e 9. Which summary sequence correctly tracks electrons through the overall process of aerobic cellular respiration? a. glucose -> electron carriers -> electron transport chain -> water b. glucose -> pyruvate -> acetyl-CoA -> carbon dioxide c. glucose -> pyruvate -> acetyl-CoA -> carbon dioxide -> ATP d. NAD+ and FAD -> NADH and FADH2 -> ATP ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 10. Some tissue types, such as brain tissue, use glucose exclusively as an energy source, whereas other tissues utilize a diversity of sources, including fats and proteins. a. true b. false ANSWER: a 11. In aerobic cellular respiration, the approximate yield of ATP molecules from the full oxidation of a molecule of glucose is: a. 2. b. 6. c. 12. d. 32. e. 64. ANSWER: d 12. During the important redox reactions of electron transport chain, the transfer of electrons is typically accompanied by transfer of: a. phosphate groups. b. oxygen atoms. c. water molecules. d. carbon atoms. e. protons. ANSWER: e 13. Which molecule has the greatest chemical potential energy? a. pyruvate b. glucose c. fructose 1,6 bisphosphate d. NADH e. ATP ANSWER: c 14. How many reactions in glycolysis directly generate ATP by substrate-level phosphorylation? a. 4 b. 1 c. 3 d. 2 e. 5 ANSWER: d 15. In eukaryotic cells, the oxidation of pyruvate occurs in: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 a. the matrix of the mitochondria. b. the nucleus. c. the cytoplasm. d. the endoplasmic reticulum. e. vacuoles. ANSWER: a 16. In 1937, two German biochemists published a paper proposing these reactions as part of glucose oxidation: citrate → isocitrate → α-ketoglutarate → succinate → fumarate → malate → oxaloacetate. Adding succinate, fumarate, or malate to thin slices of tissue increased oxygen consumption, supporting the hypothesis that these molecules are intermediates in the process. However, they were puzzled by the observation that these intermediates were still present in the reaction mixture at the end of the experiment. They had thought that intermediates would be consumed as they were converted to the next molecule in the pathway. What explains the observation that these intermediates were still present? a. Succinate, fumarate, and malate increase metabolism and therefore oxygen consumption, but they are not directly part of the glucose oxidation pathway. b. Succinate, fumarate, and malate are not reactants, but catalysts, and catalysts are not consumed in the process. c. The pathway is a cycle, constantly regenerating intermediates as glucose is broken down. ANSWER: c 17. The citric acid cycle begins when acetyl-CoA reacts with _____ to form _____ and ______. a. citrate; isocitrate; ATP b. malate; oxaloacetate; CoA c. oxaloacetate; malate-CoA; acetate d. pyruvate; citrate; NADH e. oxaloacetate; citrate; CoA ANSWER: e 18. Certain complexes of the mitochondrial electron transport chain pump protons. Protons are pumped across the _____ mitochondrial membrane, from the _____ to the _____. a. inner; matrix; intermembrane space b. outer; cytoplasm; intermembrane space c. outer; intermembrane space; cytoplasm d. inner; intermembrane space; matrix ANSWER: a 19. If oxygen is unavailable, predict what will happen to the citric acid cycle. a. It will continue because none of the reactions in the citric acid cycle require oxygen. b. It will stop because ADP levels will increase in the absence of oxygen. c. It will stop because the supplies of NAD+ and FAD will become depleted. d. It will continue because ATP levels will be low, and low ATP activates enzymes of the cycle. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 ANSWER: c 20. In human cells, such as those in muscle tissue, the product of fermentation is: a. pyruvate. b. lactic acid. c. acetic acid. d. FADH2. e. ethanol. ANSWER: b 21. Glycogen stored in muscles can be "mobilized" to supply metabolic energy by hydrolyzing individual glucose subunits from the polymer. What other organ has a major function of storing glycogen? a. the large intestine b. the brain c. the stomach d. the liver ANSWER: d 22. The liver plays an important role in modulating blood glucose levels by removing glucose from the blood during periods of abundance or secreting it into the blood when blood glucose concentration is low. However, the liver doesn't simply stockpile glucose. Rather, it polymerizes it into glycogen, which it can then hydrolyze back into glucose monomers as needed. Why would storage as glycogen be better than simply storing the glucose? a. Glycogen is easier to transport across the cell membrane. b. Glycogen increases the available energy yield from glucose. c. Glucose molecules would create potential osmotic damage. d. Glycogen creates more osmotic pressure on the cell membrane. ANSWER: c 23. Why is the storage of glycogen in the liver an important function in terms of aerobic cellular respiration? a. Glycogen is a polymer of glucose, and hydrolysis to monomers supplies glucose to the blood in times of high energy needs. b. Glycogen is like cellulose and can form structural supports for cells. c. Glycogen is an important enzyme that metabolizes ethanol that may flow through the liver. d. Glycogen is transported directly into mitochondria where it is oxidized in the electron transport chain. ANSWER: a 24. The breakdown of one glucose molecule during glycolysis results in two molecules of pyruvate, each of which is then oxidized to a molecule of acetyl-CoA, and these, in turn, are substrates for the citric acid cycle. If all three fatty acids of a triacylglycerol molecule (a fat molecule) are palmitic acid (C16, having 16 carbons), how many acetyl-CoA molecules would result from beta-oxidation of all of these fatty acids? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 a. 4 b. 8 c. 16 d. 24 e. 32 ANSWER: d 25. Which regulatory mechanism is activated when the overall energy availability of a cell is high? a. ATP inhibits phosphofructokinase-1. b. ADP up-regulates phosphofructokinase-1. c. ADP inhibits phosphofructokinase-1. d. AMP up-regulates phosphofructokinase-1. e. ATP up-regulates phosphofructokinase-1. ANSWER: a 26. Indicate which step in glycolysis is considered the first committed step. Refer to the figure shown.

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Chapter 7

a. step 3 b. step 1 c. step 5 d. step 7 e. step 9 ANSWER: a 27. PFK-1 is _____ by ATP and _____ by ADP. a. inhibited; inhibited b. activated; inhibited c. activated; activated d. inhibited; activated ANSWER: d 28. When oxygen is depleted, the citric acid cycle stops. What could we add to the system to restore citric acid cycle activity (other than oxygen)? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 a. ethanol and lactic acid b. acetyl-CoA c. ADP and Pi d. NAD+ or FAD ANSWER: d 29. Acetate is the starting point for synthesis of a cell's: a. RNA b. carbohydrates c. DNA d. lipids e. proteins ANSWER: d 30. Oxidation of FADH2 to FAD by _____ results in fewer _____ being pumped across the inner membrane, and therefore yields fewer _____ molecules when compared to oxidation of NADH to NAD+ by _____. a. complex II; Co Q; ATP; complex III b. complex III; protons; water; complex II c. CoQ; electrons; ATP; cytochrome c d. complex III; CoQ; ATP; complex II e. complex II; protons; ATP; complex I ANSWER: e 31. Which choice correctly lists the approximate number of ATP produced by (1) metabolism of a 16-carbon fatty acid, (2) fermentation of a single glucose, and (3) aerobic respiration of a single glucose, in that order? For this calculation, you will need to know that each 2-carbon unit cleaved from the fatty acid yields one NADH and one FADH2, and the 2-carbon unit is added to coenzyme A to become acetyl-CoA, which is further metabolized by the citric acid cycle. a. 40, 10, 20 b. 75, 10, 25 c. 50, 10, 20 d. 108, 2, 32 e. 100, 1, 10 ANSWER: d 32. In oxidation-reduction reactions (redox reactions), ________ are _______ during reduction. a. neutrons; gained b. electrons; lost c. protons; gained d. neutrons; lost Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 e. electrons; gained ANSWER: e 33. Why do some marathon runners attempt to "carbo load" (that is, eat a lot of pasta) before a big race? a. They will feel full longer. b. The bonds in carbohydrates have high potential energy. c. The bonds in carbohydrates have low potential energy and can thus be broken down and readily consumed. d. Through the process of anabolism, the athlete will break down the carbohydrates into smaller components, including ATP. ANSWER: b 34. Which example represents the reduced forms of the two major electron carriers? a. NADH and FAD b. NAD+ and FADH2 c. NADH and FADH2 d. NAD+ and FAD ANSWER: a 35. The most stable and least reactive form of carbon is: a. carbon dioxide. b. fructose 1,6-bisphosphate. c. glucose. d. pyruvate. e. ethanol. ANSWER: a 36. Glycolysis results in the partial oxidation of glucose to pyruvate. This means that: a. the electron carriers donate electrons to proteins in the mitochondria that in turn produce ATP. b. glucose is broken down partially to ATP in the cytoplasm. c. glucose combines with oxygen in the cytoplasm to get partially oxidized. d. glycolysis consists only of exergonic reactions so that ATP can be made from the release of energy. e. in the process of the conversion of glucose to pyruvate, some potential energy is transferred to NADH and ATP. ANSWER: e 37. Even though the full oxidation of glucose is exergonic, some of the reactions in cellular respiration are endergonic. a. true b. false ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 38. When considering the transfer and capture of potential energy derived from glucose during cellular respiration, which molecule carries the smallest amount of that potential energy? a. acetyl-CoA b. NADH c. ATP d. pyruvate e. FADH2 ANSWER: c 39. If an energy source is available, the citric acid cycle can run in reverse in some organisms. a. true b. false ANSWER: a 40. In which stage of aerobic cellular respiration is the most energy transferred from chemical bonds in the fuel molecule to bonds in other molecules? a. glycolysis b. the conversion of pyruvate to acetyl-CoA c. the citric acid cycle d. the electron-transport chain ANSWER: c 41. What is the final electron acceptor in the electron transport chain? a. NAD+ b. glucose c. ATP d. ADP e. oxygen ANSWER: e 42. What would happen if complexes I-IV of the electron transport chain pumped protons in the opposite direction? a. Too much ATP would be synthesized. b. No ATP would be synthesized. c. There would be too many electrons in the mitochondrial matrix. d. ATP synthase would operate in reverse. ANSWER: b 43. The outer surface of the F0 subunit of ATP synthase must be _____ to be embedded in the membrane. a. hydrophobic Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 b. hydrophilic c. negatively charged d. positively charged ANSWER: a 44. Animals breathe in air containing oxygen and breathe out air containing less oxygen. The consumed oxygen is used: a. in photosynthesis. b. in the glycolysis pathway. c. in the conversion of pyruvate to acetyl Co-A. d. as an electron acceptor in the respiratory electron transport chain. e. in the citric acid cycle. ANSWER: d 45. The energy from the movement of electrons through the electron transport chain is directly used to synthesize ATP. a. true b. false ANSWER: b 46. In order for a pathway to produce its products, it must have sufficient inputs. Which input that results from fermentation makes glycolysis possible in anoxic (no oxygen) conditions? a. ADP b. glucose (or other sugar) c. NAD+ d. ATP e. NADH ANSWER: c 47. Kangaroo rats live in the deserts of the southwestern United States. Kangaroo rats have many adaptations to minimize water loss. They obtain a small amount of water from seeds that they eat. However, the rest of the water they obtain is from cellular respiration. a. This could be true, as water is produced in glycolysis. b. This cannot be true, as water is actually consumed in cellular respiration. c. This cannot be true, as cellular respiration doesn't really "produce" water. d. This could be true, as water is produced in cellular respiration. e. This could be true, as water is produced in the citric acid cycle. ANSWER: d 48. When carbohydrates are metabolized as cellular fuel, the C—H and C—C bonds of the carbohydrate are oxidized to C=O bonds of carbon dioxide. Oxidation is defined as a loss of electrons, but carbon does not become positively charged in the process. Why, then, is this considered oxidation? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 a. The phosphate groups of ATP are ionized and carbons donate those electrons. b. Electrons in the C=O bonds are higher energy than the electrons in the C—H bonds. c. C=O bonds in CO2 are double bonds, and C—H bonds are single bonds. d. The shared electrons in C—O bonds spend less time close to the carbon nucleus than the shared electrons in C—H or C—C bonds. ANSWER: d 49. Which statement best summarizes cellular respiration? a. The chemical potential energy stored in organic molecules is converted to chemical energy that can be used to do the work of the cell. b. Organic molecules, such as carbohydrates, are converted to chemical energy that can be used to do the work of the cell. c. Chemical potential energy in the bonds of ADP is transferred to the chemical potential energy in the bonds of ATP. ANSWER: a 50. An organism that carries out cellular respiration in its mitochondria: a. could be a cell from a terrestrial (land) plant. b. could be a prokaryotic cell from the domain Archaea. c. could be a bacterium. d. could be any kind of cell. e. None of these choices is correct. ANSWER: a 51. Complete oxidation of glucose to CO2 involves two different mechanisms for synthesizing ATP: oxidative phosphorylation and substrate-level phosphorylation. Which is true of substrate-level phosphorylation? a. Most of the ATP generated in cellular respiration is generated by substrate-level phosphoylation. b. An enzyme catalyzes the transfer of a phosphate group from an organic molecule to ADP to form ATP. c. ATP is generated indirectly through the transfer of high-energy electrons from electron carriers to an electron transport chain. d. ATP is generated by release of energy from the electron carriers NADH and FADH2. ANSWER: b 52. The energy in organic molecules is released in a series of steps, rather than a single step, because: a. it is not possible to release it in a single step. b. more total energy is released in multiple steps than would be released in a single step. c. less total energy is released in multiple steps than would be released in a single step. d. the amount of energy released would be too much to capture in one reaction. e. only a single electron can be moved at a time in a cellular reaction. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 53. Glycolysis is a series of chemical reactions (endergonic and exergonic) by which the cell can obtain ATP. NAD+ plays a crucial role in the reactions of glycolysis by: a. donating electrons to ADP to make ATP. b. donating electrons to pyruvate when glucose becomes partially oxidized. c. converting endergonic reactions to exergonic reactions so that there is an output of energy to make ATP. d. accepting electrons during glycolysis, with the overall result that glucose is partially oxidized to pyruvate. ANSWER: d 54. You are trying to find the maximum source of energy for an organism, and you are limited to 0.1 moles of a molecule from a compound. Which compound would you choose to provide 0.1 moles for maximum energy? a. ATP b. oxygen c. glucose d. pyruvate e. NADH ANSWER: c 55. If one follows the respiratory energy transfers and transformations sequentially from glucose through glycolysis, pyruvate oxidation, and the citric acid cycle, but before the electron transport chain and oxidative phosphorylation, most of the free energy captured from the original glucose molecule is found in: a. NADH. b. acetyl-CoA. c. CO2. d. pyruvate. e. ATP. ANSWER: a 56. Each round of the citric acid cycle begins when the four-carbon molecule oxaloacetate is converted to the six-carbon molecule citrate. As the cycle progresses, two carbons are eliminated to regenerate the oxaloacetate. The added carbon is supplied by _____ and the two eliminated carbons are released as _____. a. ATP; acetyl-CoA b. CO2; pyruvate c. acetyl-CoA; CO2 d. CO2; NADH e. CO2; acetyl-CoA ANSWER: c 57. Atractyloside is a poison that inhibits the transport of ADP from the cytosol across the mitochondrial Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 membranes and into the mitochondrial matrix. The direct effect of this drug is to stop ATP synthesis because: a. the poison prevents proton pumping across the inner mitochondrial membrane. b. ADP is a necessary substrate for the reactions catalyzed by ATP synthase. c. the poison prevents electron transfer to O2. d. the poison prevents electron transfer from NADH to complex I. ANSWER: b 58. Oligomycin is an antibiotic that binds ATP synthase, blocking the flow of protons through the enzyme's proton channel. In addition to preventing synthesis of ATP, what additional effect might you expect in response to the presence of oligomycin? a. a buildup of protons in the mitochondrial matrix b. higher pH in the intermembrane space c. lower pH in the intermembrane space ANSWER: c 59. Very low concentrations of detergent make membranes leaky to small molecules and ions without damaging proteins. In isolated mitochondria exposed to detergent, the molecules of the electron transport chain and of ATP synthase remain intact. Do you expect ATP synthesis to continue in the presence of low concentrations of detergent? a. No, because with a leaky membrane, the proton gradient cannot be maintained. b. Yes, because all enzymes and electron carriers remain intact. c. No, because leaky membranes do not allow NADH and FADH2 to donate their electrons to the electron transport chain. d. No, because leaky membranes inhibit glycolysis. ANSWER: a 60. The pH in the intermembrane space of the mitochondria should be _____ compared with the matrix, due to the _____ concentration of protons in the intermembrane space. a. higher; higher b. lower; higher c. higher; lower d. lower; lower ANSWER: b 61. Brown fat is a specialized tissue found especially in infants and hibernating mammals. Brown fat mitochondria have proton channels located in their inner membranes that allow protons to flow from the intermembrane space into the mitochondrial matrix without passing through ATP synthase. What does the flow of protons through these channels mean for the organism? a. There would be no effect. b. Channels would contribute to the formation of the proton electrochemical gradient. c. ATP production would be reduced, because these organisms do not need as much energy. d. Rather than producing ATP, much more of the energy of metabolism would be lost as heat. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 ANSWER: d 62. A research group has discovered an organism with cells that contain a previously undescribed organelle. Researchers perform some tests on the isolated organelle to see if it is involved in any major metabolic reactions. To do this, they incubate these organelles for a period of time and determine changes in the amount of various substances in the suspending solution. The results are shown in the table.

Based on this analysis, which metabolic process would you conclude could be taking place in this organelle? a. pyruvate oxidation b. citric acid cycle c. glycolysis d. electron transport chain/oxidative phosphorylation ANSWER: b 63. A research group has discovered an organism with cells that contain a previously undescribed organelle. Researchers perform some tests on the isolated organelle to see if it is involved in any major metabolic reactions. To do this, they incubate these organelles for a period of time and determine changes in the amount of various substances in the suspending solution. The results are:

Based on this analysis, which metabolic process would you conclude could be taking place in this organelle? a. glycolysis b. citric acid cycle c. pyruvate oxidation d. electron transport chain/oxidative phosphorylation ANSWER: a 64. A research group has discovered an organism with cells that contain a previously undescribed organelle. Researchers perform some tests on the isolated organelle to see if it is involved in any major metabolic reactions. To do this, they incubate these organelles for a period of time and determine changes in the amount of various substances in the suspending solution. The results are:

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Based on this analysis, which metabolic process would you conclude could be taking place in this organelle? a. glycolysis b. citric acid cycle c. pyruvate oxidation d. electron transport chain/oxidative phosphorylation ANSWER: c 65. Which of these reactions summarizes the overall reactions of cellular respiration? a. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + energy b. 6 CO2 + 6 H2O + energy → C6H12O6 + 6 O2 c. C6H12O6 + 6 O2 + energy → 6 CO2 + 12 H2O d. 6 CO2 + 6 O2 → C6H12O6 + 6 H2O e. H2O → 2 H+ + 1/2 O2 + 2eANSWER: a 66. Most of the ATP produced during cellular respiration is generated through: a. oxidative phosphorylation. b. substrate-level phosphorylation. c. glycolysis. d. fermentation. e. pyruvate oxidation. ANSWER: a 67. Which molecule would you expect to act as allosteric activator of an enzyme in glycolysis? a. pyruvate b. NADH c. ADP d. ATP ANSWER: c 68. In the absence of carbohydrates, which of the answer choices can maintain production of NADH and FADH2 by the citric acid cycle? a. β-oxidation of fatty acids Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 b. fermentation of glucose c. hydrolysis of glycogen ANSWER: a 69. When fats are used as an energy source, the fatty acids are broken down to acetyl-CoA. That means that fats bypass the reactions of _____ and enter the respiratory pathway at _____. a. oxidative phosphorylation; fermentation b. fermentation; glycolysis c. the citric acid cycle; oxidative phosphorylation d. the citric acid cycle; glycolysis e. glycolysis; the citric acid cycle ANSWER: e 70. The emperor penguins of Antarctica live on a diet of fish and crustaceans obtained from the cold Antarctic seawaters. During their annual breeding cycle, however, they migrate across the frozen continent to their breeding grounds 50 miles away from the sea (and 50 miles away from their source of food). For over two months, the male emperor penguins care for and incubate eggs, while the females return to the sea to feed. During this time, a male penguin can lose up to 50% of its biomass (by dry weight). Where does this biomass go? a. It is converted to heat and then lost to the environment. b. It is converted to CO2 and H2O and then released. c. It is converted to ATP molecules. d. It leaves the penguins as excrement. ANSWER: b 71. Which statement is true concerning glycogen? a. Glycogen is made by animal cells and is primarily an energy storage molecule. b. Glycogen is made by plant cells and is primarily an energy storage molecule. c. Glycogen is made by plant cells and provides structural support for the cell. d. Glycogen is made by animal cells and provides structural support for the cell. ANSWER: a 72. Which statement is true concerning starch? a. Starch is made by animal cells and provides structural support for the cell. b. Starch is made by animal cells and is primarily an energy storage molecule. c. Starch is made by plant cells and provides structural support for the cell. d. Starch is made by plant cells and is primarily an energy storage molecule. ANSWER: d 73. Given that the free energy available from the full oxidation of glucose is -686 kcal/mole, which ratio represents the ratio of the free energy captured in ATP by aerobic cellular respiration to the free energy in glucose as the starting fuel molecule? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 a. 234:686 b. 7.3:686 c. 530:686 d. 53:234 e. 7.3:234 ANSWER: a 74. The NADH yield from oxidation of glucose up through and including the citric acid cycle is 10 NADH per glucose (10 moles NADH per mole glucose). The ΔG for of oxidation of NADH to NAD+ is -53 kcal per mole, and the yield of ATP per NADH from oxidative phosphorylation is about 2.5 ATP per NADH. What is the approximate efficiency of converting the energy captured in NADH into potential energy in ATP? a. 7.3% b. 53% c. 34.4% d. 13.7% ANSWER: c 75. In the electron transport chain, the energy from high-energy electrons is transformed into _____ before being used to generate ATP. a. a proton gradient b. bonds in phosphate c. C-C and C-H bonds in carbohydrates d. O2 to form H2O ANSWER: a 76. The diagram shown represents the four stages of aerobic cellular respiration as four simple boxes. Many of the "inputs" and "outputs" are shown as arrows. Open arrows represent "energy management molecules" (ATP and electron carriers), and closed arrows represent other inputs and outputs, such as substrates and products. Use this diagram to answer the questions.

Which arrow in the diagram shown could represent NADH? a. A b. B Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 c. C d. D e. E ANSWER: a 77. The diagram shown represents the four stages of aerobic cellular respiration as four simple boxes. Many of the "inputs" and "outputs" are shown as arrows. Open arrows represent "energy management molecules" (ATP and electron carriers), and closed arrows represent other inputs and outputs, such as substrates and products. Use this diagram to answer the questions.

Which arrow in the diagram shown could represent ATP? a. A b. B c. C d. D e. E ANSWER: c 78. The diagram shown represents the four stages of aerobic cellular respiration as four simple boxes. Many of the "inputs" and "outputs" are shown as arrows. Open arrows represent "energy management molecules" (ATP and electron carriers), and closed arrows represent other inputs and outputs, such as substrates and products. Use this diagram to answer the questions.

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Chapter 7 Which arrow in the diagram shown could represent acetyl-CoA? a. A b. B c. C d. D e. E ANSWER: d 79. The diagram shown represents the four stages of aerobic cellular respiration as four simple boxes. Many of the "inputs" and "outputs" are shown as arrows. Open arrows represent "energy management molecules" (ATP and electron carriers), and closed arrows represent other inputs and outputs, such as substrates and products. Use this diagram to answer the questions.

Which arrow in the diagram shown could represent O2? a. A b. B c. C d. D e. E ANSWER: b 80. The diagram shown represents the four stages of aerobic cellular respiration as four simple boxes. Many of the "inputs" and "outputs" are shown as arrows. Open arrows represent "energy management molecules" (ATP and electron carriers), and closed arrows represent other inputs and outputs, such as substrates and products. Use this diagram to answer the questions.

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Which arrow in the diagram shown could represent CO2? a. A b. B c. C d. D e. E ANSWER: e 81. Why do you need to breathe in oxygen? a. It is required as the final electron acceptor for the electron transport chain. b. It is required for creating CO2 during the oxidation steps of cellular respiration. c. It is required for the pyruvate oxidation stage of cellular respiration. d. It used to oxidize glucose during glycolysis. e. It is used as an input to the pathway that produces glycogen for energy storage. ANSWER: a Multiple Response 82. What are direct sources of electrons for the electron transport chain? Select all that apply. a. FAD b. FADH2 c. NADH d. NAD+ e. ATP ANSWER: b, c 83. During which stages of cellular respiration is carbon dioxide released? Select all that apply. a. pyruvate oxidation b. the citric acid cycle c. electron transport chain d. glycolysis Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 e. oxidative phosphorylation ANSWER: a, b 84. Which of these components of the electron transport chain directly move protons across the inner mitochondrial membrane? Select all that apply. a. complex I b. complex III c. complex IV d. complex II e. cytochrome c f. CoQ ANSWER: a, b, c 85. Which of the processes can pyruvate undergo as a result of fermentation? Select all that apply. a. It can be reduced to ethanol. b. It can be oxidized to ethanol. c. It can be reduced to acetyl-CoA. d. It can be reduced to lactic acid. e. It can be oxidized to pyruvic acid. ANSWER: a, d 86. Which regulatory mechanisms are important in keeping glycolysis and the citric acid cycle in relative balance to each other? Select all that apply. a. Citrate inhibits phosphofructokinase-1. b. ADP up-regulates phosphofructokinase-1. c. ATP inhibits phosphofructokinase-1. d. AMP up-regulates phosphofructokinase-1. ANSWER: a, b, c, d 87. Which one of the answer choices must be cleaved into monosaccharides before being transported into cells? Select all that apply. a. lactose b. maltose c. sucrose d. fructose ANSWER: a, b, c 88. The phosphorylation of glucose during glycolysis serves to: Select all that apply. a. destabilize the molecule, making it easier to cleave. b. trap imported glucose inside the cell. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 c. reduce an electron carrier, storing energy for later. d. trap imported glucose inside the mitochondria. ANSWER: a, b 89. Pyruvate can be used to produce: Select all that apply. a. sugars. b. acetyl-CoA. c. alanine. ANSWER: a, b, c 90. Which of the answer choices is a net products of the citric acid cycle for each molecule of acetyl-CoA generated by pyruvate oxidation? Select all that apply. a. 1 ATP b. 3 NADH c. 1 FADH2 d. 2 NADH e. 2 ATP f. 2 FADH2 ANSWER: a, b, c 91. When oxygen is depleted, the citric acid cycle stops. Which of the answer choices would you need to add to the system to restore activity? Select all that apply. a. NADH b. NAD+ c. FAD d. glucose e. acetyl-CoA f. FADH2 ANSWER: b, c 92. When oxygen is depleted, the electron transport chain stops. What else happens when oxygen is depleted? Select all that apply. a. The citric acid cycle would stop. b. The citric acid cycle would not change. c. ATP synthase would stop. d. ATP synthase would not change. e. Fermentation would start. ANSWER: a, c, e 93. Which of the answer choices is true about ATP synthase? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 a. ATP synthase is an integral membrane protein. b. ATP synthase is a transport protein. c. ATP synthase makes a total of four ATP molecules in glycolysis. d. Under aerobic conditions, the F1 portion of ATP synthase catalyzes a catabolic reaction. e. ATP synthase is an important enzyme in the citric acid cycle. f. ATP synthase relies on an electrochemical gradient of sodium ions to catalyze the formation of ATP. ANSWER: a, b 94. Which of the answer choices is a mobile electron carrier in the electron transport chain? Select all that apply. a. complex II b. complex III c. ATP synthase d. CoQ e. cytochrome c f. complex I g. complex IV ANSWER: d, e 95. Which of the answer choices is an electron carrier in the electron transport chain? Select all that apply. a. CoQ b. cytochrome c c. ATP synthase d. phosphofructokinase e. pyruvate dehydrogenase ANSWER: a, b 96. The coronary arteries supply oxygenated blood to heart muscle. Following a heart attack, which of the events would occur in heart tissue due to a block in the coronary arteries? Select all that apply. a. Pyruvate would accumulate. b. The production of ATP would decrease. c. Phosphofructokinase, an intermediary enzyme of glycolysis, would be activated. d. Lactic acid would accumulate in the tissue due to glycolysis and fermentation. ANSWER: b, c, d 97. How would you explain why fermentation yields so much less ATP than the yield from aerobic cellular respiration? Select all that apply. a. The end products of fermentation retain much more of the starting free energy than CO2. b. The products of fermentation have no more free energy. c. The end products of fermentation are only partially oxidized, compared to the end product of aerobic Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 cellular respiration (CO2). d. The process of glycolysis and fermentation can make ethanol and lactic acid. ANSWER: a, c 98. Which of the answer choices are products of the breakdown of triacylglycerols? Select all that apply. a. glycerol b. NADH c. FADH2 d. acetyl-CoA e. pyruvate f. glucose ANSWER: b, c, d 99. What glycolysis products are transported into the mitochondria? Select all that apply. a. NADH b. pyruvate c. ATP d. glucose ANSWER: a, b 100. DNP (2,4-dinitrophenol) is an effective weight loss agent that was used in diet pills in the 1930s. It has since been removed from the market because of serious side effects such as fever, cataracts, rashes, and sometimes death. DNP inserts into the inner mitochondrial membrane and allows the flow of protons from the intermembrane space to the matrix. Based on this information, which of the answer choices might you predict? Select all that apply. a. decreased difference in pH between the matrix and intermembrane space b. reduced ATP production c. dissipation of the proton gradient d. increased ATP production e. increased hydrolysis of ATP ANSWER: a, b, c 101. In order for a pathway to produce its products, it must have sufficient inputs. Which of the answer choices must be directly supplied to the citric acid cycle for it to proceed? Select all that apply. a. acetyl-CoA b. NAD+ c. FAD d. ADP e. NADH f. ATP g. pyruvate Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 h. glucose (or other sugar) ANSWER: a, b, c, d 102. In order for a pathway to produce its products, it must have sufficient inputs. Which of the answer choices must be available in sufficient supply for glycolysis to proceed? Select all that apply. a. glucose (or other sugar) b. NAD+ c. ADP d. ATP e. NADH f. FAD g. pyruvate h. acetyl-CoA ANSWER: a, b, c, d 103. Which of the answer choices are products of pyruvate oxidation? Select all that apply. a. ATP b. pyruvate c. acetyl Co-A d. CO2 e. NADH f. FADH2 ANSWER: c, d, e 104. Which of the answer choices is/are true regarding redox reactions? (Note that in redox reactions, the molecule that "causes" another to gain or lose electrons is referred to as the agent.) Select all that apply. a. Oxidizing agents accept electrons. b. If a molecule accepts electrons, it has been reduced. c. Redox reactions may involve the transfer of hydrogen ions (H+). d. A molecule that has gained H atoms is said to be reduced. e. Oxidizing agents may accept H+ ions. f. Reducing agents may accept H+ ions. ANSWER: a, b, c, d, e 105. In the reactions of glycolysis, pyruvate oxidation, and the citric acid cycle, chemical energy is transferred to: Select all that apply. a. bonds in ATP. b. electrons in electron carriers. c. C-C and C-H bonds in fats. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 d. a proton gradient. ANSWER: a, b 106. The reaction diagrammed: Select all that apply.

a. is one possible pathway in the fermentation of pyruvate. b. occurs in the mitochondrial matrix. c. occurs twice for each glucose oxidized. d. is the final reaction of glycolysis. e. shows synthesis of the substrate that enters the citric acid cycle. f. occurs in the cytosol. ANSWER: b, c, e 107. In step 6 of the citric acid cycle, succinate + FAD → fumarate + FADH2. Which statements concerning this reaction are true? Refer to the figure shown. (Note that in redox reactions, the molecule that "causes" another to gain or lose electrons is referred to as the agent.) Select all that apply. a. Fumarate is more oxidized than succinate. b. FAD is an oxidizing agent in the reaction. c. FAD is more reduced than FADH2. d. This is not an oxidation/reduction reaction; that is, no electrons have moved. ANSWER: a, b 108. Complete oxidation of glucose to CO2 involves two different mechanisms for synthesizing ATP: oxidative phosphorylation and substrate-level phosphorylation. Substrate-level phosphorylation: Select all that apply. a. requires an electron transport chain. b. requires activity of the enzyme ATP synthase. c. occurs in the cytosol. d. occurs across the inner mitochondrial membrane. e. occurs in the mitochondria. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 ANSWER: c, e 109. How would you characterize the process of fermentation? Select all that apply. a. a limited way of deriving energy without oxygen, as the amount of ATP generated in fermentation is low b. a useful process in the production of alcohol c. a useful process in the production of cheese and yogurt d. a useful process for organisms that were present when there was very little atmospheric oxygen e. a useful process for making glucose f. a very useful way of deriving energy without oxygen, as the amount of ATP generated in fermentation made is high ANSWER: a, b, c, d 110. Which of the answer choices would you expect to observe in response to an increase in the concentration of cellular ADP, provided that the cell has plenty of glucose and oxygen? Select all that apply. a. an increase in phosphofructokinase activity b. an increase in ATP production by ATP synthase c. an increase in CO2 production d. an increase in electron transport chain activity e. an increase in ATP production by substrate-level phosphorylation f. a decrease in acetyl-CoA production g. an increase in the pH of the mitochondrial matrix h. a decrease in the reduction of oxygen to form water ANSWER: a, b, c, d, e 111. Phosphofructokinase (PFK-1) is an allosteric enzyme that catalyzes a key regulatory step in glycolysis. What statements are true about this enzyme? Select all that apply. a. Decreased levels of citrate in the cytoplasm inhibit PFK-1 activity. b. ADP and AMP are allosteric activators of PFK-1 activity. c. ATP acts as an allosteric inhibitor of PFK-1 activity. d. Cells switch to β-oxidation of fatty acids when PFK-1 activity is inhibited. e. ADP can bind PFK-1's active site to block substrate binding. ANSWER: b, c 112. Which of the answer choices can yield a significant amount of metabolic energy via cellular respiration? Select all that apply. a. candy bar b. egg c. bread d. multivitamins e. oil Copyright Macmillan Learning. Powered by Cognero.

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Chapter 7 f. fiber g. coffee ANSWER: a, b, c, e 113. An individual's ability to excel in certain types of sports may be due to different muscle fiber types. Certain muscle types allow some people, like marathoners, to achieve sustained muscle activity, whereas other muscle types allow people, like sprinters, to make use of a rapid burst of muscle activity, although the muscles fatigue quickly. The sustained activity of muscles in marathon runners is due to the higher yield of ATP per glucose. What differences would you predict for marathoners' muscles compared to sprinters'? Select all that apply. a. None of the other answer options is correct. b. Marathoners' muscles require greater oxygen delivery to the muscle cells. c. Marathoners' muscles have a greater number of mitochondria. d. Marathoners' muscles use a different set of enzymes for cellular respiration. e. Marathoners' muscles rely more on oxidative phosphorylation. ANSWER: b, c, e 114. Based on your interpretation of the model shown of phosphofructokinase-1 (PFK-1) activity, which alterations might result in an increase in PFK-1 activity? Select all that apply.

a. a mutation that decreases binding affinity for ATP b. a mutation that increases binding affinity for AMP c. a mutation that increases binding affinity for fructose 6-phosphate d. a mutation that increases binding affinity for ATP e. a mutation that decreases binding affinity for ADP f. a mutation that decreases binding affinity for fructose 6-phosphate ANSWER: a, b, c

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Chapter 8 Multiple Choice 1. Which of the sequences best represents the flow of energy through living systems? a. H2O → photosynthesis → carbohydrate → cellular respiration → ATP → H2O b. H2O → photosynthesis → carbohydrate → cellular respiration → ATP → CO2 c. light → photosynthesis → carbohydrate → cellular respiration →ATP → heat d. light → photosynthesis → carbohydrate → cellular respiration → ATP → H2O e. CO2 → photosynthesis → carbohydrate → cellular respiration → ATP → heat ANSWER: c 2. Photosynthesis in green plants is comprised of the _____, in which H2O is oxidized, and _____, in which CO2 is reduced. a. CO2 reduction; NADPH oxidation b. Calvin cycle; photosynthetic electron transport chain c. photosystem II; photosystem I d. photosystem I; photosystem II e. photosynthetic electron transport chain; the Calvin cycle ANSWER: e 3. The experiment shown in the figure demonstrates that the percentage of 18O2 increases when the:

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a. H2O in the experiment contains 18O. b. CO2 in the experiment contains 18O. c. H2O and CO2 in the experiment both contain 18O. ANSWER: a 4. In which lettered location in the figure would you find electrons moving between large protein complexes?

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a. K b. J c. L d. M e. N ANSWER: c 5. In which lettered location in the figure would you find carbohydrate synthesis taking place?

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a. J b. K c. L d. M e. N ANSWER: e 6. In which lettered location in the figure is light initially captured?

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a. J b. K c. L d. M e. N ANSWER: c 7. A large agricultural chemical company is testing a new experimental pesticide. An unfortunate side effect in plants treated with this new product is a decrease in NADPH production in the chloroplasts. (Interestingly, reduction of NAD+ to NADH in the mitochondria is unaffected.) Given this observation, which of the answer choices would you expect to observe in the chloroplasts of these plants? a. an increase in 3-phosphoglycerate levels and a decrease in RuBP levels b. a decrease in 3-phosphoglycerate levels and an increase in RuBP levels c. an increase in glyceraldehyde-3-phosphate levels d. an increase in 3-phosphoglycerate levels and an increase in RuBP levels ANSWER: a 8. The most abundant protein on Earth is thought to be: a. cellulose. b. rubisco. c. chlorophyll. d. hemoglobin. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 e. keratin. ANSWER: b 9. How many CO2 molecules must enter the Calvin cycle for each molecule of triose phosphate exported from the chloroplast? a. 1 b. 3 c. 5 d. 6 e. 12 ANSWER: b 10. The Calvin cycle is composed of 15 discrete biochemical reactions. How many of these are associated with the regeneration phase? a. 3 b. 5 c. 9 d. 12 e. 14 ANSWER: d 11. The regeneration phase of the Calvin cycle regenerates _____ molecules of _____ from _____ molecules of _____. a. 5; ATP; 5; triose phosphate b. 3; triose phosphate; 5; RuBP c. 5; RuBP; 5; triose phosphate d. 3; RuBP; 5; triose phosphate e. 5; NADPH; 3; RuBP ANSWER: d 12. Incorporation of each molecule of CO2 into the Calvin cycle requires the energy of 3 ATP and 2 NADPH. If the regeneration phase of the Calvin cycle did not use ATP, what would be the requirement for incorporation of a CO2 molecule? a. 2 ATP and 2 NADPH b. 1 ATP and 1 NADPH c. 0 ATP and 2 NADPH d. 2 ATP and 1 NADPH e. 2 ATP and 3 NADPH ANSWER: a 13. Carbohydrates are stored primarily as starch in plants. This is important because: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 a. the space carbohydrates occupy is needed for other macromolecules. b. if carbohydrates were left to accumulate as individual monosaccharides, H2O would enter the cell by osmosis and damage the cell. c. energy is needed in the nucleus, and only starch is capable of being transported into the nucleus. d. carbohydrates would otherwise be utilized as energy sources, even if such energy was not needed at that time. e. All of these choices are correct. ANSWER: b 14. Which of the outcomes would you expect if the pH of the thylakoid lumen increases from 4 to 6? a. The rate of ATP synthesis by ATP synthase will decrease. b. More NADP+ will be reduced to NADPH. c. Less O2 will be produced. d. Proton transport through ATP synthase from the thylakoid lumen to the stroma will increase. e. The change in pH will not have a significant effect on the rate of photosynthesis. ANSWER: a 15. The figure shown illustrates a part of the Calvin-Benson experiments that led to our understanding of the Calvin cycle. What is happening here?

a. They are testing whether artificial light from a candle can replace sunlight in photosynthesis. b. They are extracting the chlorophyll and accessory pigments in hot alcohol. c. They are filtering the cells into a solution of H218O so that they can trace the heavy isotope of oxygen as it is incorporated into the cell mass. d. They are dripping photosynthetic cells into hot alcohol to quickly stop the reactions of photosynthesis. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 16. In this interpretation of a photosystem, the item labeled 2 is a(n) _____ and its main function is to _____.

a. reaction center chlorophyll; make ATP and NADPH b. reaction center chlorophyll; donate excited electrons c. antenna chlorophyll; capture light energy from the sun d. reaction center; capture light energy from the sun e. antenna chlorophyll; donate excited electrons ANSWER: b 17. In this interpretation of a photosystem, the item labeled 1 is a(n) _____ and its main function is to _____.

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a. antenna chlorophyll; donate excited electrons b. reaction center chlorophyll; make ATP and NADPH c. reaction center chlorophyll; donate excited electrons d. reaction center; capture light energy from the sun e. antenna chlorophyll; capture light energy from the sun ANSWER: e 18. In this interpretation of a photosystem, excited electrons can flow from 1 to 2.

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a. true b. false ANSWER: b 19. If the photosystem shown is photosystem II, what will happen to the excited electron from the part labeled 2?

a. It will oxidize an electron acceptor. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 b. It will reduce an antenna chlorophyll molecule. c. It will be passed to an electron acceptor. d. It will be transferred to H2O. e. It will normally return to a ground state with release of heat. ANSWER: c 20. If the photosystem shown is photosystem II, where will molecule 2 obtain replacement electrons after donating an electron to an electron acceptor?

a. from CO2 b. from H2O c. from Pc (plastocyanin) d. from antenna chlorophyll e. from O2 ANSWER: b 21. If the photosystem shown is photosystem I, where will molecule 2 obtain replacement electrons after donating electrons to an electron acceptor?

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a. from antenna chlorophyll b. from CO2 c. from H2O d. from Pc (plastocyanin) e. from O2 ANSWER: d 22. After several weeks of no rain during the summer, you observe that leaves of some plants are wilted. Which of the answer choices would you expect to observe at the cellular level with respect to photosynthesis? a. The rate of the reduction of chlorophyll molecules will be decreased. b. The rate of the oxidation and reduction of the thylakoid electron transport chain will be greater than normal. c. More CO2 will be incorporated into carbohydrate. d. More ATP will be synthesized by ATP synthase in the chloroplasts. e. More glucose will be produced to compensate for the absence of H2O. ANSWER: a 23. Which of the answer choices could result if a plant cell is exposed to a toxin that causes the outer chloroplast membrane to be less permeable to H2O? a. The synthesis of carbohydrate in the chloroplast would increase. b. The synthesis of ATP in the chloroplast would increase. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 c. There would be an increase in the amount of O2 produced. d. The cytoplasm of plant cells would become hypertonic. e. There would be a decrease in the amount of O2 produced. ANSWER: e 24. In a thylakoid membrane, absorbed light energy is passed from one antenna chlorophyll molecule to another until it is: a. dissipated as heat. b. released as fluorescence. c. passed to a reaction center. d. both dissipated as heat and released as fluorescence. ANSWER: c 25. The electrochemical gradient in a chloroplast can have a concentration of protons 1000 times greater on one side of a thylakoid membrane than the other. a. true b. false ANSWER: a 26. Which of the answer choices lists the correct order of steps taken by electrons through the light dependent reactions of photosynthesis? a. H2O → photosystem II → cytochrome b6f → photosystem I → NADPH b. H2O → photosystem I → cytochrome b6f → photosystem II → NADPH c. NADPH → photosystem I → cytochrome b6f → photosystem II → H2O d. NADPH → photosystem II → cytochrome b6f → photosystem I → H2O ANSWER: a 27. As electrons flow through the light-dependent reactions, they become associated with several molecules and molecular complexes. When do the electrons have the greatest amount of energy? a. when they are associated with photosystem II b. when they are associated with cytochrome b6f c. when they are associated with NADPH d. when they are associated with H2O e. when they are associated with photosystem I ANSWER: e 28. Which of the answer choices is an advantage of having two slightly different photosystems in the chloroplasts? a. The electrons can be elevated to a higher energy level than is possible with a single photosystem. b. Cells are able to use light energy at twice the maximum efficiency predicted for cells with a single Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 photosystem. c. Cells are able to use both blue and green wavelengths of light for photosynthesis. d. Twice as many electrons can be excited by a given amount of light energy. ANSWER: a 29. Approximately what percentage of the sun's usable energy, arriving at the surface of a leaf, does the photosynthetic electron transport chain capture? a. 40% b. 24% c. 8% d. 60% e. 2% ANSWER: b 30. The estimate of the actual percentage of the solar energy reaching a leaf that gets converted into carbohydrate products is approximately: a. 24%. b. 40-50%. c. 8-12%. d. 50-60%. e. 1-2%. ANSWER: e 31. When NADP+ is in short supply, excited electrons from the electron transport chain of photosynthesis have no "safe" place to go. What is the result? a. carbohydrate synthesis stops b. electron transport stops c. light absorption stops d. reactive oxygen species e. All of these choices are correct. ANSWER: d 32. It is possible that the earliest reaction centers used light energy to drive the movement of electrons from an electron donor in the surrounding medium to an electron acceptor within the cell. a. true b. false ANSWER: a 33. Which of the answer choices do scientists think might have been the first electron donor to arise? a. Fe2+ b. 18O2 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 c. Mg2+ d. NADPH e. 35S ANSWER: a 34. Carbon enters the Calvin cycle in the form of: a. triose phosphates. b. glucose. c. CO2. d. RuBP. e. 3-PGA. ANSWER: c 35. Protons in a chloroplast flow through an ATP synthase from the: a. stroma to the thylakoid lumen b. stroma to the intermembrane space. c. cytoplasm to the intermembrane space. d. thylakoid lumen to the stroma. e. intermembrane space to the stroma. ANSWER: d 36. The ATP generated by the photosynthetic electron transport chain is used in the _____ phase(s) of the Calvin cycle. a. carboxylation b. carboxylation and reduction c. carboxylation and regeneration d. reduction and regeneration e. carboxylation, reduction, and regeneration ANSWER: d 37. Light energy is converted to chemical energy in the: a. thylakoid membrane. b. central vacuole. c. nucleus. d. mitochondrial membrane. e. cytoplasm. ANSWER: a 38. The enzyme called rubisco: a. connects a 3-carbon molecule to a 5-carbon molecule. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 b. breaks a 3-carbon molecule into 1-carbon and 2-carbon molecules. c. breaks a 1-carbon molecule off of a 6-carbon molecule. d. connects two 3-carbon molecules into a 6-carbon molecule. e. connects a 1-carbon molecule to a 5-carbon molecule. ANSWER: e 39. The terminal electron acceptor of a photosynthetic electron transport chain is: a. NADPH. b. NADP+. c. NAD+. d. O2. e. H2O. ANSWER: b 40. The primary reducing agent in the Calvin cycle is: a. NADP+. b. H2O. c. NADPH. d. O2. e. NAD+. ANSWER: c 41. In plants, the initial electron donor in photosynthesis is: a. H2O. b. O2. c. CO2. d. ATP. e. NADPH. ANSWER: a 42. Which one of the statements best represents the relationship between cellular respiration and photosynthesis? a. Cellular respiration stores energy in organic molecules, whereas photosynthesis releases it. b. Photosynthesis occurs only during the day, and cellular respiration occurs only at night. c. Cellular respiration occurs only in animals, and photosynthesis occurs only in plants. d. Photosynthesis stores energy in organic molecules, whereas cellular respiration releases it. e. Photosynthesis reverses the biochemical pathways of cellular respiration. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 43. The product of the Calvin cycle is a triose-phosphate sugar that is either exported from the chloroplast or used to regenerate RUBP. How many times must each of the reactions in the Calvin cycle take place in order to complete the synthesis of one 3-carbon triose phosphate sugar molecule? a. 6 b. 3 c. 2 d. 1 ANSWER: b 44. The reactions of the Calvin cycle used to be referred to as the "dark reactions" of photosynthesis, implying that light was not required for this pathway to incorporate CO2 into carbohydrate. However, in the absence of light, the Calvin cycle shuts down. Which parts of the Calvin cycle would be impacted first by the absence of light energy? a. the synthesis of hexose sugars b. the carboxylation of RuBP c. the regeneration of RuBP d. the reduction of 3-phosphoglycerate ANSWER: d 45. Which of the answer choices is estimated to be the most abundant protein on Earth? a. rubisco b. ATP synthetase c. NADP+ reductase d. cytochrome complex ANSWER: a 46. What is the initial carbon input in the Calvin cycle? a. C6H12O6 b. triose phosphate c. CO2 d. rubisco ANSWER: c 47. Why is the absorption and capture of light energy by chlorophyll molecules in plants much more efficient than its absorption and capture by a solution of chlorophyll molecules in the lab? a. In the plant, the energy is transferred between molecules. b. An artificial light source is used in the lab. c. In the plant, other accessory proteins are produced. d. Electrons are more readily available in solution. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 48. Energy captured during the "photo" part of photosynthesis is stored in _____ during the "synthesis" part of the process. a. ionic bonds b. carbon atoms c. covalent bonds d. hydrogen bonds ANSWER: c 49. What result would you predict if a plant was exposed to a toxin that made the thylakoid membranes permeable to protons? a. The proton gradient across the thylakoid membrane would be eliminated. b. The reduction of NADP+ in the chloroplasts would no longer be possible. c. The transport of electrons by the protein complexes in the thylakoids would be prevented. d. A reversal of the proton gradient across the thylakoid membrane would result in a higher pH in the thylakoid lumen than in the surrounding stroma. ANSWER: a 50. If plants are exposed to a toxin that makes the thylakoid membrane leaky to protons, what would you predict about the rate of ATP production by the chloroplasts? a. ATP production will increase. b. ATP production will decrease. c. ATP production does not change. ANSWER: b 51. Which of the molecules are reduced during photosynthesis? a. NADPH and 3-phosphoglycerate b. ADP and 3-phosphoglycerate c. NADP+ and 3-phosphoglycerate d. NAD+ and glyceraldehyde-3-phosphate ANSWER: c 52. When photosynthesis is fully operational under conditions of intense light, the pH of the thylakoid lumen may be as much as three full pH units lower than the pH of the stroma of the chloroplast. That means that there are _____ protons in the thylakoid lumen compared to the stroma. a. 1000 times more b. 100 times more c. 3 times more d. 3 times fewer ANSWER: a 53. Electrons removed from H2O molecules are transported through the photosystems and photosynthetic Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 electron transport chain and are ultimately used to reduce NADP+ to NADPH. At what point do the electrons have the greatest amount of potential energy? a. after capturing photon energy in the reaction center of photosystem II b. after capturing photon energy in the reaction center of photosystem I c. just after being removed from the H2O molecule d. when they are associated with NADPH e. when they are transferred to the cytochrome complex of the electron transport chain ANSWER: b 54. If a mutation occurs such that photosystem I no longer functions as efficiently as it should, which of the answer choices would be most directly affected? a. the amount of O2 produced b. the rate at which NADPH is produced c. the rate at which protons are transported into the thylakoid d. the rate of ATP synthesis by ATP synthase in the chloroplast ANSWER: b 55. If a mutation occurs such that photosystem II no longer functions as efficiently as it should, which of the answer choices would be most directly affected? a. the amount of O2 produced b. the rate at which NADPH is produced c. the rate at which protons are transported into the thylakoid d. the rate of ATP synthesis by ATP synthase in the chloroplast ANSWER: a 56. How does the plant cell protect itself from harmful reactive oxygen species formed by the interaction of high-energy electrons with O2 molecules? a. by increasing the amount of e- donated to ferredoxin b. by increasing the amount of NADP c. by decreasing the production of ATP d. by producing antioxidants ANSWER: d 57. Of the total solar radiation that reaches the surface of the earth, photosynthetically active radiation (PAR, those wavelengths of light that can be absorbed by photosynthetic pigments) comprises about 40%. Approximately what proportion of this PAR do the photosynthetic electron transport systems capture? a. 2/40 b. 8/40 c. 24/40 d. 16/40 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 e. 35/40 ANSWER: c 58. Plants that possess high levels of xanthophylls are better suited to growth in bright sunlight. a. true b. false ANSWER: a 59. In the evolution of photosynthesis, which of these processes happened first? a. endosymbiosis b. mitochondrial membrane structure c. chloroplast membrane structure d. the acquisition of two photosystems by cyanobacteria ANSWER: d 60. Some of the intermediates in the biosynthesis of chlorophyll are themselves capable of absorbing light energy. a. true b. false ANSWER: a 61. During photosynthesis, _____ is reduced to _____. a. CO2; O2 b. H2O; CO2 c. O2; H2O d. sugar; O2 e. CO2; sugar ANSWER: e 62. The ratio of ATP to NADPH used by the Calvin cycle is approximately 3:2 (3 moles ATP used for each 2 moles NADPH). What is the minimum number of these molecules required to synthesize one mole of glucose? a. 18 moles ATP + 12 moles NADPH b. 30 moles ATP + 20 moles NADPH c. 12 moles ATP + 8 moles NADPH d. 6 moles ATP + 4 moles NADPH e. 3 moles ATP + 2 moles NADPH ANSWER: a 63. Two petri dishes of radish seeds are prepared as follows: A disk of absorbent, but otherwise inert, filter paper is placed in the bottom of each dish. Next, 5 mL of distilled water are added to each disk. Then, 2 grams of dry radish seeds are spread over each moist disk. One dish is put in a dark cupboard, and the other is put on a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 sunny windowsill. The investigator checks the dishes daily and adds water equally as needed. Seeds in both treatments germinate within 2 days. At the end of 10 days, the plant material on each petri plate is dried and weighed. The results are: - starting dry weight 2.00 g - sunny windowsill 2.25 g - dark cupboard 1.75g What is the most likely explanation for why the dark-grown seedlings lost biomass? a. The metabolism in the cells of the seedlings increased to compensate for the lack of light energy. b. Cellular respiration in the cells of the seedlings consumed the starch and oils present in the seeds as sources of energy. c. Cellular respiration and ATP synthesis needed for growth does not occur as efficiently in plants in the dark as in plants exposed to light. ANSWER: b 64. Two petri dishes of radish seeds are prepared as follows: A disk of absorbent, but otherwise inert, filter paper is placed in the bottom of each dish. Next, 5 mL of distilled water are added to each disk. Then, two grams of dry radish seeds are spread over each moist disk. One dish is put in a dark cupboard, and the other is put on a sunny windowsill. The investigator checks the dishes daily and adds water equally as needed. Seeds in both treatments germinate within 2 days. At the end of 10 days, the plant material on each petri plate is dried and weighed. The results are: - starting dry weight 2.00 g - sunny windowsill 2.25 g - dark cupboard 1.75g How did the increased biomass in the sunlit dish arise (that is, where did the additional mass come from, and by what process)? a. The additional mass came from the carbon and oxygen atoms present in the CO2 it took up during photosynthesis, which it used to synthesize organic molecules. b. The additional mass came from the hydrogen and oxygen atoms present in the H2O it took up during photosynthesis, which it used to synthesize organic molecules. c. The additional mass came from the extra H2O present in the plant tissues due to greater photosynthetic activity. ANSWER: a 65. A new experimental pesticide is being tested by a large agricultural chemical company. An unfortunate side effect in plants treated with this new product is a decrease in ATP production in the chloroplasts. (Interestingly, production of ATP in the mitochondria is unaffected.) Which of the answer choices would you predict to be directly affected by the new pesticide? a. the carboxylation of RuBP by rubisco b. the transport of electrons along the photosynthetic electron transport chain c. the synthesis of triose phosphates from 3-phosphoglycerate d. the oxidation of RuBP by rubisco ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 66. A new experimental pesticide is being tested by a large agricultural chemical company. An unfortunate side effect in plants treated with this new product is a decrease in ATP production in the chloroplasts. (Interestingly, production of ATP in the mitochondria is unaffected.) Which of the answer choices would you expect to be directly affected by the new pesticide? a. the rate of photorespiration b. the transport of electrons along the photosynthetic electron transport chain c. the carboxylation of RuBP by rubisco d. the oxidation of RuBP by rubisco ANSWER: a 67. What is the principle product of the Calvin cycle that is exported from the chloroplast for use by the plant cell? a. NADPH b. sucrose and glucose c. RuBP d. triose phosphates e. ATP ANSWER: d 68. During photosynthesis, _____ is reduced to _____. a. CO2; O2 b. H2O; CO2 c. O2; H2O d. triose phosphate; O2 e. CO2; triose phosphate ANSWER: e 69. Beginning with the step catalyzed by rubisco, select the correct order of phases in the Calvin cycle. a. carboxylation, regeneration, reduction b. reduction, regeneration, carboxylation c. regeneration, carboxylation, reduction d. carboxylation, reduction, regeneration e. reduction, carboxylation, regeneration ANSWER: d 70. The reducing agent during the Calvin cycle is: a. NADH. b. ATP. c. O2. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 d. NADPH. e. FADH2. ANSWER: d 71. A new experimental pesticide is being tested by a large agricultural chemical company. An unfortunate side effect in plants treated with this new product is the partial inhibition of ferredoxin-NADP+ reductase. Given this observation, which of the answer choices would you expect to observe in the photosynthetic cells of these plants? a. decreased rate of triose phosphate production b. decreased levels of NADP+ in the chloroplast c. decreased levels of ATP in the chloroplast d. decreased rate of cyclic electron transport ANSWER: a 72. The pH in the stroma of the chloroplast should be _____ compared with the thylakoid lumen, due to the _____ concentration of protons in the thylakoid lumen. a. higher; lower b. lower; higher c. higher; higher d. lower; lower ANSWER: c 73. In the chloroplast, cytochrome b6f is an important complex in the photosynthetic electron transport chain because it accepts electrons from photosystem II and donates them to photosystem I. Another vital function of cytochrome b6f is that, during its reduction-oxidation cycle, it moves _____ from the _____. a. electrons; stroma into the thylakoid lumen b. protons; stroma into the thylakoid lumen c. protons; thylakoid lumen into the stroma d. electrons; thylakoid lumen into the stroma ANSWER: b 74. Suppose you discovered a mutant strain of spinach in which the thylakoid membranes were slightly permeable to protons, thus allowing a slow leakage. (Remember that normal membranes are not permeable to protons at all.) How might this defect affect the yield of ATP or NADPH from the light reactions? a. ATP yield would decrease. b. NADPH yield would increase. c. NADPH yield would decrease. d. ATP yield would increase. e. There would be no change. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 75. As electrons flow from one photosystem to another, the amount of energy present in the electrons decreases. How does the plant cell use the energy lost by the electrons? a. The energy is used to increase the pH of the thylakoid lumen. b. The energy is used to reduce NADP+ to NADPH. c. The energy is used to create and maintain a proton gradient. d. The energy is used directly by ATP synthase to make ATP. ANSWER: c 76. Suppose a mutation caused photosystem I to function at only 10 percent of its normal capacity. Which of the answer choices would you predict? a. The rate of O2 production would decrease. b. The pH of the thylakoid lumen would be higher than normal. c. NADP+ reductase activity would increase. d. ATP production would increase. ANSWER: a 77. Which of the answer choices is the primary advantage to eukaryotic photosynthetic organisms of having two photosystems? a. Twice as much ATP can be made, compared to cells with a single photosystem. b. Twice as much NADPH can be made, compared to cells with a single photosystem. c. Twice as much O2 can be made, compared to cells with a single photosystem. d. H2O can be used as a source of electrons. e. ATP can be synthesized by ATP synthase. ANSWER: d 78. Some bacteria carry out a type of photosynthesis that uses H2S in place of H2O. If we assumed that the process is otherwise similar to green plant photosynthesis, which of the answer choices could represent the overall reaction? a. 6 CO2 + 12 H2S + sunlight → C6H12O6 + 6 H2O + 12 S b. 6 CO2 + 12 H2S + sunlight → C6H12S6 + 6 H2S + 6 O2 c. 6 CO2 + 12 H2S + sunlight → C6H12O6 + 6 H2S + 6 O2 d. 6 CO2 + 12 H2S + sunlight → C6H12S6 + 6 H2O + 6 SO2 ANSWER: a 79. Which of the sequences accurately represents the flow of electrons through living organisms? a. H2O → photosynthesis → carbohydrate → cellular respiration → ATP b. CO2 → H2O → chlorophyll → carbohydrate → cellular respiration → ATP c. O2 → cellular respiration → carbohydrate → photosynthesis → CO2 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 d. CO2 → photosynthesis → carbohydrate → cellular respiration → ATP e. H2O→ photosynthesis → carbohydrate → cellular respiration → H2O ANSWER: e 80. Leaves absorb the least amount of light in the _____ range of the visible spectrum. a. orange b. green c. blue d. yellow e. violet ANSWER: b 81. Photorespiration in chloroplasts resembles cellular respiration in mitochondria in that: a. CO2 is produced in both pathways. b. NADH is used in both pathways as the source of high-energy electrons. c. they both store energy in the form of ATP. d. both pathways are beneficial to the cell. e. both employ the same reactions, but one occurs only in the light. ANSWER: a 82. Photorespiration in chloroplasts differs from cellular respiration in mitochondria in that: a. CO2 is produced in mitochondrial respiration, whereas O2 is produced in photorespiration. b. NADH supplies high-energy electrons in mitochondrial respiration, whereas RuBP supplies the electrons necessary for photorespiration. c. Photorespiration supplies plant cells with ATP, whereas mitochondrial respiration supplies animal cells with ATP. d. ATP is available for cellular processes in cellular respiration only. ANSWER: d 83. Normally, xanthophylls are present in the thylakoid membrane and are associated with photosystem II. If a mutation occurred that reduced the amount of xanthophylls in the thylakoid membrane, which of the answer choices would you predict? a. The levels of harmful, reactive oxygen species in the cell would increase. b. NADPH production would increase under conditions of intense light. c. Levels of reactive oxygen species in the thylakoid lumen would drop dramatically. d. The rate of cyclic electron transport would increase. ANSWER: a 84. Suppose you discovered a mutant strain of spinach in which the thylakoid membranes were slightly permeable to protons, thus allowing a slow leakage. (Remember that normal thylakoid membranes are not permeable to protons at all.) What change in the reactions of photosynthesis might occur in compensation for Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 this defect? a. Noncyclic electron transport would increase. b. Cyclic electron transport would increase. c. O2 production would decrease. d. Cyclic electron transport would decrease. e. Noncyclic electron transport would decrease. ANSWER: b 85. Photorespiration results in the: a. production of ATP. b. loss of O2. c. inactivation of rubisco. d. consumption of ATP and the loss of CO2. e. All of these choices are correct. ANSWER: d 86. How much of the solar radiation that reaches a leaf is captured by the photosynthetic electron transport chain? a. close to 100% b. about 60% c. between 20% and 25% d. between 5% and 10% ANSWER: c 87. In nature, only about 1-2% of the sun's usable energy is ultimately stored in carbohydrate. Which of the answer choices contributes to this apparent inefficiency of the photosynthetic process? a. The presence of accessory pigments in chloroplasts decreases the amount of light energy available to excite chlorophyll molecules. b. PS II and PS I are not completely synchronized because they absorb light energy at slightly different wavelengths. c. Rubisco is able to use both CO2 and O2 as a substrate. ANSWER: c 88. Which of the answer choices is the hypothesis that may explain the evolutionary origin of the presence of two photosystems in photosynthetic eukaryotes? a. An ancient species of photosynthetic bacteria with a single photosystem took up residence in a eukaryotic host and eventually evolved a second photosystem. b. An ancient species of photosynthetic bacteria with two photosystems took up residence in a primitive eukaryotic host and exist today as chloroplasts. c. Two different ancient photosynthetic bacteria took up residence in a primitive eukaryotic host and are present today as chloroplasts. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 ANSWER: b 89. Over evolutionary time, photosynthesis has, in effect, introduced challenges to itself by producing an O2 atmosphere because: a. rubisco is an oxygenase and O2 is a strong reductant. b. rubisco is a carboxylase. c. H2O is a strong oxidant. d. rubisco is a dehydrogenase. e. rubisco is an oxygenase and O2 is a strong oxidant. ANSWER: e 90. Paraquat is an herbicide that blocks the transfer of electrons from photosystem I to NADPH. Mechanistically, paraquat is effective because it: a. increases the formation of reactive oxygen species. b. prevents the formation of reactive oxygen species. c. prevents the reduction of NADPH. d. absorbs light energy that cannot then be used for photosynthesis. e. increases cyclic electron transport, which does not produce NADPH. ANSWER: a 91. Low temperatures slow the activity of rubisco but enhance its ability to select CO2 over O2. Low temperatures thus: a. increase photorespiration rates. b. decrease photorespiration rates. c. increase the production of reactive oxygen species. d. increase photosynthesis rates. ANSWER: b 92. High temperatures increase reaction rates but lower rubisco selectivity for CO2 versus O2. High leaf temperatures thus: a. decrease photorespiration rates. b. increase photorespiration rates. c. increase respiration. d. decrease RuBP carboxylation. e. increase RuBP carboxylation. ANSWER: b 93. The reactions of photosynthesis are an important part of the energy conversion pathways in plants. Suppose you are conducting research on photosynthesis with these conditions in the plants you are studying: • There are high concentrations of NADPH in the chloroplasts. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 • There are low concentrations of ATP in the chloroplasts. • There are low intracellular concentrations of glyceraldehyde-3-phosphate (G3P). • The plant is growing in full sunlight. Which of the answer choices would you predict? a. Levels of antioxidants in the cells will be elevated. b. Carbohydrate production will be elevated. c. Cellular metabolic processes will slow due to decreased ATP levels in the chloroplasts. ANSWER: a 94. The only known "solution" to the loss of efficiency caused by the oxygenation of RuBP by rubisco is to raise the concentration of CO2 in the cell using energy from ATP. What would you predict of such plants? a. They would not produce O2. b. They would be tolerant of shade. c. They would often be found in bright, sunny habitats. d. They would not produce reactive O2 species. e. They would require high CO2 levels. ANSWER: c 95. In the lab, you isolate a chlorella mutant that is unable to switch xanthophyll pigments into the inactive, non-heat-dissipating form. If you released that mutant into a small pond shaded by trees, would the mutation spread, and why? a. No, because xanthophyll dissipates absorbed light as heat, and in low light environments, such dissipation would decrease photosynthesis and therefore growth. b. No, because the xanthophylls prevent the production of reactive oxygen species. c. Yes, because the xanthophylls prevent the production of reactive oxygen species. d. Yes, because the heat released by the xanthophylls would warm up the cells and enhance photosynthesis. e. No, because mutations never spread in nature. ANSWER: a Multiple Response 96. Which answer choices are products when photosystem II oxidizes a molecule of H2O? Select all that apply. a. O2 b. protons c. electrons d. CO2 e. hydrogen gas (H2) ANSWER: a, b, c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 97. Under which types of conditions is cellular damage in photosynthetic cells by reactive oxygen species likely to occur? Select all that apply. a. a decrease in the amount of xanthophylls associated with photosystem II b. cold, bright sunny conditions c. the loss of the proton gradient across the thylakoid membrane d. a decrease in the amount of CO2 available to the cell e. an increase in the amount of NADP+ in the chloroplast f. an increase in the amount of antioxidants like ascorbate and beta-carotene g. the loss of antenna chlorophyll molecules in the photosystems ANSWER: a, b, c, d 98. Which answer choices are true about photorespiration? Select all that apply. a. CO2 and ATP are produced. b. Photorespiration increases the efficiency of photosynthesis. c. It happens because rubisco can add either O2 or CO2 to RuBP. d. It happens because there is so much more CO2 in the atmosphere than O2. e. It happens because 3-phosphoglycerate cannot be reduced to form triose phosphate sugars. f. ATP is consumed by photorespiration. ANSWER: c, f 99. Which of the answer choices is true concerning eukaryotic photosynthesis? Select all that apply. a. Electrons excited by photosystem I have more energy than electrons excited by photosystem II. b. The electrons that reduce cytochrome-b6f complex provide energy to move protons across the thylakoid membrane. c. Photosystem II provides electrons with sufficient energy to reduce NADP+ to NADPH. d. Excited electrons associated with photosystem I may be used to reduce NADP+, or NADP+ may be recycled through the electron transport chain. e. Photosystem I accepts electrons from H2O and allows them to enter the electron transport chain. f. The production of ATP by ATP synthase depends, in part, on the movement across the thylakoid membrane of protons that were once part of H2O molecules. ANSWER: a, b, d, f 100. The proton gradient across the thylakoid membrane is required to: Select all that apply. a. reduce NADP+ to NADPH. b. provide the energy for the reaction ADP + Pi → ATP. c. provide a source of protons needed to synthesize carbohydrate in the Calvin cycle. d. provide a source of protons to combine with molecular O2 to make H2O. e. power ATP synthase. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 8 ANSWER: b, e 101. Which of the answer choices could be true if a plant cell is exposed to a toxin that makes the thylakoid membrane freely permeable to protons? Select all that apply. a. There would be no ATP available to drive carbohydrate synthesis in the Calvin cycle. b. The amount of NADPH in the chloroplast would increase. c. There would be no ATP needed to split H2Omolecules early in the light-dependent pathway. d. ATP generated by the mitochondria would be used to power photosynthesis, so plant cells would not have enough ATP left to supply the many other processes for which they require ATP. e. The amount of carbohydrate produced would increase. f. There would be no ATP produced in the chloroplasts to power photosynthesis in the absence of light energy. ANSWER: a, b

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Chapter 9 Multiple Choice 1. Which statement is true about G proteins? a. All of these choices are correct. b. Some G proteins are composed of three subunits. c. G proteins release GDP and bind GTP when associated with an activated receptor. d. G proteins can deactivate themselves by catalyzing the hydrolysis of bound GTP to produce GDP and inorganic phosphate. e. G proteins are a component of the signal-transduction pathways associate with a G protein-coupled receptor. ANSWER: a 2. A protein on a cell surface that binds to a signaling molecule is an example of which element of cellular communication? a. a signaling cell b. a signaling molecule c. a receptor protein d. a responding cell e. None of the other answer options is correct. ANSWER: c 3. Platelet-derived growth factor (PDGF) is a signaling molecule released by platelets at wound sites to cause nearby fibroblasts to divide and contribute to wound healing. Which of the answer choices best describes PDGF signaling? a. endocrine signaling b. contact-dependent signaling c. autocrine signaling d. paracrine signaling e. All of these choices are correct. ANSWER: d 4. When T cells in your immune system recognize a pathogen, they release Interleukin 6 (IL6), which binds back to receptors on the T cell surface. Which of the answer choices best describes this kind of IL6 signaling? a. autocrine signaling b. paracrine signaling c. contact-dependent signaling d. endocrine signaling e. All of these choices are correct. ANSWER: a 5. Kohler and Lipton first discovered platelet-derived growth factor (PDGF) by observing that fibroblasts: a. grew better in cell culture blood plasma without the proteins released by platelets. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 b. grew better in cell culture containing blood serum containing proteins released by platelets during clotting. c. grew at the same rate in cell culture containing either blood plasma or serum. ANSWER: b 6. Which statement is true about the Delta protein? a. All of these choices are correct. b. Delta is produced by embryonic stem cells as they differentiate into neurons in the brain. c. Delta directly signals to a Notch transmembrane protein in adjacent cells. d. Delta directs adjacent cells to differentiate into glial cells. ANSWER: a 7. Which of the answer choices would be considered a cell-surface receptor? a. a protein that acts as an enzyme that attaches phosphate groups to substrates b. a protein that binds a nonpolar steroid hormone and activates transcription c. a protein that causes GDP to be exchanged for GTP in a G protein d. a protein that forms a channel that allows ions to enter the cell when a ligand binds ANSWER: d 8. According to the figure, what is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

a. Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules. b. Cell-surface receptors typically bind to signaling molecules that are smaller than those bound by intracellular receptors. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 c. Cell-surface receptors bind to specific signaling molecules; intracellular receptors bind any signaling molecule. d. Signaling molecules that bind to cell-surface receptors lead to cellular responses restricted to the cytoplasm; signaling molecules that bind to intracellular receptors lead to cellular responses restricted to the nucleus. e. None of the other answer options is correct. ANSWER: a 9. Which type of receptor is involved in rapid responses of muscle cells and neurons? a. intracellular receptor b. receptor kinase c. G protein-coupled receptor d. ligand-gated ion channel ANSWER: d 10. When a ligand binds to a G protein-coupled receptor, which of the answer choices would you expect to happen before any of the others? a. Protein kinase activity increases. b. Adenylyl cyclase activity increases. c. The amount of cAMP in the cytoplasm increases. d. Phosphodiesterase activity to break down cAMP increases. e. None of the other answer options is correct. ANSWER: b 11. Why do the functions of many receptor kinases depend on the fluid nature of the plasma membrane? a. The receptor monomers must move together and dimerize to be activated. b. Binding of ligand to the receptor requires a fluid membrane. c. The generation of cAMP requires a fluid membrane. d. Phosphorylation requires a fluid membrane. ANSWER: a 12. According to the figure shown, phosphate groups play a key role in receptor kinase activation by:

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Chapter 9 a. All of these choices are correct. b. activating the dimerization of the receptor kinase proteins in the membrane. c. activating receptor kinase activity. d. activating the receptor so that the receptor is capable of binding its signal. e. providing binding sites to recruit and activate signal-transduction proteins. ANSWER: e 13. Many mutations in receptor kinases that lead to cancer allow the receptor to dimerize and become active, even in the absence of signaling molecules. An example is a mutant form of the EGF receptor kinase called Her2/neu. An antibody that prevents dimerization of Her2/neu receptor kinases is being tested for its effectiveness in stopping cancer. Which of the answer choices would be blocked if this antibody was bound to the EGF receptor? a. Receptor activation b. Signal transduction c. Cellular response d. All of these choices are correct. ANSWER: d 14. To have communication between cells, you must have a: a. All of these choices are correct. b. signaling molecule. c. responding cell. d. receptor. ANSWER: a 15. Which is a step in cell signaling? a. cellular response b. signal transduction c. receptor activation d. signal termination e. All of these choices are correct. ANSWER: e 16. Use the terms listed to fill in the blanks of each statement that follows. A. signaling cell B. ligand C. receptor D. responding cell In communication between cells, the _____ produces the signaling molecule, also known as the _____; the _____ produces the _____, to which the signaling molecule binds. a. B; C; A; D b. A; B; C; D Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 c. B; A; D; C d. D; C; B; A ANSWER: c 17. Nicotine from cigarette smoke acts as a ligand and associates with the acetylcholine receptor on specific cells in the nervous system. Nicotine eventually produces feelings of pleasure and well-being. The listed statements are descriptions of the events that happen in the cellular response to nicotine. Place the events in the correct order to describe the steps in the signaling pathway. A. The acetylcholine receptor is an ion channel, and when a ligand binds, the ion channel opens. B. An influx of ions carries the signal to the reward areas of the brain. C. Nicotine binds to the transmembrane protein that normally binds the neurotransmitter acetylcholine. D. The signal causes release of dopamine in the brain, which causes good feelings. E. Nicotine is quickly eliminated from the body (causing cravings for more cigarettes to produce good feelings). a. C; A; B; D; E b. A; D; B; C; E c. B; A; C; D; E d. C; D; B; A; E ANSWER: a 18. In which of the statements is cell signaling prevented? a. The nicotine in cigarette smoke binds to and activates the acetylcholine receptor in plasma membranes of neurons in the brain. b. Proteins on the surface of cells in taste buds called umami receptors bind to glutamates and nucleotides in food, which changes the membrane potential of these cells and tells the brain, "This is savory." c. Allergy medicines, which are called antihistamines, bind to and block histamine receptors to keep fluids in capillaries that would normally be released and cause stuffy noses and watery eyes. ANSWER: c 19. You observe a cell that has a receptor protein that binds a signaling molecule that is produced by the same cell. What kind of signaling is occurring in this cell? a. autocrine signaling b. contact-dependent signaling c. paracrine signaling d. endocrine signaling ANSWER: a 20. Of the answer choices, what is the most likely reason that paracrine signaling does not activate the cell that is producing the signaling molecule? a. The cell's receptors have a mutation. b. The neighboring cells are too far away. c. The concentration of the signaling molecule is not high enough. d. The cell does not have the proper receptor. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 ANSWER: d 21. Which type of signaling regulates the differentiation of cells in the nervous system into neurons or glial cells? a. All of these choices are correct. b. paracrine signaling c. autocrine signaling d. endocrine signaling e. contact-dependent signaling ANSWER: e 22. You strip off all proteins on the cell surface by using a protease (an enzyme that destroys proteins). Now, when you add a specific signaling molecule, the cell still responds. What is the most reasonable explanation of this? a. The receptor is a ligand-gated ion channel. b. The signaling molecule doesn't need a receptor. c. The signal can directly activate the second messenger system and does not need to activate the receptor. d. The receptor for this signal is inside the cell, and the signaling molecule is nonpolar and can diffuse into the cell. ANSWER: d 23. Which type of receptor requires a set of accessory proteins to signal? a. G protein-coupled receptors b. receptor kinases c. ligand-gated ion channels d. None of the other answer options is correct. ANSWER: a 24. How would signaling be affected if a mutation caused a G protein to replace GDP with GTP on its own without needing to be activated by the G protein-coupled receptor? a. The signal would not be affected and signaling would only occur when a ligand bound to the receptor. b. The signaling pathway would be activated even when no ligand was present. c. The signaling would be blocked because G protein would be active but unable to signal due to the lack of a bound ligand. d. The signaling would be blocked because replacement of GDP with GTP would deactivate the G protein. ANSWER: b 25. The extent and duration of a cellular response to a signal depend on the: a. binding affinity of the receptor to the signaling molecule. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 b. concentration of the signaling molecule in the vicinity of the receptor. c. level of expression of the genes encoding the signal transduction proteins. d. level of expression of the genes encoding the proteins that terminate the response. e. All of these choices are correct. ANSWER: e 26. What would happen to a signaling pathway if phosphatases had reduced levels of function? a. The response to the signal would persist longer after activation. b. The response to the signal would be shorter after activation. c. There would be no change in the persistence of the response to the pathway. ANSWER: a 27. A number of mutations have been described in the G protein Ras that have profound effects on its activity. For example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the GTPase activity of Ras, preventing it from causing the hydrolysis of GTP to GDP. Which of the scenarios would result in a persisting proliferation response to growth factor receptor activation after the ligand is no longer binding to its receptor kinase? a. a mutation that blocks the exchange of GDP with GTP b. a mutation that blocks the GTPase activity of Ras c. Both a mutation that blocks the GTPase activity of Ras and a mutation that blocks the exchange of GDP with GTP would cause the response to persist. d. Neither a mutation that blocks the GTPase activity of Ras nor a mutation that blocks the exchange of GDP with GTP would cause the response to persist. ANSWER: b 28. Which type of receptor undergoes a conformational change upon activation? a. All of these choices are correct. b. receptor kinase c. ligand-gated ion channel d. intracellular receptor e. G protein-coupled receptor ANSWER: a 29. Which of the answer choices is part of the general response of cells during cellular communication? a. All of these choices are correct. b. Receptors on the outside of the cell bind to specific signal molecules. c. A cell-surface receptor molecule becomes activated by binding to a molecular signal. d. An activated cell-surface receptor transfers the signal to the interior of the cell. e. The signal is transmitted inside the cell and amplified as a series of proteins are activated in sequence, affecting cellular activities according to the type of signal involved. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 30. Vascular endothelial growth factor (abbreviated VEGF and pronounced "Veg-F") is a peptide signaling molecule related to platelet-derived growth factor. VEGF is important in the formation of the circulatory system because its signaling pathway causes the formation of blood vessels in developing embryos during normal development. Tumors also produce and secrete VEGF, causing nearby blood vessels to branch and grow to form new blood vessels that supply these tumors. Given what you know about the different kinds of cell signaling and VEGF, which statement is true? a. VEGF is an endocrine-signaling molecule because it is released from platelets into the bloodstream and is carried throughout the body, causing widespread activation of platelet-derived growth factor receptors on cells in a variety of tissues. b. VEGF is a paracrine signaling molecule because it binds to receptors on nearby cells, at the site where new blood vessels are needed. c. VEGF is an endocrine-signaling molecule because it circulates through the bloodstream inside of platelets. ANSWER: b 31. Which type of cell signaling does not rely on the diffusion of a chemical signal molecule? a. autocrine signaling b. paracrine signaling c. contact-dependent signaling d. endocrine signaling e. All of these choices are correct. ANSWER: c 32. Communication between neurons is an example of which type of cell signaling? a. endocrine signaling b. contact-dependent signaling c. autocrine signaling d. paracrine signaling e. All of these choices are correct. ANSWER: d 33. Which of the answer choices would be found in the cytoplasm and not on the cell surface? a. a receptor that binds a nonpolar steroid hormone and activates transcription b. a receptor that allows ions to enter the cell when a ligand binds c. a receptor that causes GDP to be exchanged for GTP in a G protein d. a receptor that must form a dimer after binding the ligand to transmit a signal ANSWER: a 34. A drug designed to inhibit the response of cells to the steroid testosterone would almost certainly result in which of the answer choices? a. lower levels of cAMP b. a decrease in receptor kinase activity Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 c. a decrease in the transcription of certain genes d. an increase in cytosolic calcium levels e. a decrease in G protein activity ANSWER: c 35. Which one of the answer choices could be found bound to the regulatory region of a gene? a. protein kinase A b. MAP kinase c. adenylyl cyclase d. cAMP e. steroid hormone receptor ANSWER: e 36. How are steroid hormone receptors and cell-surface receptors similar? a. Both types undergo a conformational change when they bind to their ligand. b. Both types cause G proteins to exchange GDP for GTP. c. When bound to their ligand, both types enter the nucleus to activate transcription. d. When activated, both types carry signals across the plasma membrane. ANSWER: a 37. Ion channels can be involved in cell signaling because: a. they interact with G proteins. b. they lead to receptor phosphorylation. c. their signal is amplified in the cell by a series of phosphorylation events. d. they receive signals from other cells and lead to a cellular response. e. Ion channels are not involved in cell signaling. ANSWER: d 38. Despite their differences, steroid hormones: a. bind to cell-surface receptors. b. bind intracellular receptors to form complexes that enter the nucleus. c. have the same effect on different types of cells. d. are hydrophilic small molecules that bind to intracellular receptors. e. facilitate the initiation of translation by ribosomes. ANSWER: b 39. A bacterium releases a toxin that binds to and activates a G protein-coupled receptor in the mammalian gut. The result is high levels of cAMP in these cells. This toxin might work by: a. binding to and activating the G protein-coupled receptor in these cells. b. binding to and inhibiting the G protein-coupled receptor in these cells. c. causing the a subunit of the G protein to stay associated with the b and g subunits. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 d. causing the a subunit of the G protein to bind GDP constantly. e. causing adenylyl cyclase to dissociate from the membrane. ANSWER: a 40. Compared to cells with normal GTP, what would result if a G protein-coupled receptor was activated but treated with a nonhydrolyzable version of GTP (a form of GTP that cannot be converted to GDP) that was taken up by the cells? These treated cells would have: a. increased cAMP and increased protein kinase A (PKA) activity. b. decreased cAMP and decreased protein kinase A (PKA) activity. c. decreased cAMP and increased protein kinase A (PKA) activity. d. increased cAMP and decreased protein kinase A (PKA) activity. ANSWER: a 41. Which of the events are listed in the correct order of G protein-coupled signaling? a. cAMP activates adenylyl cyclase, which activates protein kinase A (PKA). b. Protein kinase A (PKA) phosphorylates adenylyl cyclase, which then synthesizes cAMP from ATP. c. Adenylyl cyclase catalyzes the formation of cAMP, which activates protein kinase A (PKA). ANSWER: c 42. An increased heart rate caused by the release of adrenaline from the adrenal glands, which are located just above the kidney, is an example of _____ signaling. a. autocrine b. contact-dependent c. paracrine d. endocrine ANSWER: d 43. The high variability in the cell-specific expression of different types of G protein-coupled receptors, each with a distinct affinity for specific signaling molecules allows: a. All of these choices are correct. b. signaling cells to activate only certain other cell types. c. amplification of signal response when multiple signals are present. d. specific cell types to respond to multiple different signals. ANSWER: a 44. Phosphatases are a family of enzymes that remove phosphate groups from specific proteins; these phosphate groups had been added to the proteins by protein kinases. Vanadate is an inhibitor of phosphatases in eukaryotic cells. What effect would vanadate have on the response of cells to signals received by receptor kinases? a. The signal would still bind the receptor, so there would be no effect. b. The response of the cell would be shorter than it normally would. c. The response of the cell would last longer than it normally would. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 ANSWER: c 45. Many mutations in receptor kinases that lead to cancer allow the receptor to dimerize and become activated, even in the absence of signaling molecules. An example is a mutant form of the EGF receptor kinase called Her2/neu. An antibody that prevents dimerization of Her2/neu receptor kinases is being tested for its effectiveness in preventing cancer. At which stage does this drug work? a. It prevents the receptor from bonding to the signal. b. It prevents the receptor from becoming activated. c. It prevents the signaling cell from producing the signal. d. It prevents the termination of the signal. ANSWER: b 46. The figure shows how normal signaling works with a Ras protein acting downstream of a receptor kinase. You examine a cell line in which Ras is always activated even in the absence of a signaling molecule. This causes constant activation of the kinases in the MAP kinase pathway. Which condition would be most likely to turn off this abnormally active signaling pathway?

a. The addition of a drug that prevents the final kinase from interacting with its target protein in the nucleus. b. The addition of a drug that increases the binding affinity of Ras for MAP kinase enzymes in the cytoplasm. c. The addition of a drug that prevents the dimerization of the receptor kinase. d. The addition of a drug that prevents the phosphorylation of the receptor kinase. ANSWER: a 47. Defects in cell signaling can lead to a cancerous cell as a result of: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 a. overproduction of signals that trigger cell division. b. overproduction of a receptor that triggers cell division. c. a defective receptor that stays in the activated state and triggers cell division continuously. d. a defective signal transduction protein that stays in the activated state and triggers cell division continuously. e. All of these choices are correct. ANSWER: e 48. A cellular response to a signal can be terminated by: a. All of these choices are correct. b. inactivation of intracellular signal transduction proteins. c. inactivation of proteins in the signal pathway over time. d. depletion of a second messenger. e. depletion of the signal that activates the receptor. ANSWER: a 49. The type of cellular response to a signal depends on: a. the activity of terminator proteins expressed in the cell. b. the type of signal transduction proteins expressed in the cell. c. the activity of other signaling pathways in the cell. d. All of these choices are correct. e. the type of receptor being expressed in the cell. ANSWER: d 50. Vascular endothelial growth factor (abbreviated VEGF and pronounced "Veg-F") is a peptide signaling molecule related to platelet-derived growth factor. VEGF is important in the formation of the circulatory system because its signaling pathway causes the formation of new blood vessels in developing embryos during normal development. Tumors also produce and secrete VEGF, causing the formation of new blood vessels that supply these tumors. Given what you know about the different kinds of cell signaling and about VEGF, which of the answer choices best describes VEGF signaling? a. endocrine signaling b. paracrine signaling c. autocrine signaling d. contact-dependent signaling ANSWER: b 51. Which type of receptor activates a signaling transduction pathway by transferring phosphate groups to a substrate? a. G protein-coupled receptor b. receptor kinase c. ligand-gated ion channel d. intracellular receptor Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 ANSWER: b 52. Nicotine from cigarette smoke acts as a ligand and associates with specific proteins on the surface of cells in the brain, causing feelings of pleasure and well-being. In the following four questions, select the appropriate event in the cellular response to nicotine. The opening of an ion channel as a result of nicotine binding is an example of: a. receptor activation. b. receptor binding. c. signal transduction. d. response. e. termination. ANSWER: a 53. Nicotine from cigarette smoke acts as a ligand and associates with specific proteins on the surface of cells in the brain, causing feelings of pleasure and well-being. In the following four questions, select the appropriate event in the cellular response to nicotine. Nicotine interaction with the transmembrane protein that normally binds the neurotransmitter acetylcholine is an example of: a. receptor binding. b. receptor activation. c. signal transduction. d. response. e. termination. ANSWER: a 54. Nicotine from cigarette smoke acts as a ligand and associates with specific proteins on the surface of cells in the brain, causing feelings of pleasure and well-being. In the following four questions, select the appropriate event in the cellular response to nicotine. What is it called when nicotine binding to the cell signal causes release of dopamine in the brain, which causes relaxation and feelings of well-being? a. termination b. receptor activation c. signal transduction d. receptor binding e. cellular response ANSWER: e 55. Nicotine from cigarette smoke acts as a ligand and associates with specific proteins on the surface of cells in the brain, causing feelings of pleasure and well-being. In the following four questions, select the appropriate event in the cellular response to nicotine. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 Rapid elimination of nicotine from the body is an example of: a. signal transduction. b. receptor activation. c. signal termination. d. cellular response. e. receptor binding. ANSWER: c 56. To see if a growth-promoting factor is released from blood platelets, cells called fibroblasts were placed in a dish that contained liquid cell culture medium. The medium was supplemented with either serum (the liquid component collected from blood that had been allowed to clot) or plasma (the liquid component collected from blood that has not clotted). The number of cells in each culture is shown in the graph.

Which of the answer choices is the dependent variable in this experiment? a. the number of cells b. whether the cells were cultured in serum or plasma c. the number of days the cells were cultured ANSWER: a 57. To see if a growth-promoting factor is released from blood platelets, cells called fibroblasts were placed in a dish that contained liquid cell culture medium. The medium was supplemented with either serum (the liquid component collected from blood that had been allowed to clot) or plasma (the liquid component collected from blood that has not clotted). The number of cells in each culture is shown in the graph.

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Chapter 9

How many cells are there on day three in the culture that contains serum? a. 80,000 b. 800,000 c. 40,000 d. 400,000 ANSWER: b 58. To see if a growth-promoting factor is released from blood platelets, cells called fibroblasts were placed in a dish that contained liquid cell culture medium. The medium was supplemented with either serum (the liquid component collected from blood that had been allowed to clot) or plasma (the liquid component collected from blood that has not clotted). The number of cells in each culture is shown in the graph.

At what point are there approximately seven times more cells in one treatment than in the other? a. day 4 b. day 1 c. day 3 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 d. day 2 ANSWER: d 59. To see if a growth-promoting factor is released from blood platelets, cells called fibroblasts were placed in a dish that contained liquid cell culture medium. The medium was supplemented with either serum (the liquid component collected from blood that had been allowed to clot) or plasma (the liquid component collected from blood that has not clotted). The number of cells in each culture is shown in the graph.

Which group of cells shows a decrease in the rate of cell growth from day 2 to day 3? a. Neither; the rate of both is faster. b. the cells cultured with plasma c. the cells cultured with serum ANSWER: c 60. To see if a growth-promoting factor is released from blood platelets, cells called fibroblasts were placed in a dish that contained liquid cell culture medium. The medium was supplemented with either serum (the liquid component collected from blood that had been allowed to clot) or plasma (the liquid component collected from blood that has not clotted). The number of cells in each culture is shown in the graph.

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Chapter 9 Which treatment shows an increase in number of cells over the entire duration of the experiment? a. cultured with plasma b. cultured with serum ANSWER: a 61. You are doing an experiment on a new class of G protein–coupled receptor. You make several cell lines that each express the receptor, the corresponding G protein, and adenylyl cyclase. Each cell line has different mutations in these components. You decide to measure cyclic AMP (cAMP) in the cells of each cell line to understand the effect of each of these mutations. For each of the descriptions of the mutations, indicate whether the amount of cAMP in the cells will increase, decrease, or stay the same. A mutation in the G protein–coupled receptor that prevents it from binding to the G protein. The amount of cAMP in cells will: a. increase. b. decrease. c. stay the same. ANSWER: b 62. You are doing an experiment on a new class of G protein–coupled receptor. You make several cell lines that each express the receptor, the corresponding G protein, and adenylyl cyclase. Each cell line has different mutations in these components. You decide to measure cyclic AMP (cAMP) in the cells of each cell line to understand the effect of each of these mutations. For each of the descriptions of the mutations, indicate whether the amount of cAMP in the cells will increase, decrease, or stay the same. A mutation in a subunit of the G protein prevents it from binding to adenylyl cyclase but not the receptor. The amount of cAMP in cells will: a. increase. b. decrease. c. stay the same. ANSWER: b 63. You are doing an experiment on a new class of G protein–coupled receptor. You make several cell lines that each express the receptor, the corresponding G protein, and adenylyl cyclase. Each cell line has different mutations in these components. You decide to measure cyclic AMP (cAMP) in the cells of each cell line to understand the effect of each of these mutations. For each of the descriptions of the mutations, indicate whether the amount of cAMP in the cells will increase, decrease, or stay the same. A mutation in a subunit of the G protein prevents the release of bound GDP. The amount of cAMP in cells will: a. increase. b. decrease. c. stay the same. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 64. You are doing an experiment on a new class of G protein–coupled receptor. You make several cell lines that each express the receptor, the corresponding G protein, and adenylyl cyclase. Each cell line has different mutations in these components. You decide to measure cyclic AMP (cAMP) in the cells of each cell line to understand the effect of each of these mutations. For each of the descriptions of the mutations, indicate whether the amount of cAMP in the cells will increase, decrease, or stay the same. A mutation in a subunit of the G protein prevents it from hydrolyzing bound GTP. The amount of cAMP in cells will: a. increase. b. decrease. c. stay the same. ANSWER: a 65. You are doing an experiment on a new class of G protein–coupled receptor. You make several cell lines that each express the receptor, the corresponding G protein, and adenylyl cyclase. Each cell line has different mutations in these components. You decide to measure cyclic AMP (cAMP) in the cells of each cell line to understand the effect of each of these mutations. For each of the descriptions of the mutations, indicate whether the amount of cAMP in the cells will increase, decrease, or stay the same. A mutation of the G protein–coupled receptor removes the extracellular domain of the receptor. The amount of cAMP in cells will: a. increase. b. decrease. c. stay the same. ANSWER: b 66. You are doing an experiment on a new class of G protein–coupled receptor. You make several cell lines that each express the receptor, the corresponding G protein, and adenylyl cyclase. Each cell line has different mutations in these components. You decide to measure cyclic AMP (cAMP) in the cells of each cell line to understand the effect of each of these mutations. For each of the descriptions of the mutations, indicate whether the amount of cAMP in the cells will increase, decrease, or stay the same. A mutation of the G protein–coupled receptor allows it to bind a subunit of the G protein in the absence of ligand. The amount of cAMP in cells will: a. increase. b. decrease. c. stay the same. ANSWER: a 67. A number of mutations have been described in G proteins that have profound effects on their activity. For example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the hydrolysis of GTP to GDP. For each of the statements that describe a malfunction in cellular response, indicate the type of mutation that is responsible. The MAP kinase pathway is continually activated, even in the absence of a signaling molecule. Which type of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 mutation could be responsible for this malfunction in cellular response? a. a Ras mutant with high GDP affinity b. a Ras mutant that cannot hydrolyze GTP to GDP c. Neither of the other answer options is correct. ANSWER: b 68. A number of mutations have been described in G proteins that have profound effects on their activity. For example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the hydrolysis of GTP to GDP. For each of the statements that describe a malfunction in cellular response, indicate the type of mutation that is responsible. Following ligand binding to receptor kinase, the MAP kinase pathway fails to activate. Which of the types of mutation listed is responsible for this malfunction in cellular response? a. a Ras mutant with high GDP affinity b. a Ras mutant that cannot hydrolyze GTP to GDP c. Neither of the other answer options is correct. ANSWER: a 69. A number of mutations have been described in G proteins that have profound effects on their activity. For example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the hydrolysis of GTP to GDP. For each of the statements that describe a malfunction in cellular response, indicate the type of mutation that is responsible. The receptor kinase fails to dimerize in the presence of ligand. Which type of mutation is responsible for this malfunction in cellular response? a. a Ras mutant with high GDP affinity b. a Ras mutant that cannot hydrolyze GTP to GDP c. Neither of the other answer options is correct. ANSWER: c 70. A number of mutations have been described in G proteins that have profound effects on their activity. For example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the hydrolysis of GTP to GDP. For each of the statements that describe a malfunction in cellular response, indicate the type of mutation that is responsible. Phosphorylation of the receptor kinase occurs when it dimerizes and it is bound by ligand. Some mutations can cause the receptor to be constantly phosphorylated. Which of the types of mutation could be responsible for persistent receptor phosphorylation? a. a Ras mutant with high GDP affinity b. a Ras mutant that cannot hydrolyze GTP to GDP c. Neither of the other answer options is correct. ANSWER: c 71. A number of mutations have been described in G proteins that have profound effects on their activity. For Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the hydrolysis of GTP to GDP. For each of the statements that describe a malfunction in cellular response, indicate the type of mutation that is responsible. An absence of the appropriate signaling molecule causes a change in the expression of genes. Which type of mutation is responsible for this malfunction in cellular response? a. a Ras mutant with high GDP affinity b. a Ras mutant that cannot hydrolyze GTP to GDP c. Neither of the other answer options is correct. ANSWER: b 72. A number of mutations have been described in G proteins that have profound effects on their activity. For example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the hydrolysis of GTP to GDP. For each of the statements that describe a malfunction in cellular response, indicate the type of mutation that is responsible. Target genes are not expressed, even in the presence of signaling molecule. Which type of mutation is responsible for this malfunction in cellular response? a. a Ras mutant with high GDP affinity b. a Ras mutant that cannot hydrolyze GTP to GDP c. Neither of the other answer options is correct. ANSWER: a 73. A number of mutations have been described in G proteins that have profound effects on their activity. For example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the hydrolysis of GTP to GDP. For each of the statements that describe a malfunction in cellular response, indicate the type of mutation that is responsible. A mutation causes an increase in kinase activity in the cytoplasm. Which type of mutation could be responsible for this increase in cytoplasmic kinase activity? a. a Ras mutant with high GDP affinity b. a Ras mutant that cannot hydrolyze GTP to GDP c. Neither of the other answer options is correct. ANSWER: b 74. A number of mutations have been described in G proteins that have profound effects on their activity. For example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the hydrolysis of GTP to GDP. For each of the statements that describe a malfunction in cellular response, indicate the type of mutation that is responsible. There is no increase in cytosolic kinase activity as a result of ligand binding. Which type of mutation is responsible for this malfunction in cellular response? a. a Ras mutant with high GDP affinity b. a Ras mutant that cannot hydrolyze GTP to GDP Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 c. Neither of the other answer options is correct. ANSWER: a 75. A number of mutations have been described in G proteins that have profound effects on their activity. For example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the hydrolysis of GTP to GDP. For each of the statements that describe a malfunction in cellular response, indicate the type of mutation that is responsible. There is no change in membrane potential, even in the presence of a signaling molecule. Which type of mutation is responsible for this malfunction in cellular response? a. a Ras mutant with high GDP affinity b. a Ras mutant that cannot hydrolyze GTP to GDP c. Neither of the other answer options is correct. ANSWER: c 76. A number of mutations have been described in G proteins that have profound effects on their activity. For example, some mutations greatly increase the affinity of the G protein for GDP, making it very difficult for GDP to be exchanged for GTP. Other mutations prevent the hydrolysis of GTP to GDP. For each of the statements that describe a malfunction in cellular response, indicate the type of mutation that is responsible. A receptor that is continuously activated in the absence of ligand. Which type of mutation is responsible for this malfunction in cellular response? a. a Ras mutant with high GDP affinity b. a Ras mutant that cannot hydrolyze GTP to GDP c. Neither of the other answer options is correct. ANSWER: c 77. Insulin signaling through its receptor kinase enables virtually every cell in our body to transport glucose across the plasma membrane into the cytosol. In the next set of questions, select the appropriate event in the cellular response to insulin. Phosphorylation of transmembrane proteins caused by insulin binding to those proteins is an example of: a. receptor binding. b. receptor activation. c. signal transduction and amplification. d. response. e. termination. ANSWER: b 78. Insulin signaling through its receptor kinase enables virtually every cell in our body to transport glucose across the plasma membrane into the cytosol. In the next set of questions, select the appropriate event in the cellular response to insulin. After insulin binds to the cell-surface protein, cytoplasmic proteins are phosphorylated. What process occurs Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 when those proteins are dephosphorylated? a. receptor binding b. receptor activation c. signal transduction and amplification d. response e. termination ANSWER: e 79. Insulin signaling through its receptor kinase enables virtually every cell in our body to transport glucose across the plasma membrane into the cytosol. In the next set of questions, select the appropriate event in the cellular response to insulin. The noncovalent association of insulin with specific proteins on the surface of a cell is an example of: a. receptor binding. b. receptor activation. c. signal transduction and amplification. d. response. e. termination. ANSWER: a 80. Insulin signaling through its receptor kinase enables virtually every cell in our body to transport glucose across the plasma membrane into the cytosol. In the next set of questions, select the appropriate event in the cellular response to insulin. Activated proteins in the nucleus that cause transcription (and eventually translation in the cytosol) of proteins needed for cell division is an example of: a. receptor binding. b. receptor activation. c. signal transduction and amplification. d. response. e. termination. ANSWER: d 81. Insulin signaling through its receptor kinase enables virtually every cell in our body to transport glucose across the plasma membrane into the cytosol. In the next set of questions, select the appropriate event in the cellular response to insulin. What occurs after the insulin receptors phosphorylate each other, making it possible for other proteins in the cytoplasm to bind to them? a. receptor binding b. receptor activation c. signal transduction and amplification d. response Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 e. termination ANSWER: c 82. Insulin signaling through its receptor kinase enables virtually every cell in our body to transport glucose across the plasma membrane into the cytosol. In the next set of questions, select the appropriate event in the cellular response to insulin. What occurs when the phosphorylated cytosolic proteins bind to other cytosolic proteins to activate them, and these activated cytosolic proteins in turn bind to and activate others? a. receptor binding b. receptor activation c. signal transduction and amplification d. response e. termination ANSWER: c 83. Insulin signaling through its receptor kinase enables virtually every cell in our body to transport glucose across the plasma membrane into the cytosol. In the next set of questions, select the appropriate event in the cellular response to insulin. One consequence of insulin binding its receptor is the uptake of glucose by the cell, which is an example of: a. receptor binding. b. receptor activation. c. signal transduction and amplification. d. response. e. termination. ANSWER: d Multiple Response 84. Which of the functions in termination of a cell-signaling event are initiated by a G protein-coupled receptor? Select all that apply. a. phosphatase activity b. protein kinase activity c. adenylyl cyclase activity d. phosphodiesterase activity that breaks down cAMP e. GTP activity to convert GTP to GDP ANSWER: a, d, e 85. According to the figure shown, which of the answer choices is responsible for amplifying the signal in a cell responding to adrenaline? Select all that apply.

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Chapter 9

a. One adrenaline molecule binds to change the conformation of the receptor protein. b. One molecule of adenylyl cyclase can activate many molecules of protein kinase A (PKA). c. One molecule of protein kinase A (PKA) can phosphorylate and activate many target proteins. d. One activated receptor can activate many G proteins. ANSWER: b, c, d 86. Which statements are true about G proteins? Select all that apply. a. Some G proteins are composed of three subunits. b. All G proteins associate with G protein-coupled receptors. c. G proteins release GDP and bind GTP when associated with an activated receptor. d. G proteins become deactivated when bound GTP is hydrolyzed to GDP. e. All of these choices are correct. ANSWER: a, c, d 87. Which type of receptor is membrane-associated? Select all that apply. a. G protein-coupled receptor b. receptor kinase c. ligand-gated ion channel d. intracellular receptor ANSWER: a, b, c 88. Which steps in a signaling cascade initiated by a G protein-coupled receptor would be directly affected in cells cultured in the presence of a non-hydrolyzable ATP analog? (The analog could bind to other molecules but could not be converted to ADP and Pi.) Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 a. protein phosphatase activity b. protein kinase activity c. adenylyl cyclase activity d. phosphodiesterase activity ANSWER: b, c 89. Which of the answer choices are responsible for the amplification of an intracellular signal? Select all that apply. a. protein kinase activity b. protein phosphatase activity c. phosphodiesterase activity that breaks down cAMP d. adenylyl cyclase activity ANSWER: a, d 90. Which of the answer choices correctly pair enzymes that activate and terminate the same step in a signaling event? Select all that apply. a. adenylyl cyclase (activates) and phosphodiesterase (terminates) b. protein kinase (activates) and phosphatase (terminates) c. protein kinase (activates) and phosphodiesterase (terminates) d. adenylyl cyclase (activates) and phosphatase (terminates) ANSWER: a, b 91. You are a medical student working in a cancer research laboratory. Your boss has given you samples of cancer cells and blood from a patient. She has told you that the cells have no mutations in any of their signaling molecule proteins (that is, all the proteins are normal), but that the cells' signaling transduction properties are hyperactive, causing them to divide too rapidly. Which answer choices could account for the increased cell proliferation in the patient? Select all that apply. a. None of the other answer options is correct. b. The concentration of a signaling molecule in the patient is higher than usual. c. The cells taken from the patient exhibit a reduced level of signal amplification. d. The cells taken from the patient have more growth factor receptors than normal. ANSWER: b, d 92. You are studying a newly discovered growth factor. You find that this growth factor stimulates the proliferation of cells grown in the laboratory. You have also found that the receptor that binds the growth factor is a receptor kinase that activates Ras, which activates the MAP kinase pathway. Which mutations affecting this growth factor pathway would you expect to promote uncontrolled cell proliferation? Select all that apply. a. a mutation that prevents entry into the nucleus of the last kinase in the MAP kinase pathway. b. a mutation that blocks the kinase activity of the receptor c. a mutation that inactivates the phosphatase that dephosphorylates the activated receptor d. a mutation that prevents the GTPase activity of Ras. e. a mutation that prevents dimerization of the receptor. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 9 ANSWER: c, d 93. Which statements about communication among bacteria are correct? Select all that apply. a. At high population density, a high concentration of signaling molecules involved in DNA uptake is typically observed. b. Unlike communication among eukaryotes, no receptor molecule is required in communication among bacteria. c. Small peptides can stimulate a DNA-uptake response. d. Cellular communication among bacteria is based on the same principles as cellular communication among eukaryotic cells. ANSWER: a, c, d 94. Which statements describe a part of the MAP kinase pathway? Select all that apply. a. A receptor kinase that is activated by binding to a growth factor. b. The activated receptor kinase that forms a dimer. c. The G protein that is activated by binding to GTP. d. Adenylyl cyclase that converts ATP to cAMP. e. An activated kinase that enters the nucleus to activate transcription. ANSWER: a, b, c, e 95. Complexity in cell communication from molecular crosstalk can be due to: Select all that apply. a. inhibition of one signaling pathway by a component of another signaling pathway. b. an intersection of two signaling pathways that allow two signals to enhance the cellular response. c. signaling molecules interacting with each other before binding the receptor. d. None of the other answer options is correct. ANSWER: a, b

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Chapter 10 Multiple Choice 1. Which of the choices enable(s) the movement of cells? a. microfilaments only b. microtubules and microfilaments c. intermediate filaments only d. microtubules only e. microfilaments and intermediate filaments ANSWER: b 2. Microtubules and microfilaments are said to be "dynamic" elements of the cytoskeleton. In this case, "dynamic" means that: a. microtubules and microfilaments are assembled, disassembled, and then reassembled again in a regulated manner. b. microtubules and microfilaments are constantly changing and are never in the same place for very long. c. once individual microfilaments and microtubules have formed, they can change shape and move around inside the cell. ANSWER: a 3. Microtubules increase in length: a. All of these choices are correct. b. by lengthening outward from the centrosome. c. in cycles, following rapid depolymerization. d. by adding tubulin dimers to the ends. e. more quickly at one end than the other. ANSWER: a 4. Epidermolysis bullosa is a set of rare genetic disorders that is caused by: a. All of these choices are correct. b. a microfilament gene that disrupts desmosomes, weakening connections between epidermal cell layers. c. an intermediate filament gene that disrupts hemidesmosomes, weakening connections between epidermal cell layers. d. a keratin gene that disrupts microfilaments, weakening connections between epidermal cell layers. e. a keratin gene that disrupts intermediate filaments, weakening connections between epidermal cell layers. ANSWER: e 5. Dynein motor proteins use the energy from ATP to: a. carry vesicles along a microtubule within a cell in a minus-to-plus direction. b. carry vesicles along a microfilament within a cell in a minus-to-plus direction. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 c. carry vesicles along a microtubule within a cell in a plus-to-minus direction. d. carry vesicles along a microfilament within a cell in a plus-to-minus direction. e. slide along microfilaments to contract muscle cells. ANSWER: c 6. At night time, Zebrafish embryos change the color of the melanophore cells in their skin to a darker shade by: a. using dynein motor proteins to move pigment granules outward from the center of the cell. b. using kinesin motor proteins to move pigment granules outward from the center of the cell. c. using dynein motor proteins to move pigment granules inward toward the center of the cell. d. using kinesin motor proteins to move pigment granules inward toward the center of the cell. e. using dynein motor proteins to move pigment granules inward toward the plus end of microtubules. ANSWER: b 7. Which of the statements are true of motile cilia? a. All of these choices are correct. b. Motile cilia propel the movement of cells or fluid surrounding cells. c. Motile cilia help protists such as paramecium to move. d. Motile cilia contain microtubules. e. Motile cilia wave back and forth because dynein moves microtubes that are fixed in place. ANSWER: a 8. Which of the pairs of cell junctions perform the most similar functions? a. desmosomes and plasmodesmata b. tight junctions and desmosomes c. tight junctions and adherens junctions d. desmosomes and adherens junctions ANSWER: d 9. Which one of the choices matches a type of cell junction with the correct cytoskeletal element and cell adhesion molecules? a. desmosome, intermediate filament, integrin b. adherens junction, microfilament, integrin c. hemidesmosome, intermediate filament, integrin d. hemidesmosome, intermediate filament, cadherin ANSWER: c 10. A decrease in cell-cell adhesion was caused by the introduction of an experimental substance that compromised the structural integrity of the tissue. Which type of cell junction would most likely be affected by this treatment given that the treatment caused a decrease in the strength of cell-cell adhesion? a. adherens junctions b. gap junctions Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 c. hemidesmosomes d. tight junctions ANSWER: a 11. How does an adherens junction differ from a desmosome? a. Whereas both adherens junctions and desmosomes connect cells using cadherin proteins, adherens junctions connect to microfilaments in the cytoplasm and desmosomes connect to intermediate filaments. b. An adherens junction connects cells using adherin proteins, and a desmosome connects cells using cadherin proteins. c. Whereas both adherens junctions and desmosomes connect cells using cadherin proteins, only an adherens junction connects to the cytoskeleton. d. An adherens junction connects a cell to neighboring cells, and a desmosome connects a cell to the extracellular matrix. e. These two types of junctions only differ in their bandlike versus button-like architecture in the plasma membrane. ANSWER: a 12. Which of the choices correctly lists the sequence in which the plant cell wall is synthesized? a. middle lamella → secondary cell wall → primary cell wall b. secondary cell wall → middle lamella → primary cell wall c. secondary cell wall → primary cell wall → middle lamella d. primary cell wall → secondary cell wall → middle lamella e. middle lamella → primary cell wall → secondary cell wall ANSWER: e 13. Which of the choices is a factor in determining a cell's particular shape? a. cytoskeletal protein networks in the cytoplasm b. a mesh of proteins and polysaccharides in the extracellular matrix c. cytosolic proteins that assemble into cell junctions d. All of these choices are correct. ANSWER: d 14. Which of the choices is only found in the extracellular matrix of plant cells? a. elastin b. cellulose c. collagen d. laminin e. polysaccharide ANSWER: b 15. Metastatic cancer cells: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 a. are able to leave the tumor where they originated and travel to distant locations to begin new tumors at distant sites. b. have lost their adhesion to the extracellular matrix, freeing them from the original tumor. c. become malignant when they grow rapidly. d. only need to cross a single layer of capillary endothelial cells to form tumors at distant sites. e. All of these choices are correct. ANSWER: a 16. Cells differentiate into multiple tissue types because each tissue has its own unique DNA. a. true b. false ANSWER: b 17. Which of the statements about mammalian skin is incorrect? a. The epidermis is an outer layer that provides a water-resistant, protective barrier. b. The dermis is the layer below the epidermis containing connective tissue, blood vessels, and nerve endings that provides nutrients to the epidermis and a cushion layer for the body. c. Epithelial cells in the epidermis are primarily composed of keratinocytes that protect underlying tissue. d. A specialized form of extracellular matrix, called the basal lamina, underlies the dermis and separates it from tissues below it. e. Fibroblasts in the dermis produce extracellular matrix proteins to make the dermis strong and flexible. ANSWER: d 18. What would happen in a cell if its gene for α-tubulin was mutated so it was unable to bind to its β-tubulin? a. The cell would be unaffected. b. The cell would have no microfilaments. c. The cell would have no microtubules. d. The cell would have no intermediate filaments. ANSWER: c 19. What would happen to a cell that depends on motile cilia for movement if it had no actin? a. Nothing; motile cilia contain microtubules, not microfilaments. b. The cell would not be able to move. c. The cell would not be able to move because dynein and kinesin could not work. d. The cell might be able to move if it used actin and myosin to change shape. ANSWER: a 20. If cells did not have actin, they would be unable to: a. maintain shape. b. All of these choices are correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 c. connect properly to neighboring cells. d. organize proteins associated with the cytoplasmic membrane. ANSWER: b 21. Why are the ends of microfilaments and microtubules called "plus ends" and "minus ends"? a. The actin and tubulin monomers that make up microfilaments and microtubules have evolved more rapidly than most other proteins. b. Polymerization occurs at one end (the "plus end") and depolymerization occurs at the other end (the "minus end"). c. Monomers can be added to one end only (the "plus end"), and can only be removed from the other end (the "minus end"). d. Monomers are added more quickly to one end (the "plus end") than they are to the other end (the "minus end"). ANSWER: d 22. Which one of the statements about intracellular transport is true? a. Kinesin and dynein move substances along microfilaments. b. Myosin moves substances along microfilaments. c. Myosin and dynein move substances along microfilaments. d. Kinesin and myosin move substances along microtubules. ANSWER: b 23. Evidence that cytoskeletal elements have ancient origins comes from the: a. All of these choices are correct. b. presence of cytoskeletal elements in both eukaryotes and prokaryotes. c. observation that both prokaryotes and eukaryotes use cytoskeletal elements to assist cell division. d. ability to form functional hybrid microfilaments from actin monomers taken from distantly related organisms. e. sequence similarities of cytoskeletal elements when comparing distantly related organisms. ANSWER: a 24. Kinesin motor proteins use the energy of ATP to: a. carry vesicles along a microtubule within a cell in a minus-to-plus direction. b. carry vesicles along a microfilament within a cell in a minus-to-plus direction. c. carry vesicles along a microtubule within a cell in a plus-to-minus direction. d. carry vesicles along a microfilament within a cell in a plus-to-minus direction. e. slide along microfilaments to contract muscle cells. ANSWER: a 25. Motile cilia move in a(n): a. whiplike motion by the motor protein kinesin sliding microtubules past each other. b. oarlike motion by the motor protein kinesin sliding microtubules past each other. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 c. oarlike motion by the motor protein dynein sliding microtubules past each other. d. whiplike motion by the motor protein dynein sliding microtubules past each other. e. whiplike motion by the motor protein dynein sliding microfilaments past each other. ANSWER: d 26. Which cell adhesion molecule is particularly good at attaching the cell to the basement membrane? a. myosin b. cadherins c. integrins d. actin ANSWER: c 27. Which type of cell junction prevents the movement of substances through the space between cells? a. gap junctions b. adherens junctions c. desmosomes d. hemidesmosomes e. tight junctions ANSWER: e 28. Suppose there is a mutation in a laminin-binding integrin gene that causes a loss of function in the cytoplasmic domains of the integrin. Which one of the outcomes would you expect to observe as a result of this mutation? a. The strength of tissues would decrease because the integrin could no longer associate with microfilaments. b. The integrin would function normally because the cytoplasmic domain is not responsible for binding to laminin. c. The integrin would adhere to laminin and the cytoplasmic domain would adhere to microtubules instead of microfilaments. ANSWER: a 29. Which one of the statements about integrins is correct? a. Integrins are the primary cell adhesion molecule in a desmosome. b. There is a single type of integrin, capable of binding to multiple types of extracellular matrix proteins. c. Integrins indirectly connect microfilaments with the extracellular matrix. d. Adjacent integrins create channels referred to as gap junctions. ANSWER: c 30. The dermis of mammalian skin, including that of humans, contains a large amount of extracellular matrix proteins, including collagen and elastin. As a person ages, the amount of elastin in the dermis declines. Which of the cell types is likely to be the cause of this change in the dermis of the skin? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 a. fibroblasts b. keratinocytes c. melanocytes d. epidermal cells ANSWER: a 31. The extracellular matrix can influence: a. cell movement. b. cell shape. c. gene expression. d. All of these choices are correct. ANSWER: d 32. The extracellular matrix in animals: a. is attached to cells by gap junctions. b. is formed by many types of cells, primarily fibroblasts, on their cell surfaces. c. is a meshwork of insoluble fibrous proteins, primarily collagens, that assembles into a double helix. d. is located exclusively in muscle tissue. e. differs from one tissue to the next, affecting the expression of genes in the cells that it surrounds. ANSWER: e 33. Epithelial tissue is found: a. lining the lungs. b. lining the gastrointestinal tract. c. All of these choices are correct. d. lining blood vessels. e. on the outer surface of the body. ANSWER: c 34. The cytoskeleton of animal cells includes: a. microtubules, microfilaments, and intermediate filaments. b. actin, microfilaments, and intermediate filaments. c. actin, microtubules, and intermediate filaments. d. tubulin, microfilaments, and intermediate filaments. e. microtubules and microfilaments. ANSWER: a 35. Which of the cytoskeletal elements contribute to the structural integrity and strength of epithelial tissues in animals? a. intermediate filaments and microtubules only b. intermediate filaments and microfilaments and microtubules Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 c. microtubules and microfilaments only ANSWER: b 36. Microfilaments increase in length: a. more quickly at one end than the other. b. by assembling outward from the centrosome. c. in cycles of slower polymerization following rapid depolymerization. d. if free tubulin dimers are available. e. All of these choices are correct. ANSWER: a 37. Dynamic instability of microtubules: a. All of these choices are correct. b. results from a deficiency of free tubulin subunits. c. is caused by the large size of this type of cytoskeletal element. d. is shared by no other cytoskeletal elements. e. allows microtubules to perform their functions. ANSWER: e 38. If a mutation occurred in the cadherin gene so that the cytoplasmic domain no longer attached to the cytoskeleton, which of the outcomes would occur? a. Cadherins in desmosomes would no longer be anchored to intermediate filaments. b. Cadherins in adherens junctions would no longer be anchored to microtubules. c. Cadherins in hemidesmosomes would no longer be anchored to intermediate filaments. d. Cadherins in desmosomes would no longer be anchored to microfilaments. ANSWER: a 39. Which one of the choices matches a type of cell junction with its proper cytoskeletal element and cell adhesion molecule? a. desmosome, microfilament, integrin b. adherens junction, microfilament, cadherin c. tight junction, cadherin, intermediate filament d. desmosome, microtubule, integrin ANSWER: b 40. Which one of the choices properly matches a type of cell junction with its proper cytoskeletal element and a cell adhesion molecule? a. desmosome, intermediate filament, cadherin b. hemidesmosome, microfilament, cadherin c. adherens junction, microfilament, integrin d. hemidesmosome, intermediate filament, cadherin Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 ANSWER: a 41. Which one of the choices is a characteristic shared by integrins and cadherins? a. Both are peripheral membrane proteins. b. Both facilitate the adhesion of cells to extracellular matrix proteins. c. Both are present in adherens junctions, desmosomes, and hemidesmosomes. d. Both proteins have a cytoplasmic domain connected to the cytoskeleton. ANSWER: d 42. Suppose that frog embryos are genetically manipulated so that the epidermal cells express both E-cadherin and N-cadherin on their cell surface. The neuronal tissue and epidermal tissues are collected from the embryos and treated as illustrated in the figure shown. Which of the choices would you predict would happen now that N-cadherin is also present on the surface of the epidermal cells?

a. The genetically modified epidermal cells would adhere to neuronal cells and would no longer adhere to epidermal cells. b. The genetically modified epidermal cells would adhere to each other and to neuronal cells. c. The genetically modified epidermal cells would adhere to each other, and not to the neuronal cells, because they would still be epidermal cells regardless of the presence of an additional cell adhesion protein on their surface. ANSWER: b 43. In order for a tumor cell to metastasize, which of the changes would you expect to happen? a. a decrease in cadherin expression and an increase of integrin expression b. an increase of cadherin expression and a decrease in integrin expression c. an increase in both cadherin and integrin expression d. a decrease in both cadherin and integrin expression ANSWER: a 44. In a study of tadpole coloration, you noticed that a certain percentage of tadpoles in a population displayed a decreased ability to shift from dark coloration at night to light coloration during the day. These individuals were Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 studied and found to have the normal number of melanophores and these melanophores produced normal amounts of melanin pigment granules. You remember reading that pigment granule transport in melanophores involves the cytoskeleton. Which part of the cytoskeleton would you suggest investigating as a potential source of the faulty color adjustment in these tadpoles? a. intermediate filaments b. microfilaments c. microtubules ANSWER: c 45. In a study of tadpole coloration, you noticed that a certain percentage of tadpoles in a population displayed a decreased ability to shift from dark coloration at night to light coloration during the day. These individuals were studied and found to have the normal number of melanophores and these melanophores produced normal amounts of melanin pigment granules. You remember reading that pigment granule transport in melanophores involves the cytoskeleton. You investigated the cytoskeleton and found that all three elements, including the one you suspected, are completely normal. You then decide to look at the motor proteins to see if the cause of the abnormality might lie with one of them. Which one of the statements would be your hypothesis? a. The melanophores in these tadpoles have partially functional or nonfunctional dynein that would normally transport organelles along microtubules to the center of the cell. b. The melanophores in these tadpoles have partially functional or nonfunctional kinesin that would normally transport organelles along microfilaments to the center of the cell. c. The melanophores in these tadpoles have partially functional or nonfunctional kinesin that would normally transport organelles along microtubules to the center of the cell. d. The melanophores in these tadpoles have partially functional or nonfunctional dynein that would normally transport organelles along microfilaments to the center of the cell. e. The melanophores in these tadpoles have partially functional or nonfunctional myosin that would normally transport organelles along microfilaments to the center of the cell. ANSWER: a 46. Biologists spend a great deal of time studying the cytoskeleton of cells. Some of the tools they use in their research include chemicals that interfere with the structure of the cytoskeletal elements or with the activities of the motor proteins that associate with the cytoskeleton to facilitate movement. Four examples of such interfering agents include: • • • •

cytochalasins: disrupt the normal assembly and disassembly of microfilaments. colchicine: binds to tubulin and prevents the assembly of microtubules. EHNA (erythro-9-[3-2-(hydroxynonyl)] adeninE.: interferes with the function of dynein. blebbistatin: inhibits the activity of myosin.

Where would you predict cytochalasins to have the greatest effect? a. sperm cell motility b. fibroblast migration Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 c. the shortening of a muscle cell d. chromosome segregation e. Cytochalasins would likely have no effect on any of these cells or processes. ANSWER: b 47. Biologists spend a great deal of time studying the cytoskeleton of cells. Some of the tools they use in their research include chemicals that interfere with the structure of the cytoskeletal elements or with the activities of the motor proteins that associate with the cytoskeleton to facilitate movement. Four examples of such interfering agents include: • • • •

Where would you predict EHNA to have the greatest effect? a. sperm cell motility b. fibroblast migration c. the shortening of a muscle cell d. chromosome segregation e. EHNA would likely have no effect on any of these cells or processes. ANSWER: a 48. Biologists spend a great deal of time studying the cytoskeleton of cells. Some of the tools they use in their research include chemicals that interfere with the structure of the cytoskeletal elements or with the activities of the motor proteins that associate with the cytoskeleton to facilitate movement. Four examples of such interfering agents include: • • • •

Where would you predict colchicine to have the greatest effect? a. fibroblast migration b. sperm cell motility c. chromosome segregation d. the shortening of a muscle cell e. Colchicine would likely have no effect on any of these cells or processes. ANSWER: c 49. Biologists spend a great deal of time studying the cytoskeleton of cells. Some of the tools they use in their research include chemicals that interfere with the structure of the cytoskeletal elements or with the activities of the motor proteins that associate with the cytoskeleton to facilitate movement. Four examples of such Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 interfering agents include: cytochalasins: disrupt the normal assembly and disassembly of microfilaments. colchicine: binds to tubulin and prevents the assembly of microtubules. EHNA (erythro-9-[3-2-(hydroxynonyl)] adeninE.: interferes with the function of dynein. blebbistatin: inhibits the activity of myosin. Where would you predict blebbistatin to have the greatest effect? a. the shortening of a muscle cell b. sperm cell motility c. fibroblast migration d. chromosome segregation e. Blebbistatin would likely have no effect on any of these cells or processes. ANSWER: a 50. INSTRUCTIONS: Read the passage. Notice that there are numbers at the end of some of the sentences. Refer to these numbers when answering the questions. Melanoma cells metastasize by leaving the primary tumor and entering the bloodstream, which carries them to distant parts of the body. (1) The cancer cells leave the bloodstream by crossing the capillary basal lamina to enter another tissue. (2) As a cancer researcher, you think that somehow preventing the cancer cells from crossing the basal lamina of the blood vessels will inhibit metastasis. (3) You believe that if you block the integrin receptors on the melanoma cells from binding to the extracellular matrix proteins of the basal lamina, then the cells will not be able to adhere to the basal lamina and therefore will not be able to migrate across it

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Chapter 10

(4) You and your colleagues make antibodies, proteins that bind specifically to only one type of protein in the cell, and your antibodies bind specifically to the integrins on the surface of the melanoma cells, preventing them from adhering to the proteins in the basal lamina. Once you have these integrin-specific antibodies, you test them on cancer cells in culture to see if they prevent the cells from crossing an artificial basal lamina. To do this, you and you colleagues set up nine cultures of cancer cells. You treat three cultures with the integrinspecific antibody. You treat three more cultures with a nonspecific antibody that does not bind to integrins. Three cultures of cells are not treated with antibodies. In the passage, sentence number (1) is: a. a hypothesis. b. proof that their idea about blocking integrins will prevent metastasis. c. support for their idea that blocking integrin function inhibits some tumor cells from binding the basal lamina. d. a prediction. e. an observation. ANSWER: e 51. INSTRUCTIONS: Read the passage. Notice that there are numbers at the end of some of the sentences. Refer to these numbers when answering the questions. Melanoma cells metastasize by leaving the primary tumor and entering the bloodstream, which carries them to distant parts of the body. (1) The cancer cells leave the bloodstream by crossing the capillary basal lamina to enter another tissue. (2) As a cancer researcher, you think that somehow preventing the cancer cells from crossing the basal lamina of the blood vessels will inhibit metastasis. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 (3) You believe that if you block the integrin receptors on the melanoma cells from binding to the extracellular matrix proteins of the basal lamina, then the cells will not be able to adhere to the basal lamina and therefore will not be able to migrate across it

(4) You and your colleagues make antibodies, proteins that bind specifically to only one type of protein in the cell, and your antibodies bind specifically to the integrins on the surface of the melanoma cells, preventing them from adhering to the proteins in the basal lamina. Once you have these integrin-specific antibodies, you test them on cancer cells in culture to see if they prevent the cells from crossing an artificial basal lamina. To do this, you and you colleagues set up nine cultures of cancer cells. You treat three cultures with the integrinspecific antibody. You treat three more cultures with a nonspecific antibody that does not bind to integrins. Three cultures of cells are not treated with antibodies. In the passage, sentence number 2 is: a. an observation. b. support for their idea that blocking integrin function inhibits some tumor cells from binding the basal lamina. c. proof that their idea about blocking integrins will prevent metastasis. d. a hypothesis. e. a prediction. ANSWER: d 52. INSTRUCTIONS: Read the passage. Notice that there are numbers at the end of some of the sentences. Refer to these numbers when answering the questions. Melanoma cells metastasize by leaving the primary tumor and entering the bloodstream, which carries them to distant parts of the body. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 (1) The cancer cells leave the bloodstream by crossing the capillary basal lamina to enter another tissue. (2) As a cancer researcher, you think that somehow preventing the cancer cells from crossing the basal lamina of the blood vessels will inhibit metastasis. (3) You believe that if you block the integrin receptors on the melanoma cells from binding to the extracellular matrix proteins of the basal lamina, then the cells will not be able to adhere to the basal lamina and therefore will not be able to migrate across it

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Chapter 10 53. INSTRUCTIONS: Read the passage. Notice that there are numbers at the end of some of the sentences. Refer to these numbers when answering the questions. Melanoma cells metastasize by leaving the primary tumor and entering the bloodstream, which carries them to distant parts of the body. (1) The cancer cells leave the bloodstream by crossing the capillary basal lamina to enter another tissue. (2) As a cancer researcher, you think that somehow preventing the cancer cells from crossing the basal lamina of the blood vessels will inhibit metastasis. (3) You believe that if you block the integrin receptors on the melanoma cells from binding to the extracellular matrix proteins of the basal lamina, then the cells will not be able to adhere to the basal lamina and therefore will not be able to migrate across it

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Chapter 10 d. a prediction. e. support for their idea that blocking integrin function inhibits some tumor cells from binding to the basal lamina. ANSWER: e 54. INSTRUCTIONS: Read the passage. Notice that there are numbers at the end of some of the sentences. Refer to these numbers when answering the questions. Melanoma cells metastasize by leaving the primary tumor and entering the bloodstream, which carries them to distant parts of the body. (1) The cancer cells leave the bloodstream by crossing the capillary basal lamina to enter another tissue. (2) As a cancer researcher, you think that somehow preventing the cancer cells from crossing the basal lamina of the blood vessels will inhibit metastasis. (3) You believe that if you block the integrin receptors on the melanoma cells from binding to the extracellular matrix proteins of the basal lamina, then the cells will not be able to adhere to the basal lamina and therefore will not be able to migrate across it

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Chapter 10 Which of the choices is the dependent variable? a. the untreated cells b. the number of cells attached to the artificial basal lamina c. whether or not the cells were treated with antibodies d. the cells treated with integrin-specific antibodies e. the cells treated with the nonspecific antibodies ANSWER: b 55. INSTRUCTIONS: Read the passage. Notice that there are numbers at the end of some of the sentences. Refer to these numbers when answering the questions. Melanoma cells metastasize by leaving the primary tumor and entering the bloodstream, which carries them to distant parts of the body. (1) The cancer cells leave the bloodstream by crossing the capillary basal lamina to enter another tissue. (2) As a cancer researcher, you think that somehow preventing the cancer cells from crossing the basal lamina of the blood vessels will inhibit metastasis. (3) You believe that if you block the integrin receptors on the melanoma cells from binding to the extracellular matrix proteins of the basal lamina, then the cells will not be able to adhere to the basal lamina and therefore will not be able to migrate across it

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Chapter 10 from adhering to the proteins in the basal lamina. Once you have these integrin-specific antibodies, you test them on cancer cells in culture to see if they prevent the cells from crossing an artificial basal lamina. To do this, you and you colleagues set up nine cultures of cancer cells. You treat three cultures with the integrinspecific antibody. You treat three more cultures with a nonspecific antibody that does not bind to integrins. Three cultures of cells are not treated with antibodies. Which of the choices is the independent variable? a. whether or not the cells were treated with antibodies b. the cells treated with integrin-specific antibodies c. the cells treated with the nonspecific antibodies d. the number of cells attached to the artificial basal lamina e. the untreated cells ANSWER: a 56. INSTRUCTIONS: Read the passage. Notice that there are numbers at the end of some of the sentences. Refer to these numbers when answering the questions. Melanoma cells metastasize by leaving the primary tumor and entering the bloodstream, which carries them to distant parts of the body. (1) The cancer cells leave the bloodstream by crossing the capillary basal lamina to enter another tissue. (2) As a cancer researcher, you think that somehow preventing the cancer cells from crossing the basal lamina of the blood vessels will inhibit metastasis. (3) You believe that if you block the integrin receptors on the melanoma cells from binding to the extracellular matrix proteins of the basal lamina, then the cells will not be able to adhere to the basal lamina and therefore will not be able to migrate across it

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Chapter 10

(4) You and your colleagues make antibodies, proteins that bind specifically to only one type of protein in the cell, and your antibodies bind specifically to the integrins on the surface of the melanoma cells, preventing them from adhering to the proteins in the basal lamina. Once you have these integrin-specific antibodies, you test them on cancer cells in culture to see if they prevent the cells from crossing an artificial basal lamina. To do this, you and you colleagues set up nine cultures of cancer cells. You treat three cultures with the integrinspecific antibody. You treat three more cultures with a nonspecific antibody that does not bind to integrins. Three cultures of cells are not treated with antibodies. The three untreated cultures of cells were included in the experiment: a. to avoid sampling error. b. to increase the sample size. c. as a control group. d. in case extra cultures of cells were needed. ANSWER: c 57. INSTRUCTIONS: Read the passage. Notice that there are numbers at the end of some of the sentences. Refer to these numbers when answering the questions. Melanoma cells metastasize by leaving the primary tumor and entering the bloodstream, which carries them to distant parts of the body. (1) The cancer cells leave the bloodstream by crossing the capillary basal lamina to enter another tissue. (2) As a cancer researcher, you think that somehow preventing the cancer cells from crossing the basal lamina of the blood vessels will inhibit metastasis. Copyright Macmillan Learning. Powered by Cognero.

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(4) You and your colleagues make antibodies, proteins that bind specifically to only one type of protein in the cell, and your antibodies bind specifically to the integrins on the surface of the melanoma cells, preventing them from adhering to the proteins in the basal lamina. Once you have these integrin-specific antibodies, you test them on cancer cells in culture to see if they prevent the cells from crossing an artificial basal lamina. To do this, you and you colleagues set up nine cultures of cancer cells. You treat three cultures with the integrinspecific antibody. You treat three more cultures with a nonspecific antibody that does not bind to integrins. Three cultures of cells are not treated with antibodies. Which of the choices is shown in the graph of the data? a. Cells treated with antibodies do not attach to basal lamina proteins as well as untreated cells. b. The antibody treatment is shown on the y-axis. c. The independent variable is shown on the y-axis. d. The data proved that the hypothesis was correct. ANSWER: a Multiple Response 58. How does the extracellular matrix affect a cell that it surrounds? Select all that apply. a. All of these choices are correct. b. The genes that the cell expresses depend in part on the types of proteins in the extracellular matrix. c. The size that the cell grows to depends in part on the polysaccharides in the extracellular matrix. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 d. The extracellular matrix has no effect on the cells that it surrounds. e. The cell's shape depends in part on the structure and composition of the extracellular matrix. ANSWER: b, e 59. Which are examples of a motor protein? Select all that apply. a. myosin b. kinesin c. actin d. dynein ANSWER: a, b, d 60. Which of the choices are correct comparisons between gap junctions and plasmodesmata? Select all that apply. a. Both allow exchange of material between cells, but gap junctions allow larger-sized molecules to transfer between cells. b. Animal cells have gap junctions, and plant cells have plasmodesmata. c. In plasmodesmata, the plasma membranes of the connected cells are continuous, whereas in gap junctions, the plasma membranes of the connected cells remain distinct. d. Gap junctions connect cells directly at the plasma membranes, but plasmodesmata must also bridge the cell wall separating the two cells. ANSWER: b, c, d 61. Which of the statements about plant cell walls is true? Select all that apply. a. The cell wall has three layers: the middle lamella, the primary cell wall, and the secondary cell wall. b. In the late stages of cell division, the middle lamella is the first cell wall layer to be formed, serving to attach the daughter cells to each other. c. The primary cell wall is thin and flexible, formed while the cells are still growing. d. The secondary cell wall contains complex components such as lignin that are rigid. e. Like the extracellular matrix formed by animal cells, the cell wall polymers are formed in the cytoplasm and secreted to the outside of the cell. ANSWER: a, b, c, d 62. Some cytoskeletal elements are more permanent than others. Which components of the cytoskeleton are dynamic structures? Select all that apply. a. microtubules b. microfilaments c. intermediate filaments ANSWER: a, b 63. Which of the statements about intracellular transport are true? Select all that apply. a. Microtubules facilitate cellular movement. b. Microfilaments facilitate cellular movement. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 10 c. Intermediate filaments facilitate cellular movement. d. Intermediate filaments, microfilaments, and microtubules all facilitate cellular movement of one kind or another. ANSWER: a, b 64. Which of the statements about intracellular transport are true? Select all that apply. a. Kinesin moves substances toward the "plus end" of microtubules. b. Kinesin moves substances toward the "minus end" of microtubules. c. Dynein move substances toward the "plus end" of microtubules. d. Dynein moves substances toward the "minus end" of microtubules. ANSWER: a, d 65. Myosin motor proteins use the energy of ATP to: Select all that apply. a. shorten muscle fibers. b. carry vesicles along microtubules within a cell. c. carry vesicles along microfilaments within a cell. d. cause changes in cell shape. ANSWER: a, c, d 66. Which of the statements about cadherins and integrins are true? Select all that apply. a. They are found on the surfaces of most animal cells. b. They are transmembrane proteins that attach to cytoskeletal elements inside the cell. c. They are highly specific for particular proteins in the extracellular matrix. d. They are not found in embryonic cells. ANSWER: a, b, c

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Chapter 11 Multiple Choice 1. It is estimated that there are a total of 245 cells (roughly 50 trillion) in the adult the human body. Starting from a fertilized egg, how many cell divisions would be required to produce 245, assuming that every cell divides in every cycle? a. 450 divisions b. 45 divisions c. 4.5 divisions d. one division per day for nine months ANSWER: b 2. Which of the steps in prokaryotic binary fission is correct? a. All of these choices are correct. b. The two replicated chromosomes remain attached to the plasma membrane. c. The cell continues to grow outward symmetrically, separating the two chromosomes. d. Cell wall material is laid down at the midpoint to separate the two daughter cells. e. DNA is replicated bidirectionally from a single point on the circular chromosome. ANSWER: a 3. In what way is cytokinesis in plant cells similar to binary fission in a bacterium? a. Cell wall material is deposited to separate the daughter cells. b. A ring of actin filaments constricts the plasma membrane between the two nuclei to separate the daughter cells. c. A microtubule-like structure constricts the plasma membrane between the two nuclei to separate the daughter cells. d. A motor protein slides microtubules in a contractile ring at the plasma membrane between the two nuclei to separate the daughter cells. e. All of these choices are correct. ANSWER: a 4. In a dividing plant cell, a phragmoplast structure forms during telophase that directs vesicles carrying cell wall components to the middle of the cell, assembling a new cell wall called the cell plate. a. true b. false ANSWER: a 5. Which of the choices has the most similarity in nucleotide sequence? a. non-sister chromatids b. nonhomologous chromosomes c. homologous chromosomes d. sister chromatids Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 e. the X and the Y chromosomes ANSWER: d 6. In meiosis, recombination occurs: a. during both prophases I and II and involves exchange of chromosomes fragments between all four chromatids. b. during only prophase I and involves exchange between chromatids of homologous chromosomes. c. during only prophase II and involves exchange of chromosome fragments between sister chromatids. d. during both prophases I and II and involves exchange of chromosome fragments between sister chromatids. ANSWER: b 7. In fruit flies, chromosomal crossing over does not occur in meiosis in males, whereas crossing over does occur in meiosis in females. In fruit flies that are heterozygous at many genes, at what stage would cells no longer be heterozygous for any gene during the process of meiosis? a. After the first meiotic division in both males and females. b. After the second meiotic division in males, and after the first meiotic division in females. c. After the first meiotic division in males, and after the second meiotic division in females. d. After the second meiotic division in both males and females. ANSWER: c 8. A cell in prophase I of meiosis has _____ as many copies of chromosomes as each of the daughter cells following cytokinesis of meiosis II. a. half b. four times c. one quarter d. twice ANSWER: b 9. A cell in prophase I of meiosis has _____ as much DNA as each of the daughter cells following cytokinesis of meiosis II. a. four times b. half c. one quarter d. twice ANSWER: a 10. Sexual reproduction results: a. in offspring that are not genetically identical. b. from combining genetic material from two gametes. c. in a new generation that is not genetically identical to its parents. d. in a new generation with the same number of chromosomes as each parent. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 e. All of these choices are correct. ANSWER: e 11. Meiosis is thought to have appeared early in evolution because most eukaryotes perform meiosis and the steps of meiosis are the same in all eukaryotes. a. true b. false ANSWER: a 12. Which of the statements is true about gamete formation in mammals? a. In male gamete formation, only one of the four cells resulting from meiosis becomes a sperm. b. In female gamete formation, the separation of the cytoplasm is unequal during meiosis. c. In female gamete formation, all four of the cells resulting from meiosis become an egg, which upon fertilization by a sperm cell, forms a zygote. d. In female gamete formation, the polar bodies formed during meiosis can be fertilized by sperm. ANSWER: b 13. In fungi and plants, mitotic cell division of haploid cells follows meiosis. a. true b. false ANSWER: a 14. What process can account for the phenomenon wherein a normal XY male produces a sperm carrying two Y chromosomes? a. disjunction b. translocation c. second-division nondisjunction d. first-division nondisjunction e. copy-number variation ANSWER: c 15. How many chromosomes are in a human cell that is triploid? a. 92 b. 46 c. 23 d. 69 e. None of the answer options is correct. ANSWER: d 16. Which statement best explains why embryos with a missing autosome are frequently not recorded into statistics about spontaneously aborted fetuses? a. None of the other answer options is correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 b. Fertilization cannot produce embryos with a missing autosome. c. Nondisjunction produces twice as many gametes with extra autosomes as gametes with a missing autosome. d. Fetuses with a missing autosome usually do not get aborted but continue to live birth. e. Fetuses missing an autosome are usually aborted before the pregnancy is recognized. ANSWER: e 17. Which of the statements is true about Klinefelter syndrome (47, XXY)? a. Nondisjunction is the usual cause. b. Affected males can reproduce normally. c. The risk increases with the father's age. d. Most affected fetuses undergo spontaneous abortion. e. None of the other answer options is correct. ANSWER: a 18. Which of the qualities is associated with the risk of having a baby with Down syndrome? a. health of the mother b. age of the mother c. socioeconomic status of the mother d. IQ of the mother e. All of these choices are correct. ANSWER: b 19. Nondisjunction in _____ results in offspring inheriting a change in the number of chromosomes. a. meiosis b. mitosis ANSWER: a 20. The most common form of hemophilia is a defect in blood clotting factor VIII, which is caused by a mutant form of a gene on the X chromosome. Boys who inherit that mutation from their mother suffer from uncontrolled bleeding. Girls carrying one copy of this mutation have near normal blood clotting. Uncommonly, a girl is born with hemophilia even though both parents have normal phenotypes. Which of the statements might explain hemophilia in a girl born to parents with normal blood clotting phenotypes? a. Nondisjunction during sperm formation resulted in her receiving no sex chromosome from her father and an X chromosome from her mother; she is XO. b. Nondisjunction during egg formation resulted in her receiving no X chromosome from her mother but an X from her father; she is XO. c. A nondisjunction event in sperm production resulted in her receiving X and Y chromosomes from her father and an X from her mother; she is XXY. d. Nondisjunction during egg formation resulted in her receiving two X chromosomes from her mother and an X chromosome from her father; she is XXX. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 21. Autosomal trisomies are associated with the X or Y chromosome. a. true b. false ANSWER: b 22. The incidence of Down syndrome is positively correlated with: a. copy-number variation. b. the father's age. c. the mother's age. d. inversion. e. None of the other answer options is correct. ANSWER: c 23. During the formation of female gametes, nondisjunction of X chromosomes may occur during meiosis I, resulting in two types of eggs with different compositions of sex chromosomes. If normal X or Y-bearing sperm fertilize these two types of egg, which of the options are possible sex chromosome complements in the resulting fertilized egg? a. XXX and XO or XXY and YO b. XX and XY or XYY and YO ANSWER: a 24. The lethality of most monosomies and trisomies in humans shows the: a. advantage of sexual reproduction. b. deleterious effects of a variable number of tandem repeat sequences. c. importance of gene dosage (the number of copies of a gene). d. rarity of variation in the number of copies of each gene in the genome. ANSWER: c 25. The presence of an extra X chromosome is more likely to result in a viable human fetus than an extra copy of an autosome because: a. expression of genes from the X chromosome is restricted to early development of females. b. the X chromosome carries few genes, so gene dosage (the number of copies of a gene) is not a problem. c. the X and the Y chromosomes do not recombine along most of their length. d. the X chromosome is the "female" chromosome. e. female mammals inactivate all but one copy of their X chromosomes, which does not occur for autosomes. ANSWER: e 26. Diploid somatic cells of elephants have 56 chromosomes. If nondisjunction of one of an elephant's chromosomes occurs in meiosis I, the resulting sperm are expected to have the chromosome complement: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 a. 29, 29, 27, 27. b. 28, 28, 29, 27. c. 28, 28, 28, 28. d. 29, 29, 28, 28. e. 56, 56, 57, 55. ANSWER: a 27. Diploid somatic cells of elephants have 56 chromosomes. If nondisjunction of one of an elephant's chromosomes occurs in meiosis II, the resulting sperm are expected to have the chromosome complement: a. 28, 28, 28, 28. b. 28, 28, 29, 27. c. 29, 29, 27, 27. d. 29, 29, 28, 28. e. 56, 56, 57, 55. ANSWER: b 28. A normal XY male has a son with the karyotype XYY. The likely explanation is: a. dosage compensation. b. translocation. c. second-division nondisjunction. d. first-division nondisjunction. e. copy-number variation. ANSWER: c 29. How many chromosomes are there in a human cell that is tetraploid? a. 92 b. 46 c. 69 d. 23 ANSWER: a 30. Which of the processes is most important for passage through the G1 cell cycle checkpoint? a. inhibition of cyclin proteins b. changes in membrane polarization c. expression of genes encoding cyclin-dependent kinases d. activation of DNA polymerase e. activation of cyclin-dependent kinases ANSWER: e 31. The amount of _____ is fairly constant throughout the cell cycle, but the amount of _____ varies. a. DNA; cyclins Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 b. cyclin-dependent kinases; cyclins c. cyclin-dependent kinase; DNA d. cyclins; DNA ANSWER: b 32. Studies of rapidly dividing embryonic animal cells revealed: a. All of these choices are correct. b. a periodic activation of protein kinases in sync with the cell cycle. c. that increased levels of cyclin proteins are correlated with activation of cyclin-dependent kinase (CDK) enzymes. d. that inhibition of protein synthesis blocks mitosis. e. increased synthesis of certain proteins in sync with the cell cycle. ANSWER: a 33. Which of these events includes a checkpoint in the cell cycle? a. the transition from G1 to S phase b. the transition from anaphase to telophase c. the transition from S phase and the completion of DNA synthesis to G2 phase d. All of these events include a checkpoint in the cell cycle. ANSWER: a 34. The S cyclin-CDK complex: a. controls the cell cycle during the S and G2 phases. b. triggers initiation of DNA synthesis during the S phase. c. prevents initiation of DNA synthesis a second time during the S and G2 phases. d. both controls the cell cycle during the S and G2 phases and triggers initiation of DNA synthesis during the S phase. e. All of these choices are correct. ANSWER: e 35. The M cyclin-CDK complex: a. All of these choices are correct. b. initiates many events of mitosis. c. triggers phosphorylation of certain nuclear proteins, resulting in breakdown of the nuclear envelope during prophase. d. phosphorylates proteins that promote formation of the mitotic spindle. e. controls passage of the G1 checkpoint. ANSWER: a 36. Cancer-causing genes found in some viruses are called: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 a. coat protein genes. b. proto-oncogenes. c. tumor suppressor genes. d. oncogenes. ANSWER: d 37. Phosphorylated p53 is a protein that accumulates in the nuclei of cells that have damaged DNA where it functions to block the cell cycle and activate DNA repair. The p53 gene that encodes this protein is an example of a(n): a. protein kinase. b. proto-oncogene. c. oncogene. d. tumor suppressor gene. ANSWER: d 38. Which of the statements is true when a cell has a mutation in the p53 gene, such that the p53 protein is not able to be phosphorylated? a. The amount of p53 protein in the nucleus would increase in response to DNA damage. b. The cell would arrest, giving time for the DNA damage to be repaired. c. The cell would proceed through the cell cycle even in the presence of DNA damage. d. Nothing would change in the cell cycle, because the p53 protein is normally not able to be phosphorylated. ANSWER: c 39. The genetic information of daughter cells is the same as the genetic information of the parent in binary fission. a. true b. false ANSWER: a 40. A mutation acquired by a bacterium will very likely be inherited by both daughter cells. a. true b. false ANSWER: a 41. Mitosis most likely evolved from what process? a. the cell cycle b. binary fission c. cytokinesis d. meiosis ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 42. A skin cell in G2 of interphase has _____ as much DNA as it had in G1. a. four times b. half c. exactly d. one-fourth e. twice ANSWER: e 43. All of these events happen during M phase except: a. formation of the spindle. b. condensation of the chromosomes. c. synthesis of DNA. d. separation of sister chromatids. ANSWER: c 44. Paramecium is a single-cell eukaryotic organism that can reproduce by mitotic cell division. Prior to the M phase of the cell cycle, which of the events must occur? a. The cell must replicate its chromosomes. b. The cell must first be fertilized. c. The nucleus must divide. d. Sister chromatids must be separated. e. The nuclear envelope must disintegrate. ANSWER: a 45. Which of the statements is correct about cells in G1 phase in the cell cycle? a. Cells in G1 phase attach copies of their chromosomes to the cell membrane in preparation for binary fission. b. Cells at the end of G1 activate the proteins that make the mitotic spindle. c. Cells in G1 phase have a gap in protein synthesis and suppress the process of translation. d. Cells in G1 phase that do not activate the enzymes associated with DNA synthesis for long periods are said to be in G0 phase. ANSWER: d 46. Which of the choices is reproduced by binary fission? a. chloroplasts b. archaea c. All of these choices are correct. d. mitochondria e. bacteria ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 47. Actin microfilaments and tubulin-based microtubules are important components of the cytoskeleton that function during mitosis. Which process of cell division would be most affected if actin function was blocked? a. The mitotic spindle would not form. b. Cytokinesis would not occur to divide the cytoplasm. c. The sister chromatids would not separate. d. The chromosomes would not replicate. ANSWER: b 48. The correct sequence of steps in the M phase of the cell cycle is: a. prophase, prometaphase, metaphase, anaphase, telophase, cytokinesis. b. prophase, prometaphase, metaphase, anaphase, telophase, nuclear division. c. prophase, metaphase, prometaphase, anaphase, nuclear division, telophase. d. prophase, metaphase, anaphase, telophase, cytokinesis, nuclear division. ANSWER: a 49. Sister chromatids are best described as two DNA molecules that have: a. different genes in a different order with different alleles. b. the same genes in the same order with different alleles. c. different genes in the same order with different alleles. d. the same genes in the same order with the same alleles. ANSWER: d 50. Colchicine is a drug that blocks the assembly of microtubules. If dividing cells are treated with colchicine, at what stage of mitosis would you predict the arrest would occur? a. late anaphase b. metaphase c. prophase d. telophase e. G1 of interphase ANSWER: c 51. At the end of mitosis, the daughter cells are _____, whereas at the end of meiosis the daughter cells are _____. a. diploid; haploid b. haploid; diploid c. diploid; polyploid d. polyploid; haploid ANSWER: a 52. Which of the events could produce new combinations of alleles of genes along a chromosome if crossing over happened during prophase I of meiosis? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 a. crossing over between sister chromatids b. crossing over between non-sister chromatids of homologous chromosomes c. either kind of crossing over could produce new combinations of alleles ANSWER: b 53. Homologous pairs of chromosomes separate during: a. meiosis II only. b. meiosis I and mitosis. c. meiosis II and mitosis. d. mitosis only. e. meiosis I only ANSWER: e 54. During meiosis, two rounds of DNA synthesis are required to form four gametes from one parent cell. a. true b. false ANSWER: b 55. If a cell underwent mitosis, and its daughter cells were immediately exposed to chemicals that damaged the DNA, at which stage of the cell cycle checkpoint would you predict the cell would arrest? a. M checkpoint b. G2 checkpoint c. G1 checkpoint d. G1, G2, or the M checkpoints ANSWER: c 56. Which one of the choices would most likely contribute to uncontrolled cell proliferation (that is, cancer)? a. a mutant DNA synthesis mechanism causing blocked chromosome replication b. a mutant cyclin that cannot bind to its normal CDK binding partner c. a mutant enzyme needed for microtubule synthesis/polymerization d. a mutant CDK that was active in the absence of its cyclin binding partner e. a mutant kinetochore protein that causes reduced microtubule attachment ANSWER: d 57. p53 is an example of a(n): a. tumor suppressor. b. proto-oncogene. c. oncogene. d. cyclin-dependent kinase. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 58. An enzyme that catalyzes the addition of a phosphate group from ATP to another molecule is called a: a. phosphorylase. b. kinase. c. phosphatase d. cyclase. ANSWER: b 59. Which of the choices would be most likely to lead to the development of cancer? a. the activation of an oncogene and the inactivation of a tumor suppressor gene b. the activation of a proto-oncogene c. the activation of a tumor suppressor gene d. the activation of an oncogene and the activation of a tumor suppressor gene ANSWER: a 60. Cancers develop stepwise over time because: a. All of these choices are correct. b. tumor cells divide slowly and in a stepwise pattern. c. it takes time for a tumor to dissolve through its encapsulation before it can invade neighboring tissues. d. cells keep leaving tumors and are destroyed in lymph nodes. e. it takes multiple mutations of multiple genes to allow cancer cells to develop. ANSWER: e 61. Which of the options is circled in this electron micrograph?

a. one double-stranded DNA molecule b. one single strand of a DNA molecule Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 c. two double-stranded DNA molecules ANSWER: a 62. Prokaryotic cells and eukaryotic cells reproduce by cell division. Regardless of the type of cell, all cells must _____ before they divide. a. complete mitosis b. separate sister chromatids from one another c. make a copy of their genetic information d. reconstruct their nuclear envelope ANSWER: c 63. The correct sequence of steps in the eukaryotic cell cycle is: a. G1 → S phase → G2 → mitosis → cytokinesis. b. G0 → S phase → G1 → S phase → G2 → mitosis → cytokinesis. c. G0 → S phase → G1 → G2 → cytokinesis → mitosis. d. G0 → S phase → G1 → S phase → G2 → cytokinesis → mitosis. e. G1 → S phase → G2 → cytokinesis → mitosis. ANSWER: a 64. Which of the options is circled in this electron micrograph?

a. one double-stranded DNA molecule b. one single strand of a DNA molecule c. two double-stranded DNA molecules Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 ANSWER: c 65. How many copies of each gene are present in human skin cells at G1 of interphase? a. 4 b. 2 c. 8 d. 1 ANSWER: b 66. How many copies of each gene are present in human skin cells at G2 of interphase? a. 8 b. 2 c. 6 d. 4 ANSWER: d 67. Taxol is an anti-cancer drug that prevents uncontrolled cell proliferation by blocking the lengthening and shortening of microtubules, which causes arrest of the cell cycle. If dividing cells are treated with Taxol, what will occur? a. The cells will be unable to form a mitotic spindle. b. The cells will be unable to replicate their chromosomes. c. The cells will be unable to divide the cytoplasm in cytokinesis. d. The cells will be unable to increase cyclin levels required to pass the G1 checkpoint. ANSWER: a 68. In a diploid individual, there are two genes called "A" and "B" on one chromosome. One homolog of that chromosome carries A and B alleles of those genes, and the homologous chromosome carries different forms (alleles) of these same genes, a and b. If there is a single crossover between these two genes involving one pair of non-sister chromatids during metaphase I of meiosis, the resulting four gametes would be: a. Ab, Ab, aB, aB. b. AB, ab, AB, ab. c. AaBb, AaBb, AaBb, AaBb. d. AB, AB, ab, ab. e. AB, Ab, aB, ab. ANSWER: e 69. The random alignment of maternal and paternal homologous chromosomes during metaphase I is one of the ways genetic variability among gametes comes about. For example, it is possible for an organism with 4 pairs of homologous chromosomes to produce gametes with up to 16 different combinations of maternal and paternal chromosomes (24= 16). In the case of humans with 23 pairs of chromosomes, there are over 8 million possible combinations. How many possible combinations of maternal and paternal chromosomes are possible in the gametes of an organism with 6 chromosomes? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 a. 512 b. 128 c. 256 d. 64 e. 1024 ANSWER: d 70. The epithelial cells in the skin of an animal have 24 chromosomes. How many chromosomes are present in the gametes of this animal? a. 12 b. 24 c. 48 d. 96 ANSWER: a 71. Consider a diploid organism with a haploid complement of 4 chromosomes in its gametes. At meiotic prophase I, how many total chromosomes copies will be present in a cell? a. 16; 4 pairs of homologous chromosomes and 2 sister chromatids per chromosome b. 8; 4 pairs of homologous chromosomes c. 12; 4 pairs of homologous chromosomes and their haploid complement d. 4; 1 complete set of chromosomes ANSWER: a 72. The FoxP2 gene is thought to be involved in language in humans. At prophase I, how many copies of the FoxP2 gene are present in a cell? Keep in mind that humans are diploid. a. 2 copies; 1 on each homologous chromosome b. 4 copies; 1 on each sister chromatid in a pair of homologous chromosomes c. 8 copies; 1 on each sister chromatid in a pair of homologous chromosomes d. 8 copies; 1 on each strand of each chromatid per homologous pair ANSWER: b 73. Sister chromatids are separated during: a. meiosis II only. b. meiosis I and mitosis. c. mitosis only. d. meiosis I only. e. meiosis II and mitosis. ANSWER: e 74. A homologous chromosome pair is best described as 2 chromosomes that have: a. the same genes in the same order but possibly with different alleles of some of the genes. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 b. the same alleles of the same genes in the same order. c. different genes arranged in a different order with potentially different alleles of some of the genes. d. different alleles of the same genes arranged in a different order. e. identical sequences of nucleotides. ANSWER: a 75. The rise in cyclin levels correlate with passage of cell cycle checkpoints. Which event is triggered by high cyclin levels? a. polymerization of DNA b. signal transduction from receptors c. kinase activity to phosphorylate specific proteins d. ATP synthesis ANSWER: c 76. Which one of the choices is capable of phosphorylating key proteins involved in regulating the cell cycle? a. CDK alone b. cyclin alone c. phosphatases d. p53 protein e. cyclin-CDK complex ANSWER: e 77. Which major checkpoint delays the cell cycle when DNA replication is incomplete? a. M checkpoint b. G1 checkpoint c. G2 checkpoint d. G1, G2, and the M checkpoints ANSWER: c 78. CDKs are important in the regulation of the cell cycle. They carry out their function by: a. removing phosphate groups from target proteins. b. adding phosphate groups to target proteins. c. degrading cyclin proteins. d. preventing the progression of a cell from one stage of the cell cycle to the next. ANSWER: b 79. Which of the statements concerning cyclin-dependent kinases (CDKs) is true? a. All of these choices are true. b. CDKs are enzymes that attach phosphate groups to other proteins. c. CDKs are active, or "turned on," when complexed with cyclins. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 d. CDKs are present throughout the cell cycle. e. CDKs are inactive, or "turned off," in the presence of cyclins. ANSWER: a 80. Processes that regulate cell division in mammals include: a. rapid degradation of cyclins after CDK activation. b. activated cyclin-CDK complexes triggering cell cycle events. c. different cyclins and CDKs acting at different stages of the cell cycle to promote cell division. d. All of these choices are correct. e. production of cyclin proteins that activate CDK enzymes. ANSWER: d 81. Some genes have important functions in cells, but they can sometimes acquire mutations that cause negative effects, including the development of cancer. These genes are called: a. oncogenes. b. proto-oncogenes. c. tumor suppressor genes. d. viral genes. ANSWER: b 82. The first oncogene to be discovered: a. was discovered in a Rous sarcoma virus that causes cancer. b. is a gene that contributes to uncontrolled cell division or cancer. c. is a protein kinase that acts to promote cell division. d. has a less-active normal counterpart called a proto-oncogene. e. All of these choices are correct. ANSWER: a 83. Considering Douglas Hanahan and Robert Weinberg's Hallmarks of Cancer, which of the choices could block cancer progression? a. uncontrolled cell division in the absence of growth signals b. resistance to signals that slow cell division or promote cell death c. metastasis, which allows invasion of local tissues d. ability to stimulate growth of blood vessels to provide nutrients to the rapidly growing tumor e. inhibitory signals causing cell cycle arrest at the M checkpoint ANSWER: e 84. During meiosis, genetic variability is introduced during which phase? a. metaphase I b. anaphase II c. prophase II Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 d. telophase I ANSWER: a 85. In a normal cell, tumor suppressors are responsible for: a. allowing the cell to correct replication errors. b. increasing the rate of cell division. c. signaling that cells should pass a cell cycle checkpoint. d. signaling that DNA replication should begin. ANSWER: a 86. There are two stages to the cell cycle: M phase and interphase. a. true b. false ANSWER: a 87. A student was studying cell growth using cells grown in laboratory cultures. The cultures were synchronized so that all of the cells passed through the same stage of the cell cycle at the same time. The cells were examined during five different periods of time, intervals (A–E). The amount of DNA present per cell was determined for each interval. The graph shows the result of this study.

Most of the cells examined during interval A have half as much DNA as those measured in interval C. Therefore, most of the cells at interval B must have been in: a. M phase. b. S phase. c. G0 phase. d. G1 phase. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 e. G2 phase. ANSWER: b 88. A student was studying cell growth using cells grown in laboratory cultures. The cultures were synchronized so that all of the cells passed through the same stage of the cell cycle at the same time. The cells were examined during five different periods of time, intervals (A–E). The amount of DNA present per cell was determined for each interval. The graph shows the result of this study.

Most of the cells examined during interval D were in: a. S phase. b. G1 phase. c. G2 phase. d. M phase. ANSWER: d 89. A student was studying cell growth using cells grown in laboratory cultures. The cultures were synchronized so that all of the cells passed through the same stage of the cell cycle at the same time. The cells were examined during five different periods of time, intervals (A–E). The amount of DNA present per cell was determined for each interval. The graph shows the result of this study.

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Chapter 11

Most of the cells examined during interval C were in: a. S phase. b. G1 phase. c. G2 phase. d. cytokinesis. e. M phase. ANSWER: c 90. A student was studying cell growth using cells grown in laboratory cultures. The cultures were synchronized so that all of the cells passed through the same stage of the cell cycle at the same time. The cells were examined during five different periods of time, intervals (A–E). The amount of DNA present per cell was determined for each interval. The graph shows the result of this study.

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Chapter 11

Based on the information in the graph, which interval(s) correspond(s) to times at which most of the cells were at G1 of the cell cycle? a. intervals A and E b. intervals A and B c. intervals B and D d. intervals A, B, C, and D ANSWER: a 91. An experiment was performed to study the progress of cells through the mitotic cell cycle. The compounds listed were used one at a time to study their effects on the cell cycle: cytochalasin: an inhibitor of actin microfilament assembly colchicine: an inhibitor of microtubule formation aphidicolin: an inhibitor of DNA polymerase activity emetine: an inhibitor of ribosome activity (blocks protein synthesis) Which of these compounds would be most likely to arrest cells in S Phase? a. cytochalasin: an inhibitor of actin microfilament assembly b. aphidicolin: an inhibitor of DNA polymerase activity c. colchicine: an inhibitor of microtubule formation d. emetine: an inhibitor of ribosome activity (and therefore protein synthesis) ANSWER: b 92. An experiment was performed to study the progress of cells through the mitotic cell cycle. The compounds Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 listed were used one at a time to study their effects on the cell cycle: cytochalasin: an inhibitor of actin microfilament assembly colchicine: an inhibitor of microtubule formation aphidicolin: an inhibitor of DNA polymerase activity emetine: an inhibitor of ribosome activity (blocks protein synthesis) Which of these compounds would be most likely to arrest cells in prophase? a. colchicine: an inhibitor of microtubule formation b. cytochalasin: an inhibitor of actin microfilament assembly c. aphidicolin: an inhibitor of DNA polymerase activity d. emetine: an inhibitor of ribosome activity (and therefore protein synthesis) ANSWER: a 93. An experiment was performed to study the progress of cells through the mitotic cell cycle. The compounds listed were used one at a time to study their effects on the cell cycle: cytochalasin: an inhibitor of actin microfilament assembly colchicine: an inhibitor of microtubule formation aphidicolin: an inhibitor of DNA polymerase activity emetine: an inhibitor of ribosome activity (blocks protein synthesis) Which of these compounds would be most likely to arrest cells in telophase/cytokinesis? a. emetine: an inhibitor of ribosome activity (and therefore protein synthesis). b. colchicine: an inhibitor of microtubule formation c. aphidicolin: an inhibitor of DNA polymerase activity d. cytochalasin: an inhibitor of actin microfilament assembly ANSWER: d 94. An experiment was performed to study the progress of cells through the mitotic cell cycle. The compounds listed were used one at a time to study their effects on the cell cycle: cytochalasin: an inhibitor of actin microfilament assembly colchicine: an inhibitor of microtubule formation aphidicolin: an inhibitor of DNA polymerase activity emetine: an inhibitor of ribosome activity (blocks protein synthesis) Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 Which of these compounds would most likely to prevent cells from passing the G1 checkpoint? a. emetine: an inhibitor of ribosome activity (and therefore protein synthesis) b. colchicine: an inhibitor of microtubule formation c. aphidicolin: an inhibitor of DNA polymerase activity d. cytochalasin: an inhibitor of actin microfilament assembly ANSWER: a 95. An experiment was performed to study the progress of cells through the mitotic cell cycle. Consider the image. The group of cells on the left are examples of the cells at the beginning of the experiment. The cells on the right are representatives of three different experimental treatments using four different compounds.

Each of these compounds were used individually to study their effects on the cell cycle. cytochalasin: an inhibitor of actin microfilament assembly colchicine: an inhibitor of microtubule formation aphidicolin: an inhibitor of DNA polymerase activity emetine: an inhibitor of ribosome activity (and therefore protein synthesis) Which of the cells came from a culture treated with cytochalasin? a. cell C b. cell A c. cell B ANSWER: c 96. An experiment was performed to study the progress of cells through the mitotic cell cycle. Consider the image. The group of cells on the left are examples of the cells at the beginning of the experiment. The cells on the right are representatives of three different experimental treatments using four different compounds. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11

Each of these compounds were used individually to study their effects on the cell cycle. cytochalasin: an inhibitor of actin microfilament assembly colchicine: an inhibitor of microtubule formation aphidicolin: an inhibitor of DNA polymerase activity emetine: an inhibitor of ribosome activity (and therefore protein synthesis) Which of the cells came from a culture treated with colchicine? a. cell A b. cell C c. cell B ANSWER: b 97. An experiment was performed to study the progress of cells through the mitotic cell cycle. Consider the image. The group of cells on the left are examples of the cells at the beginning of the experiment. The cells on the right are representatives of three different experimental treatments using four different compounds.

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Chapter 11 Each of these compounds were used individually to study their effects on the cell cycle. cytochalasin: an inhibitor of actin microfilament assembly colchicine: an inhibitor of microtubule formation aphidicolin: an inhibitor of DNA polymerase activity emetine: an inhibitor of ribosome activity (and therefore protein synthesis) Which of the cells most likely came from a culture treated with emetine or aphidicolin? a. cell A b. cell B c. cell C ANSWER: a 98. A student was studying gamete production in male frogs using cells removed from the testes of adult frogs. These cells undergo meiotic cell division to become mature sperm cells. The cells were grown in laboratory cultures that were synchronized so that all of the cells passed through the same stage of the cell cycle at the same time. The cells were examined during five different periods of time, or intervals (1–5). The amount of DNA present per cell was determined for each interval. The graph shows the result of this study.

In which of the intervals is the number of chromosomes reduced by one-half? a. interval 5 b. interval 2 c. interval 1 d. interval 4 e. interval 3 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 ANSWER: e 99. A student was studying gamete production in male frogs using cells removed from the testes of adult frogs. These cells undergo meiotic cell division to become mature sperm cells. The cells were grown in laboratory cultures that were synchronized so that all of the cells passed through the same stage of the cell cycle at the same time. The cells were examined during five different periods of time, or intervals (1–5). The amount of DNA present per cell was determined for each interval. The graph shows the result of this study.

In which of the intervals is the amount of DNA reduced by one-half? a. intervals 3 and 4 b. interval 3 c. interval 4 d. interval 1 e. intervals 1 and 5 ANSWER: a 100. A student was studying gamete production in male frogs using cells removed from the testes of adult frogs. These cells undergo meiotic cell division to become mature sperm cells. The cells were grown in laboratory cultures that were synchronized so that all of the cells passed through the same stage of the cell cycle at the same time. The cells were examined during five different periods of time, or intervals (1–5). The amount of DNA present per cell was determined for each interval. The graph shows the result of this study.

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Chapter 11

In which of the intervals are most of the cells in meiosis II? a. interval 2 b. interval 4 c. interval 3 d. interval 1 e. interval 5 ANSWER: b 101. A student was studying gamete production in male frogs using cells removed from the testes of adult frogs. These cells undergo meiotic cell division to become mature sperm cells. The cells were grown in laboratory cultures that were synchronized so that all of the cells passed through the same stage of the cell cycle at the same time. The cells were examined during five different periods of time, or intervals (1–5). The amount of DNA present per cell was determined for each interval. The graph shows the result of this study.

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Chapter 11

In which of the intervals do the cells have twice as many chromosomes as the cells in interval 5? a. intervals 1, 2, and 3 b. interval 2 c. intervals 1 and 2 d. intervals 2 and 3 e. interval 1 ANSWER: a 102. A student was studying gamete production in male frogs using cells removed from the testes of adult frogs. These cells undergo meiotic cell division to become mature sperm cells. The cells were grown in laboratory cultures that were synchronized so that all of the cells passed through the same stage of the cell cycle at the same time. The cells were examined during five different periods of time, or intervals (1–5). The amount of DNA present per cell was determined for each interval. The graph shows the result of this study.

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Chapter 11

In which of the intervals would the addition of an inhibitor of DNA polymerase affect the progression of the cells through the cell cycle? a. interval 1 b. interval 2 c. interval 3 d. interval 4 e. interval 5 ANSWER: a 103. The M phase consists of two events: prophase and cytokinesis. a. true b. false ANSWER: b 104. Interphase is typically the shortest of the two stages of the cell cycle. a. true b. false ANSWER: b 105. There are three phases of interphase called: G1 phase, S phase and G2 phase. a. true b. false ANSWER: a 106. Cells that have fully differentiated and no longer divide are said to be in G0 phase. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 a. true b. false ANSWER: a Multiple Response 107. Which of the chromosome number abnormalities have detectable phenotypic effects? Select all that apply. a. 45, X b. 47, XXY c. 47, +21 d. 47, XYY e. 47, XXX ANSWER: a, b, c 108. Failure of cell division in anaphase in meiosis can result in a: Select all that apply. a. diploid gamete. b. triploid fertilized egg. c. haploid gamete. d. diploid fertilized egg. ANSWER: a, b 109. Which of the statements are true of both first- and second-division nondisjunction? Select all that apply. a. Homologous chromosomes fail to separate. b. Gametes are produced with missing chromosomes. c. Sister chromatids fail to separate. d. Gametes are produced with extra chromosomes. e. Half of the gametes produced are wild-type, whereas half are mutated. ANSWER: b, d 110. Which of the statements are true of mitotic cell division? Select all that apply. a. It occurs in eukaryotes but not in prokaryotes. b. It is a carefully regulated process that only occurs in certain conditions during the life of a cell. c. It is a form of sexual reproduction. d. It is a process that ends once development is complete and no longer occurs in adult organisms. e. It does not require DNA replication. ANSWER: a, b 111. Which of the statements are true about the cellular processes that occur during specific phases of the cell cycle in somatic cells? Select all that apply. a. During S phase, there is synthesis of cell components required for mitosis. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 b. During S phase, there is expression of proteins and enzymes required for DNA replication. c. During cytokinesis, separation of the cytoplasm forms two genetically identical daughter cells. d. During the second gap phase, there is separation of replicated chromosomes so that each daughter cell gets the same complement of chromosomes. e. During the first gap phase, there is increasing expression of cyclin genes if conditions favor cell division. ANSWER: b, c, e 112. Which of the options correctly match the type of cell division with the cellular events or results characteristic of that type of cell division? Select all that apply. a. Mitosis produces genetically identical daughter cells. b. Meiosis produces genetically unique daughter cells. c. Mitosis produces four daughter cells. d. In meiosis, the original cell undergoes two rounds of cell division. e. In mitosis, homologous chromosomes undergo synapsis. ANSWER: a, b, d 113. What outcomes could happen if a defect in S phase occurred during cell division? Select all that apply. a. All the organelles may not be duplicated. b. There could be too many chromosomes. c. There could be a lack of cytoplasm. d. There could be too few chromosomes. ANSWER: b, d 114. Why does meiosis result in more genetic variation than can be explained by mutation alone? Select all that apply. a. because of the random alignment of maternal and paternal homologs during metaphase of meiosis I b. because of the segregation of sister chromatids in anaphase of meiosis II c. because of crossing over between homologs during prophase I d. because not all the DNA gets replicated during S phase ANSWER: a, c 115. Which events would happen to a cell if cyclin levels were always high in the cell? Select all that apply. a. Protein substrates of CDKs would be constantly phosphorylated. b. Cyclin-dependent kinases would not be activated. c. The cell cycle would not stop at checkpoints. d. The cell would divide rapidly. ANSWER: a, c, d 116. Which choices accurately match each of the cellular processes to the stage of meiosis in which it occurs? Select all that apply. a. Microtubules attach to kinetochores at the centromere of each homolog in prometaphase I. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 11 b. Microtubules from both poles of the spindle move chromosomes to the center of the cell in anaphase I. c. Chromosomes condense to thickened structures that are visible under the microscope in prophase I. d. The centromere splits, and the sister chromatids separate in anaphase I. e. The nuclear envelope disappears in prophase I. ANSWER: a, c, e 117. Which choices accurately match each of the cellular processes to the stage of meiosis in which it occurs? Select all that apply. a. Microtubules pull homologous chromosomes toward opposite poles in anaphase I. b. Sister chromatids move toward opposite poles of the cell in anaphase II. c. Crossing over occurs between homologous chromosomes in prophase II. d. Homologous chromosomes undergo synapsis in prophase II. e. Both chromatids of homologous chromosomes migrate to the same pole in anaphase I. ANSWER: b, e 118. Which of the statements are true about the eukaryotic cell cycle? Select all that apply. a. There are two stages to the cell cycle: M phase and interphase. b. The M phase consists of two events: prophase and cytokinesis. c. Interphase is typically the shortest of the two stages of the cell cycle. d. There are three phases of interphase called: G1phase, S phase and G2 phase. e. Cells that have fully differentiated and no longer divide are said to be in G0phase. ANSWER: a, d, e

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Chapter 12 Multiple Choice 1. A new nucleotide can only be added to the _____ end of a growing DNA strand. DNA therefore always grows in the _____ direction. a. 3´; 5´ to 3´ b. 5´; 5´ to 3´ c. 3´; 3´ to 5´ d. 5´; 3´ to 5´ ANSWER: a 2. The leading strand is the daughter strand that has its _____ end pointed toward the replication fork and is therefore synthesized _____. a. None of the other answer options is correct. b. 3´; in a series of segments c. 5´; in a series of segments d. 5´; continuously e. 3´; continuously ANSWER: e 3. The lagging strand is the daughter strand that has its _____ end pointed toward the replication fork and is therefore synthesized _____. a. 3´; continuously b. 5´; in a series of segments c. 3´; in a series of segments d. 5´; continuously ANSWER: b 4. The enzyme responsible for replacing RNA primers with DNA is a type of: a. DNA ligase. b. helicase. c. topoisomerase II. d. DNA polymerase. e. DNA replicase. ANSWER: d 5. The enzyme responsible for proofreading a growing DNA strand and for replacing mismatched nucleotides is a. DNA replicase. b. helicase. c. topoisomerase II. d. DNA ligase. e. DNA polymerase. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 ANSWER: e 6. In DNA replication, each individual parent strand acts as a _____ strand for the synthesis of a _____ strand. a. duplicate; daughter b. template; daughter c. daughter; template d. template; duplicate e. daughter; duplicate ANSWER: b 7. The enzyme that catalyzes the addition of new nucleotides to a growing DNA strand is: a. DNA polymerase. b. helicase. c. topoisomerase II. d. DNA ligase. e. DNA replicase. ANSWER: a 8. As a piece of linear DNA is replicated, the leading strand will have _____ RNA primer(s) and the lagging strand will have _____ RNA primer(s). a. many; many b. one; one c. many; one d. one; many ANSWER: d 9. The enzyme responsible for joining Okazaki fragments together during DNA replication is: a. DNA ligase. b. helicase. c. topoisomerase II. d. DNA polymerase. e. DNA replicase. ANSWER: a 10. Given that the rate of DNA synthesis of a plasmid in yeast is 50 nucleotides per second and the circular plasmid replicates in 20 minutes, what is the approximate maximum size of the plasmid? a. 30,000 bp b. 120,000 bp c. 60,000 bp ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 11. Given that a chromosome is 60,000,000 bases in length, and DNA polymerase functions at a rate of 50 base pairs per second, what is the minimum number of origins of replication required on that chromosome to allow replication to complete if S phase lasts for 5.5 hours? a. about 60 b. one c. about 30 d. about 300 e. about 30,000 ANSWER: c 12. In replication of a linear double-stranded DNA molecule, one end of each strand becomes shorter in each round of replication. This happens because: a. the RNA primer cannot be replaced at the very end of a lagging DNA strand. b. DNA polymerase elongates a growing DNA strand at only the 5´ end. c. only one replicated DNA strand requires an RNA primer. d. All of these choices are correct. ANSWER: a 13. Consider the end of a DNA strand in a double-stranded molecule that has become shorter as a result of DNA replication. This same end will be shortened still further in: a. every fourth round of DNA replication. b. alternate rounds of DNA replication. c. every third round of DNA replication. d. every round of DNA replication. ANSWER: d 14. DNA replication in human cells occurs at a rate of about 50 base pairs per second, and the length of the longest human chromosome (chromosome 1) is 249 Mb (megabase, each Mb is 1,000,000 bases). If DNA replication took place continuously from one end of chromosome 1 to the other, it would take approximately: a. 2 minutes. b. 2 months. c. 2 hours. d. 2 days. e. 2 seconds. ANSWER: b 15. In contrast to linear DNA replication, circular DNA replication typically: a. does not produce Okazaki fragments. b. does not produce a replication bubble. c. occurs only at a single replication fork. d. has a single origin of replication. e. All of these choices are correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 ANSWER: d 16. The enzyme _____ compensates for chromosomes shortening at the tips after each round of DNA replication. a. DNA polymerase b. telomerase c. topoisomerase II d. DNA ligase e. DNA replicase ANSWER: b 17. If you were able to find a drug that could inhibit the reactivation of telomerase activity in cancer cells, the cancer cells would: a. slowly erode their chromosome ends. b. become less invasive. c. eventually die from lack of energy. d. stop dividing immediately. e. gradually revert to normal cells. ANSWER: a 18. In a long DNA molecule, each origin of replication produces a _____ with a _____ on each side. a. replication bubble; DNA ligase b. replication fork; replication bubble c. replication bubble; replication fork d. replication fork; DNA ligase e. None of the other answer options is correct. ANSWER: c 19. Each DNA parent strand within a replication bubble acts as a template strand that produces: a. only lagging strands. b. two leading strands or two lagging strands. c. either a leading strand or a lagging strand. d. only leading strands. e. one leading strand and one lagging strand. ANSWER: e 20. Telomerase is fully active in _____ and _____ cells, but almost completely inactive in _____ cells. a. germ; stem; somatic b. germ; somatic; stem c. stem; blood; germ d. somatic; blood; germ Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 e. somatic; germ; stem ANSWER: a 21. You are interested in cloning a fragment from the genome of an organism where the GC content of the sequence is 50%. The length of the gene you wish to study is 2 kb (kilobase). Each kb is 1000 bases and does not include cleavage sites for any of the restriction enzymes shown. For convenience in later experimental procedures, you want the average size of the fragments produced by the restriction enzyme to be approximately 4 kb. Which of the restriction enzymes would you choose to digest your genomic DNA? Consider the table of restriction enzymes and DNA.

a. Notl b. Swal c. EcoRI d. Alul ANSWER: c 22. The size of the yeast genome is 1.2 × 107, and the GC content is 38%. Approximately how many EcoR1 restriction fragments would you expect to get from cutting the yeast genome with EcoR1? a. 4000 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 b. 1000 c. 1 × 106 d. 500 e. 40,000 ANSWER: a 23. Which of these is necessary for the first step of PCR? a. original template DNA b. DNA polymerase c. the four DNA nucleotides d. DNA primers e. All of these choices are correct. ANSWER: a 24. The oligonucleotide primers used in the polymerase chain reaction are typically 20-30 nucleotides in length or longer; however, for purposes of this question, assume that 6 nucleotides is long enough. Suppose you wish to amplify the fragment shown (the raised dots indicate several kilobases of DNA sequence not shown). You decide to design primers corresponding to the regions that are underlined. What primer sequences would you use? 5´–ATGCTGAAACTTCTC···GGGATGAAATCAGTTT–3´ 3´–TACGACTTTGAAGAG···CCCTACTTTAGTCAAA–5´ a. 5´–TGAAAC–3´ and 5´–CTGATT–3´ b. 5´–GTTTCA–3´ and 5´–AATCAG–3´ c. 5´–GTTTCA–3´ and 5´–CTGATT–3´ d. 5´–TGAAAC–3´ and 5´–AATCAG–3´ ANSWER: a 25. Suppose you know the sequence of a region of DNA in an organism, but you want to know the unknown sequences that flank this known sequence. Cleverly, you cleave the DNA with a restriction enzyme at restriction sites outside the known sequence and then use DNA ligase to form a circle as shown in the figure. The dark blue portion of the circles represents double-stranded DNA whose sequence is known, and the light blue portion represents the flanking regions with unknown sequences. The numbered arrows are places where you consider designing PCR primers, with the 3´ end of each primer indicated by the arrowhead.

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Chapter 12

Which primers would you use to amplify the light blue region of the circle? a. 2 and 4 b. 1 and 2 c. 1 and 4 d. 2 and 3 e. 1 and 3 f. 3 and 4 ANSWER: e 26. If a restriction enzyme has a cleavage site consisting of six nucleotide pairs, what is the average distance in nucleotides between cleavage sites in a random sequence of double-stranded DNA where the AT content is 50%? a. 256 b. 64 c. 4096 d. 16 ANSWER: c 27. Gel electrophoresis separates DNA fragments based on: a. replication rate. b. nucleotide sequence. c. mutations. d. length. e. primers. ANSWER: d 28. DNA renaturation (hybridization) can be used to: a. determine the degree of similarity between two DNA samples. b. cleave long fragments of DNA into smaller fragments for analysis. c. amplify small samples of DNA by replicating them. d. determine the nucleotide sequence of a DNA strand. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 e. None of the other answer options is correct. ANSWER: a 29. DNA or RNA probes are used to: a. amplify small samples of DNA by replicating them. b. determine whether a particular DNA sequence is present in a sample. c. cut large segments of DNA into smaller fragments for analysis. d. determine the nucleotide sequence of a DNA strand. e. determine the degree of similarity between two DNA samples. ANSWER: b 30. The Southern blotting technique requires: a. None of the other answer options is correct. b. DNA primers. c. complementarity between the protein being expressed and the radioactive probe. d. separation of proteins by size. e. complementarity between the nucleotide sequence of the probe DNA and the DNA to be detected. ANSWER: e 31. Modern methods of massively parallel sequencing include devices for detecting fluorescence or visible light as each: a. nucleotide passes through a gel. b. nucleotide is released from a strand by DNase. c. nucleotide is incorporated during synthesis. d. dideoxynucleotide in turn terminates synthesis. ANSWER: c 32. The "$1000" genome is a catchphrase that describes a goal of reducing sequencing costs compared to the first genome ever sequenced by a factor of: a. ten. b. a hundred. c. a thousand. d. a million. e. a billion. ANSWER: e 33. The polymerase chain reaction (PCR) is used to generate: a. multiple copies of whole chromosomes. b. multiple copies of a targeted region of DNA. c. a single copy of a targeted region of DNA. d. single copies of whole chromosomes. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 e. multiple copies of whole chromosomes and a targeted region of DNA. ANSWER: b 34. One characteristic of restriction enzymes is that they cut: a. double-stranded DNA at specific sites. b. single-stranded DNA at specific sites. c. single-stranded DNA at random sites. d. double-stranded DNA at random sites. e. DNA fragments generated by gel electrophoresis. ANSWER: a 35. If you were to genetically engineer a bacterial cell to produce a human protein, you would use a bacterial plasmid as the _____, or carrier of the human DNA. a. probe b. vector c. palindrome d. shuttle e. None of the other answer options is correct. ANSWER: b 36. The transformation step in creating bacteria genetically engineered to produce human proteins involves: a. bacteria taking up the recombinant DNA in the form of the vectors. b. cleaving the donor and vector DNA so they can be bound into a single molecule. c. binding of the donor DNA to the vector DNA and ligating the two pieces. d. bacteria expressing the novel proteins encoded by the donor DNA. e. None of the other answer options is correct. ANSWER: a 37. In the technique of DNA editing by means of CRISPR, the purpose of the editing template DNA is to serve as a: a. template for CRISPR RNA. b. targeting guide for CRISPR-associated protein. c. targeting guide to direct DNA cleavage. d. repair of target DNA. e. None of the answer options is correct. ANSWER: d 38. An experiment in DNA editing using CRISPR is carried out in which the editing template DNA contains a G-A mismatch in the region to be edited. At the same nucleotide site in the resulting edited targeted DNA, the base pair is expected to be: a. C-T. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 b. G-C. c. T-A. d. G-A. e. None of the answer options is correct. ANSWER: a 39. Each end of a eukaryotic chromosome is capped by a repeating DNA sequence called the telomere. a. true b. false ANSWER: a 40. What feature of double-stranded DNA makes it necessary to have a leading strand and a lagging strand during replication? a. the antiparallel orientation of the strands b. the base stacking of the bases c. the negative charge on the sugar-phosphate backbone d. the hydrogen bonding between bases ANSWER: a 41. What would happen to the variation between organisms in a population if their DNA polymerase did not have a proofreading function? a. The amount of variation would decrease. b. The amount of variation would increase. c. The amount of variation would stay the same. ANSWER: b 42. A lack of telomerase activity limits what? a. the rate of meiosis b. the number of replication bubbles that form during replication c. the rate of DNA replication d. the number of Okazaki fragments that can be linked together e. the number of times a cell can divide ANSWER: e 43. The main function of DNA polymerase is to add new nucleotides to the 3´ end of a growing chain. What is the second role of this enzyme in DNA replication? a. synthesis of RNA primer b. unwinding of the DNA duplex c. proofreading d. stabilizing strands of DNA at replication fork e. joining Okazaki fragments Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 ANSWER: c 44. During the process of PCR, what plays the role in the test tube that helicase plays in the cell? a. dNTPs b. heat c. polymerase d. oligonucleotides ANSWER: b 45. Gel electrophoresis can distinguish between the _____ of different DNA fragments. a. sizes b. charge c. sequences ANSWER: a 46. Why do bacteria produce restriction enzymes? a. Restriction enzymes are involved in the bacterium's defense against viruses. b. Restriction enzymes are involved in the transcription of the bacterium's DNA. c. Restriction enzymes are important in the bacterium's DNA replication. d. Restriction enzymes serve no function in bacteria and are just used as a molecular tool by scientists. ANSWER: a 47. Two closely related DNA sequences renature more readily at higher temperatures than do two DNA sequences that are not as closely related. a. true b. false ANSWER: a 48. Using PCR, a student hopes to amplify a sequence of known DNA from a genomic DNA sample. The two primers are both designed to pair with sequences that have a 40% GC content. Using a particular set of denaturation temperatures, annealing temperatures, and extension temperatures, the experiment produces not the expected single DNA fragment of a single known size but a set of fragments of different sizes. Which of these factors could account for these results? a. The denaturation temperature was too high. b. The annealing temperature was too high. c. The sequences of the primers were not specific enough. d. The extension temperature was too low. ANSWER: c 49. Restriction enzymes cleave double-stranded DNA at the sites that show a particular type of symmetry; these sequences read the same on both strands and are called palindromes. Which of these sequences is not a palindrome? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 a. 5´-CCTCAGG-3´ b. 5´-CGGCCG-3´ c. 5´-GGTACC-3´ d. 5´-CCTGCAGG-3´ e. 5´-AGATCT-3´ ANSWER: a 50. The biggest remaining obstacle for personal genome sequencing is still: a. sequencing speed. b. the cost. c. gel electrophoresis. d. heat-stable DNA polymerase. e. the labor intensiveness of Southern blots. ANSWER: b 51. Which of the statements best describes the way you would engineer bacterial cells to produce a human protein? a. None of the answer options is correct. b. Use restriction enzymes to cleave the donor DNA and to shear the vector DNA into random pieces. c. Use restriction enzymes to cleave the vector DNA and to shear the donor DNA into random pieces. d. Randomly shear the donor DNA and the vector DNA. e. Use a specific restriction enzyme to cleave both the donor DNA and the vector DNA. ANSWER: e 52. DNA replication of small circular molecules usually starts at a single origin of replication and proceeds bidirectionally, that is, with two replication forks proceeding in opposite directions from the origin of replication. The time required for replication would be longer if replication of such a molecule were unidirectional rather than bidirectional. a. true b. false ANSWER: a 53. A student in a biology laboratory designs oligonucleotide primers to carry out the polymerase chain reaction. However, when ordering the sequences, she reverses the 3´ and 5´ ends. The vendor synthesizes the sequences as ordered and returns them to the student. The PCR will produce the expected DNA fragment. a. true b. false ANSWER: b 54. Which of these choices would block the process of DNA synthesis by DNA polymerase? a. All of these choices are correct. b. removal of the 5´ triphosphate from the nucleotides used by DNA polymerase Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 c. removal of the base from the nucleotides used by DNA polymerase d. removal of DNA ligase from the DNA replication process e. removal of the 3´-OH from the last nucleotide of the daughter strand ANSWER: a 55. What features of DNA make it possible to make recombinant DNA in the lab? a. Two double helices from different sources can be ligated together. b. Restriction enzymes cut DNA from all species. c. The genetic code is the same for all organisms. d. All of these choices are correct. ANSWER: d 56. In DNA editing by means of CRISPR, the function of the CRISPR-associated protein is to: Select all that apply. a. bind guide RNA b. bind target DNA. c. translate guide RNA. d. transcribe guide RNA. e. cleave target DNA. ANSWER: a 57. Suppose you add fluorescent ribonucleotides to a cell undergoing DNA replication so that the RNA primers used in DNA synthesis glow when viewed with a fluorescent microscope. You notice that, near each replication fork, one strand glows more than the other. Which strand glows more, and why? a. The leading strand glows more because it is elongated nearest the replication fork. b. The leading strand glows more because it forms the "trombone loop." c. The lagging strand glows more because its RNA primer is nearer the replication fork. d. The lagging strand glows more because it forms the "trombone loop." ANSWER: c 58. Consider a cell in which one of the proteins or enzymes involved in DNA replication is altered in a way that results in an increased rate of single-base changes in the newly synthesized DNA strand. Which function of which protein is most likely disrupted in this situation? a. the winding stress relief function of topoisomerase II b. the proofreading function of DNA polymerase c. the fragment joining function of DNA ligase d. the unwinding function of helicase e. the strand separation function of single-stranded binding protein ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 59. Which enzyme is the first to bind to the DNA sequences at the origin of replication? a. DNA primase b. DNA polymerase c. DNA helicase d. single-strand binding protein e. DNA ligase ANSWER: c 60. The fact that DNA replication occurs in virtually the same way in all organisms reflects: a. the universal genetic code. b. its multiple origins in different lineages of organisms. c. the laws of thermodynamics. d. the fact that mutations are generally harmful. e. its origin early in the history of life. ANSWER: e 61. When we say that DNA replication is semiconservative, we mean that: a. when DNA is replicated each new double helix contains one parental strand and one newly synthesized daughter strand. b. only half of an organism's DNA is replicated during each cell division. c. parental DNA stays in the parent cell and daughter DNA ends up in the daughter cell. d. None of the other answer options is correct. ANSWER: a 62. During DNA replication in a cell, RNA primase synthesizes a primer that is complementary to the region in the sequence shown in bold: 5´–CACAGCAGAAACCTACAACTCATG–3´ What is the primer sequence? a. 5´-GUUGUAGGUUUC-3´ b. 5´-GTTGTAGGTTTC-3´ c. 5´-CUUUGGAUGUUG-3´ d. 5´-CTTTGGATGTTG-3´ ANSWER: a 63. To cells that are defective in primer removal, you add fluorescent ribonucleotides when the cells are undergoing DNA replication. In this case, you observe that one strand glows more than the other not only near the replication fork but also at intervals along its length. Which strand glows in this way and why? a. The leading strand glows in this way because it is synthesized discontinuously. b. The lagging strand glows in this way because its RNA primers are required for each Okazaki fragment and are closely spaced. c. The lagging strand glows in this way because it is synthesized continuously. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 d. The leading strand glows in this way because it is synthesized continuously. ANSWER: b 64. The parallel lines shown represent the paired strands of a DNA double helix. If this molecule undergoes one round of replication, which ends are shorter in the daughter molecules than in the parental molecules?

a. y and x b. w and y c. w and x d. w and z e. y and z ANSWER: d 65. An organism is found in which the template RNA sequence in the telomerase is 5´-CCAUAA-3´. What is the DNA sequence of the organism's telomeres? a.

ANSWER: d 66. If a restriction enzyme has a cleavage site consisting of six nucleotide pairs, what is the chance that six adjacent nucleotides in a random sequence of double-stranded DNA will match the restriction site? a. (1/4)6 = 1/4096 b. (1/2)6 = 1/64 c. (1/4)4 = 1/256 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 d. (1/2)4 = 1/16 ANSWER: a 67. The polymerase chain reaction (PCR) amplifies DNA by means of repeated rounds of DNA replication. This means that the ends of the amplified fragments become shorter with each round of replication. a. true b. false ANSWER: b 68. The oligonucleotide primers used in the polymerase chain reaction are typically 20-30 nucleotides in length or longer; however, for purposes of this problem, assume that six nucleotides is long enough. You wish to amplify the fragment shown (the raised dots indicate several kilobases of DNA sequence not shown) and decide to design primers corresponding to the regions that are shown in bold. What primer sequences would you use?

a. 5´-ACTTGC-3´ and 5´-GTGGCA-3´ b. 5´-ACTTGC-3´ and 5´-TGCCAC-3´ c. 5´-GCAAGT-3´ and 5´-GTGGCA-3´ d. 5´-GCAAGT-3´ and 5´-TGCCAC-3´ ANSWER: b 69. Using PCR, a student wants to amplify a sequence of known DNA from a genomic DNA sample. The two primers are both designed to pair with sequences that have a 40% GC content. Using a particular set of denaturation temperatures, annealing temperatures, and extension temperatures, the experiment does not produce the expected single DNA fragment of a single known size, but instead a set of fragments of varied sizes. Which factor could account for these results? a. The extension temperature was too low, so many fragments were terminated before the full fragment was replicated. b. The denaturation temperature was too high, so the primers could bind with each other as well as with multiple genomic sites containing mismatches. c. The annealing temperature was too low, so the primers could bind with multiple genomic sequences that contain some mismatched bases. ANSWER: c 70. The technique of Sanger sequencing takes advantage of the fact that dideoxynucleotides (nucleotides in which the 3´ hydroxyl group is absent) act as: a. None of the other answer options is correct. b. restriction sites. c. palindromic sites. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 d. probes. e. chain terminators. ANSWER: e 71. A certain restriction enzyme X cleaves double-stranded DNA at the sequence shown, where the slash indicates where each strand is cleaved. 5´-AA/ATTT-3 3´-TTTA/AA-5´ Note that the cleavage results in a two-base pair single-stranded region at the 5´ end that allows the cleaved ends to undergo base pairing. Which of the restriction enzymes that cleave double-stranded DNA (indicated by the slash) would produce overhanging ends able to pair with those produced by enzyme X? a. 5´-GG/ATCC-3´ 3´-CCTA/GG-5´ b. 5´-AATA/TT-3´ 3´-TT/ATAA-5´ c. 5´-AAA/TTT-3´ 3´-TTT/AAA-5´ d. 5´-GGA/TCC-3´ 3´-CCT/AGG-5´ e. 5´-GGAT/CC-3´ 3´-CCTA/GG-5´ ANSWER: a 72. If you were to use CRISPR to cleave DNA at sites flanking a gene 3000 bp in length, and in this way, produce a deletion of the gene, how many guide RNAs would you need? a. 3 b. 1 c. 2 d. 4 ANSWER: c 73. Suppose you carry out a DNA-editing experiment using CRISPR in which the editing template DNA has strands labeled with the heavy nitrogen isotope 15N. The experiment is carried out in the presence of the normal light isotope 14N. Then the expected distribution of the isotope in the strands in the edited region of the target DNA would be: a. that both strands are labeled with 14N. b. that both strands are labeled with 15N. c. that only one strand is labeled with 14N and the other strand with 15N. ANSWER: a 74. Consider the restriction enzymes shown. What type of DNA ends would they produce (5´ overhangs, 3´ Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 overhangs, or no overhangs)?

a. AluI: no overhang; BAM HI: 3´ overhang; KpnI: 5´ overhang b. AluI: 5´ overhang; BAM HI: 5´ overhang; KpnI: 5´ overhang c. AluI: 3´ overhang; BAM HI: no overhang; KpnI: no overhang d. AluI: no overhang; BAM HI: 5´ overhang; KpnI: 3´ overhang ANSWER: d 75. Incorrect nucleotides are sometimes incorporated into DNA during the process of replication. In each of the two examples shown, labeled Example 1 and Example 2, the template strand is on top and the replicating daughter strand at the bottom. In each example, the nucleotide underlined in the daughter strand is incorrect because it cannot pair properly with the base in the template strand. In Example 1, the proofreading function of DNA polymerase will remove the incorrect nucleotide from the daughter strand and replace it with the correct nucleotide, but in Example 2, the incorrect nucleotide cannot be repaired.

a. true b. false ANSWER: a 76. You carry out a polymerase chain reaction (PCR) to amplify a particular sequence of DNA from a solution that initially contains exactly one double-stranded molecule containing the sequence. After 30 cycles of PCR, how many copies of the sequence are present? a. 230 b. 30 c. 215 d. 260 ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 77. A molecular biologist carries out a PCR reaction and discovers to his disappointment that no PCR amplification takes place. It turns out that the problem is one of the primers, which has this sequence: 5´–ATAACACTAAATAGCGCTATTTAGTGTTAT–3´ What is a reason why this primer does not work for PCR? a. The primer is too long. b. The last 15 nucleotides of the primer are the exact base-pair complement of the first 15 nucleotides. c. The primer contains too many As and Ts. d. The annealing temperature for the PCR to work is excessively high. ANSWER: b 78. Production of a recombinant DNA molecule makes use of which of these properties of DNA? a. Complementary single-stranded nucleic acid sequences can come together to form a duplex molecule. b. A restriction enzyme cleaves duplex DNA molecules only at the positions of certain short sequences that constitute the enzyme's restrictions site. c. DNA polymerase elongates a growing DNA strand by the addition of successive deoxynucleotides to the 3¢ end. d. DNA ligase is an enzyme that can join the ends of single-stranded DNA molecules. e. All of the choices are correct. ANSWER: e 79. Suppose that you digested the genomic DNA of D. melanogaster with EcoRI 5´-G|AATTC-3´. The vertical line indicates the position of cleavage. Then you ligate the resulting fragments into a unique MunI cloning site 5´-C|AATTG-3´ of a plasmid vector. What proportion of the cloned fragments would it be possible to isolate from the vector using MunI? a. 6.25% b. 25% c. 50% d. 100% ANSWER: a 80. The accompanying table gives hypothetical data on the average length of telomeres among groups of people that differ in the lifestyle variables of stress, level of vigorous exercise, and daily caloric intake.

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Chapter 12

To test the hypothesis that "high levels of stress have an effect on average length of telomeres," which groups would you compare? a. group 1 and group 2 b. group 1 and group 3 c. group 1 and group 4 d. group 2 and group 3 e. group 3 and group 4 ANSWER: d 81. The accompanying table gives hypothetical data on the average length of telomeres among groups of people that differ in the lifestyle variables of stress, level of vigorous exercise, and daily caloric intake.

If you were to compare group 1 and group 3, which of the hypotheses could you test? a. Caloric intake has an effect on telomere length. b. Vigorous exercise has an effect on telomere length. c. High levels of stress have an effect on telomere length. d. No single variable can be tested in this comparison. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 ANSWER: a 82. The accompanying table gives hypothetical data on the average length of telomeres among groups of people that differ in the lifestyle variables of stress, level of vigorous exercise, and daily caloric intake.

If you were to compare group 3 and group 4, which of the hypotheses could you test? a. Caloric intake has an effect on telomere length. b. Vigorous exercise has an effect on telomere length. c. No single variable can be tested in this comparison. d. A high level of stress has an effect on telomere length. ANSWER: c 83. Shown in the diagram is an abstract representation of a region of double-stranded DNA in which the vertical tick marks indicate the location of cleavage sites for the restriction enzyme BamHI. The numbers are the sizes of the restriction fragments in kilobase pairs. X, Y, and Z represent three possible probes that could be used in a Southern blot to identify one or more restriction fragments from the region.

You carry out such a Southern blot experiment, digesting the genomic DNA with BamHI and probing with one or more of the probes X, Y, or Z. In the gel diagram shown, lane 1 consists of DNA fragments of known size ranging from 1 to 10 kb.

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Chapter 12

Which of the lanes depicts what you would expect to observe when probing the Southern blot with probe X? a. lane H b. lane K c. lane L d. lane M e. lane Q ANSWER: e 84. Shown in the diagram is an abstract representation of a region of double-stranded DNA in which the vertical tick marks indicate the location of cleavage sites for the restriction enzyme BamHI. The numbers are the sizes of the restriction fragments in kilobase pairs. X, Y, and Z represent three possible probes that could be used in a Southern blot to identify one or more restriction fragments from the region.

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Which of the lanes depicts what you would expect to observe when probing the Southern blot with probe Y? a. lane H b. lane K c. lane L d. lane M e. lane Q ANSWER: b 85. Shown in the diagram is an abstract representation of a region of double-stranded DNA in which the vertical tick marks indicate the location of cleavage sites for the restriction enzyme BamHI. The numbers are the sizes of the restriction fragments in kilobase pairs. X, Y, and Z represent three possible probes that could be used in a Southern blot to identify one or more restriction fragments from the region.

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Which of the lanes depicts what you would expect to observe when probing the Southern blot with probe Z? a. lane H b. lane K c. lane L d. lane M e. lane Q ANSWER: c 86. Shown in the diagram is an abstract representation of a region of double-stranded DNA in which the vertical tick marks indicate the location of cleavage sites for the restriction enzyme BamHI. The numbers are the sizes of the restriction fragments in kilobase pairs. X, Y, and Z represent three possible probes that could be used in a Southern blot to identify one or more restriction fragments from the region.

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Which of the lanes depicts what you would expect to observe when probing the Southern blot with probes Y and Z together? a. lane H b. lane K c. lane L d. lane M e. lane Q ANSWER: d 87. Shown in the diagram is an abstract representation of a region of double-stranded DNA in which the vertical tick marks indicate the location of cleavage sites for the restriction enzyme BamHI. The numbers are the sizes of the restriction fragments in kilobase pairs. X, Y, and Z represent three possible probes that could be used in a Southern blot to identify one or more restriction fragments from the region.

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In the gel electrophoresis diagram, note that the DNA fragments in lane 1 are sorted by size and that the larger size fragments move a smaller distance from their original positions in the slots at the top of the gel. Why do smaller fragments move a greater distance than larger fragments in an electrophoresis gel? a. The smaller fragments have less difficulty getting through the pores in the gel. b. The smaller fragments are more likely to remain double stranded. c. The smaller fragments have fewer negative charges along the DNA backbone. d. The smaller fragments are less likely to get tangled up with one another. ANSWER: a 88. You find a way to attach either a red or green fluorescent dye to nucleotides. Double-stranded DNA molecules with both strands labeled red fluoresce red, those with both strands labeled green fluoresce green, and those with one strand labeled red and the other green fluoresce yellow. You grow human cells in the presence of "red" nucleotides until both DNA strands of all chromosomes fluoresce red. You then allow one round of DNA replication in the presence of "green" nucleotides. What pattern of fluorescence do you expect to see in the sister chromatids of each chromosome? a. Both chromatids will be yellow. b. One chromatid will be red, and the other will be yellow. c. One chromatid will be green, and the other will be yellow. d. One chromatid will be red, and the other will be green. ANSWER: a 89. You find a way to attach either a red or green fluorescent dye to nucleotides. Double-stranded DNA molecules with both strands labeled red fluoresce red, those with both strands labeled green fluoresce green, and those with one strand labeled red and the other green fluoresce yellow. You grow human cells in the presence of "red" nucleotides until both DNA strands of all chromosomes fluoresce red. You then allow cells with DNA labeled "red" in both DNA strands to undergo two rounds of DNA replication in the presence of "green" Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 nucleotides. What pattern of fluorescence would you expect to see in the sister chromatids of each chromosome? a. Both chromatids will be yellow. b. One chromatid will be red, and the other will be yellow. c. One chromatid will be green, and the other will be yellow. d. One chromatid will be red, and the other will be green. ANSWER: c 90. Refer to the figure shown. On the left is a diagram of double-stranded DNA of a circular plasmid from a bacterial cell. The positions X, Y, and Z are sites where the plasmid DNA is cleaved by restriction enzymes X, Y, and Z, respectively. The numbers are the size in kilobase pairs (kb) of the DNA regions between the restriction sites. On the right is a diagram of an electrophoresis gel. Lane 1 is the ladder, comprised of DNA fragments that range in size from 1–12 kb, showing the position of each band size after electrophoresis. The other lanes are DNA bands observed from digestion of the plasmid with one or more of the restriction enzymes.

Which lane contains the DNA fragments produced by digestion of the plasmid with enzyme X only? a. lane H b. lane K c. lane L d. lane M e. lane Q ANSWER: b 91. Refer to the figure shown. On the left is a diagram of double-stranded DNA of a circular plasmid from a bacterial cell. The positions X, Y, and Z are sites where the plasmid DNA is cleaved by restriction enzymes X, Y, and Z, respectively. The numbers are the size in kilobase pairs (kb) of the DNA regions between the restriction sites. On the right is a diagram of an electrophoresis gel. Lane 1 is the ladder, comprised of DNA fragments that range in size from 1–12 kb, showing the position of each band size after electrophoresis. The other lanes are DNA bands observed from digestion of the plasmid with one or more of the restriction enzymes.

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Which lane contains the DNA fragments produced by digestion of the plasmid with enzymes X and Y? a. lane H b. lane K c. lane L d. lane M e. lane Q ANSWER: d 92. Refer to the figure shown. On the left is a diagram of double-stranded DNA of a circular plasmid from a bacterial cell. The positions X, Y, and Z are sites where the plasmid DNA is cleaved by restriction enzymes X, Y, and Z, respectively. The numbers are the size in kilobase pairs (kb) of the DNA regions between the restriction sites. On the right is a diagram of an electrophoresis gel. Lane 1 is the ladder, comprised of DNA fragments that range in size from 1–12 kb, showing the position of each band size after electrophoresis. The other lanes are DNA bands observed from digestion of the plasmid with one or more of the restriction enzymes.

Which lane contains the DNA fragments produced by digestion of the plasmid with enzymes X and Z? a. lane H b. lane K Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 c. lane L d. lane M e. lane Q ANSWER: e 93. Refer to the figure shown. On the left is a diagram of double-stranded DNA of a circular plasmid from a bacterial cell. The positions X, Y, and Z are sites where the plasmid DNA is cleaved by restriction enzymes X, Y, and Z, respectively. The numbers are the size in kilobase pairs (kb) of the DNA regions between the restriction sites. On the right is a diagram of an electrophoresis gel. Lane 1 is the ladder, comprised of DNA fragments that range in size from 1–12 kb, showing the position of each band size after electrophoresis. The other lanes are DNA bands observed from digestion of the plasmid with one or more of the restriction enzymes.

Which lane contains the DNA fragments produced by digestion of the plasmid with enzymes Y and Z? a. lane H b. lane K c. lane L d. lane M e. lane Q ANSWER: c 94. Refer to the figure shown. On the left is a diagram of double-stranded DNA of a circular plasmid from a bacterial cell. The positions X, Y, and Z are sites where the plasmid DNA is cleaved by restriction enzymes X, Y, and Z, respectively. The numbers are the size in kilobase pairs (kb) of the DNA regions between the restriction sites. On the right is a diagram of an electrophoresis gel. Lane 1 is the ladder, comprised of DNA fragments that range in size from 1–12 kb, showing the position of each band size after electrophoresis. The other lanes are DNA bands observed from digestion of the plasmid with one or more of the restriction enzymes.

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Which lane contains the DNA fragments produced by digestion of the plasmid with enzymes X and Y and Z? a. lane H b. lane K c. lane L d. lane M e. lane Q ANSWER: a 95. Refer to the figure shown. On the left is a diagram of double-stranded DNA of a circular plasmid from a bacterial cell. The positions X, Y, and Z are sites where the plasmid DNA is cleaved by restriction enzymes X, Y, and Z, respectively. The numbers are the size in kilobase pairs (kb) of the DNA regions between the restriction sites. On the right is a diagram of an electrophoresis gel. Lane 1 is the ladder, comprised of DNA fragments that range in size from 1–12 kb, showing the position of each band size after electrophoresis. The other lanes are DNA bands observed from digestion of the plasmid with one or more of the restriction enzymes.

Why do the sizes of the bands in lane 2 not add up to 12 kb? a. Because this restriction digest produces two 5 kb fragments, and they form one band on the gel. b. Because one of the fragments is less than 1 kb in size and can't be seen on the gel. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 c. Because this restriction digest produces two 2 kb fragments and they form one band on the gel. ANSWER: a 96. You would like to amplify 100 bp sequence from the genome of the baker's yeast S. cerevisiae (12 × 106 bp) using polymerase chain reaction. At the end of the amplification, you want the amount of amplified DNA to represent not less than 90% of the total DNA. How many cycles of amplification will you need to run to achieve that 90% goal? a. 7 b. 14 c. 24 d. 35 ANSWER: b 97. You would like to amplify 100 bp sequence from the genome of the baker's yeast S. cerevisiae (12 × 106 bp) using polymerase chain reaction. At the end of the amplification, you want the amount of amplified DNA to represent not less than 99% of the total DNA. How many cycles of amplification you need to run if you want amplified DNA to represent 99% of total DNA? a. 7 b. 14 c. 24 d. It is impossible to achieve this goal. ANSWER: c 98. You would like to amplify 1000 bp sequence from the genome of the baker's yeast S. cerevisiae (12 × 106 bp) using polymerase chain reaction. At the end of the amplification, you want the amount of amplified DNA to represent not less than 99% of the total DNA. How many cycles of amplification you need to run if you want amplified DNA to represent 99% of total DNA? a. 14 b. 21 c. 30 d. It is impossible to achieve this goal. ANSWER: b 99. You would like to amplify 1000 bp sequence from the human genome (3 × 109 bp) using polymerase chain reaction. At the end of the amplification, you want the amount of amplified DNA to represent not less than 99% of the total DNA. How many cycles of amplification you need to run if you want amplified DNA to represent 99% of total DNA? a. 14 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 b. 29 c. 30 d. It is impossible to achieve this goal. ANSWER: b Multiple Response 100. A messenger RNA from a human gene that is 4 kb in length is used as a probe in a Southern blot to identify restriction fragments of chromosomal DNA to which the mRNA can hybridize. Restriction fragments of DNA of 4 kb and 5 kb are observed to hybridize with the mRNA probe. What could explain how a 4 kb probe can hybridize with fragments that together sum to 9 kb? Select all that apply. a. A restriction fragment may extend beyond the 5´ end of the transcription start site. b. The restriction fragments may extend beyond the 5´ or 3´ ends of the transcription stop site. c. The gene from which the mRNA probe is transcribed may contain introns. d. The gene from which the mRNA probe is transcribed has only one exon. e. The result must be some sort of mistake. ANSWER: b, c 101. As in the development of computer hardware, interest in large-scale DNA sequencing has stimulated the development of sequencing devices that: Select all that apply. a. are smaller. b. have improved capacity through automation. c. have improved capacity through open-source software development. d. require less energy to operate. e. None of the other answer options is correct. ANSWER: a, b 102. Why is DNA ligase important for a cell? Select all that apply. a. It unwinds DNA. b. It joins DNA together from different origins of replication. c. It relieves the stress of DNA unwinding. d. It joins Okazaki fragments. ANSWER: b, d 103. You are interested in studying human ß-globin and decide to produce the protein in bacterial cells. The ßglobin gene is a relatively simple gene with a single intron. You insert the entire gene, including the gene's promoter, in a plasmid and transform the recombinant plasmid into E. coli. You are disappointed to discover that your bacterial culture does not produce the correct protein. What are some possible explanations? Select all that apply. a. The bacterial cell cannot splice introns. b. The bacterial cell does not recognize the eukaryote promoter. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 12 c. The bacterial cell produces the protein, but the protein is rapidly degraded. d. The bacterial cell is missing RNA polymerase. ANSWER: a, b, c 104. Which factors make the Southern blot technique possible? Select all that apply. a. New nucleotides are added only to the 3´ end of a growing DNA strand. b. A DNA strand whose 3´ end terminates in a dideoxynucleotide cannot be elongated. c. Complementary single-stranded nucleic acid sequences can come together to form a duplex molecule. d. Duplex nucleic acid molecules can be separated by size by means of gel electrophoresis. e. Single-stranded nucleic acid molecules can be immobilized on certain types of filter paper. ANSWER: c, d, e 105. Which factors make sequencing by the Sanger chain-termination method possible? Select all that apply. a. New nucleotides are added only to the 3´ end of a growing DNA strand. b. A DNA strand whose 3´ end terminates in a dideoxynucleotide cannot be elongated. c. Complementary single-stranded nucleic acid sequences can come together to form a duplex molecule. d. Duplex nucleic acid molecules can be separated by size by means of gel electrophoresis. e. Single-stranded nucleic acid molecules can be immobilized on certain types of filter paper. ANSWER: a, b, c, d 106. Which factors make amplification of a DNA fragment by means of the polymerase chain reaction possible? Select all that apply. a. Complementary single-stranded nucleic acid sequences can come together to form a duplex molecule. b. Duplex nucleic acid molecules can be separated by size by means of electrophoresis. c. A restriction enzyme cleaves duplex DNA molecules only at the positions of certain short sequences that constitute the enzyme's restrictions site. d. DNA polymerase elongates a growing DNA strand by the addition of successive deoxynucleotides to the 3´ end. e. DNA ligase is an enzyme that can join the ends of single-stranded DNA molecules. ANSWER: a, d

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Chapter 13 Multiple Choice 1. A shotgun sequencing project yields a sequence for a strand of DNA. 5′-GGTTTGGAGTAT-3′ In assembling the genome sequence from the short fragments, you look for other fragments that overlap the sequence above. Which sequence overlaps with the given sequence? a. 5′-TTTGGAGTATGG-3′ b. 5′-AATTTGGAGTAT-3′ c. 5′-GGTTTGGAGTGG-3′ d. 5′-CATTTGGAGTGG-3′ ANSWER: a 2. Approximately what percentage of the human genome actually codes for proteins? a. 1.0% b. 2.5% c. 45.0% d. 97.5% e. 99.0% ANSWER: b 3. Sometimes a single-stranded molecule of RNA is able to fold back on itself because the nucleotide sequence on one part of the RNA is complementary to another part. This sequence motif results in a: a. negative supercoil. b. transcription factor binding site. c. chromosome scaffold. d. hairpin-shaped structure. e. positive supercoil. ANSWER: d 4. Which step comes first in shotgun sequencing? a. matching regions of overlap b. putting the sequences in the correct order c. breaking the DNA into small fragments d. sequencing the DNA e. reconstructing the long sequence of nucleotides ANSWER: c 5. A certain sequence motif consists of 5 nucleotide pairs. In a random sequence of double-stranded DNA in which the percentage of A-T nucleotide pairs equals that of G-C, what is the probability that any group of 5 adjacent nucleotides has the same sequence as the sequence motif? a. (1/4)5= 1/1024 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 b. (1/5)4= 1/625 c. (1/2)5= 1/32 d. (1/5)2= 1/25 ANSWER: a 6. Sequences of genomic DNA and the corresponding messenger RNA (mRNA) are often compared to obtain valuable information for genome annotation. Why is this comparison useful? a. The open reading frame in mRNA includes the introns from the genomic DNA. b. The exclusion of introns in mRNA reveals the intron-exon structure of many protein coding genes. c. Because mRNA is shorter than genomic DNA, it is easier to find open reading frames. d. The sequences of genomic DNA and mRNA are identical, which serves as independent validation. ANSWER: b 7. According to the phylogenetic tree of the evolutionary relationships among viruses, which virus is most closely related to the cow lentivirus?

a. a sheep lentivirus b. human lentivirus HIV1 c. a horse lentivirus d. a cat lentivirus ANSWER: b 8. Which statement best describes why genome sequencing can be complicated by repeated sequences? a. The repeated sequences are too small to gather usable sequence information from. b. Automated sequencing devices interpret repeated sequences as a single-copy sequence. c. Repeated sequences artificially inflate the genome size, leading to an exaggerated interpretation of the complexity of the organism. d. The repeated sequences are often longer than the sequences obtained by automated sequencing. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 9. Which statement represents repeated sequences commonly found in the human genome? a. All of these choices are correct. b. long sequences dispersed throughout the genome c. short sequences repeated many times in tandem d. sequences whose transcript can fold back to form a hairpin e. ong sequences repeated in tandem ANSWER: a 10. Why does the phylogenetic tree suggest that human HIV came from chimpanzees at least twice?

a. It shows that all human HIVs share one common ancestor. b. It shows that all human HIVs are more evolved than cat lentiviruses. c. It shows that not all human HIVs are clustered together in the tree. d. It shows that sheep lentiviruses are more closely related to all human HIVs than cow lentiviruses. ANSWER: c 11. The estimated number of genes in the human genome is: a. 25,000. b. 15,000. c. 40,000. d. 250,000. e. 400,000. ANSWER: a 12. The marbled lungfish has a genome size almost 50 times larger than that of the human genome. The most likely explanation is that the lungfish: a. genome is most likely polyploid. b. genome contains a lot of repetitive DNA. c. genome has a lot more genes. d. has much bigger cells and therefore needs more DNA. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 ANSWER: b 13. Various species of the flowering plant of the genus Chrysanthemum have 18, 36, 54, 72, and 90 chromosomes. The variation is likely due to: a. repetitive DNA. b. transposable elements. c. chromosome fission. d. aneuploidy. e. polyploidy. ANSWER: e 14. The C-value paradox states that genome size: a. is uncorrelated with the complexity of the organism. b. is positively correlated with the complexity of the organism. c. is negatively correlated with the complexity of the organism. d. differs in reproductive cells and nonreproductive cells. ANSWER: a 15. A retrotransposon known as LINE1 is about 1000 base pairs (bp) in length and is present in the human genome in about 516,000 copies. Approximately what percentage of the human genome is accounted for by this transposon? a. 0.17% b. 0.017% c. 17% d. 1.7% e. 0.0017% ANSWER: c 16. According to the figure, which one set accurately lists the taxonomic groups in order of increasing genome size, that is, from smallest to largest?

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Chapter 13

a. bacteria, teleosts, mammals, salamanders b. teleosts, bacteria, salamanders, mammals c. bacteria, teleosts, salamanders, mammals d. salamanders, mammals, teleosts, bacteria ANSWER: a 17. Protozoa make up the taxonomic group with the broadest range of genome sizes, including the smallest and largest genome, among all eukaryotes.

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Chapter 13

a. true b. false ANSWER: a 18. Can it be determined from the data in the table which eukaryotic species has the largest genome? Can it be determined which species has the greatest number of chromosomes?

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Chapter 13 a. yes; yes b. yes; no c. no; yes d. no; no ANSWER: d 19. Human mitochondrial DNA does not have telomeres because it is: a. too small to need a telomere. b. replicated without an RNA primer. c. circular. d. not in the nucleus. ANSWER: c 20. The diameter of a metaphase chromosome is about 1400 nm, whereas that of the DNA double helix is 2 nm. This means that DNA is coiled into a structure with a diameter 700 times larger than itself. The diameter of a human hair is about 80 micrometers (0.003 inch). If a human hair were coiled into a structure with a diameter 700 times larger than itself, the diameter of the coiled human hair would approximate that of a: a. soda can (about 2 inches). b. pencil (about 0.3 inch). c. spaghetti noodle (about 0.1 inch). d. telephone pole (about 18 inches). e. rash can (about 36 inches). ANSWER: a 21. The human genome has 23 distinct types of chromosomes. How many individual DNA molecules would be present in the nucleus of a human skin cell just before mitosis? a. roughly 23,000 b. 46 c. 69 d. 23 e. 92 ANSWER: e 22. How many pairs of chromosomes does the human genome normally have? a. 15 b. 23 c. 6 d. 38 e. 46 ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 23. Bacterial nucleoid does not contain: a. histones. b. supercoiled DNA. c. enzymes. d. proteins. ANSWER: a 24. A virus with a wide host range may infect multiple species of both animals and plants: a. true b. false ANSWER: b 25. A genetic engineer wants to modify a virus to insert a synthetic gene into human lung cells. Which modified viruses would achieve her goal? a. All of these choices are correct. b. a bacteriophage that has been modified to carry the gene c. a tobacco virus modified to carry the gene d. a rabies virus modified to infect lung cells and carry the gene ANSWER: d 26. If a genome is sequenced in such a way that any given nucleotide is present in n sequenced fragments, and if each nucleotide is equally likely to be sequenced, the expected proportion of nucleotides that are not present in any of the fragments is given by e–n, where e is the base of natural logarithms. (The value of e is approximately 2.7182.) What is the probability that a nucleotide is not sequenced if n = 1? a. 37% b. 95% c. 5% ANSWER: a 27. If a genome is sequenced in such a way that any given nucleotide is present in n sequenced fragments, and if each nucleotide is equally likely to be sequenced, the expected proportion of nucleotides that are not present in any of the fragments is given by e–n, where e is the base of natural logarithms. The value of e is approximately 2.7182. What is the probability that a nucleotide is sequenced if n = 3? a. 5% b. 37% c. 95% ANSWER: c 28. Some viruses with an RNA genome require a virus-encoded reverse transcriptase. a. true b. false ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 29. Assembling complete genomic sequences based on the overlap of sequenced fragments can be difficult because the sequenced fragments: Select all that apply. a. All of these choices are correct b. may have errors in them. c. may come from either DNA strand. d. may overlap but come from different locations in the genome. e. may actually come from bacteria, viruses, or other contaminants. f. may be relatively short (50–100 nucleotides). ANSWER: a 30. In terms of the sequence known as "the human genome," which statement is true? a. It is accessible online only for physicians and researchers. b. It represents what the genome of a perfectly healthy person would be. c. It corresponds to the genome sequence of the common ancestor of all humans. d. It includes no mutant genes. e. It does not really exist in any human being. ANSWER: e 31. A new virus is discovered that infects sheep. The viral genome does not contain a reverse transcriptase gene and can base pair with gene sequences complementary to its mRNA sequence. To which virus group could this newly discovered virus belong? Select all that apply.

a. type VI b. type III Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 c. type V d. type II e. type VII ANSWER: b 32. Which characteristic is true of all viruses? Select all that apply. a. They must infect a host cell to reproduce. b. They contain a DNA genome. c. They have a protein coat called a capsid. d. They contain a double-stranded RNA genome. e. They are surrounded by a lipid envelope. ANSWER: a 33. Tandem repeats complicate the alignment of DNA fragments in shotgun sequencing. Which action might solve that problem? a. Sequence more DNA fragments. b. Develop techniques to sequence longer DNA fragments. c. Combine sequencing with chromosome painting. d. Sequence both strands of the tandem repeats. ANSWER: b 34. Chloroplasts and mitochondrion have double-stranded RNA genomes. a. true b. false ANSWER: b 35. Repeated DNA sequences represent a special challenge in genome sequence assembly. Which would be harder to assemble correctly, assuming the number of copies of the repeat can be determined? a. sequences containing repeats longer than the DNA fragments to assemble b. sequences containing repeats shorter than the DNA fragments to assemble ANSWER: a 36. Which pair of people has the exact same genome? a. father and son b. mother and daughter c. monozygotic twins d. dizygotic twins e. None of the other answer options is correct. ANSWER: c 37. A baby is born to a biological mother and father. For conception, each parent donated a set of chromosomes and an unrelated woman donated mitochondria. How many genomes did this baby inherit? How many genomes Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 does the baby have? a. 3; 3 b. 2; 1 c. 2.5; 2.5 d. 1; 1 e. 3; 1 ANSWER: e 38. In a random sequence of double-stranded DNA with equal nucleotide frequencies, what is the average number of nucleotide pairs between two occurrences of a sequence motif consisting of four nucleotides? a. 256 b. 32 c. 64 d. 128 e. 16 ANSWER: a 39. Consider a stretch of DNA with two short sequence motifs separated by an unknown number of base pairs. Each position in the intervening DNA is equally likely to be any of the four nucleotides. If you try to match your short sequence by moving it along a much larger sequence, would you expect many matching sites to be close together and a few farther apart? Few to be close together and many farther apart? Equal spacing? A normal distribution of distances between sites? In other words, which curve do you think best approximates the distribution of the number of matching nucleotide pairs between two occurrences of a short sequence motif?

a. curve H b. curve M c. curve K d. curve L ANSWER: b 40. How can researchers distinguish exons from introns in a segment of DNA? a. Only exons contain three-base sequences that can code for amino acids. b. Primers will not bind to introns. c. Exons have a characteristic sequence. d. The sequence of exons complements mRNA molecules in the cell. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 41. Which property of transposable elements allows them to contribute to the C-value paradox? a. All of these choices are correct. b. Copies of transposable elements can appear on multiple chromosomes. c. Their replication is controlled by genes found on the transposable element itself. d. Their copy number can increase from one generation to the next. ANSWER: a 42. The complexity of an organism is proportional to the number of genes in its genome. a. true b. false ANSWER: b 43. The genome of a newly identified organism is discovered and genome assembly shows that it is a single continuous strand of DNA. This organism is more likely to be: a. a eukaryote. b. a bacterium. c. Each option is equally possible. ANSWER: b 44. How many human chromosomes could fit into a single bacterial cell? a. 92 b. 1 c. 23 d. 46 e. Zero because bacteria are too small to contain a human chromosome. ANSWER: e 45. One difference between eukaryotic and prokaryotic DNA packaging is that: a. DNA wraps around histones in eukaryotes. b. DNA molecules in eukaryotic chromosomes are circular, whereas the genome of prokaryotes is a single linear DNA molecule. c. the first level of packaging of the DNA in prokaryotes is sometimes referred to as "beads on a string", whereas DNA in eukaryotic chromosomes is supercoiled. d. chromosomes of eukaryotes are confined within the nucleus, whereas prokaryotes have their genome spread out over the volume of the cell. ANSWER: a 46. Which group lists the levels of genetic information in order from smallest to largest. a. genome; gene; chromosome; exon b. exon, gene; chromosome; genome c. exon; chromosome; gene; genome d. exon; gene; genome; chromosome Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 e. gene; chromosome; exon; genome ANSWER: b 47. Similarities between the nucleoid in bacteria and the chromosome scaffold in eukaryotes demonstrate that these structures did not evolve independently and that they use same types of protein to bind DNA to form their folded structures. a. true b. false ANSWER: b 48. In eukaryotes, genetic material is packaged in the nucleus. Which group accurately lists the levels of the packaging in order of increasing size? a. chromatin, nucleosome, chromosome b. chromosome, nucleosome, chromatin c. chromosome, chromatin, nucleosome d. nucleosome, chromatin, chromosome e. nucleosome, chromosome, chromatin ANSWER: d 49. Virus genomes can be: a. single-stranded RNA. b. single-stranded DNA. c. double-stranded DNA. d. double-stranded RNA. e. All of these choices are correct. ANSWER: e 50. DNA that does not code for proteins serves no function in the cell. a. true b. false ANSWER: b 51. In a random sequence of double-stranded DNA, the probability that there are exactly n nucleotide pairs between two successive occurrences of a short sequence motif is given by (1 – x)nx, where x is the probability of occurrence of the sequence motif along the double-stranded DNA. When x = 1/512, the mean distance between occurrences is 512 nucleotide pairs. What is the probability that the distance between two occurrences is exactly equal to this mean? a. 0.072 b. 0.72 c. 0.00072 d. 0.0072 ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 52. When an open reading frame (ORF) is identified, it may not actually correspond to the amino acid sequence of any polypeptide in the cell. a. true b. false ANSWER: a 53. Which statement explains why the size of an organism's genome is not directly related to the complexity of the organism? a. All of these choices are correct. b. Some genomes have more than two copies of each chromosome. c. Different genomes have different amounts of highly repetitive DNA sequences. d. Different genomes have different amounts of moderately repetitive DNA sequences. ANSWER: a 54. Using current DNA-sequencing technology, the sequence of an entire chromosome (for example, human chromosome 1, 250 million nucleotides) is read from one long molecule. a. true b. false ANSWER: b 55. A shotgun sequencing project yields the sequence for a strand of DNA: 5'-GGTTTGGAGTAT-3' Which sequence overlaps with the sequence? a. 5'-GAGTATCCAAAT-3' b. 5'-GTGTTGGAGCTT-3' c. 5'-GGTTTTTAGACT-3' d. 5'-GGCTTGAGGTTA-3' e. None of the sequences overlaps with this sequence. ANSWER: a 56. A shotgun sequencing project yields the sequence for a strand of DNA. 5'-ATTAGAGAAAAT-3'. Which sequence overlaps with this sequence? a. All of these choices are correct. b. 5'-GAAAATTTCAGA-3' c. 5'-TGTCACATTAGA-3' d. 5'-AAGCCTATTTTC-3' e. 5'-TCTAATGGGCCA-3' ANSWER: a 57. A shotgun genome sequencing project is designed in which the average length of a sequenced fragment will be 100 nucleotides and each nucleotide will be sequenced an average of 30 times. For the human genome, Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 which is 3.1 x 109 base pairs long, how many fragments of 100 nucleotides would need to be sequenced? a. (3.1 x 109) x 30 x 100 = 9.3 x 1011 b. (3.1 x 109) x 30 ÷ 100 = 9 x 108 c. (3.1 x 109) x 100 ÷ 30 = 1.03 x 1010 d. (3.1 x 109) ÷ (30 x 100) = 1.03 x 106 ANSWER: b 58. A shotgun genome sequencing project is designed in which the average length of a sequenced fragment will be 150 nucleotides and each nucleotide will be sequenced an average of 20 times. For a genome that is 3 x 107 base pairs long, how many fragments of 150 nucleotides would need to be sequenced? a. (3 x 107) x 20 ÷ 150 = 4 x 106 b. (3 x 107) x 150 ÷ 20 = 2.25 x 108 c. (3 x 107) ÷ (20 x 150) = 1 x 104 d. (3 x 107) x 20 x 150 = 9 x 1010 ANSWER: a 59. Why is a DNA sequence motif consisting of codons uninterrupted by a stop codon often called a "putative" open reading frame (putative ORF) instead of simply an open reading frame (ORF)? a. because the ORF could be due to chance b. because the ORF could be due to a sequencing error c. because the ORF might not be transcribed d. because the ORF could be part of an intron e. All of these choices are correct. ANSWER: e 60. In using the shotgun approach to genome sequencing, the sequenced fragments originate from: a. the beginning of a gene. b. the centromere of each chromosome. c. random sites scattered across the genome. d. targeted sites evenly space along each chromosome. e. None of the other answer options is correct. ANSWER: c 61. Sequences that are conserved, that is, similar in many different organisms, are unlikely to be functionally important. a. true b. false ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 62. Genome sequencing includes the steps: 1. putting the sequences in the correct order. 2. matching regions of overlap. 3. breaking the DNA into small fragments. 4. reconstructing the long sequence of nucleotides from smaller ordered fragments. 5. sequencing the DNA. In what order are these steps carried out? a. 3-5-2-1-4 b. 5-2-3-1-4 c. 1-2-3-4-5 d. 3-5-2-4-1 e. 5-4-3-2-1 ANSWER: a 63. Which is a feature of DNA that could allow you to distinguish between a DNA sequence that is protein coding and a DNA sequence that is not? a. Protein-coding DNA sequences are single stranded; nonprotein-coding DNA sequences are double stranded. b. Protein-coding DNA sequences contain U (uracil) and nontranscribed DNA sequences contain T (thymine). c. Protein-coding DNA sequences contain tandem repeats; nonprotein-coding DNA sequences do not. d. Protein-coding DNA sequences frequently contain long open reading frames; nonprotein-coding DNA sequences rarely do. e. Protein-coding DNA sequences form hairpin structures, in which the molecule folds back on itself; nonprotein-coding DNA sequences do not form hairpin structures. ANSWER: d 64. If you want to identify protein-coding DNA sequences, why is it more efficient to study messenger RNAs than DNA? a. because mRNAs have poly-A tails b. because mRNAs do not contain the promoter sequence c. because triplets of nucleotides in DNA do not indicate what amino acids will be present in a protein d. because processed mRNA does not contain introns ANSWER: d 65. Which would be found in DNA sequences but not in mRNA sequences? a. transcription start sites b. translation start sites c. open reading frames d. exons e. All of these choices are correct. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 66. Which statement about transposable elements is correct? a. They are sometimes described as "selfish" and called "the ultimate parasite." b. All of these choices are correct. c. They make up 45% of the DNA in the human genome. d. They can transpose via DNA replication or an RNA intermediate. e. They replicate themselves and insert themselves into new positions. ANSWER: b 67. Which statement applies to transposable elements? a. All of these choices are correct. b. Their numbers can increase from one generation to the next. c. They account for a significant fraction of the human genome. d. Some kinds can replicate via an RNA intermediate. e. Some can replicate and insert themselves into another place in the genome. ANSWER: a 68. A diploid species of plant with 22 chromosomes has a tetraploid relative with 44. How many chromosomes are present in the reproductive cells of each species? a. diploid 22, tetraploid 11 b. diploid 22, tetraploid 22 c. diploid 11, tetraploid 11 d. diploid 11, tetraploid 22 ANSWER: d 69. The C-value paradox applies to: a. bacteria. b. archaea. c. eukaryotes. d. viruses. e. All of these choices are correct. ANSWER: c 70. Two viruses with genomes of double-stranded DNA are compared, and one has a genome size two times as big as the other. This is likely because the one with the larger genome: a. has more repetitive DNA. b. has more genes. c. has more transposable elements. d. is polyploid. ANSWER: b 71. A retrotransposon known as Alu1 is about 300 base pairs in length and is present in the human genome in Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 about 1 million copies. Approximately what percentage of the human genome is accounted for by this transposon? a. 10% b. 0.1% c. 1% d. 0.03% e. 30% ANSWER: a 72. Among human embryos that are unable to complete development, some have three complete copies of the nuclear genome and others have four complete copies of the nuclear genome. How many chromosomes do cells of these embryos possess, respectively? a. 92 and 115 b. 46 and 69 c. 23 and 46 d. 69 and 92 e. None of the answer options is correct. ANSWER: d 73. The vast majority of naturally occurring polyploids have an even number of sets of chromosomes. a. true b. false ANSWER: a 74. Polyploids with an odd number of sets of chromosomes have many problems in meiosis but none in mitosis. Why? a. Crossing over occurs in meiosis. b. DNA replication is less accurate in meiosis than in mitosis. c. Homologous chromosomes pair in meiosis. d. Spindles are attached to kinetochores only in meiosis. ANSWER: c 75. Some common domesticated fruit plants, including seedless watermelon and banana, are triploids. The seeds are small because most are genetically abnormal and do not undergo complete development. Why are most of the seeds genetically abnormal? a. They have extra and/or missing chromosomes because of chromosome segregation problems in meiosis. b. They have broken chromosomes because of crossing over in meiosis. c. Their DNA replication is disrupted because of the extra copy of the genome. d. They have excessive numbers of mitochondria. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 76. Organisms that have bigger genomes are more complex than organisms with smaller genomes? a. true b. false. ANSWER: b 77. Chromosome painting, as shown in the figure, is achieved by labeling various DNA fragments with different fluorescent molecules and then hybridizing those fragments to human chromosomes spread on a microscope slide. The result, if done properly, is that each chromosome can be visualized as having a different color.

Given the ability to paint each chromosome differently, what else must be true? a. Each chromosome condenses around histones in its own way. b. No similar gene sequences are found on more than one chromosome. c. Transposable element sequences must cluster together on single chromosomes. d. Each chromosome must have at least some sequences not found on other chromosomes. e. None of the other answer choices is correct. ANSWER: d 78. Which statement is true about the chromatin packaging in eukaryotes? a. Genes can be transcribed during chromatin condensation because the major groove of the DNA molecule is always accessible to RNA polymerase complex. b. Histone proteins are positively charged, which allows them to bind to and pack negatively charged DNA. c. Histones are fast-evolving proteins, and therefore, histones from one species cannot bind and pack DNA from another species. d. Histones H2A and H3 are removed before chromosomes can undergo condensation. ANSWER: b 79. Which type of organism packages its DNA in the form of nucleosomes? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 a. eukaryotes b. bacteria c. archaea d. All of these choices are correct. ANSWER: a 80. Nucleosomes are found in: a. chloroplasts. b. mitochondria. c. the nucleus. d. All of these choices are correct. ANSWER: c 81. The ratio of the length of a metaphase chromosome to that of the fully extended and noncondensed molecule of DNA is about 1:10,000. Consider a length of string 1 mile long. If it was coiled in a ratio of 1:10,000, its length would approximate that of a: a. telephone pole, about 144 inches. b. pencil, about 7.5 inches. c. spaghetti noodle, about 20 inches. d. trash can, about 48 inches. e. soda can, about 6 inches. ANSWER: e 82. If a double-stranded DNA molecule of 125 million base pairs were scaled to the size of a human hair, it would have a diameter of 0.003 inches and be 1 mile long. If this scaled-up DNA molecule were condensed and compacted into a chromosome, into which object could that chromosome fit? a. soda can, about 6 inches b. pencil, about 7.5 inches c. spaghetti noodle, about 20 inches d. trash can, about 48 inches e. telephone pole, about 144 inches ANSWER: a 83. Which modification of a virus might allow it to infect one cell but would prevent it from releasing additional viruses that could infect other cells? a. the addition of viral mRNA to the host cell b. the removal of the viral proteins on the surface of the capsid that bind to host cell receptors c. the modification of genes for capsid proteins that render these proteins nonfunctional d. the removal of the reverse transcriptase gene e. All of these choices are correct. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 84. Viruses are grouped into broad categories according to: a. their genome type. b. the type of organism they infect. c. the evolutionary relatedness. d. their particle size. e. None of the other answer options is correct. ANSWER: a 85. A virus with a narrow host range: a. is usually highly virulent. b. infects cells of only a few species. c. is likely a retrovirus. d. infects plants and bacteria but not animals. e. All of these choices are correct. ANSWER: b 86. The genome of viruses consists of: a. double-stranded DNA. b. single-stranded DNA. c. double-stranded RNA. d. single-stranded RNA. e. All of these choices are correct. ANSWER: e 87. A repetitive DNA sequence 300 base pairs (bp) in length is present in 200 identical copies arranged end to end in one region of the genome. It would be possible for you to assemble the entire DNA sequence and determine the number of repeats from a large number of 100 bp sequence fragments. a. true b. false ANSWER: b 88. Viruses with a DNA genome require a virus-encoded DNA polymerase. a. true b. false ANSWER: b 89. Which factor contributes to the C-value paradox illustrated in the figure shown?

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a. Some genomes contain more repetitive DNA than others. b. Some genomes contain 2, 4, 8, 16, or even more complete sets of chromosomes. c. Genomes differ in the amount of coding and noncoding DNA they contain. d. Some genomes contain many more transposable elements than others. e. All these choices are correct. ANSWER: e 90. Which single-stranded DNA sequence forms a hairpin with an 8 base-pair stem and a 5 base-pair loop. a. 5'-ATGGACTTTACCTGAAATGCA-3' b. 5'-ATGGACTTATGCATACCTGAA-3' c. 5'-ATGGACTTATGCAAAGTCCAT-3' d. 5'-ATGGACTTAAGTCCATATGCA-3' e. 5'-ATGACTTATGCAAGTCCAT-3' ANSWER: c Multiple Response 91. The figure shown provides examples of various types of sequences that can be found in a segment of double-stranded DNA. What kind of sequences are most likely to make the processes of DNA sequencing and sequence assembly difficult? Select all that apply.

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a. dispersed repeat b. tandem repeat c. simple-sequence repeat d. noncoding RNA e. single-copy gene ANSWER: a, b, c 92. Knowledge of your personal genome: Select all that apply. a. identifies genetic risk factors for disease. b. allows you to make informed decisions about health. c. tells you what genetic diseases you will get and when in life they will appear. d. tells you which risky lifestyle behaviors are okay for you to indulge in. ANSWER: a, b 93. Polyploidy is widespread among plants and can arise from which processes? Select all that apply. a. duplication of a complete set of chromosomes in a single species b. hybridization between related species followed by duplication c. deletion of long stretches of noncoding DNA over time d. introduction of DNA from one species into another via insects e. expansion of regions of highly repetitive DNA ANSWER: a, b 94. Imagine a genomic researcher who is analyzing the genome of different types of cats. She finds that a particular sequence in the North American Bobcat genome is identical to a sequence found in the common house cat, whereas most other sequences between the genomes of these cats differ at many nucleotides. Bobcats and house cats diverged an estimated 6.8 million years ago, plenty of time for mutation to generate DNA sequence variation. Which statements could explain the identical sequence in these otherwise differing Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 genomes? Select all that apply. a. The sequence encodes a gene that is critical for life and that cannot retain its function if mutated. b. The sequence is contained in a retrovirus that has infected both species. c. The sequence encodes a protein critical for the production of fur color. d. The sequence is from an intron of a gene that encodes a muscle protein. ANSWER: a, b 95. Humans do not have significantly more genes than some other animals, for instance, the nematode worm C. elegans. What accounts for the diversity of cell types and functions in humans relative to C. elegans or D. melanogaster? Select all that apply. a. Most genes in the other animals are inactive. b. Complexity arises from differential gene expression. c. Complexity arises from different combinations of proteins. d. Humans cells frequently gain more genes through horizontal gene transfer. e. Many human genes can encode multiple proteins. ANSWER: b, c, e 96. Recall the analogy of genome sequencing and sentence fragments. Which scenarios would make assembling the sentence fragments into a complete sentence more difficult? Select all that apply. a. if the letters were randomly assembled and did not form known words b. if some of the words in the sentence were used several times c. if each of the words in the sentence were used only once d. if the fragments were longer ANSWER: a, b 97. In addition to noncoding sequences, most eukaryotic genomes contain not only functional genes that encode proteins but also nonfunctional pseudogenes. Pseudogenes are copies of functional genes that contain mutations, which prevent protein function coded by this gene or block either transcription or translation of the gene. Which scenarios could indicate that a gene-like DNA sequence is a pseudogene sequence and not a functioning gene? Select all that apply. a. The sequence folds back to itself to form a hairpin. b. The sequence has no putative open reading frame (ORF). c. The sequence doesn't have 5' cap and poly(A) tail. d. The sequence has no transcription-factor binding sites. ANSWER: b, d 98. Which statements provide evidence that organelles, for example, mitochondria and chloroplasts, originated billions of years ago as a result of primitive eukaryotic cells engulfing free-living bacterial cells? Select all that apply. a. Organelle DNA is usually circular. b. Organelles have a large amount of noncoding DNA. c. Organelle DNA is not associated with nucleosomes. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 13 d. Chromosomes in organelles undergo the process similar to mitosis. e. Òhe structures of the nucleoid in mitochondria and chloroplasts differ from each other. ANSWER: a, c 99. Homologous chromosomes are: Select all that apply. a. pairs of chromosomes that match in size and morphology. b. pairs of chromosomes that have the same genes in the same order. c. pairs of chromosomes that cannot undergo crossing-over. d. identical double helices of DNA that are the products of DNA replication. e. pairs of chromosomes between different species that carry approximately the same genes in the same locations. ANSWER: a, b

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Chapter 14 Multiple Choice 1. The enzyme _____ repairs breaks in the DNA sugar-phosphate backbone. a. uracyl glycosylase b. DNA polymerase c. AP endonuclease d. DNA ligase e. None of the other answer options is correct. ANSWER: d 2. A family can share a genetic risk of developing cancer if: a. the cancer is caused by somatic cell mutations. b. the cancer is caused by germ-line mutations. c. a germ-line mutation in one of the genes implicated in the cancer occurred in an ancestor. d. a somatic cell mutation in one of the genes implicated in the cancer occurred in an ancestor. e. All of these choices are correct. ANSWER: c 3. If an organism is treated with a chemical agent that doubles the rate of mutation observed in its absence, and a mutation occurs, what is the probability that it was caused by the chemical agent? a. 1/4 b. 1/8 c. 1/2 d. 3/4 ANSWER: c 4. In many organisms, mutations that replace one pyrimidine with another or one purine with another are more frequent than mutations that replace a pyrimidine with a purine, or vice versa. What does this finding imply about the accuracy of the statement "mutations are random"? a. The statement applies to all types of mutations, regardless of their chemistry. b. The statement applies only to single-nucleotide point mutations. c. The statement applies only to single-nucleotide frameshift mutations. d. The statement applies only to nucleotide changes that result in nonsense codons. e. "Random" means without regard for the needs of the organism. ANSWER: e 5. In flowering plants, somatic tissue can differentiate into the sexual organs. This means that: a. all somatic mutations in plants can affect future generations. b. some somatic mutations in plants can affect future generations. c. no somatic mutations in plants can affect future generations. d. the DNA repair mechanisms in plants are much more efficient than those in animals. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 ANSWER: b 6. In living organisms, on average, about one nucleotide in every _____ is mistakenly substituted for another during each round of DNA replication. a. 100 b. 1000 c. 100,000 d. 1 billion e. 10 billion ANSWER: e 7. According to the figure shown, the average number of new mutations that occur across an entire human genome in one generation is approximately _____ times higher than in nematodes. a. 2 b. 10 c. 100 d. 1000 e. 10,000 ANSWER: c 8. In a genome sequence, multiple copies of a transposable element scattered throughout the genome are annotated as a(n): a. direct repeat. b. inverted repeat. c. tandem repeat. d. privileged repeat. e. dispersed repeat. ANSWER: e 9. Insertions and deletions of single nucleotides: a. cause missense mutations. b. cause frameshift mutations. c. add or delete amino acids to or from the normal polypeptide. d. shorten chromosomes. e. cause cancer. ANSWER: b 10. The human genome contains a family of genes that code for different forms of myosin. How could this gene family have arisen? a. The original myosin gene was duplicated and the resulting copies have diverged. b. Two different original genes accumulated mutations, and both evolved into myosin. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 c. The original myosin gene was duplicated without divergence. d. The genes were transferred from a bacterium as a result of infection. ANSWER: a 11. The human genome contains a family of genes that code for different forms of myosin, which are expressed in different cell types and have somewhat different functions. Where in the sequence of each gene would you expect to see differences among them? a. regulatory region only b. protein-coding region only c. regulatory and protein-coding regions d. no difference in any part of the gene ANSWER: c 12. The human genome contains a family of genes that code for different forms of myosin. How different do you think the sequences of these genes are from each other? a. completely different, no identical base sequences b. completely the same, identical base sequence c. some differences in sequence, some similarities ANSWER: c 13. Chromosomal mutations that are most likely to cause serious damage to an organism are those that affect the: a. centromere. b. telomere. c. long arm of the chromosome. d. short arm of the chromosome. e. None of the other answer options is correct. ANSWER: a 14. In large genomes, most reciprocal translocations occur in: a. coding DNA. b. noncoding DNA. c. open reading frames. d. closed reading frames. e. inversion regions. ANSWER: b 15. Reciprocal translocations can affect gene dosage in offspring because: a. the genes involved in the translocation are nonfunctional. b. the two chromosomes involved in the translocation may not assort together during meiosis. c. the two chromosomes involved in the translocation always assort together during meiosis. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 d. reciprocal translocation involves the duplication of one set of genes and the deletion of another set of genes. e. None of the other answer options is correct. ANSWER: b 16. The main mutagenic effects of ultraviolet light are chemical cross-links between adjacent thymines in a DNA strand. These could be repaired by: a. DNA replication. b. mismatch repair. c. base excision repair. d. nucleotide excision repair. ANSWER: d 17. Incorrectly repaired double-stranded breaks in DNA can produce: a. duplications. b. deletions. c. inversions. d. translocations. e. All of these choices are correct. ANSWER: e 18. The enzyme _____ repairs 99% of mismatched bases immediately during DNA replication. a. DNA ligase b. DNA polymerase c. AP endonuclease d. uracyl glycosylase e. None of the other answer options is correct. ANSWER: b 19. Which kind of damage to DNA can be caused by X-rays? a. breaks in one or both of the sugar-phosphate backbones b. cross-links between adjacent pyrimidine bases c. addition of bulky side groups that hinder proper base pairing d. loss of a base from one of the deoxyribose sugars, resulting in a gap in one DNA strand ANSWER: a 20. A loss-of-function mutation is one in which the function of a gene is completely knocked out or obliterated. Frameshift mutations never result in the loss-of-function of a gene. a. true b. false ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 21. Beneficial mutations are those that increase survival or reproduction in the prevailing environment. a. true b. false ANSWER: a 22. In humans, the rate of point mutation tends to be greater in males than in females because: a. the testes are more exposed to the environment than the ovaries. b. male meiosis is more sensitive to mutagens than female meiosis. c. DNA repair in males is less efficient than that in females. d. male germ-line cells undergo many more divisions than female germ-line cells. ANSWER: d 23. Which mutations in an animal somatic cell are inherited by the next generation? a. point mutations b. synonymous mutations c. deletions d. None of the other answer options is correct. ANSWER: d 24. The relatively large number of new mutations that occur in the human genome in each generation is tolerable because: a. humans have excellent DNA repair mechanisms. b. most of the mutations occur in somatic cells, not germ cells. c. most of the human genome is noncoding DNA, so few mutations affect human proteins. d. humans have excellent protein repair mechanisms. e. compared to other organisms, changes in human proteins have relatively small effects on their cells' structures and functions. ANSWER: c 25. Somatic mutations are important to the evolutionary process; most cancers result from somatic mutations. a. true b. false ANSWER: b 26. A population of mosquitoes is exposed to the pesticide DDT for several generations. At the end of that time, most individuals in the population are resistant to DDT. The most likely reason is that: a. DDT caused the mutations that led to resistance. b. some individuals in the original population had the mutations that led to resistance. c. by chance, new mutations that led to DDT resistance arose after DDT was used. d. somatic mutations in the original population were passed on to subsequent generations. e. random mutations in each generation made mosquitoes resistant to DDT. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 ANSWER: b 27. Transposons can: a. interfere with transcription. b. cause a frameshift. c. cause errors in mRNA processing. d. All of these choices are correct. ANSWER: d 28. Which of the answer choices is most likely to result in a nonfunctional polypeptide? a. a silent mutation b. a missense mutation c. a nonsense mutation ANSWER: c 29. Cancer is usually due to a series of mutations that occur in a single lineage of somatic cells. a. true b. false ANSWER: a 30. In what situation can a harmful deletion in a chromosome persist in a population? a. if it is homozygous b. if the homologous chromosome lacks the deletion c. if the deletion is in the centromere d. if a transposon replaces the deleted region ANSWER: b 31. Which of the types of mutation is likely to be the most harmful? a. reciprocal translocation b. gene inversion c. deletion of the centromere d. gene duplication ANSWER: c 32. Single-stranded DNA breaks are more harmful to an organism than double-stranded DNA breaks. a. true b. false ANSWER: b 33. The accompanying diagram shows three gene duplication events that gave rise to a gene family consisting of four sequences X1 - X4. A researcher sequences the genes in two related species and discovers that each Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 species has two copies. One species has copies X1 and X2, and the other has copies X3 and X4. The sequence of X1 is very close to that of X2, and the sequence of X3 is very close to that of X4; however, the (X1, X2) pair is quite dissimilar from the (X3, X4) pair. The most likely interpretation of this pattern is that speciation took place in the interval labeled: a. M. b. H. c. K. d. None of the other answer options is correct. ANSWER: b 34. Genetic risk factors act independently of environment and lifestyle choices. a. true b. false ANSWER: b 35. Two independent mutants are discovered in a strain of fruit flies. Both are due to the insertion of a transposable element into a gene that is required to produce red pigment in the eyes: the eyes are yellow instead of red. When the transposon undergoes transposition to another location, the copy present in the original location is sometimes deleted. When the copy of the transposon in the gene required for red pigment is deleted, the gene regains its ability to function, and the cell with the deletion and all its descendants can produce the red pigment. In one mutant, A in the figure shown, the eyes have tiny red sectors scattered throughout, whereas in the other mutant, B, the eyes have larger red blotches. What do these patterns tell you about the transposition of the transposons in the mutants A and B? a. The transposons in A and B transpose at about the same time in development. b. The transposon in A tends to transpose at an earlier time in development than that in B. c. The transposon in A tends to transpose at a later time in development than that in B. d. There is not enough information to make a conclusion about the time of transposition. ANSWER: c 36. A chemical agent that interferes with DNA repair may be considered a mutagen. a. true b. false ANSWER: a 37. In general, which is less harmful, a duplication or a deletion? a. duplication b. deletion ANSWER: a 38. In organisms with large genomes, inversions are more likely to be tolerated if the breakpoints occur in: a. coding DNA. b. noncoding DNA. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 c. open reading frames. d. closed reading frames. e. reciprocal translocations. ANSWER: b 39. Which type of repair is a backup for the DNA polymerase proofreading function? a. mismatch repair b. base excision repair c. nucleotide excision repair d. DNA ligase ANSWER: a 40. Which of the statements most accurately explains why the average rate of mutation in humans does not accurately describe the pattern of mutations across the genome? a. Some sites in the genome (hotspots) are more mutable than others. b. Mutation rates differ between males and females. c. Mutation rates differ between germ-line cells and somatic cells. d. Mutation rates vary with age. e. All of these choices are correct. ANSWER: e 41. The human genome codes for a form of myosin that is approximately one-third the size of typical human myosin. Select all likely events that could have created this form of myosin. a. All of these choices are correct. b. a frameshift mutation in the coding region of the gene c. a point mutation in the coding sequence of the gene d. alternative splicing with no change in base sequence e. alternative splicing with a change in base sequence ANSWER: a 42. The American Cancer Society currently estimates that only about 10% of all people with melanoma have a family history of the disease. What factors might contribute to the development of melanoma in the other 90% of patients? a. exposure to solar radiation b. mutations in a melanocyte c. exposure to chemical mutagens d. All of these choices are correct. ANSWER: d 43. Any mutation that increases the risk of disease in an individual is known as a: a. transmission mutation. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 b. somatic mutation. c. genetic load. d. genetic risk factor. e. predisposition factor. ANSWER: d 44. It has been estimated that the average human gamete contains about 30 new nucleotide-substitution mutations and approximately 3 new small insertions or deletions (indels). Assuming a human genome size of 3 × 109 base pairs, what is the estimated rate of nucleotide-substitution mutations and of indel mutations per nucleotide pair per generation? a. 106; 10-7 b. 10-7; 10-8 c. 10-8; 10-9 d. 10-9; 10-10 ANSWER: c 45. In the standard genetic code shown in Table 4.1, how many codons for amino acids allow synonymous mutations in the third position? a. 19 b. 29 c. 39 d. 49 e. 59 ANSWER: e 46. If an organism is treated with a dose of radiation that quadruples the rate of mutation and a mutation subsequently occurs, what is the probability that it was caused by radiation? a. 0 b. 1/4 c. 1/2 d. 3/4 e. 1 ANSWER: d 47. Some people with blue eyes have a small sector of one eye that is brown. What kind of mutation could cause this color difference? a. a somatic mutation in a germ-line cell b. a somatic mutation very early in development c. a somatic mutation late in development d. a germ-line mutation Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 ANSWER: c 48. Why do mutations associated with cancer almost always occur sequentially instead of simultaneously? a. The mutations are extremely unlikely to occur in the same cell at the same time. b. Each mutation increases the cellular growth rate, allowing more cells to have a higher chance of the next mutation. c. Each mutation adds to the growth advantage of the ones occurring previously. d. All of these choices are correct. ANSWER: d 49. In the standard genetic code shown in Table 4.1, a nucleotide in a codon is said to be a fourfold degenerate site if its mutation to any other nucleotide results in a synonymous mutation. How many codons in the standard genetic code have a fourfold degenerate site? a. 4 b. 8 c. 12 d. 16 e. 20 ANSWER: b 50. In the standard genetic code shown in Table 4.1, what fraction of codons are nonsense codons? a. 1/64 b. 2/64 c. 3/64 d. 4/64 e. 5/64 ANSWER: c 51. A mutation consisting of a single-nucleotide deletion creates a frameshift in a long open reading frame. In the shifted reading frame, what is the expected average number of codons before a nonsense codon is encountered? Round any fraction to the nearest whole number. a. 11 b. 13 c. 16 d. 21 e. 32 ANSWER: d 52. In bacteria, many transposable elements are able to regulate their own transposition so that the probability of transposition decreases as the number of copies of the transposon in a cell increases. If a transposon invades a bacterial cell that does not currently have any transposons, and then increases in copy number as the cell divides, what curve best depicts the increase in copy number of such a self-regulating transposon? Copyright Macmillan Learning. Powered by Cognero.

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a. curve M b. curve H c. curve K d. curve L e. curve Q ANSWER: b 53. Deletions that eliminate a multiple of three nucleotides can: a. cause nonsense mutations in an open reading frame. b. cause frameshift mutations in an open reading frame. c. delete amino acids in a polypeptide chain. d. All of these choices are correct. ANSWER: c 54. Which of the statements applies to frameshift mutations? a. Frameshift mutations cause the insertion or deletion of a single amino acid from the polypeptide chain. b. Frameshift mutations create a premature stop codon at the site of the mutation. c. Frameshift mutations change the amino acid sequence downstream from the site of the mutation. d. Frameshift mutations are known risk factors in most forms of cancer, including breast and colon cancer. e. Frameshift mutations are known risk factors in breast cancer, but not colon cancer. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 55. Consider the tryptophan codon 5′ - UGG - 3′ in the standard genetic code shown in Table 4.1. Can a single base change in this codon create a synonymous mutation? Can a single base change in this codon create a nonsense codon?

A) yes; yes

a. b. yes; no c. no; yes d. no; no ANSWER: c 56. In the standard genetic code shown in Table 4.1, how many amino acids have codons that allow synonymous mutations (no change to the amino acid) in the second position?

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a. 0 b. 2 c. 8 d. 16 e. 20 ANSWER: a 57. You have identified two different mutations that inactivate the same gene. One is a point mutation (a single nucleotide substitution), and the other is a deletion. Which of the statements is true? a. The deletion will have a more severe effect than the point mutation. b. The point mutation can be reversed by another point mutation, but the deletion cannot. c. If the point mutation is heterozygous, the deletion is likely to be homozygous. d. The deletion, but not the point mutation, will be visible through a light microscope. ANSWER: b 58. The figure shown represents several different versions of human chromosome 2. In each one, the gray region of the chromosome represents the gene that codes for the light chain of the myosin protein (the diagrams are not to scale). Diagram 1 represents the nonmutant human chromosome 2. Among diagrams 2–4, which represents a mutation of chromosome 2?

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a. diagram 2 b. diagrams 3 and 4 c. diagrams 2, 3, and 4 d. None of the other answer options is correct. ANSWER: c 59. Large chromosomal inversions can cause problems in which of these processes? a. meiosis b. mismatch repair c. mitosis d. None of the other answer options is correct. ANSWER: a 60. Bacterial species that live entirely inside eukaryotic cells have deletions of genes for the synthesis of molecules that can be taken in from the host cell. This finding implies that the deletions are: a. based on the needs of the organism, and these genes are not needed. b. random, but that those that delete unneeded genes are not harmful. c. unlikely to have happened by chance. d. induced by mutagens present inside the host cell. ANSWER: b 61. Which of the statements most accurately describes the benefits of the proofreading function of DNA polymerase? a. All DNA mutations can be detected and repaired during DNA replication. b. DNA polymerase is always present in the nucleus and can repair all mutations when they occur. c. DNA polymerase can be recruited to recently mutated sites to repair mutations. d. DNA polymerase can repair most mutations as they occur during DNA replication. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 62. If a covalent bond joining adjacent nucleotides in a DNA strand is broken, the break could be repaired by: a. DNA ligase. b. AP endonuclease. c. DNA uracil glycosylase. d. DNA exonuclease. ANSWER: a 63. If a cytosine base in DNA is converted into uracil, the site is repaired by: a. DNA ligase. b. mismatch repair. c. base excision repair. d. nucleotide excision repair. ANSWER: c 64. A double-stranded DNA break is harder to repair than a single-stranded DNA break. a. true b. false ANSWER: a 65. When a chromosome has either a duplication or a deletion, the chances that the mutation will be harmful are a function of: a. its size—the smaller the duplication or deletion, the greater the chance of harm. b. its size—the larger the duplication or deletion, the greater the chance of harm. c. its position on the chromosome—if it occurs at the tip of a long arm, it is likely to be less harmful than if it occurs at the tip of a short arm. d. None of the other answer options is correct. ANSWER: b 66. Mismatch repair, base excision repair, and nucleotide excision repair are similar in that each: a. repairs a single mismatched base. b. repairs a short strand of mismatched nucleotides. c. repairs multiple mismatched or damaged bases across a region. d. uses an undamaged segment of DNA as the template to repair a damaged segment of DNA. e. None of the other answer options is correct. ANSWER: d 67. Approximately half of the small insertions or deletions that are found in coding regions of DNA have lengths that are an exact multiple of 3. If the lengths of insertions or deletions in coding regions were completely random, what proportion is expected to be an exact multiple of 3? a. 1/6 b. 1/3 c. 2/3 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 d. 1/2 ANSWER: b 68. In a population of organisms with 3 alleles, how many homozygous genotypes are possible? How many heterozygous genotypes are possible? a. 2; 1 b. 2; 3 c. 3; 2 d. 3; 3 e. 3; 4 ANSWER: d 69. The phenylthiocarbamide "taster" allele is denoted PAV and the "nontaster" allele AVI. Heterozygous genotypes are almost all tasters. The frequency of nontasters is low in West Africa but high in India. Based on this information, which allele would you expect to be more common in India? a. PAV allele b. S allele c. A allele d. PiZ allele e. AVI allele ANSWER: e 70. Figure 14.4 shows the three alleles for the beta-globin gene.

Which of the statements best describes the effect of having the S allele, which codes for sickle-cell anemia? a. Having the allele is always harmful. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 b. Having the allele is always beneficial. c. The effect depends on whether the allele occurs in its homozygous or heterozygous form. d. The effect depends on the environment (i.e., whether malaria is present). e. The effect depends on whether the allele occurs in its homozygous or heterozygous form, and the environment (i.e., whether malaria is present). ANSWER: e 71. The gel diagram shows the bands obtained for a single tandem repeat in evidence obtained at a crime scene (W) and genomic DNA from four suspects (A–E).

Which suspect cannot be ruled out as the source of the DNA in the sample? a. suspect A b. suspect B c. suspect C d. suspect D e. suspect E ANSWER: d 72. Single-nucleotide polymorphisms (SNPs) can be detected by microarrays, which are wafer-like substrates to Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 which millions of short stretches of DNA are attached. A microarray consists of many different squares, each one containing a different sequence of single-stranded DNA to which fluorescently labeled strands of DNAs from patients with different SNPs can hybridize. If one square will hybridize with an SNP having a C-G base pair at a particular site, and an adjacent square will hybridize with an SNP having a T-A base pair at the same site, which of the genotypes will hybridize with both squares and make them fluoresce? a. the homozygous C-G/C-G genotype b. the homozygous T-A/T-A genotype c. the heterozygous C-G/T-A genotype d. All of the genotypes hybridize with both squares. ANSWER: c 73. Using the polymerase chain reaction (PCR) to amplify the same region of the genome from different individuals can identify differing lengths due to different numbers of repeated sequences referred to as: a. tandem repeats. b. single-nucleotide polymorphisms. c. copy-number variations. d. first-division nondisjunctions. e. All of these choices are correct. ANSWER: a 74. Would you consider the single-nucleotide polymorphism (SNP) associated with sickle-cell hemoglobin to be a risk factor for sickle-cell anemia? a. No, because it is a nonsynonymous mutation, not an SNP. b. No, because there are many different forms of sickle-cell anemia. c. No, because sickle-cell anemia is associated with malaria. d. Yes, because individuals homozygous for the SNP have the condition. e. Yes, although lifestyle choices are also important in the condition. ANSWER: d 75. What is a single-nucleotide polymorphism (SNP)? a. any point mutation b. the presence of single-nucleotides in the DNA double helix c. a site where a restriction enzyme makes a cut d. a site where RNA polymerase binds to DNA e. a difference between two sequences at a single nucleotide position in a DNA sequence ANSWER: e 76. What is the approximate percentage of the genome that is subject to copy-number variation (CNV)? a. 1% b. 10% c. 50% d. 80% Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 e. 100% ANSWER: b 77. A mutation in the BRCA1 gene is associated with which disease? a. breast cancer b. retinal cancer c. bone cancer d. colon cancer ANSWER: a 78. Harmful mutations are often eliminated in a population because they: a. are fatal to the individual. b. are repaired. c. decrease survival and/or reproduction of the individuals. d. do not confer a benefit to the individual. ANSWER: c 79. You use the polymerase chain reaction (PCR) to amplify DNA from a specific small region of a person's genome to detect whether the individual's DNA contains a small deletion in a gene. You separate the PCR products on a gel. The bands in the gel constitute a: a. genotype. b. karyotype. c. phenotype. d. DNA type. e. diagnosis. ANSWER: c 80. Cancer can be caused by mutations. Genetic analysis of a tumor found in a patient shows that the cell proliferation was triggered by a somatic mutation in the MYC gene, causing this gene to be inappropriately activated. The patient is concerned about passing this cancer on to the children she plans to have in the future. Should she be concerned? a. Yes, she should worry because tumor growth was triggered by a genetic change, and mutations are passed on through cell divisions. b. No, this is not something to worry about because mutations in cancer cells cannot be passed on in cell division. c. No, she should not worry about her children because this did not occur in a germ-line cell. ANSWER: c 81. The nonmutant allele of the BRCA1 gene helps to suppress tumor formation in women who are heterozygous for the mutation. Women heterozygous for BRCA1 nevertheless have a 50% to 70% chance of developing breast cancer before age 70, and the usual reason is that the nonmutant allele is lost or inactivated in a lineage of cells. One possible mechanism for such "loss of heterozygosity" is: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 a. germ cells in the affected individual develop a mutation in the nonmutant allele of BRCA1. b. a somatic mutation in a breast cell inactivates the nonmutant BRCA1 allele. c. a silent mutation occurs in the nonmutant BRCA1 allele. ANSWER: b 82. When the DNA sequence of the gene that codes for the peptide hormone insulin is compared in two mammals (e.g., humans and rats), most of the sequence differences are synonymous mutations. These far outnumber sequence differences that result in amino acid substitutions. Why might this be? a. Mutations are random with respect to an organism's needs. b. Amino acid substitutions often result in proteins that have lost or compromised function and, therefore, are selected against. c. A DNA sequence is more likely to mutate if the change does not alter the amino acid sequence. d. Most amino acid substitutions result in a beneficial phenotype that improves fitness and results in a greater number of offspring. e. Synonymous mutations occur in noncoding DNA sequences. ANSWER: b 83. If a population has only one allelic form of the gene, every individual is: a. conserved. b. polymorphic. c. homozygous. d. heterozygous. e. identical. ANSWER: c 84. An individual is homozygous for a gene if: a. the mother is homozygous for the gene. b. the father is homozygous for the gene. c. they developed from a sperm and an egg that carried the same allele. d. they developed from a sperm and an egg that carried different alleles. e. None of the other answer options is correct. ANSWER: c 85. If there are 3 different alleles for a particular gene in a population of diploid organisms, how many different genotypes are possible in the population? a. 6 b. 3 c. 2 d. 9 e. 12 ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 86. Gene A exists in five forms in the human population. Each form, or allele, has a different number of tandem repeats. The alleles are amplified using PCR and then run on a polyacrylamide gel for analysis, yielding the banding pattern shown on the gel.

Which of the lanes in the gel could represent the alleles found in one individual? a. All 5 lanes are possible. b. Lanes 2, 4, and 5 are possible. c. Lanes 2, 3, 4, and 5 are possible. d. Lanes 1, 2, 3, and 4 are possible. ANSWER: b 87. Ultraviolet light is a mutagen, but humans need some exposure to it in order to synthesize vitamin D3. The amount of ultraviolet light that penetrates the skin depends on the skin's pigmentation; more melanin (skin pigment) means less penetration. Certain mutations result in decreased melanin production. Such mutations: a. are harmful only if inherited from both parents. b. are beneficial only if inherited from both parents. c. are neutral only if inherited from one parent, not the other. d. may be harmful in one environment and beneficial in another. ANSWER: d 88. An individual is heterozygous for a gene if: a. his or her mother is heterozygous for the gene. b. his or her father is heterozygous for the gene. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 c. he or she developed from a sperm and egg that carried different alleles. d. he or she developed from a sperm and egg that carried the same allele. e. only one allele is present in the population. ANSWER: c 89. A polymorphism is a nucleotide pair, gene, or region of the genome: a. that has recently undergone mutation. b. that has escaped DNA repair mechanisms sometime in the past. c. for which every individual is homozygous. d. that decreases the risk of a genetic disease. e. that increases the risk of a genetic disease. ANSWER: b 90. In a population of organisms with 4 alleles, how many homozygous genotypes are possible? How many heterozygous genotypes are possible? a. 3; 4 b. 4; 3 c. 4; 4 d. 4; 6 e. 4; 10 ANSWER: d 91. Which type of mutation is thought to be the least common? a. somatic mutations b. germ-line mutations c. neutral mutations d. harmful mutations e. beneficial mutations ANSWER: e 92. The phenotype of an individual results from an interaction between: a. allele frequency and mutation rate. b. allele frequency and genotype. c. genotype and mutation rate. d. genotype and the environment. e. allele frequency and the environment. ANSWER: d 93. If 8 alleles exist in a population, what is the maximum number of copies a normal individual is expected to have? a. 8 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 b. 5 c. 4 d. 3 e. 2 ANSWER: e 94. Imagine a region of DNA containing a tandem repeat that is found in the genome of koalas. The possible number of tandem repeat alleles in the population equals the number of: a. genes. b. base pairs. c. times a sequence is repeated in tandem along the DNA. d. restriction fragment length polymorphisms there are in the genome. e. times a sequence is repeated in the koala genome. ANSWER: c 95. When DNA with tandem repeats is visualized on a gel, the resulting fragments separate according to their: a. A-T content. b. G-C content. c. color. d. size. e. shape. ANSWER: d 96. Genomic regions with tandem repeats used in DNA typing are highly polymorphic, with many different possible genotypes existing in a population. However, matching patterns of bands for any tandem repeat between the DNA in a sample and the genomic DNA of a particular individual is not sufficient to establish that the two come from the same individual. The primary reason is that: a. the DNA sample could be contaminated. b. the gel may have been handled improperly. c. the individual in question may have a twin. d. any one tandem repeat could match in two individuals by chance. e. any one tandem repeat has multiple alleles and many possible genotypes. ANSWER: d 97. A weapon (W) left behind at a crime scene is typed for a tandem repeat. The DNA from the weapon includes bands from the victim (V) as well as those from the perpetrator.

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Which of the suspects (A–E) has tandem repeatbands that are consistent with those of the perpetrator? a. suspect A b. suspect B c. suspect C d. suspect D e. suspect E ANSWER: b 98. Are all point mutations SNPs? a. Yes, because SNPs are single base-pair changes of the DNA. b. Yes, because SNPs are random in their occurrence. c. No, because SNPs require a certain frequency in the population. d. No, because SNPs are mismatched base pairs. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 ANSWER: c 99. A typical human is heterozygous for about 3 million SNPs. This means that a typical sperm or egg differ at how many nucleotide sites? a. 1.5 million b. 2 million c. 2.5 million d. 3 million e. None of the other answer options is correct. ANSWER: d 100. There is a copy-number variation (CNV) in the human AMY1 gene that encodes a starch-degrading enzyme. If a homozygous genotype with only one copy of AMY1 in each homologous chromosome produces 1 mg/mL of salivary enzyme, what amount of enzyme would be expected in a heterozygous individual with three copies of the AMY1 gene in one chromosome and two copies of AMY1 in the homologous chromosome? a. 5.0 mg/mL b. 4.5 mg/mL c. 3.5 mg/mL d. 2.5 mg/mL e. 2.0 mg/mL ANSWER: d 101. The graphs shown depict the relative proportions of individuals affected with a certain condition (darker shaded bar) and individuals not affected (lighter bar), in individuals carrying either the A - T or the G - C allele of a single-nucleotide polymorphism (SNP). Heterozygous genotypes carry both alleles and are included in both categories.

Which graph shows a pattern that suggests that the G - C allele is a risk factor for the disease? a. graph M b. graph H c. graph K d. graph L e. graph Q Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 ANSWER: d 102. The graphs shown depict the relative proportions of individuals affected with a certain condition (darker shaded bar) and individuals not affected (lighter bar), in individuals carrying either the A - T or the G - C allele of a single-nucleotide polymorphism (SNP). Heterozygous genotypes carry both alleles and are included in both categories.

Which graph shows a pattern that suggests that the A - T allele is a factor that protects against the disease? a. graph M b. graph H c. graph K d. graph L e. graph Q ANSWER: b 103. Hairy cell leukemia is a cancer of the white blood cells that responds to treatment with drugs that inhibit DNA synthesis. A mutation in the BRAF gene, designated BRAF V600E, is associated with hairy cell leukemia. BRAF V600E differs from the nonmutant BRAF gene in a single base pair. Imagine that this change to the single base-pair added a Hae III restriction site indicated by the arrow on the gene shown. You decide to identify hairy cell leukemia with PCR and the restriction enzyme Hae III.

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Which lane on the gel represents DNA from cells heterozygous for the BRAF V600E allele? a. lane 1 b. lane 2 c. lane 3 d. lane 4 ANSWER: b 104. Hairy cell leukemia is a cancer of the white blood cells that responds to treatment with drugs that inhibit DNA synthesis. A mutation in the BRAF gene, designated BRAF V600E, is associated with hairy cell leukemia. BRAF V600E differs from the nonmutant BRAF gene in a single base pair. Imagine that this change to the single base-pair added a Hae III restriction site indicated by the arrow on the gene shown. You decide to identify hairy cell leukemia with PCR and the restriction enzyme Hae III. Copyright Macmillan Learning. Powered by Cognero.

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Which lane in the gel represents an individual who is homozygous for the nonmutant allele? a. lane 1 b. lane 2 c. lane 3 d. lane 4 ANSWER: a 105. Hairy cell leukemia is a cancer of the white blood cells that responds to treatment with drugs that inhibit DNA synthesis. A mutation in the BRAF gene, designated BRAF V600E, is associated with hairy cell leukemia. BRAF V600E differs from the nonmutant BRAF gene in a single base pair. Imagine that this change to the single base-pair added a Hae III restriction site indicated by the arrow on the gene shown. You decide to identify hairy cell leukemia with PCR and the restriction enzyme Hae III. Copyright Macmillan Learning. Powered by Cognero.

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Which lane in the gel represents an individual who is homozygous for the Hae III restriction site? a. lane 1 b. lane 2 c. lane 3 d. lane 4 ANSWER: c 106. In the figure shown, the horizontal lines represent DNA strands in a double-stranded molecule, the vertical lines mark the positions of cleavage sites for a particular restriction enzyme, the arrows show the positions of primers used in the polymerase chain reaction to amplify the region, and the numbers are the number of nucleotides between successive restriction sites. Copyright Macmillan Learning. Powered by Cognero.

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If this particular DNA molecule is amplified and then cleaved with the restriction enzymes, what size bands would be observed in a gel? a. 6kb b. 6 kb, 3 kb, 2 kb, and 1 kb c. 6 kb and 3 kb d. 3 kb, 2 kb, and 1 kb e. None of the other answer options is correct. ANSWER: d 107. In the figure shown, the horizontal lines represent DNA strands in a double-stranded molecule, the vertical lines mark the positions of cleavage sites for a particular restriction enzyme, the arrows show the positions of primers used in the polymerase chain reaction to amplify the region, and the numbers are the number of nucleotides between successive restriction sites.

For the DNA shown, the population contains restriction site polymorphisms for sites X and Y. We will use the symbol X+Y+ to denote the allele containing both sites, X+Y– for the allele containing site X but not Y, X–Y+ for the allele containing the site Y but not X, and X–Y– for the allele containing neither site. If DNA from individuals in the population were amplified with PCR, then cleaved with the restriction enzymes and run on a gel, what bands would be observed in the heterozygous genotype X–Y+/X–Y–? a. 6 kb b. 6 kb, 3 kb, 2 kb, and 1 kb c. 6 kb and 3 kb d. 3 kb, 2 kb, and 1 kb e. None of the other answer options is correct. ANSWER: c 108. In the figure shown, the horizontal lines represent DNA strands in a double-stranded molecule, the vertical lines mark the positions of cleavage sites for a particular restriction enzyme, the arrows show the positions of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 primers used in the polymerase chain reaction to amplify the region, and the numbers are the number of nucleotides between successive restriction sites.

For the DNA shown, the population contains restriction site polymorphisms for sites X and Y. We will use the symbol X+Y+ to denote the allele containing both sites, X+Y– for the allele containing site X but not Y, X–Y+ for the allele containing the site Y but not X, and X–Y– for the allele containing neither site. If DNA from individuals in the population were amplified with PCR, then cleaved with the restriction enzymes and run on a gel, what bands would be observed in the heterozygous genotype X+Y–/X–Y+? a. 6 kb b. 6 kb, 3 kb, 2 kb, and 1 kb c. 5 kb and 3 kb d. 5 kb, 3 kb, and 1 kb e. None of the other answer options is correct. ANSWER: d 109. Shown are two alleles of a single-nucletoide polymorphism (SNP), one of which is associated with a higher risk of developing high blood pressure (hypertension). Normal allele: 5' -ATTCGCGGAATTCTGG -3' 3' -TAAGCGCCTTAAGACC -5' Allele associated with hypertension: 5'-ATTCGCGGGATTCTGG- 3' 3'-TAAGCGCCCTAAGACC -5' You amplify DNA from multiple patients at a cardiologist's office and digest each with EcoRI in order to determine the patients' genotypes for this SNP. EcoRI recognizes the sequence shown and cuts each strand at the locations indicated by the arrows.

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Which lane in the gel represents a sample from an individual heterozygous for this SNP?

a. lane A b. lane B c. lane C d. lane D e. lane E ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 110. In the gel shown, which lane represents an unaffected individual? Normal allele: 5' -ATTCGCGGAATTCTGG -3' 3' -TAAGCGCCTTAAGACC -5' Allele associated with hypertension: 5'-ATTCGCGGGATTCTGG- 3' 3'-TAAGCGCCCTAAGACC -5'

a. lane A b. lane B c. lane C d. lane D e. lane E ANSWER: c 111. In the gel shown, which lane represents an individual homozygous for the hypertension allele? Normal allele: 5' -ATTCGCGGAATTCTGG -3' 3' -TAAGCGCCTTAAGACC -5' Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 Allele associated with hypertension: 5'-ATTCGCGGGATTCTGG- 3' 3'-TAAGCGCCCTAAGACC -5'

a. lane A b. lane B c. lane C d. lane D e. lane E ANSWER: a 112. In the gel shown, which lane represents an individual heterozygous for the hypertension allele? Normal allele: 5' -ATTCGCGGAATTCTGG -3' 3' -TAAGCGCCTTAAGACC -5' Allele associated with hypertension: 5'-ATTCGCGGGATTCTGG- 3' 3'-TAAGCGCCCTAAGACC -5' Copyright Macmillan Learning. Powered by Cognero.

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a. lane A b. lane B c. lane C d. lane D e. lane E ANSWER: b Multiple Response 113. Which of the answer choices may occur when a transposable element inserts into a gene? Select all that apply. a. interference with transcription b. errors in RNA processing c. disruption to the open reading frame d. the gene duplicates itself ANSWER: a, b, c 114. Which answer choices does the effect of an insertion or deletion of a small number of nucleotides depend on? Select all that apply. a. whether it occurs in a coding region or noncoding region of the DNA b. whether it is a multiple of two c. whether it occurs in a germ-line cell or somatic cell Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 d. whether it involves purine or pyrimidine bases ANSWER: a, c 115. What can a transposable element do when it inserts into a gene? Select all that apply. a. interfere with transcription b. cause errors in RNA processing c. disrupt the open reading frame d. cause the gene to duplicate itself e. cause the gene to delete itself ANSWER: a, b, c 116. Point mutations can impair a protein if they result in which of the answer choices? Select all that apply. a. synonymous codon b. nonsense codon c. nonsynonymous codon d. shift in reading frame ANSWER: b, c, d 117. It has been estimated that the average human gamete contains about 30 new nucleotide-substitution mutations and approximately 3 new small insertions or deletions (indels). Which of the answer choices is/are true? Select all that apply. a. The estimate must be wrong; nobody could live with so many mutations. b. Many of the mutations are likely to be in nonprotein-coding DNA. c. Most of the mutations have little or no effect on survival or health. d. None of the other answer options is correct. ANSWER: b, c 118. Why do data on observable mutant phenotypes underestimate the actual frequency of mutation? Select all that apply. a. Many mutations are in noncoding regions of the genome. b. Some mutations in protein-coding regions of the genome are synonymous mutations. c. Most mutations result in death of the individual or an inability to reproduce. d. The data on observable mutant phenotypes are accurate. ANSWER: a, b 119. Which of the point mutations is unlikely to change a protein's ability to function? Select all that apply. a. one that occurs in a noncoding region of DNA b. one that creates a new codon code for the same amino acid as the original codon c. one that occurs in somatic cells d. one that creates a new codon that does not code for any amino acid e. one that occurs in germ cells Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 ANSWER: a, b 120. The accompanying diagram shows four chromosomes in an individual who carries a translocation. Chromosomes 1 and 2 are structurally normal copies of two nonhomologous chromosomes, and chromosomes 3 and 4 have undergone a translocation. Which pair of chromosomes included in a sperm or egg would result in a fertilized egg with abnormal gene dosage (missing or extra pieces of a chromosome)? Select all that apply and assume that the other gamete carries structurally normal chromosomes like 1 and 2.

a. 1 and 2 b. 1 and 3 c. 2 and 3 d. 2 and 4 e. 1 and 4 f. 3 and 4 ANSWER: b, c, d, e 121. The process of gene duplication and divergence refers to which answer choices? Select all that apply. a. creating new genes by mutation in duplicates of old genes b. the creation of gene families, which are similar genes within a species c. offspring becoming increasingly different from their ancestors d. the creation of new species as new genes are created from duplicates of old genes e. None of the other answer options is correct. ANSWER: a, b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 122. An individual is charged with murder, based on DNA typing of a blood sample that is more than 20 years old. Which answer choices are legitimate arguments that the accused could make to claim that the evidence is no longer relevant? Select all that apply, even if unlikely. a. 20-year-old DNA cannot be typed reliably because it becomes chemically degraded. b. His DNA has changed because of somatic mutation. c. The chance of mislabeling an evidence sample from so long ago is too great. d. A sample that old could no longer discern the accused from his brother. e. None of the other answer options is correct. ANSWER: a, c, d 123. Which answer choices can produce copy-number variations (CNVs)? Select all that apply. a. frameshift mutation b. nonsense mutation c. inversion d. deletion e. duplication ANSWER: d, e 124. Most genetic variation in a population is: a. obvious. b. beneficial. c. harmful. d. deleterious. e. neutral. ANSWER: e 125. Harmful mutations are always eliminated from the genome in one or a few generations, because they decrease the survival and reproduction rates of the individuals that carry them. a. true b. false ANSWER: b 126. The gel diagram shown represents analysis of blood from a victim of a crime and from two suspects. Criminologists created this gel using a single tandem repeat site.

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Which suspect is definitely innocent? a. suspect 1 b. suspect 2 c. Both suspects are definitely innocent. d. Neither suspect is definitely innocent. ANSWER: b 127. In England, a man who is upset about the amount of dog feces on his lawn supposedly collected hair samples from neighborhood dogs and compared their DNA types with that of the feces, thereby identifying the offending dog (and the negligent owner). Is this possible? a. This is not only possible, it is fairly easy. b. This is ridiculous—no way. c. It might be possible, but it would cost hundreds of thousands of dollars. d. It's an urban legend, forget it. ANSWER: a 128. What kind of genetic variation can alter gene dosage? a. single-nucleotide polymorphisms b. restriction fragment length polymorphisms c. copy-number variations d. All of these choices are correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 ANSWER: c 129. It is impossible to override a genetic predisposition to a disease such as skin cancer. a. true b. false ANSWER: b 130. Beneficial mutations can: Select all that apply. a. protect from disease. b. never be reversed. c. permit an organism to become adapted to its environment. d. None of the other answer options is correct. ANSWER: a, c 131. Which statements about mutations are false? Select all that apply. a. Mutations make evolution possible. b. Mutations result in genetic variation among individuals. c. Mutations accumulate in a genome over time. d. Mutations are usually caused by exposure to radiation. e. Mutations can be harmful, beneficial, or neutral. f. Mutations occur as a direct response to an organism's attempt to change one of its traits. ANSWER: d, f 132. Which answer choices are the same in identical twins? Select all that apply. a. phenotype b. genotype c. karyotype d. DNA fingerprints e. fingerprints ANSWER: b, c, d 133. Many single-nucleotide polymorphisms (SNPs) are regarded as "neutral" in the sense they they have little or no effect on the likelihood of survival or reproduction of the different genotypes. Could you ever say for certain that a particular SNP is "neutral"? Select all that apply. a. Yes, if the polymorphism is in noncoding DNA. b. Yes, if the polymorphism is in a transposable element. c. Yes, if the polymorphism results in synonymous codons. d. No, because any effect may be too small to detect. e. No, because you can never prove a negative. ANSWER: d, e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 134. With regard to alleles that encode different forms of b(beta)-globin in humans and their relation to malaria, which answer choices are examples of phenotypes? Select all that apply. a. sickle-cell anemia b. AA, AS, SS c. susceptibility to malaria d. sickled red blood cells e. allele C ANSWER: a, c, d 135. With regard to alleles that encode different forms of the phenylthiocarbamide taste receptor in humans, which options are phenotypes? Select all that apply. a. taster b. PAV/PAV c. nontaster d. PAV/AVI e. aversion to broccoli ANSWER: a, c, e 136. With regard to alleles that encode different forms of the phenylthiocarbamide taste receptor in humans, which options are genotypes? Select all that apply. a. taster b. PAV/PAV c. nontaster d. PAV/AVI e. aversion to broccoli ANSWER: b, d 137. Which of the statements about single-nucleotide polymorphisms (SNPs) is/are correct? Select all that apply. a. For some disorders, multiple SNPs plus environmental factors are necessary in order for the disorder to present itself. b. Genotyping SNPs instead of sequencing entire genomes allows researchers to find correlations between genetic risk factors and disease. c. SNPs are useful in the diagnosis of a disease but not in the prediction of an individual's risk for acquiring that disease. d. An individual is unlikely to be heterozygous for one or more SNPs. e. The presence of a SNP in an organism's genome when compared to the "standard" genome of that organism means that a disease will develop. ANSWER: a, b, c 138. Over the past several decades, populations of Staphylococcus aureus, a type of bacteria that infects wounds, have become resistant to penicillin. You run an experiment on a population of 10 million S. aureus Copyright Macmillan Learning. Powered by Cognero.

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Chapter 14 cells that has been exposed to penicillin for many generations. You select three samples (A–C) of 100,000 cells each and expose each sample to a new antibiotic called Antibiotic X. None of the cells had previously been exposed to Antibiotic X, which has a different mechanism of action than penicillin. You measure the percent of the population that survives exposure to Antibiotic X and plot your data. The graph shows your results.

What is the probable origin of Antibiotic X resistance in these cells? a. The cells probably developed resistance to Antibiotic X by undergoing mutation when exposed to the drug and thus survived the exposure. b. A small percentage of the cells had preexisting mutations that confer resistance to Antibiotic X. These cells survived exposure to the drug. c. Either of the other answer options could be correct because both happen often in nature, but more information is needed to answer the question. ANSWER: b 139. Over the past several decades, populations of Staphylococcus aureus, a type of bacteria that infects wounds, have become resistant to penicillin. You run an experiment on a population of 10 million S. aureus cells that has been exposed to penicillin for many generations. You select three samples (A–C) of 100,000 cells each and expose each sample to a new antibiotic called Antibiotic X. None of the cells had previously been exposed to Antibiotic X, which has a different mechanism of action than penicillin. You measure the percent of the population that survives exposure to Antibiotic X and plot your data. The graph shows your results. Copyright Macmillan Learning. Powered by Cognero.

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In a different set of experiments, you identify two types of mutations in S. aureus. The first type of mutation makes the bacteria more susceptible to Antibiotic X; the second type of mutation makes it more resistant to Antibiotic X. Which type of mutation is more likely to occur in the population? a. The first type (more susceptible) is more likely to occur. b. The second type (more resistant) is more likely to occur. c. It depends if the cell population has been exposed to Antibiotic X. d. It depends if either type of mutation confers additional survival benefits to S. aureus. e. There is no difference in likelihood because the mutation is a random event. ANSWER: e 140. Over the past several decades, populations of Staphylococcus aureus, a type of bacteria that infects wounds, have become resistant to penicillin. You run an experiment on a population of 10 million S. aureus cells that has been exposed to penicillin for many generations. You select three samples (A–C) of 100,000 cells each and expose each sample to a new antibiotic called Antibiotic X. None of the cells had previously been exposed to Antibiotic X, which has a different mechanism of action than penicillin. You measure the percent of the population that survives exposure to Antibiotic X and plot your data. The graph shows your results.

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You decide to sequence the genome of your resistant S. aureus cells and compare it to the sequence for nonresistant S. aureus cells. What type of genetic difference probably causes this difference in antibiotic resistance in the two cell types? a. It must be a difference in a single base pair of a single gene because the definition of a mutation is a change to a single base pair. b. It must involve many base pairs because of the huge selective advantage the mutation confers. c. It may involve a single base pair or many base pairs; genetic changes on both scales are mutations. ANSWER: c 141. The graphs shown depict the relative proportions of individuals affected with a certain condition (darker shaded bar) and individuals not affected (the lighter bar), in individuals carrying either the A - T or the G - C allele of a single-nucleotide polymorphism (SNP). Heterozygous genotypes carry both alleles and are included in both categories.

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Which graph shows a pattern that suggests that the SNP is not a risk factor for the disease? a. graph M b. graph H c. graph K d. graph L e. graph Q ANSWER: a

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Chapter 15 Multiple Choice 1. Imagine that blending inheritance was true, and black and white rabbits mated to produce gray offspring. If the offspring show only half the intensity of black pigment after one generation, how many generations would be required for them to show 1/16 the intensity of black pigment? a. 3 b. 4 c. 5 d. 6 e. 7 ANSWER: b 2. Mendel crossed true-breeding yellow-seed plants with true-breeding green-seed plants. Given that yellow color is a dominant trait, what would the genotype of the offspring be? a. yellow seed b. green seed c. a mixture of yellow and green seed d. heterozygous for yellow and green alleles ANSWER: d 3. In Mendel's pea plants, each cell in a pea plant had a total of: a. one allele for each gene. b. two alleles for each gene. c. one allele for each gene if it was a true-breeding parent and two if it was an F1 offspring. d. one allele for each gene if it had green seeds and two if it had yellow seeds. e. between one and four alleles for each gene. ANSWER: b 4. In Mendel's pea plants, an individual that is heterozygous for seed color: a. had only one kind of allele for seed color. b. had two different alleles for seed color. c. could have either one or two alleles for seed color. d. had two different alleles for seed color and expressed the dominant allele. ANSWER: d 5. Mendel's experiments with garden peas differed from those of other plant hybridizers of his time in that: a. Mendel studied true-breeding strains instead of heterogeneous strains. b. Mendel focused on a small number of easily contrasted traits instead of a large number of more complex traits. c. Mendel quantified his results and looked for statistical patterns instead of simply noting the presence or absence of a trait among a group of offspring. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 d. Mendel performed intentional crosses between strains, whereas other hybridizers simply allowed strains to interbreed. e. All of these choices are correct. ANSWER: e 6. In the F2 generation of a homozygous round (AA) × homozygous wrinkled (aa) cross in peas, two round seeds are chosen at random. What is the probability that one is AA and the other Aa? a. (1/3)2 b. (2/3)(1/3) c. 2(2/3)(1/3) d. (2/3)2 ANSWER: c 7. In the F2 generation of a homozygous round (AA) × homozygous wrinkled (aa) cross in peas, three seeds are chosen at random. What is the probability that all three seeds are round? a. (1/4)3 b. 3(1/4)(3/4)2 c. 3(1/4)2(3/4) d. (3/4)3 ANSWER: d 8. Flower color in snapdragons is due to a gene with incomplete dominance: CRCR plants have red flowers, CRCW have pink flowers, and CWCW plants have white flowers. What types and ratios of flower color are expected among the progeny of a pink × white cross? a. 1 red:2 pink:1 white b. 1 red:1 pink c. 1 pink:1 white d. all pink e. all white ANSWER: c 9. If you crossed a true-breeding yellow-seed plant (AA) with a heterozygous yellow-seed plant (Aa), offspring: a. genotypes would be 1 AA:2 Aa. b. genotypes would be 1 Aa:1 aa. c. genotypes would be 1 AA:1 Aa. d. phenotypes would be 1/2 yellow-seed plants and 1/2 green-seed plants. ANSWER: c 10. The bands observed in a gel for a VNTR (variable number of tandem repeats) polymorphism in a father (F) Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 and mother (M) are shown in the diagram. What possible patterns of bands are expected among their offspring?

a. (1, 2), (1, 3), (2, 3), (3, 4) b. (1, 2), (1, 4), (2, 3), (3, 4) c. (1, 3), (1, 4), (2, 3), (2, 4) d. (1, 3), (2, 3), (1, 4), (3, 4) ANSWER: b 11. Which of the processes would result in gametes that violate Mendel's principle of segregation? a. dominance b. independent assortment c. epistasis d. nondisjunction ANSWER: d 12. Nondisjunction results in gametes that violate which principle? a. dominance b. segregation c. independent assortment ANSWER: b 13. A woman has her personal genome analyzed for the BRCA1 mutation after learning that her father is heterozygous and carries one mutant allele. What is her chance of inheriting the mutant allele from her father? a. 0: men cannot transmit genes affecting breast cancer. b. 25% c. 50% d. 75% Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 e. 100% ANSWER: c 14. In genetic crosses, the symbols AA and Aa refer respectively to: a. heterozygous and homozygous phenotypes. b. heterozygous and homozygous genotypes. c. homozygous and heterozygous phenotypes. d. homozygous and heterozygous genotypes. e. homozygous and heterozygous morphotypes. ANSWER: d 15. A woman and a man are both heterozygous for a recessive allele for a rare genetic disease. If they have one child, what is the probability that he or she will be affected? If they have two children, what is the probability that at least one of them will be affected? a. 7/16, 1/4 b. 1/4, 3/4 c. 3/4, 1/4 d. 3/4, 7/16 e. 1/4, 7/16 ANSWER: e 16. Suppose that in humans the ability to roll the tongue (R) is dominant to being unable to roll it (r), and having freckles (F) is dominant to having no freckles (f). If a woman heterozygous for both traits married a man with no freckles who couldn't roll his tongue, what is the probability that they would have a freckled, tongue-rolling child? a. 9/16 b. 3/16 c. 1/16 d. 3/4 e. 1/4 ANSWER: e 17. With independent assortment, the ratio of phenotypes in the F2 generation of a cross between true-breeding strains (AA bb x aa BB) can be described as 9:3:3:1 when A and B are dominant over a and b. To what phenotype do the "3"s in the ratio refer? a. recessive for both traits b. dominant for the A trait and recessive for the B trait c. dominant for one trait and recessive for the other d. dominant for both traits ANSWER: c 18. In genetics, the dash symbol (-) is a "wild card" that stands for either the dominant allele or the recessive Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 allele; for example, R- means the individual has either the genotype RR or Rr. In Duroc pigs, genotypes R- Shave red coats, R- ss and rr S- have sandy-colored coats, and rr ss pigs are white. R and S show independent assortment. What is the ratio of red : sandy : white among progeny of the cross Rr Ss x Rr Ss? a. 12:3:1 b. 9:3:1 c. 10:3:3 d. 9:6:1 e. None of the other answer options is correct. ANSWER: d 19. In genetics, the dash symbol (–) is a "wild card" that stands for either the dominant allele or the recessive allele; for example, A– means the individual has either the genotype AA or Aa. Two genes that undergo independent assortment affect fruit shape in summer squash. Each gene has two alleles, one of which is dominant for fruit shape. Genotypes of the form A– B– have disc-shaped fruit, those of the form A– bb and aa B– have sphere-shaped fruit, and genotype aa bb has long fruit. What ratio of disc : sphere : long is expected from the cross Aa Bb x Aa Bb? a. 10:3:3 b. 9:3:4 c. 12:3:1 d. 9:6:1 ANSWER: d 20. The genes C and I, which undergo independent assortment, affect feather color in leghorn chickens. Each gene has two alleles, one of which is dominant for feather color. The gene product of the dominant allele C is necessary for color in the feathers; however, the gene product of the dominant allele I is an inhibitor of color production in the feathers even when C is present. What ratio of white feathers : colored feathers is expected among the progeny of the cross Cc Ii x Cc Ii? a. 9:7 b. 10:6 c. 12:4 d. 13:3 e. 15:1 ANSWER: d 21. Mendel was the first to demonstrate: a. dominance. b. segregation. c. independent assortment. d. equivalence of reciprocal crosses. e. All of these choices are correct. ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 22. Assuming that the trait represented by the filled symbols below is a rare inherited trait with complete penetrance due to a single gene with alleles A and a, what mode of inheritance does the pedigree suggest?

a. dominant b. recessive ANSWER: a 23. Consider a gene with four alleles: A1, A2, A3, and A4. How many distinct homozygous genotypes are possible? a. 2 b. 3 c. 4 d. 5 e. 6 ANSWER: c 24. Consider a gene with four alleles: A1, A2, A3, and A4. How many distinct heterozygous genotypes are possible? a. 2 b. 3 c. 4 d. 5 e. 6 ANSWER: e 25. Consider a gene with four alleles: A1, A2, A3, and A4. If the cross A1A2 x A2A2 yields two offspring, what is the probability that they have the same genotype as each other? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 a. 1/8 b. 1/4 c. 1/3 d. 1/2 e. 3/4 ANSWER: d 26. Consider a gene with four alleles: A1, A2, A3, and A4. In the cross A1A2 x A3A4, how many offspring genotypes are possible? a. 2 b. 3 c. 4 d. 5 e. 6 ANSWER: c 27. Consider a gene with n alleles A1, A2, …, An. How many distinct homozygous genotypes are possible? a. 1 x n b. 2 x n c. 3 x n d. 4 x n e. 5 x n ANSWER: a 28. A trait with incomplete penetrance: a. is only expressed in homozygous recessive individuals. b. is only expressed in homozygous dominant individuals. c. is only expressed in some of the individuals that have the genotype for that trait. d. is expressed in only some offspring of affected individuals. e. can be expressed differently in different individuals that have the genotype for that trait. ANSWER: c 29. A trait with variable expressivity: a. is only expressed in homozygous recessive individuals. b. is only expressed in homozygous dominant individuals. c. is only expressed in some of the individuals that have the genotype for that trait d. is expressed in only some offspring of affected individuals. e. can be expressed differently in different individuals that have the same genotype for that trait. ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 30. Which of the statements is true of direct-to-consumer (DTC) genetic testing? a. DTC genetic testing can tell you exactly how you are going to die. b. DTC genetic testing is tightly regulated and always performed with expert advice available to the consumer. c. DTC genetic test results are sometimes are based on flimsy evidence connecting a particular allele to a disease. d. DTC genetic testing provides certain results that prove that you will or will not have a disease. e. None of the other answer options is correct. ANSWER: c 31. Consider a gene with four alleles: A1, A2, A3, and A4. In the cross A1A2? genotypes are possible? a. 2 b. 3 c. 4 d. 5 e. 6 ANSWER: a

A2A2, how many offspring

32. Consider a gene with four alleles: A1, A2, A3, and A4. If the cross A1A2 the probability that the offspring have the same genotype? a. 1/8 b. 1/4 c. 1/3 d. 1/2 e. 3/4 ANSWER: d

A2A2 yields two offspring, what is

33. Consider a gene with four alleles: A1, A2, A3, and A4. In the cross A1A2 genotypes are possible? a. 2 b. 3 c. 4 d. 5 e. 6 ANSWER: a

A3A3, how many offspring

34. Consider two genes that undergo independent assortment. One gene has alleles A and a with A dominant to a, and the other has alleles B and b, with B dominant to b. From the cross Aa Bb x Aa Bb, the expected ratio of genotypes A– B– : A– bb : aa B– : aa bb is 9:3:3:1. In these symbols, the dash is a "wild card": in the symbol A– , the dash means the unspecified allele is either A or a; and in the symbol B–, the dash means the unspecified Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 allele is either B or b. If the two genes are epistatic to each other, the 9:3:3:1 ratio can be modified according to the manner in which the genotypes of the A and B genes interact to affect the phenotype. There are exactly nine ways in which the 9:3:3:1 ratio can be modified. Each possibility corresponds to one row in the figure shown, where the phenotypes are color coded. In any row, two boxes of the same color indicate that the corresponding genotypes have indistinguishable phenotypes.

What phenotypic ratio results from epistasis of the type in row K? a. 10:3:3 b. 9:3:4 c. 12:4 d. 12:3:1 e. 9:6:1 ANSWER: d 35. Assuming that the trait represented by the filled symbols in the pedigree is a rare inherited trait with complete penetrance due to a single gene with alleles A and a, what mode of inheritance does the pedigree shown suggest?

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Chapter 15

a. dominant b. recessive ANSWER: b 36. In the pedigree shown, what are the genotypes of individuals C and D?

a. Both are AA. b. One is AA and the other is Aa. c. Both are Aa. d. It is not possible to determine this for certain from the information provided. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 37. In the pedigree shown, what can you say about the likely genotypes of individuals A and B?

a. Both are likely to be AA. b. Both are likely to be Aa. c. One is likely to be AA, and the other Aa. d. It is not possible to determine this for certain from the information provided. ANSWER: c 38. In 1871, Francis Galton reported experiments in which, in each of multiple generations, he transfused blood from a true-breeding strain of black rabbits into individuals of a true-breeding strain of white rabbits. Nevertheless, in each generation the progeny of the white rabbits were as white as their parents. Regarding heredity, this experiment demonstrates that: a. characteristics of blood are not hereditary. b. red blood cells (erythrocytes) have no nucleus. c. blood transfusions do not affect hereditary characteristics. d. the recipient's coat color is not affected by blood transfusions. ANSWER: c 39. The F1 generation of a cross between two individuals, one who is homozygous dominant for a trait and the other who is homozygous recessive for the trait, will yield progeny with the same phenotype. a. true b. false ANSWER: a 40. Segregation of alleles corresponds to the separation of: a. genes. b. genotypes. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 c. chromosomes. d. phenotypes. ANSWER: c 41. The phenotypic ratio of the progeny of a cross between two individuals with a trait coded for by a single gene that displays incomplete dominance is: a. 1:3. b. 1:1:1. c. 1:2. d. 1:2:1. ANSWER: d 42. What is the probability that an individual is either Aa or AA if his or her parents are both heterozygous for the trait? a. 1/4 b. 1/2 c. 3/4 d. 1 ANSWER: c 43. Mendel's principle of segregation corresponds to what part of meiosis? a. condensation of chromosomes in prophase I b. alignment of homologs in metaphase I c. separation of homologs in anaphase I d. alignment of chromosomes in metaphase II e. separation of daughter chromatids in anaphase II ANSWER: c 44. The bands observed in a gel for a VNTR (variable number of tandem repeats) polymorphism in a father (F) and mother (M) are shown in the diagram. What possible patterns of bands are expected among their offspring?

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Chapter 15

a. (1, 2), (1, 3), (2, 3), (3, 4) b. (1, 2), (1, 4), (2, 3), (3, 4) c. (1, 3), (1, 4), (2, 3), (2, 4) d. (1, 3), (2, 3), (1, 4), (3, 4) ANSWER: c 45. In the F2 generation of a homozygous round (AA) homozygous wrinkled (aa) cross in peas, two seeds are chosen at random. What is the probability that one is round and the other is wrinkled? a. (1/4)2 b. (3/4) (1/4) c. 2 (3/4) (1/4) d. (3/4)2 ANSWER: c 46. With independent assortment, the ratio of phenotypes in the F2 generation of a cross between true-breeding strains (AA bb x aa BB) can be described as 9:3:3:1 when A and B are dominant over a and b. To what phenotype does the "9" in the ratio refer? a. recessive for both traits b. dominant for the A trait and recessive for the B trait c. dominant for one trait and recessive for the other d. dominant for both traits ANSWER: d 47. For unlinked genes, the sorting of one gene pair influences the sorting of another gene pair. a. true Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 b. false ANSWER: b 48. In pea plants, flowers are either white or purple; the purple color is produced by pigments called anthocyanins. The production of anthocyanins is a two-step process: the first step is controlled by the C gene and the second by the P gene. Both genes must produce functional proteins for anthocyanin to be expressed. For each gene, the dominant (C and P) alleles produce functional proteins. You cross two pea plants, each with the genotype CcPp. What proportion of their offspring will have white flowers? a. 13/16 b. 9/16 c. 7/16 d. 3/4 e. 1/4 ANSWER: c 49. In genetics, the dash symbol (–) is a "wild card" that stands for either the dominant allele or the recessive allele; for example, A– means the individual has either the genotype AA or Aa. Two genes that undergo independent assortment affect fruit color in summer squash. Each gene has two alleles, one of which is dominant for fruit color. Genotypes of the form A– B– and A– bb have white fruit, genotypes of the form aa B– have yellow fruit, and genotype aa bb has green fruit. What ratio of white : yellow : green is expected from the cross Aa Bb x Aa Bb? a. 10:3:3 b. 9:3:4 c. 12:3:1 d. 9:6:1 ANSWER: c 50. In genetics, the dash symbol (–) is a "wild card" that stands for either the dominant allele or the recessive allele; for example, A– means the individual has either the genotype AA or Aa. Two genes that undergo independent assortment affect coat color in Duroc pigs. Each gene has two alleles, one of which is dominant for coat color. Genotypes of the form A– B– are red, those of the form A– bb and aa B– are sandy-colored, and genotype aa bb is white. What ratio of red : sandy : white is expected from the cross Aa Bb x Aa Bb? a. 10:3:3 b. 9:3:4 c. 12:3:1 d. 9:6:1 ANSWER: d 51. Mendel's principle of independent assortment corresponds to which part of meiosis? a. random pairing of homologs during prophase I b. random alignment of homologs on the metaphase plate during metaphase I c. separation of chromosomes in anaphase I d. random alignment of homologs on the metaphase plate during metaphase II Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 e. separation of daughter chromatids during anaphase II ANSWER: b 52. Suppose that in humans the ability to roll the tongue (R) is dominant to being unable to roll it (r), and having freckles (F) is dominant to having no freckles (f). A man heterozygous for both traits marries a woman heterozygous for both traits. What is the probability that they will have a child with freckles? a. 9/16 b. 3/16 c. 1/16 d. 3/4 e. 1/4 ANSWER: d 53. Mothers who consume excessive alcohol during pregnancy may have babies who are affected with a pattern of physical and mental disabilities known as fetal alcohol syndrome. Similarly, mothers who become addicted to drugs during pregnancy may have babies who show signs of addiction. Which of the statements properly describes why these are not examples that contradict the statement that traits acquired by the parents are not transmitted to the offspring? a. Maternal drug or alcohol use does not alter the genome sequences of the child. b. Though gene expression patterns in the child may be affected by maternal drug or alcohol use, those changes result from epigenetic or environmental changes and not from mutations. c. All tissues of the mother and child can be affected by exposure to drugs or alcohol as long as they pass through the placenta into the fetus. d. All of these choices are correct. ANSWER: d 54. If blending inheritance was an accurate model of transmission genetics, which statement would be true? a. Variation in natural populations would remain the same over time. b. Variation in natural populations would increase over time. c. Variation in natural populations would decrease over time. d. Variation would increase in some populations, decrease in some populations, and remain the same in some populations. e. Mutation would not contribute to variation in natural populations. ANSWER: c 55. Aristotle seems to have been the first person to point out that men who become bald have male offspring who are born with hair, which implies that acquired characteristics are not transmitted to offspring. Today we know that hereditary pattern baldness has a characteristic age of onset, a time at which the hair falls out and baldness gradually spreads. This means that: a. acquired characteristics are transmitted to the offspring. b. pattern baldness is not an acquired trait. c. Aristotle was wrong—bald males do have bald sons. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 d. None of the other answer options is correct. ANSWER: b 56. In 1891, August Weismann reported an experiment in which he cut the tails off 10 mice in each of 19 consecutive generations. The expected result is: a. after the first generation of amputation, the progeny are born with no tails. b. offspring are born with progressively shorter tails in each generation. c. offspring are born with tails that are the same length as their parents for a few generations, but subsequent generations of offspring have shorter tails. d. all generations of offspring have tails of approximately the same length. ANSWER: d 57. Imagine that blending inheritance was true, and black and white rabbits mated, resulting in gray rabbits. If the offspring show only half the intensity of black pigment after one generation, how many generations would be required for them to show 1/64th the intensity of black pigment? a. 3 b. 4 c. 5 d. 6 e. 7 ANSWER: d 58. Mendel used true-breeding strains of peas. Would true-breeding peas with the trait caused by a dominant allele or a recessive allele have been easier to produce? a. dominant allele b. recessive allele c. Both strains would be equally easy to produce. ANSWER: b 59. Mendel's experiments with garden peas differed from those of other plant hybridizers of his time in that: a. Mendel studied true-breeding strains instead of poorly defined material. b. Mendel focused on a small number of easily contrasted traits instead of a large number of more complex traits. c. Mendel quantified his results and looked for statistical patterns instead of simply noting the presence or absence of a trait among a group of offspring. d. All of these choices are correct. ANSWER: d 60. In modern terminology, a true-breeding strain is: a. hybrid. b. dominant. c. heterozygous. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 d. homozygous. ANSWER: d 61. In a reciprocal cross: a. both parents are male. b. both parents are female. c. heterozygous genotypes are crossed with homozygous genotypes. d. the crosses are set up between individuals with two traits: in one cross the male has the first trait and in the second cross the female has that trait. ANSWER: d 62. If one of the traits Mendel studied was encoded by a single gene with three alleles that produced different phenotypes, what would he have observed if he crossed all possible pairs of the three true-breeding lines in the F1 and F2 generations? a. In each of the crosses, he would have observed two phenotypes in the F1 generation. b. Only a single phenotype would have been observed in the F1 generation because only one allele can be dominant. c. No evidence would exist for the principle of segregation. d. No true-breeding plants would have been produced in the F2 generation. e. None of the other answer options is correct. ANSWER: e 63. According to the principle of segregation, a heterozygous plant with alleles Aa will produce: a. gametes with only the A allele. b. gametes with only the a allele. c. gametes in the ratio of 1 A allele:1 a allele. d. gametes in the ratio of 3 A alleles:1 a allele. e. some gametes with the A allele and some with the a allele, but in no predictable ratio ANSWER: c 64. If you crossed two heterozygous yellow-seed pea plants (genotypes Aa), the relative frequency of: a. the a allele in each parent's gametes would be 1/2. b. the A allele in each parent's gametes would be 1/2. c. green-seed plants (genotype aa) would be 1/4. d. homozygous yellow-seed plants (genotype AA) would be 1/4. e. All of these choices are correct. ANSWER: e 65. In meiosis, allele separation during _____ is the chromosomal basis of segregation. a. prophase Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 b. metaphase c. anaphase d. telophase ANSWER: c 66. Which of the ratios is associated with Mendel's discoveries? a. 3:1 b. 1:2 c. 1:2:1 d. 9:3:3:1 e. All of these choices are correct. ANSWER: e 67. In the F2 generation of Mendel's monohybrid crosses, the 3:1 ratio refers to: a. single-nucleotide polymorphisms (SNPs). b. copy-number variations (CNVs). c. variable number of tandem repeats (VNTRs). d. genotypes. e. phenotypes. ANSWER: e 68. In the F2 generation of Mendel's monohybrid crosses, the 1:2:1 ratio refers to: a. single-nucleotide polymorphisms (SNPs). b. copy-number variations (CNVs). c. variable number of tandem repeats (VNTRs). d. genotypes. e. phenotypes. ANSWER: d 69. A Punnett square is merely a graphical way to depict the expression: a. (1/2 A + 1/2 A). b. (1/2 A + 1/2 A)2. c. (1/2 A + 1/2 A)3. d. (3/4 A + 1/4 A). ANSWER: b 70. Among the progeny of a heterozygous round (Aa) homozygous wrinkled (aa) testcross, three seeds are chosen at random. What is the probability that all three seeds are round? a. (1/2)3 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 b. 2(1/2)3 c. 3(1/2)3 d. 4(1/2)3 ANSWER: a 71. Flower color in snapdragons is due to a gene with incomplete dominance: CRCR plants have red flowers, CRCW have pink flowers, and CWCW plants have white flowers. Which cross is expected to yield progeny with flower colors in a ratio of 1 red:1 pink? a. CRCR x CRCW b. CRCR x CWCW c. CRCW x CRCW d. CRCW x CWCW e. CWCW x CWCW ANSWER: a 72. Flower color in snapdragons is due to a gene with incomplete dominance: CRCR plants have red flowers, CRCW have pink flowers, and CWCW plants have white flowers. Which cross is expected to yield progeny that all have pink flowers? a. CRCR x CRCW b. CRCR x CWCW c. CRCW x CRCW d. CRCW x CWCW e. CWCW x CWCW ANSWER: b 73. With independent assortment, the ratio of genotypes in the F2 generation of a cross between true-breeding strains (AA bb x aa BB) can be described as 1:2:1:2:4:2:1:2:1. To what genotype does the "4" in the ratio refer? a. AA Bb b. Aa BB c. Aa Bb d. None of the other answer options is correct. ANSWER: c 74. With independent assortment, the ratio of genotypes in the F2 generation of a cross between true-breeding strains (AA bb x aa BB) can be described as 1:2:1:2:4:2:1:2:1. To what genotypes do the "2s" in the ratio refer? a. homozygous for both genes b. heterozygous for both genes Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 c. homozygous for one gene, heterozygous for the other d. None of the other answer options is correct. ANSWER: c 75. In genetics, the dash symbol (–) is a "wild card" that stands for either the dominant allele or the recessive allele; for example, W– means the individual has either the genotype WW or Ww. In summer squash, genotypes W– G– and W– gg are white, ww G– are yellow, and ww gg are green. W and G show independent assortment. What is the ratio of white : yellow : green among progeny of the cross Ww Gg x Ww Gg? a. 12:3:1 b. 9:3:1 c. 10:3:3 d. 9:6:1 e. None of the other answer options is correct. ANSWER: a 76. In genetics, the dash symbol (–) is a "wild card" that stands for either the dominant allele or the recessive allele; for example, W– means the individual has either the genotype WW or Ww. In summer squash, genotypes W– G– and W– gg are white, ww G– are yellow, and ww gg are green. W and G show independent assortment. What cross would yield white and yellow progeny in the ratio of 1:1? a. WW gg x ww gg b. WW Gg x Ww GG c. Ww Gg x Ww Gg d. WW gg x ww GG e. Ww Gg x ww GG ANSWER: e 77. In genetics, the dash symbol (–) is a "wild card" that stands for either the dominant allele or the recessive allele; for example, A– means the individual has either the genotype AA or Aa. Two genes that undergo independent assortment affect coat color in mice. Each gene has two alleles, one of which is dominant for coat color. Genotypes of the form A– B– have a brownish color called agouti, A– bb are black, and aa B– and aa bb are albino (white). What ratio of agouti : black : white is expected from the cross Aa Bb x Aa Bb? a. 10:3:3 b. 9:3:4 c. 12:3:1 d. 9:6:1 ANSWER: b 78. In genetics, the dash symbol (–) is a "wild card" that stands for either the dominant allele or the recessive allele; for example, A– means the individual has either the genotype AA or Aa. Two genes that undergo independent assortment affect the shape of the seed capsule in the broadleaf weed known as shepherd's purse. Each gene has two alleles, one of which is dominant for the shape of the seed capsule. Genotypes of the form A– B–, A– bb, and aa B– have triangular seed capsules, whereas the seed capsules of aa bb genotypes are ovoid. What ratio of triangular : ovoid is expected from the cross Aa Bb x Aa Bb? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 a. 9:7 b. 10:6 c. 12:4 d. 13:3 e. 15:1 ANSWER: e 79. You are examining a human pedigree for a trait. You notice that an offspring can be affected even if neither parent is affected. This immediately tells you that the trait is: a. dominant. b. recessive. c. epistatic. d. produced by multiple alleles. e. incomplete penetrance. ANSWER: b 80. Two individuals with the same genotype may not express the same phenotype because gene expression is influenced by: a. other genes alone. b. the environment alone. c. a combination of genes and environment. d. genes alone in some people and the environment alone in other people. e. None of the other answer options is correct. ANSWER: c 81. Consider a gene with four alleles: A1, A2, A3, and A4. In the cross A1A2_ genotypes are possible? a. 2 b. 3 c. 4 d. 5 e. 6 ANSWER: a

A3A3, how many offspring

82. Consider a gene with four alleles: A1, A2, A3, and A4. If the cross A1A2 the probability that both of them have the same genotype? a. 1/8 b. 1/4 c. 1/3 d. 1/2 e. 3/4

A3A3 yields two offspring, what is

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Chapter 15 ANSWER: d 83. Consider a gene with four alleles: A1, A2, A3, and A4. In the cross A1A2 genotypes are possible? a. 2 b. 3 c. 4 d. 5 e. 6 ANSWER: c

A3A4, how many offspring

84. Consider a gene with four alleles: A1, A2, A3, and A4. How many distinct genotypes are possible, taking into account both homozygous and heterozygous genotypes? a. 6 b. 7 c. 8 d. 9 e. 10 ANSWER: e 85. Consider a gene with four alleles: A1, A2, A3, and A4. If the cross A1A2 the probability that both of them have the same genotype? a. 1/8 b. 1/4 c. 1/3 d. 1/2 e. 3/4 ANSWER: b

A3A4 yields two offspring, what is

86. Consider a gene with six alleles: A1, A2, A3, A4, A5, and A6. How many distinct homozygous genotypes are possible? a. 2 b. 3 c. 4 d. 5 e. 6 ANSWER: e 87. Consider a gene with six alleles: A1, A2, A3, A4, A5, and A6. How many distinct heterozygous genotypes are possible? a. 3 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 b. 6 c. 9 d. 12 e. 15 ANSWER: e 88. Consider a gene with n alleles A1, A2, …, An. How many distinct heterozygous genotypes are possible? a. n2 b. n2/2 c. n(n-1) d. n(n-1)/2 e. n(n+1)/2 ANSWER: d 89. Consider Mendel's experiments involving the trait of round versus wrinkled peas. In some experiments, Mendel removed the anthers of the flowers from a plant that was true-breeding for round peas and applied pollen from the anthers of a plant that was true-breeding for wrinkled peas. What would have happened if Mendel missed some anthers that he was trying to remove from the true-breeding round pea plants, and then counted the peas produced by a plant that bore a mix of round and wrinkled pollen? Which of the statements is true? a. The F1 peas produced would have mixed phenotypes. b. The F1 peas produced would have a single phenotype. c. The F2 peas would have phenotypes in a 9:3:3:1 ratio. d. The F2 peas would all have the phenotype of the parental plant with the anthers not properly removed. e. The F1 peas would have phenotypes in a 1:1 ratio. ANSWER: b 90. Consider two genes that undergo independent assortment. One gene has alleles A and a with A dominant to a, and the other has alleles B and b with B dominant to b. From the cross Aa Bb x Aa Bb, the expected ratio of genotypes A– B–: A– bb:aa B– : aa bb is 9:3:3:1. In these symbols, the dash is a "wild card": in the symbol A–, the dash means the unspecified allele is either A or a; and in the symbol B–, the dash means the unspecified allele is either B or b. If the two genes are epistatic to each other, the 9:3:3:1 ratio can be modified according to the manner in which the genotypes of the A and B genes interact to affect the phenotype. There are exactly 9 ways in which the 9:3:3:1 ratio can be modified. Each possibility corresponds to one row in the figure, where the phenotypes are color coded. In any row, two boxes of the same color indicate that the corresponding genotypes have indistinguishable phenotypes.

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What phenotypic ratio results from epistasis of the type in row B? a. 9:7 b. 10:6 c. 12:4 d. 13:3 e. 15:1 ANSWER: d 91. Assuming that the trait represented by the filled symbols in the pedigree shown is a rare inherited trait with complete penetrance due to a single gene with alleles A and a, what mode of inheritance does the pedigree suggest?

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a. dominant b. recessive ANSWER: b 92. In the pedigree shown, what can you say about the likely genotypes of individuals A and B?

a. both are AA b. one is AA and the other is Aa c. both are Aa d. It is not possible to determine this for certain from the information provided. ANSWER: c 93. Consider two genes that undergo independent assortment. One gene has alleles A and a with A dominant to a, and the other has alleles B and b, with B dominant to b. From the cross Aa Bb x Aa Bb, the expected ratio of genotypes A– B– : A– bb : aa B– : aa bb is 9:3:3:1. In these symbols, the dash is a "wild card": in the symbol A– Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 , the dash means the unspecified allele is either A or a; and in the symbol B–, the dash means the unspecified allele is either B or b. If the two genes are epistatic to each other, the 9:3:3:1 ratio can be modified according to the manner in which the genotypes of the A and B genes interact to affect the phenotype. There are exactly nine ways in which the 9:3:3:1 ratio can be modified. Each possibility corresponds to one row in the figure shown, where the phenotypes are color coded. In any row, two boxes of the same color indicate that the corresponding genotypes have indistinguishable phenotypes. What phenotypic ratio results from epistasis of the type in row G?

a. 9:7 b. 10:6 c. 12:4 d. 13:3 e. 15:1 ANSWER: a 94. Assuming that the trait represented by the filled symbols in the pedigree is a rare inherited trait with complete penetrance due to a single gene with alleles A and a, what mode of inheritance does the pedigree shown suggest?

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a. dominant b. recessive ANSWER: b 95. In the pedigree shown, what are the genotypes of individuals C and D?

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Chapter 15 96. In the pedigree shown, what can you say about the likely genotypes of individuals A and B?

a. Both are likely to be AA. b. Both are likely to be Aa. c. One is likely to be AA, and the other Aa. d. It is not possible to determine this for certain from the information provided. ANSWER: c 97. Consider two genes that undergo independent assortment. One gene has alleles A and a with A dominant to a, and the other has alleles B and b with B dominant to b. From the cross Aa Bb x Aa Bb, the expected ratio of genotypes A– B–: A– bb:aa B– : aa bb is 9:3:3:1. In these symbols, the dash is a "wild card": in the symbol A–, the dash means the unspecified allele is either A or a; and in the symbol B–, the dash means the unspecified allele is either B or b. If the two genes are epistatic to each other, the 9:3:3:1 ratio can be modified according to the manner in which the genotypes of the A and B genes interact to affect the phenotype. There are exactly 9 ways in which the 9:3:3:1 ratio can be modified. Each possibility corresponds to one row in the figure, where the phenotypes are color coded. In any row, two boxes of the same color indicate that the corresponding genotypes have indistinguishable phenotypes.

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What phenotypic ratio results from epistasis of the type in row Q? a. 9:7 b. 10:6 c. 12:4 d. 13:3 e. 15:1 ANSWER: e 98. Consider two genes that undergo independent assortment. One gene has alleles A and a with A dominant to a, and the other has alleles B and b with B dominant to b. From the cross Aa Bb x Aa Bb, the expected ratio of genotypes A– B–: A– bb:aa B– : aa bb is 9:3:3:1. In these symbols, the dash is a "wild card": in the symbol A–, the dash means the unspecified allele is either A or a; and in the symbol B–, the dash means the unspecified allele is either B or b. If the two genes are epistatic to each other, the 9:3:3:1 ratio can be modified according to the manner in which the genotypes of the A and B genes interact to affect the phenotype. There are exactly 9 ways in which the 9:3:3:1 ratio can be modified. Each possibility corresponds to one row in the figure, where the phenotypes are color coded. In any row, two boxes of the same color indicate that the corresponding genotypes have indistinguishable phenotypes.

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What phenotypic ratio results from epistasis of the type in row B? a. 9:7 b. 10:6 c. 12:4 d. 13:3 e. 15:1 ANSWER: d 99. Assuming that the trait represented by the filled symbols in the pedigree shown is a rare inherited trait with complete penetrance due to a single gene with alleles A and a, what mode of inheritance does the pedigree suggest?

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a. dominant b. recessive ANSWER: b 100. In the pedigree shown, what can you say about the likely genotypes of individuals A and B?

a. both are AA b. one is AA and the other is Aa c. both are Aa d. It is not possible to determine this for certain from the information provided. ANSWER: c 101. Consider two genes that undergo independent assortment. One gene has alleles A and a with A dominant to a, and the other has alleles B and b, with B dominant to b. From the cross Aa Bb x Aa Bb, the expected ratio of genotypes A– B– : A– bb : aa B– : aa bb is 9:3:3:1. In these symbols, the dash is a "wild card": in the symbol A– Copyright Macmillan Learning. Powered by Cognero.

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a. ANSWER: Multiple Response 102. Which ratios are possible modifications of the 9:3:3:1 ratio that results from different types of epistasis? Select all that apply. a. 15:1 b. 14:2 c. 13:3 d. 12:4 e. 12:3:1 f. 12:2:2 g. 10:6 h. 10:3:3 i. 10:4:2 j. 9:4:3 k. 9:7 l. 9:6:1 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 m. 9:5:2 n. 8:8 ANSWER: a, c, d, e, g, h, j, k, l 103. Which of the choices result in a given genotype not necessarily producing the same phenotype? Select all that apply. a. incomplete penetrance b. variable expressivity c. incomplete dominance d. principle of segregation ANSWER: a, b 104. Which of the statements are benefits of genetic testing? Select all that apply. a. Newborns can be screened for treatable diseases. b. Individuals can find out whether or not they carry harmful recessive alleles. c. Adults can find out whether or not they have specific risk factors for some diseases. d. Physicians can use the results of genetic testing to engineer drugs specifically for an individual's genotype. e. None of the other answer options is correct. ANSWER: a, b, c 105. Which of the statements are accurate about direct-to-consumer (DTC) genetic tests? Select all that apply. a. DTC genetic tests can be purchased directly by consumers. b. DTC genetic tests involve the active intervention of medical professionals at each stage of the process, from initial submission of samples to final consultation with the consumer. c. DTC genetic tests may or may not be reliable. d. DTC genetic tests are the best way to determine whether or not an individual carries an allele associated with a specific disease. e. None of the other answer options is correct ANSWER: a, c 106. Suppose that in humans the ability to roll the tongue (R) is dominant to being unable to roll it (r). Having freckles (F) is dominant to having no freckles (f). A freckled tongue-roller could have which genotypes? Select all that apply. a. RRFF b. RrFf c. rrFF d. RRff and RrFf only e. All of these choices are correct. ANSWER: a, b 107. In pea plants, flowers are either white or purple: the purple color is produced by pigments called Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 anthocyanins. The production of anthocyanins is a two-step process: the first step is controlled by the C gene and the second by the P gene. Both genes must produce functional proteins for anthocyanin to be produced. For each gene, the dominant (C and P) alleles produce functional proteins. Which genotypes will produce white (no pigment) flowers? Select all that apply. a. ccPP b. CcPP c. CCpp d. CcPp e. CCPP ANSWER: a, c 108. A pea plant is heterozygous (Aa) for seed color and heterozygous (Bb) for seed shape. Which statements are correct according to Mendel's principle of independent assortment? Select all that apply. a. Each gamete will contain either a seed-color allele or a seed-shape allele, but not both. b. A gamete that contains the dominant allele for seed color must also contain the dominant allele for seed shape. c. A gamete that contains the dominant allele for seed color must also contain the recessive allele for seed shape. d. A gamete that contains the dominant allele for seed color is equally likely to contain the dominant or the recessive allele for seed shape. e. Possible gamete genotypes are AB or ab; each is equally likely to occur. ANSWER: d, e 109. In genetics, the dash symbol (–) is a "wild card" that stands for either the dominant allele or the recessive allele; for example, R– means the individual has either the genotype RR or Rr. In Duroc pigs, genotypes R– S– have red coats, R– ss and rr S– have sandy-colored coats, and rr ss pigs are white. R and S show independent assortment. What crosses are expected to result in all sandy-colored progeny? Select all that apply. a. RR SS × rr ss b. RR Ss × Rr SS c. Rs Ss × Rr Ss D) RR ss × rr SS d. e. rr ss × rr ss ANSWER: a, d 110. Consider two genes that undergo independent assortment. One gene has alleles A and a with A dominant to a, and the other has alleles B and b with B dominant to b. From the cross Aa Bb x Aa Bb, the expected ratio of genotypes A– B–: A– bb:aa B– : aa bb is 9:3:3:1. In these symbols, the dash is a "wild card": in the symbol A–, the dash means the unspecified allele is either A or a; and in the symbol B–, the dash means the unspecified allele is either B or b. If the two genes are epistatic to each other, the 9:3:3:1 ratio can be modified according to the manner in which the genotypes of the A and B genes interact to affect the phenotype. There are exactly 9 ways in which the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 15 9:3:3:1 ratio can be modified. Each possibility corresponds to one row in the figure, where the phenotypes are color coded. In any row, two boxes of the same color indicate that the corresponding genotypes have indistinguishable phenotypes.

What phenotypic ratio results from epistasis of the type in row Q? a. 9:7 b. 10:6 c. 12:4 d. 13:3 e. 15:1 ANSWER: e

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Chapter 16 Multiple Choice 1. Pairing of homologous chromosomes in prophase of meiosis I appears to be critical for proper alignment, crossing over, and subsequent separation. This pairing is facilitated by the sharing of sequence homology between homologous chromosomes. If X and Y chromosomes are so different, how can they achieve the necessary pairing? a. They do not pair because they are not homologous. b. They are only different in size, most of the DNA sequences match. c. They share a small region of homology at their respective tips. d. A special function of the spindle apparatus forces X and Y chromosomes together. ANSWER: c 2. Given equal probabilities of a boy or girl at birth, what is the probability that a group of three siblings includes: Exactly two boys? Exactly three boys? Two or more boys? a. 3/8; 1/8; 1/2 b. 1/4; 1/4; 1/2 c. 1/8; 1/8; 1/4 d. 1/2; 1/8; 5/8 ANSWER: a 3. In the human X and Y chromosomes, their alignment during meiosis I helps ensure proper: a. random distribution. b. gene dosage. c. independent assortment. d. segregation. ANSWER: d 4. Gene A exists in three forms at a site in the X chromosome: A1, A2, and A3. Each form, or allele, has a different number of tandem repeats. The alleles are amplified with PCR and then run on a polyacrylamide gel for analysis. The gel shown is the DNA fingerprint for two girls who have the same parents. What is the genotype of the mother? Of the father?

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a. A1/A2; A3 b. A1/A3; A2 c. A2/A3; A1 d. A1/A3; A3 e. A1/A3; A1 ANSWER: a 5. The pedigree shown here pertains to a trait due to a rare, X-linked recessive mutation.

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Individual 1 has an affected father, but the genotypes and phenotypes of individuals 1-4 are unknown. What is the probability that individual 1 is heterozygous? What is the probability that individual 3 is heterozygous? What is the probability that individual 4 is affected? a. 1; 1/2; 1/4 b. 1/2; 1/4; 1/2 c. 1; 1/4; 1/2 d. 1; 1/2; 1/2 e. 1/2; 1/4; 1/4 ANSWER: d 6. The pedigree shown pertains to a trait due to a rare, X-linked recessive mutation.

Individual 1 has an affected brother and husband, but the genotypes and phenotypes of individuals 1, 3, and 4 are unknown. What is the probability that individual 1 is heterozygous? What is the probability that individual 3 is affected? What is the probability that individual 4 is affected? a. 1/2; 0; 1/2 b. 1/2; 1/4; 1/4 c. 1; 1/2; 1/4 d. 1; 1/2; 1/2 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 e. 1; 1/4; 1/4 ANSWER: b 7. Mutations in the Y chromosome are inherited by males in every generation, whereas mutations in the X chromosome exhibit a crisscross inheritance pattern. Why? a. All the genes in the Y chromosome have complete linkage. b. The Y chromosome shares a high homology with the X chromosome. c. The Y chromosome determines maleness. d. The mutations are advantageous therefore they are preserved. ANSWER: c 8. In one set of experiments, Calvin Bridges crossed white-eyed female fruit flies with red-eyed males. Some of the offspring were white-eyed females. He demonstrated these individuals had the genotype _____, which occurred because of _____ during meiosis. a. XXX; nondisjunction b. XXO; independent assortment c. XXY; random segregation d. XXO; nondisjunction e. XXY: nondisjunction ANSWER: e 9. An X-linked ichthyosis is a recessive form of a family of skin diseases caused by a hereditary deficiency of the steroid sulfatase (STS) enzyme. A woman heterozygous for this mutation mates with the phenotypically normal man and produces an XXY son who suffers from the disorder. Which of the answer choices can explain this result? a. nondisjunction in meiosis II in the mother b. nondisjunction in meiosis I in the father c. nondisjunction in meiosis I in the mother d. nondisjunction in meiosis II in the father ANSWER: a 10. Recessive alleles in the X chromosome are expressed in males because: a. All of these choices are correct. b. the X chromosome is dominant in males. c. males are more sensitive to mutations than females. d. the Y chromosome does not contain the wild-type allele. ANSWER: d 11. X-linked genes show crisscross inheritance because males: a. get their X chromosome from their father. b. get their X chromosomes from their mother and transmit their X chromosomes to their daughters. c. transmit their Y chromosome to their sons. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 d. transmit their X chromosome to their sons. e. get their Y chromosomes from their father and transmit their Y chromosomes to their sons. f. get their Y chromosome from their father. ANSWER: b 12. A male fly that is homozygous for recessive alleles at two linked autosomal genes is mated with a wild-type female fly that is homozygous dominant for both alleles. All F1 offspring are then randomly mated. If the genes are so close as to have no recombination, what proportion of offspring will be homozygous for both mutations in the F2 generation? a. 1/4 b. 1/16 c. 1/2 d. 3/4 ANSWER: a 13. Which of Mendel's findings is not applicable when predicting the results of crosses involving genes that are closely linked? a. the law of segregation b. the law of independent assortment c. the occurrence of alternate forms (alleles) of genes d. complete dominance e. All of these choices are correct. ANSWER: b 14. What is the limit on how far apart two genes can be in the same chromosome to be considered linked genes? a. 100 map units b. 75 map units c. 50 map units d. 25 map units ANSWER: c 15. In a cell undergoing meiosis, when a single crossover occurs between two genes of interest, the result is: a. 4 chromosomes: 2 nonrecombinant and 2 recombinant. b. 4 chromosomes: 3 nonrecombinant and 1 recombinant. c. 4 chromosomes: all nonrecombinant. d. 4 chromosomes: 1 nonrecombinant and 3 recombinant. e. 4 chromosomes: all recombinant. ANSWER: a 16. An individual is heterozygous for two linked genes, but whether its genotype is AB/ab or Ab/aB is not known. The individual is crossed with an ab/ab individual, and among the progeny are: 16 AB/ab; 54 Ab/ab; 46 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 aB/ab; 24 ab/ab. These results imply that the genotype of the doubly heterozygous parent was AB/ab. a. true b. false ANSWER: b 17. Linked genes have recombination frequencies between: a. 0 and 75%. b. 0 and 25%. c. 0 and 10%. d. 0 and 50%. e. 0 and 100%. ANSWER: d 18. An individual is heterozygous for two linked genes, but whether its genotype is AB/ab or Ab/aB is not known. The individual is crossed with ab/ab, and among the progeny are: 62 AB/ab; 13 Ab/ab; 18 aB/ab; 51 ab/ab. These results imply that the genotype of the doubly heterozygous parent was AB/ab. a. true b. false ANSWER: a 19. In the gel diagram shown here, A and B are both X-linked restriction polymorphisms with three alleles (denoted A1, A2, A3 and B1, B2, B3, respectively). Mo and Fa show the banding patterns corresponding to the genotypes of mother and father in a mating, and Da is the banding pattern corresponding to the genotype of a daughter. What is the daughter's genotype?

a. A1B3/A3B2 b. A1B2/A3B3 c. A1B1/A3B3 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 d. A1B3/A2B2 ANSWER: a 20. Map distances and recombination frequencies are additive: a. for all distances between genes in one chromosome. b. for distances up to about 75 map units, after which recombination frequencies become higher than expected. c. for distances up to about 50 map units, at which point recombination frequencies become higher than expected. d. for distances up to about 25 map units, at which point recombination frequency becomes higher than expected. e. for distances up to about 15 map units, at which point recombination frequency becomes lower than expected. ANSWER: e 21. Map distances are not additive when sets of genes are more than about 15 map units apart. What could account for non-additive map distances? a. At distances greater than about 15 map units, the frequency of double crossover events is high enough to reduce the observed recombination. b. At distances greater than about 15 map units, nondisjunction rates increase and create new combinations not expected by recombination. c. At distances greater than 15 map units, homologs no longer pair and recombination cannot occur. ANSWER: a 22. If a pedigree involves a Y-linked trait, which of the observations are true? a. All affected males should have affected fathers. b. Males are exclusively affected. c. Females are not affected. d. Affected males have affected sons and unaffected daughters. e. All of these choices are correct. ANSWER: e 23. Except for the regions at its tips, where it is homologous to the X chromosome, genes in the Y chromosome: a. are completely linked. b. exhibit independent assortment. c. exhibit random segregation. d. may or may not be linked; it just depends on how far apart they are from one another. e. may or may not be linked; it just depends on which specific alleles of the gene are involved. ANSWER: a 24. In humans, one reason why mitochondrial inheritance is strikingly different from nuclear inheritance is that: a. mitochondrial DNA is circular. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 b. mitochondrial DNA is maternally inherited. c. mitochondrial DNA is single-stranded. d. unlike mitochondrial inheritance, only one parent contributes nuclear DNA. e. mitochondrial DNA is X-linked. ANSWER: b 25. A cell can contain many mitochondria. Assume 50% of the mitochondria in a human egg are healthy, whereas the other 50% harbor a defective gene that leads to poor function. If this egg is fertilized, then the probability of the resulting child showing signs of defective mitochondrial function: a. will be 50%. b. will depend upon whether or not the healthy version of the gene is dominant. c. cannot be determined because, unlike chromosomal separation in mitosis, mitochondria do not segregate as evenly to daughter cells. d. can be determined by using the Punnett square. e. can be determined because, like nuclear division, mitochondria segregate evenly to descendent cells. ANSWER: c 26. The gene density of the X chromosome is best described by which of the statements? a. The gene density of the X chromosome is like the gene density in an autosome. b. The gene density of the X chromosome is like the gene density in the Y chromosome. c. The gene density of the X chromosome is much higher than the gene density of any other chromosome. d. None of the answer options is correct. ANSWER: a 27. The total number of chromosomes differs between human males and females. a. true b. false ANSWER: b 28. The fact that the ratio of human male to female births is nearly 1:1 demonstrates that: a. All of these choices are correct. b. X and Y chromosomes undergo independent assortment. c. X and Y chromosomes pair along their length and undergo recombination. d. X and Y chromosomes together determine sex. e. X and Y chromosomes undergo segregation. ANSWER: e 29. The human X chromosome carries approximately 1000 genes. The Y chromosome contains only about 50 genes. How can males survive with only these 50 genes in the Y chromosome? a. The Y is paired with X chromosome, which has genes necessary for survival. b. The genes in the X chromosome are not required for survival. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 c. In males, the missing genes are found in autosomal chromosomes. d. Most of the 1000 genes in the X chromosome are only required for females ANSWER: a 30. Which of the statements is true of human sex chromosomes and the genes they contain? a. There are genes that females have that males do not have. b. There are genes that males have that females do not have. c. Sperm contain only the Y chromosome, whereas all eggs contain only the X chromosome. d. The contribution from the mother determines the sex of the child. e. A healthy child with two sex chromosomes of the same kind can be male. ANSWER: b 31. In Drosophila, the red-eye allele for eye color exhibits complete dominance over the recessive white-eye allele. In some of Thomas Hunt Morgan's initial experiments, he crossed red-eyed female flies with white-eyed males. When Morgan crossed the F1 red-eyed heterozygous females with red-eyed males, he found that half of the male progeny were white-eyed. What is the most likely explanation of this result? a. Chemical factors secreted by males changed the color of the eye. b. The gene responsible for eye color is on the Y chromosome. c. The gene responsible for eye color is on an autosome. d. The gene responsible for eye color is on the X chromosome. e. Chemical factors secreted by males mutated the eye color gene. ANSWER: d 32. A tandem repeat site on the X chromosome has three alleles, A1, A2, and A3. The gel shown is the DNA fingerprint for two prospective parents (Mo and Fa) along with DNA types that might (or might not) correspond to their offspring. What are the possible phenotypes of daughters from this mating?

a. K and L b. H and K c. H and L d. M and H e. M and L ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 33. In mammals, males inherit their X chromosome from their _____ and transmit it to their _____. a. fathers; sons b. mothers; sons c. fathers; daughters d. mothers; daughters e. mothers; sons and daughters ANSWER: d 34. Red-green color blindness is a sex-linked recessive trait in humans. If a carrier female (heterozygous for the trait) mated with an affected male, what would be the expected outcome? a. None of the daughters would be color blind. b. Half of the daughters would be color blind. c. None of the offspring would be color blind. d. All of the sons would be color blind. e. Half of the daughters would be color blind and all of the sons would be color blind. ANSWER: b 35. A woman with color vision and a color-blind brother: a. cannot carry the gene causing color blindness. b. must be heterozygous for the gene causing color blindness. c. has a 25% chance of being heterozygous for the gene causing color blindness. d. has a 10% chance of being heterozygous for the gene causing color blindness. e. has a 50% chance of being heterozygous for the gene causing color blindness. ANSWER: e 36. What process allows a gamete to carry a chromosome containing some alleles from the paternal chromosome and other alleles from the maternal chromosome? a. nondisjunction b. crossing over c. chromosome deletion d. chromosome duplication ANSWER: b 37. The frequency of recombination is _____ for genes that are closer together compared to genes that are further apart in the same chromosome. a. smaller b. larger ANSWER: a 38. Two genes that are 8.2 map units apart will have a recombination frequency of: a. 2.05% Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 b. 4.1%. c. 8.2%. d. 16.4%. ANSWER: c 39. Two genes are located close together in the same chromosome. Allele A for long wings is completely dominant to allele A for short wings. Allele B for light body color is completely dominant to allele B for dark body color. A female fly is heterozygous for both genes such that alleles A and B are present in one homolog, whereas alleles A and B are present in the other homolog. If she is crossed with a male fly that is homozygous recessive for both genes, then the majority of their progeny would: a. look just like their mother. b. have long wings and a light body or long wings and a dark body. c. have long wings and a light body or short wings and a dark body. d. have short wings and a light body or long wings and a light body. e. have short wings and a light body or long wings and a dark body. ANSWER: e 40. If a woman's husband has a deleterious mutation in a Y-linked trait, then she should: a. be concerned because her husband will definitely transmit it to all their sons. b. be concerned because there is a 50% probability her husband will transmit it to their sons. c. be concerned because there is a 50% probability she can transmit it to their sons. d. not be concerned because there is a 0% probability she can transmit it to their sons. ANSWER: a 41. The frequency of recombination during meiosis is a function of: a. the alleles of the genes: recombination happens less frequently between recessive alleles. b. the distance between genes: the closer the genes are, the more frequent the recombination between them. c. the distance between genes: the farther apart the genes are, the more frequent the recombination between them. d. whether or not the genes are sex-linked: genes in X chromosomes recombine much more frequently than do genes in the autosomes. e. whether or not the genes are sex-linked: genes in Y chromosomes recombine much more frequently than do genes in the autosomes. ANSWER: c 42. Three genes (A, B, and C) are in the same chromosome. The distance between A and B is 1 map unit, between B and C is 4 map units and between A and C is 3 map units. The correct order of genes in the chromosome is: a. B A C. b. B C A. c. C B A. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 d. A B C. e. A C B. ANSWER: a 43. Four genes (A, B, C, and D) are in this order: D, A, B, and C. The distance between A and B is 4 map units, between A and C is 6 map units, and between B and D is 5 map units. The recombination frequency between genes C and D should be: a. 6%. b. 4%. c. 5%. d. 7%. e. 2%. ANSWER: d 44. The genetic map shown here depicts the locations of five single-nucleotide polymorphisms (SNPs, designated a-e) on an autosome and the frequency of recombination (in percent) between adjacent SNPs. The region includes a genetic risk factor Q for a disease, which may be located in region M, H, K, or L along the chromosome.

percent recombination a-Q equals 20% percent recombination b-Q equals 16% percent recombination c-Q equals 4% percent recombination d-Q equals 4% percent recombination e-Q equals 20% What is the most likely position of Q in the genetic map? a. position L b. position H c. position M d. position K e. None of the answer options is correct. ANSWER: d 45. A Y-chromosome haplotype is a set of Y-linked nucleotides that: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 a. are present in noncoding DNA. b. allow one Y chromosome to be distinguished from another. c. distinguish the Y chromosome from the X chromosome. d. None of the answer options is correct. ANSWER: b 46. A trait due to a gene in the Y chromosome can be observed in both males and females in a pedigree. a. true b. false ANSWER: b 47. Because of its unique structure and pattern of inheritance, mutations in the Y chromosome: a. are accumulated within individual Y chromosome lineages. b. generally only affect male reproductive function. c. are helpful for mapping genes. d. are shared only with specific regions of the X chromosome. e. happen only rarely. ANSWER: a 48. The Y chromosome can be used to trace ancestry because: a. natural selection favors specific mutations in specific places. b. mutations in the Y chromosome occurred as people migrated around the globe. c. mutations in the Y chromosome are independent of mutations in mitochondria. d. mutations in the Y chromosome occur only in a few locations. e. nonmutant Y chromosome genotypes are eliminated from human populations. ANSWER: b 49. The mitochondrial DNA sequence of a woman should match all of these people except: a. her siblings from the same parents. b. her mother. c. her sons. d. her daughters. e. her father. ANSWER: e 50. Analysis of mitochondrial DNA can answer interesting and important questions. For which of the questions would analyzing mitochondrial DNA be futile? a. Are you my long-lost brother? b. I have poor muscle function, is there something wrong with my ability to make lots of ATP? c. Are you my father? d. From biological samplings at a crime scene, can we keep this person as our primary suspect? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 e. Which is more closely related to the great white shark, the tiger shark or the hammerhead shark? ANSWER: c 51. In humans, mitochondria (and their genomes) show: a. paternal inheritance. b. maternal inheritance. c. biparental inheritance. d. primarily maternal inheritance with some paternal inheritance. e. primarily paternal inheritance with some maternal inheritance. ANSWER: b 52. A recessive X -linked trait can be expressed in both males and females. a. true b. false ANSWER: a 53. If the genome of plant species' chloroplast was sequenced, where would you predict it to appear on the phylogeny shown, in position 1 or position 2?

a. Position 2 because the chloroplast came from a plant. b. Position 1 because chloroplasts are responsible for photosynthesis and would evolve to have similar genes to cyanobacteria. c. Position 1 because chloroplasts originated from photosynthetic cyanobacteria. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 d. Position 2 because the chloroplast would have originated with plant species a, so it should be more closely related to this plant species. ANSWER: c 54. Why didn't Mendel discover the nature of sex chromosomes in his pea plants? a. Peas, and most other plants, do not have sex chromosomes. b. He focused only on the flowers and the seeds. c. The 3:1 and 9:3:3:1 ratios of offspring did not fit the pattern of sex determination in peas. d. The sex chromosomes were too difficult to see with his rudimentary microscope. ANSWER: a 55. Given equal probabilities of the birth of a boy or girl, what is the probability that a group of four siblings includes all boys? All girls? All boys or all girls? a. 1/16; 1/8; 3/8 b. 1/16; 1/16; 1/8 c. 1/16; 1/16; 1/64 d. 1/8; 1/16; 3/8 ANSWER: b 56. Given equal probabilities of the birth of a boy or girl, if a group of four siblings consists of all girls, what is the probability that the next birth is a boy? a. 2/3 b. 3/4 c. 1/2 d. 4/5 ANSWER: c 57. The pedigree shown here pertains to a trait due to a rare, X-linked recessive mutation.

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Individual 1 has an affected grandfather, but the genotypes and phenotypes of individuals 1-4 are unknown. What is the probability that individual 1 is heterozygous? What is the probability that individual 3 is heterozygous? What is the probability that individual 4 is affected? a. 1; 1/4; 1/4 b. 1; 1/2; 1/2 c. 1/2; 1/4; ½ d. 1/2; 1/2; ½ e. 1/2; 1/4; 1/4 ANSWER: e 58. The pedigree shown here pertains to a trait due to a rare, X-linked recessive mutation.

Individual 1 has an affected brother, but the genotypes and phenotypes of individuals 1-4 are unknown. What is the probability that individual 1 is heterozygous? What is the probability that individual 3 is heterozygous? What is the probability that individual 4 is affected? a. 1; 1/2; 1/2 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 b. 1/2; 1/2; 1/2 c. 1/2; 1/2; 1/4 d. 1/2; 1/4; 1/4 e. 1; 1/4; 1/2 ANSWER: d 59. For an X-linked gene, which of the genotype options is possible? a. A female can be homozygous or heterozygous for this gene. b. A female or a male can be heterozygous for this gene. c. A female or a male can be homozygous for this gene. d. A male can be homozygous for this gene. e. A male can be heterozygous for this gene. ANSWER: a 60. Considering an X-linked dominant trait, if an unaffected woman and an affected man decide to have children, which of the answer choices is possible for their children? a. Their sons are expected to be heterozygous for the gene. b. Their daughters are expected be heterozygous for the gene. c. All their children, whether male or female, are expected to show the dominant trait. d. All of their sons are expected to show the dominant trait. e. Their daughters are not expected to show the dominant trait. ANSWER: b 61. A cross involving an autosomal gene versus a sex-linked gene would differ in that: a. the phenotype outcomes of females compared to males could be different. b. a Punnett square could not be utilized for solving a cross involving a sex-linked gene. c. both parents would have two copies of a sex-linked gene. d. only autosomal genes observe the law of segregation. e. recombination or crossing over could not take place between sex chromosomes. ANSWER: a 62. When X-linked traits are recessive, males who express the trait: a. must receive recessive alleles from both parents. b. receive the recessive allele from their fathers. c. receive the recessive allele from their mothers. ANSWER: c 63. Experimental demonstration that genes are present in chromosomes was inferred from the: a. nondisjunction of the X and Y chromosomes. b. disjunction of the X and Y chromosomes. c. segregation of the X and Y chromosomes. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 d. appearance of white-eyed males. ANSWER: a 64. Inheritance due to a rare X-linked dominant mutation can be distinguished from inheritance due to a rare autosomal dominant mutation because, in the X-linked case: a. the two possibilities cannot be distinguished. b. affected mothers do not have affected sons. c. affected fathers do not have affected daughters. d. affected mothers do not have affected daughters. e. affected fathers do not have affected sons. ANSWER: e 65. If you tracked an individual X chromosome from a male to his first-generation offspring, then to his secondgeneration offspring, what pattern would you see? a. His X chromosome would be found only in his sons and grandsons. b. His X chromosome would be found only in his daughters, then in both his granddaughters and grandsons. c. His X chromosome would be found only in his daughters and grandsons. d. His X chromosome would be found only in his daughters and granddaughters. e. His X chromosome would be found only in his sons, then in both his granddaughters and grandsons. ANSWER: b 66. A female fruit fly that is heterozygous for the white-eye mutation has red-eyed sons and white-eyed sons in equal proportions: a. regardless of the genotype of the male. b. only when mated with a white-eyed male. c. only when mated with a heterozygous male. d. only when mated with a red-eyed male. ANSWER: a 67. Red-green color blindness is due to a mutant gene in the X chromosome. An XX female with normal color vision and an XY male with normal color vision have a child with karyotype XXY who is color blind. The likely explanation is that the: a. mother was heterozygous for the color-blind mutation. b. father was heterozygous for the color-blindness mutation. c. father had CNV of the color-blind mutation. d. mother had CNV of the color-blind mutation. e. None of the other answer options is correct. ANSWER: a 68. What is the highest recombination frequency possible between two genes in the same chromosome? a. 100% Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 b. 25% c. 75% d. 50% e. The value depends on the chromosome in question. ANSWER: d 69. An individual who is heterozygous for two linked genes (with alleles A, a and B, b) is crossed with an AB/ab individual, and among the progeny are: 14 AB/ab 36 Ab/ab 34 aB/ab 16 ab/ab What is the frequency of recombination? a. 0.35 b. 0.30 c. 0.40 d. 0.60 e. 0.70 ANSWER: b 70. The genetic map shown here depicts the locations of five single-nucleotide polymorphisms (SNPs, designated a-e) in an autosome and the frequency of recombination (in percent) between adjacent SNPs. The region includes a genetic risk factor Q for a disease, which may be located at position M, H, K, or L.

Pedigree studies indicate that the frequencies of recombination between each of the SNPs and Q are: percent recombination a-Q equals 10% percent recombination b-Q equals 6% percent recombination c-Q equals 6% percent recombination d-Q equals 14% percent recombination e-Q equals 30% What is the most likely position of Q in the genetic map? a. position H b. position M c. position K d. position L e. None of the other answer options is correct. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 71. In genetic mapping, a map unit is defined as: a. 1 micrometer. b. 1 millimeter. c. the distance between genes resulting in 1% recombination. d. the physical distance between two genes. e. the distance between genes resulting in 50% recombination. ANSWER: c 72. Four genes (A, B, C, D) are in the same chromosome, but their order is not yet known. The distance between A and B is 4 map units, between A and C is 6 map units, between A and D is 1 map unit, between B and C is 2 map units, and between B and D is 5 map units. The correct order of genes in the chromosome is: a. A, B, C, D. b. D, C, B, A. c. B, A, C, D. d. C, B, A, D. e. B, D, A, C. ANSWER: d 73. Sometimes certain diseases affect all sons of an affected father, even when the mother is unaffected. What is the best explanation of this? a. The mother is a heterozygote and gives the affected gene to her son. b. The disease-related gene is in the Y chromosome. c. There is a breakdown of the crisscross pattern of inheritance. d. The corresponding genes in the Y chromosome are not expressed. e. The disease exhibits dominant inheritance. ANSWER: b 74. The pedigree shown is following a certain trait that is not Y-linked.

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Which individuals are expected to have essentially the same Y chromosome? a. II-1, III-1, and IV-1 b. II-5, III-8, and IV-3 c. I-1, II-5, III-9 d. I-1, II-7, and III-9 e. All the men in this pedigree should have identical Y chromosomes. ANSWER: a 75. If you were male and sent a biological sample to a DTC (direct-to-consumer) genetic testing service, it could tell you the possible ethnic origin of your: a. autosomes. b. X chromosome. c. female relatives' X chromosomes. d. Y chromosome. e. Y chromosome, your X chromosome, and your female relatives' X chromosomes. ANSWER: d 76. The mitochondrial DNA sequences of a large extended family were analyzed and compared to a single male in the family. Which of the relatives' mitochondrial DNA should be a match to this person? a. his children b. his paternal grandmother Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 c. his maternal grandfather d. his brother's daughter e. his sister's son ANSWER: e 77. Harmful mutations in mitochondrial DNA almost always affect: a. ATP production. b. males more seriously than females. c. glycolysis. d. females more seriously than males. ANSWER: a 78. Mitochondrial DNA haplotypes can be used to trace the ancestry of the molecule because mutations in mitochondrial DNA: a. occur randomly in time. b. accumulate over time. c. do not undergo recombination. d. All of these answer choices are correct. ANSWER: d 79. Mitochondrial DNA is useful for tracing ancestry because: a. both recombination and mutation in mitochondrial DNA are rare, but when mutations do occur, they are almost always harmful. b. mutations in mitochondrial DNA are rare; mitochondrial DNA has relatively little variation. c. mutations in mitochondrial DNA are almost always harmful. d. both recombination and mutation in mitochondrial DNA are rare. e. recombination rarely occurs. ANSWER: e 80. Mitochondria in eukaryotic cells arose from a unicellular group of organisms called alpha-proteobacteria. If you take a cell from a pine tree, a photosynthetic eukaryotic organism, and sequence the mitochondrial genome, the chloroplast genome, and the nuclear genome which of the answer choices would you expect? a. The mitochondrial and chloroplast genomes would have the highest sequence similarity to one another because they both arose from prokaryotic organisms early in eukaryotic history. b. The mitochondrial and nuclear genome would have the highest sequence similarity because mitochondria originated first, so those two genomes have been evolving together for a longer period of time. c. The chloroplast genome and the nuclear genome would have the highest sequence similarity because pine trees are photosynthetic. d. The mitochondrial, chloroplast, and nuclear genomes will have low sequence similarity because all three have separate evolutionary origins. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 81. In the pedigree shown here, male I-1 has hemophilia and males II-1, II-4, and III-1 are red-green color blind.

What must be true of individual I-2? a. She is a carrier of the color-blindness mutation. b. She is a carrier of the hemophilia mutation. c. She is a carrier of both mutations, carrying both mutations in the same chromosome. d. She is a carrier of both mutations, carrying one mutation in each of her two homologous chromosomes. ANSWER: a 82. In the pedigree shown here, male I-1 has hemophilia and males II-1, II-4, and III-1 are red-green color blind.

What must be true of individual II-2? a. She is a carrier of the hemophilia mutation. b. She is a carrier of both mutations, carrying one mutation in each of her two homologous X chromosomes. c. She is a carrier of the color-blindness mutation. d. She is a carrier of both mutations, carrying both mutations in the same chromosome. ANSWER: b 83. In the pedigree shown here, male I-1 has hemophilia and males II-1, II-4, and III-1 are red-green color blind.

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If these two genes are so far apart that they are unlinked, which of the two situations is more likely? Situation 1: II-1 and II-2 have a son with only color blindness. Situation 2: II-1 and II-2 have a son with only hemophilia. a. Both situations are equally likely. b. Situation 1 is more likely. c. Situation 2 is more likely. ANSWER: a 84. In the pedigree shown here, male I-1 has hemophilia and males II-1, II-4, and III-1 are red-green color blind.

If these two genes have a recombination frequency of 14% and individuals II-1 and II-2 have their third son, what is the chance that he will have both color blindness and hemophilia? a. 5% b. 7% c. 20% d. 40% e. 80% ANSWER: b 85. In the pedigree shown here, male I-1 has hemophilia and males II-1, II-4, and III-1 are red-green color blind.

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If these genes are far enough apart to undergo independent assortment, and II-1 and II-2 have another son, what is the probability that the child will be affected with color blindness alone, hemophilia alone, or both traits together? a. 5/8 b. 1/2 c. 3/8 d. 3/4 e. None of the other answer options is correct. ANSWER: d Multiple Response 86. To discover the location of a genetic mutation that causes a specific disease in humans, scientists use genetic mapping. Specifically, they look for genetic markers, or previously discovered DNA polymorphisms, that show statistical association with the occurrence of the disease. What does the statistical association mean? Select all that apply. a. DNA polymorphism and a gene causing the disease are linked. b. Disease causes the DNA polymorphism. c. DNA polymorphism and the disease gene are unlinked. d. DNA polymorphism causes the disease. ANSWER: a, b 87. Which of the answer choices best describes the behavior of the human X and Y chromosomes during meiosis? Select all that apply. a. The X and Y chromosomes have no regions of homology and therefore do not pair during meiosis. b. The X and Y chromosomes have short regions of homology that allow them to pair during meiosis. c. When the X and Y chromosomes pair during meiosis, crossover does occur in the short regions of homology. d. X and Y chromosomes pair with each other during meiosis the same way autosomes do. e. None of the other answer options is correct. ANSWER: b, c 88. In Bridges' experiments on Drosophila, he found that rarely, among progeny from matings between whiteCopyright Macmillan Learning. Powered by Cognero.

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Chapter 16 eyed females and red-eyed males, he would observe unusual white-eyed daughters and red-eyed sons. He further found that these white-eyed females had an XXY genotype while the red-eyed males were XO. What did Bridges conclude from these observations? Select all that apply. a. In Drosophila, sex is determined by the number of X chromosomes, and not by the presence of a Y chromosome. b. Genes for white eyes are on the X chromosome. c. The X and Y chromosome can occasionally undergo nondisjunction. d. Separation of the X and Y chromosomes is random, and does not follow the laws of segregation. ANSWER: a, b, c 89. An individual that is heterozygous for two linked genes has the genotype Ab/aB, which means that the A and b alleles are in one chromosome and the a and B alleles are on the homologous chromosome. Which of the products of meiosis listed are the recombinant types? Select all that apply. a. AB b. Ab c. aB d. ab ANSWER: a, d 90. In Jabberwocks, flame eyes (F) are dominant to blue eyes (f) and burbling (B) is dominant to whistling (b). Jabberwock geneticists suspect that the two genes are linked on one of the autosomes. They mate a truebreeding, flame-eyed, burbling female with a true-breeding, blue-eyed, whistling male. They then isolate a sample of gametes produced by one of the resulting offspring. If the genes are linked, which of the answer choices is/are a recombinant gamete genotype? Select all that apply. a. FB b. Fb c. fb d. fB e. None of the answer options is correct. ANSWER: b, d 91. Y-linked traits are: Select all that apply. a. never inherited by females. b. never transmitted by females. c. transmitted by carrier females. d. transmitted by affected fathers to their sons. e. transmitted by affected fathers to their daughters. ANSWER: a, b, d 92. Nondisjunction of X chromosomes in females that occurs in the first meiotic division results in eggs with these genotypes: Select all that apply. a. XX. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 b. O. c. XY. d. X. ANSWER: a, b 93. An XXY individual inherited three sex chromosomes due to nondisjunction during meiosis in one the parents. This person can be: Select all that apply. a. a male who received an X normally from his mom, but an X and Y from his dad due to nondisjunction. b. a male who received two X's from his mom due to nondisjunction, but a Y normally from his dad. c. a female who received an X normally from her mom, but an X and Y from her dad due to nondisjunction. d. a female who received two X's from her mom due to nondisjunction, but a Y normally from her dad. ANSWER: a, b 94. Which of the answer choices are useful in reconstructing phylogenetic histories? Select all that apply. a. Y chromosome b. mitochondrial DNA c. chloroplast DNA d. X chromosome ANSWER: a, b, c 95. Hemophilia is a sex-linked recessive trait in humans. If a carrier female (heterozygous for the trait) mated with a non-affected male, what would be the expected outcome(s)? Select all that apply. a. Half the daughters would have hemophilia. b. None of the daughters would have hemophilia. c. Half the sons would have hemophilia. d. None of the offspring would have hemophilia. ANSWER: b, c 96. Y-linked traits: Select all that apply. a. only affect males. b. affect all sons of an affected male. c. affect males and females equally. d. affect primarily males but can also affect females (depending on her parents). ANSWER: a, b 97. What patterns in a pedigree characterize a trait determined by the mitochondrial genome? Select all that apply. a. All offspring of an affected female show the trait. b. Males never transmit the trait to their offspring. c. Males are never affected. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 16 d. Males transmit the trait to their offspring. ANSWER: a, b 98. Considering a rare X-linked recessive trait, which of the statements is/are true? Select all that apply. a. An unaffected father is expected to have unaffected daughters. b. Both parents have to be affected in order to have affected daughters. c. An affected father is expected to have affected daughters. d. An affected mother is expected to have affected sons. e. An unaffected woman whose father was affected is expected to have unaffected sons. ANSWER: a, d 99. In Jabberwocks, flame eyes (F) are dominant to blue eyes (f) and burbling (B) is dominant to whistling (b). Jabberwock geneticists suspect that the two genes are linked in one of the autosomes. They mate a truebreeding, flame-eyed, burbling female with a true-breeding, blue-eyed, whistling male. They then mate a pair of the offspring. If the genes are linked, which of the F2 genotypes will be rarer than would be expected if the genes were not linked? Select all that apply. a. ff BB b. FF bb c. FF Bb d. FF BB e. ff bb ANSWER: a, b, c

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Chapter 17 Multiple Choice 1. Which of the traits are quantitative traits? a. serum cholesterol b. blood pressure c. height d. weight e. All of these choices are correct. ANSWER: e 2. In general, the phenotypes of a complex trait should exhibit a(n) _____ distribution. a. normal b. complex c. skewed d. inverted e. bimodal ANSWER: a 3. Of the students in your class, 95% are between 140 cm and 180 cm tall. Assuming that these data are normally distributed, what is the mean height of the students? a. 140 cm b. 150 cm c. 160 cm d. 170 cm e. 180 cm ANSWER: c 4. The data shown are the results of growing one strain of a crop plant in soils with different nitrogen concentrations.

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If a different strain of the same crop plant were grown in the same soil conditions, you would predict that: a. it would respond completely differently to soil nitrogen concentrations because they are different strains. b. it would respond the same way to soil nitrogen concentrations because they are the same type of crop plant. c. it could respond the same way or differently; different genotypes can, but may not always, respond differently to soil nitrogen concentrations. d. it could respond the same way or differently because the environment has too many variables to predict. ANSWER: c 5. Phenylketonuria (PKU) is a metabolic disorder that results in intellectual deficiency, seizures, and sometimes mental disorders when phenylalanine is present in the diet. It was found that putting children with this genetic condition on a special diet low in phenylalanine prevents the symptoms of the disease from developing. This is an example of: a. genotype-by-environment interaction. b. correlation between genotype and environment. c. phenotypic variation due to genetic variation. d. None of the other answer options is correct. ANSWER: a 6. Complex traits are often called _____ because they _____. a. epistatic; are influenced by many genes b. external; are influenced by the environment as well as by genes c. qualitative; come in one of several discrete (distinct) forms d. quantitative; can be measured Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 e. environmental; are influenced by the environment as well as by genes ANSWER: d 7. One reason that the effects of individual genes on complex traits can be difficult to determine is that: a. environmental effects cannot be measured. b. environmental effects are so large that genetic effects cannot be measured. c. a single phenotype may be produced by many different genotypes. d. neither environmental nor genetic effects can be measured. e. genetic effects cannot be measured, regardless of the effects of the environment. ANSWER: c 8. Inbred lines show: a. phenotypic variation due to mutation. b. phenotypic variation due to environmental variation. c. no phenotypic variation. d. phenotypic variation due to genetic variation. ANSWER: d 9. In a parent-offspring regression experiment, suggest a hypothesis that could explain the solid line in the graph.

a. The offspring were raised in a higher-quality environment than were the parents. b. The offspring were raised in a lower-quality environment than were the parents. c. The trait has a heritability that is less than 0. d. The trait has a heritability that is greater than 1. e. None of the answer options is correct. ANSWER: a 10. Which of the traits in the accompanying table has the greatest heritability? Copyright Macmillan Learning. Powered by Cognero.

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a. trait K b. trait H c. trait Q d. trait L e. trait M ANSWER: c 11. The scatterplots show the relationship between the average phenotype of parents (x-axis) and that of their offspring (y-axis) for two traits in a variety of tropical sweet corn. One trait is stalk height, which has a heritability of 75%; and the other is ear length, which has a heritability of 25%. Which of the matchups is correct?

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a. Ear length and stalk height are both trait Q b. Ear length is trait Q, and stalk height is trait K. c. Ear length and stalk height are both trait K. d. Ear length is trait K, and stalk height is trait Q. ANSWER: d 12. The scatterplots show the relation between the average phenotype of parents (x-axis) and that of their offspring (y-axis) for three traits. One trait has a heritability of 90% (comparable to that of fingerprint ridge count), another has a heritability of 50% (comparable to that of human height), and yet another has a heritability of 25% (comparable to that of human longevity).

The fact that all of the lines slope upward means that: a. some genes affect all three traits. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 b. variation in each trait is affected by only environmental differences. c. the genes affecting any one trait have no effect on any other trait. d. variation in each trait is affected by differences in heritability. e. the traits all show genotype-by-environment interaction. ANSWER: d 13. Regression toward the mean is observed because of: a. breakdown of genotype-by-environment interaction. b. segregation and recombination of alleles. c. mating between relatives. d. correlation between genotype and environment. e. variable norms of reaction. ANSWER: b 14. When the average weight of the parents is smaller than the population mean, what is the average weight of the offspring compared with their parents? Compared with the population? a. greater; smaller b. greater; greater c. smaller; smaller d. smaller; greater e. It is not possible to determine the average weight of the offspring compared to the parents' weights or the mean of the population from the information provided. ANSWER: a 15. Heritability is a measure of the: a. norm of reaction of a trait. b. proportion of a trait that is determined by genes rather than by the environment. c. extent to which a trait is passed from parent to offspring. d. extent to which a trait exhibits regression toward the mean in a population. e. proportion of total variation in a trait that is due to genetic differences among individuals within a population. ANSWER: e 16. Heritability is the proportion of the total phenotypic variation in a trait that is the result of: a. genetic variation among individuals. b. environmental variation among the individuals. c. variation in the degree of dominance of different genes. d. variation in the degree of epistasis among genes. e. variation in the frequency of recombination between genes. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 17. When heritability is 100%, the environment plays: a. no role in the expression of the trait. b. a role in the expression of a trait, but it is a small one. c. a role in the expression of a trait, but it cannot be measured. d. no role in variation in the trait among individuals. e. a role in variation in the trait among individuals, but it cannot be measured. ANSWER: d 18. In the graph of Galton's height data, which line on the graph represents the population mean?

a. line b b. line c c. line a ANSWER: b 19. In the graph of Galton's height data, which line on the graph represents the average height of the parents in a mating?

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a. line a b. line b c. line c ANSWER: a 20. In the graph of Galton's height data, which line on the graph represents the mean height of the offspring for a given average height of the mother and father?

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a. line c b. line a c. line b ANSWER: c 21. Which line on the graph demonstrates Galton's concept of regression toward the mean?

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a. line b b. line a c. line c ANSWER: a 22. In which of the complex traits shown is variation determined almost completely by the environment?

a. trait Q b. trait H c. trait K d. trait L e. trait M Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 ANSWER: e 23. Which of the traits depicted in the graph shown has the lowest heritability?

a. trait H b. trait L c. trait K d. trait M ANSWER: b 24. Herman Nilsson-Ehle's studies of seed color in wheat were important because they showed that: a. complex traits are subject to Mendelian laws of inheritance. b. inbred strains can be used to study the environmental component of complex traits. c. complex traits have a large environmental component. d. complex traits are not subject to Mendelian laws of inheritance. e. complex traits both have a large environmental component and are subject to Mendelian laws of inheritance. ANSWER: a 25. The data in the graph shown are the concordance between identical twins (monozygotic) and same-sex fraternal twins (dizygotic) for five complex traits. Based on these data, which of the traits shows the greatest effect of environment relative to heredity?

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a. trait K b. trait H c. trait L d. trait M e. trait Q ANSWER: c 26. The data in the table shown are the concordance between monozygotic twins and same-sex dizygotic twins for five complex traits. Based on these data, which of the traits shows the greatest effect of heredity relative to environment?

a. trait H b. trait K c. trait M d. trait L e. trait Q Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 ANSWER: b 27. Consider a trait determined entirely by environmental risk factors for which the chance that a random individual is affected is 25%. What is the concordance between monozygotic twins? What is the concordance between same-sex dizygotic twins? a. monozygotic concordance = 0.143; dizygotic concordance = 0.143 b. monozygotic concordance = 0.143; dizygotic concordance = 0.167 c. monozygotic concordance = 0.167; dizygotic concordance = 0.143 d. monozygotic concordance = 1.0; dizygotic concordance = 0.167 ANSWER: a 28. Consider a trait determined by a rare, simple Mendelian, autosomal dominant allele. What is the concordance between identical (monozygotic) twins? Between same-sex fraternal (dizygotic) twins? a. monozygotic concordance = 1; fraternal concordance = 0.333 b. monozygotic concordance = 1; fraternal concordance = 0.143 c. monozygotic concordance = 1; fraternal concordance = 0.250 d. monozygotic concordance = 1; fraternal concordance = 0.071 e. monozygotic concordance = 1; fraternal concordance = 0.500 ANSWER: e 29. Consider a trait determined by a rare, simple Mendelian, autosomal recessive allele. What is the concordance between monozygotic twins? Between same-sex dizygotic twins? a. monozygotic concordance = 1; dizygotic concordance = 0.071 b. monozygotic concordance = 1; dizygotic concordance = 0.143 c. monozygotic concordance = 1; dizygotic concordance = 0.250 d. monozygotic concordance = 1; dizygotic concordance = 0.333 e. monozygotic concordance = 1; fraternal concordance = 0.500 ANSWER: b 30. Consider a trait determined by a rare mutation in mitochondrial DNA. What is the concordance between monozygotic twins? Between same-sex dizygotic twins? a. monozygotic concordance = 1; dizygotic concordance = 1 b. monozygotic concordance = 1; dizygotic concordance = 1/4 c. monozygotic concordance = 1; dizygotic concordance = 1/2 d. monozygotic concordance = 1; dizygotic concordance = 3/4 e. monozygotic concordance = 1; dizygotic concordance = 0 ANSWER: a 31. Consider a trait determined by a rare mutation in the Y chromosome. What is the concordance between male monozygotic twins? Between dizygotic twins that are both male? a. monozygotic concordance = 1; dizygotic concordance = 1 b. monozygotic concordance = 1; dizygotic concordance = 1/4 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 c. monozygotic concordance = 1; dizygotic concordance = 1/2 d. monozygotic concordance = 1; dizygotic concordance = 3/4 e. monozygotic concordance = 1; dizygotic concordance = 0 ANSWER: a 32. If the genetic contribution to variation in a trait is negligible: a. then for that trait monozygotic twins will be no more similar to one another than are dizygotic twins. b. then for that trait dizygotic twins will be more similar to one another than monozygotic twins are to one another. c. then for that trait monozygotic twins will be more similar to one another than fraternal twins are to one another. d. then for that trait monozygotic twins will resemble one another more closely than do any other siblings. e. None of the other answer options is correct. ANSWER: a 33. When identical twins differ in phenotype of a complex trait, it is almost always due to: a. epistasis. b. segregation. c. the environment. d. mutation. ANSWER: c 34. BRCA1 and BRCA2 are: a. genes known to cause cancer. b. proteins necessary for normal breast cell division. c. genes related to DNA synthesis. d. proteins necessary for tumor suppression. e. genes related to DNA repair. ANSWER: e 35. Personalized medicine refers to the idea of: a. basing treatment on an individual's long-term medical history. b. using a patient's individual lifestyle and other risk factors to inform treatment for a given disease. c. using a combination of a patient's lifestyle and genotype to inform treatment for a given disease. d. using a patient's individual genotype to select treatment for a given disease. e. using a combination of a patient's lifestyle, genotype, and long-term medical history to inform treatment for a given disease. ANSWER: d 36. One hope of personalized medicine is that: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 a. it will lead to medications with fewer side effects. b. physicians will be able to use an individual's lifestyle factors to moderate harmful side effects of medication. c. physicians will be able to use an individual's genotype to avoid harmful side effects of some medications. d. physicians will be able to use an individual's lifestyle factors to identify preventative medicines that have minimal side effects. e. individuals will receive preventative medications for diseases they have not yet contracted. ANSWER: c 37. A new mutation arises that affects a complex trait. In most cases, its effect is likely to be: a. small. b. medium. c. large. ANSWER: a 38. Personalized medicine matches the treatment to the: a. hospital. b. genotype. c. disease. d. phenotype. e. insurance. ANSWER: b 39. A complex trait can be influenced by the environment. a. true b. false ANSWER: a 40. Which of the statements best reflects the extent to which we can distinguish the effects of genes and the environment on the expression of a complex trait? a. About 80% of an individual's height is determined by genes and 20% by environment. b. For identical twins raised in the same environment, height should not differ by more than 20%. c. In a group of individuals of the same sex, about 80% of the variation in height among individuals is due to genetic differences and 20% is due to environmental differences. d. If one individual is 60 inches tall and another is 70 inches tall, they differ in about 80% of their genes. ANSWER: c 41. Ten individuals of two inbred strains of mice, strain A and strain B, are fed identical low sugar diets. All have normal blood sugar as adults. The same mice are then fed diets high in sugar. Mice from strain B develop diabetes, but mice from strain A do not. Diabetes in strain B mice results from: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 a. their genotypes, which are different from strain A mice. b. the interaction of their diet with their genotype. c. their diets, which are different from strain A mice. d. their norm of reaction to sugar in their diets. e. None of the other answer choices is correct. ANSWER: b 42. Suppose four genes that display independent assortment affect a quantitative trait with no significant effects from the environment, as in the three-gene wheat seed color example shown. As in this example, each allele indicated by a capital letter adds one unit to the phenotype. What is the total number of possible phenotypes?

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Chapter 17 43. The plots shown depict several possible outcomes of an experiment comparing the mean phenotype of a behavioral trait in two strains of mice in three different environments. Strain 1 is represented by filled circles and Strain 2 is represented by open circles. Environment 1 (E<sub1) is deficient in stimulation for the behavior, environment 2 (E2) has an average level of stimulation, and environment 3 (E3) is enriched in stimulation. Which of the possible results indicates the absence of genotype-by-environment interaction?

a. result H b. result L c. result K d. result M e. None of the answer options is correct. ANSWER: b 44. Sunlight exposure has stronger effect on skin cancer risk in fair-skinned humans than in individuals with darker skin. This is an example of: a. All of these choices are correct. b. epistasis. c. genotype-by-environment interaction. d. differences in the norm of reaction. e. pleiotropy. ANSWER: a 45. Suppose two genes that display independent assortment affect a quantitative trait with no significant effects from the environment, like the three genes in the seed color example shown. As in the example, each allele indicated by a capital letter adds one unit to the phenotype. What is the total number of possible phenotypes? What is the mean phenotype among the F2 progeny of a cross between homozygous genotypes?

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Chapter 17 46. Which of the traits depicted in the graphs shown illustrates regression toward the mean?

a. trait M b. trait H c. trait K d. trait L e. trait Q ANSWER: c 47. If all variation among individuals in a population is due to differing genotypes alone, heritability is _____, and the slope of the line used to measure it is _____. a. 100%; 1.0 b. 100%; 0.5 c. 50%; 0.5 d. 50%; 1 e. 0%; 0 ANSWER: a 48. When heritability is 0%, genes play: a. no role in the expression of a trait. b. a role in the expression of a trait but only a small one. c. a role in the expression of a trait, but it cannot be measured. d. no role in variation in the trait among individuals. e. a role in variation in the trait among individuals, but it cannot be measured. ANSWER: d 49. A scientist measured heritability in crop yield per acre of a strain of wheat grown on a farm in eastern Europe and determined that it was 80%. She can therefore conclude that: a. None of the answer options is correct. b. there will be 80% of variation among individuals due to genetic differences when the same strain of wheat is grown on the same farm in eastern Europe the following year. c. when the same strain of wheat is grown on two different farms at the same time 80% of the average difference in yield per acre will be due to genetic differences among populations. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 d. 80% of the total crop yield is due to genotype and 20% to environment. e. there will be 80% of variation among individuals due to genetic differences when the same strain of wheat is grown in Nebraska. ANSWER: d 50. When heritability is 100%, the variation among individuals in the population is due to: a. phenotype. b. genotype. c. Both of these choices are correct. ANSWER: b 51. Comparisons of traits in monozygotic and dizygotic twins are useful because they: a. allow estimates of the number of genes affecting a complex trait. b. reveal the extent of the environment only in causing phenotypic variation. c. reveal the relative importance of genes and environment in causing phenotypic variation. d. None of the answer options is correct. ANSWER: c 52. Using monozygotic twins to study heritability is complicated by the fact that: a. monozygotic twins often experience a more similar uterine environment and more similar treatment during childhood than do fraternal twins. b. monozygotic twins are often treated more similarly to one another than are dizygotic twins. c. dizygotic twins often share the same uterine environment, whereas monozygotic twins do not. d. monozygotic twins often experience identical uterine environments, whereas dizygotic twins do not. e. even monozygotic twins are sometimes genetically different from one another. ANSWER: a 53. In a hypothetical twin study, investigators identify 50 monozygotic twin pairs, in which both twins in each pair share the trait in question. In another hypothetical twin study, the investigators identify 50 monozygotic pairs, in which only one twin of each pair exhibits the trait. The concordance among monozygotic twins for this trait is: a. 5%. b. 25%. c. 50%. d. 75%. e. 100%. ANSWER: c 54. The concordance among monozygotic twins for epilepsy is 37% and for dizygotic twins is 10%. From this information, we can conclude that: a. epilepsy has an important genetic component, and no environmental components factor into the disorder. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 b. epilepsy has both an important genetic component and an important environmental component. c. lifestyle and other environmental factors play an important role in epilepsy. d. epilepsy has only a minor genetic component, whereas lifestyle and other environmental factors play an important role. e. epilepsy has only a minor genetic component. ANSWER: b 55. It is a general assumption that monozygotic twins share exactly the same DNA because they arise from the same fertilized egg. However, SNP studies show that they are only about 99% identical. Where might the 1% difference have come from? a. mutations that arise in somatic cells b. mutations that arise in gametic cells ANSWER: a 56. Personalized medicine matches the treatment to the disease, not to the patient. a. true b. false ANSWER: b 57. If the concordance rate between identical twins is less than 100%, what does this imply? a. Concordance rates are estimates of genetic influence only. b. Concordance rates are estimates and can be only measured if one of the twins shows the trait. c. Concordance rates are difficult for researchers to calculate; a lack of 100% concordance reflects experimental error. d. Concordance rates less than 100% reflect environmental effects on the trait. ANSWER: d 58. One drawback of personalized medicine can be that medications tailored to a person's condition may cause side effects, because the response to the medication is also a complex trait. a. true b. false ANSWER: a 59. Which of the graphs would likely represent the distribution of a complex trait?

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a. graph A b. graph B c. graph C d. graph D ANSWER: a 60. Which of the graphs would likely represent the distribution of a simple, single gene trait?

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a. graph c b. graph a c. graph b d. graph d ANSWER: c 61. A key difference between single gene traits, such as those Mendel studied, and complex traits, such as human height, is that: a. complex traits are generally not significantly influenced by the environment, whereas single gene traits have a significant environmental component. b. single gene traits are generally not significantly influenced by the environment, whereas complex traits have a significant environmental component. c. complex traits show incomplete dominance, whereas single gene traits show complete dominance. d. single gene traits show both complete and incomplete dominance, whereas complex traits show only complete dominance. e. None of the other answer options is correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 ANSWER: b 62. A farmer in Kansas and a farmer in California each plant an acre of the same strain of corn. Which of the outcomes would you expect in regard to corn heights in each field? a. Both the average height of the corn and the variation in corn height within each field could differ between fields. b. There would be no variation in the height of the corn within each field, but the height of corn could differ between fields. c. The variation in corn height would be the same in both fields, because they are planting the same strain of corn. ANSWER: a 63. Inbred lines of animals, such as laboratory mice, are useful for studying complex traits because: a. individuals are genetically identical, so differences in traits must be due to environmental differences. b. although individuals differ both genetically and environmentally, the genetic differences are known, so differences in traits can be understood. c. animals are kept in identical environments, so any differences in traits are due to differences in their genes. d. both genetic and environmental differences between individuals are known, making differences in their traits easy to understand. e. None of the other answer options is correct. ANSWER: a 64. The phenotypes of complex traits vary with both genotype and environment. An important implication of this is that: a. you cannot predict the expression of a genotype without knowing its environment. b. you cannot predict how a given environment will affect the expression of a trait without knowing its genotype. c. no one genotype will produce the "best" phenotype in all environments. d. several different genotypes may produce equally fit phenotypes in the same environment. e. All of these choices are correct. ANSWER: e 65. Consider a simple Mendelian trait that is due to a single gene with alleles A and a. In one environment, allele A is dominant, whereas in a different environment allele a is dominant. This is an example of: a. epistasis. b. genotype-by-environment interaction. c. a complex trait. ANSWER: b 66. If all variation among individuals in a population is due to differing environments, heritability is _____, and the slope of the line used to measure it is _____. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 a. 50%; 1 b. 100%; 1 c. 50%; 0.5 d. 0%; 0 e. 100%; 0.5 ANSWER: d 67. A trait with high heritability will not respond to natural selection. a. true b. false ANSWER: b 68. In which of the complex traits shown is variation determined completely by heredity?

a. trait Q b. trait M c. trait K d. trait L e. trait H ANSWER: e 69. Which of the traits depicted in the graph has the greatest heritability?

a. trait H b. trait M c. trait K d. trait L ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 70. The data in the graph shown indicate the concordance between monozygotic (MZ) twins and same-sex dizygotic (DZ) twins for five complex traits. Based on these data, which of the traits shows the greatest effect of heredity relative to environment?

a. trait M b. trait H c. trait K d. trait L e. trait Q ANSWER: c 71. The data in the table shown indicate the concordance between monozygotic twins and same-sex dizygotic twins for five complex traits. Based on these data, which of the traits shows the greatest effect of environment relative to heredity?

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Chapter 17 a. trait L b. trait H c. trait K d. trait M e. trait Q ANSWER: a 72. Consider the table for concordance rates between fraternal (DZ) and identical (MZ) twins.

Which of the traits shows the most genetic influence? a. trait 1 b. trait 2 c. trait 3 ANSWER: a 73. Consider the table for concordance rates between fraternal (DZ) and identical (MZ) twins.

Suppose that in the general public the percentage of people showing a certain trait is 11%. What does this imply about the genetic influence on the trait? a. Genetic influence is high because the percentage is higher than fraternal twins. b. Genetic influence cannot be determined from these data. c. Genetic influence is high because over half the general public is affected. d. Genetic influence is high because this value is greater than that for identical twins. e. Genetic influence cannot be estimated because values for fraternal twins are too low. ANSWER: b 74. Consider the table for concordance rates between fraternal (DZ) and identical (MZ) twins. Copyright Macmillan Learning. Powered by Cognero.

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When the concordance rate of a trait is significantly higher in identical twins than it is in fraternal twins, we can conclude that: a. the trait has a strong norm of reaction. b. the trait has a strong environmental component. c. the trait has a strong genotype-by-environment interaction. d. the trait has a strong genetic component. e. the trait has both a strong genetic component and a strong norm of reaction. ANSWER: d 75. Sarah has always lived what many consider an active and healthy lifestyle. She makes a conscious effort to eat well and exercise daily, has never smoked, and drinks only on a few occasions each year. Many of Sarah's female relatives, including her mother and grandmother, have had breast cancer. Sarah has decided to have her genome sequenced. She is specifically interested in whether she shows mutations in BRCA1 and BRCA2 that are linked to breast cancer. What information can be drawn from the sequence data Sarah will receive? a. Sarah will know whether she is predisposed to breast cancer. b. Sarah will know if she will develop breast cancer. c. Sarah will know if and when she will develop breast cancer. d. Sarah will know the severity of her breast cancer based on the number of mutations—that is, the more mutations in those genes, the greater the severity of the disease. e. Sarah will know that she will not develop breast cancer if there are no mutations in those genes. ANSWER: a 76. Sarah has always lived what many consider an active and healthy lifestyle. She makes a conscious effort to eat well and exercise daily, has never smoked, and drinks only on a few occasions each year. Many of Sarah's female relatives, including her mother and grandmother, have had breast cancer. Sarah has decided to have her genome sequenced. She is specifically interested in whether she shows mutations in BRCA1 and BRCA2 that are linked to breast cancer. The results of Sarah's genetic testing show that she has mutations in BRCA1 that are linked to cancer. Many diseases, including breast cancer, are the result of the interaction of multiple genes and the environment. As a preventative measure, Sarah decides to have a radical bilateral mastectomy (surgical removal of the breasts). Other women in the same situation might make a different choice because women with BRCA1 do not always develop breast cancer. a. true b. false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 ANSWER: a 77. A woman with BRCA1 should first find out if the mutation is found in both of her breasts or just one, so that she can have surgery only where she will develop cancer. a. true b. false ANSWER: b 78. With BRCA1, the development and spread of cancer is inevitable, which makes the surgery ineffective. a. true b. false ANSWER: b 79. Studies have shown that 5%-10% of all breast cancers are related to mutations in BRCA1 and BRCA2. A logical conclusion from these data is that: a. at least some breast cancers are caused by the environment or by genes other than BRCA1 and BRCA2. b. most women with breast cancer must have inherited the gene from their mother. c. most women with breast cancer have mutations in one or both of these genes. d. not all mutations in BRCA1 and BRCA2 have been identified to account for the other 90%-95% of breast cancers. ANSWER: a 80. The figure shown includes the 23 human chromosomes. Regions are highlighted in different colors to identify regions of the chromosome where there is a QTL (quantitative trait locus) affecting levels of HDL ("good cholesterol") in green, LDL ("bad cholesterol") in red, and triglycerides (lipids) in blue in the blood.

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Levels of HDL, LDL, and triglycerides in the blood: a. are simple traits. b. are influenced by many different genes. c. tend to cluster in one chromosome. d. are influenced by the same set of genes. e. are X linked. ANSWER: b 81. The figure shown includes the 23 human chromosomes. Regions are highlighted in different colors to identify regions of the chromosome where there is a QTL (quantitative trait locus) affecting levels of HDL ("good cholesterol") in green, LDL ("bad cholesterol") in red, and triglycerides (lipids) in blue in the blood.

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In the figure shown, each of the regions highlighted by a colored bar: a. contains genes whose influence on the trait cannot be determined from the information given. b. contains two genes that influence the trait. c. contains three genes that influence the trait. d. contains four genes that influence the trait. e. contains one gene that influences the trait. ANSWER: a 82. The graph shown indicates how bristle number changes at different temperatures in the fly Drosophila pseudoobsura. The lines on the graph represent the average response for a population of genetically identical flies. Copyright Macmillan Learning. Powered by Cognero.

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Each of the lines is called a: a. scatter plot. b. normal curve. c. normal reaction. d. norm of reaction. e. histogram. ANSWER: d 83. The graph shown indicates how bristle number changes at different temperatures in the fly Drosophila pseudoobsura. The lines on the graph represent the average response for a population of genetically identical flies.

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In general, the graph shows that: a. as the temperature increases, bristle number increases. b. bristle number is affected by humidity. c. temperature affects bristle number in a manner that is hard to predict. d. as the temperature increases, bristle number decreases. e. bristle number is not affected by temperature. ANSWER: c 84. Consider the figure shown. The levels of HDL, LDL, and triglycerides are influenced by both by variation in genes highlighted by the colored bars and by diet. From this information, we can conclude which of the statements?

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a. If we know the levels of HDL, LDL, and triglycerides for a particular individual, we can determine the diet of the individual. b. If we know the diet of a particular individual, we can determine the levels of HDL, LDL, and triglycerides. c. If we know all of the alleles of the genes of a particular individual affecting these traits, we can determine the levels of HDL, LDL, and triglycerides. d. If we know the levels of HDL, LDL, and triglycerides for a particular individual, we can determine the alleles of the genes that affect these traits. e. It is not possible to predict the levels of HDL, LDL, and triglycerides of a particular genotype without knowing the diet of the individual. ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 85. Consider the figure shown. The levels of HDL, LDL, and triglycerides are influenced by both by variation in genes highlighted by the colored bars and by diet. From this information, we can conclude which of the statements?

The level of HDL in the blood: a. is a complex trait. b. All of these choices are correct. c. is influenced by variation in many different genes. d. is influenced by at least some genes that also affect the level of LDL in the blood. e. is a phenotype. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 86. The graph shows how bristle number changes at different temperatures in the fly Drosophila pseudoobsura. The lines on the graph indicate the average response for a population of genetically identical flies, where each line represents a different population.

In general, at what degrees is the number of bristles greatest? a. 26 degrees centigrade b. 18 degrees centigrade c. 22 degrees centigrade d. It depends on the population of flies. e. 14 degrees centigrade ANSWER: d 87. The graph shows how bristle number changes at different temperatures in the fly Drosophila pseudoobsura. The lines on the graph indicate the average response for a population of genetically identical flies, where each line represents a different population.

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The lines on the graph are all different because: a. each fly population has different alleles that respond to changes in temperature. b. they represent different species of insects. c. they represent different species of flies. d. each fly population was raised at a different temperature. e. the experiment was not well controlled. ANSWER: a Multiple Response 88. Most complex traits are: Select all that apply. a. affected by multiple genes. b. affected by environmental factors. c. affected by interactions between genes and environmental factors. d. homozygous for all alleles affecting the trait. e. heterozygous for all alleles affecting the trait. f. inherited in pedigrees showing simple Mendelian patterns. ANSWER: a, b, c 89. Which diseases are complex traits? Select all that apply. a. hemophilia Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 b. obesity c. red-green color blindness d. type II diabetes e. sickle-cell anemia ANSWER: b, d 90. Identical twins are: Select all that apply. a. always concordant. b. identical in phenotype. c. always of the same sex. d. identical in genotype. ANSWER: c, d 91. Some tentative patterns that are emerging from the study of complex traits are that: Select all that apply. a. many genes involved in complex traits are pleiotropic. b. many genes involved in complex traits show epistasis. c. most genes affecting complex traits have minor effects; few have major effects. d. genes involved in complex traits usually show pleiotropy or epistasis. e. genes involved in complex traits can show pleiotropy or epistasis, and most have major effects. ANSWER: a, b, c 92. Which of the traits are complex traits in humans? Select all that apply. a. depression b. height c. autism d. cystic fibrosis ANSWER: a, b, c 93. Which of the traits would you expect to conform to a normal distribution? Select all that apply. a. albinism b. type II diabetes c. Alzheimer's disease d. hypertension (high blood pressure) e. sickle-cell anemia ANSWER: b, c, d 94. Genetic mapping of inherited risk factors that code for metabolic components of complex traits, such as genes affecting sugar metabolism in the case of type II diabetes, sometimes implicates different genes affecting Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 two or more of the traits that are close together in a chromosome. This finding can be explained by: Select all that apply. a. linkage. b. dominance. c. segregation. d. epistasis. e. pleiotropy. ANSWER: a, e 95. Which of the examples would be good examples of genotype-by-environment interactions? Select all that apply. a. a strain of corn whose yield varies with the amount of nutrients in the soil b. two different strains of corn that differ in height due to genotype c. a line of dairy cows that increases milk yield in relation to feed amount d. two different mutants of mice that share similar fat metabolism abilities e. a line of chickens whose eggshells are extra thick regardless of feed type f. a strain of mutant mice that becomes obese regardless of feed amount ANSWER: a, c 96. The plots shown in the graphs depict several possible outcomes of an experiment comparing the mean phenotype of a behavioral trait in two strains of mice in three environments. Strain 1 is represented by filled circles and Strain 2 is represented by open circles. Environment 1 (E1) is deficient in stimulation for the behavior, environment 2 (E2) has an average level of stimulation for the behavior, and environment 3 (E3) is enriched in stimulation for the behavior. Which of the possible results indicates the presence of genotype-byenvironment interaction for one or both of the strains? Select all that apply.

a. result L b. result H c. result K d. result M e. None of the answer options is correct. ANSWER: b, c, d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 17 97. The plots shown depict several possible outcomes of experiments comparing the mean phenotype for a quantitative trait across a range of environments among four varieties of rice (M–L). Which pairs of test varieties indicate genotype-by-environment interaction in which the better-performing variety depends on the environment? Select all that apply.

a. M and L b. M and H c. H and L d. H and K e. M and K f. K and L ANSWER: d, e 98. You study a complex trait in the offspring of a cross between inbred lines, and you discover that the heritability is 0. How can this result be explained? Select all that apply. a. There is no genetic variation among offspring. b. The offspring are heterozygous for all genes in which the parents differ. c. The offspring are homozygous for all genes in which the parents differ. d. The environmental variation on the trait is huge. e. None of the other answer options is correct. ANSWER: a, b

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Chapter 18 Multiple Choice 1. Consider the levels of gene regulation. In what order do the levels of gene regulation take place? (1) post-translational modification (2) RNA processing (3) transcription (4) chromatin structure a. (4)-(3)-(2)-(1) b. (3)-(2)-(1)-(4) c. (1)-(3)-(2)-(4) d. (4)-(1)-(2)-(3) e. (4)-(2)-(3)-(1) ANSWER: a 2. Chromatin structure can increase or decrease transcription of a gene according to the: a. combination of amino acid modifications in the histone tails. b. coding sequences in the messenger RNAs for histone proteins. c. proportion of arginine and lysine amino acids in the histone proteins. d. combination of histone proteins found within the nucleosome. ANSWER: a 3. In general, when cytosine bases in CpG islands are methylated: a. transcription is repressed. b. transcription is active but slow. c. transcription is active and rapid. d. translation is active and rapid. e. translation is repressed. ANSWER: a 4. Chromatin remodeling refers to the process by which: a. mutations change DNA structure and therefore chromatin structure. b. DNA sequences are rearranged to create binding sites for proteins that carry out transcription. c. DNA strands are "unzipped" to allow access to the proteins that carry out transcription. d. methylation occurs in CpG islands. e. nucleosomes are repositioned to expose different stretches of DNA to the nuclear environment. ANSWER: e 5. Dosage compensation in mammals refers to a mechanism in which: a. genes in the Y chromosome must be expressed at higher levels than genes in the X chromosome because they are not homologous with genes on the X chromosome. b. X chromosome genes are regulated differently in males and females because females have two X chromosomes, whereas males have only one. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 c. genes in the X chromosome must be expressed at lower levels than genes in the Y chromosome because they are not homologous with genes on the Y chromosome. d. the level of expression for a given gene is directly related to the number of copies of the gene. e. None of the other answer options is correct. ANSWER: b 6. X-inactivation is caused by the accumulation of: a. noncoding RNA produced by the Xist gene, which coats the X chromosome and induces DNA methylation, histone modification, and other changes associated with preventing transcription. b. proteins translated from coding RNA produced by the Xist gene that bind to and coat the X chromosome undergoing inactivation, physically preventing it from being transcribed. c. noncoding RNA produced by the Xist gene, which coats the X chromosome and covalently crosslinks the DNA strands preventing them from being unwound, "unzipped," and transcribed. d. proteins produced by the Xist gene; these proteins induce methylation, histone modification, and other changes associated with preventing transcription. e. None of the answer choices accurately describes the process of X-inactivation. ANSWER: a 7. A woman is heterozygous for a recessive allele that causes X-linked hemophilia. This recessive allele has a mutation that makes a gene for a blood-clotting factor nonfunctional. Considering the implications of Xinactivation, the level of clotting factor in her blood is likely to be approximately _____ that of a homozygous non-mutant woman. a. 1/4 b. 3/4 c. 1.0 d. 1/2 e. None of the other answer options is correct. ANSWER: d 8. Many genes have multiple enhancer sequences. The multiple enhancer sequences allow multiple: a. options for RNA editing of the RNA transcript. b. transcription factors to control gene expression. c. options for alternative splicing of the RNA transcript. d. proteins to be made from the same protein-coding gene. e. None of the other answer options is correct. ANSWER: b 9. Enhancer sequences are bound by: a. transcription factors. b. RNA editing complexes. c. histone-modifying complexes. d. cytosine methylation enzymes. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 e. RNA splicing complexes. ANSWER: a 10. Combinatorial control refers to a regulatory mechanism in which: a. transcription is initiated by a combination of sites in the promoter. b. transcription is terminated by a combination of sites in the terminator. c. each alternatively spliced transcript has a different combination of exons. d. transcription requires a specific combination of transcription factors. e. None of the other answer options is correct. ANSWER: d 11. RNA splicing provides an opportunity for regulating gene expression because: a. methylation of the poly (A) tail controls how rapidly the primary transcript can be broken up and spliced back together again. b. the same introns may be spliced together in different sequences to produce different proteins from the same primary transcript. c. the same exons may be spliced together in different sequences to produce different proteins from the same primary transcript. d. methylation of spliceosomes controls how rapidly primary transcripts are processed and sent to the cytoplasm. e. alternative RNA processing can include different sets of exons in the final mRNA that can encode different proteins. ANSWER: e 12. The diagram shown here is part of an RNA transcript containing four open reading frames (M, H, K, and L) and three introns (1, 2, 3).

Splicing out an intron involves a spliceosome facilitating the cleavage of the primary RNA transcript at nucleotides on either side of the intron. How many processed transcripts are possible if all three 5' nucleotides (immediately before each intron) and all three 3' nucleotides (immediately after each intron) are cleaved? a. 3 b. 5 c. 1 d. 4 e. 2 ANSWER: c 13. Consider the table. One type of RNA editing converts Cs to Us, thereby affecting what amino acid is Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 translated by some codons. How many codons in the genetic code would not be affected by such a change?

a. 28 b. 25 c. 34 d. 18 e. 42 ANSWER: d 14. The enzymatic processing of a polypeptide chain is an example of: a. signal transduction. b. post-transcriptional modification. c. dosage compensation. d. epigenetic modification. ANSWER: b 15. Small regulatory RNAs work in conjunction with: a. RISC proteins. b. nucleosomes. c. RNA polymerase. d. ribosomal RNA. ANSWER: a 16. In eukaryotes, initiation of translation of a messenger RNA is determined in part by: a. the secondary structure of the 5′ untranslated region. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 b. the length of the 5′ untranslated region. c. whether or not the 5′ cap on the mRNA is present. d. the base sequence of the 3′ untranslated region. ANSWER: c 17. Positive and negative transcriptional regulation differ in that: a. in positive regulation, the binding of a regulatory protein to the DNA is necessary for transcription to occur; in negative regulation, such binding prevents transcription. b. in positive regulation, the binding of a regulatory protein to the DNA is necessary for transcription to occur; in negative regulation, no such protein is necessary. c. positive regulation requires that a promoter be present; a promoter is not necessary in negative regulation. d. in positive regulation, the absence of a regulatory protein promotes transcription; in negative regulation, the absence of a regulatory protein promotes transcription. e. None of the other answer options is correct. ANSWER: a 18. The CRP-cAMP complex binds the lactose operon when: a. glucose levels are high and cAMP levels are low. b. glucose levels are low and cAMP levels are high. c. glucose levels and cAMP levels are high. d. glucose levels and cAMP levels are low. e. None of the other answer options is correct. ANSWER: b 19. The lacZ and lacY genes are transcribed when: a. lactose is present and glucose levels are low. b. lactose is absent and glucose levels are high. c. lactose is present, regardless of the level of glucose in the cell. d. glucose levels are high and lactose levels are low. e. glucose levels are low, regardless of the level of lactose in the cell. ANSWER: a 20. It is necessary for the gene that codes for the repressor of lac operon to be near the structural genes. a. true b. false ANSWER: b 21. An activator protein combines with a small molecule and undergoes a change in shape that alters its binding affinity to DNA. This change in shape is an example of a/an: a. dosage compensation. b. induced silencing. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 c. allosteric effect. d. activator effect. e. position effect. ANSWER: c 22. Which of the genotypes produces functional lactose permease constitutively? a. I- P+ O+ Z- Y+ b. I+ P+ O+ Z+ Yc. I+ P+ O+ Z- Y+ d. I- P+ O+ Z+ YANSWER: a 23. In the lactose operon of E. coli, the lacP- mutation: a. produces the structural genes all the time. b. produces the structural genes only in the presence of the inducer. c. produces the structural genes only in the absence of the inducer. d. never produces the structural genes. ANSWER: d 24. At the molecular level, the "choice" between lytic and lysogenic pathways is determined by the: a. positive and negative regulatory effects of a small number of different bacteriophage proteins produced soon after infection. b. negative regulatory effects of a large number of different bacteriophage proteins produced soon after infection. c. positive regulatory effects of a small number of different bacteriophage proteins produced soon after infection. d. negative regulatory effects of a small number of different bacteriophage proteins produced soon after infection. e. positive regulatory effects of a large number of different bacteriophage proteins produced soon after infection. ANSWER: a 25. In bacteriophage lambda, the choice between the lytic and lysogenic pathways is often thought of as a sort of race between the production of cro protein and production of cI protein. The lysogenic pathway would likely predominate in a mutant in which cro was overexpressed relative to a wild-type virus. a. true b. false ANSWER: b 26. In bacteriophage lambda, the choice between the lytic and lysogenic pathways is often thought of as a sort of race between the production of cro protein and production of cI protein. The lysogenic pathway would likely Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 predominate in a mutant in which cro was under-expressed relative to a wild-type virus: a. true b. false ANSWER: a 27. The cellular processes that control the rate and manner of gene expression are called: a. gene recombination. b. gene conversion. c. gene regulation. d. gene incompatibility. ANSWER: c 28. A CpG island is: a. a stretch of nucleotides with many adjacent CG bases in a small region near or in the promoter site of a gene. b. a stretch of nucleotides with many adjacent CG bases found anywhere along a DNA strand. c. a stretch of nucleotides with many adjacent CG bases found near exon/intron border. d. a stretch of nucleotides with many adjacent CG bases found near centromere. ANSWER: a 29. Certain genes have apparently unusual inheritance patterns. For instance, in the mouse autosomal gene H19, only the copy inherited from the mouse's mother is expressed. Conversely, in the mouse autosomal gene Igf2, only the copy inherited from the mouse's father is expressed. The consequence of this is that imprinted genes are expressed as if they were present in only one copy, even though there are two copies of each of these genes in each cell. What process or processes are responsible for this phenomenon? a. incomplete dominance b. epigenetic inheritance c. combinatorial gene regulation d. X-chromosome inactivation e. None of the other answer options is correct. ANSWER: b 30. In prokaryotes, gene expression is regulated at the level of post-translational modification of RNA. a. true b. false ANSWER: b 31. The repressor of lactose operon binds to the operator in the presence of lactose. a. true b. false ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 32. The binding of cAMP-CRP to DNA affects the binding of a repressor. a. true b. false ANSWER: b 33. The inversion of a promoter sequence will: a. block transcription. b. not affect gene expression. c. increase the level of transcription. d. have unpredictable effect on gene expression. ANSWER: a 34. Humans with hypohidrotic ectodermal dysplasia have defects in the formation and function of their sweat glands. Males with the condition have reduced ability to sweat over their entire body. Females with the condition lose the ability to sweat only in patches of skin. Where must the gene that encodes this trait be located? a. in the Y chromosome b. in the X chromosome c. in an autosome ANSWER: b 35. A single gene can produce different proteins. a. true b. false ANSWER: a 36. Alternative splicing may be considered a mechanism of gene regulation because it: a. is mutagenic. b. enhances RNA editing. c. results in DNA rearrangements. d. results in different protein products. ANSWER: d 37. Gene regulation can occur through: a. DNA modification. b. histone modification. c. RNA modification. d. All of these choices are correct. ANSWER: d 38. The human body contains approximately 200 major cell types. They look and function differently from one another because each: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 a. expresses a different set of genes. b. has a slightly different genome. c. expresses the same set of genes, but in different orders at different times. d. has a slightly different genome and each expresses a different set of genes. ANSWER: a 39. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

Suppose you digest genomic DNA with EcoRI and isolate the 10-kb DNA fragment corresponding to the diagram. You digest this fragment with HpaII and carry out electrophoresis. What size DNA fragments would you expect if only site 2 is methylated? a. 5 kb b. 1 kb, 4 kb, and 5 kb c. 2 kb and 8 kb d. 2 kb and 4 kb e. 4 kb and 6 kb ANSWER: b 40. Chromatin may be remodeled by chemical modification of the histone proteins. a. true b. false ANSWER: a 41. Methylation, acetylation, and other histone modifications are important because they are associated with: a. cancer genes and their regulation. b. chromatin remodeling and thereby affect gene transcription. c. the environmental and developmental cues that determine whether viral genes and genes in transposable elements are turned off or on. d. All these choices are correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 ANSWER: b 42. Gene regulation by means of epigenetic mechanisms is useful to the organism because it: a. does not require new mutations. b. can quickly be reversed. c. can respond rapidly to environmental change. d. can differ between the sexes. e. All of these choices are correct. ANSWER: e 43. Adding or removing the 5' cap on messenger RNA helps regulate gene activity at the level of translation because the 5′ cap is necessary for: a. translational initiation. b. 3′ polyadenylation. c. RNA splicing. d. translational termination. ANSWER: a 44. Half or more of human proteins may have their synthesis regulated in part by small regulatory RNA. a. true b. false ANSWER: a 45. Gene regulation is influenced by both genetic and environmental factors. a. true b. false ANSWER: a 46. In the lactose operon of E. coli, which of the genotypes transcribes the lacZ and lacY genes constitutively? a. I+ P+ O+ Z- Y+ b. I+ P- O+ Z+ Y+ c. I- P+ O+ Z+ Y+ d. I+ P+ O+ Z+ YANSWER: c 47. In the lactose operon of E. coli, which of the genotypes produces functional β-galactosidase constitutively? a. I- P+ O+ Z+ Yb. I+ P+ O+ Z- Y+ c. I+ P+ O+ Z+ Yd. I- P+ O+ Z- Y+ Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 e. I+ P+ O- Z+ YANSWER: a 48. In E. coli, the production of lac repressor is constitutive. a. true b. false ANSWER: a 49. In the lactose operon of E. coli, the lacOc mutation is transcribed: a. only in the absence of the repressor. b. only in the presence of the inducer. c. only in the absence of the inducer. d. all the time. ANSWER: d 50. In the lactose operon of E. coli, the product of the lacI gene is a(n) _____ of transcription. a. enhancer b. positive regulator c. negative regulator d. silencer ANSWER: c 51. In E. coli, if the cell contains one mutant and one normal copy lac repressor gene, the expression of lac operon is a. inducible. b. constitutive. ANSWER: a 52. A typical eukaryotic gene may be regulated by multiple enhancers and silencers of different kinds, each with one or more regulatory transcription factors that can interact with it. a. true b. false ANSWER: a 53. The methylation state of an individual CpG island: a. is fixed; such genes are permanently turned off. b. is fixed, but this has no effect on whether genes are expressed. c. is random; sometimes the cytosines are methylated and sometimes they're not, but the state is independent of the environment or cell type. d. can change over time in response to environmental cues, allowing genes to be turned on or off as needed. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 e. can change over time in response to environmental cues, but this has no effect on gene expression. ANSWER: d 54. A histone code is the: a. pattern of chemical modification of the DNA wrapped around an individual histone. b. pattern of chemical modification of the histone tails. c. nucleotide sequence of an individual histone protein's gene. d. number of amino acids in an individual histone that are methylated. e. None of the other answer options is correct. ANSWER: b 55. Modifications of histone tails can: a. affect chromatin structure. b. activate transcription of some genes. c. repress transcription of some genes. d. affect expression of some genes in response to the environment. e. All of these choices are correct. ANSWER: e 56. What are the consequences of X-chromosome inactivation in mammals? a. a mosaic expression of heterozygous autosomal genes in different female cells. b. a mosaic expression of heterozygous X-linked genes in different male cells. c. a mosaic expression of heterozygous X-linked genes in different female cells. d. a mosaic expression of heterozygous autosomal genes in different male cells. e. None of the other answer options is correct. ANSWER: c 57. In humans and other mammals, X-inactivation takes place: a. late in development, with each new cell inheriting the same inactivated X chromosome as its parent cell. b. early in development, with each new cell inheriting the same inactivated X chromosome as its parent cell. c. throughout the lifetime of an organism with each new cell division, both X chromosomes are activated, and one is then inactivated in each new cell. d. early in development, with all maternal X chromosomes being inactivated and cells maintaining active copies of only the paternal X chromosome. e. early in development, with all paternal X chromosomes being inactivated and cells maintaining active copies of only the maternal X chromosome. ANSWER: b 58. The diagram shown here is part of an RNA transcript containing four open reading frames (M, H, K, and L) and three introns (1, 2, 3). The sites labeled d1-d3 are nucleotides immediately before each intron on the 5' side Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 and those labeled a1-a3 are nucleotides immediately after each intron on the 3' side. Splicing out an intron involves a spliceosome facilitating the cleavage of the DNA at the d and a sites at either end of the intron.

How many processed transcripts are possible in which two d sites and two a sites are cleaved? a. four b. one c. two d. three e. five ANSWER: d 59. A mutation is found that prevents transcription of the HOX3A gene, which is known to be in chromosome 12. Genetic mapping of the mutation suggests that it is located in chromosome 11. Which of the answer choices could make these two observations possible? a. The mutation prevents ribosome binding by altering sequences of the HOX3A mRNA. b. The mutation changes the DNA sequence of an enhancer of the HOX3A gene. c. The mutation affects a transcription factor that binds to HOX3A gene sequences. d. The mutation increases transcription of an miRNA that has sequence homology to the HOX3A gene. ANSWER: c 60. Modification of histone tails is one mechanism of: a. post-translational modification. b. RNA processing. c. RNA editing. d. X-inactivation. e. None of the other answer options is correct. ANSWER: a 61. If you were to try to develop a strain of bacteriophage lambda that could only execute the lysogenic pathway, where would you expect to find the phage DNA in the bacterial cell? a. within independent circular molecules called plasmids b. integrated into the bacterial chromosome ANSWER: b 62. The graph here shows the results of Jacob and Monod's first experiments with β-galactosidase and lactose in E. coli. If β-galactosidase were unstable in the absence of lactose (one of the hypotheses they tested), you would expect to see: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18

a. the results shown in the graph. b. an increase in β-galactosidase after lactose is removed. c. a decrease in β-galactosidase after lactose is removed. d. an initially high level of β-galactosidase at the beginning of the experiment, with a steady decline over time. e. None of the other answer options is correct. ANSWER: c 63. In order to analyze how different components of the lac operon work, scientists created special strains of E. coli, called partial diploids. A partial diploid has one full copy of the lac operon in the bacterial chromosome plus another copy of the lac operon in a plasmid. Hence, for the lac operon (and only the lac operon) the bacterial cell is a diploid. Below is one possible genotype of a partial diploid. The genotype written to the left of the slash (/) is that of the lac operon in the bacterial chromosome, and the genotype written to the right of the slash is that of the lac operon in the plasmid. I+ P- O+ Z+/ I- P+ O+ ZFor the partial diploid genotype shown here, determine whether functional β-galactosidase is synthesized in the absence and in the presence of lactose. a. β-galactosidase is not produced whether lactose is present or absent. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 b. β-galactosidase is produced whether lactose is present or absent. c. β-galactosidase is produced when lactose is present but is not produced when lactose is absent. d. β-galactosidase is not produced when lactose is present but is produced when lactose is absent. ANSWER: a 64. In order to analyze how different components of the lac operon work, scientists created special strains of E. coli, called partial diploids. A partial diploid has one full copy of the lac operon in the bacterial chromosome plus another copy of the lac operon in a plasmid. Hence, for the lac operon (and only the lac operon) the bacterial cell is a diploid. Below is one possible genotype of a partial diploid. The genotype written to the left of the slash (/) is that of the lac operon in the bacterial chromosome, and the genotype written to the right of the slash is that of the lac operon in the plasmid. I+ P+ Oc Z+ / I- P- O+ ZFor the partial diploid genotype shown here, determine whether functional β-galactosidase is synthesized in the absence and in the presence of lactose. a. β-galactosidase is not produced when lactose is present but is produced when lactose is absent. b. β-galactosidase is not produced whether lactose is present or absent. c. β-galactosidase is produced when lactose is present but is not produced when lactose is absent. d. β-galactosidase is produced whether lactose is present or absent. ANSWER: d 65. In order to analyze how different components of the lac operon work, scientists created special strains of E. coli, called partial diploids. A partial diploid has one full copy of the lac operon in the bacterial chromosome plus another copy of the lac operon in a plasmid. Hence, for the lac operon (and only the lac operon) the bacterial cell is a diploid. Below is one possible genotype of a partial diploid. The genotype written to the left of the slash (/) is that of the lac operon in the bacterial chromosome, and the genotype written to the right of the slash is that of the lac operon in the plasmid. I+ P+ O+ Z- / I- P+ O+ Z+ For the partial diploid genotype shown here, determine whether functional β-galactosidase is synthesized in the absence and in the presence of lactose (and the accompanying allolactose that is the actual inducer). a. β-galactosidase is not produced whether lactose is present or absent. b. β-galactosidase is produced when lactose is present but is not produced when lactose is absent. c. β-galactosidase produced whether lactose is present or absent. d. β-galactosidase is not produced when lactose is present but is produced when lactose is absent. ANSWER: b 66. Small regulatory RNAs can modify gene expression by: a. All these choices are correct b. inhibiting translation. c. promoting RNA degradation. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 d. chromatin remodeling. ANSWER: a 67. A frameshift mutation that occurs at the end of one exon will affect the reading frame of the next exon. a. true b. false ANSWER: a 68. In lac operon regulation, how many proteins are bound to DNA when E. coli is grown in medium containing both glucose and lactose? a. CRP is bound to the DNA, but the lac repressor is not. b. Both CRP and the lac repressor are bound to the DNA. c. Neither CAP nor the lac repressor is bound to the DNA. d. lac repressor is bound to the DNA, but CAP is not. ANSWER: c 69. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

What size of DNA fragments would you expect if only site 3 was methylated? a. 2 kb and 4 kb b. 5 kb c. 2 kb and 8 kb d. 4 kb and 6 kb e. 1 kb, 2 kb, 3 kb, and 4 kb ANSWER: a 70. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 cleaves the DNA regardless of any chemical modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

What size of DNA fragments would you expect if sites 2 and 3 were methylated? a. 4 kb and 6 kb b. 5 kb c. 2 kb and 8 kb d. 2 kb and 4 kb e. 1 kb, 4 kb, and 5 kb ANSWER: a 71. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

Suppose you digest genomic DNA with EcoRI and isolate the 10 kb DNA fragment corresponding to the diagram. You digest this fragment with HpaII and carry out electrophoresis. Among the possible patterns of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 DNA bands you might observe are those shown in the diagram below.

Which combination of cleavage sites would you infer were methylated if you observed the pattern in lane H? a. sites 2 and 4 b. site 2 only c. site 3 only d. site 4 only e. sites 2 and 3 ANSWER: a 72. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

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Chapter 18

Which combination of cleavage sites would you infer were methylated if you observed the pattern in lane L? a. site 2 only b. site 3 only c. site 4 only Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 d. sites 2 and 3 e. sites 2 and 4 ANSWER: b 73. A researcher attempts to repeat the experiment in which genetic engineering was used to transform cells in culture with the Xist gene. When the Xist gene was targeted to chromosome 21 in cells from a person with Down syndrome (trisomy 21), transcription of the copy of chromosome 21 containing Xist was shut down, much like the inactive X chromosome in cells from a normal female. After carrying out the experiment, the researcher measures the level of transcription averaged over a sample of genes in chromosome 21 and, as a control, measures the level of transcription of a sample of genes in chromosome 6. Shown here are several possible outcomes of such an experiment.

Which bar graph suggests that the experiment succeeded (that is, that the Xist gene became inserted into one of the copies of chromosome 21)? a. Q b. H c. K d. L e. M ANSWER: e 74. A researcher attempts to repeat the experiment in which genetic engineering was used to transform cells in culture with the Xist gene. When the Xist gene was targeted to chromosome 21 in cells from a person with Down syndrome (trisomy 21), transcription of the copy of chromosome 21 containing Xist was shut down, much like the inactive X chromosome in cells from a normal female. After carrying out the experiment, the researcher measures the level of transcription averaged over a sample of genes in chromosome 21 and, as a control, measures the level of transcription of a sample of genes in chromosome 6. Shown here are several possible outcomes of such an experiment.

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Chapter 18

In one replicate of the experiment, the Xist gene was incorrectly inserted into one copy of chromosome 6 rather than into chromosome 21. Which bar graph would be expected in this case? a. K b. M c. H d. L e. Q ANSWER: c 75. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical DNA modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

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Chapter 18 c. 5 kb d. 2 kb and 8 kb e. 2 kb and 4 kb ANSWER: a 76. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical DNA modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

Suppose you digest genomic DNA with EcoRI and isolate the 10 kb DNA fragment corresponding to the diagram. You digest this fragment with HpaII and carry out electrophoresis. What size of DNA fragments would you expect if sites 2 and 4 were methylated? a. 2 kb, 3 kb, and 5 kb b. 2 kb and 8 kb c. 2 kb and 4 kb d. 5 kb ANSWER: d 77. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical DNA modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

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Chapter 18

Suppose you digest genomic DNA with EcoRI and isolate the 10 kb DNA fragment corresponding to the diagram. You digest this fragment with HpaII and carry out electrophoresis. What size of DNA fragments would you expect if sites 3 and 4 were methylated? a. 5 kb b. 4 kb and 6 kb c. 2 kb and 8 kb d. 2 kb, 3 kb, and 5 kb e. 1 kb, 4 kb, and 5 kb ANSWER: c 78. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical DNA modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

Suppose you digest genomic DNA with EcoRI and isolate the 10 kb DNA fragment corresponding to the diagram. You digest this fragment with HpaII and carry out electrophoresis. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 What size of DNA fragments would you expect if sites 2 and 3 and 4 were methylated? a. 10 kb b. 5 kb c. 2 kb and 8 kb d. 2 kb and 4 kb e. 1 kb, 4 kb, and 5 kb ANSWER: a 79. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

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Which combination of cleavage sites would you infer were methylated if you observed the pattern in lane M? a. site 2 only b. site 3 only c. site 4 only d. sites 2 and 4 e. sites 3 and 4 ANSWER: c 80. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

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Which combination of cleavage sites would you infer were methylated if you observed the pattern in lane K? a. site 2 only b. site 3 only c. site 4 only Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 d. sites 2 and 4 e. sites 3 and 4 ANSWER: e 81. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

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Which combination of cleavage sites would you infer were methylated if you observed the pattern in lane Q? a. site 2 only b. site 3 only c. sites 2 and 3 d. sites 2 and 4 e. site 4 ANSWER: a 82. The horizontal lines in the accompanying diagram represent complementary strands of a DNA double helix that includes five cleavage sites for restriction enzymes. Sites 1 and 5 are cleavage sites for EcoRI, which cleaves the DNA regardless of any chemical modifications to the DNA. Sites 2–4 are cleavage sites for HpaII, however cleavage occurs only if the CG (CpG) in the center of the site is not methylated.

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Chapter 18

Which lanes in the gel include bands composed of a mixture of two or more different DNA fragments that are the same size? a. K and L b. M and H Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 c. H and K d. H and L e. K and Q ANSWER: d 83. Suresh and Gail are students in a lab working on mice with defects in the function of their HOX3A gene. In the special mouse line they are working on, no sequence changes are present in the genomic DNA of HOX3A. However, the gene is not expressed properly and a mutant phenotype results. In another mouse line, an mRNA is produced in normal amounts, but no HOX3A protein can be detected. Suresh thinks the absence of a protein results from a change in RNA processing that prevents the inclusion of an exon in the final mRNA. Gail thinks the absence of a protein results from some second gene that produces a protein that binds to the 3'UTR of the HOX3A mRNA and prevents translation. Which of the two could be correct? a. Only Suresh could be correct. b. Only Gail could be correct. c. Both Suresh and Gail could be correct. d. Both Suresh and Gail are incorrect ANSWER: c 84. Suresh and Gail are students in a lab working on mice with defects in the function of their HOX3A gene. In the special mouse line they are working on, no sequence changes are present in the genomic DNA of HOX3A. However, the gene is not expressed properly and a mutant phenotype results. In another mouse line, an mRNA is produced in normal amounts, but no HOX3A protein can be detected. Which kinds of data could distinguish between the hypotheses posed by Suresh and Gail? a. analysis of the length of the mRNA produced by the gene b. sequencing of the genomic DNA of the gene c. analysis of ribosome binding to the mRNA d. analysis of RNA polymerase binding to the gene ANSWER: a 85. Suresh and Gail are students in a lab working on mice with defects in the function of their HOX3A gene. In the special mouse line they are working on, no sequence changes are present in the genomic DNA of HOX3A. However, the gene is not expressed properly and a mutant phenotype results. In another mouse line, an mRNA is produced in normal amounts, but no HOX3A protein can be detected. Results from analysis of the defective HOX3A gene show that the mRNA is shorter than normal. Which of the hypotheses can be ruled out? a. Gail's hypothesis b. Suresh's hypothesis c. Neither hypothesis can be ruled out. d. Both hypothesis are wrong. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 ANSWER: a 86. Suresh and Gail are students in a lab working on mice with defects in the function of their HOX3A gene. In the special mouse line they are working on, no sequence changes are present in the genomic DNA of HOX3A. However, the gene is not expressed properly and a mutant phenotype results. In another mouse line, an mRNA is produced in normal amounts, but no HOX3A protein can be detected. A second set of experiments are performed that show that ribosome binding to the mRNA is normal, but translation produces a very small protein that breaks down very quickly. Is this result consistent with Suresh's hypothesis? a. yes b. no ANSWER: a 87. A strain of E. coli is genetically engineered in which the lacZ and lacY genes are removed and replaced with a gene encoding a fluorescent protein. A copy of the operon in which the protein product fluoresces green is inserted into the chromosome, and a copy in which the protein product fluoresces red is inserted into a plasmid. Both copies carry a wild-type lacO gene, and the cells of the strain have a single copy of the lacI repressor gene. If both fluorescent proteins are expressed, the cells fluoresce yellow (because a combination of red and green fluorescence appears as yellow), and if neither of the fluorescent proteins is expressed, the cells show no fluorescence. If the lacI gene in the genetically engineered fluorescent strain were nonmutant but the lacO sequence on the chromosome mutated to lacOc, how would the cells fluoresce in the presence of inducer? In the absence of inducer? a. yellow; red b. yellow; green c. yellow; no fluorescence d. yellow; yellow e. None of the other answer options is correct. ANSWER: b 88. A strain of E. coli is genetically engineered in which the lacZ and lacY genes are removed and replaced with a gene encoding a fluorescent protein. A copy of the operon in which the protein product fluoresces green is inserted into the chromosome, and a copy in which the protein product fluoresces red is inserted into a plasmid. Both copies carry a wild-type lacO gene, and the cells of the strain have a single copy of the lacI repressor gene. If both fluorescent proteins are expressed, the cells fluoresce yellow (because a combination of red and green fluorescence appears as yellow), and if neither of the fluorescent proteins is expressed, the cells show no fluorescence. If the lacI gene in the genetically engineered fluorescent strain were nonmutant but the lacO sequence on the plasmid mutated to lacOc, how would the cells fluoresce in the presence of inducer? In the absence of inducer? a. yellow; no fluorescence b. yellow; green Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 c. yellow; red d. yellow; yellow e. None of the other answer options is correct. ANSWER: c 89. The lactose operon, as well as other operons with genes that encode enzymes for the utilization of different carbohydrates, is regulated by the concentration of CRP-cAMP in the cell. One of the consequences is that the cell can have a hierarchy of preferences for the utilization of various carbohydrates. Glucose is at the top of the hierarchy, and when glucose is present in sufficient concentration, the level of CRP-cAMP in the cell is low, and none of the other operons are fully induced even if their associated carbohydrate is present. As the glucose is depleted, the level of CRP-cAMP increases, and each operon in turn becomes fully inducible according to its ranking in the hierarchy. Suppose the order of preference for sugars were, (from most to least preferred) glucose, maltose, lactose, melibiose, trehalose, and raffinose. Which operon would require the lowest level of CRP-cAMP to be fully induced? a. lactose b. maltose c. melibiose d. trahalose e. raffinose ANSWER: b 90. The lactose operon, as well as other operons with genes that encode enzymes for the utilization of different carbohydrates, is regulated by the concentration of CRP-cAMP in the cell. One of the consequences is that the cell can have a hierarchy of preferences for the utilization of various carbohydrates. Glucose is at the top of the hierarchy, and when glucose is present in sufficient concentration, the level of CRP-cAMP in the cell is low, and none of the other operons are fully induced even if their associated carbohydrate is present. As the glucose is depleted, the level of CRP-cAMP increases, and each operon in turn becomes fully inducible according to its ranking in the hierarchy. Consider a strain of E. coli in which, after the glucose in the medium is exhausted, the order of preference for the following sugars, (from most to least preferred), was maltose, lactose, melibiose, trehalose, and raffinose. Which operon would require the highest concentration of CRP-cAMP in order to be fully induced? a. raffinose b. maltose c. lactose d. melibiose e. trahalose ANSWER: a 91. A strain of E. coli is genetically engineered in which the lacZ and lacI genes are removed and replaced with a gene encoding a fluorescent protein. A copy of the operon in which the protein product fluoresces green is inserted into the chromosome, and a copy in which the protein product fluoresces red is inserted into a plasmid. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 Both copies carry a wild-type lacO gene, and the cells of the strain have a single copy of the lacI repressor gene. If both fluorescent proteins are expressed, the cells fluoresce yellow (because a combination of red and green fluorescence appears as yellow), and if neither of the fluorescent proteins is expressed, the cells show no fluorescence. The cells would fluoresce ___in the presence of an inducer and _____in the absence of an inducer. a. yellow; green b. yellow; red c. yellow; not fluorescence d. yellow; yellow ANSWER: c 92. A strain of E. coli is genetically engineered in which the lacZ and lacI genes are removed and replaced with a gene encoding a fluorescent protein. A copy of the operon in which the protein product fluoresces green is inserted into the chromosome, and a copy in which the protein product fluoresces red is inserted into a plasmid. Both copies carry a wild-type lacO gene, and the cells of the strain have a single copy of the lacI repressor gene. If both fluorescent proteins are expressed, the cells fluoresce yellow (because a combination of red and green fluorescence appears as yellow), and if neither of the fluorescent proteins is expressed, the cells show no fluorescence. If the lacI gene in the genetically engineered fluorescent strain is mutated to an inactive form (lacI-), the cells would fluoresce _____in the presence of an inducer and _____in the absence of an inducer. a. yellow; no fluorescence b. yellow; red c. yellow; green d. yellow; yellow e. no fluorescence, no fluorescence. ANSWER: d 93. A researcher attempts to repeat the experiment in which genetic engineering was used to transform cells in culture with the Xist gene. When the Xist gene was targeted to chromosome 21 in cells from a person with Down syndrome (trisomy 21), transcription of the copy of chromosome 21 containing Xist was shut down, much like the inactive X chromosome in cells from a normal female. After carrying out the experiment, the researcher measures the level of transcription averaged over a sample of genes in chromosome 21 and, as a control, measures the level of transcription of a sample of genes in chromosome 6. Several possible outcomes of such an experiment are shown here.

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Which bar graph suggests that the experiment failed (that is, that the Xist gene failed to be inserted into chromosome 21)? a. Q b. M c. H d. K e. L ANSWER: a 94. A researcher attempts to repeat the experiment in which genetic engineering was used to transform cells in culture with the Xist gene. When the Xist gene was targeted to chromosome 21 in cells from a person with Down syndrome (trisomy 21), transcription of the copy of chromosome 21 containing Xist was shut down, much like the inactive X chromosome in cells from a normal female. After carrying out the experiment, the researcher measures the level of transcription averaged over a sample of genes in chromosome 21 and, as a control, measures the level of transcription of a sample of genes in chromosome 6. Several possible outcomes of such an experiment are shown here.

In one replicate of the experiment, the Xist gene was incorrectly inserted into one copy of chromosome 6 as well as into one copy of chromosome 21. Which bar graph would be expected in this case? a. H b. K c. L d. M Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 e. Q ANSWER: c Multiple Response 95. The first level of gene regulation occurs along the chromosome, through chemical modifications of the DNA or histones. How do these chemical modifications cause changes in gene expression? Select all that apply. a. The chemical modification to histones could alter chromatin structure and affect the ability of RNA polymerase to bind to DNA. b. The chemical modifications to DNA could activate enzymes that add chemical modifications to histones. c. The chemical modifications to DNA could prevent the removal of introns from the mRNA. d. The chemical modifications to DNA and histones could affect the binding of ribosomes to mRNA to initiate translation. ANSWER: a, b 96. Which of the statements about the methylation state of CpG islands are correct? Select all that apply. a. Cells can heavily methylate CpG islands of viral DNA sequences in order to restrict their expression. b. Cells can heavily methylate CpG islands of genes in transposable elements in order to restrict their expression. c. The methylation state of CpG islands provides a way to turn genes on or off. d. The methylation state of CpG islands can change in response to the environment. e. The methylation state of CpG islands is static and will not change over time. ANSWER: a, b, c, d 97. Epigenetic mechanisms can include: Select all that apply. a. histone modification. b. cytosine methylation. c. X-inactivation. d. chromatin remodeling. e. mutations ANSWER: a, b, c, d 98. X-inactivation is an example of: Select all that apply. a. miRNA-induced silencing. b. epigenetic gene regulation. c. dosage compensation. d. positive regulation by CRP-cAMP. e. post-transcriptional modification. ANSWER: b, c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 99. Lifestyle choices that can affect levels of gene expression include: Select all that apply. a. diet. b. exercise. c. drug abuse. d. meditation. e. inherited mutation. ANSWER: a, b, c, d 100. Consider the table. In the type of RNA editing in which C is converted to U, which codons could be converted to a stop codon? Select all that apply.

a. CAA b. CAG c. UGA d. CGA e. ACG ANSWER: a, b, d 101. Small regulatory RNA can regulate gene expression by: Select all that apply. a. altering chromatin conformation. b. inhibiting translation. c. inhibiting RNA splicing. d. degrading RNA transcripts. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 e. inhibiting rRNA synthesis. ANSWER: a, b, c 102. Suppose that the RNA sequence shown here is a target for a microRNA. 5'-UUACGAUGC-3' Which of these double-stranded precursor RNA molecules could be processed for incorporation into RISC to target the sequence? Select all that apply. a. 5'-UUACGAUGC-3' 3'-AAUGCUACG-5' b. 5'-GCAUCGUAA-3' 3'-CGUAGCAUU-5' c. 5'-AAUGCUACG-3' 3'-UUACGAUGC-5' d. 5'-CGUAGCAUU-3' 3'-GCAUCGUAA-5' ANSWER: a, b 103. The 5′ UTR and 3′ UTR of mRNA may bind with proteins that: Select all that apply. a. initiate the process of dosage compensation. b. determine the lytic vs. lysogenic pathway. c. help initiate translation. d. help control mRNA degradation. e. transport the mRNA to a particular location in the cell. ANSWER: c, d, e 104. A mutant strain of E. coli is found that produces both β-galactosidase and permease constitutively. Which of the genotypes could cause this? Select all that apply. a. lacP – b. lacI – c. lacO c d. mutation in CRP-cAMP binding site e. lacZ c and lacY c ANSWER: b, c 105. What are the functions of the cI protein? Select all that apply. a. It's activity to the lytic pathway. b. It stimulates transcription of its own coding sequence. c. It stimulates expression of cro and cII. d. It induces proteases, which leads to the degradation of cII. e. It prevents expression of cro and cII. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 ANSWER: b, e 106. In a messenger RNA, the 3′ poly a tail: Select all that apply. a. is necessary for RNA splicing. b. assists with the transport of the mRNA out of the nucleus. c. alters messenger RNA stability. d. changes the rate of transcription. e. enhances translation initiation. ANSWER: b, c, e 107. Epigenetic modifications can: Select all that apply. a. be inherited. b. be reversed. c. alter gene expression. d. change the sequence of the DNA. ANSWER: a, b, c 108. Methylation is a mechanism often used by cells to prevent the expression of: Select all that apply. a. genes in transposable elements. b. harmful (mutant) genes such as those responsible for cystic fibrosis. c. genes from viruses that have been integrated into the genome. ANSWER: a, c 109. Epigenetic mechanisms of gene regulation include: Select all that apply. a. the way in which chromosomes are organized into genomes b. chemical modifications of DNA. c. chemical modifications of histones. d. changes to DNA sequences. e. the way in which DNA is packaged within chromosomes. ANSWER: b, c, e 110. Although lifestyle choices cannot generally change our genomes, they can: Select all that apply. a. change the genetic code. b. change the metabolic pathways of glycolysis. c. decrease the rate at which some genes are transcribed. d. increase the rate at which some genes are transcribed. e. change the physiological states of our cells, which in turn affects post-translational gene regulation. ANSWER: c, d, e 111. Which process produces multiple proteins from the same primary transcript in the same cell? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 a. chromatin remodeling b. alternative splicing c. combinatorial control d. histone modification e. RNA editing ANSWER: b, e 112. A researcher notices that women who are exposed to aromatic hydrocarbons while working in the petrochemical industry may have a greater incidence of breast cancer than women without such exposure. The researcher undertakes a study of the BRCA1 promoter using the restriction enzymes MspI and HpaII. MspI cleaves double-stranded DNA at 5'-CCGG-3' regardless of whether the CG in the middle has cytosine methylation, whereas HpaII cleaves double-stranded DNA at 5'-CCGG-3' only when the CG in the middle lacks cytosine methylation. For a region of double-stranded DNA in the promoter region of BRCA1, tumor cells from the breasts of exposed women with breast cancer and nontumor cells from breasts from the same women show the illustrated patterns of bands produced by MspI and HpaII:

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Which of the statements does this evidence support? Select all that apply. a. More sites are methylated in tumor cells than in nontumor cells. b. Fewer sites are methylated in tumor cells than in nontumor cells. c. The BRCA1 gene may be epigenetically silenced in tumor cells. d. The same sites are methylated in tumor cells as in nontumor cells. e. No sites are methylated in nontumor cells. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 ANSWER: a, c 113. Which processes produce different proteins in different cells from the same primary transcript? Select all that apply. a. alternative splicing b. RNA editing c. chromatin remodeling d. histone modification e. combinatorial control ANSWER: a, b 114. The diagram shown here is part of an RNA transcript containing four open reading frames (M, H, K, and L) and three introns (1, 2, 3). The sites labeled d1–d3 are nucleotides immediately before each intron on the 5' side and those labeled a1–a3 are nucleotides immediately after each intron on the 3' side. Splicing out an intron involves a spliceosome facilitating the cleavage of the DNA at the d and a sites at either end of the intron.

A missense mutation takes place in open reading frame H. In which possible alternative splice forms of the transcript would the missense mutation not affect the polypeptide product? Select all that apply. a. d1–a1 + d2–a3 b. d1–a3 c. d1–a1 + d2–a2 + d3–a3 d. d1–a2 + d3–a3 e. None of the other answer options is correct. ANSWER: b, d 115. The diagram shown is part of an RNA transcript containing four open reading frames (M, H, K, and L) and three introns (1, 2, 3). The sites labeled d1–d3 are nucleotides immediately before each intron on the 5' side and those labeled a1–a3 are nucleotides immediately after each intron on the 3' side. Splicing out an intron involves cleaving the DNA at the d and a sites at either end of the intron.

A missense mutation takes place in open reading frame K. In which possible alternative splice forms of the transcript would the missense mutation not affect the polypeptide product? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 a. d1–a1 + d2–a2 + d3–a3 b. d1–a3 c. d1–a1 + d2–a3 d. d1–a2 + d3–a3 e. None of the other answer options is correct. ANSWER: b, c 116. Which of the changes could cause the production of a shorter-than-normal protein from a gene? Select all that apply. a. production of an siRNA b. changes in RNA editing c. chromatin remodeling d. changes in RNA processing e. nonsense mutation in the ORF ANSWER: b, d, e 117. Suppose that the RNA sequence shown here is a target for a microRNA. 5'-AAUGCUACG-3' Which of these double-stranded precursor RNA molecules could be processed for incorporation into RISC to target the sequence? Select all that apply. a. 5'-UUACGAUGC-3' 3'-AAUGCUACGb. 5'-CGUAGCAUU-3' 3'-GCAUCGUAAc. 5'-AAUGCUACG-3' 3'-UUACGAUGCd. 5'-GCAUCGUAA-3' 3'-CGUAGCAUUANSWER: b, c 118. In prokaryotes, repressor proteins: Select all that apply. a. bind with activators and prevent transcription. b. are part of the mechanism of positive regulation. c. are part of the mechanism of negative regulation. d. bind with DNA and prevent transcription. ANSWER: c, d 119. In the _____ cycle, the lambda bacteriophage's DNA _____. Select all that apply. a. lytic; recombines with bacterial DNA and is passed on to subsequent bacterial cells b. lysogenic; recombines with bacterial DNA and is passed on to subsequent bacterial cells Copyright Macmillan Learning. Powered by Cognero.

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Chapter 18 c. lytic; uses the cell to create more viruses d. lysogenic; hijacks the cell to create more viruses ANSWER: b, c

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Chapter 19 Multiple Choice 1. An adult cell that is genetically reprogrammed to be pluripotent is called a(n) _____ stem cell. a. pluripotent b. embryonic c. germ layer pluripotent d. induced pluripotent e. regenerative pluripotent ANSWER: d 2. A cell in the epithelium lining of the human gut is very different in structure and function from a white blood cell. How would you describe the genetic basis for this difference? a. Genes are gradually lost as these cell types differentiate into specialized tissues. b. These different cell types contain different sets of genes as a result of modifications during development. c. These different cell types express different sets of genes, although their genomes are identical. d. The genes evolve as the body develops the specialized tissues needed for full development. ANSWER: c 3. Differentiation refers to the process by which: a. gene regulation determines which proteins are produced in a given cell during development. b. fertilized eggs undergo multiple rounds of cell division to become embryos. c. changes in gene expression allow cells to produce the correct proteins at the correct time. d. cells become progressively more specialized during development. e. None of the other answer options is correct. ANSWER: d 4. In humans, the three germ layers are established when the: a. cells of the inner cell mass reorganize themselves into the gastrula. b. blastula forms. c. inner cell mass of the blastula forms. d. blastula implants into the uterine wall. e. gastrula implants into the uterine wall. ANSWER: a 5. Stem cells are cells that are capable of: a. producing the three germ layers. b. differentiating into multiple cell types. c. producing an entire organism. d. differentiating into a limited number of cell types. e. differentiating into any type of cell. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 ANSWER: b 6. Bicoid caudal mRNA translation. a. inhibits b. promotes c. does not affect ANSWER: a 7. In animals, homeotic genes: a. encode transcription factors. b. specify the identities of individual body parts or segments during development. c. are controlled by segment-polarity genes expressed earlier in development. d. are found in a wide variety of animals with bilateral symmetry, from fruit flies to vertebrates. e. All of these choices are correct. ANSWER: e 8. In the ABC model of floral development, flower development in Arabidopsis is regulated by _____ activities controlled by _____ genes. a. four; three b. three; three c. three; four d. four; four e. three; five ANSWER: c 9. Using the natural processes of cell growth and development to replace diseased or damaged tissue is called: a. stem cell transplantation. b. regenerative medicine. c. genetic engineering. d. cloning. e. stem cell therapy. ANSWER: b 10. The Pax6 gene acts as a master regulator of eye development. If a Pax6 gene from a mouse were engineered to be expressed in the antenna of Drosophila, what would you expect to observe? a. A Drosophila compound eye develops, only in its normal position on the head. b. A mouse eye develops on the antenna. c. Drosophila compound eye develops on the antenna. d. A mouse eye develops on the head. e. The antenna develops normally. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 11. If you were collecting fish and found that some had extra eyes, which of the genes listed would you suspect to be the cause of abnormality? a. nanos b. antennapedia c. bithorax d. bicoid e. Pax6 ANSWER: e 12. In John Gurdon's nuclear-transfer experiments, he used nuclei from tadpole intestinal cells. Do you think he would have had more success or less success if he had taken cells from a blastula? From an adult frog? a. same; same b. less; more c. more; less d. more; more e. less; less ANSWER: c 13. Which of the options correctly lists the stages of human development in order? a. fertilized egg, morula, blastocyst, gastrula b. fertilized egg, blastocyst, gastrula, morula c. fertilized egg, blastocyst, morula, gastrula d. fertilized egg, morula, gastrula, blastocyst e. fertilized egg, gastrula, morula, blastocyst ANSWER: a 14. Which of the options correctly pairs a germ layer with the cells that will form from it? a. ectoderm: inner layer of the skin and muscles. b. mesoderm: inner layer of the skin and muscles c. endoderm: epithelial and pigment cells of the skin d. ectoderm: lining of the digestive tract and lungs e. mesoderm: lining of the digestive tract and lungs ANSWER: b 15. An abnormal Drosophila larva that is missing much of its posterior end is likely deficient in _____ function. a. nanos b. bicoid c. hunchback ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 16. If a Drosophila larva is missing much of its anterior end it is likely due to a mutation in the _____ gene. a. zygotic bicoid b. maternal nanos c. maternal bicoid d. zygotic nanos ANSWER: c 17. In fruit flies (Drosophila), an embryo during gastrulation: a. forms three germ layers and becomes organized into distinct segments b. cells of the blastoderm migrate inward. c. already shows an organization of discrete parts or segments. d. All of these choices are correct. ANSWER: d 18. In a Drosophila oocyte, bicoid mRNA produced by the mother is localized at one end of the egg rather than evenly distributed throughout the egg. This is an example of: a. oocyte polarization. b. translational control of development. c. hierarchical control of gene regulation during development. d. maternal developmental effect. e. translational control of gene expression during development. ANSWER: a 19. Bicoid protein is a transcription factor that promotes transcription of the hunchback gene. This is an example of: a. translational control of gene expression during development. b. translational control of development. c. hierarchical control of gene regulation during development. d. maternal developmental effect. e. oocyte polarization. ANSWER: c 20. Which list correctly orders the sequence in which genes controlling the development of the anteriorposterior axis of Drosophila larvae are expressed? a. maternal-effect genes, gap genes, segment-polarity genes, pair-rule genes b. maternal-effect genes, gap genes, pair-rule genes, segment-polarity genes c. pair-rule genes, maternal-effect genes, segment-polarity genes, gap genes d. maternal-effect genes, segment-polarity genes, pair-rule genes, gap genes e. gap genes, pair-rule genes, segment-polarity genes, maternal-effect genes ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 21. The Bithorax mutation in Drosophila results in a pair of wings developing from what would normally develop as halteres. Bithorax is an example of a _____ mutation. a. segmentation b. maternal-effect c. gap-gene d. homeotic ANSWER: d 22. In the graph shown, the dashed line shows the level of mRNA for a certain protein, Prot2, at various positions along the anterior-posterior axis of an insect embryo. The solid line represents the level of a regulatory protein that controls translation of the Prot2 mRNA.

If the regulatory protein represses translation of the Prot2 mRNA, which graph shows the expected level of protein Prot2 across the embryo?

a. graph H b. graph M c. graph K d. graph L e. graph Q ANSWER: a 23. The reason scientists hope to develop personalized stem cell therapies as opposed to transplant stem cell therapies is that personalized therapies would avoid the problem of: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 a. bacterial contamination. b. side effects of medication. c. the very low rate of success of all transplants. d. tissue rejection. e. None of the other answer options is correct. ANSWER: d 24. Walter Gehring tested the hypothesis that the Pax6 gene acts as a master regulator of eye development. He introduced mouse Pax6 genes into the genome of a Drosophila and activated the gene in the fly's antennae. What happened, and did it support or refute his hypothesis? a. A compound eye developed on the antenna; this supports the hypothesis. b. A compound eye developed on the head; this supports the hypothesis. c. The antenna developed normally; this does not support the hypothesis. d. The antenna developed where the eye normally would be; this supports the hypothesis. e. The antenna developed where the eye would normally be; this does not support the hypothesis. ANSWER: a 25. When the Pax6 gene from a mouse is introduced into a Drosophila genome, Pax6 induces eye development. The eyes appear to be _____ eyes. a. mutant fly b. mutant mouse c. normal fly d. normal mouse e. None of the other answer options is correct. ANSWER: c 26. What would you expect to happen if you added activity B to whorl 1 in an Arabidopsis floral meristem? Refer to the figure shown.

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Chapter 19 a. You would get a flower with extra sepals but no petals. b. You would get a flower with extra petals but no sepals. c. You would get a flower with extra stamens but no sepals. d. You would get a flower with extra stamens but no petals. ANSWER: b 27. What is the consequence of inactivating the Notch receptor in the type 2 progenitor cell in C. elegans? a. The type 2 cell will be unable to cause the type 1 cell to switch on type 1 cell genes. b. The progenitor cannot signal itself. c. The progenitor cell will not complete its differentiation into a type 2 cell. ANSWER: c 28. Which of the choices is the diffusible extracellular element in the process of vulval cell differentiation in C. elegans? a. the EGF receptor b. the EGF ligand c. the Notch ligand d. the Notch receptor ANSWER: b 29. Imagine that there is a loss-of-function mutation in the gene for the Notch receptor in the progenitor cell of C. elegans. How does that mutation affect vulva development? Refer to the figure shown.

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a. The progenitor cell will become a type 1 cell. b. The progenitor cell will become a type 2 cell. c. The progenitor cell will remain a type 2 cell. d. The type 2 cell will differentiate into a type 1 cell. ANSWER: a 30. In the roundworm Caenorhabditis elegans, development of the vulva begins when the _____ cell secretes _____, which then bind(s) to _____ on a(n) _____ cell. a. anchor; EGF; the EGF receptor; progenitor b. progenitor; transcription factors; cis-regulatory regions; anchor Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 c. progenitor; EGF; the EGF receptor; anchor d. anchor; transcription factors; cis-regulatory regions; progenitor e. type 1; EGF; the EGF receptor; type 2 ANSWER: a 31. The progenitor cell that is closest to the anchor cell: a. receives the largest dose of EGF. b. activates genes for differentiation into a type 1 cell. c. prevents the adjacent cells from differentiating into a type 1 cell. d. induces the adjacent cells to differentiate into type 2 cells. e. All of these choices are correct. ANSWER: e 32. During the development of the vulva in Caenorhabditis elegans, when EGF binds an EGF receptor on a progenitor cell: a. signal transduction results in activation of a set of transcription factors. b. genes that produce the Notch ligand are transcribed. c. genes that produce the Notch receptor are inhibited. d. genes that produce the EGF receptor in the adjacent cells are inhibited. e. All of these choices are correct. ANSWER: e 33. Shown are four ligands (M, H, K, and L) and their corresponding receptors along with four genes (S, T, U, and V), whose activity the receptor controls through signal transduction. The arrows indicate gene activation, the T-bars indicate gene repression.

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Chapter 19 Which ligand acts as a signal resulting in both genes U and V being inactive? a. ligand K b. ligand H c. ligand L d. ligand M e. None of the other answer options is correct. ANSWER: c 34. Shown are four ligands (M, H, K, and L) and their corresponding receptors along with three genes (U, V, and W), whose activity the receptor controls through signal transduction. The arrows indicate gene activation, the T-bars indicate gene repression.

Activation of both gene U and gene V are required for expression of gene W. Which ligand (or ligands) results in W being active? a. ligand L b. ligand H c. ligand K d. ligand M e. None of the other answer options is correct. ANSWER: d 35. Cells differentiate through: a. the cell cycle. b. growth. c. timing. d. gene regulation. ANSWER: d 36. Multipotent cells of the mesoderm can: a. give rise to any cell type of any of the three germ layers. b. give rise to any cell type in the embryo. c. give rise to any cell type of mesodermal origin. d. become induced pluripotent stem cells. e. None of the other answer options is correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 ANSWER: c 37. As cells become differentiated, they delete the DNA for genes they no longer need. a. false b. true ANSWER: a 38. Which cells, or cell types, has the greatest developmental potential? a. fertilized egg b. inner cell mass c. gastrula d. ectoderm e. nerve cell ANSWER: a 39. Which of the choices is present at greater levels at the anterior end of a Drosophila embryo than the posterior end? a. nanos RNA b. bicoid mRNA c. hunchback mRNA d. caudal mRNA ANSWER: b 40. Mutations in the Pax6 gene cause developmental defects in the eyes of flies, mice, and humans. This suggests that: a. Pax6 may be a master regulator of eye development. b. Pax6 has been evolutionarily conserved. c. Pax6 activates multiple genes. d. All of these choices are correct. ANSWER: d 41. Which statement best describes the structural and functional relationships between the Notch ligand and the Notch receptor proteins in C. elegans? a. The Notch ligand and the Notch receptor are both transmembrane proteins. b. The Notch ligand is secreted from a cell and binds to the Notch receptor located in the plasma membrane of an adjacent cell. c. The Notch receptor is located in the nuclear envelope and Notch ligand is transported through the cytoplasm to activate the receptor. d. None of the other answer options is correct. ANSWER: a 42. A cis-regulatory element may _____ transcription. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 a. not affect b. activate c. repress d. All of the choices are correct. ANSWER: d 43. The muscle cells and nerve cells in a mouse look very different and serve very different functions in the mouse's body. These differences exist because the muscle cells and nerve cells in the mouse: a. copy different genes. b. have different chromosomes. c. use different genetic codes. d. have different genes. e. have different ribosomes. f. express different genes. ANSWER: f 44. A mutation occurs that causes a defect in the development of skeletal muscle. The mutation would likely have occurred in which kinds of cells in order to produce this defect? a. ectoderm b. mesoderm c. endoderm ANSWER: b 45. The early cell divisions of human development differ from other mitotic cell divisions in that: a. cells do not grow in size; they subdivide the cytoplasm of the fertilized egg. b. cells do not replicate all chromosomes prior to dividing; this helps the cells differentiate. c. cells move as they divide. d. they are unequal; some daughter cells are much larger than others. e. All of these choices are correct. ANSWER: a 46. The body of the human embryo develops from the _____ of the blastocyst. a. outer cell mass b. inner cell mass c. wall d. membranes e. cytoplasm ANSWER: b 47. Which list is arranged in order from describing cells with the most developmental potential to cells with the least developmental potential. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 a. totipotent, multipotent, pluripotent b. pluripotent, totipotent, multipotent c. totipotent, pluripotent, multipotent d. multipotent, pluripotent, totipotent e. multipotent, totipotent, pluripotent ANSWER: c 48. Which of these curves best represents the developmental potential of cells as they progress through development?

a. curve L b. curve H c. curve M d. curve K e. curve Q ANSWER: d 49. Which pair of terms correctly matches a cell or group of cells with its ability to differentiate into different specialized cells? a. ectoderm: totipotent b. mesoderm: pluripotent c. fertilized egg: pluripotent d. fertilized egg: multipotent e. endoderm: multipotent ANSWER: e 50. A researcher isolates stem cells from a developing embryo and finds that they are able to differentiate only into bone cells or red blood cells. The isolated stem cells are likely from the: a. ectoderm. b. gastrula. c. mesoderm. d. morula. e. endoderm. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 ANSWER: c 51. During morula development, the divisions are more rapid than in most normal adult cells and the individual cells get progressively smaller. Which stage(s) of the cell cycle are probably truncated (made shorter)? a. M phase b. S phase c. G1 and G2 d. both M phase and S phase e. All of the stages are truncated. ANSWER: c 52. A cell that has the capacity to differentiate into a limited number of related cell types is a _____ stem cell. a. omnipotent b. pluripotent c. totipotent d. unipotent e. multipotent ANSWER: e 53. The mRNA for caudal and hunchback are distributed evenly throughout the Drosophila embryo, yet the Caudal protein is found only in the posterior portion of the embryo, and the Hunchback protein is present in greater amounts anteriorly. This distribution is due to: a. translational regulation of the hunchback and caudal mRNAs in the zygote. b. transcriptional regulation of the hunchback and caudal genes in the zygote. c. transcriptional regulation of the nanos and bicoid genes in the zygote. ANSWER: a 54. Gurdon's nuclear transplantation experiments with Xenopus laevis (the African clawed toad) supports the hypothesis that: a. differentiation is the result of genes being successively deleted from the genome. b. differentiation is the result of genes being successively turned on or off. c. differentiation causes cells to become pluripotent. d. only nuclei from the blastocyst retain sufficient potency to produce an adult organism. e. All of these choices are correct. ANSWER: b 55. Gene regulation during development is _____, which means that _____. a. hierarchical; genes are turned on in a specific sequence, starting with totipotent and ending with multipotent. b. restrictive; genes lose developmental potential over time. c. restrictive; genes are successively lost over time. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 d. hierarchical; genes expressed at earlier stages of development control the expression of genes at later stages of development. e. combinatorial; genes expressed at earlier stages of development control the expression of genes at later stages of development. ANSWER: d 56. Maternal-effect genes: a. affect maternal phenotype. b. are expressed by the mother and affect offspring phenotype. c. do not affect the offspring phenotype. d. affect the willingness of females to reproduce. ANSWER: b 57. In Drosophila, absence of Bicoid protein results in larvae missing anterior segments and absence of Nanos protein results in larvae missing posterior segments. Suppose you inject Bicoid protein into the posterior region of early embryos lacking Nanos protein. The expected result would be: a. larvae with two anterior regions. b. larvae-like nanos mutants. c. larvae-like bicoid mutants. d. larvae with two posterior regions. e. None of the other answer options is correct. ANSWER: a 58. In the graph, the dashed line shows the level of mRNA for a certain protein, Prot5, at various positions along the anterior-posterior axis of an insect embryo. The solid line represents the level of a regulatory protein that controls translation of the Prot5 mRNA.

If the regulatory protein represses translation of the Prot5 mRNA, which graph shows the expected level of Prot5 across the embryo?

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a. graph Q b. graph H c. graph K d. graph M e. graph L ANSWER: e 59. In almost all organisms that have been studied, the genes in Hox clusters: a. are arranged in the chromosomes in a sequence that is independent of the order in which their products function along the anterior-posterior axis of the embryo. b. are expressed in a sequence that is independent of their linear organization along the chromosome. c. each contain a unique homeodomain, the products of which are very different among organisms. d. are arranged in chromosomes in the same order as their products function, along the anteriorposterior segments of the embryo. e. None of the other answer options is correct. ANSWER: d 60. In mammals, Hox genes: a. direct development of structures that become parts of the hindbrain, spinal cord, and vertebral column. b. specify limbs and other structures associated with each body segment. c. evolved by partial-genome duplication from ancestral genes. d. are expressed in a sequence that does not reflect their linear order in the chromosome or the regions they affect. e. All of these choices are correct. ANSWER: a 61. Evolutionarily conserved molecules: a. were likely present in the most recent common ancestor of the organisms possessing them. b. likely changed little over time because they serve a vital function. c. are molecules in which most mutations are likely to be harmful. d. are similar in sequence among distantly related organisms. e. All of these choices are correct. ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 62. When the gene for Pax6 transcription factor is transferred from a donor individual of one species into the genome of a recipient individual of a different species, it: a. induces the development of eyes similar to those in the donor species. b. represses eye development. c. induces the development of eyes similar to those in the recipient species. d. usually has no detectable effect. e. induces the development of eyes intermediate between those in the donor species and those in the recipient species. ANSWER: c 63. In plants, meristem tissue that becomes committed to develop into flowers is called floral meristem. In Arabidopsis, floral meristem is organized into four concentric whorls of cells, each of which gives rise to a specific organ of the mature flower. The floral meristem is therefore analogous to what cells or tissues in animals? a. totipotent stem cells b. cells in the morula c. the inner cell mass of the blastula d. any one germ layer of the gastrula e. None of the other answer options is correct. ANSWER: d 64. What is the most logical explanation of the observation that so many organisms have a Pax6 gene? a. All organisms with a Pax6 gene have a common ancestor. b. All organisms must be able to see. c. All organisms want to see. d. All organisms evolved a Pax6 gene. ANSWER: a 65. In the ABC model of floral development, A, B, and C stand for the activities of different: a. enzymes. b. transcription factors. c. translational regulators. ANSWER: b 66. A mutation is found in C. elegans that causes the organism to have no vulval opening cells or any vulval supporting cells. What could this mutation be? a. one that blocks expression of the Notch ligand in progenitor cells b. one that results in nonfunctional EGF receptors in progenitor cells c. one that blocks expression of the EGF receptor in anchor cells d. one that blocks the Notch receptor on progenitor cells e. one that results in constitutive expression of the EGF ligand from the anchor cell ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 67. Signal transduction is the process by which: a. an extracellular molecule activates a membrane protein, which in turn activates molecules inside the cell. b. a specific combination of transcription factors determines the developmental pathway in a cell or group of cells. c. a single master gene, or signal, activates a series of downstream genes that lead to cell differentiation. d. transcription factors bind to cis-regulatory regions of DNA and either activate or repress transcription. e. None of the other answer options is correct. ANSWER: a 68. During the development of the vulva in Caenorhabditis elegans, when the Notch ligand binds a Notch receptor on a progenitor cell: a. genes that produce the Notch receptor are inhibited. b. genes that produce the Notch ligand are transcribed. c. genes that produce the EGF receptor are inhibited. d. signal transduction results in activation of a set of transcription factors. e. None of the other answer options is correct. ANSWER: c 69. In vertebrates, the expression of different combinations of four Hox genes results in structural differences in tissues and organs along the anterior-posterior axis. For example, in the anterior part of a hypothetical organism represented in the graph shown, the expression of gene 1 alone results in the formation of tissue type W, the expression of gene 1 and gene 2 results in the formation of tissue X. At the posterior end of the organism, the expression of gene 1, gene 2, gene 3, and gene 4 causes the formation of tissue type Z.

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The influence of Hox genes on the formation of specific types of tissue can be determined by introducing gainof-function and loss-of-function mutations in specific Hox genes. Predict what would happen if the expression of gene 3 was expanded anteriorly by a gain-of-function mutation so that the pattern of gene 3 expression would be the same as that of gene 2. a. The amount of tissue X would decrease, and the amounts of the other three tissue types would all increase. b. The amount of tissue X would increase, the amount of tissue Y would decrease, and the amounts of tissue W and tissue Z would remain the same. c. The amount of tissue Y would increase, the amount of tissue X would decrease, and the amounts of tissue W and tissue Z would remain the same. d. The amount of tissue Y would decrease, and the amounts of the other three tissue types would all increase. ANSWER: c 70. In vertebrates, the expression of different combinations of four Hox genes results in structural differences in tissues and organs along the anterior-posterior axis. For example, in the anterior part of a hypothetical organism represented in the graph shown, the expression of gene 1 alone results in the formation of tissue type W, the expression of gene 1 and gene 2 results in the formation of tissue X. At the posterior end of the organism, the expression of gene 1, gene 2, gene 3, and gene 4 causes the formation of tissue type Z.

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The influence of Hox genes on the formation of specific types of tissue can be determined by introducing gainof-function and loss-of-function mutations in specific Hox genes. Predict what would happen if the pattern of gene 2 expression was restricted by a loss-of-function mutation so that the pattern of gene 2 expression would be the same as that of gene 3. a. The amount of tissue W would decrease, the amount of tissue X would increase, and the amounts of tissue Y and tissue Z would remain the same. b. The amount of tissue W would increase, the amount of tissue X would decrease, and the amounts of tissue Y and tissue Z would remain the same. c. The amount of tissue Y would increase, and the amounts of the other three tissue types would decrease. d. The amount of tissue W would decrease, and the amounts of the other three tissue types would all increase. ANSWER: b 71. Unlike tetrapods or many other reptiles, snakes have no limbs. Analysis of expression of their homeotic genes shows that for three types of homeotic genes, which for convenience we will call HoxY (yellow), HoxB (blue), and HoxR (red) in the figure, the pattern of expression is different in embryos of snakes than it is in other reptilian tetrapod embryos, including chick embryos. The arrows in the figure indicate the regions in the embryo from which certain adult structures arise.

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Which combination would be required to produce a forelimb in snakes? a. the combination of HoxY (yellow) and HoxB (blue) expressed together b. the combination of HoxY (yellow) and HoxR (red) expressed together c. the combination of HoxB (blue) and HoxR (red) expressed together d. the combination of HoxY (yellow), HoxB (blue), and HoxR (red) expressed together ANSWER: b 72. Unlike tetrapods or many other reptiles, snakes have no limbs. Analysis of expression of their homeotic genes shows that for three types of homeotic genes, which for convenience we will call HoxY (yellow), HoxB (blue), and HoxR (red) in the figure, the pattern of expression is different in embryos of snakes than it is in other reptilian tetrapod embryos, including chick embryos. The arrows in the figure indicate the regions in the embryo from which certain adult structures arise.

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What kind of evolutionary change might result in the extension of the expression pattern of the HoxB (blue) and HoxR (red) factors in both the anterior and posterior directions along the snake's body? a. a deletion of the gene encoding the HoxY (yellow) factor b. expression of the gene encoding the HoxY (yellow) factor is inhibited c. a nonsense mutation in the gene encoding the HoxB (blue) factor d. a mutation arises that prevents a repressor from binding ANSWER: d 73. In vertebrates, the expression of different combinations of four Hox genes results in structural differences in tissues and organs along the anterior-posterior axis. For example, in the anterior part of a hypothetical organism represented in the graph shown, the expression of gene 1 alone results in the formation of tissue type W, the expression of gene 1 and gene 2 results in the formation of tissue X. At the posterior end of the organism, the expression of all four genes causes the formation of tissue type Z.

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The influence of Hox genes on the formation of specific types of tissue can be determined by introducing gainof-function and loss-of-function mutations in specific Hox genes. Predict what would happen if the expression of gene 2 was expanded by a gain-of-function mutation to the anterior-most region of the organism. a. The amount of tissue X would increase, and the amounts of tissues Y and Z would remain the same. b. The amount of tissue W would increase, the amount of tissue X would decrease, and the amounts of tissues Y and Z would remain the same. c. The amount of tissue X would decrease, and the amounts of the other three tissue types would all increase. d. The amount of tissue W would decrease, and the amounts of the other three tissue types would all increase. ANSWER: a 74. In vertebrates, the expression of different combinations of four Hox genes results in structural differences in tissues and organs along the anterior-posterior axis. For example, in the anterior part of a hypothetical organism represented in the graph shown, the expression of gene 1 alone results in the formation of tissue type W, the expression of gene 1 and gene 2 results in the formation of tissue X. At the posterior end of the organism, the expression of all four genes causes the formation of tissue type Z.

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The influence of Hox genes on the formation of specific types of tissue can be determined by introducing gainof-function and loss-of-function mutations in specific Hox genes. Predict what would happen if the expression of gene 4 was prevented by a loss-of-function mutation. a. The amount of tissue W would decrease, and the amounts of the other three tissue types would all increase. b. The amount of tissue Y would increase, and the amounts of the other three tissue types would remain the same. c. The amount of tissue Y would increase, and the amounts of the other three tissue types would decrease. d. The amount of tissue Z would decrease, the amount of tissue Y would increase, and the amounts of tissues X and W would be unchanged. ANSWER: d 75. In the figure shown, there are four ligands (M, H, K, and L) and their corresponding receptors along with four genes (S, T, U, and V), whose activity the receptor controls through signal transduction. The arrows indicate gene activation, the T-bars indicate gene repression.

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Which ligand acts as a signal resulting in gene U being active and V inactive? a. ligand M b. ligand H c. ligand K d. ligand L e. None of the other answer options is correct. ANSWER: b 76. In the figure shown, there are four ligands (M, H, K, and L) and their corresponding receptors along with four genes (S, T, U, and V), whose activity the receptor controls through signal transduction. The arrows indicate gene activation, the T-bars indicate gene repression.

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Which ligand acts as a signal resulting in gene U being inactive and gene V being active? a. ligand H b. ligand M c. ligand K d. ligand L e. None of the other answer options is correct. ANSWER: c Multiple Response 77. Cellular differentiation progressively restricts cell fate because the unexpressed genes in the cell: Select all that apply. a. undergo irreversible repression. b. accumulate point mutations. c. are deleted from the genome. d. accumulate near the centromeres. e. become more densely packed with nucleosomes. ANSWER: a, e 78. When the Pax6 gene from a mouse is introduced into a Drosophila genome, it induces eye development. The eyes appear to be normal fly eyes, not mouse eyes. Why? Select all that apply. a. The Pax6 gene sequence must be very similar among animal species. b. The Pax6 gene produces a product that switches on the genes for eye formation. c. The Pax6 gene codes for fly type eyes in all organisms. d. The Pax6 gene must have evolved independently in many animal species. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 ANSWER: a, b 79. Mature cells are terminally differentiated. What could cause this permanent change? Select all that apply. a. epigenetic change b. changes in gene expression c. changes in chromosome content d. changes in cell signaling ANSWER: a, b 80. The Pax6 gene is a master regulator of eye development, but it does not result in potato eyes. Why not? Select all that apply. a. Potato eyes do not share a common evolutionary origin with animal eyes. b. The cis-regulatory elements in potato-eye development are different. c. It does not work because plant cells do not have transcription factors. d. It does not make eyes because Drosophila do not eat potatoes. e. None of the other answer options is correct. ANSWER: a, b 81. Why has the development of iPS cells been viewed as a major breakthrough in the development of stem cell therapies? Select all that apply. a. It allows us to develop pluripotent stem cells without the ethical issues involved in obtaining them from human embryos. b. It may ultimately allow us to obtain stem cells from any individual—we can match the patient to the tissue. c. It is easy and inexpensive. d. It is a proven technique that we will be able to move into a clinical trial very quickly. e. It will allow us to "grow" new organs very quickly. ANSWER: a, b 82. The goals of regenerative medicine are to: Select all that apply. a. create a bank of totipotent stem cells from which new tissues and organs can be generated. b. develop cells, tissues, and organs that can replace damaged ones with minimal or no danger of tissue rejection. c. create stem cells for patients using their own adult cells. ANSWER: b, c 83. Which cells are multipotent cell types? Select all that apply. a. cells of the ectoderm b. cells of the endoderm c. cells of the mesoderm d. cells of the inner cell mass Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 e. the fertilized egg ANSWER: a, b, c 84. Maternal-effect genes code for _____ present in the egg. Select all that apply. a. pluripotent stem cells b. totipotent stem cells c. RNA d. protein ANSWER: c, d 85. The sequence and actions of genes controlling development of the anterior-posterior axis of Drosophila illustrate: Select all that apply. a. hierarchical control of gene regulation during development. b. that genes controlling development are turned on in sequence, with each set acting on a different set of cells to specify the cells' ultimate fates. c. the effect of maternal development on the development of the embryo. d. that genes controlling development are turned on in groups, with each group refining the pattern of differentiation generated by previous groups. ANSWER: a, b 86. The developmental pathway of each whorl of a developing Arabidopsis flower is determined by the specific: Select all that apply. a. combination of noncoding RNAs present in each whorl. b. number of meristem cells in each whorl. c. combination of cell types in each whorl. d. combination of genes active in each whorl. e. combination of transcription factors present in each whorl. ANSWER: d, e 87. Mature cells are terminally differentiated. What could cause this permanent change? Select all that apply. a. epigenetic change b. changes in gene expression c. changes in chromosome content d. changes in cell signaling ANSWER: a, b 88. The inference that homeotic genes in vertebrates and their corresponding genes in invertebrates evolved from counterpart genes in a common ancestor is supported by the observations that the genes in vertebrates and invertebrates: Select all that apply. a. share similarities in DNA sequence. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 b. are arranged similarly along the chromosome. c. result in the same structures during development. d. act at the same time in development. ANSWER: a, b 89. What are present in cis-regulatory elements that regulate gene expression in response to developmental signals? Select all that apply. a. binding sites for transcriptional activator proteins b. binding sites for transcriptional repressor proteins c. operator sequences d. CRP-cAMP binding sites e. All of these choices are correct. ANSWER: a, b 90. Shown are four ligands (M, H, K, and L) and their corresponding receptors along with three genes (U, V, and W), whose activity the receptor controls through signal transduction. The arrows indicate gene activation, the T-bars indicate gene repression.

If either U or V is required to activate W, which ligands result in W being active? Select all that apply. a. ligand M b. ligand H c. ligand L d. ligand K e. None of the other answer options is correct. ANSWER: a, b, d 91. During the development of the vulva in Caenorhabditis elegans, binding of ____ ligand to a cell receptor initiates development of the cell into ____ cell. Select all that apply. a. EGF; a type 2 b. Notch; a type 1 c. EGF; an anchor Copyright Macmillan Learning. Powered by Cognero.

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Chapter 19 d. EGF; a type 1 e. Notch; a type 2 ANSWER: d, e

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Chapter 20 Multiple Choice 1. The diagram shown represents a protein electrophoresis gel with samples from six individuals tested for variants (alleles A and B) of a single protein. The darkness of the band is a reflection of the intensity of staining in the gel and corresponds to the amount of enzyme present at that band. What are the allele frequencies?

a. 33.3% A, 66.6% B b. 75% A, 25% B c. 75% B, 25% A d. 66.6% A, 33.3% B e. The frequency cannot be determined from the information provided. ANSWER: d 2. Recall that in Mendel's garden peas, the yellow gene determines flower color, with the A (yellow) allele dominant to the a (green) allele. In a population of 200 plants, the genotype frequencies are 50% AA, 25% Aa, and 25% aa. What are the allele frequencies? a. 50% A, 50% a b. 50% A, 25% a c. 87.5% A, 12.5% a d. 62.5% A, 37.5% a e. 75% A, 25% a ANSWER: d 3. A population that exhibits only one allele at a particular gene is: a. fixed for that allele. b. undergoing evolution. c. in Hardy-Weinberg equilibrium. d. experiencing natural selection. e. undergoing genetic drift. ANSWER: a 4. In a population of Mendel's garden peas, the frequency of the dominant A (yellow flower) allele is 80%. Let p represent the frequency of the A allele and q represent the frequency of the a allele. Assuming that the population is in Hardy-Weinberg equilibrium, what are the genotype frequencies? a. 64% AA, 32% Aa, 4% aa b. 50% AA, 25% Aa, 25% aa Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 c. 16% AA, 40% Aa, 44% aa d. 75% AA, 15% Aa, 10% aa e. 80% AA, 10% Aa, 10% aa ANSWER: a 5. In a population of Mendel's garden peas, the frequency of green-flowered plants (genotype aa) is 49%. The population is in Hardy-Weinberg equilibrium. What are the frequencies of the AA and Aa genotypes? a. 49% AA, 2% Aa b. 9% AA, 42% Aa c. 42% AA, 9% Aa d. 33% AA, 18% Aa e. The frequencies cannot be determined from the information provided. ANSWER: b 6. The rate differences between molecular clocks based on protein-coding genes are due largely to differences in _____. Thus, the extreme case of a fast molecular clock is derived from _____. a. the intensity of selection; histone proteins b. the size of the molecule; pseudogenes c. the intensity of selection; pseudogenes d. the size of the molecule; histone proteins e. None of the answer options is correct. ANSWER: c 7. Refer to the figure shown. If you wanted to use a molecular clock to date a relatively recent divergence event - one that occurred in the last 100 million years or so - which of the four types of genes would make the BEST clock? Why?

a. Fibrinopeptides, because they have changed the most over that time period. b. A histone gene, because it has changed relatively little over that time. c. Cytochrome c, because it has changed relatively little over that time. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 d. A histone gene, because it has changed the most over that time period. e. Hemoglobin, because of its steady rate of change. ANSWER: a 8. The allele underlying sickle-cell anemia affects the structure of the _____ molecule inside _____ blood cells. a. hemoglobin; red b. myoglobin; red c. myoglobin; white d. hemoglobin; white e. fibrinopeptide: white ANSWER: a 9. In regions where malaria is prevalent, individuals who are _____ have a selective advantage. a. heterozygous for the S (sickling) allele b. homozygous for the S (sickling) allele c. homozygous for the A (nonmutant) allele d. None of the answer options is correct. ANSWER: a 10. Which type of selection decreases overall phenotypic variation in a specific trait? a. directional selection b. disruptive selection c. natural selection d. stabilizing selection ANSWER: d 11. The black-bellied seedcracker is a species of bird found in central Africa. They live in various habitats where individuals eat either hard, large seeds or small, soft seeds. Individuals with thick, shorter beaks can eat the harder seeds. Individuals with thin, longer beaks can eat softer seeds. Individuals with beaks of intermediate thickness and length cannot really eat either type of seed very well. If you measured beak width and length in populations of black-bellied seedcrackers, what type of selection would you expect to find? a. directional selection b. disruptive selection c. natural selection d. stabilizing selection ANSWER: b 12. Both Charles Darwin and Alfred Russel Wallace acknowledged the influence of the economist Thomas Malthus in the development of their ideas about natural selection. Specifically, Malthus's ideas about geometric population growth implied that: a. resources in every generation would be limited; therefore, individuals in every generation would have to compete for those resources. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 b. only the largest and strongest individuals would survive. c. organisms should have as many offspring as possible. d. organisms should have a few, very fit offspring who can compete for scarce resources. e. All of these choices are correct. ANSWER: a 13. The term "modern synthesis" refers to: a. the synthesis of Darwin's ideas about evolution with Malthus's ideas about population growth. b. the synthesis of Darwin's and Wallace's independently developed ideas about natural selection and adaptation. c. the synthesis of Darwin's ideas about natural selection and Mendelian genetics. d. the synthesis of Darwin's ideas about natural selection and modern DNA sequencing technology. e. All of these choices are correct. ANSWER: c 14. Natural selection that increases the frequency of a favorable allele is called: a. positive selection. b. balancing selection. c. sexual selection. ANSWER: a 15. On a hike through the forest, you see two large male elk fighting each other with their antlers. Nearby, a smaller female elk (doe) without antlers watches. This difference between the sexes is likely the result of: a. intersexual selection. b. intrasexual selection. c. a trait selected for defense against predators. d. a trait selected for storage of minerals. ANSWER: b 16. If two males compete directly with one another over access to a group of females, this is a form of: a. natural selection. b. intersexual selection. c. intrasexual selection. d. genetic drift. ANSWER: c 17. Nonadaptive mechanisms of evolutionary change are: a. migration (gene flow), mutations, and genetic drift. b. migration (gene flow), balancing selection, and genetic drift. c. migration (gene flow), sexual selection, and genetic drift. d. mutation, sexual selection, and genetic drift. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 e. mutation, genetic drift, and heterozygote advantage. ANSWER: a 18. Imagine these genotype frequencies in a population: p2 = 0.49, 2pq = 0.42, q2 = 0.09. Now assume that there is nonrandom mating where individuals with one genotype will only mate with individuals that also have their genotype. Assume this pattern of mating goes on until the frequency of heterozygotes is effectively zero. In addition, there is also inbreeding depression such that individuals with the genotype represented by p2 die before they can reproduce. What will be the eventual frequency of allele q? a. 0.3 b. 0.7 c. 0.91 d. 0.49 e. 1.0 ANSWER: e 19. The correlation between the time two species have been evolutionarily separated and the amount of genetic divergence between them is known as the: a. molecular clock. b. natural selection. c. Hardy-Weinberg equilibrium. d. artificial selection. e. modern synthesis ANSWER: a 20. One of the conditions of Hardy-Weinberg equilibrium is to have a large population. Why do small populations violate the conditions that must be met for a population to be in Hardy-Weinberg equilibrium? a. small populations are more likely to go extinct b. small populations are more likely to have changes in allele frequency due to genetic drift c. small populations are more likely to split into smaller populations d. small populations are more likely to have non-random mating ANSWER: b 21. If the allele frequency for the recessive autosomal allele that causes a particular rare hair color is 0.02, how frequently would you expect to observe the hair color human populations? a. 1 in every 50 individuals b. 1 in every 2000 individuals c. 1 in every 2500 individuals d. 1 in every 5000 individuals ANSWER: c 22. In general, in a sample of n individuals, the frequency of an allele is: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 a. the number of occurrences of the allele. b. the number of occurrences of the allele divided by n. c. one half the number of occurrences n. d. the number of occurrences of the allele divided by twice the number of individuals in the sample 2n. e. twice the number of occurrences of the allele divided by n. ANSWER: d 23. How can evolution of a population occur without changing the allele frequency of a trait within the population? a. There is no mutation. b. More organisms enter the gene pool. c. The genotype frequencies of the population change over time. d. The population is small. ANSWER: c 24. Refer to the table shown. Consider the two populations and the frequencies of alleles.

Which of the populations has greater allelic variation? a. population 1 b. population 2 c. Neither, allelic diversity is the same for both populations. ANSWER: c 25. You find that a wild population of antelope is not in Hardy-Weinberg equilibrium. From this information alone, can you determine the mechanism of evolution operating on the population? a. yes b. no ANSWER: b 26. If a population is not in Hardy-Weinberg equilibrium, we can conclude that: a. nonrandom mating has occurred. b. natural selection has occurred. c. one of the assumptions of the Hardy-Weinberg equilibrium has been violated. d. evolution has occurred because one or more of the assumptions of the Hardy-Weinberg equilibrium has been violated. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 e. All of these choices are correct. ANSWER: d 27. Northern elephant seals (Mirounga angustirostris) come up on a beach during breeding season. Males arrive first and establish a territory by posturing and fighting with other males. The posturing and fighting behaviors are probably the result of _____ selection. a. intrasexual b. intersexual c. stabilizing d. directional e. balancing ANSWER: a 28. The intricate plumage of male birds of paradise has become increasingly elaborate throughout their evolution due to females mating preferentially with males that display the most impressive feathers. This is an example of _____ selection. a. artificial b. sexual c. disruptive d. stabilizing ANSWER: b 29. The bright red breasts of male robins is likely the result of _____ selection. a. intrasexual b. intersexual c. directional d. stabilizing e. disruptive ANSWER: b 30. The goldenrod gall-fly lays its eggs on the terminal buds of goldenrod plants. Larvae chew through the buds and into the stems, where their saliva induces the plant to generate a gall, or outgrowth of tissue, that then provides food and shelter for the developing larva. The larvae are prey to both parasitoid wasps and to birds; wasps selectively prey on larvae inside the smallest galls while birds selectively prey on larvae inside the largest galls. Goldenrod gall-flies are therefore subject to _____ selection. a. balancing b. stabilizing c. directional d. disruptional e. heterozygote ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 31. Sexual selection always favors: a. large males. b. ornamentation such as bright feathers in male birds. c. traits that reduce an individual's survival. d. traits that increase an individual's access to reproductive opportunities. e. All of these choices are correct. ANSWER: d 32. Which of the big sagebrush plants is the most fit, based only on the information presented? a. the individual that can survive with 10% less water than the others in its population b. the individual that produces 5% more distasteful chemicals, making it 2% less likely to be attacked by herbivores than others in its population c. the individual that produces 3% more offspring each year than others in its population d. the individual that is the most resistant to disease organisms in the population e. All of these individuals are equally fit based on the information presented. ANSWER: c 33. You are studying a population of plants and you have determined that they are all genetic clones of one another. You are interested in a particular gene found in the population that you have named LLT (low-light tolerance), that allows it to live in shady areas. What is the frequency of this allele in the population you study? a. 0.5 b. 1.0 c. 0.0 d. It is not possible to determine from the information provided. ANSWER: b 34. Imagine these genotype frequencies in a population: p2 = 0.49, 2pq = 0.42, q2 = 0.09. Now assume that there is nonrandom mating where individuals with one genotype will only mate with individuals that also have their genotype. Assume this pattern of mating goes on until the frequency of heterozygotes is effectively zero. What will the frequency of allele p be in the population? a. 0.3 b. 0.7 c. 0.91 d. 0.49 e. 1.0 ANSWER: b 35. Imagine these genotype frequencies in a population: p2 = 0.49, 2pq = 0.42, q2 = 0.09. Now assume that there is nonrandom mating where individuals with one genotype will only mate with individuals that also have their genotype. Assume this pattern of mating goes on until the frequency of heterozygotes is effectively zero. Is this population in Hardy-Weinberg equilibrium? a. Yes, because allele frequencies did not change. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 b. Yes, because there is not inbreeding depression. c. No, because there is nonrandom mating. d. No, because allele frequencies changed. ANSWER: c 36. A researcher studying two species (species 1 and species 2) sequences a short stretch of eight codons from the same gene, gene B, in each and compares them. Species 1 and species 2 had a most recent common ancestor 50 million years ago. Given the differences between the sequences of the two species' genes shown here, what evolutionary force can you predict is most likely in operation on gene B?

a. genetic drift b. positive selection c. negative selection d. directional selection e. disruptive selection f. stabilizing selection g. balancing selection ANSWER: c 37. A mutation in the DNA will result in a new allele. a. true b. false ANSWER: a 38. In a population of Mendel's garden peas, the frequency of dominant yellow-flowered plants is 50%. The population is in Hardy-Weinberg equilibrium. What is the frequency of the recessive allele in the population? a. 0.5 b. 0.25 c. 0.05 d. The frequency cannot be determined from the data provided. ANSWER: d 39. A particular gene in a given population of individuals has two alleles, A and a. The frequency of the A allele equals the frequency of the a allele. What are the expected genotype frequencies, assuming the population is in Hardy-Weinberg equilibrium? a. A = 0.5, a = 0.5 b. AA = 0.5, aa = 0.5 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 c. AA = 0.25, Aa = 0.50, aa = 0.25 ANSWER: c 40. What does it mean to say that an allele is "fixed" in the population? a. It has been repaired by typical mechanisms in the cell. b. It is an indication of no genetic variation in the population. c. It is an indication of low genetic variation in the population. d. It is an indication of high genetic variation in the population. e. It is an indication that the allele cannot undergo mutation. ANSWER: b 41. In a hypothetical population of 1000 frogs a gene exists with two alleles. 280 of the frogs are homozygous dominant (DD), and 220 are homozygous recessive (dd). What is the frequency of heterozygotes in the population? a. 0.0 b. 0.28 c. 0.22 d. 0.50 ANSWER: d 42. In a hypothetical population of 1000 frogs a gene exists with two alleles. 280 of the frogs are homozygous dominant (DD), and 220 are homozygous recessive (dd). What is the frequency of the D allele in the population? a. 0.47 b. 0.50 c. 0.53 d. The answer cannot be determined from the data provided. ANSWER: c 43. Human ABO blood groups are determined by a single gene with 3 alleles: A, B, and O. In a sample of 300 individuals, 100 are blood type A and genotype AA, 100 are blood type B and genotype BO, and 100 are blood type O and genotype OO. What are the allele frequencies? a. 33.3% A, 33.3% B, 33.3% O b. 33.3% A, 33.3% B, 66.6% O c. 33.3% A, 16.6% B, 50% O d. 25% A, 25% B, 50% O e. 50% A, 25% B, 25% O ANSWER: c 44. You have a small population of beetles. One day a large rainstorm causes flooding and wipes out 87% of the population. The remaining individuals have much lower genetic variation than the original population. Which of the answer choices would explain the lack of Hardy-Weinberg equilibrium seen after the flood? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 a. founder effect b. random mating c. small population size d. genetic bottleneck e. migration ANSWER: d 45. You have two populations of flowering plants. In these populations, floral color is controlled by one gene. The red gene (R) is dominant to the white gene (r). In population 1, 75% of the flowers are red, in population 2, 25% of the flowers are white. Assuming the populations are in Hardy-Weinberg equilibrium, which population has a greater frequency of the R allele? a. population 1 b. population 2 c. Neither; they both have the same frequency. ANSWER: c 46. In a study of genetic variation of the Graceland gene, a researcher finds that there are two alleles in a population. In a large sample (500 individuals), the frequency of heterozygotes is 0.63. Is the population in Hardy-Weinberg equilibrium? a. No. The frequency of heterozygotes would be 1 if the population was in Hardy-Weinberg equilibrium. b. No. The highest frequency of heterozygotes under Hardy-Weinberg equilibrium is 0.5. c. No. There would be no heterozygotes if the population were in Hardy-Weinberg equilibrium. d. No. The frequency of each genotype would be equal if the population were in Hardy-Weinberg equilibrium. ANSWER: b 47. Small populations of a diminutive rodent, each with 100 individuals, live on two small neighboring islands, Rack and Oon. The Rack population is fixed for the A allele at the Agility gene; the Oon population is fixed for the a allele at the Agility gene. Ten individuals from Rack get carried on a drifting fallen tree trunk to Oon. Assuming the drifting individuals arrived at the start of the breeding season and that the rodents breed every year and die once they have raised their offspring, what are the genotype frequencies in the Oon population the year after the accidental migration event assuming the population is in Hardy-Weinberg equilibrium? a. AA 0.0; Aa 0.0; aa 1.0 b. AA (1/11)2; Aa 2(1/11)(10/11); aa (10/11)2 c. AA (10/11)2; Aa 2(1/11)(10/11); aa (1/11)2 d. AA 1.0; Aa 0.0; aa 1.0 e. The genotype frequencies cannot be determined from the information provided. ANSWER: b 48. Positive selection increases the frequency of an allele until it goes to fixation and negative selection decreases the frequency of an allele until it is eliminated. What is the long-term fate of either allele of a gene Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 with two alleles in which the fitness of the heterozygote is superior to that of both homozygotes? a. One allele will eventually go to fixation because it is favored in the given selective environment. b. Both alleles will remain in the population. c. Heterozygote advantage will continue until the frequency of both alleles is equal to 0.5. d. One of the alleles will eventually be lost because selection will favor the other allele. ANSWER: b 49. In a population of Mendel's garden peas, the frequency of the dominant A (yellow flower) allele is 80%. Let p represent the frequency of the A allele and q represent the frequency of the a allele. Assuming that the population is in Hardy-Weinberg equilibrium, what are the genotype frequencies? a. 64% AA, 32% Aa, 4% aa b. 50% AA, 25% Aa, 25% aa c. 16% AA, 40% Aa, 44% aa d. 75% AA, 15% Aa, 10% aa e. 80% AA, 10% Aa, 10% aa ANSWER: a 50. Male northern elephant seals (Mirounga angustirostris) compete with one another for territories along the beach and defend these territories throughout the breeding season. The males that win the male-male competitions defend the "best" territories and have access to females that stay in that territory of the beach. The females land on the beach during mating season, choose which territory of the beach to join, and then disproportionately breed with the male that defends that area of the beach. At what point may intersexual selection occur in these elephant seals? a. when the female is swimming to the beach b. when the female lands on the beach and chooses her resting location c. when the male defends his group of females against a different male d. when the male mates with a female within his group e. when the female leaves the beach after mating with a male ANSWER: b 51. You have spent time working with a population of beetles where males range in size from 2-6 cm in length. You realize that the females only mate with males that measure less than 3 cm long. If you measured allele frequencies that contributes to overall length, would you expect this population to be in Hardy-Weinberg equilibrium from one generation to the next? a. yes b. no ANSWER: b 52. In New Mexico, large expanses of black lava create patches of unique habitat. Pocket mice with darker coat color are less likely to be seen by predators on the darker patches of habitat, and more likely to survive and reproduce. If, in every generation, selection favors the darkest colored pocket mice in those habitats because they are best hidden from predators, this would be an example of: a. balancing selection. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 b. stabilizing selection. c. directional selection. d. disruptional selection. e. heterozygote advantage. ANSWER: c 53. A hypothetical endangered species of wildflower has been reduced to a single small population in a mountain meadow. A rare early spring blizzard kills all but three of the remaining plants, one of which has a rare mutation. This is an example of: a. natural selection. b. sexual selection. c. genetic drift/bottleneck. d. genetic drift/founder event. e. None of the answer options is correct. ANSWER: c 54. The only evolutionary process that leads to adaptive change is: a. mutation. b. migration. c. natural selection. d. genetic drift. e. All of these choices are correct. ANSWER: c 55. In regions where malaria is prevalent, the S allele is beneficial because: a. it improves the ability of the hemoglobin to carry oxygen. b. it improves the ability of the immune system to fight Plasmodium, the parasite that causes malaria. c. it causes red blood cells to distort, which makes them less hospitable to Plasmodium, the parasite that causes malaria. d. it prevents mosquitoes from releasing Plasmodium, the mosquito that causes malaria, into the blood. e. None of the answer options is correct. ANSWER: c 56. The figure shown illustrates the results of a series of observations and experiments on tail length in male African widowbirds. The black bars represent manipulated tail lengths by researchers (they actually added length to the tail feathers of the experimental birds). From the black bars on the figure, we see clearly that the male birds that have been artificially manipulated to have super-long tails are very good at attracting mates, approximated by the number of active nests for each male. What most likely explains why the male tails we see in nature (open bars on the figure) are so much shorter?

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Chapter 20

a. Males with really long tails in nature will eventually evolve, the researchers were simply showing the next step in the evolution of the trait. b. The sampling of males in nature was incomplete, they probably exist but were not found. c. If males in nature had the most successful tail length from the experimental study, they would not be able to fly well enough to escape predation. d. If males in nature had really long tails, like the experimental treatment for long-tails, then, over time, there would be more male African widowbirds with really long tails. ANSWER: c 57. Hammerhead bats are a species of African fruit bat. During the breeding season, males gather in mating arenas, where they display to females who pass through and assess males before selecting individuals with which to mate. This is an example of _____ selection. a. natural b. intersexual c. intrasexual d. stabilizing ANSWER: b 58. Inbreeding depression is a serious concern in small populations of endangered organisms. Why is inbreeding depression a concern for such populations? a. Related individuals are more likely to mate with one another and thereby increase the probability that two deleterious alleles will be present in the offspring. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 b. Related individuals are more likely to mate with one another and will thereby increase the frequency of homozygous genotypes. c. Related individuals will not mate with one another; therefore, population number will decline. d. Related individuals will not mate with one another, thereby increasing the probability that two deleterious alleles will be present in the offspring when unrelated individuals mate. ANSWER: a 59. A researcher studying two species (species 1 and species 2) sequences a short stretch of eight codons from the same gene, gene A, in each and compares them. Species 1 and species 2 had a most recent common ancestor 50 million years ago. Given the differences between the sequences of the two species' genes shown here, what evolutionary force can you predict is in operation on gene A?

a. genetic drift b. positive selection c. negative selection d. directional selection e. disruptive selection f. stabilizing selection g. balancing selection ANSWER: a 60. In an analysis of evolution in a DNA sequence of a single gene in multiple vertebrate species, a researcher divides the data into two classes: differences among species at the first and second positions in codons, and differences among species at the third positions in codons. When the researcher plots the results, as shown, she finds that she has two different molecular clocks. The likely cause of these data is because third codon positions are less likely to change amino acid sequences in proteins.

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Chapter 20

a. true b. false ANSWER: a 61. In a study of the alcohol dehydrogenase (ADH) enzyme in a population of Drosophila melanogaster, a researcher finds these genotype frequencies. FF FS SS 0.81 0.18 0.01 What are the allele frequencies of F and S? a. F = 0.5, S = 0.5 b. F = 0.09, S = 0.91 c. F = 0.9, S = 0.1 d. F = 0.81, S = 0.19 ANSWER: c 62. There is a volcanic island with a single population of mice that covers the island. A volcanic eruption occurs and splits the original population in two (A1 and A2). You had collected samples of DNA from the original population A prior to the eruption, as well as the two populations that resulted after the volcanic eruption. Your descendants have fortunately inherited your fascination with rodent genetics. Fifty thousand years after the original volcanic eruption, your descendants return to the island and repeat the same sampling and DNA sequencing strategy.

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Chapter 20

Your and your descendants' data indicate that members of population A1 are capable of breeding with members of A2 and producing viable, fertile A1-A2 hybrid offspring. Compare the DNA sequences presented and select the level of variation expected between two randomly chosen individuals of population A1. a. low variation b. moderate variation c. high variation ANSWER: a 63. There is a volcanic island with a single population of mice that covers the island. A volcanic eruption occurs and splits the original population in two (A1 and A2). You had collected samples of DNA from the original Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 population A prior to the eruption, as well as the two populations that resulted after the volcanic eruption. Your descendants have fortunately inherited your fascination with rodent genetics. Fifty thousand years after the original volcanic eruption, your descendants return to the island and repeat the same sampling and DNA sequencing strategy.

Compare the DNA sequences presented and select the level of variation expected between two randomly chosen individuals of population A2. a. low variation b. moderate variation c. high variation ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 64. There is a volcanic island with a single population of mice that covers the island. A volcanic eruption occurs and splits the original population in two (A1 and A2). You had collected samples of DNA from the original population A prior to the eruption, as well as the two populations that resulted after the volcanic eruption. Your descendants have fortunately inherited your fascination with rodent genetics. Fifty thousand years after the original volcanic eruption, your descendants return to the island and repeat the same sampling and DNA sequencing strategy.

Compare the DNA sequences presented and give the level of variation expected between a randomly chosen individual from population A1 and a randomly chosen individual of population A2. a. low variation b. moderate variation c. high variation Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 ANSWER: c 65. There is a volcanic island with a single population of mice that covers the island. A volcanic eruption occurs and splits the original population in two (A1 and A2). You had collected samples of DNA from the original population A prior to the eruption, as well as the two populations that resulted after the volcanic eruption. Your descendants have fortunately inherited your fascination with rodent genetics. Fifty thousand years after the original volcanic eruption, your descendants return to the island and repeat the same sampling and DNA sequencing strategy.

What is the most likely mechanism for how the fixed differences between population A1 and population A2 arise? a. genetic drift b. natural selection Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 c. random mating d. mutation ANSWER: a 66. There is a volcanic island with a single population of mice that covers the island. A volcanic eruption occurs and splits the original population in two (A1 and A2). You had collected samples of DNA from the original population A prior to the eruption, as well as the two populations that resulted after the volcanic eruption. Your descendants have fortunately inherited your fascination with rodent genetics. Fifty thousand years after the original volcanic eruption, your descendants return to the island and repeat the same sampling and DNA sequencing strategy.

One hundred thousand years after the original event, more of your descendants return to the island and repeat Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 the sampling process. There are now twice as many fixed differences between populations A1 and A2, relative to what was found in the data reported by the previous group of descendants (data running from t = -1 to t = 50,000). Unlike last time, your descendants find that individuals from population A1 cannot reproduce with individuals from population A2. Compare the DNA sequences presented and select the amount of variation expected between two randomly chosen individuals of population A1. a. low variation b. moderate variation c. high variation ANSWER: a 67. There is a volcanic island with a single population of mice that covers the island. A volcanic eruption occurs and splits the original population in two (A1 and A2). You had collected samples of DNA from the original population A prior to the eruption, as well as the two populations that resulted after the volcanic eruption. Your descendants have fortunately inherited your fascination with rodent genetics. Fifty thousand years after the original volcanic eruption, your descendants return to the island and repeat the same sampling and DNA sequencing strategy.

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Chapter 20

Compare the DNA sequences presented and select the amount of variation expected between two randomly chosen individuals of population A2. a. low variation b. moderate variation c. high variation ANSWER: a 68. There is a volcanic island with a single population of mice that covers the island. A volcanic eruption occurs and splits the original population in two (A1 and A2). You had collected samples of DNA from the original population A prior to the eruption, as well as the two populations that resulted after the volcanic eruption. Your descendants have fortunately inherited your fascination with rodent genetics. Fifty thousand years after the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 original volcanic eruption, your descendants return to the island and repeat the same sampling and DNA sequencing strategy.

Compare the DNA sequences presented and give the amount of variation expected between a randomly chosen individual from population A1 and a randomly chosen individual of population A2. a. low variation b. moderate variation c. high variation ANSWER: c 69. There is a volcanic island with a single population of mice that covers the island. A volcanic eruption occurs and splits the original population in two (A1 and A2). You had collected samples of DNA from the original Copyright Macmillan Learning. Powered by Cognero.

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Consider the effects of time from the first time the populations were sampled to the last time the population was sampled. With the passage of time, have populations A1 and A2 become more or less genetically distinct? a. more genetically distinct b. less genetically distinct ANSWER: a 70. There are three alleles in a population of humans, alleles D1, D2, and D3 with frequencies 0.42, 0.30, and Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 0.28, respectively. Based on this information, is this statement true, false, or cannot be determined? This population is triploid. a. true b. false c. cannot be determined ANSWER: b 71. There are three alleles in a population of humans, alleles D1, D2, and D3 with frequencies 0.42, 0.30, and 0.28, respectively. Based on this information, is this statement true, false, or cannot be determined? Allele D1 is dominant. a. true b. false c. cannot be determined ANSWER: c 72. There are three alleles in a population of humans, alleles D1, D2, and D3 with frequencies 0.42, 0.30, and 0.28, respectively. Based on this information, is this statement true, false, or cannot be determined? Individuals with alleles D2 or D3 should mutate to D1 because its high frequency indicates it is advantageous. a. true b. false c. cannot determine ANSWER: b 73. There are three alleles in a population of humans, alleles D1, D2, and D3 with frequencies 0.42, 0.30, and 0.28, respectively. Based on this information, is this statement true, false, or cannot be determined? If allele D3 goes extinct in the population, the new frequencies of D1 and D2 will be 0.56 and 0.44 respectively. a. true b. false c. cannot be determined ANSWER: c 74. In a study of three proteins, M, H, and K, that have approximately equal length in humans and cows, which shared a common ancestor about 90 million years ago, a researcher finds:

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Chapter 20

Which of the graphs accurately reflects the molecular clock for each of these proteins?

a. graph 1 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 b. graph 2 c. graph 3 d. graph 4 ANSWER: d 75. In a study of three proteins, M, H, and K, that have approximately equal length in humans and cows, which shared a common ancestor about 90 million years ago, a researcher finds:

Which of the choices accurately reflects the intensity of negative selection against mutations that give rise to amino acid changes in the proteins? a. K < M < H b. K > M > H c. K < M > H d. H > M > K ANSWER: b 76. On a hike through the forest, you notice a population of 100 flowering plants that all look identical (phenotypic clones), and they span an area of approximately 100 square meters. You take some representatives from one corner of their distribution back to the lab, sequence their DNA, and conclude that they are also genetic clones. Based on this information, is this statement true or false? This population can evolve. a. true b. false ANSWER: b 77. On a hike through the forest, you notice a population of 100 flowering plants that all look identical (phenotypic clones), and they span an area of approximately 100 square meters. You take some representatives from one corner of their distribution back to the lab, sequence their DNA, and conclude that they are also genetic clones. Based on this information, is this statement true or false? This population can only evolve through genetic drift. a. true b. false ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 78. On a hike through the forest, you notice a population of 100 flowering plants that all look identical (phenotypic clones), and they span an area of approximately 100 square meters. You take some representatives from one corner of their distribution back to the lab, sequence their DNA, and conclude that they are also genetic clones. Based on this information, is this statement true or false? This population can only evolve through natural selection. a. true b. false ANSWER: b 79. On a hike through the forest, you notice a population of 100 flowering plants that all look identical (phenotypic clones), and they span an area of approximately 100 square meters. You take some representatives from one corner of their distribution back to the lab, sequence their DNA, and conclude that they are also genetic clones. You go for another hike in the forest and wind up back at the same population of 100 flowering plants. You notice that one individual of the population grows in full shade, and all the others grow in at least partial sun. You take a leaf sample from the individual in the shade, sequence its DNA, and notice that the individual is heterozygous. There is a mutation in the LLT gene (the mutant is designated LLT-s). Based on this new information, is this statement true or false? This population can evolve. a. true b. false ANSWER: a 80. On a hike through the forest, you notice a population of 100 flowering plants that all look identical (phenotypic clones), and they span an area of approximately 100 square meters. You take some representatives from one corner of their distribution back to the lab, sequence their DNA, and conclude that they are also genetic clones. You go for another hike in the forest and wind up back at the same population of 100 flowering plants. You notice that one individual of the population grows in full shade, and all the others grow in at least partial sun. You take a leaf sample from the individual in the shade, sequence its DNA, and notice that the individual is heterozygous. There is a mutation in the LLT gene (the mutant is designated LLT-s). Based on this new information, is this statement true or false? This population can only evolve through genetic drift. a. true b. false ANSWER: b 81. On a hike through the forest, you notice a population of 100 flowering plants that all look identical (phenotypic clones), and they span an area of approximately 100 square meters. You take some representatives Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 from one corner of their distribution back to the lab, sequence their DNA, and conclude that they are also genetic clones. You go for another hike in the forest and wind up back at the same population of 100 flowering plants. You notice that one individual of the population grows in full shade, and all the others grow in at least partial sun. You take a leaf sample from the individual in the shade, sequence its DNA, and notice that the individual is heterozygous. There is a mutation in the LLT gene (the mutant is designated LLT-s). Based on this new information, is this statement true or false? This population can only evolve through natural selection. a. true b. false ANSWER: b 82. On a hike through the forest, you notice a population of 100 flowering plants that all look identical (phenotypic clones), and they span an area of approximately 100 square meters. You take some representatives from one corner of their distribution back to the lab, sequence their DNA, and conclude that they are also genetic clones. You go for another hike in the forest and wind up back at the same population of 100 flowering plants. You notice that one individual of the population grows in full shade, and all the others grow in at least partial sun. You take a leaf sample from the individual in the shade, sequence its DNA, and notice that the individual is heterozygous. There is a mutation in the LLT gene (the mutant is designated LLT-s). What is the frequency of the LLT-s allele in the population? a. 1/100 b. 1/200 c. 1/1 (only one plant has it) d. It is not possible to determine from the information provided. ANSWER: b Multiple Response 83. When the conditions of Hardy-Weinberg equilibrium are met: Select all that apply. a. evolution occurs. b. evolution does not occur. c. gene frequencies in the population change over time. d. gene frequencies in the population do not change over time. ANSWER: b, d 84. Using protein gel electrophoresis to measure genetic variation in populations has some drawbacks. Which of the answer choices are problems with the technique? Select all that apply. a. It cannot detect synonymous mutations. That is, differences in DNA sequences that result in the same Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 amino acid sequence cannot be detected with this method. b. It can only detect differences in noncoding DNA. c. It can only be used with proteins that can be stained or otherwise labeled and visualized. d. It cannot detect mutations that do not affect the mobility of a protein within the gel, even if the protein's amino acid sequence is changed. ANSWER: a, c, d 85. In genetics, two individuals are part of the same population if: Select all that apply a. they are the same species. b. they are from different gene pools. c. they have the same phenotype. d. they are in the same geographic area. ANSWER: a, d 86. Why can't we measure genetic variation in a population using observable traits (phenotypes)? Select all that apply. a. Many traits are encoded by multiple genes. b. Phenotypes are not determined by genes. c. The environment can also affect phenotype. d. All traits are encoded by a single gene. ANSWER: a, c 87. A snowshoe hare produces a white coat during the winter, allowing it to better hide from predators. As a result, it has thrived and over time a majority of snowshoe hares in the population also produce white coats in the winter. Which of the statements is/are true? Select all that apply. a. The white-coated hare has higher fitness than other hares that do not change coat color. b. The white-coated hare has a competitive advantage in its environment. c. The new population of hares has adapted via genetic drift. d. The alleles for a white winter coat increased over time. ANSWER: a, b, d 88. From an evolutionary perspective, germ-line mutations are more significant than somatic mutations because: Select all that apply. a. somatic mutations affect only one or a few cells. b. somatic mutations are generally harmful. c. only germ-line mutations will appear in an individual's descendants. d. only germ-line mutations are potentially beneficial. ANSWER: a, c 89. Which of the answer choices are reasons why a population on an island might have less genetic diversity throughout the genome than a population on a nearby mainland? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 20 a. Island populations are more likely to have non-random mating than mainland populations. b. The original colonizers of the island do not contain all of the genetic diversity of the larger mainland population. c. Alleles may have been lost through random chance because not all individuals were able to mate. d. Habitats on the island are the same as the mainland, so natural selection would favor the same alleles. ANSWER: b, c

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Chapter 21 Multiple Choice 1. Hybrid populations such as those seen between towhee species complicate the biological species concept because: a. stable hybrid populations are formed when two different towhee species mate. b. hybrid individuals result from mating between two different species of towhee, but they are sterile. c. hybrid populations are unable to breed with either parent species. d. hybrid populations are in the process of forming a new species. ANSWER: a 2. In some large groups of plants, including dandelions, oaks, and willows, the biological species concept is complicated because the process of _____ allows gene flow to occur between _____ that can be easily distinguished based on appearance. a. hybridization; morphospecies b. hybridization; ecological species c. allopatric speciation; ring species d. polyploidy; evolutionary species e. polyploidy; ring species ANSWER: a 3. A large population of mice (2000 individuals) lived in an area in the desert dominated by small shrubs. When the population size got too high, a small group of 6 individuals left and colonized an area adjacent to the original population's home, but the adjacent area was primarily dominated by trees instead of shrubs. There is no gene flow between the dispersers and the original population. A researcher sequenced the genomes of representative individuals from both populations and found substantial genetic differences between them. Which of the statements would most account for the amount of genetic differences observed? a. Divergence between the two populations was caused solely by drift. b. Divergence between the two groups is high because of gene flow. c. Since the original dispersal event, the small population could have accumulated more mutations than the original population. d. The divergence between populations could only have been caused by stabilizing selection. ANSWER: c 4. Male birds of different species sing species-specific songs to attract mates. Females only mate with males that sing their species-specific song. This is an example of: a. pre-zygotic isolation. b. post-zygotic isolation. c. hybridization. d. temporal isolation. ANSWER: a 5. New species can form through allopatric or sympatric speciation. Which of the mechanisms will act more strongly on populations that are initially separated in allopatry than on those initially separated in sympatry? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 a. disruptive selection b. gene flow c. genetic drift d. reinforcement ANSWER: c 6. Consider the two phylogenies showing the relationship between five related species of moths and five species of related flowers that hold nectar as a reward for their pollinators.

The similar pattern of the phylogenies suggests the characteristics co-speciated with one another. Which of the descriptions is an accurate statement reflecting the figure shown? a. The flowers and moths experienced the same allopatric speciation events during their history. b. The flowers and moths coevolved, so that spur-length and tongue-length each evolved because of changes in the other. c. The flowers and moths cannot co-speciate because co-speciation only occurs in host-parasite relationships. d. This is not an example of co-speciation. Each phylogeny represents the variation present in each population of the flower and the moth. ANSWER: b 7. Biochemical recognition systems within the stigma of a flowering plant prevent the pollen from a different species from developing a pollen tube and fertilizing its eggs. This is an example of what kind of reproductive isolation? a. pre-zygotic, behavioral isolation b. pre-zygotic, temporal separation c. pre-zygotic, mechanical isolation d. post-zygotic, temporal separation Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 e. post-zygotic, ecological separation ANSWER: c 8. When taxonomists recognize subspecies, they are identifying: a. populations that are reproductively isolated from one another. b. populations with restricted enough gene flow to develop some population-specific traits, but that can still interbreed with members of other such populations. c. populations that will inevitably become separate species as reproductive isolation is nearly complete. d. populations with extremely restricted gene flow due to allopatric isolation. e. None of the other answer options is correct. ANSWER: b 9. In general, sympatric speciation requires the action of _____ selection acting against hybrids. a. disruptive b. stabilizing c. directional d. artificial e. ecological ANSWER: a 10. According to recent studies, the human malaria parasite Plasmodium falciparum was introduced to humans by: a. rhesus macaques. b. gorillas. c. chimpanzees. d. orangutans. e. pocket gophers. ANSWER: b 11. When visiting Hawaii on vacation, a tourist notices that there are many different species of birds that look similar to one another, but are distinct in their ecology and in the morphology (beak size and shape) that helps make them successful in different habitats on the island. The tourist supposes they are closely related and result from an adaptive radiation. Which of the statements would provide the strongest evidence they are the result of an adaptive radiation? a. Some, but not all of the species can hybridize. b. Variation in beak length within any of the species is distributed like a bell curve. c. The phylogeny of the birds shows a single common ancestor for the group. d. The species move throughout all areas of the island. ANSWER: c 12. Two individuals of the opposite sex are members of the same species if they: a. look similar. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 b. behave similarly. c. share genetic information. d. can produce fertile offspring. ANSWER: d 13. Two populations are the same species if their members can potentially breed with each other and produce fertile offspring. a. true b. false ANSWER: a 14. Unlike the biological species concept, the morphospecies concept relies on: a. physiology. b. phenotype. c. behavior. d. offspring. ANSWER: b 15. Which of the statements reflects one way that two different species can produce hybrid offspring? a. They are post-zygotically isolated, but occasionally will mate anyway. b. The two species are probably hybrids as well, so reproduction can occur between the two. c. The amount of divergence between the two species allows for production of offspring, but they are sterile. d. They are distantly related, but occupy similar habitats. ANSWER: c 16. Examine the figure shown. Which of the statements is the most accurate in reflecting levels of genetic divergence between different species of lice?

a. G. cherriei and G. costaricensis will have accumulated the most genetic differences between them relative to the other Geomydoecus species because they share the most recent common ancestor. b. G. cherriei and G. costaricensis will have accumulated the fewest genetic differences between them relative to the other Geomydoecus species because they share the most recent common ancestor. c. G. cherriei and G. costaricensis will have the same number of genetic differences as O. cherrieri and O. heterodus because they coevolved together. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 d. G. cherriei and G. costaricensis will have the same number of genetic differences from G. chapini because speciation could not occur in the lice without genetic differences. ANSWER: b 17. The longer two species have been separated, the greater the number of genetic differences between them. a. true b. false ANSWER: a 18. Two species of frog mate in the same pond. One breeds in early summer and one in late summer. This is an example of what kind of reproductive isolation? a. pre-zygotic, behavioral isolation b. pre-zygotic, temporal separation c. pre-zygotic, ecological separation d. post-zygotic, temporal separation e. post-zygotic, ecological separation ANSWER: b 19. Kingfishers in New Guinea are separated into distinct subspecies. Which two subspecies would you expect to be most genetically different from one another? a. the subspecies that are furthest away from one another in distance b. the subspecies that diverged from one another first c. the subspecies that are most closely related to the mainland species of kingfisher d. the subspecies that live in the most ecologically different habitats ANSWER: b 20. Two species of antelope ground squirrels are separated by the Grand Canyon. They are hypothesized to have descended from a common ancestor population that was divided when the Grand Canyon formed. If this hypothesis is correct, it would be an example of: a. allopatric speciation by dispersal. b. allopatric speciation by vicariance. c. peripatric speciation by vicariance. d. sympatric speciation by dispersal. e. sympatric speciation by vicariance. ANSWER: b 21. Sympatric speciation: a. can occur instantaneously through formation of polyploid offspring. b. occurs whenever a physical barrier between two species is removed (i.e., a river dries up) and the two species start to interbreed. c. whenever hybrids mate with parental species. d. only occurs in bacteria. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 ANSWER: a 22. Speciation always involves geographical separation of some kind between the two organisms. a. true b. false ANSWER: b 23. The adaptive radiation of a particular species may cause adaptive evolution in another. a. true b. false ANSWER: a 24. You are studying a species of frog in which individuals from two successive generations can no longer interbreed, suggesting that the offspring are a different species from the parents. If speciation occurred because of changes in chromosome counts between the parents and the offspring, would this be an example of speciation by natural selection? a. Yes, because all speciation events involve natural selection. b. Yes, because there is genetic variation present due to differences in chromosome numbers. c. Yes, because natural selection is involved whenever there is a difference in the genome size between the parents (species 1) and the offspring (species 2). d. No. Speciation occurred, but natural selection was not involved. e. No. Natural selection can only act on populations, not on individuals. ANSWER: d 25. Identical population-splitting events occurred simultaneously on two neighboring islands, Islands B and C, which are each initially home to different populations of rodent. The population-splitting event produced two subpopulations of rodent on each island: B1 and B2, and C1 and C2. We find that populations B1 and B2 cannot successfully interbreed with each other after 50,000 years, and that C1 and C2 can still interbreed after 100,000 years. Why does reproductive isolation arrive at different rates on these two islands? a. The islands are of different geologic age, and that affects rates of reproductive isolation. b. Mutation is a random process, so the number of accumulated mutations on each island will be different. c. There is still dispersal to each of these islands from the other island. d. None of these answer choices are correct. ANSWER: b 26. A bacteriologist is studying two asexually reproducing strains of E. coli. The two require different amounts of trace minerals for survival; on these grounds, the bacteriologist determines that they are separate species. In making his determination, he is using which species concept or concepts? a. biological species concept b. ecological species concept c. morphospecies concept d. phylogenetic species concept Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 ANSWER: b 27. Ring species are a series of related populations that occur in the shape of a ring. Adjacent populations can hybridize, but populations at the ends of the ring cannot. The figure shows a ring species complex for seven populations of salamander. The seven populations are arranged around a low valley that the salamanders cannot cross. E.e. crocreator has a very specific mating "dance" that E.e. klauberi does not recognize, so they do not hybridize. This is an example of:

a. pre-zygotic isolation. b. post-zygotic isolation. c. ecological isolation. d. temporal isolation. ANSWER: a 28. The figure shows an example of a ring species complex for seven populations of salamander. The seven populations are arranged around a low valley that the salamanders cannot cross. The ring is created from an original population that experienced sequential peripatric speciation events that resulted in related species forming a ring in regions of suitable habitat. You can assume the ring begins with E.e.escholtzii.

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Chapter 21

Which of the statements best reflects the spread of alleles between E.e. crocreatur and E.e. klauberi? a. These two populations do not share any alleles because they are geographically separated from one another. b. These two populations share alleles, but alleles in E.e. crocreatur that are present in E.e. klauberi arrived in that population by travelling through the "ring" of other species. c. These two subspecies probably arose through allopatric speciation when the valley in the center divided an original population. d. The original population probably covered the entire distribution shown in the figure, but when the valley formed the different populations speciated. ANSWER: b 29. A drawback of the biological species concept is that it cannot be applied to: a. populations of a single species living in different places. b. plants, which do not physically come in contact with each other when transferring gametes the way that animals do. c. highly polymorphic species. d. extinct and asexual organisms. e. All of these choices are correct. ANSWER: d 30. Flower A blooms in June, while flower B, a close relative of A, blooms in August. A researcher is raising a population of flower B in a controlled growth chamber. Over time the researcher manipulated the day length in the growth chamber and eventually manages to induce some individuals of flower B to bloom in June. The Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 researcher finds that she can cross these individuals with flower A individuals and they produce viable, fertile offspring. What can she conclude about flower A and flower B? a. A and B are not a good example of separate species. b. A and B are genetically incompatible. c. A and B are temporally reproductively isolated from each other. d. The fact that B's blooming time can be manipulated suggests it should not be regarded as a species distinct from A. e. A and B are not pre-zygotically isolated from one another. ANSWER: c 31. The graph shown illustrates that the extent of the adaptive radiation of the Galápagos finches is correlated with the number of islands present in the archipelago. What can be concluded from the graph?

a. Rates of speciation are correlated with the net total land area available. b. Each island hosts sympatrically speciating populations. c. Opportunities for geographic isolation are a key component of the speciation process. d. Allopatric speciation tends not to occur on islands. e. The balance between rates of sympatric and allopatric speciation is governed by mutation rates. ANSWER: c 32. When we look at islands across the planet (think about the islands of the Galápagos and Hawaii) we see a remarkable number of island endemic species - species that are found nowhere else. What is a possible explanation for this observation? a. Extinction rates are lower in islands so species that have gone extinct elsewhere persist on islands. b. Sympatric speciation on islands is easier than in mainland environments because rates of chromosomal evolution are higher on islands. c. Sympatric speciation on islands is easier than in mainland environments because levels of disruptive selection are higher on islands. d. Island isolation promotes allopatric speciation. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 e. Vicariance events are more common on islands than in mainland environments. ANSWER: d 33. A single species of fish, the three-spined stickleback, once inhabited a large lake. At some point in the lake's history, lava flowed from a nearby volcano into the lake cutting it into three completely isolated mini-lakes. Despite the heat from the lava, a few individuals from the original population of stickleback survived in each mini-lake. Three million years later, a researcher finds that each mini-lake is inhabited by its own species of stickleback. Which of the figures most closely reflects how the three species of spiny back are related to one another?

a. tree M b. tree H c. tree K d. tree L ANSWER: b 34. An ancestral population is split into two different groups, and they no longer interbreed after separation. Which graph correctly illustrates the number of genetic differences that would be seen between the two populations?

a. graph M b. graph H Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 c. graph K ANSWER: b 35. Several species of fireflies are active on the same summer nights in the same fields. Males and females of each species recognize one another by their distinctive flashing patterns. This is an example of what kind of reproductive isolation? a. pre-zygotic, behavioral isolation b. pre-zygotic, temporal separation c. pre-zygotic, ecological separation d. post-zygotic, mechanical incompatibility e. post-zygotic, ecological separation ANSWER: a 36. Horses and donkeys can interbreed, but their offspring (mules) are infertile. This is an example of what kind of reproductive isolating mechanism? a. pre-zygotic, behavioral isolation b. pre-zygotic, temporal separation c. post-zygotic, behavioral isolation d. post-zygotic, genetic incompatibility e. post-zygotic, ecological separation ANSWER: d 37. Imagine a very deep lake with multiple species of fish. The lake has many different regions (i.e., shallow water with high light penetrance, deep water with low light penetrance). A phylogeny of the fish shows that they have all descended from a common ancestor. Which type of speciation is the most likely cause of speciation in fishes in this lake? a. allopatric speciation b. peripatric speciation c. sympatric speciation ANSWER: c 38. Members of two species that are each other's closest relatives are found living sympatrically on a single island. What, if anything, can we conclude about their origins? a. One species originated sympatrically from the other. b. Both species originated sympatrically from a third species, which has gone extinct. c. Neither species originated sympatrically. d. One species originated allopatrically on the island and then gave rise to the other one via sympatric speciation. e. None of the other answer options is correct. ANSWER: e 39. Most speciation is allopatric because: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 a. gene flow between diverging populations is strong in allopatry. b. gene flow between populations prevents genetic divergence in sympatry. c. related species are always found in allopatry rather than in sympatry. d. vicariance events are common. e. disruptive selection between forms of traits in sympatric populations is always very strong. ANSWER: b 40. Disruptive selection is a key component of sympatric speciation because it acts against the homogenizing effect of gene flow between the diverging populations. Imagine a case in which a bird population is undergoing disruptive selection on bill size in response to the availability of seeds to eat. The birds' environment contains large and small seeds, as no medium seeds are available. Assume that a bird's ability to eat a seed is a direct function of its bill size (large bills are good for large seeds, and so on). Disruptive selection acts against birds with: a. large bills. b. small bills. c. large bills and birds with small bills. d. medium-sized bills. e. medium bills and birds with large bills. ANSWER: d 41. Different species of fruit flies occupy each of the islands in the Hawaiian island chain, a group of volcanic islands that formed one after the other. One hypothesis for how the different fruit fly species formed is that, after each new island was formed, fruit flies from existing islands colonized it and subsequently diverged. If this hypothesis is correct, it would be an example of: a. allopatric speciation by dispersal. b. allopatric speciation by vicariance. c. peripatric speciation by vicariance. d. sympatric speciation by dispersal. e. sympatric speciation by vicariance. ANSWER: a 42. Imagine a series of small forest patches (A-D) separated by grassland. A species of beetle can only live in forested areas, and they are unable to fly long distances. An original large population of these beetles lives in forest patch A, the largest of the forest patches, and can fly as far as the next forest patch over, patch B) Over time, each patch is successively colonized by a small group of individuals from the nearest population.

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Chapter 21 After many years, each of the habitat patches is occupied by a different species of beetle, each one related to the original species in forest patch A) Speciation of the beetles was the result of _____ speciation. a. sympatric b. allopatric c. peripatric ANSWER: c 43. Xenopus tropicalis and Xenopus laevis are two closely related species of frog. X. laevis has two times the number of chromosomes as X. tropicalis. Biologists assume that the species X. laevis arose through: a. sympatric speciation. b. polyploidy. c. reinforcement. d. coevolution. ANSWER: b 44. Imagine you are looking down from the top of a mountain. As you hike from the highest elevation to the lowest elevation, you find three different species of related mice. None of the species can survive for extended periods of time at any other elevation. The vegetation (available food) changes at each elevation level.

At the highest elevation some of the individuals of species A mate with individuals in species B that have traveled up from the middle elevation. Their offspring are unable to reproduce or survive in either elevation. This is an example of: a. hybrid inviability and infertility. b. pre-zygotic isolation. c. instantaneous speciation. ANSWER: a 45. A researcher finds two closely related species of rodent, A and B, living sympatrically on a coastal peninsula. Geological analysis suggests that higher sea levels in the past have occasionally cut off parts of the peninsula, with the lower parts of the peninsula being under water and higher parts becoming separate islands. Do you think species A and B arose sympatrically or allopatrically? a. They arose sympatrically because they are on the same peninsula. b. They arose allopatrically because they diverged when separated into two islands. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 ANSWER: b 46. The example of speciation via vicariance illustrated in the figure shown focuses on the impact of the formation of the isthmus of Panama on the populations of snapping shrimp living in the shallow waters of the West Caribbean and East Pacific. Would you expect to see a similar pattern of Pacific/Caribbean sister species in other shallow-water organisms?

a. No, because speciation patterns depend on the parent species. b. No, because speciation cannot be predicted. c. Yes, because all species that are shallow-water will be impacted the same way by the rising of the Isthmus. d. Yes, because all parent species eventually will split into two new species. ANSWER: c 47. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with unlimited resources. What process will introduce genetic variation into these populations? a. natural selection b. genetic drift c. recombination d. mutation ANSWER: d 48. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with unlimited resources. After 10,000 generations, you analyze the genome of each population. Do you expect the genomes of each population to be identical after 10,000 generations? a. Yes, because the starting populations were genetically identical. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 b. Yes, because resources were unlimited. c. Yes, because the same mutations will accumulate independently in each population. d. No, because natural selection was not a factor at any point in this experiment. e. No, because mutation and natural selection (once genetic variation is present) were factors in this experiment. ANSWER: e 49. Imagine you are at the top of a mountain looking down. As you hike from the highest elevation to the lowest elevation, you find three different species of related mice. None of them are found at the same elevation, and the vegetation changes at each elevation level.

Using species C as a reference, you find that there are more genetic differences between species A and species C than between species A and species B. Which of the statements would best reflect the mode of speciation in these mice? a. Species C arose from species A later than species B. b. Only genetic drift is responsible for the differences observed. c. Speciation was allopatric or peripatric, but would depend on the number of individuals that dispersed from the original population. ANSWER: c 50. Imagine you are at the top of a mountain looking down. As you hike from the highest elevation to the lowest elevation, you find three different species of related mice. None of them are found at the same elevation, and the vegetation changes at each elevation level.

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Chapter 21

At the highest elevation there are two dominant plant species throughout the habitat that both produce seeds the mice eat. The seeds of both plant species are very soft, but seeds from one species are large and seeds from the other species are small. Some of the mice have large enough jaws to bite the large seeds, but cannot bite them because they have smaller jaws. The plants release their seeds during the mating season for the mice, so mice will mate with other mice that are foraging for seeds of the same plant type. Gradually, two species of mice form, one with larger jaws that only eat larger seeds, and one with smaller jaws that only eat the smaller seeds. The two species are likely the result of which type of speciation? a. peripatric speciation b. allopatric speciation c. sympatric speciation ANSWER: c 51. You've likely heard of polar bears (Ursus maritimus) and grizzly bears (Ursus arctos horribilis). Have you heard of a grolar bear or a pizzly bear? These might be unfamiliar because they are the offspring of matings between polar bears and grizzlies. A handful of these animals have been born in zoos; three have been documented in nature. Due to their rarity, only a few grolars (grizzly dad and polar bear mom) or pizzlys (polar bear dad and grizzly mom) have been studied in detail. Their fur color, head shape, and ear shape are striking intermediates between the phenotypes of their parents. It is thought that polar bears originated from a population of brown bears (Ursus arctos) that became geographically isolated during a glaciation event that occurred about 150,000 years ago (Lindqvist et al., 2010). Traditionally the territories of grizzly bears and polar bears did not overlap, as grizzlies ranged from Alaska to Mexico, whereas polar bears stayed on the annual sea ice over the Arctic continental shelf and within the Arctic archipelagos. Relatively recently (in the last 50 years or so) grizzlies and polar bears have begun to come in contact in the wild. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 References: Lindqvist, C), S. C) Schuster, Y. Sun, S. L. Talbot, J. Qi, A) Ratan, L. P. Tomsho, and L. Kasson, et al. 2010. "Complete mitochondrial genome of a Pleistocene jawbone unveils the origin of polar bear." PNAS 107(11):5053–5057. doi:10.1073/pnas.0914266107. PMC 2841953. PMID 20194737. Given the existence of at least a few hybrid grolars or pizzlies in the Arctic environment, do you define Ursus maritimus and Ursus arctos horribilis as separate species based on the biological species concept? a. These are not separate species because they hybridize. b. These are not separate species because they are not geographically isolated from one another. c. These are separate species because they still retain separate physical appearances from one another. d. These are separate species because the offspring are infertile. ANSWER: c 52. You've likely heard of polar bears (Ursus maritimus) and grizzly bears (Ursus arctos horribilis). Have you heard of a grolar bear or a pizzly bear? These might be unfamiliar because they are the offspring of matings between polar bears and grizzlies. A handful of these animals have been born in zoos; three have been documented in nature. Due to their rarity, only a few grolars (grizzly dad and polar bear mom) or pizzlys (polar bear dad and grizzly mom) have been studied in detail. Their fur color, head shape, and ear shape are striking intermediates between the phenotypes of their parents. It is thought that polar bears originated from a population of brown bears (Ursus arctos) that became geographically isolated during a glaciation event that occurred about 150,000 years ago (Lindqvist et al., 2010). Traditionally the territories of grizzly bears and polar bears did not overlap, as grizzlies ranged from Alaska to Mexico, whereas polar bears stayed on the annual sea ice over the Arctic continental shelf and within the Arctic archipelagos. Relatively recently (in the last 50 years or so) grizzlies and polar bears have begun to come in contact in the wild. References: Lindqvist, C), S. C) Schuster, Y. Sun, S. L. Talbot, J. Qi, A) Ratan, L. P. Tomsho, and L. Kasson, et al. 2010. "Complete mitochondrial genome of a Pleistocene jawbone unveils the origin of polar bear." PNAS 107(11):5053–5057. doi:10.1073/pnas.0914266107. PMC 2841953. PMID 20194737. The origin of polar bears was the result of: a. allopatric speciation. b. sympatric speciation. c. peripatric speciation. ANSWER: a 53. You've likely heard of polar bears (Ursus maritimus) and grizzly bears (Ursus arctos horribilis). Have you heard of a grolar bear or a pizzly bear? These might be unfamiliar because they are the offspring of matings between polar bears and grizzlies. A handful of these animals have been born in zoos; three have been documented in nature. Due to their rarity, only a few grolars (grizzly dad and polar bear mom) or pizzlys (polar bear dad and grizzly mom) have been studied in detail. Their fur color, head shape, and ear shape are striking intermediates between the phenotypes of their parents. It is thought that polar bears originated from a population of brown bears (Ursus arctos) that became Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 geographically isolated during a glaciation event that occurred about 150,000 years ago (Lindqvist et al., 2010). Traditionally the territories of grizzly bears and polar bears did not overlap, as grizzlies ranged from Alaska to Mexico, whereas polar bears stayed on the annual sea ice over the Arctic continental shelf and within the Arctic archipelagos. Relatively recently (in the last 50 years or so) grizzlies and polar bears have begun to come in contact in the wild. References: Lindqvist, C), S. C) Schuster, Y. Sun, S. L. Talbot, J. Qi, A) Ratan, L. P. Tomsho, and L. Kasson, et al. 2010. "Complete mitochondrial genome of a Pleistocene jawbone unveils the origin of polar bear." PNAS 107(11):5053–5057. doi:10.1073/pnas.0914266107. PMC 2841953. PMID 20194737. Polar bears arose from a vicariant event. a. true b. false ANSWER: a 54. You've likely heard of polar bears (Ursus maritimus) and grizzly bears (Ursus arctos horribilis). Have you heard of a grolar bear or a pizzly bear? These might be unfamiliar because they are the offspring of matings between polar bears and grizzlies. A handful of these animals have been born in zoos; three have been documented in nature. Due to their rarity, only a few grolars (grizzly dad and polar bear mom) or pizzlys (polar bear dad and grizzly mom) have been studied in detail. Their fur color, head shape, and ear shape are striking intermediates between the phenotypes of their parents. It is thought that polar bears originated from a population of brown bears (Ursus arctos) that became geographically isolated during a glaciation event that occurred about 150,000 years ago (Lindqvist et al., 2010). Traditionally the territories of grizzly bears and polar bears did not overlap, as grizzlies ranged from Alaska to Mexico, whereas polar bears stayed on the annual sea ice over the Arctic continental shelf and within the Arctic archipelagos. Relatively recently (in the last 50 years or so) grizzlies and polar bears have begun to come in contact in the wild. References: Lindqvist, C), S. C) Schuster, Y. Sun, S. L. Talbot, J. Qi, A) Ratan, L. P. Tomsho, and L. Kasson, et al. 2010. "Complete mitochondrial genome of a Pleistocene jawbone unveils the origin of polar bear." PNAS 107(11):5053–5057. doi:10.1073/pnas.0914266107. PMC 2841953. PMID 20194737. Rarely, hybridization has occurred between polar bears and grizzlies. The presence of grolars and pizzlies: a. suggests that polar and grizzly bears are not different species based on the biological species concept. b. suggests that polar and grizzly bears are not different species because there are no barriers to reproduction. c. suggests that polar and grizzly bears are different species because neither hybrid is like the parent. d. suggests that the time since divergence may not be sufficient for complete reproductive isolation to have occurred. e. that another glaciation event is needed for polar bears to become another species. ANSWER: d 55. You've likely heard of polar bears (Ursus maritimus) and grizzly bears (Ursus arctos horribilis). Have you heard of a grolar bear or a pizzly bear? These might be unfamiliar because they are the offspring of matings Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 between polar bears and grizzlies. A handful of these animals have been born in zoos; three have been documented in nature. Due to their rarity, only a few grolars (grizzly dad and polar bear mom) or pizzlys (polar bear dad and grizzly mom) have been studied in detail. Their fur color, head shape, and ear shape are striking intermediates between the phenotypes of their parents. It is thought that polar bears originated from a population of brown bears (Ursus arctos) that became geographically isolated during a glaciation event that occurred about 150,000 years ago (Lindqvist et al., 2010). Traditionally the territories of grizzly bears and polar bears did not overlap, as grizzlies ranged from Alaska to Mexico, whereas polar bears stayed on the annual sea ice over the Arctic continental shelf and within the Arctic archipelagos. Relatively recently (in the last 50 years or so) grizzlies and polar bears have begun to come in contact in the wild. References: Lindqvist, C), S. C) Schuster, Y. Sun, S. L. Talbot, J. Qi, A) Ratan, L. P. Tomsho, and L. Kasson, et al. 2010. "Complete mitochondrial genome of a Pleistocene jawbone unveils the origin of polar bear." PNAS 107(11):5053–5057. doi:10.1073/pnas.0914266107. PMC 2841953. PMID 20194737. Based on the biological species concept, what additional information would assist you in determining whether or not polar bears and grizzlies are different species? I. whether or not the F1 is fertile II. whether or not F1 offspring are viable III. the amount of DNA sequence divergence between polar and grizzly bears IV. the geographic distribution of polar and grizzly bears a. I only b. II only c. I and II only d. I and III only e. III and IV only ANSWER: c 56. Gus and Ida are two polar bears, Ursus maritimus, that live at the Central Park Zoo in New York City. They are first-generation zoo polar bears, meaning they were caught in the wild and brought to the zoo. They have lived at the zoo since 1985 and are kept in their own area so that they are reproductively isolated from all other polar bears in the zoo. _____________ is the primary force responsible for changes in allelic frequency of deleterious mutations. _______________ is the primary force responsible for changes in allelic frequency of neutral mutations. a. Genetic drift; Natural selection b. Natural selection; Genetic drift ANSWER: b 57. Gus and Ida are two polar bears, Ursus maritimus, that live at the Central Park Zoo in New York City. They are first-generation zoo polar bears, meaning they were caught in the wild and brought to the zoo. They have lived at the zoo since 1985 and are kept in their own area so that they are reproductively isolated from all other polar bears in the zoo. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 The zoo decides to keep all of Gus and Ida's descendants under zoo conditions for the next 30,000 years. They breed only with each other. You visit the zoo in 30,000 years and see their offspring. Do you expect their offspring, which are alive in 30,000 years, to be identical in genotype and phenotype to Gus and Ida? Keep in mind, polar bears start reproducing when they are between 5 and 10 years old, and adult females give birth to two cubs every 2–3 years. a. Yes, because evolution only occurs in nature. b. Yes, because they had a small population to begin with (only two bears), so there was little genetic variation to start. c. No, because mutation would occur occasionally and there are selective pressures that exist in artificial as well as natural environments. d. Maybe, it depends on mutation rates and strength of selective pressures. ANSWER: c 58. Gus and Ida are two polar bears, Ursus maritimus, that live at the Central Park Zoo in New York City. They are first-generation zoo polar bears, meaning they were caught in the wild and brought to the zoo. They have lived at the zoo since 1985 and are kept in their own area so that they are reproductively isolated from all other polar bears in the zoo. Consider the possibility that Gus and Ida were released back into the wild in 2011, in an area that already has a population of polar bears. You observe them for many years and note that Gus and Ida only breed with each other. Does this prove that they are members of a different species than the wild polar bears? a. yes b. no c. The answer cannot be determined from the information provided. ANSWER: b 59. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with sufficient resources for the survival of all individuals. Is this statement true or false? Natural selection begins immediately operating on the populations. a. true b. false ANSWER: b 60. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with sufficient resources for the survival of all individuals. Is this statement true or false? Initial differences in the populations will be adaptations to the new environment. a. true b. false ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 61. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with sufficient resources for the survival of all individuals. Is this statement true or false? Initial differences in the populations will be the result of sexual selection. a. true b. false ANSWER: b 62. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with sufficient resources for the survival of all individuals. Is this statement true or false? Initially mutation will be the only source of variation in the population. a. true b. false ANSWER: a 63. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with sufficient resources for the survival of all individuals. Is this statement true or false? Speciation cannot occur at the start of the experiment because the two populations are identical. a. true b. false ANSWER: a 64. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with sufficient resources for the survival of all individuals. Is this statement true or false? After 100 generations both populations will still be genetically the same, having accumulated the same mutations over time. a. true b. false ANSWER: b 65. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with sufficient resources for the survival of all individuals. Is this statement true or false? Natural selection cannot occur because bacteria are haploid. a. true b. false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 ANSWER: b 66. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with sufficient resources for the survival of all individuals. Is this statement true or false? Speciation can only occur after there is enough variation in the population for natural selection to act upon. a. true b. false ANSWER: b 67. You begin an experiment with two populations of E. coli that are each composed of 100 cells. The cells are all genetically identical (i.e., they are clones). You grow these populations in flasks on a lab bench under identical conditions with sufficient resources for the survival of all individuals. Is this statement true or false? Speciation can occur between these populations through genetic drift alone. a. true b. false ANSWER: b 68. Examine the figure shown. The rectangles represent the same area before and after a mine is dug. A population of worms living in the soil is evenly distributed throughout the area. After the mine is introduced, the soil near the mine becomes contaminated with high concentrations of copper. There was an initial drop in population size after the contamination, but a few years after the mine was introduced, equal numbers of worms exist again in both soil types. The worms appear identical superficially, and they do not move between the different soil types.

Natural selection is likely acting on the individuals in the high copper soils. a. true b. false c. The answer cannot be determined from the information provided. ANSWER: a 69. Examine the figure shown. The rectangles represent the same area before and after a mine is dug. A population of worms living in the soil is evenly distributed throughout the area. After the mine is introduced, the soil near the mine becomes contaminated with high concentrations of copper. There was an initial drop in Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 population size after the contamination, but a few years after the mine was introduced, equal numbers of worms exist again in both soil types. The worms appear identical superficially, and they do not move between the different soil types.

Speciation was instantaneous with the conversion to high copper concentrations for soil near the mine. a. true b. false c. The answer cannot be determined from the information provided. ANSWER: b 70. Examine the figure shown. The rectangles represent the same area before and after a mine is dug. A population of worms living in the soil is evenly distributed throughout the area. After the mine is introduced, the soil near the mine becomes contaminated with high concentrations of copper. There was an initial drop in population size after the contamination, but a few years after the mine was introduced, equal numbers of worms exist again in both soil types. The worms appear identical superficially, and they do not move between the different soil types.

Genetic divergence between individuals in the two soil types began to rise as copper levels increased. a. true b. false c. The answer cannot be determined from the information provided. ANSWER: b 71. Examine the figure shown. The rectangles represent the same area before and after a mine is dug. A population of worms living in the soil is evenly distributed throughout the area. After the mine is introduced, the soil near the mine becomes contaminated with high concentrations of copper. There was an initial drop in population size after the contamination, but a few years after the mine was introduced, equal numbers of worms exist again in both soil types. The worms appear identical superficially, and they do not move between the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 different soil types.

If using the morphospecies concept, a researcher given a worm from each region of soil would call them the same species. a. true b. false c. The answer cannot be determined from the information provided. ANSWER: a 72. Examine the figure shown. The rectangles represent the same area before and after a mine is dug. A population of worms living in the soil is evenly distributed throughout the area. After the mine is introduced, the soil near the mine becomes contaminated with high concentrations of copper. There was an initial drop in population size after the contamination, but a few years after the mine was introduced, equal numbers of worms exist again in both soil types. The worms appear identical superficially, and they do not move between the different soil types.

Now assume that some of the worms are able to move back and forth between soil types. The two populations are able to successfully interbreed on either side. Those individuals that are able to move between soil type would enhance the process of speciation between the two populations because they would be introducing new alleles from each of the populations into each population. a. true b. false c. The answer cannot be determined from the information provided. ANSWER: b Multiple Response 73. Which of the statements are true regarding hybridization? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 21 a. Hybridization occurs in plants more often than in animals. b. Hybridization involves the transfer of genetic material between members of similar species. c. Natural selection sometimes acts against progeny that result from hybridization. d. Hybridization can result in fertile offspring. e. Hybridization cannot result in the formation of new species. ANSWER: a, b, c, d 74. When determining if asexual organisms are part of the same or different species, it is useful to use the: Select all that apply. a. morphospecies concept. b. ecological species concept. c. phylogenetic species concept. d. biological species concept. ANSWER: a, b, c 75. A paleontologist is studying a group of fossils from its first appearance in the fossil record to its extinction. She determines that the group descends from a single common ancestor and did not give rise to any new species. Its appearance is distinct from any other group of fossils. Based on this information, she decides that it is a new species. In coming to this decision, which species concept or concepts is she using? Select all that apply. a. biological species concept b. morphospecies concept c. phylogenetic species concept d. ecological species concept ANSWER: b, c

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Chapter 22 Multiple Choice 1. Refer to the image. The phylogeny shown illustrates that:

a. orangutans and gorillas are more closely related than humans and gorillas. b. humans and gorillas are more closely related than orangutans and gorillas. c. the gorilla is the most recent common ancestor to bonobos, chimps, and humans. d. the orangutan is the most recent common ancestor of all great apes. e. all great apes walk with an upright gait. ANSWER: b 2. The figure shown represents a phylogenetic tree of vertebrate animals.

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Chapter 22

According to this tree, what is the sister group to the tetrapods? a. sauropsids b. mammals c. lungfish d. amphibians e. coelacanths ANSWER: c 3. Examine the figure shown.

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Chapter 22

What is the sister group to mammals? a. sauropsids b. tetrapods c. birds d. crocodiles and alligators e. birds plus crocodiles and alligators ANSWER: a 4. Examine the figure shown.

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Chapter 22

According to the figure, the closest relatives of turtles are: a. lizards and snakes. b. sauropsids. c. tetrapods. d. crocodiles, alligators, and birds. e. hagfish. ANSWER: d 5. Examine the figure shown:

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Chapter 22

According to this figure, the amphibians are a _____ group. a. monophyletic b. paraphyletic c. polyphyletic ANSWER: a 6. Examine the figure shown.

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Chapter 22

The group "fish" includes all vertebrates except the tetrapods. The taxon "fish" is therefore a _____ group. a. monophyletic b. paraphyletic c. polyphyletic ANSWER: b 7. Examine the figure shown.

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Chapter 22

If we define the taxon Actinopterygii to include the ray-finned fish, coelacanths, lungfish, tetrapods, and their most recent common ancestor, it would be a _____ group. a. monophyletic b. paraphyletic c. polyphyletic ANSWER: a 8. Examine the figure shown.

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Chapter 22

If we defined a taxon to include only the coelacanths and lungfish but not their most recent common ancestor, it would be a _____ group. a. monophyletic b. paraphyletic c. polyphyletic ANSWER: c 9. Organisms in the same family will all be from the same genus. a. true b. false ANSWER: b 10. Individual populations cannot be included as separate taxa in a phylogenetic tree. a. true b. false ANSWER: b 11. Examine the table of characters given for four different species of flower. Based on the matrix, which tree represents the most parsimonious explanation of relatedness among these species? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22

Refer to the image.

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Chapter 22

a. phylogeny 1 b. phylogeny 2 c. phylogeny 3 ANSWER: b 12. The same intron was sequenced from five different taxa (A–E) whose evolutionary relationships are disputed. The data matrix given shows six variable sites (1–6, left column) in the DNA sequences obtained from Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 each of the five species. Taxon A serves as the outgroup for this analysis.

Which of the phylogenies explains the relatedness between these taxa with the fewest evolutionary changes? Use the image to answer the question.

a. phylogeny 1 b. phylogeny 2 c. Both phylogenetic trees have the same number of evolutionary changes. ANSWER: b 13. When selecting among multiple possible phylogenetic trees that fit our data, we commonly use the principle of _____, which means we choose the _____ possible hypothesis. In phylogenetic analysis, that means selecting the tree that represents the _____ evolutionary changes or mutations. a. parsimony; simplest; fewest b. parsimony; simplest; most c. parsimony; most; likely d. phylogenetics; simplest; fewest e. phylogenetics; simplest; most ANSWER: a 14. Examine the table of characters given for four different species of flower. Based on the matrix, which character defines a synapomorphy shared by species A and B?

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Chapter 22

a. presence of sepals b. number of petals c. arrangement of petals d. number of carpals ANSWER: b 15. Examine the table of characters given for four different species of flower. Based on the character matrix, which is most likely an ancestral trait for the group of species?

a. presence of sepals Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 b. number of petals c. arrangement of petals d. number of carpals ANSWER: c 16. Streamlined bodies are common in many aquatic organisms. Dolphins, tuna, penguins, and sharks are all organisms with streamlined bodies that reduce friction and drag. Body shape in each of these organisms would be considered: a. an analogous character. b. a homologous character. c. a similar character. d. a natural character. ANSWER: a 17. You find a fossil that you think is about 350 million years old. You decide to use 235U to date a volcanic ash bed just below the key specimen. The half-life of 235U is 700 million years. If you are correct, what approximate percentage of the original 235U would you expect to remain in your sample? a. 0 percent b. 25 percent c. 50 percent d. 75 percent e. 100 percent ANSWER: d 18. Paleontologists did not just stumble across Tiktaalik. Instead, they formed a hypothesis about where they would find a transitional form between fish and tetrapods. What might have been the logical first step in deciding where to look for a transitional form between fish and tetrapods? a. They considered regions of the world where fossilization is likely to occur. b. They looked for sedimentary rocks that were formed in shallow waters 370-380 million years ago. c. They looked for sedimentary rocks that were formed in deep waters 370-380 million years ago. d. They looked for sedimentary rocks that also contained fossils of the first land plants. ANSWER: b 19. The physical features and chemical composition of rocks that contain fossils provide information about the: a. environment in which the fossil organisms lived. b. phylogenetic relationships among the fossils present. c. manner of the organisms' death. d. other organisms that were present at the time that were not captured in the fossil record. ANSWER: a 20. Examine the figure shown. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22

According to the figure, a mass extinction occurred at or near the end of the _____ Period. a. Ordovician b. Devonian c. Permian d. Triassic e. All of these choices are correct. ANSWER: e 21. The extinction of the _____ during the end-Cretaceous mass extinction allowed mammals to diversify through the process of _____. a. ammonites; peripatric speciation b. dinosaurs; peripatric speciation c. dinosaurs; adaptive radiation d. ammonites; adaptive radiation e. cephalopods; adaptive radiation ANSWER: c 22. How can fossils provide evidence for macroevolutionary processes, such as the divergence of two species Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 from a common ancestor? a. by providing a complete record of the history of life b. by exhibiting some features of ancestral organisms and some features of more derived organisms c. by preserving the bones rather than the soft parts of ancient organisms d. by preserving a large number of organisms present in one place and one time, as is seen in Messel Shale ANSWER: b 23. How do mass extinction events influence subsequent species composition and diversity? a. by immediately increasing species diversity b. by permanently altering climate c. by reducing competition among surviving organisms d. by eliminating smaller organisms instead of larger organisms ANSWER: c 24. Why don't researchers use 14C to date fossils in rocks from the Cambrian Period? a. Cambrian fossils do not contain carbon. b. 14C only formed in more recent geologic periods. c. 14C cannot provide accurate dates. d. The amount of 14C remaining in Cambrian specimens is far too small to be measured accurately. ANSWER: d 25. Which of the answer choices best describes the purpose of phylogenetics? a. Phylogenetics names species, genus, order, class, phylum, and kingdom. b. Phylogenetics looks for patterns of relatedness. c. Phylogenetics compares anatomical or molecular features. d. Phylogenetics looks for patterns of relatedness and compares anatomical or molecular features. ANSWER: d 26. Which phylogenetic group includes all descendants of a common ancestor and only the descendants of that ancestor? a. monophyletic b. paraphyletic c. polyphyletic d. genus ANSWER: a 27. Based on the diagram shown, you predict that the earliest fossil gorilla (which may not resemble modernday gorillas) would be _____ than the earliest fossil _____.

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Chapter 22

a. older; bonobo b. younger; bonobo c. younger; chimp d. older; orangutan e. None of the other answer options is correct. ANSWER: a 28. Which of the statements is always true about sister taxa on a phylogenetic tree? a. Sister taxa are always the result of speciation events that result in two new genera. b. Sister taxa always share a most recent common ancestor that is not shared with any other taxon on the phylogeny. c. Sister taxa are always the result of the most recent divergence event represented on a phylogeny. d. Sister taxa are always defined by shared ancestral characteristics that have been modified in all other taxa in the phylogeny. ANSWER: b 29. Traditional levels of taxonomy are nested in categories from least to most inclusive. This same relationship is also represented on a phylogenetic tree by moving from the terminal (most recent) nodes to the earliest nodes. a. true b. false ANSWER: a 30. A phylogenetic tree is a: a. hypothesis about the evolutionary history of species. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 b. guess about the evolutionary history of species. c. factual representation of the evolutionary history of species. d. detailed timeline of a species' evolutionary history. e. None of the other answer options is correct. ANSWER: a 31. A taxon that includes a single common ancestor and some but not all of its descendants is a _____ group. a. monophyletic b. paraphyletic c. polyphyletic ANSWER: b 32. Characters that are similar because of descent from a common ancestor are: a. homologous. b. analogous. c. examples of convergent evolution. d. synapomorphies. ANSWER: a 33. When comparing trees representing alternate hypotheses of evolutionary relationships among a group of animals, the tree with _____ changes would be the preferred candidate. a. the fewest b. three c. four d. the most ANSWER: a 34. Which molecular detail can be used to construct phylogenetic histories? a. individual nucleotides b. amino acids c. RNA d. All of these choices are correct. ANSWER: d 35. To conduct a phylogenetic analysis, an outgroup is needed in order to: a. decide which characters are analogous and which are homologous. b. determine which character states are ancestral and which are derived. c. determine which characteristics to include in our analysis. d. decide which molecular data to use. e. All of these choices are correct. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 36. A catastrophic drop in diversity is known as a: a. mass extinction. b. reverse evolution. c. transition period. d. climate change. ANSWER: a 37. Radiocarbon dating was an important tool used to determine the placement of Tiktaalik as an intermediary form between fish and tetrapods. a. true b. false ANSWER: b 38. Why is the fossil record of marine life more complete than that of organisms living in terrestrial ecosystems? a. Marine habitats are places where sedimentation is more likely than erosion. b. Organisms that live in marine environments don't have bones or other hard body parts. c. Organisms that live in terrestrial habitats are evolutionarily too old to fossilize. d. Fossilization cannot occur without water acting to preserve body parts from decomposition. ANSWER: a 39. You have found fossils of a new species of ape that walks upright. After 11 additional years in the field, you discover a fossil of what you think is the common ancestor of the new species of ape and all other gorillas. What types of data would give you the best information on the environment of this ancestral organism? a. gene sequencing of the new species of ape b. anatomical features of the new species of ape c. behavior of the new species of ape d. fossils of plants and animals found with the fossilized common ancestor e. anatomical features of the fossilized common ancestor ANSWER: d 40. Some recent discoveries of fossils surprisingly retain some coloration of feathers and skin. This is the result of the preservation of: a. proteins. b. DNA in the nucleus of a cell. c. lipids such as cholesterol. d. pigment molecules. ANSWER: d 41. Fossils' contributions to phylogenetic trees include: a. time calibration. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 b. records of extinct species. c. correlation between evolution and Earth's environmental history. d. All of these choices are correct. ANSWER: d 42. Which factor increases an organism's chance of becoming a fossil? a. being buried soon after death b. having soft body features c. being in an environment that facilitates decay d. All organisms have an equal chance of being fossilized. ANSWER: a 43. Mass extinctions: a. represent the loss of many species in a short time. b. allow surviving species to proliferate. c. can be reconstructed from fossil records. d. All of these choices are correct. ANSWER: d 44. 14C dating is most useful in determining the age of: a. bone and wood in the tombs of Egyptian pharaohs. b. samples older than 60,000 years. c. samples that originally contained only small amounts of carbon. d. All of these choices are correct. ANSWER: a 45. A population of rodents, called population A, lived on a large landmass. One group of the population dispersed to a nearby island. Two million years later, the island population split into two smaller, equal-sized populations when a river formed across the middle of the island. Now two new species, A1 and A2, have evolved on the island. They have replaced the population from which they were derived. Use the image to answer the question. Which diagram represents the phylogeny of the populations discussed in this scenario?

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Chapter 22

a. diagram M b. diagram H c. diagram K ANSWER: c 46. In a phylogenetic tree, a node or branching point represents: a. the common ancestor from which the descendent species diverged. b. the species in the fossil record from which the descendent species diverged. c. one of the descendent species in the phylogeny. d. the ancestral species from which all species in the phylogeny arose. e. None of the other answer options is correct. ANSWER: a 47. The diagram shown depicts three phylogenetic trees. Which of the three show the same sister group relationships among groups A, B, and C?

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Chapter 22

a. Trees 1 and 2 are equivalent. b. Trees 2 and 3 are equivalent. c. Trees 1 and 3 are equivalent. d. All three trees are equivalent. e. All three trees are different; no two are equivalent. ANSWER: a 48. A taxon that does NOT include the last common ancestor of all its members is a _____ group. a. monophyletic b. paraphyletic c. polyphyletic ANSWER: c 49. Consider the image. What types of data can be used to construct phylogenies such as the one shown?

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Chapter 22 species e. None of the other answer options is correct. ANSWER: d 50. Consider the image. What types of data can be used to construct phylogenies such as the one shown?

a. molecular data b. fossil evidence c. anatomical, physiological, and developmental studies of extant species d. molecular data; fossil evidence; and anatomical, physiological, and developmental studies of extant species e. None of the other answer options is correct. ANSWER: d 51. You discover a new species of ape that is more closely related to gorillas than to any other species of ape, but walks upright. Consider the image. How would you change the phylogeny shown?

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Chapter 22

a. The phylogeny should not be changed at all, because only molecular data are considered when studying evolutionary relationships. b. The phylogeny should not be changed at all until a fossil of the new species is found, because fossil evidence is a required component of the data set. c. Gorillas should be grouped with humans based on the synapomorphy (shared trait) of upright stance. d. The gorilla lineage should be branched and the new species sister added to gorillas. ANSWER: d 52. While taking a hike in the forest, you find some fossils in layers of sedimentary rocks whose age you later find out is said to cover a span of 100–400 million years. You decide to send the fossils out for analysis to a company that dates rocks by radioactive decay, and, some weeks later, receive a report informing you that a volcanic ash bed associated with one of the fossils has a 1:1 ratio of 235U: 207Pb. Do these data support or refute the assumed age of the rocks in which the fossil was found? (Note: The half-life of 235U is about 700 million years, and its decay product is 207Pb.) a. Yes, because the rock layers are less than 704 million years old. b. Yes, because the ratio of 235U to 207Pb represents one half-life, and that would be approximately 350 million years old. c. No, because the ratio of 235U to 207Pb represents one half-life, and that would be about 700 million years old. d. No, because the rock layers are less than 704 million years old. ANSWER: c 53. You are given a fossil and told that approximately 3 percent of the 14C originally in the sample is still there. How old is your sample? (Note: 14C has a half-life of 5730 years.) a. approximately 11,500 years b. approximately 17,000 years Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 c. approximately 23,000 years d. approximately 29,000 years ANSWER: d 54. Characters that are similar because of descent from a common ancestor are _____; characters that are similar due to convergent evolution are _____. a. homologous; analogous b. analogous; homologous ANSWER: a 55. With the advent of tools to incorporate molecular data, phylogenetic analysis can be used to: a. reconstruct the evolutionary history of a group of organisms over millions of years. b. track the spread of a pathogen, such as a fungus or virus, from place to place. c. identify the origin of invasive pest species. d. track shipments of endangered species or their products (such as elephant ivory or bushmeat). e. All of these choices are correct. ANSWER: e 56. Which is a disadvantage of using fossils to reconstruct phylogenetic history? a. Fossils are a physical record of organismal structure. b. Fossils yield information about the timing and order of events in the past. c. Fossils reveal relatedness. d. The probability that an organism will be fossilized varies among species and environments. ANSWER: d 57. Among the organisms listed, which is most likely to be fossilized? a. earthworm b. clam c. house flies d. jellyfish ANSWER: b 58. Why is the fossil record not a complete catalog of biological history? a. Not all organisms fossilize with equal probability. b. Fossilization destroys the structure of DNA. c. The process of fossilization often destroys anatomical features of the organisms being preserved. d. Fossils only preserve organisms for about 10 million years; older organisms are destroyed by geological processes. e. Only animals, not plants, are fossilized. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 59. Among the environments listed, in which are fossils most likely to form? a. desert b. tropical rainforest c. tundra d. shallow lake bed e. fast-flowing creek ANSWER: d 60. Pterosaurs are an extinct group of flying reptiles. Paleontologists accept that pterosaurs evolved flight independently of birds. What evidence best supports such a conclusion? a. Pterosaurs don't have feathers. b. Pterosaurs have teeth, but birds don't. c. Phylogenetic analyses place pterosaurs as the sister group of all dinosaurs. d. Pterosaurs first appeared during the Triassic Period, earlier than the oldest known birds. ANSWER: c 61. In what way have mass extinctions catalyzed evolutionary radiation? a. Mass extinctions wipe out poorly competing species, enabling good competitors to diversify. b. Mass extinctions eliminate ecologically dominant species, enabling survivors to diversify in environments with few competitors. c. Mass extinction events cause mutations, increasing the genetic variability of populations. d. Mass extinction events permanently alter environments, favoring species with novel character combinations. ANSWER: b 62. Lungfish and lizards both have lungs. Can we conclude from this observation that lungfish are the sister group of lizards? a. No, lungs are an ancestral trait present in all tetrapods and lungfish. b. No, lungs evolved convergently in lungfish and lizards. c. Yes, lungs are a synapomorphy that documents the close evolutionary relatedness of lungfish and lizards. d. No, lungfish do not have two pairs of walking legs and so cannot be closely related to any tetrapod animals. ANSWER: a 63. In 2027, an unmanned rover lands on Mars, scoops pieces of ancient sedimentary rock into its interior, and then makes careful chemical analyses. Back on Earth, scientists are astonished to find that the rocks contain traces of RNA, much like that in ribosomes. Some of the scientists conclude that the rover has found evidence of Martian life, whereas others interpret the presence of RNA as evidence that the rover was contaminated by microorganisms on Earth before launch. The statements list four possible outcomes of a molecular phylogenetic test of the two hypotheses. Rank the outcomes in order of increasing support for the hypothesis that life originated independently on Earth and Mars. 1. The RNA found by the rover has no close resemblance to that found in terrestrial organisms. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 2. Molecular phylogenetic analyses place the Martian RNA close to that of the bacterium E. coli. 3. Molecular phylogenetic analyses place the Martian RNA as a sister group to all known RNAs of bacteria, archaea, and eukaryotes. 4. Molecular phylogenetic analyses place the RNA near the bottom of the bacterial branch of the tree of life. a. 1, 2, 3, 4 b. 1, 2, 4, 3 c. 1, 3, 2, 4 d. 1, 3, 4, 2 e. 1, 4, 3, 2 ANSWER: d 64. The probability that an ancient species will be represented in the fossil record is a function of: a. the properties of the organisms themselves, such as whether they made hard skeletons. b. the properties of the environments in which they lived, such as whether burial was likely. c. the properties of the climate in which they lived, such as how warm or cold it was. d. the properties of the organisms themselves and the environments in which they lived. e. All of these choices are correct. ANSWER: d 65. The Burgess Shale preserves a remarkable fossil record of: a. delicate flowers and mushrooms from the early invasion of land plants. b. marine life during initial diversification of animals in the Cambrian Period, 505 million years ago. c. plants, insects, and mammals around 50 million years ago, when the age of mammals began. d. the tracks of dinosaurs walking on land. e. molecular evidence of bacteria and other organisms. ANSWER: b 66. During the end-Permian mass extinction: a. oxygen levels in the deep oceans dropped. b. global warming occurred as a result of volcanic eruptions. c. oceans were acidified as a result of volcanic eruptions. d. All of these choices are correct. ANSWER: d 67. The remarkable fossil Tiktaalik, which lived about 375 million years ago, is a beautiful example of an intermediate form, having the attributes of both fish (scales, fins) and tetrapods (flat head, mobile neck). The presence of Tiktaalik and fossils of other organisms that lived around the same time shows clear evidence that tetrapods were derived from fish, implying that the grouping "fish" is paraphyletic. Imagine now that there was no fossil record. Would it still be possible to determine whether "fish" is paraphyletic? a. Yes, because a phylogeny can be constructed using morphological and molecular characters of modern fish and tetrapods. b. No, because without a fossil record there is no evolutionary history of fish available for study. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 c. Yes, because Tiktaalik is a species of coelacanth, so phylogenetic reconstructions can be based on coelacanth characters. d. No, because modern fish and modern tetrapods share no synapomorphies. ANSWER: a 68. The concordance of the two great patterns in the history of life—the branching order of the tree of life and the sequence of forms in the fossil record—is powerful evidence in support of the theory of evolution. Which of the imaginary examples of evidence would reject the theory of evolution? a. a fossil of a mammal that is older than fossils of the first reptiles b. a fossil of a DNA molecule that is older than a fossil of the first fish c. lack of evidence of a transitional form between fish and tetrapods d. a fossil of a dinosaur footprint in 70-million-year-old rock ANSWER: a 69. While we can use molecular clocks based on molecular sequence data to estimate the times at which various lineages diverged, we ultimately need the _____ to calibrate the clocks. a. fossil record b. character states c. sister groups d. transitional forms e. DNA sequences ANSWER: a 70. The phylogeny shown represents a hypothesis for the evolutionary relationships among major groups of plants. If it is correct, in what order should we predict to find the fossils of these groups, in order from oldest to youngest?

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Chapter 22

a. flower group, conifer group, ferns, moss group b. conifer group, ferns, moss group, flower group c. ferns, moss group, conifer group, flower group d. moss group, ferns, conifer group, flower group e. ferns, conifer group, moss group, flower group ANSWER: d 71. After the end-Permian extinction event, all of the same taxonomic families of organisms were present as before the extinction event, there were just fewer individuals of each taxonomic family. a. true b. false ANSWER: b 72. The development of wings on a moth and a bird are _____ because they developed independently as adaptations to perform the common function of flying. a. analogous b. related Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 c. homologous d. identical ANSWER: a 73. After you have found a new species of ape, you spend 10 years in the field seeking a fossil of its ancestor. You do not find a fossil that can be considered ancestral. This lack of fossil data: a. is expected, because the fossil record is incomplete. b. is expected, because the species is too young to have fossilized ancestors. c. is unexpected, because the fossil record shows a complete record of all vertebrates. d. completely negates the anatomical and genetic data you have collected that indicate your new species is a type of gorilla. ANSWER: a 74. After you have found a new species of ape, you spend 10 years in the field seeking a fossil of its ancestor. What rationale would you have for spending 10 years looking for a fossil ancestor of your new species? a. to calibrate the time scale of your phylogeny based on molecular and anatomical data b. to reconstruct the environment, and possible selective pressures, on these lineages c. to better understand anatomical changes that occurred during the evolution of this lineage d. All of these choices are correct. ANSWER: d Multiple Response 75. Why might a phylogeny based only on molecular data show a different pattern of relationships than a phylogeny of the same taxa based only on morphological traits? Select all that apply. a. Gene sequences always provide more data than morphological traits. b. Morphological analyses always provide more data, because each morphological trait is the result of the expression of many genes. c. The molecular data may be based on the analysis of introns, which aren't expressed and don't contribute to the evolutionary history of a group of taxa. d. Some highly conserved genetic sequences can result in unrelated species appearing closely related in a molecular phylogeny, and do not reflect the same pattern as the morphologic phylogeny. e. Gene sequence changes may not result in morphological changes. ANSWER: d, e 76. What evidence of ancient organisms exists in the fossil record that is not composed of hardened body parts? Select all that apply. a. footprints b. DNA extracted from the oldest fossils c. proteins extracted from ancient fossils provide evidence of their metabolism d. molecular fossils from lipids such as cholesterol Copyright Macmillan Learning. Powered by Cognero.

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Chapter 22 ANSWER: a, d 77. To be useful for phylogenetic reconstruction, a taxonomic character must exhibit which properties? Select all that apply. a. It must vary among the taxa being analyzed, but not within individual taxa. b. It must have a genetic basis. c. It must be anatomical. d. It must be molecular (DNA or protein). e. None of the other answer options is correct. ANSWER: a, b

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Chapter 23 Multiple Choice 1. Examine the figure shown. Which group is monophyletic?

a. humans plus chimpanzees and bonobos b. gorillas plus chimpanzees and bonobos c. gibbons plus Old World monkeys d. gibbons plus orangutans e. gibbons plus orangutans plus gorillas ANSWER: a 2. The fossil Lucy (Australopithecus afarensis) is significant because, at _____ years old, hers was the first hominin species to _____. a. 20 million; migrate out of Africa b. 3.2 million; have hominin brow ridges c. 3.2 million; have been fully bipedal d. 600,000; have hominin brow ridges ANSWER: c 3. _____ was the first hominin to leave Africa and did so around _____ million years ago. a. Homo ergaster (H. erectus); 2 b. Ardipithecus ramidus; 4 c. Australopithecus afarensis; 3.2 d. H. neanderthalensis; 0.6 e. H. habilis; 1.7 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 ANSWER: a 4. Our species, H. sapiens, derived from: a. H. floresiensis. b. H. ergaster. c. H. habilis. d. H. rudolfensis. ANSWER: b 5. Examine the figure shown.

The time of greatest hominin radiation was approximately: a. 6-5 million years ago. b. 5-4 million years ago. c. 4-3 million years ago. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 d. 3-2 million years ago. e. 1 million years ago through the present. ANSWER: d 6. Why do we only see Neanderthal input in non-African genomes? a. Neanderthals never interbred with Homo sapiens in Africa. b. Only mtDNA can be transmitted from Neanderthal genomes. c. Only Y chromosome DNA can be transmitted from Neanderthal genomes. d. Neanderthals were adapted to a cold European climate. ANSWER: a 7. Based on the figure shown, which of the statements is true?

a. Neanderthals are more closely related to the French than to other groups. b. Neanderthals are more closely related to the Chinese than to other groups. c. Neanderthals are more closely related to the Nigerians than to other groups. d. Neanderthals are more closely related to some African groups than others. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 e. Neanderthals are equally related to all non-African groups. ANSWER: e 8. Which of the statements is accurate regarding the multiregional and out-of-Africa hypotheses for the origin of modern humans? a. The multiregional hypothesis predicts a divergence time of approximately 2 million years among African and non-African mitochondrial DNA sequences. b. The out-of-Africa hypothesis is consistent with a divergence time of approximately 1.3 million years among African and non-African mitochondrial DNA sequences. c. The multiregional hypothesis predicts that African samples would form a monophyletic group in a phylogenetic analysis including African and non-African samples. d. The out-of-Africa hypothesis predicts that African samples should form a paraphyletic group in a phylogenetic analysis including African and non-African samples. ANSWER: a 9. According to the multiregional hypothesis of human origins, modern H. sapiens evolved by _____ from _____ of H. ergaster. a. convergent evolution; multiple Old World populations b. adaptive radiation; a single Old World population c. allopatric speciation; multiple Old World populations d. allopatric speciation; a single Old World population e. peripatric speciation; a single Old World population ANSWER: a 10. According to genetic analyses, approximately _____ percent of the genome of every non-African is derived from Neanderthals. a. 0.05-0.01 b. 0.5-1 c. 1-4 d. 5-10 e. 15-25 ANSWER: c 11. Interbreeding between Neanderthals and modern H. sapiens likely took place in _____ around _____ years ago, before H. sapiens spread around the world. a. the Middle East; 60,000 b. Africa; 60,000 c. the Middle East; 200,000 d. Africa; 200,000 e. the Middle East; 1.4 million ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 12. Sequences in mtDNA and the Y chromosome provide an interesting perspective on the early migratory routes of H. sapiens. Both mtDNA and Y chromosome data are useful because these molecules do not undergo recombination. Which of the statements also provides support for their utility in estimating origins of various groups within H. sapiens? a. Mutation rates are lower in mtDNA and the Y chromosome so rates of evolution are slower than rates across the genome of H. sapiens. b. mtDNA and the Y chromosome can give information about interbreeding between different species of the genus Homo. c. mtDNA provides less information on origins of H. sapiens than Y chromosome data because it is only inherited through mothers. d. mtDNA and Y chromosome data do not hold as much data as the genome because they are so small and have a small relative number of base pairs. ANSWER: b 13. Which of the statements regarding the gene FOXP2 is not true? a. FOXP2 encodes a protein that is a member of a large family of evolutionarily conserved transcription factors that play an important role in development. b. Humans with mutations in FOXP2 often have problems with speech and language. c. A molecular clock analysis of the FOXP2 gene in chimpanzees and humans strongly suggests that the evolution of novel function in FOXP2 of humans led to the evolution of language. d. FOXP2 is involved in the development of many tissues, therefore it must have pleiotropic effects (additional functions that are independent of language ability). ANSWER: c 14. Which of the factors contributed to the evolution of distinctive human characteristics relative to great apes? a. alteration of developmental timing so that certain juvenile characteristics are retained in the sexually mature adult b. evolution of bipedalism followed by increasingly sophisticated tool use, cognitive ability, and social organization c. appearance of the FOXP2 gene associated with more sophisticated language d. evolutionary loss of a tail, allowing the adoption of a fully upright and bipedal stance ANSWER: a 15. Neoteny is invoked to help explain: a. how relatively little genetic change could lead to the significant differences we see between ourselves and chimps. b. how our large brains evolved. c. why selection favored relative hairlessness in humans as compared to our great ape-like ancestors. d. skin color variation among human populations. e. None of the other answer options is correct. ANSWER: a 16. Although the FOXP2 amino acid sequence is highly conserved, humans have two unique amino acids that Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 chimpanzees lack. These unique amino acids are highly correlated with differences in _____ seen between chimps and humans. a. speech and language b. sight and language c. sight and speech d. speech and fine motor control of the thumb e. sight and fine motor control of the thumb ANSWER: a 17. Why does hair color vary among humans? a. environment b. genetics c. both environment and genetics d. neither environment nor genetics ANSWER: c 18. Why is there less genetic variation among humans than among other species? a. Humans have a smaller population size than any other species. b. Humans have a larger population size than any other species. c. Human ancestral lineages are relatively old compared with other species. d. Human ancestral lineages are relatively young compared with other species. e. Humans have a much lower genetic mutation rate than any other species. ANSWER: d 19. Why do African populations of humans have higher levels of genetic variation as compared with nonAfrican populations? a. Humans originated in Africa, and subsequent migrations resulted in populations seeded by small groups with relatively small amounts of variation. b. Because African populations show strong patterns of nonrandom mating. c. Numerous migrations to Africa throughout early human history have increased genetic diversity there. d. High levels of UV radiation in equatorial regions of the world have increased levels of genetic mutations in Africa. ANSWER: a 20. Although we must be cautious about oversimplifying, when visible traits are markedly different among human races, a likely explanation is that the differences have been favored by: a. natural selection and/or sexual selection. b. diversifying selection and/or stabilizing selection. c. allopatric speciation and/or sexual selection. d. disruptive selection and/or natural selection. e. sympatric selection and/or sexual selection. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 ANSWER: a 21. Travel around the world has gotten much easier in the past century. It has become very common for immigrants to move to new areas, establish lives in regions other than where they were born, and find spouses in new regions. What, if any, changes to genetic diversity between populations will be the likely result of changes in the mobility of humans? a. The number of differences among human populations will decrease. b. The number of differences among human populations will increase. c. The number of differences among human populations will stay the same. ANSWER: a 22. You inherited your mother's crooked nose and her recipe for chocolate chip cookies (both favorable traits). Which trait will be spread more quickly through the population? a. the nose because it is a genetically based trait b. the recipe because it is a genetically based trait c. the nose because it can be passed on to all relatives d. the recipe because it can be passed on to all relatives e. They are both advantageous traits and spread at the same rate. ANSWER: d 23. Which of the experiments would provide the best data set with which to test the hypothesis that female chimpanzees have a greater influence on the vertical transmission of culture than males? a. Measure the amount of time that young chimps spend with each of their parents and measure the rate at which they acquire cultural traits. b. Measure the amount of time that young chimps spend with their mother only and measure the rate at which they acquire cultural traits. c. Measure the amount of time that young chimps spend with their father only and measure the rate at which they acquire cultural traits. d. Train a wild chimp to use a new food source and document how quickly this knowledge spreads to the rest of the population. e. Introduce a new food source to a population and see which chimps learn how to use it first. ANSWER: a 24. The data in the graph shown compares the abundance of cultural traits with the number of adult female chimpanzees in different chimpanzee communities (Lind & Lindenfors, 2010).

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Chapter 23

The K group and the M group of this data set are from the same geographic area, called Mahale. The number of female chimpanzees in the K and M groups varied by about _____ percent, but the number of cultural traits varied by about _____ percent. a. 100; 20 b. 20; 100 c. 300; 20 d. 20; 300 e. 20; 20 ANSWER: c 25. It is relatively easy to think of examples of regional "culture" in humans. Other species also have their own culture, but it is the same across all populations of the species. a. true b. false ANSWER: b 26. The convergent evolution of lactose tolerance in Europe and Africa is evidence that _____ can drive Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 biological evolution. a. cultural evolution b. mutation c. genetic drift d. recombination ANSWER: a 27. As a human, 99 percent of my _____ is (are) the same as a chimpanzee's, and 60 percent of my _____ is/are the same as a banana's, but only 50 percent of my _____ is/are the same as my biological offspring. a. alleles; alleles; alleles b. alleles; genes; genes c. nucleotide sequence; nucleotide sequence; nucleotide sequence d. nucleotide sequence; genes; alleles e. genes; alleles; nucleotide sequence ANSWER: d 28. Researchers use the denaturation temperatures of hybrid double helices to compare evolutionary relatedness. Complementary strands denature at 95 degrees Celsius. A researcher engineers a hybrid double helix from two species, and it denatures at 93 degrees Celsius. What does this mean? a. The hybrid double helices are more complementary than the non-hybrid. b. The hybrid double helices are less complementary than the non-hybrid. c. The species are more closely related than any other two species. d. The hybrid strands' base pairs had fewer mismatches. ANSWER: b 29. Hominins are defined as: a. members of species where one lineage led to humans and others went extinct. b. members of species having a common ancestor and leading to humans. c. human ancestors that survived to the present day (no human ancestors are extinct). d. None of the other answer options is correct. ANSWER: a 30. Our ancestors were bipedal by at least _____ million years ago, as evidenced by the Lucy skeleton. a. 1.2 b. 2.4 c. 3.2 d. 4.5 ANSWER: c 31. Which was a major trend during the evolution of modern humans? a. body size increase Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 b. cranium size increase c. thumb dexterity increase d. All of these choices are correct. ANSWER: d 32. Which of the properties of mitochondrial DNA is not helpful in inferring the geographic origin and timing of the evolution of modern humans? a. Mitochondrial DNA is uniparentally inherited from the mother. b. Mitochondrial DNA incurs mutations at a constant rate over time. c. In rare cases, some cells contain more than one mitochondrial genotype. d. Mitochondrial DNA in animals does not undergo recombination. e. There are a large number of mitochondria in a typical cell. ANSWER: c 33. If the out-of-Africa hypothesis of human origins is correct, which of the statements best describes the pattern of hominin migration from Africa? a. H. ergaster migrated 2 million years ago; H. sapiens migrated 60,000 years ago. b. H. sapiens migrated once 2 million years ago and again 60,000 years ago. c. H. ergaster migrated once 2 million years ago and again 60,000 years ago. d. H. neanderthalensis migrated once 1.3 million years ago and H. sapiens migrated 60,000 years ago. e. H. heidelbergensis migrated 2 million years ago and H. sapiens migrated 60,000 years ago. ANSWER: a 34. Which of the statements about the evolutionary trends from ape-like ancestors to modern humans is true? a. The evolution of larger brain size facilitated the evolution of bipedal posture. b. The evolution of bipedal posture facilitated long-distance travel, since walking on two legs allows for a greater distance to be covered in a shorter amount of time. c. The evolution of bipedal posture would facilitate tool use by freeing up the hands. d. Many primates are bipedal. ANSWER: c 35. Which characteristic is an example of neoteny? a. relatively little body hair on adult humans b. bipedal locomotion of adult humans c. dyeing graying hair blond d. hair turning gray with age e. opposable thumbs ANSWER: a 36. Complex tool use, spoken language, and complex foraging strategies were all important hallmarks of human evolution. Current thinking is that each of these was a consequence of: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 a. bipedalism and freeing the hands from locomotion. b. increasing body size and release from predation. c. increasing brain size and the development of culture. d. increasing body size and nomadic lifestyle. e. None of the other answer options is correct. ANSWER: a 37. We can conclude that natural selection must have acted in favor of large brains in humans because: a. we have large brains in proportion to our body size compared to other mammals. b. large brains have allowed us to dominate the planet. c. large brains are metabolically expensive to produce and maintain. d. we would have large brains whether or not natural selection favored them. e. None of the other answer options is correct. ANSWER: c 38. Based on King and Wilson's findings of the similarities in the genomes of chimps and humans, as well as genome sequencing studies that have found that all of our genes are present in chimpanzees, what do researchers currently think is driving the most significant evolution along the hominin lineage? a. changes in gene regulation b. increase in gene number c. decrease in gene number d. loss of mitochondrial DNA ANSWER: a 39. Neoteny is the acquisition of sexual maturity in an otherwise _____ state. a. juvenile b. less-evolved c. fully grown d. malnourished ANSWER: a 40. Which two factors are the main causes of differences among humans? a. genetic variation and differences in brain size b. genetic variation and environmental differences c. differences in brain size and environmental differences d. genetic variation and differences in metabolism ANSWER: b 41. What evolutionary force could inhibit development of phenotypic differences between races? a. genetic drift b. natural selection Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 c. sexual selection d. gene flow ANSWER: d 42. Based on comparisons of many human DNA sequences, the amount of DNA variation among individuals in the human population is _____ percent. a. 0.1 b. 0.5 c. 1 d. 10 ANSWER: a 43. When we compare levels of genetic variation in a contemporary African population with that of a contemporary non-African population, we find there is: a. more variation in the African population. b. more variation in the non-African population. c. equal variation in both populations. d. no consistent pattern; sometimes we find more variation in the African population and sometimes we find more in the non-African population. e. None of the other answer options is correct. ANSWER: a 44. When we compare levels of genetic variation in a contemporary African population with that of a contemporary non-African population, we find that there is more variation in the African population. The reason we see this pattern is that: a. African populations experience higher rates of natural selection than do non-African populations. b. non-African populations have high mutation rates. c. non-African populations experience higher rates of natural selection than do African populations. d. non-African populations are the products of founder events. e. African populations experience high rates of genetic drift. ANSWER: d 45. Approximately 85 percent of human genetic variation occurs between individuals within a population (like the Yoruba in West Africa), whereas _____ percent occurs between races. a. 0 b. 7 c. 25 d. 85 ANSWER: b 46. What do you expect to be the relationship between the number of populations and number of cultural traits? a. The larger the number of populations and geographical distribution, the larger the number of cultural Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 traits. b. The smaller the number of populations, the larger the number of cultural traits. c. There is no relationship between number of populations and the number of cultural traits. ANSWER: a 47. Which of the answer choices is a trait that is unique to humans? a. culture b. language c. consciousness d. None of the other answer options is correct. ANSWER: d 48. Recent research found that, in chimpanzees, offspring spend their first 8 years largely with their mother, and it is the mother who typically teaches the offspring to use tools to collect food (Lind & Lindenfors, 2010). These observations suggest that vertical cultural transmission by females in chimpanzees is _____ that by males. a. equal to b. greater than c. less than ANSWER: b 49. Which of the examples or elements of culture is uniquely human? a. an individual learning by imitation from another member of its own species b. an individual teaching another individual of its own species by tailoring the information available to the individual being taught c. regional variation in tool use d. language e. None of the other answer options is correct. ANSWER: e 50. Cultural evolution and biological evolution: a. always evolve separately. b. always evolve together. c. can sometimes affect each other. d. work on different factors, as cultural evolution is not quantifiable. ANSWER: c 51. Gibbons and orangutans are from Asia. Gorillas and chimpanzees are from Africa. How was Charles Darwin able to hypothesize that humans evolved in Africa? a. He understood that humans and great apes were closely related. b. He understood that humans and gorillas/chimpanzees are more closely related than humans and orangutans, based on anatomical criteria. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 c. He understood that humans and orangutans are more closely related than humans and gorillas, based on anatomical criteria. ANSWER: b 52. Consider the diagram shown.

The divergence in the tree seen at 30 million years ago represents the: a. split between New and Old World monkeys. b. split between Old World monkeys and apes. c. divergence between chimpanzees and humans. d. speciation event caused by a split in the common ancestor. ANSWER: b 53. Consider the diagram shown.

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Chapter 23

Which of the answer choices represents a characteristic that is useful in separating apes from Old World monkeys? a. loss of tail b. closely spaced nostrils c. prehensile thumb d. All of these choices are correct. ANSWER: a 54. The table shows the melting temperatures of DNA helices created by hybridizing the DNA from four species of ape:

Which of the phylogenies correctly reflects the relationships among species M, H, K, and L, based on the melting temperatures of their hybrid DNA? Copyright Macmillan Learning. Powered by Cognero.

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a. phylogeny 1 b. phylogeny 2 c. phylogeny 3 d. phylogeny 4 ANSWER: b 55. Consider the figure showing the arrangement of homologous genes in several great apes relative to human chromosome 2. Red and blue lines between chromosomes connect homologous regions. Do these data reject or support the sister group relationship between humans and chimps?

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a. These data reject the sister taxa relationship, because humans only have a single chromosome and all other great apes have two; these data cannot be used to infer relatedness. b. These data reject the sister taxa relationship, because all of the genes that are connected by the blue lines are the same in all great apes. c. These data support the sister taxa relationship, because all genes indicated by red and blue lines in chimpanzees are also present in humans. d. These data support the sister taxa relationship, because it shows how the human chromosome was divided in two to give rise to all other great ape taxa. ANSWER: c 56. Evidence on chromosomal fusion, homology of genes, and gene order on chromosome 2 in humans is often used to support their sister taxa relationship to chimpanzees. Based on these lines of evidence, does the phylogeny shown represent the most parsimonious explanation of the evolution of the number of chromosomes in humans and other apes?

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Chapter 23

a. yes b. no ANSWER: b 57. Why do we think that male Neanderthals mated with non-African ancestral Homo sapiens? a. Because Neanderthal mtDNA sequences are present in the human gene pool, and Neanderthal nuclear DNA sequences are not present. b. Because neither Neanderthal mtDNA sequences nor nuclear DNA sequences are present in the modern human gene pool. c. Because Neanderthal nuclear DNA sequences are present in the human gene pool, and Neanderthal mtDNA sequences are not present. d. Because both Neanderthal mtDNA sequences and nuclear DNA sequences are present in the modern human gene pool. ANSWER: c 58. Where did Homo sapiens originate? a. Asia b. the Middle East c. Africa d. independently in multiple regions of the world e. North America Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 ANSWER: c 59. Consider the phylogenetic tree shown. Geographic labels indicate the origin of humans whose mtDNA were sampled for sequence analysis. The chimp branch serves as the outgroup for the phylogeny.

Which of the statements is an accurate interpretation of the phylogenetic tree? a. The tree supports the out-of-Africa hypothesis for the origin of humans because it implies that the common ancestor of all the humans is of African origin. b. The tree does not reveal anything about the geographic origin of humans. c. The tree supports the multiregional hypothesis because there are four independent populations. d. The closest relative to the Kenyans are the English. e. The Chinese share a more recent common ancestor with the Ethiopians than with the Kenyans. ANSWER: a 60. Consider the mtDNA sequences obtained from chimpanzees and from humans in the given geographic regions.

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Which of the statements is not correct? a. These data depict a higher level of sequence divergence between human and chimp than is to be expected, based on overall levels of genetic similarity between humans and chimps. b. These data are consistent with the multiregional hypothesis for the origin of humans because the Asian and European sequences are more closely related to each other than either is to the African sequences. c. These data are consistent with the out-of-Africa hypothesis for the origin of humans because the Kenyan and Ethiopian samples exhibit an ancestral character state at position 7. d. The African sequences represent older evolutionary lineages than the European sequences because they have one polymorphic site. ANSWER: b 61. Which of the statements is not a piece of genetic evidence that supports the out-of-Africa hypothesis for the origin of humans? a. mtDNA sequences from African and non-African samples indicate a divergence time of approximately 200,000 years. b. Genetic diversity of mtDNA sequences among African samples is greater than genetic diversity of mtDNA sequences among non-African samples. c. mtDNA alleles present in Europe are a subset of mtDNA alleles derived from those present in Africa. d. mtDNA alleles among samples from Africa are more similar than mtDNA alleles among samples from Europe. ANSWER: d 62. All of the statements are evidence that evolution of the FOXP2 gene may have contributed to the evolution of language in humans except: a. Humans with mutations in FOXP2 have problems with speech and language. b. Differences in the FOXP2 gene between humans and chimps are mostly due to synonymous mutations, whereas differences between mice and chimps are mostly due to nonsynonymous mutations. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 c. Studies of Neanderthal DNA have shown that Neanderthals possessed the modern human version of FOXP2. d. FOXP2 encodes an evolutionarily conserved transcription factor that plays an important role in development, including brain development. ANSWER: c 63. Which of the statements most accurately describes the inference that can be made regarding the possible functional significance of FOXP2 in the evolution of human language? a. FOXP2 encodes a highly conserved transcription factor that is a "language gene." Changes in the function of this transcription factor in the human lineage were the key event that allowed language to evolve. b. FOXP2 is a developmental gene with pleiotropic effects. Because this gene is highly conserved among a group of organisms capable of acoustic communication, it must be involved in the evolution of human language. c. FOXP2 encodes a highly conserved transcription factor. Because songbirds whose FOXP2 gene have been knocked out are less capable of learning new songs, this gene is likely involved in the evolution of human language. d. FOXP2 is a developmental gene that affects acoustic communication in a range of animal species. Because there were more nonsynonymous than synonymous mutations in the human lineage, changes to the function of this gene in humans may have contributed to the evolution of language. ANSWER: d 64. What is the definition of neoteny? a. retention of juvenile characteristics in an adult b. reaching reproductive maturity as a juvenile c. the acquisition of complex language skills in an adult d. acquisition of adult characteristics in a juvenile e. accelerated aging in the adult ANSWER: a 65. Consider the figure shown.

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Chapter 23

From these data, you can infer: a. Braincase volume is correlated with body size in the Homo lineage. b. Each species in the Homo lineage has larger brain size than Australopithecines. c. The extant members of the Homo lineage show the greatest amount of range in brain size. d. The Homo lineage, on average, shows very little increase in brain size. ANSWER: c 66. Why is the advent of bipedalism considered such a significant event in human evolution? a. Only bipedal organisms have large brains, indicating that they are capable of becoming intelligent. b. Bipedalism initiated a cascade of evolutionary events that resulted in the evolution of our large brains. c. Standing upright helped our ancestors evade predators. d. The morphological adjustments required for bipedalism resulted from genetic changes that also resulted in cranial evolution, allowing for a larger braincase. ANSWER: b 67. Based on the phylogeny, does FOXP2 provide an accurate molecular clock for the timing of divergence events between mammals and primates and between humans and chimps?

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a. Yes, because we can plot the amino acid changes on the phylogeny, which shows the relationship between mice and all other primates, including humans and chimps. b. Yes, because there is one change on average for every three branches on the tree. c. No, because the changes in amino acids do not occur at regular intervals. d. No, because we do not know if Neanderthals had spoken language. ANSWER: c 68. Which human populations have the highest levels of genetic variation? a. European populations b. African populations c. Asian populations d. Melanesian populations e. North and South American populations ANSWER: b 69. Skin color is an example of a human trait influenced by natural selection. What factor has directly shaped variation in skin color? a. temperature b. disease c. diet d. intensity of sunlight Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 ANSWER: d 70. How does the level of genetic variation in humans compare to the level of genetic variation in other species? a. The level of genetic variation is higher in humans than in other species. b. The level of genetic variation is lower in humans than in other species. c. The level of genetic variation in humans is about the same as other species. d. We are not sure, because we have been unable to accurately measure genetic variation in humans and other species. ANSWER: b 71. Consider a hypothetical study in which groups of women from different regions of the world are shown the same batch of photos of men from different regions of the world. When asked to select who they would choose as a mate, nine times out of ten, women choose men from their own regions, who looked more similar to them than the men from other regions. The most likely explanation is: a. In both regions A and B, individuals of both sexes are equally likely to migrate among populations. b. In region A, males migrate more than females; in region B, females migrate more than males. c. In region A, females migrate more than males; in region B, males migrate more than females. d. In region A, males migrate more than females; in region B, males also migrate more than females. ANSWER: c 72. Refer to the figure shown.

A researcher decides to measure the total amount of genetic variation in four different human populations: Australia, Iraq, Malaysia, and Sri Lanka (listed in no particular order). He finds evidence of repeated founder events as non-African populations moved ever farther away from Africa after the initial out-of-Africa emigration event. Which of the answer options correctly lists the four populations in order of decreasing levels Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 of population genetic variation, based on known patterns of human evolution? a. Sri Lanka > Australia > Iraq > Malaysia b. Australia > Iraq > Sri Lanka > Malaysia c. Iraq > Sri Lanka > Malaysia > Australia d. Malaysia > Iraq > Australia > Sri Lanka ANSWER: c 73. Imagine that you travel to Mars and find two "races" of little green men, the Aquamarines and the Turquoises. The two groups live on different "continents", which you cleverly name Aquamarine and Turquoise after their inhabitants. Fortunately, these little fellows have DNA, so you can carry out a DNA-based phylogenetic analysis. You find that the Aquamarines are a paraphyletic group of taxa within which the Turquoises are nested. What can you conclude about the evolution of the Aquamarines and Turquoises? a. The Turquoises gave rise to the Aquamarines on the Turquoise continent. b. The Turquoises gave rise to the Aquamarines on the Aquamarine continent. c. The Aquamarines gave rise to the Turquoises on the Aquamarine continent. d. The Aquamarines gave rise to the Turquoises on the Turquoise continent. ANSWER: c 74. One reason we see relatively little genetic variation among human regional populations is that: a. extensive migration among human populations has homogenized gene pools. b. extensive convergent evolution has homogenized gene pools. c. genetic drift has reduced variation in most regional populations over time. d. humans migrated out of Africa so recently that we have not had enough time for regional variation to accumulate. e. None of the other answer options is correct. ANSWER: d 75. Statistical analysis of genetic variation within and among human populations, including those we identify as different races, reveals: a. more genetic variation between different groups than we see within groups. b. more genetic variation within groups than we see between different groups. c. no clear pattern of genetic variation; in some regions, we see more variation within than among groups, and in some regions, we see the opposite. d. a correlation between the amount of genetic variation between groups and the length of time for which the groups have been isolated from one another. e. None of the other answer options is correct. ANSWER: b 76. Cultural traits are transmitted horizontally and vertically through learning. Which of the examples shows the horizontal transmission of culture? a. A mother teaches her son how to ride a bike. b. A female chimpanzee teaches her offspring to capture termites using a stick. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 c. A son inherits the recessive form of a gene from his father. d. A teenager teaches her brother how to text. e. None of the answers is an example of horizontal transmission of culture. ANSWER: d 77. Other animals, including chimpanzees, have culture, but their culture is limited relative to the human version. Why? a. Cultural transmission serves no particular evolutionary advantage in chimpanzees. b. We are more adept at learning and imitation than other species. c. Cultural evolution cannot occur in species other than our own. d. Chimpanzees are limited to forest habitats where cultural improvement is disadvantageous. ANSWER: b 78. Which of the answer choices are uniquely human traits? a. the ability to learn b. the ability to count c. the ability to imitate d. consciousness e. None of the other answer options is correct. ANSWER: e 79. Recent studies have examined the frequency of various transcription factors and how they differ between chimps and humans. Each of the transcription factors examined is responsible for up-regulation of gene expression in specific tissues. Where would you expect to find the greatest differences in the action of these transcription factors between humans and chimps? a. in the kidney b. in the brain c. in the intestine d. in the skin ANSWER: b 80. Recent studies have examined the frequency of various transcription factors and how they differ between chimps and humans. Each of the transcription factors examined is responsible for up-regulation of gene expression in specific tissues. Would you expect to see more or fewer differences in the function of these transcription factors between humans and gorillas or between humans and chimps? a. The number of differences between humans and gorillas would be the same as the number of differences between humans and chimps because chimps and gorillas are more closely related than chimps and humans. b. The number of differences between humans and gorillas would be greater than the number of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 differences between humans and chimps because humans and chimps are more closely related than humans and gorillas. c. The number of differences between humans and gorillas would be the same as between humans and chimps, but the transcription factors that are different between humans and chimps would not be the same set as those that are different between humans and gorillas. d. The number of differences between humans and gorillas would be fewer than the number of differences between humans and chimps because humans and gorillas are more closely related than humans and chimps. ANSWER: b 81. Data in the graph shown compares the abundance of cultural traits with the number of adult female chimpanzees in different chimpanzee communities (Lind and Lindenfors, 2010).

Which statement reflects the general trend shown by the data in the graph? a. Greater numbers of cultural traits result in greater numbers of chimps. b. Greater numbers of offspring are correlated with greater numbers of cultural traits. c. Greater numbers of cultural traits result in greater numbers of adult females. d. Greater numbers of adult females are correlated with greater numbers of cultural traits. e. Fewer numbers of adult females are correlated with greater numbers of cultural traits. ANSWER: d 82. Data in the graph shown compares the abundance of cultural traits with the number of adult female chimpanzees in different chimpanzee communities (Lind and Lindenfors, 2010).

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If the Gombe and Tai population data had not been available, would the trend seen in the graph be stronger or weaker? a. stronger b. weaker c. unaffected ANSWER: b 83. Data in the graph shown compares the abundance of cultural traits with the number of adult female chimpanzees in different chimpanzee communities (Lind and Lindenfors, 2010).

If any of the answer choices are possible, which would be the best way to improve the data set? a. Increase sample size at each site. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 b. Observe more males. c. Estimate the relatedness of individuals in the population. d. Sequence the genomes of an individual from each population. e. Sample populations from additional locations. ANSWER: e Multiple Response 84. In addition to the shift to bipedalism, what major trend or trends do we see in hominin evolution as we look across the entire available fossil record? Select all that apply. a. increase in body size b. increase in brain size c. significant increase in the difference in size between males and females d. increase in tooth size e. None of the other answer options is correct. ANSWER: a, b 85. The shift to bipedalism required changes to the architecture of the: Select all that apply. a. skull. b. spine. c. pelvis. d. legs. e. hands. ANSWER: a, b, c, d 86. Which of the statements supports the hypothesis that neoteny contributed to the evolution of modern humans from our common ancestor with chimpanzees? Select all that apply. a. Juvenile great apes have large heads relative to their body size and are less hairy than adult great apes. b. Neoteny involves changes in the developmental timing of gene regulation over the course of evolution, consistent with the observation that the genome sequence of human and chimp are 99 percent identical. c. It has been suggested that the human mentality, characterized by questioning and playfulness, is characteristic of juvenile great apes. d. Sexual maturity occurs earlier in great apes than in humans. ANSWER: a, b, c 87. Examine the figures shown.

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According to the figures, humans are members of the monophyletic _____. Select all that apply. a. primate group b. monkey group, which includes both New World and Old World monkeys and their relatives c. ape group d. great ape group e. prosimians group ANSWER: a, c, d 88. Which characteristics of humans supports the neoteny hypothesis? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 23 a. large heads b. relative hairlessness c. a foramen magnum positioned at the base of the skull d. relatively flat faces and small noses e. None of the other answer options is correct. ANSWER: a, b, c 89. Human language differs from language in nonhuman animals in that: Select all that apply. a. human infants strive to learn grammatical language and nonhuman animals do not. b. human language is more complex than nonhuman language. c. only human language includes distinct words, or specific sets of sounds that can identify specific objects in the environment. d. None of the other answer options is correct. ANSWER: a, b 90. If the groups we recognize as different human races represented populations that have been genetically isolated for a long time, we would expect to see: Select all that apply. a. high levels of genetic variation between groups, relative to what we see within groups. b. high levels of genetic variation within groups, relative to what we see between groups. c. no clear pattern of genetic variation - in some regions, we would see more variation within than among groups, and in some regions, we would see the opposite. d. a correlation between the amount of genetic variation between groups and the length of time the groups have been isolated from one another. e. None of the other answer options is correct. ANSWER: a, d 91. It is empirically observed that some 85 percent of all human variation resides within populations. This means that relatively little genetic variation distinguishes different populations. This may be because: Select all that apply. a. there may have been high levels of gene flow among human populations. b. human populations have only recently spread out from Africa. c. natural selection prohibits genetic divergence of human populations. d. sexual selection tends to focus on observable traits. ANSWER: a, b

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Chapter 24 Multiple Choice 1. A researcher in a lab has sequenced three separate genes for six different bacterial species. Two of the genes (Genes 1 and 2) show the same pattern of evolutionary relatedness as the phylogeny generated from the whole genome sequence, but the third gene (Gene 3) shows a different pattern of evolutionary relatedness. Species A, D, and E all live in similar environments. Examine the two phylogenies to answer the question.

The phylogeny generated from Gene 3 could be the result of horizontal gene transfer. a. true b. false ANSWER: a 2. Bacteria are sensitive to certain antibiotics like chloramphenicol and streptomycin, which kill the bacteria by interfering with translation through mechanisms that affect the ribosome's ability to function. If a researcher were to culture populations of Archaea and then treat them with either of the previously mentioned antibiotics, what would be the predicted results? a. Both will have an effect on Archaeon, but the results may not be identical. b. Neither antibiotic will have an effect on Archaeon populations because they don't have peptidoglycan in their cell walls. c. Both will have an effect on Archaeon because they have ribosomes. d. Neither antibiotic will have an effect on Archaeon populations because their ribosomes are different from bacterial ribosomes. ANSWER: d 3. The figure shows stratification of microbial layers in a Winogradsky column (a method used to simulate environments similar to microbial mats observed in nature).

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Chapter 24

If you were to sequence the genomes of representative genera from each layer and construct a phylogenetic tree, which outcome would you expect to find? a. There would be a single monophyletic group formed by the cyanobacteria. Genera from the other layers would form paraphyletic groupings. b. There would be a single monophyletic group formed by all photosynthetic bacteria. c. Genera from the other layers would form paraphyletic groupings. d. There would be four monophyletic groups, one for each layer represented in the column. e. No monophyletic groups would be seen in the phylogeny. ANSWER: a 4. You are a student in a lab that is growing populations of bacteria in culture. Every three months you sample the population and calculate the genetic diversity in the population.

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Which of the figures would match your prediction for changes in genetic diversity over time in the population? a. Fig. A b. Fig. B c. Fig. C ANSWER: a 5. During oxygenic photosynthesis, water is _____ and loses electrons (i.e., it serves as an electron donor). a. converted b. reduced c. oxidized d. transformed ANSWER: c 6. An organism that acquires energy from the sun and uses C6H12O6 as a carbon source would be classified as a: a. photoautotroph. b. chemoautotroph. c. photoheterotroph. d. chemoheterotroph. ANSWER: c 7. A major constituent of Earth's atmosphere is nitrogen gas (N2), which makes up 79% of the air we breathe. Primary producers such as plants and algae cannot assimilate this form of nitrogen into biomolecules. Instead they rely on Bacteria and Archaea to convert N2 to ammonia (NH3). This important biological reaction is called: a. nitrification. b. nitrogen fixation. c. denitrification. d. anammox. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 ANSWER: b 8. Which of the statements is a reason why categorizing using the biological species concept for Bacterial and Archaeal species is difficult? a. Many Bacteria and Archaea live in environments where oxygen is absent. b. Archaea and Bacteria have circular chromosomes. c. Bacteria and Archaea do not have life cycles characterized by meiosis, fertilization, and reproductive isolation. d. Prokaryotic organisms are too small to track in natural environments. ANSWER: c 9. A researcher knows that bacteria are much more prevalent at the ocean's surface, compared to thaumarchaeotes. How does the ratio of bacteria to thaumarchaeotes compare at lower depths (5000 meters under the ocean's surface)? a. Bacteria are still much more prevalent. b. Thaumarchaeotes are much more prevalent. c. There are roughly equal numbers of bacteria and thaumarchaeotes. d. Neither bacteria nor thaumarchaeotes can survive at that ocean depth. ANSWER: c 10. Scientists studying ancient sedimentary rocks on Mars have found that 3 billion years ago, extremely salty water was present in an area called Meridiani Planum. Refer to Figure 24.18.

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Which group of Archaea contains strains most likely to survive in such an environment? a. Euryarchaeota b. Cyanobacteria c. Thaumarchaeota d. Korarchaeota ANSWER: a 11. Oxygenic photosynthesis is made possible by the accumulation of oxygen in Earth's atmosphere and oceans. The primary producers of oxygen for 800 million years and those that power the modern carbon cycle are mostly: a. anoxygenic phototrophs. b. algae and plants. c. cyanobacteria. d. purple bacteria. ANSWER: b 12. Nodules found on the roots of leguminous plants like soybeans harbor bacteria that fix nitrogen. These Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 nodules are an example of the intimate and specialized partnerships that have been established over time between prokaryotes and eukaryotes. Coevolution maintains these types of relationships because both participants benefit. Plants receive biologically useful forms of nitrogen needed for growth, and the bacteria reside in an oxygen-free environment required by enzymes involved in the nitrogen-fixing process. Humans also maintain intimate associations with beneficial bacteria and even archaea. In fact, the prokaryotic cells in and on human bodies outnumber the eukaryotic human cells by tenfold, according to some estimates. These microbial inhabitants help humans digest food, provide essential vitamins absorbed by the intestines, and even impact the immune system. The microbes that coevolve with plants arrive by infection of root tissue. The microbiota that coevolve with humans colonize human tissues after birth. Consider the chart showing the distribution of the gut microbiota in a human female.

Panels (a) and (b) shown depict two scenarios for the distribution of the gut microbiota in her son. The son maintains a similar diet in both scenarios and has not undergone recent antibiotic treatment, which would decimate and/or alter the entire gut community.

Which of the two scenarios supports the hypothesis that microbiota in the son arrive by infection? Which suggests the microbiota are inherited by offspring? a. Panel (a) supports infection, and panel (b) supports inheritance. b. Panel (a) supports inheritance, and panel (b) supports infection. c. Neither scenario suggests inheritance. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 13. The Department of Energy is very interested in investigating which organisms compose the gut microbiome of cows, especially in the rumen. Why would the microbes within a cow's rumen be studied by scientists in the Department of Energy? a. Scientists want to derive biofuels from methane. b. Scientists want to derive biofuels from cellulose. c. Scientists want to derive biofuels from the lactose in cow's milk. ANSWER: b 14. A researcher in a lab has sequenced three separate genes for six different bacterial species. Two of the genes (Genes 1 and 2) show the same pattern of evolutionary relatedness as the phylogeny generated from the whole genome sequence, but the third gene (Gene 3) shows a different pattern of evolutionary relatedness. Species A, D, and E all live in similar environments. Examine the two phylogenies to answer the question.

The monophyletic group formed by Species A, D, and E is the result of a shared derived character that originated in the common ancestor for that group. a. true b. false ANSWER: b 15. A botanist notices that in one area of her yard, the plants have a yellow color throughout their leaves and are not growing very well compared to the same plants in other areas of the yard. She checks the soil and determines that there is adequate nitrogen. She later determines that the plants are suffering from a protein deficiency. In order to combat this problem, she should inoculate the soil with bacteria that can reduce sulfate. a. true b. false ANSWER: b 16. Consider the image.

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Chapter 24

If nitrogen fixation were to stop, then the entire nitrogen cycle would eventually come to a halt as well. a. true b. false ANSWER: a 17. The group Archaea are thought to have originated in very harsh environments. Evidence for this is shown in the Archaeal phylogeny by the presence of thermophilic representatives in all major groups of Archaea. a. true b. false ANSWER: b 18. Plasmid DNA generally contains genetic information critical for the survival of a bacterial cell. a. true b. false ANSWER: b 19. Many antibiotics that target bacteria are effective against Archaea. a. true b. false ANSWER: b 20. A researcher discovers a new single-celled bacterium very similar to Thiomargarita namibiensis. At first, she is confused because this organism is quite large compared with other bacteria. She also knows that because of diffusion, most bacterial cells don't exceed 17 μm in size. What might the researcher notice about the internal structure of her new bacterium? a. It contains several membrane-bound organelles, facilitating the diffusion of molecules. b. It contains several large histones, facilitating the diffusion of molecules. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 c. It is completely filled by cytoplasm, limiting the distance molecules need to diffuse. d. It contains a large vacuole, limiting the distance molecules need to diffuse. ANSWER: d 21. Anoxygenic and oxygenic photosynthetic organisms both absorb the same wavelengths of light. a. true b. false ANSWER: b 22. What is one advantage of fermentation over cellular respiration in certain environments? a. Fermentation breaks down organic molecules. b. Fermentation does not require O2. c. Fermentation yields more energy. d. None of the other answer options is correct. ANSWER: b 23. Photoheterotrophs rely on sunlight and organic molecules obtained from the environment. a. true b. false ANSWER: a 24. Based on the energy and carbon sources used by humans, in what way would humans be classified? a. as chemoheterotrophs b. as chemoautotrophs c. as photoheterotrophs d. as photoautotrophs ANSWER: a 25. In what part of the ocean is denitrification likely to be most important? a. deep within marine sediments b. shallow waters rich in oxygen c. surface waters of the open ocean d. subsurface water masses that contain little or no oxygen ANSWER: d 26. Why is genomic information often more useful than phenotypic information in the bacterial phylogenetic tree? a. Most bacteria cannot be cultured. b. Most bacteria have similar phenotypic traits. c. Most bacteria have the same genes. d. Genomic information is always accurate. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 ANSWER: a 27. Photosynthetic bacteria are all found on one branch of the bacterial phylogenetic tree. a. true b. false ANSWER: b 28. Whereas the gene encoding the small subunit of ribosomes was used to generate the tree of life, many genes are inappropriate choices for this type of analysis because these genes have been subject to horizontal gene transfer for billions of years. a. true b. false ANSWER: a 29. Constructing phylogenies with genes associated with nitrogen fixation may mask actual relatedness between bacterial species because: a. these genes are not typically exchanged between bacteria by horizontal gene transfer. b. these genes are often exchanged between bacteria by horizontal gene transfer. c. their sequences are similar to those of rRNA genes. d. every bacterium contains nitrogen fixation genes. ANSWER: b 30. It is commonly thought that life originated in hot waters near marine hydrothermal vents. Which statement provides evidence in support of this hypothesis? a. Euryarchaeota are extremely diverse. b. Thaumarchaeota have high abundance in deep ocean waters. c. Early diverging branches on the archaeon tree gave rise to Thaumarchaeota. d. Some bacterial lineages also have thermophilic members. ANSWER: d 31. Archaeons are often found in environments where exploitable energy sources cannot maintain eukaryotes but can maintain bacteria. a. true b. false ANSWER: b 32. The totality of microbes that reside in/on humans and environmental interactions between the microbes and their human host are known as: a. microbiology. b. symbiosis. c. an enterotype. d. the human microbiome. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 ANSWER: d 33. Humans provide ecological niches for bacterial populations. a. true b. false ANSWER: a 34. The reasoning behind the use of fecal transplants from healthy individuals is that: a. C. difficile feeds on other bacteria that are introduced with the fecal transplant. b. the fecal matter from healthy patients does not have C. difficile. c. the competition from the introduction of other bacteria into the gastrointestinal tract keeps C. difficile numbers down. d. antibiotics are administered with the fecal transplant, thus controlling populations of C. difficile. ANSWER: c 35. The enterotype of a human depends primarily on: a. the predominant microbial genus of organisms found in the gut. b. gender. c. nationality. d. body mass index. ANSWER: a 36. Clostridium difficile is a bacterium that is often found in the gastrointestinal tract of healthy individuals. It is also a common cause of hospital-acquired infections. Most individuals who become ill with C. difficile do so following antibiotic treatment. This information suggests that C. difficile: a. is not pathogenic (disease-causing) in healthy people. b. is resistant to antibiotics and antiseptics used in the hospital setting. c. is adapted to the human gastrointestinal tract. d. rebels when other organisms are eliminated from their environment. ANSWER: b 37. Cattle production for meat is an important industry in certain areas of the United States. If you were to sample the air around regions with high cattle densities, you would expect to find high concentrations of methane produced by symbiotic Archaeons. a. true b. false ANSWER: a 38. Why is the presence of stromatolites important to our understanding of the evolution of life early in Earth's history? a. Stromatolites only appear in old rocks from South Africa and Australia, the oldest land masses on Earth. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 b. Stromatolites are found deep in the ocean and resemble modern structures formed by microbial communities. c. Isotopic composition of carbon in stromatolites can only be explained by the presence of carbon cycling through organisms. ANSWER: c 39. If scientists were able to infect strains of malaria-transmitting mosquitoes with Wolbachia, this could reduce malaria infections worldwide because a. Wolbachia infection of malaria-transmitting mosquitoes could be transmitted to humans, and Wolbachia in humans would make them resistant to infection by the parasite that causes malaria. b. Wolbachia infection of malaria-transmitting mosquitoes may inhibit infection of the mosquito by the parasite that causes malaria. c. Wolbachia infection of malaria-transmitting mosquitoes may alter the reproductive cycle of the mosquitoes so that only females are born. d. Wolbachia infection of malaria-transmitting mosquitoes would kill the mosquito, so they would no longer be able to transmit malaria to humans. ANSWER: b 40. Consider the chemical reactions of the sulfur cycle shown in Figure 24.11.

Reduction of CO2 to generate carbohydrates can occur when coupled to the oxidation of reduced sulfur compounds (e.g., H2S). This latter process is carried out by what type of microorganism(s)? a. purple bacteria b. cyanobacteria ANSWER: a 41. A researcher discovers a new unicellular organism that contains a single circular chromosome and no membrane-bound organelles. Upon closer observation, the researcher notices that this organism can also fix nitrogen and produce methane. How should this organism be classified? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 a. as an archaeon b. as a bacterium c. as a eukaryote ANSWER: a 42. Why would streptomycete bacteria be of particular interest to a pharmaceutical company? a. Streptomycete bacteria have cell walls made of lignin and not peptidoglycan. b. All streptomycete bacteria are antibiotic-resistant. c. Streptomycete bacteria produce antibiotic or antifungal compounds. d. Streptomycete bacteria change color when exposed to antibiotics. ANSWER: c 43. A researcher synthesizes a new drug that specifically targets and degrades peptidoglycan. What type of infections could this drug treat? a. infections caused by bacteria b. infections caused by archaeons c. infections caused by viruses d. infections caused by eukaryotes ANSWER: a 44. Autotrophs produce carbohydrates by reducing CO2, whereas heterotrophs synthesize carbohydrates by ingesting smaller preformed organic compounds like glucose. a. true b. false ANSWER: a 45. Fermentation yields an amount of energy equivalent to respiration because both processes achieve full oxidation of carbon compounds to CO2. a. true b. false ANSWER: b 46. An organism that oxidizes H2S to gain energy and utilizes CO2 as a carbon source to generate carbohydrates would be classified as a: a. photoautotroph. b. chemoautotroph. c. photoheterotroph. d. chemoheterotroph. ANSWER: b 47. A researcher wants to determine if a unicellular organism he discovered is an autotroph or a heterotroph. He Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 radioactively labels the carbon in CO2 and C6H12O6 and exposes one culture of his organism to the labeled CO2 and another culture to the labeled C6H12O6. What would happen if his organism is an autotroph? a. Labeled carbon would be seen in carbohydrates of both cultures. b. Labeled carbon would be seen in the carbohydrates of organisms exposed to C6H12O6. c. Labeled carbon would not be seen in the carbohydrates of either culture. d. Labeled carbon would be seen in the carbohydrates of organisms exposed to CO2. ANSWER: d 48. Consider the chemical reactions of the sulfur cycle shown in Figure 24.11.

Oceans constitute the most significant reservoir of sulfur in the biosphere, and sulfate (SO42–) reduction is the predominant form of anaerobic respiration in marine environments. In such cases, oxidized sulfur compounds like SO42– are the electron acceptors instead of what? a. oxygen b. water c. hydrogen d. carbon dioxide ANSWER: a 49. Consider the chemical reactions of the sulfur cycle shown in Figure 24.11.

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Chapter 24

Reduction of CO2 to generate carbohydrates can occur when coupled to the oxidation of reduced sulfur compounds (e.g., H2S). This latter process is carried out by what type of microorganism(s)? a. chemoautotrophs b. heterotrophs ANSWER: a 50. N2 is an end product of _____; NH3 is an end product of: _____. a. anammox; nitrogen fixation. b. anammox and denitrification; nitrogen fixation. c. nitrogen fixation; anammox. d. denitrification; anammox and nitrogen fixation. ANSWER: b 51. Molecular features that distinguish major groups of Bacteria evolved billions of years ago. The DNA sequences continued to change within each group, potentially masking inherited sequence similarities from a common ancestor that otherwise would have defined the group. a. true b. false ANSWER: a 52. Why are scientists unable to grow the majority of Bacteria and Archaea in pure culture in a laboratory setting? a. Cells with unusual morphological characteristics cannot be grown in the laboratory. b. Most microorganisms reside in extreme environments that are toxic to human life, so samples cannot be collected for cultivation of the microbial inhabitants in the laboratory. c. DNA sequencing technology was not sufficiently advanced to give scientists the ability to identify prokaryotic species in nature, a requirement for growing microbes in the laboratory. d. Many species have metabolic requirements that are difficult to fulfill in the laboratory. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 53. If the genes used to generate molecular phylogenies were subject to horizontal gene transfer, they will not reflect the evolutionary history of the organisms in which they occur. a. true b. false ANSWER: a 54. Which characteristic is not a requirement for metabolic activity seen in the Archaea that enable them to live in hydrothermal vents, acidic or salty water, and the rumens of cows? a. acid tolerance b. temperature preference c. presence of oxygen d. methane production e. salt tolerance ANSWER: c 55. A high-school student in Nepal is accepted to — and eventually attends — a college in the United States. His diet changes from one that is rich in vegetables and grains to one that primarily consists of meat and sugars. What will happen to the bacteria living in his intestines? a. There will be more Firmicutes inhabiting his intestinal tract. b. There will be more Bacteroidetes inhabiting his intestinal tract. c. Bacteria will no longer inhabit his intestines. d. Nothing will happen; there is no relationship between bacteria in the intestines and diet. ANSWER: a 56. For the first 2 billion years of its history, Earth's atmosphere lacked or had very little oxygen. Which organisms produced much of the oxygen that now makes up the present-day atmosphere of Earth? a. anoxygenic photosynthetic bacteria b. hyperthermophiles c. cyanobacteria d. Euryarchaeota ANSWER: c 57. It has been suggested that cyanobacteria are the most important organisms to have evolved on the Earth. What evidence justifies this view? a. Cyanobacteria form microbial mats, enabling them to spread across lake bottoms and shallow seafloors. b. Some cyanobacteria can fix nitrogen, which plants need in order to produce oxygen, making them unique among bacteria. c. Some cyanobacteria differentiate to form multiple cell types, making them the ancestors of plants and animals. d. Cyanobacteria evolved the ability to use water as an electron donor in photosynthesis, generating the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 24 oxygen gas found in the atmosphere. ANSWER: d

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Chapter 25 Multiple Choice 1. Which of the statements describes characteristics of eukaryotic cells but not of prokaryotic cells? a. relatively large genome, dynamic cytoskeleton, compartmentalized metabolic processes b. two or more circular chromosomes, dynamic membrane system, compartmentalized metabolic processes c. linear chromosomes, endomembrane system, nucleus, diverse means of harvesting and utilizing energy under anaerobic conditions d. two or more linear chromosomes, dynamic membrane system, diverse means of harvesting and utilizing energy under anaerobic conditions ANSWER: a 2. Many single-celled eukaryotic organisms have evolved complex life cycles that include sexual reproduction. This means that: a. diploid cells must undergo mitotic cell division in order to reproduce. b. two cells, at some point, must fuse to become diploid. c. haploid cells must undergo meiotic divisions to form gametes. d. complex mating behaviors likely evolved at the same time as sexual reproduction. ANSWER: b 3. Which of the organelles is not part of the endomembrane system? a. Golgi apparatus b. endoplasmic reticulum c. mitochondria d. nuclear envelope ANSWER: c 4. The diversity of shapes found in eukaryotic cells is made possible by the: a. nucleus. b. mitochondria. c. cytoskeleton. d. plasma membrane. ANSWER: c 5. Which of the processes introduces genetic diversity and contribute to the creation of daughter cells that are genetically unique from one another and from the parent cell? a. recombination b. independent assortment c. meiotic cell division d. fusion of gametes e. All of these choices are correct. ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 6. Many single-celled eukaryotes that normally exist in a haploid state reproduce asexually by: a. meiotic cell division. b. binary fission. c. mitotic cell division. d. fusion. ANSWER: c 7. Eukaryotic cells and bacteria can both gain nutrition by phagocytosis of other cells. a. true b. false ANSWER: b 8. Evidence that supports the origination of chloroplasts from photosynthetic bacteria includes: a. similar organization of photosynthetic membranes. b. the use of two linked photosystems to capture electrons from water. c. similarities in the DNA sequence of the chloroplast chromosome and bacterial chromosome. d. All of these answer choices are correct. ANSWER: d 9. Mitochondria and chloroplasts are thought to have arisen from: a. endosymbiotic bacteria. b. parasitic interactions. c. normal organelle development. d. None of the other answer options is correct. ANSWER: a 10. A cell with chloroplasts that have four membranes (not counting the thylakoid membranes) likely arose by _____ endosymbiotic events. a. one b. two c. four d. eight ANSWER: b 11. Which of the statements is true regarding protists? a. They may have cell walls. b. They may be photosynthetic. c. They are usually aerobic, though some are not. d. They are a monophyletic group of organisms. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 12. Refer to the figure shown. Which taxonomic groups are most closely related?

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Chapter 25 a. diatoms and dinoflagellates b. fungi and dictyostelid slime molds c. red algae and brown algae ANSWER: a 13. Which of the seven superkingdoms contains nearly all photosynthetic organisms? a. Archaeplastida b. green algae and land plants c. Stramenopila ANSWER: a 14. To which superkingdom do corn and wheat belong? a. Excavata b. Alveolata c. Archaeplastida d. Opisthokonta e. Rhizaria ANSWER: c 15. To which superkingdom do the giant kelps belong? a. Stramenopila b. Alveolata c. Excavata d. Amoebozoa e. Rhizaria ANSWER: a 16. Refer to the figure shown. To which superkingdom do the paleontologically important Foraminifera and Radiolaria belong?

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Chapter 25

a. Stramenopila Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 b. Alveolata c. Excavata d. Amoebozoa e. Rhizaria ANSWER: e 17. The most diverse eukaryotic superkingdom is: a. Archaeplastida. b. Alveolata. c. Opisthokonta. d. Amoebozoa. ANSWER: c 18. The term protist lacks phylogenetic accuracy because it represents a paraphyletic group. Which of the statements reflects the paraphyly of protist taxa? a. Each protist lineage is derived from a different common ancestor. b. Some eukaryotic lineages evolved from protists, and they are not included in the group called protists. c. Protist lineages are defined by the fact that they are not other eukaryotic lineages. d. Protist lineages are too old to determine descent from a single common ancestor. ANSWER: b 19. The first eukaryotic cell is thought to have evolved at least _____ million years ago. a. 1800 b. 1200 c. 3800 d. 4500 ANSWER: a 20. Which of the seven existing superkingdoms includes the earliest eukaryote fossil related to living protists? a. Archaeplastida b. Stramenopila c. Opisthokonta d. Excavata ANSWER: a 21. What components of eukaryotic cells leave a fossil record? a. organelles b. nuclear membranes c. cell walls d. None of the other answer options is correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 ANSWER: c 22. Which of the options is a way that genetic diversity is promoted in eukaryotes? a. conjugation b. sexual reproduction c. asexual reproduction d. transduction ANSWER: b 23. Many single-celled eukaryotic organisms have evolved complex life cycles that include sexual reproduction. This means that: a. genetic material from two haploid cells is combined. b. diploid cells must undergo mitotic cell division in order to reproduce. c. haploid cells must undergo meiotic divisions to form gametes. d. complex mating behavior likely evolved at the same time as sexual reproduction. ANSWER: a 24. A dynamic cytoskeleton enables eukaryotic cells to: a. move. b. engulf other cells. c. change shape. d. All of these choices are correct. ANSWER: d 25. What factors contribute to the high diversity observed in eukaryotic organisms? a. Their genomes contain diverse regulatory sequences. b. They reproduce sexually. c. Their dynamic cytoskeleton and membrane systems can generate cells of widely varying size and shape. d. All of these choices are correct. ANSWER: d 26. What allows multicellular eukaryotic cells to generate multiple, interacting cell types during development? a. a dynamic cytoskeleton b. membrane systems c. gene regulation d. the ability to perform photosynthesis ANSWER: c 27. Which of the statements is evidence supporting the hypothesis that eukaryotic chloroplasts originated from cyanobacteria? a. Chloroplasts and cyanobacteria have similar DNA sequences for shared genes. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 b. Chloroplasts and cyanobacteria both have cell walls. c. Chloroplasts can leave the cell and function freely on their own. d. Chloroplasts and cyanobacteria have the same number of genes. ANSWER: a 28. What is the main difference between the two hypotheses for the origin of the eukaryotic cell? a. One attributes the formation of the endoplasmic reticulum and nuclear envelope to infolding of the plasma membrane. b. Only one involves the conversion of a proteobacterium to a mitochondrion. c. One hypothesizes that eukaryotic cells evolved from an archaeon, rather than a bacterium. d. One hypothesizes that engulfment of a proteobacterium occurred after the formation of the nuclear envelope, rather than before. ANSWER: d 29. Which of the statements explains why photosynthesis is widely but discontinuously distributed throughout the eukaryotic tree? a. Photosynthesis was acquired early in evolution but then lost. b. Photosynthesis was acquired multiple times through endosymbiosis. c. All eukaryotes are capable of photosynthesis, but it is repressed in some cases. d. Some eukaryotes evolved photosynthetic pathways separate from chloroplasts. ANSWER: b 30. There are two competing theories put forth to explain the origin of eukaryotic cells. Which of the answer choices correctly states one of these theories? a. Ancient eukaryotic cells resembling archaeal cells acquired symbiotic bacteria that eventually took up permanent residence to become mitochondria. b. Two types of prokaryotic cells, one of which was photosynthetic, fused to become the first ancient eukaryotic cell. c. A symbiotic relationship between a cyanobacteria and an archaeon created the first ancient eukaryotic cell. d. Ancient eukaryotic cells resembling archaeal cells acquired symbiotic cyanobacteria that eventually took up permanent residence to become chloroplasts. ANSWER: a 31. Which of the statements is true regarding protozoa? a. All protozoa have cell walls. b. Most protozoa are aerobic. c. Most protozoa features have evolved only once. d. Protozoa are all photosynthetic. ANSWER: b 32. What causes amoeboid feeding cells of cellular slime molds to aggregate into a large, multicellular "slug" Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 form? a. low moisture level b. proximity to other cells of the same species c. low nutrient levels d. increased sunlight ANSWER: c 33. The phylogenetic groupings in the eukaryotic tree of life are based primarily on: a. DNA sequence data. b. physical characteristics shared among members of the different groups. c. whether the organisms in question are heterotrophic or autotrophic. d. whether the organisms are single-celled or multicellular organisms. ANSWER: a 34. Some single-celled organisms live inside other cells, lack many of the organelles that most eukaryotes have, and have a relatively small number of genes in their genomes. Which of the statements is true regarding these single-celled organisms? a. They have a streamlined simplicity that is possible because they rely on their hosts for many cell functions. b. They are primitive cells compared to the cells in which they live. c. They represent a fourth category of cells that exists in addition to archaea, bacteria, and eukaryotic cells. d. They are all parasitic cells. ANSWER: a 35. To which superkingdom does the organism that causes malaria belong? a. Stramenopila b. Alveolata c. Excavata d. Amoebozoa e. Rhizaria ANSWER: b 36. To which superkingdom do dogs and cats belong? a. Archaeplastida b. Alveolata c. Excavata d. Opisthokonta e. Rhizaria ANSWER: d 37. Photosynthesis spread through most of the eukaryotic superkingdoms by repeated episodes of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 endosymbiosis of photosynthetic eukaryotic organisms. a. true b. false ANSWER: a 38. How old are the earliest fossils of eukaryotes that can be clearly linked to a specific superkingdom? a. 200 million years b. 800 million years c. 1050 million years d. 1800 million years ANSWER: c 39. What do scientists rely on to estimate when extinct, ancient eukaryotes were present on Earth? a. DNA and the fossilized remains of cell walls b. the fossilized remains of cell walls c. DNA and the fossilized remains of chloroplasts d. the fossilized remains of nuclear membranes and other organelles ANSWER: b 40. The origin of eukaryotic, membrane-bound organelles other than mitochondria and chloroplasts is a contested issue. Which of the statements describes why it is difficult to determine how structures like the nuclear membrane originated? a. The first eukaryotic cells likely arose before the first fossil eukaryotes are found in the fossil record. b. We do not understand how membranes could fold inward to form novel structures within the cell. c. No living cells possess intermediate structures that could tell us how these features originated. d. All of these choices are correct. ANSWER: c 41. Diatoms are important photosynthetic organisms in marine environments. This group evolved relatively recently, during the Mesozoic Era. Diatoms also play an important role in the transfer of silica to ocean sediments. What is the importance of the recent evolution of these organisms to the marine environment in which they live? a. Because they are photosynthetic, they increase the concentration of oxygen in the atmosphere. b. Because they have silica exoskeletons, they do not deplete calcium needed by foraminiferans. c. More recently evolved groups of protists continue to alter marine environments, much like earlier groups of protists. d. More recently evolved groups of protists continue to alter marine environments in response to the evolution of the ability to ingest other protists. ANSWER: c 42. Which one of the options correctly lists the features of eukaryotic cells that distinguish them from prokaryotic cells? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 a. a dynamic cytoskeleton, dynamic system of membranes, relatively limited metabolic capabilities b. a dynamic cytoskeleton, dynamic system of membranes, flexible and wide-ranging metabolic capabilities c. a dynamic cytoskeleton that permits cellular movement, the ability to carry out photosynthesis, and the ability to switch from haploid to diploid stages in their life cycle ANSWER: a 43. Molecular motors associated with the cytoskeleton allow molecules to move through the cell faster than would be possible by diffusion. a. true b. false ANSWER: a 44. Phagocytosis provides a means by which heterotrophic protists can take in food particles. In what way do photosynthetic protists reflect phagotrophy? a. Photosynthetic protists do not reflect phagotrophy, as these cells fix carbon dioxide into sugars and do not require particulate matter as a source of carbon. b. Phagotrophy is the process by which cyanobacteria were introduced into protistan cells to become endosymbionts and, eventually, chloroplasts. c. Evolutionary loss of the ability to take in particles by phagotrophy forced some protists to evolve a capacity for photosynthesis. d. Phagotrophy introduces carbon dioxide into the cells of photosynthetic algae, where it is reduced to form sugars. ANSWER: b 45. Predation by ingesting other cells first arose in prokaryotic organisms, increasing the complexity of interactions among organisms through their ability to engulf particulate food. a. true b. false ANSWER: b 46. Which of the features enabled eukaryotes to evolve complex life cycles and processes of multicellular development? a. binary fission b. complex patterns of gene regulation c. a dynamic cytoskeleton d. sexual reproduction ANSWER: b 47. Which of the statements is an advantage of the relatively small size of the prokaryotic genome? a. It allows for more diverse regulation mechanisms. b. It allows for noncoding genetic material. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 c. It allows for increased diversity. d. It allows prokaryotes to reproduce quickly. ANSWER: d 48. Mitotic cell division in unicellular eukaryotes can result in a cell that is: a. 1n. b. 2n. c. Either 1n or 2n, depending on whether the organism predominantly exists as a haploid cell or a diploid cell. ANSWER: c 49. Animals inherited a sexual life cycle from their protistan ancestors. In what way have animals modified the life cycle characteristic of most protists? a. In animals, the haploid cells undergo multiple rounds of mitotic cell division before sexual fusion takes place. b. In animals, the diploid zygote undergoes multiple rounds of mitotic cell division before sexual fusion takes place. c. Animals have no haploid phase in their life cycle. d. Animals do not form zygotes as part of their life cycle. ANSWER: b 50. Chloroplasts have an outer and an inner membrane that separate the stroma and thylakoid membrane from the cytoplasm. What is believed to be the origin of the outer membrane of the chloroplast? a. It is the remnant of the ancient host cell's plasma membrane, following endocytosis of the cyanobacterium. b. It is a nonfunctional remnant resulting from the amplification of the inner membrane of the chloroplast to increase its surface area. c. It is an adaptation of the symbiotic cyanobacteria to protect it from fusing with lysosomes present in the cytoplasm of the ancient "host" cell. ANSWER: a 51. Which of the statements is not a line of evidence supporting the endosymbiotic theory for the origin of chloroplasts? a. Chloroplasts have a single circular chromosome. b. Chloroplasts use two linked photosystems for photosynthesis. c. Chloroplasts use exclusively chlorophyll a and b. d. Chloroplasts in some algae are surrounded by two membranes. ANSWER: c 52. What data convinced biologists that photosynthesis in Paulinella chromatophora originated independently of the endosymbiotic events that gave rise to chloroplasts in green algae? a. Nucleotide sequences of genes in the chloroplasts of Paulinella show that these organelles are most closely related to cyanobacteria different from those from which green algal chloroplasts arose. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 b. Paulinella belongs to a different superkingdom than the green algae, and each superkingdom with photosynthetic members arose from a different endosymbiotic event. c. Green algae have cell walls made of cellulose, but Paulinella does not. d. Paulinella has a chloroplast that arose by means of a second endosymbiotic event, incorporating a photosynthetic eukaryote; the presence of multiple membranes in these chloroplasts supports its independent origin from chloroplasts in green algae. ANSWER: a 53. Why do mitochondria and chloroplasts have small genomes? a. because their bacterial ancestors did b. because they do not need many genes to function c. because they lack noncoding DNA d. because, over time, much of their DNA has migrated to the host cell nucleus ANSWER: d 54. Why have more groups within the archaeplastids not developed complex multicellularity? a. Many of the groups are recently evolved; with more time, they will all evolve members with complex multicellularity. b. Many of the groups are marine, and there are a limited number of marine niches that require complex multicellularity. c. Evolution is not goal-oriented, resulting in complex multicellularity; each group is successful with its current morphology. d. Only the most diverse groups have enough species for complex multicellularity to evolve. ANSWER: c 55. Chlorarachniophyte and cryptophyte algae contain three sets of genetic material: the nuclear genome, the chloroplast genome, and a set of genes found in the nucleomorph. What, in terms of the endosymbiotic hypothesis, is represented by the nucleomorph genes? a. genes introduced by viral infection b. genes that migrated from the host's nucleus into the symbiont c. genes that migrated from the chloroplast into the host's genome d. remnant genes from the nucleus of a eukaryotic algal endosymbiont ANSWER: d 56. Among eukaryotes, some multicellular organisms lack both chloroplasts and mitochondria. a. true b. false ANSWER: b 57. Which organism is your closest relative? a. a slime mold b. an apple tree Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 c. a mushroom d. the malaria parasite ANSWER: c 58. To which superkingdom do humans belong? a. Archaeplastida b. Alveolata c. Excavata d. Opisthokonta e. Rhizaria ANSWER: d 59. Which of the statements is not true regarding protists? a. They may have cell walls. b. They may be photosynthetic. c. They may be prokaryotic. d. They are usually aerobic, though some are not. e. Some are unicellular. ANSWER: c 60. Refer to the figure shown. Of all the green algae, which are most closely related to land plants?

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Chapter 25

a. streptophytes b. chlorophytes c. glaucocystophytes d. red algae ANSWER: a 61. Refer to the figure. Which of the seven superkingdoms are thought to share a most recent common ancestor?

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Chapter 25

a. Opisthokonta, Amoebozoa, Archaeplastida Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 b. Opisthokonta, Choanoflagellates, Fungi c. Archaeplastida, Stramenopila, Alveolata d. Stramenopila, Alveolata, Rhizaria ANSWER: d 62. To which superkingdom does the organism that causes amoebic dysentery belong? a. Stramenopila b. Alveolata c. Excavata d. Amoebozoa e. Rhizaria ANSWER: d 63. Refer to the image. Phylogenies based on sequences from organellar genomes and nuclear genomes from the same group of species yield phylogenetic trees with different branching patterns. How is this possible?

a. Nuclear and organellar genes evolved at different rates, obscuring evolutionary relationships. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 b. For organelles like chloroplasts, no mutations can occur, or photosynthesis would stop and the organism would die. The nuclear genome can accumulate mutations, so it can continue to evolve. c. Endosymbiosis makes this possible. The genome of the chloroplast is more closely related to members of the group in which it originally evolved, whereas the nuclear genome of the engulfing organism has its own evolutionary trajectory. d. Endosymbiosis makes this possible. Transfer of genetic material from the genome of the endosymbiont to the host results in two different phylogenies, host and endosymbiont. The host genome continues to accumulate mutations, but the endosymbiont genome cannot. ANSWER: c 64. Which of the superkingdoms contains at least one member that is heterotrophic, one member that has cilia, and one member that is parasitic? a. Alveolates b. Stramenopiles c. Rhizaria d. Excavata ANSWER: a 65. Some slime molds make structures called plasmodia that: a. contain several forms of chloroplasts. b. provide movement for the slime molds. c. contain many nuclei in one large cell. d. are part of the reproductive structures. ANSWER: c 66. What biological technique has provided evidence used in establishing the eukaryotic superkingdoms? a. molecular sequence comparisons b. bacterial culturing c. the identification of land plants d. visualizing cell shape ANSWER: a 67. Consider the phylogenetic tree. The branching pattern of this phylogeny suggests that all of the superkingdoms evolved photosynthetic species simultaneously.

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a. true b. false ANSWER: b 68. Which phylogenetic tree shows the correct phylogenetic relationships among these organisms: you, a mushroom, an oak tree, a diatom, and the parasite that causes malaria?

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a. Phylogeny 1 b. Phylogeny 2 c. Phylogeny 3 d. Phylogeny 4 ANSWER: a Multiple Response 69. What is the possible significance of the finding that genes related to bacteria and archaea occur in the eukaryotic genome? Select all that apply. a. The ancestor of the modern eukaryotic cell may have been a primitive, mitochondria-free cell with a nucleus. b. An archaeon may have engulfed a bacterium to form a eukaryotic cell. c. The presence of genes related to bacteria and archaea is a byproduct of meiosis. d. These genes may be the result of horizontal gene transfer. ANSWER: b, d 70. Which features are found in eukaryotic cells but not in prokaryotic cells? Select all that apply. a. dynamic cytoskeleton b. cell walls c. anaerobic metabolism Copyright Macmillan Learning. Powered by Cognero.

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Chapter 25 d. nuclear envelope ANSWER: a, d 71. Which organelles contain DNA? Select all that apply. a. the nucleus b. the ribosome c. mitochondria d. chloroplasts ANSWER: a, c, d 72. Chloroplasts are thought to have evolved from cyanobacteria. Which observations support this hypothesis? Select all that apply. a. Both chloroplasts and cyanobacteria have similar internal membranes that organize the light reactions of photosynthesis. b. Both chloroplasts and cyanobacteria have small, circular DNA genomes. c. Both chloroplasts and cyanobacteria have cell walls. d. Both chloroplasts and cyanobacteria have the capability of movement through flagella. ANSWER: a, b 73. Which types of organisms are found in the superkingdom Opisthokonta? Select all that apply. a. animals b. plants c. fungi d. protists ANSWER: a, c, d 74. The members of some protist groups have a mitosome instead of functional mitochondria. It is hypothesized that the mitosome is modified from a functional mitochondrion present in the ancestor. Based on this information, what would also be expected in groups exhibiting a mitosome? Select all that apply. a. The nuclear genome of mitosome-bearing eukaryotic cells may contain genes derived from mitochondria. b. The organism is not a eukaryote. c. The mitosome has two membranes. d. The engulfed symbiont was not a proteobacterium. ANSWER: a, c

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Chapter 26 Multiple Choice 1. One advantage that simple multicellular eukaryotes have over single-celled organisms is the ability: a. of the cells in the multicellular organism to communicate with one another. b. of the cells in a group of cells to avoid predation. c. of autotrophic cells in the multicellular organism to supply heterotrophic cells with nutrients. d. to respond the external environment more efficiently. ANSWER: b 2. Imagine a hypothetical habitat where there are many simple and complex multicellular organisms of roughly the same size. A new predator is introduced into the habitat that feeds by taking bites of the organisms, rather than consuming them entirely. Given this, which of the statements is most likely? a. Complex multicellular organisms will suffer more, because vital differentiated cells that the rest of the organism requires will be lost. b. Complex multicellular organisms will suffer less because their ratio of surface area to volume will increase, and they will absorb more nutrients. c. Simple multicellular organisms will suffer more, because exterior cells responsible for nutrient absorption will be lost. d. Simple multicellular organisms will suffer more, because the greater ratio of surface area to volume will decrease rates of diffusion. ANSWER: a 3. Each of the seven superkingdoms of eukaryotes contains one member that is a complex multicellular organism. a. true b. false ANSWER: b 4. Coenocytic organization is a shared derived character of simple multicellular organisms exhibiting the trait. a. true b. false ANSWER: b 5. Which of the options is a property of simple multicellularity? a. extensive communication between cells. b. cells are highly differentiated and specialized. c. most cells retain a full range of functions. d. some cells are not in direct contact with the external environment. ANSWER: c 6. A marine biologist discovers a new species of green algae, which, under the microscope, appears to be multinuclear. Which of the options is likely true of these algae? a. It is a complex multicellular organism. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 b. It harbors parasites. c. It is coenocytic. d. It lives suspended in the water column. ANSWER: c 7. Which of the choices correctly explains features shared by simple multicellular eukaryotes? a. Cells adhere to one another, and most of the cells are capable of a wide range of functions, including reproduction. The loss of one or more cells usually does not lead to the death of the organism. b. Cells adhere to one another and engage in extensive communication with one another, and most of the cells are capable of a wide range of functions, including reproduction. The loss of one or more cells usually does not lead to the death of the organism. c. Cells adhere to one another and most of the cells are capable of a wide range of functions, except the ability to reproduce, which means that the loss of certain cells usually leads to the death of the organism. ANSWER: a 8. Cells in the interior of complex multicellular organisms are able to respond to environmental signals through: a. receptors designed to directly detect changes in the external environment. b. diffusion of environmental cues through the outer layers of the body. c. receptors for signals sent from exterior cells that relay information about environmental changes. d. pH, temperature, or osmolarity receptors in the membranes of interior cells. ANSWER: c 9. Which of the statements is a selective advantage of simple multicellularity over single-celled organisms? a. Multicellularity helps organisms avoid predation. b. Multicellularity results in fundamental changes in metabolic processes. c. Multicellularity helps organisms utilize more food sources. d. Multicellularity helps organisms enhance efficiency of reproduction. ANSWER: a 10. Which of the organisms has coenocytic organization? a. humans b. amphibians c. plants d. some green algae ANSWER: d 11. Like complex multicellularity, coenocytic morphologies evolved multiple times in the eukaryotic tree. a. true b. false ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 12. Which of the statements about the movement of oxygen in complex multicellular organisms is true? a. Oxygen is moved into and out of cells by specific transport proteins. b. The movement of oxygen by diffusion into and out of cells is limited to just a few layers of cells. c. Bulk flow systems are the only mechanism used to transport oxygen. ANSWER: b 13. Which of the statements concerning complex multicellular organisms is true? a. Plants and animals use bulk flow systems to move gasses and nutrients, but fungi and algae do not. b. Bulk flow systems require specialized organs, like a heart, to pump nutrients and dissolved gasses deep into tissues that are not in contact with the external environment. c. Bulk flow systems move substances faster than simple diffusion. ANSWER: c 14. In humans and many other animals, what organ system is involved in bulk transport of nutrients, oxygen, and signaling molecules? a. respiratory system b. circulatory system c. immune system d. digestive system ANSWER: b 15. The passive movement of ions or molecules from a point of high concentration to a point of low concentration is known as: a. solubility. b. active transport. c. endocytosis. d. diffusion. ANSWER: d 16. What is the main limitation on cell size? a. bulk flow b. diffusion c. oxygen d. cell membrane strength ANSWER: b 17. Which of the features is an adaptation, shared by virtually all complex multicellular organisms, that allows organisms to receive the oxygen and nutrients they need? a. amplification of surface areas b. active transporters for oxygen and glucose c. restricted body size that corresponds to the environment d. tissues or organs that function as pumps Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 ANSWER: a 18. The evolution from simple multicellular organisms to complex multicellular organisms required: a. a mechanism of cell-cell adhesion. b. an ability to respond to the environment. c. the development of a system that enabled bulk flow. d. cell-surface receptors. ANSWER: c 19. The oxygen needed for respiration moves from the lungs into the capillaries by simple diffusion. Once in the capillaries, what molecule or other component of blood is responsible for transporting oxygen in the circulatory system? a. There is no transport molecule; oxygen remains free in the blood. b. white blood cells c. cadherins d. hemoglobin ANSWER: d 20. Why is bulk flow a necessary condition of complex multicellularity? a. Complex multicellularity reflects cell adhesion between cells, and this cannot happen without bulk flow of signaling molecules from the environment. b. Organisms exhibiting complex multicellularity are large and have cells/tissues that are not in contact with their environment. Systems that can overcome the limits of diffusion must evolve for basic physiologic processes to occur. c. Bulk flow is necessary to move large amounts of liquid through the organism, depending on physiologic needs. The tissues can all receive necessary nutrients from the environment, but bulk flow speeds the rate of diffusion. d. Complex multicellularity is related to the increase in oxygen in the environment. Because bulk flow systems are only related to the delivery of oxygen to tissues, higher environmental oxygen selected for individuals with bulk flow systems. ANSWER: b 21. Organisms that need oxygen for cellular respiration depend on which of the processes to move oxygen into cells? a. simple diffusion b. active transport c. endocytosis d. facilitated diffusion ANSWER: a 22. Which of the statements regarding plasmodesmata is true? a. They provide points of structural support between adjacent animal cells. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 b. They are channels, located in the plasma membrane of plant cells and made of transmembrane proteins, that allow ions and small molecules to pass between adjacent cells. c. They are channels, located in the plasma membrane of animal cells and made of transmembrane proteins, that allow ions and small molecules to pass between adjacent cells. d. They provide points of structural support between adjacent plant cells. e. They are holes in the cell wall through which the plasma membrane and endoplasmic reticulum of adjacent cells are connected. ANSWER: e 23. Which of the molecules helps maintain cell-cell adhesion in plants? a. cadherins b. pectins c. integrins d. cellulose ANSWER: b 24. Choanoflagellates are: a. unicellular organisms. b. simple multicellular organisms. c. complex multicellular organisms. d. prokaryotes. ANSWER: a 25. What induces multicellularity in a choanoflagellate species? a. the presence of predators in the environment b. the presence of the preferred prey/food source in the environment c. the developmental signal molecules that induce differentiation d. the cell migrations that take place during the developmental phases of growth ANSWER: b 26. The vegetative cells of the simple multicellular organism Volvox function in: a. reproduction. b. photosynthesis. c. oxygen uptake. d. All of these choices are correct. ANSWER: b 27. Animals are to cadherins as plants are to: a. integrins. b. epithelia. c. adhesion. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 d. pectins. ANSWER: d 28. What structures are used in animal cells to move molecules between cells? a. plasmodesmata b. gap junctions c. cadherins d. plasma membranes ANSWER: b 29. Which of the statements regarding gap junctions is true? a. They are holes in the cell wall through which the plasma membrane and endoplasmic reticulum of adjacent cells are connected. b. They are channels, located in the plasma membrane of plant cells and made of transmembrane proteins, that allow ions and small molecules to pass between adjacent cells. c. They are channels, located in the plasma membrane of animal cells and made of transmembrane proteins, that allow ions and small molecules to pass between adjacent cells. d. They provide points of structural support between adjacent plant cells. e. They provide points of structural support between adjacent animal cells. ANSWER: c 30. Which of the signals is unlikely to cause a change in gene expression in complex multicellular organisms? a. oxygen levels b. nutrients in the environment c. temperature d. All of these choices are correct. ANSWER: d 31. Which of the molecules are adhesion proteins that are present in both choanoflagellates and animals? a. actins b. pectins c. cadherins d. hemoglobins ANSWER: c 32. How can two cells of a complex multicellular organism have the same genome but markedly different structures and functions? a. During differentiation, as cells become specialized, they lose portions of their genome that are no longer needed. b. Different genes are expressed in different cells. c. Part of the process of differentiation of cells is to create mutations in the genome. d. Once differentiation begins, the genome no longer controls cellular functions. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 ANSWER: b 33. The lion's mane jellyfish can be up to 8 feet wide. In spite of its large size, it has no structures for bulk flow to deliver oxygen to its cells. How can it be so large and yet lack a mechanism for bulk flow? a. The jellyfish is surrounded by water, so the concentration gradient between the water and the jellyfish is steep, making the rate of diffusion across tissues very high. b. The jellyfish is filled with tissue made of structural cells that are not metabolically active. Therefore, a mechanism to transport oxygen to its interior is not necessary. c. The jellyfish body is actually a series of intricate folds, so the ratio of surface area to volume is maximized, and diffusion occurs as it would across a tissue one or two cells thick. d. The jellyfish does not have receptors for sensing the environment, so does not need a mechanism for transmitting signal molecules throughout the body. ANSWER: b 34. Choanoflagellates are single-celled eukaryotes that feed on bacteria. In the presence of their food bacteria, some choanoflagellates develop multicellular structures in which the individual choanoflagellate cells are attached to one another. Based on this observation, it may be assumed that: a. the bacterial cells, or a chemical released by the bacteria, bind to a receptor on the plasma membrane of the choanoflagellate cells. b. the choanoflagellate cells use cadherins to attach to each other. c. the choanoflagellates use integrins to attach to each other. d. choanoflagellates are complex multicellular organisms, because they alternate between single-celled and multicellular states. ANSWER: a 35. Cell-cell signaling is limited to complex multicellular organisms composed of organs and organ systems. a. true b. false ANSWER: b 36. Complex multicellular organisms have a genome similar in scope and size to their single-celled predecessors. a. true b. false ANSWER: b 37. Which of the statements is a characteristic of complex multicellularity? a. All cells are exposed to an exterior surface. b. The interior cells are more tolerant to lower oxygen levels than exterior cells. c. Interior cells are exposed to a different physical and chemical environment than exterior cells. d. All cells of complex multicellular organisms express the same genes. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 38. Cardiac muscle cells have gap junctions between adjacent cells. What is the function of those gap junctions? a. They maintain isotonic conditions. b. They facilitate cell-cell adhesion. c. They are involved in cell-cell signaling of the cells of the heart. d. They make nutrients, like glucose, readily available to the muscle cells of the entire heart. ANSWER: c 39. Some species of choanoflagellates can form multicellular structures in the presence of their desired food bacterium; however, not all choanoflagellate species will form multicellular structures. Multicellularity is induced in some species of choanoflagellates because of signals released by the bacterium. This variation in the ability to develop multicellular structures is most likely the result of: a. changes in patterns of gene expression resulting in multicellular aggregations in some, but not all, choanoflagellates. b. changes in the pattern of integrin expression in choanoflagellates forming multicellular aggregations. c. the evolution of genes for cell-adhesion molecules only in choanoflagellates that form multicellular aggregations. d. the evolution of differential responses to environmental signals. ANSWER: d 40. Three superkingdoms—the opisthokonts, archaeplastids, and stramenopiles—all have members with complex multicellularity. What must be true about the members with complex multicellularity in these groups? a. They have tissues that are not in direct contact with their environment. b. They form aggregations in the presence of specific signals. c. They are predatory. d. They all produce integrins and cadherins. ANSWER: a 41. Meristems enable plants to develop complex anatomical structures despite the fact that: a. plant cells cannot move. b. plants have flagella only in root cells. c. plants have vascular tissue. d. plants need to capture sunlight for energy. ANSWER: a 42. Unlike plant cells, animal cells move during development. Which of the terms describes the reorganization of undifferentiated cells that results in the formation of an embryonic structure with multiple layers? a. blastula formation b. gastrula formation c. organogenesis d. cleavage ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 43. Most plants are firmly rooted in place and are not mobile. Which of the options is an adaptation to this condition? a. growth by mitosis b. the presence of hairs, spines, and toxins c. cellular migration during gastrulation d. cell-cell signaling e. lack of a cell wall in some cells ANSWER: b 44. Gastrulation in animal cells is able to take place because: a. animal cells do not have cell walls. b. mitosis stops so that cells are able to move around. c. cells in the meristem are able to migrate. ANSWER: a 45. In which of the tissues are undifferentiated cells of plants most likely to be present? a. leaves b. transport systems (xylem and phloem) c. root tips d. flowers ANSWER: c 46. An inhibitor of mitosis would most likely have the greatest effect on the _____ of a plant. a. meristems b. presence of cell walls c. cells of the vascular tissue d. gastrula ANSWER: a 47. Which of the terms describes the process by which a fertilized egg develops into a multicellular organism with many different cell types, each with different structures and functions? a. adhesion b. development c. communication d. embryology ANSWER: b 48. What is the term for a ball of undifferentiated cells that results from repeated mitosis of a fertilized egg? a. cleavage b. blastula c. gastrula Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 d. endoderm ANSWER: b 49. Plant cell division is confined to: a. flowers. b. vascular tissues. c. meristems. d. photosynthetically active tissues. ANSWER: c 50. What features of complex multicellular organisms are also present in their closest single-celled relatives? a. cell differentiation b. cell adhesion molecules c. receptors for cell-cell signaling d. regulation of gene expression e. All of these choices are correct. ANSWER: e 51. In which order are the key features of complex multicellularity likely to have evolved? a. cell adhesion, cell communication, regulated growth and development b. cell communication, cell adhesion, regulated growth and development c. regulated growth and development, cell communication, cell adhesion d. regulated growth and development, cell adhesion, cell communication ANSWER: a 52. What enabled vascular plants to spread across the land? a. the increase in atmospheric oxygen b. the coevolution of coenocytic cells c. the presence of bulk flow systems d. the evolution of new photosynthetic pathways ANSWER: c 53. Refer to the figure shown. Increased oxygen levels in the atmosphere occurred approximately 600 million years ago. This coincided with the simultaneous evolution of animals and terrestrial plants.

a. true b. false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 ANSWER: b 54. Which of the options is believed to be the major event that led to the appearance of large complex animals on Earth? a. the appearance of bulk flow systems b. the diversification of seaweeds in the ocean c. the appearance of plants on land d. the increase in atmospheric oxygen ANSWER: d 55. Refer to the figure shown. The first complex multicellular organisms appeared approximately _____ million years ago.

a. 50 b. 580 c. 1800 d. 2400 ANSWER: b 56. The acquisition of traits necessary for complex multicellularity occurred independently in plants and animals, but took place in a specific order. In what order did these processes evolve? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 a. a mechanism of communication, bulk flow, cell adhesion molecules b. bulk flow, a mechanism of communication, cell adhesion molecules c. cell adhesion molecules, a mechanism of communication, bulk flow d. a mechanism of communication, cell adhesion molecules, bulk flow ANSWER: c 57. The early evolution of land plants occurred in moist lowland habitats until about 400 million years ago, when they moved to the interiors of continents. Which of the characteristics likely contributed to the colonization of continental regions further from standing water? a. absorption of water through diffusion on photosynthetic tissue b. ability to perform photosynthesis c. development of vascular tissue for bulk flow d. cell-cell adhesion, which provides structural support for multicellularity ANSWER: c 58. Plants, animals, and some fungi are complex multicellular organisms that we encounter all the time. In which order did these groups of organisms likely appear on the planet? a. animals, then land plants, then complex fungi b. complex fungi, then land plants, then animals c. complex fungi, then animals, then land plants d. land plants, then animals, then complex fungi e. land plants, then fungi, then animals ANSWER: a 59. In the relatively new discipline of evolutionary developmental biology, or evo-devo, scientists aim to: a. integrate comparative morphology and the fossil record to reconstruct evolutionary history. b. integrate physiology and anatomy to assess the evolution of function. c. integrate genetics, comparative morphology, and phylogeny to understand the genetic basis of evolutionary change. d. integrate developmental biology and ecology to understand the distribution of species populations among environments. ANSWER: c 60. Which of these is key to the evolution of complex multicellularity? a. cell adhesion b. structures such as plasmodesmata or gap junctions for targeted signaling between cells c. regulation of gene expression within and between cells d. tissues and organs that use fluids to transport nutrients, oxygen, and molecular signals through the body e. No one of these is sufficient without the others; complex multicellularity reflects the evolutionary accumulation of multiple characters. ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 61. Sulfate, nitrate, and ferric iron are readily available, exist in appreciable amounts, and act as terminal electron acceptors in anaerobic respiratory pathways. If multiple electron acceptors existed before 575 million years ago, why do we not see evidence of complex multicellularity earlier in Earth's history? a. These compounds are very stable as gasses, so they do not readily act as electron acceptors. b. The characteristics necessary for complex multicellularity did not evolve until 575 million years ago. c. The ability of those compounds to accept electrons is lower than that of oxygen, so insufficient amounts of energy are liberated. d. The ability of these compounds to accept electrons is higher than that of oxygen, but they are scarce in the environment and thus could not support the evolution of complex multicellularity. ANSWER: b 62. The discovery that single-celled sister groups of animals synthesize proteins known to play roles in animal development tells us that: a. the single-celled sister groups are derived from more complex multicellular ancestors. b. the single-celled sister groups use the proteins in question for functions other than multicellular development. c. these proteins are found ubiquitously through the eukaryotic tree of life. d. All of these choices are correct. ANSWER: b 63. Imagine that a clade of sulfate-respiring heterotrophs in the ocean evolved complex multicellularity. How likely is this new clade to colonize the land surface successfully? a. very likely, as animals, plants, and fungi all became successful on land b. not very likely, as the land surface is already saturated with multicellular organisms c. very likely, because colonists that exploit a resource (sulfate, in this case) not used much by existing land organisms will prosper d. not very likely, as air does not contain sulfate ANSWER: d 64. Refer to the figure shown. The adhesion molecules in plants are most similar to those of:

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Chapter 26

a. animals. b. choanoflagellates. c. green algae. d. cyanobacteria. ANSWER: c 65. Within the evolutionary history of animals, there have been multiple duplications of the regulatory homeotic genes found in the hypothetical animal ancestor. How does this relate to the evolution of millions of complex multicellular animals? a. Each copy of the regulatory genes can evolve to alter patterns of development in the animal, resulting in many different morphologies. b. Each copy of the regulatory gene can evolve independently, but different copies of the genes are lost, so that only one of each type remains in different lineages. This results in different animal morphologies. c. Each copy of the regulatory gene is specific to one lineage and results in different animal morphologies. ANSWER: a 66. Consider the image shown. At which point would you expect the appearance of plasmodesmata?

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Chapter 26

a. Point A b. Point B c. Point C ANSWER: b Multiple Response 67. Which of the answer choices is a characteristic of complex multicellular organisms but not simple multicellular organisms? Select all that apply. a. the presence of subsets of cells (tissues) with specialized functions b. the use of cell-adhesion molecules to allow cells to adhere to one another c. the presence of a single cell with multiple nuclei d. the presence of cells that are not in direct contact with the external environment ANSWER: a, d 68. Differentiation of multiple cell types occurs in which organisms? Select all that apply. a. animals. b. fungi. c. protists. d. bacteria. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 26 e. choanoflagellates. f. green algae. ANSWER: a, b, c, d 69. How do the cells within your body become differentiated? Select all that apply. a. The cells have different genes. b. The cells' pattern of gene expression is regulated. c. The cells are in different local environments. d. Cells differentiate as they divide by mitosis. ANSWER: b, c 70. What is a possible mechanism for the immense diversity observed in complex multicellular organisms? Select all that apply. a. mutations in regulatory genes b. mutations in essential genes c. interactions with the environment d. a lack of cell walls ANSWER: a, c 71. In complex multicellular organisms, only a few cells become differentiated as reproductive cells. Given this observation, which statements are true? Select all that apply. a. Nutrients for reproductive cells must be taken in by other cells and transferred through the body. b. The loss of reproductive potential in cells differentiated for other functions severely limits the evolutionary potential of complex multicellular organisms. c. Reproductive cells gain nutrients by diffusion from the environment, and so must lie at the surface of complex multicellular organisms. d. Signaling molecules from surrounding tissues can influence the development of eggs and embryos. ANSWER: a, d 72. What properties of oxygen make it a critical environmental requirement for multicellularity? Select all that apply. a. Oxidation by O2 can provide a large amount of energy. b. Oxygen does not easily diffuse. c. Oxygen is present at low levels at the ocean floor. d. Oxygen is a gas. ANSWER: a, d

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Chapter 27 Multiple Choice 1. One of the earliest evolutionary adaptations that helps plants retain water in a terrestrial environment is the: a. leaf. b. xylem. c. phloem. d. root. e. cuticle. ANSWER: e 2. The main tissue types found in all vascular plants include the _____, which provides the main interface with the environment of the plant; the _____, which is/are specialized for transport of water or carbohydrates; and the _____, which includes all the other cells. a. leaves; xylem; parenchyma b. epidermis; vascular tissues; ground tissue c. parenchyma; vascular tissues; ground tissue d. cuticle; xylem; ground tissue ANSWER: b 3. If you were asked to interpret the evolution of the waxy cuticle and stomata of leaves, which of the factors would be in your strongest argument? a. carbon dioxide gain and water loss b. structural support and water gain c. reproduction and dispersal d. protection from excessive sunlight and dispersal ANSWER: a 4. What explains the higher density of microorganisms near roots? a. Microorganisms can take advantage of the much higher O2 levels that result from xylem transport. b. Microorganisms are attracted by the same mineral nutrients that the roots absorb. c. A fraction of the carbohydrates transported to the roots leaks into the soil. d. The microorganisms are plant pathogens that are attempting to infect the roots. ANSWER: c 5. What is the role of the PEP carboxylase enzyme during both CAM and C4 photosynthesis? a. to capture CO2 in the form of HCO3 b. to capture CO2 in the form of 4-carbon organic acids c. to capture CO2 in the form of 3-carbon organic acids d. to create ATP ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 6. Which of the plants would likely be the first to colonize a newly formed volcanic island (an environment of bare rock with no developed soil)? a. gymnosperms b. lycophytes c. vascular plants d. bryophytes ANSWER: d 7. Assuming the stomata are open to the same degree, the rate of transpiration should _____ on a rainy day compared with a sunny day, because the relative size of the H2O concentration gradient from the inside to the outside of the leaf would _____. a. decline; increase b. decline; decrease c. remain the same; not change d. increase; decrease e. increase; increase ANSWER: b 8. CAM plants do NOT operate the Calvin cycle at night because: a. it requires ATP and NADPH made only during daylight. b. PEP carboxylase is only active in darkness. c. CAM plants capture HCO3-, but the Calvin cycle requires CO2. d. CAM plants close their stomata during the day. ANSWER: a 9. CAM photosynthesis improves the CO2: H2O exchange ratio because: a. organic acids are decarboxylated at night. b. organic acids are decarboxylated in daylight. c. CO2 capture occurs at night when the gradient for diffusion of H2O out of the leaf is smaller. d. CO2 capture occurs at night, when the gradient for diffusion of CO2 into the leaf is larger. ANSWER: c 10. If you were evaluating varieties of rice, normally a C3 plant, in the hope of identifying a new variety capable of C4 photosynthesis, what level(s) of enzyme expression would you consider the most promising? a. Both enzymes are highly expressed in both cell types. b. Both enzymes are higher in bundle-sheath cells than in mesophyll cells. c. Rubisco levels are lower in bundle-sheath cells than in mesophyll cells. d. PEP carboxylase levels are higher in mesophyll cells than in bundle-sheath cells. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 11. C4 plants are usually _____ in shady habitats, because C4 photosynthesis has _____ ATP requirements than the C3 pathway. a. rare; lower b. common; higher c. rare; higher d. common; lower ANSWER: c 12. Recall that C4 plants tend to have higher rates of photosynthesis when compared with C3 plants (under normal 21 percent O2 atmospheric conditions). If photosynthetic rates are higher, why are all plants not C4 plants? a. Because of leaf veins, C4 photosynthesis requires more H2O than C3 photosynthesis. b. Because of rubisco, C4 photosynthesis consumes more O2 than C3 photosynthesis. c. Because of mesophyll cells, C4 photosynthesis requires more chlorophyll than C3 photosynthesis. d. Because the regeneration of PEP consumes ATP, C4 photosynthesis consumes more energy than C3 photosynthesis. ANSWER: d 13. Plants developed both cuticles and crassulacean acid metabolism (CAM) in an attempt to limit: a. excess photosynthesis. b. water loss. c. PEP carboxylase loss. d. photorespiration. ANSWER: b 14. Assuming the same pressure gradient, flow through a xylem conduit with a diameter of 100 mm is _____ times faster than a similar conduit with a 25-mm diameter. a. 16 b. 64 c. 256 d. 4 ANSWER: c 15. In which direction does phloem sap flow if there is a greater concentration of sucrose molecules in the root phloem as compared with the leaf phloem? a. in the opposite direction to xylem b. from leaf to root c. There would not be enough turgor pressure to move the sap. d. flow from root to leaf ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 16. What would happen to the cells in a leaf if roots could not exert selective control over the uptake of dissolved minerals from the soil? a. Evaporation from the cell surfaces inside the leaf would stop and the leaf would overheat. b. Sodium would accumulate in the shoot to the point that cells became damaged. c. They would become enriched in mineral nutrients and attract herbivores. d. The roots would accumulate salts, causing the xylem to reverse flow and thereby drying out the cells in the leaf. ANSWER: b 17. Which of the organisms are classified as land plants, but not as vascular plants? a. algae b. angiosperms c. lycophytes d. gymnosperms e. liverworts ANSWER: e 18. CAM plants: a. specifically avoid photorespiration. b. store CO2 in bundle-sheath cells. c. perform the Calvin cycle at night. d. capture CO2 at night. e. are very efficient at photorespiration. ANSWER: d 19. Rubisco's affinity for CO2 versus O2 is inversely related to temperature, resulting in _____ as temperatures rise. a. a slowing rate of enzymatic reaction b. increased photosynthesis c. increased photorespiration d. higher dissolved O2 levels in the chloroplast ANSWER: c 20. Water transport in xylem depends on: a. osmosis across the vessel plasma membranes. b. turgor pressure accumulating in the roots. c. hydrogen bonding between H2O molecules. d. ATP-driven pumps in root cells. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 21. Suppose a mutation arises in a population of sunflowers that results in xylem vessels twice as large in diameter as the wild-type vessels. These mutant xylem vessels will have _____ water transport capacity, and will be _____ susceptible to cavitation from freezing. a. higher; less b. higher; more c. lower; more d. lower; less ANSWER: b 22. The ability of water to support tension in the xylem is a function of the: a. hydrogen bonds between water molecules. b. hydrogen bonds within a water molecule. c. partial negative charge on the H atoms of a water molecule. d. nonpolar covalent bonds between the H and O atoms within a water molecule. ANSWER: a 23. When a blooming rose stem is cut from a rose bush, phloem sap continues to ooze out of the cut end of the stem. This is due to: a. turgor pressure. b. cavitation pressure. c. torsion pressure. d. air pressure. e. atmospheric pressure. ANSWER: a 24. Grapes growing on a vine are observed to shrink slightly during the day and increase in size at night. This is because: a. in leaves, phloem flow to growing sinks only occurs when the stomata are closed, because the transport capacity of the xylem is too small to supply water for both transpiration and phloem flow. b. developing fruits, such as grapes, transpire a lot of water to stay cool. c. photosynthesis to produce sugar occurs in the light, but respiration and, therefore, the growth of new cells occurs mostly in the dark. d. during the day, transpiration pulls on the water in the whole plant, causing the grapes to shrink slightly. ANSWER: d 25. Why is flooding harmful to many plant species? a. Flooding impairs the function of symbiotic bacteria. b. Water-logged soils have little CO2. c. Flooding reduces soil pH. d. Water-logged soils have little O2. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 ANSWER: d 26. Before the development of the Haber-Bosch process, NH3 could be produced from N2 only by: a. fungi. b. plants. c. prokaryotes. d. respiration. ANSWER: c 27. Nitrogen fixation: a. converts N2 to ammonia, making it available for incorporation into amino acids and nucleotides. b. incorporates molecular nitrogen (N2) taken from the air into amino acids and nucleotides. c. releases N2 gas to the atmosphere. d. is necessary for the electron-transport chain, because nitrogen acts as the final electron acceptor. e. is performed by biochemical pathways in the plant cells. ANSWER: a 28. A researcher discovers a mutant pea plant that is missing the Casparian strip. What process will be affected in this mutant? a. the ability of the plant to perform CAM b. the diffusion of CO2 into the leaves c. the selectivity of mineral nutrients absorbed by roots d. the opening and closing of stomata e. both the diffusion of CO2 into the leaves of the plant and the opening and closing of stomata ANSWER: c 29. Consider the three soil water profiles (A, B, and C) in the figure shown. You would expect bryophytes instead of vascular plants to be the dominant type of plant in environments for which soil water profile?

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Chapter 27

a. is constantly C b. varies spatially between A and C c. alternates temporally between B and C d. alternates temporally between A and B ANSWER: d 30. In many plants, it is possible to peel off the lower epidermis of a leaf without disturbing the photosynthetic mesophyll cells, thereby exposing the mesophyll directly to the air. What would you expect the immediate (seconds to minutes) response of photosynthesis to be? What would you expect the longer-term (minutes to hours) response of photosynthesis to be? a. immediate increase; long-term decrease b. immediate decrease; long-term increase c. no change immediately; long-term decrease d. no change immediately; long-term increase ANSWER: a 31. Suppose you set up an experiment, as illustrated in the figure shown, in which you compare the photosynthesis in C3 and C4 plants in two different growth conditions.

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Chapter 27

You would predict the _____ to have a photosynthetic advantage in cool, dim conditions and the _____ to have an advantage in hot, bright conditions. a. C3 plants; C4 plants b. C4 plants; C3 plants c. C3 plants; C3 plants d. C4 plants; C4 plants ANSWER: a 32. Suppose you set up an experiment, as illustrated in the figure shown, in which you compare the photosynthesis in C3 and C4 plants in warm temperatures and bright light, but with controlled atmospheres with different oxygen concentrations (normal air is about 21 percent oxygen). Carbon dioxide levels in both experiments are the same (0.04 percent).

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Chapter 27

You would predict the _____ to have a photosynthetic advantage in 1 percent oxygen and the _____ to have an advantage in 21 percent oxygen. a. C4 plants; C4 plants b. C4 plants; C3 plants c. C3 plants; C3 plants d. C3 plants; C4 plants ANSWER: d 33. Suppose you set up an experiment, as illustrated in the figure shown, in which you compare the photosynthesis in C3 and C4 plants in cool temperatures and dim light, but with controlled atmospheres with different oxygen concentrations (normal air is about 21 percent oxygen).

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You would predict the _____ to have a photosynthetic advantage in 1 percent oxygen and the _____ to have an advantage in 21 percent oxygen. a. C3 plants; C4 plants b. C4 plants; C3 plants c. C3 plants; C3 plants d. C4 plants; C4 plants ANSWER: c 34. In C4 plants, the Calvin cycle is performed in _____; CO2 is captured during the C4 cycle in _____. a. bundle-sheath cells; mesophyll cells b. mesophyll cells; bundle-sheath cells c. the day; the night d. the night; the day ANSWER: a 35. Openings between _____ cells create pores in the leaf epidermis that play a critical role in photosynthesis by allowing diffusion of atmospheric CO2 into a leaf's interior. a. companion b. bundle-sheath c. guard d. mesophyll ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 36. During photorespiration, which of the molecules acts as a substrate for rubisco? a. N2 b. CO2 c. H2O d. O2 ANSWER: d 37. Plants lose a great deal of water from their leaves in a process referred to as: a. transduction. b. transformation. c. transpiration. d. transfection. ANSWER: c 38. Measurements of the diameter of a plant stem during a long drought show a gradual decrease over time and then a sudden return to almost its original diameter, even as the drought continues. Which of the hypotheses is the most likely explanation for the increase in stem diameter? a. Air entry and cavitation released the tension in the stem xylem. b. Water flow from the soil increased due to greater stomatal opening. c. Water flow from the soil increased due to the use of ATP to drive the selective uptake of nutrients, leading to water uptake by osmosis. d. Redistribution of water from the leaves swelled the stem xylem. ANSWER: a 39. Gymnosperms, such as spruces, often dominate high-latitude boreal forests, whereas angiosperms dominate rainforests. This is, in part, explained by the fact that tracheids: a. do not freeze as easily as vessels, maintaining flow to support photosynthesis at colder temperatures, which is not an advantage in the tropics. b. are less susceptible to freezing-induced cavitation than vessels but less efficient for moving large volumes of water. c. mature more quickly than vessels, allowing gymnosperms to make more efficient use of the shorter growing seasons of boreal forests. d. better withstand the large tensile forces required to extract water from frozen soils than vessels, but offer no advantage in warm tropical soils. ANSWER: b 40. What happens to the tension in the xylem when the stomata close and evaporation stops? a. It drops to a value closer to that of the water-filled pores in the soil. b. It remains constant the moment the stomata close. c. It goes to zero once evaporation ceases. d. It continues to increase until the living cells of the leaf are fully hydrated. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 ANSWER: a 41. From personal experience, you may have noticed that cut flowers placed in a vase with tap water continue to live for several days. What is the most likely explanation as to why cut flowers don't immediately wither and die? a. Cavitation allows water to continue to be transported through the flower stems. b. The tap water in the vase is an excellent source of nitrogen. c. Cells in the flower petals continue to perform photosynthesis. d. The plant can still take up water from the vase via an "evaporative pump" or evaporative loss. ANSWER: d 42. If water molecules (H2O) suddenly stopped forming hydrogen bonds with each other, how would water transport in vascular plants change, if at all? a. It would remain the same, because water transport depends primarily on osmosis. b. It would be reduced, because water transport relies on H2O molecules being connected by hydrogen bonds. c. It would remain the same, because H2O molecules also form oxygen bonds. d. It would increase, because hydrogen bonds inhibit water transport. ANSWER: b 43. Which of the processes can occur without the plant having to expend metabolic energy? a. movement of water through the xylem b. DNA replication c. movement of ions into guard cells d. nutrient transport across endodermal cells ANSWER: a 44. How are sieve plates and inter-vessel pits similar in their effect on transport through phloem and xylem conduits? a. Both prevent organelles from being transported from cell to cell. b. Both support membranes that use ATP to drive flow through conduits. c. Both increase resistance to flow but allow compartmentalization of injury. d. Both act as filters that limit the viscosity of the transported fluid. ANSWER: c 45. Grapes growing on a vine are observed to shrink during the day and increase in size at night. You decide to do an experiment wherein you freeze a stem supplying a cluster of grapes with a copper clamp cooled to just a few degrees below zero and then thaw it. The rationale for this treatment is that freezing the large-diameter xylem vessels of a vine leads to the formation of air bubbles, but has no long-term effect on phloem transport. In the days following this treatment, you expect to see, for the experimental cluster, that: a. grape size remains constant. b. grape size continues to shrink during the day and increase at night. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 c. the grapes dry out and shrivel due to freeze-thaw cavitation. d. grape size increases during both the day and the night. ANSWER: d 46. Imagine that you are studying the turgor pressure at two different ends of a sieve tube. One end of the sieve tube is located within a mature leaf, and the other end is located within the roots. When comparing the turgor pressure within the sieve tube at these two locations, what would you expect to find? a. Turgor pressure will not exist at either location. b. The turgor pressure will be greater at the root end. c. The turgor pressure will be the same at the leaf and root ends. d. The turgor pressure will be greater at the leaf end. ANSWER: d 47. Roots elongate continuously in order to: a. mine new regions of soil for nutrients. b. act as a carbohydrate sink. c. obtain CO2. ANSWER: a 48. Some angiosperms produce cluster roots, which are roots that have high metabolic rates and exude large quantities of organic acids into the soil. What would you expect the most useful benefit of these organic acids to be for the plant? a. They lower soil pH to liberate tightly bound minerals. b. They provide energy for symbiotic microbes. c. They raise soil pH to liberate tightly bound minerals. d. They provide energy for free-living soil microbes. ANSWER: a 49. Which of the statements is not a reason that mycorrhizae are beneficial for plants? a. They provide additional resistance to pathogens. b. Their thin cells can access regions of soil that plant roots cannot. c. They can neutralize phosphoric acid. d. They can enhance the release of nutrients from the soil. ANSWER: c 50. Applying fungicides to plants may result in the plants showing signs of phosphorus deficiency. What is the most likely explanation for this observation? a. The fungicide destroys phosphorus transporters in the plasma membrane of root cells. b. After fungicide is sprayed, plant cell walls impede the uptake of phosphorus to limit the amount of fungicide that enters the plant. c. The fungicide blocks ATP synthesis, so phosphorus cannot be absorbed into plant cells. d. The fungicide inhibits nitrogen fixation. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 e. The fungicide destroys mycorrhizal symbionts. ANSWER: e 51. If the two guard cells of a stoma increase their concentration of potassium (K+) ions, what happens to the corresponding pore? a. The pore opens, allowing water vapor to diffuse into the leaf. b. The pore closes, preventing CO2 from diffusing out of the leaf. c. The pore closes, preventing water vapor from diffusing out of the leaf. d. The pore opens, allowing CO2 to diffuse into the leaf. ANSWER: d 52. Which of the structures are produced by roots in order to increase surface area, allowing greater access to nutrients in the soil? a. root parenchyma b. root nodules c. root hairs d. root Casparian strips ANSWER: c 53. A research team determines that a plant has a symbiotic relationship with fungi. Upon closer examination, they notice that fungal cells form a sheath enclosing the root tip, but no arbuscules are present. This is an example of: a. root nodules. b. endomycorrhizae. c. ectomycorrhizae. ANSWER: c 54. Imagine that you are comparing the nutrient content of phloem sap from three plants: one with root nodules, one with mycorrhizae, and one with neither (bare-root). Which of the three phloem samples would you expect to have the highest phosphorus content? a. Phloem sap from the mycorrhizae sample will have the most phosphorus. b. Phloem sap from the bare-root sample will have the most phosphorus. c. Phloem sap from the sample with root nodules will have the most phosphorus. d. The samples with root nodules and mycorrhizae will have a higher amount of phosphorus than the bare-root sample. ANSWER: a 55. _____ are the structures formed by roots that house nitrogen-fixing bacteria. a. Root hairs b. Mycorrhizae c. Root nodules Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 d. Root caps ANSWER: c Multiple Response 56. What important characteristics unique to land plants are thought to be critical evolutionary advances that contributed to their success on land? Select all that apply. a. cuticle b. stomata c. xylem d. chlorophylls a and b e. vacuoles ANSWER: a, b, c 57. Which of the statements support the hypothesis that plants and green algae evolved from a common ancestor? Select all that apply. a. Both plants and green algae use chlorophylls a and b in photosynthesis. b. Both plants and green algae have evolved complex multicellularity. c. Both plants and green algae have membrane-bound vacuoles in their cells. d. Both plants and green algae have specialized cells for water transport. e. Both plants and green algae have cellulose in their cell walls. ANSWER: a, c, e 58. Bryophytes are widely distributed geographically, but vascular plants account for most terrestrial photosynthesis. Why? Select all that apply. a. Vascular plants are larger CO2 sinks than bryophytes. b. Vascular plants can maintain the hydration of photosynthetic cells using deep soil water. c. Bryophyte photosynthesis depends on surface water, which is not present in as many different habitats. ANSWER: b, c 59. What advantages would a plant gain from having C4, rather than C3, photosynthesis? Select all that apply. a. reduced use of water in photorespiration b. reduced loss of carbohydrates during photorespiration c. reduced use of oxygen during photorespiration d. reduced loss of metabolic energy to photorespiration e. reduced need for CO2 for the Calvin cycle ANSWER: b, d 60. When stomata are open and a plant is transpiring normally, how does water move from the soil into the root xylem? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 a. due to the action of ATP-driven pumps in the endodermis b. without the plant doing any additional work on the fluid c. due to a tension that pulls the water upward d. driven by evaporation from the leaves e. driven by pressure generated by roots pumping water upward ANSWER: b, c, d 61. Which actions are mechanisms by which plants can increase the availability of phosphorus from the soil? Select all that apply. a. acidifying the rhizosphere b. opening stomata to increase the transpiration rate, thus increasing the flow of soil water and dissolved nutrients to the roots c. transferring more carbohydrates to symbiotic mycorrhizae d. activating phosphorus transporters in the root cap ANSWER: a, c 62. Which of the structures transport a very dilute solution of mineral nutrients within a vascular plant? Select all that apply. a. sieve tubes b. companion cells c. guard cells d. tracheids e. vessels ANSWER: d, e 63. Why do bryophytes need to be able to tolerate desiccation? Select all that apply. a. They dry out with the environment. b. They are restricted to sunny habitats. c. They have little water storage capacity. d. They are restricted to saline habitats. ANSWER: a, c 64. Which of the organisms would be classified as a vascular plant? Select all that apply. a. peat moss b. a mushroom c. a rosebush d. an evergreen tree ANSWER: c, d 65. Why do plants lose several hundred H2O molecules for each CO2 that is fixed by photosynthesis? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 a. The concentration gradient is many times larger for H2O than for CO2. b. H2O molecules are lighter than CO2 molecules and thus diffuse faster. c. H2O molecules are consumed during photosynthesis. d. The evaporation of so many H2O molecules is necessary to keep photosynthesis at its optimum temperature. ANSWER: a, b 66. Which of the statements are true of CAM plants? Select all that apply. a. The Calvin cycle is active during the day. b. Organic acids are decarboxylated at night. c. Organic acids are decarboxylated in daylight. d. CO2 capture and fixation occur at different times. e. PEP carboxylase is most active at night. ANSWER: a, c, d, e 67. How is C4 photosynthesis able to achieve higher concentrations of CO2 in the vicinity of rubisco than photosynthesis in C3 plants can? Select all that apply. a. by separating carbon capture from carbon fixation b. by suppressing photorespiration c. by closing stomata during the day d. because the C4 cycle is faster than the Calvin cycle ANSWER: a, d 68. Which properties of guard cells allow them to open and close a stoma? Select all that apply. a. radially organized cellulose fibers b. tangentially organized cellulose fibers c. two guard cells being anchored together at their ends d. the ability of guard cells to change their ion concentration e. the lack of cytoplasm in guard cells ANSWER: a, c, d 69. Which of the statements are true regarding the movement of water through the xylem of vascular plants? Select all that apply. a. Water is pulled through the xylem by an evaporative force. b. Movement of water depends on differences in turgor pressure in the leaves versus roots. c. Movement of water requires an input of metabolic energy in the form of ATP. d. Water travels through the hollow centers of cells that lack cytoplasm and membranes. e. Water is pushed up the xylem by pumps in the root cells. ANSWER: a, d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 70. We typically think of phloem as transporting carbohydrates throughout vascular plants. What other substances can travel through the phloem? Select all that apply. a. DNA b. RNA c. amino acids d. ions e. hormones ANSWER: b, c, d, e 71. Which of the structures are considered a carbohydrate sink in vascular plants? Select all that apply. a. roots b. flowers c. mature leaves d. stems ANSWER: a, b, d 72. Mineral nutrients are actively transported across the endodermis. Of what does this process require roots to have a continuous supply? Select all that apply. a. CO2 b. carbohydrates c. a nutrient-rich flow of soil water d. O2 ANSWER: b, d 73. Which of the answer choices are accurate statements about both root hairs and mycorrhizae? Select all that apply. a. Both increase the surface area for nutrient uptake. b. Both are symbiotic relationships. c. Both increase the volume of soil accessible to the plant. d. Both enhance the binding of the roots to the soil. ANSWER: a, c 74. Why does the presence of high concentrations of salts in soil pose a problem for plants? Select all that apply. a. Salts can form insoluble compounds with organic forms of nitrogen that are then unavailable to the plant. b. If the salts are not filtered from the transpiration stream, they can damage the plant. c. If excluded by the root, the salts can make it harder for the root to extract water from the soil. d. If taken up by the shoot, salts make the plant more palatable to large herbivores. ANSWER: b, c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 75. Which of the methods can plants use to acquire immobile nutrients (e.g. zinc) or nutrients that are not in readily-accessible forms (e.g. the nitrogen in N2)? Select all that apply. a. Plants can introduce acid-like compounds into the soil. b. Plants can form symbiotic relationships with bacteria. c. Plants can form symbiotic relationships with fungi. d. Plants can decompose dead plant matter in the soil to recover nutrients. ANSWER: a, b, c 76. Which of the factors could contribute to slower plant growth in saline soils? Select all that apply. a. the need to pull harder to obtain water from soil containing high concentrations of salts, potentially limiting both photosynthesis and cell expansion b. higher rates of respiration in roots that have to use energy to exclude salts from entering the xylem c. higher salt concentrations in leaves, potentially shortening their life-span d. lower rates of respiration in roots due to reduced oxygen availability ANSWER: a, b, c 77. Which of the statements are true regarding CO2 capture and the Calvin cycle in CAM plants? Select all that apply. a. CO2 capture occurs at night, and the Calvin cycle takes place during the day. b. CO2 is captured in mesophyll cells, and the Calvin cycle occurs in bundle-sheath cells. c. CO2 capture and the Calvin cycle both occur in the same cell. d. CO2 capture and the Calvin cycle both occur during the day, in sunlight. ANSWER: a, c 78. Which of the cell structures are present in mature sieve tube cells, but not in mature vessel elements? Select all that apply. a. nucleus b. mitochondria c. cell wall d. cytoplasm e. smooth endoplasmic reticulum ANSWER: b, d, e 79. In many soils, there is more than enough phosphorus to support plant growth, but it is tightly bound to soil particles and thus largely unavailable. Which of the answer choices are potential solutions to this problem for vascular plants? Select all that apply. a. They can form symbiotic relationships with mycorrhizae. b. They can form symbiotic relationships with bacteria. c. They can pump H+ into the soil. d. They can raise soil pH. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 27 ANSWER: a, c 80. Why would a farmer allow a field to be taken over by weedy, non-crop members of the bean family? Select all that apply. a. Members of this family form symbioses with N2-fixing bacteria. b. Members of this family form symbioses with mycorrhizae. c. Weedy beans are very efficient photosynthesizers. d. It could reduce the need for commercial fertilizers in subsequent seasons. ANSWER: a, d 81. A wheat farmer is concerned that his land has been depleted of nitrogen. What steps can he take to ensure that his wheat crops have enough nitrogen for the next season? Select all that apply. a. He can treat his land with phosphate fertilizers. b. He can treat his land with man-made fertilizers, thanks to Haber and Bosch. c. He can continue to plant his wheat crops, because wheat can form root nodules. d. He can plant alfalfa on his land and let it serve as "green manure." e. He can plant soy beans in alternate years. ANSWER: b, d, e

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Chapter 28 Multiple Choice 1. Which of the outcomes would occur if a moss had a mutation that inhibited the production of sporopollenin? a. The spores would dry out more quickly. b. The moss would not be able to produce pollen. c. Sperm would not be able to reach eggs for fertilization. d. Spores would not be released from the sporangia. e. The likelihood of outcrossing would decrease. ANSWER: a 2. All of the types of plants demonstrate an alternation of generations except: a. Chara and Choleochaete. b. bryophytes. c. lycophytes. d. gymnosperms. e. angiosperms. ANSWER: a 3. What major challenge did early land plants face, and bryophytes continue to face, in adapting an ancestral aquatic reproductive cycle to land? a. Successful fertilization by free-swimming sperm is rare, because it generally requires continuous water films connecting male and female gametophytes. b. Few spores can survive dispersal in air. c. Plants dependent on surface water may have dried out and died before completing their life cycles. d. Gametophytes need environmental water to undergo meiosis. ANSWER: a 4. Why is it significant that pollen contains a multicellular male gametophyte instead of just male gametes? a. The non-gamete cells of the male gametophyte provide nutrition that supports the development of the embryo. b. The non-gamete cells of the male gametophyte control the growth and development of the pollen tube. c. The non-gamete cells of the male gametophyte undergo meiosis. d. The non-gamete cells of the male gametophyte provide protection from desiccation and UV radiation. ANSWER: b 5. You have discovered a plant new to science and observe what appear to be reproductive structures. On one part of the plant, you observe organ A that appears to break open under turgor pressure to release tiny cells. On another part of the plant, organ B flings out tiny cells into the air by a mechanism driven by cavitation brought on by changes in humidity. You hypothesize that organ A belongs to the gametophyte and releases _____, while organ B belongs to the _____ and releases _____." Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 a. egg cells; sporophyte; sperm b. sperm; sporophyte; spores c. spores; gametophyte; sperm d. egg cells, gametophyte; sperm ANSWER: b 6. Which has the greater degree of genetic uniformity: all of the gametes produced by a single moss gametophyte, or all of the spores produced by a single moss sporophyte? a. Genetic uniformity would be higher among gametes. b. Genetic uniformity would be higher among spores. c. Genetic uniformity would be similar between spores and gametes. d. Both spores and gametes are sexual products and highly genetically diverse. ANSWER: a 7. What would be the result if dispersal took a gamete to an environment not previously colonized by that species? a. development of a new haploid individual b. development of a new diploid individual c. failure to colonize the new environment d. development of a clonal population ANSWER: c 8. In both bryophytes and vascular plants, meiosis occurs in the dominant or most conspicuous phase of the life cycle. a. true b. false ANSWER: b 9. In bryophytes and some vascular plants, the sperm swims to the egg. a. true b. false ANSWER: a 10. Diploid (2n) cells in the _____ of mosses, ferns, gymnosperms, and angiosperms divide meiotically and form haploid (1n) spores. a. sporangium b. gonad c. zygote d. gametophyte ANSWER: a 11. Ferns and lycophytes differ from bryophytes in that: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 a. the diploid phase of the life cycle is photosynthetically independent at maturity. b. fertilization relies on environmental water. c. products of meiosis are dispersed as spores. d. the sporophyte is adapted for spore dispersal. ANSWER: a 12. A major evolutionary constraint on the height growth of gametophytes in ferns, lycophytes, and bryophytes is thought to be imposed by: a. the reliance of swimming sperm on environmental water. b. a lack of genes encoding developmental pathways for a vascular system. c. insufficient photosynthetic ability to support growth in height. d. the dependence of the gametophyte on the sporophyte for nutrition. ANSWER: a 13. Which of the options play the same functional role of establishing new individuals physically distant from the parent plant? a. liverwort spores and gymnosperm pollen b. moss sperm and angiosperm pollen c. gymnosperm pollen and angiosperm seeds d. fern spores and gymnosperm seeds e. lycophyte sperm and fern spores ANSWER: d 14. We generally think of sexual reproduction as producing offspring whose genome is related, but not identical, to the parent plant. In which of the options does fertilization result in offspring that are genetically identical to the parent? a. self-fertilization in ferns (sperm and egg came from same gametophyte) b. self-pollination in a pine (pollen and ovule came from the same tree) c. self-pollination in a flowering plant (pollen and ovule came from the same flower) d. self-pollination in a flowering plant (pollen and ovule came from different flowers on the same plant) e. None of the other answer options is correct. ANSWER: a 15. Given what you know about pollination and fertilization in pines, what is a reasonable hypothesis for both the survival of pollen and the growth of the pollen tube inside the ovule? a. There is a substantial transfer of signals and nutrients between male and female gametophytes. b. The male gametophyte and pollen tube are nutritionally independent of both the female sporophyte and female gametophyte. c. There is a substantial transfer of signals and nutrients between the male gametophyte and female sporophyte. d. There is a substantial transfer of signals and nutrients between the male sporophyte and female sporophyte. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 ANSWER: c 16. In pines, complete development of the female gametophyte depends upon successful: a. fertilization. b. pollination. c. pollination and fertilization. d. dispersal and germination of the female spore. ANSWER: b 17. In adaptation of plants to life on land, did spores or gametes evolve as the reproductive products for longdistance dispersal? Why? a. gametes, because they are motile and can swim long distances b. gametes, because they can grow into a new gametophyte, establishing a new individual c. spores, because they are smaller and lighter than gametes d. spores, because a single spore, not a single gamete, can establish a new individual ANSWER: d 18. A reasonable hypothesis for the evolution of a delay in complete development of the female gametophyte in pines until after pollination is that: a. complete development of the female gametophyte requires fusion of gametes. b. many seeds fail to be successfully dispersed, and thus, investing resources to complete the development of female gametophytes before pollination would be costly. c. development of the female gametophyte is continuous, without any pause. d. many ovules fail to be successfully pollinated, and thus, investing resources to complete the development of female gametophytes before pollination would be costly. ANSWER: d 19. You are on an expedition in a tropical rainforest. You notice a tall plant with large, long, conical purple flowers that produce nectar. When you ask your guide about the plant, she pulls out a large pink fruit from her bag and explains that it is from the plant with the purple flowers. After tasting the fruit, you decide to bring the plant back home and cultivate it in a greenhouse. You remember seeing a particular butterfly on the purple flowers during your trip to the rainforest. After obtaining a few of these insects, you release them into the greenhouse. A few weeks later, the pink fruit plant is growing! In the process, though, a few butterflies have escaped. Is this an environmentally safe situation? a. Yes, the butterflies eat only nectar, so the ecosystem will not be disturbed. b. Yes, the butterflies will aid in the pollination of several flower species, thus benefiting the ecosystem. c. No, the butterflies may not have a natural predator in this environment; thus, they could displace native species that consume nectar. d. No, the butterflies could spread the pollen of the purple flowers to native plant species. ANSWER: c 20. The flower of an angiosperm that relies on wind to disperse pollen will be: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 a. large. b. odiferous. c. small. d. colorful. ANSWER: c 21. In which generation is a recessive deleterious allele likely to negatively affect the fitness of the individual expressing it? a. gametophyte b. sporophyte c. equally likely in both sporophyte and gametophyte d. highly improbable in both generations ANSWER: a 22. You are on an expedition in a tropical rainforest. You notice a tall plant with large, long, conical purple flowers that produce nectar. When you ask your guide about the plant, she pulls out a large pink fruit from her bag and explains that it is from the plant with the purple flowers. After tasting the fruit, you decide to bring the plant back home and cultivate it in a greenhouse. You suspect that the plants need a pollinator for reproduction. You place a beehive in the middle of the greenhouse. A few weeks later, no fruit is produced. What might be a reasonable hypothesis for why this method failed? a. The bees consume the fruit. b. The seeds of the plant are too big for the bees to carry. c. The bees cannot reach far enough into the flower to obtain pollen. d. Bees only pollinate red flowers. ANSWER: c 23. Dissection of the style shows that numerous pollen tubes have grown down from the surface of the stigma and terminate well before reaching the ovules. Their failure to reach the ovules is most likely due to: a. the presence of inferior alleles controlling pollen growth in the population. b. exhaustion of growth factors for pollen tube growth due to inadequate pollen provisioning. c. exhaustion of growth factors for pollen tube growth provided by the style. d. the presence of a self-incompatibility system controlled by S gene alleles. ANSWER: d 24. Which of the options is not diploid (2n)? a. an angiosperm seed b. pollen c. rose petals d. tree bark e. the roots of a pine tree ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 25. If a pollen grain lands on a stigma that expresses the same S gene alleles as the pollen grain itself, what would happen? a. Pollen would not germinate, or if it did, the tubes would grow slowly and die before reaching the ovules. b. Pollen would germinate quickly and the pollen tubes would grow rapidly to the ovules. c. Pollen would germinate and the pollen tubes would rapidly reach the ovules, but the sperm nuclei would fail to fuse with the egg cells. d. Fertilization would occur, but the fruit would abort. ANSWER: a 26. Endosperm is a tissue that is present in: a. the eggs of green algae. b. the spores of bryophytes. c. the spores of lycophytes. d. the seeds of gymnosperms. e. the seeds of angiosperms. ANSWER: e 27. You are on an expedition in a tropical rainforest. You notice a tall plant with large, long, conical purple flowers that produce nectar. When you ask your guide about the plant, she pulls out a large pink fruit from her bag and explains that it is from the plant with the purple flowers. After tasting the fruit, you decide to bring the plant back home and cultivate it in a greenhouse. You are successful in producing a crop of the pink fruit. You save some of the seeds to plant the following year. When the next growing season starts, you plant the seeds in the same greenhouse. To your surprise, no plants grow. What is a possible reason for this? a. The seeds do not germinate because the flowers were not pollinated by butterflies. b. The seeds must pass through the digestive tract of a seed dispersal organism to germinate. c. Because the plants self-pollinate, the endosperm does not develop properly, and thus the seeds do not germinate. ANSWER: b 28. You are studying a genus of tropical plants pollinated by lowland birds. These plants grow in low density and are widely separated from one another. However, one species in the genus has adapted to high alpine meadows, where it is the most abundant plant over large areas. The alpine species has flowers that are small, dull, and unisexual. You infer that: a. adaptation to an alpine environment required a shift from bird to wind pollination. b. reproduction is more efficient in the alpine environment than in the lowland environment. c. the alpine environment is likely nutrient-poor, leading to small, dull flowers and poor pollination success. d. the alpine species is most likely bee-pollinated, because bee pollination is only efficient at high densities of individual plants. ANSWER: a 29. If you were to remove all of the anthers from a self-incompatible plant, how would this affect the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 fertilization of its female gametes? a. Fertilization would continue, because self-incompatible plants rely on pollen from other individuals. b. Fertilization would stop, because self-incompatible plants are fertilized with pollen they produce. c. Fertilization would continue, because self-incompatible plants are fertilized with pollen they produce. d. Fertilization would stop, because self-incompatible plants rely on pollen from other individuals. ANSWER: a 30. Which of the processes occurs in angiosperms when a sperm cell merges with a diploid cell within the future seed to form the endosperm, and a second sperm fuses with a haploid (1n) egg? a. double fertilization b. polyspermy c. endospermy d. ovulation ANSWER: a 31. A truck driver is transporting a load of unripe bananas in an airtight vehicle and decides to stop and eat an apple after checking on his cargo. He opens the back of his truck and, while walking around the unripe bananas, he decides the apple is overripe and drops it in the truck. When the bananas are finally delivered days later, he is surprised to find that all of his bananas have ripened. Why? a. Opening the back of the truck exposed the bananas to CO, which accelerated their ripening. b. Exposure of the bananas to the apple (and its ethylene) instigated their ripening. c. The bananas would naturally ripen over time, even in an airtight environment. d. Opening the back of the truck exposed the bananas to light of red wavelengths. ANSWER: b 32. A strawberry farmer's crop has been devastated by a fungal infection. The farmer searches for any fungiresistant strawberry plants in his fields, but finds none. Why? a. Due to vegetative reproduction, his plants are genetically diverse. b. Due to vegetative reproduction, his plants are not genetically diverse. c. Due to sexual reproduction, his plants are not genetically diverse. d. Due to sexual reproduction, his plants are genetically diverse. ANSWER: b 33. Your lab group collects genetic samples from half of the stems in a forest, but is unable to detect any sequence differences across dozens of genes. This observation supports the conclusion that: a. the population is not evolving. b. your group sampled a range of species. c. the sampled species is capable of asexual reproduction. d. the sampled species is self-compatible. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 34. Review the figure shown. Which of the shared derived characters separates ferns and horsetails from the gymnosperms and angiosperms?

a. pollen and seeds b. alternation of generations c. xylem and phloem ANSWER: a 35. Review the figure shown. Which of the shared derived characters distinguishes algae from the land plants? a. pollen and seeds b. alternation of generations c. xylem and phloem

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ANSWER: b 36. Review the figure shown. Which of the shared derived characteristics is present in gymnosperms but absent in mosses?

a. pollen and seeds b. alternation of generations c. xylem and phloem ANSWER: c 37. Indicate which group of plants the statement refers to. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 Pollen is the male gametophyte. a. true for angiosperms only b. true for gymnosperms only c. true for both gymnosperms and angiosperms d. true for neither gymnosperms nor angiosperms ANSWER: c 38. Indicate which group of plants the statement refers to. Fruits aid in seed dispersal. a. true for angiosperms only b. true for gymnosperms only c. true for both gymnosperms and angiosperms d. true for neither gymnosperms nor angiosperms ANSWER: a 39. Indicate which group of plants the statement refers to. Fertilized mature ovules are known as seeds. a. true for angiosperms only b. true for gymnosperms only c. true for both gymnosperms and angiosperms d. true for neither gymnosperms nor angiosperms ANSWER: c 40. Indicate which group of plants the statement refers to. The nutritive tissue is produced before fertilization. a. true for angiosperms only b. true for gymnosperms only c. true for both gymnosperms and angiosperms d. true for neither gymnosperms nor angiosperms ANSWER: b 41. Indicate which group of plants the statement refers to. The male parent contributes genes to the nutritive tissue. a. true for angiosperms only b. true for gymnosperms only c. true for both gymnosperms and angiosperms d. true for neither gymnosperms nor angiosperms Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 ANSWER: a 42. Select the appropriate description for the term. sporophyte a. multicellular haploid b. unicellular haploid c. multicellular diploid d. unicellular diploid ANSWER: c 43. Select the appropriate description for the term. gametophyte a. multicellular haploid b. unicellular haploid c. multicellular diploid d. unicellular diploid ANSWER: a 44. Select the appropriate description for the term. spore a. multicellular haploid b. unicellular haploid c. multicellular diploid d. unicellular diploid ANSWER: b 45. Select the appropriate description for the term. sperm a. multicellular haploid b. unicellular haploid c. multicellular diploid d. unicellular diploid ANSWER: b 46. Select the appropriate description for the term. egg a. multicellular haploid Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 b. unicellular haploid c. multicellular diploid d. unicellular diploid ANSWER: b 47. Select the appropriate description for the term. zygote a. multicellular haploid b. unicellular haploid c. multicellular diploid d. unicellular diploid ANSWER: d 48. At which point on the phylogenetic tree shown did airborne spores evolve?

a. 1 b. 2 c. 3 d. 4 e. 5 F) 6 G) 7 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 H) 8 I) 9 ANSWER: b 49. At which point on the phylogenetic tree shown did airborne seeds evolve?

a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 9 ANSWER: b 50. At which point on the phylogenetic tree shown did airborne spores evolve?

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a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 9 ANSWER: g 51. At which point on the phylogenetic tree shown did alternation of generations evolve?

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a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 9 ANSWER: b 52. At which point on the phylogenetic tree shown did ovules evolve?

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a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 9 ANSWER: g 53. At which point on the phylogenetic tree shown did fruits evolve?

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a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 9 ANSWER: h 54. At which point on the phylogenetic tree shown did dominant sporophyte generation evolve?

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a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 9 ANSWER: e 55. At which point on the phylogenetic tree shown did pollen evolve?

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a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 9 ANSWER: g 56. At which point on the phylogenetic tree shown did multicellular sporophytes evolve?

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a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 9 ANSWER: b 57. At which point on the phylogenetic tree shown did flowers evolve?

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a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 7 h. 8 i. 9 ANSWER: h Multiple Response 58. How did the evolution of alternation of generations aid the successful establishment of a new gametophyte generation? Select all that apply. a. It allowed fertilization to occur without free-swimming sperm. b. It replaced gametophytes prone to desiccation with resistant sporophytes. c. It produced new tissues specially adapted for aerial dispersal. d. It amplified the number of spores produced per fertilization event. ANSWER: c, d 59. In which of the plants does the sporophyte eventually become self-sustaining (i.e. is not dependent on the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 gametophyte for nutrients)? Select all that apply. a. gymnosperms b. hornworts c. angiosperms d. ferns ANSWER: a, c, d 60. Why is spore dispersal an important part of reproduction? Select all that apply. a. It allows for escape from pathogens and parasites. b. It reduces competition for resources with closely related individuals. c. It can lead to increased outcrossing and gene flow in a population. d. It allows a single successful genotype to spread over more geographic area. ANSWER: a, b, c 61. In which of the plants do male gametes (sperm) require water to travel to and fertilize nearby eggs? Select all that apply. a. gymnosperms b. hornworts c. ferns d. mosses e. angiosperms ANSWER: b, c, d 62. In the life cycles of mosses and ferns, which of the structures are diploid? Select all that apply. a. sporophyte b. gametophyte c. zygote d. sperm e. eggs ANSWER: a, c 63. In ferns, both the gametophyte and sporophyte are physiologically capable of surviving on their own. Which of the statements is a reasonable hypothesis for why vascular tissues are present only in the sporophyte generation? Select all that apply. a. There is no advantage for the gametophyte to grow tall, because gametes must be produced near the ground, where the water needed for their free-swimming sperm is most likely to be found. b. Haploid organisms are incapable of developing vascular tissues, such as xylem and phloem. c. There is no advantage for the gametophyte to grow tall, because fern gametophytes typically produce only male or female gametes at any one time. d. Vascular tissues allow the sporophyte to grow to a greater height, which increases the airborne dispersal of the spores. ANSWER: a, d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 64. Which of the groups of plants has a life cycle that includes a multicellular body composed of diploid cells? Select all that apply. a. Chara b. liverworts c. horsetails d. ferns e. angiosperms ANSWER: b, c, d, e 65. The pollen of seed-bearing plants and the spores of seedless vascular plants are both transported by air. How do these structures differ? Select all that apply. a. Spores are diploid individuals, whereas pollen grains are haploid. b. Pollen contains a mature male gametophyte. c. Spores are produced by the sporophyte stage of the life cycle, whereas pollen is produced by the female gametophyte stage of the life cycle. d. Spores are unicellular, whereas pollen grains are multicellular. ANSWER: b, d 66. Which of the qualities is a major difference between all spore-dispersing and seed-dispersing plants? Select all that apply. a. The sporophyte of seed plants is photosynthetically independent. b. The gametophytes of seed plants are never exposed to the external environment. c. Seed plants disperse gametes and seeds instead of spores. d. In seed plants, dispersed spores contain a multicellular gametophyte. ANSWER: b, d 67. Compared to large seeds, small seeds are: Select all that apply. a. more likely to persist in the soil. b. more likely to exhibit some form of dormancy. c. less likely to be produced by weedy species. d. more likely to disperse long distances. ANSWER: a, b, d 68. In ferns, some leaves are "fertile" in that they bear sporangia. What are the structures in gymnosperms that are analogous to a leaf and sporangia? Select all that apply. a. cone scales and ovules b. cones and seeds c. seed coat and embryo d. cone scale and pollen-producing sporangia ANSWER: a, d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 69. Both gymnosperms and angiosperms form seeds that can travel over long distances via wind or water. Which of the options are benefits of seed production and dispersal? Select all that apply. a. Seeds can introduce offspring into environments with ample light. b. Seeds can introduce offspring into pathogen-free environments. c. Seeds can introduce offspring into nutrient-rich environments. d. Seeds can introduce offspring into environments beyond the range of spores. ANSWER: a, b, c 70. What allows fertilization in seed plants to occur without male gametes being exposed to the external environment? Select all that apply. a. aerial transport of the male gametophyte b. growth of the male gametophyte into the tissues of the ovule-bearing sporophyte c. growth of the female gametophyte within the tissues of the ovule-bearing sporophyte d. double fertilization ANSWER: a, b 71. In which of the structures are spores formed? Select all that apply. a. pollen cones b. ovule cones c. stamens d. carpels e. None of the other answer options is correct. ANSWER: a, b, c, d 72. Which of the statements are true about plants? Select all that apply. a. The sporophyte always produces spores. b. The gametophyte always produces gametes. c. Fertilization always produces diploid spores. d. The gametophyte generation is always the result of the germination of spores. ANSWER: a, b, d 73. Which of the statements are true regarding the pollen of gymnosperms and angiosperms? Select all that apply. a. Pollen grains from both of these plants consist of four cells. b. A single angiosperm produces more pollen than a single gymnosperm. c. Pollen from both types of plants form pollen tubes that play a role in fertilization. d. Pollen from both types of plants contain (or will form) haploid (1n) male gametes. ANSWER: c, d 74. Which of the statements are true about double fertilization? Select all that apply. a. It occurs in both gymnosperms and angiosperms. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 b. It refers to the fact that two sperm enter each ovule. c. It produces twin embryos, only one of which usually survives. d. It links formation of the endosperm to formation of the embryo. ANSWER: b, d 75. In ferns, some leaves are "fertile" in that they bear sporangia. What are the analogous structures to such fertile fronds in angiosperms? Select all that apply. a. carpels b. sepals c. stamens d. petals ANSWER: a, c 76. Which of the structures are part of the carpel of a flower? Select all that apply. a. ovary b. anther c. style d. stigma e. stamen ANSWER: a, c, d 77. Which of the structures constitute the stamen of a flower? Select all that apply. a. stigma b. anther c. ovary d. style e. filament ANSWER: b, e 78. Which of the features are used by flowers to attract pollinators? Select all that apply. a. nectar b. fruit c. burrs d. odors ANSWER: a, d 79. Which of the animals can serve as pollinators for some types of angiosperm? Select all that apply. a. birds b. butterflies c. bats d. moths Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 e. fish ANSWER: a, b, c, d 80. Recall that angiosperms spend a great deal of energy and resources attracting pollinators. What are some of the methods flowers use to lure pollinators? Select all that apply. a. Flowers can mimic the appearance and/or pheromones of the females in an insect species. b. Flowers can be brightly colored and produce nectar. c. Flowers can produce chemicals used by insects to synthesize pheromones. d. Flowers can produce scents similar to those emitted by dung or animal carcasses. e. Flowers can produce high-frequency sounds due to air flow through specialized floral parts to attract nocturnal pollinators such as bats. ANSWER: a, b, c, d 81. You are on an expedition in a tropical rainforest. You notice a tall plant with large, long, conical purple flowers that produce nectar. When you ask your guide about the plant, she pulls out a large pink fruit from her bag and explains that it is from the plant with the purple flowers. After tasting the fruit, you decide to bring the plant back home and cultivate it in a greenhouse. You remember seeing a particular butterfly on the purple flowers during your trip to the rainforest. You decide that the pollinating butterflies are too costly and a potential danger to the ecosystem. Instead, you are going to genetically modify the plant so that it is a selfpollinating, self-compatible plant. What are some traits, both phenotypic and genotypic, that you would look for in the modified plants? Select all that apply. a. The plants would have a mutated, non-functional S gene. b. The anther and style would be in close proximity. c. The plants would be shorter. d. The plants could be pollinated by bees. ANSWER: a, b 82. Which of the statements is true of the seeds of both gymnosperms and angiosperms? Select all that apply. a. Seeds from both plants contain a triploid (3n) endosperm. b. Seeds from both plants contain a diploid (2n) embryo. c. Seeds from both plants contain a diploid (2n) seed coat. d. Seeds from both plants contain a haploid (1n) male gametophyte. ANSWER: b, c 83. Which of the statements regarding the fruits produced by angiosperms is true? Select all that apply. a. Fruits often ripen in response to ethylene. b. Fruits are often derived from the wall of the ovary. c. Fruits are typically derived from the seed coat. d. Fruits ripen in response to far-red light. ANSWER: a, b 84. How do seeds compare to spores as dispersal units? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 28 a. They both establish a new diploid individual. b. Seeds are multicellular and contain resources to support establishment and growth. c. Spores are adapted for wind dispersal, but seeds are not. d. Seeds are larger and heavier and require extra investment in tissues that aid dispersal. ANSWER: b, d 85. In flowering plants, the style separates the site of pollen reception from the ovules to be fertilized by a distance of up to 10 centimeters. What might be the adaptive significance of this distance? Select all that apply. a. It allows pollen tube competition to play out over distances requiring significant growth and interaction with the style. b. It keeps ovules safely away from fungal spores and other pathogens that may also germinate on the style and attempt to access the ovaries. c. It may elevate the receptive surface away from the stamens to prevent self-fertilization. d. It allows the maternal sporophyte means by which self-fertilization can be prevented. ANSWER: a, b, c 86. Why might asexual reproduction typically result in local colonization, while the products of sexual reproduction are typically dispersed? Select all that apply. a. The variation resulting from meiosis and the union of gametes means some individuals will have a chance of colonizing diverse distant environments successfully. b. Vegetative propagation is more efficient than sexual reproduction when the current generation is well-adapted to the local environment. c. Asexual reproduction is only a "last ditch" alternative for plants when sexual reproduction fails. d. Sexual reproduction is only a "last ditch" alternative by which plants can "escape" the condition of being poorly adapted to the local environment. ANSWER: a, b

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Chapter 29 Multiple Choice 1. Recall that leaves form in distinct arrangements (whorled, opposite, or alternate) on the stems of vascular plants. How might chemical factors cause new leaves to form in these particular patterns? a. Leaf primordia drain auxin from neighboring cells, with new primordia forming where auxin in the meristem surface is at the highest concentration. b. Leaf primordia secrete chemical activators that promote the formation of leaves nearby. c. Mature leaves secrete chemical inhibitors that prevent the formation of leaves nearby. d. Mature leaves secrete chemical activators that promote the formation of leaves nearby. ANSWER: a 2. Which of the statements about axillary, floral, and shoot apical meristems is correct? a. When a shoot apical meristem (but not an axillary bud) converts to a floral meristem, further development of the whole plant ceases. b. Axillary buds can only grow out to become new shoots once the shoot apical meristem has been converted into a floral meristem. c. Shoot apical meristems can revert to vegetative meristem identity after having converted to a floral meristem and producing a flower, but axillary buds that give rise to flowers cannot. d. In shoot apical meristems that develop into flowers, the arrangement of floral organs is the same as that of the leaves produced earlier, while the floral organs of flowers that develop from axillary buds are free to follow a different arrangement. e. None of the other answer options is correct. ANSWER: e 3. Within the stem of a vascular plant, where would you expect to find the cells that are the most mature? a. within the shoot apical meristem at the tip of the stem b. within the zone of elongation directly beneath the shoot apical meristem c. at the base of the stem near the soil ANSWER: c 4. What is responsible for most of the growth in height of a plant? a. changes in cell number, but not cell dimension, in the apical meristem b. changes in cell dimension, but not cell number, in the elongation zone c. both changes in cell number and cell dimension, roughly equally d. thickening of the cell walls in the zone of maturation ANSWER: b 5. You are growing plants hydroponically (rooted in an aqueous solution rather than soil). The solution for the control group of plants contains normal levels of salts and nutrients, while the solution for the treatment group has a high concentration of large molecular weight solutes that are excluded by roots. You find that the plants in the treatment group are, on average, smaller than those in the control group, but there is no significant difference between the groups in the amount of CO2 being fixed by the average plant. What differences between groups, at the cellular level, would you expect to make an important contribution to the difference in size of the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 shoots? a. The total number of cells in the treatment group shoots would be smaller. b. The girth of cells in the elongation and mature zones would be larger in the control group. c. The length of cells in the elongation and mature zones would be larger in the control group. d. Cellulose molecules would be more randomly oriented in mature cell walls in the treatment group. e. Auxin levels and cell wall extensibility would be lower in the treatment group. ANSWER: c 6. An early experiment in which polar transport of auxin was demonstrated used agar blocks containing 14C labeled auxin placed at one end of a cut stem and a block of agar without 14C labeled auxin at the other end (assay block). In which of the treatments would you predict that the 14C labeled auxin would be detected in the assay block of agar? a. A and B b. C and D c. A and C d. B and C e. A and D f. A, B, C, and D ANSWER: e 7. A researcher creates a mutant pea plant in which cytokinins are overexpressed and gibberellic acid is underexpressed. What is the most likely phenotype of this mutant pea plant? a. The pea plant would be shorter with more branches compared to wild-type plants. b. The pea plant would be taller with more branches compared to wild-type plants. c. The pea plant would be shorter with fewer branches compared to wild-type plants. d. The pea plant would be taller with fewer branches compared to wild-type plants. ANSWER: a 8. A researcher discovers a pea plant mutant with PIN proteins that are located on the apical (not basal) sides of cells in the shoot apical meristem. What processes would this affect? a. polar transport of auxin b. the development of new xylem c. the development of new phloem d. the establishment of procambial cells e. All of these choices are correct. ANSWER: e 9. You are studying a mutant in which cellulose molecules are always deposited in the walls with random orientations and in which gibberellic acid is overexpressed. You find that the: a. two mutations have opposing effects on cell elongation, and so the plant appears similar to the wild Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 type. b. plants become long and spindly as the effects on elongation of the two mutations are additive. c. plants are small in stature, but with an overproliferation of branches. d. stem internodes are larger in diameter, but compressed in length, relative to wild-type plants. ANSWER: d 10. Recall that, in many plants, if the shoot apical meristem is removed, axillary buds become active and new lateral branches form. Of what is this an example? a. auxin dominance b. axillary dominance c. meristem dominance d. apical dominance ANSWER: d 11. A plant has undergone a robust phase of growth; however, the roots can no longer provide enough nutrients to the plant. What signal(s) can the roots send to the rest of the plant that will prevent new branches from forming? a. cytokinins b. gibberellic acid c. strigolactone d. ethylene ANSWER: c 12. Which cells transport auxin out of immature leaves and eventually become the framework for new xylem and phloem in vascular plants? a. procambial cells b. axillary bud cells c. shoot meristem cells d. cortical cells ANSWER: a 13. In general, when a cell or tissue type has to serve multiple functions that each contribute to fitness, natural selection is constrained from optimizing that cell type or tissue for any one function. With this concept in mind, choose the correct terms to complete the sentence. "In angiosperms, you would expect a correlation between the diameter of the water-conducting conduits and the mechanical strength of the wood. In gymnosperms, you would expect the correlation to be: _____." a. poor, negative b. poor, positive c. good, negative d. good, positive ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 14. Which structures allow oxygen to diffuse through the outer bark of woody plants to supply the respiratory needs of the underlying cells? a. guard cells b. stomata c. lenticels d. vascular cambium cells ANSWER: c 15. Which structure is the source of new cells that allows mature stems to grow in diameter? a. pith b. shoot meristem c. cork cambium d. vascular cambium ANSWER: d 16. As trees grow larger and the number of leaves increases, transpiration rates rise and demand for water increases. How do mature plant stems meet this added demand for water? a. Cells of the xylem continue to divide and produce more cells that transport water. b. The vascular cambium continues to divide, and cells of its inner surface differentiate to form additional secondary xylem cells. c. Cells of secondary phloem adjacent to the vascular cambium differentiate into xylem cells. d. Cells present in the pith closest to the xylem differentiate into xylem cells. ANSWER: b 17. The wood of gymnosperms is composed of _____, and the wood of angiosperms (e.g., a cherry tree) can consist of _____. a. vessel elements and fibers; tracheids b. fibers and tracheids; vessel elements c. tracheids; fibers and vessel elements d. tracheids and vessel elements; fibers ANSWER: c 18. A researcher examines a piece of oak tree bark under a microscope. He notices many small holes or spaces within his bark sample. What are these areas? a. stomata b. suberins c. pericycles d. lenticels ANSWER: d 19. You are comparing samples of wood from two trees. Sample A sinks in water, while sample B floats. Which Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 sample would you expect to have a higher proportion of living cells in their wood? a. sample A b. sample B ANSWER: b 20. You are comparing samples of wood from two trees. Sample A sinks in water, while sample B floats. Which sample would you expect to support larger branch systems? a. sample A b. sample B ANSWER: a 21. Vascular cambium is one of two lateral meristems; the other is cork cambium. A plant grows in diameter primarily through divisions of the vascular cambium. If you pull a small piece of bark off of a tree and look at the bark's inside surface, what tissue are you looking at? Ignore any remaining cells of vascular cambium that may be left. a. cork cambium b. cork c. xylem d. phloem ANSWER: d 22. Suppose you invented a way to genetically transform all of the cells expressing meristematic identity in a plant so that they would fluoresce green. Suppose that this transformation was also heritable by daughter cells, so that they would also fluoresce as long as they continued to express meristematic identity, but would lose the ability to fluoresce irreversibly upon differentiating to a nonmeristematic cell type. You then genetically transform all the cells of a tree and, soon after, examine a cross section of a branch stem. A few years later, you examine a cross section of another branch stem. What do you expect to see in comparing the two cross sections? a. a continuous green ring in the first sample, but multiple discontinuous rings in the later sample b. a discontinuous green ring surrounding a continuous green ring in the first sample, but multiple continuous green rings in the later sample c. multiple discontinuous green rings in the first sample, but multiple continuous green rings surrounding a discontinuous green ring in the later sample d. a continuous green ring surrounded by multiple discontinuous green rings in the first sample, but a single continuous green ring in the later sample e. a continuous green ring in the first sample, and two continuous rings in the later sample ANSWER: e 23. A botanist is studying the growth rings of a redwood tree. She notices that three concentric rings are very thin compared to the surrounding growth rings. What can she deduce from the presence of these thin growth rings? a. The tree developed more apical shoot meristems during that time period. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 b. The tree may have experienced a drought or had limited access to nutrients during that time period. c. The tree developed more secondary xylem during that time period. d. The tree had ample access to water and other resources during that time period. ANSWER: b 24. In your vegetable garden, you are growing yams. Yams are modified roots. What is the function of the yam for the plant? a. It stores water for the plant in times of drought. b. It stores starch for the plant. c. It provides greater surface area for water absorption. d. It provides greater surface area for the growth of additional root meristems. ANSWER: b 25. What is responsible for most of the uptake of water and nutrients performed by roots? a. the meristem and root cap at the root tip b. the first vascular tissues formed in the elongation zone c. the first tissues of the zone of cell maturation, which have epidermal root hairs and a developed Casparian strip d. the older portions of the root, with the highest density of branching ANSWER: c 26. Which statement is true regarding plant roots? a. All roots exhibit only primary growth. b. Roots have only lateral meristems, not apical meristems. c. New roots form at the endodermis. d. New roots form at the pericycle. ANSWER: d 27. If a stem is severed from its root system: a. the shoot will die because stems cannot produce new root meristems. b. auxin traveling down through the stem in the phloem will accumulate at the cut end and trigger the initiation of root meristem. c. axillary buds near the cut end will take on root meristem identity due to the disruption of strigolactone transport and, because root meristems are not repressed by apical dominance, they will begin to grow. d. branches at the cut end will become positively gravitropic in response to the loss of an auxin sink in the root. Their apical meristems will take on root meristem identity once they contact the soil due to ethylene accumulation. ANSWER: b 28. Recall that several vascular bundles (containing phloem and xylem) are located around the circumference of a vascular plant stem. In contrast, a root contains one vascular bundle running down its center. What might Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 account for this different arrangement of vascular bundles in stems and roots? a. Stems need to counteract the force of gravity (and support leaves), but roots do not. b. Roots need to quickly transport nutrients to the rest of the plant, but stems do not. c. Roots require fewer nutrients compared to stems. d. None of the other answer options is correct. ANSWER: a 29. In order for a root to branch, new apical meristem must form from the _____, a single layer of cells inside the endodermis of the root. a. pericycle b. cortex c. epidermis d. endodermis ANSWER: a 30. A stem segment several internodes long (Stem 1 in the figure shown) is placed upside down (apical end down) in a humid chamber. After several days, the stem looks like Stem 2.

What letter indicates the location on Stem 2 of the release of apical dominance due to auxin depletion? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 a. A b. B c. C d. D ANSWER: c 31. A stem segment several internodes long (Stem 1 in the figure shown) is placed upside down (apical end down) in a humid chamber. After several days, the stem looks like Stem 2.

What letter indicates the location of positive gravitropism on Stem 2? a. A b. B c. C d. D ANSWER: b 32. What letter indicates the location, on Stem 2, of the formation of new root meristems at sites of auxin Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 accumulation?

a. A b. B c. C d. D ANSWER: d 33. What letter indicates the location, on Stem 2, of negative gravitropism?

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Chapter 29

a. A b. B c. C d. D ANSWER: a 34. Which structures are starch-containing organelles present in root cells that sink toward one side of the cell in response to gravity (and thus play a role in gravitropism)? a. statoliths b. phytochromes c. otoliths d. phytoliths ANSWER: a 35. Turgor pressure, which causes cells to expand, results in a force applied uniformly to the entire cell wall. How is it that plant cells are often observed to expand in a particular direction, such as increasing their length but not their girth? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 a. Cell shape follows the shape of the vacuole, which occupies up to 90% of the cell volume. b. The expansion of the cell wall is controlled only where new material is deposited internally, thereby giving the cell precise control of its shape. c. Cellulose molecules individually resist extension, so the cell wall stretches more readily in directions perpendicular to the dominant orientation of cellulose molecules. d. A cell's cytoskeleton imposes its shape on the flexible cell wall "skin," so elongation of the cell follows the extensive growth of the cytoskeleton. ANSWER: c 36. A houseplant is growing on your windowsill. Every Sunday, without fail, your roommate turns the plant 180 degrees. When you ask your roommate why she does that, her reply is simply, "I want straight plants." Does this method produce straight plants? a. No. The statoliths in the plant cells do not respond to turning; the plant will grow straight whether or not your roommate turns the plant. b. No. The auxins don't respond to turning, only to light. c. Yes. Rotating the plant will even out the growth of the plant toward the light by changing where the light hits the plant every week. d. Yes. Rotating the plant redistributes the statoliths, so they remain in the bottom of each cell. ANSWER: c 37. In order to germinate, seeds of many temperate-zone species require a prolonged exposure to temperatures below some critical threshold. This prevents germination: a. during warm days in autumn. b. beneath a canopy of plants. c. during long days. d. in high-latitude populations, but not in low-latitude populations. ANSWER: a 38. You have a very leafy houseplant that completely shades the soil in the pot where it is growing. You notice that the tips of some roots are starting to break through the surface of the soil (these are likely new roots growing from the pericycle of an existing root just beneath the surface). If you do nothing, what would you expect to happen to the growth of these roots and why? a. The roots will continue to grow up because the root meristem tissue is arranged exactly like apical meristem tissue. b. The roots will continue to grow up because auxin is not being redistributed in the shady environment on the soil surface. c. The roots will grow downward, back into the soil, because the statoliths are being redistributed due to gravity, and that will change the direction of growth. d. The roots will grow downward because water stimulates root growth and the root will be moister on the side that is touching the soil. ANSWER: c 39. A high school student is carrying out a science project. She plants alfalfa seeds and encloses an area planted Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 with one set of alfalfa seeds within a box covered in red cellophane (which reflects light of red wavelengths). She leaves another set of planted alfalfa seeds uncovered. What do you expect she will find in these two alfalfa populations when her science project is over? a. Germination rates will be the same in both seed sets. b. The germination rate will be lower in the uncovered seed set. c. The germination rate will be higher in the uncovered seed set. d. None of the other answer options is correct. ANSWER: c 40. A woman is growing an African violet on her windowsill. After several weeks, she notices that the plant is leaning toward the right, in the direction of the window. In the stem(s) of her African violet, where would you expect to find most of the auxin? a. Auxin will be on the left (shaded) side of the stems, facing away from the window. b. Auxin will be on the right (lit) side of the stems, facing toward the window. c. Auxin will be equally distributed between the right and left sides of the stems. d. Gibberellic acid, not auxin, will be in the stems of the African violet. ANSWER: a 41. Upon which of these factors does photoperiodism depend? a. the activation of photoreceptors b. daily fluctuations of auxin c. the accumulation and stabilization of auxin d. the inhibition of phytochrome ANSWER: a 42. Axillary buds and leaves are found at: a. petioles. b. internodes. c. nodes. d. lateral meristems. ANSWER: c 43. Plant meristems consist of cells that undergo cell division to give rise to new cells. Which of the statements about leaf meristem cells and shoot apical meristem cells is true? a. Leaf meristem cells divide for a limited period of time, while shoot apical meristem cells can divide indefinitely. b. Leaf meristem cells are distributed uniformly throughout leaves, while shoot meristem cells are located in a single location at the tip of a shoot. c. Leaf meristem cells undergo meiotic cell division, and shoot meristem cells undergo mitotic cell division. d. Leaf meristem cells are found evenly distributed along the edges of the leaves of most plants, while shoot meristem cells are located in a single location at the tip of a shoot. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 ANSWER: a 44. Vascular bundles in the stems of seed plants are found between: a. the cortex and the endodermis. b. the cortex and the pith. c. the pith and the epidermis. d. the xylem and the phloem. ANSWER: b 45. Which of the factors permit the directional flow of auxin, also known as polar transport, down the shoot from the shoot apical meristem toward the root? a. the specific localization of PIN proteins in the basal membranes and the fact that auxin is ionized (charged) in the cytoplasm of the cells b. the specific localization of PIN proteins in the apical membranes and the fact that auxin is ionized (charged) in the cytoplasm of the cells c. the specific localization of PIN proteins in the basal membranes and the fact that auxin is not ionized (uncharged) in the cytoplasm of the cells d. the specific localization of PIN proteins in the apical membranes and the fact that auxin is not ionized (uncharged) in the cytoplasm of the cells ANSWER: a 46. The formation of branches in a root allow the root to maximize surface area for the absorption of water and nutrients. How is it that new branches may begin transporting resources to the plant almost as soon as the branch is formed? a. The new branch forms from the cortex, which is in contact with the pericycle of the root. b. The new branch forms from the cortex, which is in contact with the vascular cylinder of the root. c. The new branch forms from the pericycle, which is in contact with the vascular cylinder of the root. d. The new branch forms from the pericycle, which is in contact with the central cortex of the root. ANSWER: c 47. If plant seedlings are moved from full sun to a location that is shaded by other plants, what would you expect to happen? a. More of the phytochrome present in the leaves would be in the inactive Pr form, and internode elongation would be slowed. b. Less of the phytochrome present in the leaves would be in the inactive Pr form, and internode elongation would be accelerated. c. Less of the phytochrome present in the leaves would be in the inactive Pr form, and internode elongation would be slowed. d. More of the phytochrome present in the leaves would be in the inactive Pr form, and internode elongation would be accelerated. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 48. Two trees of the same species have been growing in a forest for ten years. One of the trees is located on a valley floor. The other tree is located near the top of a windswept ridge. What would you expect to be true of these trees? a. The tree on the ridge is taller and has a thicker trunk than the tree on the valley floor. b. The tree on the ridge is shorter and has a thicker trunk than the tree on the valley floor. c. The tree on the ridge is shorter and has a thinner trunk than the tree on the valley floor. d. The tree on the ridge is taller and has a thinner trunk than the tree on the valley floor. ANSWER: b 49. A mutant version of mouse-ear cress (Arabidopsis thaliana) has been engineered to convert it from a longday plant to a short-day plant. What is responsible for triggering the development of flowers in this plant? a. Day length increases to a critical length, causing an appropriate increase in Pfr : Pr ratio and increased expression of genes required for flowering. b. Day length decreases back to a critical length, causing an appropriate decrease in Pfr : Pr ratio and increased expression of genes required for flowering. c. Day length decreases back to a critical length, causing an appropriate decrease in Pr : Pfr ratio and increased expression of genes required for flowering. d. Day length increases to a critical length, causing an appropriate increase in Pr : Pfr ratio and increased expression of genes required for flowering. ANSWER: b Multiple Response 50. A researcher discovers a mutant carnation plant that lacks any A-class homeotic genes, which control the development of flowers. Which structures would be malformed or absent in this mutant? Select all that apply. a. stamens b. carpels c. petals d. sepals ANSWER: c, d 51. In which of the structures of a vascular plant would you expect to find totipotent cells? Select all that apply. a. mature leaves b. shoot apical meristem c. cell elongation zone d. axillary buds ANSWER: b, d 52. Cells directly beneath the shoot apical meristem begin to elongate and form the stem. Which events or structures contribute to the elongation of these cells? Select all that apply. a. the formation of a large vacuole in these cells b. the orientation of cellulose molecules around these cells Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 c. the presence of meristem identity genes in these cells d. the loss of cytoplasm from these cells ANSWER: a, b 53. If auxin failed to lose a proton (and so never became negatively charged) in a cell's cytoplasm, how would this affect auxin transport and the development of new xylem and phloem? Select all that apply. a. Auxin would remain sequestered in the cells where it was produced. b. Auxin might move between cells (via diffusion), but not in a directed manner. c. New xylem and phloem would continue to form normally. d. New xylem and phloem would be malformed or not form at all. ANSWER: b, d 54. Consider the figure shown. What phenotypic effects would you expect to observe if a mutation led to the loss of the plant cell's ability to control where PIN proteins are inserted in the plasma membrane? Select all that apply.

a. an irregular pattern of leaf primordia initiation b. a disruption of the pattern of veins in the leaves c. reduced branching d. all cells elongating more than usual ANSWER: a, b, c 55. You purchase two identical houseplants and place them side by side on your windowsill. You water both plants equally. You leave plant A alone, but you pinch off the top of the growing stem of plant B, effectively removing the apical meristem. Which results would you expect to occur? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 a. Both plants will look exactly the same, except that plant B will be shorter. b. Plant A will be much bushier, with growth of many lateral branches. c. Plant B will be much bushier, with growth of many lateral branches. d. The level of auxin in plant B will be higher than in plant A. e. The level of cytokinins will be higher in the axillary buds of plant B. f. The level of cytokinins will be the same in both plants, but these cytokinins will only affect growth in plant B. ANSWER: c, e 56. In which of these plant processes do hormones play a role? Select all that apply. a. branch development b. root development c. fruit ripening d. gravitropism e. root branching ANSWER: a, b, c, d 57. Which cell types can become the meristem cells of a vascular cambium? Select all that apply. a. shoot apical meristem cells b. procambial cells c. epidermal cells d. parenchymal cells ANSWER: b, d 58. Looking out your window, you notice that the bark of a nearby oak tree has a mosaic appearance, consisting of deep fissures and relatively flat areas. What accounts for this mosaic appearance? Select all that apply. a. the formation of new cork cambia as old cork cambia grow away from the phloem b. the formation of new secondary phloem in distinct, discontinuous patches c. the formation of new cork cambia in distinct, discontinuous patches d. the formation of new vascular cambia as old cork cambia grow away from the phloem ANSWER: a, c 59. Which of the statements is true regarding secondary growth in vascular plants? Select all that apply. a. Secondary growth depends only on the vascular cambium. b. Secondary growth depends only on the cork cambium. c. Secondary growth depends on both the vascular and cork cambia. d. Secondary growth causes a plant to grow in diameter (not height). e. Secondary growth causes a plant to grow in height (not diameter). ANSWER: c, d 60. Why can fibers in angiosperm wood can support large mechanical loads? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 a. Their cell walls are very dense compared to vessels or tracheids. b. Their cell walls are very thick, almost filling the cell. c. Their diameter is much larger than their length. d. They can make up a large fraction of the wood due to the efficiency with which vessels can conduct water. ANSWER: b, d 61. What is produced by the vascular cambium? Select all that apply. a. secondary xylem b. cork c. secondary phloem d. wood e. bark ANSWER: a, c, d 62. What are the advantages of root meristems originating at the pericycle rather than at other root structures (e.g. the epidermis)? Select all that apply. a. New roots form only at the tips of preexisting roots, reaching deep water sources. b. New roots are directly connected to the xylem and phloem of preexisting roots. c. New roots can sprout from any location (not just nodes) on preexisting roots. ANSWER: b, c 63. Which plant hormones promote elongation of either the stems or roots in vascular plants? Select all that apply. a. ethylene b. cytokinins c. auxin d. gibberellic acid e. abscisic acid ANSWER: c, d, e 64. How would shoot and root growth be affected by a mutation that caused plants to lose the ability to polymerize glucose into starch granules? Select all that apply. a. Lateral branch shoots would grow more upright and be less able to maintain a horizontal orientation. b. Lateral branch roots fully embedded in soil would grow randomly upward and downward. c. Lateral branch shoots would grow more horizontally and have less of a tendency to turn upward. d. Only shoots would be affected. e. Roots breaking the soil surface would grow upward. ANSWER: a, b 65. Collectively, what did the work of Charles Darwin, Francis Darwin, and Frits Went demonstrate about phototropism in plants? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 a. A chemical signal is produced in shoot tips in response to light. b. A chemical signal is produced in shoot stems in response to light. c. Cells exposed to a chemical signal produced by light will grow toward the light. d. Cells exposed to a chemical signal produced by light will not grow toward the light. ANSWER: a, c 66. Which statements are true of phytochrome? Select all that apply. a. Pr is the active form. b. Pfr is the active form. c. Pfr is converted to Pr by red light. d. Pfr converts to Pr in the dark. e. Pr converts to Pfr in the dark. ANSWER: b, d 67. A researcher is studying the distribution of auxin in roots and stems exposed to sunlight. He notices that more auxin collects in the sides of stems and roots that are not exposed to light; however, the effects of this distribution are very different. Why? Select all that apply. a. Auxin prevents cell elongation in the shaded cells of stems. b. Auxin (via ethylene) promotes cell elongation in the shaded cells of roots. c. Auxin (via ethylene) prevents cell elongation in the shaded cells of roots. d. Auxin promotes cell elongation in the shaded cells of stems. ANSWER: c, d 68. How does vernalization alter the genetic material of plants? Select all that apply. a. through chromatin remodeling b. through mutations in certain genes c. through DNA methylation d. through RNAi ANSWER: a, c 69. Which statement is true about roots? Select all that apply. a. Root caps are supplied with cells from a meristem. b. Cells of the root cap are protected by a covering of root hairs. c. Root caps are supplied with new cells from the pericycle. d. Root caps protect the root apical meristem. ANSWER: a, d 70. Which events occur in plants as a result of decreased photoperiod? Select all that apply. a. flowering b. formation of bud scales Copyright Macmillan Learning. Powered by Cognero.

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Chapter 29 c. germination of seeds d. removal of plugs in plasmodesmata ANSWER: a, b

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Chapter 30 Multiple Choice 1. A researcher comes across a plant that has been infected by a pathogen. As a result of the infection, the plant has undergone a systemic acquired resistance (SAR) response and has also generated several siRNAs. Based on this response, what type of pathogen has infected the plant? a. a bacterium b. a virus c. a fungi d. an oomycete e. a parasitic plant ANSWER: b 2. A botanist is studying the effects of different oomycete species on potato plants. She exposes one set of plants to an oomycete species having several avirulence (AVR) proteins. She exposes another set of plants to an oomycete species having a single AVR protein. What do you expect will be the results of her experiment? a. Mortality will be higher in plants exposed to the oomycete species having several AVR proteins. b. Mortality will be higher in plants exposed to the oomycete species having a single AVR protein. c. Mortality rates will be the same in both plant sets. ANSWER: a 3. During an experiment, you find that exposing leaves to methyl salicylic acid vapor prior to exposing them to a pathogen results in an avirulent infection, whereas exposing untreated control leaves to the same pathogen results in virulent infection. This is likely an example of: a. systemic acquired resistance. b. a hypersensitive response. c. a small interfering RNA response. d. phloem signaling. ANSWER: a 4. How will ingesting plant alkaloids affect an herbivore? a. The herbivore would be unable to digest certain proteins. b. The herbivore's enzymes would be unable to function. c. Unusable amino acids would be incorporated into the herbivore's proteins. d. The herbivore's nervous system would be damaged. e. The herbivore's jaws would become stuck together. ANSWER: d 5. Bitter cassava is often preferred as a crop over sweet cassava because it produces higher yields (more pounds of cassava per hectare planted) in the presence of herbivores. Which of the statements poses a hypothetical relationship? a. Plants produce chemicals to deter herbivores. Bitter cassava stores toxic compounds in its tissues. Bitter cassava should incur less damage and therefore have higher yields. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 b. Plants produce chemicals to deter herbivores. Bitter cassava stores toxic compounds in its roots. Because they are below ground, the roots are not eaten and the presence of toxic compounds does not contribute to higher yields. c. Plants produce chemicals to deter herbivores. Bitter cassava expends energy producing toxic compounds. The more toxic compounds a plant produces, the lower its yields. d. Plants produce chemicals to deter herbivores. Bitter cassava requires additional nitrogen to make these compounds. The additional nitrogen that is used reduces the amount of nitrogen available for plant growth and decreases yields. ANSWER: a 6. For an ecological study, you monitor herbivore-plant interactions in a rainforest and notice that mammalian herbivores avoid the leaves of a particular species of plant. However, a number of tiny herbivorous insect generalists that feed by piercing individual plant cells with their mouthparts appear capable of completing their entire life cycle feeding on this same species. You hypothesize that in this species the chemical defenses are: Select all that apply. a. stored in the cytoplasm of epidermal leaf cells. b. stored in the vacuole in an inactive form and are activated by enzymes in the cytoplasm. c. contained in a latex stored in vascular canals. d. nonexistent. ANSWER: c 7. Which of these defensive compounds produced by plants can harm an herbivore's nervous system? a. terpenes b. alkaloids c. tannins d. phenols ANSWER: b 8. The high silica content in grasses and the thick enamel layer on horse teeth are examples of: a. a trade-off between growth and defense. b. the "escape and radiate" hypothesis. c. a symbiotic relationship. d. co-evolution. ANSWER: d 9. Chemical defenses are an important part of plant-herbivore interactions. The production of these chemical defense compounds results in an energetic cost to the organism. Which of the statements explains how natural selection resulted in the evolution of these chemical compounds? a. The production of chemical compounds requires nitrogen which cannot be used to make proteins. b. The energetic cost of producing chemical compounds is reduced because only small amounts of each chemical are needed to have their desired effects. c. Individuals that produce chemical compounds are better protected from herbivores and are able to produce more offspring than individuals that do not produce the compounds. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 d. Plants produce amino acids that are not incorporated into their own proteins, but herbivorous insects do incorporate these amino acids into their own proteins, rendering the herbivores' proteins nonfunctional. ANSWER: c 10. Recall that the bullhorn acacia forms a symbiosis with P. ferruginea ants. A researcher takes an acacia seed and plants it in an area without P. ferruginea ants. What will happen to this acacia plant? a. It will immediately form a symbiotic relationship with ants native to the area. b. It will begin producing alkaloids and terpenes to deter herbivores. c. It will begin to produce hairs on its leaves to deter herbivores. d. Without P. ferruginea, the plant won't form any new defenses and will be eaten by herbivores. ANSWER: d 11. You may have heard that monarch butterflies and their caterpillars are "poisonous" to many predators. What makes these insects so toxic? a. Monarch caterpillars produce cardenolides in their digestive tracts. b. Monarch caterpillars use cardenolides from milkweed to synthesize toxins. c. Monarch caterpillars ingest milkweed latex, which "chokes" their predators. d. Monarch caterpillars store cardenolides from milkweeds. e. Monarch caterpillars secrete latex from their bodies. ANSWER: d 12. Nicotiana attenuata is subject to damage by both generalist and specialist herbivores. The production of nicotine is not effective against one specialist, M. sexta. If an individual of N. attenuata is being eaten by both M. sexta and a generalist herbivore, what would you expect to happen? a. The production of nicotine will stop because it is costly and ineffective against M. sexta. b. The production of nicotine will increase, and the individual will also produce chemicals that attract insects that prey on different life stages of M. sexta. c. The production of nicotine will decrease because energy must be devoted to the growth of spines to defend against M. sexta. d. The production of nicotine will continue because its production is stimulated by the generalist herbivore. The energetic cost of producing nicotine makes it impossible for the individual to mount any other defenses against specialist herbivores. ANSWER: b 13. Which of the senses is not analogous to a sensory pathway by which plants can detect herbivores? a. taste b. smell c. sight ANSWER: c 14. Because ant-plants must continually provide sugars, protein, and fats to ensure a large enough population of defending ants when herbivores attack, ant defense could be considered a(n) _____ form of defense. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 a. mechanical b. inducible c. constitutive d. chemical ANSWER: c 15. In response to being eaten by M. sexta caterpillars, the coyote tobacco plant produces volatile chemicals that can travel through the air to: a. signal to other tobacco plants that caterpillars are in the area. b. produce an odor that will deter other caterpillars from eating its leaves. c. attract insect predators that will eat the caterpillars. d. signal to other parts of the plant to produce more nicotine. ANSWER: c 16. When a coyote tobacco plant releases volatile signals in response to M. sexta caterpillars, it is an example of a(n) _____. The presence of hairs on the leaf of a milkweed plant is an example of a(n) _____. a. constitutive defense; inducible defense b. SAR response; basal response c. basal response; SAR response d. inducible defense; constitutive defense ANSWER: d 17. A seed from a yew tree germinates on the forest floor. During the first few months of its life, the sapling dedicates most of its resources to increasing in height rather than producing the defensive compound taxol. This allocation of resources is an example of: a. a symbiosis. b. a trade-off. c. a constitutive defense. d. an inducible defense. e. a basal response. ANSWER: b 18. Consider the experimental design of the study conducted by Baldwin and Schulze in which they investigated the effects of plants that have been attacked by herbivores on undamaged plants. Review the graph shown. What is the most important difference between the treatment group of plants and the control group of plants?

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Chapter 30

a. There are more plants in the treatment group of plants than in the control group of plants. b. The plants in the control group are not exposed to air that has passed over damaged leaves. c. The plants in the experimental group start out with damaged leaves. d. The plants in the control group are exposed to more air than the plants in the experimental group ANSWER: b 19. Consider the experimental design of the study conducted by Baldwin and Schulze in which they investigated the effects of plants that have been attacked by herbivores on undamaged plants. Review the graph shown. Which is the independent variable in the study?

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Chapter 30

a. the presence or absence of damaged plants in the path of air flow b. the amount of tannins produced by the damaged plants c. the amount of tannins produced by the undamaged experimental plants d. the amount of tannins produced by the undamaged control plants ANSWER: a 20. Consider the study in which researchers investigated the trade-off between growth and defense. Review the graph shown. Which of the statements is supported by the data in the graph?

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Chapter 30

a. All plants exhibited an increase in leaf surface area over time. b. Plants of clay origin exhibited the greatest increased in leaf surface area under all conditions. c. Plants of sand origin exhibited similar increases in leaf surface area when grown in clay regardless of whether the area they were grown in was protected or unprotected. d. Plants of sand origin exhibited greater increase in leaf surface area in protected sand habitat compared to plants of clay origin in the same sandy habitat. ANSWER: c 21. Consider the study in which researchers investigated the trade-off between growth and defense. Review the graph shown. What is the dependent variable in the study?

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Chapter 30

a. the change in leaf surface area b. the type of soil the plants were grown in c. whether the plants were protected or unprotected. ANSWER: a Multiple Response 22. Which of these locations are points of entry through which viral and bacterial pathogens can infect plant tissues? Select all that apply. a. sites of herbivore damage b. open stomata c. the cuticle d. leaf hairs ANSWER: a, b 23. Which of the statements is true regarding the systemic acquired resistance (SAR) response in plants? Select all that apply. a. It is pathogen-specific. b. It relies on salicylic acid. c. It is only effective against biotrophic pathogens. d. Strong SAR responses are associated with tissue necrosis. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 ANSWER: b, d 24. Recall that plants produce physical barriers (e.g., a cuticle or bark) to prevent pathogen infections. How can a pathogen enter, and ultimately infect, its host plant? Select all that apply. a. through the stomata b. through wounds made by insects c. through wounds made by farming equipment d. by being drawn up from the soil with water ANSWER: a, b, c 25. Which of these organisms can act as plant pathogens? Select all that apply. a. mistletoe b. nematodes c. tobacco plants d. bacteria e. P. ferruginea ants ANSWER: a, b, d 26. Which of these plant responses targets a specific organism rather than a general response? Select all that apply. a. a basal immune response b. R proteins recognizing AVR proteins c. a systemic acquired resistance response d. siRNA ANSWER: b, d 27. You are screening tomato varieties for plants that will resist a novel biotrophic pathogen. You should choose those with: Select all that apply. a. a strong hypersensitive response. b. fewer plasmodesmata connecting cells. c. a high diversity of AVR proteins. d. a high diversity of R proteins. ANSWER: a, d 28. The oomycetes that infected and killed potato plants during the Irish Potato Famine can be classified as a(n) _____ pathogen. Select all that apply. a. virulent b. necrotrophic c. avirulent d. biotrophic ANSWER: a, d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 29. Which of the statements is true regarding the RNA of plant viruses? Select all that apply. a. Outside of replication, the RNA of plant viruses is typically single-stranded. b. Outside of replication, the RNA of plant viruses is typically double-stranded. c. Plants cleave viral double-stranded RNA to form small interfering RNAs. d. Plants cleave viral single-stranded RNA to form small interfering RNAs. e. Plant viruses contain only single-stranded DNA, not RNA. ANSWER: a, c 30. You are screening tomato varieties for plants that will resist a novel necrotrophic pathogen. You should choose plants with: Select all that apply. a. a strong hypersensitive response. b. fewer plasmodesmata connecting cells. c. thicker and stronger cell walls. d. a high diversity of R proteins. ANSWER: a, c 31. Which of the defenses would you expect to be effective in combating infections by pathogens similar to Rhizobium radiobacter? Select all that apply. a. basal resistance b. R proteins c. hypersensitive response d. systemic acquired resistance e. siRNA ANSWER: a, b, c, d 32. Which of these plant and/or pathogen proteins are involved in highly specific immune responses? Select all that apply. a. flagellin receptors on the plasma membrane b. avirulence (AVR) proteins formed by pathogens c. resistance (R) proteins formed by plants d. chitin receptors on the plasma membrane ANSWER: b, c 33. The ability of plants to wall off infections through the coordinated death of a cordon of uninfected cells provides a first line of defense against: Select all that apply. a. viruses. b. fungi. c. oomycetes. d. bacteria. e. herbivores. ANSWER: a, b, c, d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 34. Removing the ants from ant-plants growing in the wild would be expected to have: Select all that apply. a. little impact on ant-plant growth because, while loses to herbivores would increase, the plant would save the resources normally consumed by the ants. b. a large impact on ant-plant growth, because ant-plants invest less than other plants in chemical defense, so they will be more palatable to herbivores. c. a positive impact on ant-plant growth, because the ants in general consume more plant resources than they protect from herbivory. d. a large impact on ant-plant growth, because the ants would not be suppressing the growth of neighboring plants that compete with the ant-plants for light and nutrients. ANSWER: b, d 35. Why do ant-plants invest in nectaries and protein/lipid food bodies on their young leaves? Select all that apply. a. It encourages the ants to visit the plant tissues most vulnerable to herbivory. b. It helps keep the ants away from floral nectaries where they might attack pollinators. c. It moves the sources of nutrition away from the parts of the plant that provide the ants' physical shelter, encouraging them to patrol the intervening stems. d. Leaves of all plants have foliar nectaries and feeding bodies to some degree and, therefore, the foliar location may not have adaptive significance. ANSWER: a, b, c 36. Plants can defend themselves with highly reactive compounds that cause very general toxic effects because: Select all that apply. a. they can sequester these compounds in specialized tissues. b. they can store these compounds in subcellular compartments separate from their activating enzymes, so the compounds are activated only when the cell is mechanically damaged. c. their metabolisms are so different from animals that there is no overlapping sensitivity to toxins between plants and herbivores. d. plant cell walls form a metabolically inactive space within the plant that can store toxins without poisoning the metabolically active cytoplasm inside the cells. ANSWER: a, b 37. Grasses first evolved in: Select all that apply. a. response to the spread of large grazing mammals. b. forested environments. c. coevolution with herbivorous dinosaurs. d. response to fire and or grazing by small browsers. ANSWER: b, d 38. What methods do grasses employ to deter herbivores and survive heavy grazing? Select all that apply. a. Grass shoot apical meristems are located at the tips of leaves. b. Grass shoot apical meristems remain close to the ground. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 c. Grasses contain silica plates. d. Grasses contain chitin plates. ANSWER: b, c 39. Which of these are chemical compounds produced by plants that interfere with an herbivore's ability to break down proteins? Select all that apply. a. protease inhibitors b. phenols c. alkaloids d. terpenes ANSWER: a, b 40. Which of these medicines are derived from defensive compounds originally made by plants? Select all that apply. a. aspirin (a pain-reliever) b. quinine (an antimalarial drug) c. vincristine (a chemotherapy drug) d. taxol (a chemotherapy drug) e. acetaminophen (a pain-reliver) ANSWER: a, b, c, d 41. What are possible mechanisms of action for defensive compounds produced by plants? Select all that apply. a. to prevent premature molting in insects b. to prevent proteins from being digested c. to interfere with cell division d. to interfere with an herbivore's nervous system ANSWER: b, c, d 42. How do grass plants elevate their leaves to compete for light without exposing their shoot apical meristems to grazing animals? Select all that apply. a. They limit internode elongation between leaves. b. They grow their leaves from persistent zones of cell division and elongation at the leaf base. c. They use axillary meristems to produce leaves. d. They never elevate the shoot apical meristem. ANSWER: a, b 43. Nitrogen is required to synthesize which of these defensive compounds (which are, as a result, very costly for plants to produce)? Select all that apply. a. alkaloids b. menthol c. phenols Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 d. nicotine e. terpenes ANSWER: a, d 44. Plants produce a variety of defensive compounds that can seriously damage herbivores. Why aren't these compounds also detrimental to the plants themselves? Select all that apply. a. They can be sequestered in vacuoles of plant cells. b. They can diffuse from leaves into the phloem of plants. c. They are restricted to "canals" in the plant. d. They can diffuse quickly between plant cells before they have an effect. ANSWER: a, c 45. Milkweed plants produce a variety of defenses, such as leaf hair and latex-filled canals, to deter herbivores. How can monarch caterpillars circumvent these obstacles? Select all that apply. a. by "mowing" hairs off of milkweed leaves b. by cutting the midvein that transports latex through leaves c. by creating channels to divert the flow of latex d. by metabolizing chemical compounds (cardenolides) produced by milkweeds ANSWER: a, b, c 46. Nonprotein amino acids produced by plants: Select all that apply. a. are plant amino acids that are not incorporated into animal protein. b. interfere with protein synthesis in herbivores. c. are not used by plants in protein synthesis. d. are used to produce nitrogen-rich chemical defenses such as alkaloids. ANSWER: b, c 47. Which of the features contribute to the ability of grasses to tolerate both grazing and fire disturbances? Select all that apply. a. persistent zones of cell division at the leaf base b. compressed internodes c. silica plates d. scattered vascular bundles in the stem ANSWER: a, b 48. Constitutive defenses differ from inducible defenses in that: Select all that apply. a. the genetic program to produce the former is always on, while the genetic program to produce the latter must be switched on by the sensing of herbivore presence. b. the latter is more efficient when the target herbivore may be absent for much of the growth of the plant. c. the former is always less efficient than the latter because inducible defenses are only turned on when needed. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 d. the former are more likely to confer broad forms of resistance to multiple herbivores, while the latter are more likely to be specific to particular herbivores. ANSWER: a, b, d 49. Plants can detect herbivore attacks on nearby plants due to: Select all that apply. a. volatile normal cell contents released when exposed to air by mechanical damage. b. synthesis of volatile compounds induced by mechanical damage. c. volatile compounds induced by detection of specific herbivores. d. mechanical vibrations arising from insect feeding that travel through the air. e. change of leaf color caused by mechanical damage. ANSWER: a, b, c, d 50. Which of the factors likely contributed to the evolution of inducible defenses against herbivory? Select all that apply. a. unpredictable fluctuations in local herbivore populations b. differing strategies for overcoming plant defenses among herbivores c. the cost of allocating resources to defense d. consistent fitness benefits of high allocations to defense ANSWER: a, b, c 51. Consider an experiment in which two species of plants, A and B, are grown in separate pots in the same enclosure, either with an herbivorous insect that eats only plant A (treatment) or with no herbivores (control). How do you expect growth for species B will differ between the two treatments? Select all that apply. a. If species B relies on defenses like hairs or spines, there will be no difference. b. If species B relies on chemical defenses, its growth will be higher in the treatment. c. If species B has defenses induced by volatile signals, its growth will be higher in the control. d. If species B relies on constitutive defenses, its growth will be higher in the treatment. ANSWER: a, c 52. In what type of environment would you expect to find a plant that spends most of its resources on defenses, rather than growth? Select all that apply. a. in the nutrient-poor soil of a tundra b. in the clay soil of a rainforest c. on well-fertilized farmland d. in the sandy soil of a rainforest ANSWER: a, d 53. Many plants produce volatile, airborne, compounds when attacked by herbivores. In what ways do these compounds aid the plant in protecting itself? Select all that apply. a. They increase the production of defensive compounds in neighboring plants, thereby reducing the population density of herbivores. b. They induce the production of defensive compounds in distant parts of the plant. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 c. They inhibit the metabolism and growth of herbivores by interfering with their respiratory pathways. d. They attract animals that can prey upon or parasitize the herbivores. e. They provide no protection and are simply the result of chewing damage. ANSWER: b, d 54. Volatile chemicals that escape from plants into the air: Select all that apply. a. assist some parasitic plants in locating their host. b. can attract insect predators to an herbivore-damaged plant. c. may play a role in systemic acquired resistance. d. are never simply released due to tissue damage with no adaptive function. ANSWER: a, b, c 55. Which of the statements is evidence for an arms race between plants and their antagonists? Select all that apply. a. R genes are one of largest gene families in plants. b. Chemical defenses are highly diverse. c. Diversification of herbivorous insects and plants occurs at the same time over evolutionary time. d. Specific and inducible defenses are more common than basal and constitutive defenses. ANSWER: a, b, c 56. Fractions of a field of Bt crops planted with non-Bt-expressing varieties: Select all that apply. a. ensure that herbivore genes conferring susceptibility to Bt are well represented in the next generation. b. ensure that selection for Bt resistance is not too intense within herbivore populations. c. impede the complete eradication of important agricultural pests. d. lead to higher losses to pests in the short term than if no fractions were planted with non-Btexpressing varieties, but maintain the effectiveness of Bt in protecting crops in the long term. ANSWER: a, b, d 57. Which of the statements are consistent with the "escape and radiate hypothesis" of Ehrlich and Raven? Select all that apply. a. A species that evolves a novel, highly effective form of defense will be able to successfully invade new habitats. b. New defenses that release a species from the intense selective pressures exerted by herbivores will lead to higher genetic diversity. c. Populations that "escape" herbivory by invading new habitats are free to evolve novel forms of defense that result in new species. d. Interactions between plants and animals lead to higher diversity than one might expect based only on animal-animal or plant-plant competition. ANSWER: a, b, d 58. Examine the figure shown. Which of the statements indicates one reason why seeds dispersed far from other Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 plants of the same species have a higher chance of survival? Select all that apply.

a. There is a smaller chance of being eaten by a seed predator because seed predators are more likely to find seeds that are in high concentration. b. There is a smaller chance of competition between a rare seed and the other plants in the area because a small plant doesn't require the same resources as larger, already established plants. c. The seed is more likely to survive because there is a smaller chance of it being exposed to pathogens that may have become established near other plants of the same species. d. Seeds that colonize new areas have unlimited resources and can undergo a population explosion, dominating an area that was previously devoid of plants. ANSWER: a, c 59. Over 380,000 species of angiosperms exist on Earth. Which of these factors contributed to this diversity of flowering plants? Select all that apply. a. pollination by different insects b. infections by different bacteria Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 c. predation by different herbivores d. infections by different fungi e. genetic engineering ANSWER: a, b, c, d 60. A farmer has adopted a new program of integrative pest management. What will this program entail for the farmer and his crops? Select all that apply. a. The farmer will assess which pests are consuming his crops. b. The farmer will use industrial pesticides as a first line of defense. c. The farmer will introduce natural predators for the pests. d. The farmer will use industrial pesticides as a last line of defense. ANSWER: a, c, d 61. You are studying a group of closely related plant species that are fed upon by numerous herbivores. One of these herbivores is a caterpillar species that has evolved the ability to sequester the alkaloid that is this plant group's principal defense against all herbivores. You then construct a phylogeny of the plant species. You expect to find that the most recently radiated species in the group that co-occur with the caterpillar: Select all that apply. a. are found in resource-rich environments. b. are found in resource-poor environments. c. have lost the ability to produce the sequestered alkaloid. d. produce a novel volatile molecule that attracts predators of the caterpillars. ANSWER: a, c, d 62. What are the components of the Janzen-Connell hypothesis? Select all that apply. a. Seeds located close to plants of the parent will have a higher survival rate. b. Seeds located far away from the parent will have a higher survival rate. c. Seeds from rare plants will have a higher survival rate. d. Seeds from common plants will have a higher survival rate. ANSWER: b, c 63. B. thuringiens bacteria is a remarkable resource for researchers and farmers alike when it comes to pest control because: Select all that apply. a. it contains a Ti plasmid. b. it produces insect toxins. c. it is responsible for crown gall disease and can be used to kill parasitic plants. d. its genes can be used to genetically engineer plants that are resistant to herbivores. e. it produces cardenolides. ANSWER: b, d 64. The phylogeny of a group of species you are studying shows that fast growth and low investment in defense is ancestral in the group, while a recently diverged subgroup containing 10% of the group's species is Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 characterized by slow growth and epidermal hairs filled with a caustic agent. You hypothesize that: Select all that apply. a. hairs were an evolutionary innovation that allowed these plants to invade lower resource environments than the ancestors of the group occupied. b. hairs were an evolutionary innovation that allowed the plants to invade all types of environments. c. hairs are an evolutionary innovation that allowed these plants to invade higher resource environments than the ancestors of the group occupied. d. the evolution of leaf hairs in this group does not support the "escape and radiate" hypothesis. ANSWER: a, d 65. According to the Janzen-Connell hypothesis, in order for pathogens or herbivores to contribute to the maintenance of diversity: Select all that apply. a. the infections must be avirulent. b. the pathogens/herbivores must be host-specific. c. the pathogens/herbivores must result in lower seedling mortality close to adult plants of the same species. d. mortality of seedlings due to competition between plants must be less important than mortality due to herbivores/pathogens. ANSWER: b, d 66. Consider the experimental design of the study conducted by Baldwin and Schulze in which they investigated the effects of plants that have been attacked by herbivores on undamaged plants. Review the graph shown. Which is the dependent variable in the study? Select all that apply.

a. the presence or absence of damaged plants in the path of air flow Copyright Macmillan Learning. Powered by Cognero.

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Chapter 30 b. the amount of tannins produced by the damaged plants c. the amount of tannins produced by the undamaged experimental plants d. the amount of tannins produced by the undamaged control plants ANSWER: b, c, d 67. Consider the study in which researchers investigated the trade-off between growth and defense. What are the independent variables in the study? Select all that apply.

a. the change in leaf surface area b. the type of soil the plants were grown in c. whether the plants were protected or unprotected. ANSWER: b, c

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Chapter 31 Multiple Choice 1. Review the phylogenetic tree shown. Which group of bryophytes appears to be ancestral to the others?

a. Liverworts b. Mosses c. Hornworts d. Lycophytes e. The relationship between the bryophytes remains uncertain. ANSWER: e 2. Review the phylogenetic tree shown. A specialized term in the construction and interpretation of phylogenetic trees is 'synapomorphy' which means the changed character that unites branches of the tree. What synapomorphy is associated with the label C in the figure shown?

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Chapter 31

a. Seeds and pollen b. Xylem and phloem c. Flowers d. Alternation of generations e. Parallel veins ANSWER: a 3. Review the phylogenetic tree shown. One of the major hurdles in the evolution of land plants was adapting to the relative scarcity of water when compared to the habitat occupied by their green algal ancestors. One such adaptation was the origin of cell types that were specialized for transporting water from one part of the plant to other parts. Which letter on this phylogenetic tree indicates this evolutionary step?

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Chapter 31 a. A b. B c. C d. D e. E ANSWER: b 4. Review the phylogenetic tree shown. One of the major events in the evolution of plants was the encapsulation of the sporophyte embryo with a significant supply of stored nutritional sources within protective cell layers. Which letter on this phylogenetic tree indicates this evolutionary step?

a. A b. B c. C d. D e. E ANSWER: c 5. Approximately 40% of living ferns are tropical epiphytes, a fact that supports the hypothesis that: a. the origin and eventual domination of angiosperms in tropical forests both displaced and created new opportunities for other plant lineages. b. plant lineages are conservative in their basic ecology and tend to diversify in the same environments in which they originated. c. most fern diversity is in temperate forests. d. epiphytic ferns are all that remain of an ancient diversification of ferns that preceded the evolution of angiosperms. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 6. Based on a lack of _____ in their fossil record, the earliest land plants are thought to have only been able to maintain photosynthesis intermittently. a. stomata b. cuticle c. vascular tissues d. sporopollenin ANSWER: c 7. Which of these occurrences is thought to be directly associated with the establishment of modern rainforests? a. the formation in the forest understory of dense mats of mosses and extensive beds of ferns b. the increase in water uptake by vascular plants c. the increase in photosynthetic oxygen production by vascular plants d. the evolution of the ability of angiosperms to support high transpiration rates ANSWER: d 8. Which taxonomic group of land plants, still in existence today, was the dominant group of land plants when the angiosperms began their rapid increase in speciation and distribution? a. bryophytes b. ferns and horsetails c. lycophytes d. gymnosperms ANSWER: d 9. Which of the statements is true for cycads and ferns, but not for conifers or ginkgo? a. Their extant diversity consists of a reduced set of "relict species" that have survived the appearance of the angiosperms by growing in environments where angiosperms do not have a clear advantage. b. Their extant diversity is confined to the same environments in which their peak historical diversity once existed. c. Their diversity has increased due to recent radiations into new environments, such as modern tropical rain forests created by angiosperms. d. Their total diversity has declined since its peak, even as the extant diversity consists of recent radiations coinciding with the diversification of angiosperms. ANSWER: d 10. The sperm cells of bryophytes are dispersed by the wind. a. true b. false ANSWER: b 11. The multicellular gametophytes of bryophytes are haploid. a. true Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 b. false ANSWER: a 12. Which of the different types of bryophytes has the sporophyte generation that lives the longest? a. mosses b. hornworts c. quillworts d. horsetails e. liverworts ANSWER: b 13. Which of the vascular plants have swimming sperm and rely on liquid water in the environment to achieve fertilization? a. gymnosperms and lycophytes b. lycophytes and gymnosperms c. ferns, horsetails, and gymnosperms d. ferns, horsetails, and lycophytes ANSWER: d 14. Fossil evidence demonstrates that the earliest vascular plants had: a. leaves. b. phloem. c. stomata. d. deep root systems. e. vascular cambium. ANSWER: c 15. Although many types of plants decreased in diversity as angiosperms gained dominance, polypod ferns radiated in conjunction with the evolution of angiosperms. Why? a. Angiosperms created new environments that ferns could inhabit. b. Angiosperms and ferns could interbreed, increasing fern diversity. c. Herbivores that once preyed on ferns began to target angiosperms. d. In response to angiosperms, ferns developed secondary growth. e. All of these choices are correct. ANSWER: a 16. Coal is what remains of ancient forests after millions of years of exposure to subterranean high temperatures and pressures. What group of land plants is thought to have contributed most to this fossil energy source? a. lycophytes b. sphagnum mosses in peat bogs c. ferns and horsetails Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 d. bryophytes e. conifer and angiosperm trees ANSWER: a 17. Although we typically think of angiosperms when we consider double fertilization, in which of these gymnosperms does double fertilization occur? a. cycads b. conifers c. ginkgo d. gnetophytes ANSWER: d 18. Which of the four major groups of gymnosperms currently has only a single species? a. ginkgos b. cycads c. conifers d. gnetophytes ANSWER: a 19. Many dinosaur movies depict huge tree-sized plants with woody trunks. These trees would have most likely belonged to which group? a. lycophytes b. cycads c. angiosperms d. bryophytes ANSWER: b 20. The xylem of angiosperms is composed of ____, allowing these plants to thrive in tropical climates and have high rates of photosynthesis. Gymnosperms are successful in cold climates because their xylem is composed of ___, which reduce the risk of cavitation due to freezing. a. vessels, tracheids b. tracheids, vessels c. cambium, cuticle d. sporopollenin, leptosporangia ANSWER: a 21. Which of the challenges faced by plants in the early colonization of land was solved by a change in the life cycle to "alternation of generations"? a. water transport and dispersal of gametes b. fertilization of the egg and dispersal of offspring c. wind pollination and dispersal of zygotes d. water uptake and dispersal of female gametophytes Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 ANSWER: b 22. Annual plants are defined as plants that: a. may overwinter as bulbs or corms. b. do not have a vascular or bark cambium. c. flower every year. d. persist through a portion of the year only as seeds. ANSWER: d 23. About two-thirds of human caloric intake consists of wheat, rice, and corn. What tissues constitute the majority of the nutritional content of these plant-based foods? a. fruits and cones b. seed embryos and ovules c. endosperm and female gametophytes d. starch granules and sporopollenin ANSWER: c 24. What characteristic arose in angiosperms that enhances animal dispersal of seeds? a. packaging of seeds within the mature ovary b. showy flower parts c. pollen with three apertures d. double fertilization ANSWER: a 25. Angiosperms only provide energy-rich nutritive materials to ovules that have been successfully fertilized, whereas gymnosperms typically provide those resources even to unfertilized, and therefore sterile, ovules. What accounts for this angiosperm advantage? a. double fertilization b. animal pollinated flowers c. multicellular xylem vessels d. seeds that are nutritious for animals e. fruits that are nutritious for animals ANSWER: a 26. Which of the choices is thought to contribute to higher photosynthetic potential in angiosperms when compared to gymnosperms? a. multicellular xylem vessels b. double fertilization c. evolution of vascular cambium d. loss of vascular cambium ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 27. Review the figure shown. What was actually measured in order to create part C of the graph?

a. the ratio of 12C:13C carbon isotopes occurring in grass-eating horse teeth fossils b. the amount of C4 carbon in fossilized grass remains from different periods of history c. the amount of 14C in fossilized pollen associated with open grasslands d. the ratio of C3 to C4 carbon compounds found in fossilized remains of grass-eating mammals e. relative proportions of microscopic SiO2 particles that are characteristic of different plant groups ANSWER: a 28. Review the figure shown. What was actually measured in order to create part B of the graph?

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Chapter 31

a. the ratio of 12C:13C carbon isotopes occurring in grass-eating horse teeth fossils b. the amount of C4 carbon in fossilized grass remains from different periods of history c. the amount of 14C in fossilized pollen associated with open grasslands d. the ratio of C3 to C4 carbon compounds found in fossilized remains of grass-eating mammals e. the relative proportions of microscopic SiO2 particles that are characteristic of different plant groups ANSWER: e 29. Review the figure shown. The conclusion drawn from the data shown in these graphs is that grasslands expanded in North America as a result of:

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Chapter 31

a. decreasing atmospheric CO2 concentrations and a drier climate. b. increasing atmospheric CO2 concentrations and a moister climate. c. evolution of horses and other mammals that could thrive on grasslands. d. a change in the atmospheric ratio of 12C:13C carbon isotopes. e. herbivorous mammals like horses can adapt to different kinds of plants in their diet. ANSWER: a Multiple Response 30. What are challenges that plants faced in colonizing land from aquatic origins? Select all that apply. a. mechanical support b. hydration c. avoidance of herbivores d. dispersal of sperm and offspring e. protection from ultraviolet radiation ANSWER: a, b, d, e 31. Which groups of plants were present on Earth 200 million years ago? Select all that apply. a. cycads b. ferns and horsetails c. conifers d. angiosperms Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 ANSWER: a, b, c 32. Which group(s) have existing diversity that mostly radiated in conjunction with the rise of angiosperms? Select all that apply. a. gnetophytes b. polypod ferns c. cycads d. mosses ANSWER: b, c, d 33. If angiosperms never existed, how would the environmental conditions of today's rainforest regions probably be different? These regions would be: Select all that apply. a. wetter. b. drier. c. colder. d. hotter. e. windier. ANSWER: b, d 34. Diversity is higher today than in the Cretaceous period (145 to 166 million years ago) for: Select all that apply. a. angiosperms. b. conifers. c. ferns. d. gnetophytes. ANSWER: a, b 35. Which of these characteristics do land plants share with their closest algal relatives? Select all that apply. a. plasmodesmata connecting the cytoplasm of cells in the multicellular plants b. an external waxy skin separating the plant body from its environment c. multicellular haploid and diploid generations d. populations of apical cells that maintain their meristematic identity ANSWER: a, d 36. The last common ancestor of gymnosperms and angiosperms possessed which of these characteristics? Select all that apply. a. pollen b. flowers c. seeds d. vascular tissues e. double fertilization Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 ANSWER: a, c, d 37. Which of these true statements could contribute to a plausible hypothesis for why the alternation of generations evolved in the earliest land plants? Select all that apply. a. It allowed one generation to specialize in fertilization and the other in dispersal. b. Gametophytes lack the ability to form rigid (lignified) vascular tissues. c. The biopolymer sporopollenin first evolved to protect the diploid stage of green algae. d. Many gametophytes tolerate desiccation. ANSWER: a, c 38. Species from which of these groups died off as angiosperms diversified and became more abundant? Select all that apply. a. ferns b. cycads c. lycophytes d. ginkgos e. green algae ANSWER: a, b, c, d 39. Which of the choices are characteristics of bryophytes? Select all that apply. a. a free-living sporophyte b. stomata c. desiccation resistance d. water-conducting cells ANSWER: b, c, d 40. Which of the characteristics of bryophytes likely contribute to their small size relative to vascular plants growing in the same environment? Select all that apply. a. The male gametes require water to swim to the egg for fertilization. b. The leaves of bryophytes are only a few cell layers thick. c. They lack the specialized cells needed for water transport to elevated parts of the plant. d. The ecologically dominant generation is haploid. ANSWER: a, c 41. Which of the choices are examples of convergent evolution? Select all that apply. a. the water-conducting tissues of some bryophytes and that of ferns b. the vascular cambium of quillworts (Isoetes) and that of angiosperms c. the roots of bryophytes and those of lycophytes d. the cuticle of bryophytes and that of gymnosperms ANSWER: a, b 42. Based on the fossil record of the Rhynie cherts, what structures were present in the earliest vascular plants? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 Select all that apply. a. stomata b. cuticles c. extensive root systems d. multi-veined leaves e. sporangia ANSWER: a, b, e 43. What novel innovations did the evolution of the seed plant life cycle involve compared with other groups of land plants? Select all that apply. a. dispersal through air b. fertilization in the absence of environmental water c. an ecologically dominant sporophyte generation d. dispersal of multicellular sporophytes e. dispersal of multicellular gametophytes f. dispersal by animals ANSWER: b, d, e 44. Which of these groups of land plants have evolved large tree-like forms? Select all that apply. a. ferns and horsetails b. hornworts c. lycophytes d. seed plants ANSWER: a, c, d 45. Based on your knowledge of how different plant groups have evolved arborescent (treelike) forms, which of the pairs of plant groups have large, tree-sized plant forms that evolved independently? Select all that apply. a. monocots and eudicots b. gymnosperms and eudicots c. lycophytes and ferns d. ferns and seed plants ANSWER: a, c, d 46. In perfectly still air, you are studying the spore dispersal of two fern species found in your area. You find that one species has an average dispersal distance many times that of the other, which is much larger than would likely be explained by chance. A reasonable hypothesis would be that your two species consist of a: Select all that apply. a. whisk fern and polypod fern. b. whisk fern and horsetails. c. marattioid and leptosporangiate fern. d. leptosporangiate and polypod fern. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 ANSWER: a, c 47. Which of the choices was an evolutionary innovation that contributed to the successful colonization of land by the first terrestrial plants? Select all that apply. a. leaves b. sporophyte generation c. roots d. gametophyte generation e. cuticle ANSWER: b, e 48. Most cycad populations are small and often vulnerable to extinction. Which of these explanations is thought to have contributed to the survival of cycad species to the present day? Select all that apply. a. the pollination of cycads by insects b. their thick, protective bud scales at the apical meristem c. their association with nitrogen-fixing bacteria d. their fragmented geographical distribution ANSWER: a, b, c 49. Gymnosperms include plants that demonstrate which of these characteristics? Select all that apply. a. fleshy seeds b. double fertilization c. pollen-containing cones d. vascular cambium e. swimming sperm f. free-living gametophytes ANSWER: a, b, c, d 50. The fact that early diverging angiosperms with vessels did not conduct water at higher rates than early diverging angiosperms that lacked vessels, supports the hypothesis that: Select all that apply. a. differences in xylem structure relative to gymnosperms did not contribute in an important way to the success of early angiosperms. b. the change in angiosperms from wood composed of tracheids to wood composed of vessels and fibers did not have much evolutionary impact on angiosperms until the diversification of monocots and eudicots. c. the growth-form flexibility conferred by xylem with fibers was more important for the ecological success of early angiosperms than the potentially high rates of transpiration made possible by vessels. d. angiosperms first evolved under shady and moist conditions. ANSWER: c, d 51. Before DNA sequence data became available, phylogenies constructed from anatomical and morphological Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 studies of living and fossil plants often supported gnetophytes as the sister group to the angiosperms. If this hypothesis were true, and the rest of the phylogeny as depicted in your text stayed the same, which of these statements would be true? Select all that apply. a. Angiosperms would be paraphyletic. b. The phylogeny would support a single origin for vessels. c. Cycads would be the sister group to all other seed plants. d. The phylogeny would support independent multiple origins of double fertilization. e. Seed plants as a whole would be monophyletic. ANSWER: b, c, e 52. Which of these fruits, vegetables, or grains are produced by eudicots? Select all that apply. a. wheat b. blueberries c. rice d. sunflower seeds e. carrots ANSWER: b, d, e 53. What can scientists study in the fossil record to learn about the distribution of grasses in the past? Select all that apply. a. herbivore teeth b. herbivore hooves c. stromatolites d. phytoliths e. the ratio of different carbon isotopes in fossils ANSWER: a, d, e 54. Which of the choices likely contributed to the diversification of angiosperms? Select all that apply. a. wind-based pollination b. animal-based pollination c. xylem composed of vessels and fibers d. greater C3 photosynthetic efficiency ANSWER: b, c 55. Which of the statements are true regarding angiosperms? Select all that apply. a. They tend to have higher maximum rates of photosynthesis compared with other plant groups in the same environment. b. They tend to have higher water transport capacities compared with other plant groups in the same environment. c. They tend to have higher rates of CO2 uptake compared with other plant groups in the same environment. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 d. They tend to have lower rates of photosynthesis compared with other plant groups in the same environment. e. They tend to have lower rates of CO2 uptake compared with other plant groups in the same environment. ANSWER: a, b, c 56. If angiosperms contained tracheids instead of xylem vessels, but kept overall transport rates the same, which of the statements would be true? Select all that apply. a. Stems of angiosperms would be much larger in cross section. b. Water transport in angiosperms would require more of the cross-sectional area of the stem. c. Angiosperms would become more resistant to freeze-thaw cavitation. d. Angiosperms could grow stems with the same overall mechanical and transport properties just as quickly. ANSWER: b, c 57. Which statements are likely to be true of the earliest angiosperms? Select all that apply. a. The presence of vessels allowed them to form a tall, emergent layer in tropical forests. b. The evolutionary innovation of flowers allowed them to be much more diverse than other seed plants radiating at this time. c. Animal pollination made possible by flowers allowed them to survive at low density in tropical forest understories. d. They evolved in shady and wet conditions, growing toward the side to exploit light gaps in the canopy overhead. ANSWER: c, d 58. Which of the characteristics do monocots demonstrate? Select all that apply. a. vascular bundles organized in a ring b. a vascular cambium that produces secondary xylem c. new roots that consistently form at nodes along the stems d. nodes giving rise to single leaves ANSWER: c, d 59. Palm trees and cycads have converged on a similar growth form, yet they actually grow and develop quite differently. What are some of these similarities and differences? Select all that apply. a. Cycad stems are composed mostly of a mantle of roots. b. Palms lack a vascular cambium, but cycads do not. c. Cycads have sheathing leaf bases that protect their apical buds, whereas palm apical meristems are protected by bud scales. d. Neither produces vegetative branches as part of normal growth. ANSWER: b, d 60. Which of the choices are characteristics of annual plants? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 31 a. They persist through a portion of the year only as seeds. b. They are herbaceous plants that complete their life cycle in less than a year. c. They can persist through unfavorable conditions as seeds. d. They are rarely animal pollinated. e. They can regrow each year from underground stems. f. They are unique to angiosperms and the majority are eudicots. ANSWER: a, b, c, f 61. Nikolai Vavilov recognized that the places where the wild relatives of crop plants are most diverse are also likely to be: Select all that apply. a. the best places to grow domesticated crops. b. the places where the crops were first domesticated. c. a source of cultivars that will produce higher yields than current commercial cultivars. d. a source of genetic variation that may be useful for breeding programs. ANSWER: b, d 62. The conservation of genetic diversity in crops and their wild relatives is important because: Select all that apply. a. modern crop cultivars have lost the ability to reproduce themselves due to artificial selection. b. genetically uniform agricultural populations may be highly susceptible to diseases and pests. c. genetically uniform agricultural populations act as strong selecting agents for the evolution of novel diseases and pests. d. domestication and selection for yield tends to reduce the genetic variation within crops over time. ANSWER: b, c, d

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Chapter 32 Multiple Choice 1. Which plant-fungal symbionts exchange nutrients with their host plant most efficiently? a. ectomycorrhizal symbionts b. endomycorrhizal symbionts c. endophytic symbionts ANSWER: b 2. Imagine that you are on a nature walk. Your guide points out a lichen growing on a tree trunk and comments that lichens are actually a type of fungi. You know this description is not correct. Why? a. Lichens are actually vascular plants that produce hyphae instead of roots. b. Lichens are actually composed of both fungi and bryophytes. c. Lichens are actually composed of both fungi and green algae (or cyanobacteria). d. Lichens are actually bryophytes that produce hyphae instead of roots. ANSWER: c 3. If all of the fungi on Earth ceased to exist, what would happen to the local carbon cycle? a. Because fungi are photosynthetic, less carbon dioxide would be removed from the atmosphere. b. Because fungi are decomposers, less carbon dioxide would be released back into the atmosphere. c. Because fungi are photosynthetic, more carbon dioxide would be removed from the atmosphere. d. Because fungi are decomposers, more carbon dioxide would be released back into the atmosphere. ANSWER: b 4. A researcher has been studying lichens in his hometown. Since he began his study, he has noticed several lichens dying in the vicinity of a new factory. What is most likely causing the decrease in lichen populations? a. the diversion of water sources to the factory causing the lichens to dry out b. an increase in air temperature, which is the result of carbon dioxide emissions from the factory c. the blocking of sun by the factory preventing the lichens from getting enough sunlight d. an increase in industrial pollutants in the area ANSWER: d 5. The majority of an individual fungus lives "in" its food source. What characteristics of the fungal growth form make this a necessity? a. Fungi have a high ratio of surface area to volume and are highly susceptible to desiccation. b. Fungi have a low ratio of surface area to volume and are highly susceptible to desiccation. c. Fungi require the stimulus of "pushing" through the substrate to stimulate growth. d. Fungi require more nutrients than other organisms, and large amounts of nutrients can only be acquired by living in a host. ANSWER: a 6. Saprophytic fungi, which live on dead and decaying material, are essential to life on Earth because they: a. grow in mycorrhizal association with the roots of most plants, providing minerals to the plants. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 32 b. speed the recycling of nutrients. c. help to digest sap in phloem of dead and dying trees. d. recycle inorganic compounds into organic compounds. e. respire like plants, taking in carbon dioxide and decreasing greenhouse gases. ANSWER: b 7. Consider the figure shown. The bars show changes in primary productivity with and without arbuscular-like mycorrhizal associations for a group of plants that evolved early in the history of terrestrial (land) plants.

Which of the statements supports the data shown in the figure? a. Endomycorrhizal associations do not increase primary productivity in these plants. b. Endomycorrhizal associations increase primary productivity in these plants. c. Endomycorrhizal associations only help these plants at high CO2 concentrations. d. Endomycorrhizal associations only help these plants at low CO2 concentrations. ANSWER: b 8. People can get an infection called ringworm, which is actually a fungus, not a worm. The characteristic growth pattern is a circle on the infected portion of the skin. The edges of the circle tend to appear redder, or more irritated, than the center. Where did the infection begin? a. It began at the margins of the circle where it is most irritated, and the fungus grew around the margin of the circle. b. It began at the center of the circle, and the fungus grew out from the initial point of infection. c. Multiple fungal spores landed in the shape of the circle and led to infection. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 32 ANSWER: b 9. In some of the earliest plant fossils, there is evidence of mycorrhizal associations. Fungi may have been instrumental in plant colonization of land by: a. colonizing areas as lichens and facilitating dispersal by terrestrial plants. b. increasing nutrient uptake by early plant species. c. increasing levels of organic matter in the soil. d. releasing antibiotics to kill bacteria in the soil where plants would colonize. ANSWER: b 10. Many antibiotics work by inhibiting the synthesis of peptidoglycan, a component of the bacterial cell wall. What is a possible mechanism of action for antifungal drugs? a. inhibiting the synthesis of peptidoglycan in fungi b. inhibiting the synthesis of cellulose in fungi c. inhibiting the synthesis of chitin in fungi d. inhibiting the synthesis of lignin in fungi ANSWER: c 11. How do fungi in the soil of rainforests encourage plant diversity? a. Fungi serve as pathogens for particular plant species, favoring the establishment of less common plant species. b. Fungi serve as pollinators, introducing plant diversity into new regions. c. Fungi form mycorrhizae with plants, introducing plant diversity into new regions. ANSWER: a 12. Recall that the phloem sap of vascular plants moves through the phloem as a result of turgor pressure. How are nutrients transported through fungi? a. through sieve tubes, as a result of capillary action b. through sieve tubes, as a result of turgor pressure c. through vessel elements, as a result of turgor pressure d. through hyphae, as a result of capillary action e. through hyphae, as a result of turgor pressure ANSWER: e 13. A researcher is looking at different petri dishes that contain the yeast C. albicans. In the first dish she evaluates, she notices that the yeast do not form hyphae. Is this a mutant strain of yeast? a. Yes, because the lack of hyphae denotes pathogenic yeast. b. Yes, because yeast typically form hyphae. c. No, because yeast do not normally form hyphae. d. No, because yeast always form rhizoids and not hyphae. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 32 14. Imagine that fungal hyphae were as thick as a human hair. How might this increased thickness affect the ecological role of fungi? a. Decomposition rates of dead organisms would decrease because the fungi's ratio of surface area to volume would decrease. b. Decomposition rates of dead organisms would increase because the fungi's ratio of surface area to volume would increase. c. There would be no effect on the rates of decomposition of dead organisms. ANSWER: a 15. Fungi feed through extracellular digestion and then absorption. How would the rates of decomposition of dead organisms be altered if fungicide, a compound that kills fungi, were sprayed on the forest floor? a. Decomposition rates would lower because fungi would be destroyed. b. Decomposition rates would increase because bacteria would take over the role of decomposition, and they reproduce at faster rates than fungi. c. Decomposition rates would be the same because fungi's hyphae is below ground and therefore would not be destroyed. d. Decomposition rates would lower because fungi would move to grow in an area without fungicide. ANSWER: a 16. The filaments that most fungi produce to absorb nutrients are called: a. root hairs. b. ectomycorrhizae. c. hyphae. d. chitins. e. fruiting bodies. ANSWER: c 17. Fungi include: a. yeasts. b. decomposers. c. pathogens. d. smuts. e. All of these choices are correct. ANSWER: e 18. Why was the development of septa in the hyphae of fungi an important evolutionary step? a. If a hypha is damaged, fungi can seal septal pores to prevent the cytoplasm from leaking out. b. If a hypha is damaged, septa can secrete chemicals to attract symbiotic cyanobacteria. c. Septa can produce fungal spores during both sexual and asexual reproduction. d. Septa can produce different fungal mating types that aid in sexual reproduction. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 32 19. Which of the fungal structures aid in spore dispersal? a. fruiting bodies b. septa c. mycelia d. hyphae ANSWER: a 20. In a parasexual fungal species, genetic diversity is the result of: a. meiotic recombination. b. mitotic recombination. c. recombination between different mating-type alleles. d. plasmogamy recombination. e. hyphae recombination. ANSWER: b 21. In the sexual life cycle of fungi, genetic variation is generated: a. during karyogamy, when nuclei become diploid. b. during plasmogamy, when hyphae become heterokaryotic. c. during germination, when spores begin to grow. d. during meiosis, when spores are formed. ANSWER: d 22. _____ is the process by which the nuclei within a heterokaryotic fungal cell fuse. a. Karyogamy b. Plasmogamy c. Mitosis d. Meiosis ANSWER: a 23. When plasmogamy occurs between two fungi of compatible mating types, a dikaryotic cell is formed in the hyphae. For some groups of fungi, only one or a few cells are dikaryotic. In other groups of fungi, a majority of the cells that make up the fungal hyphae are dikaryotic. What might be an advantage of dikaryotic hyphae? a. Dikaryotic hyphae have more nuclei and thus can achieve higher rates of transcription and translation, as well as make more digestive enzymes per cell. b. Dikaryotic hyphae are able to divide more rapidly than hyphae with only one nucleus. c. Karyogamy can occur in many cells, allowing more fruiting bodies to be produced and thereby increasing the number of spores produced by meiosis. d. Karyogamy can occur more rapidly when a dikaryotic hypha encounters a compatible mating type. ANSWER: c 24. Imagine that sexual reproduction could occur between hyphae that were the same mating type. How might this affect rates of evolution in fungi? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 32 a. Rates of evolution would decrease, because genetic variation would reduce. b. Rates of evolution would increase, because more novel genetic combinations are generated through sexual reproduction. c. Rates of evolution would stay the same, because as long as sexual reproduction is occurring, it does not matter whether it is between individuals of the same or different mating type. ANSWER: a 25. Spore production in fungi is an important part of reproduction. What is the adaptive significance of the protective coat found around fungal spores? a. The protective coat inhibits spore growth until other fungi in the area die. b. The protective coat keeps the spore viable until environmental conditions allow fungal growth. c. The protective coat increases the distance of dispersal from the parental fungus. d. The protective coat inhibits fungal growth wherever the spore lands. ANSWER: b 26. Which type of reproduction is typical in many molds and yeasts? a. sexual reproduction through plasmogamy of diploid parental mycelia, and then production of spores b. sexual reproduction through plasmogamy of haploid parental mycelia, and then production of spores c. asexual reproduction of haploid spores from a diploid parental organism d. asexual reproduction of haploid spores from a haploid parental organism ANSWER: d 27. During which of the processes does the cytoplasm of two hyphal cells fuse to form a dikaryotic (n + n) cell? a. karyogamy b. heterokaryogamy c. plasmodesmatogamy d. plasmogamy ANSWER: d 28. A researcher isolates a species of fungus that only reproduces sexually. He introduces individual hyphae, all of the same mating type (i.e. with the same mating-type alleles), into a petri dish. What do you expect will happen with the fungi in this dish? a. Fungi in the dish will mate, forming spores by means of sexual reproduction. b. Fungi in the dish will not mate, but will form spores by means of asexual reproduction. c. No spores will be formed by the fungi in the petri dish. ANSWER: c 29. Close observation of the fruiting bodies of cup fungi (ascomycetes) shows that when asci of cup fungi forcibly eject their spores into the air, they tend do so in groups (many asci at once) instead of one at a time. Furthermore, spores released during one of these group events tend to go higher into the air than do spores released from a single ascus. This observation suggests that the simultaneous spore release by multiple asci is likely an adaptation that: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 32 a. enhances the ability of spores to penetrate the layer of stagnant or unmoving air above the fruiting body. b. satiates spore predators, increasing the likelihood that some spores will not be eaten. c. increases the likelihood that multiple spores will arrive at the same food resource, thus enhancing the likelihood of successful colonization. d. allows for the formation of new asci by allowing new regions of the fruiting body to undergo karyogamy. ANSWER: a 30. A researcher discovers a new species of fungus that forms hyphae, but not septa. This fungus also lacks multicellular fruiting bodies and does not appear to form either endomycorrhizae or ectomycorrhizae with plants. How should this fungus be classified? a. as a zygomycete b. as a chytrid c. as an ascomycete d. as a glomeromycete e. as a basidiomycete ANSWER: a 31. Ug99 is a type of _____ and is a fungus (rust) that has decimated wheat crops across the globe. a. ascomycete b. basidiomycete c. chytrid d. glomeromycete e. zygomycete ANSWER: b 32. Which of the types of fungi are typically unicellular, lack heterokaryotic cells, and occasionally produce rhizoids? a. ascomycetes b. basidiomycetes c. chytrids d. glomeromycetes e. zygomycetes ANSWER: c 33. Consider the phylogeny shown. Based on the information provided in the chapter, which type of data helped the most in constructing this tree?

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Chapter 32

a. fossils from the group glomeromycetes b. morphological characters associated with fruiting bodies c. chromosome numbers of the different species d. molecular sequence data e. the presence or absence of a continuous dikaryotic life stage ANSWER: d 34. Which of the traits distinguishes (separates) reproduction in an ascomycete compared to reproduction in a basidiomycete? a. the presence of haploid spores b. the presence of dikaryotic hyphae c. only four spores are produced following meiosis d. spores are produced by meiosis e. haploid spores undergo mitosis ANSWER: e 35. Consider the phylogeny shown. Complex, multicellular fruiting bodies are an example of:

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a. a homology. b. an analogy. c. a shared, ancestral trait. d. a shared, derived trait. e. None of the other answer choices is correct. ANSWER: b 36. Consider the phylogeny shown. Hyphae are present in all of the groups except:

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a. chytrids. b. zygomycetes. c. glomeromycetes. d. basidiomycetes. e. ascomycetes. ANSWER: a 37. Consider the phylogeny shown. The sister group to fungi is:

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a. chytrids. b. zygomycetes. c. glomeromycetes. d. Dikarya. e. animals. ANSWER: e 38. Consider the phylogeny shown. Hyphae evolved in the common ancestor of zygomycetes, glomeromycetes, basidiomycetes, and ascomycetes. Why is it, then, that some of species of fungi in these groups do not have hyphae?

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a. The phylogenetic tree is incorrect. b. Hyphae were lost in the evolution of some species of fungi. c. Hyphae were gained independently in different groups of fungi. d. Hyphae are present in these groups but are difficult to detect. ANSWER: b 39. Why can yeasts only live in nutrient-rich environments, whereas many other fungi can live a wider range of environments? a. Yeasts are ascomycetes. b. Yeasts only reproduce asexually. c. Yeasts only have one mating type. d. Yeasts do not produce hyphae. e. Yeasts are unicellular. ANSWER: d 40. Which of the answer choices characterize the "gills" of common mushrooms (basidiomycetes)? a. regular, precise spacing b. vertical orientation c. a large surface area for spore production d. dispersal by gravity and the wind e. All of the answer choices are correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 32 ANSWER: e 41. Consider the map illustrating the origin and spread of the fungus Ug99.

Where did the fungus originate? a. Europe b. The Middle East c. Central Africa d. Southern Africa e. India ANSWER: c 42. Consider the map illustrating the origin and spread of the fungus Ug99.

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Why is it concerning that Ug99 might spread to South Asia and India? a. It might spread from India to Europe. b. These areas are regions where wheat production is high. c. These areas are regions where wheat production is low. d. It might spread from India to the Americas. e. All of these answer options are correct. ANSWER: b 43. Consider the figure shown of the parasexual cycle in fungi.

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What type of cell is the one at the top of the figure? a. haploid b. diploid c. aneuploid d. dikaryotic e. tetraploid ANSWER: d 44. Consider the figure shown of the parasexual cycle in fungi.

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What process occurs at step 1? a. fertilization b. crossing over c. karyogamy d. chromosome loss e. chromosome gain ANSWER: c 45. Consider the figure shown of the parasexual cycle in fungi.

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What process occurs at step 2? a. fertilization b. crossing over c. karyogamy d. chromosome loss e. chromosome gain ANSWER: b 46. Consider the figure shown of the parasexual cycle in fungi.

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What process occurs at step 3? a. fertilization b. crossing over c. karyogamy d. chromosome loss e. chromosome gain ANSWER: d Multiple Response 47. How do the two species that make up a lichen benefit from their symbiotic association? Select all that apply. a. The fungi receive carbohydrates from the algae, while the algae receive nitrogen from the fungi. b. Both partners are able to grow on substrates on which each would be unable to grow on its own. c. The algae receive carbon as a result of the fungi's ability to degrade lignin and cellulose, while the fungi receive nitrogen from the algae. d. The fungi receive carbohydrates and, in some cases, nitrogen from the algae, while the algae gain a "home" that anchors them on the substrate. ANSWER: b, d 48. How do fungi digest their food sources? Select all that apply. a. by engulfing food particles and digesting them in vacuoles b. by extending hyphae into a food source (e.g. an animal carcass) c. by secreting enzymes externally, directly into their surrounding environment Copyright Macmillan Learning. Powered by Cognero.

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Chapter 32 d. by funneling food particles through septa, which break down larger molecules ANSWER: b, c 49. How does the size of fungal hyphae contribute to their ability to decompose dead and decaying organisms? Select all that apply. a. Fungal hyphae have a low ratio of surface area to volume, so they are able to absorb more nutrients. b. Fungal hyphae have a high ratio of surface area to volume, so they are able to absorb more nutrients. c. Fungal hyphae have a low ratio of surface area to volume, so they are able to secrete more digestive enzymes into the surrounding environment to break down dead and decaying organisms. d. Fungal hyphae have a high ratio of surface area to volume, so they are able to secrete enzymes into a greater area of soil to break down dead and decaying organisms. ANSWER: b, d 50. Which of the compounds found in wood are difficult to degrade (or break apart)? Select all that apply. a. cellulose b. lignin c. starch d. protein ANSWER: a, b 51. Which of the statements is true regarding fungi as a group? Select all that apply. a. Many fungi demonstrate both asexual and sexual life cycles. b. Fungi are heterotrophs. c. Fungi are autotrophs. d. Fungi only reproduce asexually (for example, by budding). e. Many fungi are decomposers. ANSWER: a, b, e 52. Which of the answer choices is considered a fruiting body? Select all that apply. a. portobello mushrooms in the grocery store b. toadstools in your backyard c. shiitake mushrooms growing on the trunk of a tree d. truffles growing underground e. spores released from a mushroom ANSWER: a, b, c, d 53. Which of the statements is true regarding fungal spores? Select all that apply. a. They can be formed by means of asexual reproduction. b. They can be formed by means of sexual reproduction. c. They are often shaped to resist drag (similar to an airplane wing). d. They are often shaped to promote drag (similar to a parachute). Copyright Macmillan Learning. Powered by Cognero.

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Chapter 32 e. They can be dispersed by fruiting bodies. ANSWER: a, b, c, e 54. Which of the statements is true in parasexual fungi? Select all that apply. a. Crossing over occurs during mitosis. b. Crossing over occurs during meiosis. c. Chromosomes are lost until haploid (1n) cells are achieved. d. Chromosomes are added until haploid (1n) cells are achieved. e. The parasexual fungi produce dikaryotic (n + n) cells. ANSWER: a, c 55. On a hike through the forest, you notice a circle of "mushrooms," each with a stalk, cap, and gills on the underside of the cap. If you dig into the soil below the ring, which of the answer choices might you find? Select all that apply. a. a net of fungal mycelia b. remnants of a decaying tree c. low-nutrient soil d. moist soil ANSWER: a, b, d 56. Which of the statements is true regarding the sexual life cycle of basidiomycetes? Select all that apply. a. Following karyogamy, the diploid nucleus undergoes meiosis and mitosis to form spores. b. Spore precursor cells only undergo meiosis to form spores. c. Spore precursor cells only undergo mitosis to form spores. d. Fruiting bodies are composed of both haploid (1n) and dikaryotic (n + n) cells. e. Fruiting bodies are composed solely of dikaryotic cells (n + n). ANSWER: b, e 57. Using only the information in the phylogeny shown, indicate which of the statements could explain why the group Dikarya has more species than all other fungal taxa on the tree combined. Select all that apply.

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a. The amount of available habitat is greater for members of the Dikarya. b. There has not been enough research done on chytrids, so there may be many more undescribed species. c. The evolution of hyphae allowed for rapid speciation of four of the five major groups of fungi. d. The dikaryotic condition provides more genetic variation and opportunities for evolution of new species. e. Dikaryotic fungi, with greater genetic variation, may have lower rates of extinction than other groups of fungi. ANSWER: d, e 58. Which of the statements is/are true regarding the sexual life cycle of ascomycetes? Select all that apply. a. Diploid nuclei undergo meiosis, followed by mitosis, to form spores. b. Diploid nuclei only undergo meiosis to form spores. c. Diploid nuclei only undergo mitosis to form spores. d. Fruiting bodies are composed of both haploid (1n) and dikaryotic (n + n) cells. e. Fruiting bodies are only composed of dikaryotic cells (n + n). ANSWER: a, d 59. When moving between resource patches in the soil or building fruiting bodies up into the air, why do fungi use bulk flow rather than diffusion to move materials to the growing tip? Select all that apply. a. Diffusion is useful over long distances. b. Diffusion is rapid. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 32 c. Bulk flow is more useful over long distances. d. Bulk flow is rapid. e. Diffusion is energetically expensive. ANSWER: c, d

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Chapter 33 Multiple Choice 1. In male fiddler crabs both claws are the same size when they are juveniles. As the males grow their body parts all grow at the same rate except one of the claws which is ~3 times the size of the other claw. Which line on the graph shown would you expect if you were to compare growth in body size to growth in the larger claw from juvenile to adult?

a. line M b. line H c. line K ANSWER: a 2. A patient in the hospital is told that their pancreas has stopped secreting digestive enzymes into the intestine and is no longer able to digest a large portion of the food they eat. The pancreas is a/an _____________ in the body. a. tissue b. organ Copyright Macmillan Learning. Powered by Cognero.

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Chapter 33 c. organ system ANSWER: b 3. In maintaining homeostasis of body temperature, what role does the hypothalamus play? a. the response b. the sensor c. the effector d. the stimulus e. organ system ANSWER: d 4. Comparative embryology helped clarify the relationships among major groups of animals. Bilaterians, for example, are all: a. diploblastic. b. triploblastic. c. coelomates. d. acoelomates. e. protostomes. ANSWER: b 5. As in many other animals, the endoderm in cnidarians functions in formation of: a. the respiration system. b. prey capture mechanisms. c. the digestive system. d. the locomotory system. ANSWER: c 6. The fossil record details extinct fauna no longer present on Earth. Certain areas of the planet have always had high species diversity, as evidenced from both fossils and catalogs of their current species diversity. Which of the processes likely contributed most to the patterns of animal diversity we see? a. volcanic eruptions b. the appearance of new land masses over time c. changes in atmospheric oxygen d. extinction events (both local and global) ANSWER: d 7. The fossil remains of animals with body plans recognizable as arthropods, echinoderms, mollusks, and other bilaterians first appeared during the _____ Period, around _____ million years ago. a. Ediacaran; 579 b. Cambrian; 541 c. Devonian; 420 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 33 d. Devonian; 360 e. Permian; 100 ANSWER: b 8. Animals colonized land after plants did. The first animals to colonize land were _____, about _____ million years ago. a. mammals; 542 b. arthropods; 420 c. reptiles; 579 d. dinosaurs; 360 ANSWER: b 9. Tetrapods colonized land at about the same time as the early _____ radiations around _____ million years ago. a. insect; 360 b. chordate; 542 c. arthropod; 420 d. plant; 579 e. plant; 100 ANSWER: a 10. Macroscopic fossils of organisms thought to be animals first appeared in rocks deposited around _____ million years ago. a. 575 b. 542 c. 420 d. 360 e. 100 ANSWER: a 11. Isometric growth in animals is seen as they get larger to accommodate changes in surface area to volume ratio that occur with larger size. a. true b. false ANSWER: b 12. High body temperature in mammals reflexively causes a change in blood flow and sweating, which act to reduce the temperature. This is a _____ feedback loop because response to the signal results in a(n) _____ of that signal. a. negative; increase b. negative; decrease c. positive; decrease Copyright Macmillan Learning. Powered by Cognero.

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Chapter 33 d. positive; increase ANSWER: b 13. According to the phylogenetic tree shown, bilaterians (animals with bilateral symmetry) are most closely related to:

a. sponges. b. cnidarians. c. the group containing sponges and cnidarians. d. the group containing choanoflagellates, sponges, and cnidarians. e. None of the answer options is correct. ANSWER: b 14. According to the phylogenetic tree shown, which of the answer choices correctly lists the sequence in which multicellularity, tissues, and bilateral symmetry arose, from the oldest characteristic to the most recent?

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a. multicellularity, tissues, bilateral symmetry b. multicellularity, bilateral symmetry, tissues c. tissues, bilateral symmetry, multicellularity d. tissues, multicellularity, bilateral symmetry e. That information cannot be determined from the phylogeny. ANSWER: a 15. In the figure shown, which line represents isometric growth?

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a. line M b. line H c. line K ANSWER: b 16. Early scientists used morphological characteristics to create a phylogeny of animals, but recently, DNA sequence data have helped scientists revise phylogenies of taxa within the larger groups (that is, within the bilaterians, sponges, cnidarians). Even given the new DNA sequence data, morphological characteristics like "fate of the blastopore" or "body symmetry" remain relatively good at predicting larger-scale branching patterns on phylogenies. Why are morphological characteristics still good predictors of phylogenies? a. The development of these characteristics is controlled genetically, which accounts for similarities in DNA sequence among these taxa. b. Characteristics like those described are necessary for living, therefore all organisms must have those traits. c. All animals are descended from a common ancestor that had the described characteristics. d. These features reflect common adaptations to life on land. ANSWER: a 17. What is one of the primary reasons given to explain why there are very few fossils from the earliest (preEdiacaran) animals? a. The earliest animals were distributed in high-oxygen environments and decomposition occurred too rapidly for fossilization to occur. b. The area of the Earth covered with water was much smaller in the past than at present; therefore, Copyright Macmillan Learning. Powered by Cognero.

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Chapter 33 there were fewer places for fossilization to occur on the ocean floor. c. The area of the Earth covered by land, where early animals lived, was small; therefore, there were fewer opportunities for sedimentary rocks to form fossils. d. The earliest animals were small and soft-bodied and did not have body parts that could be fossilized easily. ANSWER: d 18. Mass extinctions have occurred five times in Earth's history. The Permian and Cretaceous extinctions removed a large percentage of organisms from the planet. How did these extinctions contribute to the patterns of biodiversity we see today? a. Species that remained after the extinction were unable to speciate. Therefore, the number of species on Earth today is lower than the number of species present just before either extinction. b. Species that have gone extinct were able to re-evolve from the ancestors that survived the extinction. c. Species that remained after the extinction were able to radiate, new adaptations arose, and these adaptations produced the diversity seen today. d. Species that remained after the extinction represented all of the lineages that were present before the extinction event. Therefore, extinction did not change the diversity of lineages. ANSWER: c 19. According to the phylogenetic tree shown, the house finch and rabbits are more closely related than are lizards and rabbits.

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Chapter 33 a. true b. false ANSWER: b 20. Homeostatic regulation relies on: a. positive feedback. b. negative feedback. c. both negative and positive feedback. d. the magnitude of the stimulus and response. ANSWER: b 21. All organisms with bilateral symmetry also show segmentation. a. true b. false ANSWER: a 22. Blood consists of red blood cells, white blood cells, and plasma proteins among some other components all in a fluid matrix. Based on this information, blood is classified as a connective tissue. a. true b. false ANSWER: a 23. Which of the characteristics would you expect to find in all members of the bilateria? a. segmentation b. specialized head region c. coelom d. two tissue layers (diploblastic) ANSWER: b 24. What role has mass extinction played in animal evolution? a. Mass extinction can remove ecologically dominant groups, paving the way for new or previously minor groups to diversify. b. Mass extinction selectively removes poorly adapted animals. c. Mass extinction eliminates many marine species, forcing survivors to colonize land to find food. d. All of these choices are correct. ANSWER: a 25. Which of the answer choices is an example of negative feedback altering homeostatic control? a. A person's heart rate remains elevated over the course of a long run. b. A person with a bacterial infection runs a fever and his body temperatures rises and stays elevated for several days. c. The production of oxytocin (a hormone) increases in a pregnant woman's body as oxytocin levels in Copyright Macmillan Learning. Powered by Cognero.

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Chapter 33 her body rise. d. The concentration of salt in a person's urine increases after that person eats a large bag of salty chips. ANSWER: d 26. Which of the answer choices is an example of the effector's role in maintaining homeostasis? a. increased sweating on a hot summer day b. vasodilation on a cold winter day c. increased body temperature during a workout d. decrease in body temperature on a cold day ANSWER: a 27. Some researchers have suggested that obesity is due to a change in "set point" in the brain that is related to the number of calories a person needs. People whose set point has increased eat more than they need and gain weight. Permanently raising the set point would involve a permanent change in which of the answer choices? a. the effector b. the stimulus c. the sensor d. the response ANSWER: c 28. While spending a day at the beach, you find an organism that looks like a typical cnidarian. In the lab you are able to watch embryonic development of this organism and you observe that distinct muscle tissue develops. Would you still classify this organism as a cnidarian? a. No, because it has muscle tissue. b. Yes, because it was found at the ocean. c. Yes, because it looks like a typical cnidarian. d. No, because it has radial symmetry. ANSWER: a 29. Skeletal muscle tissue appears striated under a microscope because of the arrangement of actin and myosin proteins in the filaments. If there was a mutation in the gene that produced actin such that it could not bind myosin, which of the answer choices would you expect to observe? a. The muscle's ability to contract does not change. b. Muscle shortening does not occur. c. Muscle shortening occurs, but contractions are weak. ANSWER: b 30. Individuals with type 2 diabetes produce insulin, but target cells do not respond to insulin by taking in sugar. In these individuals, blood glucose levels remain elevated. Which component of the feedback system is broken in these individuals? a. the sensor b. the effector Copyright Macmillan Learning. Powered by Cognero.

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Chapter 33 c. the stimulus d. the response ANSWER: b 31. The Cambrian explosion: a. refers to a burst of adaptive radiation that gave rise to many new animal phyla. b. refers to the rapid diversification of life following the extinction of the dinosaurs. c. produced large earthquakes that caused major species loss. d. resulted from an asteroid impact that caused the dinosaurs to die off. ANSWER: a 32. Cephalization is generally associated with all of the following except: a. a sedentary existence. b. a concentration of sensory structures at the anterior end. c. a brain. d. bilateral symmetry. ANSWER: a 33. Which of the characteristics is common to both arthropods and annelids and suggests that the two groups shared common ancestry? a. a segmented exoskeleton b. jointed appendages c. a segmented body plan d. two body layers in the embryo ANSWER: c 34. The innovation that allowed animals to break their last tie to the aquatic environment was: a. legs. b. the amniotic egg. c. the bony endoskeleton. d. the hard exoskeleton. e. impermeable skin. ANSWER: b 35. In the animal kingdom, the evolution of the coelom is important because it: a. allowed organisms to evolve muscles and a circulatory system. b. was required for the evolution of appendages adapted for locomotion. c. allowed animals to move onto land with an internal supply of extra body fluid. d. cushions the body against hard blows to the body and enables the body to turn without twisting these organs. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 33 36. According to the phylogenetic tree shown, bass and shark are not as closely related to each other as are bass and the house finch.

a. true b. false ANSWER: a 37. According to the phylogenetic tree shown, bass and rabbits are not as closely related as bass and the house finch, because the latter pair has more branching events between them.

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a. true b. false ANSWER: b Multiple Response 38. Compared with sponges, cnidarians have a wider array of cell types, permitting more sophisticated tissue functions. These functions include: Select all that apply. a. the ability to communicate within the organism using hormones. b. muscle contraction/movement. c. the ability to sense the environment via a network of nerve cells. d. secretion of digestive enzymes. ANSWER: b, c, d 39. What are the advantages of having a true coelom? Select all that apply. a. A true coelom cushions internal organs from hard blows. b. A true coelom allows the body to move independently of the movement of internal organs. c. A true coelom allowed for evolution of cephalization. d. None of the answer options is correct. ANSWER: a, b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 33 40. Phylogenetic trees help us understand: Select all that apply. a. the genetic relationships among a species. b. the evolutionary relationships among a species. c. the relationships of symmetry among related species. d. the level of complexity among related species. ANSWER: b, d 41. An environmental signal causes one neuron to send a signal to a second neuron, which then causes a muscle to contract. Which of the answer choices would you predict could be the problem if the muscle no longer contracted in response to the environmental signal. Select all that apply. a. Neuron one could be malfunctioning. b. Chemical release occurred at the synapse. c. Neuron two could be malfunctioning. d. The muscle could be damaged. ANSWER: a, c, d 42. Which of the physiological parameters in the body are maintained homeostatically by negative feedback? Select all that apply. a. body temperature b. blood sugar level c. protein levels in the blood d. blood pressure e. heart rate ANSWER: a, b, d, e 43. Which of the physiological parameters in the body are most likely to vary from normal homeostatic levels during heavy exercise? Select all that apply. a. body temperature b. blood sugar level c. cellular pH d. blood pressure e. heart rate ANSWER: a, d, e 44. Animals are distinguished from other groups of eukaryotic organisms in that they: Select all that apply. a. are multicellular heterotrophs lacking cell walls. b. have embryos that include a gastrula stage. c. produce collagen. d. are mobile. ANSWER: a, b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 33 45. Which of the characteristics can be found in organisms other than animals? Select all that apply. a. multicellularity b. chitin c. collagen d. gastrula ANSWER: a, b, c 46. Which of the statements regarding the history of life on Earth is/are correct? Select all that apply. a. The history of planet Earth includes alternations between colder periods ("icehouses") and warmer periods ("greenhouses") on the surface of the planet. b. The rise of more complex organisms such as eukaryotes has resulted in a reduction in diversity of more "primitive" organisms such as bacteria. c. There have been repeated incidents of mass extinction in the history of the planet in which species diversity has been reduced by 50% or more. d. Increases in atmospheric oxygen during the Edicarian Period helped support the evolution of larger more active animal species. ANSWER: a, c, d 47. Traits used to group animals in a simple phylogenetic tree include: Select all that apply. a. the presence of collagen. b. body symmetry. c. photosynthetic capabilities. d. complex organs. ANSWER: b, d 48. You are examining a group of cells that are growing in culture under a microscope in the lab. In order for you to classify these cells as part of the same tissue, what would you expect to observe? Select all that apply. a. The cells look similar in shape and size. b. The cells operate independently of one another. c. The cells perform similar functions. d. Individual cells within the group respond to different signals than others in the group. ANSWER: a, c 49. According to the phylogenetic tree shown, bilateral symmetry: Select all that apply.

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a. evolved before radial symmetry in the animal phyla. b. is a characteristic found in all animal phyla. c. is associated with cephalization. d. is found in animals with and without a coelom. ANSWER: c, d 50. A very general trend in animal evolution has been toward increasing structural complexity. What is the reason behind this? Select all that apply. a. Structurally complex organisms are necessarily derived from structurally simpler ancestors. b. There is an evolutionary 'drive' toward greater structural complexity. c. Natural selection tends to favor more structurally complex organisms over less structurally complex ones. d. Evolution can only modify or build on what already exists. ANSWER: a, d

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Chapter 34 Multiple Choice 1. When a gazelle senses a predator, which action is under voluntary control? a. running b. faster heartbeat c. dilation of blood vessels supplying the muscles ANSWER: a 2. Squid are predatory mollusks: they need to be able to detect both sedentary and highly mobile prey. Which feature would you expect is not a part of a squid's nervous system?

a. cephalization b. nerve net c. sense organs d. ganglia ANSWER: b 3. Which statement is true? a. All animals have a nervous system. b. All animals sense and respond to the environment. c. It is necessary to have a nervous system to sense and respond to the environment. d. All animals have a nervous system, which is necessary to sense and respond to the environment, but not all animals sense and respond to the environment. e. All animals have a nervous system and sense and respond to the environment, and a nervous system is necessary to sense and respond to the environment. ANSWER: b 4. Echinoderms (for instance, starfish or sea urchins) are descended from bilaterians, but many members of the group lack a brain, have no specialized sense organs, and have radial nerves that extend down each "arm" from a central nerve ring that surrounds the gut. What was a likely selective pressure in the evolution of this type of nervous system? a. ability to move forward b. role as a predator in the ecosystem Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 c. multicellularity d. possession of an endoskeleton of calcium carbonate e. None of the other answer options is correct. ANSWER: e 5. When an environmental stimulus is received, the signal is usually transmitted through three types of nerve cells. In which order is the signal transmitted through these cells? a. sensory neurons, motor neurons, interneurons b. motor neurons, sensory neurons, interneurons c. sensory neurons, interneurons, motor neurons d. interneurons, sensory neurons, motor neurons e. motor neurons, interneurons, sensory neurons ANSWER: c 6. All multicellular organisms have a nervous system. a. true b. false ANSWER: b 7. Which group contains animals with a nerve net? a. sponges b. cnidarians c. bilaterians ANSWER: b 8. Voltage-gated ion channels underlie the function of electrically excitable cells, such as nerve and muscle cells. Which statement is true about voltage-gated ion channels? a. Voltage-gated ion channels open and close in response to changes in membrane potential. b. Voltage-gated ion channels vary in terms of how rapidly they respond to changes in membrane potential. c. Voltage-gated ion channels involve a conformational change of the transmembrane protein, which occurs in response to membrane voltage that changes the channel's permeability to ion flow through the channel. d. All of these choices are correct. ANSWER: d 9. Consider the figure of a motor neuron shown in the image. Which areas of the motor neuron are myelinated?

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a. the dendrites b. the axon terminals c. the axon before (to the left of) the cell body d. the axon after (to the right of) the cell body e. the entire axon ANSWER: e 10. Consider the image.

Where on the neuron are nerve impulses summed? a. at letter A b. at letter B c. at letter C d. at letter D e. None of the other answer options is correct. ANSWER: e 11. An important function of myelin is to: a. increase the size of the synaptic cleft. b. decrease the size of the synaptic cleft. c. increase the speed of nerve signal transmission along the axon. d. decrease the speed of nerve signal transmission along the axon. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 e. None of the answer options is correct. ANSWER: c 12. Consider the diagram.

What aspect(s) of a neuron is/are consistent in the diagram? a. the shape of the cell body b. the branching patterns of dendrites and axon terminals c. the presence of axons and dendrites d. the length of the axon e. the number of potential connections between a neuron and its neighbor neurons ANSWER: c 13. A neuron that responds to the environment is a(n): a. interneuron. b. motor neuron. c. homeostatic neuron. d. sensory neuron. ANSWER: d 14. Neuronal stimuli are received by: a. dendrites. b. axons. c. neurotransmitters. d. cell nuclei. ANSWER: a 15. Consider the image.

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The neuron labeled number _____ in the figure is a motor neuron; its function is a. 1; connecting different types of neurons b. 2; receiving incoming information or "sensing" c. 2; stimulating muscles or glands d. 3; stimulating muscles or glands e. 3; connecting different types of neurons ANSWER: d 16. Consider the image.

In the figure, arrow B is pointing at the _____ in the _____. a. nucleus; glial cell b. nucleus; cell body c. vesicle; dendrite d. neurotransmitter; synaptic cleft e. synaptic cleft; neuron ANSWER: b 17. Consider the image.

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In the figure, which arrow points to the axon hillock? a. arrow A b. arrow C c. arrow E d. arrow F e. None of the answer options is correct. ANSWER: e 18. Ganglia were the evolutionary precursor to the centralized concentration of neurons that we now call a brain. a. true b. false ANSWER: a 19. The area within the neuron where stimuli are summed is called the: a. dendrites. b. cell body. c. axon terminal. d. axon hillock. e. synapse. ANSWER: d 20. What type of cells produce myelin for sensory neurons? a. melanocytes b. astrocytes c. glial cells d. pyramidal neurons ANSWER: c 21. The resting membrane potential of a neuron is determined by: a. movement of potassium ions relative to other ions. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 b. negatively charged proteins on the outside of the cell. c. activation of voltage-gated sodium channels. d. diffusion of potassium ions through sodium ion channels. ANSWER: a 22. Neurons can synapse with many different types of cells including muscle cells, secretory cells, and other neurons. a. true b. false ANSWER: a 23. During synaptic transmission: a. voltage-gated calcium channels open in the postsynaptic membrane. b. neurotransmitters are absorbed by the postsynaptic membrane. c. neurotransmitter-filled vesicles fuse with the postsynaptic membrane. d. receptors that bind neurotransmitters trigger a change in postsynaptic membrane potential. ANSWER: d 24. Beginning at the synapse of a neuron, place the events in neuronal signaling in the correct sequence. 1. Ion channels bind the ligand and open. 2. Na+ is pumped out of the cell, and the membrane potential is restored. 3. Acetylcholinesterase breaks down acetylcholine. 4. Na+ enters the postsynaptic cell, and the membrane potential changes. 5. Acetylcholine is released into the synapse. 6. Na+ ion channels close. a. 1, 2, 3, 4, 5, 6 b. 3, 4, 6, 1, 5, 2 c. 6, 4, 3, 1, 2, 5 d. 5, 1, 4, 3, 6, 2 e. 4, 5, 1, 2, 6, 3 ANSWER: d 25. Consider the figure.

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In the accompanying figure, which letter indicates when voltage-gated Na+ channels close? a. A b. B c. C d. D e. E ANSWER: c 26. Which channel is responsible for helping to maintain the resting membrane potential? a. ligand-gated sodium channels b. voltage-gated sodium channels c. voltage-gated potassium channels d. potassium leak channels ANSWER: d 27. What types of synaptic inputs can postsynaptic nerve cells receive? a. excitatory b. inhibitory c. both excitatory and inhibitory d. neither excitatory nor inhibitory ANSWER: c 28. What event immediately precedes the nerve firing an action potential? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 a. depolarization is initiated b. the threshold potential is reached c. voltage-gated Na+ channels close d. voltage-gated K+ channels open e. the membrane becomes hyperpolarized ANSWER: b 29. In neurons, Ca2+ ion channels are found: a. only at the axon terminal membrane. b. only along the terminal ends of the dendrites. c. throughout the length of the axon. d. throughout the entire neuron. e. only at the axon hillock. ANSWER: a 30. Inhibitory postsynaptic potentials (IPSPs) are associated with membrane depolarization, whereas excitatory postsynaptic potentials (EPSPs) are associated with hyperpolarization. a. true b. false ANSWER: b 31. Postsynaptic neurons have many connections with multiple other presynaptic neurons through their dendrites. Some of these connections are inhibitory, while others are excitatory. Inhibitory neurotransmitters have their effect by opening ligand-gated _____ channels on the dendrites they interact with. a. Na+ b. Ca+ c. ClANSWER: c 32. Some people have low levels of calcium circulating in the blood, a condition known as hypocalcemia. While for many this disorder has little to no effect, for some it can be life-threatening. How could low levels of calcium harm an individual? a. Low levels of calcium would cause sustained depolarization of the presynaptic cell. b. Low levels of calcium would result in fewer signals sent between the presynaptic and postsynaptic cell. c. Low levels of calcium would not have any effect on synaptic transmission. ANSWER: b 33. Which factor is not associated with saltatory propagation of action potentials? a. myelination Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 b. nodes of Ranvier c. variable concentration of voltage-gated Na+ and K+ channels along the axon d. faster signal transmission e. extremely low threshold potential ANSWER: e 34. Which structure is found in the synapse? a. nucleus b. neurotransmitter c. cell body d. myelin sheath ANSWER: b 35. Consider the image.

Where in the plasma membrane of these cells would you find voltage-gated Ca2+ channels? a. at letter A b. at letter B c. at letter C d. at letter D e. at letter E ANSWER: d 36. Consider the image.

Atropine is a poison that blocks nerve action by binding to acetylcholine (Ach) receptors. Where would you expect to find atropine bound on the figure? a. at letter C Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 b. at letter D c. at letter E d. at letter F e. at letter C, letter D, letter E, and letter F ANSWER: c 37. Imagine you genetically engineered a neuron to produce voltage-gated Na+ and K+ channels that opened at the same time in response to a change in voltage. How would that change the recording shown in the figure?

a. The peak voltage would be higher. b. The peak would occur over a longer period of time. c. The period of hyperpolarization would be longer. d. No action potential would be generated. e. Threshold values would increase. ANSWER: d 38. Which excitatory neurotransmitter is directly responsible for muscle contraction in vertebrates? a. dopamine b. norepinephrine Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 c. nitrous oxide d. glutamate e. acetylcholine ANSWER: e 39. The magnitude of the action potential is correlated with the strength of the stimulating input. a. true b. false ANSWER: b 40. The nervous systems of all animals are organized into peripheral and central components. a. true b. false ANSWER: b 41. All neurons have either a sensory or a motor function. a. true b. false ANSWER: b 42. All nervous system responses involve input from the brain. a. true b. false ANSWER: b 43. Reciprocal inhibition is responsible for: a. the knee-extension reflex. b. the coordinated movement of arms and legs while walking. c. voluntary motor behavior controlled by the brain. ANSWER: a 44. Consider the diagrams.

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Chapter 34

Which diagram best represents the anatomical system responsible for a spinal reflex circuit? a. Diagram A b. Diagram B c. Diagram C ANSWER: c 45. Homeostatic regulation relies on: a. negative feedback. b. positive feedback. c. both negative and positive feedback. d. the magnitude of the stimulus and response. ANSWER: a 46. When the dentist gives you a shot of anesthetic before drilling into a tooth, she is numbing a nerve in your mouth, which is a: a. cranial nerve, part of the peripheral nervous system. b. cranial nerve, part of the central nervous system. c. spinal nerve, part of the peripheral nervous system. d. spinal nerve, part of the central nervous system. e. None of the answer options is correct. ANSWER: a 47. The most commonly used dental anesthetic is lidocaine. It blocks voltage-gated Na+ channels. Where along a facial nerve would you expect lidocaine to be active? a. along the cell body b. at the dendrite c. along the axon d. at the axon terminals e. at the synaptic cleft ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 48. The dental anesthetic lidocaine blocks voltage-gated Na+ channels. How would exposure to lidocaine affect nerve function? a. Resting membrane potential values would increase. b. Resting membrane potential values would decrease. c. It would take longer to reach peak depolarization. d. Nerve depolarization would last for a longer period of time. e. None of the answer options is correct. ANSWER: e 49. A cranial nerve called the facial nerve controls facial expression muscles and conveys taste from the tongue, which means the facial nerve is part of both the _____ and _____ components of the nervous system. a. somatic; voluntary b. somatic; involuntary c. autonomic; voluntary d. autonomic; involuntary ANSWER: b 50. What is the adaptive benefit of some simple reflex circuits bypassing the brain? a. The response is slower. b. The response is faster. c. The response is more intense. d. The response is less intense. e. None of the answer options is correct. ANSWER: b 51. Which receptor carries out sensory transduction? a. a CO2 chemoreceptor in a mosquito b. a pheromone receptor in a moth's antennae c. hair cell in the ear of a gazelle d. thermoreceptors in the skin of vertebrates e. All of these choices are correct. ANSWER: e 52. The charge difference between the inside of the cell membrane and the outside of the cell membrane is known as: a. an action potential. b. the membrane potential. c. depolarization. d. repolarization. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 53. Which term is the chemical substance that allows the transmission of an impulse from a sensory neuron across a synapse to another neuron? a. receptor b. hormone c. sodium d. neurotransmitter ANSWER: d 54. What is the firing rate of a neuron? a. It is the magnitude of a depolarization. b. It is the magnitude of an action potential. c. It is the number of action potentials generated over a given period of time. d. It is related to the resting membrane potential. ANSWER: c 55. How is a nerve impulse generated when most sensory receptors are activated? a. The action potential is converted to a binary code. b. The membrane becomes hyperpolarized. c. The membrane becomes repolarized. d. The membrane becomes depolarized. ANSWER: d 56. Sensory receptors convert _____ into _____ by signal transduction. a. neurotransmitters; synapses b. physical or chemical stimuli; nerve impulses c. physical stimuli; sound d. neurons; sensory organs ANSWER: b 57. Chemoreceptor is to taste as mechanoreceptor is to: a. smell. b. pain. c. temperature. d. sound. ANSWER: d 58. Multiple sensory receptors can synapse on a single neuron. The receiving neuron has a firing rate proportional to the number of signals received. Of which process is this an example? a. temporal summation b. spatial summation Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 c. action potential d. hyperpolarization ANSWER: b 59. Multiple excitatory postsynaptic potentials (EPSPs) from temporal summation and spatial summation can result in: a. an action potential. b. filtering out of unimportant background signals. c. adaptation. d. hyperpolarization. ANSWER: a 60. Under conditions of resting membrane potential, the inside of the cell is _____ charged with respect to the outside, but when an action potential is generated, the inside becomes _____ with respect to the outside of the cell. a. negatively; even more negative b. negatively; positive c. positively; negative d. positively; hyperpolarized ANSWER: b 61. Which receptor is a type of sensory receptor that senses changes in blood pressure? a. chemoreceptor b. mechanoreceptor c. proprioceptor d. nociceptor ANSWER: b 62. Mechanoreceptors present in human skin are involved in tactile sensation. Where are the cell bodies of these mechanoreceptors located? a. in the brain b. in ganglia near the spinal cord c. in the spinal cord d. in the dermal layer of the skin near the sensory endings of the mechanoreceptors ANSWER: b 63. An interneuron may receive multiple stimuli from the same sensory neuron over a very short period of time. The firing rate of the receiving neuron is proportional to the number of signals received from the sensory neuron over time. Of which process is this an example? a. temporal summation b. spatial summation c. action potential Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 d. hyperpolarization ANSWER: a 64. The perceived flavor of food depends on both the sense of taste and the sense of smell. The sensory receptors of olfaction are _____ and those of taste buds are _____. a. chemoreceptors; thermoreceptors b. mechanoreceptors; chemoreceptors c. nociceptors; thermoreceptors d. chemoreceptors; chemoreceptors ANSWER: d 65. Which statement is true of olfaction and gustation? a. Gustation is chemosensory, whereas olfaction is not. b. Humans have more olfactory receptor proteins than gustation receptor proteins. c. Gustation and olfaction signals are processed by the same cranial nerves supplying the brain. d. Gustation is more sensitive to stimuli than olfaction. ANSWER: b 66. When you perceive the "taste" of something you are eating, it means that: a. a single sensory cell has been depolarized and the signal has been transmitted to the area of the brain that perceives the taste. b. a single taste bud has been depolarized and the signal has been transmitted to the area of the brain that perceives the taste. c. multiple sensory cells have been depolarized by a sufficient amount of a food item and their combined excitatory postsynaptic potentials (EPSPs) have been summed to transmit a signal to the brain. d. multiple sensory cells have been depolarized by a variety of food items. ANSWER: c 67. When someone detects a specific odor, a signal is sent to the brain through the olfactory nerve. Which statement describes how the brain is able to perceive the smell? a. The odorant molecule has to bind to more than 50% of olfactory neuron dendrites in order for the smell to be perceived. b. The odorant molecule binds specifically to its receptor to generate an excitatory postsynaptic potential (EPSP) that is transmitted through the olfactory nerve to the brain. c. The odorant molecule binds to specific receptors; summed excitatory postsynaptic potentials (EPSPs) transmit an action potential to general interneurons; and the signal is transmitted to the brain. d. The odorant molecules bind to specific receptors; summed excitatory postsynaptic potentials (EPSPs) transmit an action potential to the odorant-specific interneuron; and the signal is transmitted to the brain. ANSWER: d 68. What type of sensory receptor is present in taste buds? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 a. mechanoreceptors b. nociceptors c. chemoreceptors d. thermoreceptors ANSWER: c 69. How many different types of chemoreceptors can be found in the taste buds? a. hundreds b. thousands c. 40 to 50 d. 5 ANSWER: d 70. Refer to the figure shown.

The transduction of sound waves to changes in membrane potential takes place: a. within the tectorial membrane as it is stimulated by the hair cells. b. when stereocilia bend against the tectorial membrane, causing hair cell depolarization. c. in the basilar membrane as it vibrates at different locations. d. in the oval window, which vibrates at the same frequency as the original sound. e. as the vibrations received by the outer ear cause the eardrum to vibrate. ANSWER: b 71. Which type of sensory receptors function in balance and sensing gravity? a. mechanoreceptors b. photoreceptors Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 c. nociceptors d. chemoreceptors ANSWER: a 72. Which of the receptors are the mechanoreceptors that detect motion and sense gravity? a. tectorial membrane b. stereocilia c. hair cells d. semicircular canals ANSWER: c 73. Which structure is the gravity-sensing organ found in lobsters that provides information about body position and orientation? a. lateral line system b. statocyst c. semicircular canal d. vestibular system ANSWER: b 74. The human vestibular system is made up of: a. the malleus, stapes, and incus. b. two statocyst chambers and three semicircular canals. c. the cochlea. d. the pinna and the tympanic membrane. ANSWER: b 75. The frequency of sound waves reaching the ear determines: a. intensity of sound. b. loudness. c. pitch. d. tone. ANSWER: c 76. The tympanic membrane of various organisms functions to: a. amplify vibrations. b. sense gravity. c. detect body motion. d. convert sound waves to noise. ANSWER: a 77. The tympanic membrane of mammals is also known as the: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 a. oval window. b. cochlea. c. vestibular system. d. eardrum. ANSWER: d 78. In which area of the brain are sound waves processed? a. visual cortex b. auditory cortex c. cerebellum d. somatosensory cortex ANSWER: b 79. The flexibility of the basilar membrane varies along its length. Activation of different regions of the basilar membrane enables perception of sounds of different: a. pitch. b. amplitude. c. intensity. d. loudness. ANSWER: a 80. Hair cells function by: a. firing action potentials directly. b. releasing neurotransmitters. c. stabilizing homeostasis. d. using chemosensory receptors. ANSWER: b 81. What senses do the three semicircular canals in the mammalian inner ear provide? a. a sense of gravity and body orientation b. a sense of sound pitch and amplitude c. a sense of angular motion and balance d. a sense of orientation to the source of sound ANSWER: c 82. Bats are able to hear frequencies of 100,000 cycles per second. The maximum firing rate of action potentials is a few hundred per second. How is this possible? a. Action potentials are not used to transmit auditory information. The output of the sensory neuron to the brain is a graded potential. b. The auditory system is special. Its neurons conduct signals very quickly and have no refractory period. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 c. Sound wave frequency information is coded by differences in the location of vibrations along the basilar membrane, not in action potential frequency. ANSWER: c 83. Fish and sharks sense vibrations in the water that may be caused by prey. The lateral line system of these fish and sharks detects this motion through the action of mechanoreceptors known as: a. hair cells. b. statocysts. c. semicircular canals. d. stereocilia. ANSWER: a 84. The three bones of the mammalian inner ear function: a. in the sense of balance. b. to amplify sound waves that strike the tympanic membrane. c. to detect motion. d. in proprioception. ANSWER: b 85. The steps necessary for hearing include (1) amplification of sound waves, (2) transfer of sound vibrations to fluid pressure waves, and (3) mechanoreception by hair cells of the cochlea. How are the sound vibrations transferred to fluid pressure waves? a. by the bones of the middle ear b. through the oval window c. by the tympanic membrane d. by the action of the vestibular system ANSWER: b 86. In the figure shown, panel a depicts a photoreceptor and its postsynaptic cell in the dark and panel b depicts both cells in the light.

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Chapter 34

What causes the change observed in this figure? a. Hyperpolarization of the rod cell in the dark causes release of a neurotransmitter. b. Hyperpolarization of the rod cell in the light inhibits release of a neurotransmitter. c. Depolarization of the rod cell in the light inhibits release of neurotransmitter. d. This figure is incorrect; dark and light are reversed. ANSWER: b 87. Which term is the photosensitive protein that converts the energy of photons into electrical signals in the receptor cell? a. opsin b. retinal c. chlorophyll d. organ of Corti ANSWER: a 88. Which statement is evidence that all animal vision has evolved from a common origin? a. All animals use the same light-sensitive organs. b. All animals use the photopigment opsin to sense light. c. All animals have eyecups. d. All animals have ommatidia. ANSWER: b 89. Light-sensing organs are thought to have evolved first in: a. cnidarians. b. vertebrates. c. bacteria. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 d. roundworms. ANSWER: a 90. Many insects can sense ultraviolet light. What is the evolutionary advantage of this ability? a. the ability to detect predators and avoid predation b. the ability to locate certain flowers for nectar, the insects' food source c. the ability to avoid other flying insects that compete for food d. the ability to sense balance and position during flight ANSWER: b 91. Cone cells most likely evolved from: a. rhodopsin. b. retinal. c. rod cells. d. ommatidia. ANSWER: c 92. The conformational change of retinal, from the cis to the trans configuration, indirectly: a. depolarizes the photoreceptor by closing Na+ channels. b. depolarizes the photoreceptor by opening Na+ channels. c. hyperpolarizes the photoreceptor by closing Na+ channels. d. repolarizes the photoreceptor by closing Na+ channels. ANSWER: c 93. Why do flatworms move away from light sources? a. to avoid skin cancer b. to avoid the mutagenic effects of direct sunlight c. to hide from predators d. to hunt for prey, which are always found in dark environments ANSWER: c 94. Which eye type produces images with the greatest resolution? a. eyecups b. compound eyes c. single-lens eyes d. multiple sets of compound eyes ANSWER: c 95. Which portion of the forebrain regulates the endocrine system and body temperature? a. cerebrum Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 b. thalamus c. limbic system d. hypothalamus ANSWER: d 96. Where is the visual cortex of the brain found? a. frontal lobe of the cerebrum b. the occipital lobe c. the parietal lobe d. the cerebellum ANSWER: b 97. Which brain region controls drives, instincts, and emotion? a. thalamus b. cerebrum c. limbic system d. cerebellum ANSWER: c 98. Which structures are the deep crevices of the brain surface that separate the different lobes of the brain? a. topographical features b. cortexes c. sulci d. thalamus ANSWER: c 99. Consider the image.

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Chapter 34

This map of the somatosensory cortex shows distorted body parts because: a. that is the only way an illustrator can fit all the parts into the picture. b. it reflects the relative amount of sensory cortex devoted to pressure and touch sensations from a body part. c. it reflects the number of muscle fibers that are devoted to stimulating a particular body part. d. it reflects differences in different people depending on how that person perceives his or her body parts. ANSWER: b 100. Which region of the brain is larger in humans and primates than in other vertebrates? a. cerebellum b. midbrain c. forebrain d. cerebral cortex ANSWER: d 101. This map of the somatosensory cortex shows distorted body parts to illustrate the amount of sensory cortex devoted to pressure and touch sensations from a specific body part. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34

What can we assume about the body parts that are shown on the figure? a. There is a direct relationship between size of the body part and the area on the somatosensory cortex. b. There is an inverse relationship between size of the body part and the area on the somatosensory cortex. c. There is a direct relationship between the number of receptors on the body part and the area on the somatosensory cortex. d. There is an inverse relationship between the number of receptors on the body part and the area on the somatosensory cortex. ANSWER: c 102. Sometimes after falling backward and hitting their heads, people say that they are "seeing stars." Which area of the brain was affected by the fall? a. the frontal lobe b. the parietal lobe c. the occipital lobe d. the cerebellum ANSWER: c 103. H.M. was a patient whose hippocampus was surgically removed in 1953 in an attempt to cure his epilepsy. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 His case was studied until his death in 2008 and played an important role in the development of theories linking brain structures and functions. What impairment would you expect from a hippocampal deficit? a. loss of anger control b. an inability to establish memories that could later be retrieved c. loss of short-term memory d. loss of motor skills like riding a bicycle e. inability to repeat words that are spoken to him ANSWER: b 104. Which portion of the limbic system is involved in forming memories? a. thalamus b. hypothalamus c. hippocampus d. amygdala ANSWER: c 105. White matter of the cerebrum contains: a. neuronal cell bodies. b. myelinated axons. c. the respiratory centers. d. midbrain structures. ANSWER: b 106. Referring to the graph shown, indicate whether the statement is true or false.

Resting potential is generated mainly by the outward movement of K+ ions from inside the cell. A) true a. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 b. false ANSWER: a 107. Referring to the graph shown, indicate whether the statement is true or false.

The voltage differences shown in the figure are measured across the plasma membrane. a. true b. false ANSWER: a 108. Referring to the graph shown, indicate whether the statement is true or false.

The changes in voltage seen at point 2 and point 4 in the figure are caused by similar movements, in the same direction, of the same ions across the membrane. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 a. true b. false ANSWER: b 109. Referring to the graph shown, indicate whether the statement is true or false.

The trace (blue) line in the figure is following the voltage change as it moves down the axon. a. true b. false ANSWER: b 110. Referring to the graph shown, indicate whether the statement is true or false.

The voltage change shown in the figure is an all-or-nothing response. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 a. true b. false ANSWER: a 111. Referring to the graph shown, indicate whether the statement is true or false.

Different ion channels are responsible for generating the voltage change seen at point 2 and point 4 in the figure. a. true b. false ANSWER: a 112. Referring to the graph shown, indicate whether the statement is true or false.

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Chapter 34 All voltage changes in a neuron's plasma membrane have the characteristic shape shown in the figure. a. true b. false ANSWER: a 113. Referring to the graph shown, indicate whether the statement is true or false.

The changes in voltage shown in the figure are due to ligand-gated ion channels opening and closing along the axon. a. true b. false ANSWER: b 114. Referring to the graph shown, indicate whether the statement is true or false.

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Chapter 34

An excitatory postsynaptic potential (EPSP) of sufficient strength to reach threshold occurred at point 1 on the figure. a. true b. false ANSWER: a 115. Spinal cord injury can lead to paralysis of certain regions of the body. If a person's spinal cord is damaged at the level of their last thoracic (T12) and first lumbar vertebrae (L1), would this statement be true or false?

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Chapter 34

The segmental nature of the spinal cord means that the person's spinal cord injury is likely to affect only those regions of the body controlled by spinal nerves that exit or enter the spinal cord below the site of injury, at T12L1 vertebrae. a. true b. false ANSWER: a 116. Spinal cord injury can lead to paralysis of certain regions of the body. If a person's spinal cord is damaged at the level of their last thoracic (T12) and first lumbar vertebrae (L1), would this statement be true or false?

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Chapter 34

The segmental nature of the spinal cord means that the person's spinal cord injury at this level is likely to affect all regions of the body below the head. a. true b. false ANSWER: b 117. Spinal cord injury can lead to paralysis of certain regions of the body. If a person's spinal cord is damaged at the level of their last thoracic (T12) and first lumbar vertebrae (L1), would this statement be true or false?

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Chapter 34

A person's leg function is controlled by spinal nerves of the lumbosacral plexus (lumbar 2 spinal nerve through sacral 3 spinal nerve, or L2-S3). A person would suffer only limited sensory and motor deficits of their legs following this level of spinal cord injury. a. true b. false ANSWER: b 118. Spinal cord injury can lead to paralysis of certain regions of the body. If a person's spinal cord is damaged at the level of their last thoracic (T12) and first lumbar vertebrae (L1), would this statement be true or false?

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Chapter 34

A person would suffer nearly complete paralysis and sensory loss of their legs following this level of spinal cord injury. a. true b. false ANSWER: a 119. Spinal cord injury can lead to paralysis of certain regions of the body. If a person's spinal cord is damaged at the level of their last thoracic (T12) and first lumbar vertebrae (L1), would this statement be true or false?

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Chapter 34

The person's control of heart rate, as when they get excited and their sympathetic nervous system is activated, would be unaffected. a. true b. false ANSWER: a 120. Indicate whether the statement is true or false. The peripheral nervous system consists only of sensory neurons. a. true b. false ANSWER: b 121. Indicate whether the statement is true or false. All neuronal cell bodies are either in the brain or spinal cord. a. true b. false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 ANSWER: b 122. Indicate whether the statement is true or false. The organization of the human nervous system is completely different from that of other vertebrates. a. true b. false ANSWER: b 123. Consider the figure shown, and indicate whether the statement is true or false.

Chemoreceptors in the nose have multiple dendrites. a. true b. false ANSWER: b 124. Consider the figure shown, and indicate whether the statement is true or false.

Interneuron dendrites are directly connected to the olfactory sensory neuron. a. true Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 b. false ANSWER: a 125. Consider the figure shown, and indicate whether the statement is true or false.

Molecules of a single odorant type can bind to more than one receptor. a. true b. false ANSWER: a 126. Indicate whether the statement is true or false with respect to the evolution of light reception and image perception by animal eyes. All photoreceptors use the protein opsin as a photopigment to convert light energy into a biochemical and electrical change in membrane potential. a. true b. false ANSWER: a 127. Indicate whether the statement is true or false with respect to the evolution of light reception and image perception by animal eyes. Opsin evolved solely for photoreception in multicellular animals. a. true b. false ANSWER: b 128. Indicate whether the statement is true or false with respect to the evolution of light reception and image perception by animal eyes. Opsin evolved for sensing light and other physical features of the environment. a. true b. false ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 129. Indicate whether the statement is true or false with respect to the evolution of light reception and image perception by animal eyes. All image-forming animal eyes function using the same structural components and organization of a lens, pupil, and retina. a. true b. false ANSWER: a 130. Indicate whether the statement is true or false with respect to the evolution of light reception and image perception by animal eyes. The fact that image-forming eyes evolved independently in invertebrate and vertebrate animals provides strong evidence that natural selection operates on previously existing structures to converge on a similar function. a. true b. false ANSWER: a 131. Indicate whether the statement is true or false with respect to the evolution of light reception and image perception by animal eyes. Plant phytochromes that guide plant development and response to light share similar functional roles as animal photoreceptors, but phytochromes and photoreceptors evolved independently and have distinct biochemical structures. a. true b. false ANSWER: a 132. Indicate whether the statement is true or false with respect to how the functions of different brain regions are studied and understood. Patients with damage to a particular brain region can be studied to identify cognitive deficits linked to the damaged region, because these deficits indicate the function of the brain region in healthy individuals. a. true b. false ANSWER: a 133. Indicate whether the statement is true or false with respect to how the functions of different brain regions are studied and understood. Brain imaging shows which brain regions are activated when a human subject performs a particular cognitive task. a. true b. false ANSWER: a 134. Indicate whether the statement is true or false with respect to how the functions of different brain regions Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 are studied and understood. Electrical recordings of nerve cell activity in animals are used to investigate how changes in particular stimuli affect nerve cell firing patterns in a particular brain region. a. true b. false ANSWER: a Multiple Response 135. Suppose that whenever the doorbell rings your cat's pupils dilate, and she runs under the couch. When you reach under the couch to pick up the cat, you also notice that her heart rate has increased. Which responses in your cat are the result of her autonomic nervous system? Select all that apply. a. the dilation of her pupils b. the movement of her limb muscles as she runs under the couch c. her increased heart rate ANSWER: a, c 136. Which type(s) of eye can form an image? Select all that apply. a. eyecup b. compound eye c. single-lens eye ANSWER: b, c 137. Which of the structures is/are thought to be adaptive predatory characteristics? Select all that apply. a. specialized sense organs in the head b. jaws c. teeth d. a tongue ANSWER: a, b, c 138. Ligand-gated ion channels are found within the postsynaptic neuron's cell membrane. Why are ligandgated ion channels critical to how synapses communicate information? Select all that apply. a. Ligand-gated ion channels enable specific neurotransmitters released by presynaptic neurons to exert either excitatory or inhibitory effects on the postsynaptic cell. b. Ligand-gated ion channels open more rapidly than voltage-gated ion channels. c. Ligand-gated ion channels allow the postsynaptic cell to control, through intracellular signaling, which neurotransmitters the postsynaptic cell responds. d. Ligand-gated ion channels are found within a neuron's dendrites but not its axon. ANSWER: a, d 139. Which statement is true of astrocytes? Select all that apply. a. They are a specific type of glial cell. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 b. They surround blood vessels in the brain. c. They contribute to the blood-brain barrier. d. They myelinate axons in the peripheral nervous system. ANSWER: a, b, c 140. Consider the image.

Where would you expect to find ligand-gated ion channels in the plasma membranes of these cells? Select all that apply. a. at letter C b. at letter D c. at letter E d. at letter F e. at letter A ANSWER: c, e 141. Imagine you created a toxin such that binds to the sodium-potassium pump. The toxin binds immediately to the sodium-potassium pump at the peak of the action potential but does not alter the function of sodium and potassium channels. Which of the processes would the toxin prohibit in the neuron? Select all that apply. a. maintaining resting potential b. returning to resting potential after the hyperpolarization phase of an action potential c. the depolarization phase of an action potential d. the hyperpolarization phase of an action potential ANSWER: a, b, d 142. Which of the statements are true about the resting membrane potential? Select all that apply. a. It results from the sodium-potassium pump moving more Na+ ions out of the cell than K+ ions into the cell. b. It results from K+ ions diffusing out of the cell. c. It results from voltage-gated sodium channels remaining open for long periods of time. ANSWER: a, b 143. Which of the statements is true of chemical synapses? Select all that apply. a. They are more common than electrical synapses. b. Vesicles fuse with the presynaptic membrane to release neurotransmitters into the synaptic cleft. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 c. Once released from postsynaptic membrane receptors, neurotransmitter molecules may be actively returned to the presynaptic cell. d. Ligands must also bind to ligand-gated ion channels on the presynaptic membrane of the synapse. ANSWER: a, b, c 144. Which of the scenarios will most likely trigger an action potential? Select all that apply. a. multiple excitatory postsynaptic potentials (EPSPs) arriving close in time at a single synapse (temporal summation) on the postsynaptic cell b. single excitatory postsynaptic potentials (EPSPs) arriving simultaneously at several different synapses (spatial summation) on the postsynaptic cell c. an excitatory postsynaptic potential (EPSP) and an inhibitory postsynaptic potential (IPSP) arriving simultaneously on the postsynaptic cell (cancellation) ANSWER: a, b 145. Inhibitory synapses are important for: Select all that apply. a. facilitating activation of the postsynaptic neuron. b. filtering out unimportant information. c. coordinating movement. d. increasing the number of action potentials fired. ANSWER: b, c 146. Which of the statements is true of electrical synapses? Select all that apply. a. They enable rapid communication. b. They are found in the mammalian brain. c. They release neurotransmitter molecules through exocytosis. ANSWER: a, b 147. In vertebrates, sympathetic nerves, such as those responsible for the "fight-or-flight" response, are part of the: Select all that apply. a. autonomic (involuntary) nervous system. b. peripheral nervous system (PNS). c. central nervous system (CNS). ANSWER: a, b 148. You are a doctor examining a patient who fell and was injured. You conduct the patellar reflex test by tapping on his right patellar tendon. The patient's leg does not move in response to your tap. What can you conclude from this test? Select all that apply. a. The patient has definitely sustained a brain injury. b. The patient may have a motor nerve injury. c. The patient may have a sensory nerve injury. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 d. The patient may have a spinal cord injury. ANSWER: b, c, d 149. In vertebrates, which of the processes are controlled by cranial nerves? Select all that apply. a. eye movement b. facial expression c. speech d. simple reflex arc ANSWER: a, b, c 150. Which of the receptors are mechanoreceptors? Select all that apply. a. olfactory receptors b. auditory hair cells c. stretch receptors in muscle cells d. electroreceptors ANSWER: b, c 151. Sensory cells and sensory neurons allow multicellular animals to sense physical and chemical cues from their environment. What key properties of these cells enable them to perform this function? Select all that apply. a. Sensory cells and sensory neurons have electrically excitable membranes that change in charge potential in response to binding an environmental signaling molecule. b. Sensory neurons have myelinated dendrites that serve as nerve endings. c. All sensory cells and sensory neurons fire action potentials when they bind a signaling molecule. d. Sensory cells and sensory neurons have protein receptors linked to intracellular or membrane-based signaling pathways that alter ion channel permeability. ANSWER: a, d 152. How does the auditory system distinguish the volume of different sounds? Select all that apply. a. Louder sounds stimulate different parts of the auditory system than softer sounds. b. Louder sounds cause action potentials with a higher peak voltage than softer sounds. c. Louder sounds result in a higher frequency of action potentials reaching the auditory centers of the brain than softer sounds. d. Louder sounds cause the stereocilia of hair cells to bend more than softer sounds. e. Louder sounds cause vibration in one part of the basilar membrane, and softer sounds cause vibration in a different part of the basilar membrane. f. Louder sounds cause the release of more neurotransmitters from hair cells than softer sounds. ANSWER: c, d, f 153. The function of the vestibular system of humans is: Select all that apply. a. to detect the head's orientation with respect to gravity. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 34 b. to sense angular rotation of the head. c. to sense the position of the appendages. d. in hearing. ANSWER: a, b 154. Learning and memory involve changes in neural circuits within specific regions of the brain. At the cellular level, repeated stimulation of a presynaptic neuron may result in learning if: Select all that apply. a. the stimulated cell produces action potentials with greater maximum voltage. b. the postsynaptic cell produces action potentials with greater maximum voltage. c. molecular or structural changes occur at the synapse. d. the amount of neurotransmitter released from postsynaptic vesicles of the stimulated cell decreases. e. the postsynaptic cell becomes more responsive to subsequent stimulation. ANSWER: c, e

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Chapter 35 Multiple Choice 1. Imagine that you travel back in time 600 million years and are looking at marine organisms. What types of skeletons would you observe in these creatures? a. endoskeletons b. exoskeletons c. hydrostatic skeletons d. only hydrostatic skeletons and exoskeletons e. endoskeletons, exoskeletons, and hydrostatic skeletons ANSWER: d 2. Consider the image.

Which letter(s) in the diagram of a muscle shown refer(s) to an individual muscle cell? a. letter c only b. letters c, e, f, and i c. letter e only d. letters f and i e. letter i only ANSWER: c 3. Consider the image.

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Chapter 35

Which letter(s) in the diagram of a muscle shown refer(s) to skeletal muscle structures that are multicellular? a. letter c only b. letters a, c, e, f, and i c. letters a, b, c, and d d. letters f, g, h, and i e. letter i only ANSWER: c 4. Cnidarians, such as _____, have muscle fibers. a. clams b. earthworms c. jellyfish d. sea corals ANSWER: c 5. Which option is a muscle tissue that moves one bone with respect to another, enabling movement of an organism? a. skeletal muscle b. cardiac muscle c. smooth muscle d. cartilage ANSWER: a 6. The sliding filament model of muscle contraction hypothesizes that: a. tropomyosin must be removed from the actin molecule before it can bind to myosin. b. muscle contraction requires energy in the form of GTP. c. the sarcomere stretches to create contraction. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 d. actin filaments slide past myosin, causing the shortening of the muscle fiber. ANSWER: d 7. The cross-bridge power stroke corresponds to which event in muscle contraction? a. myosin detachment from actin b. sliding of actin with respect to myosin c. cocking of the myosin head d. binding of the myosin head to actin ANSWER: b 8. Which option is the neurotransmitter that is released at the neuromuscular junction? a. epinephrine b. dopamine c. norepinephrine d. acetylcholine ANSWER: d 9. What is the function of troponin in muscle contraction? a. Troponin slides past myosin, causing muscle shortening. b. Troponin forms the cross-bridges between actin and myosin. c. Troponin moves tropomyosin from actin binding sites, allowing myosin to bind to actin. d. Troponin has no function in muscle contraction. ANSWER: c 10. Skeletal muscle contraction is largely dependent on calcium ions that exit the: a. extracellular space. b. sarcoplasmic reticulum. c. extensions of the plasma membrane. d. Z disc. ANSWER: b 11. The hormone oxytocin stimulates contraction of: a. the smooth muscles in the walls of blood vessels. b. the muscles of the stomach wall. c. uterine muscles. d. skeletal muscles. ANSWER: c 12. What is the basic contracting unit of a skeletal muscle? a. sarcomere b. myofibril Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 c. actin filament d. myosin ANSWER: a 13. Myosin from a rabbit will bind to actin from an amoeba. What does this suggest about the evolution of the structures of actin and myosin? a. The binding sites of actin and myosin are strongly conserved. b. The binding sites of actin and myosin are highly variable. c. The binding sites of actin and myosin are not subject to evolution. d. There is no genetic basis to the structure of actin and myosin. e. None of the other answer options is correct. ANSWER: a 14. ATP hydrolysis allows for what component of skeletal muscle contraction? a. the myosin head to bind to actin b. the actin head to bind to tropomyosin c. calcium levels in the cytoplasm to rise d. the reorientation of tropomyosin and troponin e. cocking of the myosin head to its high-energy position ANSWER: e 15. You take a human smooth muscle cell and block the release of calcium from the sarcoplasmic reticulum. What effect does that have on contraction of that smooth muscle cell, and why? a. Contraction is completely blocked because calcium binding to troponin is required for contraction. b. Contraction is completely blocked because calcium binding to calmodulin is required for contraction. c. Contraction still occurs because Ca2+ can enter the cell directly through Ca2+ channels in the plasma membrane and bind to troponin. d. Contraction still occurs because Ca2+ can enter the cell directly through Ca2+ channels in the plasma membrane and bind to calmodulin. e. Contraction still occurs because contraction in smooth muscle is completely independent of Ca2+ levels. ANSWER: d 16. If a skeletal muscle is no longer able to make enough ATP, then: a. the muscle will be unable to shorten. b. actin and myosin in the sarcomeres are in the unbound state. c. there will be low levels of acetylcholine at the motor endplate. d. actin and myosin in the sarcomeres will remain bound. ANSWER: d 17. What is a muscle fiber? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 a. a single cell of a muscle b. a group of muscle cells that make up the same muscle group c. the myosin that makes up the contractile unit of a muscle cell d. the connective tissue outer covering of a muscle ANSWER: a 18. Which type of muscle cells are active in vasoconstriction and vasodilation of blood vessels? a. skeletal muscle cells b. cardiac muscle cells c. endocardium cells d. smooth muscle cells ANSWER: d 19. What is the function of the protein titin? a. It binds to myosin heads, enabling shortening of the muscle fiber. b. It attaches the myosin thick filament to the Z disc at the end of the sarcomere. c. It transports calcium into and out of the cell. d. It regulates contraction of muscle cells. ANSWER: b 20. What stimulates a skeletal muscle cell to contract? a. an impulse from a sensory neuron b. an impulse from a motor neuron c. auto-depolarizing cells in the membrane of muscle cells d. hormones ANSWER: b 21. The binding of the neurotransmitter to receptors on a vertebrate muscle cell causes: a. an influx of sodium ions, causing a spike in depolarization. b. an influx of potassium ions. c. an efflux of sodium ions. d. a graded depolarization of the muscle cell. ANSWER: a 22. Calcium is necessary to initiate muscle contraction. Which molecule binds calcium? a. myosin b. actin c. troponin d. tropomyosin ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 23. The combination of a nerve impulse, neurotransmitter release at the neuromuscular junction, calcium release from the sarcoplasmic reticulum, and subsequent muscle contraction is known as: a. an action potential. b. threshold potential. c. a motor event. d. excitation-contraction coupling. ANSWER: d 24. When a skeletal muscle contracts, what is happening at the level of the muscle proteins? a. Thick filaments shorten. b. Thin filaments slide relative to thick filaments. c. Thin filaments depolymerize to shorten. d. Calcium ions bind to tropomyosin, which allows actin to bind to myosin. e. Tropomyosin binds to troponin, which allows calcium to bind to actin. ANSWER: b 25. You measure levels of Ca2+ in various locations within a motor neuron and a skeletal muscle fiber when the motor neuron is not depolarized, and the muscle fiber is at rest. Where do you expect to find high levels of Ca2+? a. in vesicles within the motor neuron b. in the cytosol of the motor neuron terminus c. in the synaptic cleft d. within the sarcoplasmic reticulum of the muscle fiber ANSWER: d 26. What is the function of tropomyosin in muscle cells? a. Tropomyosin binds to actin molecules and brings about shortening of the muscles. b. Tropomyosin covers the myosin binding sites on the actin filaments, preventing contraction from occurring. c. Tropomyosin is the contractile unit of the muscle cell. d. Tropomyosin stores calcium. ANSWER: b 27. Cross-bridges form between the contractile proteins of the muscle cell. The movement of cross-bridges shortens the muscle fiber. What two proteins participate in cross-bridge formation? a. actin and myosin b. troponin and titin c. titin and myosin d. tropomyosin and actin ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 28. What event results from the hydrolysis of ATP to ADP by the myosin head? a. cross-bridge formation b. shortening of the muscle fiber c. cocking of myosin head d. the power stroke ANSWER: c 29. Which option describes the motor endplate? a. the terminal membrane of the motor neuron where neurotransmitter is released b. the portion of the membrane of the muscle cell that has postsynaptic receptors that respond to the neurotransmitter released by the motor neuron c. the membrane of the sarcoplasmic reticulum from which Ca2+ is rapidly released upon depolarization d. the Z disc where two sarcomeres adjoin ANSWER: b 30. Calmodulin is involved in the regulation of: a. skeletal muscle contraction. b. propagation of a nerve impulse across a synapse. c. contraction of smooth muscle. d. propagation of heart muscle contraction to the entire myocardium. ANSWER: c 31. Curare is a poison that was used by indigenous peoples of South America on their arrowheads. Curare blocks the action of the neurotransmitter acetylcholine. How did the action of curare make hunting easier? a. Blocking acetylcholine shut down cellular respiration and killed organisms. b. The prey were paralyzed. c. The hunted animals bled to death. d. Curare interfered with the spontaneous depolarization of heart muscle cells. ANSWER: b 32. You measure levels of Ca2+ in various locations within a motor neuron and a skeletal muscle fiber when the motor neuron is depolarized, and the muscle fiber is actively shortening. Where do you expect to find higher levels of Ca2+? a. in vesicles within the motor neuron b. binding to the receptors at the motor endplate of the muscle cell c. diffusing into the axon terminus of a motor neuron through specialized channels in the plasma membrane d. within the sarcoplasmic reticulum e. bound to troponin f. bound to tropomyosin g. bound to myosin Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 ANSWER: e 33. Rattlesnakes contract their tail shaker muscles at frequencies up to 90 hertz (90 times a second) to create their rattling sound; they move their body muscles much more slowly (about 5 hertz) to slither around. What contractile differences do you expect to find in the tail shaker muscles that result in their high contraction rate? a. Other proteins, not actin and myosin, probably form the bulk of the muscle. b. Fewer infoldings of the plasma membrane are present, which will increase the speed of response to the initial muscle depolarization. c. Calcium probably binds directly to tropomyosin rather than troponin. Skipping that extra molecule allows a faster response. d. The myosin molecules have a high rate of ATP hydrolysis, which allows for a faster rate of crossbridge cycling. e. There is probably a structural difference between the shaker and slithering muscles, not a physiological difference. ANSWER: d 34. Rattlesnakes contract their tail shaker muscles at frequencies up to 90 hertz (90 times a second) to create their rattling sound. You compare the contraction frequency of rattlesnake tail muscles at 20°C and 35°C. Do you expect to see an effect of temperature on contraction frequency in this experiment? Why or why not? a. No, because protein structure is not affected by temperature. b. No, because neither of these temperatures is seen in the natural environment of a rattlesnake. c. Maybe—it would depend on whether the snakes are male or female. d. Yes, because muscle contraction is an enzymatic process, and thus will be temperature dependent. e. Yes, because all biological structures show changes when the temperature changes from 20°C to 35°C. ANSWER: d 35. Muscle groups that produce similar motion, or work synergistically, at a joint are known as: a. agonists. b. antagonists. c. synergists. d. pro-agonists. ANSWER: a 36. What happens when action potentials stimulate a muscle at a rate that does not allow relaxation between individual action potentials? a. The strength of contraction decreases with subsequent stimuli. b. The strength of contraction increases with subsequent stimuli and reaches a steady plateau. c. The strength of contraction does not change. d. More motor units are recruited. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 37. Muscles that contract slowly and use less ATP to generate their force: a. contain primarily small motor units. b. contain primarily fast-twitch fibers. c. contain primarily slow-twitch fibers. d. obtain their energy primarily through anaerobic glycolytic processes. ANSWER: c 38. Consider the image.

Muscle contractions have _____ force at slower shortening velocities compared to higher shortening velocities. a. increased b. decreased c. equal d. less stable ANSWER: a 39. When muscle cells are stimulated at such rapid rates that there is no relaxation between individual nerve impulses, a greater force is produced. This property of muscle contraction is called: a. hyperpolarization. b. force summation. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 c. tetanus. d. agonist. ANSWER: b 40. Consider the graphs. Which graph best depicts the relationship between the amount of overlap between actin and myosin, force (y-axis), and sarcomere length (x-axis)?

a. graph A b. graph B c. graph C d. graph D e. graph E ANSWER: d 41. Maximum force is generated by a muscle contraction when: a. overlap between actin and myosin is low. b. there is excessive overlap between actin and myosin. c. the sarcomere is at intermediate length before contraction begins. d. intracellular calcium levels are low. ANSWER: c 42. A muscle can exert force while remaining the same length (e.g., when you hold a heavy weight in a fixed position). This is known as: a. isometric contraction. b. geometric contraction. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 c. isometric relaxation. d. shortening contraction. ANSWER: a 43. Motion that causes two bones to move further apart from each other is known as: a. flexion. b. extension. c. lengthening. d. rotation. ANSWER: b 44. Which option is the type of skeletal muscle fiber associated with stamina and endurance? a. smooth muscle b. glycolytic fast-twitch fibers c. oxidative slow-twitch fibers d. isometric motor units ANSWER: c 45. Consider the figure shown.

Which of the points represents the ratio of shortening velocity to load at which the load on the muscle is equal to the force, or tension, that the muscle is able to achieve? a. point A b. point B c. point C ANSWER: c 46. When an earthworm shortens its body, it is contracting its _____ muscles and relaxing its _____ muscles, Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 which _____. a. longitudinal; circumferential; are antagonists b. circumferential; longitudinal; are agonists c. exoskeleton; muscles; are internal to its cuticle d. muscles; endoskeleton; are formed of protein and chitin e. None of the other answer options is correct. ANSWER: a 47. The cuticle of insects is composed of _______ and forms its exoskeleton. a. structural proteins b. pectin c. chitin d. waxy lipids ANSWER: c 48. Which option is the portion of the vertebrate skeleton that includes the skull, vertebrae, and ribs? a. the axial skeleton b. the central skeleton c. the appendicular skeleton d. the trunk ANSWER: a 49. A(n) _____ exoskeleton made of chitin protects internal organs and limits water loss but has limited ability to grow and repair. a. invertebrate b. vertebrate c. mammalian d. hydrostatic ANSWER: a 50. Why are muscles always associated with some type of skeleton? a. The skeleton transmits the force generated by the muscles. b. The skeleton is always necessary to support the body against gravity. c. The skeleton provides a storage area for calcium ions used by muscle. d. The skeleton is necessary for animals to grow. e. All of these choices are correct. ANSWER: a 51. Walking along a beach, you find a section of an adult grey whale's vertebral column with three vertebrae. Sandwiched between two of the whale vertebrae is a disc of material. You poke it and the vertebrae with your finger. What do you expect, and why? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 a. The bone will be harder than the intervertebral disc, because the bone is mineralized and the disc is a wall of connective tissue surrounding a core of "jelly." b. The bone and the intervertebral disc will be the same hardness, because they are both skeletal composite materials. c. The bone and the intervertebral disc will be the same hardness, because their having the same physical properties allows the skeleton to work as a functional unit. d. The bone will be softer than the intervertebral disc, because it is the type of bone formed first as cartilage and retains its cartilaginous "framework" even in adult animals. ANSWER: a 52. The hydroxyapatite that makes up the bulk of the bony endoskeleton of vertebrates is found in: a. osteocytes. b. the extracellular matrix surrounding osteocytes. c. the collagen. d. osteoblasts. e. the extracellular matrix surrounding osteoblasts. ANSWER: b 53. Tendons function: a. to attach muscles to bones. b. to attach two separate bones. c. as the site of bone growth. d. as an organic component of bone. ANSWER: a 54. Hinge joints allow movement in two dimensions (flexion and extension). Which joint type is an example of a hinge joint? a. the hip joint b. the joints between bones of the skull c. the jaw joint d. the shoulder joint ANSWER: c 55. Which option is a main function of cartilage? a. forming much of the embryonic vertebrate skeleton b. forming growth plates within bones for rapid growth in adolescence c. forming the bone's articular surfaces at a joint d. All of these choices are correct. ANSWER: d 56. A ball-and-socket joint: a. has a greater range of motion than a hinge joint. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 b. is less likely to become dislocated than a hinge joint. c. has fewer muscles controlling it than a hinge joint. d. All of these choices are correct. ANSWER: a 57. What type of bone forms a mesh and supports the bone's ends? a. spongy b. compact c. cartilage d. long ANSWER: a 58. Walking along a beach, you find a section of an adult gray whale's vertebral column with three vertebrae. You saw one of the vertebra in half, revealing a hard outer layer of _____ forming the wall, and a network of _____ with _____ in the center. a. compact bone; spongy bone; trabeculae b. spongy bone; compact bone; trabeculae c. blood vessels; spongy bone; compact bone d. collagen; osteocytes; blood vessels e. collagen; hydroxyapatite; osteoblasts ANSWER: a 59. The diagram shown is a lever system.

The purple box is the load; the red circle is a hinge joint. The dotted and dashed lines are artificial muscles, or actuators. The solid lines are skeletal elements, labeled #1 and #2. Say you add a 1 kg load, cause the flexor actuator to shorten, and measure the actuator's force output. You then increase the load to 4 kg and repeat the experiment. You expect the actuator's force output to _____ with the increased load. a. increase Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 b. stay the same c. decrease ANSWER: a 60. Which option represents bone-forming cells? a. lacunae b. osteoblasts c. osteoclasts d. chondroblasts ANSWER: b 61. Dense mineralized bone tissue that forms the walls of the shafts of long bones is known as: a. condensed bone. b. spongy bone. c. compact bone. d. trabeculae. ANSWER: c 62. You reach down to pick up a bucket full of water. When you begin your arm is straight and once you have lifted the bucket your arm is bent (forearm touching your bicep). Picking up the bucket requires certain muscles to perform work. Which option best describes the action caused by the force of the muscle contraction? a. The force of muscle contraction is transmitted throughout the endoskeleton. b. The force of the muscle contraction flexes the elbow joint. c. The force of the muscle contraction extends the shoulder joint. ANSWER: b 63. The contractile machinery of muscles is only found in the most recently diverged animal groups. a. true b. false ANSWER: b 64. If a muscle contracts, all of the muscle fibers within that muscle must be contracting. a. true b. false ANSWER: b 65. Muscles produce higher forces when lengthening than when shortening. a. true b. false ANSWER: a 66. Muscles only produce force when shortening. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 a. true b. false ANSWER: b 67. To increase force production from a muscle, fewer muscle fibers can be stimulated. a. true b. false ANSWER: b 68. Increasing the frequency of stimulation from low to higher levels will increase muscle force output. a. true b. false ANSWER: a 69. Indicate whether the statement is true or false with respect to the figure of the human skeleton shown.

All of the labeled structures (A–E) are made of composite materials. a. true b. false ANSWER: a 70. Indicate whether the statement is true or false with respect to the figure of the human skeleton shown.

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Chapter 35

All of the labeled structures (A–E) have many cells and little extracellular matrix. a. true b. false ANSWER: b 71. Indicate whether the statement is true or false with respect to the figure of the human skeleton shown.

All of the labeled structures (A–E) are components of the appendicular skeleton. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 a. true b. false ANSWER: b 72. Indicate whether the statement is true or false with respect to the figure of the human skeleton shown.

Structures A and D are components of the axial skeleton. a. true b. false ANSWER: b 73. Indicate whether the statement is true or false with respect to the figure of the human skeleton shown.

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Chapter 35

All of the labeled structures (A–E) are bones that formed first as cartilage. a. true b. false ANSWER: b 74. Indicate whether the statement is true or false with respect to the figure of the human skeleton shown.

Like all skeletons, the human skeleton enables muscles to transmit forces that cause joint rotation. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 a. true b. false ANSWER: a 75. Hinge joints are inherently more stable than ball-and-socket joints. The primary reason is that a larger number of muscles are associated with movement around a hinge joint than around a ball-and-socket joint. a. true b. false ANSWER: b 76. Individuals with the congenital disorder osteogenesis imperfecta are deficient in the production of type I collagen. Which condition might you expect to find in an individual with osteogenesis imperfecta? Bones, particularly leg or arm bones, would break more easily than normal. a. true b. false ANSWER: a 77. Individuals with the congenital disorder osteogenesis imperfecta are deficient in the production of type I collagen. Which condition might you expect to find in an individual with osteogenesis imperfecta? Joints would be stiff and difficult to move. a. true b. false ANSWER: b 78. You reach down to pick up a bucket full of water. When you begin your arm is straight and once you have lifted the bucket your arm is bent (forearm touching your biceps). As you are picking up the bucket of water, you stop when your humerus and ulna are at a 90-degree angle, thereby holding the bucket in the same place. Indicate whether the statement is true or false. There is an increase in recruitment of motor units for contraction. a. true b. false ANSWER: b 79. You reach down to pick up a bucket full of water. When you begin your arm is straight and once you have lifted the bucket your arm is bent (forearm touching your biceps). As you are picking up the bucket of water, you stop when your humerus and ulna are at a 90-degree angle, thereby holding the bucket in the same place. Indicate whether the statement is true or false. Sarcomere length continues to shorten the longer you hold onto the bucket. a. true b. false ANSWER: b 80. You reach down to pick up a bucket full of water. When you begin your arm is straight and once you have lifted the bucket your arm is bent (forearm touching your biceps). As you are picking up the bucket of water, Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 you stop when your humerus and ulna are at a 90-degree angle, thereby holding the bucket in the same place. Indicate whether the statement is true or false. Ca2+ is bound to troponin. a. true b. false ANSWER: a 81. You reach down to pick up a bucket full of water. When you begin your arm is straight and once you have lifted the bucket your arm is bent (forearm touching your biceps). As you are picking up the bucket of water, you stop when your humerus and ulna are at a 90-degree angle, thereby holding the bucket in the same place. Indicate whether the statement is true or false. The force generated by the muscle increases with time (i.e., the longer you hold the bucket the more force is generated). a. true b. false ANSWER: b 82. You reach down to pick up a bucket full of water. When you begin your arm is straight and once you have lifted the bucket your arm is bent (forearm touching your biceps). As you are picking up the bucket of water, you stop when your humerus and ulna are at a 90-degree angle, thereby holding the bucket in the same place. Indicate whether the statement is true or false. The force generated by the muscle stays the same. a. true b. false ANSWER: a 83. You reach down to pick up a bucket full of water. When you begin your arm is straight and once you have lifted the bucket your arm is bent (forearm touching your biceps). As you are picking up the bucket of water, you stop when your humerus and ulna are at a 90-degree angle, thereby holding the bucket in the same place. Indicate whether the statement is true or false. Your biceps muscle is exerting a smaller force than your triceps muscle. a. true b. false ANSWER: b 84. To increase force production from a muscle, more muscle fibers can be stimulated. a. true b. false ANSWER: a 85. You control your posture with fast-twitch muscles. a. true b. false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 ANSWER: b Multiple Response 86. Which option can be associated with striated, multinucleated muscle tissue? Select all that apply. a. cardiac muscle cell b. calmodulin c. skeletal muscle cell d. smooth muscle cell ANSWER: a, c 87. Which roles are played by calcium ions in coordinating the activation of a muscle by motor neurons? Select all that apply. a. Calcium ions enter the motor endplate of the muscle fiber, causing depolarization of the muscle cell. b. Calcium ions enter the motor neuron axon terminus to stimulate vesicle fusion and neurotransmitter release into the neuromuscular synapse. c. Calcium ions are released from the sarcoplasmic reticulum and bind with troponin to open actin binding sites for myosin, leading to force generation. d. Calcium ions diffuse into the sarcoplasmic reticulum to open actin binding sites for myosin, leading to force generation. e. Calcium ions are actively pumped into the sarcoplasmic reticulum to cause muscle relaxation by allowing tropomyosin to block actin binding sites. ANSWER: b, c, e 88. Striated muscles appear striated or banded under a microscope. Striated muscle fibers include which muscle types? Select all that apply. a. skeletal muscle cells b. muscle fibers that close the shells of clams c. smooth muscle cells d. cardiac muscle cells ANSWER: a, d 89. Usain Bolt, a world champion sprinter, can run a 10-meter split in 0.82 seconds. Say you obtain a quadriceps muscle sample from Usain Bolt and compare it to one from a 25-year-old male long-distance runner. What differences do you expect to observe between those two muscle samples? Select all that apply. a. more fast-twitch fibers in Usain Bolt's leg muscles b. more mitochondria in Usain Bolt's leg muscles c. red and white muscle fibers of larger cross-sectional area in Usain Bolt's leg muscles ANSWER: a, c 90. The interaction of two proteins—actin and myosin—allows the muscles of diverse animal species to generate force and produce movement. Given this, which options are true statements? Select all that apply. a. The evolution of muscle contractile function is a highly conserved process. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 b. The evolution of muscle contractile function shows great diversity in a number of different animal groups. c. The force that animal muscles can produce depends on the amount of overlap between actin and myosin protein filaments because this overlap affects the number of actin cross-bridges that can form with myosin binding sites. d. The force that animal muscles can produce depends on the amount of overlap between actin and myosin protein filaments, because this overlap affects the number of myosin cross-bridges that can form with actin binding sites. ANSWER: a, d 91. If increased, which options increase muscle force? Select all that apply. a. motor neuron firing rate b. the number of activated motor units c. the distance between muscle fibers d. increasing the strength of an isomorphic contraction ANSWER: a, b 92. Which statements describe fast-twitch muscle fibers? Select all that apply. a. They consume more ATP than slow-twitch fibers but have fewer mitochondria. b. They obtain energy mainly through glycolysis. c. They have less resistance to fatigue than slow-twitch fibers. d. They have a higher ratio of actin to myosin. e. They tend to be smaller in diameter than slow-twitch fibers. ANSWER: a, b, c 93. The fewer muscle fibers present in a motor unit and the greater the number of motor units in a muscle as a whole: Select all that apply. a. the weaker the contractions of muscle fibers in that motor unit. b. the greater the control over the movement(s) that result from stimulation of different motor units in the muscle. c. the less likely an action potential will be generated in the muscle fibers of that motor unit. d. the less likely these muscle fibers will fatigue. ANSWER: a, b 94. Bone with insufficient or abnormal collagen would _____ compared with healthy bone. Select all that apply. a. fracture more easily b. be more flexible c. contain less hydroxyapatite d. have less range of motion ANSWER: a, c 95. Which options are true statements regarding animal skeletons? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 a. Animal skeletons rely on rigid tissues, or on water, for support and movement. b. Rigid endoskeletons and exoskeletons rely on flexible joints to allow movement in certain directions. c. An advantage of endoskeletons is that they can be repaired if broken. d. Bony endoskeletons are found only in vertebrate animals. e. Exoskeletons are found in arthropods, mollusks, and cnidarians. ANSWER: a, b, c, d 96. An evolutionary advantage of the endoskeleton is that it: Select all that apply. a. allows growth of fairly large organisms. b. allows repair of damaged bony tissue. c. provides a structure on which muscles can perform work to generate movement. d. makes the process of reproduction more efficient. ANSWER: a, b, c 97. Consider the image.

Which structure(s) in the figure shown have at least three sets of muscle antagonists to control motion? Select all that apply. a. structure A b. structure B c. structure C d. structure D ANSWER: a, c 98. You reach down to pick up a bucket full of water. When you begin your arm is straight and once you have lifted the bucket your arm is bent (forearm touching your bicep). As you begin to pick up the bucket, which option represents what is occurring within your biceps muscle? Select all that apply. a. Troponin is no longer bound to actin. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 35 b. Ca2+ is released from the sarcoplasmic reticulum. c. Myosin heads form repeated cross-bridges with actin filaments. d. Ca2+ binds to myosin. e. The distance between Z-lines stays the same. f. The infoldings of the plasma membrane are depolarized. ANSWER: a, b, c

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Chapter 36 Multiple Choice 1. In the late 1960s, Carroll Williams and Karel Salma discovered that, when grown in jars with paper towels made from balsam fir, the insect Pyrrhocoris apterus underwent several extra larval molts and finally died without completing metamorphosis. Further investigation led to the conclusion that the fir trees synthesize a hormonal analog that acts as an insecticide. Which hormone is this natural insecticide mimicking? a. ecdysone b. juvenile hormone c. PTTH (prothoracicotropic hormone) ANSWER: b 2. Consider the figure shown.

If you removed the prothoracic glands from the insect larva, the larva would: a. molt into a larger larva. b. pupate and then emerge as a normal adult. c. never molt again. ANSWER: c 3. Type 1 and type 2 diabetes mellitus have different underlying causes. Type 1 diabetes is characterized by a loss of the insulin-producing cells of the pancreas, leading to insulin deficiency. Type 2 diabetes, at least in its early stages, is characterized by a loss of sensitivity to the hormone insulin by cells that normally respond to the hormone. Which step labeled in the diagram shown is inhibited in both type 1 and type 2 diabetes (either directly or indirectly)?

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a. step a: glycogen synthesis b. step b: hormone release c. step c: sensory input to the pancreas d. step d: hormone release e. step e: glycogen hydrolysis ANSWER: a 4. In the late 1960s, Carroll Williams and Karel Salma discovered that the insect Pyrrhocoris apterus underwent several extra larval molts and finally died without completing metamorphosis. Further investigation led to the conclusion that the fir trees synthesize a hormonal analog that acts as an insecticide. The molecule is actually an analog of juvenile hormone. Furthermore, the juvenile hormone analog discovered by Salma and Williams has no effect on most other metamorphosing insects, suggesting that: a. many insects reproduce as larvae. b. some evolutionary divergence in the structure of juvenile hormone and in the juvenile hormone receptors of different insects has occurred. c. not all insects use juvenile hormone to stimulate metamorphosis. d. other insects have evolved methods to detect the juvenile hormone analog. e. these insects avoid balsam firs. ANSWER: b 5. In which process are hormones involved? a. an insect shedding its exoskeleton b. a starfish regrowing one of its arms c. a lion sleeping during the day and hunting at night d. a human maintaining her blood sugar levels e. All of these choices are correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 ANSWER: e 6. Which option is an example of positive feedback involving hormones? a. the maintenance of calcium levels in the blood b. the maintenance of glucose levels in the blood c. the increased intensity of uterine contractions during childbirth d. the maintenance of calcium and glucose levels in the blood e. All of these choices are correct. ANSWER: c 7. Calcium levels in the blood are primarily maintained through the interactions of two different hormones, calcitonin and parathyroid hormone. Based on your understanding of homeostasis through feedback mechanisms, we could also describe the action of these hormones as: a. exerting the same influence on the concentration of blood calcium. b. exerting opposite influences on the concentration of blood calcium. c. positive feedback. ANSWER: b 8. In response to low sugar levels in the blood, the hormone glucagon is released by the pancreas. In a negative feedback system, the pancreas would represent the _____ of the system. a. stimulus b. sensor c. effector d. response ANSWER: b 9. When a doctor suspects that a patient may have diabetes, she will often have the patient take a glucose challenge test. After the patient drinks a large amount of sugary solution, his blood is drawn, and the circulating levels of glucose in the blood are determined. Review the graph shown.

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Which curve do you expect from a patient with diabetes? a. Curve A b. Curve B c. Curve C ANSWER: b 10. A researcher is studying several different Rhodnius mutants. One of her mutants molts in a normal fashion until the second nymph stage. After that, the insects continue to molt, but keep demonstrating body plans of the second nymph stage. What is the likely effect of this mutation in Rhodnius? a. The juvenile hormone is not expressed. b. The ecdysone hormone is not expressed. c. The prothoracicotropic hormone (PTTH) is not expressed. d. The juvenile hormone is overexpressed. e. The ecdysone hormone is overexpressed. f. The prothoracicotropic hormone (PTTH) is overexpressed. ANSWER: d 11. Blood glucose levels are regulated by _____, in which an increase or decrease in glucose signals the pancreas to stop producing glucagon or insulin, respectively. a. negative feedback b. positive feedback Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 ANSWER: a 12. A type 1 diabetic male gives his wife an injection pen and asks her to inject him with its contents if he ever passes out due to low blood sugar. What hormone does this pen likely contain? a. insulin b. oxytocin c. glucagon d. glycogen ANSWER: c 13. Aldosterone is a hormone secreted by the adrenal cortex, and insulin is a peptide secreted by the pancreas. Refer to the figures shown. When aldosterone contacts a target cell, it binds to an intracellular receptor and migrates to the nucleus. Insulin binds to extracellular receptors on the plasma membrane.

What is the most likely reason for this difference? a. Aldosterone is hydrophilic and, therefore, must enter the aqueous environment of the cytoplasm to have an effect. b. Aldosterone is lipid-soluble and, therefore, easily crosses the phospholipid bilayer of the plasma membrane. c. Aldosterone is too small to bind extracellular receptors on the plasma membrane. d. Insulin is too large to interact chemically with DNA. ANSWER: b 14. A new hormone is discovered that appears to play a role in bone development. The hormone is hydrophilic and composed of several amino acids. How should this hormone be classified? a. as a steroid hormone b. as a peptide hormone Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 c. as an amine hormone d. as a pheromone ANSWER: b 15. Growth hormone is a protein hormone that has effects on a variety of cells. For example, it increases glucose uptake in muscles. If you could inject growth hormone molecules directly into muscle cells, what do you expect the effect would be? a. The effect on glucose uptake would be amplified, because a hormone molecule inside the cell is always more effective than one on the surface. b. The injected growth hormone molecules would have no effect on glucose uptake. c. Glucose uptake by muscle cells would decrease. d. Glucose uptake by muscle cells would increase. ANSWER: b 16. In your textbook, there is a discussion of the experiment by Wigglesworth, which demonstrated that PTTH (prothoracicotropic hormone) triggers molting and the release of ecdysone. In 1984, Nagasawa et al. published a paper entitled "Amino-terminal amino acid sequence of silkworm prothoracicotropic hormone: homology with insulin." The similar amino acid sequence between PTTH and insulin suggests that: a. a common ancestral peptide evolved into peptides with different functions in insects and vertebrates. b. PTTH must be involved with glucose metabolism in insects. c. insulin causes molting in vertebrates. d. if PTTH were given to vertebrates, it would likely trigger glucose uptake by cells. ANSWER: a 17. An amine hormone contains: a. cholesterol. b. a single amino acid chain. c. an aromatic amino acid. d. an enzyme. e. both cholesterol and an aromatic amino acid. ANSWER: c 18. Where would you find a receptor of a target cell for a peptide hormone? a. in the nucleus b. in the cytoplasm c. in the ribosome d. on the cell surface e. in the mitochondria ANSWER: d 19. Recall that a small amount of releasing factor from the hypothalamus can cause the adrenal cortex to secrete a much greater amount of cortisol. The action of cortisol, in turn, can result in a high final concentration of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 glycogen in the liver. This is an example of: a. hormone signal denaturation. b. hormone signal amplification. c. hormone signal accumulation. d. hormone signal degradation. ANSWER: b 20. Which option about hormones and their receptors is true? a. All cells have the same hormone receptors, but only a target cell will bind the hormone. b. Cells with different hormone receptors typically have different gene expression patterns. c. All cells express the same hormone receptor, and that receptor will only bind hormones that cause a response by the cell. d. All hormone receptors are found embedded in the cell membrane. e. All hormone receptors are found in the cytoplasm. ANSWER: b 21. A patient with a thyroid tumor has her thyroid gland removed. Without a thyroid gland and without Synthroid, a thyroid hormone replacement drug, which of these results is most likely to occur? a. Her blood TSH levels will be abnormally high, and her metabolic rate will be high. b. Her blood TSH levels will be abnormally low, and her energy level will be low. c. Her blood TSH levels will be abnormally low, and her metabolic rate will be high. d. Her blood TSH levels will be abnormally high, and her metabolic rate will be low. ANSWER: d 22. Which of these hormones regulates, in large part, the diurnal or nocturnal behavior of animals? a. calcitonin b. serotonin c. melatonin d. oxytocin e. epinephrine ANSWER: c 23. A pediatrician sees a patient who is exhibiting symptoms of gigantism and rapid growth. Which of these endocrine glands is likely affected in this individual? a. the thyroid gland b. the adrenal gland c. the parathyroid gland d. the anterior pituitary gland e. the posterior pituitary gland ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 24. Many animals communicate through the use of pheromones. What is one disadvantage of pheromones as a communication system? a. Most pheromones are only recognized by members of the same species. b. Most pheromones dissipate quickly in the environment. c. Most pheromones elicit a specific response from an individual of the same species. ANSWER: b 25. Goiters are often produced in the _____, a gland that plays a key role in metabolism by secreting the thyroxine and triiodothyronine hormones. a. the pituitary gland b. the thyroid gland c. the parathyroid glands d. the adrenal glands e. the pineal gland ANSWER: b 26. Recall that the pancreas helps to maintain normal blood glucose levels. What endocrine organ plays a role in stabilizing blood calcium levels? a. the pineal gland b. the posterior pituitary gland c. the adrenal cortex d. the parathyroid gland ANSWER: d 27. Kisspeptin is a protein that, in humans, has an important role in initiating secretion of the releasing factor gonadotropin-releasing hormone (GnRH). Endocrinologists are finding that the protein kisspeptin and its receptor are central to sexual maturation at puberty. If kisspeptin directs the release of GnRH, you can expect that the receptors for kisspeptin will be found in cells of the: a. anterior pituitary. b. posterior pituitary. c. hypothalamus. d. gonads. ANSWER: c 28. When people have a few drinks containing alcohol, they often feel the need to urinate. This response suggests interplay between alcohol and antidiuretic hormone (ADH). Which of these statements could explain how alcohol and ADH interact? a. Alcohol increases the release of ADH. b. Alcohol decreases the release of ADH. c. Alcohol facilitates the binding of ADH to receptors in the brain, so individuals "feel" the need to urinate. d. Alcohol inhibits the binding of ADH to receptors in the brain, so individuals "feel" the need to Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 urinate. ANSWER: b 29. A woman visits a doctor. After hearing her symptoms and running some blood work, the doctor determines that she has an underactive pineal gland. Which of these is likely a symptom she is experiencing? a. insomnia (due to low melatonin levels) b. muscle soreness and inflammation (due to low cortisol levels) c. increased urination (due to high glucose levels) d. constipation (due to low gastrin levels) ANSWER: a 30. ____, such as thyroid-stimulating hormone (TSH), signal organs of the endocrine system to secrete new, additional hormones. a. Tropic hormones b. Releasing hormones c. Stimulating hormone ANSWER: a 31. Kisspeptin is a protein, coded for by the KISS-1 gene, that, in humans, has an important role in initiating secretion of gonadotropin-releasing hormone (GnRH), a hormone that stimulates the anterior pituitary gland to release other hormones related to reproduction. Endocrinologists are finding that the protein kisspeptin and its receptor are central to sexual maturation at puberty. The receptor for kisspeptin is coded for by the gene GPR54. Given the information above, the product of GPR54 is a(n): a. steroid. b. intracellular protein. c. extracellular protein. d. amine. e. transmembrane protein. ANSWER: e 32. Kisspeptin is a protein, coded for by the KISS-1 gene, that, in humans, has an important role in initiating secretion of gonadotropin-releasing hormone (GnRH), a hormone that stimulates the anterior pituitary gland to release other hormones related to reproduction. Endocrinologists are finding that the protein kisspeptin and its receptor are central to sexual maturation at puberty. The receptor for kisspeptin is coded for by the gene GPR54. In 2003, a French family, in which three of four brothers had undeveloped testes, sparse pubic hair, and a lack of sexual development, was studied. All three affected men were homozygous for a genetic mutation that prevented puberty. That mutation could possibly be a defect in any of these genes except the: a. GPR54 gene. b. KISS-1 gene. c. GnRH gene. d. gonadotropin coding genes (LH and FSH). e. testosterone gene. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 ANSWER: e 33. Within a mammalian body, which of these signals would travel the longest distance to reach its target cell? a. a hormone signal b. a paracrine signal c. an autocrine signal d. a neurotransmitter e. a pheromone ANSWER: a 34. A scientist is searching for chemical compounds involved in embryonic limb development. He discovers a new compound that is transiently expressed in the limbs and only seems to affect cells in its immediate vicinity (up to a few cell lengths away). How should the scientist classify this chemical compound? a. as an amine hormone b. as a peptide hormone c. as a steroid hormone d. as a paracrine factor e. as a neurotransmitter ANSWER: d 35. You call an exterminator to your house because you have noticed many nymphs of a specific insect everywhere in your cupboards. The exterminator proudly tells you that they are using biological concepts to get rid of these pests, and he generously sprays a solution of juvenile hormone all over your house. Assuming there are no breeding adults and only nonreproductive nymphs, which of these results would you expect in the days after the visit from the exterminator. a. The number of nymphs will increase immediately after the exterminator sprays. b. The number of adults will increase immediately after the exterminator sprays. c. The number of nymphs will stay the same after the exterminator sprays. d. The number of adults will decrease after the exterminator sprays. ANSWER: c 36. Refer to the graph shown.

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A student wakes up at 5:00 a.m., eats breakfast, and goes to class. She has no time to eat for the rest of the day, but she does take a quick jog around lunchtime. Afterwards, she heads back to class and then returns home to study for the rest of the night. Based on the information given, indicate which point(s) on the graph reflect when the student's insulin levels were low. a. A b. B c. C d. D ANSWER: a 38. Consider the positive feedback that regulates mammalian childbirth. In this process, the pituitary gland is _____, oxytocin is _____, and a uterine contraction is _____. a. the effector; the stimulus; the response b. the sensor; the effector; both the stimulus and response c. both the stimulus and response; the effector; the sensor d. both the stimulus and response; the sensor; the effector ANSWER: a 39. Pyrrhocoris apterus, or the firebug, is native to Europe and China and eats the seeds of lime trees and mallow trees. What is the most likely explanation for the fact that a native North American species, the fir tree, synthesizes a compound that disrupts development of this European insect? a. The fir tree benefits from this compound, because its effect on the firebug decreases competition for resources in its natural environment. b. The fir tree benefits from this compound, because its effect on the firebug increases the survival of fir tree offspring. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 c. The fir tree benefits from this compound, because its effect on the firebug increases predation pressure on its competition. d. The fir tree benefits from this compound, because its effect on the firebug increases the survival of other, related members of the same species. e. There is likely a strong similarity in the hormone receptors of different insects, so that a compound made by a tree native to North America has an effect on an insect native to Europe. ANSWER: e 40. A researcher is studying the function of a peptide hormone in rabbits and humans. He notices that this hormone has a different effect in these two mammals, even though the hormone receptor has the same sequence and structure in both animals. What accounts for this difference in hormone function? a. sequence differences in the peptide hormone b. sequence differences in components of the associated signal transduction pathway c. sequence differences in the hormone's target genes d. sequence differences in cholesterol ANSWER: b 41. Please study the table shown.

Each pair of phrases in the same row shows a valid comparison between the anterior and posterior pituitary except: a. A. b. B. c. C. d. D. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 42. Honeybee workers spend their first few weeks as young adults tending the colony's brood. Later, they shift jobs to forage for food outside the colony. When researchers fed the young bees candy spiked with the molecule ethyl oleate, the young bees continued to babysit and delayed the job shift to foraging. What would be required to demonstrate that ethyl oleate normally acts as a pheromone, not as an endocrine or paracrine signal? a. It would require demonstrating that ethyl oleate binds to a receptor and initiates a series of responses. b. It would require demonstrating that ethyl oleate is released into the air by another individual of the same species. c. It would require demonstrating that ethyl oleate is released by a predator and causes a behavioral change in the prey. ANSWER: b 43. Foraging bees secrete a chemical that delays the behavioral change of other female bees in the hive from nest-tenders to foragers. Such molecules are: a. paracrine signals. b. autocrine signals. c. pheromones. ANSWER: c 44. A beekeeper opens one of his hives. Initially, only one bee gets under his suit and stings him. He is then swarmed by the rest of the bees and ends up with several stings. Why did most of the hive attack him? a. The first bee produced alarm pheromones. b. The first bee produced trail pheromones. c. The first bee produced amine hormones. d. The first bee produced steroid hormones. e. The first bee produced paracrine signals. ANSWER: a 45. Which of the statements is true regarding neurotransmitters, synaptic signaling, and hormones? a. All neurotransmitters are also hormones. b. Most synaptic signaling is rapid, compared with hormones. c. Most synaptic signaling occurs after nerve cells are stimulated by hormones. d. Similar to hormones, neurotransmitters are released into the bloodstream. ANSWER: b 46. The figure shown illustrates epithelial cells in the lungs. When the epithelial monolayer is damaged by mechanical injury, breaching of the barrier allows epithelial cells to sense injury and initiate a repair pathway. Cells on the apical side of the epithelium release the molecule heregulin, which binds only to receptors on the basal side of those cells. When the ligand, heregulin, binds its receptor, it triggers a signaling cascade that stimulates cell division. Mechanical damage allows molecules from apical and basal surfaces to mix.

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The signaling shown in the figure is best described as: a. paracrine. b. autocrine. c. endocrine. d. synaptic. ANSWER: b 47. The figure shown illustrates epithelial cells in the lungs. When the epithelial monolayer is damaged by mechanical injury, breaching of the barrier allows epithelial cells to sense injury and initiate a repair pathway. Cells on the apical side of the epithelium release the molecule heregulin, which binds only to receptors on the basal side of those cells. When the ligand, heregulin, binds its receptor, it triggers a signaling cascade that stimulates cell division. Mechanical damage allows molecules from apical and basal surfaces to mix.

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The ligand is: a. hydrophilic. b. hydrophobic. c. hydrolytic. ANSWER: a 48. High blood pressure in mammals reflexively causes a decrease in heart rate, which results in lowered blood pressure. This is a _____ feedback loop because response to the signal results in a(n) _____ of that signal. a. negative; decrease b. negative; increase c. positive; decrease d. positive; increase ANSWER: a 49. Hormones that help maintain homeostasis typically operate through a negative feedback system. a. true b. false ANSWER: a 50. Positive and negative feedback systems have all the same components: stimulus, sensor, effector, and response. The primary difference between the two is that the response in a positive feedback system stimulates the sensor. a. true b. false ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 51. Imagine you are out camping. While sleeping in your tent, you are awakened by a rustling sound outside that sound like a wild animal. In getting ready to respond to what may be a dangerous encounter with the animal, your body raises its set point for heart rate and blood flow to prepare for the potential encounter. a. true b. false ANSWER: a 52. A mouse senses that a cat is nearby, and its body begins to produce epinephrine as part of the fight-or-flight response. Which of the mouse's glands secretes epinephrine? a. the pituitary gland b. the thyroid gland c. the parathyroid glands d. the adrenal glands E. the pineal gland ANSWER: d 53. A small mouse is out looking for food just after waking up. It notices a hawk flying overhead and searches for an escape. Evaluate the statement about what is happening in the mouse's body as true or false. The release of norepinephrine will decrease digestive activity in the mouse. a. true b. false ANSWER: a 54. A small mouse is out looking for food just after waking up. It notices a hawk flying overhead and searches for an escape. Evaluate the statement about what is happening in the mouse's body as true or false. The heart rate of the mouse will decrease below normal homeostatic levels; in other words, the set point will change. a. true b. false ANSWER: b 55. A small mouse is out looking for food just after waking up. It notices a hawk flying overhead and searches for an escape. Evaluate the statement about what is happening in the mouse's body as true or false. Arteriole diameter in the skeletal muscles of the mouse will increase. a. true b. false ANSWER: a 56. A small mouse is out looking for food just after waking up. It notices a hawk flying overhead and searches for an escape. Evaluate the statement about what is happening in the mouse's body as true or false. All cells in the body of the mouse have receptors for both norepinephrine and epinephrine. a. true b. false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 ANSWER: b 57. A small mouse is out looking for food just after waking up. It notices a hawk flying overhead and searches for an escape. Evaluate the statement about what is happening in the mouse's body as true or false. Glucagon levels in the mouse will increase over minutes to hours. a. true b. false ANSWER: b Multiple Response 58. Which statement regarding hormones is true? Select all that apply. a. Hormones can act over long distances in the body. b. Hormones can act as neurotransmitters. c. Hormones can act as transcription factors in conjunction with proteins. d. Hormones can produce effects with very small concentrations. e. Hormones can be transported through the bloodstream. f. Hormones only act on one organ. g. Hormones are only released into the bloodstream. ANSWER: a, b, c, d, e 59. In which processes in humans do hormones play a role? Select all that apply. a. circadian rhythms b. regulating blood glucose levels c. sexual arousal d. childbirth e. metabolism f. sense of taste g. sense of smell ANSWER: a, b, c, d, e 60. In the human body, which structures contain cells that are part of the endocrine system? Select all that apply. a. the pancreas b. the stomach c. the intestines d. the brain e. the ovaries f. the skeletal muscles ANSWER: a, b, c, d, e 61. Which options are accurate comparisons of the endocrine and nervous systems? Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 a. The nervous system responds quickly, over short time periods, to environmental stimuli. The endocrine system responds more slowly and over longer time periods. b. Nervous systems rely exclusively on electrical communication between adjacent nerve cells. Endocrine systems rely on chemical signals carried in the bloodstream. c. The endocrine system requires electrical stimulation through sensory input from the nervous system in order to respond. d. The endocrine system responds to electrical signals from the nervous system, as well as chemical signals received from different organs. e. Chemical communication by the endocrine system uses the bloodstream to carry signals and is therefore general. Chemical communication within the nervous system is highly localized and specific. f. The tissue targets of endocrine signals are determined by the presence of receptors expressed by cells within those tissues. The targets of nervous signals are based on specific circuits and synapses that the axons of neurons make with the dendrites of other neurons. ANSWER: a, d, e, f 62. Which statements regarding animal endocrine systems are true? Select all that apply. a. Animal endocrine systems transmit chemical signals to coordinate the physiological activity of different organ systems within the animal's body. b. Animal endocrine systems transmit chemical signals to the environment that can be sensed by other members of the same species, affecting their behavior. c. Hormones may share common chemical structures in diverse animals, but often evolve new functional roles. d. Steroid hormones evolve by changing their structures through natural selection. e. Steroid hormones evolve through changes in the expression of their receptors within different cells through natural selection. ANSWER: a, b, c, e 63. It takes very few molecules of a hormone to cause changes in a target cell. This may be explained by the fact that: Select all that apply. a. activation of transcription produces many copies of mRNA, and each copy is translated many times. b. the mechanism of hormonal action involves the replication of the hormone within the target cell to quickly magnify the hormone's effect. c. hormones are large molecules that persist for years and can repeatedly stimulate the same cell. d. the mechanism of hormonal action involves an enzyme cascade that amplifies the response to a hormone. e. the mechanism of hormonal action involves memory cells that have had prior contact with the hormone and immediately respond to its presence. ANSWER: a, d 64. Which statements regarding steroid hormones are true? Select all that apply. a. They are hydrophilic. b. They are hydrophobic. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 c. They have structures similar to cholesterol. d. They are composed of single aromatic amino acids. e. They bind to cell-surface receptors. f. They bind to receptors in the cytoplasm. ANSWER: b, c, f 65. Which of these traits does an amine hormone possess? Select all that apply. a. It is made from a single aromatic amino acid. b. Its chemical structure resembles that of cholesterol. c. It only contains amino acid chains. d. It is hydrophilic. e. It is hydrophobic. ANSWER: a, d 66. Neurosecretory cells underlie nervous system communication with the endocrine system. Which of these statements reflects this coordinated function between the nervous and endocrine systems? Select all that apply. a. The vertebrate hypothalamus plays a role that is similar to neurosecretory cells located in the brain of insects, which release peptide hormones that act on the corpora allata. b. Neurosecretory cells within the vertebrate hypothalamus release peptide hormones that act directly on the anterior pituitary gland. c. Neurosecretory cells within the vertebrate hypothalamus release steroid hormones that act directly on the posterior pituitary gland. d. Neurosecretory cells within the insect brain release peptide hormones that act directly on the corpora allata and prothoracic gland. e. Neurosecretory cells within the insect brain release steroid hormones that act directly on the corpora allata and prothoracic gland. f. Neurosecretory cells within the vertebrate hypothalamus secrete hormones into blood vessels that are transported to cell receptor targets located within the anterior pituitary gland. ANSWER: a, b, d, f 67. Which of the statements regarding the pituitary gland is true? Select all that apply. a. It is composed of two glands. b. It serves as a "control center" for the vertebrate endocrine system. c. It produces hormones that can target the kidneys. d. It receives information from the hypothalamus. e. It could be considered part of the nervous system. f. It serves to regulate feelings of hunger by the body. ANSWER: a, b, c, d, e 68. Which of the statements is true regarding neurosecretory cells of the hypothalamus? Select all that apply. a. They secrete releasing factors. b. They secrete thyroid-stimulating hormone (TSH). Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 c. Their cell bodies reside in the hypothalamus. d. Their cell bodies reside in the (posterior) pituitary gland. e. They do not produce hormones, because they are neurons and not endocrine cells. ANSWER: a, c 69. Which of these hormones would be classified as a tropic hormone? Select all that apply. a. thyroid-stimulating hormone (TSH) b. oxytocin c. adrenocorticotropic hormone (ACTH) d. estrogen e. somatostatin ANSWER: a, c 70. Which of the statements is true regarding the anterior pituitary gland? Select all that apply. a. It houses neurosecretory cell axons. b. It is derived from neural tissue. c. It is the target of releasing factors. d. It is derived from epithelial tissue. e. It produces tropic hormones. ANSWER: c, d, e 71. The figure shown illustrates epithelial cells in the lungs. When the epithelial monolayer is damaged by mechanical injury, breaching of the barrier allows epithelial cells to sense injury and initiate a repair pathway. Cells on the apical side of the epithelium release the molecule heregulin, which binds only to receptors on the basal side of those cells. When the ligand, heregulin, binds its receptor, it triggers a signaling cascade that stimulates cell division. Mechanical damage allows molecules from apical and basal surfaces to mix.

The ligand/receptor complex likely acts by: Select all that apply. a. directly regulating transcription. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 36 b. activating enzymes within the cell. c. initiating signaling cascades. ANSWER: b, c 72. Refer to the graph shown.

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Chapter 37 Multiple Choice 1. Which of the structures have lamellae? a. lungs of fish b. lungs of mammals c. tracheoles of insects d. gills of fish ANSWER: d 2. One of the functions of the respiratory system is to rid the body of CO2. Where does the CO2 come from? a. CO2 is a breakdown product of the carbohydrates or lipids oxidized in cellular respiration. b. CO2 is a breakdown product of the carbohydrates or lipids reduced in cellular respiration. c. CO2 is produced when inhaled oxygen combines with carbon atoms from carbohydrates or lipids. d. CO2 is produced when the oxygen atoms of water combine with carbon atoms. ANSWER: a 3. Organisms that use bulk flow to move oxygen are able to: a. move more quickly through their environment than organisms that only use diffusion. b. achieve larger size than organisms that only use diffusion. c. remove carbon dioxide waste from their tissues, whereas organisms that use diffusion cannot. d. live in aquatic environments, whereas organisms that use diffusion cannot. ANSWER: b 4. A constant supply of oxygen is essential to animal survival. Why? Select all that apply. a. Without oxygen, cells could not generate sufficient ATP for survival. b. Oxygen is necessary for cells to release waste CO2. c. Oxygen is necessary for lactic acid fermentation. ANSWER: a 5. Chemoreceptors in the circulatory system detect changes in circulating pCO2. If CO2 concentrations get too high, the rate of ventilation increases. Why does this make sense? a. Actually, it does not make sense. There is no relationship between CO2 and ventilation rate. b. It makes sense because CO2 is formed from O2 and, therefore, CO2 is a sign that oxygen is being consumed. c. It makes sense because tissues that are more actively working produce more CO2 and, therefore, need more O2. d. It makes sense because hemoglobin carries CO2 and, therefore, blocks oxygen binding to the heme site, creating an oxygen deficiency. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 6. To aid inhalation, both the diaphragm and intercostal muscles are used. By expanding the thoracic cavity, these muscles create a space with _____ atmospheric pressure. a. lower pressure than b. higher pressure than c. pressure equal to ANSWER: a 7. At the doctor's office, you may have noticed that when you are asked to "breathe deeply," your chest rises and falls and expands more dramatically than when you are breathing normally. Why? a. Your intercostal muscles are being used to expand your chest cavity. b. Your spiracles are being used to expand your chest cavity. c. Your air sacs are being used to expand your chest cavity. d. Your bronchioles are being used to expand your chest cavity. ANSWER: a 8. Birds are able to extract more oxygen than mammals from an equal volume of air. Which of the statements reflects why birds are able to do so? a. Birds have air sacs that enable them to hold their breath for longer, thereby increasing the time for oxygen to diffuse. b. Birds have a greater surface area for diffusion because they have air sacs in addition to lungs. c. Birds do not have air mixing in their lungs; air flows in a unidirectional manner across the lung surface for increased diffusion. d. Birds have capillaries that are in close contact with the lungs. ANSWER: c 9. A man visits a doctor and discovers that, due to tumors, he needs to have both of his carotid bodies removed. How will their removal affect his body? a. His body will not be able to sense the oxygen concentration of brain-bound blood. b. His body will not be able to sense the CO2 concentration of brain-bound blood. c. His body will not be able to sense the oxygen concentration of body-bound blood. d. His body will not be able to sense the CO2 concentration of body-bound blood. ANSWER: a 10. The respiratory system of which organisms relies on air sacs to move air through the lungs? a. birds b. mammals c. fish d. amphibians e. reptiles ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 11. Imagine that a young girl suffers a horseback riding injury and one of her ribs punctures her diaphragm. How will this injury affect the girl's breathing? a. The injury will make it easier to increase negative air pressure in her chest cavity, making inhalation easier. b. The injury will make it more difficult to increase negative air pressure in her chest cavity, making inhalation more difficult. c. The injury will make it more difficult to decrease negative air pressure in her chest cavity, making inhalation more difficult. d. The injury will not affect her at all, as intercostal muscles are responsible for breathing. ANSWER: b 12. The figure shown is of tubes illustrating O2 uptake at the fish gill. The values shown in the tubes indicate percent fluid saturation of O2. Which answer choice identifies the fluid and the direction of flow correctly?

a. A b. B c. C d. D e. E ANSWER: e 13. Which of the statements is true regarding myoglobin? a. It transports O2 in the bloodstream. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 b. It possesses four heme groups. c. It is found within muscle tissue. d. It is not found in marine mammals. e. It irreversibly binds to O2. ANSWER: c 14. At arrow A in the accompanying graph on oxygen saturation of hemoglobin, the hemoglobin is only at 20% O2 saturation, and at arrow B, it is at 97% O2 saturation. The difference in O2 saturations is due to the fact that:

a. oxygen binds irreversibly to hemoglobin in the lungs. b. the pH of blood differs in muscle capillaries versus pulmonary capillaries. c. oxygen binds reversibly to hemoglobin with varying affinity due to cooperative binding. ANSWER: c 15. If the partial pressure of oxygen (pO2) in a man's blood plasma decreased below that in his red blood cells (RBCs), what would be the result? a. More oxygen would diffuse into his red blood cells and be transported throughout his body. b. Oxygen would diffuse out of his red blood cells, and less oxygen would be transported throughout his body. c. The concentration of oxygen would be the same in red blood cells and blood plasma. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 ANSWER: b 16. A genetic mutation causes an individual to produce hemoglobin molecules that do not change their binding affinity for oxygen, even after oxygen binds to a single heme site. How will this mutation affect the transport of oxygen through the individual's bloodstream? a. More oxygen will be transported by hemoglobin to tissues throughout the body. b. Oxygen will not be bound or released from hemoglobin as readily, limiting O2 delivery to the tissues. c. The same amount of oxygen will be transported by hemoglobin to tissues throughout the body. ANSWER: b 17. Referring to the accompanying graph, when the partial pressure of oxygen (pO2) is 38 mmHg, the % saturation of hemoglobin is expected to be:

a. 98%. b. 70%. c. 50%. d. 30%. ANSWER: b 18. In systemic tissue fluids, the enzyme carbonic anhydrase catalyzes the reaction CO2 + H2O ® H2CO3 (which then can dissociate into H+ and HCO3-). In fact, CO2 released from cells is converted to HCO3- ions and travels in that form in the bloodstream. HCO3- is reconverted to CO2 + H2O in the pulmonary capillaries by the same enzyme, and there the CO2 is exhaled. How is it possible for the same enzyme to catalyze reverse reactions? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 a. The direction of a reversible reaction is influenced by the concentrations of reactants and products. In pulmonary circulation, the low CO2 concentration favors the formation of CO2 and H2O. b. There must be two forms of carbonic anhydrase. One form catalyzes the forward reaction CO2 + H2O ® H2CO3 and the other catalyzes the reverse reaction H2CO3 ® CO2 + H2O. c. The lungs contain an allosteric inhibitor that prevents the formation of carbonic acid. ANSWER: a 19. Bisphosphoglyceric acid (BPG) is a byproduct of glycolysis released into the bloodstream when an animal's supply of oxygen is low. Like protons, it decreases hemoglobin's affinity for O2. The effect of high BPG levels would be that: a. hemoglobin picks up more O2 in the lungs than it would without BPG. b. hemoglobin holds on to more of its O2 than it would without BPG. c. hemoglobin loses its cooperative binding of oxygen. d. hemoglobin releases more of its bound O2 than it would without BPG. e. hemoglobin now binds more oxygen at low partial pressures than at high partial pressures. ANSWER: d 20. Which of the relationships best describes the partial pressure of oxygen (pO2) in the lung (alveolar air), red blood cells, and blood plasma? a. pO2 lung < pO2 blood plasma < pO2 red blood cell b. pO2 blood plasma < pO2 lung < pO2 red blood cell c. pO2 red blood cell < pO2 lung < pO2 blood plasma d. pO2 red blood cell < pO2 blood plasma < pO2 lung ANSWER: d 21. The red curve in the graph shown describes O2 saturation in normal conditions and the blue curve describes O2 saturation in someone with carbon monoxide poisoning. You can determine from this curve that carbon monoxide:

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Chapter 37

a. decreases the amount of O2 in the air. b. decreases the amount of hemoglobin. c. decreases the amount of O2 that can bind to hemoglobin. ANSWER: c 22. Given the properties of myoglobin described in the O2 saturation curve shown, what advantage does myoglobin confer to contracting muscles?

a. It pumps oxygen from hemoglobin in order to bind oxygen. b. It contributes to the dark color of flight muscle in birds. c. It increases the effectiveness of glycolytic muscles used in short bursts. d. It releases bound oxygen at lower pO2 conditions than hemoglobin does. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 e. It binds oxygen at pO2 values at which hemoglobin is releasing its bound O2, facilitating O2 delivery to muscle mitochondria. ANSWER: e 23. A mammalian fetus possesses a mutation that causes it to synthesize only adult hemoglobin. How will this mutation affect the fetus? a. The fetus will derive more oxygen from its mother's bloodstream. b. The fetus will derive less oxygen from its mother's bloodstream. c. The fetus will derive the same amount of oxygen from its mother's bloodstream as if it produced fetal hemoglobin. d. The fetus will not derive oxygen from its mother's bloodstream, as oxygen is provided by the amniotic fluid. e. The fetus will not derive oxygen from its mother's bloodstream, as the blood of the fetus does not come into contact with maternal blood. ANSWER: b 24. Much of the fluid that moves across capillary walls and into the tissues moves as a result of filtration, forced by blood pressure. O2, on the other hand, travels from the capillaries into the tissues because: a. of a concentration gradient between O2 in the tissues and O2 in the blood. b. of high CO2 concentration in the tissues, created as a by-product of cellular respiration. c. osmotic pressure causes a net flow of O2 from the tissues into the blood. d. O2 is pumped into the tissues where it is needed. ANSWER: a 25. Unlike open circulatory systems, closed circulatory systems are more efficient at delivering oxygenated blood directly to tissues according to their metabolic and energetic demands. a. true b. false ANSWER: a 26. Which of the statements explains why there is an elastic layer found in arteries but not veins? a. Valves present in veins provide a mechanism for withstanding high blood pressure flow going through veins. b. Blood pressure is higher in arteries than veins, and the elastic layer helps maintain the structure of the artery. c. Arteries are thicker than veins, and the elastic layer is necessary to support the additional weight of arteries. d. The total length of arteries in the body is more than the total length of veins, and the elastic layer in arteries helps push the blood over the longer length. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 27. Which of the types of blood vessels would have the greatest resistance to flow? a. veins b. capillaries c. venules d. arteries e. arterioles ANSWER: b 28. In the figure shown, the cell is _____ to the solution. If the membrane is permeable to water, but not to solutes, net movement of water will be _____ the cell.

a. hypertonic; into b. hypertonic; out of c. hypotonic; into d. hypotonic; out of ANSWER: d 29. Baroreceptors are pressure receptors found in the major arteries. If these receptors detect a fall in blood pressure, signals go to the: a. sympathetic neurons that synapse on smooth muscles surrounding the arterioles of the limbs, stimulating them to contract. b. parasympathetic neurons that synapse on smooth muscles surrounding the arterioles of the limbs, stimulating them to contract. c. sympathetic neurons that synapse on smooth muscles surrounding the arterioles of the limbs, stimulating them to relax. d. parasympathetic neurons that synapse on smooth muscles surrounding the arterioles of the limbs, stimulating them to relax. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 ANSWER: a 30. The figure shown diagrams osmotic (solid red arrows) and blood pressures (dashed green arrows) as a function of distance along a capillary. This illustration explains why:

a. diffusion of oxygen takes place in the capillaries. b. blood flow slows down in the capillaries. c. protein molecules do not leave the blood vessels. d. fluid enters the capillary at the arteriole end and leaves at the venule end. e. fluid leaves the capillary at the arteriole end and reenters at the venule end. ANSWER: e 31. Captopril (a drug derived from the venom of the pit viper) interferes with the production of angiotensin, which is a potent vasoconstrictor. Captopril is used clinically to treat high blood pressure. Given its antiangiotensin activity, can you explain why captopril reduces blood pressure? a. It blocks cardiac stimulation by the sympathetic nervous system. b. It causes an increase in the resistance to blood flow. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 c. It increases blood volume. d. It blocks stimulation of the cardiovascular system by the parasympathetic nervous system. e. It causes a decrease in the resistance to blood flow. ANSWER: e 32. In which animals does blood flow directly from respiratory organs to tissues without first returning to the heart? a. amphibians b. birds c. fish d. mammals e. reptiles ANSWER: c 33. The initiation of the mammalian heart contraction: a. takes place at the SA node. b. requires neural input. c. takes place at the AV node. d. takes place at the SA node and requires neural input. e. takes place at the AV node and requires neural input. ANSWER: a 34. BP = CO × PR expresses the relationship between blood pressure (BP), cardiac output (CO), and peripheral resistance (PR), which is the resistance to flow offered by systemic blood vessels. All of the changes would result in an increase in blood pressure except: a. stroke volume increase. b. heart rate increase. c. contraction of smooth muscle around the arterioles. d. parasympathetic input to the pacemaker. e. reduction in arteriole diameter. ANSWER: d 35. Consider the mammalian heart. Why is the muscular wall of the left ventricle thicker than that of the right ventricle? a. This difference stems from the fact that frogs have a thicker left ventricular muscle mass. b. The left ventricle must contract with more force in order to send blood to the body's extremities. c. The left ventricle has more nerve endings for contraction, and more muscle mass must be there to accommodate the extra neurons. d. The ventricle wall must be thicker to inhibit diffusion between the right and left ventricle. ANSWER: b 36. Which of the statements describes a similarity between fish and amphibian hearts? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 a. Hearts in both groups function to pump blood to the respiratory surface. b. Hearts in both groups contain mixed oxygenated and deoxygenated blood. c. Hearts in both groups are a bridge between the systemic and pulmonary circulatory pathways. d. Hearts in both groups have two atria. ANSWER: a 37. Mammals and birds have four-chambered hearts and fully divided pulmonary and systemic circulations. The circulatory systems of reptiles and amphibians are sometimes presented as "inferior" because pulmonary and systemic circulations are not completely separate. What might be an advantage to incompletely separated circuits in these organisms? a. Incompletely separated circuits take less energy to function. b. In incompletely separated circuits, the distribution of blood flow can be adjusted, depending on whether or not the animal is breathing air through its lungs. c. Incompletely separated circuits allow for higher blood pressure and therefore increased blood flow. ANSWER: b 38. What would happen to a mammalian heart if the nerves leading to it were severed? a. The heart muscle would stop contracting. b. The heart muscle would contract randomly. c. The heart muscle would continue to contract but its contraction rate may change. d. Only the atria would be able to contract. e. Systolic contraction would still occur, but diastolic relaxation could not occur. ANSWER: c 39. A veterinarian receives a heart to dissect, but he has no information regarding the origin of the heart. He notices that the heart has three chambers. What can he deduce? a. The heart was taken from a fish. b. The heart was taken from an amphibian or reptile. c. The heart was taken from a bird. d. The heart was taken from a mammal. e. The heart was taken from a fish or a reptile. ANSWER: b 40. Cells in which regions of the heart produce action potentials, thus orchestrating a "regular" heartbeat? a. aortic valve b. atrioventricular valve c. atrioventricular node d. pulmonary artery e. sinoatrial node ANSWER: e 41. In the heart of a fish, where would you expect to find oxygenated blood? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 a. in the right atrium b. in the left atrium c. in the ventricle d. in a single atrium (fish only have one). e. Nowhere; oxygenated blood does not pass through a fish's heart. ANSWER: e 42. The random thermal motion of molecules results in movement from a region of higher concentration to a region of lower concentration, and this is called diffusion. The rate of diffusion across a barrier is defined by Fick's law of diffusion: Rate of diffusion = k × A × (C2 – C1) L k = diffusion coefficient, which depends on solubility and temperature A = surface area for exchange C2 – C1 = difference in partial pressure of gas on either side of the barrier L = thickness of the barrier to diffusion Consider the structure and function of respiratory membranes. Respiratory membranes 1 and 2 represent membrane barriers between the outside medium and the inside of the organism.

Which would have the higher rate of diffusion, respiratory membrane 1 or respiratory membrane 2? a. respiratory membrane 1 b. respiratory membrane 2 ANSWER: b 43. The random thermal motion of molecules results in movement from a region of higher concentration to a region of lower concentration, and this is called diffusion. The rate of diffusion across a barrier is defined by Fick's law of diffusion: Rate of diffusion = k × A × (C2 – C1) L k = diffusion coefficient, which depends on solubility and temperature Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 A = surface area for exchange C2 – C1 = difference in partial pressure of gas on either side of the barrier L = thickness of the barrier to diffusion Consider the structure and function of respiratory membranes. Respiratory membranes 1 and 2 represent membrane barriers between the outside medium and the inside of the organism.

Which term in the equation is different between respiratory membranes 1 and 2? a. k b. A c. C2 – C1 d. L ANSWER: b 44. Referring to the figure shown and parts A and B, which would have the faster rate of diffusion of O2 molecules (blue hexagons) across the semipermeable membrane (vertical purple line)?

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Chapter 37

a. A b. B ANSWER: a 45. The random thermal motion of molecules results in movement from a region of higher concentration to a region of lower concentration, and this is called diffusion. The rate of diffusion across a barrier is defined by Fick's law of diffusion: Rate of diffusion = k × A × (C2 – C1) L k = diffusion coefficient, which depends on solubility and temperature A = surface area for exchange C2 – C1 = difference in partial pressure of gas on either side of the barrier L = thickness of the barrier to diffusion

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Chapter 37

Which term in the equation is different between figures A and B? a. k b. A c. C2 – C1 d. L ANSWER: c 46. Carotid bodies are important sensors used in the homeostatic regulation of breathing. Indicate whether the statement is true or false with respect to alterations in breathing. A decrease in arterial blood pH sensed in carotid bodies will result in increased blood O2 levels. a. true b. false ANSWER: a 47. Carotid bodies are important sensors used in the homeostatic regulation of breathing. Indicate whether the statement is true or false with respect to alterations in breathing. An increase in the animal's activity will result in stimulation of chemoreceptors in the carotid bodies. a. true b. false ANSWER: a 48. A decrease in arterial CO2 levels sensed in carotid bodies will result in increased blood O2 levels. a. true b. false ANSWER: b 49. Consider the figure shown of an electrocardiogram (EKG) trace and the typical action potential of cells in Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 the left ventricle.

Indicate whether the statement is true or false. Ventricular diastole occurs at the interval labeled 2 on the action potential figure. a. true b. false ANSWER: a 50. Consider the figure shown of an electrocardiogram (EKG) trace and the typical action potential of cells in the left ventricle.

Indicate whether the statement is true or false. Atria are emptying at the interval labeled 4 on the action potential figure. a. true b. false ANSWER: a Multiple Response Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 51. Excluding movement of blood between the lungs and the heart, which of the vessels carry CO2-rich (and oxygen-poor) blood back to the heart? Select all that apply. a. venules b. arterioles c. veins d. arteries ANSWER: a, c 52. We breathe more quickly and deeply when we exercise. Why does this make sense? Select all that apply. a. Because when we exercise, we consume more CO2 and must increase ventilation to supply the CO2. b. Because when we exercise, we use more ATP, and additional O2 is necessary to generate sufficient ATP. c. Because when we exercise, we hydrolyze more ATP to ADP and Pi, and O2 is necessary for the hydrolysis, so we increase our intake of oxygen. d. Because when we exercise, we produce more CO2 and increased ventilation is necessary to rid ourselves of CO2. ANSWER: b, d 53. The respiratory and cardiovascular systems of vertebrate animals work in tandem to transport oxygen to metabolizing tissues and to eliminate carbon dioxide as a waste gas. Which of the statements accurately describes how these systems interact for gas exchange? Select all that apply. a. The rate of oxygen transport by diffusion across the lungs must generally match the rate of transport by the circulatory system. b. The rate of gas exchange by diffusion need not match rates of ventilation or circulatory transport. c. Bulk flow of gases and solutes works well over long distances, whereas diffusion is effective only over very short distances. d. Bulk flow of gases and solutes works well over short distances, whereas diffusion is effective over longer distances. ANSWER: a, c 54. Simple multicellular organisms that are small can meet their gas exchange needs by means of: Select all that apply. a. a flattened body shape that increases their surface area. b. a flattened or thin body shape that reduces the distance between cells inside their body and the external environment. c. by living in air that provides abundant oxygen, even though they are at risk of losing water and drying out. d. by relying on anaerobic respiration (glycolysis) for long time periods, so that they don't need to use oxygen for aerobic respiration. ANSWER: a, b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 55. Which of the statements is true? Select all that apply. a. The volume of blood flow in the arteries and arterioles is greater than the volume of blood flow through the capillaries. b. Blood travels at a higher velocity in arteries than in veins. c. Capillaries have a higher resistance to flow than either arteries or veins. d. Blood pressure is higher in arteries than in veins. e. The total cross-sectional area of the capillaries is greater than the total cross-sectional area of the venules and veins. ANSWER: b, c, d, e 56. One of the functions of the respiratory system is to rid the body of CO2. What is the specific biochemical pathway that is responsible for CO2 production? Select all that apply. a. electron transport chain b. the citric acid cycle c. oxidation of pyruvate to acetyl co-A d. glycolysis ANSWER: b, c 57. What are desirable characteristics for a gas exchange surface, such as the endothelial cells lining the inside of a lung? Select all that apply. a. a large surface area b. a small surface area c. a thickness of 100 micrometers d. a thickness under 10 micrometers ANSWER: a, d 58. During certain stressful moments, some individuals will begin to hyperventilate—that is, their breathing will be very shallow and quick. A by-product of hyperventilation is much lower levels of CO2 in the blood. Which effect would you also expect to find? Select all that apply. a. decreased blood pH, less than 7.2. b. increased blood pH, greater than 7.2 c. decreased stimulation of chemoreceptors in the brainstem d. increased stimulation of chemoreceptors in the brainstem ANSWER: b, c 59. If water passed over the gills of fish in the same direction as blood flows through the capillaries in lamellae, how would gas exchange be affected? Select all that apply. a. More oxygen would diffuse from the water into the fish's bloodstream. b. Less oxygen would diffuse from the water into the fish's bloodstream. c. More carbon dioxide would diffuse out of the fish's bloodstream into the water. d. Less carbon dioxide would diffuse out of the fish's bloodstream into the water. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 e. Gas exchange would remain the same, regardless of the direction of water flow. ANSWER: b, d 60. Which of the statements is true regarding respiration in birds? Select all that apply. a. Bird lungs inflate due to negative pressure in the chest cavity. b. Bird lungs deflate due to negative pressure in the chest cavity. c. Oxygen enters the bloodstream via crosscurrent exchange. d. Oxygen enters the bloodstream via countercurrent exchange. e. Respiration depends on the movement of air through air sacs. f. Respiration depends on the movement of air through spiracles. ANSWER: c, e 61. In contrast to vertebrates, insects lack interacting respiratory and circulatory systems. Accordingly, which of the statements are true? Select all that apply. a. Insects can breathe by means of gills or trachea that supply oxygen directly to cells in their body. b. Insects breathe by means of trachea that supply oxygen directly to cells in their body. c. Insect circulatory systems are open and therefore do not directly link gas transport by trachea with supply of nutrients and other solutes to metabolizing tissues. d. Insect circulatory systems benefit from a discrete network of vessels, but these do not connect with the trachea that supply their cells with oxygen. e. Because insect respiration and circulation are not linked, insects do not regulate their rates of gas exchange to match cellular respiration. f. Even though insect respiration and circulation are not linked, insects do regulate their rates of gas exchange to match cellular respiration. ANSWER: b, c, f 62. Which of the choices presents a challenge in gas exchange for fish but not for terrestrial animals? Select all that apply. a. Water is more dense and more viscous than air. b. Water contains much less O2 than air per unit volume. c. Gills have less surface area than lungs. d. Unidirectional flow of water is less efficient than tidal ventilation. e. Diffusion occurs more slowly in water than in air. ANSWER: a, b, e 63. Which of the statements is true regarding respiration in mammals? Select all that apply. a. Air moves in a unidirectional fashion through the lungs. b. Gas exchange occurs in the bronchioles. c. Gas exchange occurs in the alveoli. d. Respiration depends on changes in air pressure in the chest cavity. e. Respiration depends on contractions of the diaphragm. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 ANSWER: c, d, e 64. Which of the statements is true regarding respiration in fish? Select all that apply. a. Gas exchange occurs at the alveoli. b. Gas exchange occurs at the gill lamellae. c. Oxygen enters the bloodstream via countercurrent exchange. d. Oxygen enters the bloodstream via crosscurrent exchange. e. Oxygen moves through spiracles. f. Oxygen moves through bronchioles. g. CO2 leaves the bloodstream via countercurrent exchange. ANSWER: b, c, g 65. Which of these factors would affect the amount of oxygen released from hemoglobin in the tissues? Select all that apply. a. a decrease in the acidity or pH of blood b. an increase in CO2 in the bloodstream c. the formation of lactic acid by muscles d. an increase in aerobic respiration e. a decrease in pO2 in the lungs ANSWER: b, c, d 66. In a mammal's bloodstream, where would you expect to find oxygen? Select all that apply. a. dissolved in blood plasma b. attached to the heme group of myoglobin in red blood cells c. attached to the heme groups of hemoglobin in red blood cells d. attached to the heme group of myosin in red blood cells e. dissolved in hemolymph ANSWER: a, c 67. In what organisms would you find oxygen bound to hemoglobin? Select all that apply. a. earthworms b. horses c. spiders d. sponges e. frogs ANSWER: a, b, c, e 68. A lioness begins her nightly hunt after spending the day beneath a tree in the Serengeti. What happens to the blood vessels in her body as she goes from resting to hunting? Select all that apply. a. Arterioles supplying her stomach and intestines constrict. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 b. Arterioles supplying her stomach and intestines expand. c. Arterioles supplying her limb muscles constrict. d. Arterioles supplying her limb muscles expand. ANSWER: a, d 69. Imagine that during her hunt, a lioness wounds a gazelle. The gazelle is able to escape, but its blood pressure is rapidly decreasing as it loses blood. What can happen in the gazelle's body to increase blood pressure? Select all that apply. a. The pituitary gland can secrete vasopressin. b. The sympathetic nervous system can signal specific arterioles to constrict. c. The sympathetic nervous system can signal specific arterioles to expand. d. The pituitary gland can secrete thyroid stimulating hormone (TSH). ANSWER: a, b 70. An aneurysm, an enlargement of the artery due to breakdown of the arterial wall, is likely the result of which changes? Select all that apply. a. the deterioration of elastin and collagen in arteries b. the deterioration of elastin and collagen in veins c. pressure pulses in arteries d. pressure pulses in veins ANSWER: a, c 71. As fluid moves through a vessel, which factors determine the flow resistance? Select all that apply. a. the viscosity of the fluid b. the length of the vessel c. the radius of the vessel d. the pressure exerted by the heart ANSWER: a, b, c 72. Which statements are true regarding lymph and the lymphatic system? Select all that apply. a. Lymph is rich in proteins filtered out of the bloodstream. b. Lymph contains water filtered out of the bloodstream. c. The lymphatic system is composed solely of the same vessels and arteries forming the circulatory system. d. The lymphatic system is composed of its own distinct set of vessels, through which blood does not travel. ANSWER: b, d 73. Cardiac output depends on or is influenced by which factors? Select all that apply. a. heart rate b. the amount of adrenaline in the blood c. stroke volume Copyright Macmillan Learning. Powered by Cognero.

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Chapter 37 d. the amount of blood entering the heart e. depolarization of the SA node ANSWER: a, b, c, d 74. Muscular hearts serve as pumps to push blood or hemolymph through an animal's circulatory system. In doing so, hearts have which properties? Select all that apply. a. Hearts can pump blood back and forth depending on the needs of the animal for breathing air or water. b. Hearts have one-way valves that ensure that blood flows in only one direction. c. Hearts have specialized cardiac muscle cells in electrical contact, which ensures that they contract in synchrony. d. Hearts display "autorhythmicity," or the ability to contract cyclically on their own. e. Hearts depend on regular input via action potentials from the nervous system to contract and pump blood. f. Hearts have to pump harder, generating greater pressure, when the smooth muscle surrounding arterioles contracts to reduce their diameter. g. Hearts pump more easily, with less pressure, when the smooth muscle surrounding arterioles contracts to reduce their diameter. ANSWER: b, c, d, f 75. In the heart of a reptile or amphibian, where would you expect to find oxygenated blood? Select all that apply. a. in the right atrium b. in the left atrium c. mixed with deoxygenated blood in a common ventricle d. mixed with deoxygenated blood in the right ventricle e. Nowhere; oxygenated blood does not travel through the heart of these animals. ANSWER: b, c

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Chapter 38 Multiple Choice 1. Refer to the figure shown. If it takes 160 grams of sugar to grow 10 grams of bacteria anaerobically, how many grams of sugar would be required to grow 10 grams of bacteria aerobically?

a. 160 grams: there should be no difference in the amount of sugar required b. 80 grams: aerobic respiration is 2 times more efficient than anaerobic respiration c. 10 grams: aerobic respiration is 16 times more efficient than anaerobic respiration d. 5 grams: aerobic respiration is 32 times more efficient than anaerobic respiration e. 2 grams: aerobic respiration is 80 times more efficient than anaerobic respiration ANSWER: c 2. Refer to the figure shown. Animals rely mainly on which of the molecules for long-term energy supply?

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a. glycogen b. glucose c. fats d. proteins e. glycogen and fats ANSWER: c 3. Which of the statements about metabolic rate is false? a. The metabolic rate per gram of tissue is greater in large endotherms than in small endotherms. b. The metabolic rate per gram of tissue is greater in endotherms than in ectotherms. c. The metabolic rate per gram of tissue in ectotherms is dependent on external temperature. d. Metabolic rate may be measured by determining the rate of oxygen consumption. e. An animal's overall metabolic rate increases with the animal's mass. ANSWER: a 4. Which of the statements best describes how cheetahs are adapted to deal with the heat generated by their extreme sprint speeds? a. They can store up to 60 times the heat generated at rest, but only for short periods. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 b. They can store up to 60 times the heat generated at rest for use in long chases. c. They can store about twice the heat generated at rest for use in long chases. d. They can store up to 100 times the heat generated at rest, but only for short periods. e. They can store about 20 times the heat generated at rest for use in long chases. ANSWER: a 5. Reactions that break down food sources to fuel the energy needs of a cell are called anabolic, whereas those that result in net energy storage are called catabolic. a. true b. false ANSWER: b 6. Assuming that aerobic metabolism starts with glucose as its fuel, at least _____% of the energy of the starting molecules will be captured in a useful form. The remaining energy will be converted to _____ and will be either lost or used to _____. a. 34; heat; warm the animal b. 33; electricity; keep the nervous system functioning c. 66; heat; warm the animal d. 75; bond energy; carry out cellular respiration e. 42; heat; carry out cellular respiration ANSWER: a 7. Endotherms usually, though not always, maintain a _____ body temperature that is _____ that of the environment. a. constant; higher than b. constant; lower than c. constant; identical to d. fluctuating; higher than e. fluctuating; identical to ANSWER: a 8. Most proteins consumed in the diet: a. are absorbed and maintained as functional proteins if the body needs that particular protein. b. are broken down into amino acids and absorbed, and may be re-polymerized into the proteins the body needs. c. are digested into nucleic acids and absorbed, and may be fed into cellular respiration pathways. d. are digested into amino acids and then enter into cellular respiration pathways to be reduced so that they provide energy to the organism. ANSWER: b 9. An amino acid that cannot be synthesized by cellular biochemical pathways, and therefore must be obtained from the diet, is called a(n) _____ amino acid. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 a. necessary b. essential c. important d. extra e. critical ANSWER: b 10. Suction feeding is an important adaptation that is seen in many aquatic organisms with jaws. The figure shown plots mandible (lower jaw bone) length versus width in several species of whales.

In the study from which this figure was taken, researchers were testing the hypothesis that species of whales with short mandible lengths are more often likely to capture prey with their jaws instead of using suction feeding. Filled symbols represent suction feeders. These data support the hypothesis of the authors. a. true b. false ANSWER: b 11. The most reliable food source containing all of the essential amino acids for humans is: a. meat. b. beans. c. corn. d. sunflower seeds. e. flavored corn chips. ANSWER: a 12. Chemical elements other than carbon, hydrogen, oxygen, and nitrogen that are required in the diet and must be obtained from the food an animal eats are otherwise known as: a. vitamins. b. essential amino acids. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 c. non-essential amino acids. d. minerals. e. fats. ANSWER: d 13. The jaws of vertebrates are thought to have been derived by the modification of which of the structures from jawless fishes? a. scales b. the vertebrae c. the notochord d. cartilage that supported the gills e. bones in the skull ANSWER: d 14. _____ is the most common form of food capture by aquatic animals. a. Suspension filter feeding b. Suction feeding c. Active swimming d. "Sit-and-wait" feeding e. None of the other answer options is correct. ANSWER: a 15. In the figure shown, digestion in the small intestine is aided by secretions from the liver, gallbladder, and pancreas.

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Secretin, a hormone produced by cells lining the duodenum, stimulates the pancreas to release bicarbonate ions. a. true b. false ANSWER: a 16. In the figure shown, carbohydrates, fats, and proteins in the duodenum are digested by enzymes produced by the liver.

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a. true b. false ANSWER: b 17. Refer to the figure shown. If you found a mammal skull that had large, pointed canines and relatively narrow, blade-shaped molars and premolars, you could infer that it came from an animal specialized for feeding on:

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a. primarily meat. b. primarily tough plant material. c. primarily soft plant material, such as fruit. d. a combination of meat and tough plant material. e. a combination of meat, soft plant material such as fruit, and tough plant material. ANSWER: a 18. Glucose enters the epithelial cells of the small intestine against its concentration gradient. Select the best explanation from the statements. a. Glucose is transported through a glucose pump in the apical membrane that hydrolyzes ATP to ADP and Pi. b. Glucose is co-transported with Na+, which moves down its concentration gradient into the cell. c. Glucose is co-transported in the Na+/K+ pump. d. Glucose follows the water that is pumped into the cell. e. Glucose enters by diffusion. ANSWER: b 19. Lactose is a disaccharide. For animals to use lactose as an energy source, it is: a. hydrolyzed into monosaccharides like glucose, which are transported into intestinal epithelial cells and then into the blood to be distributed to cells around the body. b. absorbed directly into the intestinal epithelial cells, transported across the internal space of those cells, and then transported into the blood to be distributed and fed into the glycolysis pathway. c. polymerized in the intestine, then absorbed by intestinal epithelial cells and transported via the blood to the liver, where it is stored and used as needed. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 20. Aside from whatever digestion has happened before the food reaches the small intestine, the bulk of digestion takes place in the: a. jejunum. b. ileum. c. colon. d. appendix. e. duodenum. ANSWER: e 21. The pancreas secretes substances into ducts that connect to the _____ of the digestive system. a. stomach b. duodenum c. jejunum d. ileum e. colon ANSWER: b 22. Which layer of the digestive tract secretes enzymes and absorbs nutrients? a. lumen b. submucosa c. longitudinal smooth muscles d. serosa e. mucosa ANSWER: e 23. In muscle and liver tissue, glucose consumed in the diet, but not immediately used, is stored as: a. fatty acids. b. fats. c. ATP. d. glycogen. ANSWER: d 24. You jogged across campus and arrived at your class on time. As you walk to your seat, you notice that you are still breathing more heavily than usual. This is because you are experiencing the recovery metabolism necessary to reestablish the resting metabolic rate of your cells. a. true b. false ANSWER: a 25. Protists and some other unicellular animals use ___ digestion. a. intracellular Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 b. extracellular c. both intra- and extracellular d. neither intra- nor extracellular ANSWER: a 26. Most animals rely on _____, in which food is isolated and broken down in a body compartment rather than inside individual cells. a. intracellular digestion b. extracellular digestion c. both intra- and extracellular digestion d. neither intra- nor extracellular digestion ANSWER: b 27. The gizzard carries out the same function in birds, alligators, and earthworms as does the ___ in insects. a. esophagus b. stomach c. mandible d. small intestine e. large intestine ANSWER: c 28. Why does it make sense to include temperature regulation in the same chapter as nutrition and digestion? a. The rate of energy consumption is, in part, a function of body temperature. b. Ectotherms receive most of their body heat from surroundings. c. Ectothermy and endothermy are extremes of a continuum of thermoregulatory mechanisms. d. Thermoregulation in ectotherms is energetically costly. ANSWER: a 29. _____ reactions result in net energy storage within an organism's cells. a. Catabolic b. Anabolic c. Anaerobic d. Aerobic ANSWER: b 30. An animal's metabolic rate can be affected by its: a. activity level. b. body size. c. body temperature. d. All of these choices are correct. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 31. An individual in positive energy balance takes in more energy than he or she expends. In positive energy balance, the highest-energy form of energy storage is: a. phosphocreatine in muscle tissue. b. glucose in the blood and cells. c. fat in adipose tissue. d. glycogen in the liver. ANSWER: c 32. You realize that you are late for class and begin to run across campus, which is equivalent to 15 minutes of very hard exercise. After your muscles use their stores of phosphocreatine, they will next provide themselves with energy using: a. oxidative phosphorylation. b. aerobic glycolysis. c. lactic acid fermentation. d. aerobic respiration. e. oxidative fermentation. ANSWER: c 33. An elephant has a higher total resting metabolic rate than a mouse. a. true b. false ANSWER: a 34. Endotherms have higher metabolic rates than ectotherms and are therefore able to be active over a narrower range of environmental temperatures. a. true b. false ANSWER: b 35. Essential amino acids: a. cannot be synthesized biochemically and must be taken in through the diet. b. can only be found in plants. c. include minerals such as calcium and iron. d. provide essential vitamins. ANSWER: a 36. Most vegetarians have to watch their diet closely and eat certain foods in combination, like lentils and corn, to obtain all of their essential amino acids. Which of the statements is the best explanation of why? a. Lentils are a food high in protein, therefore eating lentils is a good alternative to eating meat. b. Eating meat provides all eight essential amino acids; because vegetarians do not eat meat, they must combine foods in a way that provides those essential amino acids. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 c. Lentils and corn provide precursors to the eight essential amino acids, and when they are eaten together, the body can produce all of the essential amino acids. d. Eating meat only provides some of the essential amino acids, so it is not a problem that vegetarians do not eat meat. ANSWER: b 37. Some people who contract malaria remain weak during recovery and cannot exert themselves for any length of time, because the parasitic infection has caused the rupture of a large percentage of blood cells. One suggestion is for these people to eat rare steak while in recovery. Why? a. Eating rare steak helps increase the iron content in their bodies, which can be used in forming new red blood cells. b. Eating rare steak supplies all eight essential amino acids, which are important for building new muscle. c. The steak is high in protein and provides many calories without too much fat. d. Digesting the protein in the steak takes time, and therefore, energy is provided slowly in the hours after the meal. ANSWER: a 38. Humans are capable of synthesizing 12 of the 20 naturally occurring amino acids without having to obtain them from their diets. a. true b. false ANSWER: a 39. Tadpoles of the African genus of frogs Hymenochirus are predatory. They track their prey, chase it, and then capture it using a suction mechanism that draws the prey into their mouth. Although suction feeding is the most common method of prey capture among fish, the closest phylogenetic relatives of these frogs all have tadpoles that are suspension filter feeders. Suction feeding in Hymenochirus tadpoles and fish is an example of: a. an adaptive radiation. b. convergent evolution. c. homology. d. synapomorphies. ANSWER: b 40. Suspension filter feeding, the most common form of food capture, is: a. possible only in aquatic environments. b. an example of convergent evolution. c. used by scallops, oysters, and baleen whales. d. All of these choices are correct. ANSWER: d 41. Among the multiple types of teeth, _____ are used predominantly for crushing and grinding. a. incisors Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 b. canines c. molars and premolars d. wisdom teeth ANSWER: c 42. An animal's foregut includes the: a. large intestine. b. small intestine. c. esophagus. d. All these choices are correct. ANSWER: c 43. Which part of the digestive tract is the predominant location for water and mineral absorption? a. foregut b. midgut c. hindgut d. crop ANSWER: c 44. Which chemical or enzyme is secreted by the stomach when an organism is eating? a. hydrochloric acid (HCl) b. pepsin c. gastrin d. All of these choices are correct. ANSWER: d 45. The _____ is the first section of the small intestine, where food enters from the stomach. a. duodenum b. jejunum c. ileum d. gizzard ANSWER: a 46. Since many herbivores or ruminants lack cellulase to break down cellulose, how do they digest plants? a. They have gut bacteria that produce cellulase. b. They avoid food containing cellulose. c. They lick rocks to obtain essential minerals. d. They have a larger liver to compensate. ANSWER: a 47. The colon absorbs water from food; without the colon, water from drinking must compensate for the loss of Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 water from food. a. true b. false ANSWER: a 48. Which layer of the digestive tract covers and protects the gut? a. lumen b. mucosa c. submucosa d. longitudinal muscles e. serosa ANSWER: e 49. Horses are large herbivores that lack the four-chambered stomach of foregut fermenters, such as cows. You would, therefore, expect the equine (horse) gut to have an enlarged _____, relative to omnivores of similar size. a. stomach b. small intestine c. liver d. pancreas e. cecum ANSWER: e 50. The digestive tract is well-adapted for digestion and absorption. For most nutrients, once material is in the digestive tract, the digestive system does not vary its rate of absorption according to body needs. a. This is an example of a homeostatic positive feedback mechanism. b. This is an example of a homeostatic negative feedback mechanism. c. This response is not homeostatic. ANSWER: c 51. The digestive tracts of cows and cats are structurally different. Cows have a fermentation chamber in their foregut that houses bacteria. Why might this anatomical structure benefit cows? a. Cows are larger than cats; therefore, the fermentation chamber is necessary because of the size difference. b. Cows are herbivores. The fermentation chamber houses bacteria that produce cellulase and make more of the nutrients in the cows' food available for absorption. c. Cats are omnivores, and the bacteria that live in the cecum increase the amount of nutrition that cats can absorb from the plant material they eat. d. Cows are herbivorous and need the fermentation chamber to acquire the extra ATP from anaerobic respiration that is carried out there by the bacteria. ANSWER: b 52. Metabolic rate increases with animal mass raised to the 3/4 power. If, instead, metabolic rate scaled Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 proportionally to mass, then an animal twice the mass of another animal would have a metabolic rate: a. more than twice that of the other animal. b. less than twice that of the other animal. c. twice that of the other animal. d. equal to the other animal. e. four times that of the other animal. ANSWER: c 53. The approach most commonly used to measure metabolic rate is to measure the rate of O2 consumption. Although there are small errors inherent in this method, it is easy and provides a good estimate of metabolic rate. Why is the rate of oxygen consumption an appropriate measure of metabolic rate? a. Metabolic rate is the amount of energy used per unit time. The production of usable energy in the form of ATP depends largely on aerobic respiration. b. Metabolic rate is the amount of energy used per unit time and can be measured by the rate of hydrolysis of ATP to ADP and Pi. c. Metabolic rate is the amount of energy released in anabolic processes, and anabolic processes are all aerobic. ANSWER: a 54. Given what you have learned about metabolic rates in endotherms and ectotherms, which of the statements is true? a. Endotherms can be active for much longer periods than ectotherms. b. Ectotherms must forage longer to acquire enough food to keep warm and be active, whereas endotherms can survive longer periods without food. c. Ectotherms have an overall higher metabolic rate than endotherms. d. Endotherms require less food than ectotherms of the same size. ANSWER: a 55. Both excess carbohydrates and proteins consumed in the diet during a net positive energy balance are stored as adipose tissue. a. true b. false ANSWER: a 56. You are literally running late to class. After your initial panic, you realize that you can make it on time at your regular jogging pace. You have been jogging for a few minutes and have a few minutes left before you arrive. At this point, how are your muscles providing themselves with energy? a. anaerobic glycolysis b. aerobic glycolysis c. lactic acid fermentation d. aerobic respiration e. oxidative fermentation Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 ANSWER: d 57. Essential amino acids must be supplied in the diet because: a. only essential amino acids can be oxidized when necessary for ATP synthesis and cellular respiration. b. they are necessary for protein biosynthesis, but we do not have the biochemical pathways to synthesize those amino acids. c. proteins cannot be synthesized without amino acids and, therefore, all amino acids are essential. ANSWER: b 58. The term "empty calorie" is sometimes used in dietary terminology to denote food lacking in vitamins, minerals, dietary fiber, etc. Which of the statements is accurate? a. This terminology makes sense because a calorie unaccompanied by vitamins is not useful as a source of energy. b. This terminology makes sense because a calorie may be empty or full, depending on whether it comes from a carbohydrate, protein, or fat. c. This terminology is misleading because a calorie is a measure of energy, and a calorie has a defined energy content. d. This terminology is misleading because a calorie is the amount of energy necessary to raise the temperature of 1 gram of water by 1°C, which is irrelevant to living things. ANSWER: c 59. The jaws of vertebrates are thought to have evolved from the modification of: a. the tooth-like structures found in lampreys. b. folds of skin at the anterior region of jawless ancestors. c. the cartilage that supports the fins in ancestral cartilaginous fishes (sharks). d. the cartilage that supports the gills in jawless fishes. ANSWER: d 60. One of the effects of the hormone secretin is to stimulate the release of bicarbonate ions into the duodenum, which neutralizes the acid that enters the duodenum with food from the stomach. One consequence of neutralizing the pH in the intestine is that: a. carbohydrates are not as easily broken down in this neutral environment. b. enzymes that catalyze the hydrolysis of carbohydrates are active. c. enzymes that catalyze the hydrolysis of carbohydrates are denatured. d. the enzymatically catalyzed hydrolysis of carbohydrates that began in the stomach is halted in the duodenum. ANSWER: b 61. Rabbits produce two types of droppings: fecal pellets, which are waste products of digestion, and cecotropes, which they re-ingest. Ruminants do not re-ingest any droppings. Why is re-ingestion required in rabbits but not in ruminants? a. Ruminants use microorganisms to digest cellulose and release cellular contents into the stomach to Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 be digested; rabbits secrete cellulose-digesting enzymes into their hindgut so products of digestion can be absorbed. b. Rabbits use microorganisms in their hindgut to digest cellulose; ruminants secrete cellulose-digesting enzymes into their stomachs. c. Rabbits are hindgut fermenters; therefore, nutrients would be lost if the products of fermentation in the cecum were not eaten. Ruminants are foregut fermenters, and the products of digestion are absorbed in the small intestine. d. Rabbits are hindgut fermenters and consume a carnivorous diet; ruminants are foregut fermenters and have a cellulose-based diet. The digestive systems of rabbits and ruminants are adaptations to these two dietary specializations. ANSWER: c 62. The duct that connects the pancreas to the duodenum can sometimes become blocked. What is the result? a. The digestive enzymes produced by the pancreas still enter the duodenum because they travel by the blood to reach the duodenum, not by a duct. b. The digestive enzymes produced by the pancreas enter the large intestine instead, and digestion proceeds as normal. c. The digestive enzymes back up into the pancreas and may start to digest the pancreas. d. The digestive enzymes are not affected because they are stored in the gallbladder until they are needed in the duodenum. e. Nothing happens, because the pancreas does not produce digestive enzymes. ANSWER: c 63. Food poisoning can sometimes result from ingestion of the bacteria Salmonella in meats and eggs. When Salmonella is ingested, it multiplies in the lumen of the gut, where it can lead to inflammation of the gut lining and symptoms like diarrhea and abdominal cramps. Occasionally, the bacteria enter the bloodstream. The route the bacteria take to get to the bloodstream is: a. lumen, mucosa, submucosa, inner circular muscle layer, outer longitudinal muscle layer, serosa. b. serosa, outer longitudinal muscle layer, inner circular muscle layer, submucosa, mucosa, lumen. c. lumen, mucosa. d. lumen, mucosa, submucosa. e. lumen, mucosa, submucosa, serosa. ANSWER: d 64. Which of the choices is not a specialized function of one or more regions of the digestive tract? a. food storage b. aerobic metabolism c. chemical breakdown d. absorption of nutrients e. elimination of waste products ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 65. Most nutrient absorption takes place in the: a. small intestine. b. large intestine. c. stomach. d. pancreas. e. liver. ANSWER: a 66. Because most herbivores do not produce cellulase, they have specialized compartments in their digestive tracts that: a. provide appropriate temperature and pH for breaking down plant material. b. house large populations of bacteria that do produce cellulase. c. grind up the plant material using muscular gizzards or other structures. d. secrete alternative enzymes that carry out the same function as cellulase. e. ferment nutrients in the hindgut. ANSWER: b 67. Referring to the figure shown, the movement of glucose is from the lumen of the duodenum through an interstitial cell, out to the extracellular fluid, and into a capillary. What would happen if the transporters on the interstitial cell were reversed so that the Na+/glucose cotransporters faced the capillary in the extracellular fluid, and the glucose transport proteins faced the lumen?

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a. Nothing would differ: the movement of glucose would be the same as before the reversal. b. Glucose would move from the extracellular fluid into the interstitial cell and into the lumen of the duodenum. c. Glucose would not be able to move into the interstitial cells from the lumen of the duodenum. ANSWER: b 68. Referring to the figure shown, the movement of glucose is from the lumen of the duodenum through an interstitial cell, out to the extracellular fluid, and into a capillary. What would happen if the action of the Na+– K+ pumps was reduced?

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a. Nothing would differ: the movement of glucose would be the same as before the reversal. b. The rate of flow of glucose into the interstitial cells would be reduced. c. The rate of flow of glucose into the interstitial cells would be increased. d. The direction of flow of glucose would be reversed. ANSWER: b 69. Which of the molecules has higher potential energy?

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a. molecule A b. molecule B ANSWER: b

70.

Molecule B has more potential energy than molecule A because: a. molecule B is an acid, and molecule A is an aldehyde. b. the carbon atoms in molecule B are more reduced than the carbon atoms in molecule A. c. electrons shared with O have higher potential energy than electrons shared with H. d. molecule A is hydrophilic, and molecule B is hydrophobic. ANSWER: b 71. The microvilli of the cells lining the small intestine secrete enzymes that break disaccharides (such as lactose) into their subunits, which can be absorbed. More than 65% of adults in the world stop producing the enzyme lactase after early childhood and are thus lactose intolerant. Those who do continue to digest lactose in Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 adulthood are termed lactose tolerant, and the gene continues to be expressed into adulthood. Lactose tolerance originally arose by mutation and is associated with one of several autosomal dominant alleles. It is likely that the mutations conferring lactose tolerance are: a. mutations in regulatory DNA associated with the lactase-coding gene. b. mutations within the protein-coding sequence of the gene. c. gene duplications. ANSWER: a 72. The microvilli of the cells lining the small intestine secrete enzymes that break disaccharides (such as lactose) into their subunits, which can be absorbed. More than 65% of adults in the world stop producing the enzyme lactase after early childhood and are thus lactose intolerant. Those who do continue to digest lactose in adulthood are termed lactose tolerant, and the gene continues to be expressed into adulthood. Lactose tolerance originally arose by mutation and is associated with one of several autosomal dominant alleles. In certain populations where domesticated animals supplied milk, such as in northern Europe and east Africa, adults have the ability to digest lactose. This suggests that: a. the mutation that allowed continued production of lactase in adults conferred a selective advantage in those cultures. b. when some individuals saw the others able to drink milk, they realized the advantage and made the change also. c. drinking milk into adulthood causes a genetic change in a person's lactase gene. d. because the mutation conferring lactose tolerance is dominant, it is, by definition, the most common allele. ANSWER: a 73. It is final exam week and you are feeling stressed. You decide to eat a candy bar and drink a big bottle (32 ounces) of soda in the hope that it will make you feel better. Indicate whether the statement below is true or false, based on the scenario outlined. The carbohydrate molecules in the candy bar will yield a higher amount of ATP than an equal number of fat molecules. a. true b. false ANSWER: b 74. It is final exam week and you are feeling stressed. You decide to eat a candy bar and drink a big bottle (32 ounces) of soda in the hope that it will make you feel better. Indicate whether the statement below is true or false, based on the scenario outlined. After eating the candy bar, your circulating levels of insulin will increase. a. true b. false ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 75. It is final exam week and you are feeling stressed. You decide to eat a candy bar and drink a big bottle (32 ounces) of soda in the hope that it will make you feel better. Indicate whether the statement below is true or false, based on the scenario outlined. Concentrations of pyruvate in your cells will initially be very high, but will decrease as you sleep. a. true b. false ANSWER: a 76. It is final exam week and you are feeling stressed. You decide to eat a candy bar and drink a big bottle (32 ounces) of soda in the hope that it will make you feel better. Indicate whether the statement below is true or false, based on the scenario outlined. Gastrin secretion will be maintained at the same level until all the food is out of your stomach. a. true b. false ANSWER: b 77. It is final exam week and you are feeling stressed. You decide to eat a candy bar and drink a big bottle (32 ounces) of soda in the hope that it will make you feel better. Indicate whether the statement below is true or false, based on the scenario outlined. Secretin released by the duodenum will cause the pancreas to decrease production of digestive enzymes. a. true b. false ANSWER: b 78. It is final exam week and you are feeling stressed. You decide to eat a candy bar and drink a big bottle (32 ounces) of soda in the hope that it will make you feel better. Indicate whether the statement below is true or false, based on the scenario outlined. Bicarbonate production in the pancreas will increase. a. true b. False ANSWER: a 79. It is final exam week and you are feeling stressed. You decide to eat a candy bar and drink a big bottle (32 ounces) of soda in the hope that it will make you feel better. Indicate whether the statement below is true or false, based on the scenario outlined. Glycogen levels in your liver will increase. a. true b. false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 ANSWER: a 80. It is final exam week and you are feeling stressed. You decide to eat a candy bar and drink a big bottle (32 ounces) of soda in the hope that it will make you feel better. Indicate whether the statement below is true or false, based on the scenario outlined. CCK released in the duodenum will stimulate the release of bile from the gallbladder. a. true b. false ANSWER: a 81. It is final exam week and you are feeling stressed. You decide to eat a candy bar and drink a big bottle (32 ounces) of soda in the hope that it will make you feel better. Indicate whether the statement below is true or false, based on the scenario outlined. Lipases will be activated in the stomach to break down the fats in the candy bar. a. true b. false ANSWER: a 82. It is final exam week and you are feeling stressed. You decide to eat a candy bar and drink a big bottle (32 ounces) of soda in the hope that it will make you feel better. Indicate whether the statement below is true or false, based on the scenario outlined. Chemical digestion of the carbohydrates will begin in the stomach. a. true b. false ANSWER: b Multiple Response 83. Which of the choices describes animals that produce most of their own heat as a byproduct of metabolism? Select all that apply. a. endotherms b. active over a broad range of external temperatures c. likely to respond to cool temperatures with peripheral vasodilation d. unable to carry out fermentation in anaerobic conditions e. most active in the heat of the daytime ANSWER: a, b 84. What function does the low pH of the stomach serve? Select all that apply. a. It allows for the activation of pepsinogen into pepsin. b. Acidic conditions are less damaging for the cells lining the stomach than basic conditions. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 c. The acidic conditions break down foods directly. d. It allows gastric enzymes to work. ANSWER: a, c, d 85. Which of the choices describe(s) animals that obtain most of their heat from the environment? Select all that apply. a. ectotherms b. cold-blooded c. more likely to use behavioral means to regulate body temperature than physiological means d. unable to maintain a constant core body temperature e. most active at night in order to generate as much heat as possible ANSWER: a, b, c, d 86. Many organisms that filter feed are sedentary. Which of the choices is a characteristic that increases the amount of food obtained through filter feeding? Select all that apply. a. increasing the size of the mouth b. increasing the volume of water taken in through the mouth per unit of time c. decreasing the volume of water taken in through the mouth per unit of time d. decreasing the size of the mouth ANSWER: a, b 87. Although cellulose is a carbohydrate made of polymerized glucose molecules, the usual enzymes (such as amylase) that break down carbohydrates in your mouth and small intestine do not have any effect on cellulose. For the most part, animals do not produce the enzyme cellulase, which catalyzes cellulose hydrolysis. How, then, do herbivorous organisms manage to live on plant material composed primarily of cellulose? Select all that apply. a. They may have specialized chambers in the foregut that harbor populations of symbiotic anaerobic bacteria that break down cellulose. b. They may have a cecum that houses symbiotic anaerobic bacteria that break down cellulose. c. They have a multichambered stomach that has a very low pH and a particularly active form of pepsin for digesting plant proteins. d. They may have food take longer to pass through the intestine so more nutrients can be absorbed. ANSWER: a, b 88. Which of the statements is/are true of the hindgut? Select all that apply. a. It includes the large intestine. b. It includes the rectum. c. It is where most water and minerals are reabsorbed. d. It includes the duodenum, jejunum, and ileum. e. It is supplied with blood vessels and nerves by the submucosa. ANSWER: a, b, c, e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 89. In humans, which of the enzymes does the stomach secrete? Select all that apply. a. pepsin, an enzyme that digests proteins b. enzymes that digest lipids c. hydrochloric acid d. amylase e. cellulase ANSWER: a, b, c 90. Which of the statements about villi are correct? Select all that apply. a. Villi can be found in both the jejunum and the ileum. b. Villi serve to increase surface area. c. Villi allow for enhanced nutrient absorption. d. Villi move food particles through the digestive tract. e. Villi secrete multiple types of digestive enzymes. ANSWER: a, b, c 91. Which of the choices is a source of energy that animals can oxidize in order to synthesize ATP? Select all that apply. a. O2 b. carbohydrates c. CO2 d. lipids e. proteins ANSWER: b, d, e 92. Which of the statements is/are correct about aerobic metabolism? Select all that apply. a. It occurs in the mitochondria. b. It provides energy for longer-term, sustainable activity, in contrast to anaerobic metabolism. c. It leads to the production of energy-rich NADH and FADH2 molecules. d. It leads to the production of lactic acid in muscle cells. e. It utilizes oxygen at the end of the electron transport chain, releasing water. ANSWER: a, b, c, e 93. Which of the structures does the foregut include? Select all that apply. a. the mouth b. the esophagus c. the stomach or crop d. the small intestine e. the large intestine ANSWER: a, b, c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 94. Which of the choices is/are secretions produced by the pancreas? Select all that apply. a. lipase for fat digestion b. trypsin for further protein digestion c. bicarbonate ions to neutralize stomach acid d. methane from fermentation e. gastrin to stimulate HCl production ANSWER: a, b, c 95. Which of the statements illustrate limitations on a cheetah's ability to be an effective predator? Select all that apply. a. Cheetahs must rest for long periods between sprints. b. Cheetahs cannot run when the temperature is above about 105oF. c. Cheetahs cannot successfully prey on goats, which can release excess heat as they run. d. Cheetahs lose incredible amounts of heat during a sprint, causing their muscles to tighten after a short period of time. e. Cheetahs have a top speed significantly lower than that of most of their desired prey. ANSWER: a, b 96. Which of the statements is/are true about anaerobic metabolism? Select all that apply. a. It occurs in the cytosol in the absence of oxygen. b. It provides rapid, short-term energy. c. It breaks glucose down into carbon dioxide. d. It leads to the maximum possible yield of ATP in a cell. e. It occurs in the mitochondria in the presence of oxygen. ANSWER: a, b 97. Which of the statements accurately describe ectotherms as compared to endotherms of similar size at similar body temperatures? Select all that apply. a. Ectotherm metabolic rates are about 25% of those of endotherms. b. Ectotherms cannot sustain prolonged activity for as long as endotherms can. c. Ectotherms can survive much longer periods without food than endotherms can. d. Ectotherm metabolic rates decrease as body temperature increases. e. Ectotherms never reach the activity levels of endotherms. ANSWER: a, b, c 98. Which of the statements are true of suction feeding? Select all that apply. a. The fish's oral cavity expands slowly during suction feeding. b. It allows its users to be "sit and wait" predators. c. It is used by aquatic salamanders as well as fish. d. It can be used by some insects and young mammals. ANSWER: b, c, d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 38 99. The ability to digest lactose into adulthood results from either of two separate mutations that evolved independently of each other in different human populations. Of what are these two mutations an example? Select all that apply. a. convergent evolution b. a homology c. an analogy d. an ancestral character e. coevolution ANSWER: a, c 100. Which of the statements are true of the midgut? Select all that apply. a. It includes the small intestine. b. It includes the large intestine. c. Most nutrient absorption occurs in the midgut. d. The midgut is the primary site of food and nutrient storage. e. Glucose passively diffuses into cells in the midgut. ANSWER: a, c

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Chapter 39 Multiple Choice 1. Examine the figure. Which of the answer choices best describes conditions at equilibrium?

a. No water molecules are moving across the barrier between the two compartments. b. A net movement of water molecules continues to add water to the right-hand compartment. c. A net movement of water molecules begins to add water to the left-hand compartment. d. Water molecules move randomly across the barrier with no net addition of water to either side. e. None of the answer options is correct. ANSWER: d 2. In organisms, the ability to control osmosis depends on the properties of the plasma membranes of individual cells. Based on their ability to permit the free movement of water and some solutes, these membranes are described as: a. impermeable. b. permeable. c. selectively permeable. d. diffusible. e. impervious. ANSWER: c 3. Osmoconformers tend to live in environments with variable solute concentrations. a. true b. false ANSWER: b 4. The figure shows a tube with walls that are only permeable to water. The solute concentration of fluid entering the tube is different from that exiting the tube. There is a net flow of water out of the tube. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39

a. true b. false ANSWER: a 5. The figure shows a tube with walls that are only permeable to water. The solute concentration of fluid entering the tube is different from that exiting the tube. The fluid surrounding the tube must be hypotonic to the fluid within the tube.

a. true b. false ANSWER: b 6. The figure shows a tube with walls that are only permeable to water. The solute concentration of fluid entering the tube is different from that exiting the tube. If the tube were a portion of a nephron, it would be the ascending portion of the loop of Henle.

a. true b. false ANSWER: b 7. Fish in a marine environment must maintain a relatively constant ion concentration in their tissues and blood. How do the chloride cells in the gills aid fish in the removal of excess ions? a. The chloride cells create a countercurrent system where ions are lost from the body and water is absorbed to dilute other ions. b. The chloride cells create a countercurrent system where water is lost from the body and chloride ions are absorbed. c. The chloride cells actively move ions into the body and water follows by osmosis. d. The chloride cells actively remove ions from the body and water follows by osmosis. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 ANSWER: d 8. Which large group of animals consists largely of osmoconformers? a. marine invertebrates b. bony fish c. mammals d. insects e. amphibians ANSWER: a 9. Osmosis is the movement of water: a. from lower water concentration to higher water concentration across a selectively permeable membrane. b. from higher water concentration to lower water concentration across a selectively permeable membrane. c. between equal water concentration levels across a selectively permeable membrane. d. from warmer areas to cooler areas across a selectively permeable membrane. ANSWER: b 10. Osmoregulators _____ internal solute concentrations compared to their external environment. a. have the same b. have different c. always have lower d. always have higher ANSWER: b 11. Freshwater fish are in a hypotonic environment. In which direction do their gill chloride cells move chloride ions in this environment? a. Freshwater fish have chloride cells that actively move chloride ions out of the gills into the surrounding water, with sodium ions following. b. Freshwater fish have chloride cells that actively move chloride ions out of the gills into the surrounding water, with sodium ions moving in the opposite direction. c. Freshwater fish have chloride cells that actively move chloride ions into the gills, with sodium ions following. d. Freshwater fish have chloride cells that actively move chloride ions into the gills, with sodium ions moving in the opposite direction. ANSWER: c 12. What would happen if you placed a trout (a freshwater fish) into a saltwater tank at the aquarium and made some observations? a. The fish would take on water quickly. b. The fish would lose water. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 c. The fish would take on water initially, but could reverse the direction of Cl- ion movement to deal with excess water. d. The fish would lose water initially, but could reverse the direction of Cl- ion movement to deal with excess water. ANSWER: b 13. Despite being an osmoconformer, a shark still spends energy to regulate the concentrations of individual ions such as potassium, chloride, and sodium. a. true b. false ANSWER: a 14. When does water stop moving by osmosis between two areas of different solute concentration? a. when all solutes are dissolved in water in both areas b. when hydrostatic pressure is equal in both areas c. when osmotic and hydrostatic pressure are equal in both areas d. when the volume of water in both areas is equal ANSWER: c 15. One of the most extensive animal radiations is that of terrestrial arthropods, including insects. They excrete uric acid as their form of nitrogenous waste. Why does excreting uric acid allow them to conserve more water than a mammal that has extensive kidney tubules for water reabsorption? a. The high solubility of uric acid allows the excretory system to reabsorb water. b. Uric acid levels stay high in the blood, maintaining a strong osmotic gradient, which allows water to be reabsorbed. c. After precipitation, uric acid no longer exerts osmotic pressure; water returns to the hypertonic tissues. d. After precipitation, uric acid no longer exerts osmotic pressure; water leaves the hypertonic tissues. ANSWER: c 16. You discover a new type of marine organism that has the kidneys of a fish but no gills. What effect would the absence of gills have on excretion of nitrogenous waste? a. It would have no effect. Fish kidneys excrete all nitrogenous waste. b. Nitrogenous excretion decreases because no nitrogen uptake is occurring at the gills. c. Nitrogenous excretion increases at the kidneys because the waste cannot be lost at the gills. d. It would have no effect. Nitrogenous wastes diffuse across the skin. ANSWER: c 17. List the three forms of nitrogenous waste in order from that which requires the most water to eliminate it to that which requires the least water to eliminate it. a. uric acid, urea, ammonia b. uric acid, ammonia, urea Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 c. urea, uric acid, ammonia d. ammonia, uric acid, urea e. ammonia, urea, uric acid ANSWER: e 18. List the three forms of nitrogenous waste in order from least energetically expensive to produce to most energetically expensive to produce. a. uric acid, urea, ammonia b. uric acid, ammonia, urea c. urea, uric acid, ammonia d. ammonia, uric acid, urea e. ammonia, urea, uric acid ANSWER: e 19. Multicellular animals with pressurized circulatory systems use a three-step process to isolate and eliminate wastes. Arrange these steps in order from earliest to latest. a. filtration, reabsorption, secretion b. secretion, filtration, reabsorption c. filtration, secretion, reabsorption d. reabsorption, filtration, secretion e. secretion, reabsorption, filtration ANSWER: a 20. In unicellular organisms and simple multicellular organisms, wastes are isolated in a cellular compartment called a _____ and are eliminated by exocytosis. a. contractile vacuole b. protonephridia c. Malpighian tubule d. nephron ANSWER: a 21. Uric acid produced by insect Malpighian tubules does not influence the movement of water from the hindgut because: a. it is eliminated by the tubules before water reaches the hindgut. b. it precipitates due to the low pH of the fluid and therefore does not contribute to osmotic pressure. c. water moves only in response to the osmotic pressure created by proteins and sugars in the hindgut. d. uric acid is equally concentrated in the hindgut and in the body fluids. e. None of the other answer options is correct. ANSWER: b 22. Marine fish have impermeable renal tubules, allowing them to reabsorb scarce electrolytes without also reabsorbing water they are trying to eliminate. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 a. true b. false ANSWER: b 23. A cockroach has a very "salty" meal. Which of the answer choices would you expect to occur? a. Chloride ions will be pumped into the intestinal lumen. b. The cockroach will drink more water than usual. c. Excess water will be moved from the intestinal lumen to the hemolymph. d. Sodium ions will be moved into the intestinal lumen. ANSWER: c 24. Most mammals excrete _____; most aquatic animals excrete _____; birds, insects, and many reptiles excrete _____. a. uric acid; urea; ammonia b. uric acid; ammonia; urea c. urea; uric acid; ammonia d. urea; ammonia; uric acid e. ammonia; urea; uric acid ANSWER: d 25. Freshwater flatworms and earthworms are similar in that their excretory systems reabsorb electrolytes and eliminate concentrated urine. a. true b. false ANSWER: b 26. When animals excrete nitrogenous waste, what form does it take? a. ammonia b. urea c. uric acid d. All of these choices are correct. ANSWER: d 27. Uric acid is: a. often produced by fish. b. the least energy-expensive way to dispose of nitrogen waste. c. the least toxic form of nitrogen waste. d. associated with high water loss. ANSWER: c 28. Which process is responsible for creating most of an animal's nitrogenous waste? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 a. protein breakdown during metabolism b. carbohydrate breakdown during metabolism c. respiration d. toxins in the environment ANSWER: a 29. What organ helps concentrate and secrete nitrogenous waste in insects? a. Malpighian tubules b. kidneys c. cloaca d. protonephridia ANSWER: a 30. Which of the answer choices is an advantage of producing very concentrated urine? a. minimizing water loss b. reducing the need to drink c. ability to live in drier environments d. All of these choices are correct. ANSWER: d 31. Placing a freshwater flatworm such as Planaria in a dilute salt solution results in: a. more dilute urine. b. more concentrated urine. c. no change in urine concentration. ANSWER: b 32. In an insect, most of the reabsorption of water into body tissues occurs in the: a. Malpighian tubules. b. rectum. c. intestine. d. Malpighian tubules and intestine. e. Malpighian tubules and rectum. ANSWER: b 33. All excretory organs go through the processes of filtration, reabsorption, and secretion. a. true b. false ANSWER: b 34. Which of the answer choices correctly lists the three forms of nitrogenous waste in order from least to most toxic? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 a. uric acid, urea, ammonia b. uric acid, ammonia, urea c. urea, uric acid, ammonia d. urea, ammonia, uric acid e. ammonia, urea, uric acid ANSWER: a 35. Desert reptiles excrete nitrogenous waste in the form of uric acid, which allows them to reabsorb most of the water from their urine. a. true b. false ANSWER: a 36. The excretory organs of different organisms are extremely varied. What do all excretory organs have in common? a. a filtration system that results in a filtrate that is more concentrated than the body fluids b. a filtration system that filters nitrogenous waste from the body fluids c. a filtration system that relies on active transport to alter initial concentration of the filtrate d. a filtration system that relies on extensive tubules with a high ratio of surface area to volume to remove wastes ANSWER: b 37. You have 1 liter of 2 millimolar (mM) uric acid. Conditions change so that all the uric acid precipitates out of solution. What is the final concentration of uric acid in your solution? a. 2 mM b. 1 mM c. 1/2 mM d. 0 mM ANSWER: d 38. The figure shown illustrates that, as water flows into a region with high solute concentration, it exerts an opposing force called _____ pressure, which eventually prevents additional water from moving into the compartment, bringing the system into equilibrium.

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Chapter 39

a. hydrostatic b. osmotic c. concentration ANSWER: a 39. When proteins and nucleic acids are broken down, the toxic products are collectively called _____ waste. a. nitrogenous b. chemical c. soluble ANSWER: a 40. In the vertebrate kidney, blood filtration happens through the walls of special capillaries that form a tufted loop called a _____, which is surrounded by a capsule. a. glomerulus b. proximal convoluted tubule c. loop of Henle d. distal convoluted tubule ANSWER: a 41. The blood vessels surrounding the loop of Henle are collectively called the _____ and are arranged in a countercurrent organization, just like the loop of Henle. a. glomerulus b. Nephron c. vasa recta ANSWER: c 42. In addition to osmoregulation, the kidneys participate in regulating blood pressure. Specialized cells of the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 efferent arteriole, called the juxtaglomerular apparatus, secrete the hormone renin in response to drops in blood pressure. This, in turn, triggers additional responses that help raise blood pressure. a. true b. false ANSWER: a 43. Sometimes while traveling, people are infected with a protist parasite, Giardia. One of the side effects of Giardia infection is diarrhea. When someone has diarrhea, they lose excessive water and salt from the body. The result is loss of Na+, dehydration that leads to decreased extracellular volume and plasma volume, and decreased arterial blood pressure. Given this, the vasopressin secretion of a person infected with Giardia: a. increases. b. decreases. c. stay the same. ANSWER: a 44. The decreased solute concentration of the filtrate after passing through Bowman's capsule is caused by: a. diffusion of NaCl out of the proximal convoluted tubule. b. diffusion of HCO3- out of the proximal convoluted tubule. c. active transport of NaCl out of the proximal convoluted tubule. d. active transport of HCO3- out of the proximal convoluted tubule. ANSWER: c 45. One of the roles of the kidneys is to help buffer body fluids so that they are not too acidic or too basic. The cells of the renal tubule secrete H+ into the tubule lumen and absorb bicarbonate (HCO3-), passing it into the tissue fluid. Consider the reaction catalyzed by the enzyme carbonic anhydrase, shown in the accompanying figure.

During heavy exercise, the reaction shifts to the right in the tubule lumen of the nephron. a. true b. false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 ANSWER: a 46. Your kidneys filter your blood about _____ times per day and reabsorb _____% of the water entering the renal tubules, _____% of the electrolytes in the blood, and _____% of the glucose and amino acids that were filtered out of the blood in the glomerulus. a. 60; 99.9; 99.9; 100 b. 100; 50; 50; 99 c. 75; 100; 60; 90 d. 40; 99.9; 100; 75 e. 10; 80; 25; 70 ANSWER: a 47. In the presence of ADH, urine becomes more concentrated. a. true b. false ANSWER: a 48. The mammalian kidney has an outer layer called the _____ and an inner layer called the _____. a. pelvis; glomerulus b. epithelium; endothelium c. medulla; pelvis d. glomerulus; cortex e. cortex; medulla ANSWER: e 49. The loops of Henle create a concentration gradient in the interstitial fluid surrounding the loop, with the concentration highest in the _____ and lowest in the _____ of the kidney. a. outer medulla; inner medulla b. outer medulla; cortex c. cortex; inner medulla d. inner medulla; cortex e. cortex; outer medulla ANSWER: d 50. By the time the filtrate leaves the proximal convoluted tubule, about 75% of the water and most of the electrolytes and solutes have been reabsorbed into the bloodstream. a. true b. false ANSWER: a 51. The main solute in the filtrate as it enters the distal convoluted tubule is: a. electrolytes. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 b. glucose. c. urea. d. amino acids. e. nucleotides. ANSWER: c 52. One of the roles of the kidneys is to help buffer body fluids, to keep fluids from becoming too acidic or too basic. The cells of the renal tubule wall secrete H+ into the tubule lumen and absorb bicarbonate, passing it into the tissue fluid. Consider the reaction catalyzed by the enzyme carbonic anhydrase, shown in the accompanying figure.

During heavy exercise, the reaction shifts to the right in the renal tubule cell of the nephron. a. true b. false ANSWER: b 53. Which part of the kidney creates a concentration gradient? a. microvilli b. loop of Henle c. renal pelvis d. ureter ANSWER: b 54. An increase of _____ increases the permeability of the collecting duct to water. a. ADH (antidiuretic hormone) b. caffeine c. alcohol d. diuretic drug ANSWER: a 55. The basic unit of the kidney, the nephron, contains: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 a. the glomerulus. b. renal tubules. c. the collecting duct. d. All of these choices are correct. ANSWER: d 56. The majority of fluid that enters the nephron: a. moves into the loop of Henle. b. is reabsorbed by the proximal convoluted tubule. c. exits the nephron through the collecting duct. d. is reabsorbed by the descending loop of Henle. ANSWER: b 57. Water levels are adjusted to meet the osmoregulatory needs of the organism in the _____ of the nephron under the control of _____, also called vasopressin. a. distal convoluted tubule; antidiuretic hormone b. loops of Henle; antidiuretic hormone c. collecting ducts; antidiuretic hormone d. distal convoluted tubule; diuretic hormone e. collecting ducts; diuretic hormone ANSWER: c 58. Which of the structures is responsible for holding urine until it is excreted from the body? a. kidney b. collecting ducts c. ureter d. bladder ANSWER: d 59. When a person experiences dehydration, water can be absorbed from the bladder wall to replenish needed water back to the body. a. true b. false ANSWER: b 60. What is the significance of the fact that the loop of Henle extends into the medulla of the kidney with sections that run antiparallel to one another? a. The loop of Henle maintains the concentration gradient in the kidney. b. The loop of Henle generates the concentration gradient in the kidney. c. The loop of Henle contains more aquaporins. d. The loop of Henle is permeable to urea on one side, but not the other. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 ANSWER: b 61. Antidiuretic hormone (ADH) controls how diluted or concentrated urine is by controlling the permeability of the: a. collecting ducts to water. b. collecting ducts to electrolytes. c. collecting ducts to urea. d. loops of Henle to water. e. loops of Henle to electrolytes. ANSWER: a 62. The first part of the renal tubule is called the _____ and is specialized for _____. a. proximal convoluted tubule; reabsorption of electrolytes and other nutrients b. loop of Henle; creation of a concentration gradient in the interstitial fluid c. distal convoluted tubule; secretion of additional wastes into the urine d. distal convoluted tubule; reabsorption of water e. collecting duct; reabsorption of water ANSWER: a 63. Refer to the figure shown. The organization of the vasa recta around the loop of Henle allows:

a. urea to be concentrated in the distal convoluted tubule. b. water to be maximally reabsorbed from the distal convoluted tubule. c. the concentration gradient established by the loop of Henle to be maintained. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 d. urea to be concentrated in the distal convoluted tubule and the concentration gradient established by the loop of Henle to be maintained. e. water to be maximally reabsorbed from the distal convoluted tubule and the concentration gradient established by the loop of Henle to be maintained. ANSWER: c 64. The figure shows a tube with walls that are only permeable to water. The solute concentration of fluid entering the tube is different from that exiting the tube. The concentration of the solution outside the tube is always higher than what is inside the tube for the entire length of the tube. The volume of fluid entering the tube is _____ the volume exiting the tube.

a. greater than b. less than c. the same as ANSWER: a 65. The figure shows a tube with walls that are only permeable to water. The solute concentration of fluid entering the tube is different from that exiting the tube. The concentration of the solution outside the tube is always higher than what is inside the tube for the entire length of the tube. The number of solute molecules in the fluid entering the tube is _____ the number of solute molecules in the fluid exiting the tube.

a. greater than b. less than c. the same as ANSWER: c 66. The brine shrimp Artemia lives in salt lakes, such as the Great Salt Lake of Utah, where it usually matches its internal solute concentration to that of the outside environment. However, it sometimes encounters exceptionally high salt concentrations. In this case, it is able to adjust its internal concentration so that it is hypotonic relative to its environment. Artemia is best described as an: a. osmoconformer. b. osmoregulator. c. osmoconformer that acts as an osmoregulator in extreme environments. d. osmoregulator that acts as an osmoconformer in extreme environments. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 e. None of the other answer options accurately describe Artemia. ANSWER: c 67. The brine shrimp Artemia lives in salt lakes, such as the Great Salt Lake of Utah, where it usually matches its internal solute concentration to that of the outside environment. However, it sometimes encounters exceptionally high salt concentrations. In this case, it is able to adjust its internal concentration so that it is hypotonic relative to its environment. Based on what you know about how animals adjust to high-salt environments, how do you think Artemia survives in very high-salt environments? a. They drink water. b. They absorb excess salt across their gills. c. They absorb water across their gills. d. They excrete excess salt across their gills. e. They excrete nitrogenous waste in the form of uric acid. ANSWER: d 68. The brine shrimp Artemia lives in salt lakes, such as the Great Salt Lake of Utah, where it usually matches its internal solute concentration to that of the outside environment. However, it sometimes encounters exceptionally high salt concentrations. In this case, it is able to adjust its internal concentration so that it is hypotonic relative to its environment. Artemia sometimes encounter low-salt environments, which it tolerates by adjusting its internal concentration to be hypertonic relative to its environment. Based on what you know about how animals adjust to low-salt environments, how do you think Artemia survive in very low-salt environments? a. They drink water. b. They absorb excess salt across their gills. c. They absorb water across their gills. d. They excrete excess salt across their gills. e. They excrete nitrogenous waste in the form of uric acid. ANSWER: b 69. The brine shrimp Artemia lives in salt lakes, such as the Great Salt Lake of Utah, where it usually matches its internal solute concentration to that of the outside environment. However, it sometimes encounters exceptionally high salt concentrations. In this case, it is able to adjust its internal concentration so that it is hypotonic relative to its environment. Artemia can tolerate low-salt environments but not freshwater. Based on what you know about osmoregulation, why do you think this is the case? a. It cannot absorb enough water to maintain osmotic balance. b. It cannot absorb enough salt to maintain osmotic balance. c. It cannot excrete enough water to maintain osmotic balance. d. It cannot excrete enough salt to maintain osmotic balance. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 e. None of the other answer options provides an explanation for the observation that Artemia cannot survive in freshwater. ANSWER: b 70. You have just finished a really long workout at the gym. You are sweating quite a bit, and feel thirsty. To conserve water and counter its loss, Na+ reabsorption will increase in the distal convoluted tubule. a. true b. false ANSWER: a 71. You have just finished a really long workout at the gym. You are sweating quite a bit, and feel thirsty. The number of aquaporins in the collecting ducts will increase. a. true b. false ANSWER: a 72. You have just finished a really long workout at the gym. You are sweating quite a bit and feel thirsty. You have lost a lot of water and some ions while sweating. In response to decreased Na+ in body fluids (Na+ depletion), your kidneys will _____ renin secretion. a. increase b. decrease c. stop ANSWER: a 73. You have just finished a really long workout at the gym. You are sweating quite a bit and feel thirsty. You have lost a lot of water and some ions while sweating. In response to increased solute concentration of body fluids, your kidneys will _____ vasopressin secretion. a. increase b. decrease c. stop ANSWER: a 74. You have just finished a really long workout at the gym. You managed to sweat quite a bit and now feel thirsty. In response, your posterior pituitary gland is secreting vasopressin. Increased vasopressin (ADH) secretion _____ urine output. a. increases b. decreases c. has no effect on ANSWER: b 75. Consider the figure shown. Indicate whether the statement is true or false.

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Chapter 39

The concentration of NaCl is higher in F than G. a. true b. false ANSWER: b 76. Consider the figure shown. Indicate whether the statement is true or false.

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Chapter 39

The concentration of NaCl is higher in A than B. a. true b. false ANSWER: b 77. Consider the figure shown. Indicate whether the statement is true or false.

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Chapter 39

The concentration of NaCl is higher in A than C. a. true b. false ANSWER: b 78. Consider the figure shown. Indicate whether the statement is true or false.

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Chapter 39

The concentration of urea is higher in B than C. a. true b. false ANSWER: b 79. Consider the figure shown. Indicate whether the statement is true or false.

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Chapter 39

The concentration of urea is higher in C than D. a. true b. false ANSWER: b 80. Consider the figure shown. Indicate whether the statement is true or false.

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Chapter 39

The osmotic pressure is higher in A than D. a. true b. false ANSWER: b 81. Consider the figure shown. Indicate whether the statement is true or false.

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Chapter 39

The osmotic pressure is higher in B than C. a. true b. false ANSWER: b 82. Consider the figure shown. Indicate whether the statement is true or false.

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Chapter 39

The osmotic gradient is higher in F than D. a. true b. false ANSWER: b 83. Assume that blood pressure drops in a person because of an injury, and given the drop in blood pressure, indicate whether the statement on the body's responses is true or false. Circulating levels of aldosterone will increase. a. true b. false ANSWER: a 84. Assume that blood pressure drops in a person because of an injury, and given the drop in blood pressure, indicate whether the statement on the body's responses is true or false. Circulating levels of renin will decrease. a. true b. false ANSWER: b 85. Assume that blood pressure drops in a person because of an injury, and given the drop in blood pressure, indicate whether the statement on the body's responses is true or false. The sympathetic nervous system stimulates the adrenal glands. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 a. true b. false ANSWER: a 86. Assume that blood pressure drops in a person because of an injury, and given the drop in blood pressure, indicate whether the statement on the body's responses is true or false. Water absorption by the distal convoluted tubule will decrease. a. true b. false ANSWER: b 87. Assume that blood pressure drops in a person because of an injury, and given the drop in blood pressure, indicate whether the statement on the body's responses is true or false. Circulating levels of angiotensin II will increase. a. true b. false ANSWER: a Multiple Response 88. Depending on their environments, animals may gain water: Select all that apply. a. by drinking. b. from food. c. during cellular respiration. d. through their lungs. e. during excretion. ANSWER: a, b, c 89. Water molecules move by: Select all that apply. a. random molecular motion. b. active transport. c. osmosis. d. facilitated diffusion. ANSWER: a, d 90. Depending on their environments, animals may gain electrolytes: Select all that apply. a. from food. b. from seawater. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 c. across gills and skin. d. through cellular respiration. e. from exposure to ultraviolet light. ANSWER: a, b, c 91. Freshwater fish: Select all that apply. a. lose electrolytes across gills. b. gain water across gills. c. gain electrolytes across gills. d. gain electrolytes from drinking. e. lose water across gills. ANSWER: a, b 92. Water passes through lipid bilayers: Select all that apply. a. directly by diffusion. b. through aquaporins by facilitated diffusion. c. through protein pumps and active transport. d. through gap junctions. e. through hemidesmosomes. ANSWER: a, b 93. Depending on their environments, animals may lose water: Select all that apply. a. in urine and feces. b. through their lungs. c. by sweating. d. through their gills. e. during cellular respiration. ANSWER: a, b, c, d 94. Marine fish: Select all that apply. a. lose electrolytes across the gills. b. gain water across the gills. c. gain electrolytes across the gills. d. gain electrolytes from drinking. e. lose water across the gills. ANSWER: c, d, e 95. Depending on their environments, animals may lose electrolytes: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 Select all that apply. a. across the gills. b. in sweat. c. in urine. d. in feces. e. by cellular respiration. ANSWER: a, b, c, d 96. Because ammonia is very toxic, it is kept at low levels in the body of many different kinds of organisms. Ammonia is kept at low levels through: Select all that apply. a. elimination. b. conversion to urea. c. storage away from body tissues. d. conversion to uric acid. e. conversion to amino acids. ANSWER: a, b, d 97. In fish, their: Select all that apply. a. nitrogenous waste is excreted through the gills. b. kidneys are primarily organs of water and electrolyte balance. c. kidneys are proportionally smaller than they are in mammals. d. kidneys are unsegmented. e. renal tubules are permeable to water, regardless of the type of fish. ANSWER: a, b 98. Reabsorption is an important process in waste removal by excretory organs. Which of the answer choices occur during reabsorption in vertebrate kidneys? Select all that apply. a. Energy is expended. b. Water reenters the blood stream. c. Nitrogenous compounds reenter the blood stream. d. All electrolytes are removed from the filtrate. ANSWER: a, b 99. Which of the answer choices are components of a vertebrate kidney? Select all that apply. a. glomerulus b. Bowman's capsule c. renal tubules d. collecting ducts e. Malpighian tubules Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 ANSWER: a, b, c, d 100. Excreting ammonia directly into the environment: Select all that apply. a. is energetically inexpensive. b. requires high volumes of water. c. is energetically expensive. d. requires little water. e. converts it into a less toxic form. ANSWER: a, b 101. The walls of the proximal convoluted tubule: Select all that apply. a. are composed of epithelial cells with microvilli on their surfaces. b. possess numerous mitochondria. c. use facilitated diffusion to reabsorb glucose and amino acids. d. are where electrolytes are reabsorbed into the blood. e. have a brush border that increases the surface area of the tubule. ANSWER: a, b, d, e 102. A person's blood pressure falls. Which of the answer choices would you expect to occur? Select all that apply. a. Circulating concentrations of renin will increase. b. Circulating concentrations of angiotensin II will decrease. c. Water uptake by the distal convoluted tubule will increase. d. Circulating concentrations of aldosterone will decrease. ANSWER: a, c 103. Which of the answer choices are found in Bowman's space? Select all that apply. a. water b. glucose c. wastes d. solutes from the blood e. large proteins ANSWER: a, b, c, d 104. Which of the answer choices best describes the characteristics of the ascending limb of the loop of Henle? Select all that apply. a. It is permeable to water. b. It actively transports electrolytes. c. It is impermeable to water. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 39 d. Its filtrate is mainly urea. e. Its filtrate becomes more and more concentrated. ANSWER: b, c, d 105. In the glomerulus, blood is filtered under pressure through a filtration barrier made up of: Select all that apply. a. capillary endothelial cells. b. a basal lamina. c. a layer of podocytes. d. Bowman's capsule. e. loops of Henle. ANSWER: a, b, c 106. Which of the answer choices best describes the characteristics of the descending limb of the loop of Henle? Select all that apply. a. It is permeable to water. b. It actively transports electrolytes. c. It is impermeable to water. d. Its filtrate is mainly glucose. e. Its filtrate becomes more and more concentrated. ANSWER: a, e 107. Depending on the external solute concentration, the internal solute concentrations of osmoregulators are: Select all that apply. a. similar to the solute concentration of the external environment. b. higher than the solute concentration of the external environment. c. lower than the solute concentration of the external environment. ANSWER: b, c 108. Water moves from regions of: Select all that apply. a. high water concentration to regions of low water concentration. b. low water concentration to regions of high water concentration. c. high solute concentration to regions of low solute concentration. d. low solute concentration to regions of high solute concentration. e. low water concentration to regions of low solute concentration. ANSWER: a, d

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Chapter 40 Multiple Choice 1. One of the advantages of sexual reproduction is the production of novel arrangements of genes on chromosomes due to recombination during meiosis. This effect is further magnified upon fertilization because offspring: a. will have more alleles than either parent. b. will have higher fitness than either parent because of the union of gametes. c. may have allele combinations of genes that are not seen in either parent. d. will have more chromosomes than either parent. ANSWER: c 2. Daphnia alter their reproductive mode based on food availability. Why might it be adaptive for Daphnia to reproduce asexually in the spring? a. When food is in excess, natural selection cannot act on variation, because all individuals can get the food they need. b. When food is in excess, asexual reproduction allows for more rapid rates of reproduction. c. When food is in excess, asexual reproduction results in variable offspring. d. When food is in excess, reproductive rates will slow so that more energy can be put into growth by the individual Daphnia. ANSWER: b 3. In general, sexual reproduction is more likely found in species that live in: a. environments with constant conditions, as opposed to species that live in environments with variable conditions. b. environments with variable conditions, as opposed to species that live in environments with constant conditions. ANSWER: b 4. If the pattern seen in Daphnia is representative of other species that combine both sexual and asexual reproduction, we would expect asexual reproduction in other species to be most common: a. when environmental conditions are favorable for that species. b. when environmental conditions are unfavorable for that species. c. in the summer. d. in the winter. e. in the fall. ANSWER: a 5. Which of the choices is a form of asexual reproduction in which a new individual arises by the splitting of one organism into pieces? a. budding b. fragmentation c. parthenogenesis Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 ANSWER: b 6. In general, external fertilization is found in animals living in stable, predictable environments. a. true b. false ANSWER: b 7. In mammals such as humans, the chorion and allantois fuse to form the _____, an organ that allows the embryo to obtain nutrients directly from the mother. a. amnion b. yolk c. shell d. placenta e. embryo ANSWER: d 8. Most birth control pills contain both progestin (biologically similar to progesterone) and estrogen. By taking a pill over consecutive days, the amount of each hormone is maintained at a constant concentration in the circulating blood, inhibiting ovulation. Based on this information, what is the most likely mechanism for how birth control pills affect ovulation? a. They inhibit growth of the uterine lining so that implantation cannot occur. b. They slow egg maturation so that only immature eggs will be released. c. They inhibit maintenance of the corpus luteum after ovulation. d. They inhibit the growth of follicles so that ovulation does not occur. ANSWER: d 9. Spermatogenesis differs from oogenesis in that the cytoplasm from the precursor gametic cell is equally divided among the sperm. a. true b. false ANSWER: a 10. Consider the events that occur during the follicular phase of the menstrual cycle. 1. The lining of the uterus thickens. 2. Granulosa cells secrete estradiol. 3. FSH acts on granulosa cells. 4. LH levels surge rapidly, then decline. 5. Ovulation occurs. Select the answer that lists those events in chronological order. a. 3 → 2 → 1→ 4 → 5 b. 3 → 1 → 2 → 4 → 5 Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 c. 1 → 2 → 3 → 4 → 5 d. 2 → 4 → 3 → 1 → 5 e. 4 → 3 → 2 → 1 → 5 ANSWER: a 11. Which result would you expect if a tumor growing in the seminal vesicles rendered them nonfunctional? a. Sperm production would be reduced. b. Sperm would be blocked from exiting the penis. c. Ejaculate would have lower volume. d. Ejaculate would be more acidic. ANSWER: c 12. Men who have a vasectomy have their vas deferens cut and the ends sealed. This procedure prevents the production of sperm, so they will no longer be able to father children. a. true b. false ANSWER: b 13. Consider the figure.

Identify the structure indicated by the number 1 in the figure shown. a. epididymis b. seminal vesicle c. testis d. vas deferens e. prostate gland ANSWER: c 14. Consider the figure.

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Chapter 40

Identify the structure indicated by the number 2 in the figure shown. a. epididymis b. seminal vesicle c. testis d. vas deferens e. prostate gland ANSWER: a 15. Consider the figure.

Identify the structure indicated by the number 3 in the figure shown. a. epididymis b. seminal vesicle c. testis d. vas deferens e. prostate gland ANSWER: d 16. Consider the figure.

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Chapter 40

Identify the structure indicated by the number 4 in the figure shown. a. epididymis b. seminal vesicle c. testis d. vas deferens e. prostate gland ANSWER: e 17. Consider the figure.

Identify the structure indicated by the number 5 in the figure shown. a. epididymis b. seminal vesicle c. testis d. vas deferens e. prostate gland ANSWER: b 18. Where are sperm formed? a. epididymis b. seminal vesicle Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 c. testis d. vas deferens e. prostate gland ANSWER: c 19. Some women choose to have their fallopian tubes "tied" in order to prevent future pregnancies. How does this procedure prevent pregnancy from occurring? a. Increased acidic secretions from the damaged fallopian tubes make it unlikely for sperm to survive. b. Eggs will not be released from the ovaries, so fertilization does not occur. c. Eggs are prevented from passing through the fallopian tubes to be fertilized by sperm. d. Progesterone levels will increase, inhibiting the action of estrogen and preventing the release of eggs. ANSWER: c 20. Cervical stenosis is a condition in which the cervix swells and no longer maintains an opening to the vagina. This condition affects menstruation because menstrual fluid is unable to exit the uterus. a. true b. false ANSWER: a 21. Consider the figure.

Identify the structure indicated by the number 1 in the figure shown. a. uterus b. cervix c. vagina d. ovary e. fallopian tube or oviduct ANSWER: c 22. Consider the figure.

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Chapter 40

Identify the structure indicated by the number 2 in the figure shown. a. uterus b. cervix c. vagina d. ovary e. fallopian tube or oviduct ANSWER: d 23. Consider the figure.

Identify the structure indicated by the number 3 in the figure shown. a. uterus b. cervix c. vagina d. ovary e. fallopian tube or oviduct ANSWER: e 24. Consider the figure.

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Chapter 40

Identify the structure indicated by the number 4 in the figure shown. a. uterus b. cervix c. vagina d. ovary e. fallopian tube or oviduct ANSWER: b 25. Consider the figure.

Identify the structure indicated by the number 5 in the figure shown. a. uterus b. cervix c. vagina d. ovary e. fallopian tube or oviduct ANSWER: a 26. Which structure carries sperm to the ejaculatory duct? a. epididymis b. seminal vesicle c. testis d. vas deferens e. prostate gland Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 ANSWER: d 27. In response to FSH and LH, ovaries secrete estrogen and progesterone. a. true b. false ANSWER: a 28. Where are oocytes produced? a. uterus b. cervix c. vagina d. ovary e. fallopian tube or oviduct ANSWER: d 29. The inner cell mass of the blastocyst becomes the embryo, whereas the outer cell layer of the blastocyst forms part of the placenta. a. true b. false ANSWER: a 30. Once the plasma membranes of the egg and sperm fuse, the egg completes meiosis II and undergoes a series of changes that further prevent _____, or fertilization by more than one sperm. a. polyploidy b. aneuploidy c. polyspermy d. cleavage e. gastrulation ANSWER: c 31. The developing embryo implants into the uterine lining about 5 days after fertilization. At this stage, it is a fluid-filled ball of cells called a: a. morula. b. blastula. c. gastrula. d. epiblast. e. hypoblast. ANSWER: b 32. Which layer of cells in the human blastocyst migrates inwards during gastrulation to form all three developmental tissue layers? a. the epiblast Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 b. the hypoblast c. the inner cell mass ANSWER: a 33. During which developmental stage or process do cells move to new positions in the developing embryo? a. fertilization b. blastula c. gastrulation d. morula ANSWER: c 34. The developing fetus is most vulnerable to toxins, drugs, and infections during the first trimester of pregnancy because all of the basic structures and organ systems develop during this time. a. true b. false ANSWER: a 35. After fertilization, the zygote undergoes rounds of cell divisions in a process called: a. cleavage. b. gastrulation. c. organogenesis. d. implantation. e. origination. ANSWER: a 36. One of the disadvantages of sexual reproduction, as compared to asexual reproduction, is that: a. individuals have lower relatedness to their offspring than asexually reproducing species. b. new genotypes can be introduced into the population. c. deleterious mutations can be removed from the population more easily. d. there is greater variation introduced in each generation in sexually reproducing species. ANSWER: a 37. Each of the major hypotheses proposed to explain why sexual reproduction is more widespread than asexual reproduction ultimately bases its explanation on the adaptive benefit of: a. forming gametes of different sizes. b. forming genetically distinct offspring. c. forming large populations. d. forming relatively slow-growing populations. e. reducing mutation rates. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 38. Which of the choices are forms of asexual reproduction? a. budding b. fragmentation c. parthenogenesis d. All of these choices are correct. ANSWER: d 39. Meiosis is a form of cell division that: a. halves the chromosome number. b. doubles the chromosome number. c. allows asexual reproduction. ANSWER: a 40. Which of the statements concerning asexual reproduction is false? a. Asexual reproduction does not involve finding a mate. b. Asexual reproduction allows rapid reproduction and population increases. c. Asexual reproduction increases genetic diversity within the population. d. Asexual reproduction can occur despite low population density. ANSWER: c 41. Which of the species is most likely a K-strategist? a. butterfly b. frog c. fish d. chimpanzee ANSWER: d 42. Which of the animals is not oviparous? a. fish b. bird c. platypus d. mouse ANSWER: d 43. Which of the structures is responsible for containing waste in the amniotic egg? a. amnion b. chorion c. allantois d. yolk sac e. shell ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 44. Imagine a mutation that occurs in the gene for the progesterone receptor and prevents it from binding progesterone. What is the effect of such a mutation? a. The uterine lining is maintained throughout the menstrual cycle. b. The uterine lining does not increase to a level that can support pregnancy. c. The uterine lining is shed continuously. d. The uterine lining is shed as usual during the menstrual cycle. ANSWER: b 45. The release of luteinizing hormone (LH) is preceded by a: a. release of GnRH. b. surge of estradiol. c. surge of progesterone. d. surge of FSH. ANSWER: b 46. Which structure is also known as the birth canal? a. uterus b. cervix c. vagina d. ovary e. fallopian tube or oviduct ANSWER: c 47. Consider the events that occur if the oocyte is not fertilized during the luteal phase of the menstrual cycle. 1. The lining of the uterus is shed. 2. The corpus luteum secretes progesterone. 3. Follicle cells are converted to the corpus luteum. 4. The corpus luteum degenerates. 5. Estrogen and progesterone levels fall. Select the answer that places those events in chronological order. a. 3 → 2 → 4 → 5 → 1 b. 1 → 2 → 3 → 4 → 5 c. 2 → 4 → 3 → 1 → 5 d. 4 → 3 → 2 → 1 → 5 e. 5 → 2 → 3 → 4 →1 ANSWER: a 48. Male gametes (sperm) develop inside the: a. seminiferous tubules. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 b. vas deferens. c. urethra. d. seminal vesicles. ANSWER: a 49. Which hormone(s) do ovaries secrete? a. estrogen b. testosterone c. progesterone d. estrogen and progesterone ANSWER: d 50. Fertilization by more than one sperm would result in a polyploidy and an inviable zygote. What is the first line of defense against polyspermy? a. species-specific recognition of proteins on the outer coat of the egg b. changes in the membrane potential of the egg prohibiting additional sperm from entering c. products released from the egg changing the zona pellucida of the egg and prohibiting additional sperm from entering d. capacitation in the sperm fertilizing the egg ANSWER: b 51. Bones of a developing fetus develop from the: a. endoderm. b. mesoderm. c. ectoderm. ANSWER: b 52. Which of the choices correctly pairs a germ layer with the adult tissues it forms? a. ectoderm: bone b. mesoderm: pancreas c. endoderm: the liver and the lining of the respiratory and digestive tracts d. ectoderm: kidney e. mesoderm: connective tissue ANSWER: c 53. One of the important outcomes of gastrulation is the establishment of germ layers. If gastrulation failed to occur, then: a. only skin and the digestive tract would develop. b. only skin would develop. c. tissues and organs would not develop, and the pregnancy would not be viable. d. only skin and muscle would develop. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 ANSWER: c 54. In humans, sex determination depends on the presence or absence of the _____ gene residing on the _____ chromosome. This gene produces a protein called _____, which causes the developing gonad to differentiate. a. SRY; Y; TDF b. SRY; X; TDF c. TDF; X; SRY d. TDF; Y; SRY e. SRY; largest; TDF ANSWER: a 55. In humans, fertilization usually occurs in the: a. vagina. b. cervix. c. uterus. d. fallopian tubes. e. ovaries. ANSWER: d 56. Oocytes are the largest cells by volume in the human body, reflecting the presence of large amounts of cytoplasm containing: a. molecules that direct early cell divisions. b. the developing placenta. c. twice the amount of DNA of other cells in the body. d. polar bodies. ANSWER: a 57. In human development, implantation in the uterus occurs during: a. the first cleavage. b. the morula stage. c. the blastula stage. d. gastrulation. ANSWER: c 58. Which germ layer(s) form as a result of gastrulation? a. ectoderm b. mesoderm c. endoderm d. All of these choices are correct. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 59. Human pregnancy lasts _____ weeks after fertilization and is divided into three trimesters. a. 24 b. 38 c. 56 d. 72 ANSWER: b 60. Which of the statements about gametogenesis is true? a. In human males, spermatogenesis begins at puberty. b. In human females, primary oocytes form during puberty. c. In human females, the egg that is released during ovulation is a primary oocyte. d. In human females, primary oocytes arrest in metaphase II. e. In human males, spermatogenesis results in one full-sized sperm and three polar bodies. ANSWER: a 61. Komodo dragons (Varanus komodoensis) are an endangered species. One way that conservationists work to maintain their numbers is through captive breeding programs in zoos. Typically, females are housed alone, while males are transported between zoos to mate with females and produce offspring. In some cases, females produce offspring even though they are not in contact with a male. Which term describes the production of offspring by female Komodo dragons that lack contact with males? a. sexual reproduction b. fission c. budding d. parthenogenesis ANSWER: d 62. Komodo dragons (Varanus komodoensis) are an endangered species. One way that conservationists work to maintain their numbers is through captive breeding programs in zoos. Typically, females are housed alone, while males are transported between zoos to mate with females and produce offspring. In some cases, females produce offspring even though they are not in contact with a male. When females do not interact with males but still produce offspring, reproduction is asexual. a. true b. false ANSWER: a 63. A male and female from each of two species, species A and species B, are placed in an experimental pond. Both species have the same generation time of two weeks, produce one offspring with each round of reproduction, are the same relative size, and have the same energetic requirements. You come back after one year and notice that there are 20 times more individuals of species A than species B) Which of the two species is likely breeding asexually? a. species A b. species B c. They both reproduce asexually; species A is simply faster at reproducing. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 d. Neither; they both reproduce sexually. Species A is simply faster at reproducing. ANSWER: a 64. There are a few species of whiptail lizard in the wild that can reproduce either sexually or via parthenogenesis. In which types of environments would you expect these whiptail lizards to reproduce by parthenogenesis, rather than sexual reproduction? a. in habitats where environmental variables are relatively stable b. in habitats where environmental variables change rapidly and have little predictability ANSWER: a 65. Which of the statements accurately characterizes the early cleavage divisions in most animals? a. They do not involve an increase in the size of the zygote. b. They are directed by the zygote's genome. c. They are relatively slow. d. They do not occur at the same time, so at any given moment, the zygote may have an odd number of cells. e. All of these choices are correct. ANSWER: a 66. The amniotic egg, which is present in reptiles, birds, and mammals, is a good example of: a. coevolution. b. convergent evolution. c. homology. d. analogy. e. horizontal gene transfer. ANSWER: c 67. Salamanders are amphibians with very interesting mating rituals. The males perform species-specific dances that attract females to mate. Male salamanders deposit spermatophores (mucous packets filled with sperm) on leaves, twigs, or the ground and then guide females over them, so that the females can take them up into their cloaca. Mating dances like this result in internal fertilization in many salamander species. However, salamander reproduction is not entirely terrestrial because: a. salamanders must remain wet for gas exchange to occur. b. salamanders must lay their eggs on land near the water in which they live to be able to protect them. c. salamanders do not produce amniotic eggs; therefore, their eggs must be laid in water for protection from desiccation. ANSWER: c 68. The amniotic egg was a key adaptation that allowed vertebrates to make the transition to life on land, where diffusion to/from a watery environment did not allow a developing embryo to meet its metabolic needs. The amniotic egg's three membranes are the _____, which surrounds the embryo in a watery sac; the _____, which sequesters metabolic waste away from the developing embryo; and the _____, which surrounds the embryo and other sacs. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 a. amnion; chorion; allantois b. chorion; amnion; allantois c. chorion; allantois; amnion d. amnion; allantois; chorion e. allantois; chorion; amnion ANSWER: d 69. Many birth control pills release a constant amount of synthetic estradiol and progesterone for 21 days, followed by 7 days during which no hormones are ingested. Which of the statements describes the effects of birth control pills? a. Birth control pills inhibit the development of the uterine lining so that implantation of a fertilized oocyte cannot occur. b. Birth control pills maintain constant levels of estrogens and/or progesterone so that there is no surge in FSH and ovulation does not occur. c. The 7 days without hormones prevent the uterine lining from thickening to the point that a fertilized oocyte can implant. ANSWER: b 70. Which structure stores motile sperm? a. epididymis b. seminal vesicle c. testis d. vas deferens e. prostate gland ANSWER: a 71. A key difference between hormonal regulation of the reproductive systems of human males versus human females is that: a. LH and FSH are released only in females. b. LH and FSH are released only in males. c. LH and FSH are released cyclically in females, but nearly continuously in males. d. LH is released in males and FSH is released in females. e. LH is released continuously in males and FSH is released cyclically in females. ANSWER: c 72. Which structure in the male reproductive system produces an alkaline fluid that helps maintain sperm motility and counteracts the acidity of the female reproductive tract? a. epididymis b. seminal vesicle c. testis d. vas deferens e. prostate gland Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 ANSWER: e 73. Through which structure in the female reproductive system do oocytes travel as they leave the site where they developed? a. uterus b. cervix c. vagina d. ovary e. fallopian tube ANSWER: e 74. The head of a sperm cell contains a specialized organelle called the _____, which contains enzymes used to traverse the outer coating of the egg. a. nucleus b. cytoplasm c. acrosome d. flagellum e. mitochondrion ANSWER: c 75. Which structure in the female reproductive system is at the end of the uterus and produces different kinds of mucus capable of either blocking or guiding sperm through the opening into the uterus? a. fallopian tube b. cervix c. vagina d. ovary e. fallopian tube ANSWER: b 76. Which of the structures does not add secretions to semen? a. prostate b. bulbourethral glands c. epididymis d. seminal vesicles ANSWER: c 77. A woman is taking an oral contraceptive pill. During the time when she is taking the pill, and not the placebo, her GnRH levels are: a. high. b. low. c. increasing. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 d. decreasing. e. surging. ANSWER: b 78. A human fetus that has the genotype XO is phenotypically female, because female sex is determined by: a. the presence of a gene in the X chromosome. b. the absence of TDF, a protein that is produced by a gene on the fetus's Y chromosome. c. only one X chromosome. d. the presence of TDF, a protein produced by a gene in the fetus's X chromosome. ANSWER: b 79. Some children are born with congenital defects. In a child born without a kidney on one side of the body, it is likely that something went wrong during the _____ trimester of pregnancy. a. first b. second c. third d. The answer cannot be determined from the information provided. ANSWER: a 80. A pregnant woman is exposed to a teratogen (something that causes defects in the developing embryo) that inhibits the ability of the epiblast to undergo invagination. Which of the effects will occur? a. Only two tissue layers, the epiblast and the hypoblast, will form, and development will arrest with the bilaminar embryo. b. Only one tissue layer, the ectoderm, will form, and development will arrest with the unilinear embryo. c. Three tissue layers will form, but not from invagination of the epiblast. d. None of the other answer options is correct. ANSWER: a 81. Gastrulation results in the formation of a: a. bilaminar embryo consisting of the epiblast and the hypoblast. b. trilaminar embryo consisting of the epiblast, the mesoblast, and the hypoblast. c. gastrula consisting of a trilaminar embryo. d. trilaminar embryo consisting of ectoderm, endoderm, and mesoderm. e. bilaminar gastrula consisting of the epiblast and the hypoblast. ANSWER: d 82. The inner cell mass of the blastula forms a: a. bilaminar embryo consisting of the epiblast and the hypoblast. b. trilaminar embryo consisting of the epiblast, the mesoblast, and the hypoblast. c. gastrula consisting of a trilaminar embryo. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 d. trilaminar embryo consisting of ectoderm, endoderm, and mesoderm. e. bilaminar gastrula consisting of the epiblast and the hypoblast. ANSWER: a 83. The benefit of asymmetric meiotic division during oogenesis is that it allows most of the cytoplasm to be retained in a single cell. a. true b. false ANSWER: a 84. The key hormone of childbirth is _____, which is released from the posterior pituitary gland. It stimulates uterine contractions which, in turn, stimulate more of the hormone to be released. a. GnRH b. FSH c. LH d. estrogen e. oxytocin ANSWER: e 85. The graphs shown represent relationships between circulating levels of luteinizing hormone (LH) and testosterone in normal men. Which of the graphs indicate negative feedback regulation of testosterone on LH?

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Chapter 40

a. Graph 1 b. Graph 3 c. Graphs 1 and 2 d. Graphs 2 and 3 e. Graph 4 ANSWER: d 86. Consider the graphs.

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Chapter 40

Which graph best represents the amount of circulating testosterone in the blood of a male over the course of several weeks? a. Graph A b. Graph B c. Graph C d. Graph D e. Graph E ANSWER: e 87. Consider the graphs.

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Chapter 40

Select the graph that best illustrates the change in circulating progesterone concentrations in the luteal phase of the human menstrual cycle from just prior to ovulation until the middle of the luteal phase. a. Graph A b. Graph B c. Graph C d. Graph D e. Graph E ANSWER: a 88. Consider the graphs.

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Hypothetically, graph C represents the change in blood LH levels from 2 days prior to ovulation to 2 days after ovulation. Which ovarian hormone must have experienced a rise 2 days before ovulation? a. progesterone b. estradiol c. FSH d. GnRH ANSWER: b 89. A woman is taking oral contraceptive pills containing both progesterone and estrogen. When she is taking these pills, her GnRH levels are: a. high. b. low. c. not affected. d. surging. ANSWER: b 90. During the follicular phase of a woman's menstrual cycle, a woman is diagnosed with estrogen-responsive breast cancer. Her doctor immediately places her on an aromatase inhibitor (AI). The AI prevents an enzyme from producing estradiol anywhere in her body. During the next month, what will happen to her menstrual cycle and her reproductive hormone levels? a. She will miss her next menstrual period, and estradiol, FSH, and LH levels will all be very high (higher than normal). b. She will miss her next menstrual period, and estradiol, FSH, and LH levels will all be very low (lower than normal). Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 c. She will miss her next menstrual period, and her estradiol levels will be low (lower than normal), while her FSH and LH levels will be very high (higher than normal). d. She will miss her next menstrual period, and her estradiol levels will be high (higher than normal), while her FSH and LH levels will be very low (lower than normal). ANSWER: c 91. Consider the graphs.

Which graphs best illustrates the change in size of individual cells of a sea urchin embryo from the time of zygote formation to the end of cleavage? a. 1 b. 2 c. 3 d. 4 ANSWER: c 92. Consider the graphs.

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Which graph best describes the change in the number of cells in an embryo from fertilization to gastrulation? a. 1 b. 2 c. 3 d. 4 ANSWER: a 93. The production of eggs in females is a coordinated process that involves multiple regions in the body generating and releasing hormones. Is the statement true or false regarding hormones of the female reproductive system? The release of FSH is preceded by neurons of the hypothalamus releasing acetylcholine. a. true b. false ANSWER: b 94. The production of eggs in females is a coordinated process that involves multiple regions in the body generating and releasing hormones. Is the statement true or false regarding hormones of the female reproductive system? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 The corpus luteum secretes both estradiol and progesterone. a. true b. false ANSWER: a 95. The production of eggs in females is a coordinated process that involves multiple regions in the body generating and releasing hormones. Is the statement true or false regarding hormones of the female reproductive system? The surge in LH is preceded by a surge in estradiol. a. true b. false ANSWER: a 96. The production of eggs in females is a coordinated process that involves multiple regions in the body generating and releasing hormones. Is the statement true or false regarding hormones of the female reproductive system? Failure of fertilization causes a decrease in progesterone and estradiol and signals the end of the follicular phase of the menstrual cycle. a. true b. false ANSWER: a 97. A man with prostate cancer has his prostate removed. Indicate whether the statement is true or false following his prostate removal. His fertility will decrease because his ejaculate will have a less basic pH and be less able to counteract the acidic environment of the female vagina. a. true b. false ANSWER: a 98. A man with prostate cancer has his prostate removed. Indicate whether the statement is true or false following his prostate removal. His fertility will decrease because there will not be enough seminal fluid produced to carry the sperm from the epididymis into the vas deferens. a. true b. false ANSWER: b 99. A man with prostate cancer has his prostate removed. Indicate whether the statement is true or false Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 following his prostate removal. Fluid from the bulbourethral glands will not be added to the ejaculate, because it is only released after input of fluid from the prostate. a. true b. false ANSWER: b Multiple Response 100. The ectoderm of a human embryo goes on to form all of the choices except the: Select all that apply. a. outer layer of skin. b. pigment cells. c. nerve cells in the brain. d. muscle cells. e. red blood cells. ANSWER: d, e 101. Which of the statements about budding are true? Select all that apply. a. Budding is a form of asexual reproduction. b. Budding is a form of reproduction found in fungi, plants, and some animals. c. Budding leads to a mother cell and a daughter cell that are genetically very different. d. Budding occurs when one organism is divided into many pieces called buds. ANSWER: a, b 102. Which of the statements about parthenogenesis are true? Select all that apply. a. Unfertilized eggs will develop into adults. b. Adults may be haploid or diploid. c. Adults may develop via mitosis or meiosis. d. The parent may be either maternal or paternal. e. Eggs may be fertilized or unfertilized. ANSWER: a, b 103. Which of the statements regarding external fertilization among vertebrates are true? Select all that apply. a. External fertilization is common among mammals. b. External fertilization cannot occur in marine (saltwater) conditions. c. External fertilization is found in some fish and amphibians. d. External fertilization involves releasing gametes directly into a wet environment. e. Among vertebrates, external fertilization is a form of asexual reproduction. ANSWER: c, d 104. In females, estrogen and progesterone: Select all that apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 a. direct development of female reproductive structures during embryogenesis. b. direct development of female secondary sexual characteristics during puberty. c. regulate the menstrual cycle and support pregnancy. d. maintain adult health. e. stimulate gastrulation in the developing embryo. ANSWER: a, b, c, d 105. Which of the choices is a role of the placenta during pregnancy? Select all that apply. a. nutrient exchange between maternal and fetal blood b. gas exchange between maternal and fetal blood c. waste exchange between maternal and fetal blood d. estrogen secretion to maintain the uterine lining e. progesterone secretion stimulates ovulation ANSWER: a, b, c, d 106. Which of the events occurs during childbirth? Select all that apply. a. oxytocin release b. descent and delivery of the baby c. delivery of the placenta d. widening of the cervix e. lengthening of the cervix ANSWER: a, b, c, d 107. Upon reaching the oocyte, what layers must a human sperm cell must pass through before it can fuse with the plasma membrane of the egg? Select all that apply. a. the corona radiata b. the zona pellucida c. the vitelline membrane d. the cell wall e. the chorion ANSWER: a, b 108. Which of the choices occur when a species produces gametes of different sizes? Select all that apply. a. Smaller gametes are designated as male and are called sperm, or spermatozoa. b. Larger gametes are designated as female and are called eggs, or ova. c. Smaller gametes are designated as female and are called eggs, or ova. d. Larger gametes are designated as male and are called sperm, or spermatozoa. ANSWER: a, b 109. Which of the animals may be viviparous? Select all that apply. a. mammals Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 b. reptiles c. sharks d. insects e. birds ANSWER: a, b, c 110. In males, testosterone: Select all that apply. a. directs the development of male reproductive structures during embryogenesis. b. directs the development of male secondary sexual characters during puberty. c. helps maintain adult health. d. acts on Sertoli cells. e. acts on Leydig cells. ANSWER: a, b, c, d 111. Bacteria can take up DNA from the environment, providing a way for bacteria to increase genetic variation and acquire new genes and DNA sequences. Similarly, bdelloid rotifers can incorporate DNA from the environment into their genomes. These similar means of achieving genetic diversity in these two groups of organisms is a good example of: Select all that apply. a. coevolution. b. convergent evolution. c. homology. d. analogy. e. horizontal gene transfer. ANSWER: b, d, e 112. Compared to sexual reproduction, asexual reproduction: Select all that apply. a. requires less investment in attracting mates. b. allows for much more rapid population growth. c. is rarely found as the sole mechanism of reproduction. d. results in the production of genetically identical cells or individuals. e. involves meiosis and fertilization as core processes. ANSWER: a, b, c, d 113. How can prokaryotes reproducing by binary fission increase genetic variation? Select all that apply. a. through mutations b. through DNA transfer during conjugation c. by obtaining DNA from the environment d. by obtaining DNA through viral infection e. via the exchange of genetic material between homologous chromosomes ANSWER: a, b, c, d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 114. Sexual reproduction produces genetically unique individuals by: Select all that apply. a. recombination during gamete formation. b. independent assortment during gamete formation. c. mutation during gamete formation. d. the combination of genetically unique gametes to form a new individual. e. DNA transfer during conjugation. ANSWER: a, b, c, d 115. Viviparity is present in some sharks, snakes, and mammals. As a result, viviparity is a good example of: Select all that apply. a. coevolution. b. convergent evolution. c. homology. d. analogy. e. horizontal gene transfer. ANSWER: b, d 116. In which of the animals can oviparity and amniotic eggs be found? Select all that apply. a. birds b. reptiles c. mammals d. amphibians e. insects ANSWER: a, b, c 117. Vertebrates with external fertilization produce eggs that: Select all that apply. a. have a yolk to provide nutrients to the developing embryo. b. obtain oxygen by diffusion from the environment. c. eliminate wastes by diffusion into the environment. d. obtain water by diffusion from the environment. e. have hard shells that prevent them from drying out on land. ANSWER: a, b, c, d 118. In what groups of vertebrates can you find reproduction that primarily occurs by internal fertilization? Select all that apply. a. mammals b. reptiles c. birds d. fish e. frogs ANSWER: a, b, c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 119. Various hormonal and nonhormonal contraceptive pills are being investigated and developed for men. Which of the choices would be an effective way that these pills might prevent fertilization? Select all that apply. a. blocking sperm production in the testes b. interfering with the process in the epididymis that leads to increased sperm motility c. blocking capacitation d. interfering with action of the enzymes in the acrosome e. preventing the production of estrogen and progesterone ANSWER: a, b, c, d 120. Food shortages during World War II forced many families in Holland to eat tulip bulbs, which contain phytoestrogens. Phytoestrogens are molecules similar to estrogens that mimic the normal estrogens produced by ovaries in animals. Refer to the figure shown.

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Which of the effects might have happened to a Dutch woman who consumed large numbers of tulip bulbs? Select all that apply. a. Progesterone levels would increase, thereby stimulating ovulation. b. Luteinizing hormone would be at a higher level during the cycle, because there would be more phytoestrogen in the woman's system from tulip bulbs. c. Ovulation would occur; however, no corpus luteum would develop. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 40 d. The higher levels of estrogen would suppress ovulation. e. The higher levels of estrogen would mimic the effect of birth control pills. ANSWER: d, e 121. Which of the events occurs during gametogenesis in both human males and females? Select all that apply. a. Diploid stem cells are the precursors to haploid gametes. b. Stem cells form during embryonic development and divide by mitosis. c. All primary spermatocytes and primary oocytes divide by meiosis. d. Secondary spermatocytes and secondary oocytes form by meiosis from primary spermatocytes and primary oocytes. e. Secondary spermatocytes and secondary oocytes undergo both meiotic divisions in quick succession to produce mature gametes. ANSWER: a, b, d 122. Which of the choices are among the changes that sperm may undergo once they enter the female reproductive tract? Select all that apply. a. changes in the fluidity of the plasma membrane b. loss of surface membrane proteins c. changes in membrane potential d. decreased motility e. release of enzymes from the acrosome ANSWER: a, b, c, e

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Chapter 41 Multiple Choice 1. A helper T cell is found bound to an antigen-presenting cell and has released cytokines into the bloodstream. Indicate whether the statement is true or false. The helper T cell will destroy the cell through phagocytosis. a. true b. false ANSWER: b 2. Innate immunity is different from adaptive immunity in that innate immunity: a. only works after exposure to foreign antigens. b. only stimulates responses through mast cells. c. does not require prior exposure to a pathogen. d. does not stimulate cell-mediated immunity pathways. ANSWER: c 3. Which of the answer choices composes the immune system? a. cells. b. organs. c. proteins. d. All of these choices are correct. ANSWER: d 4. You are a doctor who examines a swollen and warm puncture wound on the hand of a patient. Next door, a vet examines a puncture wound on the foot of a dog. Do you and the vet expect to see the same immune response in the dog and human patient? a. Yes, because dogs have innate immune systems but not adaptive immune systems. b. Yes, because dogs have adaptive immune systems but not innate immune systems. c. No, because the immune response is completely unique to humans. d. Yes, because vertebrates have both adaptive immune systems and innate immune systems. e. No, because the foot of a dog is a completely different structure than the hand of a patient. ANSWER: d 5. Which of the processes are considered components of the complement system? a. major histocompatibility complex (MHC) molecules on the surface of cells b. membrane attack complexes (MACs) c. histamine produced by mast cells ANSWER: b 6. What are the components that make up the tertiary structure of an antibody molecule? Refer to the figure shown. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41

a. one heavy chain and one light chain b. two heavy chains only c. two heavy chains and two light chains d. one heavy chain and two light chains ANSWER: c 7. Refer to the figure. The structure in the figure shown is:

a. only found circulating in the blood. b. attached to B cells plasma membrane or circulating in the blood. c. contained in the lysosome of B cells. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 d. found on all types of leukocytes. e. contained in the lysosome of all types of leukocyte. ANSWER: b 8. Refer to the figure. Which area(s) of the molecule shown is/are variable, and what is the consequence of its/their variability?

a. area V; each individual antibody binds many different antigens b. area V; each individual antibody binds a unique antigen c. area C of the heavy chain; each individual antibody binds many different antigens d. area C of the light chain; each individual antibody binds a unique antigen e. Both area V and area C are variable, which allows the body to produce many different antigens using a relatively small genome. ANSWER: b 9. After clonal selection, B cells become either _____ cells that secrete _____, or _____ cells with membranebound antibodies. a. memory; antibodies; plasma b. plasma; antibodies; memory c. memory; antigens; plasma d. plasma; antigens; memory ANSWER: b 10. You are a doctor who examines a swollen and warm puncture wound on the hand of a patient, which was caused by a rusty nail. You vaccinate her for tetanus. In her body's response to the vaccine, how does one of her B cells produce antibodies against C. tetani? a. Each antigen is coded for by a unique gene in the human genome; the correct gene is expressed to produce the antigens against C. tetani. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 b. Rearrangement of different gene segments in B cells occurs after exposure to C. tetani, which creates a unique gene. The cell then produces a large amount of that specific antibody. c. Post-translational modification of a generic antibody protein changes it from a generic form to a specific form against C. tetani. d. Rearrangement of different gene segments in maturing B cells creates a unique gene. After exposure to C. tetani, the B cell with a particular arrangement of DNA produces antibodies that recognize C. tetani antigens and the B cell is stimulated to produce lots of that specific antibody. ANSWER: d 11. A researcher discovers a new antibody. Upon evaluating its genomic sequence, she determines that the antibody is not the result of genomic rearrangement alone. What other processes could have produced this specific antibody? a. germ cell hypomutation b. RNA editing c. alternative mRNA splicing d. hybridization ANSWER: c 12. What is the primary reason it is important to get a new influenza (flu) shot every year? a. The immune system is unable to make memory B cells against influenza viruses. b. The immune system is unable to respond quickly to antigenic shift in viruses. c. The immune system is only able to recognize one viral antigen each year. d. The immune system is only able to devote a small number of B cells to influenza so that it can combat other infections in the body in the same year. ANSWER: b 13. Individuals who have had a transplant are often prescribed immunosuppressant medications. As a result, they often suffer from frequent infections and in some cases tumor formation. Why does this occur? a. Immunosuppressants destroy mast cells in the tissues. b. Immunosuppressants increase recombination in stem B cells. c. Immunosuppressants decrease T cell activity. d. Immunosuppressants activate immunoglobulins. ANSWER: c 14. Autoimmune disorders result from T cells that: a. produce alternate T cell receptors (TCRs) while circulating in the bloodstream. b. are activated by major histocompatibility complex (MHC) molecules displaying self antigens. c. fail to recognize MHC proteins on cell membranes. d. fail to produce TCRs. ANSWER: b 15. T cell receptors (TCRs): Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 a. are generated through genetic rearrangement similar to antibody formation in B cells. b. are antibodies. c. have the same constant regions but different variable regions. d. only bind major histocompatibility complex (MHC) class II proteins on antigen-presenting cells. ANSWER: a 16. Mature T cells that have been released into the circulatory system have undergone: a. only positive selection. b. only negative selection. c. both positive and negative selection. ANSWER: c 17. Why is it useful for T cells that do not recognize self MHC (major histocompatibility complex) molecules to be removed from the maturing T cell population? a. T cells unable to recognize self MHC molecules will be unable to bind to cells that present foreign antigens on MHC class I proteins. b. T cells unable to recognize self MHC molecules will bind and potentially attack, and kill, all the cells of the body. c. T cells unable to recognize self MHC molecules cannot be stimulated by cytokines released in areas of infection. ANSWER: a 18. A loss-of-function mutation in the gene for CD8 on a cytotoxic T cell results in the T cell: a. switching to a helper T cell. b. being unable to bind major histocompatibility complex (MHC) class I proteins. c. being able to bind to antigens on bacteria to trigger phagocytosis. d. causing the development of an autoimmune disorder. ANSWER: b 19. _____ are cells of the adaptive immune system that can activate B cells by binding to major histocompatibility complex (MHC) class II protein-antigen complexes. a. Helper T cells b. Cytotoxic T cells c. Plasma B cells d. Memory B cells e. Phagocytes ANSWER: a 20. As a person ages, his or her thymus shrinks. How does this change affect the effectiveness of vaccinations? a. Vaccinations will have no effect because the entire immune system shuts down when the thymus shrinks to a certain size. b. Vaccinations will be more effective because at birth your thymus made all the T cells you'll need for Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 your entire life. c. Vaccinations will be less effective because B cells will not produce appropriate amounts of antibodies. d. Vaccinations will be less effective because the rate of T cell maturation will be reduced. e. Vaccinations will be as effective in aging patients as in young patients because T cell development occurs in the bone marrow. ANSWER: d 21. You are a doctor examining a patient whose thymus never developed because of a medical condition called DiGeorge syndrome. Which of the consequences of a missing thymus would you expect DiGeorge syndrome individuals to have? a. These individuals would have a compromised innate immune system and an immune system unable to cause inflammation. b. These individuals would to be able to produce antibodies from their B cells. c. These individuals' T cell receptors would not undergo positive and negative selection. d. These individuals would have an enhanced adaptive immune response. ANSWER: c 22. It is difficult to treat tuberculosis infections because the bacterium: a. has no antigens on its cell surface. b. infects a region of the body with little contact from immune cells in the circulatory system. c. has developed resistance to many antibiotics. d. it is small and attaches itself to the surface of cells in the lungs. ANSWER: c 23. After infection with the malaria-causing parasite, P. falciparum, the human liver and circulatory system (especially red blood cells): a. both generate immune cells that express the PfEMP1 T cell receptor. b. are both places in which P. falciparum resides during some stage of its life cycle. c. both have immune cells that can effectively keep up with P. falciparum's antigenic variation. d. are both places where granulomas form in response to P. falciparum infection. ANSWER: b 24. A person inhales a potentially pathogenic bacterium. What defenses will the bacterium encounter as it travels through the human body? a. hairs in the nasal cavity b. mucus and cilia in the respiratory tract (if the bacterium enters the respiratory tract) c. digestive enzymes in the stomach (if the bacterium enters the gastrointestinal tract) d. phagocytes (if the bacterium enters the bloodstream) e. All of these choices are correct. ANSWER: e Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 25. Which of the statements are true regarding histamine? a. It can be produced by mast cells. b. It contributes to inflammation. c. It causes blood vessels to dilate. d. It changes the permeability of blood vessels. e. All of these choices are correct. ANSWER: e 26. Refer to the graph shown indicating relative antibody response. At which point on the curve would you expect circulating, novel viral concentrations to be highest?

a. A b. B c. C d. D e. E ANSWER: a 27. Innate immunity activation depends on: a. previous exposure to a foreign antigen. b. diversity of antibodies in the blood stream. c. major histocompatibility complex (MHC) class II proteins. d. memory B cells. e. None of the other answer options is correct. ANSWER: e 28. Lupus is an autoimmune disorder. Which of the qualities would you expect to be a symptom of lupus? a. an increase in ciliary action of the esophagus b. increased phagocytosis of bacterial pathogens in the body Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 c. B cells binding to antigens on host cells and targeting their degradation d. increased blood clotting if the skin is cut ANSWER: c 29. Skin is an important part of the immune response because: a. it is relatively porous. b. the presence of white blood cells on the surface provides protection to the host. c. it acts as a barrier to keep out pathogens. d. bacteria on the surface kill viruses that touch the skin. ANSWER: c 30. You are a doctor who examines a swollen and warm puncture wound on the hand of a patient. Which was not a response of the patient's immune system to this injury when it occurred? a. Antibodies were produced as part of the adaptive immune response. b. Phagocytes attacked foreign cells that entered the wound. c. The innate immune system triggered an inflammation response. d. Cell division was inhibited at the wound site. ANSWER: d 31. A man is scratched by his cat. A phagocyte near the scratch site recognizes and engulfs a bacterium. Shortly thereafter, more phagocytes arrive in the tissue surrounding the scratch. How are the additional phagocytes recruited to the site of the scratch? a. by antigens secreted by the bacteria b. by cytokines produced by the bacteria c. by antigens secreted by the initial phagocyte d. by cytokines secreted by the initial phagocyte ANSWER: d 32. Your phagocytes have transmembrane proteins called toll-like receptors (TLR) on their surface, but your mast cells do not because: a. phagocytes carry the genetic information to make TLRs, but mast cells do not. b. phagocytes run transcription and translation, but mast cells only run transcription. c. some body's cells contain ribosomes but most do not. d. phagocytes express some genes that are not expressed in mast cells (and vice versa). ANSWER: d 33. Immunodeficiency is a general term that refers to the loss of: a. one or more components of the immune system. b. functional mast cells in the body. c. immune system function with age. d. immune system function because of genetic mutations. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 ANSWER: a 34. The complement system refers to: a. proteins present on macrophages that recognize foreign proteins. b. proteins circulating in the blood that are activated by opsonization. c. proteins circulating in the blood that are activated by antibodies or molecules on pathogens. d. proteins that are activated when histamine levels increase. ANSWER: c 35. Tetanus is a bacterial infection that is commonly acquired from puncture wounds. Why do you need to be vaccinated against the flu every year, but you only need a virus booster against tetanus every 10 years? a. Antibody production against viruses drops off about 10 times faster than antibody production against bacterial infections, so you need to be vaccinated 10 times more frequently against the flu. b. The flu is everywhere but bacterial infections are incredibly rare, so you don't need to be as careful in maintaining your immunity against them. c. There is a higher rate of mutation in viruses than in bacteria, so you need to change the antibodies you are producing (through vaccination) every year for the flu. d. Infections that enter the body through the bloodstream (like tetanus) are much less toxic than infections that enter the body through binding to epithelial cells (like the flu). ANSWER: c 36. You are a doctor who examines a swollen and warm puncture wound on the hand of a patient. Because the wound was made by a rusty nail, you are concerned about infection by Clostridium tetani bacteria, which cause tetanus. You are relieved to see that the patient received a tetanus vaccine at her last checkup a year ago. The vaccine: a. exposed the patient to C. tetani so that she contracted a mild version of tetanus. b. exposed the patient to deactivated C. tetani, so that her B cells would produce antibodies to the bacteria with no risk of contracting the disease. c. stimulated the production of mast cells by her bone marrow. d. exposed the patient to other types of bacteria, so that her immune system would be prepared for infection by prokaryotes. ANSWER: b 37. The large number of antibodies that can be produced in a single individual is a result of: a. making multiple unique antibodies by genomic rearrangement. b. being able to produce unique antibodies from both the maternal and paternal alleles. c. making a unique antibody from all other B cells by genomic rearrangement. d. secreting multiple antibodies that will be presented on the surface of other B cells. ANSWER: c 38. All antibodies in a single individual are similar in that they all contain: a. the same C segment but different V segments. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 b. C and V segments on their heavy and light chains. c. the same C segments on their heavy chains, but different C segments on their light chains. d. None of the answer options is correct. ANSWER: b 39. Which type of cell produces antibodies? a. helper T cells b. cytotoxic T cells c. mast cells d. B cells e. natural killer cells ANSWER: d 40. When a B cell binds an antigen, the B cell undergoes clonal selection, and B cells that do not present the appropriate antibody die. a. true b. false ANSWER: b 41. Cells infected with Plasmodium falciparum evade being filtered from the bloodstream by the spleen by: a. being too large to enter vessels of the spleen. b. inhibiting extravasation so they do not enter the spleen. c. producing adhesion proteins that cause them to stick to the sides of blood vessels. d. destroying major histocompatibility complex (MHC) proteins on the surface of red blood cells. ANSWER: c 42. Cells with class I major histocompatibility complex (MHC) proteins that contain a foreign antigen will be destroyed after: a. being bound by helper T cells. b. being bound by cytotoxic T cells. c. stimulating the membrane attack complex (MAC). d. releasing histamine. ANSWER: b 43. Some populations of cheetahs in Africa have been shown to be highly inbred. Individuals in the population are so closely related to each other that they do not reject skin transplants from other individuals in the population. Which of the statements would also be true based on this information? a. Individuals in this population lack cytotoxic T cells. b. Individuals in this population have the same major histocompatibility complex (MHC) proteins. c. Individuals in this population have MHC proteins that are unable to bind antigens. d. Individuals in this population lack class II MHC proteins. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 ANSWER: b 44. T cell receptors are only able to bind antigens that are presented with major histocompatibility complex (MHC) proteins. a. true b. false ANSWER: a 45. Virus-infected cells are detected and destroyed by which cell-mediated immune response? a. B cells b. helper T cells c. cytotoxic T cells d. macrophages ANSWER: c 46. Which of the answer choices is not part of the structure of an antibody? a. a major histocompatibility complex b. an antigen-binding site c. heavy chains d. a hypervariable region e. light chains ANSWER: a 47. Which of the answer choices is required to activate a helper T cell, which in turn can stimulate antibody production by B cells? a. an antigen alone b. an antibody c. a major histocompatibility complex alone d. an antibody displayed by a major histocompatibility complex protein e. an antigen displayed by a major histocompatibility complex protein ANSWER: e 48. Consider the graph. Which letter under the graph shown corresponds to increased mitotic division in memory B cells?

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Chapter 41

a. A b. B c. C d. D e. E ANSWER: d 49. A small child enters a clinic for the third time in one month with another bad skin infection filled with pus (tissue fluid combined with large numbers of bacteria growing in the fluid). Decreased numbers, or loss of function, in which of the cell types could account for these symptoms? a. eosinophils b. neutrophils c. T cells d. B cells ANSWER: b 50. You are a doctor and have a patient who has been injured by a rusty nail. As a precaution, you vaccinate her against tetanus. In response to the vaccine, how does her body produce antibodies against C. tetani and prevent future illness due to tetanus? a. Every B cell in her body produces antibodies against C. tetani. b. Every cell in her immune system produces antibodies against C. tetani. c. The B cell with the appropriate antibody is stimulated to divide, producing plasma cells that make antibodies to C. tetani, and memory cells that "remember" C. tetani. d. The B cell that produces the appropriate antibody undergoes genomic rearrangement in order to produce other cells that produce the same antibody. ANSWER: c 51. B cells undergo _____, where a specific cell will divide in response to its associated antibody recognizing an antigen. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 a. clonal selection b. genomic recombination c. autoimmune disease d. inflammation e. positive selection ANSWER: a 52. You are a doctor examining a patient's blood test reports. You are looking at the ratio of helper T cells to cytotoxic T cells. Helper T cells _____; cytotoxic T cells _____. You can tell them apart by the type of protein they _____. a. activate other immune system cells; kill altered host cells; secrete b. activate other immune system cells; kill altered host cells; express on their plasma membrane c. kill altered host cells; activate other immune system cells; secrete d. kill altered host cells; activate other immune system cells; express on their plasma membrane e. activate antigens; destroy antigens; digest ANSWER: b 53. You are a doctor examining a patient's blood-test results. The patient's T cell count is low. You are concerned that the patient will: a. be susceptible to infection by all types of bacteria. b. have low antigen levels from decreased B-cell activity. c. have strong allergies. d. have a reduced ability to kill host cells if she becomes infected. e. develop cancer. ANSWER: d 54. Diseased organs can sometimes be replaced by healthy organs from donors. A long-standing problem with organ transplants is rejection of the organ by the recipient. Rejection would not be expected if the transplanted organ came from a(n): a. sibling. b. person with the same blood type. c. fraternal (dizygotic) twin. d. identical (monozygotic) twin. e. closely related species. ANSWER: d 55. Sometimes an organ can be replaced by moving it from one part of the body to another. This can be done, for example, to replace damaged skin or joints. In these cases: a. rejection of the transplanted organ would still be expected. b. rejection of the transplanted organ would not be expected. c. immunosuppressant drugs would be required to prevent transplant rejection. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 d. antibodies would be generated against the transplanted organ. e. antigens would be generated against the transplanted organ. ANSWER: b 56. A new technique to replace diseased organs is harvesting stem cells from the patient's own body and using them to grow a new organ that is then transplanted into the body. In this case: a. rejection of the transplanted organ would still be expected. b. rejection of the transplanted organ would not be expected. c. immunosuppressant drugs would be required to prevent transplant rejection. d. antibodies would be generated against the transplanted organ. e. antigens would be generated against the transplanted organ. ANSWER: b 57. You are a doctor who examines a swollen and warm puncture wound on the hand of a patient. Does the fact that the patient's immune system responded normally to this wound tell you that every aspect of his or her immune system is functioning normally? a. Yes, because the response of swelling and warmth at a wound site shows a complete response by both the innate and adaptive immune systems of animals. b. No, because this response does not include the response of inflammation. c. No, because this response does not include the complement response or adaptive immune response. d. None of the answer options is correct. ANSWER: c 58. A mother may be able to donate blood to one of her offspring but may not be a match for a kidney transplant because major histocompatibility complex (MHC) proteins are: a. not presented on red blood cells (RBCs), so the child will not reject the blood but could reject the kidney because of MHC incompatibility. b. presented on RBCs, so the child's immune system will recognize those as self but could reject the kidney because of MHC incompatibility. c. not presented on RBCs, so the child will not reject the blood but could reject the kidney because it is too large for the child's body. d. presented on RBCs, so the child's immune system will recognize those as self but genetic recombination of MHC alleles in organs will result in rejection of the kidney. ANSWER: a 59. If an individual carries a mutation that results in a complete lack of major histocompatibility complex (MHC) class I proteins, how would the individual's immune response be affected? a. Memory cell antibodies would not be able to attach to antigens. b. Cytotoxic T cells would not be able to target (and destroy) infected cells. c. Helper T cells would not be able to interact with antigen-presenting cells. d. Plasma cell antibodies would not be able to attach to antigens. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 60. What is the purpose of getting a flu shot every year? a. It lets your body get used to having the flu so that you won't feel as sick when you actually do get the flu. b. It exposes you to antigens present on the flu virus specific to this year infectious strain so that your immune system can make antibodies to the virus. c. It exposes you to antibodies present on the flu virus so that your immune system can make antigens. d. It exposes your immune system to the virus specific to this year infectious strain and causes genomic rearrangement in memory B-cells. ANSWER: b 61. A helper T cell is found bound to an antigen-presenting cell and has released cytokines into the bloodstream. Indicate whether the statement is true or false. The helper T cell is bound to the antigens on an invading bacterial cell. a. true b. false ANSWER: b 62. A helper T cell is found bound to an antigen-presenting cell and has released cytokines into the bloodstream. Indicate whether the statement is true or false. The variable region of the T cell receptor has recognized an antigen on a macrophage with major histocompatibility complex (MHC) class I proteins. a. true b. false ANSWER: b 63. A helper T cell is found bound to an antigen-presenting cell and has released cytokines into the bloodstream. Indicate whether the statement is true or false. The helper T cell was positively selected in the thymus. a. true b. false ANSWER: a 64. A helper T cell is found bound to an antigen-presenting cell and has released cytokines into the bloodstream. Indicate whether the statement is true or false. Cytokines will be released into the circulatory system. a. true b. false ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 65. A helper T cell is found bound to an antigen-presenting cell and has released cytokines into the bloodstream. Indicate whether the statement is true or false. The helper T cell will destroy the cell through phagocytosis. a. true b. false ANSWER: b 66. You are a doctor who examines a swollen and warm puncture wound on the hand of a patient. Indicate whether the statement is true or false. The wound is swollen and red due to vasoconstriction. a. true b. false ANSWER: b 67. You are a doctor who examines a swollen and warm puncture wound on the hand of a patient. Indicate whether the statement is true or false. By breaking the skin, the wound breached one part of the patient's innate immune system. a. true b. false ANSWER: a 68. You are a doctor who examines a swollen and warm puncture wound on the hand of a patient. Indicate whether the statement is true or false. Healing of the wound may require both the innate and the adaptive immune system. a. true b. false ANSWER: a 69. You are a doctor who examines a swollen and warm puncture wound on the hand of a patient. Indicate whether the statement is true or false. The wound is swollen and red due to vasodilation. a. true b. false ANSWER: a 70. You are a doctor who examines a swollen and warm puncture wound on the hand of a patient. Indicate whether the statement is true or false. Cells in the wound secrete antihistamines. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 a. true b. false ANSWER: b 71. You are bitten by a mosquito that carries Plasmodium. Indicate whether the statement is true or false. Histamine is released at the site of the bite and swelling occurs. a. true b. false ANSWER: a 72. You are bitten by a mosquito that carries Plasmodium. Indicate whether the statement is true or false. There is no response because single-cell eukaryotes have no effect on humans. a. true b. false ANSWER: b 73. You are bitten by a mosquito that carries Plasmodium. Indicate whether the statement is true or false. Cytotoxic T cells recognize and kill Plasmodium. a. true b. false ANSWER: b 74. You are bitten by a mosquito that carries Plasmodium. Indicate whether the statement is true or false. The complement system will lyse the infected cells. a. true b. false ANSWER: b 75. You are bitten by a mosquito that carries Plasmodium. Indicate whether the statement is true or false. There will be a strong secondary response since everyone is vaccinated for malaria at birth. a. true b. false ANSWER: b 76. A pathogen has been recognized and bound by a free antibody. Indicate whether the statement is true or false. Cytotoxic T cells will bind the pathogen and destroy it. a. true Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 b. false ANSWER: b 77. A pathogen has been recognized and bound by a free antibody. Indicate whether the statement is true or false. Inactive circulating membrane attack complex (MAC) proteins will become activated. a. true b. false ANSWER: a 78. A pathogen has been recognized and bound by a free antibody. Indicate whether the statement is true or false. Opsonization and phagocytosis will occur. a. true b. false ANSWER: a 79. A pathogen has been recognized and bound by a free antibody. Indicate whether the statement is true or false. Histamine levels in the blood could increase. a. true b. false ANSWER: a Multiple Response 80. Phagocytic cells are an important part of the innate immune system because of their ability to: Select all that apply. a. engulf many different pathogens. b. present foreign antigens that stimulate adaptive immune responses. c. attack cells infected by a virus. d. produce multiple antibodies. ANSWER: a, b 81. A researcher purifying antibodies in patients' blood samples discovers that one patient has an abnormally high level of IgE. He goes back to check the patient's chart. What are the likely symptoms of this patient? Select all that apply. a. persistent infections of the mucous membranes b. persistent viral infections c. severe allergies Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 d. severe asthma ANSWER: c, d 82. Which of the statements is/are true regarding antibody heavy (H) chains? Select all that apply. a. Antibodies contain two heavy chains. b. Heavy chains contain variable and constant regions. c. Heavy chains are formed by the recombination of V, J, C, and D gene segments. d. Heavy chains contain only one hypervariable region. ANSWER: a, b, c 83. Which of the cells are part of the adaptive immune system? Select all that apply. a. memory cells b. plasma cells c. mast cells d. T cells e. B cells ANSWER: a, b, d, e 84. T cell maturation in the thymus requires selection of T cells: Select all that apply. a. with adequate variability to bind foreign antigens. b. that do not bind tightly to self antigens. c. that recognize major histocompatibility complex (MHC) class I and II. d. None of the choices is correct ANSWER: b, c 85. Which of the statements is (are) true regarding bacteria in the human body? Select all that apply. a. Human bodies may contain pathogenic bacteria, but these are often kept in check by other (nonpathogenic) species. b. Human bodies contain bacteria that are beneficial to certain processes (for example, digestion). c. Cholera is caused by a bacterium that infects and disables part of the immune system. ANSWER: a, b 86. What are possible mechanisms of action of phagocytes? Select all that apply. a. They engulf and digest pathogens with lysosome-enclosed enzymes. b. They destroy pathogens by the production of reactive nitrogen species. c. They specifically target and kill virus-infected cells. d. They form membrane attack complexes (MACs). ANSWER: a, b 87. You are a doctor examining a patient whose thymus is underdeveloped because of a medical condition called DiGeorge syndrome. What concerns do you have about this patient's immune system? Select all that Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 apply. a. The level of antigens in the blood will be low. b. Cell-mediated immunity will be weakened. c. The possibility of autoimmune diseases will have increased because T cell maturation will be affected. ANSWER: b, c 88. What are some of the ways that viruses can bypass an animal's immune system? Select all that apply. a. antigenic disappearance b. antigenic display c. antigenic shift d. antigenic drift ANSWER: c, d 89. Recall Edward Jenner's experiment, where a boy was inoculated with cowpox and, as a result, did not contract smallpox when he was exposed to the virus. Why didn't the boy get smallpox? Select all that apply. a. because of immune "memory" provided by plasma cells b. because of immune "memory" provided by memory cells c. because of a primary immune response d. because of a secondary immune response ANSWER: b, d 90. Which of the statements is/are true regarding the secondary immune response? Select all that apply. a. It occurs more slowly than the primary immune response. b. It occurs faster than the primary immune response. c. It produces more antibodies than the primary immune response. d. It produces fewer antibodies than the primary immune response. ANSWER: b, c 91. You are a doctor examining a patient's blood-test results. The patient's T cell count is low. You examine scans of the patient's thymus. It is underdeveloped, which means the T cells in the blood will: Select all that apply. a. be present in lower numbers because the thymus is where T cells are formed. b. be present in lower numbers because the thymus is where T cells mature. c. not have undergone selection against self-recognition, and the patient will be more susceptible to autoimmune diseases. ANSWER: b, c 92. Which of the answer choices are unique characteristics or components of the adaptive—but not innate— immune system? Select all that apply. a. immune "memory" b. opsonization Copyright Macmillan Learning. Powered by Cognero.

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Chapter 41 c. natural killer cells d. phagocytosis e. specificity ANSWER: a, e 93. Which of the statements is true regarding T cell receptors? Select all that apply. a. They recognize antigen-MHC complexes. b. They contain variable and constant regions. c. They are composed of two heavy (H) chains and two light (L) chains. d. They are formed by the rearrangement of V, J, C, and D gene segments. e. They can be found in the bloodstream independent of T cells. ANSWER: a, b, d 94. A woman goes to the doctor and is diagnosed with rheumatoid arthritis. What factors led to her developing this disease? Select all that apply. a. Some of her T cells were not positively selected. b. Some of her T cells were not negatively selected. c. Some of her T cells are recognizing and responding to her self antigens. d. Some of her T cells are not recognizing and associating with her major histocompatibility complex (MHC) proteins. ANSWER: b, c 95. Which of the cell types are part of the adaptive immune system and not the innate immune system? Select all that apply. a. eosinophils b. plasma cells c. mast cells d. dendritic cells e. helper T cells ANSWER: b, e 96. Which of the characteristics of the malaria parasite Plasmodium listed reflect the evolution of mechanisms to evade the immune system? Select all that apply. a. The fact that the malaria parasite is a eukaryote. b. The fact that infected cells remain in the blood vessels and out of the spleen because the malaria parasite expresses a sticky protein in its plasma membrane. c. The fact that the malaria parasite has a nucleus and membrane-bound organelles. d. The fact that the malaria parasite produces proteins. e. The fact that the malaria parasite shows antigenic variation in its cell surface proteins. ANSWER: b, e

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Chapter 42 Multiple Choice 1. As in many other animals, the endoderm in ctenophores functions in: a. respiration. b. prey capture. c. digestion. d. locomotion. ANSWER: c 2. Comparative embryology helped clarify the relationships among major groups of animals. Bilaterians, for example, are: a. diploblastic. b. triploblastic. c. coelomates. d. acoelomates. e. protostomates. ANSWER: b 3. In sponges, the cells responsible for generating water currents that facilitate gas exchange and nutrition are the: a. choanocytes. b. spicules. c. mesohyl. d. sperm. e. symbionts. ANSWER: a 4. In sponges, cells differentiate into sperm or eggs in the _____. Only _____ are released into the water. a. mesohyl; sperm b. mesohyl; choanocytes c. mesohyl; spicules d. cavity; sperm e. cavity; choanocytes ANSWER: a 5. Cnidarians share a common body plan that takes two forms. The free-floating form is called a medusa, whereas the sessile form is called a: a. polyp. b. larva. c. ctenophore. d. mesoglea. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 ANSWER: a 6. In cnidarians, the mouth of both the medusa and polyp forms opens into an internal _____ where _____ digestion takes place. a. gastric cavity; extracellular b. gastric cavity; intracellular c. mesoglea; extracellular d. mesoglea; intracellular e. None of the answer options is correct. ANSWER: a 7. Arthropods are an incredibly successful group of animals. This is thought to be primarily due to the diversity in feeding appendages found in groups of arthropods. How did the diversity in feeding appendages arise? a. modification of walking appendages into feeding appendages in anterior-most segments b. modification of feeding appendages into walking appendages in body segments behind the head c. modification of portions of the exoskeleton to make room for feeding appendages d. modification of patterns of metamorphosis to allow development of feeding appendages ANSWER: a 8. Which of these groups includes almost half of all animal phyla? a. ecdysozoans b. cnidarians c. lophotrochozoans d. annelids ANSWER: c 9. Insects exchange gases through pores called _____, which connect to an internal system of tubes called _____, which are in turn connected to respiring tissues. a. spiracles; tracheae b. tracheae, spiracles c. spiracles; radulae d. radulae; tracheae e. None of the answer options is correct. ANSWER: a 10. _____, such as sea squirts, share features of the chordate body plan during early development. a. Tunicates b. Echinoderms c. Chordates d. Cephalochordates ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 11. The _____, which include the sharks, skates, and rays, are unique in possessing skeletons made of _____. a. chondrichthyes; cartilage b. coelacanths; cartilage c. osteichthyes; calcium carbonate d. chondrichthyes; silica e. osteichthyes; cartilage ANSWER: a 12. Macroscopic fossils of organisms thought to be animals first appeared in rocks deposited around _____ million years ago (mya). a. 579 b. 542 c. 420 d. 360 e. 100 ANSWER: a 13. The fossil remains of animals with body plans recognizable as arthropods, echinoderms, mollusks, and other bilaterians first appeared during the _____ Period, around _____ million years ago. a. Ediacaran; 579 b. Cambrian; 541 c. Devonian; 420 d. Devonian; 360 e. Permian; 100 ANSWER: b 14. Animals colonized land after plants did. The first land animals were _____, which first colonized land about _____ million years ago. a. insects; 579 b. hemichordates; 542 c. chelicerates; 420 d. insects; 360 e. myriapods; 100 ANSWER: c 15. The protostomes are divided into two monophyletic groups, the: a. Lophotrochozoa and Ecdysozoa. b. Protostomia and Deuterostomia. c. Chordata and Hemichordata. d. Eumetazoa and Protozoa. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 16. Mammals are divided into three major groups based on: a. ability to produce milk. b. presence of hair. c. where early development of the young occurs. d. presence or absence of a pouch. ANSWER: c 17. Animals are distinguished from other groups of eukaryotic organisms in that they: a. are multicellular heterotrophs lacking cell walls. b. have embryos that include a gastrula stage. c. produce collagen. d. are mobile. ANSWER: a 18. Many biologists were surprised by molecular data supporting the hypothesis that ctenophores are the sister group to all other animals. What anatomical or morphological features of ctenophores are conventionally thought to support a close relationship with bilaterian animals? a. the rudimentary one-way gut (mouth to anal pore) b. the presence of what many think to be a rudimentary mesoderm c. the presence of an outer epithelium d. the presence of gelatinous material in the interior of the ctenophore body ANSWER: a 19. Examine the phylogenetic tree shown.

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Chapter 42

According to this phylogeny, bilaterians (animals with bilateral symmetry) are most closely related to: a. cnidarians. b. sponges. c. the group containing sponges and cnidarians. d. the group containing choanoflagellates, sponges, and cnidarians. e. None of the answer options is correct. ANSWER: a 20. Examine the phylogenetic tree shown.

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Chapter 42

According to this phylogeny, which of the answer choice correctly lists the sequence in which multicellularity, tissues, and bilateral symmetry arose, from the oldest characteristic to the most recent? a. multicellularity, tissues, bilateral symmetry b. multicellularity, bilateral symmetry, tissues c. tissues, bilateral symmetry, multicellularity d. tissues, multicellularity, bilateral symmetry e. The sequence cannot be determined from the phylogeny. ANSWER: a 21. Early scientists used morphological characteristics to create a phylogeny of animals, but recently, DNA sequence data have helped in the revision of taxa phylogenies within the larger groups (that is, bilaterians, sponges, cnidarians). Why are morphological characteristics like "fate of the blastopore" or "body symmetry" relatively good at predicting larger-scale branching patterns on phylogenies? a. The development of these characteristics is controlled genetically, therefore similarities in DNA sequence would be expected in more closely related taxa. b. Characteristics like those described are necessary for living, therefore all organisms must have those traits. c. All animals are descended from a common ancestor that had the traits described above. d. These features reflect adaptations to life on land. ANSWER: a 22. In the past, species of sponges, cnidarians, and ctenophores were referred to as "lower" or "primitive" animals because of their relatively simple body plans and lack of any sophisticated organ differentiation. What suggest that the terms "lower" or "primitive" are not useful as descriptors for these organisms? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 a. Many members of these groups are active predators and feed on other organisms. b. These groups have been around for millions of years; therefore, their evolutionary path has been successful in spite of limited complexity. c. These groups are not primary producers and therefore should not be classified as "lower" because they occur higher on the trophic pyramid. ANSWER: b 23. Sponges move huge volumes of water through their bodies every day. Although this serves the function of filter feeding for the organism, how might this also provide an "ecosystem" service (something that affects and/or benefits other organisms in their environment)? a. They can filter small poisonous fish out of the water that may harm other organisms. b. They create small currents in their microenvironment that help to circulate and clean water locally for other organisms. c. They produce large numbers of gametes that will become food for other sponges. d. Their body chambers serve as places where smaller organisms can live and be protected from certain predators. ANSWER: b 24. Multicellularity allows sponges to: a. reproduce. b. access nutrients. c. form a protective skin. d. None of the answer options is correct. ANSWER: c 25. Cnidarians form true epithelial tissue. a. true b. false ANSWER: a 26. Unlike cnidarians, a ctenophore's (comb-jelly's) anterior-posterior axis allows for: a. directed gas exchange. b. more efficient nutrient digestion. c. differentiated muscle cells. d. tentacles. ANSWER: b 27. Within insects, the main evolutionary dividing line is between those that: a. fly and those that do not fly. b. have lost wings that were present in their common ancestor and those that still fly. c. shed their wings as they molt and those that do not. d. undergo metamorphosis and those that don't. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 e. have eggshells and those that don't. ANSWER: d 28. Examine the phylogenetic tree shown.

According to this phylogeny, the sister group to the Hemichordates is the: a. echinoderms. b. chordates. c. cephalochordates. d. group containing tunicates and vertebrates. e. group containing echinoderms and chordates. ANSWER: a 29. Compared to the poriferans and cnidarians, bilaterians were able to develop more specialized organs because they have: a. radial symmetry. b. bilateral symmetry. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 c. two germ layers. d. three germ layers. ANSWER: d 30. Examine the phylogenetic tree shown.

According to this phylogeny, the group marked "bony fish" is: a. monophyletic. b. paraphyletic. c. polyphyletic. d. diphyletic. e. None of the answer options is correct. ANSWER: b 31. What is one of the primary reasons given for why there are very few fossils from the earliest (pre-Ediacaran fauna) animals? a. The area covered by land, where early animals lived, was small; therefore, there were fewer opportunities for sedimentary rocks to form fossils. b. The area covered with water was much smaller than present day; therefore, there were fewer places for fossilization to occur on the ocean floor. c. The earliest animals were small and soft-bodied and did not have body parts that could be fossilized easily. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 d. The earliest animals were distributed in high oxygen environments and decomposition occurred too rapidly for fossilization to occur. ANSWER: c 32. Mass extinctions have occurred five times in Earth's history. The end Permian and Cretaceous extinctions were responsible for removing a large percentage of organisms from the planet. How do these extinctions contribute to the biodiversity we see today? a. Species that have gone extinct are able to re-evolve from the ancestors that survived the extinction. b. Species that remain after the extinction are able to radiate, new adaptations arise, and these produce the diversity seen today. c. Species that remain after the extinction are unable to speciate; therefore, the number of species on Earth today is lower than just before either the end of the Permian or the Cretaceous extinctions. d. Species that remain after an extinction represent all of the lineages that were present before the extinction event; therefore, diversity of lineages is not changed by extinction. ANSWER: b 33. Which of these characteristics would you expect to find in a member of the Bilateria? a. specialized head region b. segmentation c. coelom d. two tissue layers (diploblastic) ANSWER: a 34. An avid naturalist diver has decided to vacation on a remote island, possibly never visited by a human before, with a rocky intertidal zone. While out exploring, she comes across what looks like a jellyfish that has bilateral symmetry. She considers what she knows of the simple phylogeny of animals and would like to place this species into the group Bilateria. Name one other characteristic that would lend support to placing this organism in the group Bilateria. a. the presence of collagen in its tissues b. the presence of an oral cavity c. the presence of digestive tissues d. the presence of three different tissue layers ANSWER: d 35. Cnidarians and ctenophores are included in the group animals. Based on our current understanding of the evolution of these groups, neither is classified as protostome or deuterostome. Which statement is the most likely explanation? a. These groups do not undergo gastrulation; therefore, the timing of mouth development cannot be determined. b. These groups do not have an anus; therefore, the timing of mouth development (first or second) cannot be determined. c. These groups only have two tissue layers; currently only triploblastic organisms are classified based on the fate of the blastopore. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 d. These two groups do not have an axis of orientation for their bodies; they are radially symmetric. ANSWER: c 36. If, as some molecular data suggest, comb-jellies turn out to be the sister group of all other animals, all of these statements would be true except: a. choanoflagellates are the sister group to animals. b. triploblastic animals evolved from diploblastic ancestors. c. cnidarians and comb-jellies share a common ancestor that had a gut and nervous system. d. sponges evolved their simple structure by losing features of complex anatomy. ANSWER: c 37. Bilaterian animals are divided into two monophyletic groups, the _____ and the _____, which are supported by patterns of comparative anatomy, comparative embryology, and molecular sequence data. a. Protostomia; Deuterostomia b. Lophotrochozoa; Ecdysozoa c. Chordata; Hemichordata d. Eumetazoa; Protozoa ANSWER: a 38. The most diverse group of arthropods is the: a. insects. b. chelicerates. c. myriapods. d. crustaceans. e. hemichordates. ANSWER: a 39. The most numerous animals (in terms of abundance) on Earth are the: a. nematodes. b. mollusks. c. vertebrates. d. cnidarians. e. placozoans. ANSWER: a 40. Pollinating insects are in part responsible for _____ diversity. a. bird b. flowering plant c. predatory insect d. herbivore e. None of the answer options is correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 ANSWER: b 41. Tardigrades, also called waterbears, are very small bilaterally symmetric organisms with paired appendages along their bodies and a reduced coelom. Knowing that tardigrades molt their surface cuticle, what is the most likely interpretation of their appendages? Refer to the phylogenetic tree shown.

a. They are homologous with the limbs of tetrapods. b. They are homologous with the paired appendages of polychaete worms. c. They are homologous with the paired appendages of arthropods. d. They are a unique feature, not closely similar to appendages in any other animal. ANSWER: c 42. Insects and tetrapods have both diversified tremendously on land. Which convergently evolved features help to explain the evolutionary success of these two groups? a. internal skeleton Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 b. a flowthrough gut, with mouth and anus at either end c. lungs d. jointed limbs ANSWER: d 43. When referring to the sister group of the cephalochordates, why do some biologists use the name craniata instead of the name vertebrata? a. Hagfish are too different from cephalohordates to be included in that group. b. Hagfish, which are included in the group, have a cranium but no vertebrae. c. Hagfish are not represented by very many species and therefore cannot be given their own taxonomic grouping. d. Hagfish are understudied and therefore can only be grouped with other taxa based on their lack of characteristics. ANSWER: b 44. Examine the phylogenetic tree shown.

According to this phylogeny, the tetrapods are: a. monophyletic. b. paraphyletic. c. polyphyletic. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 d. diphyletic. e. None of the answer options is correct. ANSWER: a 45. Animals transitioned from water to land through a series of significant morphological and physiological changes. Which of these adaptations did not contribute to the transition? a. the ability to breathe air b. the evolution of appendages that could support the weight of the body c. the evolution of jaws for better prey capture and feeding ANSWER: c 46. Ediacaran fossils are generally interpreted as animals, but most show no evidence of a mouth, gut, or other organ systems. Which group of living animals best helps us to understand how Ediacaran organisms might have fed, exchanged gases, and reproduced? a. cnidarians b. placozoans c. sponges d. annelid worms ANSWER: b 47. What role has mass extinction played in animal evolution? a. Mass extinction can remove ecologically dominant groups, paving the way for new or previously minor groups to diversify. b. Mass extinction selectively removes poorly adapted animals. c. Mass extinction eliminates many marine species, forcing survivors to colonize land to find food. d. All of these choices are correct. ANSWER: a 48. The Ecdysozoa include: a. annelid worms. b. mollusks. c. sea anemones. d. arthropods. e. vertebrates. ANSWER: d 49. Examine the phylogenetic tree shown.

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Chapter 42

Among the vertebrates, the different groups of fish are: a. polyphyletic. b. paraphyletic. c. monophyletic. ANSWER: b 50. Which list represents only phyla that include at least some species which are terrestrial? a. Nematoda, Mollusca, Annelida, Arthropoda b. Porifera, Mollusca, Annelida, Chordata c. Cnidaria, Mollusca, Arthropoda, Chordata d. Nematoda, Annelida, Arthropoda, Echinodermata e. Cnidaria, Annelida, Arthropoda, Chordata ANSWER: a 51. A key characteristic found in all reptiles, birds, and mammals that allowed them to become fully terrestrial is: a. four walking limbs. b. a bony vertebral column. c. an amniotic egg. d. lungs. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 ANSWER: c Multiple Response 52. Echinoderms are unique in their: Select all that apply. a. five-fold symmetry. b. skeletons made of interlocking plates of protein. c. water vascular system. d. skeletons made of interlocking plates of calcite. e. filter feeding. ANSWER: a, c, d 53. Animals with bilateral symmetry typically have: Select all that apply. a. distinct front (head) and back (tail) ends. b. specialized sensory organs clustered at the front, or head, end. c. specialized appendages along both sides for locomotion. d. specializations for rapid locomotion in all directions. e. complex behavior. ANSWER: a, b 54. The phylum Cnidaria includes: Select all that apply. a. sponges. b. jellyfish. c. corals. d. placozoans. e. comb-jellies. ANSWER: b, c 55. Comb-jellies are similar to bilaterians in that they: Select all that apply. a. have a "flowthrough" gut with both a mouth and an anus. b. have a well-defined nervous system. c. have an anterior–posterior body axis. d. None of the answer options is correct. ANSWER: a, c 56. If ctenophores are sister to all animal taxa, excluding cnidarians and placozoans, then: Select all that apply. a. the anatomical complexity of ctenophores arose independently of what is found in other eumetazoan Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 groups. b. the nervous system of ctenophores evolved independently of other animals. c. ancestral sponges exhibited morphologic complexity seen in other eumetazoans. d. None of the answer options is correct. ANSWER: a, c 57. The amniotes include: Select all that apply. a. sharks. b. amphibians. c. reptiles. d. birds. e. mammals. ANSWER: c, d, e 58. The fossil record is a testament to extinct fauna no longer present on Earth. Certain areas of the planet have always had high species diversity (evident from fossils and catalogs of current species diversity). Which of the processes listed likely contributed to the patterns of animal diversity we see? Select all that apply. a. the appearance of new land masses over time b. changes in atmospheric oxygen c. extinction events (both local and global) d. volcanic eruptions ANSWER: b, c 59. Compared with sponges, cnidarians have a wider array of cell types permitting more sophisticated tissue functions. These functions include: Select all that apply. a. secretion of digestive enzymes. b. muscle contraction/movement. c. the ability to sense the environment via a network of nerve cells. d. the ability to communicate within the organism using hormones. e. None of the answer options is correct. ANSWER: a, b, c 60. Not all sponges have skeletons. If they do, their skeletons may be made of: Select all that apply. a. silica. b. calcium carbonate. c. bone. d. protein. ANSWER: a, b, c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 61. Which of these statements accurately describe characteristics of placozoans? Select all that apply. a. Placozoans are tiny animals containing only a few thousand cells. b. Placozoans are simple animals, consisting of only two tissue layers that sandwich an interior fluid layer. c. Placozoans are, in spite of their morphological simplicity, genetically complex, sharing many genes for transcription factors and signaling molecules with both cnidarians and bilaterian animals. d. Placozoans are a highly diverse group, consisting of many species occupying a wide range of habitats. e. Placozoans are large radially symmetrical animals superficially similar to cnidarians and ctenophores. ANSWER: a, b, c 62. The Lophotrochozoa include: Select all that apply. a. annelid worms. b. mollusks. c. sea anemones. d. arthropods. e. chordates. ANSWER: a, b 63. Annelids are characterized by: Select all that apply. a. cylindrical, segmented bodies. b. extensive nervous systems. c. radula for feeding. d. hydrostatic skeletons. ANSWER: a, b, d 64. Mollusks are distinguished from other animal groups by the presence of: Select all that apply. a. a mantle that plays a role in breathing and excretion. b. a toothlike radula (not present in all). c. a segmented body with a hydrostatic skeleton. d. ecdysis. e. None of the answer options is correct. ANSWER: a, b 65. Members of the Ecdysozoa: Select all that apply. a. secrete an external cuticle. b. secrete a cuticle made of calcium carbonate. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 c. molt during growth. d. have trochophore larvae. e. have lophophores. ANSWER: a, c 66. Chelicerates: Select all that apply. a. include spiders, scorpions, and their relatives. b. include many species known for their venoms. c. were among the first animals to invade land more than 400 million years ago. d. have pincer-like claws, antennae, and wings. e. are ecologically diverse, particularly in their food habits, with different groups specializing on a wide variety of plant and animal foods. ANSWER: a, b, c 67. Which of the relationships between groups are correct? Select all that apply. a. Cephalopods are a group of annelids. b. Crustaceans are a group of arthropods. c. Insects are a group of hemichordates. d. Sea stars are a group of echinoderms. e. Tunicates are a group of chordates. ANSWER: b, d, e 68. Vertebrates are distinguished from other chordates by: Select all that apply. a. well-developed vertebrae. b. a protective cranium. c. a bony jaw. d. paired appendages. e. a mineralized skeleton made of calcium phosphate. ANSWER: a, b 69. Osteichthyes, or bony fish, make up about half of vertebrate species diversity. Their success is due to a combination of adaptations that include: Select all that apply. a. mobile jaws for specialized feeding. b. kidneys that allow fine regulation of water balance. c. lungs modified from pharyngeal gill slits. d. metamorphosis allowing them to occupy distinct niches as juveniles and adults. e. paired, fleshy pectoral and pelvic fins. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 ANSWER: a, b 70. Birds are one of two groups of flying vertebrates (bats are the other). Their adaptations for flight include: Select all that apply. a. hollow bones. b. heavy jaws. c. a method of breathing that extracts a majority of the oxygen from each breath of air. d. efficient kidneys that remove metabolic wastes more quickly than is the case in other vertebrates. e. None of the answer options is correct. ANSWER: a, c 71. One characteristic found only in chordates is the presence of pharyngeal slits. In the evolution of chordates, the pharyngeal gill slits gave rise to which features? Select all that apply. a. respiratory surface (gills) b. support of the head or cranium c. lungs d. part of the jaw ANSWER: a, d 72. What is not a characteristic exclusively found in animals? Select all that apply. a. multicellularity b. chitin c. collagen d. gastrula ANSWER: a, b 73. If continuing analyses confirm that ctenophores are the sister group to sponges plus eumetazoans, which of these conclusions would be supported? Select all that apply. a. Nervous systems evolved convergently in ctenophores and eumetazoans. b. One-way guts (mouth to anus) evolved convergently in ctenophores and eumetazoans. c. Nematocysts evolved convergently in ctenophores and eumetazoans. d. Sexual reproduction evolved convergently in ctenophores and eumetazoans. ANSWER: a, b 74. Traits used to group animals in a simple phylogenetic tree include: a. presence of collagen. b. body symmetry. c. photosynthetic capabilities. d. complex organs. ANSWER: a, b, d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 75. Insects are likely so successful because of which adaptations that allow them to live in a wide range of habitats? Select all that apply. a. desiccation-resistant eggs b. wings c. metamorphosis d. diverse feeding appendages e. unique respiratory system of spiracles and tracheae ANSWER: a, b, c, e 76. Which of the listed relationships between groups is correct? Select all that apply. a. Gastropods are a group of mollusks. b. Myriapods are a group of insects. c. Chelicerates are a group of arthropods. d. Sea urchins are a group of echinoderms. e. Hemichordates are a group of chordates. ANSWER: a, c, d 77. Comb-jellies, or ctenophores, are similar to cnidarians in that they: Select all that apply. a. are radially symmetrical. b. have two tissue layers surrounding a gelatinous interior. c. have a simple nerve net. d. digest prey within their gastric cavity. e. None of the answer options is correct. ANSWER: a, b, c, d 78. The Protostomia include: Select all that apply. a. annelid worms. b. mollusks. c. sea anemones. d. arthropods. e. chordates. ANSWER: a, b, d 79. The major defining characteristics of arthropods are: Select all that apply. a. jointed appendages. b. segmented bodies. c. complex mouthparts. d. trochophore larvae. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 42 e. chitin. f. metamorphosis. ANSWER: a, b, c, e 80. The Deuterostomia include: Select all that apply. a. echinoderms. b. cnidarians. c. chordates. d. placozoans. e. arthropods. ANSWER: a, c 81. Chordates all possess, at least during embryological development, which of these features? Select all that apply. a. a notochord b. a dorsal nerve cord c. a postanal tail d. a bony skeleton e. jaws ANSWER: a, b, c 82. The advantages of the amniotic egg include: Select all that apply. a. relatively long development times supported by a large yolk. b. the ability to resist desiccation in dry terrestrial habitats. c. improved protection from predators. d. improved ability to withstand wide temperature fluctuations. e. reduced energetic investment by the mother relative to amphibians. ANSWER: a, b 83. Bilateral symmetry: Select all that apply. a. is a characteristic of all the lophotrochozoans. b. most likely evolved before radial symmetry. c. is associated with cephalization. d. is associated with a sessile existence. ANSWER: a, c 84. Examine the phylogenetic tree shown.

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Chapter 42

The bony fish: Select all that apply. a. include the ray-finned fishes and the lobe-finned fishes. b. are a monophyletic group. c. include species with lungs. d. all have a cranium and jaws. ANSWER: a, c, d

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Chapter 43 Multiple Choice 1. Male peacocks have tail feathers that make up 60% of their body length. During the mating season they fan their tails and shake these feathers in front of females. Females evaluate the tail-shaking behavior as part of selecting their mate. Which behavior is most likely to have been the original behavior that, through natural selection, resulted in this mate choice behavior of tail shaking? a. flight b. sound production with the larynx c. grooming feathers with the beak d. involuntary muscle contraction such as shivering e. head movement to make eye contact ANSWER: d 2. Male Australian bowerbirds build and decorate elaborate structures, called bowers, out of grasses and other vegetation. If we want to understand how this behavior promotes a male bowerbird's ability to survive and reproduce, we want to understand its: a. causation. b. development. c. adaptive function. d. evolutionary history. e. genetic basis. ANSWER: c 3. Questions of causation and development are _____ explanations of behavior. a. mechanistic b. evolutionary c. innate d. learned ANSWER: a 4. The example of the honeybee dance given in the chapter text is an example of an ________ behavior. a. mechanistic b. evolutionary c. innate d. learned ANSWER: c 5. During their lifetime, many female birds establish a pair bond with a single male with whom they build a nest and rear young. Many of those females also copulate with other males outside the pair bond; the young they raise are fathered by more than one male. If we want to understand the role of genes and the environment in shaping this behavior, we want to understand its: a. causation. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 b. development. c. adaptive function. d. evolutionary history. e. genetic basis. ANSWER: b 6. Male peacocks have tail feathers that make up about 60% of their body length. During the mating season they shake these feathers in front of females. Longer feathers produce a larger volume of rustling sound for a given shaking rate, but they are also heavier, and thus require more energy to shake at a given rate than shorter feathers. This suggests that: a. selective pressures will continue selection for longer and longer peacock tails. b. selective pressures will always result in shorter peacock tails to reduce the energetic cost of shaking. c. feather length is a compromise between the conflicting selective pressures against too long or too short of a tail. d. tail length is completely random because there are selective pressures both for and against longer tails. ANSWER: c 7. Which terms best describe the class of questions that Insel and Young were addressing when they studied the effect of antidiuretic hormone on monogamous voles by experimentally increasing the number of receptors in the brain? a. ultimate; mechanistic b. ultimate; survival-value/adaptation c. proximate; survival-value/adaptation d. ultimate; developmental/ontogenetic e. proximate; mechanistic ANSWER: e 8. Male Australian bowerbirds build and decorate elaborate structures, called bowers, out of grasses and other vegetation. If we want to understand the physiological mechanism behind this behavior, we want to understand its: a. causation. b. development. c. adaptive function. d. evolutionary history. e. causation and development. ANSWER: a 9. Some types of fish have specialized cells called electrocytes that generate an electric field around their body, with which they can communicate with other individuals of their own species. The electrocytes become functional when specialized Na+ channels are inserted into their plasma membrane. Because sodium channels are proteins, we can conclude that the behavior of generating an electrical field for communication: Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 a. has a genetic basis because proteins are produced through gene expression. b. is a learned behavior. c. is dependent on environmental conditions. d. does not change regardless of input from the environment. e. does not have a genetic basis because proteins are produced independent of the genome. ANSWER: a 10. The cuckoo is a type of bird found in Europe, Asia, Africa, and Australia. Cuckoo chicks manipulate the host parent through a number of strategies, one of which involves producing a call that mimics the call of an entire clutch of nestlings; this causes the host parent to feed it more. In this context, the call of the cuckoo is best described as a(n): a. feature detector. b. supernormal stimulus. c. fixed action pattern. d. associatively learned trait. e. ancestral trait. ANSWER: b 11. _____ behavior corresponds to an animal's "nature"; _____ behavior corresponds to its "nurture." a. Innate; learned b. Learned; innate c. Innate; imprinted d. Imprinted; innate ANSWER: a 12. Which statement best describes the genetic basis of behavior? a. All behaviors are influenced by many genes. b. Most behaviors are strongly influenced by a single gene. c. Most behaviors have no genetic basis at all. d. Most behaviors are influenced by many genes, and a few behaviors are strongly influenced by a single gene. e. A few behaviors are influenced by many genes, and most behaviors are strongly influenced by a single gene. ANSWER: d 13. Some types of fish have specialized cells called electrocytes that generate an electric field around their body, which they use to communicate with other individuals of the same species and stun prey. Researchers hypothesized that electrocytes produce the electric field in response to blood levels of a hormone called ACTH, and that it was a positive relationship (higher levels of hormone results in a stronger electric field). Which experiment would be the most direct test of this hypothesis? a. While measuring the strength of a fish's electric field, inject it with ACTH. b. Take a blood sample from a fish that is producing a strong electric field and inject the blood into a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 second fish and measure its electric field strength. c. Take a blood sample from a fish that is producing a strong electric field and measure the ACTH levels. d. Remove the pituitary gland (which secretes ACTH) from a fish and measure its electric field strength. e. While measuring the strength of a fish's electric field, inject it with a saline solution. ANSWER: a 14. What characteristic associated with male Anolis lizards caused a change in behavior in female Anolis lizards? a. testosterone only b. testosterone that causes courting behavior in males c. testosterone that causes males to fight with each other d. the presence of males, regardless of their behavior ANSWER: b 15. Displays are patterns of behavior that: a. are species-specific. b. are highly repeatable. c. function as signals. d. are species-specific and function as signals. e. are species-specific, are highly stereotyped, and function as signals. ANSWER: e 16. A fixed action pattern is a sequence of behaviors that, once triggered, is followed to completion. a. true b. false ANSWER: a 17. Many squirrel species cache food for consumption later. Even young squirrels are able to dig and bury food, even if they have never watched another member of their species bury food. They are able to retrieve food that is buried very well, and it is thought they use landmarks around the burial site of the food to return to that spot and find their food. Based on this description, the burial of food is a(n) _____ behavior, and the retrieval of food is _____. a. innate; innate b. learned; learned c. innate; learned d. learned; innate e. innate; a conditioned response ANSWER: c 18. Males of each species of field crickets have specific songs that differ from songs of other species. Females Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 typically only respond to males exhibiting their species-specific calls. However, hybrids can form: hybrid females will choose hybrid males that sing a hybrid song that has components of the songs from both parental species. Females of either parent species will not choose males that sing a hybrid song. Which statement is the most likely reason for differential responses to calls in non-hybrid and hybrid females? a. The songs have a genetic basis and females respond to any calls they have heard while growing up. b. The songs have a genetic basis and females respond to the calls based on specialized sensory receptors that result in recognition of the call of other members of the same species. c. The songs have no genetic basis and females respond to the calls they have learned from being around other members of the same species. d. The songs have no genetic basis and females respond to the calls based on specialized sensory receptors that result in recognition of the call of other members of the same species. ANSWER: b 19. You observe a chimpanzee using a stick to dig termites from a nest. The chimpanzee then eats the termites. This type of behavior is an example of _____ and the termites are the _____. a. non-associative learning; stimulus b. habituation; signal c. sensitization; second stimulus d. classical conditioning; neutral stimulus e. operant conditioning; reward ANSWER: e 20. You observe that chimpanzees use sticks to collect termites for food. Which observation would indicate that the use of the stick is a learned behavior? a. The chimpanzee is the only member of his or her population that exhibits this behavior. b. The chimpanzee prefers eating termites to other types of insects. c. Food is abundant in the chimpanzee's environment. d. Chimpanzees raised in isolation do not use sticks. ANSWER: d 21. The enhancement of a response to a stimulus following a novel pre-stimulus is called: a. sensitization. b. habituation. c. imitation. d. imprinting. ANSWER: a 22. In the 1950s, Japanese scientists studying a group of macaques habituated the animals to their presence by providing them with sweet potatoes. A young female macaque, who was given the name Imo by the scientists, began washing the sand and grit off her sweet potatoes by rinsing and rubbing them in a nearby stream. Now, 60 years later, all the macaques in the group wash their sweet potatoes—only now they wash them in the ocean. This is likely an example of learning by: a. classical conditioning. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 b. imitation. c. non-association. d. habituation. e. sensitization. ANSWER: b 23. Associative learning, also called conditioning, occurs when: a. an animal learns that two events are correlated. b. an animal learns a behavior without reward or punishment. c. an animal's behavior is lessened or eliminated through repeated exposure to the stimulus. d. an animal's response to a stimulus is enhanced when a strong or novel stimulus is presented first. ANSWER: a 24. Each morning before work a cat owner makes a lot of noise when she opens the utensil drawer in the kitchen, grabs the can opener, and then shuts the drawer. Each morning her new cat runs into the kitchen to be fed. After a few weeks, the cat owner notices that the cat runs into the kitchen whenever she hears the utensil drawer close. This is an example of: a. operant conditioning. b. classical conditioning. c. habituation. d. sensitization. ANSWER: b 25. Imprinting is most likely to occur: a. during any point in an individual's lifetime. b. when an organism is most at danger. c. during a critical point in development. d. when an organism is surrounded by conspecific individuals. ANSWER: c 26. You observe a chimpanzee using a stick to dig termites from a nest. How could you determine if this digging behavior is learned or innate? a. Take the stick away and see if the chimpanzee picks up another one. b. Show the chimpanzee termite mounds of different sizes and see which mound it chooses for digging. c. Provide the chimpanzee with lots of different foods and see if it still digs for termites. d. Raise a group of chimpanzees in isolation and see if they still exhibit the behavior. e. Raise chimpanzees in groups and see if they still exhibit the behavior. ANSWER: d 27. Habituation and sensitization are forms of _____ learning, meaning that they occur in the absence of a particular outcome, such as a reward or punishment. a. non-associative Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 b. associative c. conditioned d. imitated ANSWER: a 28. Many farmers place scarecrows in their fields to keep crows away from the crops. Initially the crows will stay out of the field when they see the "person" in the middle of the field. After a few weeks, a farmer notices crows are perching in his cornfield. Why didn't placing the scarecrow in the field work over a long period of time? a. Crows are too smart to be fooled by a scarecrow. b. Crows will land on anything as long as it doesn't move. c. Crows have become habituated to the scarecrow and no longer avoid it. d. Crows have become sensitized to the scarecrow and no longer avoid it. ANSWER: c 29. Imprinting is a common phenomenon in ducks and geese. Assume that a baby duck sees a small dog during a critical period after hatching and thinks it is its mother. The duck follows the dog everywhere. What, if anything, could be done to help the duck recognize its real mother instead of the dog? a. The dog should be left inside the farmhouse, and the duckling should be placed with an adult duck so it will start following the duck around. b. The dog should be left inside the farmhouse, and the duckling should be placed with other ducklings so it can learn normal duckling behavior. c. The duckling should be shown an adult duck so it will follow something that looks just like the duckling. d. Nothing can be done because the duckling has already imprinted on the dog. ANSWER: d 30. Salmon that have grown to maturity in the ocean are well known for their ability to find their way back to the streams where they were born in order to breed. For this kind of navigation, they likely: a. use magnetotaxis only. b. use map information only. c. rely on kinesis only. d. use a variety of environmental cues. ANSWER: d 31. Some types of dinoflagellates (single-celled eukaryotes) are photosynthetic. They have flagella and can swim through water. If you grow them in a container that is lit from the right side, the cells will swim to the right side of the container. This result suggests that dinoflagellates: a. sense light. b. sense Earth's magnetic field. c. sense a heterotrophic food source. d. swim in random directions. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 e. swim in response to temperature gradients. ANSWER: a 32. At a mechanistic level, learning involves a change in the: a. strength of connections between neurons. b. tissues to which responding neurons are synapsed. c. structure of neurons. d. firing mechanism of neurons. ANSWER: a 33. Mediterranean slipper lobsters (Scyllarides latus) are thought to be active at night. A group of Mediterranean slipper lobsters were brought into the lab and exposed to a day–night cycle of 12 hours of day, and 12 hours of night. Then they were placed in 24 hours of dark. Their activity patterns did not change even when they were exposed to only dark. These results suggest their activity patterns are governed by: a. a lunar clock. b. a circadian clock. c. a response to photoperiod. d. orientation. ANSWER: b 34. Male salamanders do a species-specific dance that lures females to follow them. The male then leads the female to walk over a spermatophore (sperm packet) left by the male, resulting in fertilization of the females' eggs. This kind of species-specific mating dance is likely the result of: a. imprinting. b. ritualization. c. sensitization. d. conditioning. ANSWER: b 35. Male crickets create sounds that we call "chirps" by rubbing one wing against the other. If another male presents himself to the singing male, the two males will fight. First they interlock antenna, then they spread their jaws, and finally they wrestle. _____ is/are examples of communication _____. a. All of the actions; because all involve information being sent and received by the two crickets b. Interlocking antenna and spreading jaws, but not wrestling; because communication is no longer needed when the two crickets are physically in contact c. The initial song and the response of the second male cricket arriving in the area; between the two crickets d. The initial song; because it is an intraspecific signal ANSWER: a 36. A hummingbird is attracted to the red, tubular flowers of a trumpet vine. It drinks nectar from the flower and, in doing so, collects pollen that it transfers to the next flower. In this form of communication, who is the sender, who is the receiver, and what information is being transferred? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 a. The hummingbird is the sender, the flower is the receiver, and the availability of pollen is the information. b. The hummingbird is the sender, the flower is the receiver, and the availability of the hummingbird to transfer pollen is the information. c. The flower is the sender, the hummingbird is the receiver, and the availability of nectar is the information. d. The flower is the sender, the hummingbird is the receiver, and the need of the flower to have its pollen transferred is the information. e. The flower is the sender, the pollen is the receiver, and the presence of the hummingbird is the information. ANSWER: c 37. Two forager bees return to the hive. Forager A does a dance that goes in rapid circles for approximately two minutes. Forager B does a dance that moves in a straight line before it circles back and starts again. The whole time he is waggling his abdomen. The second forager waggles for more than five minutes. Which forager is conveying a food source that is close to the hive? a. forager A b. forager B c. both foragers A and B ANSWER: a 38. If group selection—the differential survival and reproduction of groups—is an evolutionarily stable strategy, then: a. family members of groups behave altruistically to one another but not to unrelated members of the group. b. the characteristics of the group must be heritable. c. a small portion of the group must sacrifice their own fitness for the good of the group. d. groups are only formed by related individuals. ANSWER: b 39. In many species of lizards, the males have elaborate displays that communicate their health condition during mating season. This is targeted to other males that may be looking to expand the size of their own territories into the displaying male's territory. What prediction could you make about such lizard display behaviors in the field? a. Males will only display when females are present. b. The display will be the same to any male that arrives in their territory. c. The display will vary depending on the size of the territory. d. Males will display continually during the mating season even in the absence of other lizards. ANSWER: b 40. Meerkats are small desert mammals that live in groups of 20–50 individuals. Which behavior would you categorize as examples of reciprocal altruism if you saw them while observing a group of meerkats? a. Two meerkats take turns grooming each other, each spending about the same amount of time Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 grooming the other. b. One meerkat grooms a second meerkat, who is sleeping. c. One meerkat spends time standing upright on a stump, looking around, while other meerkats forage for food. d. One meerkat spends time sleeping in the burrow after eating, while other meerkats forage for food. e. One meerkat stands in the sun, while a second spends time digging out the burrow. ANSWER: a 41. Meerkats are small desert mammals that live in groups of 20–50 individuals. You observe a female meerkat standing upright on a stump, looking around. Eight of her young nieces and nephews are foraging for food around her. You hypothesize that this is an example of kin selection. Which additional observation about the meerkats' behavior would support your hypothesis? a. All adults in the group, including those that are not related to the young meerkats, are standing upright and looking around. b. The female meerkat in the question takes food from the foraging nieces and nephews when they find it. c. The female meerkat in the question does not stand "on guard" when she is surrounded by more distantly related juveniles. d. Other aunts and uncles of the young meerkats are foraging and do not spend any time standing up and looking around. e. All adults in the group, including those who are not related to the young meerkats, spend energy in all the activities that involve caring for the young meerkats. ANSWER: c 42. Meerkats are small desert mammals that live in groups of 20–50 individuals. You observe one meerkat standing upright on a stump, looking around, while other meerkats forage for food. You hypothesize that this is an example of kin selection. What additional information would you need to know to evaluate your hypothesis? a. the body size of this meerkat relative to other members of the group b. the gender of this meerkat as well as that of all other members of the group c. the degree of genetic relatedness of this meerkat to all members of the group d. all other activities of this meerkat as well as those of all other members of the group e. the foraging success of this meerkat as well as that of all other members of the group ANSWER: c 43. For reciprocal altruism to work, individuals must be able to: a. recognize one another. b. remember previous interactions. c. calculate the fitness value of individual interactions. d. recognize one another and remember previous interactions. e. recognize one another, remember previous interactions, and calculate the fitness value of individual interactions. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 44. Group selection is not an evolutionary stable strategy because: a. natural selection operates at the level of individual fitness. b. populations can go extinct for many reasons. c. intersexual selection is stronger than group selection. d. behaviors are learned and not innate. ANSWER: a 45. Group selection is not typically seen as an evolutionarily stable strategy because: a. selfish behaviors are more likely to increase individual fitness. b. it cannot explain behaviors for species with solitary individuals. c. average relatedness is very low between individuals in a population; therefore, individual fitness cannot increase. d. None of the other answer options is correct. ANSWER: a 46. An entomologist finds two closely related species of fly, one (P) of which appears to reproduce parthenogenetically while the other (S) reproduces sexually. In the same location, the entomologist finds a bizarre third kind of insect, a sterile defensive form with enlarged mandibles. Because it is sterile, the entomologist concludes that this form must be produced by either P or S. Which do you think is most likely the producer of the sterile defensive individuals? a. P, because all individuals would be clones of the parent b. P, because they could all produce clones of individuals from the colony c. S, because a different type could only be produced through sexual reproduction d. S, because by protecting other members of the group they are related to, they still have reproductive success ANSWER: d 47. A neutral stimulus in classical conditioning: a. elicits a response to a behavior unrelated to the typical stimulus. b. elicits a response only when paired with the typical stimulus. c. elicits a response without exposure to the typical stimulus. d. elicits a response only during a critical period of conditioning. ANSWER: a 48. A dog sometimes puts its paw in the air when its owner walks in the door. The owner makes sure that she always has treats in her pocket when she comes home and if the dog puts its paw in the air, the owner rewards it with a treat. The owner does not give the dog a treat if it doesn't put its paw in the air. Over time, the dog always puts its paw in the air. This is an example of: a. operant conditioning b. classical conditioning ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 49. Many species of gulls feed their chicks by regurgitating food. When parents return from foraging, the gull chicks peck at a red spot on the parent's bill, which causes the parents to open the bill and regurgitate their food. Chicks will peck at a red dot painted on any long, slender object. For the chicks, pecking is a ________ and the red dot is a __________. a. fixed action pattern; key stimulus b. learned behavior; supernormal stimulus c. key stimulus; fixed action pattern d. supernormal stimulus; learned behavior e. supernormal stimulus; fixed action pattern ANSWER: a 50. Many species of gulls feed their chicks by regurgitating food. When parents return from foraging, the gull chicks peck at a red spot on the parent's bill, which causes the parents to open the bill and regurgitate their food. Chicks will peck at a red dot painted on any long, slender object. For the adult gulls, the chicks pecking at their bills is a ________ and the regurgitating food is a ________. a. fixed action pattern; key stimulus b. learned behavior; supernormal stimulus c. key stimulus; fixed action pattern d. supernormal stimulus; learned behavior e. supernormal stimulus; fixed action pattern ANSWER: c Multiple Response 51. According to the Dutch behavioral biologist Niko Tinbergen, which of the terms defines—and helps researchers understand—an animal's behavior? Select all that apply. a. conditioning b. causation c. development d. evolutionary history e. habituation ANSWER: b, c, d 52. In an expedition to a remote part of South America, you find two very closely related rodent species that build very different burrows: one is shaped like a corkscrew and the other is straight. Which scenario describes an experiment you could do to determine whether or not this burrowing pattern is a learned or an innate behavior? Select all that apply. a. Observe juveniles as they dig burrows. b. Raise juveniles of each species in isolation and upon release determine if they dig their speciesspecific burrow shape. c. Raise juveniles of one species with adults of the other species and observe if juveniles dig their species-specific burrow or the burrow of the species they were raised with. d. Raise juveniles with other members of their species, but allow them to watch adults of the other Copyright Macmillan Learning. Powered by Cognero.

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Chapter 43 species dig their burrows. Then observe to see which type of burrow they dig. ANSWER: b, c, d 53. Which scenario would be considered a type of communication? Select all that apply. a. cardinals singing b. the plumage of a bird of paradise c. a lion marking its territory d. a duckling imprinting on a human e. a goose rolling an egg back into the nest ANSWER: a, b, c 54. The "waggle dance" of honeybees conveys what information about a food source to other foragers in the hive? Select all that apply. a. direction b. distance from the hive c. quality of the food d. quantity of food e. presence of predators ANSWER: a, b

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Chapter 44 Multiple Choice 1. You are studying a population of monarch butterflies that overwinters in Mexico and migrates to Northern California. The figure shows the Thanksgiving counts from the past 20 years. For the purposes of this question, assume carrying capacity is at the population size recorded in 1998.

Growth rate is lowest from 2009-2010. a. true b. false ANSWER: b 2. You are studying a population of monarch butterflies that overwinters in Mexico and migrates to Northern California. The figure shows the Thanksgiving counts from the past 20 years. For the purposes of this question, assume carrying capacity is at the population size recorded in 1998.

The growth rate is higher from 2015-2016 than 2000-2001. a. true b. false ANSWER: b 3. Prairie dogs are rodents that live in colonies. Black-tailed prairie dog (Cynomys ludovicianus) colonies typically have around 12 adult animals per hectare. What aspect of the population ecology of the black-tailed prairie dog does this statement describe? a. population density b. population size c. population growth Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 d. life history ANSWER: a 4. The figures shown represent three hypothetical populations; each circle is an individual. Which figure depicts a pattern that illustrates a scenario in which food is most abundant near waterholes in the desert?

a. diagram M b. diagram H c. diagram K ANSWER: b 5. The figures shown represent three hypothetical populations; each circle is an individual. Which figure depicts a pattern that illustrates a scenario in which individuals establish exclusive territories to secure access to resources?

a. diagram M b. diagram H c. diagram K ANSWER: a 6. The figures shown represent three hypothetical populations; each circle is an individual. Which figure depicts a pattern that illustrates a scenario in which individuals are distributed randomly within the environment?

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Chapter 44

a. diagram M b. diagram H c. diagram K ANSWER: c 7. The figures shown represent three hypothetical populations; each circle is an individual. Which figure depicts a pattern that illustrates a social species?

a. diagram M b. diagram H c. diagram K ANSWER: b 8. Consider the Lime Swallowtail, Papilio demoleus, which lives in citrus groves on the island of Hispaniola. Which type of spatial distribution would you expect for this species? a. random b. clustered (clumped) c. uniform ANSWER: b 9. The growth curve shown depicts growth projections for a single population. What would happen if the birth rate were to decline? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44

a. The curve would be less steep and shift to the right. b. The curve would be steeper and shift to the left. c. The curve would be steeper and shift to the right. d. The curve would be less steep and shift to the left. e. The curve would stay the same. ANSWER: a 10. A population of 20 squirrels, with 10 males and 10 females, colonizes a new area with no predators and enough plant species to support a squirrel population of 200 individuals. Each year, a female produces 2 offspring, one male and one female, that survive to reproduce. Within the first 2 years in the new area, the growth curve for this squirrel population will show: A. logistic growth because the squirrel population will reach maximum population size immediately after colonization. B. logistic growth because there is a carrying capacity in the newly colonized area. C. exponential growth because the colonizing population size is far below the carrying capacity. D. exponential growth because only a few individuals will have offspring. a. b. c. ANSWER: c 11. Consider the logistic growth equation: Nt = N1 + rN1(K - N1K). Which of the choices will result in an increase in population size? a. increasing the death rate b. decreasing the birth rate c. decreasing the carrying capacity d. increasing the carrying capacity Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 ANSWER: d 12. Consider a population undergoing logistic growth. When are the most individuals added to the population per unit time?

a. (K - N)/K = 1 b. (K - N)/K = 0.75 c. (K - N)/K = 0.5 d. (K - N)/K = 0.25 e. (K - N)/K = 0 ANSWER: c 13. Consider the data in the histograms shown that illustrate the population age structures of France and India. In many parts of the world, the average age of reproductive maturity is increasing as humans move from urban to rural environments. Suppose that the current average age of first reproduction is 15 years old in India and 20 years old in France. How would a 5-year increase in the average age of reproduction in both countries be expected to affect the per capita birth rate, and would this effect be greater in France or in India?

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Chapter 44

a. increase in per capita birth rate, with a greater effect in India b. increase in per capita birth rate, with a greater effect in France c. decrease in per capita birth rate, with a greater effect in India d. decrease in per capita birth rate, with a greater effect in France ANSWER: d 14. If having large numbers of offspring results in high fitness, why do females of all species not have very large numbers of offspring? a. In some species, the number of eggs a female will have in her lifetime is finite, developing before she reaches reproductive age. b. A female, regardless of species, would die before they could raise all of the offspring they could produce. c. Female reproduction is limited by the amount of energy a female can devote to reproduction in any one season. d. Females are limited by the number of males they mate with in a single season. ANSWER: c 15. When sea turtle eggs hatch, predators gather, and mortality among the hatchlings is extremely high. Once turtles reach adult size, they can live for decades. This is an example of a Type _____ survivorship curve. a. I b. II c. III ANSWER: c 16. Habitat fragmentation occurs frequently with current logging practices. The figure shown represents a fragmented set of patches in a logged area. Which of the subpopulations is most likely to go extinct?

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a. population J b. population K c. population L d. population M ANSWER: b 17. Habitat fragmentation occurs frequently with current logging practices. The figure shown represents a fragmented set of patches in a logged area. At equilibrium, which forest fragment is likely to hold the largest number of species?

a. population J b. population K c. population L d. population M Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 ANSWER: a 18. In 1987, 18 black-footed ferrets, the last known individuals of this species, were captured and brought into a captive breeding program in Wyoming. In 1989, the total black-footed ferret population, still in captivity, was 120 animals. These 120 animals in 1989 represented: a. N, the total population size. b. ΔN, the change in population size from 1987 to 1989. c. ΔN/Δt, the rate of change in population size. d. r, the per capita change in population size. e. r, the exponential rate of growth of the population. ANSWER: a 19. Population size can fluctuate because of many different factors. Which of the choices would not affect population size? a. a change in the primary predators of the population b. a change in the resources available to the population c. a change in the birth rate of the population d. a change in the death rate of the population e. All of these choices would affect population size. ANSWER: e 20. When the number of young produced at birth exceeds the number of adults that can be supported by available resources, this leads to natural selection. a. true b. false ANSWER: a 21. As a population approaches its carrying capacity, its growth rate: a. increases. b. decreases. c. stays the same. d. stops. ANSWER: b 22. Only density-dependent factors impact population growth. a. true b. false ANSWER: b 23. Giant clams synchronize their reproduction with the phases of the moon. During these reproductive bouts, a single clam can release approximately 500 million eggs to be fertilized in the water column by sperm released from other clams. However, the population sizes of giant clams are very low, because many of the fertilized Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 eggs and larvae will be eaten by fish in the local area. Based on this information, what kind of reproductive strategist is the giant clam? a. It is an r-strategist. b. It is a K-strategist. ANSWER: a 24. In a population affected mainly by density-dependent factors, would you expect natural selection to favor an r-selected or K-selected life history? a. r-selected b. K-selected ANSWER: b 25. A population that has a high mortality earlier in the life cycle than later in the life cycle would have a Type _____ survivorship curve. a. I b. II c. III ANSWER: c 26. In any one place, multiple species of the lizard genus Anolis can coexist because they hunt food in different ways and in different parts of the vegetation. a. true b. false ANSWER: a 27. Population size can be influenced by: Select all that apply. a. births. b. deaths. c. immigration. d. emigration. e. All of these choices are correct. ANSWER: e 28. What is the importance of mark-recapture methods in determining population size? a. Population size differs from year to year for most species. Repeated mark-recapture allows researchers to determine the actual population size for a specific year. b. Population size does not stay the same from year to year for most species. Mark-recapture provides an estimate of population size for species that are difficult to study. c. Population size is relatively constant for most species. Mark-recapture gives the actual population size. d. Population size is relatively constant for most species. Mark-recapture provides density-independent Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 factors affecting population size. ANSWER: b 29. A group of field biologists goes out to mark beetles living in an area around a local pond in order to estimate the population size of the beetles. The biologists capture 500 beetles on the first day. They return the following day and catch a total of 400 individuals, 200 of which are marked. What is the estimated population size for the beetles? a. 10,000 b. 1000 c. 500 d. 250 ANSWER: b 30. The coast redwood Sequoia sempervirens occurs naturally in a strip of habitat about 450 miles long and 535 miles wide, from southwest Oregon along the coast to just south of Monterey, California. What aspect of the population ecology of the coast redwood does this statement describe? a. geographic range b. population density c. population size d. distribution ANSWER: a 31. Consider a population undergoing logistic growth. When (K - N)/K = 0, what does that imply about the birth rate b and the death rate d? Refer to the figure shown.

a. b/d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 b. b < d c. b = d d. It is not possible to make an inference about b and d from this information. ANSWER: c 32. The equation ΔN/Δt = rN(1 - N/K) represents an alternate way to express the logistic growth equation. (It is equivalent to the equation offered in your text.) What will occur in the population as the ratio N/K is increased? a. Population growth will increase. b. Population growth will decrease. c. The intrinsic rate of population growth will increase. d. The intrinsic rate of population growth will decrease. ANSWER: b 33. In 1987, 18 black-footed ferrets, the last known individuals of this species, were captured and brought into a captive breeding program in Wyoming. In 1989, the total ferret population, still in captivity, was 120 animals. Given r = (ΔN/Δt)/N1, and based on the numbers provided, the per capita growth rate (ferrets/year) from 1987 to 1989 was: a. 102. b. 60. c. 51. d. 25.5. e. 2.83. ANSWER: e 34. During exponential growth, the number of individuals added per unit time increases, while the _____ stays the same. a. population size b. carrying capacity c. per capita growth rate d. death rate ANSWER: c 35. As a population approaches its carrying capacity, how does its growth change? a. The growth rate slows until N is 0. b. The growth rate slows until N is close to K. c. The growth rate slows until N is close to r. d. The growth rate stays the same. e. r changes until it is close to K. ANSWER: b 36. The graph shown illustrates a life-history trade-off between survivorship and reproduction in red deer from Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 the Isle of Rhum in northern Scotland. At what age is this trade-off minimized (that is, at what age is the cost of reproduction lowest)?

a. 4 years b. 6 years c. 8 years d. 10 years ANSWER: c 37. Natural populations are affected by both density-dependent and density-independent factors. In a population affected mainly by density-independent factors, would you expect natural selection to favor an r-selected or Kselected life history? a. r-selected b. K-selected ANSWER: a 38. In the 1940s, biologist Adolph Murie studied survival in Dall sheep in what is now Denali National Park in Alaska. He found that Dall sheep that survive their first year of life have a high probability of surviving until they are around 9 years old. After that, they rapidly become easy prey for wolves and die at a very high rate. This pattern best fits a Type _____ survivorship curve. a. I b. II c. III ANSWER: a 39. If you were a conservation biologist in charge of the recovery plans for an endangered species, which would you prefer to find within that species: an even distribution of age classes or a pyramid-shaped age distribution? Why? a. an even distribution, because then the population is stable Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 b. a pyramid-shaped distribution, because then the population is stable c. an even distribution, because this means the population will grow d. a pyramid-shaped distribution, because this means the population will grow ANSWER: d 40. A population that exhibits density-dependent growth is described in the graph shown.

The most rapid growth in population size is indicated at: a. point M. b. point H. c. point K. d. point L. ANSWER: c 41. A population that exhibits density-dependent growth is described in the graph shown.

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Chapter 44

The carrying capacity for the population is indicated at: a. point M. b. point H. c. point K. d. point L. ANSWER: d 42. A population that exhibits density-dependent growth is described in the graph shown.

Which point on the graph would change if more food resources were available to the population? a. point M Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 b. point H c. point K d. point L ANSWER: d 43. Consider the data in the table and the graph on the survivorship of the Lime Swallowtail butterfly. As indicated in the table, the survivorship from the egg to the larval stage is 68%.

What is the survivorship from the egg to the adult stage? a. 0.28 b. 0.32 c. 0.68 d. . 0.93 ANSWER: a 44. Consider the data in the table and the graph on the survivorship of the Lime Swallowtail butterfly. As indicated in the table, the survivorship from the egg to the larval stage is 68%.

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What is the per capita death rate during the egg to the larval stage? a. 0.28 b. 0.32 c. 0.68 d. . 0.93 ANSWER: b 45. Examine the life table for a species of rodent that lives in the southwestern deserts of the United States.

According to the life table, at the end of 3 years, the population goes extinct. a. true b. false ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 46. Examine the life table for a species of rodent that lives in the southwestern deserts of the United States.

According to the life table, which age class contributes the most to the population growth of these rodents? a. females of age class 1 b. females of age class 2 c. females of age class 3 ANSWER: a Multiple Response 47. A researcher is trying to calculate the population density of bobcats in Colorado. What information must the research team have in order to calculate this value? Select all that apply. a. the size of the bobcat population within the United States b. the size of the bobcat population within Colorado c. the geographic range of the bobcat population in the United States d. the geographic range of the bobcat population in Colorado ANSWER: b, d 48. An r-strategist will typically: Select all that apply. a. produce many offspring. b. produce few offspring. c. produce small offspring. d. produce large offspring. e. provide abundant parental care. f. provide little parental care. ANSWER: a, c, f 49. Although all Anolis lizards feed on insects and other invertebrates, they have evolved different feeding strategies that are reflected in: Select all that apply. a. behavior. b. leg morphology. c. skull morphology. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 d. habitat preference. e. growth rate. f. age at reproductive maturity. ANSWER: a, b, c, d 50. Species diversity on ecological islands reflects the rate of: Select all that apply. a. new species arrival. b. population growth. c. growth toward the carrying capacity. d. colonist species extinction. ANSWER: a, d 51. Which of the important factors do ecologists use to understand past changes and predict future changes in populations? Select all that apply. a. the population's birth rate b. expected longevity of individuals in the population c. the proportion of individuals in the population able to reproduce d. the ratio of males to females ANSWER: a, b, c 52. As populations approach carrying capacity, crowding can affect: Select all that apply. a. birth rates through resource limitation. b. death rates through resource limitation. c. death rates by increasing the transmission of pathogens and parasites. d. birth and death rates through closer access to mates. e. death rates due to climatic factors. ANSWER: a, b, c 53. What are some factors that keep a population under its carrying capacity? Select all that apply. a. predation b. increased birth rate c. decreased death rate d. parasitism ANSWER: a, d 54. Which of the choices are ecological islands? Select all that apply. a. a freshwater pond in a meadow b. a black basalt lava flow in the middle of the desert Copyright Macmillan Learning. Powered by Cognero.

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Chapter 44 c. an individual cluster of pines in the middle of a pine forest d. a deep sea hydrothermal vent ANSWER: a, b, d 55. A K-strategist will typically: Select all that apply. a. produce a small number of offspring. b. produce a large number of offspring. c. produce relatively large offspring. d. produce relatively small offspring. e. provide abundant parental care. f. provide minimal parental care. ANSWER: a, c, e 56. To determine whether a population is getting larger or smaller, what variables must be known? Select all that apply. a. births b. deaths c. immigration d. emigration e. carrying capacity ANSWER: a, b, c, d

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Chapter 45 Multiple Choice 1. Toucans are frugivorous (fruit-eating) birds that inhabit tropical forests in Central and South America where they can potentially consume any fruit that is available (generalists). The realized niches of toucans are determined by competition in the consumption of different species of fruits (niche overlap). White-throated toucans are the largest species and very strong competitors who easily scare the mid-sized Cuvier's and Ivorybilled toucans off the fruiting trees, altering access to the resources among the three species. The graphs shown represent the occupation of niche space at two different sites, where two or three species coexist, in different abundances.

Which of the statements best reflects why all three species of toucans are found in the same area and exploit similar resources? a. Each species evolved from a different ancestor so their bill sizes are different. b. The three species have evolved from a common ancestor. c. There was convergent evolution of adaptations associated with their bills to exploit fruit. ANSWER: c 2. Toucans are frugivorous (fruit-eating) birds that inhabit tropical forests in Central and South America where they can potentially consume any fruit that is available (generalists). The realized niches of toucans are determined by competition in the consumption of different species of fruits (niche overlap). White-throated toucans are the largest species and very strong competitors who easily scare the mid-sized Cuvier's and Ivorybilled toucans off the fruiting trees, altering access to the resources among the three species. The graphs shown represent the occupation of niche space at two different sites, where two or three species coexist, in different abundances. a. ANSWER: 3. When would a researcher be most likely to observe resource partitioning? a. between two species, one predator and one prey b. between two sympatric species that eat similar-sized seeds Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 c. between two sympatric species, one herbivore and one carnivore d. between two allopatric populations that eat the same thing ANSWER: b 4. Banner-tailed kangaroo rats, Dipodomys spectabilis, are rodents that feed on seeds and are endemic to (found only in) the deserts of the southwestern United States. Which of the statements is the best description of their niche? a. seed-eating rodents that can tolerate the intense heat and dryness in the southwest deserts b. rodents found in parts of Arizona, New Mexico, Texas, and Mexico c. rodents found primarily in Sonoran and Chihuahuan deserts d. rodents found only in arid habitats e. rodents found primarily in areas with relatively heavy soils ANSWER: a 5. _____ are close interactions between species that have evolved over long periods of time. When these interactions enhance the reproduction and population growth of both species, they are called _____. a. Mutualisms; antagonisms b. Mutualisms; symbioses c. Symbioses; antagonisms d. Antagonisms; symbioses e. Antagonisms; mutualisms f. Symbioses; mutualisms ANSWER: f 6. Measuring its costs and benefits in terms of energy spent and/or gained, predation: a. is a lose-lose interaction. b. results in gain for both individuals. c. results in a gain for one individual and a loss for the other. d. results in a gain for one individual and neither a gain nor a loss for the other. e. None of the answer options is correct. ANSWER: c 7. In the late 1960s, Robert Paine conducted landmark studies on diversity in the rocky intertidal zone, comparing the species diversity in control plots with diversity in experimental plots from which he removed the top predator, sea stars. After 5 years, 15 species of intertidal invertebrates lived in the control plots, while the experimental plots were dominated by only two species, one mussel and one barnacle. The process most likely responsible for the loss of species diversity in the experimental plots was: a. mutualism. b. predation. c. competitive exclusion. d. parasitism. e. resource partitioning. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 ANSWER: c 8. What is the difference between evolution and coevolution? a. These concepts are not different; they only differ on time scale. b. Coevolution is the only way speciation can occur. c. Coevolution results when two or more species influence adaptation in each other. d. Evolution of one species only occurs in response to natural selection based on the changing environment that species lives in at a particular time. ANSWER: c 9. We think of termites as insects that eat wood. However, termites cannot digest the wood and rely on a variety of eukaryotic and prokaryotic gut microbes to digest cellulose. Without the microbes, the termites will still ingest wood but will starve. This, then, is an example of a(n); a. obligate mutualism. b. obligate antagonism. c. facultative mutualism. d. facultative antagonism. e. None of the answer options is correct. ANSWER: a 10. Measuring its costs and benefits in terms of energy spent and/or gained, which interaction involves one individual neither gaining nor losing? a. competition b. predation c. obligate mutualism d. facultative mutualism e. commensalism ANSWER: e 11. Tropical leafcutter ants collect leaf cuttings which they transport to special underground chambers. There, the ants chew the leaves to create nursery beds on which the ants grow a species of fungus they use for food. When ant queens disperse to establish new colonies, they carry the fungus with them, dispersing it as well (this benefits the fungus). In the ants' new nests, the fungus is at risk of being destroyed by another fungal species that is able to grow in the habitat, using the same limited resources. On their bodies, the ants carry and provide a home for bacteria that produces antibiotics the ants use to kill the newly encountered fungus and thereby protect their food supply. In this system, the relationship between the leafcutter ants and their fungal food species is: a. interspecific competition. b. mutualism. c. predation. d. parasitism. e. commensalism. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 12. You decide to follow the development of the human gut microbiota over time by sampling the solid waste of infants from birth to 24 weeks old. Which of the setups minimizes the sources of variation in your experiment? a. Sample three infants from three different families that are 1 day old, three infants that are 1 week old, three infants that are 2 weeks old, and so on. b. Sample three infants from three different families the day they are born, when they are 1 week old, 2 weeks old, and so on. c. Sample identical triplets from a single family when they are 1 day old, 1 week old, 2 weeks old, and so on. d. All of the experimental designs described here have the same level of variation. ANSWER: c 13. The bacterial species Staphylococus aureus is found on the skin and in the nasal passages of about 20% of the human population. It can survive on polyester for 3 months and can also infect cows and chickens. Thus, its relationship with humans is: a. obligative. b. facultative. c. required for its survival. d. All of these choices are correct. ANSWER: b 14. In a successional community, some kinds of organisms like fungi or beetles are present in every stage of succession. However, the particular species of fungi or beetles that may be there at the earliest stages of succession are often not the same species present in the last successional stage. This is a result of: a. changes in plant communities during succession that support different animal communities. b. biogeochemical cycles that change with each stage of succession. c. the fact that predators and herbivores are not a part of the community until the last stage of succession. d. the fact that species in an early successional community will always be in their fundamental niche, and later stages only allow species to occupy their realized niche. ANSWER: a 15. When lichens grow on bare rock, they may eventually accumulate enough organic material around them to supply a foothold for rooted vegetation. These early pioneering lichens can be said to do what for species arriving in a later successional stage? a. inhibit b. exclude c. facilitate d. tolerate e. concentrate ANSWER: c 16. You are most likely to observe succession in a terrestrial community where lichens were one of the first Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 species to establish the area when you visit a(n): a. recently burned forest. b. recently created volcanic island. c. abandoned field. d. recently plowed field. e. tropical rain forest. ANSWER: b 17. Communities change over time because species persistence in an area can change over time. As a result, a species within a community can change if it had interacted with a species that has been removed or interacts with a species introduced into the community. a. true b. false ANSWER: a 18. Once a hemlock-spruce forest is established in Glacier Bay, it tends to remain relatively unchanged unless disturbed. This makes it the _____ for the region. a. climax community b. ecosystem c. food web d. primary producer ANSWER: a 19. The diversity of our native grassland species is in decline. Restoration ecologists working to restore native grasslands are experimenting with combinations of burning and grazing to simulate natural conditions and increase species diversity. According to ecologist Joseph Connell, what level of burning and grazing should ecologists aim for? a. strong enough to limit interspecific competition, but not strong enough to limit the number of species that can tolerate the environment b. strong enough to promote interspecific competition and niche diversification, but not strong enough to promote competitive exclusion c. strong enough to promote indirect competition, but not strong enough to promote direct competition d. strong enough to selectively favor one or more keystone species, but not strong enough to favor their competitors e. None of the answer options is correct. ANSWER: a 20. A _____ affects other members of the community in ways that are disproportionate to its abundance or biomass. a. primary producer b. predator c. keystone species Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 d. detritivore ANSWER: c 21. You are surveying biodiversity on a new island chain. You have counted the number of bat species on one island already. The next island is smaller and farther from the mainland than the one you have just surveyed. According to the theory of island biogeography, the total number of its bat species should be _____ than on the current island because the rate of immigration to the new island should be _____ and the rate of extinction should be _____. Refer to the figure shown.

a. smaller; higher; lower b. smaller; lower; higher c. greater; higher; lower d. greater; lower; higher ANSWER: b 22. The realized niche of Canada geese depends on which factor? a. interspecific competition between Canada geese and other birds b. competitive exclusion of Canada geese by other organisms Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 c. the distribution of Canada geese parasites d. the climate of a given environment e. All of these choices are correct. ANSWER: e 23. Spanish moss is not a moss, but a flowering plant. Tillandsia usneoides, often seen draped from the branches of live oaks, cypress, and other large trees in the southeastern United States. The trees on which Spanish moss grows provide support but no nutrients or moisture — the plant provides those things for itself, apparently without harming its support tree. This is an example of a: a. commensalism. b. antagonism. c. mutualism. d. parasitism. e. competition. ANSWER: a 24. Tropical leafcutter ants collect leaf cuttings which they transport to special underground chambers. There, the ants chew the leaves to create nursery beds on which they grow a species of fungus they use for food. When ant queens disperse to establish new colonies, they carry the fungus with them, dispersing it as well (this benefits the fungus). In the ants' new nest, the fungus is at risk of being destroyed by another fungal species that is able to grow in the same habitat, using the same limited resources. On their bodies, the ants carry and provide a home for bacteria that produce antibiotics the ants use to kill the newly encountered fungus and thereby protect their food supply. In this system, the relationship between the two fungal species is: a. interspecific competition. b. mutualism. c. intraspecific competition. d. parasitism. e. commensalism. ANSWER: a 25. In the Sierra Nevada mountains of California, there are many populations of the checkerspot butterfly, Euphydryas editha. You notice that females of one population, population A, lay their eggs near the tip of a plant's stem. Females of another population in the same area, population B, lay their eggs at the base of the stem on a different type of plant. The young hatch as caterpillars, and they live on the host plant and eat its leaves. Is the information provided a complete description of the butterflies' niche? a. yes, because you know the butterflies live in the Sierra Nevada b. yes, because both eggs and adults are described c. yes, because eggs, young, and adults are described d. no, because I also need to know what the adults eat e. no, because a complete niche includes physical and biological factors ANSWER: e 26. In the Sierra Nevada mountains of California, there are many populations of the checkerspot butterfly, Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 Euphydryas editha. The young of the populations hatch as caterpillars, and they live on the host plant and eat the new leaves. The host plant used by one population grows over a single season, with new leaves (the most nutritious) appearing at the tip of the stem. Where would you expect females from this population to lay their eggs? a. at the base of the plant stem b. at the tip of the plant stem c. anywhere along the plant stem d. in the exact middle of the plant stem ANSWER: b 27. Lampreys have sucker-like mouths and are known to attach themselves to fish. In doing so, they rasp off a section of the fish using hard keratinized structures in the mouth and "feed" on the blood and fluid that oozes into their oral cavity. Fish can survive with a lamprey attached, but they have to expend extra energy swimming with the lamprey attached. This association would be designated a: a. mutualism. b. parasitism. c. predation. d. commensalism. ANSWER: b 28. When two species compete for similar resources, resource partitioning can alter the niche of the species that compete. The graphs shown provide an example of this.

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Resource partitioning results in: a. individuals of each species sharing the resources that they both exploit. b. individuals of each species hybridizing in the region of overlap. c. a move from the fundamental niche to the realized niche for both species. d. a move from the realized niche to the fundamental niche for both species. ANSWER: c 29. Measuring its costs and benefits in terms of energy spent and/or gained, competition among individuals results in: a. a loss for both individuals. b. a gain for both individuals. c. a gain for one individual and a loss for the other. d. a gain for one individual and neither a gain nor a loss for the other. e. None of the answer options is correct. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 30. Which of the answer choices best describes the kinds of resources for which individuals may compete? a. food or some form of nutrient b. space or shelter c. food and mates d. food, nutrients, space, or shelter e. any resource that is limited in the environment ANSWER: e 31. In some rocky intertidal habitats, the barnacle Chthamalus can survive across the same depths as can the barnacle Balanus. However, where the two species occur together, Chthamalus adults are found only in the more stressful upper portions of the habitat where they are more able than Balanus to resist desiccation. This is an example of: a. a mutualistic symbiosis. b. intraspecific competition reducing the size of Chthamalus' fundamental niche to a smaller realized niche. c. mutualism reducing the size of Balanus' realized niche to a smaller fundamental niche. d. competitive exclusion reducing the size of Chthamalus' fundamental niche to a smaller realized niche. e. competitive exclusion expanding the size of Balanus' realized niche to a larger fundamental niche. ANSWER: d 32. Most symbiotic interactions require a long-term association between the two species. Which of the interactions is not a symbiotic association? a. photosynthetic algae living with a fungus as a lichen b. a lion that eats a small antelope in the open grasslands c. leafcutter ants that supply food to their fungal gardens d. ants that live on a host tree and chew off the branches of neighboring trees that grow too close ANSWER: b 33. Many plants have specialist pollinators. That is, the plant has evolved adaptations that only allow one type of pollinator access to nectar when pollinating. This is a result of: a. coevolution. b. adaptive radiation. c. parasitism. d. commensalism. ANSWER: a 34. Some argue that any members of the microbiota that take up space on our body surfaces and do no harm to us are actually providing us a service because they prevent the colonization of harmful species. From this point of view, the relationship between the skin microbiota and their human host is: a. mutualism. b. an antagonistic relationship. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 c. commensalism. d. predation. ANSWER: a 35. You are studying the composition of human gut microbiota. You sample the solid waste of three infants from three different families each week for 24 weeks. Base on the infants, you find that the population of bacteria in the human gut is dominated by one species for the first 12 weeks of life, and then shifts to another species. This suggests the composition of bacterial community in the human gut: a. is fixed and unchanging. b. is essential for human health. c. is harmful to humans. d. changes over time. e. is helpful to humans. ANSWER: d 36. Antibiotics are compounds that kill bacteria. If you take an antibiotic in pill form, it is distributed throughout your body. Individuals who take antibiotic pills for a bacterial infection in their throat often find they develop an "upset stomach" because the antibiotic also kills the microbial communities in their gut. This suggests that: a. gut bacteria have positive effects on the humans they inhabit. b. gut bacteria have only negative effects on the humans they inhabit. c. it is good to sterilize the gut. d. humans have a predatory relationship with their gut bacteria. e. None of the answer options is correct. ANSWER: a 37. The presence of bacteria that produce essential amino acids in the bodies of aphids is an example of: a. obligate mutualism. b. commensalism. c. facultative mutualism. d. antagonism. ANSWER: a 38. When sea otters are present in a kelp forest community, there are more than 12 other species found in the community. If sea otters are removed, there are only 5-7 other species that remain in the community. These data suggest that sea otters are: a. a keystone species. b. top predators. c. primary consumers. d. secondary consumers. ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 39. Biodiversity is more than the number of species found in a specific area. Also included in biodiversity are the number of: a. populations of species in a specific area. b. distinct evolutionary lineages in a specific area. c. symbiotic interactions in a specific area. d. trophic levels in a specific area. ANSWER: b 40. The volcanic oceanic island of Krakatoa is located in Indonesia. In 1883, this island experienced a massive volcanic explosion that killed most of the plants and animals living on the island. After a very, very long period of recovery following the eruption, which group of organisms would we expect to have the most representation among species on the island? a. lizards b. birds c. butterflies d. beetles ANSWER: d 41. Imagine you are studying a forest community that is subject to unusually intense fires and hurricanes approximately every 100 years, which is more frequent than in most environments. What would you expect of the species diversity in this community? a. The plant and animal species in the area are entirely different after each fire; therefore, every 100 years, the species present before the fire are replaced with species that have never been present in that area. b. The species present are able to tolerate fires, and the species composition never changes. c. The combination of species present before the fire are different than those present 90 years after the fire. d. The recovery after the fire takes longer for each subsequent fire that occurs in the area. ANSWER: c 42. Consider two islands located an equal distance from a mainland, which is the colonization source. One island is large, the other is small. Which island will have a higher rate of change in species composition? a. the large island b. the small island c. The rate of change in species composition will be the same on both islands. d. Species composition does not change at equilibrium ANSWER: b 43. In the Sierra Nevada mountains of California, there are many populations of the checkerspot butterfly, Euphydryas editha. You notice that females of one population, population A, lay their eggs near the tip of a plant's stem. Females of another population in the same area, population B, lay their eggs at the base of the stem on a different type of plant. The young hatch as caterpillars, and they live on the host plant and eat its leaves. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 You breed a member of population A with a member of population B. You raise the hybrid caterpillars on paper towels in the lab. You cross all the female hybrids with members of population A, and then you release the pregnant hybrid females back into the natural environment. All of the hybrid females lay their eggs at the center of the stems, not at the top or the bottom. What does this suggest about the trait of female egg placement on stems? a. It is random. b. There is no genetic component. They always lay their eggs in the same environment that they experienced as caterpillars. c. There is a genetic component to egg placement. ANSWER: c 44. In ponds where there are southern toads, eastern spadefoot toads, and spring peeper toads, all these toads eat the same food source. When red-spotted newts are also present in the ponds, they preferentially eat the tadpoles of southern toads and eastern spadefoot toads. What happens to the ecological niche of spring peeper toad tadpoles when newts are in the pond? a. Their fundamental niche expands. b. Their realized niche expands. c. Their ecological niche stays the same. ANSWER: c 45. In the Sierra Nevada mountains of California, there are many populations of the checkerspot butterfly, Euphydryas editha. You notice that females of one population, population A, lay their eggs near the tip of a plant's stem. Females of another population in the same area, population B, lay their eggs at the base of the stem on a different type of plant. The young hatch as caterpillars, and they live on the host plant and eat its leaves. What direct advantage does laying their eggs on different plants give these two populations? a. reduced competition through resource partitioning b. reduced competition through sharing a common resource c. increased speciation rate through resource partitioning d. decreased speciation rate through resource partitioning e. a larger niche in their natural environment ANSWER: a 46. Competition for resources occurs often within communities. In the figures shown, the black line represents the change in fitness for one species, and the gray line represents the change in fitness for another species. Which of the graphs shown represents the change in fitness for each species when there is competition for the same resource?

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Chapter 45

a. graph M b. graph H c. graph K ANSWER: b 47. In a specific community, there are some isolated populations of flowering plants that have evolved adaptations to their local areas within the community. Each isolated population contributes to: a. species diversity within the community. b. overall biodiversity of the community. c. species extinction within the community. d. additional trophic levels within the community. ANSWER: b 48. Similar to predators, _____ can have effects on population sizes because they can keep populations below carrying capacity and limit competitive exclusion. a. parasites b. interspecific competitors c. facultative mutualists d. commensals e. obligate mutualists ANSWER: a 49. The human body has about 10 bacterial cells for every eukaryotic cell. Bacteria coat our skin, gut, and mouth. Also present are protists, archaeans, and viruses. Collectively, these organisms are our microbiota. For most members of our microbiota, our body provides their environment, or space to live. They, in turn, have no effect on us. This is an example of: a. mutualism. b. an antagonistic relationship. c. commensalism. d. predation. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 50. The bacterial species Staphylococus aureus is found on the skin and in the nasal passages of about 20% of the human population. In healthy individuals, S. aureus is benign (does not cause harm) in these locations. However, if it is introduced into the bloodstream through, for example, a wound, it can make a person gravely ill. Thus, the ecological relationship of S. aureus with a human: a. can be commensal or antagonistic. b. can be mutualistic or antagonistic. c. is always commensal. d. is fixed or unchanging. e. is antagonistic. ANSWER: a 51. Some species of ants "farm" aphids by protecting them from predators. In return, the ants feed on a sugarrich liquid, called honeydew, secreted by the aphids. The ecological relationship between the ants and the aphids is: a. mutualism. b. parasitism. c. competition. d. None of the answer options is correct. ANSWER: a 52. In the late 1960s, Robert Paine conducted landmark studies on diversity in the rocky intertidal zone comparing the species diversity in control plots with diversity in experimental plots from which he removed the top predator, sea stars. After 5 years, 15 species of intertidal invertebrates lived in the control plots, while the experimental plots were dominated by only two species, one mussel and one barnacle. Why did species diversity most likely remain high in the presence of a predator? a. The sea star kept the mussel and barnacle populations low enough to prevent competitive exclusion from occurring. b. The sea star kept the mussels and barnacles from developing a mutualistic relationship. That, in turn, kept their populations low. c. The mussels and the barnacles parasitized the sea stars, keeping populations low enough to prevent competitive exclusion from occurring. d. The sea stars developed a mutualistic relationship with the mussels and barnacles, keeping their populations low and preventing competitive exclusion from happening. ANSWER: a 53. What is the ecological explanation for why sunflowers and grassy weeds thrive in your untended, sunny garden, but mosses, which require shade, do not? a. Communities depend on interactions among organisms. b. Communities depend on their physical environment. c. Disturbance can modify community composition. d. None of the answer options is correct. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 54. Many flowering plant species are pollinated by a single animal species. The evolution of specialist pollinators is the result of: a. low pollinator diversity in the community. b. colonization of the community at the same time by both the pollinator and the flowering plant. c. coevolution between the flowering plant and pollinator. d. local extinction of other pollinators. ANSWER: c 55. Southern California is an ecosystem in which organisms have evolved adaptations to frequent fires. Each time a fire occurs in the region a biologist should observe: a. ecological succession. b. increased growth of organisms already present. c. an assemblage of species none of which were present before the fire. ANSWER: a 56. Typically, when a keystone species is removed from a community: a. another keystone species colonizes the area and the community remains relatively unchanged b. species diversity in the community will increase. c. species diversity in the community will decrease. d. there will be species turnover and none of the original species will remain in the community. ANSWER: c 57. Glacier Bay in Alaska is a fjord whose lowlands were covered by glacial ice until around 230 years ago. As the glacier retreated, it left behind sediment that was exposed to plant colonization. Remarkably (and fortunately for generations of ecologists), the location of the tip of the retreating glacier has been frequently documented since the late 1700s. This means that scientists have been able to develop a detailed timeline of the changes in the plant community, or _____ in Glacier Bay over time. a. ecological succession b. niche diversification c. niche divergence d. competitive exclusion e. None of the answer options is correct. ANSWER: a 58. In Glacier Bay, one of the first species to colonize a newly exposed area of sediment left behind by a glacier is fireweed. You would predict that fireweed is a(n) _____-strategist. a. r b. K ANSWER: a 59. You are surveying biodiversity on a new island chain. You have counted the number of bird species on one Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 island already. The next island is larger and closer to the mainland than the one you have just surveyed. According to the theory of island biogeography, the total number of bird species on the island that is larger and closer to the mainland should be _____ than on the second island because the rate of immigration to the new island should be _____ and the rate of extinction should be _____. a. smaller; higher; lower b. smaller; lower; higher c. greater; higher; lower d. greater; lower; higher ANSWER: c 60. As a rough estimate, the species-area relationship derived from the theory of island biogeography predicts that a 90% reduction of habitat area results in a _____% loss of species diversity if no other factors are taken into account. a. 10 b. 15 c. 20 d. 25 e. 50 ANSWER: e 61. Consider a scenario in which you observe population changes among the plants in your garden. Over the winter the hibiscus plants in your garden were killed by cold winter temperatures. Much to your surprise sunflowers sprouted on their own where the hibiscus plants used to grow. The following summer, for the first time, you see lots of goldfinches (small yellow birds) visit the garden to eat the sunflower seeds. What is the ecological relationship between the goldfinches, who eat the sunflower seeds, and the sunflowers? a. competition b. commensalism c. mutualism d. predation ANSWER: d 62. Consider a scenario in which you observe population changes among the plants in your garden. You notice that two small birch trees sprout and begin to grow among the sunflowers in your garden. Birch trees are slower growers than sunflowers. Five years pass, and you find that the birch trees are starting to shade the sunflowers. In this scenario, the birch trees and the sunflowers are in competition for: a. water. b. light. c. nutrients. d. All of these choices are correct. ANSWER: d 63. Consider a scenario in which you observe population changes among the plants in your garden. You notice that two small birch trees sprout and begin to grow among the sunflowers you planted in your garden last Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 summer. Birch trees are slower growers than sunflowers. By the fifth summer after the birch trees appear, they start to shade the sunflowers. The progression of your garden from sunflowers to birch trees is an example of: a. succession. b. evolution. c. ecology. d. mutualism. e. None of the answer options is correct. ANSWER: a 64. In the Sierra Nevada mountains of California, there are many populations of the checkerspot butterfly, Euphydryas editha. You notice that females of one population, population A, lay their eggs near the tip of a plant's stem. Females of another population in the same area, population B, lay their eggs at the base of the stem on a different type of plant. The young hatch as caterpillars, and they live on the host plant and eat its leaves. Based on this description, what is the ecological relationship between the butterfly and its host plant? a. mutualism b. commensalism c. parasitism d. competition ANSWER: c 65. In the Sierra Nevada mountains of California, there are many populations of the checkerspot butterfly, Euphydryas editha. You notice that females of one population, population A, lay their eggs near the tip of a plant's stem. Females of another population in the same area, population B, lay their eggs at the base of the stem on a different type of plant. The young hatch as caterpillars, and they live on the host plant and eat its leaves. The choice of different host plants by members of these two populations of butterfly: a. increases competition between members of the two different butterfly populations. b. decreases competition between members of the two different butterfly populations. c. has no effect on competition between members of the two different butterfly populations. d. decreases competition among members of the same butterfly population. e. increases competition among members of the same butterfly population. ANSWER: c 66. Consider a scenario in which you observe population changes among the plants in your garden. You notice that two small birch trees sprout and begin to grow among the sunflowers you planted in your garden last summer. Birch trees are slower growers than sunflowers. By the fifth summer after the birch trees appear, they start to shade the sunflowers. Compared to the sunflowers, birch trees are: a. r-strategists. b. K-strategists. c. less competitive. d. more competitive. e. None of the answer options is correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 ANSWER: b 67. You observe two species of squirrels, Species A and Species B, coexisting on a mountain, and diagram the distribution of the two species as shown. In the diagram, the distribution of Species A is shaded gray and that of Species B is marked with vertical lines. You indicate in your diagrams the distributions of each species, either in the presence or absence of the other species.

Based on the diagram, which species is the stronger competitor? a. species A b. species B c. Neither species is a better competitor. ANSWER: b 68. You observe two species of squirrels, Species A and Species B, coexisting on a mountain, and diagram the distribution of the two species as shown. In the diagram, the distribution of Species A is shaded gray and that of Species B is marked with vertical lines. You indicate in your diagrams the distributions of each species, either in the presence or absence of the other species.

The area shaded gray in the figure on the left represents the _____ of species Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 a. actual niche b. realized niche c. accessible niche d. fundamental niche ANSWER: d 69. Consider the location of islands off the coast of a mainland. Determine which island is indicated in the statement and whether the statement is true or false.

Island K is expected to have the highest extinction rate. a. true b. false ANSWER: a 70. Consider the location of islands off the coast of a mainland. Determine which island is indicated in the statement and whether the statement is true or false.

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Species diversity on island M will be highest. a. true b. false ANSWER: a 71. Consider the location of islands off the coast of a mainland. Determine which island is indicated in the statement and whether the statement is true or false.

Species richness could be the same on islands L and M. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 a. true b. false ANSWER: a 72. Aphids can feed in leafy treetops or tree canopies. Ants are usually not abundant in tree canopies unless aphids are also present. You introduce aphids to an area of the tree canopy and survey the abundance of ants over time. Consider the graphs. Which of the plots shown represents your expected trend in ant abundance?

a. graph M b. graph H c. graph K d. graph L e. graph Q ANSWER: b 73. Aphids can feed in leafy treetops or tree canopies. Ants are usually not abundant in tree canopies unless aphids are also present. You introduce aphids to an area of the tree canopy and survey the abundance of ants over time. You continue your experiments by taking one of the trees in the canopy and removing all of the ants. If the ants and aphids are in a mutualistic relationship that strongly benefits them both, you expect the aphid population will _____ in response to the lack of ants. a. increase in size b. stay the same size c. decrease in size d. change species composition e. change in demographics (age distribution) ANSWER: c 74. Aphids can feed in leafy treetops or tree canopies. Ants are usually not abundant in tree canopies unless aphids are also present. You introduce aphids to an area of the tree canopy and survey the abundance of ants over time. You continue your experiments by removing all of the ants from one of the trees in the canopy. If the ants and aphids are in a mutualistic relationship that strongly benefits them both, you expect the aphid population will decrease in size in response to the lack of ants. The best control for your experiment in this scenario is a tree in which you _____ and you measure the abundance of _____ over time. a. do not remove any insects; ants b. do not remove any insects; aphids c. add a second ant species; the first ant species Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 d. Actually, the control is built into the original experiment. You do not need to set up a separate control. ANSWER: b 75. You conduct an experiment to look at the relationship between ants and other insect diversity in and around trees. You take a tree and block the ants from getting to the canopy, where there are aphids that produce sugary substances that attract ants, because the aphids are parasites of the tree and remove phloem. You then spend 3 years counting the number and types of insects on the forest floor around this tree (tree A) and in the area around a tree in which ants are not blocked from accessing the canopy (tree B). You find that the area around tree A has 25% fewer ants and 40% more beetles than the area around tree B) These data support the hypothesis that: a. there is no relationship between ant abundance and beetle abundance. b. there is a positive relationship between ant abundance and beetle abundance. c. there is a negative relationship between ant abundance and beetle abundance/ d. ecological interactions in the tree canopy have no effect on forest floor communities. e. ecological interactions in the tree canopy always have a negative effect on forest floor communities. ANSWER: c 76. You conduct an experiment to look at the relationship between ants and other insect diversity in and around trees. You take a tree and block the ants from getting to the canopy, where there are aphids that produce sugary substances that attract ants, because the aphids are parasites of the tree and remove phloem. You then spend 3 years counting the number and types of insects on the forest floor around this tree (tree A) and in the area around a tree in which ants are not blocked from accessing the canopy (tree B). As part of the experiment, you monitor the growth of tree A, where the ants have been blocked from accessing the canopy, and tree B, where the ants do have access to the canopy, over the 3-year period. You expect _____ because _____. a. higher growth in tree A; the number of aphids will fall because ants are absent, and aphids are parasites of the tree. b. higher growth in tree B; the natural environment is always optimized so any change (for example, blocking the ants from tree A will reduce the performance of all members of the community. c. no change in growth; due to the size difference between the trees and the insects, the presence or absence of insects will not have an effect on tree growth. ANSWER: a 77. You conduct an experiment to look at the relationship between ants and other insect diversity in and around trees. You take a tree and block the ants from getting to the canopy, where there are aphids that produce sugary substances that attract ants, because the aphids are parasites of the tree and remove phloem. You then spend 3 years counting the number and types of insects on the forest floor around this tree (tree A) and in the area around a tree in which ants are not blocked from accessing the canopy (tree B). As part of the experiment, you monitor the growth of tree A, where the ants have been blocked from accessing the canopy, and tree B, where the ants do have access to the canopy, over the 3-year period. Which of the factors would also affect tree growth and thus complicate the experiment described? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 a. the presence of other tree parasites. b. the distribution of nutrients in each tree's local environment. c. the amount of leaves, and thus photosynthetic production, of each of the trees. d. the ambient light conditions, and thus photosynthetic production, for each of the trees. e. All of these responses would also affect the growth of the trees, and thus complicate the experiment described. ANSWER: e 78. You are studying the island biogeography of birds in the Caribbean islands and find a species-area relationship with a constant, c = 8.1, and an exponent, x = 0.15. Recall that the species-area relationship is defined as S = cAx . How many bird species would you expect to find on an island of 240 mi2? a. 3 species b. 18 species c. 21 species d. 50,000 species ANSWER: b 79. The theory of island biogeography describes the number of species on an island as being determined by an equilibrium between the processes of immigration and extinction. Consider a small, far-offshore island and a large, near-shore island. Which island is expected to have a higher equilibrium number of species? a. far-offshore, small island b. near-shore, large island c. The equilibrium number of species is expected to be the same. d. It is not possible to make a definite prediction based on the information given. ANSWER: b 80. Consider the location of islands off the coast of the mainland.

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Island H is colonized with 30 individuals of a bird species from the mainland. Subsequently, 5 individuals of the same species leave island H and colonize island L. Indicate whether the statement is true or false. Genetic diversity will be higher on island L than H because island L is larger. a. true b. false ANSWER: b 81. Consider the location of islands off the coast of the mainland.

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Chapter 45 The bird species is more likely to go extinct on island L than island H. a. true b. false ANSWER: b 82. Toucans are frugivorous (fruit-eating) birds that inhabit tropical forests in Central and South America where they can potentially consume any fruit that is available (generalists). The realized niches of toucans are determined by competition in the consumption of different species of fruits (niche overlap). White-throated toucans are the largest species and very strong competitors who easily scare the mid-sized Cuvier's and Ivorybilled toucans off the fruiting trees, altering access to the resources among the three species. The graphs shown represent the occupation of niche space at two different sites, where two or three species coexist, in different abundances.

Consider the graphs. Which of the graphs shown best characterizes the fundamental niche of any of these species of toucans? Assume the X-axis is the niche space available to any species of toucan.

a. Graph M b. Graph H c. Graph K d. Graph L ANSWER: c 83. Toucans are frugivorous (fruit-eating) birds that inhabit tropical forests in Central and South America where Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 they can potentially consume any fruit that is available (generalists). The realized niches of toucans are determined by competition in the consumption of different species of fruits (niche overlap). White-throated toucans are the largest species and very strong competitors who easily scare the mid-sized Cuvier's and Ivorybilled toucans off the fruiting trees, altering access to the resources among the three species. The graphs shown represent the occupation of niche space at two different sites, where two or three species coexist, in different abundances.

Which of the statements best describes the competition and coexistence among toucans in Site 1? a. Ivory-billed toucans are better competitors than any of the other species b. Niche differentiation allows the two most abundant species to coexist in relatively high abundance. c. Cuvier's toucans cannot coexist with other toucan species in Site 1 because their niche overlaps with those species. d. There is no overlap in the niches of the three species present in Site 1. ANSWER: b 84. Toucans are frugivorous (fruit-eating) birds that inhabit tropical forests in Central and South America where they can potentially consume any fruit that is available (generalists). The realized niches of toucans are determined by competition in the consumption of different species of fruits (niche overlap). White-throated toucans are the largest species and very strong competitors who easily scare the mid-sized Cuvier's and Ivorybilled toucans off the fruiting trees, altering access to the resources among the three species. The graphs shown represent the occupation of niche space at two different sites, where two or three species coexist, in different abundances.

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According to the graphs, Cuvier's toucans are stronger competitors than Ivory-billed toucans. a. true b. false ANSWER: b 85. Toucans are frugivorous (fruit-eating) birds that inhabit tropical forests in Central and South America where they can potentially consume any fruit that is available (generalists). The realized niches of toucans are determined by competition in the consumption of different species of fruits (niche overlap). White-throated toucans are the largest species and very strong competitors who easily scare the mid-sized Cuvier's and Ivorybilled toucans off the fruiting trees, altering access to the resources among the three species. The graphs shown represent the occupation of niche space at two different sites, where two or three species coexist, in different abundances.

In absence of Ivory-billed toucans at Site 2, the realized niche of Cuvier's toucans expands but the realized niche of White-throated toucans remains mostly unchanged. Which of the statements is best supported by the Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 data presented in the graph? a. Competition is less intense among toucans at Site 1 compared to Site 2. b. Ivory-billed toucans are constantly migrating from Site 1 to Site 2. c. The expanded realized niche of Cuvier's toucans resulted in higher abundance of this species. d. At Site 2, the fundamental niche of White-throated toucans is reduced compared to Site 1 due to increased competition. ANSWER: c 86. Toucans are frugivorous (fruit-eating) birds that inhabit tropical forests in Central and South America where they can potentially consume any fruit that is available (generalists). The realized niches of toucans are determined by competition in the consumption of different species of fruits (niche overlap). White-throated toucans are the largest species and very strong competitors who easily scare the mid-sized Cuvier's and Ivorybilled toucans off the fruiting trees, altering access to the resources among the three species. The graphs shown represent the occupation of niche space at two different sites, where two or three species coexist, in different abundances.

The carrying capacity of Cuvier's toucans is: a. higher in Site 1 b. higher in Site 2 c. constant d. equal in both sites e. None of the other answer options is correct. ANSWER: b Multiple Response 87. Which of the answer choices could influence the pattern of succession in a recently disturbed habitat? a. the existence of fungal symbionts in the soil. b. air quality Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 c. soil composition d. the distance to the nearest undisturbed habitats e. the surviving species in the habitat ANSWER: a, c, d, e 88. During a biodiversity sampling exercise, an undergraduate found 8 species of bees in one area of campus yet calculated that there were probably 12 species of bees in that specific area of the campus. Why is there typically a discrepancy between the number of different species caught and the calculated value of how many species probably exist in an area? Select all that apply. a. Some of the species in the taxonomic group may actually belong to a different taxonomic group. b. The methods used to catch a specific taxonomic group may not be enticing to all species of the group. c. The time when traps were set out may not be when some species of the taxonomic group are active. d. Without sequencing the genomes of all the species caught, you cannot know how many there are. ANSWER: b, c 89. Species diversity on habitat islands reflects the rate of: Select all that apply. a. new species arrival. b. population growth. c. growth toward the carrying capacity. d. colonist species extinction. ANSWER: a, d 90. Which of the answer choices could be used to define the ecological niche of a plant species? Select all that apply. a. the pH of its soil b. the insects that eat it c. the depth to which the soil freezes in the winter d. the number of competitors present ANSWER: a, b, c 91. Competition occurs: Select all that apply. a. among members of the same species. b. between members of different species. c. only between individuals who share the same realized niche. d. only between individuals in populations that are close to their carrying capacity in the environment. e. None of the answer options is correct. ANSWER: a, b 92. Measuring its costs and benefits in terms of energy spent and/or gained, which interaction(s) involve at least Copyright Macmillan Learning. Powered by Cognero.

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Chapter 45 one individual gaining? Select all that apply. a. competition b. predation c. obligate mutualism d. facultative mutualism e. commensalism ANSWER: b, c, d, e

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Chapter 46 Multiple Choice 1. Examine the figure, which shows the parallel history of atmospheric CO2 levels and surface temperature over the last 400,000 years, based on measurements of air bubbles trapped in glacial ice sheets in Antarctica.

Glacial expansion correlates with _____ amounts of CO2 in the atmosphere and temperatures _____ than those measured in 1950. a. decreasing; lower b. increasing; higher ANSWER: a 2. Which of the equations is the chemical reaction for cellular respiration? a. C6H12O6 + 6O2 → 6CO2 + 6H2O b. 6CO2 + 6H2O → C6H12O6 + 6O2 c. CaSiO3 + CO2 → CaCO3 + SiO2 d. C6H12O6 + 6CO2 → 6O2 + 6H2O ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 3. Recall that sedimentary rocks, living organisms, oceans, and soil are all carbon reservoirs. If the carbon in each of these reservoirs was instantly transformed into atmospheric CO2, which reservoir would contribute the most CO2 to the atmosphere? a. oceans b. living organisms c. sedimentary rocks d. soil ANSWER: c 4. Which of the statements is an accurate reflection of photosynthesis and chemical weathering? a. Both processes release CO2 as a product; that is, they increase atmospheric CO2 levels. b. Both processes use CO2 as a reactant; that is, they decrease atmospheric CO2 levels. c. Both processes are typically associated with the long-term carbon cycle. d. Both processes are typically associated with the short-term carbon cycle. ANSWER: b 5. Which of the answer choices represents an accurate description of the movement of carbon through a food web? a. Carbon is being removed from the atmosphere at each level. b. Carbon is being added to the atmosphere at each level. c. Carbon is removed by the producers and added by the consumers. ANSWER: c 6. Primary producers are the basis for all aquatic and terrestrial food webs. How does the amount of primary production affect community structure? a. The more energy available from primary producers, the more species that could be supported at higher trophic levels in the community. b. The more energy available from primary producers, the more individuals that could be supported at higher trophic levels in the community. c. Primary production has no direct relationship to community structure. ANSWER: b 7. Consider the food web shown in the figure.

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If steelhead were removed from the web, what would you predict would happen to the numbers of tuft-weaving chironomids? a. They would increase. b. They would decrease. c. They would stay the same. ANSWER: b 8. Because energy transfer is not 100% efficient across trophic levels, an average of about _____ of energy and biomass available at one trophic level is available at the next. a. 1% b. 5% c. 10% d. 15% e. 20% ANSWER: c 9. In a pond, tadpoles eat algae and fish eat the tadpoles. Around the pond, grasshoppers eat grass and, at night, are preyed upon by bats. Other bats eat the fish that eat the tadpoles. In this community, the algae are: a. primary producers. b. primary consumers. c. secondary consumers. d. tertiary consumers. e. detritivores. ANSWER: a 10. The first law of thermodynamics states that energy can be transformed from one state to another, but it cannot be created or destroyed. Taking this into consideration, what can be said about the energy transferred Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 between levels in a trophic pyramid? a. Some of the energy in a trophic level is converted to heat and is unavailable to the next level of the trophic pyramid. b. All of the energy is transferred from one trophic level to the next. Components that have no energy, such as light, are used to create the energy in the chemical bonds of carbohydrates. c. All of the energy necessary for a trophic level is transferred to it from the trophic level below it. Because there is less biomass at each successive trophic level, less energy needs to be transferred at each successive level. d. Some of the energy is held in one trophic level, and only the energy needed to support the smaller biomass at a higher level is transferred. e. Organisms at higher trophic levels consume only the energy they need from lower trophic levels. This leaves a "bank" of energy in lower trophic levels to support higher trophic levels when food is scarce. ANSWER: a 11. The levels of CO2 in the atmosphere have been: a. increasing. b. decreasing. c. staying the same. ANSWER: a 12. The CO2 level is _____ during winter in the northern hemisphere compared to levels in the summer. a. lower b. higher c. the same ANSWER: b 13. Which correlates most closely with net increase in CO2 levels over the past two centuries? a. volcanic gases b. carbon dissolved in the oceans c. respiration d. human activities ANSWER: d 14. Because CO2 is a greenhouse gas, a rise in CO2 levels correlates with a(n): a. rise in respiration. b. rise in temperature. c. increase in the size of glaciers. d. rise in photosynthesis. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 15. The carbon cycled through a food web primarily comes from: a. primary producers. b. consumers. c. decomposers. ANSWER: a 16. Which of the choices provided best depicts the flow of carbon between primary producers, consumers, and decomposers? a. primary producers → consumers → decomposers b. consumers → primary producers → decomposers c. decomposers → consumers → primary producers ANSWER: a 17. In general, a larger biomass in primary producers for a trophic pyramid: a. supports less biomass at higher levels. This is because more energy is gained between the level of primary producer and primary consumer. b. does not affect the amount of biomass that can be supported at the level of primary consumer. c. can support more biomass at higher trophic levels. This is because more energy is available between the level of primary producer and primary customer. ANSWER: c 18. Consider the food web shown.

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The minnow fry in this system is a: a. primary producer. b. primary consumer. c. secondary consumer. d. tertiary consumer. ANSWER: c 19. It is thought that approximately 90% of plant species went extinct during the Carboniferous Period, approximately 320-290 million years ago. Which of the graphs represent the changes that likely occurred in carbon dioxide and oxygen levels? The solid black line depicts CO2 levels. The y-axis represents levels of oxygen or carbon dioxide. The dashed line represents O2 levels. The left side of the x-axis represents the time of the extinction event. As you move to the right, the x-axis represents years since the extinction event.

a. graph M Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 b. graph H c. graph K d. graph L ANSWER: b 20. According to the figure, atmospheric CO2 levels oscillate throughout the year.

How would this trend change if it were summer for an entire year? Shown are three graphs. Select the graph that represents the atmospheric CO2 levels of an Earth with a full year of summer. Assume you are in the Northern Hemisphere.

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a. graph A b. graph B c. graph C ANSWER: b 21. The Keeling curve shows that, while CO2 levels oscillate on an annual basis, overall CO2 levels increase from year to year.

What is the primary reason that this trend continued in the last 50 years? a. increasing rates of aerobic respiration associated with the rise in world population b. decreasing rates of oxygenic photosynthesis by plants and algae c. clearing of forests to create agricultural land outside the tropics d. combustion of fossil fuels ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 22. Positive feedback is one in which the products of a process act to increase the rate at which the process operates. Which of the answer choices would constitute a positive feedback response in the carbon cycle? a. increasing CO2 causing an increase in rates of photosynthesis b. increasing temperature causing an increase in rates of photosynthesis c. increasing temperature causing an increase in rates of respiration d. increasing temperature causing an increase in rates of continental weathering ANSWER: c 23. Consider the Keeling curve, which shows a steady increase in CO2 levels over the past 50+ years. This steady rise has been attributed to both natural and human activity. Imagine if everyone in the world planted a fast-growing vine that was an annual (meaning that it grows large and fast, but dies after one year). Predict the effect this would hypothetically have on the level of CO2 in the atmosphere over the next 20 years. a. The level of CO2 in the atmosphere will steadily decrease because of increased rates of photosynthesis. b. The level of CO2 in the atmosphere will steadily increase because decomposition of the annuals simply returns the CO2 back into the atmosphere. c. The level of CO2 in the atmosphere will stay the same because the growing annuals will remove the extra CO2 from human activity. ANSWER: b 24. Studies assessing the amounts of different carbon isotopes (12C, 13C, and 14C) in the atmosphere were important because they provided evidence that human activities were, in fact, adding CO2 to the atmosphere. Of what type of relationship is this an example? a. correlation b. causation c. conduction d. antagonism ANSWER: b 25. When looking at the carbon isotopes present in the atmosphere, scientists determined that 13C and 14C levels have decreased over time. What did researchers conclude from these observations? a. Volcanic activity is responsible for increased atmospheric CO2. b. The burning of forests is responsible for increased atmospheric CO2. c. The use of fossil fuels is responsible for increased atmospheric CO2. d. Increased plant/animal respiration is responsible for increased atmospheric CO2. ANSWER: c 26. In reviewing historical records and ice core data, researchers noted that atmospheric CO2 levels started to rise as humans began to burn fossil fuels. Of what type of relationship is this an example? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 a. correlation b. contradiction c. causation d. antagonism e. conduction ANSWER: a 27. Imagine that a time traveler was able to successfully travel 2 billion years back in time and modify the environment so that photosynthesis and respiration were completely coupled; that is, all carbon dioxide produced by respiration is consumed through photosynthesis, and all oxygen produced is utilized through cellular respiration. How would this affect Earth's present-day atmosphere, if at all? a. Earth's atmosphere would be unaffected. b. Atmospheric O2 levels would increase. c. Atmospheric O2 levels would decrease. d. Atmospheric CO2 levels would increase, but O2 levels would be unaffected. e. Atmospheric O2 levels would increase, but CO2 levels would be unaffected. ANSWER: c 28. How is atmospheric oxygen linked to the carbon cycle? a. An increase in burial of organic carbon in sediments results in more oxygen in the atmosphere and oceans. b. More photosynthesis leads to more oxygen in the atmosphere and oceans. c. More volcanic CO2 emission leads to more oxygen in the atmosphere and oceans. d. Longer food chains with more consumers lead to more oxygen in the atmosphere and oceans. ANSWER: b 29. What would be the environmental consequences if Earth's volcanoes all shut down tomorrow? a. CO2 levels in the atmosphere would slowly rise. b. CO2 levels in the atmosphere would slowly fall. c. Rates of continental weathering would slowly increase. d. Temperature would slowly increase. ANSWER: b 30. What would be the environmental consequences if Earth's volcanoes all shut down tomorrow? a. CO2 levels in the atmosphere would slowly rise. b. Continental weathering would decrease. c. Rates of continental weathering would slowly increase. d. Temperature would slowly increase. ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 31. CO2 is an example of a _____, a gas in the atmosphere that is transparent to solar radiation but can absorb heat emitted from Earth's surface. a. photosynthesis product b. synthetic carbon cycle product c. greenhouse gas d. None of the other answer options is correct. ANSWER: c 32. How has Earth's atmosphere been influenced by the fact that, as part of the carbon cycle, some organic carbon is ultimately stored in sedimentary rocks? a. The storing of organic carbon in sedimentary rocks allowed for Earth's atmosphere to remain the same for the last 4 billion years. b. The storing of organic carbon in sedimentary rocks allowed for oxygen to become a main component of Earth's atmosphere. c. The storing of organic carbon in sedimentary rocks kept carbon from becoming an integral part of Earth's atmosphere. d. The carbon cycle is not related to the composition of Earth's atmosphere. ANSWER: b 33. Over the last 400,000 years, how have atmospheric CO2 levels changed? a. Atmospheric CO2 levels have remained the same. b. Atmospheric CO2 levels have been steadily increasing overall. c. Atmospheric CO2 levels have been steadily decreasing overall. d. Atmospheric CO2 levels have fluctuated periodically. ANSWER: d 34. The movement of carbon from CO2 in the atmosphere to HCO3- (a byproduct of chemical weathering) on rocks to CaCO3 in coral skeletons and, finally, to CaCO3 in limestone are all steps in the _____ cycle. a. short-term carbon b. nitrogen c. sulfur d. long-term carbon ANSWER: d 35. Of the processes that release CO2 into the atmosphere, which are influenced by human technology? a. increased volcanic activity because of local earthquakes b. wildfires set by lightning strikes in remote areas of forests c. burning organic carbon that has been stored in sediments d. respiration rates dropping because of increased temperatures at the poles ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 36. Recall that over the last 400,000 years, atmospheric CO2 levels have fluctuated, with the lowest levels occurring during glacial periods and the highest levels occurring during interglacial periods. Why might atmospheric CO2 levels be lower during glacial periods? a. During glacial periods, there are more volcanic eruptions. b. During glacial periods, there are fewer photosynthesizing plants. c. During glacial periods, more carbon is stored in ocean reservoirs. d. During glacial periods, chemical weathering stops. ANSWER: c 37. The photo shown is of the white cliffs of Dover, which are made of chalk, a type of sedimentary rock formed from calcium carbonate shells of cocoliths. The cliffs, as well as the ocean and all of the life within the ocean, are carbon reservoirs. If the carbon in each of these reservoirs was instantly transformed into atmospheric CO2, which reservoir would contribute the most CO2 to the atmosphere?

a. sedimentary rocks b. living organisms c. oceans d. soil ANSWER: a 38. Which of the answer choices could provide an explanation for the rapid increase in oxygen levels roughly 2.5 billion years ago? a. Sedimentation rates were constant, photosynthetic rates did not change, and respiration rates decreased. b. Sedimentation rates increased, photosynthetic rates increased, and respiration rates were low. c. Sedimentation rates increased, photosynthetic rates increased, and respiration rates increased. d. Sedimentation rates increased, photosynthetic rates remained constant, and respiration rates were low. ANSWER: b 39. What is the difference between primary and secondary consumers? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 a. Primary consumers obtain carbon from carbon dioxide in air or dissolved in water, whereas secondary consumers gain carbon by feeding on other organisms. b. Primary consumers feed on small animals, whereas secondary consumers feed on larger animals. c. Primary consumers feed on primary producers, whereas secondary consumers feed on primary consumers. d. Primary consumers obtain carbon by ingesting organic molecules, whereas secondary consumers gain carbon by fixing CO2. ANSWER: c 40. _____ describes the rates at which carbon is transferred from one reservoir to another. a. Respiration b. Sink c. Flux d. Erosion ANSWER: c 41. In a trophic pyramid, the broader the base of primary producers, the more biomass that can be supported in upper levels. a. true b. false ANSWER: a 42. In a pond, tadpoles eat algae and fish eat the tadpoles. Around the pond, grasshoppers eat grass and, at night, are preyed upon by bats. Other bats eat the fish that eat the tadpoles. In this community, the fish-eating bats are: a. primary producers. b. primary consumers. c. secondary consumers. d. tertiary consumers. e. detritivores. ANSWER: d 43. Consider the graph shown. Which biological process accounts for the oscillating pattern observed in this dataset?

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a. photosynthesis b. respiration c. It is not possible to answer this question with the information provided. ANSWER: a 44. During what season would you expect oxygen levels to be lowest in a single year? a. summer b. fall c. winter d. spring ANSWER: d 45. Consider the figure showing atmospheric CO2 level oscillations over the course of the year.

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Which biological process accounts for maximum CO2 levels observed in spring (May) and for lowest CO2 levels observed in fall (September)? a. photosynthesis b. respiration c. More information would be needed to answer this question. ANSWER: a 46. Consider the figure showing atmospheric CO2 level oscillations over the course of the year.

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Chapter 46

In what season in the Northern Hemisphere would you expect oxygen levels to be highest? a. summer b. fall c. winter d. spring ANSWER: b 47. Oxygen (O2) makes up about 21% of Earth's atmosphere. The graph shown depicts changes in atmospheric O2 levels measured between the years 1990 and 2008 at Cape Grim Station in Australia. Note: Oxygen content is recorded as the ratio of oxygen to nitrogen gas (which is essentially invariant on short time scales) relative to a reference gas mixture.

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Chapter 46

The burning of fossil fuels produces CO2, removing O2 from the atmosphere. It has been estimated that approximately three molecules of O2 are consumed by every single molecule of CO2 produced by fossil fuel combustion. Over about 20 years, a total loss of 0.0317% of O2 from the atmosphere was calculated from these data.

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Chapter 46

If you compare the Cape Grim data to the Keeling curve shown, what can you conclude about how the trend in O2 levels is related to the trend in CO2 levels? a. Both O2 levels and CO2 levels increase over time. b. Both O2 levels and CO2 levels decrease over time. c. O2 levels decrease as CO2 levels increase. d. O2 levels increase as CO2 levels decrease. ANSWER: c 48. Oxygen (O2) makes up about 21% of Earth's atmosphere. The graph shown depicts changes in atmospheric O2 levels measured between the years 1990 and 2008 at Cape Grim Station in Australia. Note: Oxygen content is recorded as the ratio of oxygen to nitrogen gas (which is essentially invariant on short time scales) relative to a reference gas mixture. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46

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Chapter 46

The data show that decreasing O2 levels are _____ increasing CO2 levels in the atmosphere. a. caused by b. correlated with c. independent of ANSWER: b 49. The figure shows carbon sources and sinks.

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Chapter 46

Order the carbon reservoirs listed from smallest to largest. - deep ocean - organic matter in soil - surface water - sedimentary rock - atmosphere - terrestrial organisms a. atmosphere, terrestrial organisms, surface water, organic matter in soil, deep ocean, sedimentary rock b. atmosphere, terrestrial organisms, surface water, deep ocean, organic matter in soil, sedimentary rock c. atmosphere, organic matter in soil, surface water, sedimentary rock, terrestrial organisms, deep ocean d. atmosphere, surface water, terrestrial organisms, organic matter in soil, deep ocean, sedimentary rock ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 50. The figure shows carbon sources and sinks.

What is the difference in the rate of carbon exchange (in gigatons per year of carbon) between photosynthesis and respiration in the marine environment? a. 2 gigatons/year b. –2 gigatons/year c. 1 gigaton/year d. –1 gigaton/year ANSWER: a 51. The figure shows carbon sources and sinks.

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What is the difference in the rate of carbon exchange (in gigatons per year of carbon) between photosynthesis and respiration in the terrestrial environment? a. 2 gigatons/year b. –2 gigatons/year c. 1 gigaton/year d. –1 gigaton/year ANSWER: c 52. The figure shows carbon sources and sinks.

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At what rate does fossil fuel combustion contribute carbon to the atmosphere? a. 0.5 gigaton/year b. 1.0 gigaton/year c. 1.5 gigatons/year d. 5.0 gigatons/year e. 7.8 gigatons/year ANSWER: e 53. The figure shows carbon sources and sinks.

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Based on the calculations in the figure, burning fossil fuels results in _____ inputs of atmospheric carbon than the net exchange of atmospheric carbon from natural biological processes. a. smaller b. larger ANSWER: b 54. Examine the graph shown. It depicts the parallel history of atmospheric CO2 levels and surface temperature over the last 400,000 years, based on measurements of air bubbles trapped in glacial ice sheets in Antarctica. Glacial expansion correlates with _____ amounts of CO2 in the atmosphere and temperatures _____ than those measured in 1950. a. decreasing; lower Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 b. increasing; higher ANSWER: a 55. Examine the graph shown. It depicts the parallel history of atmospheric CO2 levels and surface temperature over the last 400,000 years, based on measurements of air bubbles trapped in glacial ice sheets in Antarctica. Consider the processes necessary for glacier formation. As glaciers advance, it is thought that the oceans circulate _____, causing deep ocean carbon reservoirs to _____. a. more slowly; increase b. more vigorously; decrease ANSWER: a 56. Consider the graph shown. Which point on the graph is most likely to coincide with the appearance of the first photosynthetic organisms? a. point A b. point B c. point C d. point D ANSWER: b 57. Consider the graph shown. Which point on the graph is most likely to coincide with increased formation of sedimentary rock? a. point A b. point B c. point C d. point D ANSWER: b 58. Consider the graph shown. Which point on the figure is most likely to coincide with the appearance of the first animals? a. point A b. point B c. point C d. point D ANSWER: d 59. Consider the food web shown, and answer the questions. If the tertiary consumer was removed, how would this affect the abundance of primary producers? a. They would increase. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 b. They would decrease. c. They would stay the same. ANSWER: a 60. Consider the food web shown, and answer the questions.

If the tertiary consumer was removed, how would this affect the abundance of primary consumers? a. They would increase. b. They would decrease. c. They would stay the same. ANSWER: b 61. Consider the food web shown, and answer the questions.

If the tertiary consumer was removed, how would this affect the abundance of secondary consumers? a. They would increase. b. They would decrease. c. They would stay the same. ANSWER: a 62. Think about the organization of a food web, and then classify the organism as a primary producer, a primary consumer, or a secondary consumer. More than one label may apply. Herbivore a. primary producer b. primary consumer c. secondary consumer ANSWER: b 63. Think about the organization of a food web, and then classify the organism as a primary producer, a primary consumer, or a secondary consumer. More than one label may apply. Tyrannosaurus rex a. primary producer b. primary consumer c. secondary consumer ANSWER: c 64. Think about the organization of a food web, and then classify the organism as a primary producer, a primary consumer, or a secondary consumer. More than one label may apply. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 Brown algae a. primary producer b. primary consumer c. secondary consumer ANSWER: a Multiple Response 65. Which of these processes increase(s) the amount of CO2 in the atmosphere? Select all that apply. a. photosynthesis b. volcanic eruptions c. subduction d. chemical weathering e. respiration ANSWER: b, e 66. Which of the answer choices are reservoirs of carbon? Select all that apply. a. soil b. oceans c. organisms d. the atmosphere e. burning wood ANSWER: a, b, c, d 67. Consider the representation of the biomass of a forest community. This diagram also correlates with the trophic structure of the community. The size of each rectangle represents the relative biomass at each trophic level. Based on the representation, what conclusion can be drawn about the transfer of energy between the organisms in each trophic level? Select all that apply.

a. The transfer of energy is less than 100% between trophic levels, but the rectangles representing biomass should be larger. For example, there should be more herbivorous insects, because the trophic Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 level that includes trees is large enough to support more herbivorous insects. b. The transfer of energy is incomplete between trophic levels, because insectivorous bats consume organisms from both trophic levels with insects. c. The transfer of energy is incomplete between trophic levels, because trees lose most of their energy in the form of heat, so the energy consumed is not available for higher trophic levels. d. The transfer of energy is incomplete between trophic levels, because at each trophic level, organisms consume some of the available energy in metabolic processes. The energy used for metabolism, therefore, cannot be available for the higher trophic levels. e. When the transfer of energy between levels looks like much more than 10%, it suggests that insects are especially efficient metabolizers of biomass. ANSWER: d, e 68. What can Antarctic ice core samples tell us about atmospheric CO2 levels? Select all that apply. a. Nothing can be learned about CO2 levels from Antarctic ice cores, because ice is made of H2O and not CO2. b. CO2 ice core data confirm CO2 measurements taken directly from the atmosphere. c. Ice cores can help scientists estimate atmospheric CO2 levels from hundreds of thousands of years ago. d. Ice cores can tell scientists exactly what plants lived on Earth centuries ago. ANSWER: b, c 69. There is evidence that increased levels of CO2 in the atmosphere may increase Earth's surface temperature. How might this affect the spread of mosquitoes that transmit the disease malaria? Select all that apply. a. Warmer temperatures may increase the habitat range of mosquitoes. b. Warmer temperatures may result in malaria cases in new regions. c. Warmer temperatures will have no effect on mosquitoes or the incidence of malaria. d. Warmer temperatures will kill off mosquitoes, thereby decreasing malaria cases. ANSWER: a, b 70. The process of cellular respiration consumes _____ and produces _____. Select all that apply. a. CO2; O2 b. O2; CO2 c. O2; H2O d. CO2; H2O ANSWER: b, c 71. An ecosystem includes: Select all that apply. a. the interacting organisms in the area. b. abiotic factors of the area. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 c. predators in the area. d. closely related organisms in a different area. ANSWER: a, b, c 72. If plants consume CO2 during photosynthesis, why hasn't all the atmospheric CO2 been used up? Select all that apply. a. Photosynthesis also produces CO2 as a product. b. Respiration produces CO2 as a product. c. Both photosynthesis and respiration produce CO2 as products. d. Plants do not actually use CO2 during photosynthesis; they use O2. e. Photosynthesis and respiration use one another's products as reactants. ANSWER: b, e 73. Why are carbon-based organic compounds, such as C6H12O6 (glucose), often referred to as energy molecules? Select all that apply. a. Energy is released when carbon-containing compounds are broken down. b. Energy is stored in carbon-containing organic compounds. c. Energy is required to build carbon-containing organic compounds. d. None of the other answer options is correct. ANSWER: a, b, c 74. Which of the processes are typically associated with the short-term carbon cycle? Select all that apply. a. respiration b. plate tectonics c. photosynthesis d. chemical weathering ANSWER: a, c 75. CO2 is added to the atmosphere by: Select all that apply. a. respiration. b. photosynthesis. c. deforestation. d. the weathering of rocks. ANSWER: a, c 76. When we think about the carbon cycle and human activities, it is important to differentiate between facts and hypotheses. Which of the answer choices can be considered a fact? Select all that apply. a. The amount of carbon dioxide in the atmosphere has increased since 1950. b. Increasing atmospheric carbon dioxide will cause mean global temperature to increase by 2 degrees Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 Celsius over the next century. c. The burning of fossil fuels contributes substantially to the ongoing rise of atmospheric CO2. d. In the past, atmospheric CO2 levels rose higher than those observed today. ANSWER: a, c, d 77. How can scientists determine what atmospheric CO2 levels were like hundreds or thousands of years ago? Select all that apply. a. They can examine ice core samples from Antarctica. b. They can examine CO2 levels in the outermost layer of Earth's atmosphere. c. They can examine the number of stomata in fossilized leaves. d. They can examine the number of fossilized dinosaur skeletons from different eras. ANSWER: a, c 78. Which of these things are made (in part) of carbon? Select all that apply. a. the shells of clams and other mollusks b. the glucose/sugars in fruits c. limestone at the bottom of the sea d. certain gases in the atmosphere e. water in the ocean ANSWER: a, b, c, d 79. In a food web, which of these organisms converts the carbon in animal carcasses or waste back to atmospheric CO2? Select all that apply. a. primary producers b. algae c. fungi d. decomposers ANSWER: c, d 80. Which of the answer choices is considered a carbon reservoir? Select all that apply. a. the Atlantic Ocean b. Redwood National Park c. the Great Barrier Reef d. all of the organisms on Earth e. water in the ocean ANSWER: a, b, c, d 81. Which of the processes are typically associated with the long-term carbon cycle? Select all that apply. a. respiration b. burial of carbon in sediments Copyright Macmillan Learning. Powered by Cognero.

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Chapter 46 c. photosynthesis d. volcanic eruptions ANSWER: b, d 82. Which of the processes listed naturally release CO2 into the atmosphere? Select all that apply. a. volcanism b. chemical weathering c. subduction d. biomineralization e. oxidation of fossil fuels (by bacteria) ANSWER: a, e 83. In a food web, which of the organisms are considered heterotrophs? Select all that apply. a. decomposers b. primary producers c. primary consumers d. secondary consumers ANSWER: a, c, d 84. Think about the organization of a food web, and then classify the organism as a primary producer, a primary consumer, or a secondary consumer. More than one label may apply. Venus flytrap a. primary producer b. primary consumer c. secondary consumer ANSWER: a, b

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Chapter 47 Multiple Choice 1. What physical properties of the Earth best explain why there are different seasons? a. One side of the Earth does not face the sun for half of each day. b. The Earth rotates at a slight angle on its axis. c. The elliptical orbit of the Earth causes seasons. d. Seasons are caused by the dominant primary producers in any given area. ANSWER: b 2. At the equator, warm air rises and absorbs atmospheric water. The water comes out of the air at about 30 degrees latitude. a. true b. false ANSWER: b 3. Theoretical predictions suggest that the North Pole should be cooler than it actually is. What helps to warm regions at high latitudes? a. The Earth's tilt on its axis means that for half the year the temperatures at the North Pole will increase. b. The heat-carrying capacity of water takes warm water toward the poles with ocean currents, and warms some northern regions. c. Warm air from the equator moves toward the poles before descending. d. There is a large amount of ice, and the ability of water held in the ice to resist temperature changes means it holds a lot of heat. ANSWER: c 4. Air cools as it moves from the base up the side of a mountain. What is one consequence of this movement of air? a. Moisture will drop as rain as the air moves up the side of the mountain. b. The air is deflected and circles back down to the base of the mountain. c. The air slows because cold air is less dense than warm air. d. The air picks up additional moisture that falls on the other side of the mountain. ANSWER: a 5. Consider the figure shown.

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Chapter 47

If the Earth spun on an axis that was perfectly perpendicular to the equator, how might this affect latitudinal differences in incoming solar radiation? a. There would be no latitudinal difference because the axis of the Earth would be perpendicular to the sun. b. There would be no latitudinal differences because all points on the Earth would be equidistant from the sun. c. There would still be latitudinal differences because latitude is not determined by the Earth's position on its axis. d. There would still be latitudinal differences because incoming solar radiation would be spread across a greater area at the poles than at the equator. ANSWER: d 6. Consider the figure shown. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 47

If the Earth spun on an axis that was perfectly perpendicular to the equator: a. there would be no seasonality in climate. b. seasonality would increase at high latitudes. c. seasonality would increase at low latitudes. d. summers would be warmer in the Northern Hemisphere. ANSWER: a 7. Most deserts occur at approximately 30° latitudes. Which of the statements best reflects this finding? a. Westerlies pull all moisture away from land and to the ocean at 30° latitude. b. Warm dry air descends at 30° latitude. c. Trade winds pull all moisture away from land and to the ocean at 30° latitude. d. Cool dry air descends at 30° latitude. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 47 ANSWER: d 8. Most terrestrial biomes are defined by their average annual temperature and precipitation. The range of values can be organized in such a way to produce the figure shown.

Current predictions of rainfall change in the West have indicated that rainfall may decrease by as much as 40 cm in some areas. Imagine a savannah location that experiences 100 cm of average annual rainfall and an average annual temperature of 20°C. If rainfall decreased by 40 cm per year, what would you expect to occur? a. The area within the curve for savannah would shift further to the right. b. The area within the curve for savannah would shift further up. c. The region would still be classified as savannah. d. The region would be classified as a desert. ANSWER: d 9. Most terrestrial biomes are defined by their average annual temperature and precipitation. The range of values can be organized in such a way to produce the figure shown.

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Chapter 47

Current predictions of global climate change indicate that temperatures may rise by as much as 5°C in some areas. Imagine a chaparral location that experiences 100 cm of average annual rainfall and an average annual temperature of 15°C. If there were an increase of 5°C, what would you expect? a. The area within the curve for chaparral would shift further to the left. b. The region would be classified as a Savanna. c. The region would still be classified as chaparral because rainfall did not change. d. The area within the curve for chaparral would shift further up. ANSWER: b 10. At a minimum, primary producers require water, carbon dioxide, and light. While these requirements are readily available in most ecosystems, we do not observe the same rates of net primary productivity in every region of the globe. Which of the statements best reflects one reason why rates of productivity are not the same everywhere? a. Rates of net primary productivity are actually the same everywhere; scientists just need to calculate the rates relative to plant biomass in a given ecosystem. b. Although water is limited in certain ecosystems, nutrients that are scarce actually limit net primary productivity because they are necessary for growth. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 47 c. Although water is limited in certain ecosystems, irrigation would solve the problem and equalize rates of net primary productivity in every ecosystem. d. Rates of net primary productivity are actually limited by the amount of solar radiation. If net primary productivity were calculated as a yearly average, then this would show that ecosystems have the same rate of productivity. ANSWER: b 11. Primary producers are present in every biome, including the deep sea. How can primary producers exist in the deep sea when sunlight does not penetrate to these depths? a. There are probably only two trophic levels because there is only light available from bioluminescent organisms. b. The higher trophic levels (secondary or tertiary consumers) are probably only represented by one or two species because of the low levels of energy available from primary producers. c. Primary producers in the deep sea use energy from chemical reactions to drive the reduction of CO2 to organic compounds. d. The biome is not stable because there is not a constant source of molecules available for primary producers to convert into biomass. ANSWER: c 12. Recent studies indicate that the addition of iron to areas of the ocean could increase productivity. Increased productivity could be beneficial for many reasons. What may be one of the largest negative impacts that could occur with "fertilizing the oceans with iron"? a. Increased productivity would provide more biomass for subsequent levels on the trophic pyramid. b. Increased productivity could result in higher respiration rates and regions of depleted oxygen in the ocean. c. Increased productivity would be matched by increased detritus falling to the ocean floor, burying many deep-sea organisms. d. Increased productivity could result in depletion of nitrogen available in ocean waters. ANSWER: b 13. A tundra ecosystem would include: a. all of the bryophytes inhabiting the tundra. b. all of the mammals inhabiting the tundra. c. the rainfall patterns the tundra experiences. d. the amount/duration of sunlight the tundra receives. e. All of these choices are correct. ANSWER: e 14. Is the nitrogen cycle different in marine biomes than in terrestrial biomes? a. Yes, because one is in water and one is on land. b. Yes, because the organisms living in each area are different. c. No, because nitrogen cycles between and within all biomes. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 47 d. No, because all nitrogen ultimately comes from the atmosphere. ANSWER: d 15. If rates of nitrogen fixation increased tenfold in aquatic ecosystems, would you expect a tenfold increase in primary productivity? a. yes, because nitrogen is a limiting nutrient in aquatic ecosystems b. yes, because there is an abundance of dissolved carbon dioxide in aquatic ecosystems c. no, because there are other limiting nutrients in aquatic ecosystems besides nitrogen d. no, because the increase in primary productivity would be greater than tenfold ANSWER: c 16. If rates of nitrogen fixation increased tenfold in terrestrial ecosystems, would you expect a tenfold increase in primary productivity? a. Yes, because nitrogen is a limiting nutrient in terrestrial ecosystems. b. Yes, because there is an abundance of carbon dioxide in terrestrial ecosystems. c. No, because there are other limiting nutrients in terrestrial ecosystems besides nitrogen. d. No, because the increase in primary productivity would be greater than tenfold. ANSWER: c 17. Some biologists believe that the latitudinal gradient in species diversity reflects the fact that high latitudes were recently covered by large glaciers (within the last 10,000-15,000 years). If there had been no recent ice age, would we still expect to see a latitudinal diversity gradient? a. No, without the ice ages, biomes at high latitudes would be older, with more time for the diversification of species. b. No, without polar ice (large ice formation around the pole) warm ocean currents such as the Gulf Stream would carry more heat to high latitudes, making them more productive and, thus, supporting more diversity than in the tropics. c. Yes, with or without recent glaciation of high latitudes, there is still less energy available at high latitudes to support diverse biomes. d. Yes, because nutrient availability is always lower at high latitudes. ANSWER: c 18. Refer to the figures shown.

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Chapter 47

The latitudinal diversity gradient refers to what pattern of species diversity? a. For many kinds of organisms, species diversity is greatest near the equator and lowest near the poles. b. For many kinds of organisms, species diversity is greatest near the poles and lowest near the equator.

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Chapter 47 19. Which of the statements is not a hypothesis that can adequately explain global patterns in species' diversity? a. Tropical biomes have existed for a longer time than other biomes, allowing more time for speciation to occur. b. The number of tree species limits the number of animal species in an area. c. Temperate species are fewer because it is difficult to adapt to local conditions. d. The tropics cover more land than temperate regions. ANSWER: b 20. In which biome would you expect to find the greatest diversity of insects? a. desert b. deciduous forest c. savanna d. tropical rain forest e. tundra ANSWER: d 21. Tropical biomes have evolved over _____ years. Biomes at higher latitudes have experienced an ice age within the past _____ years. a. a few million; few million b. a few million; thousand c. tens of millions of; few million d. several thousand; few million e. a few thousand; few thousand ANSWER: c 22. The hypothesis that species diversity is greater at lower latitudes than higher latitudes because low-latitude habitats are older is a hypothesis that would be supported if speciation rates were: a. the same at high and low latitudes. b. higher at low latitude than at high latitude. c. higher at high latitude than at low latitude. d. more variable at high latitude than at low latitude. e. more variable at low latitude than at high latitude. ANSWER: a 23. The growing season is the portion of the year when a plant can grow successfully. The growing season is determined by: a. mean annual temperature. b. mean annual precipitation. c. annual variation in nutrient availability. d. seasonal variations in temperature and precipitation. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 47 ANSWER: d 24. If you stand at the equator and throw a ball north (with all of your might), which direction will it deflect, relative to where you were standing when you threw the ball? a. The ball will deflect east. b. The ball will deflect west. c. The ball will move at the same speed as you and come right back to you. d. None of the answer options is correct. ANSWER: a 25. Which of the choices is one of the many contributing factors to high levels of rainfall at the equator? a. warm air holds less moisture than cool air b. high levels of solar radiation at the equator c. relatively constant temperatures ANSWER: b 26. If Earth had no oceans, would temperatures at the equator be hotter or colder than they are now? a. Temperatures would be hotter. b. Temperatures would be colder. c. Temperatures would remain the same. ANSWER: a 27. Consider the figure shown.

If the Earth spun on an axis that was perfectly perpendicular to the equator, how might this affect the light at the Arctic Circle? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 47 a. There would be fewer days where there was no solar radiation at the Arctic Circle. b. The land area where no solar radiation occurred would be reduced relative to its current area. c. Solar radiation would strike the Arctic Circle every day. ANSWER: c 28. If the Earth spun on an axis that was perfectly perpendicular to the equator, then the poles would receive the same amount of solar energy as the equator. a. true b. false ANSWER: b 29. Which of the choices contributes to high rainfall in the tropics (at the equator)? a. ability of warm air to hold more moisture than cool air b. density of air at different temperatures c. evapotranspiration rate d. All of these choices are correct. ANSWER: d 30. Estimates of temperature at any given location on Earth based solely on the intensity of solar radiation will be: a. very close to actual average annual temperature for that region. b. slightly overestimated at the poles because ice at the poles reflects heat. c. slightly underestimated at the equator because ocean currents carry heat from the equator toward higher latitudes. d. slightly overestimated at the equator and underestimated at the poles because heat moves from the equator to the poles. ANSWER: d 31. Why are savanna biomes found primarily at subtropical latitudes? a. Temperatures are warm, causing high rates of evaporation and transpiration. b. Rainfall is low because of descending air masses. c. Fires in dry grasslands inhibit the expansion of tree and shrub populations. d. All of these choices are correct. ANSWER: d 32. In western mountain ranges of North America, such as the Sierra Nevadas and the Rockies, where will you find a rain shadow? a. on the west facing slope b. on the east facing slope c. on the furthest northern tip of the mountain range d. on the furthest southern tip of the mountain range ANSWER: b Copyright Macmillan Learning. Powered by Cognero.

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Chapter 47 33. Most terrestrial biomes are defined by their average annual temperature and precipitation. The graph shown illustrates the range of values for each biome.

Based on your understanding of wind, air, and water movements, would you expect to find each of these biomes represented in both the Northern and Southern Hemispheres at roughly equal latitudes? a. no, because Earth rotates on an axis b. no, because land masses are not equally distributed on Earth c. yes, because there are circulation cells that form and move in an equal and opposite way on the Northern and Southern Hemispheres ANSWER: b 34. Why do some water masses in subsurface oceans have little or no oxygen? a. Rates of photosynthesis are low in overlying waters, limiting the availability of oxygen. b. Rates of photosynthesis are high in overlying waters, supporting high rates of respiration in waters that deplete the oxygen there. c. Waters underneath the surface oceans are cold and thus have a limited capacity to carry oxygen in solution. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 47 d. Large carnivores deplete oxygen in subsurface oceans via their high rates of oxygen metabolism. ANSWER: b 35. Consider the representation of the marine nitrogen cycle.

NH4+ (ammonium ion) plays an important role in the nitrogen cycle, and is excreted by many aquatic organisms. Why is the fixation of atmospheric nitrogen necessary if there is a readily available source of biologically usable nitrogen already present in aquatic biomes? a. Atmospheric nitrogen serves as a reservoir to be converted when other forms of nitrogen have been depleted from an area. b. Ammonium ions will eventually be oxidized to nitrogen gas (N2) that will be returned to the atmosphere. c. Nitrogen still must be fixed from the atmosphere so that global climate doesn't change. d. Nitrogen must be fixed so that terrestrial plants can utilize it for growth. ANSWER: b 36. Productivity in coastal ecosystems is higher than in other areas of the ocean. How would the removal of viruses from coastal waters affect this productivity? a. Productivity would increase because fewer organisms would be dying. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 47 b. Productivity would increase because there would be more space for organisms to grow. c. Productivity would decrease because the turnover of nutrients would be slower. d. Productivity would decrease because fewer inorganic molecules would be available. e. Productivity would remain the same. ANSWER: c 37. Patterns of primary productivity in marine environments are not delineated in the same way as in terrestrial environments. Why are there regions of high primary productivity at 60° N and 60° S in the oceans, when these are not regions of high productivity on land? a. The ocean at those latitudes is warmer, so more primary producers live there. b. There is little ocean movement at these latitudes, so primary producers can concentrate there. c. There is higher nutrient availability at these latitudes, supporting more primary producers. d. More sunlight is absorbed at these latitudes by the water than by the land. ANSWER: c 38. Consider the characteristics of terrestrial biomes. Would you expect more biomass in a trophic pyramid from the tropical rainforest compared to one from the taiga? a. Yes, because there is higher productivity at the equator than at the poles. b. Yes, because there is higher fungal diversity cycling carbon at the equator than at the poles. c. Yes, because there are higher numbers of predators at the equator than at the poles. d. No, because there are photosynthetic organisms in both biomes that support the trophic pyramids. e. No, because there are other producers at the poles that do not require sunlight to make up the difference in biomass that would be there if only sunlight were used as an energy source. ANSWER: a 39. Tropical rainforests have high species richness of trees. Which of the statements provides the most reasonable explanation why animal species richness is also high in tropical rainforests? a. Animals have had the same amount of time to evolve in rainforest areas as trees. b. Animal diversification rates are higher than those of trees. c. Animals have diversified into the vast number of niches provided by tree diversity. d. Animals are unable to adapt to climate variability in higher latitudes. ANSWER: c 40. Climate has been proposed to play a role in the latitudinal diversity gradient. Specifically, some have hypothesized that: a. the warm climate of the tropics promotes high speciation rates. b. the warm climate of the tropics promotes niche diversification. c. relatively few species are able to adapt to the harsher, more variable climates of the high latitudes. d. the harsh, more variable climates of the high latitudes reduce speciation rates. e. None of the other answer options is correct. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 47 41. High biomass of primary producers in surface waters is important: a. because it forms the basis for many aquatic food webs and supports primary consumers. b. to decrease global warming. c. because there are so many viruses that they need high numbers in order for the different species to survive. ANSWER: a 42. Refer to the figure shown.

If you were to draw one line through the map at 90 degrees W and one line at 90 degrees E, you would encounter the biomes in the same order as you move down the lines from 90 degrees N to 90 degrees S? a. true b. false ANSWER: a 43. Examine the figure shown.

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Chapter 47

If the surface of the Earth that was covered by oceans was approximately 20% smaller, then temperatures across the globe would be higher. a. true b. false ANSWER: a 44. Examine the figure shown.

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Chapter 47

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If phosphorous, in addition to iron, was a limiting nutrient in the ocean how would you expect the purple line in panel a from the figure to change? a. The purple line would follow a path closer to the red line for all points on the x-axis.

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Chapter 47 45. Examine the figure shown.

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Chapter 47

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If phosphorous, in addition to iron, was a limiting nutrient in the ocean how would you expect the purple line in panel c from the figure to change? a. The purple line would follow a path closer to the red line for all points on the x-axis.

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Chapter 47

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Chapter 48 Multiple Choice 1. Examine the figure illustrating human population growth.

What pattern of growth are we currently exhibiting? a. exponential growth b. logistic growth c. intrinsic growth d. geometric growth e. None of the other answer options is correct. ANSWER: a 2. In general, what is the relationship between a country's ecological footprint and its overall standard of living? a. It depends on the country—in developing countries, the ecological footprint and standard of living are positively correlated; in developed countries, they are negatively correlated. b. As the standard of living decreases, the ecological footprint increases. c. It depends on country—in developed countries, the ecological footprint and standard of living are positively correlated; in developing countries, they are negatively correlated. d. As the standard of living increases, so does the ecological footprint. e. None of the answer options is correct. ANSWER: d Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 3. A college student walks down New York City's Fifth Avenue at 8:00 AM. She is late for class, so she catches the bus for the last 2 miles of her commute. Did she have a larger ecological footprint when she was walking or when she was on the bus? a. on the bus, because it is using energy to move b. on the bus, because the road took more energy to build than the sidewalk c. walking, because she is using energy that must be replaced by eating foods that are not locally grown d. walking, because she is moving on a sidewalk that took energy and raw materials to build e. More information is needed about, for example, the number of people on the bus and its fuel source, to accurately determine her ecological footprint in both situations. ANSWER: e 4. The graph shown depicts observed temperature changes, natural plus anthropogenic (human) causes of temperature changes, and temperature changes due to natural causes. How would the lines in the graph be different if humans had no impact on global temperatures? a. The line for natural forces only would be the same as that for human and natural forces. b. All three lines on the graph would be identical. c. Observed data would match the zero line on the graph, and the natural forces and human and natural forces lines would fluctuate above the zero line. d. The line for natural forces would be above the zero line, and the line for natural and human forces would be below the zero line on the graph. ANSWER: a 5. A number of plant species have declined in abundance around Boston, Massachusetts, since Thoreau's time. These tend to be the plants that: a. have the earliest flowering time. b. have changed their flowering time the most. c. are least able to change their flowering time. d. have the latest flowering time. e. None of the answer options is correct. ANSWER: c 6. If you were studying the effects of climate change on the geographic ranges of a number of species living in the mountains, and your climate records showed that mean temperatures were increasing, which species would you be most concerned about from a conservation standpoint? a. a species that is a habitat generalist and lives at low elevations b. a species that is a habitat specialist and lives at the highest elevations c. a species that is a habitat specialist and lives at low elevations d. a species that is a habitat generalist and lives at middle elevations e. a species that is a habitat specialist and lives at middle elevations ANSWER: b 7. Examine the figure shown that exhibits data on predicted changes in global temperatures based on current Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 CO2 emission increases. Compare the projected change in temperature in central South America to that in central Africa. Which of the statements best describes the predicted changes? a. Central South America will warm by an average of 7-8 degrees; central Africa will warm by around 3-4 degrees. b. Central South America will warm by an average of more than 8 degrees; central Africa will warm by 1-2 degrees. c. Central South America will remain about the same; central Africa will warm by 7-8 degrees. d. Central South America will warm by 2-3 degrees; central Africa will not change. e. Both central South America and central Africa will warm by 4-5 degrees. ANSWER: a 8. As oceans acidify, carbonate ions in the ocean: a. become more abundant and more available to organisms that build calcium carbonate skeletons. b. become less abundant and less available to organisms that build calcium carbonate skeletons. c. become more abundant but less bioavailable to organisms that build calcium carbonate skeletons. d. become less abundant but more bioavailable to organisms that build calcium carbonate skeletons. e. remain as available to organisms that build calcium carbonate skeletons as they were before oceans acidified. ANSWER: b 9. Carbon dioxide combines with water to form a weak _____; this has caused the pH of seawater to go _____. a. antioxidant; up b. acid; up c. base; down d. base; up e. acid; down ANSWER: e 10. Geoengineering consists of a suite of widely debated proposals to mitigate global warming by actively managing Earth's radiation budget or greenhouse gas concentrations. Nobel laureate Paul Crutzen has proposed injecting sulfate aerosols into the stratosphere, which would reflect more incoming solar radiation away from the Earth, decreasing surface temperatures. Imagine we adopted this strategy for 100 years and then stopped. How would climate react at the end of the experiment? a. Cooler surface temperatures achieved during the experiment would be maintained over long time intervals. b. Cooler temperatures would continue, but only until CO2 regained its level in the atmosphere prior to the experiment. c. Temperature would quickly return to the increasing trajectory established before the experiment started. d. This question can't be answered without additional information. ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 11. Humans have a nearly ubiquitous influence on physical environments throughout the world. Looking back through geologic history, which type of organism can claim a comparable impact on physical environments? a. cyanobacteria b. trilobites c. dinosaurs d. birds ANSWER: a 12. Natural gas is a form of fossil fuel, but many scientists prefer we use natural gas rather than coal to meet societal needs. What advantage does natural gas have relative to coal? a. Burning natural gas does not add CO2 to the atmosphere. b. Burning natural gas releases less CO2 per kilowatt of energy obtained than burning coal does. c. Natural gas is available in unlimited supply. d. Development of natural gas fields via fracking has no potential consequences for the environment. ANSWER: b 13. Remember that, as a rule, temperatures decrease with an increase in altitude. If you were studying the effects of climate change on the geographic ranges of species living in the mountains, and your climate records showed that mean temperatures were increasing, you would predict that species ranges would show a shift toward: a. higher elevations. b. lower elevations. c. intermediate elevations. d. unpredictable elevations—the precise response would depend on individual species, with no clear trends or averages. e. None of the answer options is correct. ANSWER: a 14. You are given 100 acres of land and asked to use the land for long-term reduction of atmospheric CO2 levels. Which of the options is the best use of the land for this purpose? a. Plant a rapidly growing grass that can be harvested every year and used to make ethanol, which when added to gasoline reduces vehicle CO2 emissions. b. Plant slow-growing, bristle-cone pine trees that are known to live for thousands of years and reach a height of about 6 feet. c. Plant corn that is used to feed pigs, which, when raised, are harvested for sausage meat that is sold at the grocery store. d. Plant a regular lawn and use the land as a park to encourage people in your local community to exercise more. e. Plant trees that grow rapidly, reach a large size, and live for a long time. ANSWER: e 15. Which of the outcomes is a result of excess greenhouse gases added to the atmosphere by human activities? Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 a. an increase in Earth's surface temperature b. a decrease in the pH of Earth's oceans c. changes in Earth's precipitation patterns d. changes in the behavior of different plant/animal species e. All of these choices are correct. ANSWER: e 16. Which of the statements is true regarding the process of eutrophication? a. It is the result of the natural accumulation of nitrogen in bodies of water. b. It increases the species diversity and health of bodies of water. c. It can result in "dead zones" devoid of aquatic animal life. d. It increases the oxygen content of water by promoting bacterial growth. e. All of these choices are correct. ANSWER: c 17. Using records taken by Henry David Thoreau in the 1840s, scientists have documented what pattern of change in flowering plants around Boston, Massachusetts? a. As temperatures have become warmer, flowering is occurring earlier in many species. b. As temperatures have become warmer, flowering is occurring later in many species. c. As temperatures have become more variable, flowering is occurring earlier in many species. d. As temperatures have become more variable, flowering is occurring later in many species. e. As extreme weather events have become more common, flowering is occurring earlier in many species. ANSWER: a 18. Refer to the figure shown. According to the data in the figure, mean temperature changes from 1940-1980 and 1999-2008 have been greatest: a. in Australia. b. in southern Africa. c. near the equator. d. in South America. e. in the northern hemisphere. ANSWER: e 19. Giant sequoia trees (Sequoiadendron giganteum) are among the largest and longest-living organisms on Earth. These trees grow in weight by up to 1000 pounds of wood per year. As these trees are growing, they: a. incorporate CO2 into their biomass (wood, roots, and needles). b. are probably a net sink of CO2. c. generate some CO2 through cellular respiration. d. All of these choices are correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 ANSWER: d 20. Oceans are about _____% more acidic than they were in the 1960s. a. 10 b. 20 c. 25 d. 30 e. 40 ANSWER: d 21. About _____ of the CO2 produced by humans during the past century has been absorbed by the ocean. a. one-quarter b. one-third c. one-half d. one-eighth e. one-sixteenth ANSWER: b 22. Human impact on the nitrogen and phosphorus cycles happens primarily through our use of: a. fossil fuels. b. fertilizer. c. groundwater. d. marine resources. e. minerals. ANSWER: b 23. About how much of the nitrogen added to croplands ends up in food? a. 10% b. 25% c. 45% d. 60% e. 85% ANSWER: a 24. The source of nitrogen in nitrogen fertilizer manufacture is: a. nitrogen gas. b. ammonia. c. nitrate. d. nitrite. e. None of the answer options is correct. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 ANSWER: a 25. Nutrients from croplands are carried by the Mississippi River into the Gulf of Mexico. If nutrients are good for growth, why are scientists concerned about this nutrient influx? a. Higher productivity may support greater numbers of sharks. b. Higher productivity may support greater diversity of phytoplankton species. c. Higher productivity may result in increased respiration by bacteria feeding on sinking algal material, depleting oxygen within gulf waters. d. The nutrient runoff depletes nitrate, which cannot be made industrially. ANSWER: c 26. So-called "dead zones" in coastal oceans are associated with nutrient runoff and oxygen-depleted bottom waters. What causes the oxygen depletion? a. Nutrient runoff causes population growth of cyanobacteria and algae, which consume oxygen via aerobic respiration. b. Nutrient runoff causes population growth of cyanobacteria and algae which, in turn, support the growth of fish populations. Fish consume oxygen via aerobic respiration. c. Nutrient runoff causes population growth of algae, which consume oxygen via aerobic respiration. d. Nutrient runoff causes algae and cyanobacteria populations to grow. When they die, they are consumed by heterotrophic bacteria that consume oxygen via aerobic respiration. e. Nutrient runoff causes the waters to warm, which causes them to lose oxygen. ANSWER: d 27. An abundance of nitrogen (N) and phosphorus (P) in an aquatic environment leads to _____ levels of dissolved oxygen due to _____. a. high; the oxidation of water b. high; the abundance of anaerobic bacteria c. low; reduction of oxygen d. low; high aerobic respiration ANSWER: d 28. You track aquatic coastal levels of nitrogen and phosphorus at high latitudes and notice a sharp increase in the winter and then again in the spring. What do you expect the effect of these nutrient increases will be on phytoplankton populations? a. sharp spikes in phytoplankton population levels at essentially the same time (or, in phase) with the spikes in nutrients b. sharp spikes in phytoplankton population levels some time after (or, out of phase) with the spikes in nutrients c. a sharp spike in phytoplankton population levels in the spring, when sunlight levels are higher, but not much of a response in winter d. no effect of the changes in nutrient concentration on the phytoplankton populations in either winter or spring ANSWER: c Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 29. Dead zones have been documented in many areas. These dead zones are increasing in size each year. Which of the options would be one of the ways to decrease the size of these dead zones? a. Plant more crops so that more of the nitrogen in the fertilizers will be taken up. b. Decrease the volume of nitrogen used in agricultural areas near dead zones. c. Increase population sizes of cyanobacteria in coastal waters near dead zones. d. Increase populations of decomposers in coastal water areas near dead zones. ANSWER: b 30. Compared with nitrogen, do we need to be concerned that we may run out of phosphates? a. Yes. Nitrogen is obtained from nitrogen gas, which is in virtually infinite supply. Phosphate is mined from sedimentary rock, and deposits are limited. b. Yes. Nitrogen is obtained from ammonia, which is in virtually infinite supply. Phosphate is mined from sedimentary rock, and deposits are limited. c. No. Although phosphate is mined from sedimentary rock, deposits are extensive enough that we can consider them virtually unlimited, just like nitrogen gas. d. No. Although phosphate is mined from sedimentary rock, deposits are virtually unlimited, just like the ammonia from which we derive our nitrogen. e. No. Both phosphate and nitrogen are readily available as gas from the atmosphere. ANSWER: a 31. Invasive species have particularly devastating effects on islands because: a. islands are small and can support relatively few species to begin with. b. island species have typically evolved with few competitors and predators. c. islands include many amphibians, which are particularly susceptible to environmental disruption. d. speciation is slow on islands. e. None of the answer options is correct. ANSWER: b 32. What factors promote the recent spread of drug-resistant pathogens? a. Global warming has made pathogens more abundant. b. Biodiversity loss focuses more pathogens on fewer species. c. Air travel has introduced pathogens to areas far beyond their original distribution. d. Aggressive use of antibiotics has provided strong selective pressure favoring mutants able to tolerate drugs. ANSWER: d 33. Which of the occurrences have been the result (in part) of human activities? a. the introduction of invasive species into new environments b. the evolution of antibiotic-resistant bacteria c. the increase in the amount of greenhouse gases in the atmosphere d. eutrophication of certain bodies of water Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 e. All of these choices are correct. ANSWER: e 34. Many endangered species of birds in Hawaii are in danger of contracting avian malaria from mosquito bites. Many of these bird species live in the higher elevations of mountains in Hawaii, where it is too cold for the mosquitoes to exist. If global warming continues at the current estimated pace, what would you expect to happen to the distribution of these bird species? a. Their distribution will decrease to areas of even higher elevation. b. Their distribution will remain the same, but their population size will decrease. c. Their distribution will increase because mosquitoes will die in the warmer regions. ANSWER: a 35. Many species are introduced into new areas every day through ballast water and international shipping; however, not all of these species become established or invasive. Which of the characteristics would be the most likely in a species that would establish itself and become invasive? a. a K-strategist b. an r-strategist c. a primary producer d. a detritivore ANSWER: b 36. Wheat is a group of grass species. Cultivation of wheat originated in the eastern Mediterranean about 10,000 years ago. In comparison to closely related wild grasses, cultivated wheat has shorter stalks and fewer, larger seeds that are firmly attached to the stalk—the seeds don't easily blow away in the wind. The seed characteristics of cultivated wheat are useful for humans, but will they increase or decrease the fitness of cultivated strains of wheat in their natural environment? a. Fitness is increased because the dispersal of embryos is reduced. b. Fitness is increased because cultivated individuals produce more offspring. c. Fitness is decreased because more energy is devoted to trying to grow longer stalks. d. Fitness is decreased because the seeds are not dispersed. ANSWER: d 37. Wheat is a group of grass species. Cultivation of wheat originated in the eastern Mediterranean about 10,000 years ago. In comparison to closely related wild grasses, cultivated wheat has shorter stalks and fewer, larger seeds that are firmly attached to the stalk and are unable to disperse via wind like other grasses. Which of the statements provides an explanation as to why wheat is so successful despite the loss of seed dispersal? a. Wheat seeds are dropped below the parent plant and will grow the following season. This maintains the worldwide distribution of wheat. b. Wheat seeds are collected and dispersed by humans. The worldwide distribution of the seeds increases every time a new field of wheat is planted. c. Wheat seeds have evolved to be consumed by animals. Wheat seeds no longer rely on wind as their dispersal mechanism. d. Wheat seeds are not required for reproduction. Because wheat is a grass species, it has asexual Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 reproduction. ANSWER: b 38. Clearcutting for grazing and croplands has destroyed more than _____ of the preindustrial area of tropical rainforest, and if left unchecked, deforestation over the next century could eliminate _____ of all tree species in the Amazon forest. a. 25%; 30-40% b. 25%; 40-50% c. 30%; 75-80% d. 50%; 40-50% e. 50%; 75-80% ANSWER: d 39. _____ of all amphibian species are threatened with extinction, and more than _____% have undergone significant population declines in the last decade. a. One-quarter; 25 b. One-third; 25 c. One-half; 40 d. One-half; 25 e. One-third; 40 ANSWER: e 40. Amphibian population declines have been linked to many factors including habitat loss, pesticides, and fungal disease. a. true b. false ANSWER: a 41. One of the goals of conservation biology is to conserve biodiversity. An important part of conservation strategy is the design of reserves. Reserves are often designed to be as large as possible. How does the theory of island biogeography support biologists' efforts to utilize large reserves in order to conserve as much biodiversity as possible? a. Species area curves show that island size (that is, habitat patch) correlates to the number of species that can be supported within that area. b. Species area curves can be used to estimate species diversity to determine if several small reserves in a region will be better than one large reserve. c. Species area curves can be used to ensure that populations divided into metapopulations will have at least one large fragment. d. Species area curves can only be used to determine the number of species that may colonize an area and, in doing so, serve to help predict future species diversity in a habitat patch. ANSWER: a 42. In order for the red areas in the map shown to represent biodiversity hotspots, which of the criteria would Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 have to be true? a. There must be more bird species in the area than in another area of the same biome for that region to be considered a biodiversity hotspot. b. There must be at least 100 native bird species and 100 native plant species for the area to be a biodiversity hotspot. c. There must be at least 1500 native plant species and habitat reduction of the original area for the area to be designated a biodiversity hotspot. d. There must be high diversity in three different taxonomic groups (mammals, plants, birds, amphibians, and so forth) for the area to be a biodiversity hotspot. ANSWER: c 43. Which of these fish species is the best candidate for sustainable harvesting? a. herring (a small, fast-reproducing species) b. orange roughy (a long-lived, deep-sea fish) c. bluefin tuna (a large, predatory species) d. sturgeon (a large, long-lived, bottom-feeding species that migrates to reproduce) ANSWER: a 44. Zoos play a critical role in the conservation of biodiversity through maintaining: a. individuals of endangered species in different zoos. b. breeding populations of endangered species for potential release to the wild. c. populations of endangered species in zoos for people to see. d. breeding populations that can survive in the captive environment of zoos. ANSWER: b 45. The figure shown depicts arrival of migratory hummingbirds at their spring breeding grounds from 18801969 in the map on the left and from 1997-2010 in the map on the right. Which of the statements reflects changes over approximately the past 130 years in hummingbirds and their utilization of the spring breeding grounds? a. Hummingbirds are arriving at spring breeding grounds earlier in the past 20 years or so than 100 years ago. b. Hummingbirds have expanded their spring breeding grounds in the south, areas depicted in orange on the map. c. Hummingbirds are no longer breeding in the easternmost region on the map, area in dark green. d. In the past 20 years, hummingbirds are arriving at spring breeding grounds later than they were 100 years ago. ANSWER: a 46. A biological reserve that is designed for a single species is an important tool in maintaining biodiversity because a reserve: a. that is designed for a single species will also protect other species in the reserve. b. will not change over time so all the species will be conserved as long as the reserve exists. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 c. will generate income when people visit to see the biodiversity protected in the reserve. d. that is designed to protect a species will always be maintained as a reserve. ANSWER: a 47. While some may consider that the goal of conservation biology is to preserve all species on the planet, which of the statements is a more accurate statement of conservation biology's goals? a. Conservation biology is most concerned with preserving biodiversity hotspots so that the most species on the planet can be preserved. b. Conservation biology concentrates on preserving levels of species diversity in communities to maintain ecosystem services provided by the species in that community. c. Conservation biology only concentrates on regions where species diversity is declining faster than other areas on the planet. d. Conservation biology only responds to dire circumstances and does not serve to predict areas that may need conserving in the future. ANSWER: b 48. Biologists who study biodiversity describe new species and monitor species of which we are aware. Why is the study of biodiversity important in the Anthropocene? a. The study of biodiversity is important because it provides information about the number of species that go extinct every 1000 years. b. The study of biodiversity is important because it helps quantify the effects of human actions on changes in the traits of many populations (size, density, and distribution). c. The study of biodiversity is important because many politicians are biologists and they can influence government policy. d. The study of biodiversity is important because if we don't count all the species on the planet, we won't be able to identify more potential sources of food. ANSWER: b 49. Most human-dominated ecosystems have lower diversity than pristine ecosystems in a similar environment. What ecosystem services are diminished in human-dominated ecosystems? a. resilience of the ecosystem to disturbance. b. candidate organisms for new sources of food or medicine. c. levels of primary production. d. All of these choices are correct. ANSWER: d 50. Why do corridors increase the effectiveness of conservation reserves? Select all that apply. a. They increase the effective area for resident species, facilitating the conservation of species that require a large habitat area to survive. b. They provide access to a greater diversity of habitat types, providing greater possibilities for a good match between species and environment. c. They facilitate hunting because important games species traverse corridors, where they are more easily encountered. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 d. They mitigate the effects of invasive species. ANSWER: a 51. Which of the outcomes are suspected environmental impacts of fracking? Select all that apply. a. methane leakage into the atmosphere b. increased incidence of earthquakes c. water table contamination d. increasing levels of water in streams ANSWER: a 52. Indicate whether the statement is true or false with regard to the carbon cycle. Animals are always sinks for CO2. a. true b. false ANSWER: b 53. Indicate whether the statement is true or false with regard to the carbon cycle. Plants are always sinks for CO2. a. true b. false ANSWER: b 54. Indicate whether the statement is true or false with regard to the carbon cycle. Cellular respiration releases CO2 into the environment. a. true b. false ANSWER: a 55. Indicate whether the statement is true or false with regard to the carbon cycle. Photosynthesis removes CO2 from the environment. a. true b. false ANSWER: a 56. Indicate whether the statement is true or false with regard to the carbon cycle. Plants are permanent sinks for CO2. a. true b. false ANSWER: b 57. Indicate whether the statement is true or false with regard to the carbon cycle. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 Fossil fuels are stores of CO2. a. true b. false ANSWER: a 58. Indicate whether the statement is true or false with regard to the carbon cycle. The ocean is both a source and a sink of CO2. a. true b. false ANSWER: b 59. Indicate whether the statement is true or false with regard to the carbon cycle. Humans only contribute to the carbon cycle by the burning of fossil fuels. a. true b. false ANSWER: b 60. Methane is an important greenhouse gas, and levels of methane in the atmosphere are rising. Indicate whether the statement about why atmospheric methane is rising is true or false. Atmospheric methane is rising because methane is generated by archaeons living in the guts of cattle and beef production is increasing. a. true b. false ANSWER: a 61. Methane is an important greenhouse gas, and levels of methane in the atmosphere are rising. Indicate whether the statement about why atmospheric methane is rising is true or false. Atmospheric methane is rising because methane is generated by archaeons living in waterlogged rice paddies and rice production is increasing. a. true b. false ANSWER: a 62. Methane is an important greenhouse gas, and levels of methane in the atmosphere are rising. Indicate whether the statement about why atmospheric methane is rising is true or false. Atmospheric methane is rising because methane is released when permafrost at high latitudes melts and more permafrost is melting. a. true b. false ANSWER: a Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 63. Methane is an important greenhouse gas, and levels of methane in the atmosphere are rising. Indicate whether the statement about why atmospheric methane is rising is true or false. Atmospheric methane is rising because methane is generated by archaeons living in the guts of cattle and in rice paddies and both beef and rice production are increasing. a. true b. false ANSWER: a 64. Methane is an important greenhouse gas, and levels of methane in the atmosphere are rising. Indicate whether the statement about why atmospheric methane is rising is true or false. Atmospheric methane is rising because methane is released through leaks in gas wells and gas pipelines. a. true b. false ANSWER: a 65. Annual average temperature in the spring months has been steadily increasing in the eastern United States. Along with this change in temperature, larvae of many insect species have begun to emerge up to 2 weeks earlier over the past 2 decades. In the past, arrival of many songbird species has coincided with peak larval emergence and is correlated with reproductive success of bird species feeding on caterpillars. Over the past 2 decades, many songbird species have not shown a shift in arrival date. The graph shown indicates changes in caterpillar biomass peak and great tit, a species of songbird that relies on caterpillars for food, arrival times.

Which of the statements could reflect changes in reproductive success of great tits over the 20-year period from 1984–2004? a. Reproductive success of great tits increased over the 20-year interval. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 b. Reproductive success of great tits decreased over the 20-year interval. c. Reproductive success of great tits stayed the same over the 20-year interval. ANSWER: b 66. Annual average temperature in the spring months has been steadily increasing in the eastern United States. Along with this change in temperature, larvae of many insect species have begun to emerge up to 2 weeks earlier over the past 2 decades. In the past, arrival of many songbird species has coincided with peak larval emergence and is correlated with reproductive success of bird species feeding on caterpillars. Over the past 2 decades, many songbird species have not shown a shift in arrival date. The graph shown indicates changes in caterpillar biomass peak and great tit, a species of songbird that relies on caterpillars for food, arrival times.

Assume that researchers continue tracking great tit arrival times at the spring breeding grounds. If great tit populations were showing adaptations to alterations in peak caterpillar biomass, how would you expect the graph to change for great tits? a. Great tit arrival time at the spring breeding grounds should be earlier in the year. b. Great tit arrival time at the spring breeding grounds should be later in the year. c. Great tit arrival time at the spring breeding grounds should remain the same as in past years. ANSWER: a Multiple Response 67. The ecological footprint is an attempt to quantify the amount of land required to provide _____ for each person in a given country or region or in the world. Select all that apply. a. energy b. food c. materials d. services Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 e. None of the answer options is correct. ANSWER: a, b, c, d 68. Energy use by individuals varies among countries primarily because of differences in: Select all that apply. a. standard of living. b. the need to heat buildings in countries at high latitude. c. amount of travel. d. the type of food consumed. ANSWER: a, b, c 69. There is a giant sequoia Sequoiadendron giganteum named Washington (after George Washington) that was believed to be the second largest tree in the world (at 253.7 ft. in height!) until it was struck by lightning in 2003. The lightning strike caused a fire at the top of the tree that reduced Washington's height to about 229 feet. Carbon released from the tree due to that fire could have: Select all that apply. a. been incorporated into the wood of a nearby sequoia through the process of photosynthesis. b. been incorporated into the wood of a nearby sequoia through the process of cellular respiration. c. been incorporated into the cells of a human through the process of cellular respiration. d. traveled through the atmosphere and eventually dissolved into the ocean. ANSWER: a, d 70. Alternatives to fossil fuel–based energy—wind, sun, tidal energy, and nuclear power—offer what potential benefits? Select all that apply. a. They can reverse the effects of global climate change. b. They conserve finite resources. c. They help contain the cost of energy production. d. They directly conserve biodiversity. e. They reduce the amount of CO2 emitted per kilowatt hour of energy produced. ANSWER: b, c, e 71. Potential costs associated with increasing crop area include: Select all that apply. a. decreasing the biological storage capacity for carbon as forests are converted to cropland. b. loss of biodiversity as natural habitat is converted to cropland. c. the use of genetically modified organisms and the risks associated with them. d. increased food costs as land prices escalate. e. None of the answer options is correct. ANSWER: a, b 72. Potential costs associated with increasing crop yield are that it requires: Select all that apply. a. increasing the use of fertilizer. Copyright Macmillan Learning. Powered by Cognero.

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Chapter 48 b. increased food costs as land prices escalate. c. losing biological diversity as more natural habitat is converted to cropland. d. increasing the use of fossil fuels to power machinery. ANSWER: a, c, d 73. Humans have changed the selective landscape of Plasmodium, the infectious agent that causes malaria, by developing drugs that kill this parasite. Which drugs have been identified as resistant to strains of Plasmodium? Select all that apply. a. drugs in the quinine family b. atrazine c. artemesins d. All of these choices are correct. ANSWER: a, c 74. Many biologists consider amphibians to be the "canary in a coal mine" for the global environment. What features of their biology make amphibians especially sensitive to or indicative of environmental disruption? Select all that apply. a. They have narrow environmental tolerances. b. They exchange gasses through their skin. c. They are preferred prey species in both their aquatic and terrestrial life stages. d. They are found only in remote habitats, far from direct human disruption. ANSWER: a, b 75. Humans have changed the selective landscape of Plasmodium, the infectious agent that causes malaria, by developing drugs that kill this parasite. Plasmodium is resistant to many of these drugs. Which of the factors is a likely characteristic of Plasmodium that contributes to its ability to develop drug resistance? Select all that apply. a. Plasmodium has not changed in response to human efforts to eradicate it. b. The parasite has a short generation time and high mutation rates. c. The parasite populations are constantly evolving in anticipation of the next drug. d. It occurs in a region where medical treatments are not regularly monitored. ANSWER: b, d 76. Based on the figure shown, which of the statements is/are a direct reflection of the data presented? Select all that apply.

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Chapter 48

a. Increasing ivory prices always increases the amount of illegal killing of elephants. b. The natural mortality rate of elephants is higher than the illegal killing rate of elephants between 1998 and 2006. c. After 2008, increasing ivory prices are coincident with increases in illegal killing of elephants. d. In 2010, illegal killing rates of elephants surpassed natural mortality rates of elephants. ANSWER: c, d

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