Cell Molecular Approach 8th Edition by Cooper Test Bank by QuentinAxel

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 1: An Overview of Cells and Cell Research TEST FILE QUESTIONS Multiple Choice 1. RNA is believed to have been the original genetic system because it can a. form a stable double helix with a complementary nucleic acid strand. b. catalyze the polymerization of nucleotides into another RNA strand. c. form ribosomes. d. transfer amino acids to ribosomes. Answer: b Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain how the first cell originated. 2. The initial importance of membrane-enclosing, self-replicating RNA molecules and associated proteins was that they a. maintained these molecules as a unit capable of reproduction and evolution. b. provided sites for proteins to function. c. transported materials in and out of the compartment. d. kept other molecules out of the compartment. Answer: a Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain how the first cell originated. 3. Stanley Miller applied electric sparks to a reducing atmosphere of methane, ammonia, water, and hydrogen, and the resulting products showed that under these conditions, a. amino acids can form. b. nucleotides can form. c. amino acids can polymerize into polypeptides. d. nucleotides can polymerize into nucleic acids. Answer: a Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Describe the major steps in evolution of metabolism.

4. Which energy-producing process is thought to have come first during cellular evolution? a. Photosynthesis b. Glycolysis c. Oxidative phosphorylation d. Proteolysis Answer: b Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Describe the major steps in evolution of metabolism. 5. At the time life arose on Earth, Earth’s atmosphere contained abundant amounts of all of the following except a. H2. b. N2. c. O2. d. H2S. Answer: c Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Describe the major steps in evolution of metabolism. 6. Organisms that evolved the ability to use H2O as a donor of electrons and hydrogen for the photosynthetic conversion of CO2 to organic compounds radically changed Earth by producing a. sugar. b. cellulose. c. H2. d. O2. Answer: d Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Describe the major steps in evolution of metabolism. 7. The feature that most clearly distinguishes eukaryotes from prokaryotes is the presence of _______ in eukaryotic cells. a. ribosomes b. oxidative phosphorylation c. RNA molecules d. a nucleus Answer: d Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structures of eukaryotic and prokaryotic cells. 8. Cytoplasmic organelles are

a. absent in prokaryotic cells and present in eukaryotic cells. b. present in both prokaryotic and eukaryotic cells. c. present in prokaryotic cells and absent in eukaryotic cells. d. absent in both prokaryotic and eukaryotic cells. Answer: a Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Illustrate the structures of eukaryotic and prokaryotic cells. 9. The eukaryotic nucleus contains _______ DNA molecule(s). a. a single linear b. a single circular c. multiple linear d. multiple circular Answer: c Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structures of eukaryotic and prokaryotic cells. 10. The bacterial genome is located in a portion of the cell called the a. nucleus. b. nucleolus. c. mesosome. d. nucleoid. Answer: d Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structures of eukaryotic and prokaryotic cells. 11. Eukaryotic cell nuclei contain genes that are a. primarily of bacterial origin. b. primarily of archaebacterial origin. c. partly archaebacterial and partly bacterial in origin. d. all of eukaryotic origin. Answer: c Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Illustrate the structures of eukaryotic and prokaryotic cells. 12. Organelles such as mitochondria and chloroplasts are thought to have originated in eukaryotic cells via a process called a. phagocytosis. b. endosymbiosis. c. endocytosis. d. exocytosis. Answer: b

Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 13. Mitochondria and chloroplasts resemble bacteria in that they a. have their own DNA. b. have their own ribosomes. c. reproduce by simple division into two. d. All of the above Answer: d Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 14. Which of the following is a colonial organism closely related to the evolutionary precursors of present-day plants? a. Paramecium b. Dictyostelium discoideum c. Volvox d. Arabidopsis thaliana Answer: c Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 15. The human body is composed of more than _______ different types of cells. a. 100 b. 200 c. 400 d. 2,000 Answer: b Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 16. E. coli is a useful model system for molecular biology studies because a. it has a small genome. b. it reproduces rapidly. c. mutants can easily be isolated from culture dishes. d. All of the above Answer: d Textbook Reference: Experimental Models in Cell Biology

Bloom’s Category: 2. Understanding Learning Objective: Explain the advantages of E. coli for studying basic concepts of molecular biology. 17. A yeast cell divides in culture about every a. 20 minutes. b. 40 minutes. c. 2 hours. d. 12 hours. Answer: c Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Contrast yeast with E. coli as a model system. 18. How many genes does the haploid yeast nuclear genome contain? a. 1,000 b. 6,000 c. 10,000 d. 20,000 Answer: b Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Contrast yeast with E. coli as a model system. 19. The adult nematode worm Caenorhabditis elegans consists of _______ somatic cells. a. 95 b. 959 c. 1,500 d. 9,590 Answer: b Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the simple models for studying plant and animal development. 20. The simplicity and clarity of Caenorhabditis elegans allowed researchers to a. trace the developmental lineages of all cells in the adult. b. identify the genes involved in differentiation of each cell type. c. observe the process of fertilization better than had been possible in the past. d. follow the process by which cell aggregation forms a multicellular organism. Answer: a Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the simple models for studying plant and animal development.

21. Drosophila melanogaster was an organism of choice for the study of a. the process of fertilization. b. the process of cell aggregation to form a multicellular organism. c. the process of development. d. mammalian genetics. Answer: c Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the simple models for studying plant and animal development. 22. Studies on which model organism led to the initial discovery of important mechanisms controlling the development of the animal body plan? a. Mice b. Drosophila melanogaster c. Xenopus laevis d. Yeast Answer: b Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the simple models for studying plant and animal development. 23. Arabidopsis thaliana is a model organism for studying the molecular biology of a. plants. b. fungi. c. fruit flies. d. vertebrates. Answer: a Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the simple models for studying plant and animal development. 24. Which fish is proving to be a useful model organism for the study of vertebrate development? a. Zebrafish b. Goldfish c. Salmon d. Guppy Answer: a Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the simple models for studying plant and animal development.

25. Cultures grown from cells of a dissociated tissue are called a. primary cell cultures. b. transformed cell lines. c. normal cell lines. d. secondary cell cultures. Answer: a Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 2. Understanding Learning Objective: Summarize the principles of animal cell culture. 26. Most laboratories make use of cultured cells to study human disease. What makes these cultured cells such useful models? a. Cells only function when they are not organized into tissues. b. Cells cultured in the lab behave exactly the same as cells in the human body. c. Cells are very inexpensive to grow. d. They provide a continuous and uniform source of new cells. Answer: d Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 3. Applying Learning Objective: Summarize the principles of animal cell culture. 27. Viruses are useful model systems for studying a. tooth development. b. cell signaling. c. DNA replication. d. regulation of cell cycle control. Answer: c Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 2. Understanding Learning Objective: Explain how viruses can be used to study cell biology. 28. The light microscope was used to observe the cellular structure of cork by _______, who named the chambers “cells.” a. Schleiden and Schwann b. Hooke c. Virchow d. Leeuwenhoek Answer: b Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 1. Remembering Learning Objective: Summarize the uses and limitations of the light microscope. 29. The diffraction limited resolution of a standard light microscope is determined by which equation? a. R = 0.61λNA b. R = 0.61NA / λ

c. R = 0.61λ / NA d. R = 0.61 / λNA Answer: c Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 1. Remembering Learning Objective: Summarize the uses and limitations of the light microscope. 30. The diffraction limited resolving power of a standard light microscope is approximately a. 0.2 mm. b. 0.2 μm. c. 0.2 nm. d. 2 Å. Answer: b Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 1. Remembering Learning Objective: Summarize the uses and limitations of the light microscope. 31. Living cells are commonly visualized by means of _______ microscopy. a. phase-contrast b. bright-field c. fluorescence d. electron Answer: a Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 1. Remembering Learning Objective: Summarize the uses and limitations of the light microscope. 32. Fluorescent-labeled antibodies are used on cells primarily to locate a specific a. DNA sequence. b. RNA sequence. c. protein. d. carbohydrate. Answer: c Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 2. Understanding Learning Objective: Explain how fluorescence microscopy is used to visualize specific proteins. 33. A common use of green fluorescent protein (GFP) is to a. label antibodies. b. visualize proteins in living cells. c. photobleach other fluorescent proteins. d. label DNA sequences. Answer: b Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation

Bloom’s Category: 1. Remembering Learning Objective: Describe how GFP can be used to study proteins in living cells. 34. Which of the following experimental approaches would be useful for studying protein–protein interactions within a cell? a. Differential interference-contrast microscopy b. Differential centrifugation c. Fluorescence pulse field gel electrophoresis (FPFGE) d. Fluorescence resonance energy transfer (FRET) Answer: d Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 2. Understanding Learning Objective: Describe how GFP can be used to study proteins in living cells. 35. In fluorescence photobleaching experiments, recovery time is defined as the amount of time it takes the bleached area to regain its fluorescence. The recovery time for a cell with fluorescently labeled membrane proteins is a. longer than the recovery time for a cell with fluorescently labeled lipids because lipids diffuse faster than proteins. b. shorter than the recovery time for a cell with fluorescently labeled lipids because proteins diffuse faster than lipids. c. equal to the recovery time for a cell with fluorescently labeled lipids because proteins and lipids diffuse at the same rate. d. shorter than the recovery time for a cell with fluorescently labeled lipids because lipids diffuse faster than proteins. Answer: a Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 4. Analyzing Learning Objective: Describe how GFP can be used to study proteins in living cells. 36. Super-resolution light microscopes extend the resolution limits of light microscopy to a. 0.2 to 0.5 mm. b. 0.2 to 0.5 μm. c. 20 to 100 nm. d. 20 to 100 Å. Answer: c Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 1. Remembering Learning Objective: Explain super-resolution microscopy. 37. Electron microscopes have a resolution advantage over light microscopes because a. electron microscopes allow the viewer to examine living cells. b. electron microscope lenses have a larger numerical aperture. c. electron microscopes are easier to use. d. the wavelength of electrons is shorter than that of light. Answer: d

Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 2. Understanding Learning Objective: Compare electron microscopy and light microscopy. 38. Specific proteins can be localized in the transmission electron microscope by use of a. positive staining. b. negative staining. c. fluorescent-labeled antibodies. d. gold-labeled antibodies. Answer: d Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 2. Understanding Learning Objective: Compare electron microscopy and light microscopy. 39. The method by which cell components are separated by centrifugation at progressively higher speeds is called _______ centrifugation. a. differential b. velocity c. density-gradient d. equilibrium Answer: a Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 1. Remembering Learning Objective: Summarize procedures for isolation of subcellular organelles. 40. Which of the following cell components will be concentrated in the first pellet produced by differential centrifugation? a. Mitochondria b. Nuclei c. Endoplasmic reticulum d. The cytosol Answer: b Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 3. Applying Learning Objective: Summarize procedures for isolation of subcellular organelles.

Fill in the Blank 1. It is generally believed that Earth’s original cell enclosed a self-replicating _______ molecule. Answer: RNA Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Explain how the first cell originated.

2. RNA, unlike DNA, is capable of _______ a number of different chemical reactions. Answer: catalyzing Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Describe the major steps in evolution of metabolism. 3. The common source of electrons in the conversion of CO 2 into organic molecules in photosynthesis is _______. Answer: water Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Describe the major steps in evolution of metabolism. 4. The two major groups of prokaryotic cells are the _______ and the _______, which diverged early in evolution. Answer: archaebacteria; bacteria Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structures of eukaryotic and prokaryotic cells. 5. The DNA in a bacterium is located in a region termed the _______. Answer: nucleoid Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structures of eukaryotic and prokaryotic cells. 6. The cell lineage that eventually became the plants acquired _______before acquiring _______. Answer: mitochondria; chloroplasts Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 7. The amoeba _______ exists as a unicellular organism in abundant nutrient conditions and forms multicellular fruiting bodies in starvation conditions. Answer: Dictyostelium discoideum (slime mold) Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 8. _______ cells cover the surface of internal organs in the body, such as the intestine. Answer: Epithelial Textbook Reference: The Origin and Evolution of Cells

Bloom’s Category: 1. Remembering Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 9. Short generation time is one of the major factors that support using yeast or _______ for many fundamental molecular genetics experiments. Answer: E. coli Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Contrast yeast with E. coli as a model system. 10. Model systems serve as simplified examples in which the properties and development of cells can be examined easily. For mammals, a common model system is _______, and _______ is a common model for plants. Answer: mice; Arabidopsis thaliana Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the simple models for studying plant and animal development. 11. Model systems such as Caenorhabditis elegans, Drosophila melanogaster, and zebrafish are especially useful for studying problems in cell differentiation and developmental biology because, in contrast to single cells such as E. coli and yeast, they are _______ systems. Answer: multicellular (organismic) Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the simple models for studying plant and animal development. 12. _______ cells, the most commonly used cell line, were derived from a biopsy of a tumor carried by Henrietta Lacks. Answer: HeLa Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the principles of animal cell culture. 13. Viruses have proven to be useful tools with which to explore the properties of cells and complex processes such as cancer because viruses are dependent on _______ processes for much of their replication and metabolism. Answer: host cell Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Explain how viruses can be used to study cell biology. 14. _______ is a light microscope technique for studying the interaction of proteins.

Answer: Fluorescence resonance energy transfer (FRET) Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fraction Bloom’s Category: 1. Remembering Learning Objective: Explain how fluorescence microscopy is used to visualize specific proteins. 15. In the _______ microscope, imaging is restricted to a single focal plane by the use of an aperture pinhole to reject out of focus light; whereas in the _______ microscope, the same outcome is achieved through the use of light wavelengths such that excitation of the fluorescent dye requires the simultaneous absorption of two or more photons. Answer: confocal; multiphoton excitation Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fraction Bloom’s Category: 1. Remembering Learning Objective: Explain how fluorescence microscopy is used to visualize specific proteins. 16. In transmission electron microscopy, objects are imaged by their differences in _______ density. Answer: electron Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fraction Bloom’s Category: 1. Remembering Learning Objective: Compare electron microscopy and light microscopy.

True/False 1. Similarities in basic metabolic mechanisms indicate that all present-day cells on Earth descended from a single primordial ancestor. Answer: T Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain how the first cell originated. 2. Most of the genes of the bacterial symbionts that evolved into mitochondria are now found in the nucleus. Answer: T Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Explain how the first cell originated. 3. Eukaryotes contain a mixture of genes from eubacteria and archaebacteria. Answer: T

Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain how the first cell originated. 4. Photosynthesis first evolved in the green algae. Answer: F Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Describe the major steps in evolution of metabolism. 5. Compared to humans, E. coli has about a thousand times less DNA. Answer: T Textbook Reference: Cells as Experimental Models Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structures of eukaryotic and prokaryotic cells. 6. Higher plants and higher animals are similar in that they contain about the same number of different cell types. Answer: F Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 7. Epithelial cells form sheets that cover the surface of the animal body and line internal organs. Answer: T Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 8. In good culture conditions, E. coli typically divides every two hours. Answer: F Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the simple models for studying plant and animal development. 9. Yeasts are simple eukaryotes. Answer: T Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the simple models for studying plant and animal development.

10. Caenorhabditis elegans is an important organism for the study of plant development. Answer: F Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the simple models for studying plant and animal development. 11. Studies on Arabidopsis thaliana have indicated that the mechanisms that control development in plants are completely different from those that control development in animals. Answer: F Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the simple models for studying plant and animal development. 12. The frog Xenopus laevis is useful for studying early development because its eggs are small and clear. Answer: F Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the simple models for studying plant and animal development. 13. Xenopus laevis, an amphibian vertebrate, is useful for developmental studies because when its eggs are fertilized, the embryos develop outside its body. Answer: T Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the simple models for studying plant and animal development. 14. A single somatic animal cell in culture is capable of forming an entire animal. Answer: F Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the principles of animal cell culture. 15. Normal human fibroblasts can be grown only for 50–100 doublings in culture, after which they stop growing and die. Answer: T Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the principles of animal cell culture.

16. Immortal cell lines are cells that can continue to proliferate in culture for an indefinite number of generations. Answer: T Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the principles of animal cell culture. 17. A single immortalized fibroblast in culture is capable of forming an entire human. Answer: F Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the principles of animal cell culture. 18. With appropriate stimulation, mammalian stem cells in culture can give rise to individual mammalian cell types and hence may provide a source of patient-specific replacement cells. Answer: T Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 3. Applying Learning Objective: Summarize the principles of animal cell culture. 19. Viruses are complexes of nucleic acid and protein that can replicate on their own. Answer: F Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Explain how viruses can be used to study cell biology. 20. Viruses can have either RNA or DNA for their genetic material. Answer: T Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Explain how viruses can be used to study cell biology. 21. Synthesis of DNA from RNA was first observed using retroviruses. Answer: T Textbook Reference: Tools of Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Explain how viruses can be used to study cell biology. 22. Robert Hooke first coined the term “cell” following his observation, with a light microscope, of the pattern of cell walls in cork. Answer: T Textbook Reference: Tools of Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the uses and limitations of the light microscope.

23. Because the wavelength of electrons is 100,000 times shorter than the wavelength of visible light, the resolution of the electron microscope is 100,000 times better than the resolution of the light microscope. Answer: F Textbook Reference: Tools of Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Compare electron microscopy and light microscopy. 24. Scanning electron microscopy is usually used for a three-dimensional view of the surface of cells. Answer: T Textbook Reference: Tools of Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Compare electron microscopy and light microscopy. 25. Freeze fracture splits lipid bilayers and allows examination of the distribution of proteins that span the bilayer. Answer: T Textbook Reference: Tools of Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Compare electron microscopy and light microscopy. 26. All implementations of the light microscope are absolutely limited in their resolution by the diffraction properties of the illuminating light. Answer: F Textbook Reference: Tools of Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Compare electron microscopy and light microscopy. 27. In velocity centrifugation, a suspension is layered on top of a sucrose density gradient and centrifuged until the density of the particles matches the density of the sucrose solution. Answer: F Textbook Reference: Tools of Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize procedures for isolation of subcellular organelles.

Short Answer 1. List three significant differences that distinguish prokaryotic cells from eukaryotic cells. Answer: Any three of these four answers are correct: (1) Prokaryotes have no nucleus, whereas eukaryotes have a nucleus or nuclear envelope; (2) prokaryotes have small circular DNA molecules, whereas eukaryotes have large linear DNA molecules; (3)

prokaryotes have no organelles, whereas eukaryotes do; (4) prokaryotes have small cells compared to eukaryotes. Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Illustrate the structures of eukaryotic and prokaryotic cells. 2. The formation of a phospholipid bilayer membrane around a set of macromolecules was an important step in the origin and early evolution of life. What two advantages of such a system are thought to have allowed the first cells to grow and evolve? Answer: (1) The membrane forms a barrier between the interior of the cell and its environment, and (2) it allows a cell to grow and evolve as a unit, instead of existing as isolated molecules. Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 3. Which technique would be best for studying whether cells from different tissues of the body have different nutritional requirements? Answer: Use of primary animal cell culture (since established cell lines are already adapted to the medium in which they are grown) Textbook Reference: Tools of Cell Biology Bloom’s Category: 3. Applying Learning Objective: Summarize the simple models for studying plant and animal development. 4. Which approach provides a good model for studying the effect of a gene on the development of mammals? Answer: Use of genetically engineered mice in which a mutant gene has replaced the wild type gene Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 3. Applying Learning Objective: Summarize the simple models for studying plant and animal development. 5. Why is plant cell culture important to agriculture? Answer: New genes can be introduced into plant cells in culture, and then individual genetically engineered cells can be grown into entire plants with new characteristics. Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the simple models for studying plant and animal development. 6. Embryonic stem cells are important because they have the potential to do what? Answer: Differentiate into all the cell types present in an adult Textbook Reference: Experimental Models in Cell Biology

Bloom’s Category: 2. Understanding Learning Objective: Summarize the principles of animal cell culture. 7. Studies of certain viruses have led to the discovery that DNA can be synthesized from an RNA template. What are these viruses called? Answer: Retroviruses Textbook Reference: Tools of Cell Biology Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Explain how viruses can be used to study cell biology. 8. What technique would be best for initial examinations of the molecular arrangement of actin subunits in actin filaments? Answer: Negative staining and transmission electron microscopy Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 3. Applying Learning Objective: Compare electron microscopy and light microscopy. 9. What technique would reveal the morphological pattern of proteins that span the plasma membrane of red blood cells? Answer: Freeze fracture and transmission electron microscopy Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 3. Applying Learning Objective: Compare electron microscopy and light microscopy. 10. What technique would be best for observing the detailed structure of the surface of single-celled eukaryotes like Paramecium? Answer: Scanning electron microscopy Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 3. Applying Learning Objective: Compare electron microscopy and light microscopy. 11. What technique would be best for separating rough endoplasmic reticulum derived membrane vesicles from smooth endoplasmic reticulum membrane vesicles? Answer: Equilibrium centrifugation Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 3. Applying Learning Objective: Summarize procedures for isolation of subcellular organelles. 12. Arrange the following in the order they will sediment out as centrifugal speed increases from low to high during differential centrifugation. Lysosomes Microsomes Nuclei Ribosomes (free cytosolic) Answer: Nuclei, lysosomes, microsomes, ribosomes Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation

Bloom’s Category: 3. Applying Learning Objective: Summarize procedures for isolation of subcellular organelles.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. The original cell was thought to have arisen from the enclosure of self-replicating _______ by a phospholipid membrane. a. DNA b. carbohydrate c. protein d. RNA Answer: d Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Explain how the first cell originated. Feedback A: Incorrect. Although DNA is the normal genetic material, it does not have the catalytic properties to be self-replicating. Feedback B: Incorrect. Carbohydrate, whether a small sugar or a polysaccharide, does not have the information-encoding or catalytic properties to be self-replicating. Feedback C: Incorrect. Protein, although often catalytic, does not have the informationencoding properties to be self-replicating. Feedback D: Correct! RNA has the capacity to encode information and catalytic properties that DNA does not have. Therefore, RNA can be self-replicating. 2. The source of the atmospheric oxygen necessary for the development of oxidative metabolism is thought to have been a. glycolysis. b. the formation of Earth. c. the breakdown of ATP. d. photosynthesis. Answer: d Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Describe the major steps in evolution of metabolism. Feedback A: Incorrect. Glycolysis does not generate oxygen. Feedback B: Incorrect. Oxygen was very low in the atmosphere of the newly formed Earth. Feedback C: Incorrect. Energy is released by the breakdown of ATP, but it does not yield oxygen. Feedback D: Correct! Photosynthesis harnesses energy from the sun to generate glucose and oxygen; it is thought to have provided the oxygen necessary for the evolution of oxidative metabolism. 3. The original atmosphere of Earth is thought to have been rich in a. CO2, N2, O2, Rn, and Ne. b. CO2, N2, O2, Ar, and Ne. c. CO2, N2, H2, H2S, and CO.

d. CO2, N2, H2, Kr, and Xe. Answer: c Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Describe the major steps in evolution of metabolism. Feedback A: Incorrect. Oxygen (O2) was not at all abundant in Earth’s original atmosphere. It became abundant later because of photosynthesis. Feedback B: Incorrect. These are the five most abundant gases in the present atmosphere of Earth. Feedback C: Correct! These gases are thought to have been the most abundant gases in the original atmosphere of Earth. Feedback D: Incorrect. Kr and Xe are not thought to have been among the most abundant gases in the original Earth's atmosphere. 4. The feature that most clearly separates eukaryotes from prokaryotes is the presence of _______ in eukaryotic cells. a. ribosomes b. oxidative phosphorylation c. DNA molecules d. a nucleus Answer: d Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structures of eukaryotic and prokaryotic cells. Feedback A: Incorrect. Both eukaryotes and prokaryotes have ribosomes. Feedback B: Incorrect. Both eukaryotic and prokaryotic cells can carry out oxidative phosphorylation. Feedback C: Incorrect. Both eukaryotic and prokaryotic cells possess DNA. Feedback D: Correct! Prokaryotic cells lack a nucleus. 5. Chloroplasts are thought to have originated from endosymbiosis of _______ by a large host cell. a. anaerobic bacteria b. photosynthetic eubacteria such as cyanobacteria c. fungi such as yeast d. nonphotosynthetic aerobic bacteria Answer: b Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. Feedback A: Incorrect. Anaerobic bacteria do not carry out photosynthesis. Feedback B: Correct! Cyanobacteria can carry out photosynthesis and are photosynthetic eubacteria. Chloroplasts are thought to have originated from them via endosymbiosis. Feedback C: Incorrect. Fungi, including yeast, do not carry out photosynthesis. Feedback D: Incorrect. Mitochondria, not chloroplasts, are thought to have originated

from nonphotosynthetic aerobic bacteria. 6. The genome of eukaryotes consists of genes derived from a. archaebacteria alone. b. bacteria alone. c. both archaebacteria and bacteria. d. neither archaebacteria nor bacteria. Answer: c Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. Feedback A: Incorrect. The eukaryotic genome is a mosaic with respect to its origin. Feedback B: Incorrect. The eukaryotic genome is a mosaic with respect to its origin. Feedback C: Correct! Individual eukaryotic genes appear to have originated from either archaebacteria or bacteria, with basic cellular metabolism genes typically coming from eubacteria. Hence, the eukaryotic genome is a mosaic with respect to its origin. Feedback D: Incorrect. The eukaryotic genome is a mosaic with respect to its origin. 7. Which of these organisms is not a unicellular eukaryote? a. Saccharomyces cerevisiae b. Paramecium c. Methanococcus d. Chlamydomonas Answer: c Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. Feedback A: Incorrect. This yeast is a unicellular fungus—a eukaryote. Feedback B: Incorrect. This ciliated protozoan is a unicellular eukaryote. Feedback C: Correct! Archaebacteria are prokaryotes. Feedback D: Incorrect. This is a unicellular eukaryote; however, some green algae, such as Volvox, can form colonies of cells. 8. Solid tissues in animals include epithelial tissue, _______ tissue, nervous tissue, and muscle. a. blood b. connective c. ground d. epidermal Answer: b Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 1. Remembering Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms.

Feedback A: Incorrect. Blood is considered a tissue, but mature blood cells are typically circulating cells rather than solid tissue. Feedback B: Correct! Connective tissue underlies epithelial cell layers and is composed of fibroblasts as the major cell type. Feedback C: Incorrect. Ground tissue is a tissue type found in plants that is composed of parenchyma cells that carry out most of the metabolic reactions of the plant. Feedback D: Incorrect. Epidermal tissues are found in plants and cover the surface of the plant. 9. How many genes does an E. coli have? a. 4,200 b. 6,000 c. 14,000 d. 20,000–25,000 Answer: a Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Explain the advantages of E. coli for studying basic concepts of molecular biology. Feedback A: Correct! This is the number of genes present in the bacteria E. coli. Feedback B: Incorrect. This is the number of genes present in the yeast Saccharomyces cerevisiae. Feedback C: Incorrect. This is the number of genes present in Drosophila melanogaster. Feedback D: Incorrect. This is the number of genes present in humans. 10. An E. coli cell under well-defined laboratory conditions divides about every a. 20 minutes. b. 2 hours. c. 12 hours. d. 24 hours. Answer: a Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Explain the advantages of E. coli for studying basic concepts of molecular biology. Feedback A: Correct! On rich laboratory medium, a typical E. coli cell will divide every 20 minutes. Feedback B: Incorrect. This is the approximate doubling time for Saccharomyces cerevisiae cells grown on rich medium. Feedback C: Incorrect. E. coli cells double more quickly than this, unless they possess a mutation that slows their growth and division. Feedback D: Incorrect. This is more typical of a mammalian cell, such as a fibroblast, grown in culture. 11. Which of the following is the most commonly used mammal for genetic studies? a. Human

b. Xenopus laevis c. Mouse d. Cat Answer: c Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the simple models for studying plant and animal development. Feedback A: Incorrect. Aside from the ethical considerations prohibiting the use of humans, humans reproduce and develop slowly and are not easily manipulated genetically. Feedback B: Incorrect. This is a type of frog and is thus an amphibian, not a mammal. Feedback C: Correct! Although not as easily manipulated genetically as organisms such as C. elegans and D. melanogaster, mice are the most commonly used mammals for genetic studies. Feedback D: Incorrect. Cats are rarely used for genetic studies. 12. Embryonic stem cells are different from primary cell cultures or permanent cell lines derived from a tissue in that they are capable of a. generating many cell types. b. growing and dividing. c. synthesizing their own DNA. d. transporting nutrients across their plasma membrane. Answer: a Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the principles of animal cell culture. Feedback A: Correct! Only embryonic stem cells are capable of generating all of the cell types present in the adult organism. Feedback B: Incorrect. All cases cited are capable of growing and dividing. Feedback C: Incorrect. All cases cited are capable of synthesizing their own DNA. Feedback D: Incorrect. All cells must transport nutrients across their membranes. 13. In contrast to yeast or bacterial cells, animal cell cultures are grown on fairly complex media, with added amino acids, vitamins, and/or hormones. This is primarily because animal cells a. typically live in association with other cells. b. are less capable of transporting small molecules across their membranes. c. come from organisms that have specialized cell types. d. have too many genes to be able to keep track of the biosynthesis of needed metabolites. Answer: c Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the principles of animal cell culture. Feedback A: Incorrect. Animal cells typically are part of various solid tissues. However, bacteria, for example, can associate with and signal to one another.

Feedback B: Incorrect. Animal cells must have an extensive set of transporters in their membranes to be able to transport, for example, amino acids or vitamins into cells. Feedback C: Correct! Because of cell specialization and, as part of this, the existence of a circulatory system, animal cells can be specialized and derive nutrients from diet and vitamins and derive hormones from other cell types. Feedback D: Incorrect. All cells have strong abilities to integrate their metabolic networks. 14. Approximately how many doublings can normal human fibroblasts undergo in culture? a. 5 b. 50 to 100 c. They will go on proliferating indefinitely. d. They will not double at all. Answer: b Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the principles of animal cell culture. Feedback A: Incorrect. They can divide more than five times. Feedback B: Correct! After 50 to 100 doublings, the cells will stop growing and die. Feedback C: Incorrect. Immortalized cell lines, which are derived from tumors, will proliferate indefinitely, but normal cells will divide only a limited number of times. Feedback D: Incorrect. Normal human fibroblasts will continue to proliferate for a number of generations after being removed from an organism. 15. The Rous sarcoma virus (RSV) is a. a strain of virus that causes neurodegeneration in chickens. b. a virus which has been useful in the development of human gene therapy. c. the causative agent of the disease chicken pox. d. the first cancer causing animal virus to be identified. Answer: d Textbook Reference: Experimental Models in Cell Biology Bloom’s Category: 1. Remembering Learning Objective: Explain how viruses can be used to study cell biology. Feedback A: Incorrect. Rous sarcoma virus (RSV) causes cancer, specifically sarcomas, in chickens. Feedback B: Incorrect. The RSV genome and mechanism of action is not suitable as a gene delivery tool in human. Feedback C: Incorrect. Chicken pox is caused by the varicella zoster virus (VZV). Feedback D: Correct! In 1911, Peyton Rous found that sarcomas (cancers of connective tissues) in chickens could be transmitted by a virus, now known as Rous sarcoma virus, or RSV. 16. What is the theoretical diffraction limit of resolution of a light microscope used to look at a sample through oil? a. 0.22 μm

b. 0.2 nm c. 1 mm d. 0.305 μm Answer: a Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 1. Remembering Learning Objective: Summarize the uses and limitations of the light microscope. Feedback A: Correct! This would be the value if an oil-immersion lens were used. Resolution is calculated from the following formula: Resolution = 0.61λ/NA, where λ is the wavelength of light (approximately 0.5 μm) and NA (numerical aperture) represents the size of the cone of light that enters the lens, which in the case of oil has a maximal value of 1.4. Feedback B: Incorrect. This is the approximate resolving power of an electron microscope. Feedback C: Incorrect. Objects this far apart could be resolved by eye; a light microscope would not be necessary. Feedback D: Incorrect. This is the maximal resolution for a dry, non-oil-immersion objective. 17. In fluorescence microscopy, the immediate source of the light detected is light that has been _______ the sample. a. absorbed by b. emitted by c. exciting d. scattered from Answer: b Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 2. Understanding Learning Objective: Explain how fluorescence microscopy is used to visualize specific proteins. Feedback A: Incorrect. The absorbed light is detected only indirectly. The light emitted by the sample is the detected light. Feedback B: Correct! The light detected is emitted by the sample. In fluorescence, the excitation light is absorbed by the fluorescent group (electrons excited to a higher energy level), and light is then emitted as the electrons fall back to their basal energy level. Feedback C: Incorrect. It is the light emitted by the sample that is detected. Feedback D: Incorrect. There must be light emitted by the sample for fluorescence detection. 18. Which of the following is a high-resolution light microscope technique for detecting interactions between proteins? a. FRAP (fluorescence recovery after photobleaching) b. FRET (fluorescence resonance energy transfer) c. MPEM (multi-photon excitation microscopy) d. Confocal microscopy Answer: b

Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 2. Understanding Learning Objective: Explain how fluorescence microscopy is used to visualize specific proteins. Feedback A: Incorrect. FRAP is used to measure the diffusion rate of proteins. Feedback B: Correct! A FRET signal requires that individual fluorescent dyes carried by two separate proteins be close together in a complex. Energy transfer between fluorescent dyes is efficient only when the two fluorescent dyes are brought close together. Feedback C: Incorrect. MPEM is a technique for imaging a single plane of a sample. It is an alternative to confocal microscopy. Feedback D: Incorrect. TIRF is an optical technique for visualizing fluorescence in a zone close to the cell membrane. 19. Transmission electron microscopy is used to a. study the shapes of whole cells in single, thin sections. b. view living cells in three dimensions. c. view fluorescently labeled proteins in cells. d. observe subcellular organelles and macromolecules. Answer: d Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 1. Understanding Learning Objective: Compare electron microscopy and light microscopy. Feedback A: Incorrect. Many thin sections pass through an individual cell. Light microscopy and scanning electron microscopy, with their lower resolutions, are more commonly used to study whole cells. Feedback B: Incorrect. However, light microscopy can be used to obtain images of living cells in three dimensions. Cells must be fixed for electron microscopy. Feedback C: Incorrect. In electron microscopy, proteins are visualized by means of staining with salts and heavy metals, not by fluorescent labeling. Feedback D: Correct! With the use of salts and heavy metals to provide contrast, the electron microscope can be used to visualize subcellular structures in some detail. 20. What is the smallest number of ultracentrifugation steps necessary to separate nuclei from ribosomes in a cellular lysate? a. 1 b. 2 c. 4 d. 7 Answer: a Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 2. Understanding Learning Objective: Summarize procedures for isolation of subcellular organelles. Feedback A: Correct! A single low-speed ultracentrifugation step will sediment nuclei and whole cells and leave the ribosomes in the supernatant. Neither the nuclei nor the ribosomes would be pure, but they would be separated from each other. Feedback B: Incorrect. However, two ultracentrifugation steps can be used to separate

organelles such as mitochondria and chloroplasts from the plasma membrane. Feedback C: Incorrect. A separation between the two can be achieved with fewer steps. Feedback D: Incorrect. Only the soluble portion of the cytosol remains by a fourth highspeed ultracentrifugation step, so further centrifugation would not yield further purification of cellular substructures.

Essay 1. Why is RNA thought to have arisen on Earth prior to proteins and DNA? Answer: Unlike proteins, RNA molecules contain the information required for their own replication. Thus, base pairing could direct the synthesis of the complementary RNA strand, and the synthesis of the complement would in turn regenerate the original RNA molecule. In the early 1980s it was shown that, like proteins, some RNA molecules possess catalytic activity. In addition, RNA molecules capable of directing the synthesis of a new RNA strand from an RNA template have been described. Therefore, the unique abilities of RNA, both to store the information necessary for its own replication and to catalyze chemical reactions, make it a likely candidate as the basis for the evolution of the first living system. DNA does not have similar catalytic properties. Proteins possess catalytic activity but cannot replicate themselves. Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain how the first cell originated. 2. Why must the development of photosynthesis have preceded the development of oxidative phosphorylation? Answer: The major source of oxygen in Earth's atmosphere is from photosynthesis. Without atmospheric oxygen, oxidative phosphorylation is not possible. Therefore, photosynthesis must have preceded the development of oxidative phosphorylation. Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 2. Understanding Learning Objective: Describe the major steps in evolution of metabolism. 3. Mitochondria and chloroplasts are both thought to have become organelles within eukaryotic cells through endocytosis. Mitochondria are thought to have evolved from aerobic eubacteria, while chloroplasts can be readily related to cyanobacteria. What present-day traits of these organelles provide evidence of these origins? Answer: A free-living organism has its own DNA that can be transcribed and translated into proteins within the organism. Both mitochondria and chloroplasts contain their own DNA genome that is transcribed within the organelle into RNA and translated therein into proteins. The ribosomes and proteins made can be related by their size and sequence to present-day aerobic bacteria and cyanobacteria, respectively. Textbook Reference: The Origin and Evolution of Cells Bloom’s Category: 3. Applying Learning Objective: Outline the evolution of eukaryotic cells and multicellular

organisms. 4. How good an indicator is DNA content of the cellular complexity of an organism? Give examples to explain your answer. Answer: DNA content is a poor indicator of the cellular complexity of an organism. Chlamydomonas, a single-celled organism, has nearly the same haploid DNA content as the multicellular plant Arabidopsis thaliana, with about 120 million base pairs, and the fruit fly (180 million base pairs). In fact, the multicellular animal, Caenorhabditis elegans, at 97 million base pairs has fewer than Chlamydomonas does. In contrast, multicellular plants and animals have multiple cell types and tissues. Textbook Reference: Cells as Experimental Models Bloom’s Category: 3. Applying Learning Objective: Outline the evolution of eukaryotic cells and multicellular organisms. 5. Totipotency is a term that is often applied to cells in a plant callus tissue in culture and might by extension be applied to embryonic stem cells from animals. With appropriate manipulation of nutrients and growth regulatory molecules, an entire plant can be regenerated from a single cell within a callus. Taking this as an example of the developmental potential of a single plant cell, define the term “totipotency.” Answer: Totipotency is the ability of a somatic cell to give rise to a whole organism, with all its differentiated tissues. Differentiated animal cells do not exhibit totipotency, but many plant cells do. This allows them to repair themselves when damaged. For example, after a wound to the stem of a plant, cells in the vicinity of the wound will begin proliferating again, and some will differentiate into xylem cells, such that vascularization is maintained throughout the plant. For a number of plants, including petunias and carrots, a single cell in culture can be made to give rise to an entire plant if it is treated with the appropriate nutrients and growth factors. Similarly, cells derived from animal embryos—embryonic stem cells—can give rise to many, and perhaps all, of the cell types in the human body. Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 3. Applying Learning Objective: Summarize the principles of animal cell culture. 6. Why have viruses been so useful in the elucidation of cellular processes? Answer: Unlike bacteria, which can grow and divide in growth medium, viruses cannot proliferate outside of cells. They rely on many host cell processes to replicate and transmit themselves to other host cells, and for this reason they have been extremely useful to cell biologists. They divide rapidly and in many cases infection of the host cell has dramatic consequences that are easy to observe. For example, poliovirus, which subverts a cell’s translation machinery for its own purposes, has yielded information about the normal process of translation in the cell. Similarly, a number of viruses that carry oncogenes, which are mutant versions of normal cellular genes, have been used to elucidate mechanisms of cell growth and division. Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 2. Understanding

Learning Objective: Explain how viruses can be used to study cell biology. 7. In visualizing protein localization within a cell, what are the relative advantages and disadvantages of tagging proteins with green fluorescent protein (GFP) versus using a fluorescent antibody specific to the protein of interest (immunofluorescence)? Answer: The main advantage of using GFP-tagged proteins is that fixing, which kills cells and can cause artifactual results, is not required. This makes it possible to observe subcellular localization in live cells. The localization of GFP-tagged proteins can be observed in real time in response to various inducers by infusing them into the sample while on the microscope. One potential disadvantage of using GFP-tagged proteins is that the presence of the GFP tag (which is approximately 30 kilodaltons) can alter the structure of the protein and sometimes produces artifactual results. Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 5. Evaluating Learning Objective: Describe how GFP can be used to study proteins in living cells. 8. Imagine you are studying an inducible transcription factor called X. You make a cellular lysate and carry out a series of centrifugation steps at increasing speeds to localize X. You find that prior to induction, X is in the supernatant after a fourth, very high speed ultracentrifugation step; however, after induction it is found in the pellet after only one, relatively low speed ultracentrifugation step. What might this tell you about how X is regulated? What other techniques might you carry out to confirm your results? Answer: Protein X is probably regulated by nuclear localization: In uninduced cells it is found in the supernatant after a very high speed centrifugation step, indicating a cytosolic localization. And after induction it is pelleted after a single, relatively low speed ultracentrifugation step, indicating a nuclear localization. To test this, you could use fluorescence microscopy and an antibody or GFP-fuse X to observe it under both inducing and noninducing conditions. Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 4. Analyzing Learning Objective: Describe how GFP can be used to study proteins in living cells. 9. Diffraction is considered to be a fundamental principle that limits the resolution of a light microscope. Super-resolution light microscopes are indeed light microscopes. In achieving higher resolution than a conventional light microscope, does a super-resolution light microscope prove the diffraction limit to be false? If not, how does a superresolution light microscope achieve higher resolution? Answer: Super-resolution light microscopy neither breaks fundamental laws of optics nor proves these rules to be false. Rather, super-resolution methods build upon known principles to extend the observed resolution. For example, in the STORM (stochastic optical reconstruction microscopy) technique cited in the textbook, resolution is extended by “painting” the image point by point by mapping the center of probability for each diffracted point under conditions of low excitation incidence for each point. At high light intensity, the ability to resolve each point is still the diffraction limit. However, the calculated placement of each point in the STORM approach results, after many points are collected, in a higher resolution outcome—a super-resolution outcome.

Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 5. Evaluating Learning Objective: Explain super-resolution microscopy. 10. Microscopy, be it light or electron microscopy, is a major research tool in a molecular approach to the study of cells. What does this tell us about the level of resolution needed to solve many problems in cell biology? Answer: Microscopy provides a good indication of the scale at which cell biology attempts to answer questions about cell structure and function. These include questions at the subcellular level, cellular level, and tissue level. At the subcellular level, genes or protein machines must be understood. At the tissue to organism level, processes as complicated as nerve function and memory are a goal for our understanding. For example, much of protein localization work in cells can be done within the 0.22 μm limit of light microscopy. Textbook Reference: Tools of Cell Biology: Microscopy and Subcellular Fractionation Bloom’s Category: 4. Analyzing Learning Objective: Compare electron microscopy and light microscopy.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 2: Molecules and Membranes TEST FILE QUESTIONS Multiple Choice 1. The most abundant molecules in cells are a. proteins. b. carbohydrates. c. lipids. d. water. Answer: d Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Explain the properties of different types of chemical bonds. 2. Water is an ideal solvent in cells because it a. has low heat of vaporization. b. is a polar molecule that can form hydrogen bonds with itself and with other polar molecules. c. dissolves nonpolar molecules. d. contracts when it freezes. Answer: b Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain the properties of different types of chemical bonds. 3. In polysaccharides, sugars are linked together by means of _______ bonds. a. phosphodiester b. peptide c. glycosidic d. hydrophobic Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of a simple carbohydrate. 4. Sugars can cyclize if they contain _______ or more carbons. a. four

b. five c. six d. seven Answer: b Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of a simple carbohydrate. 5. A few sugars joined together are called a(n) a. glycoside. b. oligosaccharide. c. polysaccharide. d. starch. Answer: b Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of a simple carbohydrate. 6. The major bonds in glycogen are _______ glycosidic bonds. a. α(1→4) b. α(1→6) c. β(1→4) d. β(1→6) Answer: a Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of a simple carbohydrate. 7. Cell membranes are composed principally of a. carbohydrates. b. nucleic acids. c. phospholipids. d. proteins. Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 8. Lipids play a major role in cells as a. a form of energy storage. b. components of cell membranes. c. part of the cell signaling function. d. All of the above Answer: d Textbook Reference: The Molecules of Cells

Bloom’s Category: 2. Understanding Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 9. Fatty acids are stored in fat droplets in the form of a. triacylglycerols. b. phospholipids. c. cholesterol. d. glycolipids. Answer: a Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 10. Which of the following molecules stores the most chemical energy per unit of weight? a. Carbohydrates b. Lipids c. Proteins d. Nucleic acids Answer: b Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 11. Sphingomyelin contains two hydrocarbon chains linked to a. glycerol. b. choline. c. serine. d. glycine. Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 12. An amphipathic molecule is a. water-soluble. b. water-insoluble. c. part water-soluble and part water-insoluble. d. hydrophilic. Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering

Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 13. Steroid hormones are derivatives of a. cholesterol. b. phospholipids. c. amino acids. d. sugars. Answer: a Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 14. RNA in cells can serve as a a. regulator of gene expression. b. carrier of information from the nucleus to the cytoplasm. c. catalyst or enzyme. d. All of the above Answer: d Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Contrast the structures of RNA and DNA. 15. The pyrimidine bases in DNA are a. adenine and guanine. b. adenine and cytosine. c. cytosine and thymine. d. cytosine and guanine. Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Contrast the structures of RNA and DNA. 16. Complementary base pairs form _______ bonds to direct replication of DNA. a. hydrogen b. phosphodiester c. glycosidic d. covalent Answer: a Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Contrast the structures of RNA and DNA. 17. Which of the following is not a way in which DNA differs from RNA? a. DNA contains deoxyribose sugars.

b. DNA is usually a double-stranded molecule. c. DNA contains thymine as one of its bases. d. DNA can form hydrogen bonds with complementary sequences. Answer: d Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Contrast the structures of RNA and DNA. 18. Nucleotides function in cells as all of the following except a. building blocks of nucleic acids. b. carriers of chemical energy. c. intracellular signal molecules. d. defenders against infection. Answer: d Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Contrast the structures of RNA and DNA. 19. In DNA, a. G pairs with T and A pairs with C. b. G pairs with A and C pairs with T. c. G pairs with C and A pairs with T. d. G pairs with C and U pairs with A. Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Contrast the structures of RNA and DNA. 20. Polymerization of nucleotides to form nucleic acids involves the formation of _______ bonds. a. peptide b. phosphodiester c. glycosidic d. hydrogen Answer: b Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Contrast the structures of RNA and DNA. 21. The main difference between double-stranded DNA (dsDNA) and single-stranded DNA (ssDNA) is that a. dsDNA has bases that include thymine (T), while ssDNA has bases including uracil (U). b. dsDNA has strands oriented in an antiparallel fashion, while ssDNA does not. c. dsDNA cannot undergo translation while ssDNA can leave the nucleus to undergo translation.

d. dsDNA has 5’ to 3’ polarity while ssDNA does not. Answer: a Textbook Reference: The Molecules of Cells Bloom’s Category: 4. Analyzing Learning Objective: Contrast the structures of RNA and DNA. 22. A researcher is trying to determine the contents of a viral genome. Upon chemical analysis, the nucleic acid is found to contain 27% cytosine, 27% adenine, 23% uracil, and 23% guanine. Based on this data, the viral genome most likely consists of a. single-stranded DNA. b. double-stranded DNA. c. single-stranded RNA. d. double-stranded RNA. Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 3. Applying Learning Objective: Contrast the structures of RNA and DNA. 23. Serine, threonine, asparagine, and glutamine are all _______ amino acids. a. basic b. acidic c. polar d. nonpolar Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of the different groups of amino acids. 24. Which of the following is not a basic amino acid? a. Arginine b. Glutamine c. Histidine e. All of the above are basic amino acids. Answer: b Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of the different groups of amino acids. 25. Proteins are polymers of how many different amino acids? a. 16 b. 20 c. 24 d. 36 Answer: b Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding

Learning Objective: Summarize the properties of the different groups of amino acids. 26. In the primary structure of a protein, amino acids are joined together by _______ bonds. a. peptide b. phosphodiester c. glycosidic d. hydrophobic Answer: a Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of the different groups of amino acids. 27. The scientist who first determined the complete amino acid sequence of a protein (insulin) was a. Christian Anfinsen. b. Frederick Sanger. c. Linus Pauling. d. John Kendrew. Answer: b Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of the different groups of amino acids. 28. Anfinsen’s experiments on denatured ribonuclease showed that a. protein denaturation is irreversible. b. proteins can renature to regain their activity only with the assistance of specialized enzymes. c. proteins have unique amino acid sequences. d. the conformation of the folded protein is determined by its amino acid sequence. Answer: d Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of noncovalent bonds in protein folding. 29. Which of the following are involved in forming the tertiary structure of proteins? a. H bonds b. Hydrophobic interactions c. Ionic bonds d. All of the above Answer: d Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of noncovalent bonds in protein folding.

30. The interaction of two α and two β subunits to form a functional hemoglobin molecule is an example of _______ structure. a. primary b. secondary c. tertiary d. quaternary Answer: d Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of noncovalent bonds in protein folding. 31. Fully folded proteins typically have polar side chains on their surfaces, where electrostatic attractions and hydrogen bonds can form between the polar group on the amino acid and the polar molecules in the solvent. In contrast, some proteins have a polar side chain in their hydrophobic interior. Which of the following would not occur to help accommodate an internal, polar side chain? a. A hydrogen bond forms between two polar side chains. b. A hydrogen bond forms between a polar side chain and the protein backbone. c. A hydrogen bond forms between a polar side chain and an aromatic side chain. d. Hydrogen bonds form between polar side chains and a buried water molecule. Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 3. Applying Learning Objective: Explain the roles of noncovalent bonds in protein folding. 32. Like other catalysts, enzymes a. increase the rate of reactions without being consumed in reactions. b. shift the chemical equilibrium from more reactants to more products. c. do not alter the chemical equilibrium between reactants and products. d. Both a and c Answer: d Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Explain why enzymes affect the kinetics of chemical reactions without changing the equilibrium between reactants and products. 33. Enzymes act by a. lowering the overall free energy change of a reaction. b. decreasing the distance reactants must diffuse to find each other. c. increasing activation energy. d. decreasing activation energy. Answer: d Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Explain why enzymes affect the kinetics of chemical reactions without changing the equilibrium between reactants and products.

34. A reaction in which the substrate glucose binds to the enzyme hexokinase and the configuration of both molecules changes is an example of a. a lock-and-key mechanism. b. an induced fit mechanism. c. competitive inhibition. d. allosteric inhibition. Answer: b Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 3. Applying Learning Objective: Summarize the mechanisms of enzymatic catalysis. 35. Because of the central role that one amino acid plays in the mechanism by which proteins are cleaved by the enzymes trypsin and chymotrypsin, these enzymes are called _______ proteases. a. histidine b. lysine c. arginine d. serine Answer: d Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms of enzymatic catalysis. 36. The trypsin substrate binding site contains an aspartate residue, which is able to form an ionic bond with which amino acids in its polypeptide substrates? a. Lysine or arginine b. Glutamate or glutamine c. Leucine or phenylalanine d. Serine or threonine Answer: a Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 3. Applying Learning Objective: Summarize the mechanisms of enzymatic catalysis. 37. Coenzymes are a. enzymes in the same pathway. b. proteins that form dimeric enzymes. c. small molecules that work with an enzyme to enhance reaction rate. d. small molecules that allosterically regulate enzymes. Answer: c Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Distinguish between enzymes and coenzymes.

38. Which statement describes the most common relationship of the inhibitor molecule to the allosteric enzyme in feedback inhibition of enzyme activity? a. The inhibitor is the substrate of the enzyme. b. The inhibitor is the final product of the metabolic pathway. c. The inhibitor is the product of the enzyme-catalyzed reaction. d. The inhibitor is a metabolically unrelated signal molecule. Answer: b Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Explain why regulating the activity of enzymes is important to cell function. 39. In allosteric regulation, binding of a small regulatory molecule to an enzyme _______ enzyme activity. a. inhibits b. stimulates c. may stimulate or inhibit d. neither stimulates nor inhibits Answer: c Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Explain why regulating the activity of enzymes is important to cell function. 40. Proteins can be covalently modified by the addition of phosphate groups to all but which of the following amino acids? a. Arginine b. Threonine c. Serine d. All of the above Answer: a Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Explain why regulating the activity of enzymes is important to cell function. 41. The fundamental building block of cellular membranes is a. the glycolipid. b. the phospholipid. c. cholesterol. d. protein. Answer: b Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers.

42. Common factors affecting membrane fluidity include a. length of phospholipid fatty acid chains. b. temperature. c. number of double bonds in the fatty acid chains. d. All of the above Answer: d Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers. 43. Cholesterol affects membrane fluidity by a. increasing membrane fluidity at all temperatures. b. decreasing membrane fluidity at all temperatures. c. decreasing membrane fluidity at high temperatures and increasing membrane fluidity at low temperatures. d. increasing membrane fluidity at high temperatures and decreasing membrane fluidity at low temperatures. Answer: c Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers. 44. Phospholipids in a membrane commonly a. move laterally in the plane of the bilayer. b. rotate within the bilayer. c. move from one bilayer to the other. d. Both a and b Answer: d Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers. 45. The role of double bonds in the fatty acid tails of membrane phospholipids is to a. help stabilize the membrane. b. react with adjacent double bonds. c. increase membrane fluidity. d. interact with membrane proteins. Answer: c Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers.

46. The currently accepted model of membrane structure is called the _______ model. a. lipid bilayer b. unit membrane c. lipid raft d. fluid mosaic Answer: d Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers. 47. The fluid mosaic model of cell membranes was proposed by a. Frye and Edidin. b. Singer and Nicolson. c. Gorter and Grendel. d. Watson and Crick. Answer: b Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers. 48. The unique functions of different membranes are primarily due to their a. proteins. b. phospholipids. c. cholesterol molecules. d. glycolipids. Answer: a Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between integral and peripheral membrane proteins. 49. The mitochondrial inner membrane is about _______ protein. a. 0% b. 10% c. 42% d. 75% Answer: d Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between integral and peripheral membrane proteins. 50. Integral membrane proteins are those that

a. directly associate with membrane lipids. b. associate with the membrane indirectly. c. do not span the lipid bilayer. d. None of the above Answer: a Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between integral and peripheral membrane proteins. 51. Transmembrane proteins that span the membrane are a. peripheral membrane proteins. b. covalently linked to membrane lipids. c. integral membrane proteins. d. active transporters. Answer: c Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between integral and peripheral membrane proteins. 52. Transmembrane proteins can span the lipid bilayer as a. α helices. b. β turns. c. unstructured chains. d. Both a and c Answer: a Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between integral and peripheral membrane proteins. 53. Membrane proteins can be anchored to the cytosolic face of the plasma membrane by a. sugar groups of glycolipids. b. prenyl groups. c. sugar groups of glycoproteins. d. None of the above Answer: b Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between integral and peripheral membrane proteins. 54. Phospholipid bilayers are permeable only to molecules that are _______ and _______. a. large; uncharged

b. large; charged c. small; uncharged d. small; charged Answer: c Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Distinguish molecules that can diffuse through a lipid bilayer from those that require transporters to cross a membrane. 55. Passive transport molecules a. allow small molecules across membranes. b. are peripheral proteins. c. can transport against a concentration gradient. d. use the energy of ATP to transport molecules. Answer: a Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Distinguish molecules that can diffuse through a lipid bilayer from those that require transporters to cross a membrane. 56. Molecules that traverse a membrane against their concentration gradient do so by _______ transport. a. active b. passive c. carrier-mediated d. channel-mediated Answer: a Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Distinguish molecules that can diffuse through a lipid bilayer from those that require transporters to cross a membrane. 57. Channels form pores through which molecules of appropriate size and charge can cross a membrane. By contrast, carrier proteins a. actively transport molecules. b. selectively bind the molecule to be transported, change configuration, and release it on the other side. c. require ATP. d. transport a molecule against its concentration gradient. Answer: b Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Distinguish molecules that can diffuse through a lipid bilayer from those that require transporters to cross a membrane.

Fill in the Blank 1. Plants store glucose in the form of _______, and animals store glucose in the form of _______. Answer: starch; glycogen Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of a simple carbohydrate. 2. Because phospholipids consist of hydrophobic hydrocarbon chains and hydrophilic head groups, they are _______ molecules. Answer: amphipathic Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 3. Steroid hormones are derivatives of the membrane lipid _______. Answer: cholesterol Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 4. RNA contains uracil in place of the _______ found in DNA. Answer: thymine Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Contrast the structures of RNA and DNA. 5. Polymerization of DNA and RNA always occurs in the _______ direction. Answer: 5ʹ to 3ʹ Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Contrast the structures of RNA and DNA. 6. The purine base adenine is found in DNA and in the principal form of chemical energy in cells, _______. Answer: adenosine 5ʹ-triphosphate (ATP) Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Contrast the structures of RNA and DNA. 7. The side chains of basic amino acids such as lysine and arginine are _______ charged while those of acidic amino acids such as aspartic acid and glutamic acid are _______ charged at normal cellular pH.

Answer: positively (+); negatively (‒) Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of the different groups of amino acids. 8. In most proteins, α helices and β sheets fold into globular domains with _______ on the inside. Answer: hydrophobic amino acids Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Explain the roles of noncovalent bonds in protein folding. 9. The two polypeptide chains in insulin are held together by _______ bonds. Answer: disulfide Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of noncovalent bonds in protein folding. 10. Proteins associated with a membrane by protein-protein interactions are called _______. Answer: peripheral membrane proteins Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between integral and peripheral membrane proteins. 11. Under the mild conditions of temperature and pressure that are compatible with life, most biological reactions are so slow that they would not occur in the absence of _______. Answer: enzymes (enzymatic catalysis) Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Explain why enzymes affect the kinetics of chemical reactions without changing the equilibrium between reactants and products. 12. Enzymes increase reaction rates without either being _______ themselves or altering the _______ of the reaction. Answer: consumed (altered permanently, used up); equilibrium Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Explain why enzymes affect the kinetics of chemical reactions without changing the equilibrium between reactants and products. 13. Enzymes reduce the _______ energy required to reach the _______ state in a chemical reaction. Answer: activation; transition

Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Explain why enzymes affect the kinetics of chemical reactions without changing the equilibrium between reactants and products. 14. Coenzymes serve as _______ of several types of chemical groups. Answer: carriers Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Distinguish between enzymes and coenzymes. 15. In allosteric regulation of enzyme activity, the regulatory molecules bind to a site that is _______ from the active site of the protein, producing a _______ change in the protein that affects substrate binding to the active site. Answer: distal (distant, far); conformational (shape) Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Explain why regulating the activity of enzymes is important to cell function. 16. The formation of cell membranes is based on the capacity of _______ to form a _______. Answer: lipids (phospholipids, amphipathic lipids); bilayer (phospholipid bilayer) Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers. 17. Integral membrane proteins typically possess _______ segments of 20−25 amino acids that traverse the membrane and are rich in _______ amino acids. Answer: alpha-helical; hydrophobic Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between integral and peripheral membrane proteins. 18. Cell membranes are _______ permeable to small, uncharged molecules, while transport of larger polar or charged substances through membranes occurs via _______ or _______ proteins. Answer: selectively; channel (carrier); carrier (channel) Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Distinguish molecules that can diffuse through a lipid bilayer from those that require transporters to cross a membrane. True/False

1. Most proteins interact with other proteins through covalent protein–protein interactions. Answer: F Textbook Reference: The Molecules of Cells Bloom’s Category: 3. Applying Learning Objective: Explain the properties of different types of chemical bonds. 2. The function of glycogen, starch, and cellulose is to store glucose. Answer: F Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of a simple carbohydrate. 3. The glucose molecules in cellulose are joined by β(1→4) bonds. Answer: T Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of a simple carbohydrate. 4. Cholesterol is an amphipathic molecule. Answer: T Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 5. Each membrane phospholipid contains 3 fatty acid chains. Answer: F Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 6. Sphingomyelin is a serine-based phospholipid. Answer: T Textbook Reference: The Molecules of Cells. Bloom’s Category: 1. Remembering Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones 7. RNAs can be enzymes. Answer: T Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Contrast the structures of RNA and DNA.

8. Christian Anfinsen and colleagues showed in the 1950s that proteins contain all the information to spontaneous fold into an active enzyme. Yet sixty years later, we can still not predict the three-dimension shape of a protein from its amino acid sequence. Answer: T Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of noncovalent bonds in protein folding. 9. Enzymes accelerate reactions and affect the end equilibrium concentration of reactants and products. Answer: F Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 4. Analyzing Learning Objective: Explain why enzymes affect the kinetics of chemical reactions without changing the equilibrium between reactants and products. 10. Some enzymes participate directly in the chemical reactions they catalyze. Answer: T Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms of enzymatic catalysis. 11. Coenzymes function in conjunction with enzymes to carry chemical groups between substrates. Answer: T Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Distinguish between enzymes and coenzymes. 12. Approximately 50% of the chemical reactions in cells are catalyzed by enzymes. Answer: F Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Explain why regulating the activity of enzymes is important to cell function. 13. The activity of enzymes can be controlled by the binding of small molecules, by interaction with other proteins, and by covalent modifications. Answer: T Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Explain why regulating the activity of enzymes is important to cell function.

14. One conclusion derived from the Singer and Nicolson fluid mosaic model of membrane structure is that integral membrane proteins are freely soluble in water. Answer: F Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between integral and peripheral membrane proteins. 15. The sugar groups of glycolipids and glycoproteins are found on the outer surface of the plasma membrane. Answer: T Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between integral and peripheral membrane proteins.

Short Answer 1. How is the information in DNA or RNA conveyed? Answer: By the order of the bases in the polynucleotide chain Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Contrast the structures of RNA and DNA. 2. Provide the name or three-letter code of one amino acid to fit each of the following descriptions: a) It is normally positively charged at pH 7.0. b) Its R group can form ionic bonds with lysine in a protein. c) It has a hydrophobic R group. d) It commonly forms covalent bonds between its R group and another identical R group. Answer: a) Arginine (Arg), histidine (His), or lysine (Lys); b) Aspartic acid (Asp) or glutamic acid (Glu); c) Alanine (Ala), valine (Val), isoleucine (Ile), leucine (Leu), methionine (Met), phenylalanine (Phe), tyrosine (Tyr), or tryptophan (Trp); d) Cysteine (Cys) Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of the different groups of amino acids. 3. Identify the amino acids below as positively (+) charged, negatively (–) charged, able to form disulfide bonds (SS), and/or having OH groups that can be phosphorylated (P) by protein kinases at pH 7.0. Place the appropriate symbols (+, –, SS, or P) in the blanks beside the three-letter codes. (Hint: Some amino acids have none of these properties.) Ala ______ Leu ______ Arg ______ Lys ______ Asn ______ Met ______

Asp ______ Cys ______ Glu ______ Gln ______ Gly ______ His ______ Ile ______ Answer: Ala ______ Arg + Asn ______ Asp – Cys SS Glu – Gln ______ Gly ______ His + Ile ______

Phe Pro Ser Thr Trp Tyr Val

______ ______ ______ ______ ______ ______ ______

Leu Lys Met Phe Pro Ser Thr Trp Tyr Val

______ + ______ ______ ______ P P ______ P ______

Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of the different groups of amino acids. 4. What kind of bonds hold together the two chains of insulin? Answer: Disulfide bonds Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of the different groups of amino acids. 5. What is the name of the technique commonly used to determine the three-dimensional structure of proteins? Answer: X-ray crystallography Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Explain the roles of noncovalent bonds in protein folding. 6. How is it possible for serine proteases, which all have different substrates, to share a similar chemical mechanism? Answer: The enzymes of the serine protease family each have a differently sized and shaped substrate binding pocket near their active site that accepts a specific type of amino acid. Their active sites all hydrolyze peptide bonds using the amino acid that fits in their active site. Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms of enzymatic catalysis.

7. How does NAD+ act as a chemical carrier? Answer: NAD+ can accept a hydrogen ion and two electrons from one substrate, forming NADH, which can then donate these electrons to a second substrate, re-forming NAD+ Thus, NAD+ transfers electrons from the first substrate (which becomes oxidized) to the second substrate (which becomes reduced). Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Distinguish between enzymes and coenzymes. 8. What is the relationship between feedback inhibition and allosteric inhibition of enzyme activity? Answer: Feedback inhibition is one example of allosteric inhibition. In the example given in the text, isoleucine, the end product of a metabolic pathway, inhibits threonine deaminase, the first enzyme in the pathway. Isoleucine is different in size and shape from the other substrate of threonine deaminase. Rather than binding to the active site as a competitor, it binds to a distal site as an allosteric regulator. Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Explain why regulating the activity of enzymes is important to cell function. 9. State the two main ways that cells regulate the activity of their enzymes and provide one example of each. Answer: (1) Allosteric or feedback inhibition or activation of enzyme activity (e.g., amino acids such as isoleucine feedback and inhibit the first enzyme in their pathway of synthesis); (2) Phosphorylation or other covalent modification of an enzyme (e.g., phosphorylation of glycogen phosphorylase takes place); or proteolytic activation, a variation on covalent modification (e.g., a pro-enzyme or zymogen like chymotrypsin or trypsin is activated by proteolysis). Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 3. Applying Learning Objective: Explain why regulating the activity of enzymes is important to cell function. 10. On which face of the plasma membrane are glycolipids and glycoproteins found? Answer: The extracellular face (or outside face) Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between integral and peripheral membrane proteins.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. The most abundant molecule in cells is a. aspartic acid. b. DNA. c. sucrose. d. water. Answer: d Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Explain the properties of different types of chemical bonds. Feedback A: Incorrect. Aspartic acid is a very important amino acid, but is a very small portion of the total molecules in cells. Feedback B: Incorrect. DNA is the essential genetic material in cells, but it is only a small portion of cellular mass and an even smaller fraction of the total molecules in cells. Feedback C: Incorrect. Sucrose is stored in some plants and can sometimes be abundant. However, sucrose is not a normal cell constituent in animals. Feedback D: Correct! Water is the most abundant molecule in cells, accounting for 70% of cellular mass, and considering its small size, an even higher fraction of the total molecules in cells. 2. Which of the following is not one of the four major classes of organic molecules in the cell? a. Carbohydrates b. Lipids c. Water d. Proteins Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Explain the properties of different types of chemical bonds. Feedback A: Incorrect. Carbohydrates, including simple sugars and polysaccharides, are one of the major classes. Feedback B: Incorrect. Lipids are important organic components of the cell, not only for their structural role in membranes, but also as signaling molecules and in energy storage. Feedback C: Correct! Although water is a major cell constituent, accounting for 70% of the cell's mass, it is not an organic molecule. Nucleic acids are the fourth major class of organic molecules in the cell. Feedback D: Incorrect. Proteins are critical components of cells, carrying out a variety of functions. 3. What is the major carbohydrate-storage molecule in plants? a. Starch b. Cellulose

c. Glycogen d. Deoxyribonucleic acid Answer: a Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of a simple carbohydrate. Feedback A: Correct! Starch is a polysaccharide composed of glucose residues in the  configuration. The principal linkage is between carbon 1 of one glucose and carbon 4 of a second glucose. Feedback B: Incorrect. Cellulose is a plant polysaccharide, but it has a structural role rather than a role in energy storage. Feedback C: Incorrect. Glycogen is the major carbohydrate-storage molecule in animals. Feedback D: Incorrect. The structure of deoxyribonucleic acid (DNA) includes sugars, but it is a nucleic acid, not a carbohydrate, and its role is in the storage of genetic information. 4. Molecules that are partly water-soluble and partly water-insoluble are a. hydrophilic. b. amphipathic. c. hydrophobic. d. allosteric. Answer: b Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. Feedback A: Incorrect. Hydrophilic molecules are either charged or polar and are watersoluble. Feedback B: Correct! An example of an amphipathic molecule is a phospholipid, which has a long hydrocarbon chain (the water-insoluble part) attached to a polar head group (the water-soluble part). Feedback C: Incorrect. Hydrophobic molecules, such as triacylglycerols, do not mix with water and form insoluble droplets in aqueous medium. Feedback D: Incorrect. This is a type of regulation used to control a protein's activity. 5. Cholesterol, a membrane lipid in animals, has a chemical structure similar to a. estradiol. b. phosphatidylinositol. c. thymine. d. triacylglycerol. Answer: a Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. Feedback A: Correct! Cholesterol is a steroid and a precursor to steroid hormones such as

estradiol. Feedback B: Incorrect. Phosphatidylinositol is a glycerol phospholipid. Feedback C: Incorrect. Thymine is a pyrimidine base found in DNA molecules. Feedback D: Incorrect. Triacylglycerol is a fat—three fatty acids linked to a glycerol molecule. 6. Which of the following bases is not found in DNA? a. Adenine b. Cytosine c. Thymine d. Uracil Answer: d Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Contrast the structures of RNA and DNA. Feedback A: Incorrect. Adenine is a purine base found both in DNA and RNA. Feedback B: Incorrect. Cytosine is a pyrimidine base found both in DNA and RNA. Feedback C: Incorrect. Thymine is a pyrimidine base found in DNA, but not in RNA. Feedback D: Correct! Uracil is a pyrimidine base found in RNA, but not in DNA. 7. How many amino acids are commonly incorporated into proteins? a. 5 b. 10 c. 20 d. 25 Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of the different groups of amino acids. Feedback A: Incorrect. Twenty amino acids are commonly found in proteins. Feedback B: Incorrect. Twenty amino acids are commonly found in proteins. Feedback C: Correct! Feedback D: Incorrect. Twenty amino acids are commonly found in proteins. 8. A disulfide bond is formed between _______ residues. a. cysteine b. glycine c. methionine d. tyrosine Answer: a Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of the different groups of amino acids. Feedback A: Correct! Each cysteine (Cys, C) has an SH (sulfhydryl) group. Following oxidation, two cysteine residues are then linked to form a disulfide bond. Feedback B: Incorrect. Glycine (Gly, G) is the shortest amino acid. Its side chain consists

of a hydrogen (H) atom. Feedback C: Incorrect. Methionine (Met, M) also contains a sulfur (S) atom. However, the sulfur is internal in the side chain and cannot form a disulfide bond. Feedback D: Incorrect. Tyrosine (Tyr, Y) has an aromatic, hydrophobic side chain that contains no sulfur. 9. Which of the following classes of amino acids is buried within the folded structure of the protein? a. Acidic b. Basic c. Nonpolar d. Polar Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of the different groups of amino acids. Feedback A: Incorrect. Acidic amino acids are hydrophilic and hence are on the surface of proteins, where they can interact with water. Feedback B: Incorrect. Basic amino acids are hydrophilic and hence are on the surface of proteins, where they can interact with water. Feedback C: Correct! Nonpolar amino acids are hydrophobic and are in the interior of proteins. Feedback D: Incorrect. Polar amino acids are hydrophilic and hence are on the surface of proteins, where they can interact with water. 10. The three-dimensional structure of a protein is analyzed most definitively by a. electron microscopy. b. light microscopy. c. subcellular fractionation. d. X-ray crystallography. Answer: d Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of noncovalent bonds in protein folding. Feedback A: Incorrect. Electron microscopy does not have the resolution to provide information on the arrangement of atoms in a protein. Feedback B: Incorrect. The resolution of light microscopy is about 0.2 m and hence is insufficient to detect even the overall shape of a protein. Feedback C: Incorrect. Subcellular fractionation is an approach for separating organelles, membranes, and soluble molecules, either in individual molecules or in molecular complexes. Feedback D: Correct! X-ray crystallography is a high-resolution technique that is capable of defining the position of atoms within a protein. 11. The  (alpha) helix is an example of which level of protein structure? a. Primary

b. Quaternary c. Secondary d. Tertiary Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Explain the roles of noncovalent bonds in protein folding. Feedback A: Incorrect. Primary structure refers simply to the amino acid sequence of a protein. Feedback B: Incorrect. Quaternary structure refers to interactions between different polypeptide chains. Feedback C: Correct! Secondary structure refers to the arrangement of amino acids within a localized region, and an  helix is a coiled structure of often relatively limited length. Feedback D: Incorrect. Tertiary structure refers to the overall structure of a relatively large portion of a protein. 12. Proteins must have more than one _______ to have a quaternary structure. a.  helix b.  sheet c. polypeptide chain d. transmembrane segment Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of noncovalent bonds in protein folding. Feedback A: Incorrect. The  helix is a feature of the secondary structure of a protein. Feedback B: Incorrect. The  sheet is a feature of the secondary structure of a protein. Feedback C: Correct! The quaternary structure of a protein is the association of multiple polypeptide chains with one another. Feedback D: Incorrect. A transmembrane segment of a protein is typically an  helix, a secondary feature of the protein. 13. Enzymes act by a. lowering the overall change in free energy of a reaction. b. decreasing the distance that reactants must diffuse to find one another. c. increasing activation energy. d. decreasing activation energy. Answer: d Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Explain why enzymes affect the kinetics of chemical reactions without changing the equilibrium between reactants and products. Feedback A: Incorrect. The total change in free energy for a reaction is not altered by an enzyme.

Feedback B: Incorrect. Enzymes do not affect the distance that reactants must diffuse toward one another. Feedback C: Incorrect. Increasing activation energy would slow down the reaction. Feedback D: Correct! Enzymes decrease the energy required to reach the transition state (the activation energy), thus speeding up a reaction. 14. Enzymes affect the transition state of a chemical reaction by a. binding to substrate(s). b. providing a surface on which reactions converting substrate to product can occur more rapidly. c. altering the conformation of substrate(s) to approach that of the transition state. d. All of the above Answer: d Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms of enzymatic catalysis. Feedback A: Incorrect. Enzymes do bind to substrates, but the other answer choices are correct as well. Feedback B: Incorrect. Enzymes do provide a surface on which the reactions can occur more rapidly, but the other answer choices are correct as well. Feedback C: Incorrect. Enzymes do alter the conformation of the substrate to approach that of the transition state, but the other answer choices are correct as well. Feedback D: Correct! Because A, B, and C are true, D is the best and therefore the correct answer. 15. Chymotrypsin, trypsin, elastase, and thrombin are all members of the serine protease family because each a. has similar charge and shape properties in its substrate insertion pockets. b. can form a heptahedral complex with its substrate. c. uses the same catalytic mechanism involving the same key amino acids. d. can be modified by the cell-wall degrading enzyme chitinase. Answer: c Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms of enzymatic catalysis. Feedback A: Incorrect. If each had similar substrate binding properties, then each would cleave proteins at the same sequence. This is not the case. For example, chymotrypsin cleaves next to a hydrophobic amino acid and trypsin cleaves next to a basic amino acid. The insertion pocket of chymotrypsin is hydrophobic, while that of trypsin is negatively charged. Feedback B: Incorrect. Carbon atoms form tetrahedral complexes, not heptahedral complexes. Feedback C: Correct! A similar tetrahedral transition complex is formed between a serine residue in the active site of the enzyme and the substrate amino acid N-terminal of the cleavage site, hence the name serine protease. A carbon atom forms a tetrahedral

complex. A proton is transferred from the serine to a histidine as a part of the overall catalytic mechanism. Feedback D: Incorrect. All four examples are enzymes found in mammals. Mammals do not have cell walls. 16. The coenzymes NAD+ and NADP+ are structurally related to the vitamin a. riboflavin (B2). b. niacin. c. pantothenate. d. pyridoxal (B6). Answer: b Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering Learning Objective: Distinguish between enzymes and coenzymes Feedback A: Incorrect. Riboflavin (B2) is structurally related to the coenzyme FAD. Feedback B: Correct! Feedback C: Incorrect. Pantothenate is structurally related to Coenzyme A Feedback D: Incorrect. Pyridoxal (B6) is structurally related to the coenzyme pyridoxal phosphate 17. Which of the following is not true of coenzymes? a. They are branched amino acids. b. They serve as carriers of chemical groups. c. They transfer specific chemical groups among a wide range of substrates. d. They work together with enzymes to enhance chemical reactions without being irreversibly altered. Answer: a Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Distinguish between enzymes and coenzymes. Feedback A: Correct! Coenzymes are not amino acids but are small molecules related to a number of vitamins. Feedback B: Incorrect. Coenzymes do act as carriers for several different types of chemical groups. Feedback C: Incorrect. The enzyme, not the coenzyme, determines the specificity of the substrate. Feedback D: Incorrect. Coenzymes are regenerated in a cyclical manner. 18. Coenzymes are chemically related to a. amino acids. b. inorganic phosphate. c. glucose. d. vitamins. Answer: d Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 1. Remembering

Learning Objective: Distinguish between enzymes and coenzymes. Feedback A: Incorrect. Coenzymes are not derived metabolically from amino acids, so they have little chemical similarity to amino acids. Feedback B: Incorrect. Coenzymes are organic compounds. Some, such as NADH, are dinucleotides that include phosphate within their overall structure, but this is not as free inorganic phosphate. Feedback C: Incorrect. Glucose is a very important metabolic sugar, but the overall structure of coenzymes is that of a sugar. Feedback D: Correct! Many coenzymes are closely related to vitamins that contribute part or all of the structure of the coenzyme. 19. All of the following are ways in which enzyme activity can be regulated except a. by binding to an allosteric site. b. through feedback inhibition. c. by modulation of intracellular sucrose concentrations. d. through phosphorylation. Answer: c Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 2. Understanding Learning Objective: Explain why regulating the activity of enzymes is important to cell function. Feedback A: Incorrect. Allosteric regulation, the outcome of binding to an allosteric site, can regulate enzyme activity. Feedback B: Incorrect. Feedback inhibition by the end product of an early step in a metabolic pathway can regulate enzyme activity. Feedback C: Correct! Typically, sucrose is not an intracellular metabolite; hence, this cannot regulate enzyme activity. Feedback D: Incorrect. Phosphorylation is a common regulator of enzyme activity. 20. About 50% of the mass of most biological membranes consists of lipids and about 50% consists of proteins. Therefore, a. the membranes contain fewer molecules of lipid than of protein. b. the membranes contain equal numbers of lipid and protein molecules. c. the membranes contain more molecules of lipid than of protein. d. only a few membrane proteins are exposed at the cell surface. Answer: c Textbook Reference: Cell Membranes Bloom’s Category: 3. Applying Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers. Feedback A: Incorrect. Lipids are much lower in molecular weight than membrane proteins are, and hence in terms of molecular abundance there are many more molecules of lipid than of protein. Feedback B: Incorrect. Lipids are much lower in molecular weight than membrane proteins are, and hence in terms of molecular abundance there are many more molecules of lipid than of protein.

Feedback C: Correct! Lipids are much lower in molecular weight than membrane proteins are, and hence in terms of molecular abundance there are many more molecules of lipid than of protein. Feedback D: Incorrect. Many of these proteins are transmembrane proteins and hence are exposed at the cell surface. 21. Introducing a double bond into a fatty acid puts a(n) _______ into the conformation of the molecule. a. amino acid bulge b. kink c. reverse spiral d. branch Answer: b Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers. Feedback A: Incorrect. There are no amino acids in fatty acids. Feedback B: Correct! The double bond introduces a kink in the hydrocarbon chain. Feedback C: Incorrect. In the absence of a double bond, the hydrocarbon chain of a fatty acid is fairly straight. Feedback D: Incorrect. The hydrocarbon chain of a fatty acid has either single or double carbon-carbon bonds holding it together. 22. Phospholipids consist of a 3-carbon core to which two fatty acids and a phosphate group are linked. The most common 3-carbon core is a. dihydroxyacetone. b. glyceraldehyde. c. glycerol. d. serine. Answer: c Textbook Reference: The Molecules of Cells Bloom’s Category: 1. Remembering Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers. Feedback A: Incorrect. Dihydroxyacetone is a 3-carbon simple sugar. Feedback B: Incorrect. Glyceraldehyde is a 3-carbon simple sugar. Feedback C: Correct! Glycerol has three hydroxyl groups and is the 3-carbon core to which phosphate and two fatty acids are added in the four most common phospholipids. Feedback D: Incorrect. Serine is an amino acid and is the 3-carbon core to which phosphate and two fatty acids are added in one class of phospholipids—sphingomyelin. Serine is also a core component of glycolipids. 23. Lipids with unsaturated fatty acids a. decrease fluidity of membranes. b. increase the charge associated with the inner face of a membrane.

c. increase fluidity of membranes. d. are present only on the inner side of the plasma membrane. Answer: c Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Illustrate the hydrophobic and hydrophilic interactions that lead to the formation of lipid bilayers. Feedback A: Incorrect. However, cholesterol can decrease the fluidity of membranes. Feedback B: Incorrect. Unsaturated fatty acids do not affect the charge associated with a membrane. Feedback C: Correct! Double bonds produce kinks in fatty-acid chains, causing them to pack irregularly, which increases the fluidity of membranes. Feedback D: Incorrect. Unsaturated fatty acids can be located on either side of the membrane. 24. In the fluid mosaic model of biological membrane structure, transmembrane proteins are a. embedded nearly randomly in the lipid bilayer. b. almost completely surrounded by membrane lipid. c. segregated into large protein clusters or rafts. d. weakly held in place on the surface of the lipid bilayer. Answer: a Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between integral and peripheral membrane proteins. Feedback A: Correct! Typically, only small, -helical portions of the protein are embedded in the membrane, and individual proteins are distributed fairly randomly within the membrane bilayer. Feedback B: Incorrect. Typically, only small, -helical portions of the protein are embedded in the membrane. Feedback C: Incorrect. Typically, transmembrane proteins form at most small polypeptide clusters (e.g., dimers). Feedback D: Incorrect. This is a characteristic of peripheral membrane proteins. 25. What is the effect of a  barrel on the permeability of a membrane? a. It decreases permeability. b. It increases permeability. c. It has no effect. d.  barrels are peripheral membrane proteins. Answer: b Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between integral and peripheral membrane proteins. Feedback A: Incorrect. A  barrel is a protein structural feature that spans a membrane,

creating a pore-like opening and increasing permeability. Feedback B: Correct! A  barrel is a protein structural feature that spans a membrane, creating a pore-like opening. The presence of this feature therefore increases the permeability of a membrane. Feedback C: Incorrect. A  barrel is a protein structural feature that spans a membrane, creating a pore-like opening and increasing permeability. Feedback D: Incorrect. A  barrel is a protein structural feature that spans a membrane, creating a pore-like opening. 26. Which class of molecule accelerates transport across biological membranes? a. Carbohydrates b. Lipids c. Nucleic acids d. Proteins Answer: d Textbook Reference: Cell Membranes Bloom’s Category: 1. Remembering Learning Objective: Distinguish molecules that can diffuse through a lipid bilayer from those that require transporters to cross a membrane. Feedback A: Incorrect. Carbohydrates can be part of proteins, but it is the polypeptide chain that forms the transporter. Feedback B: Incorrect. Lipids form the permeability barrier of biological membranes. Feedback C: Incorrect. Nucleic acids encode the proteins that form transporters in membranes. Feedback D: Correct! Transporters in membranes are proteins of either the channel or carrier protein class. 27. While small, uncharged molecules can diffuse through the hydrophobic core of a phospholipid bilayer, larger polar molecules such as glucose must enter cells by binding to a. a nonphospholipid such as cholesterol. b. the carbohydrate portion of glycolipids. c. peripheral membrane proteins located on the inner side of the membrane. d. carrier proteins that facilitate the passage of specific molecules across membranes. Answer: d Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Distinguish molecules that can diffuse through a lipid bilayer from those that require transporters to cross a membrane. Feedback A: Incorrect. Cholesterol is a minor portion of membrane lipids. Binding of a polar glucose to the hydrophobic portions of cholesterol would be energetically unfavorable. Feedback B: Incorrect. The carbohydrate portions of glycolipids are exposed at the cell surface. Therefore, binding to these would not cause the glucose to be transported into the cell. Feedback C: Incorrect. Peripheral membrane proteins located on the inner side of the

membrane are inaccessible from the outside of the cell. Binding of glucose to these proteins is physically impossible. Feedback D: Correct! Through binding to a carrier protein, a transporter, glucose is placed in a hydrophilic environment and not exposed to the hydrophobic membrane lipids. 28. Passive transport across a membrane refers to a. transport into the interior of a cell. b. transport out of a cell. c. transport in the energetically favorable direction. d. simple diffusion across membranes, without the help of proteins such as channels or carriers. Answer: c Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Distinguish molecules that can diffuse through a lipid bilayer from those that require transporters to cross a membrane. Feedback A: Incorrect. Passive transport can be directed toward either the inside or the outside of a cell. Feedback B: Incorrect. Passive transport can be directed toward either the inside or the outside of a cell. Feedback C: Correct! Passive transport refers to the transport of a molecule in the energetically favorable direction, taking into consideration concentration and electrochemical gradients via membrane channels or carrier proteins. Feedback D: Incorrect. The term “passive transport,” by convention, is restricted to energetically favorable transport across membranes via membrane channels and carrier proteins. Hence, the term does not include simple diffusion.

Essay 1. Can RNA be considered a polysaccharide? Answer: Ribose is a five-carbon sugar, and RNA has many riboses linked together through phosphodiester bonds. Hence, both RNA and polysaccharides are polysugars. In the case of polysaccharides, however, the sugars are linked by glycosidic bonds rather than by phosphodiester bonds. Because of this difference, RNA is not considered a polysaccharide. Textbook Reference: The Molecules of Cells Bloom’s Category: 3. Applying Learning Objective: Explain the properties of different types of chemical bonds. 2. Why is phosphatidylethanolamine considered a neutral phospholipid, while phosphatidylinositol is considered an acidic phospholipid? Answer: Both phosphatidylethanolamine and phosphatidylinositol are glycerol-based phospholipids. At physiological pH (approximately 7), each has one negative charge contribution from the phosphate group. In addition, phosphatidylethanolamine has a

positive charge contribution from the amino group of ethanolamine. Since the sum of –1 and +1 is 0, phosphatidylethanolamine is a neutral phospholipid. The inositol of phosphatidylinositol is a sugar alcohol and carries no charge contribution. Since the sum of –1 and 0 is –1, phosphatidylinositol is an acidic phospholipid (i.e., it has a net negative charge). Textbook Reference: The Molecules of Cells Bloom’s Category: 4. Analyzing Learning Objective: Compare the structures of fatty acids, phospholipids, and steroid hormones. 3. Water is a polar molecule and can form hydrogen bonds. How is this property of water an important factor in determining protein structure? Answer: Hydrophilic amino acids tend to be exposed on the surface of proteins, where their hydrophilic side chains can both form hydrogen bonds and have charged interactions with water. Textbook Reference: The Molecules of Cells Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of noncovalent bonds in protein folding. 4. Like ribonuclease, insulin is a small protein. However, upon renaturation, insulin is much less efficient in resuming its native conformation. Why is this? Answer: Insulin is composed of two polypeptide chains that are joined by disulfide bonds. When completely denatured, the two polypeptide chains are separated. Upon renaturation, chain A might bond to chain B or to chain A, and the efficiency of correct renaturation is quite low. During the normal biosynthesis of insulin, the protein is synthesized as a single larger polypeptide that is subsequently cleaved to give two polypeptide chains. The correct disulfide bonds are formed before the cleavage. Textbook Reference: The Molecules of Cells Bloom’s Category: 3. Applying Learning Objective: Explain the roles of noncovalent bonds in protein folding. 5. Suppose you have completed a series of polyacrylamide gel electrophoresis experiments and found that under both denaturing and non-denaturing conditions, purified hemoglobin migrates as one molecular species. However, the apparent mass of the observed species under denaturing conditions is approximately one-quarter of that observed under non-denaturing conditions. What is the most likely explanation for these results? Answer: If native hemoglobin had a quaternary structure consisting of four polypeptide chains of near identical mass, then the observed experimental outcomes would be exactly the expected outcomes. Hemoglobin is a protein consisting of four very similar polypeptide chains; two  and two  chains. The  and  chains of hemoglobin arose from gene duplication. Textbook Reference: The Molecules of Cells Bloom’s Category: 3. Applying Learning Objective: Explain the roles of noncovalent bonds in protein folding.

6. Suppose you are studying the change in reaction rate for lactate dehydrogenase as the concentration of lactate is increased. You find that initially the reaction rate increases linearly with lactate concentration, but as you continue to increase lactate concentration, there is less and less increase in the reaction rate. Eventually there is almost no increase. Why? Answer: Lactate, as the substrate of lactate dehydrogenase, forms a complex with the enzyme. In any reaction mixture there is a limited amount of enzyme. As the concentration of substrate, in this case lactate, is increased, more and more of the binding sites of the enzyme molecules are occupied by substrate. The increase in reaction rate becomes less tightly linked to substrate concentration. At a sufficiently high substrate concentration, the enzyme is saturated with substrate and there is no further increase in reaction rate. Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 3. Applying Learning Objective: Summarize the mechanisms of enzymatic catalysis. 7. A rat’s liver is respiring actively, and you supply the liver with glucose labeled with carbon-14. Which of the following will rapidly become radioactively labeled: water, carbon dioxide, or NADH? Answer: One must consider here the inputs to, and the outputs from, glycolysis, the citric acid cycle, and the electron transport chain, as well as the chemical structures of the compounds named in the question. Water does not become labeled, as it consists of only hydrogen and oxygen atoms, with no carbons. NADH does not become labeled, since it is formed by the addition of hydrogen, not carbon, to NAD+. The carbon dioxide does become labeled, since the carbons of glucose are all released as carbon dioxide. Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 4. Analyzing Learning Objective: Distinguish between enzymes and coenzymes. 8. Cells can both make cholesterol and utilize cholesterol from the diet. High levels of cholesterol depress cholesterol biosynthesis by cells. What is the biochemical term applied to this type of enzyme regulation, and what are the implications of lowered dietary cholesterol for the rate of cellular cholesterol biosynthesis? Answer: The inhibition of product biosynthesis by the binding of the end product of a metabolic pathway to an early enzyme is termed feedback inhibition. In the case of cholesterol biosynthesis, reduction of dietary cholesterol could affect the level of cellular cholesterol such that feedback inhibition of the biosynthetic pathway is decreased and the rate of cellular cholesterol biosynthesis is increased. Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 3. Applying Learning Objective: Explain why regulating the activity of enzymes is important to cell function. 9. Suppose that an organic chemist wishes to synthesize an inhibitor of an important enzyme in pyrimidine biosynthesis. What two different kinds of molecules should be considered the most effective inhibitors?

Answer: Pyrimidines are heterocyclic bases found as part of DNA and RNA. Biosynthesis of pyrimidines is essential for cell multiplication. One chemical approach to inhibiting an enzyme in pyrimidine biosynthesis would be to synthesize a chemical analog of the enzyme substrate that might bind more tightly to the enzyme’s active site or be unable to be reacted on by the enzyme. A second chemical approach would be to synthesize a chemical analog of an allosteric effector of the enzyme. Some enzymes are regulated by metabolic products of the enzyme pathway that bind to a second site on the enzyme that is distant from the active site of the enzyme. A tight binding allosteric effector analog would also inhibit enzyme activity. Textbook Reference: Enzymes as Biological Catalysts Bloom’s Category: 3. Applying Learning Objective: Explain why regulating the activity of enzymes is important to cell function. 10. Transmembrane proteins are water-insoluble. Why? Answer: Hydrophobic portions of transmembrane proteins are “dissolved” in the lipid bilayer of the membrane. These portions are not water-soluble; hence the whole protein is water-insoluble. To solubilize a transmembrane protein, detergent must be used. The binding of detergent with the transmembrane domains of the protein solubilizes the protein. Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between integral and peripheral membrane proteins. 11. The only protein structure known to span a biological membrane, other than  helix, is the  barrel, formed by the folding of  sheets into a barrel-like structure. Why does the presence of  barrel proteins in the outer membrane of chloroplasts and mitochondria make lipid bilayers permeable to small molecules and ions? Answer: A -barrel structure in a membrane creates a porous opening in the lipid bilayer. Therefore, the lipid bilayer, in this case the outer membrane of a chloroplast or mitochondrion, will be permeable to small molecules and ions. Textbook Reference: Cell Membranes Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between integral and peripheral membrane proteins. 12. Suppose you are studying the transport of glucose into red blood cells and find that as you increase the concentration of glucose outside of the cells, a concentration is reached at which there is no further increase in the rate of accumulation of glucose in the cells. What is the explanation for this? Answer: Glucose is too polar a molecule to diffuse through a biological membrane by simple diffusion. Instead, glucose is transported into cells by glucose transporters. The red blood cell membrane has a fixed number of transporters. At a certain concentration of extracellular glucose, glucose is bound to all of the transporters and no further increase in

the rate of accumulation with increased concentration is possible. The number of transporter molecules limits the rate of passive transport. Textbook Reference: Cell Membranes Bloom’s Category: 4. Analyzing Learning Objective: Distinguish molecules that can diffuse through a lipid bilayer from those that require transporters to cross a membrane.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 3: Bioenergetics and Metabolism TEST FILE QUESTIONS Multiple Choice 1. If the equilibrium constant for the reaction A  B is 0.5, and the initial concentration of A is 30 mM and of B is 15 mM, then the reaction a. will proceed in the direction it is written, producing a net increase in the concentration of B. b. will produce energy, which can be used to drive ATP synthesis. c. will proceed in the reverse direction, producing a net increase in the concentration of A. d. is at equilibrium. Answer: d Textbook Reference: Metabolic Energy and ATP. Bloom’s Category: 3. Applying Learning Objective: Interpret the first and second laws of thermodynamics. 2. The free-energy change of a reaction is determined by the a. intrinsic properties of reactants and products. b. concentrations of reactants and products. c. temperature of reactants and products. d. All of the above Answer: d Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Explain how changes in Gibbs free energy determine the direction of chemical reactions. 3. The G°′ for hydrolysis of ATP to ADP and P i is _______ kcal/mole. a. +14 b. +7.3 c. –7.3 d. –0.5 Answer: c Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of ATP in cell physiology.

4. The hydrolysis of ATP to ADP and phosphate allows glucose-6-phosphate to be synthesized from glucose and phosphate because the a. heat produced from ATP hydrolysis drives glucose-6-phosphate synthesis. b. enzymatic coupling of these two reactions allows the energy of ATP hydrolysis to drive the synthesis of glucose-6-phosphate. c. energy of glucose phosphorylation drives ATP splitting. d. All of the above Answer: b Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Summarize the role of ATP in cell physiology. 5. Glycolysis yields a net gain of _______ ATP molecules per molecule of glucose. a. 2 b. 4 c. 32 d. 36 Answer: a Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 6. Under aerobic conditions, the end product of glycolysis is a. ethanol. b. lactate. c. phosphoenolpyruvate. d. pyruvate. Answer: d Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 7. Glycolysis produces four molecules of ATP from each molecule of fructose-1,6-bisphosphate because a. two molecules of ATP are required to convert glucose to fructose-1,6-bisphosphate. b. fructose-1,6-bisphosphate is converted into two molecules of 1,3-bisphosphoglycerate, each of which produces two ATPs when metabolized to pyruvate. c. fructose-1,6-bisphosphate is converted into four molecules of 1,3-bisphosphoglycerate, each of which produces one ATP when metabolized to pyruvate. d. Both a and b Answer: b Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 8. In glycolysis, the key control enzyme that is inhibited by excess ATP is a. hexokinase. © 2019 Oxford University Press

b. phosphofructokinase. c. alcohol dehydrogenase. d. fructose-1,6-bisphosphate. Answer: b Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 9. Under anaerobic conditions, NADH produced by glycolysis is recycled by a reaction that produces NAD+ and a. ethanol. b. lactate. c. pyruvate. d. ethanol or lactate. Answer: d Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 10. Which of the following is a product of glycolysis that is transported into the mitochondria? a. Acetate as acetyl CoA b. Pyruvate c. Ethanol d. Lactic acid Answer: b

Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 11. In eukaryotic cells, glycolysis takes place in the a. cytosol. b. nucleus. c. mitochondria. d. endoplasmic reticulum. Answer: a Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 12. In eukaryotic cells, the citric acid cycle occurs in the a. cytosol. b. nucleus. c. mitochondria. d. chloroplasts. Answer: c Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering © 2019 Oxford University Press

Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 13. The citric acid cycle consists of the oxidation of _______ to produce _______. a. pyruvate; CO2, NADH, and FADH2 b. acetyl CoA; CO2, NADH, and FADH2 c. pyruvate; CO2 d. acetate; CO2 Answer: b

Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 14. Through the oxidation process, fatty acids produce approximately _______ energy per gram compared to carbohydrates. a. the same b. 50% more c. 2.5 times more d. 5 times more Answer: c Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Describe the breakdown of lipids. 15. Fatty acids are broken down in a stepwise process, _______ carbon(s) at a time. a. one b. two c. four d. six Answer: b Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Describe the breakdown of lipids. 16. Protons are freely permeable across a. both mitochondrial membranes. b. the inner, but not the outer, mitochondrial membrane. c. the outer, but not the inner, mitochondrial membrane. d. neither mitochondrial membrane. Answer: c

Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 17. The inner mitochondrial membrane contains proteins that a. synthesize ATP. b. pump protons. © 2019 Oxford University Press

c. transport pyruvate and fatty acids. d. All of the above Answer: d

Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Explain chemiosmotic coupling. 18. The G° for the transfer of two electrons from NADH to O2 through the electron transport chain is _______ kcal/mol. a. +52.5 b. –52.5 c. +7.5 d. –7.5 Answer: b

Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Explain chemiosmotic coupling. 19. When electrons from the cofactors reduced by breakdown of one glucose are transferred to oxygen in mitochondria, the process is coupled to the formation of _______ molecules of ATP. a. 2 b. 10–12 c. 22–24 d. 32–34 Answer: d

Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 2. Understanding Learning Objective: Explain chemiosmotic coupling. 20. Electron transport and oxidative phosphorylation are performed by protein complexes in the mitochondrial a. outer membrane. b. intermembrane space. c. inner membrane. d. matrix. Answer: c

Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Explain chemiosmotic coupling. 21. Coenzyme Q carries electrons from complex _______ to complex _______. a. I; II b. II; III c. I; III d. III; IV Answer: c

Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Explain chemiosmotic coupling. 22. Peter Mitchell received the Nobel Prize in 1978 for his revolutionary hypothesis of oxidative phosphorylation, which is called the _______ hypothesis. a. electron transport b. substrate-level phosphorylation c. endosymbiotic d. chemiosmotic Answer: d

Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Explain chemiosmotic coupling. 23. Photosynthesis uses light energy to chemically convert a. CO and H2O into CH4 and O2. b. CH3OH into C6H12O6. c. C6H12O6 and O2 into CO2 and H2O. d. CO2 and H2O into C6H12O6 and O2. Answer: d Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Explain the role of chlorophyll in harvesting energy from sunlight. 24. Photosynthesis takes place in two stages, referred to as a. light reactions and dark reactions. b. oxygen requiring and oxygen releasing reactions. c. H2O splitting and sugar synthesizing reactions. d. chlorophyll dependent and independent reactions. Answer: a Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Explain the role of chlorophyll in harvesting energy from sunlight. 25. In photosynthesis, all of the following reactions are directly dependent on light except a. electron transport. b. synthesis of ATP. c. nitrogen fixation. d. removal of electrons from H2O. Answer: c Textbook Reference: Photosynthesis. Bloom’s Category: 3. Applying Learning Objective: Explain the role of chlorophyll in harvesting energy from sunlight. 26. In the light reactions of photosynthesis, a. sugars are synthesized from CO2 and water. © 2019 Oxford University Press

b. H2O is converted to O2 and NADP+ is reduced to NADPH. c. oxidative phosphorylation produces ATP. d. All of the above Answer: b

Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 27. Light-dependent generation of ATP in photosynthesis occurs in the a. stroma. b. inner membrane. c. thylakoid membrane. d. thylakoid lumen. Answer: c

Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 28. The electron carrier that transports electrons from cytochrome bf to photosystem I where NADP+ is reduced to NADPH is a. cytochrome c. b. plastocyanin. c. plastoquinone. d. QA. Answer: b

Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 29. ATP is synthesized in photosynthesis by which of the following means? a. Light energy stored by chlorophyll of photosystem I is directly transferred to ATP synthase, which uses the energy to synthesize ATP. b. Light energy absorbed by photosystems I and II generate energetic electrons that cause the cytochrome bf complex to pump protons across the thylakoid membrane; these protons drive ATP synthase to synthesize ATP. c. Light energy absorbed by photosystems I and II generate energetic electrons that cause the cytochrome bf complex to pump protons across the inner chloroplast membrane; these protons drive ATP synthase to synthesize ATP. d. Light energy is used to synthesize glucose, which is then metabolized to produce ATP. Answer: b

Textbook Reference: Photosynthesis Bloom’s Category: 2. Understanding

Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 30. The dark reactions occur in the a. outer membrane. b. inner membrane. c. thylakoid membrane. d. stroma. Answer: d

Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of the Calvin cycle. 31. The Calvin cycle can occur a. in the absence of CO2. b. only in the presence of light. c. in the absence of light. d. only in the absence of oxygen. Answer: c

Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of the Calvin cycle. 32. In the synthesis of glycogen, glucose-1-phosphate reacts with a nucleotide triphosphate to form _______-glucose. a. ADP b. GDP c. CDP d. UDP Answer: d Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between gluconeogenesis and glycolysis. 33. Gluconeogenesis can produce glucose beginning with a. lactate. b. pyruvate. c. fatty acids. d. All of the above Answer: d Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between gluconeogenesis and glycolysis. 34. The production of one glucose from two pyruvates requires _______ more than is obtained through glycolysis. a. 2 NADH © 2019 Oxford University Press

b. 2 ATP and 2 GTP c. 2 NADH and 2 ATP d. 4 ATP Answer: b Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between gluconeogenesis and glycolysis. 35. The most energetically favorable reactions in glycolysis are catalyzed in reverse in gluconeogenesis by a. shifting their equilibrium. b. the glycolytic enzymes using ATP. c. the use of different enzymes coupled with energy from ATP or NADH. d. accumulation of high substrate concentrations. Answer: c Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between gluconeogenesis and glycolysis. 36. The fixation of atmospheric nitrogen can be carried out by a. a few species of bacteria. b. bacteria, fungi, and plants. c. plants. d. all eukaryotes. Answer: a Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. 37. For humans, _______ essential amino acids must be provided by the diet. a. 2 b. 4 c. 9 d. 18 Answer: c Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. 38. The incorporation of each amino acid into a protein is coupled to the hydrolysis of _______ molecule(s). a. one ATP b. two ATP c. one ATP and one GTP d. one ATP and two GTP Answer: d Textbook Reference: The Biosynthesis of Cell Constituents © 2019 Oxford University Press

Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. 39. 6-Mercaptopurine is a potent inhibitor of a. protein translation. b. DNA synthesis. c. conversion of pyruvate to oxaloacetate in gluconeogenesis. d. glycogen biosynthesis. Answer: b Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathways of nucleic acid synthesis. 40. A nucleotide is a a. base found in nucleic acids. b. purine or pyrimidine base plus a five-carbon sugar. c. purine or pyrimidine base, a five-carbon sugar, and one or more phosphate groups. d. purine or pyrimidine base, a five-carbon sugar, and three phosphate groups. Answer: c Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathways of nucleic acid synthesis.

Fill in the Blank 1. In anaerobic cells, the oxidation of glucose yields _______ molecules of ATP per molecule of glucose. Answer: two Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 2. Under anaerobic conditions, pyruvate can be converted to either lactic acid or _______. Answer: ethanol Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 3. The citric acid cycle involves the addition of two carbons from the compound _______ to the four-carbon oxaloacetate to produce the six-carbon compound _______. Answer: acetyl CoA; citrate ( citric acid) Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle.

4. The acceptor of acyl groups from pyruvate is _______. Answer: coenzyme A Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 5. In addition to the three molecules of NADH produced per molecule of acetate introduced into the citric acid cycle, one molecule of an additional reduced cofactor called _______ is produced. Answer: FADH2 Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 6. The citric acid cycle produces two reduced cofactors, NADH and _______. Answer: FADH2 Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 7. During oxidation of a fatty acid, the fatty acid is joined to the molecule _______ and is broken down by stepwise removal of _______-carbon units as part of the molecule _______, which can then enter the citric acid cycle to be metabolized. Answer: coenzyme A (CoA-SH); two; acetyl CoA Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Describe the breakdown of lipids. 8. In aerobic cells, the complete oxidation of glucose via glycolysis and the citric acid cycle yields between _______ and _______ molecules of ATP per molecule of glucose, depending on how the electrons from NADH enter the mitochondria. Answer: 36; 38 Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 9. ATP is produced in mitochondria by a process called _______. Answer: oxidative phosphorylation Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 10. The reduced cofactors NADH and FADH2 transfer their electrons to O2 to produce H2O via a set of four membrane-bound complexes collectively called the _______. These complexes pump _______ across the _______ membrane. Answer: electron transport chain; H+ (or protons); inner mitochondrial © 2019 Oxford University Press

Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 11. The mobile carrier that transfers electrons from Complex I or II to Complex III of the electron transport chain is called _______. Answer: coenzyme Q (ubiquinone) Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 12. The ultimate acceptor of electrons from the electron transport chain is _______. Answer: oxygen Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 13. The only complex in the mitochondrial electron transport chain that is not involved directly in the transport of protons is complex _______. Answer: II Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 2. Understanding Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 14. The _______ produced by the electron transport chain drives a flow of these ions through a complex of proteins called _______ to produce ATP from ADP and P i. Answer: proton gradient (electrochemical gradient, proton motive force); ATP synthase

(F0-F1 proton pump) Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Explain chemiosmotic coupling. 15. The light reactions in photosynthesis produce _______ and _______ while the dark reactions produce _______. Answer: ATP (NADPH); NADPH (ATP); carbohydrates Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 16. In chloroplasts, the proton gradient that drives ATP production is established across the _______ membrane.

Answer: thylakoid Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 17. In photosystem II, water is split to produce _______ and _______, while in photosystem I the movement of electrons is coupled to the reduction of _______. Answer: O2 (oxygen); H+ (protons); NADP+ Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 18. Photosystem II is also involved in cyclic electron flow in which electrons are transferred to _______. This electron transfer is coupled to the generation of a _______ gradient and ultimately to the generation of _______. Answer: cytochrome bf; H+ (proton); ATP Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 19. The ATP and NADPH produced by photosynthesis are used to synthesize _______ from _______ and _______. Answer: carbohydrates; carbon dioxide (water); water (carbon dioxide) Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of the Calvin cycle. 20. In the dark reactions, _______ is synthesized in the chloroplast stroma by the _______ cycle. Answer: glucose; Calvin Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of the Calvin cycle. 21. Plants store glucose in the form of _______, and animals store glucose in the form of _______. Answer: starch; glycogen Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between gluconeogenesis and glycolysis. 22. _______ differs from a reverse of glycolysis in that energetically unfavorable reverse steps of glycolysis are replaced with energetically favorable reactions catalyzed by different enzymes. Answer: Gluconeogenesis Textbook Reference: The Biosynthesis of Cell Constituents © 2019 Oxford University Press

Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between gluconeogenesis and glycolysis. 23. Amino acids are synthesized from intermediates in _______ and from the _______. Answer: glycolysis; citric acid cycle Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 2. Understanding Learning Objective: Summarize how amino acids and proteins are synthesized. 24. In the process of protein synthesis, _______ serves as a template for translation. Answer: messenger RNA Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. 25. The reduction of atmospheric N2 to NH3 by certain species of bacteria is referred to as _______. Answer: nitrogen fixation Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. 26. Polymerization of DNA and RNA is driven by _______ as activated precursors. Answer: nucleoside triphosphates Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathways of nucleic acid synthesis. 27. The compound 6-mercaptopurine has been used successfully to treat _______. Answer: leukemia Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathways of nucleic acid synthesis.

True/False 1. The amount of free energy released when bonds are broken during a reaction is higher when the molecule has more electronegative atoms. Answer: F Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 3. Applying Learning Objective: Explain how changes in Gibbs free energy determine the direction of chemical reactions. 2. ATP serves to transfer energy from energy-producing to energy-requiring reactions. © 2019 Oxford University Press

Answer: T Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Summarize the role of ATP in cell physiology. 3. Glycolysis can act on glucose or fatty acids as initial substrates. Answer: F Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 4. Per unit weight, carbohydrates are more efficient energy storage molecules than lipids. Answer: F Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Describe the breakdown of lipids. 5. Most of the ATP derived from the breakdown of glucose in aerobic cells is derived from glycolysis. Answer: F Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 6. Most of the ATP derived from the breakdown of glucose in anaerobic cells is derived from glycolysis. Answer: T Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 7. The oxidation of NADH via the electron transport chain is a highly efficient process with little or no energy loss to heat. Answer: T Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 2. Understanding Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 8. The electron transport chain consists of four complexes, I through IV. Answer: T Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. © 2019 Oxford University Press

9. The final electron acceptor in oxidative phosphorylation is O2. Answer: T Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 2. Understanding Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 10. The potential energy stored in the proton gradient that drives oxidative phosphorylation is totally chemical in nature. Answer: F Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 2. Understanding Learning Objective: Explain chemiosmotic coupling. 11. The mode of action of mitochondrial ATP synthase involves mechanical coupling between protein subunits and hence can be considered an example of a rotary motor. Answer: T Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Explain chemiosmotic coupling. 12. In photosynthesis, energy from sunlight is used to drive the synthesis of ATP and NADPH. Answer: T Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 13. Photosystems I and II share the property of absorbing light and transferring energy to chlorophyll. Answer: T Textbook Reference: Photosynthesis Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 14. Plastoquinone (PQ) can carry electrons from both photosystems I and II to cytochrome bf. Answer: T Textbook Reference: Photosynthesis Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 15. Photosystems I and II both contribute electrons to cytochrome bf to generate a proton gradient that powers ATP generation by the chloroplast ATP synthase. Answer: T © 2019 Oxford University Press

Textbook Reference: Photosynthesis Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 16. The dark reactions of photosynthesis are directly coupled to chlorophyll and light energy. Answer: F Textbook Reference: Photosynthesis Bloom’s Category: 2. Understanding Learning Objective: Summarize the reactions of the Calvin cycle. 17. The Calvin cycle in plants, in which glucose is synthesized from CO 2, is an example of a light reaction. Answer: F Textbook Reference: Photosynthesis Bloom’s Category: 2. Understanding Learning Objective: Summarize the reactions of the Calvin cycle. 18. The gluconeogenesis pathway in animal cells directly reverses glycolysis to synthesize glucose. Answer: F Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between gluconeogenesis and glycolysis. 19. Lipids are synthesized from acetyl CoA, which is produced by reversing the citric acid cycle. Answer: F Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 2. Understanding Learning Objective: Describe the synthesis of lipids. 20. Amino acids are synthesized from intermediates in glycolysis and in the citric acid cycle. Answer: T Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 2. Understanding Learning Objective: Summarize how amino acids and proteins are synthesized. 21. Ammonia is incorporated into organic molecules primarily during the synthesis of the amino acids glutamate and glutamine. Answer: T Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. 22. Many bacteria and plants can synthesize all 20 amino acids found in proteins. Answer: T Textbook Reference: The Biosynthesis of Cell Constituents © 2019 Oxford University Press

Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. 23. Polymerization of amino acids to form proteins requires additional energy in the form of ATP and GTP. Answer: T Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. 24. Nucleic acid synthesis occurs by polymerization of nucleotide monophosphates into DNA or RNA. Answer: F Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathways of nucleic acid synthesis. 25. DNA synthesis requires the presence of a complementary nucleotide strand, whereas RNA synthesis does not. Answer: F Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathways of nucleic acid synthesis.

Short Answer 1. What is the definition of Gibbs free energy (G)? Answer: G is the free energy of a reaction (in other words, the energy change resulting from a reaction). The energy change is a sum of the changes in enthalpy, the heat released or absorbed during a chemical reaction, and the changes in entropy, the degree of disorder resulting from the reaction. Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Explain how changes in Gibbs free energy determine the direction of chemical reactions. 2. What is the definition of free-energy change? Answer: Free-energy change is the energy required or produced when a reaction occurs. This is the G of the reaction. Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Explain how changes in Gibbs free energy determine the direction of chemical reactions.

3. In anaerobic reactions, pyruvate is converted to either lactate or ethanol. In both of these anaerobic reactions, the cofactor NADH is oxidized to NAD +, which then can be used to perform what function? Answer: NAD+ is a required electron acceptor to keep glycolysis going by providing the cofactor to reduce glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate. Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 4. Where in “typical” eukaryotic cells does glycolysis occur? Be specific if the process or event occurs in only one part of an organelle. If you are not sure, give the location of the enzymes that catalyze the process. Answer: The cytosol Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 5. Where in “typical” eukaryotic cells does the citric acid cycle occur? Be specific if the process or event occurs in only one part of an organelle. If you are not sure, give the location of the enzymes that catalyze the process. Answer: The mitochondrial matrix

Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 6. Where in “typical” eukaryotic cells does oxidative phosphorylation of ATP occur? Be specific if the process or event occurs in only one part of an organelle. If you are not sure, give the location of the enzymes that catalyze the process. Answer: The inner membranes of mitochondria Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 7. In which location does cytochrome c carry electrons from complex III to complex IV? Answer: In the intermembrane space (or on the outer surface of the inner membrane)

Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 8. How does ATP synthesis in oxidative phosphorylation differ from ATP synthesis in glycolysis? Answer: In oxidative phosphorylation, the protons generated by the electron transport chain flow down their gradient through ATP synthase and drive synthesis of ATP. In glycolysis, a phosphate is transferred to ADP from a higher energy phosphate of a substrate molecule to directly produce ATP. © 2019 Oxford University Press

Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 3. Applying Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 9. Oxidation of one molecule of NADH leads to the synthesis of how many molecules of ATP? Answer: Three Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 2. Understanding Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 10. Light is absorbed by photosystems I and II. Based on the known pigments in a chloroplast photocenter, would you expect the isolated chloroplasts to show a single component light absorption spectra? Answer: Since photocenters contain multiple pigments, one would expect each to have its own individual absorption spectrum; therefore the absorption spectrum for isolated chloroplasts should be a multi-component composite. Textbook Reference: Photosynthesis Bloom’s Category: 3. Applying Learning Objective: Explain the role of chlorophyll in harvesting energy from sunlight. 11. The ATP synthase in chloroplast thylakoid membranes is oriented in the opposite direction within the membrane to that of the mitochondrial ATP synthase. How can the chloroplast synthase still be functional in ATP production? Answer: In chloroplasts, the proton gradient is oriented in the opposite direction. Hence both the ATP synthase and the proton gradient are oriented in proper manner for function in chloroplasts. Textbook Reference: Photosynthesis Bloom’s Category: 4. Analyzing Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 12. What is the role of light with respect to photosystem I (PS I) and NADPH generation? Answer: Light input into PS I is used to put electrons into a higher energy state. This provides the reducing power for NADP reductase to convert NADP to NADPH. Textbook Reference: Photosynthesis Bloom’s Category: 2. Understanding Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 13. Plastoquinone (PQ) is involved as an electron carrier from either PS II or PS I to cytochrome bf. In transferring electrons from PS I, the plastoquinone is part of cyclic electron flow, while in the case of transfer from PS II it is not. What is the key difference between the two processes? Answer: In the case of PS I, the electrons transferred by PQ are carried back at a lower energy state to PS I by plastocyanin (i.e., a cyclic flow), while in the case of PS II that is not true. Textbook Reference: Photosynthesis

Bloom’s Category: 4. Analyzing Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts. 14. Synthesis of glucose from CO 2 in the Calvin cycle requires energy input. What are the sources of energy for the Calvin cycle? Answer: ATP and NADPH Textbook Reference: Photosynthesis Bloom’s Category: 2. Understanding Learning Objective: Summarize the reactions of the Calvin cycle. 15. Gluconeogenesis involves the conversion of pyruvate to glucose. In a formal sense, this is essentially the reversal of glycolysis. However, the actual reactions are different. Explain this. Answer: Glycolysis is not an equilibrium set of reactions. There is first energy input in the form of ATP and later the generation of net ATP in subsequent reactions. Non-equilibrium reactions cannot simply be reversed. Different mechanisms, reactions, are needed. Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between gluconeogenesis and glycolysis. 16. How is nitrogen incorporated into organic molecules in animal cells? Answer: Ammonia is attached to the 5-carbon molecules -ketoglutarate from the citric acid cycle to form the amino acids glutamate and glutamine. Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 2. Understanding Learning Objective: Summarize how amino acids and proteins are synthesized. 17. DNA and RNA contain deoxyribose or ribose in repeated linkages. Yet the biosynthesis of DNA and RNA differs greatly from that of long polysaccharides such as starch or glycogen. Describe this difference. Answer: The biosynthesis of DNA or RNA is a template process in which base pairing ensures a constancy in sequence between starting material and product and produces a new molecule of both defined sequence and length. Starch and glycogen are both polymers of glucose, a singlesugar unit. Any difference in the length of individual polysaccharide molecules is without informational consequences. There is no informational content to these molecules. Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 3. Applying Learning Objective: Summarize the pathways of nucleic acid synthesis.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. What are the energetic consequences of coupling a reaction with a + G to one with a larger – G? a. An outcome in which the overall G is positive b. An outcome in which both reactions are energetically unfavorable overall c. A G that is favorable overall and drives the reaction with a + G in the forward direction d. An outcome in which the overall G is unfavorable Answer: c Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 3. Applying Learning Objective: Explain how changes in Gibbs free energy determine the direction of chemical reactions. Feedback A: Incorrect. The overall sum of the reactions has a favorable net –G. Feedback B: Incorrect. The overall sum of the reactions has a favorable net –G. Feedback C: Correct! Feedback D: Incorrect. The overall sum of the reactions has a favorable net –G. 2. What is the change in free energy (G) for the hydrolysis of ATP to ADP and P i in the cell? a. +12 kcal/mol b. –1.2 kcal/mol c. –7.3 kcal/mol d. –12 kcal/mol Answer: d Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of ATP in cell physiology. Feedback A: Incorrect. The hydrolysis of ATP is an energetically favorable reaction, so energy is released and the change in free energy has a negative value. Feedback B: Incorrect. The phosphate bonds in ATP are high-energy bonds and release a large amount of energy when broken. Feedback C: Incorrect. This is the standard change in free energy for the reaction, and thus represents the free-energy change under standard conditions. The conditions in the cell differ from standard conditions and hence a different amount of energy is released upon ATP hydrolysis. Feedback D: Correct! Because of the large amount of energy released, the hydrolysis of ATP is used to drive many energetically unfavorable reactions in the cell. 3. What is the net yield of ATP molecules per molecule of glucose from the process of glycolysis? a. 2 b. 4 c. 10 d. 38

Answer: a Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. Feedback A: Correct! Four molecules are generated, but two are consumed during glycolysis, resulting in a final net yield of two ATP molecules. Feedback B: Incorrect. Four molecules of ATP are generated from glycolysis, but some are also used up in the process. Feedback C: Incorrect. This is the number of NADH molecules generated from the oxidation of glucose. Feedback D: Incorrect. This is the total number of molecules generated from the oxidation of glucose. 4. Under aerobic conditions, the NADH formed during glycolysis a. has no metabolic advantage to the cell. b. is a waste product of glycolysis. c. provides additional energy by donating its electrons to the electron transport chain. d. converts acetaldehyde to ethanol and hence serves as the basis of many fermentation processes. Answer: c Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. Feedback A: Incorrect. NADH is a coenzyme and can donate electrons to many different reactions. Under aerobic conditions, it can donate electrons to the electron transport chain to generate energy. Feedback B: Incorrect. NADH is an important coenzyme and supports enzymatic processes even under anaerobic conditions. Feedback C: Correct! Under aerobic conditions NADH can donate electrons to the electron transport chain and additional ATP can be generated. Feedback D: Incorrect. NADH is a coenzyme for the conversion of acetaldehyde to ethanol, but this only occurs under anaerobic conditions. 5. The end product of the anaerobic metabolism of glucose may be any of the following except a. acetaldehyde. b. ethanol. c. lactate. d. methanol. Answer: d Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. Feedback A: Incorrect. Acetaldehyde can be a normal anaerobic product of glucose metabolism. Feedback B: Incorrect. Ethanol can be a normal anaerobic product of glucose metabolism. Feedback C: Incorrect. Lactate can be a normal anaerobic product of glucose metabolism. Feedback D: Correct! Methanol is not a normal end product of glycolytic metabolism of glucose under anaerobic conditions.

6. Together, glycolysis and the citric acid cycle (also known as the Krebs cycle) directly generate two molecules of ATP and two molecules of GTP. Most of the energy derived from the oxidation of glucose comes from the oxidation of the _______ generated from _______. a. FADH2; the citric acid cycle b. NADH; glycolysis c. NADH; the citric acid cycle d. NADPH; the citric acid cycle Answer: c Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. Feedback A: Incorrect. Two molecules of FADH2 are generated from the citric acid cycle. This yields four molecules of ATP. More ATP comes from the oxidation of NADH generated from the citric acid cycle. Feedback B: Incorrect. Two molecules of NADH are generated from glycolysis, while six are generated from the citric acid cycle. Feedback C: Correct! Six molecules of NADH are generated from the citric acid cycle, four more than from glycolysis. Eighteen molecules of ATP are formed from the oxidation of NADH through the electron transport chain. Feedback D: Incorrect. No NADPH is generated from the citric acid cycle. 7. How many carbons does the acetate group of acetyl CoA have? a. 1 b. 2 c. 3 d. 4 Answer: b Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. Feedback A: Incorrect. Feedback B: Correct! Feedback C: Incorrect. Feedback D: Incorrect. 8. Which of the following is a key molecule in the electron transport chain in mitochondria? a. Citrate b. Cytochrome c c. Pyruvate d. Tetrahydrofolate Answer: b Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. Feedback A: Incorrect. Citrate is the first six-carbon acid in the citric acid cycle.

Feedback B: Correct! Cytochrome c is one of the cytochromes involved in the electron transport chain. Feedback C: Incorrect. Pyruvate is produced by the glycolytic metabolism of glucose. Feedback D: Incorrect. Tetrahydrofolate is a coenzyme that acts as a donor in the transfer of onecarbon groups. 9. NADH produced in glycolysis yields fewer ATP molecules than that produced in the citric acid cycle because a. it must be converted to FADH 2 for transport into the mitochondrial matrix. b. its electrons are shuttled across the mitochondrial inner membrane to complex II. c. its electrons are donated to complex I on the cytoplasmic side, so it pumps no protons out of the matrix. d. ATP molecules are hydrolyzed to transport the NADH across the mitochondrial inner membrane. Answer: b Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. Feedback A: Incorrect. NADH is not converted to FADH2. Feedback B: Correct! Since electrons are donated to complex II, no protons are pumped out of the matrix by complex I, and fewer ATP molecules are made. Feedback C: Incorrect. Complex I does not accept electrons on the cytoplasmic side of the mitochondrial inner membrane. Feedback D: Incorrect. Transport of NADH across the mitochondrial inner membrane is not driven by ATP hydrolysis. 10. The role of cytochrome c in the electron transport chain is to a. transfer electrons from complex III to complex IV. b. transfer the electrons to molecular oxygen. c. transfer electrons from complex I to complex III. d. couple the passage of protons down the gradient to ATP production. Answer: a Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Explain chemiosmotic coupling. Feedback A: Correct! Cytochrome c is a peripheral inner-membrane protein that receives electrons from cytochrome b in complex III and transfers them to complex IV. Feedback B: Incorrect. This function is carried out by complex IV. Feedback C: Incorrect. This function is carried out by coenzyme Q. Feedback D: Incorrect. This occurs in complex V. 11. The ATP synthases of mitochondria and chloroplasts are examples of _______ proteins in which polypeptide rotation provides a mechanical coupling to ATP synthesis. a. symport b. Na+-K+ ATPase c. proton pump © 2019 Oxford University Press

d. motor Answer: d Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Explain chemiosmotic coupling. Feedback A: Incorrect. The ATP synthase is not a symport that moves two substances in the same direction. Feedback B: Incorrect. The Na +-K+ ATPase is a plasma membrane protein whose generation of Na+ and K+ gradients drives much of transport across the plasma membrane. Feedback C: Incorrect. The ATP synthase is not a proton pump. The movement of protons here drives the generation of ATP by the ATP synthase. Feedback D: Correct! Rotation of the polypeptide, a property of a motor protein, leads to mechanical coupling necessary for ATP synthesis. 12. What is the major site of ATP production in human cells? a. The mitochondrial matrix b. The cytoplasm c. The outer mitochondrial membrane d. The inner mitochondrial membrane Answer: d Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 1. Remembering Learning Objective: Explain chemiosmotic coupling. Feedback A: Incorrect. The citric acid cycle takes place in the mitochondrial matrix and generates some energy, but not the major portion of it. Feedback B: Incorrect. Glycolysis, which takes place in the cytoplasm, is only the first step in the metabolism of glucose, and it produces only a small fraction of the total energy. Feedback C: Incorrect. Oxidative phosphorylation is the process that produces the majority of energy in the cell, but it does not take place at the outer mitochondrial membrane. Feedback D: Correct! The inner mitochondrial membrane is the site of oxidative phosphorylation, which produces close to 90% of the ATP derived from glucose metabolism. 13. In plants, the molecule that captures energy from sunlight is a. NADPH. b. chlorophyll. c. chloroplast. d. chromatin. Answer: b Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Explain the role of chlorophyll in harvesting energy from sunlight. Feedback A: Incorrect. NADPH is a coenzyme and is generated during photosynthesis. Feedback B: Correct! Chlorophylls absorb photons from visible light in the light reaction of photosynthesis. This causes an electron to be transferred to a higher energy orbital, which eventually results in the generation of usable energy in the forms of ATP and NADPH. Feedback C: Incorrect. Chloroplasts are organelles within plant cells where photosynthesis is carried out. © 2019 Oxford University Press

Feedback D: Incorrect. This refers to a complex between DNA and histones. 14. Each photocenter in a chloroplast consists of hundreds of antenna pigment molecules that absorb light and transfer energy to a reaction center a. carotenoid. b. chlorophyll. c. cytochrome bf. d. plastoquinone. Answer: b Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Explain the role of chlorophyll in harvesting energy from sunlight. Feedback A: Incorrect. Feedback B: Correct! Feedback C: Incorrect. Feedback D: Incorrect. 15. Light is captured by _______ different photosystems associated with the thylakoid membranes of chloroplasts. a. two b. three c. five d. ten Answer: a Textbook Reference: Photosynthesis Bloom’s Category: 1. Remembering Learning Objective: Explain the role of chlorophyll in harvesting energy from sunlight. Feedback A: Correct! The two photosystems are photosystem I and photosystem II. Feedback B: Incorrect. There are two photosystems, photosystem I and photosystem II. Feedback C: Incorrect. There are two photosystems, photosystem I and photosystem II. Feedback D: Incorrect. There are two photosystems, photosystem I and photosystem II. 16. Which statement correctly describes the difference between the proton gradient across the inner mitochondrial membrane and the proton gradient across the thylakoid membrane of chloroplasts? a. The first creates an electrochemical gradient, while the second produces a mostly chemical gradient. b. There is no proton gradient across the thylakoid membrane; magnesium ions create the gradient instead. c. It is generated by an electron transport system in the first case but not the second. d. It drives the synthesis of ATP in the first case but not in the second. Answer: a Textbook Reference: Photosynthesis Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanisms used for generation of ATP and NADPH in chloroplasts.

Feedback A: Correct! The thylakoid membrane is permeable to chloride and magnesium ions, so electrical neutrality is maintained. Despite the smaller electrical component, the same amount of energy is stored in the gradient, due to the higher pH differential across it: 3.5 pH units, as opposed to 1.4 pH units for the inner mitochondrial membrane. Feedback B: Incorrect. Protons, not magnesium ions, create the gradient across the thylakoid membrane. Feedback C: Incorrect. In both cases the gradient is established by an electron transport system. Feedback D: Incorrect. ATP is produced in both cases. 17. The Calvin cycle is involved in the synthesis of _______ from _______. a. fructose; carbon dioxide b. glucose; carbon dioxide c. glucose; water d. maltose; malt Answer: b Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 1. Remembering Learning Objective: Summarize the reactions of the Calvin cycle. Feedback A: Incorrect. Glucose is the product of the Calvin cycle. Feedback B: Correct! The Calvin cycle converts carbon dioxide (CO 2) into glucose. Feedback C: Incorrect. The Calvin cycle converts carbon dioxide (CO2) into glucose. Feedback D: Incorrect. The Calvin cycle converts carbon dioxide (CO 2) into glucose. 18. Animal cells can synthesize glucose (gluconeogenesis) from all of the following except a. lactate. b. amino acids. c. glycerol. d. fatty acids. Answer: d Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between gluconeogenesis and glycolysis. Feedback A: Incorrect. Lactate derived from anaerobic glycolysis can be used in gluconeogenesis. Feedback B: Incorrect. Amino acids generated from the breakdown of proteins can be used in gluconeogenesis. Feedback C: Incorrect. Glycerol generated from lipid breakdown can be used in gluconeogenesis. Feedback D: Correct! Plants, but not animals, can use fatty acids to synthesize glucose. 19. Gluconeogenesis normally starts from a. amino acids. b. glycerol. c. lactate. d. All of the above Answer: d Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 2. Understanding © 2019 Oxford University Press

Learning Objective: Explain the difference between gluconeogenesis and glycolysis. Feedback A: Incorrect. This answer choice is correct, but the other answer choices are correct as well. Amino acids derived from the breakdown of proteins are only one of the normal starting molecules for gluconeogenesis. Feedback B: Incorrect. This answer choice is correct, but the other answer choices are correct as well. Glycerol derived from the breakdown of lipids is only one of the normal starting molecules for gluconeogenesis. Feedback C: Incorrect. This answer choice is correct, but the other answer choices are correct as well. Lactate produced by anaerobic glycolysis is only one of the normal starting molecules for gluconeogenesis. Feedback D: Correct! All of the above are normal starting molecules for gluconeogenesis. 20. Because fatty acids are synthesized by the stepwise addition of acetyl CoA derivatives, they differ in length by units of _______ carbon(s). a. 1 b. 2 c. 3 d. 4 Answer: b Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Describe the synthesis of lipids. Feedback A: Incorrect. The acetate group of acetyl CoA is a two-carbon unit. Therefore, the difference in length between various fatty acids is a multiple of two carbons. Feedback B: Correct! The acetate group of acetyl CoA is a two-carbon unit. Therefore, the difference in length between various fatty acids is a multiple of two carbons. Feedback C: Incorrect. The acetate group of acetyl CoA is a two-carbon unit. Therefore, the difference in length between various fatty acids is a multiple of two carbons. Feedback D: Incorrect. The acetate group of acetyl CoA is a two-carbon unit. Therefore, the difference in length between various fatty acids is a multiple of two carbons. 21. All organisms can use _______ as a nitrogen source. a. nitrate b. atmospheric nitrogen c. ammonia d. nitrite Answer: c Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. Feedback A: Incorrect. Only bacteria, fungi, and plants can use nitrate. Feedback B: Incorrect. Only nitrogen-fixing bacteria can use atmospheric nitrogen. Feedback C: Correct! Feedback D: Incorrect. Nitrite is chemically related to nitrate. 22. The carbon skeletons of amino acids are derived from a. the Calvin cycle. © 2019 Oxford University Press

b. gluconeogenesis. c. intermediates in glycolysis and the citric acid cycle. d. photosynthesis. Answer: c Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. Feedback A: Incorrect. The Calvin cycle in plants uses carbon dioxide to synthesize glucose. Feedback B: Incorrect. Gluconeogenesis uses a number of starting molecules to synthesize glucose. Feedback C: Correct! The 20 common amino acids are all derived from intermediates in glycolysis or the citric acid cycle. Feedback D: Incorrect. Photosynthesis is the process of capturing light energy. The relationship of photosynthesis to amino acid biosynthesis is quite distant. 23. In humans, how many amino acids must be taken up in the diet? a. 0 b. 9 c. 19 d. 20 Answer: b Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize how amino acids and proteins are synthesized. Feedback A: Incorrect. Not all amino acids can be synthesized by humans; some must be obtained from the diet. Feedback B: Correct! Humans lack the enzymes necessary to synthesize 9 of the 20 amino acids. Feedback C: Incorrect. Humans have the biosynthetic enzymes necessary to synthesize more than just a single amino acid. Feedback D: Incorrect. Some amino acids can be made in humans via biosynthetic reactions, and thus not all of them need to be obtained from the diet. 24. Antimetabolite chemotherapy succeeds as a treatment for childhood leukemia because a. cancer cells are dividing rapidly. b. cancer cells are qualitatively different from normal cells. c. normal cells are totally unaffected by the antimetabolite. d. normal cells are impermeable to the antimetabolite. Answer: a Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathways of nucleic acid synthesis. Feedback A: Correct! Cancer cells are often quantitatively different from normal cells (e.g., faster growing). Feedback B: Incorrect. Unfortunately, this is not the case. If it were, cancer chemotherapy would be much simpler. Feedback C: Incorrect. Unfortunately, this is not the case. Secondary side effects on normal cells do occur. © 2019 Oxford University Press

Feedback D: Incorrect. Normal cells and cancer cells typically have the same permeability properties. 25. Nucleotide chains are elongated in the 5-to-3 direction. The 5 and 3 notations refer to the chemical positions on the a. base. b. phosphate. c. ribose or deoxyribose. d. coenzyme that participates in polynucleotide biosynthesis. Answer: c Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathways of nucleic acid synthesis. Feedback A: Incorrect. The 5 and 3 refer to the positions of hydroxyl groups on the ribose or deoxyribose sugar. Feedback B: Incorrect. The 5 and 3 refer to the positions of hydroxyl groups on the ribose or deoxyribose sugar. Feedback C: Correct! The 5 and 3 refer to the positions of hydroxyl groups on the ribose or deoxyribose sugar. Feedback D: Incorrect. Coenzymes are not directly important to the biosynthesis of polynucleotides.

Essay 1. The conversion of A into B has a G of +13 kcal/mol, and the conversion of C into D has a G of –25 kcal/mol. If the two reactions are coupled through an enzyme, such that A + C goes to B + D, what is the free energy change, and in which direction will the reaction proceed? Answer: When two reactions are coupled, the change in free energy is the sum of the changes in free energy for each reaction. Thus, the G for the coupled reaction is –12 kcal/mol, and because the value is negative, the reaction proceeds as written (A + C goes to B + D), in the energetically favorable direction. Enzymes are often used to couple reactions in this way, such that the synthesis of substances that require energy occur at the expense of reactions that release energy, such as the hydrolysis of ATP. At physiological conditions, the hydrolysis of ATP to ADP + P i releases about –12 kcal/mol. Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 3. Applying Learning Objective: Explain how changes in Gibbs free energy determine the direction of chemical reactions. 2. Dihydroxyacetone is interconvertible with glyceraldehyde-3-phosphate. If this were not the case, what would be the net ATP yield from glycolysis? Answer: Although both dihydroxyacetone and glyceraldehyde-3-phosphate are products of glycolytic metabolism of glucose, only glyceraldehyde-3-phosphate is metabolized further in glycolysis. In total, for each glucose molecule metabolized through glycolysis, one molecule of dihydroxyacetone and one molecule of glyceraldehyde-3-phosphate are formed. The dihydroxyacetone is metabolized further through glycolysis only if converted to glyceraldehyde© 2019 Oxford University Press

3-phosphate. If this conversion did not occur, only the two molecules of ATP generated directly from the further glycolytic metabolism of glyceraldehyde-3-phosphate would be formed. If only two molecules of ATP were generated from glycolysis, then the net formation of ATP from glucose would be zero, as two molecules of ATP are consumed in the first portion of glycolysis. It is only because dihydroxyacetone is converted that four molecules of ATP are generated in the second half of glycolysis to give a net formation of two molecules of ATP per glucose. Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 4. Analyzing Learning Objective: Summarize the reactions of glycolysis and the citric acid cycle. 3. To harvest energy in the absence of oxygen, a cell breaks down glucose at a steady and rapid rate. If oxygen becomes available, the rate of glucose breakdown will decrease and be maintained at a much lower rate than in the absence of oxygen. Explain why this occurs. Answer: In the absence of oxygen, the cell ferments glucose to lactate, using the glycolysis pathway to generate ATP. In the presence of oxygen, the cell switches to oxidative phosphorylation, which generates ATP much more efficiently than glycolysis. Therefore, less glucose is needed to supply ATP at the same rate. Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 4. Analyzing Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 4. At low pH, the chemical 2,4-dinitrophenol (DNP) is neutral and can diffuse freely across membranes, including those of mitochondria. At high pH, it gives off a proton, becomes negatively charged, and can no longer diffuse across membranes. What effect would DNP have on ATP production by mitochondria? Answer: DNP would have the effect of dissipating the proton gradient across the inner membrane. It would be neutral on the outside and thus pass freely into the mitochondrial lumen, where it would encounter a high pH environment and release its proton. Thus the overall effect would be proton transport into the mitochondrial matrix. This would result in a halt in ATP production via oxidative phosphorylation, which depends on the proton gradient, and would drastically decrease the overall ATP production of the cell. Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 4. Analyzing Learning Objective: Explain chemiosmotic coupling. 5. Traditionally, it was assumed that ATP synthesis in mitochondria would be via high-energy phosphate group transfer intermediates, but none have ever been found. How does the chemiosmotic theory explain their absence? Answer: According to the chemiosmotic theory, the movement of protons down an electrochemical gradient across the inner mitochondrial membrane drives the formation of ATP. This occurs because the proton movement causes electrically driven shape changes in the F0 subunit of the ATP synthase. The F0 and F1 subunits are coupled. The F1 subunit of the motorlike ATP synthase catalyzes the synthesis of ATP from ADP and P i. The mechanism is entirely different from that of a high-energy phosphate intermediate. Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 4. Analyzing © 2019 Oxford University Press

Learning Objective: Explain chemiosmotic coupling. 6. Suppose you are visiting a local greenhouse and the greenhouse manager excitedly points out “the cool, white corn plant” nestled among the green ones. How large is the white plant likely to be? Explain your answer. Answer: The plant is very small. This is likely a mutant plant that lacks chlorophyll. Therefore, it cannot perform photosynthesis and is doomed to starve to death. Such albino mutants, if left to themselves, rarely grow to more than a few centimeters in height. Textbook Reference: Photosynthesis Bloom’s Category: 3. Applying Learning Objective: Explain the role of chlorophyll in harvesting energy from sunlight. 7. ATP synthase is composed of two complex proteins, F 0 and F1. What is the function of each protein complex, and where is each found in mitochondria? Answer: F0 is a proton channel, and the F1 complex is an ATP synthase. F0 is found in the mitochondrial inner membrane, and F1 is associated with F0 on the matrix face of the inner mitochondrial membrane. Textbook Reference: Glycolysis and Oxidative Phosphorylation Bloom’s Category: 3. Applying Learning Objective: Compare the mechanisms of ATP formation during glycolysis and oxidative phosphorylation. 8. Although none of the reactions in the Calvin cycle directly requires light to produce carbohydrates from CO2 and H2O, the cycle ceases in the dark. What is the reason? Answer: The Calvin cycle uses the ATP and NADPH synthesized in the light reactions to drive the synthesis of carbohydrates from CO2 and H2O, and thus depends on light for its continuation. Though the Calvin cycle can occur in the so-called “dark reaction,” its duration of operation is very limited. Textbook Reference: Metabolic Energy and ATP Bloom’s Category: 2. Understanding Learning Objective: Summarize the reactions of the Calvin cycle. 9. Pyruvate is a product of glycolytic metabolism of glucose. Pyruvate is also a major intermediate in glucose formation through gluconeogenesis. Why should gluconeogenesis not be considered the reverse of glycolysis? Answer: Though gluconeogenesis (illustrated in Figure 3.16 of the textbook) reverses many of the individual reactions of glycolysis, the overall process is not simply a reversal of glycolysis. Some of the reactions in glycolysis yield large quantities of energy and are thus essentially irreversible. The reversal of these reactions in gluconeogenesis requires the input of energy from ATP and/or consumption of NADH. These are not equilibrium reactions, and they are catalyzed by different enzymes than those of glycolysis. Hence, gluconeogenesis cannot be considered the reverse of glycolysis. Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between gluconeogenesis and glycolysis.

10. Suppose that in a cell culture experiment you add glucose uniformly labeled with carbon-14 to the culture medium. You then assay for the kinetics of carbon-14 incorporation into various amino acids. You find that carbon-14 is incorporated into histidine before it is incorporated into aspartate. Why? Answer: Amino acids derive their carbon skeletons from intermediates in glycolysis and the citric acid cycle. Histidine is derived from glucose-6-phosphate and aspartate is derived from oxaloacetate. Glucose-6-phosphate is separated from glucose by only one reaction, while oxaloacetate is separated by several reactions. Hence, the carbon-14 label from glucose is incorporated first into serine and later into aspartate. Textbook Reference: The Biosynthesis of Cell Constituents Bloom’s Category: 3. Applying Learning Objective: Summarize how amino acids and proteins are synthesized.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 4: Fundamentals of Molecular Biology TEST FILE QUESTIONS Multiple Choice 1. Gregor Mendel discovered the principles of genetics in his early experiments on a. peas. b. fruit flies. c. mice. d. humans. Answer: a Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Explain the relationship between genes and chromosomes. 2. During formation of eggs and sperm, one member of each chromosome pair is transmitted to each progeny cell in a type of cell division called a. mitosis. b. gametogenesis. c. cytokinesis. d. meiosis. Answer: d Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 2. Understanding Learning Objective: Explain the relationship between genes and chromosomes. 3. Animal cells and most cells of higher plants are a. haploid. b. diploid. c. euploid. d. tetraploid. Answer: b Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Explain the relationship between genes and chromosomes. 4. Eggs and sperm are a. haploid. b. diploid.

c. euploid. d. tetraploid. Answer: a Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Explain the relationship between genes and chromosomes. 5. Our fundamental knowledge about mutation, genetic linkage, and the relationships between genes and chromosomes was acquired largely from experiments on the a. bacterium Pneumococcus. b. fruit fly Drosophila. c. plant maize. d. pea plant. Answer: b Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Explain the relationship between genes and chromosomes. 6. DNA was identified as the genetic material in experiments done by a. Mendel. b. Chargaff. c. Avery, MacLeod, and McCarty. d. Watson and Crick. Answer: c Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Summarize the experiment that established DNA as the genetic material. 7. Deduction of the structure of DNA was based on X-ray diffraction patterns obtained by a. Avery, MacLeod, and McCarty. b. Pauling. c. Meselson and Stahl. d. Wilkins and Franklin. Answer: d Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of DNA. 8. DNA is a double helix with _______ base pairs per complete turn of the helix. a. 3.4 b. 10 c. 20 d. 34 Answer: b Textbook Reference: Heredity, Genes, and DNA

Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of DNA. 9. In the base pairing within a DNA double helix, a. purines pair with purines. b. A pairs with U and G pairs with C. c. A pairs with G and C pairs with T. d. A pairs with T and G pairs with C. Answer: d Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of DNA. 10. When thymidine and adenosine interact within a DNA molecule, they form a. two hydrogen bonds. b. three hydrogen bonds. c. nucleotide triphosphates. d. two phosphodiester bonds. Answer: a Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of DNA. 11. What is the complementary sequence to the following bases? 5ʹ-TCAAGG-3ʹ a. 5ʹ-AGTTCC-3ʹ b. 5ʹ-TCAAGG-3ʹ c. 5ʹ-CCTTGA-3ʹ d. 5ʹ-AGUUCC-3ʹ Answer: c Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 3. Applying Learning Objective: Diagram the structure of DNA. 12. DNA replicates a. conservatively. b. semiconservatively. c. liberally. d. by hybridization. Answer: b Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the experimental evidence for semiconservative replication.

13. In Meselson and Stahl’s experiment, cells were grown in media containing 15N in place of 14N, and then the cells’ DNA was separated by equilibrium centrifugation. This experiment demonstrated that DNA a. replication requires DNA polymerase. b. replication is conservative. c. replication is semiconservative. d. forms double helices by means of hydrogen bonding between base pairs. Answer: c Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the experimental evidence for semiconservative replication. 14. RNA molecules that serve as templates for protein synthesis are called _______ RNAs. a. messenger b. transfer c. ribosomal d. small nuclear Answer: a Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of mRNA, tRNA, and rRNA in protein synthesis. 15. Proteins are polymers of amino acids. How many different amino acids are found in proteins? a. 16 b. 20 c. 24 d. 36 Answer: b Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental evidence for a triplet code. 16. You have identified an unknown protein of approximately 1,500 amino acids. The minimum length of its mRNA is approximately a. 1,500 nucleotides. b. 4,500 nucleotides. c. 500 nucleotides. d. 750 nucleotides. Answer: c Textbook Reference: Expression of Genetic Information Bloom’s Category: 3. Applying Learning Objective: Summarize the experimental evidence for a triplet code.

17. A codon is a a. region of DNA coding for one protein. b. sequence of three nucleotides on a tRNA that binds to an mRNA. c. sequence of three nucleotides on an mRNA that binds to specific tRNAs. d. sequence of three nucleotides on the coding strand of DNA. Answer: c Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental evidence for a triplet code. 18. Because the genetic code has more than one codon for most amino acids, the code is said to be a. redundant. b. wobbly. c. a template. d. degenerate. Answer: d Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental evidence for a triplet code. 19. How many different RNA triplets specify amino acids? a. 36 b. 60 c. 61 d. 64 Answer: c Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental evidence for a triplet code. 20. Which of the following deletions will not change the reading frame of a gene? a. One nucleotide b. Two nucleotides c. Three nucleotides d. Four nucleotides Answer: c Textbook Reference: Expression of Genetic Information Bloom’s Category: 2. Understanding Learning Objective: Predict the effects of specific mutations on the amino acid sequence of an encoded protein. 21. Reverse transcriptase is an enzyme that catalyzes _______ synthesis. a. DNA-directed RNA b. RNA-directed DNA c. DNA-directed DNA

d. RNA-directed RNA Answer: b Textbook Reference: Expression of Genetic Information Bloom’s Category: 2. Understanding Learning Objective: Summarize the experimental evidence for reverse transcription. 22. The restriction enzyme NotI recognizes a specific DNA sequence of eight base pairs and cleaves at that site. Statistically, in any genome there should be one NotI recognition site for every _______ base pairs. a. 38 b. 83 c. 48 d. 84 Answer: c Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Predict the average sizes of DNA fragments produced by cleavage with a restriction endonuclease with a known recognition site. 23. Restriction endonucleases are enzymes that a. cleave DNA only at specific sequences. b. act only on the ends of DNA strands. c. act only in a single species of bacteria. d. cleave only nuclear DNA. Answer: a Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Summarize how a fragment of host DNA is cloned in a plasmid vector. 24. The first enzymatic step in making a cDNA uses a. Taq polymerase. b. DNA polymerase. c. RNA polymerase. d. reverse transcriptase. Answer: d Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Summarize how a fragment of host DNA is cloned in a plasmid vector. 25. To clone genes, a plasmid vector should contain all of the following except a. foreign DNA. b. an origin of replication. c. a restriction nuclease cut site. d. an antibiotic resistance gene.

Answer: a Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Summarize how a fragment of host DNA is cloned in a plasmid vector. 26. Which technique would be used to clone intact pieces of human DNA that are 400 kb pairs long? a. An E. coli plasmid b. Bacteriophage  c. Yeast artificial chromosomes d. Polymerase chain reaction Answer: c Textbook Reference: Recombinant DNA Bloom’s Category: 3. Applying Learning Objective: Summarize how a fragment of host DNA is cloned in a plasmid vector. 27. The most common method of DNA sequencing is based on premature termination of DNA synthesis by inclusion of a. dideoxynucleotides. b. endonuclease. c. ampicillin. d. puromycin. Answer: a Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Describe how DNA is sequenced with dideoxynucleotides. 28. Which vector would be used to express a human protein in E. coli? a. An expression plasmid containing cDNA made from mRNA for the human protein b. An expression plasmid containing an EcoRI-cut piece of the gene for the human protein c. The human gene for the protein inserted into E. coli by a bacteriophage  expression vector d. A yeast artificial chromosome containing the human gene for the protein Answer: a Textbook Reference: Recombinant DNA Bloom’s Category: 3. Applying Learning Objective: Identify the key features of a vector used to express cloned genes. 29. Vectors for expressing a eukaryotic protein in a bacterium must contain a. a eukaryotic promoter. b. antibody recognition sequences. c. an origin of replication. d. EcoRI restriction enzyme recognition sites.

Answer: c Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Identify the key features of a vector used to express cloned genes. 30. Expression of a human protein in a bacterial cell is best done with human a. cDNA in a bacterial expression vector. b. genomic DNA in a bacterial expression vector. c. cDNA in a virus-derived eukaryotic expression vector. d. genomic DNA in a virus-derived eukaryotic expression vector. Answer: a Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Identify the key features of a vector used to express cloned genes. 31. Taq polymerase is used in the polymerase chain reaction because it a. is stable in organic solvents. b. is stable at high temperatures. c. recognizes specific primer sequences. d. can make DNA sequences from RNA sequences. Answer: b Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 2. Understanding Learning Objective: Explain how DNA is amplified by the polymerase chain reaction (PCR). 32. Which pair of primers will amplify the template DNA sequence shown below in a PCR? 5ʹ-AGGCTGCGAT...//...CCGTTAGATT-3ʹ 3ʹ-TCCGACGCTA...//...GGCAATCTAA-5ʹ a. 5ʹ-AGGCTGCGAT-3ʹ and 5ʹ-CCGTTAGATT-3ʹ b. 5ʹ-AGGCTGCGAT-3ʹ and 5ʹ-AATCTAACGG-3ʹ c. 5ʹ-TCCGACGCTA-3ʹ and 5ʹ-GGCAATCTAA-3ʹ d. 5ʹ-TCCGACGCTA-3ʹ and 5ʹ-CCGTTAGATT-3ʹ Answer: b Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 3. Applying Learning Objective: Explain how DNA is amplified by the polymerase chain reaction (PCR). 33. Heat-denatured DNA will renature when cooled to about a. 95C. b. 65C. c. 37C. d. 20C. Answer: b

Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 1. Understanding Learning Objective: Explain how DNA is amplified by the polymerase chain reaction (PCR). 34. Nucleic acid hybridization is the a. mating of two genetically different organisms. b. formation of hybrid nuclei when two cells are fused. c. formation of a double-stranded molecule from the interaction of two complementarysequence single-stranded molecules. d. recombining of DNA molecules from two different organisms. Answer: c Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used to separate and detect fragments of DNA or molecules of RNA. 35. The technique by which fragments of DNA are separated by gel electrophoresis, denatured, blotted on nitrocellulose paper, labeled with a complementary radioactive probe, and detected by autoradiography is called _______ blotting. a. Southern b. Northern c. Western d. Eastern Answer: a Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 1. Remembering Learning Objective: Summarize the methods used to separate and detect fragments of DNA or molecules of RNA. 36. Which method would be used to determine if drug treatment of a cell culture led to increased expression of mRNA? a. Southern blotting b. Northern blotting c. Western blotting d. Eastern blotting Answer: b Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 3. Applying Learning Objective: Summarize the methods used to separate and detect fragments of DNA or molecules of RNA. 37. The technique by which a particular nucleic acid sequence is localized with labeled probes within a cell, tissue, or isolated chromosome is called _______ hybridization. a. in vivo b. in vitro

c. in situ d. in toto Answer: c Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used to separate and detect fragments of DNA or molecules of RNA. 38. To identify the protein against which a given sample has an antibody, the best technique to use would be _______ blotting. a. Southern b. Northern c. Western d. Eastern Answer: c Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 3. Applying Learning Objective: Describe how antibodies are used to detect proteins. 39. SDS-polyacrylamide gel electrophoresis (SDS-PAGE) separates proteins by a. negative charge. b. positive charge. c. charge: size ratio. d. size. Answer: d Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 2. Understanding Learning Objective: Describe how antibodies are used to detect proteins. 40. Retroviral vectors can be advantageous over liposomes for stable introduction of DNA into animal cells because a. retrovirus infected cells are more readily selectable. b. one step in the viral life cycle is integration of its DNA into the host genome. c. they allow insertion of larger DNA molecules. d. All of the above Answer: b Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Distinguish transient expression from stable transformation of animal cells in culture. 41. The experimental introduction of DNA into eukaryotic cells is called a. infection. b. transfection. c. transient expression. d. transcription.

Answer: b Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Distinguish transient expression from stable transformation of animal cells in culture. 42. DNA can be introduced into cells by a. microinjection into the nucleus of a cell. b. fusion of DNA-containing liposomes with cells. c. infection with engineered retroviruses. d. All of the above Answer: d Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Distinguish transient expression from stable transformation of animal cells in culture. 43. Most eukaryotic cells transfected with DNA using liposomes will only express the introduced genes for several days because a. foreign DNA will be targeted by CRISPR/Cas. b. foreign RNA will be targeted by antisense RNAs. c. the introduced DNA will not be replicated with the cells’ chromosomes. d. miRNAs will block translation of the foreign mRNA. Answer: c Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Distinguish transient expression from stable transformation of animal cells in culture. 44. Transgenic mice are usually produced by a. infecting mice with viruses carrying the gene. b. injecting DNA containing the gene into the pronucleus of a fertilized egg. c. injecting nuclei from embryonic stem cells into enucleated eggs. d. injecting RNA coding for a human protein into the nucleus of a fertilized egg. Answer: b Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used to introduce mutations into homologous cellular genes. 45. Mutations may be introduced into selected genes of living cells by a. homologous recombination with an altered cloned gene. b. hybridization with antisense nucleic acid sequences. c. the polymerase chain reaction. d. ultraviolet laser irradiation. Answer: a

Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Summarize the methods used to introduce mutations into homologous cellular genes. 46. The CRISPR/Cas system has revolutionized the efficiency with which scientists can mutate a gene in vivo to investigate the functional role of the protein it encodes. Which component is not essential in the CRISPR/Cas system? a. Guide RNA complementary to gene of interest b. Access to target mRNA c. Cas9 nuclease d. Access to target gene Answer: b Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Describe the CRISPR/Cas system. 47. When introduced into living cells, antisense RNA molecules bind a. to complementary DNA sequences and block RNA synthesis. b. to the sense strand of DNA and block RNA synthesis. c. to complementary mRNA sequences and block protein synthesis. d. proteins called RISC and cleave the complementary mRNA. Answer: c Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between antisense RNA and RNA interference. 48. In RNA interference, expression of a gene is most effectively inhibited by injection of a. antisense RNA complementary to its mRNA. b. sense RNA complementary to its mRNA. c. double-stranded DNA. d. double-stranded RNA. Answer: d Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between antisense RNA and RNA interference. Fill in the Blank 1. RNA is copied from DNA through a process called _______. Answer: transcription Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of mRNA, tRNA, and rRNA in protein synthesis.

2. Proteins are synthesized from RNA templates through a process called _______. Answer: translation Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of mRNA, tRNA, and rRNA in protein synthesis. 3. In addition to mRNA, the two types of RNA necessary for protein synthesis are _______ and _______. Answer: tRNA; rRNA Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of mRNA, tRNA, and rRNA in protein synthesis. 4. The central dogma of molecular biology is _______  _______  _______. Answer: DNA; RNA; protein Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of mRNA, tRNA, and rRNA in protein synthesis. 5. The three codons that do not specify amino acids serve as _______. Answer: stop codons Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental evidence for a triplet code. 6. A poly-U RNA template codes only for the amino acid _______. Answer: phenylalanine Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental evidence for a triplet code. 7. RNA viruses that replicate via a DNA intermediate are called _______. Answer: retroviruses Textbook Reference: Expression of Genetic Information Bloom’s Category: 2. Understanding Learning Objective: Summarize the experimental evidence for reverse transcription. 8. The enzyme discovered by Temin, Mizutani, and Baltimore, which synthesizes DNA from an RNA template, is called _______ Answer: reverse transcriptase (or RNA-dependent DNA polymerase) Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental evidence for reverse transcription.

9. A schematic representation of a length of DNA showing the locations of cleavage sites for multiple different restriction endonucleases is called a _______. Answer: restriction map Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Predict the average sizes of DNA fragments produced by cleavage with a restriction endonuclease with a known recognition site. 10. The DNA made from an mRNA is called _______. Answer: cDNA Textbook Reference: Recombinant DNA Bloom’s Category: 1. Remembering Learning Objective: Summarize how a fragment of host DNA is cloned in a plasmid vector. 11. Methods of sequencing DNA that have made it possible to sequence whole genomes faster and less expensively than the Sanger dideoxy method are collectively referred to as . Answer: next-generation sequencing Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Describe how DNA is sequenced with dideoxynucleotides. 12. Eukaryotic genes expressed in bacteria produce products that differ from the eukaryotic protein in that they do not have _______ modification. Answer: posttranslational Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Identify the key features of a vector used to express cloned genes. 13. A single species of antibodies produced from a cultured clone of B lymphocytes is called a(n) _______. Answer: monoclonal antibody Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 1. Remembering Learning Objective: Describe how antibodies are used to detect proteins. 14. When foreign DNA is injected into the nucleus of a fertilized mouse egg and integrated into the mouse genome, the egg can develop into a(n) _______ mouse. Answer: transgenic Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used to introduce mutations into homologous cellular genes. 15. A cell that can give rise to an entire animal is called a(n) _______ cell.

Answer: embryonic stem Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used to introduce mutations into homologous cellular genes. 16. Site-specific mutations can be introduced into cultured animal cells by a technique that relies on _______ recombination. Answer: homologous Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Summarize the methods used to introduce mutations into homologous cellular genes. 17. CRISPR were discovered in bacteria, where they are used as a defense against foreign genetic elements, such as _______. Answer: viruses Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Describe the CRISPR/Cas system. 18. The use of double-stranded RNA to inhibit expression of a gene with a complementary sequence is called _______. Answer: RNA interference Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between antisense RNA and RNA interference.

True/False 1. Genes that do not normally separate from one another during meiosis are on the same chromosome. Answer: T Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 2. Understanding Learning Objective: Explain the relationship between genes and chromosomes. 2. Meiosis results in a change of nuclear DNA from haploid to diploid. Answer: F Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 2. Understanding Learning Objective: Explain the relationship between genes and chromosomes.

3. Avery, MacLeod, and McCarty showed that one type of bacteria could be transformed into another type by uptake of purified RNA. Answer: T Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Summarize the experiment that established DNA as the genetic material. 4. Erwin Chargaff found that in DNA, the amount of A is equal to that of T and the amount of G is equal to that of C. Answer: T Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of DNA. 5. DNA contains the base uracil, whereas RNA contains the base thymine. Answer: F Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of DNA. 6. Vectors usually contain an antibiotic-resistance gene so that cells that incorporate the vector can be selected by their resistance to the antibiotic. Answer: T Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Summarize how a fragment of host DNA is cloned in a plasmid vector. 7. cDNA made from genomic DNA is used to create cDNA libraries. Answer: F Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 1. Remembering Learning Objective: Explain how molecular cloning allows a unique fragment of DNA to be isolated from a mixture. 8. One method of DNA sequencing involves incorporation of dideoxynucleotides during synthesis of new DNA. Answer: T Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Describe how DNA is sequenced with dideoxynucleotides. 9. A cDNA is especially useful for bacterial expression of a eukaryotic protein. Answer: T Textbook Reference: Detection of Nucleic Acids and Proteins

Bloom’s Category: 2. Understanding Learning Objective: Identify the key features of a vector used to express cloned genes. 10. The polymerase chain reaction is used to amplify specific fragments of DNA in vitro. Answer: T Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 1. Remembering Learning Objective: Explain how DNA is amplified by the polymerase chain reaction (PCR). 11. Embryonic stem cells introduced into early mouse embryos can give rise to cells in all of the tissues of the mouse, including germ cells. Answer: T Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used to introduce mutations into homologous cellular genes. 12. The CRISPR/Cas system can knock out the expression of a gene in vivo and/or introduce a new mutation into a gene. Answer: T Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Describe the CRISPR/Cas system.

Short Answer 1. Considering Mendel’s findings on the genetics of peas, identify the genotypes of two yellow parent plants that produced a green offspring. Answer: Both parents would have to be heterozygous carriers of the recessive allele, and both of their genotypes must be Yy. Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 3. Applying Learning Objective: Explain the relationship between genes and chromosomes. 2. Consider Mendel’s experiments on the genetics of peas. What is the probability that two green parent pea plants would also have green offspring? Answer: The probability is 100%. All would be green because green is recessive, so the genotype of both parents must be yy; no dominant allele (Y) can be present. Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 3. Applying Learning Objective: Explain the relationship between genes and chromosomes.

3. Consider this statement: An increase in the levels of a protein within a cell indicates that expression of the gene encoding it was also increased. Why is the statement not necessarily true? Answer: The statement may be true, but a number of factors can give rise to an increase in the level of a protein. For example, as stated, the expression of the gene could be responsible, but stabilization of the mRNA could increase the protein as well. Think about the central dogma and other factors that can increase protein levels. Textbook Reference: Expression of Genetic Information Bloom’s Category: 3. Applying Learning Objective: Describe the roles of mRNA, tRNA, and rRNA in protein synthesis. 4. Addition or deletion of one or two nucleotides in the coding part of a gene produces a nonfunctional protein, whereas addition or deletion of three nucleotides often results in a protein with nearly normal function. Explain. Answer: The addition or deletion of one or two nucleotides would alter the reading frame, resulting in a completely different amino acid sequence, whereas a change of three nucleotides affects only one amino acid, and this frequently yields a protein that functions normally. Textbook Reference: Expression of Genetic Information Bloom’s Category: 2. Understanding Learning Objective: Predict the effects of specific mutations on the amino acid sequence of an encoded protein. 5. What is unique about RNA tumor viruses (retroviruses) that allows their RNA genome to be replicated and inserted as DNA into host cells? Answer: Retroviruses encode a very unique enzyme, called reverse transcriptase, that uses an RNA template to synthesize DNA copies. This is widespread knowledge now, but at the time it was controversial because it flew in the face of the then accepted model of the “central dogma” in which the flow of information in genetic material is from DNA to RNA to protein. Textbook Reference: Expression of Genetic Information Bloom’s Category: 3. Applying Learning Objective: Summarize the experimental evidence for reverse transcription. 6. EcoRI restriction mapping on a sample of a large piece of DNA resulted in six distinct bands following agarose gel electrophoresis. Similar mapping on an identical sample that was subjected to high intensity ultraviolet light resulted in only four distinct bands. What might best explain the reduction in the number of detectable bands in the UV light-treated sample? Answer: EcoRI specifically cleaves DNA at GAATTC sequences. In the first experiment, there must have been five EcoRI sites. Following UV treatment, in which only four bands were detectable, there were only three EcoRI sites. Therefore, the most likely explanation is that the UV light causes mutations in at least two of the GAATTC EcoRI sites. If any one of the six nucleotides in the recognition sequence becomes mutated, that site will not be recognized by the restriction enzyme. Textbook Reference: Recombinant DNA

Bloom’s Category: 4. Analyzing Learning Objective: Predict the average sizes of DNA fragments produced by cleavage with a restriction endonuclease with a known recognition site. 7. Why are dideoxynucleotides used in DNA sequencing? Answer: Deoxyribonucleotides have hydrogen at the 2ʹ position and a hydroxyl group on the 3ʹ position of the ribose sugar. The hydroxyl group is absolutely required to form the covalent bond with the incoming nucleotide. With the dideoxy, there is no 3ʹ hydroxyl, so the growth of the chain is terminated. Textbook Reference: Recombinant DNA Bloom’s Category: 3. Applying Learning Objective: Describe how DNA is sequenced with dideoxynucleotides. 8. If a plasmid does not contain an origin of replication, what would happen if you were to transfect it into bacteria? Answer: Without an origin of replication, it cannot be replicated. It would get into the bacteria successfully, but as the bacteria reproduced, the plasmid would get diluted out in progeny until no new bacteria contained the plasmid. Textbook Reference: Recombinant DNA Bloom’s Category: 4. Analyzing Learning Objective: Identify the key features of a vector used to express cloned genes. 9. Suppose you have a cell line that expresses a protein whose structure you know but not the function. Describe an experiment that would help you determine the protein’s function. Answer: There are several possible experiments. For example, you could develop or obtain an antibody to the protein and inject it in cells and observe changes in the cells after injection. Alternatively, you could use antisense RNA or RNA interference to block expression of the protein from the mRNA encoding the protein and observe changes. Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 4. Analyzing Learning Objective: Explain the difference between antisense RNA and RNA interference. 10. Suppose you have a cell line that expresses a protein whose structure you know but not the function. You use RNA interference to block expression of the protein. How could you confirm that the interference was effective? Answer: Since you know the structure of the protein, you can procure an antibody. So the easiest way to confirm efficacy of the RNA interference would be to do a Western blot on the cellular proteins isolated from two samples, treated and untreated cells, and evaluate whether the level of the protein is lower in cells subjected to RNA interference. Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 4. Analyzing Learning Objective: Explain the difference between antisense RNA and RNA interference.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. Which statement correctly represents the distinction between the terms “genotype” and “phenotype”? a. Phenotype is the genetic composition of an organism, and genotype is the physical appearance. b. There is no difference between a genotype and a phenotype. c. A phenotype is dominant, and a genotype is recessive. d. The genotype is the genetic composition of an organism, and the phenotype is its physical appearance. Answer: d Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Explain the relationship between genes and chromosomes. Feedback A: Incorrect. The opposite is true. The genotype is the genetic composition, and the phenotype is the physical appearance. Feedback B: Incorrect. Genotype and phenotype are different descriptors. Feedback C: Incorrect. Alleles are described as dominant or recessive, based on inheritance patterns. Feedback D: Correct! For example, a gene specifying seed color can be denoted (Y) for yellow and (y) for green. Yellow (Y) is said to be dominant, and green (y) is said to be recessive. In this example, (YY) and (Yy) are both genotypes that confer a yellow phenotype. The genotype (yy), in contrast, confers a green phenotype. 2. In Gregor Mendel’s experiments on pea genetics, he determined that yellow seed color, Y, was dominant over green seed color, y. In a cross between two plants, one with yellow seeds and the other with green seeds, all the seeds produced from the offspring were yellow. What was the genotype of the parent plant with yellow seeds? a. YY b. Yy c. yY d. yy Answer: a Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 3. Applying Learning Objective: Explain the relationship between genes and chromosomes. Feedback A: Correct! To be yellow, the offspring need only one dominant Y allele. For all offspring to be yellow, the yellow parent would have to have two copies of the dominant allele (homozygous). Feedback B: Incorrect. If the yellow parent plant had one of each allele, the offspring would be expected to be 50% yellow and 50% green. Feedback C: Incorrect. If the yellow parent plant had one of each allele, the offspring would be expected to be 50% yellow and 50% green.

Feedback D: Incorrect. If both alleles are recessive, y, the parent plant could only produce green seeds. 3. The two strands of DNA in the double helix are held together by a. interactions between the sugar residues of each chain. b. interactions between phosphate residues of each strand. c. hydrogen bonds between the bases of each strand. d. covalent bonds between the bases of each strand. Answer: c Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of DNA. Feedback A: Incorrect. The sugar residues lie on the outside and do not contribute to bonding between the two DNA strands. Feedback B: Incorrect. The phosphate residues are negatively charged and repel each other. This repulsion is counterbalanced by a stronger attractive force. Feedback C: Correct! Hydrogen bonds between adenines and thymines and between cytosines and guanines of opposite strands hold the double helix together. Feedback D: Incorrect. Covalent bonds are formed within a strand of DNA but not between the complementary strands. 4. The process by which proteins are made from RNA templates is called a. transcription. b. translation. c. transposition. d. interpretation. Answer: b Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of mRNA, tRNA, and rRNA in protein synthesis. Feedback A: Incorrect. Transcription is the synthesis of RNA from DNA templates. Feedback B: Correct! Translation is the process by which an RNA sequence (made up of nucleotides) is translated into a protein sequence (made of up amino acids). Feedback C: Incorrect. Transposition refers to the movement of DNA within a genome. Feedback D: Incorrect. This is not an official process in cell biology. 5. Which molecule serves as an adaptor between amino acids and mRNA during translation? a. rRNA b. DNA c. tRNA d. siRNA Answer: c Textbook Reference: Expression of Genetic Information Bloom’s Category: 2. Understanding Learning Objective: Describe the roles of mRNA, tRNA, and rRNA in protein synthesis.

Feedback A: Incorrect. Ribosomal RNA (rRNA) is a component of ribosomes. Feedback B: Incorrect. DNA serves as a template for synthesis of RNA molecules. Feedback C: Correct! Transfer RNAs (tRNAs) serve as adaptor molecules that align amino acids along the mRNA template. Each amino acid is attached by a specific enzyme to its appropriate tRNA, which contains a recognition sequence that base pairs to a complementary sequence on the mRNA. In this way, the tRNA directs its attached amino acid to its correct position on the mRNA template. Feedback D: Incorrect. Short interfering RNAs (siRNAs) are derived from cleavage of double-stranded RNA and are involved in RNA interference rather than translation. 6. The enzyme that synthesizes DNA using RNA templates is called a. DNA polymerase. b. RNA polymerase. c. ribozymes. d. reverse transcriptase. Answer: d Textbook Reference: Expression of Genetic Information Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental evidence for reverse transcription. Feedback A: Incorrect. DNA polymerases synthesize DNA using a DNA template. Feedback B: Incorrect. RNA polymerases synthesize RNA from DNA templates. Feedback C: Incorrect. Ribozymes are enzymes made up of RNA. Feedback D: Correct! Reverse transcriptase is found in RNA tumor viruses and catalyzes the synthesis of a DNA copy of the RNA genome. The DNA copy can then be stably integrated into the host genome, an event that can initiate the development of cancer. 7. A restriction enzyme with a four-base recognition site would cleave DNA with a statistical frequency of once every a. 4 base pairs. b. 256 base pairs. c. 4,096 base pairs. d. 65.5 kilobases. Answer: b Textbook Reference: Recombinant DNA Bloom’s Category: 3. Applying Learning Objective: Predict the average sizes of DNA fragments produced by cleavage with a restriction endonuclease with a known recognition site. Feedback A: Incorrect. Because the bases in DNA occur in an irregular order, the recognition sequence would be encountered less frequently than every four base pairs. Feedback B: Correct! Given that the genetic code makes use of four bases, the sequence would randomly occur every 4 × 4 × 4 × 4 base pairs (every 256 base pairs). Feedback C: Incorrect. This would be the frequency for a six-base recognition site. Feedback D: Incorrect. This would be approximately the frequency for an eight-base recognition site. 8. Why can’t a digest of human genomic DNA be resolved by gel electrophoresis?

a. Once cut by the restriction enzyme, the fragments are unstable b. A typical digest of the human genome produces too many fragments to resolve c. The fragments will be too large to move through the gel d. The fragments will all be the same size Answer: b Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 2. Understanding Learning Objective: Predict the average sizes of DNA fragments produced by cleavage with a restriction endonuclease with a known recognition site. Feedback A: Incorrect. In the absence of exonucleases, DNA fragments are very stable. Feedback B: Correct! Digestion of the human genome with a restriction enzyme that recognizes a 6 base sequence will produce roughly 500,000 fragments, which will produce a smear of DNA in the gel rather than discrete bands. Feedback C: Incorrect. Restriction enzymes target sequences are randomly located within the genome, yielding fragments that span a large range of different sizes. Feedback D: Incorrect. Restriction enzymes target sequences are randomly located within the genome, yielding fragments that span a large range of different sizes. 9. The characterization of restriction endonucleases was a key step in the development of recombinant DNA technology. What is the function of these enzymes? a. Synthesis of complementary strands of DNA from a DNA template b. Synthesis of RNA from a DNA template c. Ligation of two digested DNA strands d. Cleavage of DNA at specific sequences Answer: d Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Summarize how a fragment of host DNA is cloned in a plasmid vector. Feedback A: Incorrect. DNA polymerase directs the incorporation of nucleotides into a complementary DNA molecule. Feedback B: Incorrect. RNA is transcribed from DNA by RNA polymerase. Feedback C: Incorrect. This is an important step in the generation of recombinant DNA molecules, but it is carried out by DNA ligase, which seals breaks in DNA strands. Feedback D: Correct! These enzymes, identified in bacteria, cleave DNA at distinct recognition sites four to eight base pairs in length. They can be used to cleave a DNA molecule at unique sites. 10. Which character is not necessary in a plasmid vector to facilitate successful cloning? a. Origin of replication b. Gene conferring resistance to antibiotic c. Restriction site d. Strong bacterial promoter Answer: d Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding

Learning Objective: Summarize how a fragment of host DNA is cloned in a plasmid vector. Feedback A: Incorrect. An origin of replication is a DNA sequence that signals the host cell DNA polymerase to replicate the DNA molecule. An origin of replication is required on all plasmid vectors to ensure that the vector, along with the gene of interest, is replicated within the host cell. Feedback B: Incorrect. Antibiotic resistance genes are required and included as a means of selecting bacteria that carry the plasmid. Selection is accomplished by plating bacteria on media with the antibiotic to which resistance was conferred. Only bacteria that have the plasmid are resistant to the antibiotic and will survive, forming colonies. Feedback C: Incorrect. Restriction sites are required for insertion of foreign DNA. Feedback D: Correct! A strong bacterial promoter is needed for protein expression, but not for cloning of a DNA sequence. 11. A cDNA molecule is a. DNA synthesized from an RNA template using the enzyme reverse transcriptase. b. the nucleic acid portion of a ribosome. c. a DNA molecule capable of independent replication in a host cell. d. RNA or single-stranded DNA, complementary to the mRNA of a gene of interest, that hybridizes with the mRNA and blocks its translation. Answer: a Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Explain how molecular cloning allows a unique fragment of DNA to be isolated from a mixture. Feedback A: Correct! Reverse transcriptases catalyze the synthesis of DNA from an RNA template. These enzymes can be used experimentally to synthesize a DNA copy of an RNA sequence. The resulting molecule is named cDNA because it is complementary to the RNA template. Feedback B: Incorrect. Ribosomal RNA (rRNA) is a component of ribosomes. Feedback C: Incorrect. A vector is a DNA molecule capable of independent replication in a host cell. Examples of vectors are plasmids, cosmids, and artificial chromosomes. Feedback D: Incorrect. This describes antisense nucleic acids. 12. A PCR reaction does not require a. a heat stable DNA polymerase. b. DNA containing the sequence to be amplified. c. short single stranded DNA primers complementary to each end of the target sequence. d. sodium dodecyl sulfate. Answer: d Textbook Reference: Recombinant DNA Bloom’s Category: 2. Understanding Learning Objective: Explain how DNA is amplified by the polymerase chain reaction (PCR). Feedback A: Incorrect. A heat stable DNA polymerase is required to synthesize new DNA under PCR conditions

Feedback B: Incorrect. DNA containing the sequence of interest is required so that this sequence can be copied. Feedback C: Incorrect. DNA polymerases require these sequences to target them to the correct location in template DNA and give them a free 3ʹ hydroxyl group to attach new nucleotides to. Feedback D: Correct! SDS is useful for purifying DNA template, but it inhibits polymerase, and thus interferes with PCR reactions. 13. How many molecules would be produced from two molecules of DNA following five rounds of PCR amplification? a. 10 b. 32 c. 64 d. One billion Answer: c Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 3. Applying Learning Objective: Explain how DNA is amplified by the polymerase chain reaction (PCR). Feedback A: Incorrect. PCR amplification is exponential; more than 10 molecules would be produced. Feedback B: Incorrect. This would be the number produced starting with one molecule of DNA. Feedback C: Correct! Each starting molecule of DNA would double during each round of PCR [(2 molecules)  2  2  2  2  2 = 64; or twice (1 molecule)  2  2  2  2 2 = 32]. Feedback D: Incorrect. This would be approximately the number produced from 30 rounds of PCR starting with a single DNA molecule. 14. What technique would you use to determine if the level of a specific mRNA had been increased in response to an inducer? a. Northern blotting b. Southern blotting c. Western blotting d. Immunoprecipitation Answer: a Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 3. Applying Learning Objective: Summarize the methods used to separate and detect fragments of DNA or molecules of RNA. Feedback A: Correct! Northern blotting uses a radioactive probe to detect cellular RNAs separated out by size on a blot. The amount of radioactivity reflects the amount of RNA in the sample, so samples can be compared to determine the effect of the inducer. Feedback B: Incorrect. Southern blotting is used to detect genes in cellular DNA. Feedback C: Incorrect. Western blotting is used to detect proteins, not RNA molecules. Feedback D: Incorrect. This is used in the detection of proteins, not RNAs.

15. The SDS in SDS-polyacrylamide gel electrophoresis (SDS-PAGE) of a protein is used to a. neutralize the charge of a protein so that charge is not a factor in its rate of migration. b. denature the protein and give it an overall negative charge. c. cross-link the protein’s structure so that it does not become denatured during electrophoresis. d. digest the protein into smaller fragments that are more easily separated by electrophoresis. Answer: b Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 2. Understanding Learning Objective: Describe how antibodies are used to detect proteins. Feedback A: Incorrect. SDS does not neutralize charges. Feedback B: Correct! This eliminates the differences in shape between proteins as a factor in their migration rates so that they migrate according to size. Feedback C: Incorrect. SDS is not a cross-linking agent. Feedback D: Incorrect. SDS is not a protease and thus does not digest a protein into smaller fragments. 16. Transgenic mice carry a foreign gene a. only in their germ line cells. b. only in their somatic cells. c. in some cells (both germ line and somatic) but not in others—transgenic mice are chimeras. d. in all of their cells. Answer: d Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used to introduce mutations into homologous cellular genes. Feedback A: Incorrect. The foreign gene is present in the germ line cells of transgenic mice, but not exclusively so. Feedback B: Incorrect. The foreign gene is present in the somatic cells of transgenic mice, but not exclusively so. Feedback C: Incorrect. Transgenic mice are not chimeras. Feedback D: Correct! When a transgenic mouse is made, a foreign gene is introduced into a fertilized egg, which then goes on to develop into an entire mouse, and thus all cells in the mouse carry the gene. 17. Transgenic plants are easier to produce than transgenic animals because a. plants can be grown more easily from single cultured cells into which recombinant DNA has been introduced. b. plant DNA can be cloned more easily. c. plant cells can be transformed by bacterial infection. d. DNA passes more readily through the plant cell wall than through the animal cell

membrane. Answer: a Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used to introduce mutations into homologous cellular genes. Feedback A: Correct! Plant cells exhibit the ability to completely regenerate a new plant from single cells much more frequently than animal cells do, so it is relatively easy to generate an entire plant from a single cell. Feedback B: Incorrect. Plant DNA cannot be cloned more easily than any other kind of DNA. Feedback C: Incorrect. This is true, but it does not explain the relative ease with which transgenic plants are generated. Feedback D: Incorrect. Plant cells possess both a cell wall and a cell membrane, and thus DNA passes less readily into the plant cell than into an animal cell. 18. Which technique is used to inactivate a gene by altering the DNA that encodes it? a. Homologous recombination b. Antisense nucleic acid blocks c. Antibody microinjection d. Use of siRNA Answer: a Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used to introduce mutations into homologous cellular genes. Feedback A: Correct! Mutant versions of a gene can be made to replace the endogenous, wild-type copy by homologous recombination. Feedback B: Incorrect. This technique targets a gene’s mRNA and does not alter gene structure. Feedback C: Incorrect. This technique targets the protein encoded by the gene and does not affect gene structure. Feedback D: Incorrect. siRNA functions by causing degradation of specific mRNAs. 19. The CRISPR/Cas system is a fast and powerful method to introduce specific mutations into cellular genomes via homologous recombination to study the function of genes and the proteins they encode. Which reagents are necessary for CRISPR/Cas? a. Antibody to DNA nuclease, Cas9 b. Mutant copy of target gene, guide RNA, and Cas9 c. Antisense RNA complementary to target gene, Cas9 d. siRNA complementary to the target mRNA, Cas9 Answer: b Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Describe the CRISPR/Cas system. Feedback A: Incorrect. Antibodies can be injected into cells to inhibit proteins, but not to

alter genes. Feedback B: Correct! These three reagents work together with intracellular nucleases to cut out and replace sequences of DNA. Feedback C: Incorrect. Antisense RNA can inhibit translation of mRNAs but does not have any effect on the gene encoding the mRNA. Feedback D: Incorrect. siRNAs are another way to inhibit expression of mRNAs but not genes. 20. Which sequence correctly outlines the process of RNA interference, beginning with a double-stranded RNA molecule? a. Unwinding of siRNA; cleavage by Dicer; association with RISC; pairing with target mRNA; mRNA cleavage b. Reverse transcription of RNA to cDNA; addition of oligonucleotide linkers containing restriction endonuclease cleavage sites; cDNA ligation to appropriate vector; introduction of recombinant molecule into E. coli c. Cleavage by Dicer; association with RISC; unwinding of siRNA; pairing with target mRNA; mRNA cleavage d. Reverse transcription of RNA to cDNA; double-stranded DNA heated to separate strands; cooling to allow primers to bind to each strand of DNA; synthesis of new DNA molecules by Taq polymerase Answer: c Textbook Reference: Gene Function in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Explain the difference between antisense RNA and RNA interference. Feedback A: Incorrect. Double-stranded RNA is first cleaved by the enzyme Dicer into short interfering RNAs (siRNAs). The siRNAs then associate with the RNA-induced silencing complex (RISC). Within this complex, the two strands of siRNA are unwound, and the strand complementary to the mRNA guides the complex to the target mRNA by complementary base pairing. Feedback B: Incorrect. This answer choice correctly outlines the process of cloning an RNA sequence but does not outline the process of RNA interference. Feedback C: Correct! Double-stranded RNAs are cleaved into short interfering RNAs by the enzyme Dicer. The siRNAs associate with the RNA-induced silencing complex (RISC) and are unwound. The antisense strand of siRNA then targets RISC to a homologous mRNA, which is cleaved by one of the RISC proteins. Feedback D: Incorrect. This answer choice correctly outlines the process of amplifying an RNA sequence using PCR. This process is not related to RNA interference.

Essay 1. If adenine makes up 22% of the bases in a given double-stranded DNA molecule, what percentage is made up of guanines? Answer: If adenine makes up 22%, then thymine would also make up 22% because of complementary base pairing. That would leave 56% of the DNA to cytosines and

guanines, which must be present in equal amounts, again due to the complementarity between them. Thus, the percentage of guanines would be 56/2, or 28%. Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 3. Applying Learning Objective: Diagram the structure of DNA. 2. In experiments using polymerase chain reactions (PCR), it is often more difficult to amplify through regions of DNA that are high in GC content versus those regions that are either lower in GC content or are AT-rich. Based on your knowledge of DNA structure, explain why. Answer: In base pairing between strands of DNA, there are three hydrogen bonds between each G-C base pair and only two hydrogen bonds between A-T base pairs. In PCR, heat is used to denature the hydrogen bonds between strands of DNA. The energy required to denature regions of DNA high in GC are significantly higher than those that are either less so or are AT-rich. Textbook Reference: Heredity, Genes, and DNA Bloom’s Category: 4. Analyzing Learning Objective: Diagram the structure of DNA. 3. In the study of molecular biology one frequently hears references to the “central dogma” of molecular biology. What is meant by this term? Answer: The “central dogma” refers to the critical flow of information from DNA to protein. It includes the molecular processes regulating the transcription of DNA into RNA and similarly the regulation of RNA into protein. Although it is a simple concept, understanding these processes allows one to think critically about mechanisms that can go awry and lead to alterations, mutations, and disease. Textbook Reference: Expression of Genetic Information Bloom’s Category: 2. Understanding Learning Objective: Describe the roles of mRNA, tRNA, and rRNA in protein synthesis. 4. Imagine you are a student trying to decode the genetic code. An artificial mRNA molecule consisting of poly-CA (5ʹ... CACACACACACACAC..3ʹ) yields a polypeptide consisting solely of histidines and threonine, and another artificial mRNA consisting of poly-CAA (5ʹ...CAACAACAACAACAA...3ʹ) yields three different polypeptides: polythreonine, polyglutamine, and polyasparagine. On the basis of this information, which codons can you assign to which amino acids? Answer: Poly-CA contains two alternating codons, CAC and ACA, one of which codes for histidine and the other for threonine. Depending on where translation begins, polyCAA can be read as repeating CAA, repeating AAC, or repeating ACA. The common amino acid is threonine, and the common codon is ACA. Thus ACA codes for threonine and CAC must therefore code for histidine. That leaves CAA and AAC, one of which codes for glutamine and the other for asparagine. More data are needed to determine which codes for which. Textbook Reference: Expression of Genetic Information Bloom’s Category: 3. Applying Learning Objective: Summarize the experimental evidence for a triplet code.

5. The genetic code is said to be degenerate. What does “degenerate” mean in this context? Answer: The genetic code is based on multiples of three nucleotides, and it is a triplet code. In other words, once initiated, every three RNA bases encode a different amino acid or stop codon. There are four possible bases: A, U, G, and C. Therefore, the maximum number of possible codon sequences of three bases would be 43, or 64 possible triplets. Three codons that are referred to as “stop codons” signal termination of the growing polypeptide strand, leaving a possible 61 codons that can encode amino acids. Since there are only 20 amino acids, most are specified in the genetic code by more than one codon. For example, CGU, CGC, CGA, and CGG all encode the amino acid arginine. This property of multiple codons per amino acid is what is meant by “degenerate.” Textbook Reference: Expression of Genetic Information Bloom’s Category: 3. Applying Learning Objective: Summarize the experimental evidence for a triplet code. 6. According to the “central dogma” of molecular biology, information flows from DNA to RNA to protein. In its original version, the dogma stated that information does not flow in the reverse direction. To what extent is this true? Answer: The reproductive cycle of retroviruses includes a step, catalyzed by reverse transcriptase, in which a DNA copy of the viral RNA genome is synthesized. Thus, DNA is made from an RNA template, contradicting the original version of the dogma. However, it has not yet been found that a nucleic acid can be made using a protein template, so it appears that information does not flow backward from proteins. Textbook Reference: Expression of Genetic Information Bloom’s Category: 2. Understanding Learning Objective: Summarize the experimental evidence for reverse transcription. 7. EcoRI and NotI are two DNA restriction enzymes that recognize the specific sequences GAATTC and GCGGCCGC, respectively. Which of these restriction enzymes would cleave DNA most frequently, and why? Answer: EcoRI. The specific sequences recognized by EcoRI would be expected to occur once in every 46 base pairs, based on four possible nucleotides in a specific sequence of six. NotI would occur less frequently—only once in 48 base pairs, based on four possible nucleotides in a specific sequence of eight. Textbook Reference: Recombinant DNA Bloom’s Category: 4. Analyzing Learning Objective: Predict the average sizes of DNA fragments produced by cleavage with a restriction endonuclease with a known recognition site. 8. Suppose you are studying a mammalian transcription factor that you would like to express in bacteria so that you can purify large quantities of it to use in biochemical studies. You introduce the cDNA encoding the transcription factor into an expression plasmid and transform E. coli with the new recombinant vector. You find no expression. What might be occurring?

Answer: Lack of expression could result from a number of factors. The plasmid must have appropriate unique restriction sites, an origin of replication, and a selective antibiotic resistance marker. If any of these are perturbed during cloning, there will be no expression of the gene of interest. Textbook Reference: Recombinant DNA Bloom’s Category: 4. Analyzing Learning Objective: Summarize how a fragment of host DNA is cloned in a plasmid vector. 9. A common method of DNA sequencing is based on the premature termination of DNA synthesis that accompanies the use of dideoxynucleotides in the PCR reactions. Why does elongation of a growing strand cease when a dideoxynucleotide is incorporated? Answer: RNA, or ribonucleic acid, uses ribose and has hydroxyl groups at both the 2ʹ and 3ʹ positions. DNA is deoxynucleic acid—deoxy refers to one less hydroxyl group, at the 2ʹ position. Both molecules depend on the 3ʹ hydroxyl for elongation because it is required for a phosphodiester bond to forms between it and 5ʹ phosphate group of the next nucleotide. Dideoxynucleotides have only hydrogens at both the 2ʹ and 3ʹ positions (i.e., both hydroxyl groups are gone at these positions). Therefore when dideoxynucleotides are incorporated into growing strands of DNA, elongation ceases because there is no 3ʹ hydroxyl with which to form a phosphodiester bond. Textbook Reference: Recombinant DNA Bloom’s Category: 4. Analyzing Learning Objective: Describe how DNA is sequenced with dideoxynucleotides. 10. Why are E. coli DNA polymerases not used in the polymerase chain reaction (PCR)? Answer: E. coli DNA polymerases cannot be used in PCR because they are not stable at the temperatures required to denature the DNA strands in the first step of each PCR cycle—the enzyme would rapidly be inactivated, and polymerization would cease. Instead, polymerases from bacteria that have evolved to live at high temperatures (such as Thermus aquaticus) are used, since they are stable at high temperatures. Textbook Reference: Detection of Nucleic Acids and Proteins Bloom’s Category: 4. Analyzing Learning Objective: Explain how DNA is amplified by the polymerase chain reaction (PCR).

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 5: Genomics, Proteomics, and Systems Biology TEST FILE QUESTIONS Multiple Choice 1. In 1995, the first complete genome of _______ was sequenced by Dr. Craig Venter. a. Escherichia coli b. Saccharomyces cerevisiae c. Haemophilus influenzae d. Caenorhabditis elegans Answer: c Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. 2. Analysis of the H. influenza and E. coli genomes revealed that _______% of each genome encodes for protein. a. 25 b. 50 c. 75 d. 90 Answer: d Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. 3. The simplest eukaryotic genome is found in a. Saccharomyces cerevisiae. b. Caenorhabditis elegans. c. Drosophila melanogaster. d. Arabidopsis thaliana. Answer: a Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. 4. Analysis of the first bacterial genome to be sequenced, H. influenzae, revealed _______ rRNA genes, _______ different tRNA genes, and _______ potential proteincoding genes.

a. 6; 54; 1,743 b. 54; 6; 174 c. 54; 6; 1,743 d. 6; 54; 174 Answer: a Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. 5. The first eukaryotic genome to be completely sequenced was a. the protist Dictyostelium discoideum. b. the yeast Saccharomyces cerevisiae. c. the fruit fly Drosophila melanogaster. d. the human genome. Answer: b Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. 6. Which of the following is not a stop codon? a. UAA b. UCA c. UAG d. UGA Answer: b Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. 7. Which statement below best explains why there are 64 possible codons? a. There are only four nucleotides and with a triplet code, the total number of possibilities would be 43. b. There are 21 amino acids and each amino acid had three codons. c. There are only four nucleotides and with a triplet code, the total number of possibilities would be 34. d. There are only four nucleotides, yet in the double-stranded form, there are eight, thus there are 82 possibilities. Answer: a Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 3. Applying Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. 8. Stretches of sequenced nucleotides that code for polypeptides are recognized by the a. presence of a start codon.

b. absence of chain-terminating codons. c. presence of closed-reading frames. d. absence of open-reading frames. Answer: b Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. 9. Potential protein-coding regions of genomic DNA containing long stretches of nucleotide sequences that can encode polypeptides and contain no stop codons are called a. introns. b. exons. c. pseudogenes. d. open-reading frames. Answer: d Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. 10. Sequencing the genomes of Caenorhabditis elegans and Drosophila melanogaster led to the discovery of many new genes for a. enzymes of cell metabolism. b. proteins regulating the cell cycle. c. proteins regulating mitosis. d. proteins regulating animal development. Answer: d Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. 11. The Arabidopsis genome, like many plant genomes, contains more genes than many animals. This is due largely to a. fewer stop codons. b. larger exons. c. larger introns. d. gene duplication. Answer: d Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. 12. The human genome contains approximately _______ billion base pairs.

a. 1.5 b. 2 c. 3 d. 6 Answer: c Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 13. Surprisingly, after sequencing many different species, what organism stands out as having the largest number of genes? a. Arabidopsis b. Zebrafish c. Chimpanzee d. Apple Answer: d Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 14. The human genome is estimated to contain about _______ genes. a. 10,000–15,000 b. 20,000–25,000 c. 100,000–150,000 d. 200,000–250,000 Answer: b Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 15. To sequence the human genome, the International Human Genome Sequencing Consortium a. used the shotgun approach to sequence fragments and assemble them in order using overlaps between the sequences. b. sequenced cloned DNA and then mapped it by FISH. c. used BAC clones as substrates for sequencing. d. used massively parallel sequencing. Answer: c Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Outline the basic approach used in next-generation sequencing. 16. To sequence the human genome, the Celera Genomics team led by Craig Venter used

a. the shotgun approach to sequence fragments and assemble them in order using overlaps between the sequences. b. BAC clones as substrates for sequencing. c. sequence-cloned DNA and then mapped it by FISH. d. massively parallel sequencing. Answer: a Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Outline the basic approach used in next-generation sequencing. 17. What percent of genes are similar between mice, rats, and humans? a. 10% b. 30% c. 70% d. 90% Answer: d Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Explain why studies of other vertebrates are useful in understanding human genomics. 18. Which method revealed that there were many more RNAs transcribed than are accounted for by the protein-coding genes in human cells? a. DNA microarray b. Next-generation sequencing c. RNA-seq d. cDNA hybridization Answer: c Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Describe the global methods used to study gene expression. 19. There are more distinct proteins in eukaryotic cells than there are number of genes. Which statement best explains this? a. Many genes can give rise to multiple RNAs through alternate splicing. b. Pseudogene expression often undergo transcription of truncated mRNAs. c. Many genes undergo amplification and produce higher levels of mRNA. d. Some mRNAs cross-hybridize and get translated into fusion proteins. Answer: a Textbook Reference: Proteomics Bloom’s Category: 2. Understanding Learning Objective: Describe the global methods used to study gene expression. 20. What is a common method for testing blood samples of athletes for the presence of performance enhancing drugs? a. Next-generation sequencing

b. Mass spectrometry c. Two-dimensional gel electrophoresis d. RNA-seq Answer: b Textbook Reference: Proteomics Bloom’s Category: 1. Remembering Learning Objective: Describe the global methods used to study gene expression. 21. Mass spectrometers are used to a. transport a molecule against its concentration gradient. b. couple the energy of ATP hydrolysis to a configurational change. c. transport H+ across a membrane against a pH gradient. d. identify proteins by mass/ratio peptide analysis as part of proteomic procedures. Answer: d Textbook Reference: Proteomics Bloom’s Category: 1. Remembering Learning Objective: Explain how proteins are identified by mass spectrometry. 22. In tandem mass spectrometry the unique second phase is characterized by a. the further isolation of the first phase samples by two-dimensional gel electrophoresis. b. the purification of the first phase samples by column chromatography. c. the addition of a collision based fragmentation step to the first phase sample to determine amino acid sequence. d. running simultaneous samples to compare and confirm amino acid identity. Answer: c Textbook Reference: Proteomics Bloom’s Category: 1. Understanding Learning Objective: Explain how proteins are identified by mass spectrometry. 23. Which cellular structure has the largest number of unique proteins? a. Plasma membrane b. Endoplasmic reticulum c. Mitochondria d. Golgi apparatus Answer: a Textbook Reference: Proteomics Bloom’s Category: 1. Remembering Learning Objective: Summarize the approaches used for analysis of the proteome of subcellular organelles. 24. Which method uses an antibody that recognizes cellular proteins to investigate protein–protein interactions? a. Immunoprecipitation b. The yeast two-hybrid system c. Protein array d. Electrophoretic mobility shift assay

Answer: a Textbook Reference: Proteomics Bloom’s Category: 1. Remembering Learning Objective: Describe the approaches used for identification of protein interactions. 25. Which method requires the experimental construction of two fusion proteins to assess protein–protein interactions? a. Immunoprecipitation b. The yeast two-hybrid system c. Protein array d. Electrophoretic mobility shift assay Answer: a Textbook Reference: Proteomics Bloom’s Category: 2. Understanding Learning Objective: Describe the approaches used for identification of protein interactions. 26. The new field of science called _______ lies at the interface between biology and computer science. It is focused on developing the computational methods needed to extract useful biological information from the sequence of billions of bases of DNA. a. systems biology 100 b. genome wide association studies c. microarray d. bioinformatics Answer: d Textbook Reference: Systems Biology Bloom’s Category: 1. Remembering Learning Objective: Contrast the approaches of traditional biological experimentation and systems biology. 27. Which phrase best defines systems biology? a. The use of global experimental data for quantitative modeling of integrated cellular processes b. The systematic analysis of individual molecules and pathways c. The comparative analysis of large datasets from multiple cell types, tissues, and/or organs d. The computational overlay of data from both genomic and proteomic experiments Answer: d Textbook Reference: Systems Biology Bloom’s Category: 1. Remembering Learning Objective: Contrast the approaches of traditional biological experimentation and systems biology.

28. The experimental inactivation of a gene by homologous recombination with an inactive mutant to assess function of the normal genes in intact organisms is often referred to as what type of experiment? a. Transgenic b. Hybrid c. Knockout d. Gain-of-function Answer: c Textbook Reference: Systems Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the methods used for large-scale screens of gene function. 29. In RNA interference experiments, siRNAs homologous to known mRNAs inhibit expression of the proteins by what mechanism? a. They form double-stranded RNA molecules that get degraded in the cytoplasm. b. They form double-stranded RNA molecules that block the complete mRNA form traversing through the ribosomal complex. c. They form double-stranded RNA molecules in pre-spliced primary RNA and prevent passage from the nucleus to the cytoplasm. d. They form double-stranded RNA molecules in pre-spliced primary RNA and prevent the splicing out of introns. Answer: a Textbook Reference: Systems Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used for large-scale screens of gene function. 30. Five percent of the genome is conserved among all mammals, yet only ~1% directly encodes proteins. What is the likely role for the remaining 4%? a. Pseudogene duplication among species b. Regulation of expression for the 1% that encodes proteins c. Redundant open-reading frames d. Positioning of genes into transcriptional active zones Answer: b Textbook Reference: Systems Biology Bloom’s Category: 2. Understanding Learning Objective: Explain the approaches used to identify gene regulatory sequences 31. Why is it thought that regulatory elements occur in clusters? a. Redundancy to ensure stimulation of the gene b. Allows gene expression to move in both directions on separate strands of the DNA c. Allows hairpin formation to enhance protein binding d. Allows the interaction of multiple transcription factors Answer: d Textbook Reference: Systems Biology

Bloom’s Category: 2. Understanding Learning Objective: Explain the approaches used to identify gene regulatory sequences 32. The large-scale project called ENCODE used RNA-seq to characterize all RNAs in 147 human cell types. The surprise result from analysis of the data revealed that even though there is only 1% of the genome that encodes for protein, as much as _______% of the genome is actually transcribed. a. 30 b. 50 c. 70 d. 90 Answer: b Textbook Reference: Systems Biology Bloom’s Category: 1. Remembering Learning Objective: Explain the approaches used to identify gene regulatory sequences. 33. In the upstream promoter of some human genes, a regulatory sequence of eight nucleotides, TGACGTCA, has been shown to regulate a number of cAMP responsive genes. Based on this knowledge, you conduct a genome search and identify 250 of these identical sequences that occur. Could you then conclude that these sequences coincide with genes that are cAMP responsive? a. Yes, because each of those regulatory regions containing that sequence would only be found near genes. b. Yes, because this sequence would be unlikely to occur randomly. c. No, because this sequence, in the context of 3 billion base pairs, would occur fairly regularly and may or may not be associated with a gene. d. No, because there must be other regulatory sequences present as well. Answer: c Textbook Reference: Systems Biology Bloom’s Category: 4. Analyzing Learning Objective: Explain the approaches used to identify gene regulatory sequences. 34. What is the term used in cell biology to describe the transmission of information from the environment, such as the presence of a hormone, to targets within the cell? a. Crosstalk b. Networks c. Hierarchical clustering d. Signaling pathway Answer: d Textbook Reference: Systems Biology Bloom’s Category: 1. Remembering Learning Objective: Illustrate the types of interactions between pathways in a regulatory network. 35. Consider a hypothetical signaling pathway where the initiating signal is A, the terminal response is E, and there are three intermediate components, B, C, and D, such

that the complete pathway is A-B-C-D-E. Experimentally, you determine that E acts to simulate A. This is an example of a. negative feedback. b. positive feedback. c. feedforward relay. d. stimulatory crosstalk. Answer: b Textbook Reference: Systems Biology Bloom’s Category: 3. Applying Learning Objective: Illustrate the types of interactions between pathways in a regulatory network. 36. Consider two signaling pathways: A-B-C-D-E and 1-2-3-4-5-6-7. Which response would be representative of a feedforward relay? a. B stimulates E b. B stimulates 6 c. D stimulates A d. 2 stimulates E Answer: a Textbook Reference: Systems Biology Bloom’s Category: 3. Applying Learning Objective: Illustrate the types of interactions between pathways in a regulatory network. 37. Consider two signaling pathways: A-B-C-D-E and 1-2-3-4-5-6-7.Which response would be representative of stimulatory crosstalk? a. B stimulates E b. B stimulates 6 c. D stimulates A d. 2 stimulates 7 Answer: b Textbook Reference: Systems Biology Bloom’s Category: 3. Applying Learning Objective: Illustrate the types of interactions between pathways in a regulatory network. 38. Which of the following would not necessarily be part of a cellular signaling network? a. Crosstalk b. Feedback loops c. Synthetic genetic toggle switches d. Feedforward relays Answer: c Textbook Reference: Systems Biology Bloom’s Category: 2. Understanding Learning Objective: Illustrate the types of interactions between pathways in a regulatory network.

39. What is best described as an engineering approach to understanding and manipulating biological systems? a. Synthetic biology b. Biomedical engineering c. Systems biology d. Theranostics Answer: a Textbook Reference: Systems Biology Bloom’s Category: 1. Remembering Learning Objective: Define synthetic biology. 40. In 2010, scientists developed a series of overlapping oligonucleotides that extended a full 1.08 million base pairs and used it to develop the first complete organism using a synthetic genome. The newly created cells had normal growth and morphology. What organism did they generate in these groundbreaking experiments? a. Escherichia coli b. Saccharomyces cerevisiae c. Mycoplasma mycoides d. Haemophilus influenza Answer: a Textbook Reference: Systems Biology Bloom’s Category: 1. Remembering Learning Objective: Define synthetic biology. 41. In 2010, scientists developed a series of overlapping oligonucleotides that extended a full 1.08 million base pairs and used it to develop the first complete organism using a synthetic genome. The newly created cells had normal growth and morphology. In addition to the synthetic genome, why did the investigators also include a gene that confers resistance to tetracycline? a. By including an exogenous gene, they could confirm propagation of the genome by PCR. b. The tetracycline resistance gene increases the genomic stability of genomes in prokaryotes. c. The tetracycline resistance gene destabilizes the genome of the host DNA allowing the novel synthetic genome to survive. d. The tetracycline resistance gene allows the selection of only the cells that are resistant to tetracycline and thus cells that survive selection carry the novel synthetic genome. Answer: d Textbook Reference: Systems Biology Bloom’s Category: 3. Applying Learning Objective: Define synthetic biology.

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1. The first complete sequence of a cellular genome was of the bacterium _______. Answer: H. influenza Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. 2. In comparing the genomes of two yeasts, S. pombe and S. cerevisiae, researchers found that S. pombe contains more _______. Answer: introns Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. 3. There are three chain termination or stop codons: UAA, UAG, and _______. Answer: UGA Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. 4. In comparing genomes between the plant Arabidopsis thaliana and humans, it was found that _______ had the largest number of protein-coding genes. Answer: Arabidopsis thaliana Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. 5. A long DNA sequence with no stop codons is a(n) _______. Answer: open-reading frame Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 6. The genome of the fruit fly Drosophila contains approximately _______ as many genes as the human genome. Answer: two-thirds (2/3) Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 7. Remarkably, approximately _______% of the predicted human proteins are related to proteins in simpler sequenced eukaryotes, including yeast, Drosophila, and C. elegans. Answer: 40 (acceptable of anything between 30–50)

Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 8. The human genome contains approximately _______ base pairs. Answer: 3 billion (3  109) Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 9. Remarkably, only _______% of the human genome corresponds to protein-coding sequences. Answer: 1.2 Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 10. From 2001 to 2014, the cost of sequencing an individual’s genome has dropped from several million dollars to several _______ dollars. Answer: thousand Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Outline the basic approach used in next-generation sequencing. 11. A more common name for the high-throughput DNA sequencing method of massively parallel sequencing is _______. Answer: next-generation sequencing Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Outline the basic approach used in next-generation sequencing. 12. The source of cDNA used in microarray analyses is _______. Answer: mRNA Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Describe the global methods used to study gene expression. 13. A new method that has the ability to assess every RNA expressed in a cell is _______. Answer: RNA-seq Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Describe the global methods used to study gene expression.

14. Mass spectrometry measures the _______ ratio of small peptides that have been proteolytically cleaved. Answer: mass-to-charge Textbook Reference: Proteomics Bloom’s Category: 1. Remembering Learning Objective: Explain how proteins are identified by mass spectrometry. 15. A reasonable method to determine if two proteins differed by only a single amino acid would be to use _______. Answer: tandem mass spectrometry Textbook Reference: Proteomics Bloom’s Category: 3. Applying Learning Objective: Explain how proteins are identified by mass spectrometry. 16. The use of antibodies to determine protein-protein interactions is known as _______. Answer: immunoprecipitation Textbook Reference: Proteomics Bloom’s Category: 1. Remembering Learning Objective: Describe the approaches used for identification of protein interactions. 17. When an element in one signaling pathway either stimulates or inhibits an element of a second pathway it is referred to as _______. Answer: crosstalk Textbook Reference: Systems Biology Bloom’s Category: 2. Understanding Learning Objective: Illustrate the types of interactions between pathways in a regulatory network.

True/False 1. An open-reading frame is a long stretch of nucleotide sequence that contains no chainterminating (stop) codons. Answer: T Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. 2. A common characteristic of most bacterial genomes is that ~90% of the genome encodes for proteins. Answer: T Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes.

3. The human genome has more genes than plant genomes have. Answer: F Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. 4. An open-reading frame of 1,200 base pairs long would contain 300 codons. Answer: F Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 3. Applying Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 5. Surprisingly, the size of genomes among a wide variety of organisms does not correlate with complexity of the organisms. Answer: T Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Explain why studies of other vertebrates are useful in understanding human genomics. 6. There is only a 1% variance between the genomes of human, Neandertals, and chimpanzees. Answer: T Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Explain why studies of other vertebrates are useful in understanding human genomics. 7. The 20,000–25,000 human genes may encode more than 100,000 distinct mRNAs and proteins. Answer: T Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Describe the global methods used to study gene expression. 8. About half of the human genes identified by sequencing can be assigned a function by homology to previously characterized genes, and the other half represent new genes of unknown function. Answer: T Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Describe the global methods used to study gene expression.

9. More than 50% of the human genome is composed of various types of repetitive and duplicated DNA sequences. Answer: T Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Describe the global methods used to study gene expression. 10. Hybridization of fluorescent labeled cDNAs to microarrays of thousands of gene sequences allows gene expression to be assayed. Answer: T Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Describe the global methods used to study gene expression. 11. Mass spectrometric identification of individual proteins in a complex mixture would be impossible in the absence of large-scale protein sequence databases. Answer: T Textbook Reference: Proteomics Bloom’s Category: 2. Understanding Learning Objective: Explain how proteins are identified by mass spectrometry. 12. Hybridization of fluorescent labeled cDNAs to microarrays of thousands of gene sequences allows gene expression to be assayed. Answer: T Textbook Reference: Systems Biology Bloom’s Category: 1. Remembering Learning Objective: Summarize the methods used for large-scale screens of gene function. 13. The computational analysis comparing the expression of all mRNAs between two different cells is an example of synthetic biology. Answer: F Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Define synthetic biology.

Short Answer 1. If an organism has a 2 million base genome and 90% of it encodes proteins, and the average gene is 300 codons, how many genes does it have? Answer: 2,000 (2,000,000  0.9/900) Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 3. Applying Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis.

2. Why would approximately 40% of the predicted human proteins be related to proteins in simpler organisms? Answer: Even though humans are vastly different from simpler organisms, we still share many basic cellular processes such as metabolism, DNA replication and repair, transcription, translation, and protein trafficking. Because of this, many of the genes encoding proteins that regulate these functions are very similar across many species. Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 3. Applying Learning Objective: Explain why studies of other vertebrates are useful in understanding human genomics. 3. Explain the likely effect that the sequencing of the human genome will have on medical science. Answer: Knowing the sequence of the human genome will help clarify the genetic basis of many human diseases by providing a means of comparison of gene sequences from different individuals (e.g., those with and those without a certain disease). Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Outline the basic approach used in next-generation sequencing. 4. What is the most significant difference between microarray analyses and RNA-seq analyses? Answer: Microarray identifies only the mRNAs in the sample that are expressed and that hybridize to oligonucleotides on the microarray that represent genes in the cells, whereas RNA-seq identifies all the RNA (rRNA, tRNA, etc.) that is expressed in a cell, via nextgeneration sequencing. Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 3. Applying Learning Objective: Describe the global methods used to study gene expression. 5. Even though there are only about 20,000 different protein-coding genes in humans, it is estimated that these give rise to more than 100,000 different proteins. What are two mechanisms that can account for this dramatic difference? Answer: 1) Alternative splicing; 2) Protein modifications, such as the addition of phosphate groups, carbohydrates, and lipids. Textbook Reference: Proteomics Bloom’s Category: 3. Applying Learning Objective: Explain how proteins are identified by mass spectrometry 6. When carrying out immunoprecipitation experiments to investigate protein–protein interactions, why is it important to isolate the proteins under gentle conditions? Answer: Protein–protein interactions are due to a combination of hydrogen bonding and ionic bonding. Unlike very strong covalent bonds, they are subject to dissolution in high salts, detergents, and strong acids and bases, so these in vitro reactions should be done under conditions very similar to those inside the cell.

Textbook Reference: Proteomics Bloom’s Category: 3. Applying Learning Objective: Describe the approaches used for identification of protein interactions. 7. You have two proteins you suspect of interacting with each other, but have an antibody to only one of them. Describe an experiment that might allow you to determine if the two proteins do indeed interact with each other. Answer: Armed with only one antibody, an immunoprecipitation experiment could be performed in which cell extracts are incubated with the one antibody, then the sample could be run on a mass spectrometer. The results of the mass spectrometer could then be analyzed for evidence of the second protein. Textbook Reference: Proteomics Bloom’s Category: 3. Applying Learning Objective: Describe the approaches used for identification of protein interactions. 8. You have identified a unique 8-base sequence in the promoter of a gene that controls growth. Describe an experiment that may address whether this sequence is capable of directing transcription of other genes. Answer: You could synthesize this sequence in an oligonucleotide flanked by restriction sites and clone it into a plasmid that contains a reporter gene. Then transfect the plasmid into cells and look for expression of the reporter gene. Textbook Reference: Systems Biology Bloom’s Category: 3. Applying Learning Objective: Explain the approaches used to identify gene regulatory sequences. 9. You have a cell signaling cascade in which a cell signal initially targets and activates protein 1 which in turn activates protein 2 which then activates protein 3 and finally activates protein 4. You are given a cell line that never expresses protein 2, yet in every experiment in which you give the initiating signal, you get activation of protein 4. How could you explain these results? Answer: Somehow, the signal is being appropriately transferred, so there must be an alternate signaling path. There may be a feedforward relay in which either the initiating signal or protein 1 bypass protein 2 and directly activate either protein 3 or protein 4. Textbook Reference: Systems Biology Bloom’s Category: 3. Applying Learning Objective: Illustrate the types of interactions between pathways in a regulatory network. 10. In the pathway through which epinephrine signals the breakdown of glycogen to glucose in muscle cells, there are at least six different proteins involved before glycogen phosphorylase acts to cleave a glucose off of glycogen. Additional administration of epinephrine will increase the pathway. In a cell line in which phosphorylase kinase is mutated, what will happen if epinephrine is used to treat the cells?

Answer: There would be no effect. In the signaling cascade (see textbook Figure 5.18) epinephrine is upstream of the phosphorylase kinase where the mutation is, so the cascade would be blocked at that point. Textbook Reference: Systems Biology Bloom’s Category: 2. Understanding Learning Objective: Illustrate the types of interactions between pathways in a regulatory network.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. In which of the following organisms is the highest percentage of its genome used to encode protein? a. Haemophilus influenzae b. Saccharomyces cerevisiae c. Drosophila melanogaster d. Caenorhabditis elegans Answer: a Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. Feedback A: Correct! Remarkably, 90% of this bacterial genome encodes protein. Feedback B: Incorrect. Yeast is among the highest among eukaryotes at 70%, but still is less than in prokaryotes. Feedback C: Incorrect. The protein coding component of the fruit fly genome is 10%. Feedback D: Incorrect. The protein coding component of the roundworm genome is 25%. 2. The main difference between the genomes of the bacteria Haemophilus influenzae and E. coli is that a. the genome of H. influenzae is a circular molecule and that of E. coli is linear. b. the H. influenzae genome encodes a larger number of genes. c. the genome of E. coli is much larger than the H. influenzae genome. d. a higher percentage of the H. influenzae genome contains genes encoding proteins. Answer: c Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. Feedback A: Incorrect. Both genomes are encoded in circular molecules. Feedback B: Incorrect. The H. influenzae genome encodes approximately one-half the number of genes as E. coli. Feedback C: Correct! The genome of E. coli is 2.5 times the size of the H. influenzae genome. Feedback D: Incorrect. Approximately 90% of the genomes of both E. coli and H. influenzae encode proteins.

3. The eukaryote Caenorhabditis elegans has been used especially in the study of a. animal development. b. photosynthesis. c. flagellar movement. d. immunology. Answer: a Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. Feedback A: Correct! The developmental pattern of each cell, from the embryonic stage to the full-grown organism, has been determined. Feedback B: Incorrect. Photosynthesis is a process that takes place in plants, and C. elegans belongs to the animal kingdom. Feedback C: Incorrect. Flagellar movement is a means of locomotion used by some unicellular organisms such as Chlamydomonas; C. elegans is multicellular. Feedback D: Incorrect. Mice are the most commonly used organism for immunological studies. 4. The plant Arabidopsis thaliana contains significantly more genes than either C. elegans or Drosophila. Which statement is true? a. Arabidopsis has a much larger number of genes due to duplication of large segments of the Arabidopsis genome. b. The large number of genes reflects a greater diversity of proteins than in the other two organisms. c. Very few Arabidopsis genes are unique to plants. d. Arabidopsis is the only plant for which the genome has been sequenced. Answer: a Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Summarize the contributions of coding and noncoding sequences to the genomes of Drosophila, C. elegans, and Arabidopsis. Feedback A: Correct! The number of distinct protein-coding genes in Arabidopsis is about the same as in C. elegans and Drosophila. Feedback B: Incorrect. The number of distinct protein-coding genes in Arabidopsis is about the same as in C. elegans and Drosophila. Feedback C: Incorrect. About one-third of the Arabidopsis genes are not found in yeast or animal genomes. Feedback D: Incorrect. Draft sequences of two important strains of rice were published less than two years after the sequencing of Arabidopsis. 5. An estimated 40% of the human genome encodes proteins that are related to proteins found in a number of other species. These proteins are involved in which cellular processes? a. Replication

b. Transcription c. Translation d. All of the above Answer: d Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Explain why studies of other vertebrates are useful in understanding human genomics. Feedback A: Incorrect. This is a correct answer, but it is not the only correct answer. Feedback B: Incorrect. This is a correct answer, but it is not the only correct answer. Feedback C: Incorrect. This is a correct answer, but it is not the only correct answer. Feedback D: Correct! All of these processes, plus others such as basic metabolism and the functioning of signaling pathways, are often very similar among species. 6. Neandertals are thought to be our closest evolutionary relatives, having diverged from humans 300,000 to 400,000 years ago, and their genome has been sequenced. Which statement about humans and Neandertals is false? a. The genomes of modern humans and Neandertals are 99.9% identical. b. Neandertal and chimpanzee genomes share about 99% identity. c. The Neandertal genome was surprisingly GC-rich. d. The coding sequences of only 90 genes in Neandertals are different from those of modern humans. Answer: c Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Explain why studies of other vertebrates are useful in understanding human genomics. Feedback A: Incorrect. This is a true statement. Feedback B: Incorrect. This is a true statement. Feedback C: Correct! This was not found to be the case and thus the incorrect statement. Feedback D: Incorrect. This is a true statement. 7. In 2001, two groups of researchers announced their successful completion of draft sequences of the human genome. Which statement about these draft sequences is true? a. The publicly funded team used a whole-genome shotgun sequencing approach. b. The Celera Genomics team used an approach that had also been used to sequence the yeast and C. elegans genomes. c. Both groups reported sequences amounting to five megabases. d. Both groups presented draft sequences that cover only the euchromatin portion of the genome. Answer: d Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Outline the basic approach used in next-generation sequencing. Feedback A: Incorrect. The whole-genome approach was used by the Celera group. Feedback B: Incorrect. That approach was used by the publicly funded team.

Feedback C: Incorrect. The reported euchromatin sequences were about 2.9 billion bases of DNA. Feedback D: Correct! The heterochromatin repeat-rich portion of the genome remained unsequenced. 8. Which of the following steps comes first in next-generation sequencing? a. Amplification of the DNA sample by polymerase chain reaction (PCR). b. Fragmentation of the DNA sample into smaller pieces. c. Addition of adapters to the ends of the DNA sample to allow amplification by PCR. d. Separation of the DNA sample by electrophoresis to ensure uniform lengths. Answer: b Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Outline the basic approach used in next-generation sequencing. Feedback A: Incorrect. PCR is performed after fragmentation, adapter addition, anchoring, and setting up the reaction with primers and labeled nucleotides. Feedback B: Correct! This is done to allow many small pieces to be amplified by PCR simultaneously, in parallel. Feedback C: Incorrect. This step follows fragmentation. Feedback D: Incorrect. Unlike older methods for sequencing, gel electrophoresis is not used in next-generation sequencing. 9. A collection of all the RNAs that are expressed at any given time in a cell are referred to as the a. proteome. b. genome. c. metabolome. d. transcriptome. Answer: d Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 1. Remembering Learning Objective: Describe the global methods used to study gene expression. Feedback A: Incorrect. This refers to all the proteins present in a cell. Feedback B: Incorrect. This refers to all the genes in a cell. Feedback C: Incorrect. This refers to all the metabolic intermediates in a cell. Feedback D: Correct! RNA is generated by the process of transcription, thus the collection of RNAs is called the transcriptome. 10. Which methods can accurately quantify the amount of mRNA expressed in a given cell or tissue sample? a. Immunoprecipitation and mass spectrometry b. DNA microarray and RNA-seq c. Next-generation sequencing and tandem mass spectrometry d. RNAi and CRISPR/Cas Answer: b Textbook Reference: Genomes and Transcriptomes

Bloom’s Category: 2. Understanding Learning Objective: Describe the global methods used to study gene expression. Feedback A: Incorrect. These methods measure proteins. Feedback B: Correct! Both of these are large scale methods to measure RNA presence and abundance in a variety of cells and tissue samples. Feedback C: Incorrect. These are DNA sequencing and protein analysis methods, respectively. Feedback D: Incorrect. These are methods to assess gene function. 11. Which of the following methods would be best for direct determination of the relative level of all mRNAs in a given cell or tissue? a. Mass spectrometry b. Immunoprecipitation c. DNA microarray d. Two-dimensional gel electrophoresis Answer: c Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Describe the global methods used to study gene expression. Feedback A: Incorrect. This is a method to analyze proteins. Feedback B: Incorrect. This is a method to isolate specific proteins using an antibody. Feedback C: Correct! Microarray may rapidly assess all mRNAs present in a cell or tissue and provide quantitative data on the amount of each mRNA. Feedback D: Incorrect. This is a method to analyze proteins. 12. Next-generation DNA sequencing is not an acceptable method for detecting differences in gene expression during development. Which statement most likely describes the reason why? a. The amount of DNA in each cell changes during development, making such analysis difficult. b. Proteins of varying size and charge would not be distinguishable using this approach. c. The genome remains constant throughout development, so DNA would not be an appropriate molecule to analyze. d. The approach would be cost prohibitive. Answer: c Textbook Reference: Proteomics Bloom’s Category: 3. Applying Learning Objective: Describe the global methods used to study gene expression. Feedback A: Incorrect! The amount of DNA in a cell remains constant over time (although the somatic cells are “2n” and mature gametes are “n.” Feedback B: Incorrect. Next-generation sequencing is a method that allows the assessment of DNA sequences, it does not provide information on post-translational modification of proteins. Feedback C: Correct! All cells contain the same DNA; gene expression must be measured thorough RNA, cDNA, or protein analysis. Feedback D: Incorrect. Next-generation DNA sequencing is a highly cost-effective

method, and the entire genome of an individual can be sequenced for about $1,000. However, this method does not provide information on gene-expression. 13. Humans have about 20,000 genes that encode proteins. It is estimated that through alternative splicing and protein modifications, these genes can give rise to how many different proteins? a. 40,000 b. 60,000 c. 80,000 d. 100,000 Answer: d Textbook Reference: Proteomics Bloom’s Category: 1. Remembering Learning Objective: Explain how proteins are identified by mass spectrometry. Feedback A: Incorrect. Feedback B: Incorrect. Feedback C: Incorrect. Feedback D: Correct! This is one of the reasons why the regulation of these posttranscriptional processes is so important. 14. Which method would best determine posttranslational modifications to a protein, such as phosphorylation? a. Tandem mass spectrometry b. Mass spectrometry c. Two-dimensional gel electrophoresis d. Immunoprecipitation Answer: a Textbook Reference: Proteomics Bloom’s Category: 2. Understanding Learning Objective: Explain how proteins are identified by mass spectrometry. Feedback A: Correct! The second mass spectrometer in this method analyzes smaller fragments that differ by single amino acids, and the sequence of the peptide can be deduced based on molecular weight. Phosphorylation can be detected because protein modifications change the weight of the amino acid. Feedback B: Incorrect. Simple mass spectrometry determines the mass-to-charge ratio of a peptide and identifies it based on comparison with known protein spectra in a database. Feedback C: Incorrect. The location of a phosphorylated protein may well be different, but it would be very difficult to determine if it was due to phosphorylation. Feedback D: Incorrect. After a protein has been identified as being phosphorylated, a phospho-specific antibody could be generated and used in immunoprecipitations, but it would not work to make the initial identification. 15. Which method for investigating protein–protein interactions uses gene expression as the detection component? a. Yeast two-hybrid b. Immunoprecipitation

c. Mass spectrometry d. Tandem mass spectrometry Answer: a Textbook Reference: Proteomics Bloom’s Category: 2. Understanding Learning Objective: Describe the approaches used for identification of protein interactions. Feedback A: Correct! This method couples two yeast proteins that act as a transcription factor to express an identifiable target gene and works in vivo. Feedback B: Incorrect. This method is used solely for looking at protein–protein interactions using antibodies. Feedback C: Incorrect. This method enzymatically cleaves proteins and then analyzes them against a large database. Feedback D: Incorrect. This method goes a step further than mass spectrometry, breaking the peptides into smaller fragments and analyzes individual amino acids. 16. Bioinformatics is a. an electronic copy of genomic sequences that is exchanged between scientists across the world. b. a quantitative understanding of integrated dynamic behavior of complex biological systems and processes. c. a field of biology that is focused on developing the computational methods to analyze and extract useful biological information from the sequences of DNA. d. the systematic inactivation of each gene in the genome by homologous recombination. Answer: c Textbook Reference: Systems Biology Bloom’s Category: 1. Remembering Learning Objective: Contrast the approaches of traditional biological experimentation and systems biology. Feedback A: Incorrect. This does not describe what bioinformatics encompasses. Feedback B: Incorrect. This describes a new field of biology known as systems biology. Feedback C: Correct! This is the definition of bioinformatics. Feedback D: Incorrect. This describes “gene knockout” experiments. 17. In RNAi screens, siRNAs are incubated with cells in culture to assess functional consequences of the targeted gene. Which statement regarding siRNA and its action is true? a. The siRNA is homologous to and binds intronic gene sequences, preventing appropriate RNA splicing. b. The siRNA is homologous to and binds promoter sequences in the gene, preventing expression of the gene. c. The siRNA is homologous to and binds exonic gene sequences, giving rise to a truncated nonfunctional protein. d. The siRNA is homologous to and binds the mRNA, leading to degradation of the mRNA. Answer: d

Textbook Reference: Systems Biology Bloom’s Category: 2. Understanding Learning Objective: Summarize the methods used for large-scale screens of gene function. Feedback A: Incorrect. The siRNA is not directed toward any of the gene sequences. Feedback B: Incorrect. The siRNA is not directed toward any of the gene sequences. Feedback C: Incorrect. The siRNA is not directed toward any of the gene sequences. Feedback D: Correct! The siRNA is thought to bind the mRNA in the cytoplasm where it cannot be used for translation and subsequently gets degraded. 18. Epinephrine increases available glucose by binding to a cell surface receptor that activates a G protein, which then activates cAMP-dependent protein kinase, then phosphorylase kinase, and finally glycogen phosphorylase, which breaks down glycogen into glucose-1-phosphate. This is an example of a. a network. b. a signaling pathway. c. crosstalk. d. a feedback loop. Answer: b Textbook Reference: Systems Biology Bloom’s Category: 2. Understanding Learning Objective: Illustrate the types of interactions between pathways in a regulatory network. Feedback A: Incorrect. Networks are multiple pathways working together. Feedback B: Correct! This is a classic example of a single signaling pathway from the initiating event to the terminal effect. Feedback C: Incorrect. Crosstalk is when activity in one signaling pathway leads to an effect in another signaling pathway. Feedback D: Incorrect. Feedback loops are when a downstream effector within a pathway either positively or negatively affects an upstream activity. 19. Epinephrine increases available glucose by binding to a cell surface receptor that activates a G protein, which then activates cAMP-dependent protein kinase, then phosphorylase kinase, and finally glycogen phosphorylase, which breaks down glycogen into glucose-1-phosphate. Epinephrine breaks down glucose by activating five different proteins, stimulating glycogen synthase. Experimentally, you discover that in addition to stimulating the G protein, the epinephrine receptor can also directly stimulate glycogen synthase, bypassing the intermediate proteins. This would be an example of a. a network. b. a feedforward relay. c. crosstalk. d. positive feedback. Answer: b Textbook Reference: Systems Biology Bloom’s Category: 3. Applying

Learning Objective: Illustrate the types of interactions between pathways in a regulatory network. Feedback A: Incorrect. Networks are multiple pathways working together. Feedback B: Correct! It is considered feedforward because it jumps one or more steps in the multistep signaling pathway. Feedback C: Incorrect. Crosstalk is when activity in one signaling pathway leads to an effect in another signaling pathway. Feedback D: Incorrect. This is a downstream component within a signaling pathway stimulates an upstream component of the same pathway. 20. The treat malaria, pharmaceutical company Sanofi developed a. a genetic toggle switch that induced apoptosis genes in Plasmodium falciparum. b. a method to extract high yields of artemisinin from the sweet wormwood plant. c. a sterile strain of Plasmodium falciparum. d. a new yeast strain that produces artemisinic acid. Answer: d Textbook Reference: Systems Biology Bloom’s Category: 1. Remembering Learning Objective: Define synthetic biology. Feedback A: Incorrect. However, this is an example of synthetic biology. Feedback B: Incorrect. This was the old method for obtaining artemisinin and was almost prohibitively expensive and very limiting. Feedback C: Incorrect. However, this could be an example of synthetic biology. Feedback D: Correct! The purified artemisinic acid can easily be converted to artemisinin.

Essay 1. The genome of the yeast S. cerevisiae has been completely sequenced and has been predicted to contain 6,241 genes. Why is this only an estimate? Answer: When sequencing a genome that has not been fully characterized, any openreading frame of a given arbitrary length, usually 100 codons (300 base pairs), is considered to be a gene. However, it is possible that there are genes, smaller than this cutoff value, that were missed, resulting in an underestimate. Alternatively, randomly occurring stretches of DNA that are 300 base pairs long and lack a stop codon may have been counted as genes, resulting in an overestimate. An exact count must await characterization of the coding portion of the genome. Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Compare the numbers of genes in bacterial and yeast genomes. 2. Prior to sequencing the complete genomes of a variety of species, it was thought that the number of protein-encoding genes in an organism was proportional to the complexity of the organism. What are some examples that have proved this idea to be invalid? Answer: We now know that there is no relationship between gene number and an

organism’s complexity. By sequencing and analyzing the complete genomes of multiple organisms, we have a good approximation of the number of genes. This analysis has shown that the roundworm C. elegans has a gene number of 19,000, but Drosophila, a much more complex organism, has only 14,000. Humans have about 12,000 genes, yet the plant Arabidopsis thaliana has 26,000. While it is true that some less complex organisms, such as E. coli, have simple genomes with only 4,200 genes, there isn’t a reliable correlation. Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 3. Sequencing the human genome revealed that 40% of the predicted proteins are related to proteins in a remarkably wide variety of eukaryotic organisms, such as yeast, Drosophila, and C. elegans. Considering that shared proteins suggest shared functions, what functions would you expect to be shared? Answer: We first think about what the organisms have in common, such as the “central dogma.” Many of shared genes are involved in replication, transcription, and translation. These organisms also share many basic cellular processes, such as those involved in metabolism and respiration, so the shared genes likely have functions such as these. Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 2. Understanding Learning Objective: Compare the approximate size and amount of protein-coding sequence in the human genome with that of Drosophila. 4. Describe how a DNA microarray experiment could help identify genes that are differentially expressed in a cancer cell and a normal cell. Answer: DNA microarray is a method to rapidly examine all the mRNAs within a given cell or tissue sample using a glass or silicone chip on which oligonucleotides representative of all the human genes are printed. It is presumed that the transcriptome in the cancer cell and normal cell is different. Initially the mRNA from cultures of both cells would be isolated and converted into cDNA using reverse transcriptase. Each of the samples would be hybridized to the microarray chip and analyzed. Comparison of the two samples will identify genes that are unique to the cancer cell. Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 3. Applying Learning Objective: Describe the global methods used to study gene expression. 5. You have analyzed two identical samples by both RNA-seq and DNA microarray to analyze the transcriptome of a human cancer cell. DNA microarray showed expression of 8,500 unique mRNAs, while RNA-seq results showed that there were 11,000 unique mRNAs. What could explain this difference? Answer: Both methods convert the mRNA into cDNA using reverse transcriptase. But in microarray these cDNA are hybridized to short oligonucleotides representative of all 21,000 human genes. In RNA-seq, the cDNAs are actually subjected to complete nextgeneration sequencing. Therefore, if a particular gene gave rise to two different mRNAs

through alternative splicing, each one of those would hybridize to the short oligonucleotide on the microarray slide giving a single signal. Sequencing the two mRNAs by RNA-seq would reveal that they are actually two distinct mRNAs. Textbook Reference: Genomes and Transcriptomes Bloom’s Category: 4. Analyzing Learning Objective: Describe the global methods used to study gene expression. 6. An interesting observation that has arisen out of systems biology in the study of the genome and proteome is that humans have 21,000 genes, yet we have identified over 100,000 proteins. What are some of the contributors to this almost five-fold difference? Answer: One mechanism is that of alternate splicing of primary pre-RNA transcripts. In some cases, a single pre-RNA from a single gene can yield dozens of different mRNAs, each of which encode a unique protein. Other mechanisms include posttranslational modifications such as phosphorylation and the addition of carbohydrates or lipids. Textbook Reference: Proteomics Bloom’s Category: 3. Applying Learning Objective: Explain how proteins are identified by mass spectrometry. 7. In analyzing DNA sequences, why are regulatory regions so much more difficult to identify than protein coding regions? Answer: Protein-coding regions follow certain rules, such as open-reading frames, termination codons, intron-exon boundaries, etc., and they are often quite large. They can also be confirmed by analyzing mRNAs. On the other hand, regulatory regions are often nontranscribed and very small. Most are around 10 nucleotides long. Being so short, with only four base possibilities, the odds of the sequence occurring randomly are high. The mere fact that a sequence occurs does not mean it has a function, so once a potential regulatory region is identified, it must be assessed for functionality in gene regulation. Textbook Reference: Systems Biology Bloom’s Category: 2. Understanding Learning Objective: Contrast the approaches of traditional biological experimentation and systems biology. 8. Based on major technological innovations since the late 1990s, the relatively new field of bioinformatics has been growing significantly. Describe the need for and the uses of bioinformatics. Answer: Recent advances in sequencing technology, microarray analysis, and large-scale small molecule screening have resulted in the accumulation of huge amounts of experimental data. The human genome project, for example, was a massive effort, sequencing 3 × 109 bases. Simply reading the primary amino acid sequences from exons of genes, much less being able to use the data in an applied manner, would take years of human processing. These huge datasets have motivated scientists to develop methods to analyze the data. Bioinformatics lies at the interface of biology and computer science and focuses on developing the computational methods necessary to analyze and extract useful biological information from these large datasets in a timely manner. Textbook Reference: Systems Biology Bloom’s Category: 2. Understanding

Learning Objective: Contrast the approaches of traditional biological experimentation and systems biology. 9. Bioinformatics is not limited to genomic analyses. Its approaches can be used to analyze any of the new methods of experimentation that give rise to datasets. One application is in mass spectrometry, in which protein samples are partially cleaved, giving rise to a series of shorter peptides of about 20 amino acids each. These are then ionized and placed in a mass spectrometer that measures the unique mass-to-charge ratio of each peptide. What is the bioinformatic strategy for subsequent identification of the individual peptides? Answer: Incredibly powerful high-speed, and now relatively inexpensive, computers compare the data from the mass spectrometry of the protein sample with a theoretical dataset stored as a “library” in the computer. There are billions of possibilities, but with modern information technology and hardware, the comparison is easily accomplished. So, when an experiment is conducted, and the spectra are obtained for the sample in question, it is computationally compared to the library until a match to known protein is identified. Today this bioinformatic identification can be accomplished in minutes to hours, whereas in the not too distant past, it would have taken months or even lifetimes to make such comparisons. Textbook Reference: Systems Biology Bloom’s Category: 3. Applying Learning Objective: Summarize the methods used for large-scale screens of gene function. 10. Approximately 5% of mammalian genomes are conserved, suggesting that these sequences are important. However, after the human genome was sequenced, it was determined that only about 1.2% of the total DNA corresponds to protein-coding sequences. What important roles may the other 3.8% of these conserved sequences serve? Answer: The other conserved sequences are predominantly associated with regulating expression of the protein-encoding sequences. Textbook Reference: Systems Biology Bloom’s Category: 3. Applying Learning Objective: Explain the approaches used to identify gene regulatory sequences.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 6: Genes and Genomes TEST FILE QUESTIONS Multiple Choice 1. The genomes of salamanders contain ten times more DNA than the genomes of humans because salamanders a. have ten times more genes than humans have. b. need more DNA so they can regenerate new limbs. c. have more noncoding DNA than humans have. d. are more complex than humans. Answer: c Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. 2. A gene can be defined as a segment of DNA that codes for a. a protein. b. a functional product. c. messenger RNA. d. messenger RNA or ribosomal RNA. Answer: b Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. 3. Introns in mRNA-coding genes are the transcribed sequences a. that code for proteins. b. that regulate mRNA translation. c. that are removed by nucleases. d. between protein-coding sequences. Answer: d Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans.

4. In the nucleus, introns are removed from transcripts by a. restriction nucleases. b. splicing. c. exonucleases. d. endonucleases. Answer: b Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. 5. Introns were discovered when mRNA for adenovirus protein expressed in mammalian cells was hybridized to single-stranded virus DNA coding for that mRNA. When the complexes were observed under the electron microscope, a. a completely double-stranded hybrid was seen. b. a partial hybrid with loops of mRNA extending from the hybrid regions was seen. c. a partial hybrid with loops of DNA extending from the hybrid regions was seen. d. no hybrid was seen, because the introns had been removed. Answer: c Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. 6. Introns can encode a. small nucleolar RNAs. b. microRNAs. c. sequences that control gene expression. d. All of the above Answer: d Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. 7. Introns are found a. commonly in eukaryotic genes and rarely in prokaryotic genes. b. only in eukaryotic genes. c. commonly in both eukaryotic and prokaryotic genes. d. only in prokaryotic genes. Answer: a Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans.

8. Histone genes have a. a single long intron. b. no introns. c. larger introns than exons. d. larger exons than introns. Answer: b Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. 9. Introns constitute about what percentage of the average human gene? a. 3% b. 40% c. 60% d. 90% Answer: d Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. 10. A gene family is a a. set of related but slightly different genes present in one individual. b. family of individuals with the same gene. c. set of slightly different genes present as one copy each in a set of individuals. d. family of individuals in which each has an identical sequence of the same gene. Answer: a Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. 11. What is the term used to describe functional protein encoding regions found completely within an intron? a. Pseudogene b. Polymorphism c. Nested gene d. Alternative splicing Answer: c Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Explain how an intron can encode a functional protein. 12. Synthesis of different proteins from the same gene is due to

a. intron shuffling. b. exon shuffling. c. alternative splicing. d. exon splicing. Answer: c Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Show how alternative splicing can generate multiple different proteins from a single gene. 13. During the processing of miRNAs, which of the following cleaves the 5ʹ and 3ʹ tails away from the hairpin structure in the primary miRNA transcript? a. Drosha b. Dicer c. RNAse d. Integrase Answer: a Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Distinguish miRNAs from lncRNAs. 14. The consequence of the action of many miRNAs is the blocking or reduction of the eventual expression of proteins. What is the mechanism of action by which miRNAs accomplish this? a. miRNAs complementary base pair with mRNAs and inhibit translation and stimulate mRNA degradation. b. miRNAs complementary base pair to regulatory regions of genes and inhibit the transcription of genes. c. miRNAs complementary base pair to the active site on rRNAs and prevent the mRNA from interacting with ribosomes. d. miRNAs complementary base pair with intron-exon boundary sequences and prevent correct mRNA processing. Answer: a Textbook Reference: Noncoding Sequences Bloom’s Category: 2. Understanding Learning Objective: Distinguish miRNAs from lncRNAs. 15. Simple-sequence repeats a. are not transcribed. b. are present in tandem arrays of thousands of copies. c. account for approximately 10% of the human genome. d. All of the above Answer: d Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 2. Understanding Learning Objective: Describe the different types of repetitive DNA sequences.

16. Pseudogenes are a. genes that code for an RNA but do not code for a protein. b. nonfunctional gene copies. c. inactive genes. d. genes containing variant sequences. Answer: b Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Describe the different types of repetitive DNA sequences. 17. Retrotransposons and retroviruses share similar mechanisms by encoding which of the following enzymes? a. Reverse transcriptase and RNA polymerase b. Reverse transcriptase and integrase c. RNA polymerase and integrase d. DNA polymerase and integrase Answer: b Textbook Reference: Noncoding Sequences Bloom’s Category: 2. Understanding Learning Objective: Explain how transposable elements can affect gene expression. 18. Most pseudogenes are thought to have originated via a. duplication of genes that then became nonfunctional through mutation. b. reverse transcription of an mRNA and integration of the cDNA into a new chromosomal site. c. genome-wide duplication and the inactivation of one homolog by mutation. d. fusion of a prokaryotic cell with a eukaryotic cell. Answer: b Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Distinguish processed pseudogenes from pseudogenes that arose by DNA duplication. 19. Which of the following is the most abundant form of chromatin in the interphase nucleus of a cell? a. Euchromatin b. Nucleoli c. Chromosomes d. Heterochromatin Answer: a Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 20. During metaphase in actively dividing cells, DNA is in

a. heterochromatin. b. nucleoli. c. chromosomes. d. euchromatin. Answer: c Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 21. During which phase of mitosis does the nuclear membrane reform? a. Telophase b. Metaphase c. Anaphase d. Prophase Answer: a Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 22. The extended length of all the DNA in one human cell is about two a. millimeters. b. centimeters. c. meters. d. kilometers. Answer: c Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 23. Each human has _______ pairs of chromosomes. a. 21 b. 22 c. 23 d. 24 Answer: c Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 24. The chromosomes of prokaryotes differ from those of eukaryotes in that prokaryotic chromosomes are a. linear. b. multiple. c. complexed with histones. d. circular. Answer: d

Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 25. The DNA of eukaryotic cells is wrapped around histones to form structures called a. nucleoli. b. nuclear matrices. c. nucleosomes. d. centromeres. Answer: c Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 26. Which histone is not part of the nucleosome core particle? a. Histone H1 b. Histone H2A c. Histone H2B d. Histone H3 Answer: a Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 27. Nuclease digestion of chromatin occurs at sites separated by approximately 200 base pairs because a. an AT-rich region occurs every 200 base pairs. b. nucleosomes are spaced 200 base pairs apart. c. a restriction nuclease site occurs every 200 base pairs. d. two turns of the DNA around the nucleosome consist of 200 base pairs. Answer: b Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of chromatin. 28. Heterochromatin consists of a. DNA associated with nucleosomes. b. 10-nm chromatin fibers. c. decondensed, transcriptionally active chromatin. d. highly condensed, transcriptionally inactive chromatin. Answer: d Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of chromatin. 29. Euchromatin consists of

a. decondensed, transcriptionally active chromatin. b. 100-nm chromatin fibers. c. DNA associated with nucleosomes. d. highly condensed, transcriptionally inactive chromatin. Answer: a Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of chromatin. 30. A centromere is a a. region of euchromatin devoid of histones. b. sequence at the end of chromatids. c. region where proteins bind to form kinetochores. d. specific, repeated DNA sequence. Answer: c Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of centromeres and their epigenetic transmission.

31. Kinetochores are the a. sites of spindle fiber attachment to chromosomes. b. regions where two chromosomes remain attached during mitosis. c. same as centromeres. d. structures at the base of cilia and flagella. Answer: a Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of centromeres and their epigenetic transmission. 32. During mitosis, highly condensed chromosomes consist of two sister chromatids joined at the a. centrosome. b. centromere. c. kinetochore. d. spindle fiber. Answer: b Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of centromeres and their epigenetic transmission. 33. The transfer of information from parent to progeny that is not based on DNA sequences is known as

a. epigenetic inheritance. b. non-Mendelian genetics. c. sex-linked inheritance. d. Mendelian genetics. Answer: a Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of centromeres and their epigenetic transmission. 34. Telomeres are a. microtubule binding sites in the center of chromosomes. b. sites at the ends of chromosomes where DNA replication begins. c. chromosome end structures required for complete replication of linear chromosomes. d. sites at the ends of chromosomes where microtubules bind. Answer: c Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of telomeres in chromosome maintenance. 35. The function of telomeres is to a. degrade the ends of chromosomes. b. provide repetitive sequences in circular DNA molecules. c. join two sister chromatids. d. provide a site for replication of chromosome ends. Answer: d Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of telomeres in chromosome maintenance.

Fill In The Blank 1. Introns are removed by _______. Answer: RNA splicing Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. 2. Different proteins may be produced from the same gene through the process of _______. Answer: alternative splicing Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 2. Understanding

Learning Objective: Show how alternative splicing can generate multiple different proteins from a single gene. 3. Regulation of gene expression by short double-stranded RNAs is not only an important experimental method but is also normally used by cells to control mRNA _______ and _______. Answer: translation; degradation Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Distinguish miRNAs from lncRNAs. 4. The endogenous noncoding RNAs that mediate RNA interference are _______. Answer: microRNAs (miRNAs) Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Distinguish miRNAs from lncRNAs. 5. SINEs and LINEs are types of _______ DNA sequences. Answer: repetitive Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Describe the different types of repetitive DNA sequences. 6. In some breeds of dogs, the premature termination of bone growth results in short legs. In these breeds, the mRNA encoding by the gene responsible for leg length undergoes _______ and gets inserted into a LINE where its expression is impaired. Answer: retrotransposition Textbook Reference: Noncoding Sequences Bloom’s Category: 2. Understanding Learning Objective: Explain how transposable elements can affect gene expression. 7. Electron microscopy of chromatin revealed that chromatin has a beaded appearance. The beads were later identified to be _______. Answer: nucleosomes Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 8. Eukaryotic DNA wrapped around a core of eight histones forms a _______. Answer: nucleosome Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 9. The chromatin that is decondensed and contains most of the active genes is called _______.

Answer: euchromatin Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 10. The chromatin that is highly condensed and contains mostly inactive genes is called _______. Answer: heterochromatin Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 11. Prior to undergoing mitosis, dividing cells replicate DNA during _______. Answer: interphase Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 12. The specialized regions of eukaryotic chromosomes that serve as sites for association with sister chromatids are called _______. Answer: centromeres Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of centromeres and their epigenetic transmission. 13. The specialized sequences at the ends of eukaryotic chromosomes are called _______. Answer: telomeres Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of telomeres in chromosome maintenance. 14. The enzyme that replicates the sequences at the end of eukaryotic chromosomes is called _______. Answer: telomerase Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of telomeres in chromosome maintenance. 15. Cancer cells have high levels of the enzyme _______, allowing them to maintain the ends of their chromosome through indefinite divisions. Answer: telomerase Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of telomeres in chromosome maintenance.

16. The protein complex that protects the ends of chromosomes is _______. Answer: shelterin Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of telomeres in chromosome maintenance.

True/False 1. The ENCODE Project revealed that as much as 75% of the human genome is transcribed. Answer: T Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. 2. Introns are nonfunctional sequences between exons. Answer: F Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 1. Remembering Learning Objective: Explain how an intron can encode a functional protein. 3. A single miRNA can target up to 100 different mRNAs. Answer: T Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Distinguish miRNAs from lncRNAs. 4. miRNAs often lie within introns of protein-coding genes. Answer: T Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Distinguish miRNAs from lncRNAs. 5. The mature miRNA contains a double-stranded RNA incorporated into the RISC complex. Answer: F Textbook Reference: Noncoding Sequences Bloom’s Category: 2. Understanding Learning Objective: Distinguish miRNAs from lncRNAs. 6. During retrotransposition, reverse transcription occurs before integration. Answer: T Textbook Reference: Noncoding Sequences

Bloom’s Category: 2. Understanding Learning Objective: Explain how transposable elements can affect gene expression. 7. Transposable elements have played a major role in stimulating gene rearrangements that have contributed to the generation of genetic diversity. Answer: T Textbook Reference: Noncoding Sequences Bloom’s Category: 2. Understanding Learning Objective: Explain how transposable elements can affect gene expression. 8. Many of the genetic alterations responsible for inherited diseases may be due to mutations in functional noncoding regions rather than in protein-coding sequences. Answer: T Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Explain how transposable elements can affect gene expression. 9. Partial digestion of eukaryotic chromatin with micrococcal nuclease (an endonuclease) results in DNA fragments that are 200 (or multiples of 200) base pairs long because nucleosomes are spaced 200 base pairs apart. Answer: T Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of chromatin. 10. Heterochromatin gets its name from the fact that it contains a mixture of active and inactive genes. Answer: F Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 11. Telomeres are required for the complete replication of linear chromosomes. Answer: T Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 2. Understanding Learning Objective: Summarize the role of telomeres in chromosome maintenance. 12. Epigenetic inheritance is the transfer from parent to progeny of information that is not encoded in DNA sequences. Answer: T Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 2. Understanding Learning Objective: Describe the functions of centromeres and their epigenetic transmission.

Short Answer 1. A gene containing three exons would have how many introns? Answer: Since exons are distinctly separated by introns, a three-exon gene would have to contain at least two introns. Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 3. Applying Learning Objective: Explain how an intron can encode a functional protein. 2. How many possible mRNAs could be derived from a gene with three exons (exon 1, exon 2, and exon 3)? Answer: Three mRNAs could be produced through alternative splicing with exons 1‒2, 1‒3, and 2‒3. Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 3. Applying Learning Objective: Explain how an intron can encode a functional protein. 3. What explains how humans are able to express over 85,000 proteins when we are reported to only have about 21,000 genes? Answer: This is due to alternative splicing. Many genes can give rise to six or more unique mRNAs via alternative splicing, each of which can encode a different protein. Textbook Reference: The Structure of Eukaryotic Genomes Bloom’s Category: 2. Understanding Learning Objective: Show how alternative splicing can generate multiple different proteins from a single gene. 4. Why would a deleterious genetic mutation in a gene on the X chromosome more likely affect males than females? Answer: Females have two X chromosomes, and the gene on one may functionally mask the mutation on the other. Males only have one X chromosome, so there is no chance another gene can cover the mutated gene’s functions. Textbook Reference: Noncoding Sequences Bloom’s Category: 2. Understanding Learning Objective: Distinguish miRNAs from lncRNAs. 5. What was the very important outcome of the ENCODE Project? Answer: Prior to ENCODE, it was thought that the vast majority of the DNA that did not encode protein was nonfunctional “junk” DNA. The results of ENCODE have suggested that the vast majority of our genome is functional, and our views of mammalian genomes, particularly with respect to the prevalence of noncoding RNAs, have changed. Textbook Reference: Noncoding Sequences Bloom’s Category: 2. Understanding Learning Objective: Distinguish miRNAs from lncRNAs.

6. Describe the mechanisms by which retrotransposons such as SINEs and LINEs get moved around within the genome. Answer: Their transposition is mediated by reverse transcription. An RNA copy of a SINE or LINE is converted to DNA by reverse transcriptase within the cell, and the new DNA copy is integrated at a new site in the genome. Textbook Reference: Noncoding Sequences Bloom’s Category: 3. Applying Learning Objective: Explain how transposable elements can affect gene expression. 7. When chromatin DNA is partially digested with micrococcal nuclease (an enzyme that degrades DNA) and subsequently run on an agarose gel, it is found to yield DNA fragments approximately 200 base pairs long. What is the explanation for this? Answer: The binding of proteins to DNA in chromatin protects regions of the DNA from nuclease digestion so that the enzyme can attack DNA only at sites separated by approximately 200 base pairs. Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of chromatin. 8. Give two reasons why a specific telomere structure is required at the end of a eukaryotic chromosome. Answer: (1) The telomere sequence is required for the complete replication of the ends of linear chromosomes. (2) The telomere protects the ends of chromosomes from being degraded. Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 2. Understanding Learning Objective: Summarize the role of telomeres in chromosome maintenance. 9. What is the difference between telomerase in cancer cells and telomerase in normal somatic cells of an adult, and what is the significance of that difference? Answer: Cancer cells contain high levels of telomerase, which allows them to maintain the ends of their chromosomes through indefinite divisions. Normal adult somatic cells lack telomerase activity and do not divide indefinitely. Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 3. Applying Learning Objective: Summarize the role of telomeres in chromosome maintenance. 10. What is epigenetic inheritance? Answer: The transfer of information from parent to progeny that is not based on DNA sequence. An example could be information carried by histones. Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 3. Applying Learning Objective: Summarize the role of telomeres in chromosome maintenance.

DASHBOARD QUIZ QUESTIONS

Multiple Choice 1. Even though its complete genome is significantly smaller than that of humans, the flowering plant Arabidopsis thaliana has about _______ genes, which is about the same number as humans have. a. 4,000 b. 6,000 c. 26,000 d. 100,000 Answer: c Textbook Reference: The Structure of Eukaryotic Genes Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. Feedback A: Incorrect. The bacterium E. coli contains 4,000 genes. Feedback B: Incorrect. The yeast S. cerevisiae contains 6,000 genes. Feedback C: Correct! Feedback D: Incorrect. This number is incorrect, though prior to the genome project this was one estimate of the number of human genes. 2. Which statement about introns is true? a. Though they are part of a gene, they are absent from the corresponding mRNA. b. They are sequences within genes that are not transcribed. c. They make up only a small fraction of the DNA of a mammalian gene. d. Prokaryotes do not have introns in their genes. Answer: a Textbook Reference: The Structure of Eukaryotic Genes Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. Feedback A: Correct! Introns lie within genes but are not present in the mRNA encoded by the gene. Feedback B: Incorrect. Introns are transcribed but are later spliced out of the RNA transcript. Feedback C: Incorrect. In higher eukaryotes, approximately 90% of a gene consists of introns, on average. Feedback D: Incorrect. A few prokaryotic genes do have introns, though such genes are rare. 3. Which statement about exons is false? a. An exon may contain a 5ʹ untranslated region. b. An exon may contain sequences for small nucleolar RNAs. c. An exon may contain a 3ʹ untranslated region. d. Exons are separated by introns. Answer: b

Textbook Reference: The Structure of Eukaryotic Genes Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. Feedback A: Incorrect. This is true of exons. Feedback B: Correct! However, snRNA sequences may be found in introns. Feedback C: Incorrect. This is true of exons. Feedback D: Incorrect. This is true of exons. 4. Which statement about introns in the yeast Saccharomyces cerevisiae is true? a. S. cerevisiae genes do not contain introns. b. A small percentage of S. cerevisiae genes contain introns, and these introns are usually located near the beginning of the gene. c. Most genes in S. cerevisiae contain introns. d. Few S. cerevisiae genes contain introns, but those that do contain several. Answer: b Textbook Reference: The Structure of Eukaryotic Genes Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. Feedback A: Incorrect. Some S. cerevisiae genes contain introns. Feedback B: Correct! Only about 4% of S. cerevisiae genes contain introns, and they are usually located at the beginning of the gene. Feedback C: Incorrect. Only about 4% of the genes of S. cerevisiae contain introns. Feedback D: Incorrect. The S. cerevisiae genes that contain introns usually contain only one. 5. The percentage of the average human gene that encodes proteins is approximately a. 3%. b. 25%. c. 70%. d. 90%. Answer: a Textbook Reference: The Structure of Eukaryotic Genes Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. Feedback A: Correct! The human genome contains many introns and repetitive sequences that do not encode proteins, hence the low value for protein-encoding DNA. Feedback B: Incorrect. This is the approximate value for the C. elegans genome. Feedback C: Incorrect. This is the approximate value for the S. cerevisiae genome. Feedback D: Incorrect. This is the approximate value for the E. coli genome. 6. The human genome is distributed among _______ chromosomes. a. 20 b. 23

c. 30 d. 39 Answer: b Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of typical genes in bacteria, yeast, and humans. Feedback A: Incorrect. Mice have 20 chromosomes. Feedback B: Correct! The human genome contains 23 pairs of chromosomes. Feedback C: Incorrect. Cows have 30 chromosomes. Feedback D: Incorrect. Dogs have 39 chromosomes. 7. If a human gene is found to contain five introns, the mature mRNA encoded by that gene would have how many exons? a. Four exons b. Five exons c. Six exons d. There could be multiple mRNAs that contain between one and six introns. Answer: d Textbook Reference: The Structure of Eukaryotic Genes Bloom’s Category: 3. Applying Learning Objective: Explain how an intron can encode a functional protein. Feedback A: Incorrect. It is possible to have 4 exons, but d is the best answer. Feedback B: Incorrect. It is possible to have 5 exons, but d is the best answer. Feedback C: Incorrect. It is possible to have 6 exons, but d is the best answer. Feedback D: Correct! Due to alternative splicing multiple mature mRNAs containing from 1 to 6 exons could be possible from such a gene, since each exon is separated by an intron. 8. The human genome contains only 20,000 to 25,000 actual gene sequences, yet the human genome can produce up to 100,000 gene products. Which of the following processes accounts for this difference? a. Alternative splicing b. Spacer sequences c. Satellite DNA d. Short interspersed elements (SINEs) Answer: a Textbook Reference: The Structure of Eukaryotic Genes Bloom’s Category: 2. Understanding Learning Objective: Show how alternative splicing can generate multiple different proteins from a single gene. Feedback A: Correct! The presence of introns allows the exons of a gene to be joined in different combinations, resulting in the synthesis of different proteins from the same gene. Feedback B: Incorrect. Spacer sequences are noncoding DNA in eukaryotes that lie between genes.

Feedback C: Incorrect. Satellite DNAs are repeated sequences that make up about 10% of eukaryotic DNA. They are not transcribed and do not carry functional genetic information. They may, however, play important roles in chromosome structure. Feedback D: Incorrect. SINEs are 100–300 base pairs long. About 1.5 million such sequences are dispersed throughout the genome. Although SINEs are transcribed into RNA, they do not encode proteins, and their function is unknown. 9. The ENCODE project revealed that as much as 75% of the genome in humans is actually transcribed into RNA. Most of this is non‒protein-coding RNA. Which of the following include two new classes of noncoding RNA discovered by the ENCODE project? a. mRNA and tRNA b. rRNA and snRNA c. Pre-mRNA and hnRNA d. MicroRNA and lncRNA Answer: d Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Distinguish miRNAs from lncRNAs. Feedback A: Incorrect. mRNAs encode proteins, and we have known about tRNAs role in translation many years. Feedback B: Incorrect. The ribosomal and small nucleolar RNAs are indeed noncoding but were discovered years ago. Feedback C: Incorrect. These RNA species are intermediate forms of RNA found during the process of mRNA processing. Feedback D: Correct! MicroRNAs and long noncoding RNAs were unknown and a direct result of the ENCODE project. 10. Which statement best describes the mechanism of action of miRNAs in blocking expression of proteins? a. Single-stranded miRNA, in association with the RISC complex, binds to the complementary sequence in the 3ʹ UTR of mRNAs, blocking translation and inducing degradation of the mRNA. b. The miRNA duplex, in association with the RISC complex, binds to the complementary sequence in the 3ʹ UTR of mRNAs, blocking translation and inducing degradation of the mRNA. c. Single-stranded miRNA, in association with the RISC complex, binds to the complementary sequence in the 5ʹ UTR of mRNAs, blocking translation and inducing degradation of the mRNA. d. Single-stranded miRNA, in association with the Drosha and Dicer complex, binds to the complementary sequence in the 3ʹ UTR of mRNAs, blocking translation and inducing degradation of the mRNA. Answer: a Textbook Reference: Noncoding Sequences Bloom’s Category: 2. Understanding Learning Objective: Distinguish miRNAs from lncRNAs.

Feedback A: Correct! Feedback B: Incorrect. After association with RISC, the duplex RNA unwinds into a single stranded form. Feedback C: Incorrect. The complex interacts with complementary sequences on the 3ʹ end of the mRNA. Feedback D: Incorrect. Drosha and Dicer are involved in earlier steps that convert the hairpin structure into the duplex miRNA that then interacts with RISC. 11. Which of the following transposable elements accounts for ~21% of the human genome? a. Long interspersed elements (LINEs) b. Short interspersed elements (SINEs) c. Retrovirus-like elements d. DNA transposons Answer: a Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Explain how transposable elements can affect gene expression. Feedback A: Correct! LINEs are the most abundant repetitive elements accounting for 21% of the human genome. Feedback B: Incorrect. SINEs have the highest copy number, but are only 13% of the genome. Feedback C: Incorrect. Retrovirus-like elements represent 8%. Feedback D: Incorrect. DNA transposons make up about 3% of the genome. 12. Which statement about LINEs is false? a. Transposition of LINEs often carries adjacent sequences of DNA along with them into the new sites of integration. b. Transposition of LINEs may carry regulatory elements of one gene into a region of another gene. c. Some LINEs contain a gene encoding reverse transcriptase. d. Mutations induced by LINEs are exclusively harmful. Answer: d Textbook Reference: Noncoding Sequences Bloom’s Category: 2. Understanding Learning Objective: Explain how transposable elements can affect gene expression. Feedback A: Incorrect. This is true. LINEs have a very wide and highly variable influence on remodeling the genome. Feedback B: Incorrect. This is true. Feedback C: Incorrect. This is true. Feedback D: Correct! This is false. The transposition of some LINEs have been shown to induce beneficial mutations. 13. In dogs, the Fgf4 gene encodes a protein that regulates bone growth, and in its normal chromosomal location, Fgf4 yields dogs with long legs. Which statement best explains the mechanism leading to a short-legged phenotype in dogs?

a. During retrotransposition, the Fgf4 gene loses two exons, resulting in a shorter mRNA. b. A LINE is retrotransposed into the third exon of the Fgf4 gene, leading to a less functional protein that regulates bone growth. c. The Fgf4 gene is retrotransposed into the middle of a LINE that contains mutated regulatory sequences that give abnormal expression of Fgf4. d. During reverse transcription of the Fgf4 mRNA, an intron is included that inhibits expression of the Fgf4 gene. Answer: c Textbook Reference: Noncoding Sequences Bloom’s Category: 4. Analyzing Learning Objective: Explain how transposable elements can affect gene expression. Feedback A: Incorrect. This might be a possible mechanism but unlikely. Feedback B: Incorrect. This is a very possible mechanism since LINEs can be reinserted almost anywhere in the genome. Feedback C: Correct! This is the correct mechanism associated with leg length in dogs. Feedback D: Incorrect. An intron would not be present in the mRNA, so it could not be reverse transcribed. 14. A pseudogene is a a. second copy of a gene that functions when the original copy becomes damaged. b. gene that is unrelated in sequence to another gene but has the same function. c. gene that arose through gene duplication, but after acquiring mutations became nonfunctional. d. gene that evolved by gene duplication and the acquisition of mutations to yield a gene product that has a slightly different function from that of the original gene product. Answer: c Textbook Reference: Noncoding Sequences Bloom’s Category: 1. Remembering Learning Objective: Distinguish processed pseudogenes from pseudogenes that arose by DNA duplication. Feedback A: Incorrect. Pseudogenes do not functionally substitute for other genes. Feedback B: Incorrect. A pseudogene is related in sequence to another gene in the genome. Feedback C: Correct! A pseudogene is related in sequence to another gene in the genome but is unable to function, either because it is not transcribed or because the mRNA transcript does not yield a functional protein. Feedback D: Incorrect. A pseudogene does not have a function related to that of the original gene product. 15. Partial digestion of chromatin with micrococcal nuclease was found to yield DNA fragments approximately 200 base pairs long. What accounts for this phenomenon? a. The low pH environment of the nucleus renders the nuclease able to cut only in 200base-pair increments. b. This result suggests that the binding of proteins to DNA in chromatin protects regions of DNA from nuclease digestion. c. This is a random phenomenon, and its significance is unknown.

d. The nuclease cleaves at specific DNA sequences that occur every 200 base pairs in the genome. Answer: b Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 4. Analyzing Learning Objective: Diagram the structure of chromatin. Feedback A: Incorrect. The low pH environment of the nucleus does not affect the nuclease because the experiment is conducted with purified chromatin isolated from the nucleus. Feedback B: Correct! The DNA helix wraps itself around an octamer histone complex that gives chromatin fibers a beaded appearance, with beads spaced at intervals of approximately 200 base pairs. Feedback C: Incorrect. This phenomenon is well studied and is critical to our current understanding of chromosomal structures. Feedback D: Incorrect. The nuclease cleaves nonspecifically at random sequences. 16. Extensive digestion of chromatin with micrococcal nuclease was found to yield particles called nucleosome core particles that appear as beads when viewed by electron microscopy. Which of the following is not found in the nucleosome core particle? a. Two molecules of H2A protein b. Two molecules of H3 protein c. Two molecules of H4 protein d. One molecule of H1 protein Answer: d Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 4. Analyzing Learning Objective: Diagram the structure of chromatin. Feedback A: Incorrect. The nucleosome core particle contains two molecules of H2A protein. Feedback B: Incorrect. The nucleosome core particle contains two molecules of H3 protein. Feedback C: Incorrect. The nucleosome core particle contains two molecules of H4 protein. Feedback D: Correct! The H1 protein is not part of the nucleosome core particle. It is bound to the DNA as it enters each nucleosome core particle. 17. A centromere is a region of the chromosome that a. is located at the ends of the chromosomes and plays a critical role in chromosome replication and maintenance. b. is relatively decondensed and distributed throughout the nucleus. c. is very highly condensed and resembles the chromatin of cells undergoing mitosis. d. plays a critical role in ensuring the correct distribution of duplicated chromosomes to daughter cells during mitosis. Answer: d Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering

Learning Objective: Describe the functions of centromeres and their epigenetic transmission. Feedback A: Incorrect. This describes a telomere. Feedback B: Incorrect. This describes euchromatin. Feedback C: Incorrect. This describes heterochromatin. Feedback D: Correct! Centromeres serve both as the sites of association of sister chromatids and as the attachment sites for microtubules of the mitotic spindle. 18. Which statement describes centromeric sequences in humans? a. They consist of two short conserved sequences, CDEI/III, separated by 78–86 base pairs. b. They consist of a central core of unique-sequence DNA, flanked by tandem repeats of three repetitive sequence elements. c. They consist of A/T-rich -satellite DNA, spanning 1–5 million base pairs. d. They consist of two satellite sequences, transposable elements, and nonrepetitive DNA, spanning hundreds of kilobases. Answer: c Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of centromeres and their epigenetic transmission. Feedback A: Incorrect. This is the centromeric configuration for S. cerevisiae. Feedback B: Incorrect. This is the centromeric configuration for S. pombe. Feedback C: Correct! Remarkably, centromeric sequences are highly variable among species. Feedback D: Incorrect. This is the centromeric configuration for Drosophila. 19. In the chromatin around centromeres, nucleosomes are present that have a unique structure in which the core protein _______ is replaced by _______, and this variant of the nucleosome is required for protein assembly at the kinetochore. a. histone H3; CENP-A b. histone H3; histone H5 c. histone H1; CENP-A d. histone H1; histone H5 Answer: a Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of centromeres and their epigenetic transmission. Feedback A: Correct! Feedback B: Incorrect. There is no histone H5. Feedback C: Incorrect. Given the role of histone H1 locking the complex together this makes sense, but is not correct. Feedback D: Incorrect. 20. The function of telomeres is to

a. condense the chromosomes in preparation for mitosis. b. ensure chromosome replication and maintenance. c. hold the sister chromatids together and bind the spindle during mitosis. d. provide a binding site for DNA polymerase to initiate replication. Answer: b Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of telomeres in chromosome maintenance. Feedback A: Incorrect. The mechanism by which chromosomes are condensed during mitosis is not well understood but is not thought to involve telomeres. Feedback B: Correct! Telomeres stabilize chromosomes and assist in the replication of their ends. Feedback C: Incorrect. This is the function of centromeres. Feedback D: Incorrect. This is an origin of replication.

Essay 1. Suppose you are working on a set of experiments using Western blotting. Using an antibody that recognizes amino sequences encoded by exon 3 of a five-exon gene, you have demonstrated that a protein is highly expressed in a particular cell line. To assess the mRNA, you design reverse transcriptase polymerase chain reaction (rtPCR) primers to exons 2 and 4, which flank exon 3. Despite all your efforts, you cannot amplify an mRNA. You know that it is indeed expressed, since the protein is detectable. You also know that your controls are working and your experimental procedure is sound, because you are able to detect an mRNA for the other cell lines. Given what you know about the “central dogma” and gene structure, how might you explain this inability to identify the mRNA encoding this exon 3 protein? Answer: It is likely that of the five exons, the experimental cell line does not express exons 2 and/or 4; if even one of the exons is missing in the mRNA, there will be no PCR amplification. For example, the cell line may only express an mRNA which, by alternative splicing, uses exons 1, 3, and 5. In this case the protein would be detected by Western blot analysis because the antibody recognizes protein sequences in exon 3, but the primers designed to hybridize to exons 2 and 4 have no target to which they could anneal. Textbook Reference: The Structure of Eukaryotic Genes Bloom’s Category: 4. Analyzing Learning Objective: Show how alternative splicing can generate multiple different proteins from a single gene. 2. Briefly describe two molecular mechanisms that give rise to gene duplication. Answer: One is the duplication of a segment of DNA, which can result in the transfer of a block of DNA sequence to a new location in the genome. Alternatively, genes can be duplicated by reverse transcription of an mRNA, followed by integration of the cDNA copy into a new chromosomal site (retrotransposition). Textbook Reference: Noncoding Sequences

Bloom’s Category: 3. Applying Learning Objective: Distinguish processed pseudogenes from pseudogenes that arose by DNA duplication. 3. Processed pseudogenes do not have introns. Why not? Answer: Processed pseudogenes are cDNA that come from the reverse transcription of mRNAs. Since the mRNA templates have no introns, there are no intronic sequences available to be reverse transcribed. Textbook Reference: Noncoding Sequences Bloom’s Category: 3. Applying Learning Objective: Distinguish processed pseudogenes from pseudogenes that arose by DNA duplication. 4. Arabidopsis, and many other plants, have more protein coding genes than humans. What molecular mechanism is likely responsible for an organism so much less complex than humans having so many protein-coding genes? Answer: Gene duplication. For reasons unknown, plants have a much higher rate of gene duplication than humans, and in fact, it is thought that in the case of Arabidopsis, the entire genome has undergone at least two full-genome duplications. Textbook Reference: Noncoding Sequences Bloom’s Category: 4. Analyzing Learning Objective: Distinguish processed pseudogenes from pseudogenes that arose by DNA duplication. 5. A “complete” sequence of the Drosophila genome has recently been obtained; that is, the part of the genome that lies within euchromatin (and only that part) has been sequenced. Why would sequencing be focused on euchromatin, and why is the acquisition of the Drosophila genome sequence now considered to be virtually complete? Answer: Euchromatin is the part of the genome that consists of relatively decondensed chromatin, and active genes are located within it. Researchers focused on euchromatin because knowledge of the genes encoded by a genome yields the most information about the cell biology of an organism. In contrast, the useful contribution from heterochromatin, with its noncoding repetitive sequences, would most likely be very small. Hence, the sequencing of the genome is considered virtually complete. Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of chromatin. 6. Which genes are actively undergoing transcription during mitosis? Answer: None. There are essentially no active transcriptional processes occurring during mitosis because as cells enter mitosis, the chromosomes become highly condensed so they can be distributed to daughter cells. In mitotic cells at metaphase, the DNA will have been condensed approximately ten-thousand-fold. This highly condensed DNA can no longer serve as a template for RNA synthesis. and transcription ceases during mitosis. Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 2. Understanding

Learning Objective: Diagram the structure of chromatin. 7. What are two important functions of centromeres? Answer: Centromeres ensure the correct distribution of duplicated chromosomes to daughter cells during mitosis. Metaphase chromosomes consist of two identical sister chromatids held together at the centromere. Centromeres also serve as the site of attachment of microtubules of the mitotic spindle, whereby specific proteins bind to the centromere, forming the kinetochore, which is the site of spindle fiber attachment. Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 2. Understanding Learning Objective: Describe the functions of centromeres and their epigenetic transmission. 8. Why is it thought that the DNA sequences of centromeres are not critical to their function, whereas the chromatin structure is? Answer: There are two primary reasons: The first is that the sequences of centromeres among species are incredibly variable and share virtually no sequence identity. The other reason is that the substitution of histone H3 with CENP-A in nucleosome around centromeric regions occurs in all species in which it has been studied, and these unique nucleosomes are required for the assembly of kinetochore proteins, suggesting it is the chromatin structure, and not the DNA sequence, that is important. Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 4. Analyzing Learning Objective: Describe the functions of centromeres and their epigenetic transmission. 9. Because telomeres are longer in the cells of infants than they are in older people, it has been hypothesized that telomere length determines a cell’s life expectancy. Thus, some believe that finding a way to extend the length of telomeres in older people would extend their life expectancy. What is a possible problem with this approach? Answer: As cells age, the lengths of their telomeres gradually decrease until the chromosomes, and subsequently the cells themselves, are destroyed. In some cancer cells, however, this process reverses itself and the telomeres start to extend again. If telomere length really does extend a cell’s life span, there is a danger that cells will start dividing uncontrollably, as in cancer, and kill the organism. Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 4. Analyzing Learning Objective: Summarize the role of telomeres in chromosome maintenance. 10. What is the rationale for developing anti-cancer agents that inhibit telomerase? Answer: Normal somatic cells lack telomerase activity and do not divide indefinitely. However, cancer cells have high levels of telomerase activity, allowing them to maintain the ends of their chromosomes through indefinite divisions. Inhibiting telomerase would effectively limit the number of divisions a cancer cell could undergo. Textbook Reference: Chromosomes and Chromatin Bloom’s Category: 3. Applying

Learning Objective: Summarize the role of telomeres in chromosome maintenance.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 7: Replication, Maintenance, and Rearrangements of Genomic DNA Multiple Choice 1. DNA polymerases can synthesize DNA a. de novo by catalyzing the polymerization of free dNTPs. b. by adding dNTPs to complementary dNTPs on a single-stranded DNA. c. by adding dNTPs to a hydroxyl group on the end of a growing polynucleotide chain hydrogen-bonded to a strand of RNA. d. by adding dNTPs to a hydroxyl group on the end of a growing polynucleotide chain hydrogen-bonded to a strand of DNA. Answer: d Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Compare the roles of DNA polymerases in E. coli with those in mammalian cells. 2. In addition to synthesizing DNA, DNA polymerase I has a second catalytic activity: it can a. synthesize short RNA sequences. b. synthesize short polypeptide sequences. c. remove RNA primers. d. ligate short segments of DNA together. Answer: c Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare the roles of DNA polymerases in E. coli with those in mammalian cells. 3. Short segments of newly synthesized DNA on the lagging strand of DNA are called a. Okazaki fragments. b. replicons. c. origins of replication. d. lagging fragments. Answer: a Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering

Learning Objective: Contrast the mechanisms of synthesis of the leading and lagging strands of DNA. 4. The short fragments of DNA produced during DNA replication are joined together by a. RNA polymerase. b. DNA polymerase. c. DNA helicase. d. DNA ligase. Answer: d Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Contrast the mechanisms of synthesis of the leading and lagging strands of DNA. 5. Which eukaryotic DNA polymerase replicates the leading strand in the 5ʹ to 3ʹ direction? a. α b. ε c. δ d.  Answer: b Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Contrast the mechanisms of synthesis of the leading and lagging strands of DNA. 6. Which eukaryotic DNA polymerase replicates the lagging strand in the 3ʹ to 5ʹ direction? a. α b. γ c. δ d. None of the above; the lagging strand is replicated in the 5ʹ to 3ʹ direction. Answer: d Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Contrast the mechanisms of synthesis of the leading and lagging strands of DNA. 7. Primase synthesizes short sequences of _______ complementary to the _______ strand. a. DNA; leading b. RNA; lagging c. DNA; lagging d. All of the above Answer: b Textbook Reference: DNA Replication

Bloom’s Category: 2. Understanding Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. 8. In eukaryotic cells, RNA primers are removed by the combined action of 5ʹ to 3ʹ exonucleases and a. RNase A. b. RNase H. c. DNA polymerase I. d. DNA polymerase α. Answer: b Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. 9. Eukaryotic DNA polymerase ε a. is the polymerase for leading-strand replication. b. is the polymerase for lagging-strand replication. c. fills in the gaps between Okazaki fragments. d. functions primarily in repair of DNA damage. Answer: a Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. 10. In E. coli, the major enzyme responsible for DNA replication is DNA polymerase a.  b.  c.  d. α. Answer: c Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. 11. Proliferating cell nuclear antigen (PCNA) is a(n) _______ in eukaryotes. a. DNA polymerase b. sliding-clamp protein c. single-stranded DNA binding protein d. origin-of-replication binding protein Answer: b Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering

Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. 12. Free rotation of one cut DNA strand around one uncut strand is the primary function of a. topoisomerase I. b. topoisomerase II. c. DNA helicase. d. DNA polymerase. Answer: a Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. 13. The function of topoisomerase II is to a. resolve DNA tangles. b. allow DNA to swivel and unwind. c. allow daughter chromatids to separate in anaphase. d. All of the above Answer: d Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. 14. During DNA replication, the overall error frequency is 1 in _______ base pairs. a. 105 b. 106 c. 108 d. 109 Answer: d Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms that ensure accurate DNA replication. 15. The proofreading property of DNA polymerase is due to its _______ activity. a. 3ʹ to 5ʹ exonuclease b. 5ʹ to 3ʹ exonuclease c. excision repair d. endonuclease Answer: a Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms that ensure accurate DNA replication.

16. Origins of replication are the a. sites where DNA transcription starts. b. binding sites for the protein complex that initiates DNA synthesis. c. loops with two replication forks seen in replicating DNA. d. forks where DNA replication is occurring. Answer: b Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare origins of replication in bacteria and mammalian cells. 17. Autonomously replicating sequences are a. yeast plasmids. b. yeast telomeres. c. bacterial plasmids. d. yeast origins of replication. Answer: d Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare origins of replication in bacteria and mammalian cells. 18. Telomeres are the a. midpoints of chromosomes. b. microtubule attachment points on chromosomes. c. end-sequences of chromosomes. d. enzyme complexes that complete DNA replication. Answer: c Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Summarize the action of telomerase. 19. Telomerase is a. a reverse transcriptase. b. the enzyme that adds a random sequence to the ends of chromosomes. c. an enzyme first discovered in Thermus aquaticus. d. An enzyme that breaks down DNA from the ends. Answer: a Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Summarize the action of telomerase. 20. Pyrimidine dimers a. block DNA replication and transcription. b. can be repaired by photoreactivation. c. can be repaired by nucleotide-excision repair. d. All of the above Answer: d

Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. 21. The human disease in which affected individuals are extremely sensitive to sunlight and develop multiple skin cancers on exposed areas is called a. melanoma. b. zero pigment disease. c. xeroderma pigmentosum. d. Cockayne’s syndrome. Answer: c Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. 22. Cultured cells from xeroderma pigmentosum patients were unable to carry out a. base-excision repair. b. nucleotide-excision repair. c. synthesis of melanin. d. DNA synthesis. Answer: b Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. 23. The most common cause of skin cancer is damage to DNA by a. infrared light. b. ultraviolet light. c. γ radiation. d. β particle radiation. Answer: b Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Explain why defects in DNA repair lead to cancer. 24. The human genes that convey a susceptibility to hereditary nonpolyposis colorectal cancer are genes coding proteins involved in the DNA repair mechanism called a. mismatch repair. b. nucleotide-excision repair. c. photoreactivation. d. transcription-coupled repair. Answer: a Textbook Reference: DNA Repair

Bloom’s Category: 2. Understanding Learning Objective: Explain why defects in DNA repair lead to cancer. 25. Patients with hereditary nonpolyposis colorectal cancer genes should have a. frequent colonoscopies. b. a colectomy. c. gene-replacement therapy. d. an intraperitoneal injection of the normal enzyme. Answer: a Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Explain why defects in DNA repair lead to cancer. 26. E. coli DNA polymerase V a. is induced in response to high temperatures. b. recognizes thymine dimers and inserts nucleotides on the opposite strand. c. has a low frequency of errors. d. is activated during transcription. Answer: b Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Describe translesion DNA synthesis. 27. Double-stranded breaks are repaired by a. direct reversal of DNA damage. b. translesion repair. c. excision repair. d. recombinational repair. Answer: d Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms cells use to repair double-strand breaks. 28. The gene responsible for inherited breast cancer (BRCA2) encodes a protein that is involved in a. initiation of cell death by apoptosis. b. regulation of cell proliferation. c. repair of double-strand DNA breaks by homologous recombination. d. error-prone repair. Answer: c Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms cells use to repair double-strand breaks. 29. Site-specific recombination a. occurs in randomly occurring DNA sequences.

b. is mediated by proteins that recognize specific DNA target sequences. c. leads to programmed cell death. d. is also referred to as homologous recombination. Answer: b Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Describe the rearrangements in immunoglobulin heavy and light chains. 30. Site-specific recombination occurs commonly during a. mitosis of somatic cells. b. meiosis of germ cells. c. development of immune-system cells. d. prokaryotic cell division. Answer: c Textbook Reference: DNA Rearrangements Bloom’s Category: 1. Remembering Learning Objective: Describe the rearrangements in immunoglobulin heavy and light chains. 31. Recombination of DNA strands is important because it can a. rearrange DNA sequences to change gene expression during development. b. inactivate the repair of damaged sequences. c. inhibit diversity in the next generation. d. All of the above Answer: a Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Describe the rearrangements in immunoglobulin heavy and light chains. 32. Which of the following regions make up a mature immunoglobulin heavy chain? a. VDJC b. VJC c. VDJ d. VDJR Answer: a Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Describe the rearrangements in immunoglobulin heavy and light chains. 33. Which region is not part of the immunoglobulin light chain? a. V region b. J region c. D region

d. C region Answer: c Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Describe the rearrangements in immunoglobulin heavy and light chains. 34. In the mouse, the total number of heavy chains that can be generated by site-specific recombination events is about a. 600. b. 7,200. c. 105. d. 106. Answer: b Textbook Reference: DNA Rearrangements Bloom’s Category: 1. Remembering Learning Objective: Describe the rearrangements in immunoglobulin heavy and light chains. 35. A major difference between immunoglobulin heavy chains and light chains is that heavy chains contain _______ regions. a. V b. D c. J d. C Answer: d Textbook Reference: DNA Rearrangements Bloom’s Category: 1. Remembering Learning Objective: Describe the rearrangements in immunoglobulin heavy and light chains. 36. Mutations in the genes for RAG1 and RAG2 would most likely have an effect on a. VDJ recombination in the immune system. b. mismatch repair. c. excision repair. d. translesion DNA synthesis. Answer: a Textbook Reference: DNA Rearrangements Bloom’s Category: 3. Applying Learning Objective: Explain how nonhomologous end joining of double-strand breaks and cytosine deamination contribute to immunoglobulin diversity. 37. T cell receptors a. are found in the circulation. b. contain variable and regulatory regions. c. are tetrameric complexes that contain α, β, , and  polypeptide chains.

d. bind antigens present on the surface of other cells. Answer: d Textbook Reference: DNA Rearrangements Bloom’s Category: 1. Remembering Learning Objective: Explain how nonhomologous end joining of double-strand breaks and cytosine deamination contribute to immunoglobulin diversity. 38. Gene amplification is a. seen in ribosomal genes of developing amphibian oocytes. b. seen in the genes for muscle proteins of developing muscle cells. c. seen in the genes for immunoglobulins during B lymphocyte development. d. never seen in eukaryotic cells, since the amount of DNA per cell is always constant, except during S phase of the cell cycle. Answer: a Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Explain why DNA amplification increases gene expression.

Fill in the Blank 1. A mutation in the enzyme _______ will block the separation of daughter chromatids in mitosis. Answer: topoisomerase II Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. 2. On one typical eukaryotic chromosome there are, in number, _______ telomere(s) and _______ origin(s) of replication. Answer: two (or 2); many (several hundred to a thousand) Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare origins of replication in bacteria and mammalian cells. 3. On a typical prokaryotic chromosome the number of telomeres and origins of replication is _______ telomere(s) and _______ origin(s) of replication. Answer: zero; one Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare origins of replication in bacteria and mammalian cells. 4. The BRCA genes responsible for inherited breast cancer are involved in the repair of _______ DNA breaks by homologous recombination. Answer: double-stranded (double-strand)

Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms cells use to repair double-strand breaks. 5. Repair of double-strand breaks in DNA caused by ionizing radiation often leads to the loss of _______ around the site of damage. Answer: bases Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms cells use to repair double-strand breaks. 6. The genes that encode the immunoglobulin light chain consist of regions that are called _______, _______, and _______. Answer: variable (V); constant (C); joining (J) Textbook Reference: DNA Rearrangements Bloom’s Category: 1. Remembering Learning Objective: Describe the rearrangements in immunoglobulin heavy and light chains. 7. IgM antibodies contain _______ constant regions in their heavy chains. Answer: C Textbook Reference: DNA Rearrangements Bloom’s Category: 1. Remembering Learning Objective: Describe the rearrangements in immunoglobulin heavy and light chains. 8. T cell receptors bind to _______ on the surface of other cells during immune responses. Answer: antigens Textbook Reference: DNA Rearrangements Bloom’s Category: 1. Remembering Learning Objective: Explain how nonhomologous end joining of double-strand breaks and cytosine deamination contribute to immunoglobulin diversity. 9. T cell receptors resemble _______ in many ways. Answer: immunoglobulins Textbook Reference: DNA Rearrangements Bloom’s Category: 1. Remembering Learning Objective: Explain how nonhomologous end joining of double-strand breaks and cytosine deamination contribute to immunoglobulin diversity. 10. _______ can be beneficial to organisms such as amphibians to accommodate the generation of large pools of rRNAs, while it can be deleterious in some forms of cancer by dramatically increasing the number of genes driving proliferation. Answer: Gene amplification Textbook Reference: DNA Rearrangements

Bloom’s Category: 2. Understanding Learning Objective: Explain why DNA amplification increases gene expression.

True/False 1. All known DNA polymerases synthesize DNA in the 5ʹ to 3ʹ direction. Answer: T Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare the roles of DNA polymerases in E. coli with those in mammalian cells. 2. Mitochondria have their own unique DNA polymerase. Answer: T Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare the roles of DNA polymerases in E. coli with those in mammalian cells. 3. Eukaryotic cells have many different DNA polymerases, one or more of which function primarily in repair of damaged DNA. Answer: T Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding Learning Objective: Compare the roles of DNA polymerases in E. coli with those in mammalian cells. 4. The leading and lagging strands at a replication fork are synthesized in opposite directions, but both are synthesized in a continuous manner. Answer: F Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Contrast the mechanisms of synthesis of the leading and lagging strands of DNA. 5. Both topoisomerase I and topoisomerase II allow DNA to relieve torsion. Answer: T Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. 6. Some DNA polymerases have a nuclease activity that allows them to remove mismatched nucleotides and repair a sequence. Answer: T

Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanisms that ensure accurate DNA replication. 7. Proofreading is done by DNA polymerases. Answer: T Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanisms that ensure accurate DNA replication. 8. DNA replication starts at sites called replication forks. Answer: F Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Compare origins of replication in bacteria and mammalian cells. 9. Yeast origins of replication are autonomously replicating sequences. Answer: T Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare origins of replication in bacteria and mammalian cells. 10. The rate of DNA replication in mammalian cells is tenfold faster than in prokaryotic cells. Answer: F Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare origins of replication in bacteria and mammalian cells. 11. Telomere sequences form loops at the ends of eukaryotic chromosomes. Answer: T Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Summarize the action of telomerase. 12. Telomerase is a reverse transcriptase. Answer: T Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Summarize the action of telomerase. 13. Telomerase carries its own template DNA. Answer: F Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Summarize the action of telomerase.

14. Telomerase extends the ends of linear chromosomes by making a copy that is complementary to the other strand of DNA. Answer: F Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Summarize the action of telomerase. 15. DNA glycosylase can give rise to apyrimidinic and apurinic sites by cleaving bases from their deoxyribose on the DNA backbone. Answer: T Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. 16. DNA polymerases cannot replicate DNA across from a site of DNA damage. Answer: F Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Describe translesion DNA synthesis. 17. Somatic hypermutation is focused in the diversity region of the immunoglobulin gene. Answer: F Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Explain how nonhomologous end joining of double-strand breaks and cytosine deamination contribute to immunoglobulin diversity.

Short Answer 1. How do DNA polymerases differ from RNA polymerases in terms of primer requirement? Answer: DNA polymerases require a primer, whereas RNA polymerases do not. Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Compare the roles of DNA polymerases in E. coli with those in mammalian cells. 2. Why does the synthesis of Okazaki fragments require an RNA primer? Answer: An RNA primer is needed because RNA polymerase can initiate an RNA primer on a single strand of DNA, and DNA polymerases can only elongate a double strand from an existing double strand–single strand boundary. Textbook Reference: DNA Replication

Bloom’s Category: 2. Understanding Learning Objective: Contrast the mechanisms of synthesis of the leading and lagging strands of DNA. 3. Why does DNA polymerase synthesize DNA only in the 5ʹ to 3ʹ direction? Answer: In the 5ʹ to 3ʹ direction, the energy for polymerization comes from hydrolysis of the 5ʹ triphosphate of the free dNTP as it is added to the 3ʹ hydroxyl group of the existing chain. If the direction were 3ʹ to 5ʹ, proofreading repair could not be performed because the removal of the terminal nucleotide would leave no 5ʹ triphosphate to provide the energy to bond to the 3ʹ hydroxyl group of the incoming nucleotide triphosphate. Textbook Reference: DNA Replication Bloom’s Category: 3. Applying Learning Objective: Describe the mechanisms that ensure accurate DNA replication. 4. Explain how a prokaryotic chromosome can replicate and divide normally with no telomeres and only one origin of replication. Answer: Because a prokaryotic chromosome is circular, it has no ends, so no telomeres are required. Because prokaryote genomes are a single small circle and the rate of prokaryotic chromosome replication is faster than that of eukaryotes, a single origin of replication is sufficient. Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Compare origins of replication in bacteria and mammalian cells. 5. What are the functions of origins of replication? Answer: Origins of replication initiate DNA synthesis at those sites on a chromosome. Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Compare origins of replication in bacteria and mammalian cells. 6. What are the functions of telomeres? Answer: Telomeres form or maintain the ends of linear chromosomes so that they can be completely replicated with the aid of telomerase. Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Summarize the action of telomerase. 7. Why is telomerase important in cancer cells? Answer: Abnormal increases in telomerase activity can allow tumor cells to replicate telomeres and keep proliferating indefinitely. Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Summarize the action of telomerase. 8. What type of DNA damage would you expect to occur more frequently in populations as they near the equator?

Answer: Given the greater intensity of sunlight as you approach the equator, one would expect to see more pyrimidine dimers in populations closer to the equator due to the elevated UV light. This is actually true, and in fact the incidence of melanoma increases among populations closer to the equator. Textbook Reference: DNA Repair Bloom’s Category: 4. Analyzing Learning Objective: Explain why defects in DNA repair lead to cancer. 9. Explain Tonegawa’s key conclusion about the development of immunoglobulin genes. Answer: During development, immunoglobulin genes undergo recombination so that each gene encoding a heavy or a light chain consists of a different combination of V, J, and C regions. Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Describe the rearrangements in immunoglobulin heavy and light chains.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. In E. coli, the major enzyme responsible for DNA replication is DNA polymerase a. I. b. II. c. III. d. . Answer: c Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare the roles of DNA polymerases in E. coli with those in mammalian cells. Feedback A: Incorrect. DNA polymerase I was the first to be isolated, but it is not the major polymerase responsible for DNA replication. Feedback B: Incorrect. E. coli lacking DNA polymerase II grow normally; its role is unknown. Feedback C: Correct! E. coli lacking DNA polymerase III activity are unable to replicate their DNA and this, together with other data, has identified it as the major DNA replication enzyme. Feedback D: Incorrect. This is a major replication enzyme in eukaryotic cells, not in E. coli. 2. Which statement is true of all known DNA polymerases? a. They synthesize DNA in the 5ʹ to 3ʹ direction, and they require a preformed primer hydrogen-bonded to the template. b. They synthesize DNA in the 5ʹ to 3ʹ direction, and they possess primase activity.

c. They require a preformed primer, and they possess helicase activity. d. They synthesize DNA in the 3ʹ to 5ʹ direction, and they possess exonuclease activity. Answer: a Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Compare the roles of DNA polymerases in E. coli with those in mammalian cells. Feedback A: Correct! They can synthesize DNA only from 5ʹ to 3ʹ, and they cannot begin polymerization de novo (i.e., in the absence of a primer). Feedback B: Incorrect. Primase is a separate enzyme that synthesizes short RNA molecules. Feedback C: Incorrect. Helicase action is involved in DNA replication, but it is performed by other enzymes. Feedback D: Incorrect. DNA is not synthesized in the 3ʹ to 5ʹ direction, and not all polymerases have exonuclease activity. 3. DNA polymerase requires a primer and cannot initiate synthesis de novo. What serves as a primer for DNA replication? a. Short fragments of DNA complementary to the template strand b. A protein with a free OH group c. Short fragments of RNA complementary to the template strand d. Double-stranded hairpins at the end of the DNA molecule that are formed by a looping of the DNA Answer: c Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare the roles of DNA polymerases in E. coli with those in mammalian cells. Feedback A: Incorrect. DNA polymerase cannot initiate synthesis de novo, so synthesizing a DNA primer would present the same problem as initiating synthesis of Okazaki fragments without a primer. Feedback B: Incorrect. While some viruses do employ proteins as primers, prokaryotic and eukaryotic DNA replication require an RNA primer. Feedback C: Correct! In contrast to DNA synthesis, the synthesis of RNA can be initiated de novo, and an enzyme called primase synthesizes short fragments of RNA complementary to the template strand at the replication fork. Feedback D: Incorrect. This method is employed by some viruses, but prokaryotic and eukaryotic DNA replication requires an RNA primer. 4. Which statement concerning elongation of DNA at the replication fork is false? a. The leading strand is synthesized continuously in the direction of replication fork movement. b. The lagging strand is synthesized in Okazaki fragments backward from the overall direction of replication. c. The Okazaki fragments are joined by the action of DNA ligase. d. Both strands are synthesized continuously at the replication fork.

Answer: d Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Contrast the mechanisms of synthesis of the leading and lagging strands of DNA. Feedback A: Incorrect. This statement is true. Only one strand is synthesized continuously, and this strand is called the leading strand, since its elongation in the direction of replication fork movement exposes the template used for the synthesis of Okazaki fragments. Feedback B: Incorrect. This statement is true. The lagging strand is formed from short, discontinuous pieces of DNA that are synthesized backward with respect to the direction of movement. The small pieces of DNA are named Okazaki fragments after their discoverer. Feedback C: Incorrect. This statement is true. DNA ligase joins Okazaki fragments, forming a new, intact DNA strand. Feedback D: Correct! This statement is false. The two strands of double-helical DNA run in antiparallel directions. Continuous synthesis of two new strands at the replication fork would require that one strand be synthesized in the 3ʹ to 5ʹ direction, but DNA polymerase synthesizes DNA only in the 5ʹ to 3ʹ direction. 5. The twisting of the parental DNA strands around each other ahead of a replication fork is relieved by enzymes called a. DNA helicases. b. topoisomerases. c. DNA ligases. d. DNA polymerases. Answer: b Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. Feedback A: Incorrect. DNA helicases unwind the DNA ahead of the replication fork, thus exposing the template strands for daughter strand synthesis. Feedback B: Correct! Type I topoisomerase cleaves a single strand and type II topoisomerase cleaves both parental strands, but they carry out the same function, which is to relieve the twisting ahead of the replication fork by allowing the DNA to swivel at the breakpoint. Feedback C: Incorrect. DNA ligases join the DNA fragments of the lagging strand together to make a single molecule. Feedback D: Incorrect. DNA polymerases do not carry out this function. 6. During replication, _______ stabilize the unwound DNA template so it can be copied by the DNA polymerase. a. Single-stranded DNA-binding proteins b. Helicases c. PCNAs

d. Topoisomerases Answer: a Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. Feedback A: Correct! Single-stranded DNA-binding proteins stabilize the newly exposed single strand of DNA. Feedback B: Incorrect. These unwind the DNA ahead of the replication fork. Feedback C: Incorrect. This is a sliding clamp that helps load the polymerase on the template. Feedback D: Incorrect. Topoisomerases relax the torsional constraints caused by the helicases in unwinding. 7. Estimates of mutation rates for a variety of genes indicate that the frequency of errors during replication is much lower than would be predicted on the basis of complementary base pairing. What accounts for the higher degree of fidelity? a. Conformational changes in DNA polymerase b. 3ʹ to 5ʹ exonuclease activity of DNA polymerase c. Requirement of a primer for DNA synthesis by DNA polymerase d. All of the above Answer: d Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanisms that ensure accurate DNA replication. Feedback A: Incorrect. This is correct, but the other answer choices are correct as well. Structural studies of several DNA polymerases indicate that the binding of correctly matched dNTPs induces conformational changes in DNA polymerase that lead to the incorporation of the nucleotide into DNA. Feedback B: Incorrect. This is correct, but the other answer choices are correct as well. This enzyme operates in the reverse direction of DNA synthesis and participates in proofreading newly synthesized DNA. Feedback C: Incorrect. This is correct, but the other answer choices are correct as well. Proofreading is effective because DNA polymerase requires a primer. Because a mispaired base is not hydrogen-bonded to the template strand, this mismatch at the 3ʹ terminus of the growing chain is recognized and excised by the 3ʹ to 5ʹ exonuclease activity of DNA polymerase, which requires a primer hydrogen-bonded to the template strand in order to continue synthesis. Feedback D: Correct! All three of the answer choices correctly identify mechanisms that contribute to the high degree of fidelity of DNA replication by DNA polymerase. 8. How many replication origins are there in the human genome? a. 1 b. 350 c. 30,000 d. 350,000

Answer: c Textbook Reference: DNA Replication Bloom’s Category: 1. Remembering Learning Objective: Compare origins of replication in bacteria and mammalian cells. Feedback A: Incorrect. The E. coli genome has a single origin of replication. Feedback B: Incorrect. This is the approximate number in the yeast genome. Feedback C: Correct! The large size of the human genome requires many origins for replication to be accomplished in a reasonable amount of time. Feedback D: Incorrect. This is too many. This would mean that origins would be spaced approximately every 8 kb in the human genome. 9. Which statement about pyrimidine dimers is false? a. They are lesions in DNA caused by UV radiation. b. They are formed between adjacent pyrimidines on a DNA strand. c. Their formation blocks DNA replication and transcription. d. They can be repaired by photoreactivation in human cells. Answer: d Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. Feedback A: Incorrect. This is a true statement. The major type of damage to DNA caused by UV light is the formation of pyrimidine dimers. Feedback B: Incorrect. This is a true statement. Pyrimidine dimers are formed between residues on the same strand of DNA. Feedback C: Incorrect. This is a true statement. Both these processes are inhibited by pyrimidine dimers. Feedback D: Correct! Photoreactivation returns the pyrimidine bases to their original state, and although it functions in a variety of organisms, human cells lack this mechanism of DNA repair. 10. How does nucleotide-excision repair differ from base-excision repair? a. Base-excision repair recognizes and removes single damaged bases, whereas nucleotide-excision repair is more general, recognizing many different kinds of lesions that distort the DNA molecule. b. Nucleotide-excision repair reverses the chemical reaction that caused the lesion, whereas base-excision repair removes the damaged bases and replaces them with normal ones. c. Only the base is removed in base-excision repair, whereas the entire nucleotide is removed in nucleotide-excision repair. d. Base-excision repair requires no protein components and can occur by simple absorption of UV light, whereas nucleotide-excision repair requires several enzymes. Answer: a Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding

Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. Feedback A: Correct! In addition, base-excision repair removes single damaged bases, whereas nucleotide-excision repair removes a larger region of DNA encompassing the damaged site. Feedback B: Incorrect. Both mechanisms involve excision followed by replacement of the damaged bases. Feedback C: Incorrect. The first step in base-excision repair is removal of the base, but the entire nucleotide is subsequently removed. Feedback D: Incorrect. Both processes require several enzymes. 11. During mismatch repair in E. coli, the parental strand is recognized by a. single-stranded breaks. b. glycosylated adenines. c. methylated adenines. d. methylation of the O6 position of guanine residues. Answer: c Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. Feedback A: Incorrect. This is the mechanism for identification in eukaryotes. Feedback B: Incorrect. Adenines in DNA are not glycosylated. Feedback C: Correct! Methylation of adenine residues in the GATC sequence occurs after replication, and therefore the mismatch-repair machinery can distinguish the newly replicated (unmethylated) strand from the parental (methylated) strand. Feedback D: Incorrect. This causes a lesion in DNA and is not a normal part of DNA structure. 12. In nucleotide-excision repair of thymine dimers, what is the correct order of enzymatic activity, following damage recognition? a. Helicase  nuclease  ligase  polymerase b. Nuclease  helicase  ligase  polymerase c. Nuclease  helicase  polymerase  ligase d. Polymerase  helicase  ligase  nuclease Answer: c Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. Feedback A: Incorrect. Just after damage recognition, nucleases cleave a few bases on either side of the dimers. Feedback B: Incorrect. After helicase unwinds the DNA, DNA polymerase fills the gap left by the excised piece. Feedback C: Correct! Following damage recognition, nucleases cleave a few bases on either side of the dimers and then helicase unwinds the DNA, after which DNA

polymerase fills the gap left by the excised piece and ligases seal the ends. Feedback D: Incorrect. Just after damage recognition, nucleases cleave a few bases on either side of the dimers. 13. Deamination of cytosine results in the conversion of cytosine to a. hypoxanthine. b. adenine. c. 7-methyl guanosine. d. uracil. Answer: d Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. Feedback A: Incorrect. Hypoxanthine is derived from the deamination of adenine. Feedback B: Incorrect. This is incorrect, though adenine is also capable of being deaminated. Feedback C: Incorrect. This is the cap structure for mRNAs. Feedback D: Correct! Removal of the amino group of cytosine converts it to uracil. 14. Mutation in the MutS gene is most commonly associated with a. colorectal cancer. b. melanoma. c. squamous cell carcinoma of the skin. d. small-cell carcinoma. Answer: a Textbook Reference: DNA Repair Bloom’s Category: 1. Remembering Learning Objective: Explain why defects in DNA repair lead to cancer. Feedback A: Correct! Mutations in MutS are responsible for about half of all HNPCC cases. Feedback B: Incorrect. Melanoma is more commonly associated with excision repair genes, such as those defective in xeroderma pigmentosum. Feedback C: Incorrect. Squamous cell carcinoma is more commonly associated with excision repair genes, such as those defective in xeroderma pigmentosum. Feedback D: Incorrect. Small-cell carcinoma of the lung is most commonly associated with smoking. 15. When a thymine dimer is encountered on the DNA template strand, what mechanism is responsible for correcting it on the newly synthesized strand? a. Mismatch repair b. Base-excision repair c. Translesion DNA repair d. Transcription-coupled repair Answer: c Textbook Reference: DNA Repair

Bloom’s Category: 2. Understanding Learning Objective: Describe translesion DNA synthesis. Feedback A: Incorrect. Mismatch is part of proofreading and correcting an incorrectly incorporated base. Feedback B: Incorrect. Base excision removes modified bases. Feedback C: Correct! When the DNA polymerase encounters thymine dimers, a specialized DNA polymerase (pol V) recognizes the dimers and synthesizes the new adenines, and the dimers are subsequently repaired. Feedback D: Incorrect. This repair mechanism is used in genes that are actively being transcribed. 16. Which of the following does not contribute to the large variety of antigen-binding specificities found among the immunoglobulins? a. Recombination between different versions of V, D, and J segments b. Somatic hypermutation c. Imprecise joining of immunoglobulin segments d. Retrotransposons Answer: d Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Explain how nonhomologous end joining of double-strand breaks and cytosine deamination contribute to immunoglobulin diversity. Feedback A: Incorrect. V(D)J recombination is a major contributor to the diversity of immunoglobulin heavy chains. Recombination between V and J segments of the light chains also contributes to diversity. Feedback B: Incorrect. This is a contributor to the genetic diversity among immunoglobulins. Feedback C: Incorrect. This increases the genetic diversity among immunoglobulins by about 100 times. Feedback D: Correct! The DNA rearrangements that occur to generate immunoglobulin genes involve site-specific recombination, but they do not involve an RNA intermediate. 17. Which statement regarding somatic hypermutation is false? a. The enzyme activation-induced deaminase (AID) is a key player in somatic hypermutation. b. Somatic hypermutation is thought to be the result of a high frequency of errors during DNA repair. c. Somatic hypermutation is thought to control the proliferation of B lymphocytes by rendering their genome nonreplicable. d. Somatic hypermutation substantially increases affinity for antigen. Answer: c Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Explain how nonhomologous end joining of double-strand breaks and cytosine deamination contribute to immunoglobulin diversity. Feedback A: Incorrect. This is a true statement. AID is a key player in somatic

hypermutation. It is expressed only in B lymphocytes, and it catalyzes the deamination of cytosine to form uracil in DNA. Feedback B: Incorrect. This is a true statement. Somatic hypermutation is thought to be the result of a high frequency of errors during repair of CU mutations. Feedback C: Correct! This is false regarding somatic hypermutation, as it does not have any role in the regulation of B cell proliferation during the immune response. Feedback D: Incorrect. This statement is true. Somatic hypermutation increases affinity for antigen and is an important contributor to an effective immune response. 18. Which of the following enzymes serves an important role in both somatic hypermutation and class switch recombination? a. Helicase b. Exonuclease c. DNA glycosylase d. Activation-induced deaminase Answer: d Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Explain how nonhomologous end joining of double-strand breaks and cytosine deamination contribute to immunoglobulin diversity. Feedback A: Incorrect. Helicase unwinds double-stranded DNA. Feedback B: Incorrect. Exonucleases cleave nucleotides from one strand of doublestranded DNA. Feedback C: Incorrect. DNA glycosylase can be involved in removing uracils from DNA in base-excision repair. Feedback D: Correct! Deaminating C to U in many locations throughout the variable and switch regions leads to the production of immunoglobulins with increased affinity for antigens. 19. Which statement about gene amplification is false? a. Gene amplification is responsible for amplification of ribosomal RNA genes in amphibian oocytes. b. Amplified DNA sequences can be found as free extrachromosomal molecules. c. Gene amplification occurs as an abnormal event in cancer cells. d. Amplified DNA sequences are a common occurrence in virally infected cells. Answer: d Textbook Reference: DNA Rearrangements Bloom’s Category: 2. Understanding Learning Objective: Explain why DNA amplification increases gene expression. Feedback A: Incorrect. This is a true statement. Gene amplification is seen in amphibian oocytes because amphibian oocytes are about one million times larger in volume than typical somatic cells and must support large amounts of protein synthesis. Feedback B: Incorrect. This is a true statement. Amplified DNA sequences can be found as free extra-chromosomal molecules or as tandem arrays of sequences within a chromosome. Feedback C: Incorrect. This is a true statement. Gene amplification is seen as an

abnormal event in malignant cells. Such gene amplification was first recognized in cancer cells that had become resistant to methotrexate, a commonly used chemotherapy drug. Feedback D: Correct! This statement is false. Gene amplification is not associated with cells infected with viruses.

Essay 1. Why does the proofreading activity of DNA polymerases rely on synthesis in the 5ʹ to 3ʹ direction as opposed to the 3ʹ to 5ʹ direction? Answer: In 5ʹ to 3ʹ synthesis, the energy for polymerization comes from cleavage of the 5ʹ triphosphate of the dNTP about to be incorporated into the chain. Therefore, if a mismatched nucleotide has to be replaced, it is simply removed and the correct nucleotide is incorporated in its place. If synthesis were in the 3ʹ to 5ʹ direction, the energy for incorporation would be derived from the 5ʹ triphosphate of the last nucleotide incorporated into the chain. Replacing it due to a mismatch would mean its removal and the loss of the energy source needed for the addition of the next nucleotide in the chain. Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Compare the roles of DNA polymerases in E. coli with those in mammalian cells. 2. Why do E. coli chromosomes not have telomeres? Answer: The E. coli chromosome is circular and therefore does not have ends. It is thus not necessary to extend the end of the template strand in lagging strand synthesis as it is for the replication of linear DNA molecules. Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Contrast the mechanisms of synthesis of the leading and lagging strands of DNA. 3. Why are topoisomerases required in replication? Answer: Topoisomerases are needed because as DNA unwinds, or as the two strands of the double helix get pulled apart, the helix ahead of the unwinding helix can get stuck (much like the strands of a twisted rope get stuck as one tries to separate them). Topoisomerases catalyze reactions that break the strand, allowing the helix to relax and then rejoin the strand. This is repeated over and over and prevents the torsional strain that would prevent unwinding. Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Identify the proteins found at replication forks of bacteria and mammalian cells. 4. Explain why accuracy is more important for DNA replication than for transcription. Answer: Mistakes made during DNA replication are heritable and thus are passed on to generations of cells in a lineage. The accumulation of such mistakes would eventually

lead to a breakdown of essential processes in the cell and cell death. In contrast, mistakes made in an RNA transcription are not heritable, and the cell can afford a few aberrant RNA transcripts as long as the majority encode the wild-type protein. Hence, an elaborate system of proofreading has evolved for DNA replication but not for RNA transcription. Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanisms that ensure accurate DNA replication. 5. Why would chromosomes shorten with each replication cycle in the absence of telomerase? Answer: The synthesis of the lagging strand involves the continual synthesis of RNA primers, which are then elongated by DNA polymerase, as the replication fork moves along a DNA molecule. The enzyme that synthesizes the primers, primase, is templatedirected, meaning that it cannot synthesize a primer de novo but instead needs a template. This is problematic once the fork reaches the end of the chromosome because the template runs out. This problem is solved by telomerase, which extends the template strand by a repeat unit, thus providing a template for primase and, consequently, replication of the lagging strand to its full length. Textbook Reference: DNA Replication Bloom’s Category: 2. Understanding Learning Objective: Summarize the action of telomerase. 6. Describe the two major forms of spontaneous DNA damage. Answer: One form of spontaneous damage is simple deamination. There are three bases with amino groups (cytosine, adenine, and guanine), and all three of these could be deaminated. The other form of spontaneous damage is depurination, which occurs to both adenine and guanine nucleotides when the complete purine base is lost after cleavage of the phosphodiester bond. Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. 7. UV irradiation often induces DNA damage in the form of pyrimidine dimers. Briefly describe the two mechanisms that cells have developed to repair this damage. Which mechanism is lacking in humans? Answer: The most direct process is photoreactivation, which uses energy from visible light to break the cyclobutane ring structure formed by the two adjacent bases. Photoreactivation is not observed in humans. The other process is nucleotide excision repair, in which DNA on either side of the dimer is cleaved by 3ʹ and 5ʹ nucleases and the short fragment containing the dimer is excised out. The resulting gap is filled in by DNA polymerase. Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair.

8. Describe the AP site that is generated by the action of DNA glycosylase. Answer: DNA glycosylase initiates the repair mechanism to replace a uracil that was inadvertently placed instead of thymine or a uracil that resulted from the deamination of cytosine. It cleaves the bond linking the nitrogenous base to its deoxyribose sugar on the DNA backbone. So, instead of the entire nucleoside being removed and the backbone getting cleaved, the base is removed, creating an apyrimidinic (AP) site. Subsequently, deoxyribosephosphodiesterase is needed to cleave off the deoxyribose. Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast direct repair of DNA damage with the different types of excision repair. 9. Pyrimidine dimers cannot be copied by the normal action of DNA polymerases. Therefore, DNA replication would stop when the polymerase encounters one of these dimers. What mechanism have cells developed to deal with this damage when it is encountered during replication? Answer: This type of repair is called translesion DNA synthesis. When these dimers are encountered during replication, special DNA polymerases are recruited to the site, and new nucleotides are added to the new strand that are complementary to the dimerized bases. Thus, thymine dimers would have complementary adenines added to the new strand by these special polymerases. It should be noted, however, that these special polymerases are error-prone, in that they have low fidelity (meaning they often insert incorrect bases), and that they do not have proofreading capability. After synthesis of the new strand, the dimers are recognized and repaired by normal nucleotide excision repair mechanisms. Textbook Reference: DNA Repair Bloom’s Category: 2. Understanding Learning Objective: Describe translesion DNA synthesis. 10. Though not a common mechanism, gene amplification is used by cells to rapidly give rise to multiple copies of mRNAs to meet large demands for particular proteins. How is gene amplification accomplished? Give an example of when a cell may activate this process. Answer: Gene amplification occurs through repeated rounds of replication of a gene. The result is many copies of a single gene are available for transcription. A common example of normal gene amplification occurs in the development of Drosophila, in which the genes that encode egg shell proteins are amplified to meet the high demands for the protein in the ovarian cell. Unfortunately, gene amplification is also used in some cancer cells. A classic example is cancer that has become resistant to methotrexate treatment. Methotrexate is a chemotherapeutic agent that inhibits dihydrofolate reductase (DHFR), a critical enzyme needed for DNA synthesis (and thus for cancer cell growth). Cancer cells may develop methotrexate resistance through gene amplification of the DHFR gene, which effectively outcompetes the drug. Other cancer cells may directly amplify genes critical to the cell cycle, thus constitutively activating cell growth. Textbook Reference: DNA Rearrangements

Bloom’s Category: 3. Applying Learning Objective: Explain why DNA amplification increases gene expression.

Test Bank to accompany

The Cell: A Molecular Approach, Seventh Edition Geoffrey M. Cooper

Chapter 8: RNA Synthesis and Processing TEST FILE QUESTIONS Multiple Choice 1. The prokaryotic cells most commonly used to study transcription and translation are from a. E. coli. b. B. subtilis. c. S. cerevisiae. d. D. discoideum. Answer: a Textbook Reference: Transcription in Bacteria Bloom’s Category: 1. Remembering Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 2. RNA polymerase differs from DNA polymerase in that it a. synthesizes new strands of RNA in a 3ʹ to 5ʹ direction. b. is a monomeric protein. c. can synthesize a complementary strand without the two strands of DNA being separated. d. does not require a primer to initiate synthesis of RNA. Answer: d Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 3. The DNA sequence to which an RNA polymerase binds to initiate transcription of a gene is called a(n) a. enhancer. b. promoter. c. polymerase-binding element. d. origin of transcription. Answer: b Textbook Reference: Transcription in Bacteria Bloom’s Category: 1. Remembering Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 4. The regions of the DNA where RNA polymerase binds can be identified by

a. restriction mapping. b. PCR. c. inhibited transcription following mutagenesis in the –35 and –10 promoter regions. d. polymerase mapping. Answer: c Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 5. The role of the sigma ( ) factor in prokaryote transcription is to a. unwind the DNA during transcription. b. direct RNA polymerases to bind to different promoter regions. c. recognize the transcription initiation site. d. terminate transcription. Answer: b Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 6. Transcription of a bacterial gene is initiated a. with RNA polymerase in the closed-promoter complex. b. after release of the sigma ( ) factor. c. with RNA polymerase in the open-promoter complex. d. Both b and c Answer: c Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 7. The sigma () subunit of bacterial RNA polymerase remains associated for the addition of _______ during transcription. a. the first ten nucleotides b. the first half of the nucleotides c. all of the nucleotides d. the last ten nucleotides Answer: a Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 8. Transcription is _______-dependent _______ synthesis. a. DNA; DNA b. DNA; RNA c. RNA; DNA d. RNA; protein Answer: b

Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Describe the processes of transcriptional elongation and termination. 9. Termination of transcription in E. coli is signaled by a. formation of a stem-loop structure in the RNA. b. binding of Rho protein to the end of the mRNA. c. binding of a sigma () factor to the end of the mRNA. d. an inverted GC-rich sequence followed by seven A residues. Answer: d Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Describe the processes of transcriptional elongation and termination. 10. Eukaryotic RNA polymerase I genes code for a. mRNAs. b. tRNAs. c. small nuclear RNAs and small cytoplasmic RNAs. d. ribosomal RNAs. Answer: d Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of different eukaryotic RNA polymerases. 11. Mitochondrial genes are transcribed by a. RNA polymerase I. b. RNA polymerase II. c. RNA polymerase III. d. a separate mitochondrial RNA polymerase. Answer: d Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of different eukaryotic RNA polymerases. 12. The large multi-subunit complex that links the general transcription factors to the gene-specific transcription factors is called a. the transcription complex. b. Mediator. c. the operator. d. TBP. Answer: b Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Distinguish between the binding of bacterial and eukaryotic RNA polymerases to promoters.

13. A major difference between eukaryotic and prokaryotic RNA polymerases is that eukaryotic polymerases a. use a set of transcription factors to bind to and initiate transcription. b. use sigma ( ) factors to initiate transcription. c. start from promoters. d. start from origins of replication. Answer: a Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Distinguish between the binding of bacterial and eukaryotic RNA polymerases to promoters. 14. The TATA box is similar to the _______ in E. coli. a. –35 promoter sequence b. –10 promoter sequence c. +1 transcription start site d. None of the above Answer: b Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Distinguish between the binding of bacterial and eukaryotic RNA polymerases to promoters. 15. The first step in the formation of a transcription complex for mRNA transcription is the binding of _______ to the TATA box. a. TFIA b. TFIIA c. TFIIIA d. TFIID Answer: d Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II. 16. Release of RNA polymerase II to initiate transcription appears to be the direct result of the a. binding of TAFs to the polymerase. b. unwinding of the DNA by helicases. c. phosphorylation of RNA polymerase by a protein kinase. d. removal of the nucleosome occupying the promoter site. Answer: c Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II.

17. All of the following are required for eukaryotic transcription in vitro except a. TAF. b. TFIIE. c. TBP. d. Mediator. Answer: d Textbook Reference: Eukaryotic RNA Polymerase and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II. 18. Processing of pre-tRNAs to produce tRNAs involves a. cleavage of the pre-tRNA. b. CCA addition to the 5ʹ end. c. phosphodiesterase activity to cleave the base and deoxyribose. d. the removal of one intron. Answer: a Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. 19. Processing of RNA transcripts occurs a. only in eukaryotic cells. b. only with mRNA transcripts. c. only with rRNA and mRNA transcripts. d. with tRNA, rRNA, and mRNA transcripts. Answer: d Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. 20. Self-splicing was discovered in studies of the a. pre-mRNAs of Tetrahymena. b. pre-mRNAs of E. coli. c. pre-rRNA of Tetrahymena. d. pre-rRNA of E. coli. Answer: c Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. 21. rRNA modifications are mediated by a. general transcription factors. b. miRNAs. c. snoRNPs.

d. Mediator. Answer: c Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. 22. Eukaryote mRNA processing occurs in a. the cytoplasm. b. the nucleus after completion of transcription. c. a complex with RNA polymerase. d. the Golgi apparatus. Answer: c Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Diagram mRNA processing. 23. Fertilization stimulates a. addition of 7-methylguanosine caps to pre-mRNAs. b. the lengthening of poly-A sequences to stored mRNAs in eggs. c. removal of introns from pre-mRNAs. d. the degradation of early developmental mRNAs. Answer: b Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Diagram mRNA processing. 24. The 7-methylguanosine cap on mNRA is required for a. protecting an mRNA from exonuclease attack. b. protecting an mRNA from endonuclease attack. c. initiation of translation of the mRNA. d. the addition of the poly-A tail. Answer: c Textbook Reference: RNA Processing and Turnover Bloom’s Category: 3. Applying Learning Objective: Diagram mRNA processing. 25. A poly-A tail is added to an mRNA by a. RNA polymerase, which reads a string of complementary dTs at the end of the gene. b. poly-A polymerase, which adds A’s sequentially to the end of the transcript. c. poly-A transferase, which adds a pre-made poly-A sequence to the end of the transcript. d. poly-A correctase, which recognizes an AAUAAA signal at the end of the message and changes the U to an A. Answer: b Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding

Learning Objective: Diagram mRNA processing. 26. The signal for the addition of a poly-A tail to pre-mRNA is a. UUUUUUU. b. TTTTTTTT. c. AAUAAA. d. GCGCUGC. Answer: c Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Diagram mRNA processing. 27. During splicing, pre-mRNAs go through an intermediate stage when they are shaped like a. a lariat. b. a circle. c. the Greek letter theta (). d. a cross. Answer: a Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Diagram mRNA processing. 28. Spliced mRNAs have protective mechanisms, but introns are rapidly degraded in the a. nucleus. b. cytoplasm. c. lysosomes. d. nucleoli. Answer: a Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Diagram mRNA processing. 29. The RNA components of the spliceosome are five different a. small cytoplasmic RNAs. b. small nuclear RNAs. c. microRNAs. d. siRNAs. Answer: b Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of snRNAs in mRNA splicing. 30. Which is not part of the spliceosome? a. U1 b. U2

c. U3 d. U5 Answer: c Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of snRNAs in mRNA splicing. 31. Which snRNA is responsible for recognition of the 5′ splice site consensus sequence in mRNA splicing? a. U1 b. U4 c. U5 d. U6 Answer: a Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of snRNAs in mRNA splicing. 32. A process called alternative splicing of mRNA transcripts can produce mRNAs with _______ from the same gene. a. one different exon b. more or fewer exons c. completely different exons d. All of the above Answer: d Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Illustrate patterns of alternative splicing. 33. The form of apolipoprotein B (Apo-B) produced by the intestine is shorter than the Apo-B produced by the liver. The intestine produces a smaller protein by a. proteolytic cleavage of the forming polypeptide chain. b. a base alteration that changes a codon for glutamine to a termination codon. c. shortening the mRNA from the 3 end after translation. d. including fewer exons in the mRNA by alternative splicing. Answer: b Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Describe RNA editing. Fill in the Blank 1. The structure that is formed during termination of transcription due to an inverted GCrich inverted repeat is known as a _______. Answer: stem-loop Textbook Reference: Transcription in Bacteria

Bloom’s Category: 1. Remembering Learning Objective: Describe the processes of transcriptional elongation and termination. 2. The predominant RNA synthesized by eukaryotic RNA polymerase II is _______. Answer: mRNA Textbook Reference: Transcription in Bacteria Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of different eukaryotic RNA polymerases. 3. Eukaryotic RNA polymerase III synthesizes _______. Answer: tRNA Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of different eukaryotic RNA polymerases. 4. A core promotion element located just upstream from the transcription start site in eukaryotic genes is an A- and T-rich region called the _______. Answer: TATA box Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Distinguish between the binding of bacterial and eukaryotic RNA polymerases to promoters. 5. The subunit of TFIID that is responsible for binding specifically to the TATA box is known as _______. Answer: TATA-binding protein (TBP) Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II. 6. During transcriptional initiation in eukaryotes, _______ is responsible for phosphorylating RNA polymerase II C-terminal domain (CTD). Answer: TFIIH Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II. 7. When Tetrahymena 28S rRNA is incubated in the absence of proteins, it undergoes _______. Answer: self-splicing Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Summarize the events involved in processing rRNAs and tRNAs.

8. Processing of the 5′ end of pre-tRNAs involves cleavage by an enzyme called _______. Answer: RNase P Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. 9. RNA splicing takes place in large complexes called _______. Answer: spliceosomes Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Diagram mRNA processing. 10. The RNA components of spliceosomes are five types of _______. Answer: snRNAs (small nuclear RNAs) Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of snRNAs in mRNA splicing. 11. Through the process of _______, cells can produce different proteins from the same gene. Answer: alternative spicing Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Illustrate patterns of alternative splicing. 12. RNA processing events (other than splicing) that alter the protein-coding sequences of some RNAs are called _______. Answer: RNA editing Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Describe RNA editing. 13. Unstable mRNAs with very short half-lives are frequently _______ proteins. Answer: regulatory Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Explain how mRNA degradation can be regulated by the environment. 14. Cytoplasmic degradation of most eukaryotic mRNA is initiated by shortening of their _______. Answer: poly-A tails Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding

Learning Objective: Explain how mRNA degradation can be regulated by the environment.

True/False 1. Sigma () subunits of bacterial RNA polymerase initiate binding to the DNA and are released from the polymerase at the end of transcription. Answer: F Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 2. The  subunit of E. coli RNA polymerase is necessary for transcriptional elongation. Answer: F Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 3. An internal channel between the  and ′ subunits contains the bacterial RNA polymerase active site. Answer: T Textbook Reference: Transcription in Bacteria Bloom’s Category: 1. Remembering Learning Objective: Describe the processes of transcriptional elongation and termination. 4. In eukaryotes, transcription of mitochondrial genes uses typical nuclear RNA polymerases. Answer: F Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of different eukaryotic RNA polymerases. 5. tRNA genes are transcribed by RNA polymerase III. Answer: T Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of different eukaryotic RNA polymerases. 6. Nucleosomes are an impediment to transcription in eukaryotic cells. Answer: T Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II.

7. A minimum of six general transcription factors are required for the recruitment of RNA polymerase II. Answer: F Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II. 8. The genes for the 5S, 5.8S and 28S rRNAs are found in tandem and transcribed as a unit. Answer: F Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Summarize the organization of promoters transcribed by RNA polymerases I and III. 9. The processing of pre-rRNA requires proteins and RNAs that are localized to the nucleolus. Answer: T Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. 10. Processing of tRNAs involves addition of a CCA 5′ terminus. Answer: F Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. 11. RNA editing refers to RNA processing events (other than splicing) that alter the protein-coding sequences of some mRNAs. Answer: T Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Describe RNA editing. 12. Tissue-specific RNA editing can occur. Answer: T Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Describe RNA editing. 13. Messenger RNAs have varying half-lives in the cytoplasm, and those differences are usually due to differences in the sequences near the 3ʹ end of the mRNA. Answer: T Textbook Reference: RNA Processing and Turnover

Bloom’s Category: 2. Understanding Learning Objective: Explain how mRNA degradation can be regulated by the environment.

Short Answer 1. What are the 5 subunits that make up bacterial RNA polymerase, and which one(s), if any, are weakly bound to the complex? Answer: The 5 subunits are , , ’, , and . The  subunit is weakly bound. Textbook Reference: Transcription in Bacteria Bloom’s Category: 1. Remembering Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 2. What are two reasons why eukaryotic transcription is considered to be more complex than bacterial transcription? Answer: First, in bacteria, a single RNA polymerase is responsible for transcription, but in eukaryotes, three nuclear RNA polymerases transcribe distinct classes of genes. Second, eukaryotic RNA polymerases need to interact with various additional proteins to initiate transcription. Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Summarize the roles of different eukaryotic RNA polymerases. 3. Identify two functions of capping and polyadenylation of mRNA. Answer: Capping and polyadenylation affect mRNA translation and stability. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Diagram mRNA processing. 4. What are the two signals for polyadenylation of mRNA? Answer: A highly conserved hexanucleotide located 10 to 30 nucleotides upstream of polyadenylation and a U- or GU-rich downstream sequence. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Diagram mRNA processing. 5. What happens to the lariat-like structure that is excised during mRNA splicing? Answer: It is linearized and degraded within the nucleus. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Diagram mRNA processing. 6. Why are some RNAs considered to be self-splicing? Answer: They can catalyze the removal of their own introns in the absence of other protein or RNA factors.

Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Diagram mRNA processing. 7. What two types of molecules make up snoRNPs? Answer: RNA (snoRNA) and proteins (8–10) Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of snRNAs in mRNA splicing. 8. What are the five types of small nuclear RNAs that make up the spliceosome? Answer: The five snRNAs that make up the spliceosome are U1, U2, U4, U5, and U6. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of snRNAs in mRNA splicing. 9. Why is dystrophin absent in Duchenne muscular dystrophy? Answer: This is due to mutations in the DMD gene that lead to premature termination of the translation. These mutations are usually deletions of multiple exons that cause frameshifts in the spliced mRNA. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Illustrate patterns of alternative splicing. 10. How was RNA editing first discovered? Answer: It was first discovered in mitochondrial mRNAs in which U residues were added and deleted at multiple sites along the pre-mRNA. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Describe RNA editing. 11. What is the most common form of nuclear RNA editing in mammals? Answer: Deamination of adenosine to inosine is the most common. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Describe RNA editing. 12. What is the general mechanism for degrading mature mRNAs in the cytoplasm? Answer: The poly-A tails are shortened (deadenylated) and then degraded by a 3ʹ to 5ʹ nuclease, or their 5ʹ cap is removed and it is degraded by a 5ʹ to 3ʹ nuclease. Either or both of these mechanisms may be used on a single mRNA. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Explain how mRNA degradation can be regulated by the environment.

13. Explain why bacterial mRNAs need to be very unstable, usually having half-lives of about 2 to 3 minutes. Answer: This rapid turnover allows the cell to respond quickly to changes in its environment, such as the availability of different nutrients. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Explain how mRNA degradation can be regulated by the environment. 14. What feature is often found on rapidly degraded mRNAs? Answer: Rapidly degraded mRNAs often contain specific AU-rich sequences near their 3′ ends that serve as binding sites for proteins that can stabilize these mRNAs or target them for degradation. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Explain how mRNA degradation can be regulated by the environment.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. In prokaryotes,  factors are necessary for a. specific binding to certain promoters. b. low-affinity binding upstream from promoters. c. elongation of the RNA strand to its end. d. All of the above Answer: a Textbook Reference: Transcription in Bacteria Bloom’s Category: 1. Remembering Learning Objective: Explain how E. coli RNA polymerase initiates transcription. Feedback A: Correct! A  factor directs the polymerase to the promoter by binding to its –35 and –10 regions. Feedback B: Incorrect. The RNA polymerase core, which lacks the  subunit, binds with low affinity to DNA. Feedback C: Incorrect. Following the synthesis of the first few nucleotides, the  factor dissociates from the core enzyme. Feedback D: Incorrect. Only one of the answers is correct. 2. According to the central dogma of molecular biology, transcription of genetic information occurs via a. DNA-dependent DNA synthesis. b. DNA-dependent RNA synthesis. c. RNA-dependent DNA synthesis. d. RNA-dependent protein synthesis.

Answer: b Textbook Reference: Introduction Bloom’s Category: 1. Remembering Learning Objective: Describe the processes of transcriptional elongation and termination. Feedback A: Incorrect. This is called replication. Transcription is a different process. Feedback B: Correct! Transcription produces the three types of RNA in the cell—mRNA, tRNA, and rRNA—using DNA as a template. Feedback C: Incorrect. RNA-dependent DNA synthesis is a process that occurs in retroviruses. It is catalyzed by reverse transcriptase. Feedback D: Incorrect. This is called translation. 3. Which statement about transcriptional termination in prokaryotes is false? a. Termination is signaled by transcription of a GC-rich interval. b. Transcription terminates when the RNA polymerase dissociates from its DNA template. c. The ribosome comes to a UAA, UAG, or UGA stop codon, and transcription ceases. d. A segment of RNA forms a stable stem-loop structure by complementary base pairing. Answer: c Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Describe the processes of transcriptional elongation and termination. Feedback A: Incorrect. This is a true statement. Feedback B: Incorrect. This is a true statement. Feedback C: Correct! These stop codons terminate translation, not transcription. Feedback D: Incorrect. This is a true statement. 4. Small nuclear RNAs (snRNAs) and small cytoplasmic RNAs (scRNAs) are transcribed by RNA polymerase a. I. b. II. c. III. d. II and III. Answer: d Textbook Reference: Eukaryotic RNA polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of different eukaryotic RNA polymerases. Feedback A: Incorrect. RNA polymerase I transcribes rRNAs. Feedback B: Incorrect. This is partially correct; RNA polymerase III also transcribes snRNA and scRNA. Feedback C: Incorrect. This is partially correct; RNA polymerase II also transcribes snRNA and scRNA. Feedback D: Correct! Some snRNAs and scRNAs are transcribed by RNA polymerase II and others by RNA polymerase III. 5. Which of the following RNAs is RNA polymerase II not responsible for transcribing? a. mRNA

b. miRNA c. tRNA d. lncRNA Answer: c Textbook Reference: Eukaryotic RNA polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of different eukaryotic RNA polymerases. Feedback A: Incorrect. RNA polymerase II is responsible for transcribing mRNA. Feedback B: Incorrect. RNA polymerase II is responsible for transcribing miRNA. Feedback C: Correct! RNA polymerase III is responsible for transcribing tRNA. Feedback D: Incorrect. RNA polymerase II is responsible for transcribing lncRNA. 6. All of the following are general transcription factors used in eukaryotic transcription except a. TFIIE b. TFIIH c. TBP d. BRE Answer: d Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II. Feedback A: Incorrect. TFIIE is a transcription factor used in eukaryotic transcription. Feedback B: Incorrect. TFIIH is a transcription factor used in eukaryotic transcription. Feedback C: Incorrect. TBP is part of TFIID, which is a transcription factor used in eukaryotic transcription. Feedback D: Correct! BRE is a recognition element in promoter sequences of eukaryotic genes. 7. Which of the following is not part of the transcription complex? a. TBP b. TAFs c. TFIIH d. RFC Answer: d Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II. Feedback A: Incorrect. TBP is TATA-binding protein and recognizes the TATA sequences. Feedback B: Incorrect. TAFs are TBP-associated factors and interact with TBP. Feedback C: Incorrect. TFIIH is a complex of proteins that have phosphorylase and helicase activity. Feedback D: Correct! RFC is replication factor C and is associated with the DNA

replication machinery. 8. A yeast mutant shows decreased expression of the 5.8S rRNA, the 5S rRNA, and protein-encoding mRNAs. If only one gene is mutated, which of the following genes is most likely affected? a. TATA-binding protein (TBP) b. Upstream binding factor (UBF) c. RNA polymerase III d. TFIIB Answer: a Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 4. Analyzing Learning Objective: Summarize the organization of promoters transcribed by RNA polymerases I and III. Feedback A: Correct! TBP has been found to be necessary for the transcription of genes transcribed by all three RNA polymerases. The 5.8S rRNA is transcribed by RNA polymerase I, the 5S rRNA by RNA polymerase III, and protein-encoding mRNAs by RNA polymerase II. Hence all three would be affected by a mutation in the TBP gene. Feedback B: Incorrect. UBF is a transcription factor specific for RNA polymerase I, and so only the 5.8S, 18S, and 28S rRNAs would be affected. Feedback C: Incorrect. RNA polymerase III transcribes only tRNAs, 5S rRNA, and some other small RNAs, so it would not affect the transcription of 5.8S rRNA or the proteinencoding mRNAs. Feedback D: Incorrect. TFIIB is a basal transcription factor that functions as RNA polymerase II promoter, so only the protein-encoding mRNAs would be affected. 9. Which statement about transcription of ribosomal RNA is false? a. Ribosomal RNA is transcribed by RNA polymerase I. b. A single pre-RNA gives rise to the 18S, 5.8S, and 28S rRNAs. c. TBP binds the TATA sequences in the promoter of the ribosomal gene. d. Two transcription factors, UBF and SL1, bind the RNA gene promoter and recruit polymerase I to form the initiation complex. Answer: c Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Summarize the organization of promoters transcribed by RNA polymerases I and III. Feedback A: Incorrect. This is a true statement. Feedback B: Incorrect. This is a true statement. Feedback C: Correct! This is a false statement. Ribosomal genes do not contain a TATA box. TBP does bind the promoter, but not at TATA sequences. Feedback D: Incorrect. This is a true statement. 10. A ribozyme is defined as an enzyme a. in which an RNA molecule is responsible for the catalytic activity. b. that catalyzes the cleavage of RNA.

c. that is involved in translation. d. that catalyzes the addition of ribose moieties to RNA. Answer: a Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. Feedback A: Correct! A prototypical example of a ribozyme is RNase P, in which the RNA component of the enzyme catalyzes the 5′ end cleavage of pre-tRNAs during their processing. Feedback B: Incorrect. An enzyme that cleaves RNA is called a ribonuclease. Feedback C: Incorrect. The ribosome carries out translation; ribozymes play no part in translation. Feedback D: Incorrect. An enzyme that adds ribose moieties to RNA is an RNA polymerase. 11. Approximately what percentage of the bases in tRNAs are altered to produce a variety of modified nucleotides within the tRNA molecule? a. 2% b. 10% c. 50% d. 90% Answer: b Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. Feedback A: Incorrect. This percentage is too low. Feedback B. Correct! About 10% of nucleotides are modified in a tRNA molecule. Feedback C: Incorrect. This percentage is too high. Feedback D: Incorrect. This is the percentage of nucleotides that are not modified. 12. The 5′ end of messenger RNA is modified by a. polyadenylation. b. ribothymidine addition. c. methylation. d. capping with 7-methylguanosine. Answer: d Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Diagram mRNA processing. Feedback A: Incorrect. This is added to the 3′ end of mRNA. Feedback B: Incorrect. This is a modified base found in tRNA. Feedback C: Incorrect. Methyl groups alone are not added to the 5′ end of mRNAs. Feedback D: Correct! This is also known as the 5′ cap. 13. Which process both stabilizes and increases the efficiency of translation of an mRNA?

a. Editing b. Splicing c. Addition of a 7-methylguanosine cap d. Addition of the CCA sequence to the 3ʹ end Answer: c Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Diagram mRNA processing. Feedback A: Incorrect. Editing usually changes bases within the coding sequence of an mRNA transcript and does not typically alter either the stability or the efficiency of translation. Feedback B: Incorrect. Splicing is the process whereby introns are removed from the premRNA, leaving the correct mRNA for export to the cytoplasm. Feedback C: Correct! The cap structure is recognized and bound by the ribosome, thus facilitating translation. In addition, the cap structure protects the RNA from nucleases. Feedback D: Incorrect. This modification occurs only on tRNAs and serves as the site of attachment for the amino acid. 14. Suppose that a gene has three exons and two introns in the following order: 5ʹ exon 1 – intron 1 – exon 2 – intron 2 – exon 3 3ʹ Which of the following could not result from alternative splicing in this gene? a. 5ʹ exon 1 – exon 2 – exon 3 3ʹ b. 5ʹ exon 2 – exon 3 3ʹ c. 5ʹ exon 2 – exon 1 3ʹ d. 5ʹ exon 1 – exon 3 3ʹ Answer: c Textbook Reference: RNA Processing and Turnover Bloom’s Category: 3. Applying Learning Objective: Illustrate patterns of alternative splicing. Feedback A: Incorrect. This is the prototypical splicing reaction in which the introns are removed from the pre-mRNA and all exons are retained. Feedback B: Incorrect. Removal of both introns and loss of the first exon would be a possible product of alternative splicing. Feedback C: Correct! Although the process of alternative splicing can change the identity of the exons in the mature mRNA, it does not alter the order of exons. Feedback D: Incorrect. Deletion of an exon is a possible outcome of alternative splicing. 15. All of the following are considered to be RNA editing events that occur except a. addition of the 5′ cap. b. deamination of cytosine to uridine. c. addition and deletion of U residues. d. deamination of adenosine to inosine. Answer: a Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Describe RNA editing.

Feedback A: Correct! This is considered an mRNA processing event and does not alter the protein coding sequence of the mRNA. Feedback B: Incorrect. This occurs during RNA editing. Feedback C: Incorrect. This occurs during RNA editing. Feedback D: Incorrect. This occurs during RNA editing. 16. Which step is considered the final aspect of processing an RNA molecule? a. Addition of the 5′ cap b. RNA editing c. Degradation in the cytoplasm d. Splicing Answer: c Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Explain how mRNA degradation can be regulated by the environment. Feedback A: Incorrect. This is not the final RNA processing event. Feedback B: Incorrect. This is not the final RNA processing event. Feedback C: Correct! This is the last processing event that occurs. Feedback D: Incorrect. This is not the final RNA processing event. 17. Which of the following RNAs is least stable and has the shortest half-life? a. Bacterial mRNAs b. rRNAs c. tRNAs d. Eukaryotic mRNAs Answer: a Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering Learning Objective: Explain how mRNA degradation can be regulated by the environment. Feedback A: Correct! Bacterial mRNAs are very unstable, usually having half-lives of only 2 to 3 minutes. Feedback B: Incorrect. rRNAs are very stable and have long half-lives. Feedback C: Incorrect. tRNAs are very stable and have long half-lives. Feedback D: Incorrect. Eukaryotic mRNAs have half-lives from 30 minutes to 20 hours. 18. mRNAs are degraded in the cytoplasm by a. shortening of the poly-A tail and degradation from the 3′ end by nucleases. b. removal of the 5′ cap and degradation from the 5′ end by nucleases. c. Both a and b d. Neither a nor b Answer: c Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering

Learning Objective: Explain how mRNA degradation can be regulated by the environment. Feedback A: Incorrect. This is one method, but mRNA can also be degraded from the 5′ end. Feedback B: Incorrect. This is one method, but mRNA can also be degraded from the 3′ end. Feedback C: Correct! Both the poly-A tail and the 5′ cap can be removed from mRNAs; mRNAs can then be degraded by nucleases from either end. Feedback D: Incorrect. Both the poly-A tail and the 5′ cap can be removed from mRNAs; mRNAs can then be degraded by nucleases from either end. 19. Which of the following regulates degradation of mRNAs within a cell? a. Growth factors b. miRNAs c. Hormones d. All of the above Answer: d Textbook Reference: RNA Processing and Turnover Bloom Category: 2. Understanding Learning Objective: Explain how mRNA degradation can be regulated by the environment. Feedback A: Incorrect. This does regulate degradation of mRNAs, but there are other correct answers. Feedback B: Incorrect. This does regulate degradation of mRNAs, but there are other correct answers. Feedback C: Incorrect. This does regulate degradation of mRNAs, but there are other correct answers. Feedback D: Correct! Growth factors, hormones, and miRNAs are all responsible for regulating mRNA degradation.

Essay 1. Explain both the closed- and open-promoter complex that is formed during the initiation of transcription in E. coli. Answer: RNA polymerase initially binds nonspecifically to DNA and moves along until it binds to the –10 and –35 promoter sequences. This forms the closed-promoter complex. The polymerase unwinds the DNA at the initiation site to form the open-promoter complex, and transcription is initiated. Textbook Reference: Transcription in Bacteria Bloom’s Category: 2. Understanding Learning Objective: Explain how E. coli RNA polymerase initiates transcription. 2. In vitro experiments have shown that formation of the preinitiation transcription complex on RNA polymerase II promoters requires the dephosphorylated form of the RNA polymerase II large subunit, whereas the phosphorylated form is associated with

elongating complexes. Describe how the timing of phosphorylation might influence gene transcription. Answer: If phosphorylation occurred before the formation of the preinitiation complex, then RNA polymerase II would not bind the promoter, and transcription would be inhibited. In contrast, phosphorylation after the formation of the preinitiation complex would favor the elongating form of the complex, and transcription would be stimulated. Thus, the timing of activation of kinases and phosphatases can be a factor in the regulation of gene transcription. Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Summarize the roles of different eukaryotic RNA polymerases. 3. Explain what TFIIH is, including its function in initiating transcription. Answer: TFIIH is a multisubunit factor that has two roles: The first is to act as a helicase to unwind DNA around the initiation site. The second role is to act as a protein kinase that phosphorylates repeated sequences on the CTD of RNA polymerase II. Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II. 4. Explain the role of Meditator in transcription. Answer: Mediator is required for transcription within the cell (it is not required for in vitro transcription). Mediator interacts with both the general transcription factors and with RNA polymerase to stimulate basal transcription and regulate gene-specific transcription. Textbook Reference: Eukaryotic RNA Polymerases and General Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Describe the functions of the general transcription factors for RNA polymerase II. 5. What important information was revealed by the discovery of catalytic RNAs? Answer: Biologists had long puzzled over the need for nucleic acids to encode (protein) enzymes and the reciprocal need for enzymes to catalyze the replication of the nucleic acids. Which came first, the nucleic acid or the protein? The discovery of catalytic RNAs suggested that the nucleic acids came first. This discovery also demonstrated that some RNAs are good catalysts, and it was already well known that some RNAs function as the genetic material in some viruses. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. 6. What role do the modified bases in the tRNA molecule play? Answer: It is believed that some of these modified bases play a role in protein synthesis by altering the base-pairing properties of the tRNA molecule. Textbook Reference: RNA Processing and Turnover

Bloom’s Category: 2. Understanding Learning Objective: Summarize the events involved in processing rRNAs and tRNAs. 7. Explain why the lariat structure that is formed as an intermediate in RNA splicing could not be formed in a DNA molecule. Answer: The first step of splicing is cleavage at the 5ʹ splice site and joining of the 5ʹ end of the intron to an adenine nucleotide within the intron to form the lariat. This reaction occurs by means of an attack on the 2ʹ-OH of the adenylate residue on the phosphate residue at the intron-exon border. DNA, unlike RNA, lacks a 2ʹ-OH, and therefore this reaction could not occur. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 2. Understanding Learning Objective: Diagram mRNA processing. 8. Explain how the 5′ cap is formed in messenger RNA. Answer: The 5′ cap is formed by the addition of a guanosine triphosphate (GTP) in the reverse orientation to the 5′ end of the mRNA. This added nucleotide is then methylated, and methyl groups are added to the first one or two nucleotides in the mRNA. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering. Learning Objective: Diagram mRNA processing. 9. Explain the two steps involved in pre-mRNA splicing. Answer: First, the pre-mRNA is cleaved at the 5′ splice site, and the 5′ end of the intron is joined to an adenine nucleotide within the intron. Second, there is simultaneous cleavage at the 3′ splice site and ligation of the two exons. Textbook Reference: RNA Processing and Turnover Bloom’s Category: 1. Remembering. Learning Objective: Diagram mRNA processing. 10. Explain how alternative splicing of transformer (tra)in Drosophila leads to sex determination. Answer: In males, the first exon of tra pre-mRNA is joined to a 3′ splice site that yields a second exon that contains a translation termination codon, so no tra protein is expressed. In females, SXL protein binds to block the 3′ splice site, resulting in the use of an alternative splice site further downstream in exon 2. This alternative splice site is downstream of the translation termination codon, so a functional tra protein is produced. Textbook Reference: RNA Processing and Turnover Bloom Category: 2. Understanding Learning Objective: Illustrate patterns of alternative splicing. 11. In eukaryotic cells, different mRNAs have different half-lives, and therefore are degraded at different rates. What is the usefulness of this wide range of half-lives? Answer: Unstable mRNAs frequently code for regulatory proteins, and the levels of these proteins within the cell vary rapidly in response to environmental stimuli. mRNAs

encoding metabolic enzymes or structural proteins generally have longer half-lives, and the levels of these proteins in the cell do not fluctuate frequently. Textbook Reference: RNA Processing and Turnover Bloom Category: 2. Understanding Learning Objective: Explain how mRNA degradation can be regulated by the environment.

Test File to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 9: Transcriptional Regulation and Epigenetics TEST BANK QUESTIONS Multiple Choice 1. The lac operon in E. coli is regulated by allolactose, which _______ a(n) _______ of transcription. a. activates; activator b. inactivates; activator c. activates; repressor d. inactivates; repressor Answer: d Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 2. Understanding Learning Objective: Explain how lactose regulates transcription of the lac operon. 2. The lac operator consists of approximately 20 base pairs of DNA located a. 80–100 base pairs upstream of the transcription initiation site. b. 20–40 base pairs upstream of the transcription initiation site. c. in a position overlapping a few bases of the initiation site and extending downstream. d. downstream of the entire operon. Answer: c Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 1. Remembering Learning Objective: Explain how lactose regulates transcription of the lac operon. 3. If E. coli is in an environment that is rich in lactose, a metabolite of lactose binds to a. a repressor, releasing it from the operator site and activating transcription of the genes involved in lactose breakdown. b. RNA polymerase, activating it to transcribe the genes involved in lactose breakdown. c. an activator protein, which binds to RNA polymerase and stimulates transcription of the genes involved in lactose breakdown. d. the operator site of the lac operon, stimulating transcription of the genes involved in lactose breakdown. Answer: a Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 3. Applying Learning Objective: Explain how lactose regulates transcription of the lac operon. © 2019 Oxford University Press

4. A mutation in the i gene encoding the lac operon repressor protein causes this protein to lose its ability to bind to the operator site. Which would you expect to observe in an E. coli population having this mutation? a. The cells would die if lactose were the only energy source available. b. The cells would require lactose as their only energy source. c. The cells would not be able to express lac genes when lactose is present. d. The cells would express lac genes when lactose is absent. Answer: d Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 4. Analyzing Learning Objective: Explain how lactose regulates transcription of the lac operon. 5. E. coli regulates the biosynthesis of the amino acid tryptophan through the tryptophan (trp) operon. When tryptophan levels are high inside the cell, tryptophan binds to a regulatory protein, which then binds to the operator site of the operon. This shuts down transcription of the genes encoding enzymes needed for tryptophan biosynthesis, and tryptophan biosynthesis is halted. The trp operon is an example of a. positive control because binding of a repressor blocks transcription. b. negative control because binding of a repressor blocks transcription. c. positive control because binding of an activator stimulates transcription. d. negative control because binding of an activator stimulates transcription. Answer: b Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 3. Applying Learning Objective: Distinguish between positive and negative control. 6. Which information would be helpful in determining whether a particular bacterial gene is under positive or negative transcriptional control? a. The gene is part of an operon. b. The gene is regulated by the level of a metabolite inside the cell. c. The gene codes for an enzyme that has an important metabolic purpose. d. The gene is not transcribed when a repressor protein binds to DNA. Answer: d Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 2. Understanding Learning Objective: Distinguish between positive and negative control. 7. A researcher hypothesizes that expression of a particular gene is under positive transcriptional control. What evidence would support his hypothesis? a. A repressor protein is discovered that blocks transcription of this gene. b. The gene is only expressed when catabolite activator protein binds to an upstream DNA site. c. The gene is no longer regulated when an upstream DNA region undergoes mutation. d. Mutation of the gene itself does not affect its regulation. Answer: b Textbook Reference: Gene Regulation in E. coli © 2019 Oxford University Press

Bloom’s Category: 3. Applying Learning Objective: Distinguish between positive and negative control. 8. Regulation of the lac operon by lactose involves a(n) _______ that inhibits transcription, while regulation by glucose involves a(n) _______ that stimulates transcription. a. activator; repressor b. repressor; activator c. promoter; promoter d. operator; operator Answer: b Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 1. Remembering Learning Objective: Explain why repressors inhibit but activators stimulate transcription. 9. How do repressors differ from activators of transcription in bacteria? a. Repressors, but not activators, can act on one chromosome only. b. Activators are proteins, repressors are not. c. Repressors inhibit transcription, and activators stimulate transcription. d. Activators bind to DNA, and repressors bind to RNA polymerase. Answer: c Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 2. Understanding Learning Objective: Explain why repressors inhibit but activators stimulate transcription. 10. Two bacterial populations were prepared in a laboratory. Population A expresses the wildtype regulator protein that controls the transcription of gene X. Population B contains a mutation in the gene for this regulator protein and does not express any form of the protein. Which statement provides a possible hypothesis to explain the gene X expression data shown in the table below?

a. The regulator protein is an activator that is required for stimulating RNA polymerase to transcribe gene X. b. The regulator protein binds to RNA polymerase to prevent it from binding to the promoter site and transcribing gene X. c. The regulator protein is a repressor that binds to an operator site and blocks RNA polymerase from carrying out transcription. d. The regulator protein is an inhibitor that binds to RNA polymerase and prevents it from binding to the promoter site. Answer: a Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 4. Analyzing © 2019 Oxford University Press

Learning Objective: Explain why repressors inhibit but activators stimulate transcription. 11. One way that a promoter differs from an enhancer is that it a. is not a sequence of DNA. b. does not bind a transcription factor. c. is not located far away from the gene it regulates. d. does not bind to an RNA polymerase. Answer: c Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast promoters and enhancers. 12. TATA boxes are found a. in enhancers only. b. in promoters only. c. in both enhancers and promoters. d. in neither enhancers nor promoters. Answer: b Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast promoters and enhancers. 13. An investigation was carried out to identify different sections of mammalian DNA as either promoters or enhancers of gene Y transcription. Wild-type mammalian cells functioned normally, while mutant cells lacked the ability to produce functional cohesion protein. Based on the data in the table below, which conclusion is valid?

a. Both A and B are enhancers. b. Both A and B are promoters. c. A is an enhancer, and B is a promoter. d. A is a promoter, and B is an enhancer. Answer: d Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 4. Analyzing Learning Objective: Compare and contrast promoters and enhancers. 14. Looping of DNA has been observed by investigators using electron microscopy techniques. The presence of which feature in DNA might be indicated by such observations? a. Promoters b. Enhancers © 2019 Oxford University Press

c. Base-pair repeats d. DNA methylation Answer: b Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Compare and contrast promoters and enhancers. 15. A reporter gene is one that a. causes other genes to be expressed. b. can produce an easily detectable product when expressed under the control of a regulatory sequence to which it has been ligated and cloned. c. is always on. d. is turned on by the products of other genes. Answer: b Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. 16. A researcher wants to isolate regulatory sequences of DNA bound to their transcription factors. She then wants to sequence the DNA fragments that form the binding sites that the transcription factors recognize and bind. Which technique should the researcher use for her first step? a. Gene transfer assay b. Electrophoretic-mobility shift assay c. Chromatin immunoprecipitation d. DNA affinity chromatography Answer: c Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. 17. An electrophoresis assay used to identify the sequences of DNA to which specific regulatory proteins bind is called a. an electrophoretic-mobility shift assay. b. DNA affinity chromatography. c. Western blotting. d. Southern blotting. Answer: a Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA.

18. Kadonaga and Tjian were able to isolate the transcription factor Sp1 by exploiting its ability to bind a. the sequence TATA. b. the sequence AAAUAAA. c. the sequence GGGCGG. d. RNA polymerase. Answer: c Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. 19. Eukaryotic gene repressor proteins are thought to act by binding to a. DNA sites in competition with activating proteins. b. specific activating proteins, preventing their binding to DNA. c. basal transcription factors, inhibiting transcription. d. All of the above Answer: d Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Describe how activators and repressors affect transcription. 20. Many transcriptional activators are composed of two or more independent _______, each with its own separate function. a. domains b. genes c. enhancers d. proteins Answer: a Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Describe how activators and repressors affect transcription. 21. Suppose a transcriptional activator binds to a specific DNA site near a promoter region. Which is a protein that is then stimulated to form a functional transcription complex at this site? a. Transcriptional factors b. Mediator proteins c. RNA polymerase d. All of the above Answer: d Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Describe how activators and repressors affect transcription.

22. Investigators discovered two eukaryotic regulatory proteins that control expression of the same gene. Their findings are summarized in the table below. Which hypothesis is most consistent with these findings?

a. Both regulatory proteins block the binding of activators to DNA. b. Both regulatory proteins affect transcription by binding to mediator proteins. c. Regulatory protein I blocks the binding of activators to DNA, whereas regulatory protein II affects transcription by binding to mediator proteins. d. Regulatory protein I affects transcription by binding to mediator proteins, whereas regulatory protein II blocks the binding of activators to DNA. Answer: d Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 4. Analyzing Learning Objective: Describe how activators and repressors affect transcription. 23. What causes RNA polymerase to pause just after transcription is initiated? a. Phosphorylation of the C-terminal domain (CTD) of RNA polymerase b. Phosphorylation of NELF and DSIF c. Binding of NELF and DSIF to RNA polymerase d. Production of P-TEFb Answer: c Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Explain how transcriptional elongation is controlled. 24. Suppose a mutation renders a cell incapable of producing functional P-TEFb. What effect would the mutation have on this cell? a. RNA polymerases that are paused would not be able to resume elongation. b. No transcription initiation complexes would be able to form. c. No serine residues in the RNA polymerase CTD would be phosphorylated. d. The cell would lose the ability to pause any of its transcription initiation complexes. Answer: a Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Explain how transcriptional elongation is controlled. 25. What is the role of the transcription factor c-Myc in human embryonic stem cells? © 2019 Oxford University Press

a. It acts to halt elongation of transcription. b. It functions to speed up the formation of transcription initiation complexes. c. It prevents NELF and DSIF from binding to transcription initiation complexes. d. It recruits P-TEFb to release RNA polymerase from its paused state and continue transcript elongation. Answer: d Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Explain how transcriptional elongation is controlled. 26. Eukaryotic gene regulation is facilitated by a. acetylation of histones. b. phosphorylation of histones. c. methylation of histones. d. All of the above Answer: d Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Describe the effects of different histone modifications on transcription. 27. Methylation of histones results in a. activation of transcription. b. repression of transcription. c. activation or repression of transcription, depending on the site. d. neither activation nor repression of transcription, only condensation of chromatin. Answer: c Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Describe the effects of different histone modifications on transcription. 28. Some types of cancer are able to progress because certain normally active genes are repressed. An anticancer drug was developed to counteract this by functioning as a histone deacetylase inhibitor. Such an agent will _______ and thus _______ the expression of genes. a. halt acetylation of histones; repress b. stimulate acetylation of histones; activate c. halt removal of acetyl groups on histones; activate d. stimulate removal of acetyl groups on histones; repress Answer: c Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Describe the effects of different histone modifications on transcription. 29. Which characteristic of chromatin would you expect to see increase when incubated with histone acetyltransferase enzymes? a. Condensation to heterochromatin b. Sensitivity to DNAse I © 2019 Oxford University Press

c. Decreased gene expression d. Repression of transcription Answer: b Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Describe the effects of different histone modifications on transcription. 30. Which molecule provides a source of energy used by chromatin remodeling factors to alter DNA−histone interactions? a. NADH b. FADH2 c. ATP d. Glucose Answer: c Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 1. Remembering Learning Objective: Summarize the action of chromatin remodeling factors. 31. Which is not a mechanism of histone change that is mediated by chromatin remodeling factors? a. Sliding of histones along DNA b. Proteolytic breakdown of histone proteins c. Conformational change of histone tertiary structure d. Dissociation of histones from DNA Answer: b Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Summarize the action of chromatin remodeling factors. 32. Which enzyme would be most useful in an experimental investigation of the effects of chromatin remodeling factors on a particular section of chromatin? a. DNAse I b. Histone acetyltransferase c. Histone deacetylase d. Histone methyltransferase Answer: a Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Summarize the action of chromatin remodeling factors. 33. DNA methylation patterns in parental cells are established and maintained in progeny cells by which of the following mechanisms? a. Enzymes put a methyl group on cytosine residues of newly replicated CpG sequences basepaired with G-methyl-C sequences. b. Methyl-CTP is incorporated into DNA during replication across from G-methyl-C sequences only. © 2019 Oxford University Press

c. When its gene is activated, a methyl group is added to certain CpG sequences in the promoter region. d. Methylation of genes on the X chromosome only. Answer: a Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Explain epigenetic inheritance based on histone modifications and on DNA methylation. 34. A skin cell and a muscle cell taken from a single mouse differ in appearance and metabolism. If you were to examine the DNA in these two cells, how would they compare? a. The DNA sequences are arranged in different orders in the two cells. b. DNA sequences differ in various parts of the genomes of the two cells. c. The cells contain the same DNA sequences but have different promoters and enhancers. d. The same DNA sequences are present but different regions of chromatin are modified. Answer: d Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Explain epigenetic inheritance based on histone modifications and on DNA methylation. 35. Genomic imprinting is thought to result from a. acetylation of histones. b. methylation of cytosine residues in DNA. c. binding of transcription factors to promoters. d. binding of Xist RNA to genes. Answer: b Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 1. Remembering Learning Objective: Explain epigenetic inheritance based on histone modifications and on DNA methylation. 36. How are methylated sites in DNA maintained through multiple cycles of DNA replication during development? a. DNA polymerase incorporates a 5-methyl cytosine next to a G bound to an existing 5-methyl cytosine. b. DNA polymerase adds a methyl group to a cytosine next to a G bound to an existing 5-methyl cytosine. c. A methylating enzyme adds a methyl group to a cytosine next to a G bound to an existing 5methyl cytosine. d. A methylating enzyme adds a methyl group to a guanosine next to a C bound to an existing 5methyl guanosine. Answer: c Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding

Learning Objective: Explain epigenetic inheritance based on histone modifications and on DNA methylation. 37. The Xist lncRNA assists with X chromosome inactivation by binding which protein(s) to facilitate histone modification? a. Histone acetylase b. Polycomb proteins c. DNA demethylase d. All of the above Answer: d Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 1. Remembering Learning Objective: Describe the action of lncRNAs in gene repression and activation. 38. How do long noncoding RNAs (lncRNAs) function in regulating gene expression? a. They bind to complementary DNA sites and interfere with RNA polymerase binding. b. They form scaffolds to help stabilize protein complexes that modify chromatin. c. They assist with DNA loop formation between enhancers to facilitate transcription initiation. d. They catalyze the modification of histones to stimulate chromatin condensation. Answer: b Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Describe the action of lncRNAs in gene repression and activation. 39. The table below provides information about RNA strands that were found to be involved in gene regulation in mammalian cells. Which strand(s) would be classified as lncRNAs?

a. I and IV b. II and III c. I only d. I, II, III, and IV Answer: a Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 4. Analyzing Learning Objective: Describe the action of lncRNAs in gene repression and activation. 40. Which statement concerning lncRNAs is true? a. lncRNAs are extracellular signals that stimulate activation of specific genes. © 2019 Oxford University Press

b. lncRNAs catalyze the chemical modification of histones. c. lncRNAs function in both activation and repression of gene expression. d. lncRNAs are formed once RNA polymerase becomes active after having been paused. Answer: c Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Describe the action of lncRNAs in gene repression and activation. Fill in the Blank 1. Low concentrations of lactose in a bacterial cell lead to a _______ binding to the operator site of the lac operon and blocking transcription. Answer: repressor Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 2. Understanding Learning Objective: Explain how lactose regulates transcription of the lac operon. 2. A bacterial operon is said to be under _______ control if it involves an activator protein that functions to stimulate RNA polymerase to initiate transcription of its genes. Answer: positive Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 1. Remembering Learning Objective: Distinguish between positive and negative control. 3. The lac operon is under negative control by a repressor that is sensitive to lactose concentrations and is also under positive control by the activator protein _______, which is sensitive to glucose concentrations. Answer: CAP Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 1. Remembering Learning Objective: Explain why repressors inhibit but activators stimulate transcription. 4. The diagram below illustrates how two _______-acting elements are positioned near the gene that they regulate.

Answer: cis Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Compare and contrast promoters and enhancers.

5. DNA is able to form _______, which enable transcription factors at enhancers to interact with the transcription initiation complexes at promoters. Answer: loops Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast promoters and enhancers. 6. The results from several gene transfer assays carried out to locate regulatory DNA sequences are shown in the table below. Three different transcription factors were used in the assay as probes. Luciferase was used as the reporter protein, and production of light is noted with + signs. No production of light is noted with a 0. The results indicate that _______ of the five DNA sequences tested may be have some sort of gene regulatory role.

Answer: three Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 4. Analyzing Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. 7. A researcher has several transcription factors in purified form as well as a DNA fragment in purified form. He suspects the DNA fragment binds to one of the transcription factors. He can use the _______ technique to determine which transcription factor the DNA fragment binds to. Answer: electrophoretic-mobility shift assay Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. 8. Chromosomes are organized into multiple loop structures that maintain specificity of enhancers for their cognate genes by the proteins _______ and _______. Answer: CTCF, cohesin Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Describe how activators and repressors affect transcription. 9. Some transcriptional activators have two domains: a _______ domain and an activation domain, each of which carries out specific functions. Answer: DNA binding Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding © 2019 Oxford University Press

Learning Objective: Describe how activators and repressors affect transcription. 10. P-TEFb contains a _______ that phosphorylates NELF, DSIF, and serine 2 in the C-terminal domain of RNA polymerase, all of which lead to re-activation of RNA polymerase and transcriptional elongation. Answer: protein kinase Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Explain how transcriptional elongation is controlled. 11. Acetylation of lysine residues on histones causes these proteins to lose _______ charges, which allows the chromatin structure to relax. Answer: positive Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Describe the effects of different histone modifications on transcription. 12. An investigation identified whether histones nearest specific genes were acetylated or deacetylated. The results are summarized in the table below. Based on these results, it can be predicted that genes _______ would be actively expressed, while the others would not.

Answer: B and E Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Describe the effects of different histone modifications on transcription. 13. _______ use the energy of ATP hydrolysis to alter and restructure nucleosomes. Answer: Chromatin remodeling factors Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 1. Remembering Learning Objective: Summarize the action of chromatin remodeling factors. 14. Islet cells in the pancreas are the only cells in the human body that produce the hormone insulin. The table below compares the extent of methylation within the promoter region upstream of the gene encoding the insulin protein in three types of human cells. The results provide an example illustrating that methylation of DNA is correlated with transcriptional _______.

Answer: repression Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 4. Analyzing Learning Objective: Explain epigenetic inheritance based on histone modifications and on DNA methylation. 15. Polycomb proteins are important in repressing gene transcription during development and differentiation. They act by _______ lysine residues in histones. Answer: methylating Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 1. Remembering Learning Objective: Explain epigenetic inheritance based on histone modifications and on DNA methylation. 16. The molecule providing the scaffolding in the figure below to allow multiple chromatin modifications is a(n) _______.

Answer: long noncoding RNA (lncRNA) Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Describe the action of lncRNAs in gene expression and activation.

Textbook Reference: Gene Regulation in E.coli Bloom’s Category: 2. Understanding Learning Objective: Explain how lactose regulates transcription of the lac operon. 2. Some bacterial genes are regulated by transcriptional activators rather than by repressors. Answer: T Textbook Reference: Gene Regulation in E.coli Bloom’s Category: 2. Understanding Learning Objective: Distinguish between positive and negative control. 3. For each protein-coding gene, there is one promoter and one enhancer that act as regulatory elements for that gene. Answer: F Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Compare and contrast promoters and enhancers. 4. DNA-affinity chromatography is a useful method for purifying individual transcription factors from cell extracts, even though these proteins are present in extremely low quantities. Answer: T Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. 5. Some corepressors bind to transcription factors at transcription initiation sites and recruit deacetylases that remove acetyl groups from histone lysine groups, leading to condensation of chromatin and repression of transcription. Answer: T Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Describe how activators and repressors affect transcription. 6. The light micrograph image below shows a section of a Drosophila chromosome. The arrows indicate regions containing histones that have been deacetylated.

Answer: F Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 4. Analyzing Learning Objective: Describe the effects of different histone modifications on transcription. 7. Chromatin remodeling factors are responsible for shuffling and rearranging DNA sequences in chromatin to make them either more or less accessible for transcription. Answer: F Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Summarize the action of chromatin remodeling factors. 8. In eukaryotic cells, genes with methylated DNA tend to be more active than unmethylated genes. Answer: F Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Explain epigenetic inheritance based on histone modifications and on DNA methylation. 9. Transmission of information that is not contained within the sequence of DNA to daughter cells at cell division is called epigenetic inheritance. Answer: T Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding © 2019 Oxford University Press

Learning Objective: Explain epigenetic inheritance based on histone modifications and on DNA methylation. 10. lncRNAs assist with changes in chromatin. In embryonic stem cells some of these molecules can act in a repressive role to suppress differentiation and maintain the stem cell state, while others can act in an activating role to allow differentiation to proceed. Answer: T Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Describe the action of lncRNAs in gene repression and activation.

Short Answer 1. A mutation in the i gene of the lac operon produces a repressor that retains its DNA binding function but loses its ability to bind to lactose. Will bacteria with this mutant repressor be able to survive in a medium containing lactose as the only carbon source? Explain. Answer: No, bacteria with this mutation will not survive because the repressor has lost its ability to shift to a state in which it does not bind to the operator site. This shift depended on the repressor’s ability to bind lactose. Having lost this ability, it will continue binding to the operator site, even though lactose signals are present that would normally cause it to be released. Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 3. Applying Learning Objective: Explain how lactose regulates transcription of the lac operon. 2. Suppose a culture of E. coli is growing in media containing only glucose. The cells eventually deplete this glucose supply. Just as the last bit of glucose is used up, lactose and other sugars become available. How does both the loss of glucose and the addition of lactose affect expression of the lac operon in these cells? Answer: The lac operon is repressed when lactose is absent and glucose is present. When lactose becomes available, it forms a metabolite that binds to the repressor, causing its removal from the operator site on the DNA. At the same time, glucose levels drop, which induces CAP to activate RNA polymerase. The combined loss of the repressor and activation of RNA polymerase causes rapid transcription of the lac operon genes. Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 3. Applying Learning Objective: Explain how lactose regulates transcription of the lac operon. 3. The L-arabinose operon in an E. coli strain contains genes that code for enzymes used to break down the sugar L-arabinose. A regulatory gene located upstream from the operon encodes the regulatory protein AraC. AraC undergoes a conformational change when it binds to L-arabinose, and this form of AraC binds to the operator site of the operon, acting as an activator, to stimulate RNA polymerase to begin transcription. Explain whether this is an example of positive or negative control and how this compares to the control mechanism of the lac operon. Answer: The L-arabinose operon is under positive control because the regulatory element causes transcription to be stimulated, not repressed. This differs from the lac operon, which is an © 2019 Oxford University Press

example of negative control, in which the regulatory element (a repressor protein) causes transcription to be repressed. Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 3. Applying Learning Objective: Distinguish between positive and negative control. 4. Prokaryotes are single-celled organisms and do not undergo differentiation. Why is it important that these cells have gene regulatory mechanisms, some of which repress transcription and others that stimulate transcription? Answer: Prokaryotes must use their available energy wisely and not waste it generating gene products that are not essential at any given point in time. The cells maintain efficiency by producing only the proteins that will help it survive in whatever environment they find themselves in. As their environment changes, the cells need to shift from expressing one set of genes to another to take advantage of new food sources or to protect themselves against a sudden change in temperature, for example, and the gene regulatory mechanisms provide the means for doing that. Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 3. Applying Learning Objective: Explain why repressors inhibit but activators stimulate transcription. 5. What are three ways in which an enhancer’s function differs from that of a promoter? Answer: An enhancer can (1) act at great distance from the initiation site it regulates; (2) be either upstream or downstream from the initiation site; and (3) function in either orientation (forward or backward). Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast promoters and enhancers. 6. Describe an electrophoretic mobility shift assay. Answer: It is a method to identify binding sites of transcriptional regulatory proteins within the promoter or enhancer of a gene. A radiolabeled DNA fragment is incubated with a purified protein of interest and subjected to electrophoresis through a nondenaturing gel. Protein binding is detected as a decrease in electrophoretic mobility of the DNA fragment, since its migration through the gel is slowed by the bound protein. Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. 7. If you had genomic DNA and a purified transcription factor, how might you use that to determine the DNA sequence to which the transcription factor binds? Answer: You could do a chromatin precipitation assay by incubating the purified protein with chromatin and treating with formaldehyde to crosslink the protein to the DNA. Then, using an antibody to the protein, you could isolate the protein and sequence the DNA that was crosslinked to the protein. Textbook Reference: Transcription Factors in Eukaryotes © 2019 Oxford University Press

Bloom’s Category: 3. Applying Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. 8. What two different mechanisms do transcription factors use to regulate transcription of eukaryotic genes? Answer: Transcription factors regulate transcription of eukaryotic genes by (1) binding to mediator proteins and/or general transcription factors to recruit and facilitate the action of a transcription complex at promoters; and (2) modifying chromatin structure. Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Describe how activators and repressors affect transcription. 9. A research lab working with mouse cells characterized three mutant cell lines summarized in the table below. Which of these is most likely to contain a mutated form of P-TEFb? Explain.

Answer: Mutant cell line X most likely carries a mutated form of P-TEFb because this protein is responsible for releasing paused RNA polymerase from its paused state, thereby allowing it to continue elongating messenger RNA transcripts. Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 4. Analyzing Learning Objective: Explain how transcriptional elongation is controlled. 10. Three amino acids, present in histone tails, that are chemically modified as part of gene regulatory processes are lysine, arginine, and _______. Answer: serine Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 1. Remembering Learning Objective: Describe the effects of different histone modifications on transcription. 11. Refer to the figure below.

Explain how HDACs function to regulate gene expression. Answer: HDAC stands for histone deacetylase, which is an enzyme that removes acetyl groups from lysine residues in histones. Removal of acetyl groups causes chromatin to condense and its DNA to become less accessible to transcription factors and RNA polymerase, which prevents genes in this region of the DNA from being transcribed. Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Describe the effects of different histone modifications on transcription. 12. Methylation of certain lysine residues in histones leads to gene silencing, whereas methylation of other lysine residues in the same histones leads to transcriptional activation. How can methylation lead to these two different outcomes? Answer: Methylation changes which regulatory proteins interact with histones. The specific change depends on the locations of the lysines that are methylated. In some locations, methylation may cause an activating protein to bind to a histone, which leads to relaxation of chromatin and activation of transcription. In other locations, methylation may cause a repressor protein to bind to a histone, which leads to tightening of the chromatin and repression of transcription. Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Describe the effects of different histone modifications on transcription. 13. Some chromatin remodeling factors use energy from ATP hydrolysis to slide histones along a chromosome. Explain how this type of activity can regulate gene expression. Answer: When DNA is wound around histones, it is less accessible to RNA polymerase and other transcription factors and is not easily transcribed. This DNA could become more accessible to the transcription machinery when chromatin remodeling factors slide its histones to another region of the chromosome. In this way, a previously repressed set of genes can shift to being actively expressed. Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding © 2019 Oxford University Press

Learning Objective: Summarize the action of chromatin remodeling factors. 14. Long noncoding RNAs (lncRNAs) are often described as having a scaffolding role in the regulation of gene expression. Explain what that means. Answer: LncRNAs are elements that bind together multiple proteins that modify chromatin. They also interact with chromatin binding proteins in ways that bind the chromatin remodeling proteins to specific DNA regions and orient them so that they can carry out their modification functions at specific locations in the chromatin. Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Describe the action of lncRNAs in gene expression and activation.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. The lac operon in E. coli is regulated by lactose, which a. activates an activator of transcription. b. inactivates an activator of transcription. c. activates a repressor of transcription. d. inactivates a repressor of transcription. Answer: d Textbook Reference: Gene Regulation in E.coli Bloom’s Category: 1. Remembering Learning Objective: Explain how lactose regulates transcription of the lac operon. Feedback A: Incorrect. That would be a form of positive regulation. Feedback B: Incorrect. There is no activator of transcription in this operon. Feedback C: Incorrect. The repressor is already active in the absence of lactose. Feedback D: Correct! Lactose binds to the repressor, encoded by i, and prevents it from binding to the operator and inhibiting transcription. 2. A research group working with E. coli isolated several lac operon mutants. The group measured -galactosidase activity under different environmental conditions, as shown in the table below. Which statement draws a plausible conclusion about one of these mutants?

a. Mutant I has an operator region that no longer binds repressor. b. Mutant II has a repressor that no longer binds to the operator. c. Mutant III has a promoter site that no longer binds RNA polymerase. d. Mutant IV has a mutation in the -galactosidase gene, rendering its gene product inactive. Answer: b Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 4. Analyzing Learning Objective: Explain how lactose regulates transcription of the lac operon. Feedback A: Incorrect. If the operator no longer binds repressor, the -galactosidase gene would be transcribed with high enzyme activity observed under both lactose and glucose conditions. Feedback B: Correct! If repressor no longer binds to the operator, there is nothing to stop transcription of the -galactosidase gene under any conditions. Feedback C: Incorrect. If the promoter site no longer binds RNA polymerase, there would be no expression of the -galactosidase gene and no enzyme activity under any conditions. © 2019 Oxford University Press

Feedback D: Incorrect. If the -galactosidase gene produces not active product, there can be no enzyme activity under any conditions. 3. Suppose an E. coli cell is in an environment that is rich in glucose and poor in lactose. Which statement best explains gene expression of enzymes used in lactose breakdown? a. A repressor binds to the operator site of the lac operon and inhibits transcription of the genes involved in lactose breakdown. b. The high levels of glucose cause a cascade that ends with CAP binding to RNA polymerase, which activates transcription of the genes involved in lactose breakdown. c. Glucose binds to a repressor protein, which binds to the operator site and blocks RNA polymerase from transcribing the genes involved in lactose breakdown. d. Glucose binds to the operator site of the lac operon, repressing transcription of the genes involved in lactose breakdown. Answer: a Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 3. Applying Learning Objective: Explain how lactose regulates transcription of the lac operon. Feedback A: Correct! When lactose is absent, transcription of lac operon genes is repressed since they would not be used. Feedback B: Incorrect. High levels of glucose have the opposite effect on CAP. Feedback C: Incorrect. Glucose does not bind to the lac repressor protein. Feedback D: Incorrect. Only the repressor protein binds to the operator site. 4. A student made the following table to summarize what he had learned about bacterial transcriptional control mechanisms. A fellow classmate pointed out that the table is not accurate. Which statement describes the problem?

a. The terms listed for the types of regulatory proteins for both positive and negative control systems are incorrect. b. The wild-type regulatory protein in positive control systems causes transcription of the operon to be turned off. c. The mutated nonfunctional regulatory protein in negative control systems causes transcription of the operon to be turned on. d. Both the mutated nonfunctional regulatory protein in positive and negative control systems cause transcription of the operon to be turned on. Answer: c Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 4. Analyzing Learning Objective: Distinguish between positive and negative control. © 2019 Oxford University Press

Feedback A: Incorrect. The correct terms are used for each system. Feedback B: Incorrect. The chart correctly shows that in positive control systems, the operon is turned on in response to an activator. Feedback C: Correct! In negative control systems, a repressor that has lost its repressor function can no longer block RNA polymerase from binding to the promoter and initiating transcription. Feedback D: Incorrect. The chart correctly shows that in positive control systems, the operon is turned off in response to a mutated nonfunctional activator. 5. The lac operon is regulated by both a positive transcriptional control system and a negative transcriptional control system. Under what conditions will both systems operate to stimulate transcription of the lac operon? a. High glucose and high allolactose b. Low glucose and low allolactose c. High glucose and low allolactose d. Low glucose and high allolactose Answer: d Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 3. Applying Learning Objective: Distinguish between positive and negative control. Feedback A: Incorrect. The positive system is not set for transcription in this case. High glucose will not stimulate CAP binding to RNA polymerase to activate it for transcription. Feedback B: Incorrect. The negative system is not set for transcription in this case. Low allolactose will cause the lac repressor to remain bound to the operator site, blocking RNA polymerase from carrying out transcription. Feedback C: Incorrect. Neither the positive system nor the negative system is set for transcription in this case. Feedback D: Correct! Both the positive and negative control systems are set for transcription. Low glucose leads to CAP binding to RNA polymerase to activate it for transcription and high allolactose will cause the lac repressor to be released from the operator site, enabling RNA polymerase to proceed with transcription. 6. Some bacterial promoter regions are regulated by _______ that stimulate RNA polymerase binding and by _______ that block RNA polymerase binding. a. inducers; suppressors b. triggers; arrestors c. activators; repressors d. promoters; operators Answer: c Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 1. Remembering Learning Objective: Explain why repressors inhibit but activators stimulate transcription. Feedback A: Incorrect. These are not the correct terms used to identify the regulatory proteins that stimulate and block transcription, respectively. Feedback B: Incorrect. These are not the correct terms used to identify the regulatory proteins that stimulate and block transcription, respectively. Feedback C: Correct! The terms activator and repressor are used to identify the regulatory © 2019 Oxford University Press

proteins that stimulate and block transcription, respectively. Feedback D: Incorrect. These are not the correct terms used to identify the regulatory proteins that stimulate and block transcription, respectively. 7. Cells regulate gene expression to conserve which resource? a. Water b. Energy c. Food d. Air Answer: b Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 2. Understanding Learning Objective: Explain why repressors inhibit but activators stimulate transcription. Feedback A: Incorrect. Cells use other mechanisms to conserve water. Feedback B: Correct! Cells conserve energy by using regulatory control systems to inhibit transcription of the genes they don’t need, and to stimulate transcription of the genes they do need, at any given moment in time. Feedback C: Incorrect. Cells do not have a way to conserve food. Feedback D: Incorrect. Cells do not conserve air. 8. Which statement about cis-acting elements is true? a. They are specific DNA sequences that control the transcription of adjacent genes. b. Various proteins specifically recognize and bind to these cis-acting sequences. c. They may be directly adjacent to the gene they control or far away. d. All of the above Answer: d Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast promoters and enhancers. Feedback A: Incorrect. This is a true statement, but other answer choices are true as well. Feedback B: Incorrect. This is a true statement, but other answer choices are true as well. Feedback C: Incorrect. This is a true statement, but other answer choices are true as well. Feedback D: Correct! All of the statements are true. 9. Which rationale best describes the ability of an enhancer to mediate transcription from very distant sites? a. Looping of the DNA can occur, allowing the transcription factor to come into proximity of the RNA polymerase. b. When needed, enhancers are spliced into a region closer to the promoter. c. They are recognized by RNA polymerase, which binds to the enhancers and then slides down the promoter toward the gene. d. All of the above Answer: a Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast promoters and enhancers. © 2019 Oxford University Press

Feedback A: Correct! This allows for enhancer regulation from sites located literally kilobases away. Feedback B: Incorrect. This doesn’t occur, unless by chance. Feedback C: Incorrect. RNA polymerase does not bind enhancers. Feedback D: Incorrect. 10. A reporter gene is used to a. identify regulatory sequences from the upstream regions of other genes. b. determine if a protein binds to a given sequence element. c. determine if a gene contains introns. d. determine the stability of a protein. Answer: a Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. Feedback A: Correct! A reporter gene encodes a protein whose level can be measured easily. When DNA sequences are ligated upstream of the reporter, their ability to stimulate or repress transcription can be determined by measuring the level of the protein product. Feedback B: Incorrect. An assay such as the electrophoretic mobility shift assay would be used to assay for binding. Feedback C: Incorrect. A reporter gene plays no such role. Feedback D: Incorrect. A reporter gene is not used to measure the properties of the protein produced by a gene. 11. Which of the following experimental approaches is least likely to be used to identify transcription factor binding sites on DNA? a. Electrophoretic-mobility shift assay b. DNA-affinity chromatography c. A reporter gene assay d. Chromatin immunoprecipitation Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. Answer: b Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Feedback A: Incorrect. This is routinely used in assessing transcription factors. Feedback B: Correct! DNA-affinity chromatography is used to purify transcription factors using DNA sequences known to be binding sites. Feedback C: Incorrect. This approach could be used to identify DNA sequences that respond to transcription factors. Feedback D: Incorrect. With the use of specific antibodies to transcription factors, this could be used to identify binding sites. 12. Eukaryotic transcriptional activators have activation domains that interact with a. DNA binding domains. © 2019 Oxford University Press

b. enhancers and promoters. c. mediator proteins and transcription factors. d. histones. Answer: c Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Describe how activators and repressors affect transcription. Feedback A: Incorrect. Activation domains of transcriptional activators do not bind to DNA binding domains of themselves or other transcriptional activators. Feedback B: Incorrect. Activation domains of transcriptional activators bind to other proteins, not to DNA. Feedback C: Correct! Activation domains of transcriptional activators bind to Mediator proteins and general transcription factors to facilitate assembly of the transcription initiation complex. Feedback D: Incorrect. Activation domains of transcriptional activators are not involved in modifications of histones. 13. A eukaryotic regulatory protein controls expression of a gene. Some data regarding this protein are listed in the table below. Which hypothesis about the mechanism of this protein can be ruled out?

a. This protein blocks an activating transcription factor from binding to the promoter. b. This protein prevents RNA polymerase from binding to the promoter. c. This protein competes with a transcription activator for binding to an enhancer. d. This protein binds to Mediator proteins in the transcription initiation complex. Answer: d Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 4. Analyzing Learning Objective: Describe how activators and repressors affect transcription. Feedback A: Incorrect. This is a plausible mechanism of repression by a regulatory protein that consists of one DNA binding domain. Feedback B: Incorrect. This is a plausible mechanism of repression by a regulatory protein that consists of one DNA binding domain. Feedback C: Incorrect. This is a plausible mechanism of repression by a regulatory protein that consists of one DNA binding domain. Feedback D: Correct! The regulatory protein has only one domain and it is a DNA binding domain. It does not have a regulatory domain that can interact with other proteins. 14. Which molecule has a critical role in transcription elongation? a. P-TEFb b. HAT c. TET © 2019 Oxford University Press

d. lncRNA Answer: a Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 1. Remembering Learning Objective: Explain how transcriptional elongation is controlled. Feedback A: Correct! P-TEFb has protein kinase activity necessary to phosphorylate NELF and DSIF as well as the CTD of RNA polymerase to stimulate elongation of transcription. Feedback B: Incorrect. HAT is a histone acetyltransferase enzyme important in histone modification. Feedback C: Incorrect. TET is an enzyme that removes methyl groups from cytosine residues of DNA. Feedback D: Incorrect. LncRNAs assist with assembly of coactivators or corepressors in the modification of chromatin. 15. In Drosophila, genes encoding heat shock proteins are normally not expressed by are arrested at early transcription phase. However, if these organisms are exposed to sudden bursts of heat, transcription elongation is stimulated and the heat shock proteins are expressed. Which of the following is a possible mechanism that could account for this heat shock effect? a. Heat causes repression of HAT expression. b. Heat stimulates PTEFb expression. c. Heat leads to dephosphorylation of the C terminal domain of RNA polymerase. d. Heat causes expression of an inhibitor of PTEFb. Answer: b Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Explain how transcriptional elongation is controlled. Feedback A: Incorrect. HAT is a class of histone acetyltransferase enzymes that modify histones. Feedback B: Correct! PTEFb is required to phosphorylate NELF, DSIF, and the C terminal domain of RNA polymerase, which stimulates transcription elongation. Feedback C: Incorrect. The C terminal domain of RNA polymerase must be phosphorylated to become active in transcript elongation. Feedback D: Incorrect. PTEFb is required for stimulation of transcription elongation. 16. Which experimental evidence was not useful in the search for a link between histone acetylation and gene activation? a. A Tetrahymena histone acetyltransferase was found to be homologous to a known transcriptional activator in yeast. b. A mammalian histone deacetylase was found to be homologous to a known transcriptional repressor in yeast. c. The expression of a gene for a known transcriptional activator from yeast produced a protein that had high levels of histone acetyltransferase activity. d. Chemical analysis of acetylated histones showed that the acetyl groups were added to lysine residues. Answer: d Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding © 2019 Oxford University Press

Learning Objective: Describe the effects of different histone modifications on transcription. Feedback A: Incorrect. This evidence helped link histone acetylation to gene activation. Feedback B: Incorrect. This evidence helped link histone acetylation to gene activation. Feedback C: Incorrect. This evidence strongly linked histone acetylation to gene activation. Feedback D: Correct! This evidence identified the target residues of the modification events, but did not provide any information about whether the modification was linked to gene expression. 17. A research group characterized regions of DNA within chromatin from mouse cells. They also characterized the histones flanking these regions. The table below summarizes some of their results.

Which statement provides a plausible hypothesis based on these data? a. Regions I and IV are promoters. b. Region III is an enhancer. c. Region II is a promoter. d. Regions I, II, and IV are genes under active transcription. Answer: c Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 4. Analyzing Learning Objective: Describe the effects of different histone modifications on transcription. Feedback A: Incorrect. Enhancers are marked by monomethylated lysines in histones flanking these regions. Feedback B: Incorrect. Enhancer regions are DNAse sensitive. Feedback C: Correct! Promoter regions are DNAse sensitive and marked by trimethylated lysines in histones flanking these regions. Feedback D: Incorrect. Regions I, II, and IV are enhancer and promoter regions, not coding regions. 18. Which is not a type of histone modification used for controlling gene expression? a. C-terminal amidation b. Arginine methylation c. Lysine ubiquitination d. Serine phosphorylation Answer: a Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 1. Remembering Learning Objective: Describe the effects of different histone modifications on transcription. Feedback A: Correct! Histones are not modified via C-terminal amidation as a means for © 2019 Oxford University Press

controlling gene expression. Feedback B: Incorrect. Histones are modified via arginine methylation as a means for controlling gene expression. Feedback C: Incorrect. Histones are modified via lysine ubiquitination as a means for controlling gene expression. Feedback D: Incorrect. Histones are modified via serine phosphorylation as a means for controlling gene expression. 19. Which is a mechanism of histone change that is mediated by chromatin remodeling factors? a. Proteolytic breakdown of histone proteins b. Acetylation of histone proteins c. Ubiquitination of histone proteins d. Conformational change of histone tertiary structure Answer: d Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Summarize the action of chromatin remodeling factors. Feedback A: Incorrect. Chromatin remodeling factors do not break down histone proteins via proteolysis. Feedback B: Incorrect. Chromatin remodeling factors do not chemically modify histone proteins. Feedback C: Incorrect. Chromatin remodeling factors do not chemically modify histone proteins. Feedback D: Correct! Chromatin remodeling factors modify histones in several ways including changing their tertiary structure. 20. A scientist used a mammalian cell extract to see whether it contained any chromatin remodeling factors. She incubated the cell extract with and without ATP in the presence of chromatin containing promoters X, Y, and Z and tested whether RNA polymerase would bind to any of the promoters. The table below summarizes the results. Do the results suggest that a chromatin remodeling factor was present in the cell extract?

a. Yes, because Promoter Y binds RNA polymerase independent of the presence of ATP. b. Yes, because Promoter X binds RNA polymerase only when ATP is present. c. No, because Promoter Z cannot bind RNA polymerase even when ATP is absent. d. No, because there is not enough data to tell whether a chromatin remodeling factor may be present. Answer: b Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 4. Analyzing Learning Objective: Summarize the action of chromatin remodeling factors.

Feedback A: Incorrect. If ATP were absent, a chromatin remodeling factor would not be able to shift nucleosomes so that RNA polymerase can bind. Feedback B: Correct! Chromatin remodeling factors require ATP to shift nucleosomes so that RNA polymerase can bind. Feedback C: Incorrect. If ATP were present, a chromatin remodeling factor would shift nucleosomes so that RNA polymerase can bind. Feedback D: Incorrect. The data suggest that ATP is required for RNA polymerase to bind to the X promoter, which suggests that a chromatin remodeling factor may be involved. 21. Which statement about epigenetic changes is true? a. They alter the base sequences within genes. b. They cannot be passed from one generation to the next. c. They always lead to activation of gene expression. d. They can be reversed. Answer: d Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Explain epigenetic inheritance based on histone modification and on DNA methylation. Feedback A: Incorrect. Epigenetic changes do not change the base sequences within any region of DNA. Feedback B: Incorrect. Epigenetic changes can be passed from parent to offspring. Feedback C: Incorrect. Epigenetic changes can either stimulate or repress gene expression. Feedback D: Correct! There are mechanisms whereby epigenetic changes can be reversed. 22. Cancer cells repress some genes that are expressed in normal cells. Suppose you wanted to design two drugs to treat cancer cells. Which activities would you want your drugs to have? a. DNA methylase activity and histone acetyltransferase activity b. DNA demethylase activity and histone acetyltransferase activity c. DNA methylase activity and histone deacetylase activity d. DNA demethylase activity and histone deacetylase activity Answer: b Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Explain epigenetic inheritance based on histone modification and on DNA methylation. Feedback A: Incorrect. DNA methylases add methyl groups to DNA, which causes suppression of gene expression. Feedback B: Correct! A DNA demethylase will remove methyl groups from DNA and a histone acetyltransferase will add acetyl groups to histones, and both activities are associated with stimulated gene expression. Feedback C: Incorrect. DNA methylases add methyl groups to DNA and histone deacetylases remove acetyl groups from histones, both of which causes suppression of gene expression. Feedback D: Incorrect. Histone deacetylases remove acetyl groups from histones, which causes chromatin to condense and gene expression to be suppressed.

23. How can epigenetic changes explain why identical twins develop different physical characteristics later in life? a. Epigenetic changes are mutations that change the primary structures of proteins. b. Epigenetic changes are changes that occur to proteins after they have been synthesized in cells. c. Epigenetic changes are modifications of DNA and histones that change which genes are expressed. d. Epigenetic changes are additions and deletions of DNA by the action of viruses that infect cells. Answer: c Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Explain epigenetic inheritance based on histone modification and on DNA methylation. Feedback A: Incorrect. Epigenetic changes do not involve mutations of the DNA sequences in an organism. Feedback B: Incorrect. Epigenetic changes are ones that take place in DNA or histones associated with DNA that affect which genes are expressed. Feedback C: Correct! Epigenetic changes are chemical modifications of DNA and histones that affect the expression of genes. Feedback D: Incorrect. Epigenetic changes do not involve viruses or their actions. 24. How are lncRNAs similar to miRNAs? a. They contain similar numbers of nucleotides. b. They help regulate gene expression at the level of transcription. c. They function as activators of gene expression. d. They correspond to noncoding regions of the genome. Answer: d Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Describe the action of lncRNAs in gene repression and activation. Feedback A: Incorrect. lncRNAs contain more than 200 nucleotides, while miRNAs contain 20 to 30 nucleotides. Feedback B: Incorrect. miRNAs inhibit gene expression at the level of translation. Feedback C: Incorrect. lncRNAs help with both activation and repression of gene expression, while miRNAs function to repress gene expression. Feedback D: Correct! Both miRNAs and lncRNAs are noncoding, which means that their sequences are not used to produce proteins 25. lncRNAs act as scaffolds to support proteins that regulate a. chromatin modification. b. RNA polymerase binding to promoters. c. looping of DNA to bring enhancers in proximity to transcription initiation complexes. d. recruitment of Mediator to transcription complexes. Answer: a Textbook Reference: Chromatin and Epigenetics © 2019 Oxford University Press

Bloom’s Category: 2. Understanding Learning Objective: Describe the action of lncRNAs in gene repression and activation. Feedback A: Correct! lncRNAs function in assisting with chromatin modification. Feedback B: Incorrect. Transcription activators assist RNA polymerase binding to promoters. Feedback C: Incorrect. CTCF and cohesin help stabilize looped DNA. Feedback D: Incorrect. Mediator is recruited by the basal transcription complex.

Free-response 1. Suppose that gene X is induced in E. coli cells when they are treated with high levels of ethanol. A mutation in region 1 in the upstream region of gene X results in high-level expression under basal conditions, and a mutation in region 2 results in an uninducible gene. What would you expect from a strain in which both regions 1 and 2 are mutated? Answer: Because a mutation in region 1 causes high-level expression even in the absence of ethanol, region 1 must bind a transcriptional repressor and regulation of gene X must occur by relief of repression. Since a mutation in region 2 blocks inducibility of the gene, region 2 must bind a transcriptional activator: In the absence of an activator, the repressor becomes irrelevant, and the gene is inactive, regardless of conditions. Mutations in both regions 1 and 2 would have the same effect as the single mutation in region 2, because in the absence of a bound transcriptional activator, the repressor is irrelevant, and the gene is inactive regardless of the presence or absence of ethanol. Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 4. Analyzing Learning Objective: Explain how lactose regulates transcription of the lac operon. 2. Describe whether the control mechanism of the lac operon is positive or negative, and why. Answer: The control mechanism is negative because the operon is regulated by a repressor that binds to DNA, preventing RNA polymerase from carrying out transcription. If the control mechanism was positive, some molecular signal would bind to RNA polymerase and act as an activator to stimulate transcription. Textbook Reference: Gene Regulation in E. coli Bloom’s Category: 3. Applying Learning Objective: Distinguish between positive and negative control. 3. You know that there is a protein in a particular cell line that binds within the first 200 bp of a gene’s promoter to activate its transcription. You have another cell line in which the same gene is not expressed, and you suspect that this is due to the lack of binding to this protein. How could electrophoretic-mobility shift assays (EMSA) help test your hypothesis? Answer: EMSA is an excellent method for determining if cellular proteins interact on a particular piece of DNA. First, the 200 bp piece of DNA would be radiolabeled. It would then be incubated with nuclear extracts from the two cell lines and a no-extract control. The reaction with noextract or probe-only control would yield an unbound probe at the bottom of the gel. If proteins were present in the nuclear extracts that bound the probe, the protein bound to the probe would slow or retard the migration of the probe through the gel, resulting in a “shift” of the probe to a point higher in the gel than the unbound free probe. If the hypothesis is correct, the extract from © 2019 Oxford University Press

the cell line with the gene that was not expressed would result in a pattern that looks very similar to that of the probe-only control. The extract from the cell line that does express the gene would have a band that is shifted up in the gel. Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Summarize the experimental approaches used to study the binding of transcription factors to DNA. 4. You have a cell line that contains a gene that regulates cell cycle re-entry. It is normally repressed by a transcriptional repressor that has two functional domains: a DNA binding domain and a repressor domain that inhibits the activity of RNA polymerase. What would likely happen if a protein encoding only the DNA binding domain became dramatically overexpressed in these cells? Answer: Most likely, the repression would be removed, and the cells would undergo uncontrolled growth. This is because in the normal protein, the DNA binding domain brings along the repression domain that inhibits transcription. The mutant lacking the repression domain that would soon overwhelm the normal protein would be favored stoichiometrically to interact with the DNA binding site, and it would not permit the repressor domain to inhibit cell cycle reentry. Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 3. Applying Learning Objective: Describe how activators and repressors affect transcription. 5. Describe two general mechanisms of transcriptional inhibition by eukaryotic repressors. Answer: Some repressors directly bind DNA and prevent binding of activators to the DNA (i.e., they compete for binding with activators). Other repressors have active repression domains that interact with Mediator and other transcription factors. Textbook Reference: Transcription Factors in Eukaryotes Bloom’s Category: 2. Understanding Learning Objective: Describe how activators and repressors affect transcription. 6. In the 1960s, scientists discovered that acetylation of histones was correlated with increased transcription of chromatin. However, it took another 30 years before evidence was found that allowed scientists to say that acetylation was directly responsible for activating regions of chromatin for transcription. What was this evidence, and why did it provide a direct link between histone acetylation and gene activation? Answer: The evidence was from an experiment in which a gene known to be involved in transcriptional activation was cloned and expressed in a bacterial host. The protein that was expressed was studied using a gel assay. It was clear from the experiment that this transcriptional activator was a histone acetylase, providing a direct link between the acetylase activity and gene activation. Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Describe the effects of different histone modifications on transcription.

7. How are chromatin remodeling factors similar to and different from histone acetyltransferases and deacetylases? Answer: While both types of factors disrupt histone-DNA interactions, they do so by different mechanisms. Chromatin remodeling factors disrupt histone-DNA interactions, which changes histone tertiary structure to allow the histones to change position or dissociate from DNA. This occurs without any chemical modification of histones. In contrast, histone acetyltransferases and deacetylases catalyze chemical change to histones, which leads to changes in histone tertiary structure, disrupting histone-DNA interactions. Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 2. Understanding Learning Objective: Summarize the action of chromatin remodeling factors. 8. After they had germinated and sent up shoots, a group of plants was induced to take up the cytosine analog azacytosine, which cannot be methylated. Compared to control plants that were not treated with the analog, these plants produced many more flower stalks displaying flowers that had odd shapes. How can these results be explained? Answer: Methylation of DNA must be involved in repressing transcription of genes needed for producing flower stalks and for producing the flower shape normally observed on this plant. The azacytosine prevented methylation of these genes, and so they were not repressed. The result was transcription of genes that led to a change in flower stalk number and in flower shape. Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 4. Analyzing Learning Objective: Explain epigenetic inheritance based on histone modification and on DNA methylation. 9. The nucleus of a mature mouse skin cell is transplanted into an enucleated mouse embryo. The embryo develops into a blastocyst but then stalls at this stage. How can epigenetics explain this? Answer: The nucleus of a mature mouse skin cell contains chromatin that has been modified through all of the stages of development and differentiation such that it is only able to express the genes needed by a skin cell. Its genes that are required for development of the blastocyst to the next stage have been repressed via epigenetic changes, including DNA methylation and chromatin modification. Therefore, the cell cannot move past the blastocyst stage because its chromatin is unable to transcribe the genes necessary for this stage of development. Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying Learning Objective: Explain epigenetic inheritance based on histone modification and on DNA methylation. 10. How could you use chromatin immunoprecipitation to isolate and identify a lncRNA involved in chromatin modification? Answer: Use formaldehyde to crosslink proteins to lncRNA and to DNA in chromatin. Shear the DNA into smaller pieces. Then use immunoprecipitation to isolate specific proteins known to bind to lncRNA, such as polycomb proteins, histone deacetylases, or DNA methylases. Finally, reverse the crosslinks between proteins and lncRNA, and sequence the lncRNA. Textbook Reference: Chromatin and Epigenetics Bloom’s Category: 3. Applying © 2019 Oxford University Press

Learning Objective: Describe the action of lncRNAs in gene repression and activation.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 10: Protein Synthesis, Processing, and Regulation TEST FILE QUESTIONS Multiple Choice 1. Aminoacyl tRNA synthetases are enzymes that a. synthesize transfer RNAs. b. attach amino acids to specific transfer RNAs. c. connect amino acids while they are held in place on ribosomes by transfer RNAs. d. attach the terminal CCA sequence to transfer RNAs. Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Explain the role of tRNAs in translation. 2. E. coli contains about _______ different tRNAs that code for _______ different amino acids. a. 62; 40 b. 62; 20 c. 50; 20 d. 40; 20 Answer: d Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Explain the role of tRNAs in translation. 3. The capacity for some tRNAs to recognize more than one codon in mRNA is explained by a phenomenon called a. redundancy. b. wobble. c. reading frameshift. d. degeneracy. Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Explain the role of tRNAs in translation. 4. Translation always occurs on which of the following structures? a. Ribosomes © 2019 Oxford University Press

b. Endoplasmic reticulum c. Nuclear envelope d. Mitochondria Answer: a Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of ribosomes. 5. The combined sizes of prokaryotic and eukaryotic ribosomes are a. the same. b. 30S and 50S, respectively. c. 40S and 60S, respectively. d. 70S and 80S, respectively. Answer: d Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of ribosomes. 6. Which statement is true and provides evidence that a certain component of the ribosome catalyzes protein synthesis? a. Ribosomes are inactive after protease digestion. b. Ribosomes are inactive after RNase digestion. c. Structural analysis shows that proteins occupy the catalytic site where peptide bonds are formed. d. Structural analysis shows that mRNA occupies the catalytic site where peptide bonds are formed. Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and function of ribosomes. 7. The formation of peptide bonds is catalyzed by which portion of the bacterial ribosome? a. Proteins of the small subunit b. 16S rRNA c. Proteins of the large subunit d. 23S rRNA Answer: d Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and function of ribosomes. 8. Eukaryotic ribosomes recognize and initially bind to what structure on the mRNA? a. A Shine-Dalgarno sequence b. The 7-methylguanosine cap c. A TATA sequence © 2019 Oxford University Press

d. A CCAAT sequence Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. 9. The first step in the initiation of protein synthesis is the binding of _______ to the _______. a. initiation factors; initiation codon b. initiation factors; small ribosomal subunit c. the small ribosomal subunit; initiation codon d. the initiator tRNA; initiation codon Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. 10. The first amino acid that initiates the eukaryotic polypeptide is a. any amino acid. b. glutamine. c. methionine. d. N-formylmethionine. Answer: c Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. 11. In translation, mRNAs are read in the _______ direction, and polypeptide chains are synthesized from the _______ ends. a. 5ʹ to 3ʹ; carboxyl to the amino b. 5ʹ to 3ʹ; amino to the carboxyl c. 3ʹ to 5ʹ; carboxyl to the amino d. 3ʹ to 5ʹ; amino to the carboxyl Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Outline the events of initiation, elongation, and termination of translation. 12. During translation, the codons on the mRNA are recognized by complementary base pairing to the anticodon on the a. ribosome. b. transfer RNA. c. small cytoplasmic RNA. d. aminoacyl tRNA synthetase. Answer: b © 2019 Oxford University Press

Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Outline the events of initiation, elongation, and termination of translation. 13. Translation of mRNAs starts at a. the 3ʹ end of the mRNA. b. a site downstream of a 3ʹ untranslated region. c. the 5ʹ end of the mRNA. d. a site downstream of a 5ʹ untranslated region. Answer: d Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Outline the events of initiation, elongation, and termination of translation. 14. The factor that escorts the aminoacyl tRNA to the eukaryotic ribosome and then releases it with GTP hydrolysis following the correct codon–anticodon base pairing is a. eIF1. b. eRF1. c. eIF2. d. eEF1. Answer: d Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Outline the events of initiation, elongation, and termination of translation. 15. Termination of translation and release of the polypeptide chain occur when a. tRNA binds to a termination codon via a complementary anticodon but lacks an amino acid. b. a protein release factor binds to the termination codon. c. a tRNA with a complementary anticodon binds to a termination codon, and a release factor bound to the CCA end releases the chain. d. a small molecule shaped like puromycin binds to the termination codon. Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Outline the events of initiation, elongation, and termination of translation. 16. To function as a repressor-binding site, the iron response element (IRE) must be within the first 70 nucleotides of the ferritin mRNA. This suggests that protein binding to this IRE a. regulates message degradation. b. interferes with the 5ʹ cap binding to the ribosome. © 2019 Oxford University Press

c. blocks the translation start site. d. inhibits the binding of eEF2. Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that regulate translation. 17. The amount of a protein in a cell is regulated by the rate of a. translation of its mRNA only. b. transcription of its gene and translation of its mRNA only. c. translation of its mRNA and degradation of the protein only. d. transcription of its gene, translation of its mRNA, and degradation of the protein. Answer: d Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that regulate translation. 18. Oocytes store mRNAs with _______ and activate them by _______. a. no cap; adding a cap b. long poly-A tails; shortening the tails c. short poly-A tails; lengthening the tails d. bound siRNA; removing the siRNA Answer: c Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms that regulate translation. 19. Peptide bond formation in translation occurs by a. an enzyme catalyzed reaction. b. a tRNA catalyzed reaction. c. a tRNA synthase catalyzed reaction. d. an rRNA catalyzed reaction. Answer: d Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms that regulate translation. 20. siRNAs act primarily to inhibit a. transcription, by binding to specific genes. b. transcription, by formation of heterochromatin. c. translation, by blocking ribosomal attachment. d. translation, by binding to an mRNA. Answer: d Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms that regulate translation. © 2019 Oxford University Press

21. Phosphorylation of eukaryotic initiation factors 2 and 2B (eIF2 and eIF2B) a. allows them to be recycled back to the ribosome. b. allows them to initiate translation. c. blocks their exchange of bound GDP for GTP. d. blocks the addition of a terminal phosphate to the bound GDP. Answer: c Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that regulate translation. 22. The correctly folded three-dimensional configuration of a protein is determined primarily by the a. sequence of nucleotides of its gene. b. primary sequence of its amino acids. c. chaperones with which it interacts. d. pathway by which it folds. Answer: b Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Explain how chaperones facilitate protein folding. 23. Proteins that facilitate the folding of other proteins are called a. foldases. b. binding proteins. c. chaperones. d. escorts. Answer: c Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Explain how chaperones facilitate protein folding. 24. Many chaperones are called heat-shock proteins because they a. are expressed at higher levels after a heat shock than under normal growth conditions. b. cause fever in mammals. c. misfold during a heat shock. d. denature at high temperatures. Answer: a Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Explain how chaperones facilitate protein folding. 25. Transmissible spongiform encephalopathies, such as scrapie, mad cow disease, Creutzfeldt Jakob disease, and kuru, are caused by an infectious agent known as a prion. Prions are composed of a. protein only. © 2019 Oxford University Press

b. protein and RNA. c. protein and DNA. d. RNA surrounded by a lipid-protein coat. Answer: a Textbook Reference: Protein Folding and Processing Bloom’s Category: 2. Understanding Learning Objective: Give examples of diseases associated with protein misfolding. 26. The enzyme protein disulfide isomerase, which facilitates breakage and reformation of disulfide bonds, is located primarily in the a. cytosol. b. lumen of the endoplasmic reticulum. c. matrix of mitochondria. d. lumen of lysosomes. Answer: b Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Describe the reactions catalyzed by protein disulfide isomerase and peptidyl prolyl isomerase. 27. In cystic fibrosis patients, deletion of an amino acid (phenylalanine 508) in the CFTR protein a. disrupts the normal interactions of CFTR with chaperones in the endoplasmic reticulum, leading to defective protein folding and reduced levels of functional CFTR in the plasma membranes of affected cells. b. leads to misfolded proteins, which aggregate to form insoluble amyloid fibrils characterized by β-sheet structures. c. precludes CFTR phosphorylation by a protein kinase, which is necessary for activation of this protein. d. prevents ATP hydrolysis required for the active transport of Cl– by CFTR. Answer: a Textbook Reference: Protein Folding and Processing Bloom’s Category: 2. Understanding Learning Objective: Explain how proteolysis can convert an inactive precursor to an active protein. 28. Prior to N-linked glycosylation of a protein, a complex oligosaccharide is assembled in the endoplasmic reticulum on a lipid carrier called a. a prenyl group. b. myristic acid. c. dolichol phosphate. d. phosphatidylinositol. Answer: c Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering

Learning Objective: Summarize the modifications of proteins by additions of carbohydrates and lipids. 29. N-linked glycosylation attaches a complex carbohydrate onto the a. free amino group at the amino terminal end of the polypeptide. b. amino group of asparagine. c. amino group of lysine. d. carboxyl group of aspartic acid. Answer: b Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Summarize the modifications of proteins by additions of carbohydrates and lipids. 30. Which of the following groups anchors proteins to the outer surface of the plasma membrane? a. Myristate b. Farnesyl c. Palmitate d. Glycolipid Answer: d Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Summarize the modifications of proteins by additions of carbohydrates and lipids. 31. Monomeric G proteins, such as Ras and several elongation factors, are usually in the active state when a molecule of _______ is bound to them. a. GDP b. GTP c. ADP d. ATP Answer: b Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 2. Understanding Learning Objective: Explain how the binding of a small molecule can change the catalytic activity of an enzyme. 32. Nitrosylation occurs on the side chain of a. lysine. b. cysteine. c. tyrosine. d. serine and threonine. Answer: b Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 2. Understanding © 2019 Oxford University Press

Learning Objective: Explain how the binding of a small molecule can change the catalytic activity of an enzyme. 33. Proteins are often regulated by phosphorylation, which is catalyzed by enzymes called a. protein phosphatases. b. phosphoproteases. c. protein phosphorylases. d. protein kinases. Answer: d Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of kinases and phosphatases in regulating protein activity. 34. Which of the following amino acids is not commonly phosphorylated to regulate protein activity? a. Serine b. Threonine c. Tryptophan d. Tyrosine Answer: c Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of kinases and phosphatases in regulating protein activity. 35. The removal of phosphates from proteins is catalyzed by a. protein phosphatases. b. phosphoproteases. c. protein phosphorylases. d. protein kinases. Answer: a Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of kinases and phosphatases in regulating protein activity. 36. Receptor tyrosine kinases catalyze the transfer of a phosphate group from _______ to the side chains of tyrosine residues. a. cAMP b. the plasma membrane c. ATP d. GTP Answer: c Textbook Reference: Regulation of Protein Function and Stability © 2019 Oxford University Press

Bloom’s Category: 2. Understanding Learning Objective: Describe the roles of kinases and phosphatases in regulating protein activity. 37. Which of the following regulates genes by binding directly to transcription factor proteins? a. Growth hormone b. Insulin c. Steroid hormones d. Epinephrine Answer: c Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 1. Remembering Learning Objective: Explain how protein–protein interactions play regulatory roles. 38. Which protein is inhibited when bound to its regulatory subunit(s) but becomes active when free? a. Adenylyl cyclase b. cAMP-dependent protein kinase c. Protein kinase C d. Phosphorylase kinase Answer: b Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 2. Understanding Learning Objective: Explain how protein–protein interactions play regulatory roles. 39. In a major protein degradation pathway, a short polypeptide called _______ is attached to a protein to target it for destruction. a. glutathione b. ubiquinone c. ubiquitin d. KDEL Answer: c Textbook Reference: Protein Degradation Bloom’s Category: 1. Remembering Learning Objective: Summarize protein degradation by the ubiquitin-proteasome pathway. 40. A proteasome is a a. vesicle containing proteolytic enzymes. b. precursor to lysosomes. c. complex of a proteolytic enzyme and the protein that is being degraded. d. multisubunit protease complex that degrades proteins marked for destruction. Answer: d Textbook Reference: Protein Degradation Bloom’s Category: 1. Remembering © 2019 Oxford University Press

Learning Objective: Summarize protein degradation by the ubiquitin-proteasome pathway. 41. In proteasomes, ubiquitin a. is degraded. b. is released and recycled. c. is phosphorylated. d. has a GTP group added. Answer: b Textbook Reference: Protein Degradation Bloom’s Category: 1. Remembering Learning Objective: Summarize protein degradation by the ubiquitin-proteasome pathway. 42. A protein that becomes marked in a sequence called the destruction box and is degraded by proteasomes is a. tubulin. b. cyclin. c. Ras. d. cAMP-dependent protein kinase. Answer: b Textbook Reference: Protein Degradation Bloom’s Category: 1. Remembering Learning Objective: Summarize protein degradation by the ubiquitin-proteasome pathway.

Fill in the Blank 1. The adapters that align amino acids in a sequence determined by the mRNAs are called _______. Answer: tRNAs Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Explain the role of tRNAs in translation. 2. Of the 64 possible codons, _______ (number) are usually stop codons and the others code for amino acids. Answer: three (3) Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Explain the role of tRNAs in translation. 3. Protein synthesis occurs on RNA-protein complexes called _______. Answer: ribosomes Textbook Reference: Translation of mRNA © 2019 Oxford University Press

Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of ribosomes. 4. Prokaryotic ribosomes differ from eukaryotic ribosomes in that their large and small subunits are _______ and _______, respectively, rather than 60S and 40S. Answer: 50S; 30S Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of ribosomes. 5. Translation starts at a particular start codon near the _______ end of the mRNA. Answer: 5ʹ Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. 6. Initiation sequences in bacterial mRNAs are preceded by a specific sequence called a _______ sequence. Answer: Shine-Dalgarno Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. 7. Translation is initiated by the binding of the methionyl tRNA and the mRNA to the _______ subunit, after which the other subunit joins the complex. Answer: small ribosomal Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. 8. Elongation requires _______ GTP per amino acid. Answer: two (2) Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Outline the events of initiation, elongation, and termination of translation. 9. Two human diseases caused by aggregates of misfolded proteins are _______ and _______. Answer: Alzheimer’s; Parkinson’s Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Give examples of diseases associated with protein misfolding. 10. The enzyme that catalyzes the breakage and reformation of disulfide bonds between cysteine residues of proteins is _______. © 2019 Oxford University Press

Answer: protein disulfide isomerase (or PDI) Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Describe the reactions catalyzed by protein disulfide isomerase and peptidyl prolyl isomerase 11. Many cellular proteins are regulated by the level of phosphorylation of certain amino acids. The level of phosphorylation results from the opposing actions of two types of enzymes, protein _______ and protein _______. Answer: kinases; phosphorylases Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of kinases and phosphatases in regulating protein activity. 12. A protein kinase transfers a phosphate group from _______ to a protein. Answer: ATP Textbook Reference: Regulation of Protein Function Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of kinases and phosphatases in regulating protein activity.

True/False 1. Cells contain 64 tRNAs, with one anticodon for each codon. Answer: F Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Explain the role of tRNAs in translation. 2. An 80S eukaryotic ribosome is made up of one large 50S subunit and one small 30S subunit. Answer: F Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of ribosomes. 3. Eukaryotic mRNAs have a 7-methylguanosine cap on their 3ʹ end. Answer: F Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. 4. A single mRNA can code for more than one polypeptide chain. Answer: T © 2019 Oxford University Press

Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Outline the events of initiation, elongation, and termination of translation. 5. One molecule of GTP is split in the process of loading one amino acid onto its tRNA. Answer: F Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Outline the events of initiation, elongation, and termination of translation. 6. Ferritin levels can be regulated by IRE binding proteins that bind to a site near the 5ʹ end of ferritin mRNAs and block translation of the RNA. Answer: T Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that regulate translation. 7. Translation can be regulated by proteins that bind to the 3ʹ untranslated regions of mRNAs. Answer: T Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that regulate translation. 8. MicroRNAs can regulate translation by targeting proteins that can either cleave an mRNA or repress its translation. Answer: T Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that regulate translation. 9. Chaperones determine the folded conformation of many proteins. Answer: F Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Explain how chaperones facilitate protein folding. 10. Some chaperones were originally known as heat-shock proteins because they are expressed in cells subjected to high temperatures and facilitate the refolding of denatured proteins. Answer: T Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Explain how chaperones facilitate protein folding. © 2019 Oxford University Press

Short Answer 1. One unique feature of prokaryotes is that their mRNAs are often polycistronic. What does this mean? Answer: Prokaryotic bacteria are polycistronic, meaning that a single mRNA can encode multiple proteins, each with its own start and stop sites. Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. 2. Experimentally you measure the number of mRNAs being processed from a single gene at 40 copies per minute, yet based on the rate at which ribosomes scan, there should only be about three copies per minute. How might this be explained? Answer: This is most likely due to the fact that multiple ribosomal complexes can attach to a single mRNA and process polypeptides. These groups of ribosomes are referred to as polysomes. Textbook Reference: Translation of mRNA Bloom’s Category: 4. Analyzing Learning Objective: Outline the events of initiation, elongation, and termination of translation. 3. What would happen in a bacterium in which the Shine-Dalgarno sequence was mutated? Answer: The Shine-Dalgarno sequence is required for the 16S ribosomal RNA to base pair match with and direct the initiation of translation. A mutation in this sequence would prevent 16S ribosomal RNA attachment and block or greatly impair protein synthesis. Textbook Reference: Translation of mRNA Bloom’s Category: 4. Analyzing Learning Objective: Summarize the mechanisms that regulate translation. 4. Why do chaperones bind to nascent polypeptide chains while they are still being translated? Answer: To protect the nascent chain from misfolding or aggregating until an entire domain or polypeptide is completed and can fold properly. Textbook Reference: Protein Folding and Processing Bloom’s Category: 2. Understanding Learning Objective: Explain how chaperones facilitate protein folding. 5. Why are disulfide bonds, which form between proteins in the endoplasmic reticulum, unable to form between proteins in the cytosol? Answer: Disulfide bonds cannot form between proteins in the cytosol because the cytosol contains reducing agents that maintain cysteine residues (or −SH groups) in their reduced form. Textbook Reference: Protein Folding and Processing © 2019 Oxford University Press

Bloom’s Category: 2. Understanding Learning Objective: Describe the reactions catalyzed by protein disulfide isomerase and peptidyl prolyl isomerase. 6. What would be the likely result of a mutation in the enzyme protein disulfide isomerase (PDI)? Answer: There would be significant problems with protein folding. PDI is responsible for the formation of disulfide bonds between cysteine residues, important in stabilizing the folded structures of many proteins. Textbook Reference: Protein Folding and Processing Bloom’s Category: 3. Applying Learning Objective: Describe the reactions catalyzed by protein disulfide isomerase and peptidyl prolyl isomerase. 7. By what mechanism is proinsulin converted to insulin? Answer: Proinsulin is converted to insulin by proteolytic removal of an internal connecting peptide. Textbook Reference: Protein Folding and Processing Bloom’s Category: 2. Understanding Learning Objective: Explain how proteolysis can convert an inactive precursor to an active protein. 8. To what do glycosylphosphatidylinositol (GPI) anchor target proteins? Answer: GPI anchors target proteins to the outer face of the plasma membrane. Textbook Reference: Protein Folding and Processing Bloom’s Category: 2. Understanding Learning Objective: Summarize the modifications of proteins by additions of carbohydrates and lipids. 9. What function does covalent modification of proteins serve in the cell that proteolysis cannot? Answer: Both mechanisms regulate enzymes, but covalent modification permits recycling of proteins; proteolysis is irreversible. Textbook Reference: Protein Folding and Processing Bloom’s Category: 2. Understanding Learning Objective: Summarize the modifications of proteins by additions of carbohydrates and lipids.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. Which statement regarding tRNAs is false? a. tRNAs are approximately 70–80 bases long and form a cloverleaf structure. b. All tRNAs have a CCA sequence at their 3ʹ terminus. © 2019 Oxford University Press

c. tRNAs differ in sequence only at the anticodon. d. There are several modified bases present in mature tRNAs. Answer: c Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Explain the role of tRNAs in translation. Feedback A: Incorrect. This is a true statement. Feedback B: Incorrect. This is a true statement. Feedback C: Correct! This is not true. While there are many regions of shared sequence identity between tRNAs, they are not this similar. Feedback D: Incorrect. This is a true statement. 2. The function of aminoacyl tRNA synthetases is to a. covalently attach amino acids to their corresponding tRNA molecules. b. synthesize tRNA molecules. c. catalyze the formation of the aminoacyl ATP intermediate during amino acid attachment to tRNAs. d. catalyze the formation of a peptide bond between amino acids. Answer: a Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Explain the role of tRNAs in translation. Feedback A: Correct! These enzymes attach amino acids to the 3ʹ ends of tRNA molecules, the first step in the inclusion of amino acids in a polypeptide chain. Feedback B: Incorrect. tRNA molecules are synthesized by RNA polymerase III. Feedback C: Incorrect. Aminoacyl tRNA synthetases catalyze the formation of an intermediate, but it is not aminoacyl ATP. Feedback D: Incorrect. This function is carried out by the ribosome. 3. Which statement about the attachment of amino acids to tRNAs is false? a. The amino acid is first joined to AMP, forming an aminoacyl AMP intermediate. b. Two molecules of ATP are required for the process, one at each step. c. Aminoacyl tRNA synthetases catalyze the reaction. d. The amino acid is transferred to the 3ʹ end of the tRNA. Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Explain the role of tRNAs in translation. Feedback A: Incorrect. This is the first step in the process. Feedback B: Correct! Only one ATP is required, and it is used in the formation of the aminoacyl AMP intermediate. Feedback C: Incorrect. In fact, both steps in the process are catalyzed by this enzyme. Feedback D: Incorrect. This is the second step in the process. 4. The first amino acid of eukaryotic polypeptides is a. the amino acid encoded by the first 5ʹ codon. © 2019 Oxford University Press

b. valine. c. N-formylmethionine. d. methionine. Answer: d Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. Feedback A: Incorrect. In neither prokaryotic nor eukaryotic polypeptides is the first 5ʹ codon the initiator codon. Feedback B: Incorrect. The codon for valine, GUG, is occasionally used as the initiator codon in prokaryotic translation, but this is an exception. In eukaryotes, it codes for a different amino acid. Feedback C: Incorrect. This is the first amino acid of prokaryotic polypeptides. Feedback D: Correct! Translation of eukaryotic messages always begins with an AUG codon, which codes for methionine. 5. The initiator codon in prokaryotes is a. the first codon located at the 5ʹ end of the mRNA. b. recognized by scanning of the ribosome downstream of the 5ʹ 7-methylguanosine cap. c. recognized via the Shine-Dalgarno sequence. d. the first 5ʹ AUG of the mRNA. Answer: c Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. Feedback A: Incorrect. Many prokaryotic mRNAs encode more than one protein, and thus a different mechanism must exist for locating an initiator codon. Feedback B: Incorrect. This is the mechanism used in eukaryotes. Feedback C: Correct! Initiation codons in prokaryotes are preceded by a specific sequence called the Shine-Dalgarno sequence, which correctly aligns them on the ribosome for initiation of translation. Feedback D: Incorrect. Because prokaryotic mRNAs often consist of more than one open reading frame, another mechanism must exist for recognizing an initiator codon. 6. Which statement about translational initiation is false? a. In prokaryotes, ribosomes often bind the mRNA and can scan 5ʹ or 3ʹ until recognizing a Shine-Dalgarno sequence. b. Viral mRNAs contain internal ribosome entry sites that allow ribosomes to bind to an internal site of the mRNA. c. Initiation codons in prokaryotic cells are preceded by Shine-Dalgarno sequences. d. 5ʹ 7-methylguanosine caps serve as the point of recognition and binding for ribosomes in eukaryotic cells. Answer: a Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. © 2019 Oxford University Press

Feedback A: Correct! This is a false statement. Feedback B: Incorrect. This is true, and these internal ribosome entry sites are often placed in front of genes when viruses are used as a vector in gene therapy. Feedback C: Incorrect. This is a true statement. Shine-Dalgarno sequences represent a specific sequence of bases that align the mRNA on the ribosome. Feedback D: Incorrect. This is a true statement. 7. Antibiotics are powerful medications that inhibit the growth of bacteria. They work at a variety of levels, but many target the process of protein synthesis in the bacterial cell. In the developing of an antibiotic, which of the following would be an effective strategy or target for the drug? a. Inhibiting translational initiation b. Inducing premature polypeptide chain termination c. Inhibiting aminoacyl tRNA binding d. All of the above Answer: d Textbook Reference: Translation of mRNA Bloom’s Category: 3. Applying Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. Feedback A: Incorrect. This answer choice is correct, but the other answer choices are correct as well. Feedback B: Incorrect. This answer choice is correct, but the other answer choices are correct as well. Feedback C: Incorrect. This answer choice is correct, but the other answer choices are correct as well. Feedback D: Correct! All of these would be effective strategies or targets. 8. The primary function of rRNAs in the ribosome is to a. serve as a scaffold for the ribosomal proteins. b. assist in the proper positioning of tRNAs along the mRNA template. c. catalyze peptide bond formation. d. assist in the proper folding of ribosomal proteins. Answer: c Textbook Reference: Translation of mRNA Bloom’s Category: 1. Remembering Learning Objective: Outline the events of initiation, elongation, and termination of translation. Feedback A: Incorrect. This was once thought to be the case, but it is now considered inaccurate. Feedback B: Incorrect. This is the role of ribosomal proteins. Feedback C: Correct! Although this fundamental reaction in protein synthesis was once thought to be carried out by proteins, it has recently been found to be catalyzed by rRNAs. Feedback D: Incorrect. In fact, the opposite is thought to be true—that ribosomal proteins assist in the proper folding of rRNAs.

9. Ferritin expression is stimulated by iron because iron a. stimulates a protein to bind the ferritin mRNA and inhibit its degradation. b. stimulates the dissociation of a translational inhibitor from the ferritin mRNA. c. binding stabilizes the ferritin protein. d. stimulates the extension of poly-A tails of ferritin mRNAs. Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that regulate translation. Feedback A: Incorrect. This type of mechanism regulates transferrin-receptor expression. Feedback B: Correct! A translational inhibitor bound to the 5ʹ end of the ferritin mRNA is released upon binding iron. Feedback C: Incorrect. Although ferritin binds iron, this is not the mechanism by which iron stimulates ferritin expression. Feedback D: Incorrect. This mechanism of translational regulation is at work during oocyte development, but it does not apply to ferritin. 10. Which statement about translational regulation of ferritin is false? a. The iron response element is a unique sequence of amino acids near the amino terminus of the growing polypeptide. b. In the absence of iron, an iron regulatory protein binds the IRE, preventing translation. c. In the presence of iron, the ferritin protein is translated. d. The iron binding protein must bind the mRNA within about 70 bases of the 5ʹ mRNA cap. Answer: a Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that regulate translation. Feedback A: Correct! This is a false statement. The IRE is a unique sequence of bases located near the 5ʹ cap of the mRNA. Feedback B: Incorrect. This is a true statement. Feedback C: Incorrect. This is a true statement. Feedback D: Incorrect. This is a true statement. 11. Which process is not associated with translational regulation? a. Dephosphorylation of 4E-BPs that bind eIF4E and prevent its interaction with eIF4G b. Autolytic degradation of the mRNA by folding back on itself c. Cleavage of mRNA by miRNA/RISC complexes d. Phosphorylation of eIF2, which inhibits GDP/GTP exchange Answer: b Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that regulate translation. Feedback A: Incorrect. Growth factors work through these 4E-BPs. Feedback B: Correct! This is not a form of translational regulation. Feedback C: Incorrect. © 2019 Oxford University Press

Feedback D: Incorrect. A mechanism similar to the one described in answer A works on eIF2B as well. 12. The proteins shaped like a “double chamber” that are involved in protein folding are called a. Hsp70 proteins. b. protein disulfide isomerases. c. chaperonin proteins. d. peptidyl prolyl isomerases. Answer: c Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Explain how chaperones facilitate protein folding. Feedback A: Incorrect. Hsp70s are involved in protein folding during translation and transport into various subcellular compartments, but they are not shaped like double chambers. Feedback B: Incorrect. These enzymes promote correct protein folding by stimulating exchanges between paired disulfides in the ER, but they are not double-chamber shaped. Feedback C: Correct! These proteins form a “double-chambered structure” in which proteins are sequestered from the cytosol until they are properly folded. Feedback D: Incorrect. These are small enzymes that assist in protein folding by stimulating isomerization between the cis and trans isomers of proline residues in polypeptide chains. 13. Glycosylphosphatidylinositol (or GPI) anchors are found a. at the carboxy terminus of membrane proteins. b. at the amino terminus of membrane proteins. c. on the cytoplasmic side of membrane channel proteins. d. on the amino terminus of the 28S of certain ribosomal complexes associated with secretion. Answer: a Textbook Reference: Protein Folding and Processing Bloom’s Category: 1. Remembering Learning Objective: Summarize the modifications of proteins by additions of carbohydrates and lipids. Feedback A: Correct! These sequences are usually about 20 amino acids long. Feedback B: Incorrect. Feedback C: Incorrect. Feedback D: Incorrect. 14. Which of the following is not an example of posttranslational modification? a. Glycosylation b. Proteolysis c. Palmitoylation d. Self-splicing Answer: d © 2019 Oxford University Press

Textbook Reference: Protein Folding and Processing Bloom’s Category: 2. Understanding Learning Objective: Summarize the modifications of proteins by additions of carbohydrates and lipids. Feedback A: Incorrect. This answer choice is correct, but the other answer choices are correct as well. This is the process of adding carbohydrates. Feedback B: Incorrect. This answer choice is correct, but the other answer choices are correct as well. This is a classic example of the process by which preproinsulin becomes mature insulin. Feedback C: Incorrect. This answer choice is correct, but the other answer choices are correct as well. This involves addition of palmitic acid and represents one of many types of modifications in which lipids are added to proteins. Feedback D: Correct! It is important to realize that many modifications are made to proteins and affect their function. 15. Which process is not a common lipid modification to proteins? a. N-myristylation b. Prenylation c. GPI anchor addition d. Glycosylation Answer: d Textbook Reference: Protein Folding and Processing Bloom’s Category: 2. Understanding Learning Objective: Summarize the modifications of proteins by additions of carbohydrates and lipids. Feedback A: Incorrect. Addition of myristic acid to an N-terminal glycine residue occurs for many proteins associated with the inner face of the plasma membrane. Feedback B: Incorrect. Many plasma membrane proteins involved in the regulation of cell growth and differentiation (such as Ras) are prenylated. Feedback C: Incorrect. Glycosylphosphatidylinositol is often attached to proteins that are located on the outer face of the plasma membrane. Feedback D: Correct! Glycosylation is the addition of an oligosaccharide, not a lipid, to a protein. 16. cAMP activates cAMP-dependent protein kinase by a. stimulating its phosphorylation. b. stimulating the dimerization of kinase subunits. c. stimulating the release of a translational inhibitory protein bound to its mRNA. d. binding regulatory subunits and inducing their release from the catalytic subunits. Answer: d Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 1. Remembering Learning Objective: Explain how the binding of a small molecule can change the catalytic activity of an enzyme. Feedback A: Incorrect. cAMP does not act via this mechanism. Feedback B: Incorrect. Many proteins are activated by dimerization, but not this one. © 2019 Oxford University Press

Feedback C: Incorrect. This type of mechanism functions in the regulation of ferritin, but not for cAMP-dependent protein kinase. Feedback D: Correct! The regulatory subunits inhibit the activity of the catalytic subunits until they are bound by cAMP, which induces a conformational change that causes dissociation of the complex. 17. Protein phosphatases a. catalyze the addition of phosphate residues to proteins. b. catalyze the removal of phosphate residues from proteins. c. catalyze the addition of glycosylphosphatidylinositol to proteins. d. are proteins that specifically bind phosphorylated proteins. Answer: b Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of kinases and phosphatases in regulating protein activity. Feedback A: Incorrect. Phosphatases do possess an enzymatic activity, but they cannot add phosphates to proteins. Feedback B: Correct! Phosphatases can remove phosphate residues from serine, threonine, and tyrosine residues. Feedback C: Incorrect. Protein phosphatases are not involved in adding glycolipids to proteins. Feedback D: Incorrect. Protein phosphatases are characterized by an enzymatic activity, not a binding activity. 18. The half-lives of proteins in the cell vary widely, ranging from a. milliseconds to seconds. b. three to seven minutes. c. minutes to days. d. days to weeks. Answer: c Textbook Reference: Protein Degradation Bloom’s Category: 1. Remembering Learning Objective: Summarize protein degradation by the ubiquitin-proteasome pathway. Feedback A: Incorrect. Protein half-lives are typically longer than seconds. Feedback B: Incorrect. The range is broader than this. Feedback C: Correct! A regulatory protein, such as an inducible transcription factor, may have a half-life as short as a few minutes, whereas structural proteins or other proteins necessary for the day-to-day maintenance of cell physiology may have longer half-lives, up to a few days. Feedback D: Incorrect. Few, if any, proteins last as long as a few weeks. 19. A proteasome is a a. vesicle containing proteolytic enzymes. b. multisubunit protease complex that degrades proteins marked for destruction. © 2019 Oxford University Press

c. complex including one proteolytic enzyme and the protein that is being degraded. d. precursor to lysosomes. Answer: b Textbook Reference: Protein Degradation Bloom’s Category: 1. Remembering Learning Objective: Summarize protein degradation by the ubiquitin-proteasome pathway. Feedback A: Incorrect. The proteasome is not a vesicle. Feedback B: Correct! The proteasome degrades ubiquitinated proteins. Feedback C: Incorrect. The proteasome contains more than one proteolytic enzyme. Feedback D: Incorrect. Endosomes are precursors to lysosomes.

Essay 1. Sixty-four different codons translate into only 20 amino acids. How are tRNAs adapted to address this? Answer: The genetic code is redundant, meaning that most amino acids are specified by more than one codon. Three of these codons signal translation termination, which leaves 61 triplets to code for amino acids. Although some amino acids are linked to more than one tRNA, a type of base pairing called “wobble” allows a single tRNA anticodon to recognize more than one codon. Thus G, in addition to pairing with its favored partner, C, can also pair with U; while inosine (I), present in several tRNA anticodons, can pair with C, U, or A. Thus, for example in E. coli, wobble reduces the number of necessary tRNAs from 61 to 40. Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Explain the role of tRNAs in translation. 2. What is meant by “third-codon wobble”? Answer: Third-codon wobble refers to relaxed base pairing at the third position in the anticodon of tRNAs that do not always conform to the typical Watson-Crick base pairing arrangement. Most amino acids are specified by more than codon, and the third position of the anticodon is frequently modified to inosine (a modified guanosine), which can base pair with U, A, and C. In addition, at the third position, G can also base pair with C. As a result, some tRNAs are capable of binding more than one codon. Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Explain the role of tRNAs in translation. 3. What is the key difference between prokaryotic and eukaryotic mRNAs? Answer: There are several structural and other subtle differences between prokaryotic and eukaryotic mRNAs, but the major difference is that prokaryotic mRNAs are polycistronic, meaning that the mRNA encodes more than one protein. To accomplish this, each protein has its own translation initiation site. Eukaryotic mRNAs are monocistronic and encode only one protein. They also have a single translation initiation © 2019 Oxford University Press

site. Textbook Reference: Translation of mRNA Bloom’s Category: 2. Understanding Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. 4. A complex of eukaryotic initiation factors (eIF4E, eIF4G, eIF4A, and eIF4B) brings the mRNA to the 40S ribosomal subunit and is thought to unwind base-paired structures in the mRNA to allow for the codon‒anticodon interactions necessary for translation of the message. What is the most likely reason that an equivalent activity is not present in prokaryotic cells? Answer: In prokaryotic cells the transcription and translation of an mRNA occur at the same time, with the ribosome following close behind the RNA polymerase. Thus, translation of a region of mRNA occurs before secondary structures have a chance to form. In contrast, in eukaryotic cells the message is synthesized in its entirety before being transported out of the nucleus and into the cytoplasm for translation—hence the need for enzymes to unwind the secondary structures so that the mRNA can be used as a template for protein synthesis. Textbook Reference: Translation of mRNA Bloom’s Category: 3. Applying Learning Objective: Contrast the initiation of translation in bacterial and eukaryotic cells. 5. Suppose that a researcher wishes to determine the effect of shutting off production of protein Z in the cell. How could production of protein Z be shut off without altering any of Z’s transcriptional control? Answer: An excellent example is ferritin, an iron-storage protein, that is regulated by iron at the translational level. When iron is scarce, a protein binds to a sequence at the 5ʹ end of the mRNA, called the iron response element (IRE), and inhibits translation. In the presence of iron, the IRE-binding protein is released from the mRNA and translation resumes. This regulatory mechanism can be exploited for controlling protein Z’s expression by inserting an IRE in the 5ʹ region (within 70 nucleotides of the cap) of the gene. Growing the cells in media containing low levels of iron should effectively inhibit X translation, creating observable effects. Textbook Reference: Translation of mRNA Bloom’s Category: 3. Applying Learning Objective: Summarize the mechanisms that regulate translation. 6. Suppose that a single protein is significantly overexpressed in one of two different cell lines. The proteins are identical, but Northern blot analysis indicates that the mRNA for the protein in the overexpressing cell line is about 200 base pairs shorter. Subsequent reverse transcriptase PCR has revealed that the truncation occurs in the 3ʹ untranslated region. How might this lead to overexpression of the protein? Answer: The 3ʹ UTR can serve several roles in regulating mRNA translation. It is likely that there is a translational repressor that recognizes a specific sequence and interacts with eIF-4E and prevents initiation in the full-length mRNA. In the cell line with the truncated mRNA, that overexpresses the protein, the repressor is lost; the loss of repression allows the protein to be expressed continuously, as long as the mRNA is © 2019 Oxford University Press

present. Textbook Reference: Translation of mRNA Bloom’s Category: 4. Analyzing Learning Objective: Summarize the mechanisms that regulate translation. 7. The growing polypeptide chain coming off the ribosomal complex is fairly unstable. It has a tendency to fold back on itself and can aggregate with adjacent polypeptides. If these processes are allowed to occur, degraded, improperly folded, or large aggregates of nonfunctional proteins would result. How does the cell prevent this? Answer: A class of proteins called chaperones prevents these problems. As polypeptides come off the ribosomes, chaperones quickly bind to and stabilize the growing chain. They prevent improper or premature folding until the entire chain is synthesized and the completed protein can fold properly. Textbook Reference: Protein Folding and Processing Bloom’s Category: 2. Understanding Learning Objective: Explain how chaperones facilitate protein folding. 8. What is allosteric regulation of a protein? Answer: When a protein is regulated allosterically, it is regulated by the binding of a molecule at a site other than the active site. The binding induces a conformational change in the protein that alters the active site, either increasing or decreasing the enzyme’s activity. Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 2. Understanding Learning Objective: Explain how the binding of a small molecule can change the catalytic activity of an enzyme. 9. Regulation of protein function is critical to cell physiology. Phosphorylation is frequently used to regulate protein function and is often referred to as an on/off switch for proteins. How are proteins phosphorylated, and what is a unique characteristic of phosphorylation that makes it such an ideal molecular switch? Answer: Proteins are phosphorylated by protein kinases that covalently transfer a phosphate group to a protein. The phosphate group is generally supplied by ATP, which provides the energy necessary for the reaction. Proteins are phosphorylated on either serine/threonine or tyrosine amino acids by serine/threonine kinases or tyrosine kinases, respectively. Phosphorylation of proteins may activate or inactivate the function of a particular protein. The unique characteristic of phosphorylation is that it is readily reversible by protein phosphatases, so that a protein can easily be switched on or off simply by phosphorylation/dephosphorylation with no significant impact on the rest of the protein. Textbook Reference: Regulation of Protein Function and Stability Bloom’s Category: 2. Understanding Learning Objective: Describe the roles of kinases and phosphatases in regulating protein activity.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 11: The Nucleus TEST FILE QUESTIONS Multiple Choice 1. The principal difference between eukaryotic and prokaryotic cells is that eukaryotic cells have a. larger ribosomes. b. cell walls, whereas prokaryotic cells do not. c. DNA replication and transcription in the same compartment, whereas prokaryotic cells do not. d. a nucleus, whereas prokaryotic cells do not. Answer: d Textbook Reference: Introduction Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 2. The nuclear envelope is continuous with the a. mitochondrial outer membrane. b. rough endoplasmic reticulum. c. Golgi apparatus. d. plasma membrane. Answer: b Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 3. The nuclear lamina is built of proteins called a. lamins. b. laminins. c. nucleons. d. nucleoplasmins. Answer: a Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering

Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 4. Emery–Dreifuss muscular dystrophy and Hutchinson-Gilford progeria are caused by mutations in genes that code for a. importin or Ran. b. Polycomb proteins. c. NF-κB. d. A-type nuclear lamins or emerin. Answer: d Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 5. The filaments of the nuclear lamina are composed of a type of a. intermediate filament. b. microtubule. c. microfilament. d. thick filament. Answer: a Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 6. The nuclear lamina binds to a. both the inner nuclear envelope membrane and the chromatin. b. the inner nuclear envelope membrane only. c. the chromatin only. d. both the inner and outer nuclear envelope membranes. Answer: a Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 7. Lamins are associated with the inner nuclear envelope membrane via a. GPI anchors on lamins. b. lipid tails on lamins only. c. lamin-binding proteins in the nuclear envelope membrane only. d. lipid tails on lamins and lamin-binding proteins in the nuclear envelope membrane. Answer: d

Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 8. Nuclear pores are organized with _______ symmetry. a. sixfold b. eightfold c. ninefold d. tenfold Answer: b Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 9. The major pathway for molecules to pass into and out of the nucleus is through a. gap junctions between the nuclear envelope membranes. b. porin channels in the nuclear envelope membranes. c. nuclear pore complexes. d. diffusion through the membrane bilayers of the nuclear envelope. Answer: c Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 10. Small molecules (< 20 kd) enter nuclei by way of a. passive diffusion through nuclear pore complexes. b. selective transport through nuclear pore complexes. c. active transport across the nuclear envelope membranes. d. passive diffusion across the nuclear envelope membranes. Answer: a Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 11. Which of the following is not involved in the transport of RNAs out of the nucleus? a. Ran b. Expenditure of chemical energy c. Small RNAs crossing by passive diffusion

d. Exportins Answer: c Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 12. The first nuclear localization signal to be mapped is the signal responsible for the transport of a. nucleoplasmin. b. histone H1. c. lamin B. d. SV40 T antigen. Answer: d Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 13. Nuclear proteins that are stripped of their nuclear localization signal and injected into the cytoplasm of cultured cells will localize a. to the nucleus. b. in the cytoplasm. c. in a ring around the outside of the nuclear envelope. d. extracellularly. Answer: b Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 3. Applying Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 14. The nuclear localization signal is typically a(n) _______, rich in the amino acids _______. a. α helix; Pro, Lys, and Arg b. α helix; Leu, Phe, and Gly c. short sequence; Lys and Arg d. short sequence; Pro, Leu, and Gly Answer: c Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding

Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 15. The nuclear localization signal is recognized by and binds to which protein in the process of nuclear protein import? a. Ran b. Importin c. Exportin d. The outer fibril protein Answer: b Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 16. The directionality of nuclear transport is determined by _______ in the nucleus and _______ in the cytosol. a. high Ran; low Ran b. low Ran; high Ran c. high Ran/GTP; high Ran/GDP d. high Ran/GDP; high Ran/GTP Answer: c Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 17. Export of RNAs from the nucleus occurs primarily by a. passive diffusion through nuclear pore complexes. b. cotranscriptional insertion through protein pores of the nuclear envelope membrane. c. selective transport through nuclear pore complexes. d. release from the nucleus when it breaks down at mitosis. Answer: c Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 18. Which statement about snRNAs is true? a. They are synthesized in the nucleus and function in the cytoplasm. b. They are synthesized in the cytoplasm and function in the nucleus. c. They are synthesized in the nucleus and function in the nucleus without passing into the cytoplasm.

d. They are synthesized in the nucleus, move to the cytoplasm to form functional complexes with proteins, and then return to the nucleus. Answer: d Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 19. By restricting the traffic of proteins and RNA across the nuclear envelope, eukaryotic cells can regulate _______ in unique ways. a. translation and transcription b. transcription only c. translation only d. endocytosis Answer: a Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Explain how transport across the nuclear envelope can regulate gene expression. 20. Which statement describing the chromosome conformation capture (3C) technique for chromosome structural analyses is true? a. Sequences of DNA that are physically distant from each other are chemically linked together. b. Most DNA-DNA interactions are between sequences on different chromosomes. c. Sequences separated by a centromere rarely interact with each other. d. All of the above Answer: c Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Explain chromosome territories and the methods used to study the organization of chromosomes within the nucleus. 21. Highly condensed, transcriptionally inactive chromatin is called a. euchromatin. b. heterochromatin. c. a chromatin domain. d. histone-containing chromatin. Answer: b Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization.

22. Heterochromatin is normally associated with a. ribosomes. b. the nuclear envelope. c. only a few specific human chromosomes. d. mutated chromosomes. Answer: b Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. 23. Chromatin that contains sequences that are transcribed is called a. euchromatin. b. a centromere. c. heterochromatin. d. a telomere. Answer: a Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. 24. Chromatin domains are loops of at least _______ base pairs of DNA. a. 1000 b. 10,000 c. 100,000 d. 1,000,000 Answer: c Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. 25. Newly replicated DNA is located in a few hundred spots in the nucleus because a. there are a few hundred origins of replication per nucleus. b. there are a few thousand origins of replication, but only a few hundred are active at any one time. c. there are a few thousand origins of replication active at any one time, but they are located in a few hundred discrete clusters. d. there are so many charges drawing the DNA molecules together. Answer: c Textbook Reference: The Organization of Chromosomes Bloom’s Category: 2. Understanding Learning Objective: Describe replication and transcription factories.

26. Sites of newly replicated DNA can be detected with antibodies if the DNA is labeled with a. tritiated thymidine. b. tritiated uridine. c. bromodeoxyuridine. d. bromouridine. Answer: c Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Describe replication and transcription factories. 27. The nucleolus is the site where _______ is(are) assembled. a. ribosomal subunits b. nuclear pores c. chromatin d. the nuclear matrix Answer: a Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 28. The region of chromosomes that codes for most of the rRNAs is found in the a. nucleolus. b. nucleolar organizing region. c. ribosomal assembly region. d. ribosomal organizer region. Answer: b Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 29. The 5.8S rRNA associates with a. Cajal bodies. b. the large subunit of the ribosome. c. the small subunit of the ribosome. d. the interface between the large and small subunits of a functional ribosome. Answer: b Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 30. The eukaryotic 45S pre-rRNA transcript codes for the a. 28S rRNA. b. 28S and 18S rRNAs. c. 28S, 18S, and 5.8S rRNAs. d. 28S, 18S, 5.8S, and 5S rRNAs.

Answer: c Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 31. Most snoRNAs a. splice cleaved mRNAs. b. splice cleaved rRNAs. c. cleave the 45S rRNA. d. guide regions of rRNAs to modifying enzymes. Answer: b Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 32. Eukaryotic ribosomes leave the nucleus as a. intact 80S ribosomes. b. intact 70S ribosomes. c. 40S and 60S ribosomal subunits. d. 18S and 28S rRNAs. Answer: c Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 33. Cleaving of pre-rRNA produces a. the 28S rRNA. b. the 18S rRNA. c. both 28S and 18S rRNAs. d. the 28S, 18S, and 5.8S rRNAs. Answer: d Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 34. The genes encoding 5.8S, 18S, and 28S ribosome RNA are clustered together in a. tandem arrays located on one chromosome. b. tandem arrays interspersed with tandem arrays of genes encoding 5S ribosomal RNA. c. tandem arrays located on three chromosomes. d. tandem arrays located on five chromosomes. Answer: d Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli.

35. Which of the following is an example of nucleolar involvement in processing of nonrRNAs? a. Base modification within tRNAs catalyzed by snoRNPs b. Storage of pre-mRNA splicing factors c. Gene silencing d. Proteasomal proteolysis Answer: a Textbook Reference: Nuclear Bodies Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and function of nucleoli. 36. Polycomb proteins consist of _______ complex(es) and are concentrated in Polycomb bodies within the nucleus. a. 1 b. 2 c. 3 d. either 2 or 3, depending on cell type Answer: b Textbook Reference: Nuclear Bodies Bloom’s Category: 2. Understanding Learning Objective: Compare Polycomb bodies and transcription factories. 37. Polycomb bodies act on _______ regions of chromatin, silencing gene expression. a. adjacent b. distant c. both adjacent and distant d. very short Answer: c Textbook Reference: Nuclear Bodies Bloom’s Category: 2. Understanding Learning Objective: Compare Polycomb bodies and transcription factories. 38. Polycomb proteins promote heterochromatin formation and gene silencing by a. acetylating lysine 27 of histone 3. b. deacetylating lysine 27 of histone 3. c. methylating lysine 27 of histone 3. d. methylating lysine 9 of histone 3. Answer: c Textbook Reference: Nuclear Bodies Bloom’s Category: 2. Understanding Learning Objective: Compare Polycomb bodies and transcription factories. 39. Cajal bodies are thought to represent sites of a. DNA replication. b. snRNA processing and assembly. c. large ribosomal subunit assembly.

d. small ribosomal subunit assembly. Answer: b Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Summarize the functions of Cajal bodies and nuclear speckles.

Fill in the Blank 1. In eukaryotic cells, the contents of the nucleus are separated from the cytoplasm by the _______. Answer: nuclear envelope (nuclear membranes) Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 2. Entry into the nucleus is limited by the _______ and the _______ complexes. Answer: nuclear envelope (nuclear membranes); nuclear pore Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 3. Underlying the nuclear envelope is the _______, a fibrous meshwork composed of proteins called _______. Answer: nuclear lamina; lamins Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 4. Mutations in the fibrous protein underlying the nuclear envelope can cause premature _______, or progeria. Answer: aging Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 5. Depending on size, molecules can enter the nucleus by the process of _______ or _______.

Answer: passive diffusion; selective transport Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 6. Receptor-dependent protein transport into the nucleus requires the recognition of a _______ on the cargo protein. Answer: nuclear localization signal (NLS) Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 7. Proteins called _______ act as receptors for cargo protein import into the nucleus, and proteins called _______ play the corresponding role in export from the nucleus. Answer: importins; exportins Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 8. A major regulatory protein for nuclear import and export of proteins is the small GTPase _______. Answer: Ran Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 9. Transcriptionally active chromatin, which is called _______, is decondensed, while transcriptionally inactive chromatin, which is called _______, is highly condensed. Answer: euchromatin; heterochromatin Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Explain chromosome territories and the methods used to study the organization of chromosomes within the nucleus. 10. Individual chromosomes occupy distinct _______ within the nucleus. Answer: territories (regions) Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering

Learning Objective: Explain chromosome territories and the methods used to study the organization of chromosomes within the nucleus. 11. In addition to replication factories, the nucleus contains a number of discrete structures, maintained by protein–protein and protein–RNA interactions, that are referred to as _______. Answer: nuclear bodies Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Explain the similarities and differences between nuclear bodies and cytoplasmic organelles. 12. The most prominent nuclear body, the _______, is the site of ribosome production and rRNA processing and synthesis. Answer: nucleolus Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 13. In humans, pre-ribosomal RNA consists of a large 45S RNA that contains nucleotide sequences corresponding to _______ different individual rRNAs found within ribosomes. Answer: three (3) Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 14. In ribosome assembly, _______ that were synthesized in the cytoplasm are assembled into complexes on pre-ribosomal RNA molecules. Answer: proteins (ribosomal proteins) Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 15. Nuclear speckles are nuclear bodies that contain _______ and snRNPs. Answer: splicing factors Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Summarize the functions of Cajal bodies and nuclear speckles.

True/False 1. The nuclear envelope is composed of one nuclear membrane and an underlying nuclear lamina. Answer: F

Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 2. The nuclear pore complexes provide the only known routes through which molecules can travel between the nucleus and the cytoplasm of interphase cells. Answer: T Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 3. The activity of some proteins is regulated by controlling their ability to be imported into the nucleus. Answer: T Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 4. All molecules found in the nucleus have been transported there via a process that requires ATP. Answer: F Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 5. Small nuclear RNAs are synthesized in the nucleus and function there without ever leaving the nucleus. Answer: F Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 6. Messenger RNAs are transported through the nuclear pores as ribonucleoprotein particles. Answer: T

Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 7. Individual chromosomes occupy discrete territories within the nucleus. Answer: T Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Explain chromosome territories and the methods used to study the organization of chromosomes within the nucleus. 8. Inactive, condensed chromatin is called heterochromatin. Answer: T Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. 9. Chromatin domains appear to represent discrete functional units that independently regulate gene expression. Answer: T Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. 10. The nucleolus is the site of rRNA synthesis and processing and of ribosomal subunit assembly. Answer: T Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 11. Eukaryotic ribosomal RNAs are synthesized as three transcripts, with sizes of 28S, 18S, and 5.8S. Answer: F Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 12. Ribosomal proteins are synthesized in the nucleolus and then assembled onto rRNAs. Answer: F Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering

Learning Objective: Describe the structure and function of nucleoli. 13. The eukaryotic large ribosomal subunit contains the 28S, 5.8S, and 5S rRNAs, plus many different proteins. Answer: T Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli.

Short Answer 1. The nuclear envelope consists of what four major components? Answer: The inner nuclear membrane, the outer nuclear membrane, the nuclear lamina, and nuclear pore complexes Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 2. How does the nuclear envelope differ from the membrane of the endoplasmic reticulum? Answer: The outer membrane of the nuclear envelope is enriched in proteins that bind the cytoskeleton and lacks the proteins that give the ER its tubular structure. The inner nuclear membrane contains about 60 unique transmembrane proteins, including those that interact with the nuclear lamina. The endoplasmic reticulum does not have nuclear pore complexes. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. 3. How does the structure of the outer side of the nuclear pore complex differ from that of the inner side? Answer: On the outer side, eight cytoplasmic filaments, or spokes, are attached to the cytoplasmic ring. On the inner side, eight filaments extend from the pore to a ring in the nucleus, forming a nuclear basket. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex.

4. By what mechanism are most nuclear proteins specifically targeted to the nucleus for import? Answer: They have a nuclear import signal (a short basic region, such as Pro-Arg-LysArg) in their sequence that is recognized by cytosolic transport proteins (e.g., importins); these proteins then bind to the surface of the nuclear pore complex and are transported through the nuclear pore. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 5. Nuclear localization signals are usually short sequences rich in which amino acids? Answer: Lysine and arginine Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 6. What is the function of karyopherins? Answer: They are nuclear transport receptor proteins. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 7. Give one example of how regulation of nuclear protein import is involved in gene regulation. Answer: Some transcription factors that act on nuclear genes are found in the cytoplasm associated with inhibitor proteins like IB. When IB is phosphorylated and releases NFB, NF-B can enter the nucleus because its nuclear localization signals are exposed; it can then bind to and activate certain genes. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 3. Applying Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 8. Within the nucleolus, what functions are associated with the fibrillar center, dense fibrillar component, and granular component? Answer: The fibrillar center is the site of rRNA transcription. The pre-rRNAs are processed in the dense fibrillar component, and ribosomal subunits are then assembled in the granular component.

Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 9. Because of the presence of a nuclear envelope, eukaryotes can perform several functions that prokaryotes cannot. Identify one of these functions. Answer: Answers will vary, including posttranslational processing (e.g., splicing) of transcripts and regulation of RNA synthesis by regulating transport of transcription factors into the nucleus. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Explain how transport across the nuclear envelope can regulate gene expression. 10. How can fluorescence in situ hybridization be used to show that chromosomes occupy distinct domains of space within the nucleus? Answer: Fluorescently labeled probes that base pair with repetitive sequences unique to each individual chromosome are hybridized to chromosomes in permeabilized cells. Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Explain chromosome territories and the methods used to study the organization of chromosomes within the nucleus. 11. How does euchromatin differ from heterochromatin with respect to transcription activity? Answer: Euchromatin is transcriptionally active, while heterochromatin is transcriptionally inactive. Textbook Reference: The Organization of Chromosomes Bloom’s Category: 2. Understanding Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. 12. Why is most heterochromatin bound to the periphery of the nucleus? Answer: Proteins associated with heterochromatin bind to the nuclear lamina. Textbook Reference: The Organization of Chromosomes Bloom’s Category: 2. Understanding Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. 13. How have FISH approaches been applied to studying the arrangement of distribution of gene-rich chromosomes relative the nuclear periphery versus the center of the nucleus?

Answer: FISH stands for fluorescence in situ hybridization and can using gene specific probes localization in situ gene-rich chromosomal regions. Hence relative distribution within the nucleus can be determined. Textbook Reference: The Organization of Chromosomes Bloom’s Category: 3. Applying Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. 14. How has chromosome conformation capture (3C) analysis contributed to the identification of interactions within chromosomes? Answer: 3C has revealed interacting regions of chromosomes in living cells. Most interactions identified occur between regions on the same chromosome. Textbook Reference: The Organization of Chromosomes Bloom’s Category: 3. Applying Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. 15. What are LADs and NADs? Answer: LADs are lamina associated domains in chromosomes and NADs are nucleolus associated domains. Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. 16. What is the basis for referring to replication and transcription sites in the nucleus as “factories”? Answer: Replication sites, and transcription sites are clustered in the nucleus and produce multiple DNA molecules and proteins, respectively, in these clusters in a manner reminiscent of the manufacturing that occurs in factories. Textbook Reference: The Organization of Chromosomes Bloom’s Category: 2. Understanding Learning Objective: Describe replication and transcription factories. 17. How many copies of the large ribosomal RNA gene does the human genome contain? Answer: About 200 Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 18. What takes place in the granular component of the nucleolus? Answer: Ribosome assembly Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli.

19. If you stained mammalian cells with fluorescent antibodies against the three different RNA polymerases, antibodies directed against which RNA polymerase would be expected to stain the nucleoli? Answer: Antibodies directed against RNA polymerase I Textbook Reference: Nuclear Bodies Bloom’s Category: 3. Applying Learning Objective: Describe the structure and function of nucleoli. 20. What is the size (in Svedberg units) of the pre-rRNA primary transcript? Answer: 45S Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 21. What modification of nucleotides occurs during processing of pre-rRNA? Answer: Addition of methyl groups and pseudouridylation Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 22. Where are snoRNAs located in the nucleus? Answer: In the nucleolus Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 23. Where is 5S rRNA synthesized? Answer: On a chromosome outside the nucleolar organizer region, namely chromosome 1 Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 24. What guides the enzymes that modify rRNAs to their proper positions on the prerRNA? Answer: snoRNAs Textbook Reference: Nuclear Bodies Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and function of nucleoli. 25. Where do the final stages of ribosome subunit formation occur? Answer: In the cytoplasm Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli.

26. In addition to the rRNAs derived from the pre-rRNA primary transcript, what other components make up the two ribosomal subunits? Answer: 5S rRNA and multiple ribosomal proteins Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. 27. What are nuclear speckles? Answer: Nuclear speckles are storage sites for pre-mRNA splicing factors within the nucleus. Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Summarize the functions of Cajal bodies and nuclear speckles. 28. What snRNP modifications take place in Cajal bodies? Answer: The snRNA component of the snRNPs can undergo ribose methylation and pseudouridylation. Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Summarize the functions of Cajal bodies and nuclear speckles.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. The outer nuclear membrane is contiguous with the a. endoplasmic reticulum. b. nuclear lamina. c. Golgi apparatus. d. plasma membrane. Answer: a Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. Feedback A: Correct! The endoplasmic reticulum is a large membrane-bounded structure that is contiguous with the outer membrane of the nucleus. Feedback B: Incorrect. The nuclear lamina is a fibrous network of proteins located within the nucleus. Feedback C: Incorrect. The Golgi apparatus is a membrane-bounded structure in the cytoplasm, but it is not contiguous with the outer nuclear membrane. Feedback D: Incorrect. The plasma membrane is separate from the nuclear membranes. 2. The fibrous proteins underlying the inner nuclear membrane are

a. band 3s. b. collagens. c. keratins. d. lamins. Answer: d Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. Feedback A: Incorrect. Band 3 is an erythrocyte plasma membrane protein. Feedback B: Incorrect. Collagens are extracellular matrix proteins. Feedback C: Incorrect. Keratins are intermediate filament proteins. Feedback D: Correct! Lamins are the intermediate filament family proteins from which the nuclear lamina underlying the nuclear envelope is made. 3. The mass of the nuclear pore complex is estimated to be a. smaller than that of a ribosome. b. about the same as that of a ribosome. c. about five times that of a ribosome. d. about 30 times that of a ribosome. Answer: d Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Illustrate the structure of the nuclear envelope and nuclear pore complex. Feedback A: Incorrect. The mass is much larger than that of a ribosome. Feedback B: Incorrect. The mass is much larger than that of a ribosome. Feedback C: Incorrect. The mass is much larger than that of a ribosome. Feedback D: Correct! The nuclear pore complex is very large, with a diameter of about 120 nm and a mass of about 125 million daltons. 4. Protein transport into the nucleus occurs by a. diffusion through the nuclear envelope membranes. b. passive diffusion through nuclear pore complexes. c. selective transport through nuclear pore complexes. d. selective transport through individual subsets of nuclear pore complexes that are specific for individual classes of proteins. Answer: c Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. Feedback A: Incorrect. Proteins enter the nucleus through nuclear pore complexes rather

than by diffusing through the nuclear envelope. Feedback B: Incorrect. Only small proteins can diffuse through the nuclear pore. Most proteins are imported via sequence-specific binding to importins. Feedback C: Correct! The entry of most proteins into the nucleus is through sequence mediated selective binding to importins and then entry through the pore. Feedback D: Incorrect. The nuclear pore is a common carrier for protein entry and is entry-responsive to all classes of nuclear localization signals. 5. The nuclear basket protein complex is a component of the nuclear pore complex found a. inside the nucleus. b. buried in the nuclear envelope. c. on the cytoplasmic face of the pore. d. lining the central channel of the pore. Answer: a Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. Feedback A: Correct! Feedback B: Incorrect. Feedback C: Incorrect. The cytoplasmic filaments are on this side. Feedback D: Incorrect. Proteins here are involved sequentially in the movement of cargo into and out of the nucleus. 6. Typically, nuclear localization signals are composed primarily of a. hydrophobic amino acids. b. acidic amino acids. c. basic amino acids. d. sulfur-containing amino acids. Answer: c Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. Feedback A: Incorrect. Nuclear localization signals are not rich in hydrophobic amino acids. Feedback B: Incorrect. Acidic amino acids are not typically found in nuclear localization signals. Feedback C: Correct! Nuclear localization signals are often composed of basic amino acids such as arginine and lysine. Feedback D: Incorrect. Amino acids, such as cysteine and methionine, are not typically found in nuclear localization signals. 7. The directionality of nuclear protein import is determined by

a. an unequal distribution of Ran/GTP. b. importin. c. an ion gradient. d. the nuclear lamina. Answer: a Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. Feedback A: Correct! Ran/GTP is more abundant in the nucleus than in the cytoplasm, and this factor is thought to give directionality to protein import. Feedback B: Incorrect. Although importin is the cytoplasmic receptor that recognizes proteins targeted for the nucleus and transports them there, the directionality of import is determined by something else. Feedback C: Incorrect. Although ion gradients give directionality to other transport processes within the cell, ions can pass freely through nuclear pores, so an ion gradient is not involved in nuclear transport. Feedback D: Incorrect. The nuclear lamina is a structural component of the nucleus and is not involved in the import of proteins. 8. Ran GAP association with cytoplasmic filaments of the nuclear pore results in the conversion of the nucleotide that is bound to nuclear Ran into a. ADP. b. GDP. c. ATP. d. GTP. Answer: b Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. Feedback A: Incorrect. ADP is not bound to Ran. Feedback B: Correct! GDP is bound to Ran in the cytoplasm. Ran GAP activates Ran to cleave GTP to GDP as export into the cytoplasm occurs. Feedback C: Incorrect. ATP is not bound to Ran. Feedback D: Incorrect. GTP is bound to Ran in the nucleoplasm. What is Ran bound to when it is exported to the cytoplasm? 9. What is the function of karyopherins? a. They attach to chromosomes in order to activate heterochromatin. b. They coat chromosomes. c. They define the structure of Cajal bodies. d. They transport macromolecules into or out of the nucleus. Answer: d

Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. Feedback A: Incorrect. Heterochromatin activation is a specific and highly regulated process. Feedback B: Incorrect. DNA is wrapped around histones. Feedback C: Incorrect. Cajal bodies are a structural feature of the nucleus. Feedback D: Correct! Karyopherins are nuclear transport receptors that belong to the importin or exportin class; they bind to cargo and promote its import or export. 10. An important step in the import of the transcription factor NF-B into the nucleus is regulated by a a. nuclease. b. protease. c. phospholipase. d. plasma membrane receptor that directly phosphorylates I B. Answer: b Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. Feedback A: Incorrect. Nucleases cleave either RNA or DNA. NF-B and the associated IB are proteins. Feedback B: Correct! NF-B is normally associated with I B in the cytoplasm. Upon appropriate signaling, I B is phosphorylated and cleaved by a protease. Feedback C: Incorrect. A phospholipase cleaves phospholipids. NF-B and the associated IB are proteins. Feedback D: Incorrect. NF-B is associated with the polypeptide IB in the cytoplasm. Dissociation of the two requires phosphorylation of I B. However, this is a cytosolic event. 11. mRNA molecules are exported from the nucleus to the cytoplasm via a. a consensus sequence located at the 3ʹ terminus. b. the 7-methylguanosine cap structure. c. importin. d. a recruited protein complex. Answer: d Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. Feedback A: Incorrect. There is no consensus export sequence in RNA.

Feedback B: Incorrect. The cap structure functions in mRNA translation and stability but does not play a role in export. Feedback C: Incorrect. Importin is responsible for transporting nuclear proteins into the nucleus. Feedback D: Correct! mRNAs are associated with a large number of proteins during their processing and after processing an mRNA export complex is recruited. 12. Which of the following processes does not take place in the nucleus? a. DNA replication b. Translation c. RNA processing d. Transcription Answer: b Textbook Reference: Introduction Bloom’s Category: 1. Remembering Learning Objective: Explain how transport across the nuclear envelope can regulate gene expression. Feedback A: Incorrect. DNA replication takes place within structures called replication factories in the nucleus. Feedback B: Correct! Translation is carried out by ribosomes in the cytoplasm. Feedback C: Incorrect. RNA processing events like splicing and editing take place in the nucleus. Feedback D: Incorrect. Transcription uses chromosomes as templates for synthesis of RNA and takes place in the nucleus. 13. Initial steps in the assembly of functional snRNPs occur in the a. cytoplasm. b. nucleolus. c. nucleoplasm. d. PML body. Answer: a Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 1. Remembering Learning Objective: Explain how transport across the nuclear envelope can regulate gene expression. Feedback A: Correct! The RNA associates with proteins to form functional snRNPs in the cytoplasm. Feedback B: Incorrect. The nucleolus is involved in the assembly of proteins with rRNAs to form ribosome subunits. Feedback C: Incorrect. Many transcriptional events occur on chromosomes distributed about the nucleus. Feedback D: Incorrect. The PML is involved in intranuclear events such as transcriptional regulation and DNA repair. 14. Chromosomes are distributed in the nucleus

a. as cleavage fragments. b. as condensed mitotic chromosomes. c. randomly. d. in distinct territories. Answer: d Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Explain chromosome territories and the methods used to study the organization of chromosomes within the nucleus. Feedback A: Incorrect. Chromosomes must be intact to function. Feedback B: Incorrect. In many organisms, the nuclear envelope breaks down during mitosis, and condensed mitotic chromosomes are not enclosed in a nucleus. Feedback C: Incorrect. Chromosomes are not randomly distributed. Feedback D: Correct! 15. Which statement about heterochromatin is false? a. It is highly condensed chromatin. b. There are two forms—constitutive and facultative heterochromatin. c. It is transcriptionally active. d. It is largely localized to the nuclear periphery. Answer: c Textbook Reference: The Organization of Chromosomes Bloom’s Category: 2. Understanding Learning Objective: Summarize the relationship between transcriptional activity and chromatin localization. Feedback A: Incorrect. This is a true statement. As opposed to euchromatin, which is less condensed, heterochromatin is highly condensed. Feedback B: Incorrect. This is a true statement. Constitutive heterochromatin is never transcribed, whereas facultative heterochromatin is transcribed in some cells and not in others. Feedback C: Correct! This is a false statement. Heterochromatin, due to its highly condensed nature, is inaccessible to the transcriptional machinery and is thus transcriptionally inactive. Feedback D: Incorrect. This is a true statement. Heterochromatin is frequently associated with the nuclear envelope. 16. What is the average number of replication forks per DNA replication cluster in a mammalian cell nucleus? a. Approximately 20 b. Approximately 200 c. Approximately 4,000 d. Approximately 8,000 Answer: a Textbook Reference: The Organization of Chromosomes Bloom’s Category: 1. Remembering Learning Objective: Describe replication and transcription factories.

Feedback A: Correct! There are between 5 and 50 replication forks per factory. Feedback B: Incorrect. This is the average number of discrete clusters in a nucleus. Feedback C: Incorrect. This is the average number of origins of replication in a nucleus. Feedback D: Incorrect. This is the average number of replication forks in a nucleus, two per origin of replication. 17. Which of the following are not compartments in the nucleus? a. Clustered DNA replication sites b. Lipid droplets c. PML bodies d. Speckles enriched in RNA splicing components Answer: b Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Explain the similarities and differences between nuclear bodies and cytoplasmic organelles. Feedback A: Incorrect. These are normal compartments of the nucleus. Feedback B: Correct! Lipid droplets are found in the cytoplasm, not in the nucleus. Feedback C: Incorrect. These are normal compartments of the nucleus. Feedback D: Incorrect. These are normal compartments of the nucleus. 18. The most prominent nuclear body is the a. mitochondrion. b. endoplasmic reticulum. c. histone locus body. d. nucleolus. Answer: d Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. Feedback A: Incorrect. The mitochondrion, a membrane-bounded organelle involved in ATP production in the cytoplasm, is not a nuclear body. Feedback B: Incorrect. The endoplasmic reticulum, a membrane organelle of the cytoplasm that has ribosomes associated with much of its length, is not a nuclear body. Feedback C: Incorrect. The histone locus body is a specialized compartment within the nucleus, but it is not a prominent nuclear feature. Feedback D: Correct! The nucleolus is the most prominent compartment in the nucleus and can be readily observed by light microscopy. 19. The nucleolus is the site where a. ribosomal RNA is transcribed and ribosomes are partially assembled. b. DNA replication occurs. c. proteins recently imported from the cytoplasm are deposited. d. translation occurs. Answer: a Textbook Reference: Nuclear Bodies

Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. Feedback A: Correct! The nucleolus is where the many copies of the three largest RNA subunits are transcribed. They are then processed and complexed with ribosomal proteins and the smallest ribosomal RNA to form the ribosomal subunits. Feedback B: Incorrect. DNA replication occurs at a fairly small number of discrete sites in the nucleus, but not in the nucleolus. Feedback C: Incorrect. Imported nuclear proteins are released from importin upon exchange of Ran/GDP for Ran/GTP. This does not occur in a discrete location within the nucleus. Feedback D: Incorrect. Although ribosomes are assembled in the nucleolus, they are then transported to the cytoplasm, where they carry out translation. 20. Which of the following is a possible intermediate during pre-rRNA processing? a. An RNA molecule containing 18S + 5.8S rRNAs b. An RNA molecule containing 5.8S + 28S rRNAs c. An RNA molecule containing 5S + 28S rRNAs d. An RNA molecule containing 18S + 28S rRNAs Answer: b Textbook Reference: Nuclear Bodies Bloom’s Category: 3. Applying Learning Objective: Describe the structure and function of nucleoli. Feedback A: Incorrect. One of the earliest cleavage steps occurs on the 5′ side of the 5.8S rRNA, separating the 18S and 5.8S rRNAs. Feedback B: Correct! After cleavage of some noncoding 5′ sequence from the pre-rRNA, the next cleavage step occurs between the 18S and 5.8S rRNAs, yielding a molecule containing the 5.8S and 28S rRNAs as an intermediate. Feedback C: Incorrect. The 5S rRNA is transcribed separately and is not part of the prerRNA that encodes the other three rRNA molecules. Feedback D: Incorrect. One of the first cleavage steps during pre-rRNA processing is cleavage between the 18S and 5.8S rRNA, which lies between the 18S and 28S rRNA coding regions. Therefore, a molecule containing the 18S and 28S rRNAs would not be an intermediate of pre-rRNA processing. 21. Ribosomes contain one copy each of 5.8S, 18S, and 28S rRNA. The major mechanism ensuring that each is produced in equal molar amounts is the a. selective degradation of excess transcripts of the respective rRNAs. b. grouping of the DNA sequences encoding each rRNA into a single rRNA gene. c. outcome of coevolution of the separate genes. d. existence of common sequence features for transcription initiation for the genes encoding each rRNA. Answer: b Textbook Reference: Nuclear Bodies Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and function of nucleoli. Feedback A: Incorrect.

Feedback B: Correct! Because the sequences encoding the three rRNAs are grouped together into a single gene, all three are then transcribed as a single pre-rRNA that is then processed to produce each in equal amounts. Feedback C: Incorrect. Separate genes, even if they have resulted from coevolution, do not necessarily give rise to equal molar amounts of three different gene products. Feedback D: Incorrect. This could well be a contributing mechanism, but it is not the actual mechanism. 22. Which of the following factors contributes to the relative ease of determining how pre-rRNA is processed? a. The abundance of rRNA genes b. The abundance of ribosomes (5–10 million) that need to be synthesized per cell cycle c. The grouping of rRNA genes into discrete nuclear compartments in the nucleoli d. All of the above Answer: d Textbook Reference: Nuclear Bodies Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and function of nucleoli. Feedback A: Incorrect. This is a contributing factor, since the genes’ abundance facilitates the study, but it is not the only correct answer. Feedback B: Incorrect. This is a contributing factor, since the abundance of ribosomes requires a great deal of rRNA synthesis, but it is not the only correct answer. Feedback C: Incorrect. This is a contributing factor, since it is easier to study the genes and their gene products when they are grouped together, but it is not the only correct answer. Feedback D: Correct. All of these factors contribute to the ease of study. 23. Most snoRNAs function as a. catalytic RNAs. b. self-replicating RNAs. c. guide RNAs. d. self-splicing RNAs. Answer: c Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and function of nucleoli. Feedback A: Incorrect. Feedback B: Incorrect. Feedback C: Correct! Most snRNAs function as guide RNAs to direct the specific base modifications of pre-rRNA. Feedback D: Incorrect. 24. snRNPs that are responsible for pre-mRNA splicing are assembled and stored in a. Cajal bodies. b. histone locus bodies. c. speckles.

d. both Cajal bodies and speckles. Answer: d Textbook Reference: Nuclear Bodies Bloom’s Category: 1. Remembering Learning Objective: Summarize the functions of Cajal bodies and nuclear speckles. Feedback A: Incorrect. This answer choice is correct, but there is a better correct answer choice. Feedback B: Incorrect. Feedback C: Incorrect. This answer choice is correct, but there is a better correct answer choice. Feedback D: Correct! snRNPs are assembled and stored in both Cajal bodies and nuclear speckles.

Essay 1. Protein transport into the endoplasmic reticulum (ER) or mitochondria requires unfolding of the protein and threading the amino acid chain through a channel or pore into the organelle. In contrast, protein transport into the nucleus does not require unfolding of the protein. Why is unfolding unnecessary for nuclear protein import? Answer: The nuclear pore complex is a huge structure composed of about 30 different pore proteins, and its central channel has a diameter of approximately 10–40 nm, which is large enough for even very large protein complexes. For this reason, it is unnecessary to unfold the protein, which can pass through the nuclear pore in its native state. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 2. The nuclear localization signal (NLS) is not removed (cleaved off) after entry of the protein into the nucleus. In contrast, the targeting sequences of proteins destined for organelles such as the mitochondria are located at the N-terminus of the protein and are cleaved off once the proteins reach the lumen of the organelle. What are two likely reasons that the nuclear localization signal (NLS) is not cleaved off for nuclear proteins? Answer: During mitosis, the nuclear envelope disintegrates, and nuclear proteins are released into the cytoplasm. Once the envelope re-forms around the daughter nuclei, the nuclear proteins need to be re-transported into the nuclei, explaining a need for retention of NLS. In some organisms, such as yeast, the nuclear envelope remains intact during mitosis. Thus, there are likely other reasons for retention of NLS that are of equal or greater importance. A second possible reason is that nuclear localization signals are internal to the polypeptide. Cleaved targeting sequences are typically located at the Nterminus of the protein. Cleavage of an internal sequence has much more consequence for the overall structure of the protein than removal of an N-terminal sequence. In addition, many nuclear proteins shuttle in and out of the nucleus, which would be a third reason for retaining the NLS.

Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 3. What is the role of nucleotide sequence in mediating the export of ribosomal RNA from the nucleus? Answer: The recognized export sequences reside on the proteins associated with the ribosomal RNA rather than on the ribosomal RNA itself. For example, a specific exportin, Crm1, recognizes protein sequence features (i.e., amino acids) of proteins bound to the ribosomal RNA rather than the nucleotide sequence of the ribosomal RNA itself. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and RNAs are transported into and out of the nucleus. 4. Unlike eukaryotic cells, prokaryotic cells do not have a nucleus. What are some possible benefits of the evolution of a nucleus in eukaryotes? Answer: In contrast to prokaryotes, eukaryotic RNA is extensively modified by processes such as splicing and RNA editing, which are critical to the accurate transfer of genetic information from gene to protein. Sequestering the chromosomes in a separate compartment allows RNA-processing events to be more efficient, because the enzymes involved are sequestered in the nucleus and hence have less interference from unrelated enzymes. In addition, because RNA is retained in the nucleus until the processing events are complete, inappropriate translation of unprocessed RNAs does not occur. Furthermore, an additional level of control comes from the entry of regulated transcription factor into the nucleus. This would not be possible if the cytosol were not separated from the nucleus and chromosomes. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Explain how transport across the nuclear envelope can regulate gene expression. 5. In many organisms, the nuclear envelope breaks down during mitosis. As part of this breakdown, the nuclear lamina that underlies the nuclear envelope breaks down. In fact, many investigators think that the breakdown of the lamina initiates envelope breakdown. Propose a reversible mechanism for controlling the polymerization state of the proteins that make up the nuclear lamina. Answer: Reversible phosphorylation/dephosphorylation would provide a mechanism for regulating the polymerization of nuclear lamins. When phosphorylated, negative charge repulsion would lead to dissociation, and after dephosphorylation repolymerization would occur. Specific protein kinases would be involved in the phosphorylation reaction and

specific phosphatases in the dephosphorylation reaction. These events could be regulated by various signaling pathways. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 3. Applying Learning Objective: Explain how transport across the nuclear envelope can regulate gene expression. 6. What are the expected phenotypic consequences of a mutation in Ran GAP that reduces its affinity for cytoplasmic filaments of the nuclear pore? Answer: Ran is important to both protein import into and export out of the nucleus. As part of importin recycling to the cytoplasm, Ran in the GTP-bound form complexes with importin inside the nucleus. Formation of this complex results in the release of importin from cargo and the importin/Ran-GTP complex is then recycled back to the cytoplasm. During nuclear protein export in general, GTP-bound Ran binds to exportin inside the nucleus to give rise to a Ran-GFP/exportin/cargo complex that is exported as a unit through the nuclear pore into the cytoplasm. In either case, Ran functions as a GTPdependent molecular switch. Ran GAP bound to the cytoplasmic filaments of the nuclear pore complex activates the GTPase activity of Ran in the cytoplasm and bound GTP is cleaved to GDP. If Ran GAP were not bound to the cytoplasmic filaments, its effective local concentration would be diluted, and the efficiency of Ran activation would be low. Low affinity would cause a distinct decrease in the Ran GTP gradient across the nuclear pore. This would reduce the efficiency of both import into the nucleus, which is dependent on importin recycling, and export from the nucleus. Phenotypically, cell growth rates would slow and the defect might be so severe as to be lethal. Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 4. Analyzing Learning Objective: Explain how transport across the nuclear envelope can regulate gene expression. 7. The activity of transcription factors is frequently regulated by controlling their localization within eukaryotic cells, such that only in response to a given stimulus is the transcription factor transported into the nucleus, where it can activate its target genes. For a transcription factor that has a nuclear localization signal (NLS), describe two mechanisms by which regulated transport can occur. Answer: Two possible mechanisms involve masking of the NLS such that it is not recognized by importin. This masking can be: (1) intermolecular, whereby another protein binds and obscures the NLS, or (2) intramolecular, whereby the protein itself folds in such a way that the signal is masked. Upon stimulation, the masking protein dissociates or the protein structure is altered such that the NLS is exposed, and transport to the nucleus can occur. Other mechanisms are also possible. For example, the transcription factor may be tethered to another protein that is attached to a cytoplasmic structure, such as a membrane protein or a cytoskeletal protein. In this case, the nuclear localization signal is exposed, but the protein is physically retained in the cytoplasm until the appropriate stimulus liberates it, and it is free to make its way into the nucleus.

Textbook Reference: The Nuclear Envelope and Traffic between the Nucleus and the Cytoplasm Bloom’s Category: 2. Understanding Learning Objective: Explain how transport across the nuclear envelope can regulate gene expression. 8. To determine a chromosome’s distribution within the mammalian nucleus, probes were hybridized to repeated sequences rather than to a single gene. Why? Answer: When tandemly arrayed, the repeated sequence gives a locally amplified signal. Hybridizing a nucleotide probe to a single gene on a chromosome would give the equivalent of a point signal for the chromosome. Chromosomes have many genes. Moreover, repeated sequences are spread across a given chromosome. Therefore, hybridizing a probe to a repeated sequence lights up the whole chromosome. Textbook Reference: The Organization of Chromosomes Bloom’s Category: 2. Understanding Learning Objective: Explain chromosome territories and the methods used to study the organization of chromosomes within the nucleus. 9. What attributes of the locations where replication and transcription occur within the nucleus leads to these sites being designated as “factories”? Answer: Lots of DNA replication and transcription into RNA occur within clustered regions of the nucleus where large complexes of proteins involved in each particular process are concentrated. Because of the copious production in these clustered regions and the volume of protein complexes produced, these regions resemble factories. Textbook Reference: The Organization of Chromosomes Bloom’s Category: 2. Understanding Learning Objective: Describe replication and transcription factories. 10. How does the clustering and transcriptional pattern for eukaryotic ribosomal RNA genes ensure that there will be equal numbers of copies of the major RNAs available for ribosome assembly? Answer: Ribosomal genes are organized in tandem repeats that contain one copy each of the 18S, 5.8S, and 28S rRNA sequence as a single gene unit. The transcript contains all three sequences, and hence, each is produced after processing of the pre-rRNA in the same amount. All this is located in the nucleolus. Transcription of 5S rRNA occurs outside the nucleolus and may not be as tightly linked in terms of amount. Textbook Reference: Nuclear Bodies Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and function of nucleoli.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 12: Protein Sorting and Transport The ER, Golgi, and Lysosomes

TEST FILE QUESTIONS Multiple Choice 1. Using radiolabeled amino acids, Palade and colleagues defined the pathway taken by secreted proteins. Which sequence represents that pathway? a. Rough ER  smooth ER  Golgi  secretory vesicles  cell exterior b. Rough ER  smooth ER  Golgi  endosomes  cell exterior c. Rough ER  Golgi  lysosomes  cell exterior d. Rough ER  Golgi  secretory vesicles  cell exterior Answer: d Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Diagram the secretory pathway. 2. Most plasma membrane proteins in higher eukaryotic cells are synthesized on a. free ribosomes and inserted after translation into the plasma membrane. b. rough ER ribosomes and carried to the plasma membrane by vesicles that pinch off from the Golgi apparatus. c. rough ER ribosomes and carried to the plasma membrane by vesicles that pinch off from the rough ER. d. ribosomes associated with the plasma membrane and inserted into the membrane cotranslationally. Answer: b Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Diagram the secretory pathway. 3. A signal sequence in a polypeptide chain targets all but _______ proteins to the rough ER surface. a. secreted b. plasma membrane c. mitochondrial d. lysosomal Answer: c Textbook Reference: The Endoplasmic Reticulum

Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms that target proteins to the ER. 4. Free ribosomes in the cytosol and ribosomes bound to the ER membrane have a. different types of large and small subunits. b. the same types of large and small subunits. c. different types of large subunits but the same small subunits. d. different types of small subunits but the same large subunits. Answer: b Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms that target proteins to the ER. 5. When compared to polypeptides synthesized in vitro on free ribosomes, polypeptides synthesized from the same mRNA on microsome-bound ribosomes often are a. the same size. b. larger. c. smaller. d. more hydrophobic. Answer: c Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that target proteins to the ER. 6. As they emerge from the ribosome, signal sequences are recognized and bound by a. a tRNA. b. a signal peptidase. c. a signal recognition particle (SRP). d. an SRP receptor. Answer: c Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms that target proteins to the ER. 7. Which statement describes evidence that supports the signal hypothesis as an explanation for how secretory proteins are targeted to the rough ER? a. The protein is larger when synthesized in vitro on free polysomes. b. The secretory protein ends up in the cytosol when a short sequence is deleted by genetic engineering. c. A normally cytosolic protein is secreted when a specific sequence is added to it by genetic engineering. d. All of the above Answer: d Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that target proteins to the ER.

8. If secretory proteins are isolated from secretory vesicles and injected into the cytosol, they will a. be taken up into the rough ER and follow the secretory pathway. b. remain in the cytosol until they are degraded. c. be secreted through channels in the plasma membrane. d. be taken up into secretory vesicles and secreted. Answer: b Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that target proteins to the ER. 9. The signal sequence that targets a nascent polypeptide and its mRNA to the rough ER is rich in _______ amino acids. a. positively charged b. negatively charged c. hydrophilic d. hydrophobic Answer: d Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that target proteins to the ER. 10. Protein folding in the ER is assisted by a chaperone called a. BiP. b. PiP. c. Hsp60. d. Hsp90. Answer: a Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Describe protein folding and quality control in the ER. 11. Disulfide bonds within or between proteins form easily a. in the ER but not the cytosol. b. in neither the cytosol nor the ER. c. in the cytosol but not the ER. d. in both the cytosol and the ER. Answer: a Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Describe protein folding and quality control in the ER. 12. GPI-anchored proteins are synthesized a. on free ribosomes and attached to the GPI group on the outside of the ER. b. as transmembrane proteins, then are cleaved and joined to the GPI group on the cytosolic surface of the ER.

c. as transmembrane proteins, then are cleaved and joined to the GPI group on the lumenal surface of the ER. d. on free ribosomes and attached to the GPI group on the cytosolic surface of the ER. Answer: c Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Describe protein folding and quality control in the ER. 13. Proteins attached to the outer half of the plasma membrane bilayer are usually attached by a a. farnesyl tail. b. prenyl tail. c. geranyl tail. d. glycosylphosphatidylinositol (GPI) anchor. Answer: d Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Describe protein folding and quality control in the ER. 14. The unfolded protein response involves a. translation of most proteins. b. increased synthesis of chaperones. c. decreased proteasome activity. d. All of the above Answer: b Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Describe protein folding and quality control in the ER. 15. Most cellular lipids are synthesized in a. fat droplets. b. mitochondria. c. the endoplasmic reticulum. d. the Golgi apparatus. Answer: c Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Distinguish the roles of smooth and rough ER. 16. Newly synthesized membrane lipids are found in both halves of a membrane bilayer because they are a. synthesized on one surface and flipped to the other surface by flippases. b. removed and transported to these locations by lipid transport proteins. c. synthesized on one surface and flipped spontaneously to the other surface. d. synthesized on both surfaces. Answer: a Textbook Reference: The Endoplasmic Reticulum

Bloom’s Category: 1. Remembering Learning Objective: Distinguish the roles of smooth and rough ER. 17. The sequence Lys-Asp-Glu-Leu (KDEL) serves as an ER retention signal for proteins by binding to KDEL receptors that a. hold the proteins in the rough ER. b. hold the proteins in the smooth ER. c. hold the proteins in both the smooth ER and rough ER and prevent their transport to the Golgi apparatus. d. transport the proteins from the Golgi apparatus back to the ER. Answer: d Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Explain transport to and retrieval from the Golgi apparatus. 18. Vesicles enter the Golgi apparatus by fusing with _______, and they exit from the _______. a. the cis (convex) face; trans (concave) face b. the trans (concave) face; cis (convex) face c. both the cis (convex) and the trans (concave) faces; sides of cisternae d. the sides of the cisternae; cis (convex) face Answer: a Textbook Reference: The Golgi Apparatus Bloom’s Category: 1. Remembering Learning Objective: Relate the structure of the Golgi apparatus to its function. 19. N-linked oligosaccharides are added in the _______ and modified in the _______. a. ER; Golgi b. cis Golgi; medial Golgi c. medial Golgi; trans Golgi d. cis Golgi; trans Golgi Answer: a Textbook Reference: The Golgi Apparatus Bloom’s Category: 1. Remembering Learning Objective: Describe the types of protein glycosylation that take place in the Golgi. 20. Which of the following polysaccharides is synthesized in the Golgi apparatus? a. Pectin b. Cellulose c. Amylose d. Glycogen Answer: a Textbook Reference: The Golgi Apparatus Bloom’s Category: 1. Remembering Learning Objective: Describe the types of protein glycosylation that take place in the Golgi. 21. Which of the following proteins do not pass through the Golgi apparatus?

a. Lysosomal enzymes b. Cell surface proteins c. Ribosomal proteins d. Proteins secreted by exocytosis Answer: c Textbook Reference: The Golgi Apparatus Bloom’s Category: 1. Remembering Learning Objective: Diagram the routes of protein export from the Golgi. 22. Which of the following vesicle types does not bud directly from the trans-Golgi network? a. Vesicles transporting plasma membrane receptors b. Vesicles that form regulated secretory vesicles c. Vesicles carrying lysosomal enzymes d. Lysosomes Answer: d Textbook Reference: The Golgi Apparatus Bloom’s Category: 1. Remembering Learning Objective: Diagram the routes of protein export from the Golgi. 23. Lysosomal proteins are marked by the addition of a phosphate to a _______ group. a. glucose b. serine c. GDP d. mannose Answer: d Textbook Reference: The Golgi Apparatus Bloom’s Category: 1. Remembering Learning Objective: Diagram the routes of protein export from the Golgi. 24. Plasma membrane proteins of intestinal epithelial cells require separate targeting to a. one continuous plasma membrane domain. b. two plasma membrane domains: the apical and basolateral. c. three plasma membrane domains: the apical, lateral, and basal. d. four plasma membrane domains: one apical, two lateral, and a basal. Answer: b Textbook Reference: The Golgi Apparatus Bloom’s Category: 2. Understanding Learning Objective: Diagram the routes of protein export from the Golgi. 25. In plant cells, vesicles transport proteins from the Golgi apparatus to a. lysosomes. b. vacuoles. c. chloroplasts. d. mitochondria. Answer: b Textbook Reference: The Golgi Apparatus

Bloom’s Category: 1. Remembering Learning Objective: Diagram the routes of protein export from the Golgi. 26. In polarized epithelial cells, membrane proteins containing a dileucine (LL) motif in their cytoplasmic domain are a. retained in the Golgi apparatus. b. targeted to lysosomes. c. targeted to the basolateral plasma membrane. d. targeted to the apical plasma membrane. Answer: c Textbook Reference: The Golgi Apparatus Bloom’s Category: 1. Remembering Learning Objective: Diagram the routes of protein export from the Golgi. 27. Clathrin coats are bound to specific receptors by a protein called a. an adaptor protein. b. ARF. c. COPI. d. NSF. Answer: a Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 1. Remembering Learning Objective: Summarize the process of vesicle budding and cargo selection. 28. Arf function on vesicles is regulated by a. phosphorylation of serine residues. b. phosphorylation of mannose residues. c. binding of GTP. d. binding of ATP. Answer: c Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 1. Remembering Learning Objective: Summarize the process of vesicle budding and cargo selection. 29. The pinching off of budding clathrin-coated vesicles is driven by the action of a. dynamin and GTP hydrolysis. b. dynamin and ATP hydrolysis. c. GDP bound ARF. d. GTP bound ARF. Answer: a Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 2. Understanding Learning Objective: Summarize the process of vesicle budding and cargo selection. 30. Which coat protein(s) would you expect to be recruited for vesicle formation by a KDEL receptor?

a. COPI. b. COPII. c. clathrin. d. COPI and clathrin. Answer: a Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 2. Understanding Learning Objective: Explain the role of coat proteins. 31. Which coat protein directs retrograde vesicular transport from the ERGIC or Golgi to the endoplasmic reticulum? a. COPII b. COPI c. clathrin d. Both COPI and COPII Answer: b Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 1. Remembering Learning Objective: Explain the role of coat proteins. 32. Vesicles that carry proteins from the rough ER to the Golgi apparatus bud off as _______ vesicles. a. uncoated b. clathrin-coated c. COPI-coated d. COPII-coated Answer: d Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 1. Remembering Learning Objective: Explain the role of coat proteins. 33. Lysosomal proteins are initially incorporated into _______ vesicles. a. uncoated b. clathrin-coated c. COPI-coated d. COPII-coated Answer: b Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 1. Remembering Learning Objective: Explain the role of coat proteins. 34. How many Rab proteins have been shown to play a specific role in vesicular transport? a. fewer than 10 b. about 25 c. about 50 d. 60 or more

Answer: d Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism by which vesicles fuse with the correct target membranes. 35. Lysosomes digest a. proteins only. b. nucleic acids and proteins only. c. carbohydrates only. d. proteins, nucleic acids, and carbohydrates. Answer: d Textbook Reference: Lysosomes Bloom’s Category: 1. Remembering Learning Objective: Describe the function of lysosomes. 36. The pH inside lysosomes is about a. 7.0. b. 6.0. c. 5.0. d. 4.0. Answer: c Textbook Reference: Lysosomes Bloom’s Category: 1. Remembering Learning Objective: Describe the function of lysosomes. 37. Gaucher disease results from the failure of macrophage lysosomes to hydrolyze a. proteins. b. glycolipids. c. DNA. d. polysaccharides. Answer: b Textbook Reference: Lysosomes Bloom’s Category: 1. Remembering Learning Objective: Describe the function of lysosomes. 38. Transport vesicles carrying acid hydrolases fuse with a. lysosomes. b. endocytic vesicles. c. early endosomes. d. late endosomes. Answer: d Textbook Reference: Lysosomes Bloom’s Category: 1. Remembering Learning Objective: Describe the function of lysosomes.

Fill in the Blank 1. In eukaryotic cells, most protein synthesis is initiated in the _______. Answer: cytosol Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Diagram the secretory pathway. 2. The cytosolic surface of the rough endoplasmic membrane can be readily distinguished from the luminal surface by the presence of _______. Answer: ribosomes Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Diagram the secretory pathway. 3. Proteins are targeted to the endoplasmic reticulum by a _______ that binds to a(n) _______. Answer: signal sequence; SRP (signal recognition particle) Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms that target proteins to the ER. 4. Cotranslational transport of proteins across the endoplasmic reticulum membrane occurs through a protein channel called the _______. Answer: translocon Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Explain how proteins are inserted into the ER membrane. 5. Because of the oxidizing environment of the ER lumen, the formation of _______ bonds is a frequent event. Answer: disulfide Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Describe protein folding and quality control in the ER. 6. Proteins that fail to fold properly in the endoplasmic reticulum are removed through the process of _______. Answer: ERAD (ER-associated degradation) Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Describe protein folding and quality control in the ER. 7. The transcription factors that direct the unfolded protein response are _______ and _______. Answer: XBP1; ATF6 Textbook Reference: The Endoplasmic Reticulum

Bloom’s Category: 1. Remembering Learning Objective: Describe protein folding and quality control in the ER. 8. Membrane lipid biosynthesis occurs in the _______ endoplasmic reticulum. Answer: smooth Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Distinguish the roles of smooth and rough ER. 9. Cargo protein transport between different organelles in the secretory pathway occurs by _______-mediated transport. Answer: vesicle Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Explain transport to and retrieval from the Golgi apparatus. 10. In cisternal maturation, the carriers for cargo protein transport through the Golgi apparatus are the Golgi _______. Answer: cisternae Textbook Reference: The Golgi Apparatus Bloom’s Category: 2. Understanding Learning Objective: Relate the structure of the Golgi apparatus to its function. 11. During cargo protein transport through the Golgi apparatus, N-linked oligosaccharide side chains are processed from high-_______ to _______ acid-rich. Answer: mannose; sialic Textbook Reference: The Golgi Apparatus Bloom’s Category: 2. Understanding Learning Objective: Describe the types of protein glycosylation that take place in the Golgi. 12. Adaptor proteins are involved in the selection of _______ to be incorporated into coated vesicles. Answer: cargo proteins Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 2. Understanding Learning Objective: Summarize the process of vesicle budding and cargo selection. 13. The three major classes of vesicle coat proteins are _______, _______, and _______. Answer: clathrin; COPI; COPII Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 1. Remembering Learning Objective: Explain the role of coat proteins. 14. Vesicle fusion is mediated by the formation of protein complexes between vesicle and target _______. Answer: SNAREs

Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism by which vesicles fuse with the correct target membranes. 15. Proteins destined for degradation are delivered to lysosomes by two processes: _______ and _______. Answer: autophagy; endocytosis Textbook Reference: Lysosomes Bloom’s Category: 1. Remembering Learning Objective: Describe the function of lysosomes. 16. Defects in lysosomal enzyme levels can be corrected by _______ replacement therapy. Answer: enzyme Textbook Reference: Lysosomes Bloom’s Category: 1. Remembering Learning Objective: Describe the function of lysosomes.

True/False 1. The major function of the hydrophobic sequence at the amino terminals of proteins that are to be secreted from eukaryotic cells is to target those proteins to the rough ER. Answer: T Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that target proteins to the ER. 2. Polypeptides inside the ER are usually smaller than polypeptides synthesized from the same mRNA that have not entered the ER. Answer: T Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms that target proteins to the ER. 3. Transmembrane proteins in the plasma membrane are synthesized on the rough ER. Answer: T Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Explain how proteins are inserted into the ER membrane. 4. A newly synthesized polypeptide chain enters the ER through a translocon. Answer: T Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Explain how proteins are inserted into the ER membrane.

5. The initial site of N-linked glycosylation is the Golgi apparatus. Answer: F Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Describe protein folding and quality control in the ER. 6. Phospholipids are synthesized in the cytosolic half of the ER bilayer. Answer: T Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Distinguish the roles of smooth and rough ER. 7. Smooth ER is abundant in cells actively synthesizing steroid hormones. Answer: T Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Distinguish the roles of smooth and rough ER. 8. The Golgi apparatus is the main site of intracellular drug detoxification. Answer: F Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Distinguish the roles of smooth and rough ER. 9. Vesicle formation is driven by the binding of clathrin or COP proteins. Answer: T Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 1. Remembering Learning Objective: Explain the role of coat proteins. 10. Synaptic vesicles use a special mode of fusion that is completely different from regulated secretion in a typical secretory cell. Answer: F Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanism by which vesicles fuse with the correct target membranes. 11. The main method by which a cell gets rid of old, worn-out organelles is exocytosis. Answer: F Textbook Reference: Lysosomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the process of autophagy.

Short Answer For questions 1 through 5, describe the role of each of the following in targeting and processing proteins in the endoplasmic reticulum. 1. A signal sequence Answer: A signal sequence binds to the SRP, which targets the mRNA-ribosome-nascent protein complex to the rough ER. Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that target proteins to the ER. 2. A signal recognition particle (SRP) Answer: An SRP binds to the signal sequence, stops translation, and binds to the SRP receptor on the rough ER. Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that target proteins to the ER. 3. A signal peptidase Answer: A signal peptidase cleaves off the signal sequence inside the rough ER. Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Explain how proteins are inserted into the ER membrane. 4. Protein disulfide isomerase Answer: Protein disulfide isomerase catalyzes the breakage and re-formation of disulfide bonds in the ER. Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Describe protein folding and quality control in the ER. 5. A KDEL sequence Answer: A KDEL sequence binds to the KDEL receptor protein, which is transported from the Golgi apparatus back to the ER, returning “escaped” ER proteins. Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Explain transport to and retrieval from the Golgi apparatus. 6. What happens to a glycoprotein in the ER lumen if it cannot be folded correctly? Answer: It will be translocated back to the cytosol through a translocon channel, marked by ubiquitination, and degraded in a proteasome. The overall process is referred to as ERAD (ERassociated degradation). Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Describe protein folding and quality control in the ER.

7. What is the function of the trans-Golgi network? Answer: The trans-Golgi network is the site at which proteins are sorted into vesicles that will be targeted to lysosomes, the plasma membrane, or one of two secretory pathways. Textbook Reference: The Golgi Apparatus Bloom’s Category: 2. Understanding Learning Objective: Diagram the routes of protein export from the Golgi. 8. In polarized epithelial cells, proteins may be specifically targeted to what two membrane domains? Answer: The apical and basolateral domains Textbook Reference: The Golgi Apparatus Bloom’s Category: 2. Understanding Learning Objective: Diagram the routes of protein export from the Golgi. 9. What are the two families of small GTP-binding proteins that are involved in the formation of coated vesicles? Answer: Rab proteins and ARFs Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 2. Understanding Learning Objective: Summarize the process of vesicle budding and cargo selection. 10. What are Rab proteins? Answer: Rab proteins are small GTP-binding proteins that function in specific vesicle transport and provide the initial bridge between target and vesicle membranes required for vesicle docking and fusion. Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 2. Understanding Learning Objective: Summarize the process of vesicle budding and cargo selection. 11. What roles do SNAREs play in the specific binding and fusion of vesicles to a target membrane? Answer: SNAREs are pairs of proteins, one on a vesicle and one on a target membrane, that zip together, binding a vesicle to the target membrane and leading to fusion. Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanism by which vesicles fuse with the correct target membranes. 12. What deficiency causes I-cell disease? Answer: A deficiency in the enzyme that catalyzes the first step in the addition of mannose-6phosphate onto lysosomal hydrolases causes I-cell disease. Textbook Reference: Lysosomes Bloom’s Category: 2. Understanding Learning Objective: Explain how lysosomes are formed.

13. What is the likely transport pathway for lysosomal hydrolases in patients with I-cell disease? Answer: They are secreted into the extracellular space. Textbook Reference: Lysosomes Bloom’s Category: 3. Applying Learning Objective: Explain how lysosomes are formed. 14. How do lysosomes become acidic? Answer: Proton pumps are targeted to lysosomal membranes and use the energy of ATP hydrolysis to pump protons from the cytosol into the lysosomes. Textbook Reference: Lysosomes Bloom’s Category: 2. Understanding Learning Objective: Explain how lysosomes are formed. 15. How do hydrolytic enzymes get from the Golgi apparatus to lysosomes? Answer: Hydrolytic enzymes accumulate on lysosome receptors (mannose 6-phosphate receptors) in the trans-Golgi network, where they bud off as coated vesicles which, when uncoated, are called transport vesicles. These fuse with late endosomes, where the low pH dissociates the hydrolases from the receptors. The late endosomes concentrate the hydrolases and produce lysosomes. Textbook Reference: Lysosomes Bloom’s Category: 2. Understanding Learning Objective: Explain how lysosomes are formed. 16. Describe the process of autophagy. Answer: Autophagy is the process by which cellular components are enclosed in membrane from the ER and the resulting autophagosome fuses with a lysosome, and the contents are digested. Textbook Reference: Lysosomes Bloom’s Category: 2. Understanding Learning Objective: Summarize the process of autophagy.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. Which of the following is not a contiguous membrane domain? a. Rough ER b. Secretory granules c. Smooth ER d. ER exit site (ERES) Answer: b Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Diagram the secretory pathway. Feedback A: Incorrect. The rough ER is a contiguous membrane domain of the ER. Feedback B: Correct! Secretory vesicles are not part of the ER. They form by budding from the

Golgi apparatus. Feedback C: Incorrect. The smooth ER is a contiguous membrane domain of the ER. Feedback D: Incorrect. The ERES is a contiguous membrane domain of the ER. 2. In the “pulse-chase” experiment, Palade and colleagues studied the pathway taken by newly secreted proteins in pancreatic acinar cells by labeling them with radioactive amino acids, removing unused label, and determining their location within the cells at various time points. Which sequence represents the correct order in which the proteins were identified? a. Secretory vesicles  Golgi apparatus  rough ER  nucleus b. Golgi apparatus  rough ER  nucleus  secretory vesicles c. Rough ER  secretory vesicles  Golgi apparatus d. Rough ER  Golgi apparatus  secretory vesicles Answer: d Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Diagram the secretory pathway. Feedback A: Incorrect. The order of protein transport during secretion is rough ER  Golgi apparatus  secretory vesicles. Most proteins made in pancreatic acinar cells are secreted. Feedback B: Incorrect. The order of protein transport during secretion is rough ER  Golgi apparatus  secretory vesicles. Most proteins made in pancreatic acinar cells are secreted. Feedback D: Incorrect. The order of protein transport during secretion is rough ER  Golgi apparatus  secretory vesicles. Most proteins made in pancreatic acinar cells are secreted. Feedback C: Correct! 3. Proteins with a C-terminal transmembrane sequence are a. cotranslationally brought to the ER by SRP, where they are inserted into the ER lumen. b. posttranslationally brought to the ER by TRC40, where they remain in the ER membrane with their N-terminal domain facing the lumen. c. posttranslationally brought to the ER by SRP, where they remain in the ER membrane with their N-terminal domain facing the lumen. d. posttranslationally brought to the ER by TRC40, where they remain in the ER membrane with their C-terminal domain facing the lumen. Answer: b Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms that target proteins to the ER. Feedback A: Incorrect. They cannot be recruited cotranslationally because their signal sequence is not available to interact with until translation finishes. Feedback B: Correct! These proteins are posttranslationally recruited to the ER by TRC40, and oriented with their N-terminal domain inside the ER. Feedback C: Incorrect. SRP requires presence of the ribosome to efficiently interact with signal sequences. Feedback D: Incorrect. The C-terminal domains of these proteins face the cytoplasm. 4. The ribosomes of the rough endoplasmic reticulum are targeted to the cytoplasmic side of the membrane via

a. a signal sequence within the 28S ribosomal RNA. b. a sequence within the protein being synthesized. c. the S6 ribosomal protein. d. the cap sequence at the 5′ end of the mRNA being translated. Answer: b Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms that target proteins to the ER. Feedback A: Incorrect. The ribosomal RNAs do not target ribosomes to the rough ER. Feedback B: Correct! A signal sequence, typically at the amino terminus of the protein being synthesized, directs the ribosome to the rough ER. Feedback C: Incorrect. The ribosomal proteins do not target the ribosome to the rough ER. Feedback D: Incorrect. The mRNA does not play a direct role in targeting the ribosome to the rough ER. 5. As they emerge from the ribosome, signal sequences are recognized and bound by a(n) a. tRNA. b. signal peptidase. c. signal recognition particle (SRP). d. SRP receptor. Answer: c Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms that target proteins to the ER. Feedback A: Incorrect. tRNAs are involved in the process of translation, but not in targeting to the ER. Feedback B: Incorrect. The signal peptidase cleaves the signal sequence once it has reached the lumen of the ER. Feedback C: Correct! The signal recognition particle, which consists of seven polypeptides and one RNA molecule, binds the signal sequence and targets it to the ER membrane. Feedback D: Incorrect. The SRP receptor is located on the ER membrane, and although it is necessary for the import of proteins into the ER, it does not directly bind the signal sequence. 6. In yeasts, posttranslational translocation of proteins targeting the ER is a. a process affecting many newly synthesized proteins. b. much less common than in mammalian cells. c. rare. d. impossible because translocation into the ER requires the energy of protein synthesis. Answer: a Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Explain how proteins are inserted into the ER membrane. Feedback A: Correct! Many newly synthesized proteins in yeast are posttranslationally translocated into the ER. Feedback B: Incorrect. Posttranslational translocation is much more common in yeast than in mammalian cells.

Feedback C: Incorrect. Many newly synthesized proteins in yeast are posttranslationally translocated into the ER. Feedback D: Incorrect. BiP, a chaperone in the ER, can pull fully translated proteins into the ER. 7. Which statement about transmembrane proteins is true? a. They cross the membrane only once. b. The signal sequences are always cleaved off. c. They usually have one or more  helices spanning the membrane bilayer. d. They are always inserted with the amino terminus on the lumenal side and the carboxyl terminus in the cytosol. Answer: c Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Explain how proteins are inserted into the ER membrane. Feedback A: Incorrect. They can cross the membrane multiple times. Feedback B: Incorrect. In some cases the signal sequence lies within the ER membrane and serves to anchor the protein in the membrane. Feedback C: Correct! The transmembrane portions are usually  helices of approximately 20‒25 amino acids.  helices are especially stable in the nonpolar environment of the lipid bilayer. Feedback D: Incorrect. They can be in either orientation. 8. Proteins are translocated into the ER lumen a. by flippases. b. through an aqueous channel created by the Sec61 protein. c. by being pushed by translation through the lipid bilayer of the ER membrane. d. by being pulled by BiP across the lipid bilayer of the ER membrane. Answer: b Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Explain how proteins are inserted into the ER membrane. Feedback A: Incorrect. Flippases are involved in “flipping” lipids from the cytosolic leaflet of the endoplasmic reticulum, where they are synthesized to the luminal leaflet. Feedback B: Correct! The translocon Sec61 forms an aqueous channel through which proteins are translocated into the ER lumen. Feedback C: Incorrect. Translocated proteins contain hydrophilic segments that cannot be pushed through the hydrophobic lipid bilayer. Feedback D: Incorrect. Translocated proteins contain hydrophilic segments that cannot be pulled across the lipid bilayer. 9. In some proteins, the only transmembrane sequence is located C-terminally in the protein. These proteins are recognized by a. the SRP. b. the Sec61 translocon. c. TRC40 (GET3). d. the GET1-GET2 complex embedded in the ER membrane. Answer: c

Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Explain how proteins are inserted into the ER membrane. Feedback A: Incorrect. SRP is involved in the recognition of the signal sequence, typically an Nterminal sequence. Feedback B: Incorrect. The Sec61 translocon does not directly recognize proteins for translocation; rather the translocon-associated SRP receptor recognizes SRP-bound proteins, and the separate Sec62/62 complex recognizes fully translate, chaperone-bound proteins destined for translocation into the ER lumen. Feedback C: Correct! The transmembrane domains of these proteins are recognized posttranslationally by TRC40, which then escorts the transmembrane protein to the ER membrane, where it is inserted via the GET1-GET2 complex. Feedback D: Incorrect. The GET1-GET2 complex does not directly recognize proteins for insertion into the ER membrane. 10. In proteins being made on rough ER, short spans of hydrophobic amino acids that form  helices a. induce a conformational change in the translocon, allowing them to exit into the ER membrane. b. are posttranslationally inserted into the ER membrane by BiP. c. are ubiquitinated and transferred into the cytosol for proteasomal degradation. d. cause premature termination of translation, leading to the unfolded protein response. Answer: a Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Explain how proteins are inserted into the ER membrane. Feedback A: Correct! Hydrophobic  helices exit translocons laterally to form transmembrane spanning regions. Feedback B: Incorrect. BiP only acts on luminal domains. Feedback C: Incorrect. Ubiquitination is triggered by sensors of exposed hydrophobic patches on proteins in the ER lumen. Feedback D: Incorrect. Translation continues after the hydrophobic sequences exit the translocon. 11. The lumen of the ER is equivalent topologically to the a. cytoplasm. b. cytoplasmic face of the plasma membrane. c. extracellular space. d. nucleoplasm. Answer: c Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Describe protein folding and quality control in the ER. Feedback A: Incorrect. The lumen of the ER is not topologically equivalent to the cytoplasm. Feedback B: Incorrect. The lumen of the ER is not equivalent to a membrane. Feedback C: Correct! The lumen of the ER is topologically equivalent to the extracellular space,

and hence the inner leaflet of the ER membrane is topologically equivalent to the cell surface. Feedback D: Incorrect. The lumen of the ER is not topologically equivalent to the nucleoplasm. 12. N-linked glycosylation at an Asn-X-Ser/Thr consensus sequence adds _______ sugar(s) to the protein in a single step. a. 1 b. 14 c. 30 d. 100 Answer: b Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Describe protein folding and quality control in the ER. Feedback A: Incorrect. More than one sugar is added. Sugars are added as a block during Nglycosylation. Sugars are added one at a time during O-glycosylation. Feedback B: Correct! Fourteen sugars are added in one step from a preformed precursor oligosaccharide linked to a dolichol pyrophosphate. Feedback C: Incorrect. Sugars are added as a block during N-glycosylation, but not this many. Feedback D: Incorrect. Sugars are added as a block during N-glycosylation, but not this many. 13. The major site at which membrane lipids are synthesized is the a. cytosolic side of the ER membrane. b. cytosol. c. lumenal side of the ER membrane. d. cytosolic side of the Golgi membrane. Answer: a Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Distinguish the roles of smooth and rough ER. Feedback A: Correct! Because lipids are hydrophobic in nature, they are synthesized in association with already existing membranes, and the synthetic enzymes are located in the cytosol. Feedback B: Incorrect. The hydrophobic nature of lipids makes the hydrophilic environment of the cytosol an unfavorable place for their synthesis. Feedback C: Incorrect. The enzymes necessary for the synthesis of membrane lipids are located in the cytosol. Feedback D: Incorrect. Although some membrane lipids are synthesized here, it is not the primary site of synthesis. 14. Which of the following is not a destination for vesicles leaving the Golgi apparatus? a. The plasma membrane b. The exterior of the cell c. Lysosomes d. Mitochondria Answer: d Textbook Reference: The Endoplasmic Reticulum

Bloom’s Category: 2. Understanding Learning Objective: Explain transport to and retrieval from the Golgi apparatus. Feedback A: Incorrect. Plasma membrane proteins reach their destination by association with vesicles from the Golgi. Feedback B: Incorrect. Secretion of proteins occurs by fusion of Golgi-derived vesicles with the plasma membrane. Feedback C: Incorrect. Lysosomal proteins reach the lysosomal lumen via vesicles from the Golgi apparatus. Feedback D: Correct! Proteins destined for the mitochondria are transported via a mechanism that does not involve vesicular transport or the Golgi apparatus. 15. The sequence Lys-Asp-Glu-Leu (KDEL) serves to retain proteins in the ER by a. preventing their packaging into vesicles destined for the Golgi. b. binding to receptors within the membranes of the ERGIC and Golgi, which retain them or return them to the ER. c. binding the SRP receptor in the ER membrane. d. associating with the lipids in the ER membrane. Answer: b Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 1. Remembering Learning Objective: Explain transport to and retrieval from the Golgi apparatus. Feedback A: Incorrect. Even proteins containing the KDEL sequence, which tags them as ER resident proteins, are packaged into vesicles bound for the ERGIC and/or Golgi. Feedback B: Correct! The KDEL sequence does not prevent the transport of proteins to the Golgi, but it does serve to retrieve them. Feedback C: Incorrect. The SRP receptor is involved in targeting newly synthesized proteins to the ER, but it does not play a role in the retention of ER resident proteins. Feedback D: Incorrect. The KDEL sequence does not associate directly with the ER membrane. 16. Cargo proteins are transported through the Golgi apparatus by a. forward (anterograde) moving vesicles. b. backward (retrograde) moving vesicles. c. cisternal maturation in which the cisternae themselves are the carriers for cargo transport through the Golgi apparatus. d. some process that remains controversial and may include elements of both vesicular transport and cisternal maturation. Answer: d Textbook Reference: The Golgi Apparatus Bloom’s Category: 2. Understanding Learning Objective: Relate the structure of the Golgi apparatus to its function. Feedback A: Incorrect. Whether the carriers for cargo transport through the Golgi apparatus are vesicles remains controversial. Feedback B: Incorrect. The carriers for cargo transport through the Golgi apparatus cannot be retrograde moving vesicles as cargo itself would never progress forward if it were being transported in retrograde vesicles. Feedback C: Incorrect. Whether the carriers for cargo transport through the Golgi apparatus are

the cisternae themselves remains controversial. Feedback D: Correct! In this area of molecular cell biology, the correct answer remains controversial. 17. The trans-Golgi network is the a. intermediate compartment between the ER and the Golgi. b. part of the Golgi where fusion of vesicles from the ER occurs. c. exit part of the Golgi where sorting of proteins to the lysosomes, plasma membrane, and cell exterior occurs. d. network of vesicles that transport resident Golgi proteins between cisternae. Answer: c Textbook Reference: The Golgi Apparatus Bloom’s Category: 1. Remembering Learning Objective: Relate the structure of the Golgi apparatus to its function. Feedback A: Incorrect. The trans-Golgi network lies farther away from the ER than the intermediate compartment. Feedback B: Incorrect. This would be the cis-Golgi network. Feedback C: Correct! The trans-Golgi network is the last part of the Golgi through which proteins pass. It serves as a distribution center. Feedback D: Incorrect. The trans-Golgi network is composed of cisternae, and it is not a network of vesicles. . Which of the following lipids is(are) synthesized in the Golgi apparatus? a. Phospholipids b. Cholesterol c. Ceramide d. Glycolipids Answer: d Textbook Reference: The Golgi Apparatus Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of the Golgi in synthesis of membrane lipids. Feedback A: Incorrect. Phospholipids are synthesized in the ER. Feedback B: Incorrect. Cholesterol is synthesized in the ER. Feedback C: Incorrect. Ceramide is synthesized in the ER. Feedback D: Correct! Glycolipids are synthesized from ceramide in the Golgi. 19. Clathrin-coated vesicles are involved in a. retrieval of ER resident proteins from the cis Golgi or the ER-Golgi intermediate compartment. b. uptake of extracellular molecules by endocytosis and the transport of molecules from the trans-Golgi network to the lysosomes. c. transport from the ER to the Golgi. d. recycling of Golgi resident proteins during cisternal maturation. Answer: b Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 2. Understanding Learning Objective: Explain the role of coat proteins.

Feedback A: Incorrect. COPI-coated vesicles carry out this process. Feedback B: Correct! Clathrin-coated vesicles mediate both of these processes. Feedback C: Incorrect. COPII-coated vesicles carry out this process. Feedback D: Incorrect. COPI vesicles carry out this process. 20. The major model of vesicle fusion holds that actual fusion of a vesicle with its target membrane is driven by the interaction of pairs of proteins called vesicle and target a. SNAREs. b. Rab proteins. c. COPs. d. tethers. Answer: a Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism by which vesicles fuse with the correct target membranes. Feedback A: Correct! According to the hypothesis, SNAREs on vesicles bind to corresponding SNAREs on the target membrane and initiate fusion of the two. Feedback B: Incorrect. Rab proteins are necessary for providing the initial bridge between target and vesicle membranes, but other proteins are responsible for their fusion. Feedback C: Incorrect. The COPs are involved in vesicular transport, but not specifically in the fusion event. Feedback D: Incorrect. Tethers are involved in the initial recognition between vesicles and target membranes, but not in the actual vesicle-target membrane fusion process. 21. Which of the following would you expect to find at high concentrations in lysosomes? a. Proteins destined for secretion b. Glycosylation enzymes c. Degradative enzymes d. Recycling endosomes Answer: c Textbook Reference: Lysosomes Bloom’s Category: 3. Applying Learning Objective: Describe the function of lysosomes. Feedback A: Incorrect. Secretory vesicles contain proteins to be secreted. Feedback B: Incorrect. These would be present in the ER and Golgi, where glycosylation occurs. Feedback C: Correct! The function of lysosomes is to degrade, at an acidic pH, substances taken up from outside the cell and components within the cell that have outlived their usefulness. Feedback D: Incorrect. Recycling endosomes are involved in receptor recycling and have no direct relationship with lysosomes. 22. The process by which cells degrade their own components by enclosing them in ER membrane is a. autophagy. b. ER-mediated endocytosis. c. phagocytosis.

d. pinocytosis. Answer: a Textbook Reference: Lysosomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the process of autophagy. Feedback A: Correct! Feedback B: Incorrect. This process would be an internalization process and likely does not exist. Feedback C: Incorrect. Phagocytosis (cell eating) is an internalization process. Feedback D: Incorrect. Pinocytosis (cell drinking) is an internalization process.

Essay 1. Binding of SRP to the signal sequence arrests the further translation of the nascent polypeptide chain. How does the arrest of translation promote the subsequent translocation of the polypeptide chain through the Sec61 translocation complex? Answer: SRP, in binding to the nascent polypeptide chain and ribosome, arrests polypeptide elongation. Only when the SRP has become dissociated from the chain and the nascent chain is being pushed through the translocon does polypeptide elongation resume. This results in an extended conformation of the nascent chain that is being pushed through the translocon. Folding occurs in the ER lumen. If SRP was bound but did not arrest translation, chaperones would be needed in the cytosol to bind the nascent chain and keep it in an extended conformation as it continued to elongate. In yeast, where posttranslational insertion occurs, this process is fairly common. In mammalian cells, posttranslational insertion is rare, and the pool of cytosolic chaperones is presumably limited. Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 2. Understanding Learning Objective: Explain how proteins are inserted into the ER membrane. 2. A member of the heat-shock class of proteins, BiP, is necessary for the proper folding of proteins in the ER. In terms of function, what do BiP and the heat-shock proteins (HSP) have in common? Answer: The primary function of heat-shock proteins is to prevent damage to the cell upon exposure to conditions that cause protein unfolding; in their absence, unfolded proteins would associate into large aggregates that would be difficult to disentangle, once the cell returned to normal conditions. Like BiP, the inducible heat-shock proteins bind unfolded proteins and assist in their proper folding. In fact, BiP and inducible heat-shock proteins are all chaperone proteins. BiP is a member of the Hsp70 protein family. Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 3. Applying Learning Objective: Describe protein folding and quality control in the ER. 3. Suppose you are studying a mutant that you know is impaired in secretion, but you do not know the exact nature of the defect. You analyze the sugar content of some of the proteins purified from the mutant that you know should be N-glycosylated and find that fucose, galactose,

and sialic acid residues are absent. Instead, only N-acetylglucosamine and mannose residues are evident. Where do you think the defect in secretion lies? Answer: There is a defect in transport between the ER and the Golgi. In the ER, Nacetylglucosamine, mannose, and glucose residues are added to specific asparagine residues within proteins. The glucose residues are later removed, and upon reaching the Golgi, they contain only N-acetylglucosamine and mannose residues. Fucose, galactose, and sialic acid residues are added in the Golgi. Thus, a mutation that blocks some part of the transport process between leaving the ER and leaving the Golgi would explain the lack of these sugar residues in proteins that normally contain them. Textbook Reference: The Golgi Apparatus Bloom’s Category: 4. Analyzing Learning Objective: Describe the types of protein glycosylation that take place in the Golgi. 4. Imagine a transmembrane molecule that lies in the plasma membrane and acts as a receptor for an extracellular signaling molecule. When the ligand-binding domain is inserted into the ER during synthesis of this transmembrane molecule, will it lie on the lumen side of the ER or the cytoplasm side? Answer: The ligand-binding domain will lie within the lumen of the ER, passing through the Golgi in the same orientation. In the final transport step, a vesicle containing the fusion protein will leave the trans-Golgi network and fuse with the plasma membrane, thus exposing the lumen of the vesicle to the cell exterior. In addition, any oligosaccharide subunits that may have been added to the protein will be located on the extracellular domain of the protein, since glycosylation takes place in the lumen of the ER and Golgi. Textbook Reference: The Endoplasmic Reticulum Bloom’s Category: 3. Applying Learning Objective: Diagram the routes of protein export from the Golgi. 5. Describe the roles of clathrin and associated adaptor proteins in the sorting and transport of proteins to lysosomes. Answer: Proteins destined for lysosomes are recognized in the Golgi and acquire a mannose-6phosphate tag, which then binds to the mannose-6-phosphate receptor, a transmembrane protein. As part of the formation of clathrin-coated vesicles at the trans-Golgi network, an adaptor protein is recruited: Arf/GTP. The adaptor protein has a binding site for interaction with the cytosolic tail of the mannose-6-phosphate receptor, and it also serves as a binding site for assembly of the clathrin coat. Textbook Reference: The Mechanism of Vesicular Transport Bloom’s Category: 2. Understanding Learning Objective: Summarize the process of vesicle budding and cargo selection. 6. Cargo proteins are transported through the Golgi apparatus to the trans-Golgi network, where they are sorted into various vesicles that are targeted to different destinations. What is the role of vesicles in cargo protein transport through the Golgi apparatus? Answer: Two general models for Golgi function in cargo transport through the organelle are actively being considered: the “stable cisternal model” and the “cisternal maturation model.” In the stable cisternal model, vesicles are the carriers of cargo proteins between cisternae. In the cisternal maturation model, the cisternae are the actual cargo carriers. Vesicles function to

retrieve Golgi components and hence to mature the cisternae as they transport cargo. At present, each model explains some, but not all, characteristics of Golgi transport. Further research is needed. Textbook Reference: The Golgi Apparatus Bloom’s Category: 3. Applying Learning Objective: Summarize the process of vesicle budding and cargo selection. 7. Describe the selective and nonselective cellular purposes of autophagy. Answer: Most cellular proteins degraded by autophagy are taken up nonselectively, so there is a fixed rate of general protein turnover. Rates of autophagy can be increased for specific proteins by marking them with signals, such as ubiquitin, to drive developmental processes or efficiently degrade damaged organelles and misfolded proteins produced by cellular stresses. Textbook Reference: Lysosomes Bloom’s Category: 2. Understanding Learning Objective: Summarize the process of autophagy.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 13: Mitochondria, Chloroplasts, and Peroxisomes TEST FILE QUESTIONS Multiple Choice 1. Mitochondrial and chloroplast proteins are synthesized on a. ER ribosomes. b. free cytosolic ribosomes. c. the Golgi apparatus. d. peroxisomes. Answer: b Textbook Reference: Introduction Bloom’s Category: 1. Remembering Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 2. Which of the following contain their own genomes? a. Mitochondria, but not chloroplasts and peroxisomes b. Chloroplasts, but not mitochondria and peroxisomes c. Mitochondria and chloroplasts, but not peroxisomes d. Mitochondria, chloroplasts, and peroxisomes Answer: c Textbook Reference: Introduction Bloom’s Category: 1. Remembering Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 3. Which statement about mitochondria is true? a. They produce most of the ATP derived from the breakdown of lipids and carbohydrates. b. They produce all of the ATP derived from the breakdown of fatty acids. c. They produce all of the ATP derived from the breakdown of carbohydrates. d. They produce about 10% of the ATP derived from the sunlight. Answer: a Textbook Reference: Introduction Bloom’s Category: 2. Understanding Learning Objective: Illustrate the functional organization of mitochondria. 4. The infoldings of the inner mitochondrial membrane are called

a. cisternae. b. cristae. c. laminae. d. lamellae. Answer: b Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. 5. The inner compartment of mitochondria is called the a. stroma. b. intermembrane space. c. inner membrane space. d. matrix. Answer: d Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. 6. Which compound is a product of glycolysis that is transported into the mitochondria? a. Pyruvate b. Acetate as acetyl CoA c. Lactic acid d. Citric acid Answer: a Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. 7. The citric acid cycle consists of the oxidation of _______ to produce _______. a. pyruvate; CO2, NADH, and FADH2 b. acetyl CoA; CO2, NADH, and FADH2 c. pyruvate; CO2 d. pyruvate; NADH and FADH2 Answer: b Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Illustrate the functional organization of mitochondria. 8. Most small molecules are permeable across a. both mitochondrial membranes. b. the inner, but not the outer, mitochondrial membrane. c. the outer, but not the inner, mitochondrial membrane. d. neither mitochondrial membrane. Answer: c Textbook Reference: Mitochondria

Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. 9. Mitochondria can a. divide by fission. b. fuse with one another. c. be transported to areas of high energy use. d. All of the above Answer: d Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Illustrate the functional organization of mitochondria. 10. The mitochondrial outer membrane contains channels composed of proteins called a. porins. b. aquaporins. c. connexins. d. claudins. Answer: a Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. 11. The outer mitochondrial membrane contains proteins that a. synthesize ATP. b. pump protons. c. transport pyruvate and fatty acids. d. direct translocation. Answer: d Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Illustrate the functional organization of mitochondria. 12. Electron transport occurs in the mitochondrial a. outer membrane. b. intermembrane space. c. inner membrane. d. matrix. Answer: c Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. 13. The process by which mitochondria are thought to have arisen during evolution is called a. symbiosis.

b. phagocytosis. c. endosymbiosis. d. pinocytosis. Answer: c Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Describe mitochondrial genomes. 14. Most mitochondrial genomes consist of a. a single linear DNA molecule. b. several linear DNA molecules. c. several circular DNA molecules. d. a single circular DNA molecule. Answer: c Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Describe mitochondrial genomes. 15. The organisms most similar to mitochondria are a. progenotes. b. -proteobacteria. c. cyanobacteria. d. purple sulfur bacteria. Answer: b Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Describe mitochondrial genomes. 16. Mitochondria contain a. no genes of their own. b. genes for mitochondrial proteins. c. genes for mitochondrial proteins and rRNAs. d. genes for mitochondrial proteins, rRNAs, and tRNAs. Answer: d Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Describe mitochondrial genomes. 17. In what way does the mitochondrial genetic code differ from the “universal” genetic code? a. Some codons code for different amino acids. b. There are no stop codons. c. It accommodates less wobble. d. All of the above Answer: a Textbook Reference: Mitochondria

Bloom’s Category: 2. Understanding Learning Objective: Describe mitochondrial genomes. 18. The proteins encoded by the human mitochondrial genome function in a. mitochondrial ribosomes. b. mitochondrial DNA polymerases. c. respiratory complexes and oxidative phosphorylation. d. helicases. Answer: c Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Describe mitochondrial genomes. 19. Mitochondrial DNA is inherited by means of a. maternal transmission. b. paternal transmission. c. random assortment. d. Mendelian genetics. Answer: a Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Describe mitochondrial genomes. 20. Which of the following is a mitochondrial disease? a. Lou Gehrig’s disease b. Retinoblastoma c. Leber’s hereditary optic neuropathy d. Crohn’s disease Answer: c Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Describe mitochondrial genomes. 21. Most mitochondrial proteins are synthesized on a. mitochondrial ribosomes from nuclear mRNAs. b. cytoplasmic ribosomes; they are imported co-translationally as they are being synthesized. c. cytoplasmic ribosomes; they are imported after they are completely synthesized. d. mitochondrial ribosomes from mitochondrial mRNAs. Answer: c Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 22. Mitochondrial targeting presequences usually consist of a a. hydrophobic  helix.

b. hydrophobic random chain. c. negatively charged  helix. d. positively charged  helix. Answer: d Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 23. Tim and Tom are a. twin brothers with the same mitochondrial disease. b. chaperones. c. protein translocators in mitochondrial membranes. d. transporters of small molecules across the mitochondrial membranes. Answer: c Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 24. The mitochondrial protein presequence is cleaved off by a protease called a. signal peptidase. b. presequence protease. c. ubiquitin-targeted protease. d. matrix processing peptidase. Answer: d Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 25. Mitochondrial inner membrane single-pass transmembrane proteins are inserted into the inner membrane a. through a transporter called Oxa translocase. b. through Tim. c. through Tim and Tom. d. directly from the matrix into the lipid bilayer. Answer: a Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 26. Most mitochondrial phospholipids are synthesized in the a. mitochondrial matrix. b. mitochondrial intermembrane space. c. ER. d. Golgi apparatus. Answer: c Textbook Reference: Mitochondria

Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 27. Which of the following is not involved in the initial transport of a protein across the mitochondrial outer membrane from the cytosol? a. Hsp70 b. Tim23 c. Tom d. ATP Answer: b Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 28. Which of the following is not involved in targeting a protein from the intermembrane space into the inner mitochondrial membrane? a. Tim9 b. Tim10 c. Tim22 d. Tom Answer: d Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 29. The import of mitochondrial matrix proteins from the cytoplasm requires a. an electrochemical gradient across the inner membrane. b. a potassium gradient across the inner membrane. c. ATP and a proton gradient across the inner membrane. d. ATP and a potassium gradient across the inner membrane. Answer: c Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Explain the role of the proton gradient in transport across the mitochondrial membrane. 30. The energy required to drive the transport of small molecules into and out of mitochondria is provided by the a. chemiosmotic gradient. b. hydrolysis of ATP. c. electrochemical gradient. d. negative charge of small molecules. Answer: c Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering

Learning Objective: Explain the role of the proton gradient in transport across the mitochondrial membrane. 31. Which of the following phospholipids contains four fatty acid chains and is associated with a restriction in proton flow? a. Phosphatidylcholine b. Cholesterol c. Sphingolipid d. Cardiolipin Answer: d Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Explain the role of the proton gradient in transport across the mitochondrial membrane. 32. Chloroplasts differ from mitochondria in that chloroplasts a. do not generate ATP. b. do not originate by endosymbiosis. c. do not replicate by division. d. synthesize their own amino acids and fatty acids. Answer: d Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 33. Chloroplasts are similar to mitochondria in that both a. have a porous outer membrane. b. contain light sensitive pigments. c. require presequence amino acids on proteins for import. d. require folded cristae as the site of electron transport. Answer: a Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 34. Thylakoids are often arranged in stacks called a. dictyosomes. b. grana. c. plastids. d. stroma. Answer: b Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering

Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 35. Chloroplasts synthesize a. elaioplasts. b. amino acids. c. peroxisomes. d. catalase. Answer: b Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 36. The chloroplast genome contains about _______ genes. a. 20 b. 150 c. 1,500 d. 3,000 Answer: b Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Contrast chloroplast and mitochondrial genomes. 37. Chloroplast tRNAs translate a. all mRNA codons according to the universal code. b. all the amino acids according to the universal code but have different stop codons. c. some codons as amino acids that differ from the universal code but use the same stop codons. d. some codons as amino acids that differ from the universal code and use some different stop codons. Answer: a Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Contrast chloroplast and mitochondrial genomes. 38. Most chloroplast proteins are synthesized on a. free ribosomes in the cytosol. b. RER membranes in the cytoplasm. c. ribosomes bound to the outer chloroplast membrane. d. ribosomes in the chloroplast stroma. Answer: a Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Contrast chloroplast and mitochondrial genomes.

39. The most abundant protein on Earth is a. cytochrome c. b. rubisco. c. ATP synthase. d. glucose-6-phosphatase. Answer: b Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Contrast chloroplast and mitochondrial genomes. 40. The transport of proteins across the outer and inner chloroplast membranes occurs through complexes called a. Tim and Tom. b. Tic and Toc. c. Sec and Tat. d. import complexes. Answer: b Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms of protein import into chloroplasts. 41. Many proteins incorporated into the thylakoid lumen are synthesized a. in the thylakoid lumen. b. in the stroma and transported across the thylakoid membrane. c. on the outer chloroplast membrane and transported across the thylakoid membrane by way of its hydrophobic signal sequence. d. in the cytosol, imported into the stroma, and transported across the thylakoid membrane by way of a second signal sequence. Answer: d Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 2. Understanding Learning Objective: Summarize the mechanisms of protein import into chloroplasts. 42. Carotenoids are stored in a. chloroplasts. b. chromoplasts. c. amyloplasts. d. elaioplasts. Answer: b Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of other plastids. 43. All plastids, including chloroplasts, develop from a. chromoplasts.

b. etioplasts. c. elaioplasts. d. proplastids. Answer: d Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of other plastids. 44. An amyloplast is a plastid that a. stores starch. b. stores lipid. c. is arrested in chloroplast development by lack of light. d. stores pigment. Answer: a Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of other plastids. 45. The major function of peroxisomes is to a. oxidize certain organic molecules and degrade the H2O2 produced by these reactions. b. produce hydrogen peroxide for cells. c. digest old organelles. d. digest macromolecules taken up by endocytosis. Answer: a Textbook Reference: Peroxisomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of peroxisomes in animal and plant cells. 46. Peroxisomes contain the enzyme _______, which breaks down H2O2 into H2O and oxygen. a. peroxidase b. catalase c. peroxigen d. glyoxylate Answer: b Textbook Reference: Peroxisomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of peroxisomes in animal and plant cells. 47. Which of the following is the human disease caused by mutations in the proteins required for importing functional proteins into peroxisomes? a. Turner’s syndrome b. Zellweger syndrome c. I-cell disease d. Leber’s hereditary optic neuropathy Answer: b

Textbook Reference: Peroxisomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of peroxisomes in animal and plant cells. 48. Most peroxisomal proteins are synthesized on a. free ribosomes in the cytosol. b. RER membranes in the cytoplasm. c. ribosomes bound to the outer peroxisome membrane. d. ribosomes inside the peroxisome. Answer: a Textbook Reference: Peroxisomes Bloom’s Category: 1. Remembering Learning Objective: Describe the pathways responsible for peroxisome biogenesis. 49. New peroxisomes form by a. budding from the Golgi apparatus. b. budding from preexisting peroxisomes. c. de novo assembly from proteins synthesized in the cytosol. d. budding of vesicles from the ER and growth and division of preexisting peroxisomes. Answer: d Textbook Reference: Peroxisomes Bloom’s Category: 1. Remembering Learning Objective: Describe the pathways responsible for peroxisome biogenesis.

Fill in the Blank 1. The citric acid cycle produces two reduced cofactors, _______ and _______. Answer: NADH; FADH2 Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. 2. To enter the mitochondria, precursors of the -barrel proteins pass through the _______ complex into the intermembrane space; there, they are recognized and transported to a second complex, called the _______complex, and inserted into the outer membrane. Answer: Tom; SAM (sorting and assembly machinery) Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 3. Chloroplasts have three membranes: the inner membrane, the outer membrane, and the _______ membrane. Answer: thylakoid Textbook Reference: Chloroplasts and Other Plastids

Bloom’s Category: 1. Remembering Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 4. In terms of its role in the generation of metabolic energy, the thylakoid membrane of chloroplasts is equivalent to the _______ of mitochondria. Answer: inner membrane Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 5. Chloroplasts contain their own genetic system, reflecting their evolutionary origins from photosynthetic _______. Answer: bacteria Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Contrast chloroplast and mitochondrial genomes. 6. Although chloroplasts encode more of their own proteins than mitochondria, about 3,000 or 95% of chloroplast proteins are still encoded by _______ genes. Answer: nuclear Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Contrast chloroplast and mitochondrial genomes. 7. Once through the Toc complex in the outer membrane of the chloroplast, proteins are transferred to the _______ in the inner membrane. Answer: Tic complex Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms of protein import into chloroplasts. 8. Proteins can be targeted to the thylakoid membrane by at least _______ different pathways. Answer: three (3) Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms of protein import into chloroplasts. 9. In chloroplasts, the _______ pathway is related to the pathway used to translocate proteins into the endoplasmic reticulum. Answer: Sec Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms of protein import into chloroplasts.

10. In addition to _______ from the ER, new peroxisomes can be formed by the growth and division of old peroxisomes. Answer: vesicle budding Textbook Reference: Peroxisomes Bloom’s Category: 1. Remembering Learning Objective: Describe the pathways responsible for peroxisome biogenesis. True/False 1. All mitochondria use the universal genetic code. Answer: F Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Describe mitochondrial genomes. 2. Most mitochondrial proteins are coded for by the mitochondrial genome. Answer: F Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Describe mitochondrial genomes. 3. Mitochondria lacking mitochondrial DNA are soon unable to make ATP by oxidative phosphorylation. Answer: T Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Describe mitochondrial genomes. 4. Mitochondria contain about 1,500 different proteins. Answer: T Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 5. Most proteins are imported into mitochondria in their fully folded state. Answer: F Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 6. A mitochondrial matrix protein is maintained in its unfolded state by an Hsp70 chaperone protein before transport across the mitochondrial membrane. Answer: T Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding

Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 7. Some proteins destined for insertion into the inner membrane are transported from the matrix compartment. Answer: T Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 8. Mitochondrial phospholipids are delivered to mitochondria by phospholipid transfer proteins. Answer: T Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 9. In mitochondria, the pH of the matrix is higher than the pH of the intermembrane space. Answer: T Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Explain the role of the proton gradient in transport across the mitochondrial membrane. 10. Chloroplasts are only one of a much larger family of plant organelles called plastids that all contain the same genome. Answer: T Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of other plastids.

Short Answer Questions 1–4. Where in mitochondria does each of the following processes or events occur? 1. Synthesis of mitochondrial ribosomal RNAs Answer: The mitochondrial matrix Textbook Reference: Mitochondria Bloom’s Category: 3. Applying Learning Objective: Illustrate the functional organization of mitochondria. 2. Diffusion of small molecules via porin-formed channels Answer: The outer mitochondrial membrane Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering

Learning Objective: Illustrate the functional organization of mitochondria. 3. The citric acid cycle Answer: The mitochondrial matrix Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. 4. Oxidative phosphorylation of ATP Answer: The inner mitochondrial membrane Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. 5. What are some of the reasons evolutionary biologists believe mitochondria came from bacteria through endosymbiosis? Answer: Evolutionary biologists believe mitochondria evolved from an endosymbiotic relationship between bacteria and larger cells because mitochondrial genomes are usually circular like bacteria; mitochondrial genomes are similar to free-living α-proteobacteria; and mitochondria use a genetic code that is slightly different from the “universal” genetic code. Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Describe mitochondrial genomes. 6. What would be the effect of a mutation in the Tim23 complex? Answer: Tim23 directs the translocation of polypeptides from the mitochondrial inner membrane space into the mitochondrial matrix. Thus, a mutation in Tim23 would most likely cause a buildup of proteins in the inner membrane, leading to severe mitochondrial dysfunction. Textbook Reference: Mitochondria Bloom’s Category: 4. Analyzing Learning Objective: Summarize how proteins and lipids are imported into mitochondria. 7. Where does the energy for phosphate transfer across the mitochondrial inner membrane to the matrix come from? Answer: A proton gradient (electrochemical gradient) Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Explain the role of the proton gradient in transport across the mitochondrial membrane. 8. Both mitochondria and chloroplasts use a proton gradient to drive ATP synthesis, but in different locations. What is the major structural difference, that is related to the proton gradient, between mitochondria and chloroplasts, and where is ATP generated in each?

Answer: The major structural difference is that chloroplasts contain a thylakoid membrane, which is comparable to the mitochondrial inner membrane. In chloroplasts, the proton gradient is inside the thylakoid lumen, and ATP synthesis occurs in the stroma. In mitochondria, the proton gradient is in the intermembrane space, and ATP synthesis occurs in the matrix. Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 4. Analyzing Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 9. What effect would a functional mutation in the Toc complex have on chloroplasts? Answer: Toc is a protein complex found on the outer membrane of chloroplasts that mediates the transport of proteins into the intermembrane space (between the thylakoid membrane and outer membrane). A mutation in this protein would block the transport of all proteins into the chloroplasts. Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 4. Analyzing Learning Objective: Summarize the mechanisms of protein import into chloroplasts. 10. Translocase proteins such as Toc have channels that are just barely wider than a single amino acid. Since most mature proteins have significant secondary and tertiary structure, how do they fit through the narrow channels of these translocases? Answer: Maturely formed proteins are much wider than a single amino acid due to secondary structure and would not fit through these translocases. However, upon synthesis, Hsp70 binds to these proteins to keep them in an unfolded state, allowing them to be translocated through the channels. Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 4. Analyzing Learning Objective: Summarize the mechanisms of protein import into chloroplasts. 11. What are the two mechanisms by which peroxisomes can be formed? Answer: New peroxisomes can be formed by vesicle budding from the endoplasmic reticulum, followed by the import of matrix proteins, and they can also be formed by the growth and division of existing peroxisomes in a manner very similar to the mechanism used by mitochondria and chloroplasts. Textbook Reference: Peroxisomes Bloom’s Category: 2. Understanding Learning Objective: Describe the pathways responsible for peroxisome biogenesis.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. Mitochondria differ from other organelles such as lysosomes and the Golgi apparatus

in that they a. contain enzymes specific to their function. b. contain their own genomes. c. do not contain proteins that are imported from the cytosol. d. are not membrane-bounded. Answer: b Textbook Reference: Introduction Bloom’s Category: 2. Understanding Learning Objective: Illustrate the functional organization of mitochondria. Feedback A: Incorrect. All organelles contain enzymes specific to their function. Feedback B: Correct! Mitochondria contain their own genes, including genes for proteins involved in the electron transport system, as well as some rRNA and tRNA genes. Feedback C: Incorrect. Most mitochondrial proteins are synthesized in the cytosol and subsequently imported into mitochondria. Feedback D: Incorrect. Mitochondria are surrounded by a double membrane. 2. What is the major site of energy production in the form of ATP in human cells? a. The mitochondrial matrix b. The cytoplasm c. The outer mitochondrial membrane d. The inner mitochondrial membrane Answer: d Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. Feedback A: Incorrect. The citric acid cycle takes place in the mitochondrial matrix and generates some energy, but not the major portion of it. Feedback B: Incorrect. Glycolysis, which takes place in the cytoplasm, is only the first step in the metabolism of glucose, and it produces only a small fraction of the total energy. Feedback C: Incorrect. Oxidative phosphorylation is the process that produces the majority of energy in the cell, but it does not take place at the outer mitochondrial membrane. Feedback D: Correct! The inner mitochondrial membrane is the site of oxidative phosphorylation, which produces close to 90% of the ATP derived from glucose metabolism. 3. The human mitochondrial genome encodes only 22 tRNAs. This limited array of tRNAs can read the 64 possible triplet codons through extreme wobble in base pairing at the third codon position and the use of a(n) _______ genetic code. a. chloroplast-mitochondrial-specific b. nonuniversal mitochondrial c. peroxisomal-mitochondrial-specific d. universal Answer: b Textbook Reference: Mitochondria

Bloom’s Category: 2. Understanding Learning Objective: Describe mitochondrial genomes. Feedback A: Incorrect. Chloroplasts have more tRNAs and are not so restricted in reading codons. In fact, chloroplasts use the universal genetic code. Feedback B: Correct! Mitochondria use a variation on the universal genetic code that permits all 64 codon combinations to be read. Feedback C: Incorrect. Peroxisomes do not have a genome. Therefore, peroxisomes cannot read a genetic code. Feedback D: Incorrect. Mitochondria use a nonuniversal genetic code in which some codons are read differently from the way the standard, universal genetic code is read. 4. Human diseases caused by mutations in mitochondrial genomes a. are inherited from both parents. b. are inherited from the father. c. are inherited from the mother. d. do not exist, because the mutation is always complemented by the normal gene copy in the nucleus. Answer: c Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Describe mitochondrial genomes. Feedback A: Incorrect. Most of the mitochondria are inherited from a single parent. Feedback B: Incorrect. Very few of the mitochondria in an individual are derived from the father. Feedback C: Correct! The oocyte contributes many more mitochondria to the fertilized egg than the sperm does, and thus mitochondrial diseases are inherited maternally. Feedback D: Incorrect. Mitochondria have their own genes, which are not present in the nuclear genomes. Thus diseases caused by mutations in mitochondrial genes do exist. 5. Assuming that human mitochondria contain about 1,500 different proteins, approximately what percentage of the mitochondrial proteome is encoded by mitochondrial DNA? a. 0.05% b. 0.1% c. 0.5% d. 1.0% Answer: d Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Describe mitochondrial genomes. Feedback A: Incorrect. Human mitochondria encode 13 polypeptides. Feedback B: Incorrect. Human mitochondria encode 13 polypeptides. Feedback C: Incorrect. Human mitochondria encode 13 polypeptides. Feedback D: Correct! Human mitochondria encode 13 polypeptides (13/1500  100 = 0.86%).

6. Which of the following is not involved in protein transport into mitochondria? a. A positively charged presequence of 15–55 amino acids located at the N-terminus b. The Tom complex c. The proton gradient across the inner mitochondrial membrane d. Vesicular transport Answer: d Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and lipids are imported into mitochondria. Feedback A: Incorrect. This sequence is called a presequence. The first step in protein import is the binding of the presequence to a receptor on the surface of mitochondria. Feedback B: Incorrect. This protein complex mediates the transfer of proteins across the outer mitochondrial membrane. Feedback C: Incorrect. The negatively charged matrix of mitochondria assists in the import of proteins, due to their positively charged presequences. Feedback D: Correct! Vesicular transport mediates many cellular transport steps, but it does not mediate transport into the mitochondria. 7. Which of the following is not a protein translocon/translocase found in the mitochondrial inner or outer membrane? a. Oxa1 b. Tim23 c. Tom40 d. Toc75 Answer: d Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria. Feedback A: Incorrect. Oxa1 is a translocon in the inner mitochondrial membrane. It is involved in protein export from the mitochondrial matrix. Feedback B: Incorrect. Tim23 is a translocon in the inner mitochondrial membrane involved in protein import into the mitochondrial matrix. Feedback C: Incorrect. Tom40 is a translocon in the outer mitochondrial membrane involved in protein import into mitochondria. Feedback D: Correct! Toc75 is a component of the outer chloroplast membrane. 8. Where do phospholipids, such as phosphatidylcholine and phosphatidylethanolamine in mitochondrial membranes, originate? a. In the ER b. In the intermembrane space c. On the cytosolic side of the outer membrane d. On the lumenal side of the inner membrane Answer: a Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Summarize how proteins and lipids are imported into mitochondria.

Feedback A: Correct! Phosphatidylcholine and phosphatidylethanolamine are synthesized in the ER membrane and then transported to the mitochondria via phospholipid transfer proteins. Feedback B: Incorrect. Lipids are synthesized in association with an existing membrane due to their hydrophobicity. Feedback C: Incorrect. The phospholipids are synthesized in association with a membrane, but not the mitochondrial outer membrane. Feedback D: Incorrect. The phospholipids are synthesized in association with a membrane but not the mitochondrial inner membrane. 9. The amino terminal presequence on proteins imported into the mitochondrial inner membrane is a. cleaved by mitochondrial matrix processing peptidase (MPP). b. transported back to the intermembrane space by Hsp70 chaperones. c. cleaved and translocated laterally into the inner membrane. d. cleaved during transit from Tom to Tim23. Answer: a Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Summarize how proteins and lipids are imported into mitochondria. Feedback A: Correct! Feedback B: Incorrect. Hsp70 assists with folding of inner membrane proteins. Feedback C: Incorrect. Some proteins, however, are laterally inserted into the membrane. Feedback D: Incorrect. The presequence is required for transport from Tom to Tim 23. 10. If a drug blocked the activity of the mitochondrial inner membrane proteins Tim9 and Tim10, which of the following would most likely not be found in the outer mitochondrial membrane? a. SAM complex b. -helix membrane proteins c. -barrel proteins d. Tom Answer: c Textbook Reference: Mitochondria Bloom’s Category: 4. Analyzing Learning Objective: Summarize how proteins and lipids are imported into mitochondria. Feedback A: Incorrect. The SAM complex is part of the process that constructs -barrel proteins, but is not dependent on Tim9/10. Feedback B: Incorrect. -helix proteins are inserted in the membrane with assistance from Mim1. Feedback C: Correct! Tim9-Tim10 are chaperones that bring the -barrel polypeptides to the SAM complex, which then processes it into the mature transported in the outer membrane. Without Tim9-Tim10, the polypeptides would not be escorted to SAM. Feedback D: Incorrect. The polypeptide for the -barrel proteins gets into the intermembrane space through Tom prior to interacting with Tim9-Tim10.

11. A deficiency in cardiolipin would lead to a. increased proton flow across the inner mitochondrial membrane. b. decreased proton flow across the inner mitochondrial membrane. c. increased proton flow across the outer mitochondrial membrane. d. decreased proton flow across the outer mitochondrial membrane. Answer: a Textbook Reference: Mitochondria Bloom’s Category: 3. Applying Learning Objective: Summarize how proteins and lipids are imported into mitochondria. Feedback A: Correct! Cardiolipin is located in the inner mitochondrial membrane and restricts proton flow across the membrane, thus helping maintain the energy gradient. If it were deficient, the proton flow would increase. Feedback B: Incorrect. An excess of cardiolipin might decrease proton flow. Feedback C: Incorrect. Cardiolipin is found in the inner membrane. Feedback D: Incorrect. Cardiolipin is found in the inner membrane. 12. In terms of its role in the generation of metabolic energy, the inner membrane in mitochondria is equivalent to which component in chloroplasts? a. The inner membrane b. The thylakoid membrane c. The outer membrane d. The stroma Answer: b Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. Feedback A: Incorrect. The inner membranes of chloroplasts and mitochondria are similar in terms of their permeability to ions and small metabolites, but not in terms of their roles in the generation of energy. Feedback B: Correct! The electron transport system is located in the thylakoid membrane of chloroplasts and thus is analogous to the inner membrane of mitochondria. Feedback C: Incorrect. The outer membrane in chloroplasts does not play a role in the production of metabolic energy and thus is not equivalent to the inner mitochondrial membrane of mitochondria. Feedback D: Incorrect. The stroma is the space between the inner membrane and the thylakoid membrane and is not analogous to the inner mitochondrial membrane in terms of energy production. 13. Chloroplast genomes contain approximately how many genes? a. None b. 40 c. 150 d. 80,000–100,000 Answer: c Textbook Reference: Chloroplasts and Other Plastids

Bloom’s Category: 1. Remembering Learning Objective: Contrast chloroplast and mitochondrial genomes. Feedback A: Incorrect. Although approximately 90% of chloroplast proteins are encoded by nuclear genes, chloroplasts do possess their own genes. Feedback B: Incorrect. This is the approximate number of genes in human mitochondria. Feedback C: Correct! Chloroplast genes encode both protein and RNA molecules that are involved in a variety of functions. Feedback D: Incorrect. This is the approximate number of genes encoded in human nuclei. 14. _______ different translocon systems are used for protein import from the chloroplast stroma into the thylakoid lumen or membrane. a. Two b. Three c. Five d. Ten Answer: b Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Summarize the mechanisms of protein import into chloroplasts. Feedback A: Incorrect. More different translocon systems are used. Feedback B: Correct! Feedback C: Incorrect. Fewer different translocon systems are used. Feedback D: Incorrect. Fewer different translocon systems are used. 15. The carotenoids, which give many plants their yellow, orange, and red colors, are located in a. vacuoles. b. etioplasts. c. leucoplasts. d. chromoplasts. Answer: d Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of other plastids. Feedback A: Incorrect. These are where some pigments are found, but they contain no carotenoids. Feedback B: Incorrect. These represent an intermediate stage in the development of plastids. Feedback C: Incorrect. Leucoplasts are nonpigmented plastids. Feedback D: Correct! Chromoplasts are the sites of carotenoid storage; their function, however, is still unclear. 16. Which of the following plastids is a precursor to all other plastids? a. Chromoplast b. Proplastid

c. Etioplast d. Amyloplast Answer: b Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of other plastids. Feedback A: Incorrect. Chromoplasts can convert to chloroplasts but not others. Feedback B: Correct! Proplastids are precursors to all plastids. Feedback C: Incorrect. Etioplasts can convert to chloroplasts but not others. Feedback D: Incorrect. Amyloplasts can convert to chloroplasts and chromoplasts but not others. 17. Which factor most likely triggers lipid-containing etioplasts to develop into chloroplasts? a. Exposure to light b. Hormonal signaling c. Exposure to darkness d. Internal pigments Answer: a Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 4. Analyzing Learning Objective: Describe the roles of other plastids. Feedback A: Correct! Environmental factors and intrinsic cell differentiation programs control the conversion, and since chloroplasts use light, this is a likely trigger. Feedback B: Incorrect. Hormones are not part of the signaling necessary for chloroplast development. Feedback C: Incorrect. Darkness maintains the etioplast phenotype. Feedback D: Incorrect. Etioplasts contain no pigment. 18. Peroxisomes are not involved in a. the biosynthesis of plasmalogens. b. the biosynthesis of the enzyme catalase, which breaks down hydrogen peroxide. c. the biosynthesis of lipids. d. oxidative reactions leading to the production of hydrogen peroxide. Answer: b Textbook Reference: Peroxisomes Bloom’s Category: 2. Understanding Learning Objective: Summarize the roles of peroxisomes in animal and plant cells. Feedback A: Incorrect. Peroxisomes are involved in the synthesis of plasmalogens. Feedback B: Correct! Peroxisomes have no genome. Therefore, they cannot synthesize a polypeptide. Feedback C: Incorrect. Peroxisomes are involved in the biosynthesis of lipids, especially plasmalogens. Feedback D: Incorrect. Peroxisomes are involved in oxidative reactions, and hydrogen peroxide is produced in peroxisomes.

19. Zellweger syndrome is caused by a defect in a. mitochondrial protein import. b. the electron transport system in mitochondria. c. protein import into peroxisomes. d. the synthesis of peroxisomal proteins. Answer: c Textbook Reference: Peroxisomes Bloom’s Category: 1. Remembering Learning Objective: Summarize the roles of peroxisomes in animal and plant cells. Feedback A: Incorrect. Mitochondrial protein import is not affected in individuals with Zellweger syndrome. Feedback B: Incorrect. Zellweger syndrome has nothing to do with the electron transport system in mitochondria. Feedback C: Correct! Zellweger syndrome can be caused by mutations in a number of different proteins involved in peroxisome protein import and is lethal within the first ten years of life. Feedback D: Incorrect. Synthesis of peroxisomal proteins is not affected in Zellweger syndrome.

Essay 1. What are porins? Answer: Porins are transmembrane proteins that form large pores. They are found in chloroplast and mitochondrial outer membranes. They allow the passage of molecules smaller than 1,000 daltons, and thus the space between the inner and outer membranes is equivalent to the cytosol in its concentration of ions and small molecules. Textbook Reference: Mitochondria Bloom’s Category: 1. Remembering Learning Objective: Illustrate the functional organization of mitochondria. 2. Mitochondrial mRNAs have short poly-A sequences at their 3′ end. Poly-A tails are generally considered to be a feature of eukaryotic, not bacterial, mRNA. How can this observation be reconciled with an endosymbiotic origin for mitochondria? Answer: The mitochondria of today are not the mitochondria originally introduced via endosymbiosis. Present-day mitochondria have evolved from their original status of newly introduced endosymbionts. The genome of present-day mitochondria exhibits a mix of bacterial and eukaryotic traits, which can explain this eukaryotic-like trait of mitochondrial mRNA. The presumption is that the development of poly-A tails originated after the endosymbiotic introduction of mitochondria into eukaryotic cells. Textbook Reference: Mitochondria Bloom’s Category: 3. Applying Learning Objective: Describe mitochondrial genomes. 3. At low pH, the chemical 2,4-dinitrophenol (DNP) is neutral and can diffuse freely across membranes, including those of mitochondria. At high pH, it gives off a proton,

becomes negatively charged, and can no longer diffuse across membranes. What effect would DNP have on the proton gradient between the mitochondrial intermembrane space and matrix? Answer: DNP would dissipate the proton gradient across the inner membrane. DNP in the intermembrane space would be neutral in the presence of the high concentration of protons and thus pass freely into the mitochondrial lumen, where it would encounter a high pH environment and release its proton. This would neutralize the proton gradient and drastically decrease the overall ATP production of the cell. Textbook Reference: Mitochondria Bloom’s Category: 3. Applying Learning Objective: Explain the role of the proton gradient in transport across the mitochondrial membrane. 4. Explain the role of endosymbiosis in the evolution of mitochondria and chloroplasts. Answer: Because of the prokaryotic nature of the organelle and the striking similarity between the genomes of mitochondria and some bacteria (most notably Rickettsia prowazekii), it has been hypothesized that mitochondria evolved from an endocytic event in which a bacterium was endocytosed by a eukaryotic cell. The eukaryotic cell would have provided the bacterium with protection from the outside world, and the eukaryotic cell would have benefited from the bacterium’s oxidative phosphorylation system for energy production. The evolution of chloroplasts is hypothesized to have occurred in much the same way but at a somewhat later date. Textbook Reference: Mitochondria Bloom’s Category: 2. Understanding Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 5. Explain the functions of mitochondrial matrix processing peptidase (MPP) and chloroplast stromal processing peptidase (SPP). Answer: Peptidases cleave peptide bonds in polypeptides and proteins. In both mitochondrial and chloroplast transport, there are N-terminal amino acid sequences that target polypeptides and proteins to the Tom and Toc complexes in the mitochondria and chloroplast outer membranes and then to the Tim and Tic complexes of the inner membranes. In mitochondria, the targets are the 15–55 amino acid N-terminal presequences, and in chloroplasts, the targets are the 30–100 amino acid N-terminal transit peptides. Once in the inner membrane, MMP cleaves the presequence in mitochondria, and SPP cleaves the transit peptide in chloroplasts. Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 3. Applying Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 6. Consider two unlabeled centrifuge tubes, each with a pellet at the bottom. One tube contains a pellet of mitochondria and the other a pellet of chloroplasts. How could the pellets be identified? Answer: Chloroplasts are full of chlorophyll (green) and would produce a green pellet;

mitochondria are colorless and would, if pure, produce a white pellet. Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 2. Understanding Learning Objective: Compare the structural and functional organization of chloroplasts with mitochondria. 7. Presumably, the original chloroplast genome coded for many more proteins than it does today. Some of these genes must have been transferred to the nucleus, with the resulting protein product needing to be imported into chloroplasts. Other genes must have originated in the nucleus and were then adapted for import into chloroplasts. What were the likely protein adaptations that must have occurred for protein import into chloroplasts, and how might one determine whether the protein originated from a chloroplast or a nuclear-coded gene? Answer: In general, a nuclearly coded and cytosolically synthesized protein must have an N-terminal transit peptide that can be recognized by the chloroplast guidance complex. If it does not, the polypeptide will not be guided toward the translocon located in the outer chloroplast membrane. This sequence would need to be added by some kind of gene fusion. Machinery for recognition and translocation would need to be adapted in the outer chloroplast membrane. Homology comparisons could be done to determine whether the protein is more similar to a bacterial or a eukaryotic protein. If it is more similar to a bacterial protein, it presumably originated in the chloroplast, regardless of where the coding sequence resides today. Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 3. Applying Learning Objective: Summarize the mechanisms of protein import into chloroplasts. 8. All plant plastids contain the same genome as chloroplasts. However, chromoplasts, amyloplasts, and elaioplasts are clearly different from one another. What mechanism might explain the differences between these plastids that all have the same internal genes? Answer: Plastid development is under the coordinate control of genes within the plastid and the nuclear genomes. Amyloplasts and elaioplasts are forms of leucoplasts. Leucoplasts are found in nonphotosynthetic tissues. Nonphotosynthetic tissues (e.g., roots) often have little, if any, exposure to sunlight, so sunlight may be a cue for chloroplast gene activation. However, in the case of chromoplasts in flowers, sunlight as a cue for gene expression does not provide an obvious explanation. Therefore, there must also be other environmental cues regulating chloroplast gene expression. Textbook Reference: Chloroplasts and Other Plastids Bloom’s Category: 3. Applying Learning Objective: Describe the roles of other plastids. 9. Zellweger syndrome is caused by defects in genes coding for peroxisomal protein import. Why are defects in such genes more likely to be lethal than a defect in a gene encoding a single enzyme present in the peroxisomal lumen? Answer: A defect in import machinery affects the import of several proteins into the lumen of the peroxisome. A mutation in a single peroxisomal enzyme affects only that

enzyme. Therefore, the import machinery mutation would have a greater effect and probability of being lethal. Textbook Reference: Peroxisomes Bloom’s Category: 3. Applying Learning Objective: Describe the pathways responsible for peroxisome biogenesis.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 14: The Cytoskeleton and Cell Movement TEST FILE QUESTIONS Multiple Choice 1. Actin filaments are approximately _______ in diameter. a. 5 Å b. 7 nm c. 11 nm d. 25 nm Answer: b Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 2. Actin exists in cells in two major forms called a. monomers and dimers. b. -actin and -actin. c. G actin and D actin. d. G actin and F actin. Answer: d Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 3. ATP is hydrolyzed by actin a. in the process of assembly into a filament. b. after assembly but before disassembly. c. in the process of disassociation. d. after disassociation from the filament. Answer: b Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 2. Understanding Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 4. Each monomer of actin binds one molecule of the nucleotide triphosphate

a. ATP. b. GTP. c. CTP. d. UTP. Answer: a Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 5. The phenomenon that illustrates the dynamic behavior of actin filaments and is critical to regulating the structure and function of actin filaments is known as a. assembly. b. dynamic instability. c. treadmilling. d. disassembly. Answer: c Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 6. Actin filaments are stabilized by a. cofilin. b. gelsolin. c. thymosin. d. tropomyosin. Answer: d Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 7. Branching of actin filaments can be initiated by a. Arp2/3. b. formin. c. ADF/cofilin. d. fimbrin. Answer: a Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 8. Actin filaments are bound into bundles of parallel filaments by the proteins a. filamin and spectrin.

b. troponin and tropomyosin. c. profilin and thymosin. d. -actinin and fimbrin. Answer: d Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 9. Short actin filaments bind to tetramers of which protein to form the cytoskeleton of erythrocytes? a. Myosin b. -actinin c. Spectrin d. Ankyrin Answer: c Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 10. Which movement is not based on actin–myosin interactions? a. Cell migration (crawling) over surfaces b. Chromosome movement during anaphase A c. Cytokinesis of animal cells d. Phagocytosis Answer: b Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 2. Understanding Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 11. In actin filament assembly, the process in which ATP-actin monomers are added to the barbed end of the filament while ADP-actin monomers are concurrently dissociating from the pointed end of the filament is referred to as a. equilibrium. b. dynamic instability. c. treadmilling. d. recycling. Answer: c Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 12. Duchenne’s muscular dystrophy is characterized by

a. X-chromosomal inheritance. b. childhood onset. c. abnormal dystrophin function. d. All of the above Answer: d Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Illustrate the organization of actin filaments underlying the plasma membrane. 13. Actin filaments are anchored at junctions called a. adherens junctions. b. tight junctions. c. desmosomes. d. gap junctions. Answer: a Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Illustrate the organization of actin filaments underlying the plasma membrane. 14. During muscle contraction, the A band a. stays the same width, and the I bands and H zone shorten. b. and H zone stay the same width, and the I bands shorten. c. shortens, and the I bands and H zone stay the same. d. and H zone shorten, and the I bands stay the same. Answer: a Textbook Reference: Myosin Motors Bloom’s Category: 2. Understanding Learning Objective: Explain the molecular basis of muscle contraction. 15. The barbed (fast growing) end of actin filaments is located in muscle a. at the A/I junction. b. near the M line of a contracted sarcomere. c. at the inner margin of the A/I zone. d. at the Z disc. Answer: d Textbook Reference: Myosin Motors Bloom’s Category: 1. Remembering Learning Objective: Explain the molecular basis of muscle contraction. 16. Myosin _______ is present in muscle sarcomeres. a. I b. II c. III d. IV

Answer: b Textbook Reference: Myosin Motors Bloom’s Category: 1. Remembering Learning Objective: Explain the molecular basis of muscle contraction. 17. Microtubules are typically _______ in diameter. a. 7 nm b. 10–12 nm c. 25 nm d. 35 nm Answer: c Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and dynamic instability of microtubules. 18. Microtubules are assembled from a. -tubulin dimers. b. -tubulin dimers. c. alternating -tubulin dimers and -tubulin dimers. d. dimers of - and -tubulin. Answer: d Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and dynamic instability of microtubules. 19. Which nucleotide triphosphate is hydrolyzed during a cycle of microtubule assembly and disassembly? a. ATP b. TTP c. CTP d. GTP Answer: d Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and dynamic instability of microtubules. 20. The GTP bound to -tubulin hydrolyzes to GDP and Pi a. during depolymerization of the – dimer. b. during polymerization of dimers onto microtubules. c. during depolymerization of dimers from microtubules. d. following polymerization but before depolymerization. Answer: d Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and dynamic instability of microtubules.

21. The microtubule behavior in which individual microtubules alternate between cycles of growth and shrinkage is called a. an equilibrium state. b. dynamic instability. c. treadmilling. d. recycling. Answer: b Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and dynamic instability of microtubules. 22. Both colchicine and colcemid a. block microtubule organizing centers. b. block microtubule disassembly by binding to microtubule ends. c. block microtubule assembly by binding to free tubulin. d. accelerate microtubule disassembly by binding to tubulin in microtubules, causing those molecules to exit the microtubule more quickly. Answer: c Textbook Reference: Microtubules Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and dynamic instability of microtubules. 23. The anticancer drug taxol a. blocks microtubule organizing centers. b. stabilizes microtubules and thus inhibits disassembly. c. blocks microtubule assembly by binding to free tubulin. d. accelerates microtubule disassembly by binding to tubulin in microtubules, causing those molecules to exit the microtubule more quickly. Answer: b Textbook Reference: Microtubules Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and dynamic instability of microtubules. 24. Microtubules are not involved in a. movement of chromosomes during mitosis. b. transport of membranous vesicles in the cytoplasm. c. cytokinesis of animal cells. d. movement of cilia and flagella. Answer: c Textbook Reference: Microtubules Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and dynamic instability of microtubules. 25. The major microtubule-organizing center in most animal cells is the a. kinetochore. b. nucleus.

c. centrosome. d. centromere. Answer: c Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Summarize how the growth of microtubules is initiated within cells. 26. The role of the centrosome is to a. determine the center of the cell. b. determine the position of the nucleus. c. initiate microtubule growth. d. adhere to the plus ends of microtubules. Answer: c Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Summarize how the growth of microtubules is initiated within cells. 27. Rings of the protein _______ in the pericentriolar material nucleate microtubule assembly. a. centrin b. pericentrin c. -tubulin d.  -tubulin Answer: d Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Summarize how the growth of microtubules is initiated within cells. 28. At the end of interphase, the part of the microtubule farthest from the centrosome is the _______ end. a. capped b. barbed c. minus d. plus Answer: d Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Summarize how the growth of microtubules is initiated within cells. 29. Kinesin I is a motor protein molecule consisting of a. two heavy chains. b. one heavy chain and two light chains. c. two heavy chains and two light chains. d. two heavy chains and four light chains. Answer: c Textbook Reference: Microtubule Motors and Movement

Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of kinesins and dyneins. 30. The cargo carried by kinesin along microtubules binds to kinesin on which region? a. Head b. Neck c. Coiled-coil d. Tail Answer: d Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of kinesins and dyneins. 31. Kinesins are able to transport _______ along microtubules. a. membranous vesicles b. endoplasmic reticulum c. mitochondria d. All of the above Answer: d Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 2. Understanding Learning Objective: Explain how organelles and other cargo are transported on microtubules. 32. Cytoplasmic dynein plays a key role in the positioning of which organelle? a. Nucleus b. Peroxisome c. Mitochondrion d. Golgi apparatus Answer: d Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 2. Understanding Learning Objective: Explain how organelles and other cargo are transported on microtubules. 33. A male patient at a medical clinic presents with infertility due to nonmotile sperm and an inability to clear mucous from his respiratory tract. Other tissues are normal. You suspect that these symptoms may be caused by mutant a. tubulin. b. kinesin. c. dynein. d. tau protein. Answer: c Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 3. Applying Learning Objective: Contrast the structures and functions of primary and motile cilia.

34. In a motile cilium or flagellum, _______ microtubules are arranged _______. a. 13; in a circle b. 9 triplet; in a circle c. 9 doublet; in a circle around a central pair of microtubules d. 2 microtubules; perpendicular to each other Answer: c Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 2. Understanding Learning Objective: Contrast the structures and functions of primary and motile cilia. 35. The basal bodies of cilia and flagella are similar in structure to (and can form from) a. centromeres. b. kinetomeres. c. kinetochores. d. centrioles. Answer: d Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering Learning Objective: Contrast the structures and functions of primary and motile cilia. 36. Adjacent microtubule doublets in cilia and flagella produce a bending movement because a. tubulin is contracting on one side of the microtubules. b. dynein is contracting on one side of the microtubules. c. kinesin is contracting on one side of the microtubules. d. nexin links between microtubule doublets convert a sliding movement into a bending movement. Answer: d Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 2. Understanding Learning Objective: Contrast the structures and functions of primary and motile cilia. 37. The beating of cilia and flagella occurs by means of _______-based _______. a. dynein; microtubule sliding b. kinesin; microtubule sliding c. myosin; microfilament sliding d. tubulin; microtubule contraction Answer: a Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 2. Understanding Learning Objective: Contrast the structures and functions of primary and motile cilia. 38. The microtubules that overlap in the center of the mitotic spindle are called _______ microtubules. a. astral

b. minus-end c. kinetochore d. interpolar Answer: d Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering Learning Objective: Explain how different types of microtubules act during mitosis. 39. The drug taxol stabilizes microtubules so they cannot shorten. If taxol were added during anaphase of mitosis, what effect would you expect it to have on anaphase movements? a. It would stop all movements. b. It would stop anaphase A but not anaphase B. c. It would stop anaphase B but not anaphase A. d. It would have no effect. Answer: b Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 3. Applying Learning Objective: Explain how different types of microtubules act during mitosis. 40. The intermediate filaments in the nucleus are made of a. keratins. b. lamins. c. desmin. d. vimentin. Answer: b Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the types of intermediate filament proteins. 41. The desmin filaments in muscle cells connect a. actin filaments to the Z line. b. actin filaments to the plasma membrane at the ends of myofibrils. c. Z lines of adjacent myofibrils. d. myosin filaments to the Z line. Answer: c Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the types of intermediate filament proteins. 42. Keratin filaments are found in which of the following cell types? a. Fibroblasts b. Adipocytes c. Muscle cells d. Epithelial cells Answer: d

Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the types of intermediate filament proteins. 43. Vimentin is the major intermediate filament protein of _______ cells. a. epithelial b. striated muscle c. nerve d. fibroblast Answer: d Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the types of intermediate filament proteins. 44. Which of the following is not an intermediate filament protein? a. Vimentin b. Lamin c. Desmin d. Fibronectin Answer: d Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the types of intermediate filament proteins. 45. Intermediate filaments are typically _______ in diameter. a. 5–7 nm b. 10–12 nm c. 16–22 nm d. 24–26 nm Answer: b Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of intermediate filaments. 46. Intermediate filaments function in a. cell motility. b. providing mechanical strength for cells. c. nuclear pore structure. d. All of the above Answer: b Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Describe the organization and function of intermediate filaments within cells. 47. Keratin filaments are anchored to junctions called

a. adherens junctions. b. tight junctions. c. desmosomes. d. gap junctions. Answer: c Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Describe the organization and function of intermediate filaments within cells. 48. Which protein can link intermediate filaments with actin filaments and microtubules? a. -actinin b. -catenin c. Integrin d. Plectin Answer: d Textbook Reference: Intermediate Filaments Bloom’s Category: 2. Understanding Learning Objective: Describe the organization and function of intermediate filaments within cells. 49. Expression of a shortened skin keratin gene in place of the normal keratin gene in transgenic mice results in a phenotype in which mice have a. thick skin. b. no hair. c. fragile, easily blistered skin. d. white hair. Answer: c Textbook Reference: Intermediate Filaments Bloom’s Category: 2. Understanding Learning Objective: Describe the organization and function of intermediate filaments within cells.

Fill in the Blank 1. The complex network of protein filaments and microtubules that extends throughout the cytoplasm is called the _______. Answer: cytoskeleton Textbook Reference: Introduction Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 2. The three types of filaments that provide the basis of cell shape and cell motility in eukaryotes are actin filaments, intermediate filaments, and _______.

Answer: microtubules Textbook Reference: Introduction Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 3. Most myosins move along actin filaments toward the _______ end. Answer: plus Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 4. When a striated muscle is stimulated to contract, calcium is released primarily from the _______ and interacts with a group of three thin-filament proteins called _______, which in turn cause the long, thin protein _______ to move away from the cross-bridge binding sites on the _______ molecules. This allows the cross-bridge cycle to occur. Answer: sarcoplasmic reticulum; troponins; tropomyosin; actin Textbook Reference: Myosin Motors Bloom’s Category: 2. Understanding Learning Objective: Explain the molecular basis of muscle contraction. 5. The initial role of ATP in the cross-bridge cycle is to bind to _______ and cause it to _______. Answer: myosin; release from actin Textbook Reference: Myosin Motors Bloom’s Category: 2. Understanding Learning Objective: Explain the molecular basis of muscle contraction. 6. The motor protein that moves cargo toward the minus ends of microtubules is _______. Answer: dynein Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of kinesins and dyneins. 7. The motor protein that moves cargo toward the plus ends of microtubules is _______. Answer: kinesin Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of kinesins and dyneins. 8. Dynein and kinesin are similar in that they consist of heavy chains with globular heads that bind both _______ and _______. Answer: microtubules; ATP Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering

Learning Objective: Summarize the properties of kinesins and dyneins. 9. The intermediate filament protein _______ is the major protein of the human skin, hair, and fingernails. The most prominent secondary structure in this protein is the _______. The monomers of this protein associate first to form dimers and then are assembled into protofilaments, _______ (number) of which associate laterally to form an intermediate filament. Answer: keratin; α helix; eight (8) Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Diagram the structure of intermediate filaments.

True/False 1. In cells, actin filaments form a network that provides mechanical support, determines cell shape, and allows for movement of the cell surface. Answer: T Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 2. Actin exists in two forms, a globular G form and a filamentous F form. Answer: T Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 3. Formation of actin filaments requires energy in the form of ATP. Answer: T Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 4. Actin may be cross-linked into antiparallel bundles. Answer: F Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 5. In muscle, every myofibril is organized as a chain of contractile units called myocytes. Answer: F

Textbook Reference: Myosin Motors Bloom’s Category: 2. Understanding Learning Objective: Explain the molecular basis of muscle contraction. 6. Microtubules are approximately 1/3 the diameter of actin filaments. Answer: F Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and dynamic instability of microtubules. 7. The cycle of alternating growth and shrinkage of microtubules is referred to as dynamic instability, or rescue and catastrophe. Answer: T Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and dynamic instability of microtubules. 8. Kinesins and dyneins are molecular motor proteins. Answer: T Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of kinesins and dyneins. 9. Epidermolysis bullosa simplex is a skin disease caused by a mutation in the gene encoding keratin. Answer: T Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the types of intermediate filament proteins. 10. Desmosomes are specialized junctions that help hold cells together, while their counterpart hemidesmosomes prevent cell-to-cell contact. Answer: F Textbook Reference: Intermediate Filaments Bloom’s Category: 2. Understanding Learning Objective: Describe the organization and function of intermediate filaments within cells.

Short Answer 1. What is the normal function of dystrophin in muscle cells? Answer: To attach actin filaments to transmembrane proteins of the muscle cell plasma membrane Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering

Learning Objective: Illustrate the organization of actin filaments underlying the plasma membrane. 2. Nebulin filaments attach to actin filaments in muscle for what purpose? Answer: To organize and stabilize the actin filaments Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Explain the molecular basis of muscle contraction. 3. Describe the myosin II molecule and its component parts. Answer: Myosin II is composed of two heavy chains and two pairs of light chains (two regulatory light chains and two essential light chains). The heavy chains interact in their -helical regions as a coil and have two globular head regions. Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Explain the molecular basis of muscle contraction. 4. What is a sarcomere? (Include the structures that bound it in your definition.) Answer: A sarcomere is one contractile unit of a myofibril (or muscle fibril, but not muscle fiber) that runs from Z disc to Z disc and includes the thick and thin filament sets in between. Textbook Reference: Myosin Motors Bloom’s Category: 1. Remembering Learning Objective: Explain the molecular basis of muscle contraction. 5. How is contraction regulated in smooth muscle? Answer: An increase in cytosolic calcium promotes the binding of calmodulin to myosin light-chain kinase, which phosphorylates myosin light chains, allowing myosin to assemble into filaments and interact with actin filaments to produce contraction. Textbook Reference: Myosin Motors Bloom’s Category: 2. Understanding Learning Objective: Summarize the roles of contractile actin–myosin filaments in nonmuscle cells. 6. How are the plus and minus ends of microtubules arranged in the dendrites of neurons? Answer: Microtubules are oriented in both directions—some with their plus ends extending toward the nucleus and some with their plus ends pointing toward the end of the dendrite. Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Describe the structure and dynamic instability of microtubules. 7. What is the polarity of microtubule assembly in an axon of a nerve? Answer: The plus (+) ends are located outward (distal) near the synaptic terminal. Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering

Learning Objective: Summarize how the growth of microtubules is initiated within cells. 8. What MAP is the main component of the lesions found in the brains of Alzheimer’s patients? Answer: Tau Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Explain how microtubule-associated proteins regulate the organization of microtubules. 9. In which directions do kinesin and dynein transport vesicles in an axon? Answer: Dynein moves vesicles inward (toward the cell bodies), and kinesin moves vesicles outward (toward the synapse). Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of kinesins and dyneins. 10. Describe the two separate mechanisms by which the poles of the mitotic spindle move apart in anaphase B. Answer: Plus-end-directed motor proteins (kinesin-like proteins) slide overlapping interpolar microtubules apart. Minus-end-directed motor proteins (cytoplasmic dyneins) pull the spindle poles apart, toward the periphery of the cell. Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 2. Understanding Learning Objective: Diagram the mitotic spindle 11. Intermediate filament proteins vary in size but share structural features. Describe these shared features. Answer: They all have a long, central -helical rod domain flanked by head and tail domains of variable size and structure. Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the types of intermediate filament proteins.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. Which of the following is not one of the functions of the cytoskeleton? a. To provide a structural framework for the cell b. Cell locomotion c. Protein translocation into the ER d. Intracellular movement of organelles and other structures Answer: c Textbook Reference: Introduction

Bloom’s Category: 2. Understanding Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. Feedback A: Incorrect. The cytoskeleton gives the cell its shape and contributes to the organization of the cytoplasm. Feedback B: Incorrect. The cytoskeleton is involved in cell motility, such as the movement of a macrophage toward a bacterium. Feedback C: Correct! This process is carried out by the SRP and its receptor; it is not mediated by the cytoskeleton. Feedback D: Incorrect. Intracellular structures can use cytoskeletal components to move; for example, chromosomes move toward opposite sides of the cell during mitosis on a microtubular spindle. 2. The approximate diameter of an actin filament is a. 7 nm. b. 10–12 nm. c. 25 nm. d. 5 mm. Answer: a Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. Feedback A: Correct! The actin filament is the smallest type of filament in the cytoskeleton, hence its other name, the microfilament. Feedback B: Incorrect. This is the diameter of an intermediate filament. Feedback C: Incorrect. This is the size of a microtubule. Feedback D: Incorrect. This is the approximate diameter of a eukaryotic nucleus. 3. Which of the following is not true of the assembly of actin filaments? a. It begins with the formation of an aggregate of three actin monomers. b. It requires ATP. c. Polymerization occurs from both the plus and minus ends. d. Polymerization is faster from the plus end than from the minus end. Answer: b Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 2. Understanding Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. Feedback A: Incorrect. This is the first step in filament formation and is called “nucleation.” Feedback B: Correct! Although polymerization occurs more readily in the presence of ATP, it is not absolutely required. Feedback C: Incorrect. Polymerization can occur from both ends of the growing filament. Feedback D: Incorrect. Elongation from the plus end is 5 to 10 times faster than from the minus end.

4. Which of the following is not a function of actin-binding proteins? a. Filament initiation and polymerization b. End capping c. Filament severing/depolymerization d. Incorporation of microfilaments into the extracellular matrix Answer: d Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 2. Understanding Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. Feedback A: Incorrect. Arp2/3 and formin are actin-binding proteins that are involved in filament initiation and polymerization. Formin is involved in nucleation of the actin trimer, and Arp2/3 is involved in the nucleation of actin filament branching. Feedback B: Incorrect. Actin-binding proteins are involved in end-capping actin filaments. Feedback C: Incorrect. Actin-binding proteins are involved in filament severing/depolymerization. Feedback D: Correct! Actin microfilaments are intracellular protein complexes. Actin filaments may be linked through a sequence of adaptor proteins to the extracellular matrix. However, they are not incorporated into the extracellular matrix. 5. Cofilin plays a role in the a. disassembly of actin filaments. b. stimulation of actin filament formation. c. nucleation of microfilaments. d. disassembly of microtubules. Answer: a Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. Feedback A: Correct! Cofilin contributes to the disassembly of actin filaments by severing them and by enhancing the rate of dissociation of actin monomers from the minus end. Feedback B: Incorrect. Profilin stimulates actin filament formation. Feedback C: Incorrect. The Arp2/3 proteins are involved in nucleation. Feedback D: Incorrect. Cofilin is involved in the disassembly of a cytoskeletal component but not that of microtubules. 6. Which cytosolic protein in red blood cells is the link between the plasma membrane and the spectrin/actin network beneath the cell surface? a. Band 3 b. Glycophorin c. Dystrophin d. Ankyrin

Answer: d Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 2. Understanding Learning Objective: Illustrate the organization of actin filaments underlying the plasma membrane. Feedback A: Incorrect. Band 3 is the plasma membrane protein to which the connecting protein binds. Feedback B: Incorrect. Although it is part of the link between the plasma membrane and the actin network, glycophorin is a membrane protein. Feedback C: Incorrect. Dystrophin is a spectrin-like protein that functions in muscle cells. Feedback D: Correct! Ankyrin binds both spectrin and the cytoplasmic portion of band 3, a plasma membrane protein, thus linking the actin network to the plasma membrane. 7. Which of the following does not mediate the association between actin filaments and the plasma membrane? a. Catenin−cadherin links b. Myosin I−calmodulin links c. Talin−integrin links d. Direct interaction between actin and the plasma membrane Answer: d Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 2. Understanding Learning Objective: Illustrate the organization of actin filaments underlying the plasma membrane. Feedback A: Incorrect. This is exactly the case in adherens junctions. interacts with various linkers, adaptors, or motor proteins. Feedback B: Incorrect. This is exactly the case in microvilli. Feedback C: Incorrect. This is exactly the case in focal adhesions. Feedback D: Correct! Actin does not interact directly with the plasma membrane. Rather it 8. Which statement about myosin I is true? a. It is involved in muscle contraction. b. It has a long -helical tail through which it forms homodimers. c. It does not act as a molecular motor. d. It links the actin bundles to the plasma membrane in the microvilli of intestinal cells. Answer: d Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Illustrate the organization of actin filaments underlying the plasma membrane. Feedback A: Incorrect. Myosin II is involved in muscle contraction. Feedback B: Incorrect. Myosin I has a short -helical tail and does not form dimers. Feedback C: Incorrect. Myosin I contains a globular head group which, like myosin II, acts as a molecular motor. Feedback D: Correct! Myosin I is thought to move the plasma membrane along toward

the tip of the microvillus. 9. In cell movement, branched actin filament growth pushes against the cell membrane. Proteins involved in this process include all of the following except a. Arp2/3. b. profilin. c. vinculin. d. WASP proteins. Answer: c Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 1. Remembering Learning Objective: Explain how remodeling of actin filaments is responsible for cell motility. Feedback A: Incorrect. Arp2/3 is involved in initiating branched actin filament formation. Feedback B: Incorrect. Profilin is an actin escort protein in the polymerization of branched actin filaments. Feedback C: Correct! Vinculin is involved in linking actin to focal adhesions. Feedback D: Incorrect. The WASP family of proteins stimulate Arp2/3 in initiating branched actin filament formation. 10. The basis for muscle contraction is the a. rotation of myosin fibers around actin fibers. b. expansion of the sarcomere. c. sliding of myosin and actin fibers past each other. d. movement of the Z discs away from each other. Answer: c Textbook Reference: Myosin Motors Bloom’s Category: 1. Remembering Learning Objective: Explain the molecular basis of muscle contraction. Feedback A: Incorrect. Both myosin and actin fibers are involved in muscle contraction, but they do not rotate around each other. Feedback B: Incorrect. The sarcomere shrinks during muscle contraction. Feedback C: Correct! Muscle contraction involves the movement of myosin fibers along the actin fibers in a sarcomere, thus shortening them and bringing the Z discs closer together. Feedback D: Incorrect. The Z discs are brought closer together during muscle contraction. 11. The major cation responsible for regulating actin-myosin contraction is a. Ca2+. b. H +. c. K+. d. Na +. Answer: a Textbook Reference: Myosin Motors Bloom’s Category: 1. Remembering Learning Objective: Explain the molecular basis of muscle contraction.

Feedback A: Correct! An increase in Ca 2+ ion concentration in the cytosol is the signal for actin-myosin contraction. Elevated Ca 2+ results in Ca2+ binding to either troponin or calmodulin. Feedback B: Incorrect. H+ levels are not coupled to actin-myosin contraction. Feedback C: Incorrect. K+ levels are not coupled to actin-myosin contraction. Feedback D: Incorrect. Na+ levels are not coupled to actin-myosin contraction. 12. Myosin II is found in skeletal muscle and the contractile ring. It is a two-headed myosin with tails that can form thick filaments. Other myosins, such as myosin I or myosin V, do not form thick filaments and yet are still capable of producing movement along actin filaments. To what do the tail(s) of myosin I and myosin V bind? a. Cargo such as membrane vesicles or intermediate filaments b. Centrosomes c. Chromosomes d. Tubulin Answer: a Textbook Reference: Myosin Motors Bloom’s Category: 1. Remembering Learning Objective: Describe the functions of unconventional myosins. Feedback A: Correct! The tail(s) bind to cargo. Feedback B: Incorrect. Myosins do not bind to centrosomes. Feedback C: Incorrect. Myosins do not bind to chromosomes; dynein and kinesin motors do. Feedback D: Incorrect. Myosins do not bind to tubulin; dynein and kinesin motors do. 13. Whether a microtubule shrinks or grows is determined by the a. rate of GTP-bound tubulin addition relative to the rate of tubulin GTP hydrolysis. b. phosphorylation state of -tubulin. c. rate of ATP hydrolysis relative to the rate of ATP-bound tubulin addition. d. presence or absence of  -tubulin. Answer: a Textbook Reference: Microtubules Bloom’s Category: 2. Understanding Learning Objective: Describe the structure and dynamic instability of microtubules. Feedback A: Correct! The hydrolysis of GTP by -tubulin results in depolymerization; polymerization occurs via the addition of GTP-bound tubulin. Feedback B: Incorrect. The assembly of intermediate filaments can be regulated by phosphorylation of intermediate filament proteins, but a different mechanism is at work in microtubules. Feedback C: Incorrect. Polymerization of actin filaments, not microtubules, involves this kind of mechanism. Feedback D: Incorrect. -tubulin is a special kind of tubulin involved in the nucleation of microtubules in centrosomes. 14. Kinesin and dynein are a. intermediate filament proteins.

b. microtubule motor proteins. c. proteins within centrosomes that mediate nucleation of microtubules. d. microfilament motor proteins. Answer: b Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of kinesins and dyneins. Feedback A: Incorrect. Intermediate filaments are made up of a variety of different proteins, such as keratin and desmin, but kinesin and dynein are not among them. Feedback B: Correct! Kinesin moves vesicles and organelles toward the plus end of microtubules, and dynein moves them toward the minus end. Feedback C: Incorrect. The key protein in centrosomes that mediates microtubule nucleation is  -tubulin. Feedback D: Incorrect. Myosins are the major microfilament motor proteins. 15. Like myosins, kinesins and dyneins are both families of proteins. Which statement is true of all kinesins and dyneins? a. They are microtubule-dependent motors. b. They are minus-end-directed motors. c. The motor activity of the proteins resides in their light chains. d. They are plus-end-directed motors. Answer: a Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of kinesins and dyneins. Feedback A: Correct! Kinesins and dyneins are all microtubule-dependent motors. Feedback B: Incorrect. Dyneins tend to be minus-end-directed motors while kinesins tend to be plus-end-directed motors. Feedback C: Incorrect. The motor activity resides in the head domains of the heavy chains. Feedback D: Incorrect. Kinesins tend to be plus-end-directed motors, while dyneins tend to be minus-end-directed motors. 16. Which statement about cilia is false? a. They are about 10 micrometers in length, and cells that have them usually have many of them. b. They can be used to move fluids over the cell surface or to move the cell through fluids. c. They are projections of the plasma membrane supported by microfilaments. d. Their movement relies on the motor activity of axonemal dynein. Answer: c Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 2. Understanding Learning Objective: Contrast the structures and functions of primary and motile cilia. Feedback A: Incorrect. In contrast to flagella, which can be up to 200 micrometers in length and usually occur only one or two at a time, cilia are much shorter and more numerous.

Feedback B: Incorrect. Both of these activities are performed by cilia. Feedback C: Correct! Cilia are projections of the plasma membrane, but they are supported by microtubules, not microfilaments. Feedback D: Incorrect. Axonemal dynein is responsible for the sliding of outer microtubule doublets relative to one another, which is what causes the movement of cilia. 17. A centrosome is a. a cylindrical structure made up of nine triplets of microtubules. b. a chromosomal region that connects sister chromatids during mitosis and attaches them to the spindle. c. a protein structure that binds centromeres and mediates the attachment of chromosomes to the spindle. d. the major microtubule-organizing center in animal cells. Answer: d Textbook Reference: Microtubules Bloom’s Category: 1. Remembering Learning Objective: Diagram the mitotic spindle. Feedback A: Incorrect. That is a centriole. Feedback B: Incorrect. That is a centromere. Feedback C: Incorrect. That is a kinetochore. Feedback D: Correct! Microtubules emanate from the centrosome in nondividing cells, and during mitosis they are critical to the formation of the mitotic spindle. 18. Which of the following microtubules are attached to chromosomes? a. Astral microtubules b. Interphase microtubules c. Kinetochore microtubules d. Interpolar microtubules Answer: c Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 1. Remembering Learning Objective: Explain how different types of microtubules act during mitosis. Feedback A: Incorrect. Astral microtubules radiate out from the centrosomes of the mitotic spindle. They have exposed plus ends. Feedback B: Incorrect. Interphase microtubules are found in the cytoplasm. They radiate from centrosomes and are not attached to chromosomes. The nuclear envelope separates all interphase microtubules from chromosomes. Feedback C: Correct! Kinetochore microtubules radiate out from the centrosomes in mitotic cells and attach to chromosomes at their centromeres, which are associated with specific proteins to form the kinetochore. Feedback D: Incorrect. Interpolar microtubules are not attached to chromosomes. They stabilize each other by overlapping in the center of the mitotic cell. 19. The discovery that the intermediate filament protein keratin is essential for mechanical strength of epithelial cell layers was made in a. cell cultures.

b. organ cultures. c. small invertebrates. d. transgenic mice. Answer: d Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Summarize the types of intermediate filament proteins. Feedback A: Incorrect. It is difficult to create a mechanical stress situation in cell culture. The mechanical strength contribution of keratin could not be demonstrated in cell culture. Feedback B: Incorrect. It is difficult to create a mechanical stress situation in organ culture. The mechanical strength contribution of keratin could not be demonstrated in organ culture. Feedback C: Incorrect. Many small invertebrates survive well without intermediate filaments. Other cytoskeletal structures are strengthened. Feedback D: Correct! The Fuchs laboratory made the original observation in transgenic mice in which mutant keratins were expressed. The epithelial skin layer of transgenic mice was sensitive to mechanical movement. 20. Which statement about intermediate filaments is true? a. Rather than consisting of a single type of protein, they can be made up of a number of different proteins. b. They are involved in cell movement. c. The basic structure of an intermediate filament protein is a globular head and a long helical tail. d. Like microfilaments, they exhibit treadmilling. Answer: a Textbook Reference: Intermediate Filaments Bloom’s Category: 1. Remembering Learning Objective: Describe the organization and function of intermediate filaments within cells. Feedback A: Correct! There are more than 65 types of intermediate filament proteins. Feedback B: Incorrect. They do not play a direct role in cell movement but rather provide cells with mechanical strength. Feedback C: Incorrect. This corresponds to the structure of myosin II. Intermediate filament proteins typically consist of an -helical rod structure flanked by variable head and tail regions. Feedback D: Incorrect. Intermediate filaments are more static in nature than microfilaments and microtubules, which are continually being assembled and disassembled in the cell.

Essay 1. Actin filaments are a major element of the cytoskeleton. The cytoskeleton provides a framework for the cell and acts as a scaffold that both determines cell shape and positions organelles within cells. For example, the cytoskeleton provides the tracks along which

organelles move. At the same time, the cytoskeleton is subject to cleavage by proteins like cofilin. What is the apparent function of cofilin in creating a dynamic cytoskeleton? Answer: Cofilin is an actin-binding protein, and it severs actin filaments. Actin filaments must be dynamic for the cell to able to move. Hence, the ability of cofilin to sever actin filaments creates a dynamic actin cytoskeleton. Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 2. Understanding Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 2. Describe the genetic defect that causes Duchenne’s and Becker’s muscular dystrophy and the molecular processes that characterize the diseases. Answer: These diseases are caused by a mutation in the gene for dystrophin, a spectrinrelated gene that links the actin network beneath the cell surface (the cortex) to transmembrane proteins in the plasma membrane. The transmembrane proteins, in turn, are bound to components of the extracellular matrix, and thus dystrophin plays a role in linking the cortex to the extracellular matrix. This firm anchoring to the extracellular matrix stabilizes muscle cells; without it, the constant stress of contraction results in their destruction and consequently in the loss of muscle tissue that characterizes the diseases. Textbook Reference: Structure and Organization of Actin Filaments Bloom’s Category: 2. Understanding Learning Objective: Illustrate the organization of actin filaments underlying the plasma membrane. 3. How does Ca2+ regulate the activity of myosin motors in striated muscle, nonmuscle, and smooth muscle cells? Answer: In striated muscle, actin-myosin contraction is regulated by the binding of Ca 2+ to troponin. Troponin binds to actin in the absence of Ca 2+ and blocks the binding of myosin to actin. In the presence of Ca 2+, troponin changes shape, and myosin can bind to actin. In nonmuscle and smooth muscle cells, myosin activity is regulated by phosphorylation. Myosin light-chain kinase (MLCK) is the responsible enzyme. MLCK activity is regulated by the binding of Ca 2+ to calmodulin, which in turn binds to MLCK. Textbook Reference: Myosin Motors Bloom’s Category: 3. Applying Learning Objective: Summarize the roles of contractile actin–myosin filaments in nonmuscle cells. 4. What is one way in which the more rapid growth of actin filaments at one end of the cell (the plus end) compared to the other end (the minus end) is advantageous to the cell? Answer: Since the plus end grows 5 to 10 times more rapidly than the minus end, microfilaments essentially grow in one direction. Thus, correct orientation of the plus end will cause a cell to move toward an attractant without any counteracting force in the opposite direction. Textbook Reference: Microtubules Bloom’s Category: 2. Understanding

Learning Objective: Summarize the dynamics of actin filaments and the roles of actinbinding proteins. 5. The kinetic properties of myosin have been optimized such that it moves very rapidly along actin filaments. Specifically, this movement is accomplished by myosin’s very brief attachments to the microfilament along which it moves. In contrast, kinesin moves along microtubules more slowly, with longer periods of attachment between each “step.” How might these differences be explained in evolutionary terms? Answer: Myosin works in conjunction with many other myosin molecules in muscle contraction, and it is unlikely that hundreds of myosin molecules will all let go at once. Therefore, myosin can afford to sacrifice attachment in the interest of speed, which is more adaptive for muscles that need to move quickly for survival. In contrast, vesicles and organelles are moved along a microtubule by just a few molecules of kinesin, so the chances that they could all let go at the same time are higher. Therefore, kinesin takes shorter “steps” along the microtubule with longer attachment periods. This increases the chances that the cargo will make it to its final destination without falling off the microtubule. Transporting the cargo can be a slow and steady process, having relatively little need for speed for survival. Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 4. Analyzing Learning Objective: Summarize the properties of kinesins and dyneins. 6. The positioning of various organelles—for example, the Golgi apparatus in cells—is the outcome of a balance between dynein and kinesin. What is the expected distribution of the Golgi apparatus in cells in which the dynein function is inhibited? Answer: Disruption of activity would mean there is no minus-end-directed motor tugging at the Golgi apparatus. Kinesin, a plus-end-directed motor, continues to be active, and the Golgi apparatus (or fragments thereof) will be pulled toward the plus end of microtubules (i.e., the cell periphery). Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of kinesins and dyneins. 7. A human hereditary disease has two seemingly unrelated symptoms: an inability to clear mucus from the respiratory system and male sterility. What could be the cause of this disease? Answer: The failure of cilia to beat would disable the clearing of the respiratory tract. The failure of flagella to beat would render sperm immobile—one source of male sterility. This disease is likely caused by a failure to produce dynein, the motor protein that drives ciliary and flagellar motion. Textbook Reference: Microtubule Motors and Movement Bloom’s Category: 3. Applying Learning Objective: Contrast the structures and functions of primary and motile cilia. 8. Why might colchicine, a tubulin-binding drug, be used to treat cancer? Answer: Colchicine, an alkaloid derived from plants, binds tightly to tubulin and inhibits

its polymerization, thereby preventing the assembly of the mitotic spindle. The drug thus inhibits cell division, hence its usefulness as an anticancer drug. Textbook Reference: Microtubules Bloom’s Category: 2. Understanding Learning Objective: Explain how different types of microtubules act during mitosis. 9. One approach to testing whether keratin intermediate filaments are dynamic is to inject biotin-labeled keratin into living fibroblasts and then to compare, at different times, the distribution of biotin-labeled keratin with that of endogenous keratin intermediate filaments. Assuming that keratin intermediate filaments turn over dynamically with a half-time of one hour, predict the comparative distribution of biotin-labeled keratin and keratin intermediate filaments 10 minutes and four hours after microinjection. Answer: Any incorporation of the biotin-labeled keratin into keratin intermediate filaments takes time. After 10 minutes, the injected biotin-labeled keratin should be diffusely distributed in the cell. After four hours, individual keratin intermediate filaments will have turned over (with a one-hour half-time). The biotin-labeled keratin should now be incorporated into the keratin intermediate filaments, and the distribution of the biotin-labeled keratin and of endogenous keratin in the intermediate filaments should be the same. Textbook Reference: Intermediate Filaments Bloom’s Category: 4. Analyzing Learning Objective: Describe the organization and function of intermediate filaments within cells.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 15: The Plasma Membrane TEST FILE QUESTIONS Multiple Choice 1. The plasma membrane functions as a a. selective barrier to the passage of molecules. b. site for the uptake of macromolecules into the cell. c. site for cell–cell interactions. d. All of the above Answer: d Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Summarize the lipid composition of plasma membranes. 2. The plasma membrane’s barrier to passive diffusion is primarily a function of the membrane’s a. phospholipids. b. cholesterol. c. proteins. d. glycoproteins. Answer: a Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 2. Understanding Learning Objective: Summarize the lipid composition of plasma membranes. 3. Gorter and Grendel’s classic experiment allowed them to observe that the erythrocyte plasma membrane contains _______ the surface area of the erythrocytes. a. enough lipid to occupy a monolayer equal to b. enough lipid to occupy a monolayer equal to twice c. less lipid than would occupy a monolayer equal to d. enough lipid to occupy a monolayer equal to one-half Answer: b Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 2. Understanding Learning Objective: Summarize the lipid composition of plasma membranes. 4. Plasma membrane phospholipids are a. located mostly in the outer leaflet of the bilayer.

b. located mostly in the inner leaflet of the bilayer. c. symmetrically distributed between the two membrane halves. d. asymmetrically distributed between the two membrane halves. Answer: d Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Summarize the lipid composition of plasma membranes. 5. Cholesterol is present in the membranes of all a. cells. b. eukaryotic cells. c. animal cells. d. plant cells. Answer: c Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Summarize the lipid composition of plasma membranes. 6. Plasma membrane glycolipids are found a. exclusively in the inner leaflet. b. exclusively in the outer leaflet. c. equally distributed between the inner and outer leaflets. d. only on the basal surface of epithelia. Answer: b Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Summarize the lipid composition of plasma membranes. 7. Mammalian erythrocytes (red blood cells) are particularly useful for studies of the plasma membrane because a. they have few peripheral proteins. b. they have only one membrane, the plasma membrane. c. their plasma membrane is not associated with a cytoskeleton. d. their plasma membrane contains no cholesterol. Answer: b Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 2. Understanding Learning Objective: Illustrate how proteins are associated with the plasma membrane. 8. The two erythrocyte proteins glycophorin and band 3 are examples of a. peripheral proteins. b. transmembrane proteins. c. cytoskeletal proteins. d. -barrel proteins. Answer: b Textbook Reference: Structure of the Plasma Membrane

Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. 9. Clusters of _______ in membrane proteins such as band 3 of the erythrocyte can define hydrophilic segments through which ions (for example, bicarbonate and chloride) can be transported. a. oligosaccharide side chains b. lipid modifications c. transmembrane α helices d. phosphorylation motifs Answer: c Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. 10. Which of the following features of a membrane protein is never used as a membrane anchor? a. C-terminal glycosylphosphatidyl inositol (GPI) b. An N-terminal myristoyl group c. Prenyl and palmitoyl group additions d. Oligosaccharide side chains Answer: d Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. 11. Above the temperature at which lipids are fluid, membrane proteins are able to move a. from inner to outer surfaces of a membrane. b. laterally in the plane of a membrane. c. from apical to basal surfaces of intestinal epithelial cells. d. only if attached to microtubules or microfilaments. Answer: b Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. 12. A major feature included in the updated fluid mosaic model is the a. interaction of transmembrane proteins with the cytoskeleton. b. restriction of glycolipids to the extracellular face of the plasma membrane. c. presence of multiple pass transmembrane proteins. d. presence of N-linked oligosaccharide side chains on glycoproteins at the cell surface. Answer: a Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane.

13. Clusters of sphingolipids, cholesterol, and membrane proteins that move together laterally in the plane of the plasma membrane are called lipid a. boats. b. barrels. c. rafts. d. barges. Answer: c Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of plasma membrane domains. 14. If a suspension of cells is frozen and fractured, the most likely path of the fracture plane will be a. between the cell surface and the outside solution. b. between the membrane and the cytoplasm of the cells. c. through the middle of the cytoplasm. d. between the two leaflets of the cell membranes. Answer: d Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 2. Understanding Learning Objective: Explain the significance of plasma membrane domains. 15. What is the role of tight junctions in the transport of glucose across the intestinal epithelium? a. They open and allow glucose to pass between the epithelial cells. b. They keep the Na+-K+ pumps in the apical membrane only. c. They keep the Na+-glucose cotransporter in the apical membrane and the glucosefacilitated transporter in the basolateral membrane. d. They keep the glucose-facilitated transporter in the apical membrane and the Na+glucose cotransporter in the basolateral membrane. Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of plasma membrane domains. 16. Molecules that diffuse passively across the plasma membrane most rapidly are a. small. b. hydrophobic. c. small and hydrophobic. d. small and hydrophilic. Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Describe the transport of small molecules by carrier proteins. 17. Facilitated diffusion differs from passive diffusion in that facilitated diffusion is

a. mediated only by a protein channel. b. mediated only by a protein carrier. c. mediated by a protein carrier or channel. d. against the concentration gradient. Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Describe the transport of small molecules by carrier proteins. 18. The glucose facilitated diffusion transporter can transport glucose a. into the cell only. b. out of the cell only. c. into or out of the cell. d. only in the presence of ATP. Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Describe the transport of small molecules by carrier proteins. 19. The MDR ABC transporter functions in a number of animal cells to transport a. glucose into cells. b. ions into cells. c. poisons and drugs out of cells. d. amino acids across epithelia. Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Contrast ion channels and carrier proteins. 20. Which of the following has the fastest rate of transport? a. Channel-mediated diffusion b. Facilitated diffusion c. Active transport d. All of the above are equally fast. Answer: a Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Contrast ion channels and carrier proteins. 21. In a typical mammalian cell, the concentration of _______ is higher on the inside, and the concentration of _______ is higher on the outside. a. Na+ and Cl–; K+ b. Na+; K+ and Cl– c. K+; Na+ and Cl– d. K+ and Cl– ; Na+ Answer: c

Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Contrast ion channels and carrier proteins. 22. Channels that open in response to neurotransmitters or other signal molecules are called _______ channels. a. voltage-gated b. ligand-gated c. signal-gated d. ion Answer: b Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of ion channels in transmission of nerve impulses. 23. The resting potential of a typical eukaryotic cell is _______ mV. a. 0 b. –60 c. +60 d. –100 Answer: b Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of ion channels in transmission of nerve impulses. 24. The Nernst equation allows one to calculate the a. resting membrane potential. b. equilibrium potential for each ion. c. membrane potential when all channels are closed. d. membrane potential when all channels are open. Answer: b Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of ion channels in transmission of nerve impulses. 25. What would be the resting potential across an artificial membrane if all charged molecules on both sides were equally permeable? a. –60 mV b. +60 mV c. 0 mV d. –1 mV Answer: c Textbook Reference: Transport of Small Molecules

Bloom’s Category: 3. Applying Learning Objective: Summarize the role of ion channels in transmission of nerve impulses. 26. The flow of which ion makes the largest contribution to the resting potential? a. K+ b. Na+ c. H+ d. Ca2+ Answer: a Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Describe the action of the sodium/potassium pump. 27. Voltage-sensitive K+ channels are 1,000 times more permeable to K+ than to Na+ because a. K+ ions are smaller than Na+ ions. b. K+ ions have a lower charge density than Na+ ions. c. a selectivity filter removes the water molecules from K+ ions but not from Na+ ions. d. K+ ions are more concentrated inside the cell than outside the cell. Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Describe the action of the sodium/potassium pump. 28. Cystic fibrosis is a genetic disease in which thick mucus accumulates over several types of epithelial cells, eventually blocking the pulmonary airways. The molecular basis of this disease is the production of a defective a. mucin. b. chloride channel. c. Na+-K+ pump. d. Na+-Ca2+ transporter. Answer: b Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Describe the action of the sodium/potassium pump. 29. Gene therapy for cystic fibrosis involves transfer of the _______ gene into bronchial epithelia. a. CFTR b. mucin c. Na+-K+ pump d. MDR transporter Answer: a Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering

Learning Objective: Describe the action of the sodium/potassium pump. 30. Active transport is transport in an energetically a. unfavorable direction, always driven by hydrolysis of ATP. b. unfavorable direction, always coupled to another reaction or source of energy. c. unfavorable direction, driven only by the flow of another molecule across a membrane. d. favorable direction, coupled to the hydrolysis of ATP. Answer: b Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport. 31. The Na+ and K+ ion gradients across the plasma membrane are produced primarily by the a. permeability of these ions across the lipid bilayer. b. ratio of these ions in the blood. c. action of the Na+-K+ pump. d. flow of these ions through voltage-gated channels. Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport. 32. What percent of the ATP in a typical animal cell is consumed by the Na +-K+ pump? a. 1% b. 10% c. 25% d. 66% Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport. 33. Calcium levels remain low in the cytosol a. only to ensure that calcium does not precipitate inside the cells. b. thus transient calcium increases can be used as intracellular signals. c. so that calcium does not accumulate in mitochondria. d. to allow extracellular calcium to drive import of other ions. Answer: b Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport.

34. Bacteria, fungi, and plants use a gradient of _______ ions across their plasma membranes to drive transport of other molecules into the cells. a. Na+ b. K+ c. H+ d. Ca2+ Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport. 35. Coupled transport of glucose and Na+ into the intestinal epithelial cell is an example of a. facilitated diffusion. b. symport. c. antiport. d. endocytosis. Answer: b Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport. 36. The functioning of the Na+-Ca2+ transporter in the plasma membrane is an example of a. facilitated diffusion. b. symport. c. antiport. d. endocytosis. Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport. 37. Phagocytosis involves movement of the cell surface by a. actin-based motility. b. microtubule-based motility. c. clathrin-based vesicle formation. d. dynamin-based vesicle formation. Answer: a Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism of particle uptake by phagocytosis.

38. Aged red blood cells are removed from circulation by macrophages in the a. bone marrow. b. lymph nodes. c. spleen. d. kidneys. Answer: c Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism of particle uptake by phagocytosis. 39. Phagocytosis is the main function of what two types of human white blood cells? a. Macrophages and T lymphocytes b. Platelets and neutrophils c. Macrophages and neutrophils d. Eosinophils and T lymphocytes Answer: c Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism of particle uptake by phagocytosis. 40. Cholesterol is taken up into most cells of the body by a. phagocytosis. b. receptor-mediated endocytosis. c. simple diffusion. d. caveolae formation. Answer: b Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. 41. Brown and Goldstein discovered the mechanism of cholesterol uptake by studying fibroblasts from patients with which disease? a. Chronic fatigue syndrome b. Familial hypercholesterolemia c. Lupus erythematosus d. Cystic fibrosis Answer: b Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. 42. Coated pits are converted to coated vesicles by formation of rings of the protein a. caveolin. b. clathrin. c. COPI. d. dynamin.

Answer: d Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis.

Fill in the Blank 1. The currently accepted model of membrane structure, proposed by Singer and Nicolson, is called the _______ model. Answer: fluid mosaic Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. 2. The membrane-spanning portion of a single-pass transmembrane protein is usually a secondary structure called a(n) _______. Answer: α helix (alpha helix) Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. 3. The membrane-spanning portion of the protein porin (and proteins with similar structures in the outer membranes of mitochondria and chloroplasts) has a membranespanning secondary structure called a(n) _______. Answer: β barrel (beta barrel) Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. 4. The potentially oncogenic signal-transmitting proteins Src and Ras are attached to the inner surface of the plasma membrane by covalent links to _______. Answer: membrane lipids (myristoyl groups, or prenyl groups) Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. 5. Some plasma membrane proteins are immobilized by association with _______ proteins, other plasma membrane proteins, and the _______ matrix. Answer: cytoskeletal; extracellular Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of plasma membrane domains. 6. The restriction of membrane protein movement between surfaces of intestinal epithelia is primarily due to the presence of _______ junctions between the epithelial cells.

Answer: tight Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 2. Understanding Learning Objective: Explain the significance of plasma membrane domains. 7. The plasma membrane of intestinal epithelial cells is divided into two domains: the _______ domain and the _______ domain. Answer: apical; basolateral Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of plasma membrane domains. 8. During facilitated diffusion of glucose, the glucose transporter changes its _______. Answer: conformation Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Describe the transport of small molecules by carrier proteins. 9. The Na+ and K+ channels involved in the propagation of an action potential are _______-gated. Answer: voltage Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of ion channels in transmission of nerve impulses. 10. The released neurotransmitters at a synapse are sensed by _______-gated channels. Answer: ligand Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of ion channels in transmission of nerve impulses. 11. The selectivity filter in a K+ channel is able to _______ K+ but not Na+, which is smaller. Answer: dehydrate Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Describe the action of the sodium/potassium pump 12. Active transport can be driven by _______ hydrolysis or by _______ gradients. Answer: ATP; ion Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport.

13. The cytoplasmic domain of ligand-bound LDL receptor binds to _______ proteins, which in turn associate with the _______ coat. Answer: adaptor; clathrin Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. 14. The endocytic trafficking pathway of ligands and receptors in eukaryotic cells is often controlled by sequential differences in _______ between early endosomes, late endosomes, and lysosomes. Answer: pH (acidity) Textbook Reference: Endocytosis Bloom’s Category: 2. Understanding Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. 15. Some endocytic receptors are _______ from endosomes, while others are _______ in lysosomes. Answer: recycled; degraded Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis.

True/False 1. Glycolipids are the major lipids in the plasma membrane. Answer: F Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Summarize the lipid composition of plasma membranes. 2. Phosphatidylcholine is the only major phospholipid in the plasma membrane. Answer: F Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Summarize the lipid composition of plasma membranes. 3. The outer membrane of mitochondria is the major site where porins are found in animal cells. Answer: T Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane.

4. The apical surface of intestinal epithelial cells is covered by a carbohydrate coat known as the glycocalyx. Answer: T Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of plasma membrane domains. 5. Estrogen is an example of a hormone to which the plasma membrane is impermeable. Answer: F Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Describe the transport of small molecules by carrier proteins. 6. Facilitated diffusion is a lipid-mediated process. Answer: F Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Contrast ion channels and carrier proteins. 7. The major ATPase maintaining the plasma membrane potential is the Na +-K+ pump. Answer: T Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Describe the action of the sodium/potassium pump. 8. The K+-glucose transporter is an example of a symport that actively transports glucose into cells. Answer: F Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport. 9. Phagocytosis is an example of macropinocytosis. Answer: F Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism of particle uptake by phagocytosis. 10. Defects in the human LDL receptor can cause the disease hypercholesterolemia. Answer: T Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. 11. Caveolae are rich in cholesterol and other constituents of lipid rafts.

Answer: T Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis.

Short Answer 1. What are integral membrane proteins? Answer: They are proteins that can be released from the membrane only by disrupting the membrane structure. Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. 2. What are peripheral membrane proteins? Answer: They are proteins that can be removed from the membrane by treatment with solutions of high salt or a high pH that do not disrupt the membrane bilayer. Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. 3. Frye and Edidin demonstrated the mobility of membrane proteins with the use of fused human and mouse cells. Briefly describe the results of their experiment. Answer: Immediately after fusion, the human and mouse proteins were located on different sides of the hybrid cells, but after a brief (40 minutes) incubation at 37C, the human and mouse proteins were intermixed across the cell’s surface. Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 2. Understanding Learning Objective: Illustrate how proteins are associated with the plasma membrane. 4. What is one function of cell-surface carbohydrates? Answer: Cell-surface carbohydrates serve as markers for cell–cell recognition, protect the cell surface from ionic and mechanical stress, and form a barrier to invading microorganisms. Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 2. Understanding Learning Objective: Explain the significance of plasma membrane domains. 5. How does the structure of a transport protein such as a glucose transporter create an aqueous environment in which glucose may be transported across the plasma membrane? Answer: A glucose transporter has multiple transmembrane domains. These α-helical segments cluster together to surround an aqueous environment through which the glucose may bind and pass. Textbook Reference: Transport of Small Molecules

Bloom’s Category: 2. Understanding Learning Objective: Describe the transport of small molecules by carrier proteins. 6. How is rapid transport of water across the plasma membrane achieved? Answer: The presence of aquaporin in the plasma membrane provides water-specific channels for the rapid flow of water across this membrane. Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Describe the transport of small molecules by carrier proteins. 7. Explain how phagocytosis differs mechanistically from other types of endocytosis. Answer: In phagocytosis, an actin- and myosin-based movement of the cell surface surrounds and engulfs the large particulate material that is then taken into the cell. Other types of endocytosis involve the uptake of individual molecules bound to receptors or fluid. Textbook Reference: Endocytosis Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanism of particle uptake by phagocytosis. 8. Define endocytosis and explain the process by which it operates. Answer: Endocytosis is the uptake of material into a cell by surrounding it with a part of the plasma membrane that buds off into the cytoplasm, creating vesicles formed from the plasma membrane. Textbook Reference: Endocytosis Bloom’s Category: 2. Understanding Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. 9. What is the mechanism by which cells take up specific macromolecules? Answer: Cells take up specific macromolecules by the mechanism of receptor-mediated endocytosis. Textbook Reference: Endocytosis Bloom’s Category: 2. Understanding Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. 10. Where is clathrin located in cells? Answer: On coated vesicles and coated pits Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. 11. Describe a role or function for coated pits in the process of receptor-mediated endocytosis. Answer: Receptors and ligands (cargo) cluster into coated pits and are subsequently internalized and recycled. Textbook Reference: Endocytosis Bloom’s Category: 2. Understanding

Learning Objective: Summarize the pathway of clathrin-mediated endocytosis.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. What is the primary reason that mammalian red blood cells are used in the study of the plasma membrane? a. There are a lot of them and they are easily obtained. b. They contain larger plasma membranes than any other cell type. c. They are of particular interest because they are made up of a lipid monolayer. d. They lack nuclei and membrane-bounded organelles. Answer: d Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Summarize the lipid composition of plasma membranes. Feedback A: Incorrect. This is true, but it is not the primary reason for their use. Feedback B: Incorrect. Red blood cells do not contain unusually large plasma membranes. Feedback C: Incorrect. Red blood cell plasma membranes, like all other biological membranes, are made up of a lipid bilayer. Feedback D: Correct! Red blood cells are a good source of plasma membrane because the membranes can be isolated without risk of contamination from other types of membranes. 2. Two examples of membrane lipids that are present in small amounts are a. cholesterol and phosphatidylcholine. b. glycolipids and phosphatidylinositol. c. phosphatidylcholine and sphingomyelin. d. phosphatidylethanolamine and phosphatidylserine. Answer: b Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Summarize the lipid composition of plasma membranes. Feedback A: Incorrect. Cholesterol is abundant; it is present in about the same molar amounts as the phospholipids. Phosphatidylcholine and sphingomyelin are the predominant phospholipids in the outer leaflet of the plasma membrane. Feedback B: Correct! Both glycolipids and phosphatidylinositol are quantitatively minor plasma membrane lipids. Glycolipids constitute only about 1% of the lipids of most plasma membranes. Phosphatidylinositol is a minor phospholipid of the inner leaflet. Feedback C: Incorrect. Phosphatidylcholine and sphingomyelin are the major phospholipids of the outer leaflet. Feedback D: Incorrect. Phosphatidylethanolamine and phosphatidylserine are the predominant phospholipids of the inner leaflet. 3. Which of the following will not solubilize a typical peripheral membrane protein?

a. Detergent b. Extreme pH c. High salt concentration d. Mildly hypertonic saline Answer: d Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 3. Applying Learning Objective: Illustrate how proteins are associated with the plasma membrane. Feedback A: Incorrect. Detergent treatment will solubilize membrane lipids and integral membrane proteins. In the process, peripheral membrane proteins will be solubilized. Feedback B: Incorrect. Extreme pH will solubilize the typical peripheral membrane protein. The interaction of peripheral membrane proteins with membranes is frequently through ionic bonds. Feedback C: Incorrect. A high salt concentration will solubilize the typical peripheral membrane protein. The interaction of peripheral membrane proteins with membranes is frequently through ionic bonds. Feedback D: Correct! Mildly hypertonic saline will not solubilize a typical peripheral membrane protein. Rather, because of osmolarity, it will shrink membrane vesicles or cells. 4. A feature common to most transmembrane proteins is a(n) a. phosphorylated exterior domain. b. -helical region of about 20 to 25 hydrophobic amino acids. c. amino acid sequence rich in acidic residues. d. structure consisting almost exclusively of -sheets. Answer: b Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 2. Understanding Learning Objective: Illustrate how proteins are associated with the plasma membrane. Feedback A: Incorrect. Some transmembrane proteins can become phosphorylated, but more often on their cytoplasmic sides. Feedback B: Correct! At least one of these -helical domains is present in almost all transmembrane proteins and functions as the membrane-spanning domain. Feedback C: Incorrect. A domain rich in acidic residues is often found in transcription factors, but it is not typical of transmembrane proteins. Feedback D: Incorrect. Transmembrane proteins may contain -sheets, but they are not a distinguishing feature. 5. Which of the following are integral membrane proteins that do not contain a transmembrane  helix? a. Band 3 b. Glycophorins c. Glycosylphosphatidylinositol (GPI)-anchored proteins d. Spectrins Answer: c Textbook Reference: Structure of the Plasma Membrane

Bloom’s Category: 2. Understanding Learning Objective: Illustrate how proteins are associated with the plasma membrane. Feedback A: Incorrect. Band 3 is the anion transporter in the red blood cell plasma membrane and is thought to have 14 transmembrane  helices. Feedback B: Incorrect. Glycophorin is a red blood cell transmembrane protein of unknown function with one transmembrane  helix. Feedback C: Correct! GPI-anchored proteins are associated with the outer leaflet of the plasma membrane. They are anchored by the GPI lipid to the membrane with no transmembrane domain. Feedback D: Incorrect. Spectrin is a peripheral membrane protein in red blood cells. It is part of the cytoskeleton that underlies the erythrocyte plasma membrane. 6. Membrane proteins are able to move a. from the inner to the outer surface of a membrane. b. laterally within the plane of a membrane. c. from apical to basal surfaces of intestinal epithelial cells. d. only if attached to microtubules or microfilaments. Answer: b Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Illustrate how proteins are associated with the plasma membrane. Feedback A: Incorrect. Membrane proteins cannot switch sides. Feedback B: Correct! The plasma membrane is like a viscous fluid, and proteins can move laterally within it. Feedback C: Incorrect. The apical and basal surfaces are separated by tight junctions through which proteins cannot pass. Feedback D: Incorrect. Microtubules, by transporting vesicles, are involved in moving proteins to the plasma membrane, but not within it. 7. Which of the following lipids are distinctly enriched in lipid rafts? a. Cholesterol and sphingolipids b. Glycolipids and phosphatidylinositol c. Phosphatidylcholine and sphingomyelin d. Phosphatidylethanolamine and phosphatidylserine Answer: a Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of plasma membrane domains. Feedback A: Correct! Lipid rafts are found in the outer leaflet of the plasma membrane and are formed by the clustering of cholesterol and the sphingolipids (sphingomyelin and glycolipids). Feedback B: Incorrect. Although glycolipids are enriched in lipid rafts, phosphatidylinositol is not. Quantitatively, phosphatidylinositol is a minor phospholipid in the inner leaflet of the plasma membrane. Feedback C: Incorrect. Although sphingomyelin is enriched in lipid rafts, phosphatidylcholine is not. It is a major phospholipid of the outer leaflet.

Feedback D: Incorrect. Neither phosphatidylethanolamine nor phosphatidylserine is enriched in lipid rafts. These lipids are the predominant phospholipids of the inner leaflet of the plasma membrane. 8. Which statement about the glucose transporter is false? a. It transports glucose across the plasma membrane via a mechanism called facilitated diffusion. b. It has 12 -helical transmembrane segments. c. A conformational change in the transporter is involved in the transport process. d. Flow through it is unidirectional. Answer: d Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Describe the transport of small molecules by carrier proteins. Feedback A: Incorrect. This is a true statement. Glucose movement down its concentration gradient is facilitated by the glucose transporter. Feedback B: Incorrect. This is a true statement. The glucose transporter does have 12 helical transmembrane segments. Feedback C: Incorrect. This is a true statement. Once bound to glucose, the transporter undergoes a conformational change that transfers the glucose molecule to the cytoplasmic side of the membrane. Feedback D: Correct! Flow of glucose is usually from the outside to the inside of the cell, but in some cases, such as in liver cells (where glycogen breakdown can result in high intracellular concentrations of glucose), it can be transported outward. 9. Which statement about ion channels is true? a. They are opened either by the binding of ligands or by changes in electric potential across the membrane. b. They require ATP. c. They are open most of the time. d. The rate of transport is slow compared to the rate of transport via carrier proteins. Answer: a Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Contrast ion channels and carrier proteins. Feedback A: Correct! The former are called ligand-gated channels, and the latter are voltage-gated channels. Feedback B: Incorrect. Ion pumps require energy, but the flow of molecules through channels is driven simply by a concentration gradient. Feedback C: Incorrect. They are usually closed in the resting state, and only upon stimulation are they induced to open. Feedback D: Incorrect. The flow rate through ion channels can be very high—over a million per second. 10. The resting plasma membrane potential as described by the Nernst equation is the sum of the movement of many ions. However, in practice, the movement of _______, a

single ion, is the major component determining membrane potential. a. Ca2+ b. Cl– c. K+ d. Na+ Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Summarize the role of ion channels in transmission of nerve impulses. Feedback A: Incorrect. Feedback B: Incorrect. Feedback C: Correct! The resting membrane potential, –60 mV, is close to the equilibrium potential determined by intracellular and extracellular K+ concentrations, –75 mV. Feedback D: Incorrect. 11. Which statement about the relative concentrations of Na+ and K+ inside and outside of a typical mammalian cell is true? a. The concentration of Na+ and K+ outside the cell is about 30 times higher than inside the cell. b. The concentration of Na+ outside the cell is about 30 times higher than inside the cell, and the concentration of K+ inside the cell is 10 to 30 times higher than outside the cell. c. The concentration is the same for both ions, but there is a Cl– ion concentration gradient across the plasma membrane. d. The concentration of Na+ inside the cell is 30 times higher than outside the cell, and the concentration of K+ outside the cell is 10 times higher than inside the cell. Answer: b Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Describe the action of the sodium/potassium pump. Feedback A: Incorrect. The ion concentrations across the membrane are unbalanced, but not in this way. Feedback B: Correct! The maintenance of Na+ and K+ gradients across the plasma membrane drives a number of cellular processes, such as muscle contraction and nerve impulse transmission. Feedback C: Incorrect. There is a Cl– gradient across the membrane, but there are also Na+ and K+ gradients. Feedback D: Incorrect. There are ion imbalances across the membrane, but not in the way described here. 12. Although Na+ is smaller than K+, its passage through the K+ channel is blocked by the _______ filter. a. CO2 b. negativity c. positivity

d. selectivity Answer: d Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Describe the action of the sodium/potassium pump. Feedback A: Incorrect. This is an invented term in this context. Feedback B: Incorrect. This is an invented term in this context. Feedback C: Incorrect. This is an invented term in this context. Feedback D: Correct! The channel pore contains a narrow selectivity filter. The carbonyl oxygen atoms that line the selectivity filter are positioned to displace the water molecules to which K+ is bound, but are too distant from the smaller Na+ cation to do this. 13. Transport of glucose into the intestinal epithelium is driven by ion gradients established by the a. ATP pump. b. H+ pump. c. Na+-K+ pump. d. K+ channel. Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Describe the action of the sodium/potassium pump. Feedback A: Incorrect. The pump uses ATP, but it is not an ATP pump. Feedback B: Incorrect. H+ pumps do exist, but they do not establish the major ion gradients across the epithelial apical membrane. Feedback C: Correct! The Na+-K+ pump uses ATP to establish the Na+ and K+ gradients across the epithelial apical membrane. Na+ movement can then be coupled to that of glucose by a symport to drive glucose movement in the epithelium. Feedback D: Incorrect. Although K+ channels exist, they do not generate the major ion gradients across the epithelial apical membrane. 14. Active transport differs from facilitated diffusion in that a. ions are not transported via active transport, only by facilitated diffusion. b. active transport requires a protein component, whereas facilitated diffusion occurs by simple diffusion through the plasma membrane. c. active transport involves the transport of molecules up their concentration gradient. d. active transport involves a conformational change in the transport molecule. Answer: c Textbook Reference: Transport of Small Molecules Bloom’s Category: 1. Remembering Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport. Feedback A: Incorrect. A wide range of substances are transported via both active transport and facilitated diffusion, including ions. Feedback B: Incorrect. Both processes involve protein components such as channels or pumps.

Feedback C: Correct! Active transport is thus an energy-requiring process and is coupled to either ATP hydrolysis or to the passage of another substance in an energetically favorable direction. Feedback D: Incorrect. Channels and carrier proteins used in facilitated diffusion also undergo conformational changes, such as the structural change that occurs upon the binding of glucose to its transporter. 15. Which statement about ABC transporters is false? a. Each family member has an ATP-binding cassette. b. Each member of the family is a Cl– channel. c. They are the largest family of membrane transporters in humans. d. Each member of the family has shared domain structure features. Answer: b Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport. Feedback A: Incorrect. This is a true statement. Each ABC transporter does have an ATPbinding cassette (ABC). Feedback B: Correct! This is a false statement. CFTR, the transporter protein responsible for cystic fibrosis, is a member of the ABC transporter family and is a Cl– and bicarbonate channel. However, this is only one example of what ABC transporters transport. Feedback C: Incorrect. This is a true statement. ABC transporters are members of the largest family of membrane transporters in humans. Feedback D: Incorrect. This is a true statement. One of the key features of a family of proteins is conserved domain elements. 16. The ingestion of large particles by cells is a form of endocytosis known as a. fluid phase endocytosis. b. macropinocytosis. c. phagocytosis. d. transcytosis. Answer: c Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism of particle uptake by phagocytosis. Feedback A: Incorrect. Fluid phase endocytosis is the nonselective uptake of fluid that accompanies all forms of endocytosis. Feedback B: Incorrect. Macropinocytosis (cell drinking) takes up proteins and other nonparticulates by several mechanisms. Feedback C: Correct! Phagocytosis (cell eating) takes up large particles such as bacteria, yeast, and red blood cells by engulfing them with pseudopodia. Feedback D: Incorrect. Transcytosis is a process by which receptors and bound ligands can be transferred from one side of a polarized cell (e.g., an epithelial cell) to the other.

17. Which of the following is not involved in receptor-mediated endocytosis? a. Clathrin b. Adaptor proteins c. Internalization signals d. Pseudopodia Answer: d Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. Feedback A: Incorrect. Clathrin is involved in the formation of the clathrin-coated pits, in which the receptors aggregate prior to endocytosis and form clathrin-coated vesicles once they have pinched off into the cytoplasm. Feedback B: Incorrect. Adaptor proteins bind the cytoplasmic tails of receptors and target them to coated pits. Feedback C: Incorrect. Many receptors possess internalization signals in their cytoplasmic tails that target them to coated pits. Feedback D: Correct! Pseudopodia are actin-based extensions of the cell surface that are formed during phagocytosis, not receptor-mediated endocytosis. 18. LDL uptake by cells is one of the functions of a. receptor-mediated endocytosis. b. phagocytosis. c. caveolae. d. ABC transporters. Answer: a Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. Feedback A: Correct! LDL circulating in the blood plasma binds to LDL receptors on the plasma membrane and is internalized by clathrin-coated, receptor-mediated endocytosis. Feedback B: Incorrect. Phagocytosis (cell eating) is an ingestion process for large particles such as bacteria. Feedback C: Incorrect. LDL receptors do not concentrate in caveolae. Feedback D: Incorrect. ABC transporters are members of the ATP-binding cassette family of transporters and transport a range of small molecules. 19. Mutations in the internalization signal of endocytic receptors prevent their interaction with a. adaptor proteins. b. caveolin. c. clathrin. d. dynamin. Answer: a Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Summarize the pathway of clathrin-mediated endocytosis.

Feedback A: Correct! Adaptor proteins bind to internalization signal features on endocytic receptors and form a bridge between the endocytic receptor and clathrin in receptor-mediated endocytosis. Feedback B: Incorrect. Caveolin has a role in organizing caveolae and does not bind to endocytic receptors. Feedback C: Incorrect. Clathrin does not bind directly to endocytic receptors but rather complexes with endocytic receptors indirectly through an adaptor bridge. Feedback D: Incorrect. Dynamin is involved in the pinching off of clathrin-coated vesicles from the membrane. 20. The pH of endosomes and lysosomes is a. acidic. b. alkaline. c. neutral. d. unknown. Answer: a Textbook Reference: Endocytosis Bloom’s Category: 1. Remembering Learning Objective: Explain recycling of cell surface receptors. Feedback A: Correct! The pH of early endosomes is about 6.0–6.2, and that of lysosomes is about 5.0. Any pH less than 7.0 is acidic; that is, having a higher concentration of H + than OH–. Feedback B: Incorrect. The pH is not alkaline. Feedback C: Incorrect. The pH is not neutral. Feedback D: Incorrect. The pH is known.

Essay 1. In their acceptance speech for the Nobel Prize in 1985, Michael Brown and Joseph Goldstein said, “Cholesterol is a Janus-faced molecule. The very property that makes it useful in cell membranes, namely its absolute insolubility in water, also makes it lethal.” Explain this remark. Answer: Cholesterol in the plasma membranes maintains their fluidity at a fairly constant level, preventing them from becoming overly rigid at low temperatures or overly fluid at high temperatures. However, its hydrophobicity, which is the very property that makes cholesterol useful in plasma membranes fluidity, can cause it to aggregate on arterial walls when present in high concentrations in the blood stream. Aggregations result in the narrowing of blood vessels, which can cause a heart attack or a stroke. Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 3. Applying Learning Objective: Summarize the lipid composition of plasma membranes. 2. Detergents are used to solubilize membranes because they are dual molecules possessing both hydrophobic and hydrophilic portions. Why are they also used as cleaning reagents (to clean, for example, clothes and dishes)?

Answer: Detergents are used as cleansers because the hydrophobic portions of the molecules bind to hydrophobic substances, such as an oil stain, while the hydrophilic portions bind to water, and the detergent/dirt complex is subsequently washed away in the rinse. Simply trying to remove the stain with water doesn’t work because the hydrophilic water molecules do not interact with the lipid molecules in the oil stain. Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 2. Understanding Learning Objective: Illustrate how proteins are associated with the plasma membrane. 3. According to the generally accepted fluid mosaic model of membrane structure, membranes are two-dimensional fluids through which proteins can move freely in lateral directions. Provide an example of a membrane protein that is not free to move laterally within a plasma membrane. Answer: Some membrane proteins are associated with the cytoskeleton and are thus restricted in their mobility. For example, band 3 in the plasma membrane is associated with ankyrin and spectrin, which are linked to the actin meshwork beneath the cell surface. In addition, membrane proteins such as the integrins can be anchored to the extracellular matrix. Other important examples in regard to cell adhesions are described in Chapter 15. The 64 integrin binds laminin in the extracellular matrix and intermediate filaments inside the cell, and is thus anchored on both sides. Textbook Reference: Structure of the Plasma Membrane Bloom’s Category: 3. Applying Learning Objective: Explain the significance of plasma membrane domains. 4. How does water enter a cell from the extracellular space? Answer: The plasma membrane is somewhat permeable to small polar molecules like water, and thus passive diffusion is one mechanism by which water can enter a cell. Water molecules also can pass quickly into a cell through channels formed by proteins called aquaporins; this mechanism is called facilitated diffusion. Finally, a form of endocytosis called macropinocytosis (meaning “cell drinking”) results in the uptake of macromolecules and fluids from the extracellular space. Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Describe the transport of small molecules by carrier proteins. 5. Why are most ion channels gated? Answer: Most ion channels are ligand- or voltage-gated. This means that their opening and closing, and thus the flow of ions, is regulated, which is essential to the regulation of other cell processes, such as the production of an action potential, which results from reversible changes in Na+ and K+ currents. If channels were not gated most of the time (i.e., open), then the plasma membrane would be much more likely to leak water and other substances, and it could not produce action potentials. Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Contrast ion channels and carrier proteins.

6. How does the plasma membrane come to be hyperpolarized locally during action potential reversal? Answer: The development of an action potential is due to the opening of voltage-gated Na+ channels allowing the inward flow of Na+. Later, the gated Na+ channels close and the voltage-gated K+ channels open. This results in brief hyperpolarization because the outward flow of K+ due to the combination of open voltage-gated K+ channels and resting K+ channels is transiently greater than the resting “leakage” of K+ outward through resting K+ channels. Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Describe the action of the sodium/potassium pump. 7. The drug ouabain inhibits the plasma membrane Na+-K+ pump. Why does it also inhibit glucose transport across the intestinal epithelial layer? Answer: A Na+-glucose transporter is responsible for the transport of glucose into intestinal epithelial cells. Two Na+ ions are transported inward with each glucose molecule. The Na+ moves down both a concentration and voltage gradient, and the glucose moves up a concentration gradient. For glucose, there is no charge consideration. The overall transport is energetically favorable because of the energy gain from Na +. If the Na+-K+ pump that generates the favorable Na+ gradient is inhibited, then the energetics are no longer favorable for Na+-glucose transport. The ouabain effect on the transporter is indirect. The direct effect is on the pump. Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Describe the action of the sodium/potassium pump. 8. Cystic fibrosis (CF) is the most common inherited disease of Caucasians. It is due to mutations in an ABC transporter, the cystic fibrosis transmembrane conductance regulator (CFTR), which is a Cl– and bicarbonate channel. Why have efforts at gene therapy to correct CF concentrated on the lungs when CF affects other tissues as well? Answer: The cells lining the surface of the lungs are readily accessible to aerosols carrying plasmids that express wild-type CFTR. Moreover, the immediate cause of death from CF is typically from bacterial infections of the lung trapped in CF-induced mucus accumulation. Therefore, introducing wild-type CFTR into the lung is an attractive approach to treating many of the symptoms and averting the most common cause of death associated with CF. This approach does nothing to cure mucus accumulation in pancreatic ducts or problems in other tissues. Unfortunately, to date, this approach has had only short-term success in the lungs. Textbook Reference: Transport of Small Molecules Bloom’s Category: 2. Understanding Learning Objective: Explain how ion gradients across the plasma membrane can drive active transport. 9. Shibere is a temperature-sensitive mutant in the protein dynamin in the fruit fly Drosophila. Why do flies fall out of the air when shifted to the nonpermissive temperature?

Answer: Flying requires coordinated muscle movements, and nerve function is essential to this. At nerve endings, repeated rounds of neurotransmitter release from synaptic vesicles must occur, and synaptic vesicle membrane proteins must be retrieved from the nerve endings. Dynamin is essential to the pinching off phase of this transmitter retrieval. When dynamin ceases to function at the nonpermissive temperature, synaptic vesicle recycling stops, nerve function stops, muscle function ceases to be coordinated, and the flies fall from the air. Textbook Reference: Endocytosis Bloom’s Category: 3. Applying Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. 10. Why does incubation of cells with NH3 cause the selective osmotic swelling of endosomes and lysosomes and not of other organelles in the cell? Answer: NH3 can cross membranes, while NH4+, the protonated form, cannot. In an acidic compartment, such as an endosome or a lysosome, NH3 becomes protonated. NH4+ then cannot diffuse out of the endosome or lysosome, so it accumulates in those organelles, leading to higher osmolarity and a compensating influx of water. Since these are the major acidic organelles, swelling is restricted to them. Textbook Reference: Endocytosis Bloom’s Category: 2. Understanding Learning Objective: Summarize the pathway of clathrin-mediated endocytosis. 11. Why is dissociation of ligands from their receptors in endosomes an important feature of receptor-mediated endocytosis? Answer: The dissociation of ligands from their receptors in the acidic environment of endosomes allows the receptors to be reused once they have been recycled back to the plasma membrane. In addition, internalization of receptors is a mechanism often used by cells for down-regulating a signaling pathway. However, as long as a ligand is bound to its receptor it can stimulate the signaling pathway, even if it has been internalized by endocytosis; thus the importance of ligand/receptor dissociation. Textbook Reference: Endocytosis Bloom’s Category: 2. Understanding Learning Objective: Explain recycling of cell surface receptors.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 16: Cell Walls, the Extracellular Matrix, and Cell Interactions TEST FILE QUESTIONS Multiple Choice 1. The cell walls of bacteria are made of a copolymer of a. proteins and lipids. b. proteins and polysaccharides. c. polysaccharides and peptides. d. cellulose and lignin. Answer: c Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Describe the structure of bacterial cell walls. 2. Which of the following antibiotics inhibits the synthesis of bacterial cell walls? a. Penicillin b. Streptomycin c. Tetracycline d. Actinomycin D Answer: a Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Describe the structure of bacterial cell walls. 3. The polysaccharide chains of bacterial cell walls are made of alternating a. N-acetylgalactose and N-acetylmuramic acids. b. N-acetylglucosamine and N-acetylmuramic acids. c. N-acetylglucosamine and N-acetylneuraminic acids. d. N-acetyl galactose and N-acetylneuraminic acids. Answer: b Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Describe the structure of bacterial cell walls. 4. Fungal cell walls are built primarily of a. cellulose.

b. peptidoglycan. c. pectin. d. chitin. Answer: d Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Distinguish the organization of yeast and plant cell walls. 5. Plant cell walls consist of cellulose fibrils in a matrix of a. chitin and pectin. b. pectin and hemicellulose. c. Pectin and lignin. d. lignin and chitin. Answer: b Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Distinguish the organization of yeast and plant cell walls. 6. The structural polysaccharide of fungal cell walls is the same as the one found in insect a. exoskeletons. b. endoskeletons. c. eyes. d. cartilage. Answer: a Textbook Reference: Cell Walls Bloom’s Category: 2. Understanding Learning Objective: Distinguish the organization of yeast and plant cell walls. 7. Pectins form a _______ network in plant cell walls. a. fibrous b. gel-like c. solid, rigid d. waxy Answer: b Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Distinguish the organization of yeast and plant cell walls. 8. Cellulose synthase in plants is located a. in the Golgi apparatus. b. in the cytosol. c. in the plasma membrane. d. outside the cell. Answer: c Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering

Learning Objective: Explain the role of the cell wall in plant cell growth. 9. The direction of cellulose microfibril synthesis in elongating plant cells is a. parallel to the direction of cell elongation. b. perpendicular to the direction of cell elongation. c. perpendicular to the direction of microtubules under the plasma membrane. d. parallel to the endoplasmic reticulum under the plasma membrane. Answer: b Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Explain the role of the cell wall in plant cell growth. 10. The thin, sheetlike basal laminae are found a. under epithelia. b. in the cytoplasm just below the plasma membrane. c. in the extracellular space surrounding all eukaryotic cells. d. in between adjacent cells connected by gap junctions. Answer: a Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. 11. The major protein of the extracellular matrix of animal cells is a. keratin. b. fibronectin. c. collagen. d. chondroitin sulfate. Answer: c Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. 12. Collagens commonly contain the three repeating amino acids: proline, a. glutamine, and hydroxyproline. b. glycine, and hydroxyproline. c. glycine, and hydroxylysine. d. glutamine, and hydroxylysine. Answer: b Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. 13. Basal laminae are a meshwork of fibrils composed primarily of type _______ collagen. a. I b. II

c. III d. IV Answer: d Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. 14. The smallest amino acid is a. glycine. b. proline. c. cysteine. d. asparagine. Answer: a Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. 15. Vitamin C deficiency may lead to scurvy, a disease characterized by skin lesions and blood vessel hemorrhages due to weakened connective tissue. Vitamin C deficiency has this effect because the vitamin is a. a critical structural component of collagen. b. required for the enzymatic activity of prolyl hydroxylase. c. required for the enzymatic activity of cellulose synthase. d. required for the enzymatic activity of hydroxyproline. Answer: b Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Describe the types of collagen in extracellular matrix. 16. Connective tissue cells are connected to the extracellular matrix by receptors that bind to an adhesive protein called a. fibronectin. b. aggrecan. c. integrin. d. lamin. Answer: a Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Explain the functions of adhesion proteins. 17. Integrins are a. transmembrane proteins. b. peripheral membrane proteins. c. components of the extracellular matrix. d. components of a desmosome. Answer: a

Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of integrins in cell–matrix junctions. 18. Integrins bind to a. collagen, laminin, and fibronectin. b. plectin and proteoglycans. c. α-actinin, vinculin, and talin. d. All of the above Answer: d Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of integrins in cell–matrix junctions. 19. The major cell surface receptor(s) responsible for the attachment of cells to the extracellular matrix is(are) a. hemidesmosomes. b. desmosomes. c. integrins. d. laminin. Answer: c Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of integrins in cell–matrix junctions. 20. Cell adhesion molecules can be divided into four major groups. Which of the following is not one of those groups? a. Selectins b. Integrins c. Collagens d. Immunoglobulin superfamily Answer: c Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the major types of selective cell–cell adhesion. 21. The major family of calcium-mediated cell-surface adhesion molecules is a. calmodulins. b. calsequestrins. c. cadherins. d. fibronectins. Answer: c Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the major types of selective cell–cell adhesion.

22. If developing pre-nerve cells expressing only N-cadherin on their surfaces were mixed with epithelial cells expressing only E-cadherin, what would be the most likely result? a. All cells would mix and adhere to one another equally. b. Nerve cells would adhere to one another, and epithelial cells would adhere to one another. c. Nerve cells would adhere to one another in the center, and epithelial cells would surround them. d. Epithelial cells would adhere to one another in the center, and nerve cells would surround them. Answer: b Textbook Reference: Cell–Cell Interactions Bloom’s Category: 3. Applying Learning Objective: Summarize the major types of selective cell–cell adhesion. 23. Adherens junctions are linked to cytoplasmic a. actin filaments. b. intermediate filaments. c. microtubules. d. myosin filaments. Answer: a Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Summarize the major types of selective cell–cell adhesion. 24. Desmosomes are linked to cytoplasmic a. actin filaments. b. intermediate filaments. c. microtubules. d. myosin filaments. Answer: b Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Summarize the major types of selective cell–cell adhesion. 25. Which of the following molecules mediate the association of actin filaments with the cadherins in an adherens junction? a. Plakoglobulins b. Plectins c. Talins d. - and -catenin Answer: d Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Summarize the major types of selective cell–cell adhesion.

26. Desmosomes are held together by the adhesion of the two desmosomal cadherins: a. desmoglein and desmocollin. b. desmoglein and desmoplakin. c. desmoglobin and desmocollin. d. desmoglobin and desmoplakin. Answer: a Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the major types of selective cell–cell adhesion. 27. A junctional complex consists of a tight junction associated with both adherens junctions and a. desmosomes. b. gap junctions. c. hemidesmosomes. d. focal adhesions. Answer: a Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the role of tight junctions in epithelial sheets. 28. The function of gap junctions is to a. hold epithelia together. b. seal the space between cells. c. provide direct communication between cells. d. maintain the spacing of the basal laminae beneath epithelial cells. Answer: c Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 29. Which of the following can pass through gap junctions? a. cAMP and calcium ions b. Insulin and calcium ions c. cAMP and glucose d. Insulin and glucose Answer: a Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 30. All of the following can be due to a mutation in a gene for a gap junction protein except a. Charcot-Marie-Tooth disease. b. deafness. c. cataracts.

d. scurvy. Answer: d Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast gap junctions and plasmodesmata. 31. Adhesion of plant cells is mediated by a pectin-rich region of the cell wall called the a. external lamina. b. middle lamella. c. intermediate lamella. d. plasmodesmata. Answer: b Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 32. The communicating junctions between plant cells are called a. gap junctions. b. plasmodesmata. c. plasmotubules. d. lamellar junctions. Answer: b Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata.

Fill in the Blank 1. Within the plant cell wall, cellulose microfibrils are embedded in a matrix consisting of proteins and two other types of polysaccharides: _______ and _______. Answer: hemicelluloses; pectins Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Distinguish the organization of yeast and plant cell walls. 2. In plant cell walls, _______ are highly branched polysaccharides that are hydrogenbonded to cellulose microfibrils. Answer: hemicelluloses Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Distinguish the organization of yeast and plant cell walls. 3. Responsible for much of the rigidity of plant tissues, _______ is the internal hydrostatic pressure that builds up within the cell and equalizes its osmotic pressure. Answer: turgor pressure

Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Distinguish the organization of yeast and plant cell walls. 4. The hydroxyl groups on the modified amino acids of collagen form _______ that stabilize the collagen triple helix. Answer: hydrogen bonds Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. 5. The amino acid sequence of a collagen triple helix domain consists of Gly-X-Y repeats, in which X is frequently _______ and Y is frequently _______. Answer: proline; hydroxyproline Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. 6. The only glycosaminoglycan that occurs as a single long polysaccharide chain is _______. Answer: hyaluronan Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of proteoglycans. 7. Glycosaminoglycans (GAGs) consist of repeating disaccharide units. With the exception of hyaluronan, the sugars frequently contain _______. Answer: sulfate Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of proteoglycans. 8. Gap junctions allow passage of molecules that have a molecular weight of less than _______ daltons. Answer: 1,000 Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 9. The individual units of gap junctions are composed of _______ molecules. Answer: connexin Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 10. Gap junctions between nerve cells form _______ synapses.

Answer: electrical Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata.

True/False 1. Basal laminae are formed primarily from type IV collagen and other proteins. Answer: T Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. 2. The fibrous proteins of the extracellular matrix are embedded in polysaccharide gels formed of glycosaminoglycans. Answer: T Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of proteoglycans. 3. Aggrecan is a large proteoglycan consisting of about 100 hyaluronan chains extending from a core protein. Answer: F Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of proteoglycans. 4. Integrins are homodimers of two transmembrane polypeptide subunits. Answer: F Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of integrins in cell–matrix junctions. 5. Integrin was first localized to sites of cell adhesion to the extracellular matrix by immunofluorescence microscopy. Answer: T Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of integrins in cell–matrix junctions. 6. Cell adhesion mediated by selectins, integrins, and most cadherins requires divalent cations. Answer: T Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding

Learning Objective: Explain the functions of adhesion proteins. 7. Loss of E-cadherin can lead to the development of a cancer. Answer: T Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the major types of selective cell–cell adhesion. 8. Adherens junctions and desmosomes both link to intermediate filaments in cells. Answer: F Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Summarize the major types of selective cell–cell adhesion. 9. Tight junctions provide strong adhesions between cells. Answer: F Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the role of tight junctions in epithelial sheets. 10. Tight junctions involve a fusion between the outer leaflets of two membranes in the region of the junction. Answer: F Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the role of tight junctions in epithelial sheets. 11. Tight junctions are located around the entire circumference of the cell. Answer: T Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the role of tight junctions in epithelial sheets. 12. Charcot-Marie-Tooth disease is a tight junction disease. Answer: F Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 13. Plasmodesmata contain a single microtubule running from cell to cell down their center. Answer: F Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata.

14. Plasmodesmata can open and close in response to signals. Answer: F Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata.

Short Answer 1. What prevents collagen from assembling into fibrils inside the cell? Answer: Collagen is synthesized and secreted as a procollagen. Procollagen is cleaved outside the cell to produce collagen that is capable of assembly into fibrils. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. 2. Why is the term “basal lamina” preferred over the term “basement membrane”? Answer: Because this structure is a fibrous layer, not a lipoprotein membrane. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. 3. Describe the structure of laminins. Answer: Laminins are composed of three polypeptide chains—, , and —whose tails wind around one another and whose head regions contain binding sites for collagen and cells. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Describe the types of collagen in extracellular matrix. 4. What is the function of plectin? Answer: It connects intermediate filaments to desmosomes or hemidesmosomes. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Explain the functions of adhesion proteins. 5. Describe two ways in which a hemidesmosome differs from a desmosome. Answer: A hemidesmosome has integrins as its transmembrane connector proteins and connects to the extracellular matrix; a desmosome has desmosomal cadherins as its transmembrane connector proteins and connects to another cell. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Explain the functions of adhesion proteins. 6. Which proteins attach integrins to actin filaments at focal adhesions? Answer: Talin, vinculin, and -actinin

Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of integrins in cell–matrix junctions. 7. Describe the structure of an integrin, as determined from the cDNA sequence isolated from fibroblasts by Hynes et al. Answer: An integrin monomer contains a short cytosolic domain, a hydrophobic transmembrane domain, and a large extracellular domain containing multiple glycosylation sites. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the roles of integrins in cell–matrix junctions. 8. Describe the process of adhesion between leucocytes and endothelial cells that allows leucocytes to leave the circulation and enter an inflamed tissue. Answer: First, leucocyte selectins bind to endothelial carbohydrate groups; then, leucocyte integrins bind to ICAMs on the surface of the endothelial cells. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Summarize the major types of selective cell–cell adhesion. 9. What are the four classes of cell adhesion molecules? Answer: Selectins, integrins, Ig superfamily, and cadherins Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the major types of selective cell–cell adhesion. 10. What are the three cell-to-cell junctions found in the junctional complex between two intestinal epithelial cells? Answer: Tight junctions, adherens junctions, and desmosomes Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the major types of selective cell–cell adhesion. 11. What is the primary function of tight junctions? Answer: They prevent the free passage of molecules between cells of an epithelial sheet. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the role of tight junctions in epithelial sheets. 12. What is a secondary function of tight junctions? Answer: They separate the apical and basolateral membrane domains of the plasma membrane, preventing proteins and lipids from diffusing freely between them. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the role of tight junctions in epithelial sheets.

13. How many molecules of the gap junction protein form one open channel between two cells? Answer: Six connexins in each connexon in two adjacent cells, for a total of 12 molecules of connexin. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 14. How do the hexagonal units of connexin form a junction through which molecules can diffuse? Answer: They align the hexagonal cylinder of connexins in one membrane with another in the adjacent membrane so that the pores down the centers align to form a contiguous aqueous channel. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 15. What is the function of gap junctions in heart muscle? Answer: They allow ions to flow between cells, which couples and synchronizes the contractions of neighboring cells. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 16. What are the three most common disease consequences of connexin gene mutations? Answer: Deafness, cataracts, and skin disorders Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 17. Why does a mutation in a gene for the gap junction protein not affect gap junction function in all tissues that have gap junctions? Answer: Gap junctions are composed of several of the more than 21 different connexins, each of which is expressed only in certain tissues. Even in tissues that contain a mutation in a connexin, other connexins may function or compensate for the loss of that protein. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast gap junctions and plasmodesmata. 18. Why does any mutation in a gap junction protein produce defects in gap junction functions in sensitive tissues? Give two possible explanations. Answer: The remaining connexins in sensitive tissues may not be able to compensate for the loss of the mutant protein in forming functional connexins. Alternatively, the mutant protein may function in a dominant negative way that interferes with the processing, assembly, or function of the normal connexins expressed in that tissue.

Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast gap junctions and plasmodesmata. 19. What are plasmodesmata? Answer: Plasmodesmata are junctions that form between plant cells and are analogous to gap junctions. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. 20. How do plasmodesmata form? Answer: They form by incomplete separation of plant daughter cells after mitosis. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. A mutation in the bacterial gene encoding crescentin would likely have an effect on which type of bacteria? a. Spherical b. Rod-shaped c. Curved or spiral-shaped d. Filamentous Answer: c Textbook Reference: Cell Walls Bloom’s Category: 3. Applying Learning Objective: Describe the structure of bacterial cell walls. Feedback A: Incorrect. Crescentin is unique to differently shaped bacteria. Feedback B: Incorrect. Crescentin is unique to differently shaped bacteria. Feedback C: Correct! Curved or spiral-shaped bacteria contain crescentin, which is responsible for the shape of the cell wall. Feedback D: Incorrect. Crescentin is unique to differently shaped bacteria. 2. The basic structural polysaccharide of fungal cell walls, crab shells, and insect exoskeletons is a. cellulose. b. chitin. c. collagen. d. hemicellulose. Answer: b Textbook Reference: Cell Walls

Bloom’s Category: 1. Remembering Learning Objective: Distinguish the organization of yeast and plant cell walls. Feedback A: Incorrect. Cellulose is the basic structural polysaccharide of the plant cell wall. Feedback B: Correct! Chitin, a linear polymer of N-acetylglucosamine, is the basic structural polysaccharide of fungal cell walls, crab shells, and insect exoskeletons. Feedback C: Incorrect. Collagen is a major structural component of the animal extracellular matrix. Feedback D: Incorrect. Hemicellulose is a plant polysaccharide that hydrogen-bonds to the surface of cellulose microfibrils. 3. Cellulose, chitin, and hyaluronan are all deposited extracellularly by a. secretion. b. exocytosis. c. ABC transporters. d. plasma membrane enzyme complexes. Answer: d Textbook Reference: Cell Walls Bloom’s Category: 1. Remembering Learning Objective: Distinguish the organization of yeast and plant cell walls. Feedback A: Incorrect. Although ABC transporters are responsible for the transport across the plasma membrane of many chemotherapeutic drugs, they do not deposit any of these polysaccharides extracellularly. Feedback B: Incorrect. Exocytosis involves vesicles. The polysaccharides are transported extracellularly by the enzyme complex that synthesizes them. Feedback C: Incorrect. Secretion is a vesicular transport process. The polysaccharides are transported extracellularly by the enzyme complex that synthesizes them. Feedback D: Correct! Although the plasma membrane enzyme complex is different for all three, activated sugars in all cases are polymerized and spun out from the cell by the enzyme complex. 4. Animal cells are embedded in a(n) a. cell wall. b. extracellular matrix. c. hemicellulose-rich hyaluronan layer. d. pectin-rich basal lamina. Answer: b Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the types of collagen in extracellular matrix. Feedback A: Incorrect. Plant cells, for example, are surrounded by a cell wall. Feedback B: Correct! Animal cells (e.g., fibroblasts) are embedded in an extracellular matrix. Feedback C: Incorrect. Hemicellulose is a component of plant cell walls. Feedback D: Incorrect. Pectin is a component of the primary plant cell wall.

5. Which statement about collagen is false? a. It is the most abundant protein in animal tissues. b. It forms a double helix, with two collagen molecules wrapped around each other in a ropelike structure. c. It contains an amino acid called hydroxyproline. d. It is a secreted protein. Answer: b Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Describe the types of collagen in extracellular matrix. Feedback A: Incorrect. This is true. There are at 31 members of the collagen family of proteins, and they are major structural components of the extracellular matrix. Feedback B: Correct! This is false. Collagen forms a triple helix, not a double helix. Feedback C: Incorrect. This is true. Hydroxyproline is an unusual amino acid present in collagen that is thought to stabilize the helical structure by forming hydrogen bonds between polypeptide chains. Feedback D: Incorrect. This is true. Collagen is synthesized inside cells and passes through the secretory pathway from the ER to the Golgi and finally to the exterior of the cell. 6. Collagen fibrils form extracellularly and not intracellularly, even though their components are synthesized within the cell and transported to the cell surface via the Golgi. Which statement explains why this happens? a. Type I procollagen is synthesized with nonhelical segments at each end of the polypeptide chain, limiting their ability to form collagen fibrils in cells. b. Procollagen is soluble and small enough to fit within the intracellular transport machinery. c. Type I collagen consists of about 1000 amino acids or 330 Gly-X-Y repeats. d. Type I procollagen contains hydroxylated proline and lysine residues that prevent polypeptide chain formation within the cell. Answer: a Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 4. Analyzing Learning Objective: Describe the types of collagen in extracellular matrix. Feedback A: Correct! Although all of these statements are true, this is the only one that explains why procollagen does not polymerize into collagen fibrils within the cell. Feedback B: Incorrect. Although procollagen is small enough to be transported with in the transport machinery, this statement does not directly explain why fibrils are not formed within them. Feedback C: Incorrect. Although procollagen does consist of about 1000 amino acids in repeats of Gly-X-Y, this does not directly explain why fibrils are not formed within cells. Feedback D: Incorrect. Although hydroxylated proline and lysine residues stabilize collagen in fibrils outside the cell, this does not explain why they do not form within the cell. 7. Fibrous structural proteins of the extracellular matrix are embedded in gels formed

from polysaccharides called a. elastin. b. fibronectin. c. glycosaminoglycans (GAGs). d. laminin. Answer: c Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of proteoglycans. Feedback A: Incorrect. Elastin is the primary protein of elastic fibers. Elastic fibers are rich in tissues that must expand and contract (e.g., the lung). Feedback B: Incorrect. Fibronectin is the principal adhesive protein of connective tissue. Feedback C: Correct! GAGs form hydrated gels in which extracellular fibrous proteins such as collagen are embedded. Feedback D: Incorrect. Laminin is a major adhesive protein of the basal laminae. 8. The major function of adhesion proteins such as fibronectin is to a. link epithelial cell layers. b. link collagen, proteoglycans, and cells containing integrins. c. seal cell layers at tight junctions. d. support communication between adjacent cells. Answer: b Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 1. Remembering Learning Objective: Explain the functions of adhesion proteins. Feedback A: Incorrect. Multiple types of junctions link epithelial cell layers. Feedback B: Correct! Fibronectin is an extracellular adhesive protein that links collagen, proteoglycans, and cells containing integrins. Feedback C: Incorrect. Other proteins, not discussed in this chapter, seal cell layers at tight junctions. Feedback D: Incorrect. Gap junctions support communication between adjacent cells. 9. Integrins link extracellular matrix proteins, either directly or indirectly, to all of the following intracellular proteins except a. actin. b. -actinin. c. laminin. d. talin. Answer: c Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Describe the roles of integrins in cell–matrix junctions. Feedback A: Incorrect. Actin is an intracellular cytoskeletal protein that is linked to integrins through other proteins. Feedback B: Incorrect. -actinin is an intracellular protein that is linked directly to integrins.

Feedback C: Correct! Laminin is an extracellular protein. It is a major adhesive protein of the basal laminae. Feedback D: Incorrect. Talin is an intracellular protein that links directly to integrins. 10. Which statement about integrins is false? a. The subunit binds divalent cations. b. Each subunit has a large extracellular head group. c. Each subunit has a short cytoplasmic domain that interacts with intracellular proteins such as talin, -actinin, and plectin. d. Integrin has two subunits, termed  and  that dimerize with each other. Answer: a Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Describe the roles of integrins in cell–matrix junctions. Feedback A: Correct! It is the subunit that binds divalent cations. Feedback B: Incorrect. This is a property of each integrin subunit, whether  or . Feedback C: Incorrect. This is a property of integrins. Feedback D: Incorrect. The two subunits of integrin are indeed named  and  and they do dimerize with each other. 11. The cell–cell interactions mediated by the selectins, integrins, and most members of the Ig superfamily are _______ interactions in which the cytoskeletons of adjacent cells are not linked to one another. a. nonspecific b. physiologically unimportant c. stable d. transient Answer: d Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Summarize the major types of selective cell–cell adhesion. Feedback A: Incorrect. The cell–cell interactions are indeed selective for various cell types. Feedback B: Incorrect. These interactions are highly important physiologically. Feedback C: Incorrect. Stable interactions between cells are based largely on cadherins and involve linkages to cytoskeletal elements such as actin microfilaments and intermediate filaments. Integrins can indeed form stable interactions with the extracellular matrix at focal adhesions and hemidesmosomes. Feedback D: Correct! These proteins typically mediate transient interactions with no stable cytoskeletal involvement. For example, the interactions of leukocyte selectins with carbohydrate on the surface of endothelial cells are transient and slow the movement of the leukocyte through the bloodstream. 12. Homophilic interactions between cells involve interactions a. among adhesion molecules of the same molecular class. b. among adhesion molecules of different molecular classes.

c. between integrins and ICAMs. d. between cadherins and -catenin. Answer: a Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Summarize the major types of selective cell–cell adhesion. Feedback A: Correct! For example, interactions among N-CAMs lead to highly selective associations of nerve cells during development. Feedback B: Incorrect. These interactions are termed heterophilic. Feedback C: Incorrect. This is an example of a heterophilic interaction. Feedback D: Incorrect. Cadherins are cell–cell adhesion proteins and -catenin is an intracellular protein. The two interact with each other in adherens junctions. 13. Adherens junctions and desmosomes may be distinguished from each other on the basis of a. the presence of cadherins versus ICAMs. b. the linkage of cadherins either to actin filaments or to intermediate filaments, respectively. c. whether the cadherins are linked directly or indirectly to the cytoskeletal elements. d. whether the linkage is between two cells or between a cell and the extracellular environment. Answer: b Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Summarize the major types of selective cell–cell adhesion. Feedback A: Incorrect. Both contain cadherins. Desmosomal cadherins are named, for example, desmoglein and desmocollin. Feedback B: Correct! Adherens-junction linkage is with actin filaments while desmosomes link to intermediate filaments. Feedback C: Incorrect. In both adherens junctions and desmosomes, cadherins link to cytoskeletal elements and the linkages involve a set of adaptor proteins. Feedback D: Incorrect. Both adherens junctions and desmosomes are linkages between adjacent cells. Hemidesmosomes are a type of junction that links cells to extracellular matrix. 14. Specific recognition between cell types such as leukocytes and endothelial cells of blood vessels is mediated by cell-surface glycoproteins called a. selectins. b. fibronectins. c. spectrins. d. porins. Answer: a Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Summarize the major types of selective cell–cell adhesion. Feedback A: Correct! E-selectin on the surface of endothelial cells binds specific

oligosaccharides on the surface of leukocytes, and L-selectin on the surface of leukocytes binds specific oligosaccharides on the surface of endothelial cells, thus providing a specific cell–cell interaction. Such proteins are examples of a general class of sugarbinding proteins termed lectins. Feedback B: Incorrect. Fibronectin is an extracellular matrix protein. Feedback C: Incorrect. Spectrin is involved in the formation of the cell cortex, the microfilament network beneath the cell surface. Feedback D: Incorrect. Porins are a class of proteins that form pores in bacterial outer membranes. 15. A major function of tight junctions is to a. create strong cell–cell adherence. b. promote the intermixing of proteins between the apical and basolateral surfaces of cells. c. seal the space between adjacent cells to provide, for example, a barrier to glucose diffusion across an epithelial cell layer. d. sort cells on the basis of adhesion during cell migration in embryos. Answer: c Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Describe the role of tight junctions in epithelial sheets. Feedback A: Incorrect. Tight junctions do not provide a mechanically strong junction. Rather, adherens junctions and desmosomes provide adhesive strength. Feedback B: Incorrect. Tight junctions block the intermixing by diffusion of proteins between the apical and basolateral surfaces of cells. Feedback C: Correct! This is one of the major functions of tight junctions. The other is to block diffusion between the apical and basolateral surfaces of cells. Feedback D: Incorrect. This is not a major function of tight junctions. Other cell adhesion molecules perform this function. 16. The junctions between animal cells that allow the free passage of molecules smaller than approximately 1000 daltons are called a. tight junctions. b. plasmodesmata. c. gap junctions. d. hemidesmosomes. Answer: c Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. Feedback A: Incorrect. Tight junctions prevent the passage of molecules between epithelial cells within a sheet, as well as between the apical and basolateral surfaces within a single epithelial cell. Feedback B: Incorrect. Plasmodesmata are junctions between plant cells, not animal cells. Feedback C: Correct! Gap junctions, which are formed by connexins, are open channels between cells that allow the passage of ions and small molecules. Feedback D: Incorrect. Hemidesmosomes anchor epithelial cells to the basal laminae.

17.Although structurally different from gap junctions in animal cell tissues, _______ may be thought of as the functional equivalent of gap junctions in plants. a. adherens junctions b. desmosomes c. hemidesmosomes d. plasmodesmata Answer: d Textbook Reference: Cell–Cell Interactions Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast gap junctions and plasmodesmata. Feedback A: Incorrect. Adherens junctions are found in animal cells. They link two cells and involve cadherins linked through adaptors to actin filaments. Feedback B: Incorrect. Desmosomes are found in animal cells. They link two cells and involve cadherins linked through adaptors to intermediate filaments Feedback C: Incorrect. Hemidesmosomes are found in animal cells. They link cells to the extracellular matrix and involve integrins and intermediate filaments. Feedback D: Correct! Plasmodesmata are strands of cytoplasm surrounded by plasma membrane that go through the cell wall to link adjacent plant cells metabolically.

Essay 1. Suppose that a research group is conducting experiments to test how penicillin kills bacteria. Their starting point is a recent finding that penicillin inhibits the enzyme crosslinked between different strands of the peptidoglycan that forms the principal component of the bacterial cell wall. Formulate a hypothesis based on an analogy between the bacterial and plant cell wall and on the concept of turgor pressure. Briefly state the hypothesis, and propose an experiment to test it. Answer: In the absence of cross-linking due to penicillin treatment, the bacterial cell wall should be much less rigid and hence unable to resist cell swelling in response to osmotic differences. It is likely that the bacteria will lyse as a result of osmotic swelling. If so, increasing the osmotic strength of the extracellular media should prevent swelling and enable the bacteria to grow and multiply in the presence of penicillin. One test for this would be to increase external osmotic strength by adding a disaccharide to the media that cannot be metabolized. Textbook Reference: Cell Walls Bloom’s Category: 3. Applying Learning Objective: Describe the structure of bacterial cell walls. 2. Suppose you have treated plant cells in culture with a drug to disrupt microtubules. You then examine the orientation of cellulose fibrils in newly-synthesized primary cell walls that developed under these conditions. What will be the observed orientation of the cellulose fibrils with respect to one another? Answer: Movement of cellulose synthase complexes during cellulose synthesis is believed to be aligned by microtubules. Since the drug disrupted the microtubules,

removing the tracks for cellulose synthase complex alignment, the cellulose will be laid down in a random pattern rather than an organized one. Textbook Reference: Cell Walls Bloom’s Category: 3. Applying Learning Objective: Explain the role of the cell wall in plant cell growth. 3. How is proline hydroxylation in collagen encoded in the genome? What is the relationship between this encoding mechanism and scurvy? Answer: The genome does not directly specify the presence of hydroxyproline at any given position in collagen. Rather, the genome codes for both collagen and the enzyme prolyl hydroxylase. The binding properties of the enzyme are a consequence of the amino acid sequence encoded by the genome. This enzyme requires vitamin C for its activity. In the absence of vitamin C, the enzyme is inactive, hydroxylation of proline does not occur, and collagen chains do not properly associate with one another. As a result, skin and blood vessels, for example, are weakened, producing the symptoms of scurvy. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Describe the types of collagen in extracellular matrix. 4. Unlike polysaccharides such as glycogen or cellulose, glycosaminoglycans (GAGs) are highly negatively charged. What is the source and function of this negative charge? Answer: GAGs contain saccharide groups that contain negative charge contributions from both carboxylate groups and sulfate groups. Sulfate, unlike carboxyl groups, is a strong acidic group. Hence, GAGs are highly negatively charged at physiological pH. GAGs, by virtue of their negative charge, bind positively charged ions. They also trap water, producing a hydrated gel, which provides mechanical support to the extracellular matrix. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of proteoglycans. 5. Hemidesmosomes link intermediate filaments inside intestinal epithelial cells to the basal lamina, an organized zone of extracellular matrix that underlies the intestinal epithelium. What is the advantage of this to a flexible tissue? Answer: Two forms of contacts hold the intestinal epithelium together. One is cell–cell adhesions and the other is hemidesmosomes. Hemidesmosomes link integrins as transmembrane proteins to adaptors and intermediate filaments inside the cell and to laminins and the basal lamina outside the cell. Broadly, the desmosome contributes to the strength of the cell layer. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Describe the roles of integrins in cell–matrix junctions. 6. Integrins are dimeric transmembrane proteins of the plasma membrane. According to the fluid mosaic model of membrane structure (see Chapter 14), transmembrane proteins should be free to diffuse within the lipid bilayer. Yet, as shown by Tamkun and colleagues in 1986, integrins have a very restricted distribution in the plasma membrane

of fibroblasts. The distribution often appears as a series of tracks across the cell surface. How can this distribution be explained? Answer: Integrins interact with various intracellular proteins that bridge to actin filaments or intermediate filaments. These cytoskeletal elements are arranged in a linear manner and create a linear arrangement of integrin, as revealed by immunofluorescence staining. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Describe the roles of integrins in cell–matrix junctions. 7. A standard approach to detaching tissue culture cells from substratum is to incubate the culture briefly in a saline solution containing a divalent cation chelator and trypsin. How does this procedure detach the cells from the surface? Answer: Both integrins and cadherins require divalent cations for their adhesive properties. Removal of divalent cations by chelators will greatly decrease the adhesive properties of these molecules. Cell adhesion molecules are also cell-surface proteins. Trypsin will digest these and also lead to a decrease in cell adhesiveness, affecting cell detachment. Textbook Reference: The Extracellular Matrix and Cell–Matrix Interactions Bloom’s Category: 2. Understanding Learning Objective: Describe the roles of integrins in cell–matrix junctions. 8. In vivo, epithelial cells exhibit basolateral and apical surfaces. The tight junctional complex separates these two distinct cell surfaces from each other and joins the cells into an epithelial cell layer. In vivo cells showing such properties are resistant to ion flow across the cell layer. In vitro, many cultured epithelial cell systems show high electrical resistance but fail to exhibit a polarized distribution of proteins and lipids between the apical and basolateral surfaces. Present a working hypothesis to explain this observation. Answer: According to the results of this experiment, the existence of tight junction complexes between cells appears insufficient to cause polarized protein and lipid distribution. The presence of tight junctions seems, at most, to permit the generation of polarized protein distributions. Other processes must be important in creating the asymmetry. One possible hypothesis is that targeted membrane trafficking delivers proteins selectively, although this would not explain why the trafficking is targeted in the first place. Small differences in cell–cell or cell-basal lamina interactions may be important. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 4. Analyzing Learning Objective: Describe the role of tight junctions in epithelial sheets. 9. The X-linked variant of Charcot-Marie-Tooth disease is likely caused by mutations in a gene encoding a probable connexin 32. Suppose your laboratory is working with a human cell line called HeLa cells, which do not express connexins. Propose an experiment to demonstrate that the probable wild-type connexin 32 is actually a gap junction protein. Answer: Protein expression from a plasmid encoding connexin 32 would be an important functional assay. The experiment would seek to answer two questions: Do HeLa cells

gain gap-junction function with plasmid expression? And, will a microinjected small molecule now diffuse between the gap junction–coupled cells? Such an experiment would be possible because HeLa cells do not express gap junction proteins, and therefore in the absence of expression they do not form gap junctions. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 3. Applying Learning Objective: Compare and contrast gap junctions and plasmodesmata. 10. Explain how plasmodesmata in plants allow for paracrine signaling (signaling between adjacent or neighboring cells). Answer: Plasmodesmata physically link plant cells through membrane-limited strands of cytoplasm. Small molecules or signaling proteins can be exchanged between neighboring cells through plasmodesmata. This process produces the functional equivalent of paracrine signaling. Textbook Reference: Cell–Cell Interactions Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast gap junctions and plasmodesmata.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 17: Cell Signaling TEST FILE QUESTIONS Multiple Choice 1. Signaling by the steroid hormone estrogen is an example of _______ signaling. a. autocrine b. endocrine c. paracrine d. direct cell-to-cell Answer: b Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Describe the principal modes of cell signaling. 2. Signaling by neurotransmitters is an example of _______ signaling. a. autocrine b. endocrine c. paracrine d. direct cell-to-cell Answer: c Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Describe the principal modes of cell signaling. 3. Stimulation of T lymphocytes, leading to their synthesis of a growth factor resulting in T lymphocyte proliferation, is an example of _______ signaling. a. autocrine b. endocrine c. paracrine d. direct cell-to-cell Answer: a Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Describe the principal modes of cell signaling. 4. Signaling by cadherins is an example of _______ signaling. a. autocrine b. endocrine

c. paracrine d. direct cell-to-cell Answer: d Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Describe the principal modes of cell signaling 5. Which signal molecule diffuses through the plasma membrane to ultimately bind nuclear receptors and influence transcription? a. Estrogen b. Nitric oxide c. Cadherins d. Nerve growth factor Answer: a Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 1. Remembering Learning Objective: Explain how steroid hormones regulate gene expression. 6. Steroid hormones usually act via receptors that a. are coupled to G proteins that activate adenylyl cyclase. b. activate tyrosine kinases. c. bind to DNA. d. activate phospholipase C. Answer: c Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Explain how steroid hormones regulate gene expression. 7. Statins, a class of drugs that are often administered to patients with high cholesterol, inhibit the biosynthesis of cholesterol. Statin drugs would not be expected to affect the biosynthetic pathway of which of the following signal molecules? a. Thyroid hormone b. Vitamin D3 c. Retinoic acid d. Estrogen Answer: c Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Explain how steroid hormones regulate gene expression. 8. A glucocorticoid binding to its receptor stimulates a. phosphorylation of a transcription factor protein that activates a gene. b. formation of a receptor dimer that triggers an intracellular signal pathway. c. formation of a receptor dimer that binds to and activates a gene. d. binding of the receptor monomer to a gene. Answer: c

Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Explain how steroid hormones regulate gene expression. 9. Nitric oxide is a signal molecule that can a. bind to surface receptors and activate second messengers. b. bind to surface receptors and open ion channels. c. diffuse across cell membranes and bind to receptors that regulate transcription. d. diffuse across cell membranes and directly alter the activity of intracellular enzymes. Answer: d Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 1. Remembering Learning Objective: Compare the actions of different types of small signaling molecules. 10. Nitric oxide is considered a paracrine signal molecule because it a. is slow to diffuse. b. is produced in very small quantities. c. is unstable, with a short half-life. d. binds to cell surface receptors that are very plentiful. Answer: c Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 1. Remembering Learning Objective: Compare the actions of different types of small signaling molecules. 11. The medical usage of nitroglycerine in heart disease is based on its a. inhibition of adenylate cyclase and blood vessel contraction. b. conversion to NO and blood vessel dilation. c. activation of NO-synthase and muscle cell contraction. d. promotion of cyclooxygenase and prostaglandin synthesis. Answer: b Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 1. Remembering Learning Objective: Compare the actions of different types of small signaling molecules. 12. Neurotransmitters act by binding to receptors that are a. ligand-gated ion channels. b. located in the cytoplasm. c. tyrosine-kinase receptors. d. not coupled to G proteins. Answer: a Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 1. Remembering Learning Objective: Compare the actions of different types of small signaling molecules. 13. Enkephalins and endorphins bind to the same receptors in the brain as a. nerve growth factor.

b. aspirin. c. morphine. d. acetylcholine. Answer: c Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Compare the actions of different types of small signaling molecules. 14. In plants, cytokinins stimulate a. cell division. b. cell elongation. c. cell enlargement. d. fruit ripening. Answer: a Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 1. Remembering Learning Objective: Compare the actions of different types of small signaling molecules. 15. In plant cells, which signaling molecule induces cell elongation? a. Acetylcholine. b. -aminobutyric acid (GABA). c. Auxin. d. Retinoic acid. Answer: c Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 1. Remembering Learning Objective: Compare the actions of different types of small signaling molecules. 16. Which molecule stimulates fibroblasts to proliferate and thus heal a wound? a. EGF b. NGF a. NO d. PDGF Answer: d Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Give examples of polypeptide growth factors. 17. Differentiation of blood cells and control of lymphocytes during the immune response is regulated by which type of signaling molecule? a. Enkephalins b. Fibroblast growth factor c. Platelet derived growth factor d. Cytokines Answer: d Textbook Reference: Signaling Molecules and Their Receptors

Bloom’s Category: 1. Remembering Learning Objective: Give examples of polypeptide growth factors. 18. The G protein that activates adenylyl cyclase is a a. monomeric G protein in the Ras family that binds GTP. b. dimeric G protein that separates into  and  subunits. c. heterotrimeric G protein that separates into  and  subunits. d. heterotrimeric G protein that separates into  and  subunits. Answer: c Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of a G protein-coupled receptor. 19. In an active state of a G protein, the a.  subunit binds to a target protein, and the  subunit remains bound to the receptor. b.  and  subunits both can bind to target proteins. c.  and  subunits both can bind to target proteins. d.  subunit can bind to a target protein, and the  subunit remains bound to the receptor. Answer: b Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of a G protein-coupled receptor. 20. The GTP on the G protein that is linked to adenylyl cyclase is split to GDP and Pi a. in the inactive state. b. in the process of subunit separation. c. upon activation by the receptor. d. by the active subunit. Answer: d Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 2. Understanding Learning Objective: Explain how G proteins carry signals to their target enzymes. 21. Cholera toxin inhibits the ability of the  subunit of Gs to split GTP. If you treated cells with cholera toxin, the resulting effect would be _______ of adenylyl cyclase. a. stimulation b. inhibition c. molecular degradation d. increased synthesis Answer: a Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 3. Applying Learning Objective: Explain how G proteins carry signals to their target enzymes.

22. A mutation that causes a G-protein to lose its ability to hydrolyze bound GTP would be expected to have constitutively a. inactive  subunits. b. bound  subunits. c. active  subunits. d. inactive  subunits. Answer: c Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 2. Understanding Learning Objective: Explain how G proteins carry signals to their target enzymes. 23. The alpha subunit of the G protein that is associated with the epinephrine receptor, Gs, a. opens Ca2+ channels. b. closes Na+ channels. c. activates adenylate cyclase. d. inhibits adenylate cyclase. Answer: c Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Explain how G proteins carry signals to their target enzymes. 24. A major function of cAMP in animal cells is to activate a. adenylyl cyclase. b. protein kinase A. c. protein kinase C. d. tyrosine kinases. Answer: b Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of cAMP. 25. Protein kinase A is activated by a. phosphorylation of its catalytic subunit. b. phosphorylation of its regulatory subunits. c. binding of cAMP to its catalytic subunits. d. binding of cAMP to its regulatory subunits. Answer: d Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of cAMP. 26. Binding of cAMP to the _______ subunits of protein kinase A (PKA) leads to _______ of that protein kinase. a. catalytic; activation b. catalytic; inactivation

c. regulatory; inactivation d. regulatory; activation Answer: d Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of cAMP. 27. Protein kinase A regulates glycogen metabolism by phosphorylating glycogen synthase and a. phosphorylase kinase. b. glycogen phosphatase. c. glycogen phosphorylase. d. glucokinase. Answer: a Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of cAMP. 28. Protein kinase A regulates glycogen metabolism by _______ glycogen synthase and _______ glycogen phosphorylase. a. activating; activating b. inactivating; inactivating c. activating; inactivating d. inactivating; activating Answer: d Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of cAMP. 29. The action effected by protein kinase A is terminated by a. inactivation of the initial receptor. b. inactivation of the stimulatory G protein. c. degradation of cAMP. d. dephosphorylation of phosphoproteins by protein phosphatase 1. Answer: d Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of cAMP 30. During the odorant response in the cilia of olfactory neurons, the second messenger _______, leads to the opening of ion channels in the plasma membrane of olfactory neurons and the production of a nerve impulse. a. Ca2+ b. cAMP c. cGMP d. IP3

Answer: b Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of cAMP 31. Protein X is activated by protein kinase A. A mutation in protein X that replaces the protein’s only serine residue with an arginine residue would be expected to a. increase activation of protein X. b. eliminate activation of protein X. c. have no effect on activation of protein X. d. prevent inactivation of protein X. Answer: b Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of cAMP 32. CREB is activated by a. cAMP binding in the nucleus. b. phosphorylation by protein kinase A in the nucleus. c. cAMP binding in the cytoplasm. d. phosphorylation by protein kinase A in the cytoplasm. Answer: b Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Describe gene regulation by cAMP-dependent protein kinase. 33. Which statement correctly describes how protein kinase A can activate genes? a. Nuclear protein kinase A is activated by cAMP to phosphorylate general transcription factors. b. Cytosolic protein kinase A is activated by cAMP to release the catalytic subunits, which move into the nucleus and phosphorylate CREB. c. Cytosolic protein kinase A is activated by cAMP to release the catalytic subunits, which move into the nucleus and phosphorylate general transcription factors. d. Nuclear protein kinase A is activated by cAMP to phosphorylate CREB. Answer: b Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 2. Understanding Learning Objective: Describe gene regulation by cAMP-dependent protein kinase. 34. Activated CREB protein a. phosphorylates protein phosphatase 1, leading to its activation. b. binds to specific DNA sequences and influences transcription of genes involved in growth and development. c. binds to ribosomal subunits and influences translation of proteins important in growth and development. d. phosphorylates protein kinase A in the cytoplasm.

Answer: b Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Describe gene regulation by cAMP-dependent protein kinase. 35. The first step in growth factor pathway activation is a. receptor dimerization. b. receptor phosphorylation. c. Ras activation. d. the binding of SH2-containing proteins. Answer: a Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Describe signaling by receptor tyrosine kinases. 36. MAP kinase is an abbreviation for _______ protein kinase. a. microtubule-associated b. mitogen-activated c. mitosis-activating d. mitosis-associated Answer: b Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Describe signaling by receptor tyrosine kinases. 37. A researcher generated a mutant SH2-containing protein such that it binds tyrosine and phosphotyrosine with equal affinity. As a result, MEK activity would be expected to a. increase with ligand binding-induced dimerization. b. remain the same with receptor dimerization and autophosphorylation. c. decrease due to changes in Raf activation. d. decrease due to allosteric inhibition of SH2-domain binding. Answer: b Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 4. Analyzing Learning Objective: Describe signaling by receptor tyrosine kinases. 38. Binding of integrins to the extracellular matrix leads to integrin clustering and activation of the nonreceptor tyrosine kinase a. CREB. b. FAK. c. PKA. 2+ d. Ca /calmodulin-dependent kinase. Answer: b

Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast the activities of receptor and nonreceptor kinases. 39. Mutated oncogenic Ras proteins usually a. cleave GTP more rapidly than normal. b. fail to bind Raf. c. cleave GTP less rapidly than normal. d. bind GAP more tightly than normal. Answer: c Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Explain how Ras and Raf are activated downstream of tyrosine kinases. 40. Ras is a membrane-bound _______ when activated. a. small monomeric G protein that binds GTP b. dimeric G protein that separates into  and  subunits c. heterotrimeric G protein that separates into and  subunits d. heterotrimeric G protein that separates into  and  subunits Answer: a Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Explain how Ras and Raf are activated downstream of tyrosine kinases. 41. MAP kinase signal cascades are often organized into functional groups or cassettes by a. scaffold proteins. b. binding of SH2 domains to each other. c. lipid rafts. d. binding MAP kinase-responsive genes. Answer: a Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Explain how Ras and Raf are activated downstream of tyrosine kinases. 42. The RAS guanine nucleotide-binding protein is activated by a. cAMP binding. b. serine phosphorylation by protein kinase A. c. GDP binding.

d. GTP binding. Answer: d Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Explain how Ras and Raf are activated downstream of tyrosine kinases. 43. MEK is a protein kinase that phosphorylates _______ residues. a. threonine and tyrosine b. tyrosine c. histidine d. arginine and lysine Answer: a Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Give an example of transcriptional regulation by MAP kinase signaling. 44. Hydrolysis of PIP2 by phospholipase C is stimulated by a. G protein-linked receptors. b. protein-tyrosine kinase receptors. c. serine-threonine kinase receptors. d. both G protein-linked receptors and protein-tyrosine kinase receptors. Answer: d Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Summarize signaling by PI 3-kinase and mTOR. 45. The second messenger, phosphatidylinositol 3,4,5-trisphosphate (PIP3) is generated by the action of the enzyme _______ on phosphatidylinositol 4,5-bisphosphate (PIP2). a. protein kinase A b. protein kinase C c. phosphatidylinositide (PI) 3-kinase d. mTOR kinase Answer: c Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Summarize signaling by PI 3-kinase and mTOR. 46. The mTOR pathway is a central regulator of cell growth that couples the control of a. protein synthesis to the availability of growth factors, nutrients, and energy.

b. DNA replication during mitosis and during meiosis. c. transcription factors that regulate gene expression in the liver. d. transcription factors that regulate gene expression in early embryonic development. Answer: a Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Summarize signaling by PI 3-kinase and mTOR. 47. The TGF-β/Smad pathway is similar to JAK/STAT signaling in that a protein kinase associated with a receptor directly phosphorylates and activates a transcription factor. However, the receptors for transforming growth factor beta (TGF-) and related polypeptides are protein kinases that phosphorylate _______, rather than _______, residues on their substrate proteins. a. serine or threonine; tyrosine b. tyrosine; serine or threonine c. proline or aspartate; tyrosine d. tryptophan; tyrosine Answer: a Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast TGF-/Smad and JAK/STAT signaling. 48. NF-B can activate genes during the immune response by a protein kinase that a. phosphorylates the transcription factor NF-B to activate it. b. phosphorylates the inhibitory factor IB, causing it to be degraded and to release NFB. c. phosphorylates a MAP kinase to initiate its cascade. d. activates a phosphorylase that removes an inhibiting phosphate from the NF-B. Answer: b Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Describe the NF-B pathway. 49. Which of the following would lead to a termination of Wnt signaling? a. Phosphorylation of Disheveled b. Inactivation of the destruction complex c. Phosphorylation of  -catenin d. Activation of Tcf Answer: c Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of proteolysis in the Wnt and Notch pathways. 50. Phosphorylation of G protein-coupled receptors by GRK a. turns off G protein signaling.

b. hypersensitizes G protein signaling. c. blocks the dissociation of the G protein from the receptor. d. inhibits the GTPase activity of the alpha subunit of the G protein. Answer: a Textbook Reference: Signaling Dynamics and Networks Bloom’s Category: 2. Understanding Learning Objective: Explain how the dynamics of signaling can alter biological response. 51. How would inhibition of PI 3-kinase affect ERK activity in a growth factor-stimulated cell? a. ERK activity would be inhibited. b. ERK activity would be unaffected. c. ERK activity would be upregulated. d. ERK protein would be targeted for destruction by proteolysis. Answer: a Textbook Reference: Signaling Dynamics and Networks Bloom’s Category: 2. Understanding Learning Objective: Summarize the role of crosstalk in integrating cellular response to extracellular stimuli.

Fill in the Blank 1. The region of a gene that binds to a PKA phosphorylated protein is a _______. Answer: cAMP response element (CRE) Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 1. Remembering Learning Objective: Describe gene regulation by cAMP-dependent protein kinase. 2. The ras gene is mutant in about 25% of all human cancers. When ras is mutant, it typically remains active in mitogenic signal transduction because it is unable to _______ GTP. Answer: hydrolyze (cleave) Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Explain how Ras and Raf are activated downstream of tyrosine kinases. 3. The protein kinase Raf phosphorylates the protein kinase _______, which then phosphorylates (on both threonine and tyrosine residues) _______. Answer: MEK; ERK (MAP kinase) Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Explain how Ras and Raf are activated downstream of tyrosine

kinases. 4. Key targets of the JAK kinases are _______ proteins. Answer: STAT Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Explain how Ras and Raf are activated downstream of tyrosine kinases. 5. The phosphorylated ERK kinase moves into the nucleus and phosphorylates the transcription factor _______, which binds to serum response elements in a complex with _______ and turns on a set of genes called the _______ genes. Answer: Elk-1; SRF (serum response factor); immediate early Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Give an example of transcriptional regulation by MAP kinase signaling. 6. Expression of the immediate early genes triggers the expression of a battery of other downstream genes called _______. Answer: secondary response genes Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Give an example of transcriptional regulation by MAP kinase signaling. 7. In animal cells, the common growth factor, or mitogen-activated protein kinase signaling pathway, starts with phosphorylation of tyrosines on the cytosolic domains of the receptor by a process called _______. Answer: autophosphorylation Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Summarize signaling by PI 3-kinase and mTOR. 8. The extensive crosstalk between individual signal transduction pathways means that multiple pathways interact with one another to form _______ within the cell. Answer: signaling networks Textbook Reference: Signaling Dynamics and Networks Bloom’s Category: 2. Understanding Learning Objective: Explain how the dynamics of signaling can alter biological response.

9. _______ refers to the interaction of one signaling pathway with another and integrates the activities of different pathways within the cell. Answer: Crosstalk Textbook Reference: Signaling Dynamics and Networks Bloom’s Category: 2. Understanding Learning Objective: Summarize the role of crosstalk in integrating cellular response to extracellular stimuli.

True/False 1. Signaling molecules that activate cell surface receptors are essentially the same as those that activate intracellular receptors. Answer: F Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Compare the actions of different types of small signaling molecules. 2. Intracellular signaling pathways provide multiple opportunities for the amplification of a response to an extracellular signal. Answer: T Textbook Reference: G Proteins and Cyclic AMP Bloom’s Category: 2. Understanding Learning Objective: Explain how G proteins carry signals to their target enzymes. 3. Binding of integrins to the extracellular matrix stimulates the FAK protein-tyrosine kinase. Answer: T Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast the activities of receptor and nonreceptor kinases. 4. Many cancers arise as a result of a breakdown in signaling pathways that control cell proliferation. Answer: T Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Explain how Ras and Raf are activated downstream of tyrosine kinases. 5. Ras activation is mediated by guanine nucleotide exchange factors (GEFs) that stimulate the release of bound GDP and its exchange for GTP. Answer: T

Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Remembering Learning Objective: Explain how Ras and Raf are activated downstream of tyrosine Kinases. 6. TGF- signaling leads to the activation of the Smad family of transcription factors. Answer: T Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast TGF-/Smad and JAK/STAT signaling. 7. NF-B activates the gene for IB, which feeds back and inhibits NF-B activity. Answer: T Textbook Reference: Receptors Coupled to Transcription Factors - NF-B signaling Bloom’s Category: 2. Understanding Learning Objective: Describe the NF-B pathway. 8. The binding of Hedgehog to the receptor Patched releases a transcription factor from actin microfilaments that enters the nucleus and turns on genes. Answer: F Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of proteolysis in the Wnt and Notch pathways. 9. Notch is a cell-surface receptor protein that activates transcription when a piece of the Notch molecule is cleaved, enters the nucleus, and activates genes. Answer: T Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of proteolysis in the Wnt and Notch pathways. 10. Wnt is a signal molecule that binds to Frizzled/LRP, activating Dishevelled to inhibit a protein kinase. This results in longer lived -catenin, which enters the nucleus and turns on target genes. Answer: T Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of proteolysis in the Wnt and Notch pathways. 11. In signaling pathways, the duration of a single activated pathway (for example, short versus long periods of time) often leads to completely different cellular responses. Answer: T Textbook Reference: Signaling Dynamics and Networks Bloom’s Category: 2. Understanding Learning Objective: Explain how the dynamics of signaling can alter biological response.

12. -arrestin binds to phosphorylated G protein-coupled receptors, binding Raf, MEK, and ERK into the complex and linking the G protein-coupled signal pathway to the ERK signal pathway. This is an example of inhibitory crosstalk. Answer: F Textbook Reference: Signaling Dynamics and Networks Bloom’s Category: 3. Applying Learning Objective: Summarize the role of crosstalk in integrating cellular response to extracellular stimuli.

Short Answer 1. Why is it important that neurotransmitters receptors be located on the external surface of cells? Answer: Neurotransmitters are hydrophilic and unable to cross the plasma membrane of their target cells; thus external facing receptors are required. Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Describe the principal modes of cell signaling. 2. In glucocorticoid signaling, what would happen if Hsp90 was mutated such that it was always bound to the glucocorticoid receptor (GR)? Answer: Normally, Hsp90 binds GR in the cytoplasm, and upon glucocorticoid binding of the GR, Hsp90 releases GR, and unbound GR can create a dimer and translocate to the nucleus, where it functions as a transcription factor. This mutation would completely block GR function. Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 4. Analyzing Learning Objective: Explain how steroid hormones regulate gene expression. 3. Caffeine is a phosphodiesterase inhibitor. What affect would caffeine treatment have on cells growing in culture? Answer: Phosphodiesterase converts cAMP to AMP; therefore, caffeine would elevate intracellular levels of cAMP and exacerbate cAMP signaling. Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 4. Analyzing Learning Objective: Explain how G proteins carry signals to their target enzymes. 4. In the absence of cAMP, what is the status of protein kinase A? Answer: It is in the cytoplasm in a tetramer complex of two regulatory subunits associated with the two catalytic subunits and inactive. Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 3. Applying Learning Objective: Summarize the role of cAMP.

5. How would a mutation in CREB that prevents its phosphorylation affect cAMP signaling? Answer: It would prevent transcription of CREB responsive genes since phosphorylation of CREB is required to activate it. Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 4. Analyzing Learning Objective: Describe gene regulation by cAMP-dependent protein kinase. 6. What is the family of receptors for most growth factors? Answer: Receptor tyrosine kinases Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Describe signaling by receptor tyrosine kinases. 7. What are the three steps by which receptor tyrosine kinases typically convey an extracellular signal to another intracellular enzyme? Answer: (1) Dimerization of receptors in the presence of growth factor or ligand, (2) autophosphorylation of tyrosine residues on cytoplasmic domains of the receptors, and (3) binding of downstream molecules that contain SH2 domains and activation of these molecules. Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Describe signaling by receptor tyrosine kinases. 8. Describe three features shared by protein-tyrosine kinase growth factor receptors. Answer: (1) An N-terminal extracellular ligand-binding domain, (2) a single transmembrane  helix, and (3) a cytosolic (C-terminal) protein-tyrosine kinase domain. Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Describe signaling by receptor tyrosine kinases. 9. What are the first three steps in signaling from cytokine receptors? Answer: (1) Receptor dimerization, (2) cross-phosphorylation of associated nonreceptor tyrosine kinases, and (3) phosphorylation of tyrosines on the receptors. Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast the activities of receptor and nonreceptor kinases. 10. How would a mutation in the TGF- type I receptor that prevented it from being phosphorylated affect the cell?

Answer: Binding of TGF- to the TGF- receptor activates the kinase activity of the type II receptor to phosphorylate the type I receptor, which then activates Smad signaling. Therefore, if the type I receptor could not be phosphorylated, Smad signaling would be completely blocked. Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 4. Analyzing Learning Objective: Compare and contrast TGF-/Smad and JAK/STAT signaling. 11. What is the role of -arrestin in mediating crosstalk signaling between G proteincoupled receptors (GPCR) and ERK signaling? Answer: Crosstalk from G-protein receptors and ERK occurs when GRK (G proteincoupled receptor kinase) phosphorylates the GPCR. This phosphorylation allows -arrestin to bind the phosphorylated GPCR, and then serves as an adaptor that binds and brings together Raf, MEK, and ERK, thus activating the ERK pathway. Textbook Reference: Signaling Dynamics and Networks Bloom’s Category: 3. Applying Learning Objective: Summarize the role of crosstalk in integrating cellular response to extracellular stimuli.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. Which of the following is an example of endocrine signaling? a. Epinephrine release by motor neurons at the neuromuscular junction and binding to receptors on adjacent skeletal muscle cells b. Antigen stimulation of T lymphocytes, leading to the stimulation and synthesis of a growth factor that drives their own proliferation c. Insulin release by  cells in the pancreas, mediating an effect of glucose uptake by muscle cells d. None of the above Answer: c Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Describe the principal modes of cell signaling. Feedback A: Incorrect. This is an example of paracrine signaling. Feedback B: Incorrect. This is an example of autocrine signaling. Feedback C: Correct! This is classic endocrine signaling: release of a signaling molecule affecting distant target cells, typically carried through the bloodstream. Feedback D: Incorrect. Insulin release is an example of endocrine signaling. 2. The term “paracrine signaling” refers to a. signaling between cells located far from each other. b. stimulation of a cell by substances produced by the cell itself. c. signaling between cells located close to each other.

d. signaling between parenchyma cells. Answer: c Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Describe the principal modes of cell signaling. Feedback A: Incorrect. That is endocrine signaling. Feedback B: Incorrect. That is autocrine signaling. Feedback C: Correct! An example of this is transmission of a nerve impulse between nerve cells. Feedback D: Incorrect. Paracrine signaling can occur between many different cell types and is not restricted to parenchyma cells. 3. Which of the following hormones is(are) not synthesized from cholesterol? a. Testosterone b. Progesterone c. Corticosteroids d. Retinoic acid Answer: d Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 1. Remembering Learning Objective: Describe the principal modes of cell signaling. Feedback A: Incorrect. All of the steroid hormones, and even the insect hormone ecdysone, are synthesized from cholesterol. Feedback B: Incorrect. All of the steroid hormones, and even the insect hormone ecdysone, are synthesized from cholesterol. Feedback C: Incorrect. All of the steroid hormones, and even the insect hormone ecdysone, are synthesized from cholesterol. Feedback D: Correct! Retinoic acid is synthesized from Vitamin A. 4. Which steroid hormone(s) is(are) not secreted by the gonads? a. Corticosteroids b. Progesterone c. Estrogen d. Testosterone Answer: a Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 1. Remembering Learning Objective: Describe the principal modes of cell signaling. Feedback A: Correct! Corticosteroids, which include glucocorticoids and mineralocorticoids, are secreted by the adrenal gland. Feedback B: Incorrect. Progesterone is secreted by the ovaries and is considered a sex steroid. Feedback C: Incorrect. Estrogen is secreted by the ovaries and is considered a sex steroid. Feedback D: Incorrect. Testosterone is secreted by the testes and is considered a sex steroid.

5. The signaling molecule nitric oxide (NO) functions a. by binding its cell surface receptor and triggering an intracellular signaling cascade. b. by diffusing across the membrane and binding its intracellular receptor, which then activates transcription. c. on cells located far from the cells where it was synthesized. d. by diffusing across the cell membrane and changing the activities of intracellular enzymes. Answer: d Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 1. Remembering Learning Objective: Compare the actions of different types of small signaling molecules. Feedback A: Incorrect. Nitric oxide does not bind a cell surface receptor. Feedback B: Incorrect. Nitric oxide does not bind an intracellular receptor. Feedback C: Incorrect. Nitric oxide is an unstable molecule and thus acts only on cells close to those that synthesized it. Feedback D: Correct! Unlike most signaling molecules, nitric oxide does not bind a receptor but rather directly alters the activity of an enzyme. 6. What is the difference between neurotransmitters and neuropeptides? a. Neuropeptides are generated by neuronal cells but do not transmit signals. b. Neurotransmitters are small hydrophilic molecules, and neuropeptides are small proteins. c. Neurotransmitters are small protein molecules, and neuropeptides are large ones. d. Some neuropeptides can act on distant cells, whereas neurotransmitters cannot. Answer: b Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Compare the actions of different types of small signaling molecules. Feedback A: Incorrect. Neuropeptides do transmit signals. Feedback B: Correct! Both neurotransmitters and neuropeptides transmit signals, but they are in different chemical classes. Feedback C: Incorrect. Neurotransmitters are not proteins, and neuropeptides are small, not large, proteins. Feedback D: Incorrect. Both neuropeptides and neurotransmitters can act on distant cells. 7. Which statement regarding heterotrimeric G proteins in a resting state is true? a. GDP is bound to the  subunit in a complex with both the  and  subunits. b. GDP is bound to the  subunit in a complex with both the  and  subunits. c. GTP is bound to the  subunit in a complex with both the  and  subunits. d. The proteins are in a complex with G protein-coupled receptors. Answer: b Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 2. Understanding Learning Objective: Diagram the structure of a G protein-coupled receptor. Feedback A: Incorrect. GDP is bound to the  subunit in a complex with both the  and  subunits.

Feedback B: Correct! This is the state of G proteins when in a resting state. Feedback C: Incorrect. Only upon activation by a ligand-activated receptor does GTP bind the  subunit. Feedback D: Incorrect. As a rule, the proteins are not in a complex with G proteincoupled receptors, though this has occasionally been observed. 8. G protein-coupled receptors are important molecules involved in signal transduction. Which statement about G protein-coupled receptors is true? a. They are activated only by steroid hormones. b. They bind only guanine nucleotides. c. They bind both adenine and guanine nucleotides. d. They generally contain seven membrane-spanning  helices. Answer: d Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 2. Understanding Learning Objective: Explain how G proteins carry signals to their target enzymes Feedback A: Incorrect. While steroid hormones often use G protein-coupled receptors, many other cell-signaling molecules also use G protein-coupled receptors. Feedback B: Incorrect. G protein-coupled receptors interact with G proteins, which in turn bind guanine nucleotides. Feedback C: Incorrect. G protein-coupled receptors interact with G proteins, not with any nucleotides. Feedback D: Correct! This is a structural characteristic of most G protein-coupled receptors. 9. Which statement about G protein signaling is false? a. Hormone binding induces an interaction of the receptor with the G protein, stimulating the release of GDP and the exchange of GTP on the  subunit. b. Once activated, the GTP-bound  subunit dissociates from  and interacts with its target. c. Activity of the  subunit is terminated by the hydrolysis of GTP to GDP. d. The  subunit becomes deactivated when the hormone dissociates from the receptor. Answer: d Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 2. Understanding Learning Objective: Explain how G proteins carry signals to their target enzymes Feedback A: Incorrect. This is a true statement. Feedback B: Incorrect. This is a true statement. Feedback C: Incorrect. This is a true statement. This is a result of the GTPase activity of the  subunit. Once this occurs, the  subunit reassociates with the  complex and returns to the resting state. Feedback D: Correct! This statement is false. Remember that a hallmark of signal transduction is that the ligand binding to the receptor is what initiates the cascade of events and once initiated can readily dissociate without affecting downstream events. 10. Cyclic AMP (cAMP) is synthesized from ATP by the action of

a. phosphodiesterase. b. phosphorylase kinase. c. adenylyl cyclase. d. protein kinase A (PKA). Answer: c Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 1. Remembering Learning Objective: Summarize the role of cAMP. Feedback A: Incorrect. Phosphodiesterase degrades cAMP to AMP. Feedback B: Incorrect. Phosphorylase kinase is critical to glycogen synthesis. Feedback C: Correct! This is the key enzyme in cAMP synthesis using ATP as the substrate. Feedback D: Incorrect. PKA is activated by cAMP. 11. Most of the effects of cyclic AMP (cAMP) in the cell are mediated by a. protein kinase A. b. ion channels. c. protein kinase C. d. cAMP phosphodiesterase. Answer: a Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 1. Remembering Learning Objective: Describe gene regulation by cAMP-dependent protein kinase. Feedback A: Correct! Cyclic AMP activates protein kinase A, which then stimulates the breakdown of glycogen into glucose, and also alters the pattern of gene expression via activation of the transcription factor CREB. Feedback B: Incorrect. Although cAMP can regulate ion channels such as sodium channels in sensory neurons, it is not the primary means by which it exerts its effects. Feedback C: Incorrect. Protein kinase C is not activated by cAMP, but rather by diacylglycerol. Feedback D: Incorrect. cAMP phosphodiesterase is the enzyme that breaks down cAMP to AMP. 12. Which statement about protein kinase A (PKA) is false? a. In the inactive state, PKA exists as a tetramer of two regulatory (R) and two catalytic (C) subunits. b. PKA binds a total of four molecules of cAMP, one on each of the four subunits. c. PKA binds a total of four molecules, two molecules on each of the two regulatory (R) subunits. d. Once activated, the catalytic (C) subunits dissociate and activate target molecules. Answer: b Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 2. Understanding Learning Objective: Describe gene regulation by cAMP-dependent protein kinase. Feedback A: Incorrect. This is a true statement. Feedback B: Correct! This is a false statement. The catalytic (C) subunits do not bind

cAMP. Feedback C: Incorrect. This is a true statement. Feedback D: Incorrect. This is a true statement. 13. Receptor tyrosine kinases represent critical molecules involved in growth and differentiation though phosphorylation of target substrates on tyrosine residues. Which structural feature is not common among all receptor tyrosine kinases? a. An N-terminal extracellular ligand-binding domain b. A single polypeptide c. A cytosolic C-terminal domain with tyrosine kinase activity d. A single transmembrane  helix Answer: b Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Describe signaling by receptor tyrosine kinases. Feedback A: Incorrect. They all do share this. Feedback B: Correct! Many receptor tyrosine kinases are single polypeptides, yet not all are. A notable exception to this would be the insulin receptor. Feedback C: Incorrect. They all do share this. Feedback D: Incorrect. They all do share this. 14. Which of the following is not a commonly observed consequence of the binding of a signaling molecule to its cell surface receptor? a. Receptor dimerization b. Receptor phosphorylation c. Conformational changes in the receptor d. Increased synthesis of the receptor Answer: d Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Describe signaling by receptor tyrosine kinases. Feedback A: Incorrect. Dimerization of receptors often follows binding by their stimulants. Feedback B: Incorrect. Receptor phosphorylation often occurs after binding, either by the receptor itself (autophosphorylation) or by associated cytoplasmic kinases. Feedback C: Incorrect. Conformational changes in the receptor, ultimately resulting in transmission of the signal, do occur following binding. Feedback D: Correct! The synthesis of the receptor is not typically increased either at the transcriptional or translational level following binding. In fact, sometimes the degradation of the receptor is stimulated as a step toward returning the cell to the resting state. 15. SH2 domains are a. protein domains that bind phosphotyrosine-containing peptides. b. the domains on receptor tyrosine kinases that contain the phosphorylated tyrosine.

c. domains that mediate the dimerization of receptor tyrosine kinases. d. the domains on receptor tyrosine kinases that possess the kinase activity. Answer: a Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Describe signaling by receptor tyrosine kinases. Feedback A: Correct! Proteins containing SH2 domains are the first downstream targets of receptor tyrosine kinases. Feedback B: Incorrect. SH2 domains are not usually found in receptor tyrosine kinases. Feedback C: Incorrect. SH2 domains are not dimerization domains. Feedback D: Incorrect. SH2 domains do not have enzymatic activity. 16. Integrins are transmembrane proteins that connect a. the nuclear laminae to cytoplasmic kinases. b. the extracellular matrix to the cytoskeleton. c. focal adhesions to hemidesmosomes. d. microtubules to actin filaments. Answer: b Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 1. Remembering Learning Objective: Compare and contrast the activities of receptor and nonreceptor kinases. Feedback A: Incorrect. Integrins are not located in the nuclear membrane. Feedback B: Correct! Integrins span the plasma membrane and provide a communication link between the extracellular matrix and the cytoskeleton within the cell. Feedback C: Incorrect. Focal adhesions and hemidesmosomes are two kinds of cell matrix junctions, and they are not connected to one another via integrins. Feedback D: Incorrect. These are both components of the cytoskeleton within the cell, and integrins are transmembrane proteins that connect components on opposite sides of a membrane. 17. The MEK kinase (MAP kinase/ERK kinase) is unusual in that it a. is activated by a kinase. b. lies downstream of G protein-coupled receptors. c. is a dual-specificity kinase, having the ability to phosphorylate both threonines and tyrosines. d. activates a kinase. Answer: c Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Explain how Ras and Raf are activated downstream of tyrosine kinases. Feedback A: Incorrect. Many kinases are themselves activated by other kinases.

Feedback B: Incorrect. Many kinases are downstream targets of G protein-coupled receptors. Feedback C: Correct! Most kinases can either phosphorylate serines and threonines or tyrosines, but MEK can phosphorylate both a threonine and a tyrosine residue on ERK2. Feedback D: Incorrect. Many kinases function within a cascade of kinases, whereby one phosphorylates, and thus activates, the next, and so on. 18. Heterotrimeric G proteins are not the only guanine nucleotide-binding proteins. Which of the following represents a family of GTP-binding proteins that act as monomers rather than heterotrimeric compounds? a. Ras b. ERK c. Raf d. Smad Answer: a Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Give an example of transcriptional regulation by MAP kinase signaling. Feedback A: Correct! The members of the Ras family are often called small GTP-binding proteins because they are about half the size of the  subunits of heterotrimeric G proteins. Feedback B: Incorrect. ERKs do not bind GTP. Feedback C: Incorrect. Raf is a serine/threonine kinase activated by Ras. Feedback D: Incorrect. Smads are transcription factors activated by TGF receptors. 19. In unstimulated cells, NF-B proteins are maintained in an inactive state in the cytosol by interactions with a. Hedgehog. b. I c. adaptor proteins. d. the TNF receptor. Answer: b Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 2. Understanding Learning objective: Describe the NF-B pathway. Feedback A: Incorrect. Hedgehog is not part of NF-B signaling. Feedback B: Correct! In its non-phosphorylated form, I binds to NF-B and prevents its translocation to the nucleus. Feedback C: Incorrect. Adaptor proteins are coupled to the cytokine receptors. Feedback D: Incorrect. The receptor does not directly couple to NF-B. 20. An example of signaling by direct cell–cell interactions is the a. Wingless (Wnt) signaling pathway. b. JAK/STAT pathway.

c. Notch pathway. d. pathway leading to vulval development in C. elegans. Answer: c Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 3. Applying Learning Objective: Explain the roles of proteolysis in the Wnt and Notch pathways. Feedback A: Incorrect. Stimulation of the Wnt pathway occurs by the binding of a secreted protein to a cell surface receptor. Feedback B: Incorrect. The JAK/STAT pathway is stimulated by cytokines rather than by direct cell–cell interactions. Feedback C: Correct! Notch is a transmembrane protein that is stimulated by other transmembrane proteins (such as Delta) on adjacent cells. Feedback D: Incorrect. This is triggered by the binding of Lin-3, an EGF-like protein, to a receptor called Let-23 on the surface of prevulval cells. 21. Which of the following signaling pathways allows for direct cell–cell signaling by transmembrane proteins? a. NF-B b. Wnt c. Hedgehog d. Notch Answer: d Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 2. Understanding Learning Objective: Explain the roles of proteolysis in the Wnt and Notch pathways. Feedback A: Incorrect. NF-B acts within a single cell. Feedback B: Incorrect. Wnt acts within a single cell. Feedback C: Incorrect. Hedgehog acts within a single cell. Feedback D: Correct! The Notch receptor on one cell binds the Delta receptor on an adjacent cell, initiating notch signaling.

Essay 1. In what ways are steroid hormone receptors different from most other types of cellular receptors? Answer: Most cellular receptors span the plasma membrane where they are optimally positioned to bind extracellular molecules and transmit signals to the cell’s interior. In contrast, steroid hormones are small hydrophobic molecules that can slip through membranes, and their receptors are located in the interior of the cell, in both the cytoplasm and the nucleus. In addition, once activated, the steroid hormone receptors themselves act as transcription factors, whereas for receptors located in the plasma membrane, the transcription factors usually lie at the end of a signal transduction cascade that includes several components. Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding

Learning Objective: Explain how steroid hormones regulate gene expression. 2. In plants, growth can be mediated by what is frequently referred to as “auxin-induced transcription” of genes critical to growth. This terminology was derived from experiments demonstrating that the addition of auxin leads to an increase in the expression of genes responsible for growth. However, since auxin is not a transcription factor and cannot directly activate transcription, how does it regulate gene activity? Answer: Many plant growth genes are constitutively or always repressed by proteins that bind DNA and inhibit transcription. Simply removing them will enhance the rate of transcription. Auxin binds to a receptor with ubiquitin ligase activity, which leads to the ubiquitination and subsequent proteolytic degradation of the repressor. Thus, even though auxin cannot directly affect gene expression, it can still have dramatic effects on expression by regulating proteins that do directly interact with the genes. Textbook Reference: Signaling Molecules and Their Receptors Bloom’s Category: 2. Understanding Learning Objective: Compare the actions of different types of small signaling molecules. 3. What is the basis for the different responses of nerve cells versus heart muscle cells to acetylcholine? Answer: In nerve cells, the acetylcholine receptor doubles as a ligand-gated ion channel composed of five subunits. When bound by acetylcholine, the receptor opens to allow entry of sodium and exit of potassium, thus depolarizing the membrane and triggering an action potential. In contrast, in heart muscle, the acetylcholine receptor has a very different structure—it is a G protein-linked receptor of the seven-transmembrane class. The G protein directly activates a potassium channel, which causes a parasympathetic response in the cardiac muscle. This is a classic example of how a single neurotransmitter can be used for different purposes in different cell types. Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 2. Understanding Learning Objective: Explain how G proteins carry signals to their target enzymes. 4. Why do enzymes that lie downstream of a cell surface receptor in a signal transduction pathway amplify as well as propagate the signal? Answer: Extracellular signaling molecules typically stimulate only a single receptor molecule before being degraded; the receptor, however, can often activate many molecules of the enzyme to which it is directly linked. Since most signal transduction pathways are made up of several enzymes, amplification can occur at each step, resulting in very high amplification by the time the pathway reaches the ultimate component. Thanks to this type of amplification, the cell can be exquisitely sensitive even to very low concentrations of extracellular inducers. Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 2. Understanding Learning Objective: Explain how G proteins carry signals to their target enzymes. 5. As a cell biologist for a pharmaceutical company, you are charged with developing a drug to inhibit the cAMP response element binding protein (CREB) pathway. Briefly

describe the cascade leading to CREB activation and where one might target a drug to block its activation. Answer: In the presence of cAMP, the R subunit is bound by four molecules of cAMP, allowing the catalytic subunits to dissociate, translocate to the nucleus, and then phosphorylate CREB. CREB then homodimerizes, binds to the CRE (cAMP response element) in the promoter of target genes, and activates transcription. Drugs could be designed that would interrupt signaling anywhere along the pathway. For example, a drug could stabilize the R and C subunit interactions of PKA and block translation into the nucleus. A drug could prevent interaction of cAMP with the R subunit and essentially lead to the same result. Similarly, a drug could have a more downstream action that might prevent phosphorylation of CREB, thus not allowing it to become activated. Another option would be to prevent the required dimerization of CREB, thus preventing its ability to bind DNA. Textbook Reference: G Proteins and Cyclic AMP Signaling Bloom’s Category: 3. Applying Learning Objective: Describe gene regulation by cAMP-dependent protein kinase. 6. What are the hallmarks of the first steps in signaling through receptor tyrosine kinases? Answer: Receptor tyrosine kinases span the plasma membrane; upon binding of extracellular stimulants, they form dimers with one another and autophosphorylate. The cytoplasmic domain can then bind and activate intracellular proteins containing SH2 domains, which specifically bind peptides containing phosphotyrosine residues. The SH2 domain-containing proteins then go on to activate a variety of different pathways, depending on the identity of the receptor that activated them. Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Describe signaling by receptor tyrosine kinases. 7. Growth factors and cytokines both lead to tyrosine phosphorylation through receptors, but they do so through different mechanisms. What is the key difference between the receptors for these two classes of ligands in terms of their tyrosine kinase activity? Answer: The receptors for growth factors such as EGF are membrane-spanning cell surface receptors in which the cytoplasmic domain of the receptor contains the catalytic activity necessary to phosphorylate proteins on tyrosine residues and often are capable of autophosphorylation. Cytokine receptor activation also leads to tyrosine phosphorylation, but the receptor has no intrinsic catalytic activity. Instead, upon ligand binding, the cytokine receptors interact with a class of proteins called nonreceptor tyrosine kinases, which in turn do have the catalytic activity necessary for tyrosine phosphorylation. Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Compare and contrast the activities of receptor and nonreceptor kinases. 8. Cytokine receptors frequently form dimers upon ligand binding, and these can be

heterodimers or homodimers. Explain how this ability of the receptors to form different combinations of dimers allows three unique cytokines, each with a unique action, to arise from only two cytokine receptors. Answer: If dimerization is required, then the two receptors (R1 and R2) could form homodimers (R1:R1 and R2:R2) and potentially interact with two of the cytokines; the third cytokine could bind to a heterodimer of R1:R2. Given the number of cytokines and receptors and the many combinations such as these that are possible, the complexity of these signaling networks becomes evident. Textbook Reference: Tyrosine Kinases and Signaling by the MAP Kinase and PI 3Kinase Pathways Bloom’s Category: 2. Understanding Learning Objective: Give an example of transcriptional regulation by MAP kinase signaling. 9. In the signaling of cells to initiate growth, immediate early genes become activated to initiate the growth process. What would happen to the expression of these immediate early genes if Elk-1 was not able to be dephosphorylated, and why? Answer: The immediate early genes would be constitutively activated (i.e., not shut off). Immediate early genes share the serum response element in their promoters that interacts with serum response factor (SRF). Following the growth signal pathway, ERK phosphorylates and activates Elk-1, which interacts with SRF and activates transcription of the immediate early genes. If Elk-1 could not be dephosphorylated, it would remain active and continually express immediate early genes and, being a growth signal, could lead to cancer. Textbook Reference: Tyrosine Kinases and Signaling by MAP Kinase, PI 3-Kinase, and Phospholipase C/Calcium Pathways Bloom’s Category: 2. Understanding Learning Objective: Give an example of transcriptional regulation by MAP kinase signaling. 10. What would happen to NF-B signaling if IB were unable to be dephosphorylated? Answer: Most likely nothing. When IB is phosphorylated by IB kinase, it is targeted for ubiquitination and then degraded by proteasomes. If it could not be phosphorylated, it would still be targeted for ubiquitination and then degraded by proteasomes. Textbook Reference: Receptors Coupled to Transcription Factors Bloom’s Category: 4. Analyzing Learning Objective: Describe the NF-B pathway 11. Explain the concept of “feedback loops” in signaling networks. Answer: Feedback loops allow for a form of autoregulation of a particular event or activity. In signaling, there are positive feedback loops, in which a downstream element of the pathway activates an upstream element. This leads to a continual activation, even in the absence of the initiating event such as ligand binding to a receptor. The counterpart would be a negative feedback loop, in which a downstream element inhibits the activity of an upstream element, shutting down the pathway. This may occur even in the presence of the initiating event. In the feedforward relay, an upstream element may directly

activate not only the next element in the pathway but also an element farther downstream, thus “short circuiting” the elements in between. These concepts have useful applications throughout the study of biology, from the molecular level (as observed here) all the way through organ and systems biology. Textbook Reference: Signaling Dynamics and Networks Bloom’s Category: 2. Understanding Learning Objective: Diagram a feedback loop

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 18: The Cell Cycle TEST FILE QUESTIONS Multiple Choice 1. The eukaryotic cell cycle is composed of four phases in the following order: a. G1; S; G 2; M. b. G 1; G2 ; S; M. c. G1; M; G2; S. d. S; G1; G2 ; M. Answer: a Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 2. Understanding Learning Objective: Summarize the phases of the cell cycle. 2. The length of time it takes for the cells labeled during DNA synthesis to begin to enter mitosis is a good estimate of the length of the _______ phase(s). a. G1 b. S c. G2 d. M Answer: c Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 2. Understanding Learning Objective: Summarize the phases of the cell cycle. 3. In a typical cycling mammalian cell, _______ of the cell cycle is spent in interphase. a. 99% b. 95% c. 75% d. 50% Answer: b Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle. 4. Cell cycles of early embryonic animal cells are unusual because they have a. no G1 or G2. b. an elongated G1 and G2 .

c. a short S. d. a much shortened M. Answer: d Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle. 5. The number of cells in the different phases of a population can be determined most easily by using a _______ to measure the amount of DNA per cell in a large sample of cells. a. scintillation counter b. flow cytometer c. fluorescence microscope d. phase-contrast microscope after autoradiography Answer: b Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle. 6. Cells in the quiescent phase of the cell cycle are said to be in the _______ phase. a. G0 b. G 1 c. G2 d. M Answer: a Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle. 7. Which of the following represents the amount of DNA in a typical G 2 cell? a. n b. 2n c. 4n d. 8n Answer: c Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 2. Understanding Learning Objective: Summarize the phases of the cell cycle. 8. What event takes place during interphase? a. Centromeres divide. b. Metabolic processes, such as DNA replication, are carried out by the cell. c. Spindle fibers start forming. d. Centrioles begin to appear. Answer: b Textbook Reference: The Eukaryotic Cell Cycle

Bloom’s Category: 2. Understanding Learning Objective: Summarize the phases of the cell cycle. 9. Skin fibroblasts near a wound, arrested in G0 , are stimulated to enter G1 by _______ growth factor. a. epidermal b. fibroblast c. platelet-derived d. keratinocyte Answer: c Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 2. Understanding Learning Objective: Describe growth factor regulation of cell cycle progression. 10. The oocytes in a normal 7-year-old female (prepubertal) would be expected to be arrested in which stage of the cell cycle, until they are hormonally stimulated? a. G1 b. G 2 c. Interphase d. G 0 Answer: b Textbook Reference: The Eukaryotic Cell cycle Bloom’s Category: 2. Understanding Learning Objective: Describe growth factor regulation of cell cycle progression. 11. Progression through the cell cycle in the yeast Saccharomyces cerevisiae is regulated by a. external signals such as availability of nutrients and mating factors. b. internal signals such as growth factors and steroid hormones. c. internal signals such as the availability of glucose and ATP. d. simple organisms such as yeast do not have a cell cycle. Answer: a Textbook Reference: The Eukaryotic Cell cycle Bloom’s Category: 1. Remembering Learning Objective: Describe growth factor regulation of cell cycle progression. 12. The G1 checkpoint in budding yeast cells is called a. START. b. the restriction point. c. the decision point. d. the G 1/S boundary. Answer: a Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of cell cycle checkpoints.

13. The G1 checkpoint in animal cells is called a. START. b. the restriction point. c. the decision point. d. the G 1/S boundary. Answer: b Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of cell cycle checkpoints. 14. The G2 checkpoint prevents entry into _______ if DNA synthesis is incomplete and/or the DNA is damaged. a. G1 b. S c. G2 d. M Answer: c Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of cell cycle checkpoints. 15. A scientist notices that cells being grown in culture have ceased to divide. Further investigation demonstrates the cells have abnormally segregated chromosomes. Which checkpoint are these cells likely arrested at? a. Spindle assembly checkpoint b. DNA damage checkpoint c. START d. Restriction checkpoint Answer: a Textbook Reference: The Eukaryotic Cell cycle Bloom’s Category: 2. Understanding Learning Objective: Explain the significance of cell cycle checkpoints. 16. Maturation promotion factor (MPF), discovered by Masui and Markert, is the factor that a. is missing in the cdc2 mutant in yeast. b. induces entry into meiosis when injected into frog oocytes. c. fluctuates in amount during early cleavages of sea urchin eggs. d. fluctuates in amount during the cell cycle of mammalian cells. Answer: b Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Summarize the experiments that led to the discovery of MPF. 17. MPF is a a. monomeric protein kinase.

b. dimer of two molecules of Cdk1. c. dimer of Cdk1 and cyclin A. d. dimer of Cdk1 and cyclin B. Answer: d Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Summarize the experiments that led to the discovery of MPF. 18. MPF is a general regulator of the transition from a. G1 to S. b. S to G2. c. G2 to M. d. metaphase to anaphase. Answer: c Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Summarize the experiments that led to the discovery of MPF. 19. The passage of a cell through the stages of the cell cycle is controlled by protein kinases that phosphorylate many different proteins at appropriate times. What are these protein kinases called? a. Cdk activating kinases b. Cyclin-dependent kinases c. Cyclins d. Tyrosine kinases Answer: b Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Explain the role of cyclins. 20. Studies on cell cycle mutants in budding and fission yeasts demonstrated that cdc2 and cdc28 both encode a a. protein kinase. b. protein phosphatase. c. ubiquitin ligase. d. growth factor. Answer: a Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Explain the role of cyclins. 21. Two sea urchin proteins were named cyclin A and B because, during the embryonic cell cycle, they were periodically a. activated and inactivated by phosphorylation and dephosphorylation. b. synthesized and degraded. c. bound and released by another protein.

d. ubiquitylated and nonubiquitylated. Answer: b Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Explain the role of cyclins. 22. To become active, Cdk1 must a. bind cyclin. b. bind cyclin and be phosphorylated on tyrosine 15 and threonines 14 and 161. c. bind cyclin and be phosphorylated on tyrosine 15 and be dephosphorylated on threonines 14 and 161. d. bind cyclin and be phosphorylated on threonine 161 and be dephosphorylated on threonine 14 and tyrosine 15. Answer: d Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Explain the role of cyclins. 23. Cdks bind to cyclin and are activated by a. phosphorylation by Wee1. b. dephosphorylation by Cdc25 protein phosphatase. c. the binding of Ink4. d. the binding of Cip. Answer: b Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Explain the role of cyclins. 24. Which cyclin(s) is(are) required to pass the G1 restriction point in animal cells? a. Cyclin A b. Cyclin A and Cyclin B c. Cyclin D d. Cyclin D and Cyclin E Answer: d Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Explain the role of cyclins. 25. Passage of animal cells through the cell cycle is regulated primarily by a. the availability of nutrients. b. intracellular second messengers. c. extracellular growth factors. d. direct cell surface interactions. Answer: c Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering

Learning Objective: Describe the mechanism by which growth factors regulate cell cycle progression. 26. The retinoblastoma protein (Rb) binds to and inhibits a. transcription factor E2F. b. oncogenes. c. cyclin-dependent kinase. d. p53. Answer: a Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism by which growth factors regulate cell cycle progression. 27. Growth-factor stimulation of the Ras/Raf/MEK/ERK pathway initially stimulates the synthesis of cyclin a. A. b. B. c. E. d. D. Answer: d Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism by which growth factors regulate cell cycle progression. 28. Cells are restricted to one round of DNA replication per cycle by origin of replication binding proteins called a. ATMs. b. Akts. c. Oris. d. MCMs. Answer: d Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Explain how the initiation of DNA replication is controlled. 29. To prevent cells with unreplicated DNA from passing through the G 2 checkpoint, Chk1 and Chk2 _______ protein phosphatase Cdc25. a. phosphorylate and stimulate b. phosphorylate and inhibit c. dephosphorylate and stimulate d. dephosphorylate and induce the degradation of Answer: b Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding

Learning Objective: Explain how the initiation of DNA replication is controlled. 30. p21 inhibits cell cycle progression by binding to and inhibiting a. cyclin-dependent kinases. b. p53. c. ATM. d. Rb protein. Answer: a Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Explain how the initiation of DNA replication is controlled. 31. Damaged DNA is sensed by a complex of proteins that activate the protein kinase a. ATM or ATR. b. ABS. c. Akt. d. MCM. Answer: a Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Summarize the operation of DNA damage checkpoints. 32. The disease ataxia telangiectasia, which causes nervous system defects and a high frequency of cancer in affected individuals, results from a mutation in the gene for the protein a. ATM. b. ABS. c. Akt. d. MCM. Answer: a Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Summarize the operation of DNA damage checkpoints. 33. Nuclear envelope breakdown occurs at the a. beginning of prophase. b. end of prophase. c. end of prometaphase. d. beginning of anaphase. Answer: b Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Describe the stages of mitosis. 34. Entry into mitosis occurs because a. cyclin is destroyed at the beginning of mitosis.

b. cyclin is phosphorylated at the beginning of mitosis. c. sufficient cyclin-dependent kinase is synthesized to trigger mitosis. d. Cdk1/Cyclin B (MPF) get activated. Answer: d Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Describe the stages of mitosis. 35. During prophase, Cdk1/Cyclin B (MPF) directly phosphorylates all of the following except a. condensins. b. lamins. c. APC/C. d. microtubule-associated proteins. Answer: c Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Summarize the targets of mitotic kinases. 36. Phosphorylation of the Golgi matrix proteins by Cdk1 during mitosis a. mediates breakdown of the Golgi apparatus into vesicles. b. activates proteins to degrade the Golgi proteins. c. inhibits association of the Golgi with microtubules. d. causes microtubules to pull the Golgi into two parts. Answer: a Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Summarize the targets of mitotic kinases. 37. The beginning of anaphase occurs when cohesins, holding the sister chromatids together, are degraded by a. anaphase-promoting complex. b. ATM. c. proteasome. d. separase. Answer: d Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Explain the mechanism of chromosome condensation. 38. As the cell enters M phase, the condensins are activated as a result of a. proteolytic cleavage by checkpoint proteases. b. dephosphorylation by cyclin dependent and Aurora B phosphatases. c. ubiquitylation by cyclin dependent ubiquitin ligases. d. phosphorylation by both Cdk1 and Aurora B kinase. Answer: d

Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Explain the mechanism of chromosome condensation. 39. The DNA region to which proteins bind to form the kinetochore and spindle fibers attach for segregation of the chromosomes is called the a. chiasma. b. centromere. c. synaptonemal complex. d. origin. Answer: b Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Illustrate operation of the spindle assembly checkpoint. 40. Chromosomes are attached to spindle microtubules by a. centromeres. b. kinetochores. c. centrioles. d. centrosomes. Answer: b Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Illustrate operation of the spindle assembly checkpoint. 41. Polar microtubules a. overlap in the center of the spindle. b. are attached to the contractile ring. c. are attached to kinetochores. d. pull kinetochores poleward. Answer: a Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Illustrate operation of the spindle assembly checkpoint. 42. During mitosis, the rate of microtubule turnover a. decreases. b. stays the same. c. doubles. d. increases five- to tenfold. Answer: d Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Illustrate operation of the spindle assembly checkpoint.

43. During the spindle fiber checkpoint, degradation of _______ allows for separation of the sister chromatids. a. cyclin B b. separase c. cohesin d. kinetochore Answer: c Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Illustrate operation of the spindle assembly checkpoint. 44. The mechanism of cytokinesis in plants involves the formation of the a. contractile ring. b. Aurora belt. c. cell plate. d. plasma membrane bridge. Answer: c Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Contrast cytokinesis in animal and plant cells. 45. The progression from metaphase to anaphase results from a. phosphorylation mediated by Polo-like kinase. b. ubiquitin-mediated proteolysis of key regulatory proteins. c. phosphorylation mediated by Cdk2. d. activation by Wee1. Answer: b Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Contrast cytokinesis in animal and plant cells.

Fill in the Blank 1. The most variable phase of the cell cycle is _______. Answer: G 1 Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle. 2. The spindle assembly checkpoint monitors the alignment of chromosomes on the _______. Answer: mitotic spindle Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle.

3. Injection of MPF into frog oocytes induces them to enter _______. Answer: meiosis Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Summarize the experiments that led to the discovery of MPF. 4. The degradation of cyclin B during _______ occurs after it becomes modified by the addition of ubiquitin. Answer: anaphase Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Summarize the experiments that led to the discovery of MPF. 5. Separation of anaphase chromosomes begins when _______ is degraded to break the link between sister chromatids. Answer: cohesin Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Describe the stages of mitosis. 6. Cyclin B and _______ are ubiquitylated and degraded by active APC/C. Answer: securin Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Describe the stages of mitosis. 7. Cytokinesis begins during the _______ stage of mitosis. Answer: anaphase Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Describe the stages of mitosis. 8. Cell plate formation occurs only in _______ cells. Answer: plant Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Describe the stages of mitosis.

True/False 1. Interphase usually consists of G1 , S, and G2, but in some cells it consists only of S. Answer: T Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 2. Understanding

Learning Objective: Summarize the phases of the cell cycle. 2. After development, liver cells stop cycling and stay in G 1. Answer: F Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle. 3. In animal cells, Cdk2/cyclin A act to drive cells through the restriction point in G 1. Answer: F Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Explain the role of cyclins. 4. Rb is a member of the family of tumor suppressor genes. Answer: T Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 1. Remembering Learning Objective: Describe the mechanism by which growth factors regulate cell cycle progression. 5. Rb is a tumor suppressor protein that acts by inhibiting phosphorylation of Cdk. Answer: F Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanism by which growth factors regulate cell cycle progression. 6. ATM and ATR recognize damaged or unreplicated DNA and initiate signals that arrest cells at cell cycle checkpoints. Answer: T Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Explain how the initiation of DNA replication is controlled. 7. ATM recognizes single-stranded DNA breaks and ATR recognizes double-stranded DNA breaks. Answer: F Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Summarize the operation of DNA damage checkpoints. 8. Transcription ceases as chromosome condensation occurs. Answer: T Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering

Learning Objective: Describe the stages of mitosis. 9. Re-formation of the nuclear envelope occurs as the result of cyclin B destruction, which allows phosphatases to dephosphorylate nuclear envelope lamins. Answer: T Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Describe the stages of mitosis. 10. A mutant cyclin B that is resistant to degradation by protease, but is otherwise active, would block the metaphase-to-anaphase transition. Answer: F Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Describe the stages of mitosis. 11. The metaphase-to-anaphase transition in animal cells is triggered by the breakdown of cyclin B. Answer: F Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Describe the stages of mitosis.

Short Answer 1. In what phase(s) of the cell cycle does growth typically occur? Answer: G 1, S, and G2 Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle. 2. What does the spindle assembly checkpoint prevent? Answer: It prevents chromosomes from separating until all chromosomes are properly aligned on the metaphase mitotic spindle. Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle. 3. Identify the four different molecular mechanisms by which the activity of cyclindependent kinases is regulated after they are synthesized. Answer: (1) Association of a Cdk with its partner cyclin; (2) phosphorylation of an activating threonine; (3) phosphorylation and dephosphorylation of inhibiting tyrosine and threonine residues; and (4) binding of Cdk inhibitors. Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding

Learning Objective: Explain the role of cyclins. 4. To investigate the role of APC/C ubiquitin ligase in an experimental cell culture model, you are able to insert a plasmid gene encoding a mutant APC/C protein that is always active (i.e., cannot be inhibited). What would be the predicted outcome on the cells that received this mutant APC/C? Answer: The cells would not divide and would eventually die. To initiate DNA replication, Cdk2/cyclin E must first inhibit APC/C, and the inhibition of APC/C leads to the activation of DDK protein kinase, which phosphorylates and activates MCM. Thus a mutant APC/C that cannot be inhibited would constitutively repress DNA replication and the cell could not properly re-enter the cell cycle. Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 4. Analyzing Learning Objective: Explain the role of cyclins. 5. How is DNA replication prevented from recurring in a single cell cycle? Answer: Preceding the first DNA synthesis, MCM proteins bind to replication origins in G1. Once initiation of replication has occurred, the MCMs are removed and cannot be replaced until the next G 1. Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Explain how the initiation of DNA replication is controlled. 6. How is the anaphase-promoting complex/cyclosome (APC/C) activated? Answer: Attachment of all chromosomes to the metaphase spindle no longer allows Mad to inhibit Cdc20, which can then bind to and activate APC/C. Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Summarize the targets of mitotic kinases. 7. Dephosphorylation of proteins that were phosphorylated during prophase leads to what three telophase events? Answer: (1) Chromosome decondensation; (2) nuclear envelope re-formation; and (3) return of microtubules to their interphase state Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Summarize the targets of mitotic kinases. 8. What determines the location of the contractile ring of filaments that mediate cytokinesis in an animal cell? Answer: The mitotic spindle Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Contrast cytokinesis in animal and plant cells. 9. Cytokinesis occurs in a plane that passes through which area of a cell?

Answer: The metaphase plate Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Contrast cytokinesis in animal and plant cells. 10. How does cytokinesis in plant cells differ from that in animal cells? Answer: Cytokinesis of animal cells involves constriction of the cell surface around the mid-body of the spindle by a contractile ring of actin and myosin II filaments. Cytokinesis in plant cells involves the formation of a new cell membrane and cell wall by microtubule-dependent transport of Golgi-derived vesicles to the former metaphase plate, where they fuse to form a cell plate between the two daughter cells. Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Contrast cytokinesis in animal and plant cells.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. Interphase is defined as a. the G1 and G2 phases. b. the G 1, G2 , and S phases. c. G0, the quiescent phase. d. M phase. Answer: b Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle. Feedback A: Incorrect. G1 and G2 are the “gap” phases of the cell cycle during which the cell simply grows. DNA synthesis does not occur during either of these phases. Feedback B: Correct! Interphase is the part of the cell cycle in which the chromosomes are decondensed and spread throughout the nucleus. Cell growth and DNA synthesis take place during interphase, which includes G 1, S, and G2 , in preparation for mitosis (M phase). Feedback C: Incorrect. G0 represents an exit from the cell cycle, when a cell does not divide yet remains metabolically active. Feedback D: Incorrect. M phase is the phase in which nuclear division occurs, and it happens only after interphase, during which cell growth and DNA replication occur. 2. The phase of the cell cycle that corresponds to the interval between mitosis and initiation of DNA replication is referred to as the a. M phase. b. S phase. c. G1 phase. d. G 2 phase.

Answer: c Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Summarize the phases of the cell cycle. Feedback A: Incorrect. M phase is mitosis. Feedback B: Incorrect. S phase is the phase in which replication takes place. Feedback C: Correct! This is also called gap 1 phase. Feedback D: Incorrect. G2 is characterized by continued cell growth and synthesis of proteins in preparation for mitosis. 3. You perform an experiment by incubating a radiolabeled nucleotide into the medium of a culture of cells. The cells that take up the radiolabel would be expected to be in what phase of the cell cycle? a. S phase b. G 1 phase c. G2 phase d. M phase Answer: a Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 3. Applying Learning Objective: Summarize the phases of the cell cycle. Feedback A: Correct! This is the phase in which cells are actively replicating; the radiolabeled nucleotide is incorporated into DNA that is being newly synthesized. Feedback B: Incorrect. G1 occurs prior to replication in S phase. Feedback C: Incorrect. G2 occurs after S phase and replication have been completed. Feedback D: Incorrect. M phase is the period in which mitosis occurs. 4. In which of the following cell types do G 1 and G2 not take place? a. Early embryonic cells b. Budding yeast cells c. Typical proliferating human cells d. Skin fibroblasts Answer: a Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 2. Understanding Learning Objective: Summarize the phases of the cell cycle. Feedback A: Correct! Early embryonic cells divide very rapidly, alternating between M and S phases. Since G1 and G2 do not occur, the cell does not have time to grow prior to cell division, and the resulting daughter cells are smaller than the mother cells. Feedback B: Incorrect. Yeast cells divide more rapidly than mammalian cells; however, they do go through G1 and G2 . Feedback C: Incorrect. An average proliferating human cell goes through all phases of the cell cycle, including G1 and G2. Feedback D: Incorrect. Skin fibroblasts spend most of their time in G 0, the quiescent phase. However, they can reenter the cell cycle and go through all of its phases, including G1 and G2, if stimulated to do so (for example, following an injury, when they would

need to be regenerated). 5. DNA content can be found using a method of flow cytometry that can determine 2n or 4n states of cells based on cell counting and fluorescent labeling of DNA. Which of the following phases would be identified through flow cytometry as 2n? a. G1 phase only b. S phase only c. G2 phase only d. Both G2 and M phases Answer: b Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 3. Applying Learning Objective: Summarize the phases of the cell cycle. Feedback A: Correct! G1 cells have yet to replicate and have only a 2n, or diploid, amount of DNA. Feedback B: Incorrect. S phase cells would fall between 2n and 4n and in flow cytometry would be observed as the cells in the trough between 2n and 4n peaks. Feedback C: Incorrect. G2 cells would have twice the DNA content per cell and be found in the 4n peak. Feedback A: Incorrect. Both G2 and M phases would be 4n. 6. In the absence of growth factor, most animal cells will stop the cell cycle at a restriction point in what stage? a. S phase b. M phase c. G2 phase d. G 1 phase Answer: d Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Describe growth factor regulation of cell cycle progression. Feedback A: Incorrect. Growth factors are required to get to S phase. Feedback B: Incorrect. The growth factors need to exert control long before M phase. Feedback C: Incorrect. There are G2 checkpoints, but they typically are not controlled by growth factors. Feedback D: Correct! In the absence of growth factors, cells arrest in G1 and enter G0 , remaining quiescent and awaiting a signal such as a growth factor to divide, much like quiescent fibroblasts. 7. In yeast, START triggers the progression from G 1 to S. Which of the following are involved in regulating START? a. Nutrients b. Mating factors c. Cell size d. All of the above Answer: d

Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 2. Understanding Learning Objective: Explain the significance of cell cycle checkpoints. Feedback A: Incorrect. This is a correct answer but not the only correct answer. In low nutrient conditions, progression will not occur. Feedback B: Incorrect. This is a correct answer but not the only correct answer. Proteins that signal mating will allow haploid yeast cells to fuse with one another instead of progressing to S phase. Feedback C: Incorrect. This is a correct answer but not the only correct answer. Yeast cells must attain a minimum size prior to entry into S phase. Feedback D: Correct! 8. Most cells in an adult animal are a. in G0, or quiescent, phase. b. actively proliferating. c. stem cells. d. undergoing meiosis. Answer: a Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of cell cycle checkpoints. Feedback A: Correct! Most adult cells are differentiated and cease proliferating, entering the G0 phase. However, many can re-enter the cell cycle and resume proliferation if they are needed to replace damaged tissue. Feedback B: Incorrect. Unlike early embryonic cells, most cells in adult animals have ceased to proliferate. Feedback C: Incorrect. Stem cells are partially differentiated cells that can undergo mitosis to yield both differentiated cells and stem cells. Most adult cells are differentiated and have stopped dividing, except under circumstances in which tissue regeneration is necessary. Feedback D: Incorrect. Only germ cells undergo meiosis to yield the haploid cells necessary for sexual reproduction. 9. Cell cycle checkpoints ensure that complete genomes are transmitted to daughter cells. DNA damage checkpoints are found in all of the following phases except a. M phase. b. G 1 phase. c. G2 phase. d. S phase. Answer: a Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of cell cycle checkpoints. Feedback A: Correct! DNA damage checkpoints are not found in M. Feedback B: Incorrect. DNA damage checkpoints are found in G 1 . DNA damage is assessed in G1, S, and G2 phases and is mediated by ATM and ATR, which recognize

damaged or unreplicated DNA and activate a signaling pathway that arrests the cell cycle, activates DNA repair, and/or initiates apoptosis. Feedback C: Incorrect. DNA damage checkpoints are found in G2 . Feedback D: Incorrect. DNA damage checkpoints are found in S. 10. The G2 cell cycle checkpoint detects a. the binding of the MCM proteins to origins of replication. b. chromosome misalignment. c. unreplicated or damaged DNA. d. levels of p53. Answer: c Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 1. Remembering Learning Objective: Explain the significance of cell cycle checkpoints. Feedback A: Incorrect. The MCM proteins bind replication origins only during G1, ensuring that DNA is replicated only once per cycle. Feedback B: Incorrect. Chromosome misalignment is an M phase checkpoint control. Feedback C: Correct! The G 2 checkpoint ensures that the genome has been entirely replicated and that damaged DNA has been repaired prior to entry into M phase. Feedback D: Incorrect. The G1 checkpoint is mediated by the action of p53. 11. Which statement about cyclin B is false? a. It is a component of MPF. b. It accumulates throughout S and G 2. c. It activates Cdc2 protein kinase. d. It cycles between active and inactive states in the cell cycle. Answer: d Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Explain the role of cyclins. Feedback A: Incorrect. This is a true statement. Cyclin B and Cdc2 form a complex that is known as MPF. Feedback B: Incorrect. This is a true statement. Cyclin B begins to be synthesized in S phase, accumulates throughout S and G 2, and is rapidly degraded at the end of M phase. Feedback C: Incorrect. This is a true statement. Cyclin B binds to Cdc2 and activates its kinase activity. Feedback D: Correct! This is a false statement. Cyclin B protein levels oscillate throughout the cell cycle, but its activity does not change. 12. The p21 and p15 proteins are examples of a. Cdk inhibitors. b. cyclins. c. oncogenes. d. growth factors. Answer: a Textbook Reference: Regulators of Cell Cycle Progression

Bloom’s Category: 1. Remembering Learning Objective: Explain the role of cyclins. Feedback A: Correct! p15 inhibits Cdk4, 6/cyclin D complexes; p21 inhibits several Cdk/cyclin complexes to block progression of the cell cycle at G 1. Feedback B: Incorrect. Cyclins are a class of proteins whose levels oscillate with the cell cycle and that bind and activate Cdk’s. Feedback C: Incorrect. Oncogenes are genes, not proteins. Some oncogenes stimulate cell division. Feedback D: Incorrect. Growth factors are extracellular proteins that stimulate cell division. p21 and p15 act intracellularly to inhibit cell division by binding Cdk/cyclin complexes. 13. Which statement regarding cyclin D is false? a. It is required for cell cycle progression. b. It is activated by growth factors. c. It is always present and its activity is regulated by fluctuating levels of Cdk4 and Cdk6. d. It is rapidly degraded following removal of appropriate growth factors. Answer: c Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Explain the role of cyclins. Feedback A: Incorrect. This is a true statement. Cyclin D in complex with Cdk4 and Cdk6 drives cells through the restriction point. Feedback B: Incorrect. This is a true statement. Cyclin D synthesis is rapidly induced in response to growth factor stimulation through signaling pathways such as Ras/Raf/MEK/ERK. Feedback C: Correct! This is a false statement about cyclin D. It is not always present, though it does require interaction of Cdk4 and Cdk6. Feedback D: Incorrect. This is a true statement. Cyclin D protein levels fall due to proteolytic degradation when growth factors are removed. The cell cycle will arrest if this occurs prior to passage through the restriction point. 14. Which of the following is not one of the phases of mitosis? a. Prophase b. S phase c. Metaphase d. Anaphase Answer: b Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Describe the stages of mitosis. Feedback A: Incorrect. Prophase is the first stage of mitosis, and chromosome condensation occurs at this time. Feedback B: Correct! S phase, when the cell’s DNA becomes duplicated, occurs before mitosis, during interphase. Feedback C: Incorrect. Metaphase is the stage in mitosis in which the chromosomes align

themselves along the central plate of the spindle. Feedback D: Incorrect. Anaphase is the stage of mitosis in which the sister chromatids separate and move to opposite ends of the spindle. 15. The sequence of DNA on each chromosome where the sister chromatids are held together is called the a. centrosome. b. kinetochore. c. centromere. d. centriole. Answer: c Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Describe the stages of mitosis. Feedback A: Incorrect. The centrosomes are microtubule-organizing centers from which the microtubules emanate during the formation of the spindle. Feedback B: Incorrect. This is a structure consisting of several proteins bound to the centromere that mediates movement of chromosomes along the spindle. Feedback C: Correct! The centromere is a sequence of DNA where the sister chromatids are held together prior to their migration to opposite poles of the spindle. Feedback D: Incorrect. The centriole is a microtubular structure in the centrosomes of most animal cells. 16. The process during which the cytoplasm divides is called a. anaphase. b. S phase. c. telophase. d. cytokinesis. Answer: d Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Describe the stages of mitosis. Feedback A: Incorrect. During anaphase, the sister chromatids separate and move to opposite poles of the spindle. Feedback B: Incorrect. S phase is when the chromosomes are replicated, which occurs during interphase, before the onset of mitosis. Feedback C: Incorrect. During prophase, the chromosomes condense and centrosomes move to opposite sides of the nucleus, initiating formation of the mitotic spindle. Breakdown of the nuclear envelope then allows spindle microtubules to attach to the kinetochores of chromosomes. Feedback D: Correct! Mitosis ends with re-formation of nuclear envelopes and chromosome decondensation during telophase, and cytokinesis splits the cell into two interphase daughter cells. 17. Which of the following are not phosphorylated by the Cdk1/cyclin B complex during mitosis?

a. H3 histones b. Golgi matrix proteins c. Nuclear lamins d. Condensins Answer: a Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering Learning Objective: Explain the mechanism of chromosome condensation. Feedback A: Correct! Although H3 histones are phosphorylated during mitosis, the kinase that does this has not yet been identified. Feedback B: Incorrect. Golgi matrix proteins are phosphorylated by Cdk1/cyclin B and initiate Golgi breakdown. Feedback C: Incorrect. The nuclear lamins are phosphorylated by Cdk1/cyclin B, leading to the breakdown of the nuclear envelope. Feedback D: Incorrect. Condensins are involved in condensing the chromosomes during mitosis and are directly phosphorylated by Cdk1/cyclin B. 18. The spindle assembly checkpoint monitors the alignment of chromosomes in the mitotic spindle during mitosis to insure an even distribution of chromosomes to each of the daughter cells. If the chromosomes do not align properly, the spindle assembly checkpoint will arrest cells in what phase of mitosis? a. Prophase b. Metaphase c. Anaphase d. Telophase Answer: b Textbook Reference: The Events of M Phase Bloom’s Category: 2. Understanding Learning Objective: Explain the mechanism of chromosome condensation. Feedback A: Incorrect. This would be too early, as chromosomes have not aligned on the mitotic spindle during prophase. Feedback B: Correct! Chromosomes align on the mitotic spindle during metaphase, and if they are misaligned, cell cycle is arrested. Feedback C: Incorrect. This would be too late, as the chromosomes will have already begun to be separated and pulled into respective daughter cell regions in anaphase. Feedback D: Incorrect. This would be too late, as the daughter cells are fairly well defined in telophase. 19. The progression from metaphase to anaphase is triggered by a. activation of protein kinases by MPF. b. ubiquitin-mediated proteolysis of key proteins. c. binding of an inhibitory protein to MPF. d. depolymerization of the mitotic spindle. Answer: b Textbook Reference: The Events of M Phase Bloom’s Category: 1. Remembering

Learning Objective: Contrast cytokinesis in animal and plant cells. Feedback A: Incorrect. Activation of protein kinases by MPF is involved in earlier steps in mitosis. Feedback B: Correct! MPF activates the anaphase-promoting complex, which is a ubiquitin ligase. As a result, key proteins such as Scc1, which links sister chromatids together, and cyclin B are degraded. Feedback C: Incorrect. The metaphase/anaphase transition occurs primarily via ubiquitinmediated degradation of key proteins. Feedback D: Incorrect. This does not occur until telophase, once the chromosomes have segregated and cytokinesis is about to occur.

Essay 1. Normally, following division, cells grow so that daughter cells are relatively comparable in size to the parental cell before starting a new cell cycle. Embryonic cells, however, continually get smaller for some time following fertilization. Why is this? Answer: Shortly after fertilization, embryonic cells can complete a round of the cell cycle in as little as 30 minutes. This occurs in the absence of growth; thus each division effectively halves the size of the parental cell so that an eight- or sixteen-cell embryo is about the same size as the zygote from which it started. During this time these cells cycle between S and M only, and essentially skip over G 1 and G2, the stages at which growth usually occurs. Textbook Reference: The Eukaryotic Cell Cycle Bloom’s Category: 2. Understanding Learning Objective: Summarize the phases of the cell cycle. 2. There are many critical checkpoints in the cell cycle. One in particular ensures that the DNA content is replicated only once. What is the molecular mechanism regulating this process? Answer: The key molecule in this restriction of replication is MCM helicase. MCM binds DNA at origins of replication together with the other proteins that make up the origin recognition complex (ORC). The MCMs, in effect, are “licensing factors” which, along with the ORC, are absolutely required for the initiation of replication. The key is that MCM can bind DNA only during G1, and when replication is initiated, MCM is displaced from the origin of replication on the DNA. Therefore, as the cell cycle progresses through S, G2 , and M, new rounds of replication cannot be initiated because MCM can bind DNA only in G1. Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Explain the significance of cell cycle checkpoints. 3. Briefly describe how Hunt and colleagues identified cyclins and why they named them “cyclins”? Answer: Hunt et al. knew that upon fertilization, translation of new protein by maternal RNAs present in the oocyte occurs. Experimentally, they used radiolabeled methionine

that would be incorporated into newly synthesized proteins, then examined these proteins at short time intervals after fertilization. They observed that the level of a newly synthesized protein (now detectable, due to incorporated radiolabeled methionine) increased rapidly after induction of fertilization, then fell just as rapidly, in the period of about one hour. This pattern repeated itself over the next hour. These up-and-down patterns coincided with the first two- and four-cell stage divisions, and Hunt et al. postulated that these cyclins (named for the oscillatory nature of their expression pattern) were critical to cell division. Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Summarize the experiments that led to the discovery of MPF. 4. During G2 , the Wee1 protein phosphorylates tyrosine-15 of the Cdc2 protein kinase, thus inactivating it. Only once dephosphorylation occurs and Cdc2 becomes active can the cell pass from G 2 into M and proceed to cell division. Can you think of a reason why this protein was named Wee1? Answer: G 2 is a period in the cell cycle during which cell growth occurs. In Wee1 mutant cells, Cdc2 does not become phosphorylated at tyrosine-15 and thus does not acquire the inactivating phosphate. As a result, the cells enter M phase prematurely, before they have time to grow to their full size: hence the name, “wee” mutants. Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Explain the role of cyclins. 5. Many cancer researchers and tumor biologists focus their studies on regulation of the cell cycle. Why? Answer: When cancer begins to run unchecked because normal regulation fails, the affected cell, and possibly its progeny, divide out of control, producing a tumor that may become malignant. Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanism by which growth factors regulate cell cycle progression. 6. Describe the pathway by which growth factors signal passage through the restriction point in late G1. Answer: Extracellular growth factors stimulate the Ras/Raf/MEK/ERK pathway, which results in increased synthesis of the D-type cyclins. These then complex with Cdk4, 6, activating their kinase activity. Cdk4, 6/cyclin D then phosphorylates Rb, causing it to dissociate from the E2F transcription factor and allowing E2F to activate its target genes, the products of which are necessary for progression through the restriction point. Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanism by which growth factors regulate cell cycle progression.

7. Why is Rb, a product of the retinoblastoma gene, now termed a member of a family of proteins designated tumor suppressor genes? Answer: Normally, Rb exists in an underphosphorylated state and is found in complex with the E2F transcription factors that activate genes associated with the cell cycle. When bound by Rb, E2Fs are unavailable for transcription. During initiation of the cell cycle, Cdk4,6/cyclin D phosphorylates Rb, leading to a dissociation of Rb from E2F, leaving the now free E2F available to activate transcription of cell cycle genes. When mutations in Rb prevent its interaction with E2F, E2F is left unregulated; it is available to drive the cell cycle forward and, in essence, induce unrestricted growth or tumor formation. Thus, Rb was designated a tumor suppressor, because in its natural state it can be said to have an intrinsic ability to suppress tumor formation. Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 2. Understanding Learning Objective: Describe the mechanism by which growth factors regulate cell cycle progression. 8. Single-stranded DNA and unreplicated DNA are recognized by ATR, which leads to G2 arrest. What would likely occur with a single-stranded DNA break in a cell with a mutation that prevented phosphorylation of Chk1? What if the mutation prevented phosphorylation of Chk2? Answer: Cells with a Chk1 mutation would not arrest at G 2 like their normal counterparts. A mutation that prevents activation of Chk1 would also prevent inactivation of Cdc25. Thus, Cdk1 would not be blocked, and the cell cycle would continue, allowing any single-stranded damage or unreplicated DNA to be carried forward into daughter cells. A mutation in Chk2 would not affect single-stranded breaks because Chk2 is activated by ATM in response to double-stranded breaks. Thus, in response to a single-stranded break, a cell harboring a mutation in the Chk2 would be unaffected and undergo G2 arrest as expected because the ATR–Chk1–Cdc25–Cdk1–G2 arrest pathway would be intact. Textbook Reference: Regulators of Cell Cycle Progression Bloom’s Category: 3. Applying Learning Objective: Explain how the initiation of DNA replication is controlled. 9. Using the CRISPR/Cas system in a cell culture model, you mutate the gene encoding condensin so that it now encodes a nonfunctional product. What would be the predicted phenotype for these cells? Answer: The role of condensins is to direct the condensation of chromatin, leading to the formation of metaphase chromosomes in mitosis. If the condensins were mutated, then the chromosomes would not condense into the compact size needed for efficient segregation at anaphase, and resultant cells would likely have uneven copies of chromosomes. Textbook Reference: The Events of M Phase Bloom’s Category: 3. Applying Learning Objective: Explain the mechanism of chromosome condensation.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 19: Cell Death and Cell Renewal TEST FILE QUESTIONS Multiple Choice 1. Humans possess about _______ different cell types. a. 100 b. 200 c. 500 d. 1,000 Answer: b Textbook Reference: Introduction Bloom’s Category: 1. Remembering Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 2. In adult organisms, cell death must be balanced by cell a. elongation. b. differentiation. c. renewal. d. migration. Answer: c Textbook Reference: Introduction Bloom’s Category: 1. Remembering Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 3. Most cells in adult animals are in the _______ stage of the cell cycle. a. G0 b. G1 c. G2 d. S Answer: a Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 4. Fibroblasts are stimulated to exit G0 and proliferate by _______ growth factor.

a. fibroblast b. platelet-derived c. epithelial d. stem cell Answer: b Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 5. Which of the following adult cell types is arrested in G 0? a. Fibroblasts b. Endothelial cells c. Liver cells d. All of the above Answer: d Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 6. VEGF is secreted by a. cells deprived of oxygen. b. injured endothelial cells. c. injured smooth muscle cells. d. platelets. Answer: a Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 7. Which of the following is not a key property of stem cells in tissues? a. Dividing to produce one daughter cell that remains a stem cell b. Dividing to produce one cell that then differentiates c. Exhibiting self-renewal d. Dividing within each tissue to maintain a set number of cells Answer: d Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. 8. A hematopoietic stem cell may give rise to which of the following differentiated cells? a. Erythrocytes b. Macrophages c. Granulocytes

d. All of the above Answer: d Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. 9. Which of the following signaling pathways is not important to stem cell physiology? a. Wnt b. Notch c. TGF-β d. Bax Answer: d Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Summarize the key properties of stem cells. 10. Hematopoietic stem cells can give rise to a. erythrocytes. b. erythrocytes and lymphocytes. c. all blood cells and macrophages. d. all blood cells, macrophages, and skin cells. Answer: c Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. 11. Intestinal stem cells may give rise to all of the following terminally differentiated cell types in the intestinal epithelium except _______ cells. a. absorptive epithelial b. goblet c. transit-amplifying d. enteroendocrine Answer: c Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. 12. Intestinal epithelial stem cells are located _______ the crypt. a. at the top of b. at the bottom of c. at the sides of d. throughout Answer: b Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Summarize the key properties of stem cells.

13. When stem cells of the intestinal epithelium and epidermis divide, they give rise to two different cells: one stem cell and one _______ cell. a. goblet b. crypt c. G0 d. transit-amplifying Answer: d Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Summarize the key properties of stem cells. 14. The stem cells of skeletal muscle are called a. myoblasts. b. myocytes. c. satellite cells. d. myo-stem cells. Answer: c Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Summarize the key properties of stem cells. 15. Mammalian embryonic stem cells are cultured from cells taken from a. the four-cell embryo. b. the inner cell mass. c. trophoblast cells. d. amnion cells. Answer: b Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Explain the utility of embryonic stem cells in regenerative medicine. 16. Of the offspring that result from the reproductive cloning of mammals by somatic nuclear transfer, _______% are viable. a. 1–2 b. 10 c. 15–20 d. 25 Answer: a Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Describe somatic cell nuclear transfer and therapeutic cloning.

17. In 2006, Takahashi and Yamanaka demonstrated that somatic cells can be reprogrammed or converted into induced pluripotent stem cells. Subsequent studies in _______ showed that these cells, like embryonic stem cells, are capable of differentiating into virtually all cells. a. mice b. rats c. chimpanzees d. humans Answer: a Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Summarize the derivation of induced pluripotent stem cells. 18. Early experiments by Takahashi and Yamanaka showed that a set of four genes introduced by retroviral expression are capable of transforming somatic cells into induced pluripotent stem cells. The four proteins encoded by these genes were all a. G-protein coupled receptors. b. transcription factors. c. tyrosine kinases. d. calcium-binding proteins. Answer: b Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Summarize the derivation of induced pluripotent stem cells. 19. Which of the following transcription factors has the potential to act as an oncogene, resulting in a high risk of cancer? a. Oct4 b. Sox2 c. Klf4 d. c-Myc Answer: d Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Summarize the derivation of induced pluripotent stem cells. 20. Which of the following are the three core transcription factors that can reprogram adult somatic cells into induced pluripotent cells? a. Sox2, Oct4, and Nanog b. Sox2, Oct4, and Hedgehog c. Oct4, c-Myc, and Klf4 d. Sox2, c-Myc, and Nanog Answer: a

Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Summarize the derivation of induced pluripotent stem cells. 21. The reprogramming of somatic cells into other types of differentiated cells (other than induced pluripotent stem cells) is referred to as a. transduction. b. transfection. c. transdifferentiation. d. translocation. Answer: c Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Define transdifferentiation. 22. Using approaches similar to those of Takahashi and Yamanaka, scientists have now reprogrammed adult fibroblasts to become a. heart muscle cells. b. platelets. c. red blood cells. d. adipocytes. Answer: a Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 2. Understanding Learning Objective: Define transdifferentiation. 23. The distinct series of changes that characterizes programmed cell death is called a. necrosis. b. apoptosis. c. sepsis. d. execution. Answer: b Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe the events that characterize apoptosis. 24. During development, 50% of neurons die; those that survive are those that a. generated axons. b. received epithelial growth factor. c. made correct connections with their target cells. d. did not receive an extracellular cell death signal. Answer: c Textbook Reference: Programmed Cell Death

Bloom’s Category: 2. Understanding Learning Objective: Describe the events that characterize apoptosis. 25. Which of the following events is not part of apoptosis? a. Fragmenting of chromosomal DNA b. Fragmenting of the nucleus c. Fragmenting of the cell d. Lysis of cell fragments and release of their contents Answer: d Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe the events that characterize apoptosis. 26. Cells that die as the result of acute injury undergo a. lysis. b. sepsis. c. apoptosis. d. execution. Answer: a Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe the events that characterize apoptosis. 27. Apoptotic cells produce “eat me” signals, which include _______ of cell surface _______. a. loss; carbohydrates b. loss; phosphatidylserines c. gain; phosphatidylserines d. gain; phosphatidylethanolamines Answer: c Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe the events that characterize apoptosis. 28. The gene ced-3 of C. elegans codes for a proteolytic enzyme that is a a. serine protease. b. chymotrypsin. c. ubiquitin-targeted protease. d. caspase. Answer: d Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Explain the role of caspases in programmed cell death. 29. Ced-9 functions directly to a. inhibit Ced-3.

b. inhibit Ced-4. c. stimulate Ced-3. d. stimulate Ced-4. Answer: b Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Explain the role of caspases in programmed cell death. 30. An apoptosome is a complex containing multiple subunits of a. Apaf-1. b. Apaf-1 and caspase-9. c. Apaf-1 and caspase-3. d. Apaf-1, caspase-9, and cytochrome c. Answer: d Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Explain the role of caspases in programmed cell death. 31. Effector caspases cleave a. nuclear lamins. b. cytoskeletal proteins. c. DNase inhibitor. d. All of the above Answer: d Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Explain the role of caspases in programmed cell death. 32. Members of the Bcl-2 family of proteins are a. all caspases. b. either antiapoptotic or proapoptotic. c. all antiapoptotic. d. all proapoptotic. Answer: b Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Summarize the activities of Bcl-2 family members. 33. Bax and Bak act by a. interacting with Apaf-1 to initiate a proapoptotic signal. b. interacting with Ced-9 to initiate a proapoptotic signal. c. oligomerizing in the mitochondrial outer membrane, forming pores that release cytochrome c. d. oligomerizing with Apaf-1 to block an apoptotic signal. Answer: c Textbook Reference: Programmed Cell Death

Bloom’s Category: 2. Understanding Learning Objective: Summarize the activities of Bcl-2 family members. 34. Many stimuli that trigger cell death release the molecule _______ from the intermembrane space of mitochondria so they can bind to Apaf-1. a. cytochrome c b. citric acid c. ubiquinone d. ATP synthase Answer: a Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe signaling pathways that prevent and induce apoptosis. 35. The key protein involved in “deciding” whether a cell with DNA damage should delay its cell cycle or undergo apoptosis is a. Akt. b. pRB. c. p53. d. p15. Answer: c Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe signaling pathways that prevent and induce apoptosis. 36. A major pathway that signals cell survival and inhibits cell death is the _______ pathway. a. PI 3-kinase/Akt b. Hedgehog c. TNF d. Wnt Answer: a Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe signaling pathways that prevent and induce apoptosis. 37. Which of the following is not phosphorylated by Akt? a. PI 3-kinase b. p53 c. FOXO d. Bad Answer: a Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe signaling pathways that prevent and induce apoptosis.

Fill in the Blank 1. Following injury or exercise, _______ cells become activated and repair damaged skeletal muscle. Answer: satellite Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 2. Mouse embryonic stem cells were first cultured from _______. Answer: teratocarcinomas Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Explain the utility of embryonic stem cells in regenerative medicine. 3. With the birth of _______, Wilmut and colleagues became the first scientists to produce a successfully cloned mammal by somatic cell nuclear transfer. Answer: Dolly (a sheep) Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Describe somatic cell nuclear transfer and therapeutic cloning. 4. Genes for regulation and execution of programmed cell death, or _______, were originally discovered in C. elegans. Answer: apoptosis Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe the events that characterize apoptosis. 5. In addition to cytochrome c, other proapoptotic factors are released from _______. Answer: mitochondria Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe the events that characterize apoptosis. 6. The two proapoptotic proteins that enable the release of cytochrome c from the mitochondria are _______ and _______. Answer: Bax; Bak Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe the events that characterize apoptosis.

7. If the gene _______ is mutated in C. elegans so that it does not function, no cells will die during development. Answer: Ced-3 Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Explain the role of caspases in programmed cell death. 8. Cell death in cancer cells can be caused by the cytokine _______. Answer: TNF (tumor necrosis factor) Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe signaling pathways that prevent and induce apoptosis.

True/False 1. Most differentiated cell types do not themselves proliferate, but they can be replaced by the proliferation of stem cells. Answer: T Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 2. Liver cell injury induces proliferation of liver stem cells. Answer: F Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 3. Injured muscle cells regenerate by division of muscle fibers to form myoblasts that proliferate and fuse to form new muscle fibers. Answer: F Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 4. Satellite cells are located beneath the basal lamina of muscle fibers. Answer: T Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells.

5. Each of the three cell types in the adult intestinal epithelium is derived from a different stem cell. Answer: F Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. 6. Goblet cells and enteroendocrine cells have the same stem cells as absorptive epithelial cells. Answer: T Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Summarize the key properties of stem cells. 7. The clinical application of adult stem cells is limited by difficulties in isolating and culturing the appropriate stem cell populations. Answer: T Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Summarize the key properties of stem cells. 8. It is relatively easy to isolate and propagate embryonic stem cells. Answer: T Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Explain the utility of embryonic stem cells in regenerative medicine. 9. If reintroduced into early embryos, embryonic stem cells can give rise to cells in all tissues of a mouse. Answer: T Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 2. Understanding Learning Objective: Explain the utility of embryonic stem cells in regenerative medicine. 10. Reproductive cloning of humans raises ethical concerns. Answer: T Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Describe somatic cell nuclear transfer and therapeutic cloning. 11. Programmed cell death plays an important role in development. Answer: T Textbook Reference: Programmed Cell Death

Bloom’s Category: 1. Remembering Learning Objective: Describe the events that characterize apoptosis. 12. Most tissues maintain a balance between cell proliferation and cell death. Answer: T Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe the events that characterize apoptosis. 13. As a causal factor in the development of tumors, the inability of cancer cells to undergo apoptosis is as significant as their uncontrolled proliferation. Answer: T Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe the events that characterize apoptosis. 14. A mutant bcl-2 gene can be an oncogene. Answer: T Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Summarize the activities of Bcl-2 family members. 15. In normal, nonapoptotic cells, antiapoptotic members of the Bcl-2 protein family bind to proapoptotic members and inhibit their activity. Answer: T Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Summarize the activities of Bcl-2 family members.

Short Answer 1. Provide three examples of adult human cells that are typically in G 0 but can proliferate when stimulated. Answer: (1) Fibroblasts, (2) smooth muscle cells, and (3) endothelial cells are examples of adult human cells that are typically in G0 and can proliferate when stimulated. Epithelial cells of most organs, such as the liver, kidney, lung, prostate, and breast, are also good examples. Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 2. Provide three examples of normal human cells that, in their fully differentiated form, can no longer proliferate but can be replaced by means of stem cell proliferation and differentiation.

Answer: Any three of the following: blood cells, intestinal epithelial cells, skin cells sperm cells, skeletal muscle cells, neurons, lens cells, macrophages, and plasma cells Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 3. When two-thirds of the liver of a rat is cut away, what happens to the remaining cells? Answer: They re-enter the cell cycle and proliferate until the liver regains its normal size. Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 3. Applying Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. 4. Define a stem cell. Answer: A stem cell is a cell that divides to form one daughter cell that remains a stem cell and one that proliferates and then differentiates. Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Summarize the key properties of stem cells. 5. What do skeletal muscle stem cells do when activated? Answer: The satellite cells give rise to progeny cells that undergo several cell divisions and then differentiate and fuse to form new muscle fibers. Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. 6. Why are skin, hair, intestinal epithelia, and blood cells particularly problematic areas after radiation exposure or chemotherapy for cancer? Answer: These cells and tissues are dependent on continued renewal by stem cells, and radiation or chemotherapy stops their proliferation. Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. 7. What is the advantage of using skin grafts derived from the patient’s own stem cells, as opposed to cells derived from a donor? Answer: Grafts from the patient’s own stem cells are less likely to be rejected by the patient’s immune system. Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. 8. Which human disease could potentially be cured with stem cell transplantation therapy?

Answer: Parkinson’s disease (also Alzheimer’s disease, diabetes, spinal cord injuries) Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. 9. How do embryonic stem cells differ from adult stem cells? Answer: Embryonic stem cells can be grown indefinitely as stem cell populations, and they have the broadest differentiation capacity, potentially giving rise to all cell types of an adult, whereas adult stem cells are limited to certain cell lineages. Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. 10. How can the toxic effect of high-level chemotherapy on the hematopoietic system be repaired? Answer: The toxic effects of chemotherapy can be repaired by a post-treatment transplant to the patient of hematopoietic stem cells from the bone marrow of a donor or the patient (collected and stored before treatment). Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Explain the role of stem cells in bone marrow transplantation. 11. How can embryonic stem cells be directed to differentiate along specific pathways? Answer: They can be directed to differentiate along specific pathways by the addition of appropriate growth/differentiation factors to the medium. Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 3. Applying Learning Objective: Explain the utility of embryonic stem cells in regenerative medicine. 12. What mammalian species was the first to be cloned from an adult nucleus transplanted into an oocyte cytoplasm? Answer: Sheep (lamb) Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Describe somatic cell nuclear transfer and therapeutic cloning. 13. What is the difference between reproductive and therapeutic cloning? Answer: Therapeutic cloning is the production of embryonic stem cells for the purpose of treating illness; it is accomplished by transplanting the nucleus from the patient’s somatic cell into an enucleated egg. Reproductive cloning is the production of genetically identical offspring from somatic nuclei that have been transplanted into enucleated eggs. Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine

Bloom’s Category: 2. Understanding Learning Objective: Describe somatic cell nuclear transfer and therapeutic cloning. 14. What observation in C. elegans suggested that cell death occurs as a specific developmental fate? Answer: Researchers discovered that of the 1,090 somatic cells initially produced, the same (131) cells died in every developing individual. Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe the events that characterize apoptosis. 15. In multicellular organisms, why is apoptosis not problematic, whereas cell death after acute injury is? Answer: Apoptotic cell fragments are engulfed by phagocytic cells, degraded, and recycled, whereas death by acute injury results in lysis, which releases molecules that can cause inflammation. Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe the events that characterize apoptosis. 16. What is the difference between an initiator caspase and an effector caspase? Answer: An initiator caspase starts a chain reaction by cleaving another caspase, which activates it, whereas an effector caspase cleaves other noncaspase proteins, changing cell function or structure. Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Explain the role of caspases in programmed cell death. 17. Why are cells with damaged genomes dangerous and often destroyed? Answer: They may have suffered mutations that could lead to a cancer. Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe signaling pathways that prevent and induce apoptosis. 18. What determines whether DNA damage to a given cell will result in cell cycle delay or apoptosis? Answer: The extent of DNA damage determines the level of p53 induction and hence whether the cell is arrested or killed. Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe signaling pathways that prevent and induce apoptosis. 19. How does tumor necrosis factor (TNF) trigger apoptosis? Answer: TNF binds to TNF receptors in the cell membrane and induces trimerization. The cytosolic portions of the receptors bind adaptor molecules that bind caspase-8, which in some cells activates caspase-3 and -7, inducing apoptosis directly. In other cells,

caspase-8 cleaves Bid, activating it, permeabilizing mitochondria, and activating caspase9, amplifying the cascade initiated by caspase-8 activation. Textbook Reference: Programmed Cell Death Bloom’s Category: 3. Applying Learning Objective: Describe signaling pathways that prevent and induce apoptosis. 20. Name two types of cells, other than cancer cells, that are killed by tumor necrosis factor (TNF). Answer: Virus-infected cells and excess lymphocytes at the end of an immune response are two other types of cells killed by TNF. Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe signaling pathways that prevent and induce apoptosis.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. Some differentiated cells retain the ability to proliferate as needed to repair damaged tissue throughout the life of the organism. Which of the following cell types do not retain this unique ability? a. Fibroblasts b. Neurons c. Endothelial cells d. Epithelial cells Answer: b Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Give examples of tissue maintenance by the proliferation of differentiated cells. Feedback A: Incorrect. Fibroblasts are capable of proliferation throughout life. Feedback B: Correct! As a rule, neurons do not have the capacity to proliferate, even if needed. Feedback C: Incorrect. Endothelial cells lining blood vessels can proliferate continually. Feedback D: Incorrect. Epithelial cells, especially those found in organs, can proliferate readily. 2. Which statement about stem cells is false? a. Stem cells divide to form one daughter cell that remains a stem cell and a second cell that proliferates and then differentiates. b. Stem cells are a subpopulation of less differentiated, self-renewing cells present in most adult tissues. c. Stem cells in organs can only differentiate into the tissue in which they are located. d. Stem cells were first identified in the hematopoietic system.

Answer: c Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Summarize the key properties of stem cells. Feedback A: Incorrect. This is a true statement. Feedback B: Incorrect. This is a true statement. Feedback C: Correct! This is a false statement. For example, hematopoietic stem cells can differentiate into skin, lung, bone, etc. Feedback D: Incorrect. This is a true statement. 3. Which of the following is not considered a viable source of stem cells? a. Umbilical cord blood b. Bone marrow c. An embryo d. Tumor cell Answer: d Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Summarize the key properties of stem cells. Feedback A: Incorrect. This is a correct answer. Feedback B: Incorrect. This is a correct answer. Feedback C: Incorrect. This is a correct answer. Feedback D: Correct! Tumor cells can be isolated and grown in culture but are not generally thought to be sources of stem cells. 4. Which of the following cells cannot develop from hematopoietic stem cells? a. Satellite cells b. Platelets c. B lymphocytes d. Macrophages Answer: a Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Summarize the key properties of stem cells. Feedback A: Correct! Satellite cells are the stem cells in skeletal muscle that give rise to new muscle following damage or injury. Feedback B: Incorrect. Hematopoietic stem cells develop into platelets from megakaryocytes in the myeloid lineage. Feedback C: Incorrect. Hematopoietic stem cells develop into B lymphocytes in the lymphoid lineage. Feedback D: Incorrect. Hematopoietic stem cells develop into macrophages from monocytes in the myeloid lineage. 5. Bone marrow transplants are used to treat patients with a. cancer. b. hemoglobin disorders.

c. immune deficiencies. d. All of the above Answer: d Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 1. Remembering Learning Objective: Explain the role of stem cells in bone marrow transplantation. Feedback A: Incorrect. This is a correct answer, but it is not the only correct answer. Feedback B: Incorrect. This is a correct answer, but it is not the only correct answer. Feedback C: Incorrect. This is a correct answer, but it is not the only correct answer. Feedback D: Correct! All of these illnesses are currently being treated with bone marrow transplants, which in effect are hematopoietic stem cell transfers. 6. Which of the following stem cells can give rise to all the differentiated cell types of adult organisms? a. Embryonic stem cells b. Endothelial cells c. Satellite cells d. Epithelial cells Answer: a Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 1. Remembering Learning Objective: Explain the utility of embryonic stem cells in regenerative medicine. Feedback A: Correct! Currently these are the only cells known to have this capacity, although scientists continue to learn more about stem cells with other origins. Feedback B: Incorrect. These give rise to new blood vessels. Feedback C: Incorrect. These give rise to muscle cells. Feedback D: Incorrect. These predominantly give rise to the same tissue as in which they reside. 7. Which statement about cloning by somatic cell nuclear transfer is false? a. The nucleus of an adult somatic cell is transferred to an enucleated, unfertilized egg. b. Somatic cell nuclear transfer was first carried out in frogs in the 1950s. c. In the laboratory, the cloned egg is cultured to an early embryo stage and then transferred to a foster mother. d. The cloned embryo is genetically identical to the foster mother. Answer: d Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 2. Understanding Learning Objective: Describe somatic cell nuclear transfer and therapeutic cloning. Feedback A: Incorrect. This is a true statement. Feedback B: Incorrect. This is a true statement. Feedback C: Incorrect. This is a true statement. Feedback D: Correct! The embryo is genetically identical to the adult from which the somatic cell nucleus was harvested, which is usually not the foster mother.

8. Which of the following diseases would not be an ideal target for therapeutic cloning? a. HIV b. Parkinson’s disease c. Spinal cord injury d. Diabetes Answer: a Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 3. Applying Learning Objective: Describe somatic cell nuclear transfer and therapeutic cloning. Feedback A: Correct! HIV is caused by a virus and not a cellular mutation or defect. Feedback B: Incorrect. Therapeutic cloning could be useful if stem cells could be differentiated into neurons. Feedback C: Incorrect. Therapeutic cloning could be useful if stem cells could be differentiated into functional nerves. Feedback D: Incorrect. Diabetes would be an ideal target if stem cells could be differentiated into functional pancreatic  cells. 9. What is the best rationale for the use of therapeutic cloning to replace defective cells or tissues in a patient? a. The transplanted cells are identical to those of the recipient who donated the adult nucleus, so complications of immune rejection are avoided. b. Embryos do not have to be destroyed, so ethically there is little controversy. c. It is a very efficient process for generating the tissues that need to be replaced. d. We already know how to fully differentiate the embryonic stem cells into all the major cell types. Answer: a Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 3. Applying Learning Objective: Describe somatic cell nuclear transfer and therapeutic cloning. Feedback A: Correct! Without question, this is the best argument for therapeutic cloning. Feedback B: Incorrect. In fact, early-stage embryos have to be dissected, and this has generated many ethical arguments. Feedback C: Incorrect. In fact, just the opposite is true. Much work is still needed for this to become a routine methodology for therapy. Feedback D: Incorrect. While many cell types have been generated from embryonic stem cells, the process has not been perfected for many others, and it has not yet been repeated consistently in different laboratories. 10. Apoptosis, or programmed cell death, occurs in all of the following cases except a. in virus-infected cells. b. in cells damaged by injury. c. in cells with potentially cancer-causing mutations. d. during the formation of human fingers.

Answer: b Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe the events that characterize apoptosis. Feedback A: Incorrect. Apoptosis often occurs in infected cells, helping to limit the spread of infection. Feedback B: Correct! Cells damaged by injury do not die in the distinct manner of cells undergoing apoptosis but by swelling and lysing. Feedback C: Incorrect. Cells with mutations that can cause cancer may undergo apoptosis. Feedback D: Incorrect. This is a classic case in which apoptosis plays a role in human development. 11. Which of the following events is not associated with apoptosis? a. Fragmentation of chromosomal DNA b. Condensation of chromatin c. Breaking up of the cell nucleus d. Cell cycle re-entry Answer: d Textbook Reference: Programmed Cell Death Bloom’s Category: 1. Remembering Learning Objective: Describe the events that characterize apoptosis. Feedback A: Incorrect. This event is associated with apoptosis. Feedback B: Incorrect. This event is associated with apoptosis. Feedback C: Incorrect. This event is associated with apoptosis. Feedback D: Correct! Cells do not need to enter cell cycle to undergo apoptosis. 12. Which statement about caspases is false? a. They cleave after aspartic acid residues in their substrate proteins. b. They inhibit DNase which, when activated, fragments nuclear DNA. c. They predominantly break up Golgi membranes, which releases other caspases to cleave nuclear lamins and cytoskeleton proteins. d. They contain cysteine residues in the active site. Answer: c Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Explain the role of caspases in programmed cell death. Feedback A: Incorrect. This is a true statement. Feedback B: Incorrect. This is a true statement. Feedback C: Correct! Caspases directly break up nuclear lamins and cytoskeleton proteins. Feedback D: Incorrect. This is a true statement. 13. Which products would result from caspase cleavage of a protein with the amino acid sequence AVEELCSTWYYDLLCHMRTL? a. AVEELCS + TWYYDLLCHMRTL

b. AVEELCSTWYYD + LLCHMRTL c. AVEELCSTWYYDLLCH + MRTL d. AVEELC + STWYYDLLC + HMRTL Answer: b Textbook Reference: Programmed Cell Death Bloom’s Category: 3. Applying Learning Objective: Explain the role of caspases in programmed cell death. Feedback A: Incorrect. Caspases do not cleave proteins at serine residues. Feedback B: Correct! Caspases cleave proteins at aspartate residues. Feedback C: Incorrect. Caspases do not cleave proteins at histidine residues. Feedback D: Incorrect. Caspases do not cleave proteins at cysteine residues. 14. Which statement regarding Bax and Bak, two members of the Bcl-2 family, is true? a. Bax and Bak are antiapoptotic proteins that function to repress apoptosis in cells. b. Bax and Bak act as ligands to the apoptosome in activation caspases. c. When activated, Bax and Bak form oligomers in the outer membrane of mitochondria, leading to the release of cytochrome c. d. Bax and Bak directly activate caspase-9. Answer: c Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Summarize the activities of Bcl-2 family members. Feedback A: Incorrect. They are proapoptotic multidomain proteins that induce apoptosis. Feedback B: Incorrect. Cytochrome c binds to the apoptosome at Apaf-1. Feedback C: Correct! Feedback D: Incorrect. Caspase-9 is part of the apoptosome complex and is activated only when the apoptosome is in the presence of cytochrome c. 15. The following represents a signal transduction cascade initiated by double-stranded DNA damage and ending with activation of apoptosis by the BH3-only proteins PUMA and Noxa. Double-stranded DNA damage; Chk2; _______; PUMA/Noxa; apoptosis. Which protein belongs in the blank? a. Apaf-1 b. p53 c. Chk1 d. Bax Answer: b Textbook Reference: Programmed Cell Death Bloom’s Category: 3. Applying Learning Objective: Describe signaling pathways that prevent and induce apoptosis. Feedback A: Incorrect. Apaf-1 is in the mitochondrial pathway to apoptosis. Feedback B: Correct! p53 is phosphorylated and stabilized by Chk2, which then goes on

to increase transcription of PUMA and Noxa. Feedback C: Incorrect. Chk1 is in a related DNA-damage pathway that uses either singlestranded breaks or unreplicated DNA. Feedback D: Incorrect. Bax is part of the mitochondrial pathway to apoptosis. 16. The PI 3-kinase pathway is important in countering apoptosis in healthy cells and promotes cell survival. Which statement about the PI 3-kinase pathway is false? a. PI 3-kinase is activated by a receptor tyrosine kinase. b. PIP3 activates Akt, a serine/threonine kinase. c. Activated Akt phosphorylates and activates Bad. d. PI 3-kinase phosphorylates PIP2 to form PIP3. Answer: c Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe signaling pathways that prevent and induce apoptosis. Feedback A: Incorrect. This is a true statement. Feedback B: Incorrect. This is a true statement. Feedback C: Correct! Akt phosphorylates and inactivates Bad. In its nonphosphorylated state, Bad is activated and triggers apoptosis. Feedback D: Incorrect. This is a true statement. 17. TNF and related cytokines can directly induce apoptosis. Which statement about TNF-mediated induction of apoptosis is true? a. TNF binds a trimeric receptor complex on the surface of the cell membrane. b. Upon receptor binding, caspase-8 is activated through an adaptor protein. c. Caspase-8 activates Bid, which in turn activates Bax and Bak and then feeds into the mitochondrial pathway. d. All of the above statements are true. Answer: d Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe signaling pathways that prevent and induce apoptosis. Feedback A: Incorrect. This is a correct answer, but it is not the only correct answer. Feedback B: Incorrect. This is a correct answer, but it is not the only correct answer. Feedback C: Incorrect. This is a correct answer, but it is not the only correct answer. Feedback D: Correct! All of these statements combined describe the TNF apoptotic pathway.

Essay 1. Describe some of the differences between adult-derived stem cells and embryonic stem cells. Answer: The most obvious difference is the source. Adult-derived stem cells are isolated from adult tissues, most of which have populations of stem cells for replenishing damaged or worn out cells within the tissue. Embryonic stem cells are isolated from

embryos. These cells currently appear to be limited to differentiation into cells of the tissues from which they arose. Embryonic stem cells inherently have a much greater potential than adult-derived stem cells and can be differentiated into every kind of cell and tissue or even into an entire organism. Embryonic stem cells are naturally pluripotent, whereas adult somatic cells must be reprogrammed by introducing several transcription factors into the cells in culture. These new methods allow somatic cells to become induced pluripotent stem cells that resemble embryonic stem cells. Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 2. Understanding Learning Objective: Explain the utility of embryonic stem cells in regenerative medicine. 2. Imagine you have taken stem cells isolated from umbilical cord blood and successfully differentiated them into cells that look like pancreatic -cells. What would you need to know before deciding that they could be implanted into humans to treat diabetes? Answer: First, you would need to know that the cells do in fact function like pancreatic cells. Most importantly, you would need to know that the cells secrete insulin (i.e., the function that is lost in some forms of diabetes), and that the secretion of insulin is dependent on glucose levels. Another critical factor would be knowing their proliferation potential, since cancer is a risk in some cell therapies. Yet another matter to keep in mind is where the cells would need to be transplanted for proper function. This location might not be obvious—for instance, in diabetes, the liver has been proposed as a possible site because insulin, as an endocrine hormone, can function wherever it is secreted. Finally, the potential for rejection must be considered. Textbook Reference: Stem Cells and the Maintenance of Adult Tissues Bloom’s Category: 3. Applying Learning Objective: Describe somatic cell nuclear transfer and therapeutic cloning. 3. Briefly describe the steps involved in cloning using somatic cell nuclear transfer. Give an example. Answer: The nucleus of an adult somatic cell (nongonadal cell, such as an epithelial cell) is transferred to an unfertilized egg that has been enucleated (meaning the nucleus of the egg has been removed). The egg is then cultured in vitro to an early blastocyst embryo and implanted into a foster mother who was not the source of the somatic cell nucleus. The foster mother gives birth to a clone that is identical to the donor of the somatic cell. By far the most well-known example is Dolly the sheep, which was cloned by Ian Wilmut’s group in 1997. Since then, numerous species have been cloned. Textbook Reference: Pluripotent Stem Cells, Cellular Reprogramming, and Regenerative Medicine Bloom’s Category: 3. Applying Learning Objective: Describe somatic cell nuclear transfer and therapeutic cloning. 4. Mammalian cells are constantly dying, whether from apoptosis or from acute injury. What is the key difference between these two processes? Answer: Apoptotic cells and cell fragments are recognized and efficiently removed from surrounding tissues via phagocytotic macrophages with no pathophysiological consequence. It is an active and programmed process. Cells subjected to acute injury

swell, lyse, and spill into the extracellular space. This release leads to tissue inflammation. If apoptosis had similar consequences, the subsequent inflammation likely would not allow an organism to survive. Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe the events that characterize apoptosis. 5. When fragmented nuclear DNA is visualized by gel electrophoresis following induction of apoptosis, why do the fragments almost always appear as multiples of 200 base pairs? Answer: DNA exists in the nucleus as chromatin and is intimately associated with many proteins. The basic structural unit of chromatin is the nucleosome, which contains 200 base pairs of DNA wrapped around a histone core. The linker DNA, which consists of about 50 base pairs, is particularly sensitive to digestion by nucleases such as DNase, but the 200 base pairs within each nucleosome are fairly well protected from digestion. Therefore, when run on a gel that denatures the proteins, the DNA strands that are observed are multiples of 200. Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe the events that characterize apoptosis. 6. In the 1970s and 1980s, investigators in the laboratory of Robert Horvitz demonstrated the existence of genes that directly regulate a process of programmed cell death. They did their work in C. elegans, a small nematode worm. What is the likely reason that they did these experiments with C. elegans cells and not with mammalian or human cells? Answer: The investigators studied development in C. elegans because they knew that 1,090 cells are produced during the development of each worm and that each adult worm has only 959 cells. This discrepancy suggested that there are molecular mechanisms controlling the loss or death of the other 131 cells. C. elegans was an ideal study specimen because its relatively small number of cells allowed the scientists to follow the fate of each cell. Also, the worms have a very short life span and large numbers of progeny, which allows researchers to use genetic methods to identify naturally occurring mutants that appear in offspring and to identify the gene responsible. Mammalian cells are much more complex, and there is significant genetic variation among individuals. Because many genes have related counterparts in other organisms (homologs), investigators often identify functional genes in simple organisms before looking for related genes in more complex organisms. The work on C. elegans is an example of such a strategy. Shortly after the Horvitz lab identified ced 3, ced 4, and ced 9 in C. elegans, other researchers were able to identify the mammalian homologs Apaf-1 and Bcl-2. Textbook Reference: Programmed Cell Death Bloom’s Category: 3. Applying Learning Objective: Describe the events that characterize apoptosis. 7. One of the hallmark characteristics of apoptosis is fragmentation of nuclear DNA. Which protein is responsible for this fragmentation, and what is the mechanism by which it degrades the DNA?

Answer: Caspases are the apoptotic enzymes responsible for fragmentation of nuclear DNA. All cells have DNase enzymes that digest and fragment nuclear DNA. Under normal conditions, various proteins function as inhibitors of DNase. In apoptosis, caspases block the DNase inhibitor, which in effect activates DNase and initiates the fragmentation process. This happens early in apoptosis and can be observed within 1–3 hours of induction. Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe the events that characterize apoptosis. 8. Bcl-2 is an antiapoptotic protein which, under normal conditions, contributes to the balance of activity of numerous proapoptotic and antiapoptotic proteins. What would be the result of a gain-of-function mutation in the bcl-2 gene that led to a constant and significant elevation in cellular Bcl-2 protein levels? Answer: A gain-of-function mutation can occur in the bcl-2 gene, and its discovery led to the identification of the oncogene bcl-2 in some B-cell lymphoma cancers. If Bcl-2 becomes overexpressed, the balance is tipped in favor of Bcl-2 and apoptosis is blocked, even under conditions that normally would induce cell death. The effect is proliferation of unhealthy cells that normally would be eliminated by apoptosis. Textbook Reference: Programmed Cell Death Bloom’s Category: 3. Applying Learning Objective: Summarize the activities of Bcl-2 family members. 9. Viral genomes are generally quite small, usually encoding only a small number of genes. Some identified viruses encode inhibitors of apoptosis (IAPs). With such a limited genome, why is it advantageous for a virus to have an IAP as part of its molecular machinery? Answer: Once inside the host cell, viruses rely predominantly on the host cell (its proteins, enzymes, etc.) for continued growth. Mammalian cells, however, have multiple pathways that thwart viral infection. One of these pathways involves apoptosis, which frequently is activated by virus-infected cells to limit both the spread of the infection and the generation of new viral particles. In effect, the infected cells are sacrificed to prevent infection of other cells. Viruses are constantly mutating to overcome host responses such as this. Viruses that have developed a gene that inhibits apoptosis may have an improved likelihood of survival. Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe signaling pathways that prevent and induce apoptosis. 10. Signaling pathways mediate many cellular processes. Apoptosis is also regulated by multiple signaling pathways. Briefly describe two signaling pathways that regulate apoptosis. Answer: One regulator of apoptosis is the p53 pathway. Upon DNA damage, ATM and Chk2 protein kinases phosphorylate and stabilize p53, which in turn transcribes the genes encoding PUMA and Noxa. PUMA and Noxa are proapoptotic proteins that initiate apoptosis. Another is the PI 3-kinase pathway, which can be activated by either protein

tyrosine kinases or G protein-coupled receptors. Once activated, PI 3-kinase phosphorylates PIP2 to PIP3; this activates Akt, which in turn phosphorylates another BH3-only proapoptotic protein, Bad. Another pathway involves activation of caspase 9 by TNF-alpha receptors. Textbook Reference: Programmed Cell Death Bloom’s Category: 2. Understanding Learning Objective: Describe signaling pathways that prevent and induce apoptosis.

Test Bank to accompany

The Cell: A Molecular Approach, Eighth Edition Geoffrey M. Cooper

Chapter 20: Cancer TEST FILE QUESTIONS Multiple Choice 1. A tumor is a. a malignant growth. b. any abnormal growth of cells. c. a benign growth. d. a cancerous growth. Answer: b Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Describe tumor progression. 2. Cancer is usually caused by a. a cancer virus. b. an inherited oncogene. c. the stepwise breakdown of normal cell regulatory processes. d. X rays. Answer: c Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Describe tumor progression. 3. Tumor initiation occurs in a a. single protein molecule. b. single cell. c. few cells. d. single tissue. Answer: b Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Describe tumor progression. 4. A tumor of an epithelial cell is a(n) a. carcinoma. b. sarcoma. c. leukemia.

d. lymphoma. Answer: a Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of cancer cells. 5. The form of cancer with the highest mortality rate in the United States is _______ cancer. a. breast b. prostate c. lung d. colon Answer: c Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of cancer cells. 6. The mechanism by which normal cells stop proliferating as a result of reduced availability of growth factors is called a. density dependent inhibition. b. contact inhibition. c. differentiation. d. senescence. Answer: a Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. 7. Which characteristic is commonly the same in both normal cells and cancer cells? a. Density-dependent inhibition of proliferation b. Contact inhibition of migration c. Growth factor requirements d. Dependence on oxygen and nutrients Answer: d Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. 8. Compared to their normal counterparts, leukemic cells a. continue to differentiate. b. fail to proliferate. c. fail to undergo apoptosis. d. induce widespread apoptosis. Answer: c Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding

Learning Objective: Summarize the properties of cancer cells. 9. In some cases, tumors support their unrestricted proliferation producing a growth factor that they also respond to. This mechanism is called _______ signaling. a. endocrine b. juxtacrine c. autocrine d. paracrine Answer: c Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. 10. A tumor promoter is a type of carcinogen that a. induces apoptosis of healthy cells. b. decreases chromosomal stability. c. increases cell proliferation. d. immortalizes cancer cells. Answer: c Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Compare cancer induction by chemicals and viruses. 11. Ultraviolet radiation increases the likelihood of a cell becoming malignant primarily because UV radiation a. induces melanocytes to express more melanin. b. breaks DNA into fragments. c. makes RNA polymerase more error-prone. d. causes DNA damage that leads to mutations. Answer: d Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Compare cancer induction by chemicals and viruses. 12. Asbestos increases the risk of mesothelioma and lung cancer because it a. inhibits differentiation. b. causes DNA damage. c. turns on the BRCA oncogenes. d. acts as a tumor promoter by stimulating cell proliferation. Answer: d Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Compare cancer induction by chemicals and viruses. 13. Which of the following behaviors most increases the likelihood of developing cancer? a. Drinking a glass of wine a day

b. Smoking a pack of cigarettes a day c. Kissing a person who has cancer d. Breathing the sawdust from a crown gall tumor Answer: b Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Compare cancer induction by chemicals and viruses. 14. Oncogenes were first discovered in a. a chicken retrovirus. b. a human DNA virus. c. mouse cancer cells. d. human cancer cells. Answer: a Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Explain how retroviral oncogenes were identified. 15. The RNA viruses that most commonly pick up cellular oncogenes are the a. adenoviruses. b. retroviruses. c. papilloma viruses. d. HIV viruses. Answer: b Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Explain how retroviral oncogenes were identified. 16. A proto-oncogene is a. a normal gene from which an oncogene can arise. b. one that has been picked up by an oncogenic virus. c. evolving into an oncogene. d. expressed normally in tumor cells. Answer: a Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Contrast oncogenes and proto-oncogenes. 17. The viral raf oncogene produces a constitutively active protein because the a. gene is always on. b. kinase domain is mutant and cannot be competitively inhibited. c. regulatory domain has been replaced by viral sequences so the protein cannot be turned off. d. GTPase activity is deficient and the protein is always in the GTP form. Answer: c Textbook Reference: Oncogenes

Bloom’s Category: 2. Understanding Learning Objective: Contrast oncogenes and proto-oncogenes. 18. Which statement describes the first direct evidence for involvement of cellular oncogenes in human tumors? a. Normal human cells could be transformed by the src-containing Rous sarcoma viruses. b. DNA extracted from a human bladder carcinoma was able to transform recipient mouse cells in culture. c. DNA extracted from a mouse carcinoma was able to transform recipient human cells in culture. d. DNA extracted from a human carcinoma was able to transform recipient chicken cells in culture. Answer: b Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Describe the ways in which oncogenes are formed in human cancers. 19. The most frequently encountered oncogene in human tumors is a. fos. b. myc. c. ras. d. src. Answer: c Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Describe the ways in which oncogenes are formed in human cancers. 20. C-myc becomes an oncogene in Burkitt’s lymphoma and several plasmacytomas by a a. point mutation. b. duplication. c. translocation to the immunoglobulin heavy chain locus. d. deletion of a regulatory sequence. Answer: c Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Describe the ways in which oncogenes are formed in human cancers. 21. The ras proto-oncogene can become an oncogene by a single point mutation that alters its protein product to have _______ activity. a. more GTPase b. constitutive Raf-activation c. greater nucleotide exchange d. less nucleotide exchange Answer: b Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding

Learning Objective: Describe the ways in which oncogenes are formed in human cancers. 22. Translocation of a gene can produce a fusion product with uncontrolled protein kinase activity. An example of this type of fusion protein is a. Tel/PDGFR. b. Myc/Max. c. Fos/Jun. d. ErbB/Ras. Answer: a Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Describe the ways in which oncogenes are formed in human cancers. 23. What normal cellular process would be affected in a cell line in which Bcl-2 was mutated? a. Cell cycle b. Apoptosis c. cAMP signaling d. Terminal differentiation Answer: b Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Describe the ways in which oncogenes are formed in human cancers. 24. Loss-of-function mutations can be oncogenic if the mutant gene codes for a a. telomerase. b. tumor suppressor protein. c. cyclin-dependent kinase. d. DNA synthesis enzyme. Answer: b Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Contrast tumor suppressor genes and oncogenes. Questions 25–26. Suppose you were to fuse a tumor cell with a nontumor cell in culture, grow the daughter cells of the hybrid clone in culture, and then inject them into mice with defective immune systems. 25. Tumors would likely develop in _______ of the mice. a. less than 1% b. 25% c. 50% d. nearly 100% Answer: a Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 3. Applying

Learning Objective: Contrast tumor suppressor genes and oncogenes. 26. What is the reason for the likely result of the experiment described? a. Tumorigenicity is dominant. b. 50% of the cells get a tumor suppressor gene. c. Tumor suppressor genes are recessive. d. Tumor suppressor genes are dominant. Answer: d Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 3. Applying Learning Objective: Contrast tumor suppressor genes and oncogenes. 27. The first tumor suppressor gene was identified in studies of which human tumor? a. Burkitt’s lymphoma b. Retinoblastoma c. Glioblastoma d. Promyelocytic leukemia Answer: b Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Contrast tumor suppressor genes and oncogenes. 28. Nonhereditary retinoblastoma develops only when a. a somatic mutation inactivates the Rb gene. b. a somatic mutation activates the Rb gene. c. two somatic mutations inactivate two separate Rb genes. d. one germ-line mutation and one somatic mutation inactivate Rb genes. Answer: c Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Contrast tumor suppressor genes and oncogenes. 29. The product of the human papillomavirus oncogene E7 promotes cancer by a. inhibiting DNA damage repair pathways. b. binding and preventing function of Rb protein. c. ubiquitinating the tumor suppressor p53. d. enhancing expression of telomerase. Answer: b Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 30. The tumor suppressor genes Smad2 and Smad4 encode a. growth factor receptors. b. GTP-binding signal molecules.

c. transcription factors. d. inhibitors that bind to transcription factors. Answer: c Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 31. The normal function of the tumor suppressor protein Rb is to a. induce apoptosis. b. inhibit Ras. c. inhibit progression through the G1 restriction point. d. inhibit Cdk4/cyclin B activity. Answer: c Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 32. An increase in p53 activity usually results after a cell receives a. extracellular growth inhibition signals. b. DNA damage. c. heat shock. d. extracellular growth stimulating signals. Answer: b Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 33. p53 induces apoptosis by a. inactivating a growth factor receptor. b. activating transcription of a Bcl-2 family member. c. activating transcription of a caspase. d. directly activating a caspase. Answer: b Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 34. Apoptosis is induced by p53 when a. a cell is programmed to die during development. b. growth factor levels decline to zero. c. any DNA damage is detected. d. severe DNA damage is detected and cannot be repaired.

Answer: d Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 35. One function of the tumor suppressor protein p53 is to a. activate synthesis of p21, a Cdk inhibitor. b. inhibit phosphodiesterase. c. stimulate cell cycle progression. d. phosphorylate cyclin-dependent kinases on an inhibitory site. Answer: a Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 36. MicroRNAs can act as tumor suppressors by inhibiting expression of all of the following except a. Cdk. b. ras. c. myc. d. PTEN. Answer: d Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 37. Malignant colon cancers are usually the result of a. inherited mutant oncogenes. b. inherited mutant tumor suppressor genes. c. somatic mutation of an oncogene or tumor suppressor gene. d. multiple mutations activating oncogenes and inactivating tumor suppressor genes. Answer: d Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Explain the major types of genetic alterations in human cancers. 38. What is the cure rate for colon carcinomas that remain localized to their site of origin? a. 90% b. 75% c. 50% d. 25% Answer: a

Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 1. Remembering Learning Objective: Explain the importance of early diagnosis. 39. Genetic testing for mutant oncogenes or tumor suppressor genes a. helps identify high-risk individuals before a tumor develops. b. offers a definitive indication of which individuals will develop tumors. c. aids the discovery of other mutations that lead to cancer. d. is not significantly useful as a healthcare practice. Answer: a Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 1. Remembering Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs. 40. Traditional drugs used to treat cancer patients a. target oncogene function. b. kill cancer cells specifically. c. damage DNA or inhibit DNA replication. d. target tumor suppressor protein function. Answer: c Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 1. Remembering Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs. 41. Drugs that inhibit general expression or function of an oncogene will a. inhibit division of tumor cells only. b. kill tumor cells only. c. kill tumor cells as well as normal cells. d. inhibit cell division of tumor cells and normal cells. Answer: d Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 2. Understanding Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs. 42. The therapeutic use of monoclonal antibodies against oncogene proteins is limited to a. secreted targets, such as growth factors. b. extracellular targets, such as cell surface receptors. c. extracellular and cytosolic targets. d. protein targets. Answer: b Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 2. Understanding Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs. 43. Herceptin is a monoclonal antibody against the oncogene protein a. EGF.

b. Bcr/Abl. c. ErbB-2. d. Sis. Answer: c Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 1. Remembering Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs. 44. Imatinib (Gleevec) blocks proliferation of chronic myeloid leukemia cells by inhibiting the protein kinase a. Bcr/Abl. b. Cdk/cyclin. c. MAP kinase. d. ErbB. Answer: a Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 1. Remembering Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs.

Fill in the Blank 1. The process by which tumor cells whose mutations give them a selective advantage grow preferentially is called _______. Answer: clonal selection Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Describe tumor progression. 2. Cancer cells can acquire the capacity for unlimited growth by turning on the gene for _______. Answer: telomerase Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of cancer cells. 3. SV40, papillomaviruses, and adenoviruses are _______ tumor viruses. Answer: DNA Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Compare cancer induction by chemicals and viruses. 4. Herpes viruses (HSV) can induce _______ sarcoma. Answer: Kaposi’s Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering

Learning Objective: Compare cancer induction by chemicals and viruses. 5. Common human cancers caused by viruses include cervical carcinoma and _______ cancer. Answer: liver Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Compare cancer induction by chemicals and viruses. 6. Xeroderma pigmentosum is caused by mutations in _______ genes. Answer: nucleotide-excision repair Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 7. Many cases of breast cancer can be traced to the _______ and _______ genes. Answer: BRCA1; BRCA2 Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 8. _______ treatment of promyelocytic leukemia induces terminal differentiation, forcing the cancer cells to withdraw from the cell cycle. Answer: Retinoic acid Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 1. Remembering Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs.

True/False 1. Transformation can be assayed in cell culture. Answer: T Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of cancer cells. 2. Leukemias arise from cells of the immune system. Answer: F Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of cancer cells. 3. Viruses do not induce tumors in humans.

Answer: F Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Compare cancer induction by chemicals and viruses. 4. SV40, papillomaviruses, and adenoviruses induce transformation by producing proteins that interact with the tumor suppressor proteins p53 and Rb. Answer: T Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Compare cancer induction by chemicals and viruses. 5. Oncogenic mutations in ras promote accumulation of the GDP bound form of the protein. Answer: F Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Describe the ways in which oncogenes are formed in human cancers. 6. Several oncogenes encode antiapoptotic proteins that promote cell survival. Answer: T Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Summarize the functions of oncogene proteins. 7. BRCA1 and BRCA2 prevent breast cancer by maintaining the integrity of the genome. Answer: T Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 8. The cure rate for carcinomas detected early, when they have not spread from their site of origin, is 50%. Answer: F Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 1. Remembering Learning Objective: Explain the importance of early diagnosis. 9. Acute promyelocytic leukemia results from a mutated retinoic acid receptor, effectively blocking cell differentiation and allowing leukemic cells to continue proliferation. Answer: T Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 2. Understanding Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs.

10. Cancer patients with mutant p53 can be treated effectively with radiation therapy. Answer: F Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 2. Understanding Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs. 11. The sensitivity of tumors to inhibition of activated oncogenes is called oncogene addiction. Answer: T Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 1. Remembering Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs. 12. The first drug developed against a specific oncogene and approved for clinical use was a monoclonal antibody against ErbB-2. Answer: T Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 1. Remembering Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs.

Short Answer 1. How do cancer cells become able to proliferate indefinitely, while normal cells become senescent after a limited number of cell divisions? Answer: With each cell division, the chromosomes are replicated, and each time they replicate, they lose a portion of the repetitive sequence (telomere) on each end. The enzyme telomerase is able to extend the repetitive sequences but normally is not expressed at high enough levels to maintain chromosomal length indefinitely. Telomerase expression is significantly increased in most cancers, allowing these cells to divide indefinitely. Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. 2. When does autocrine induction of cell proliferation occur? Answer: When a mutation causes a growth factor and its receptor to be expressed by the same cell Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. 3. List three ways in which a proto-oncogene can be activated to become an oncogene. Answer: (1) Point mutation, (2) gene amplification, and (3) gene translocation Textbook Reference: Oncogenes

Bloom’s Category: 2. Understanding Learning Objective: Contrast oncogenes and proto-oncogenes. Questions 4–10. Explain the conditions in which the following oncogene protein products may be oncogenic. 4. The growth factor EGF Answer: When it is produced by the same cell that has active receptors for EGF, thereby creating an autocrine (positive) feedback system to stimulate cell proliferation Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Summarize the functions of oncogene proteins. 5. The transcription factors Myc, Fos, or Jun Answer: When they are expressed at high or constant levels in cells in which they normally would not be expressed Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Summarize the functions of oncogene proteins. 6. The growth factor receptor PDGFR Answer: When translocation (deletion or mutation) alters the receptor domain so that the kinase domain becomes constitutively active Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Summarize the functions of oncogene proteins. 7. The intracellular signal transducing molecule Ras Answer: When mutation of its GTPase activity occurs so that it cannot split the bound GTP and thus continually sends a signal to Raf (to activate the MAP kinase pathway) Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Summarize the functions of oncogene proteins. 8. The Cdk-activating protein cyclin D Answer: When it is expressed constitutively or at high levels Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Summarize the functions of oncogene proteins. 9. The cell-differentiation-inducing receptor ErbA (thyroid hormone receptor) or PML/RARα (retinoic acid receptor) Answer: When mutant receptors interfere (compete) with the normal differentiation receptors Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding

Learning Objective: Summarize the functions of oncogene proteins. 10. The tumor suppressor protein p53 or Rb Answer: When mutant suppressor proteins are not able to prevent cell cycle progression; or when p53 is not able to induce apoptosis when the DNA is damaged; or when the cell is infected by viruses Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. 11. Name three tumor suppressor genes. Answer: Any three of the following: BRCA1 or BRCA2, Rb, p53, INK4, PTEN, APC, TBRII, Smad2, and Smad4 Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival.

DASHBOARD QUIZ QUESTIONS Multiple Choice 1. What is the difference between a benign tumor and a malignant tumor? a. A malignant tumor is painful, and a benign tumor is not. b. A malignant tumor has the ability to spread to other tissues and to initiate tumors at secondary sites, whereas a benign tumor does not spread. c. A benign tumor will cause a less severe form of cancer than a malignant tumor. d. A malignant tumor is caused by a virus, whereas a benign tumor arises spontaneously. Answer: b Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Explain the difference between benign and malignant tumors. Feedback A: Incorrect. Pain is not the distinguishing feature between malignant and benign tumors; both can be either painless or painful. Feedback B: Correct! Benign tumors do not spread, and they are better candidates for removal by surgery. Malignant tumors, on the other hand, can spread throughout the body, invading other tissues and organs. They are far more destructive and difficult to treat. Feedback C: Incorrect. Benign tumors are not considered cancerous. Feedback D: Incorrect. Both malignant and benign tumors can arise spontaneously or be caused by viruses. 2. Which statement about cancers is false? a. Carcinomas are malignancies of epithelial cells. b. Leukemias are malignancies that arise from blood-forming cells.

c. Lymphomas are malignancies from colon tissue. d. Sarcomas are malignancies of connective tissue like muscle and bone. Answer: c Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. Feedback A: Incorrect. This is true. Feedback B: Incorrect. This is true. Feedback C: Correct! This is false. Lymphomas are malignancies that arise from cells of the immune system. Feedback D: Incorrect. This is true. 3. The type of cancer with highest mortality in the United States is cancer of the a. prostate. b. colon/rectum. c. breast. d. lung. Answer: d Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of cancer cells. Feedback A: Incorrect. Prostate cancer is the most frequently occurring cancer, but it is not the most lethal. Feedback B: Incorrect. This is the second most lethal cancer, killing approximately onethird the number of people as the most lethal cancer. Feedback C: Incorrect. This is one of the most lethal cancers, but not the most lethal. Feedback D: Correct! Deaths from lung cancer account for approximately 30% of all cancer deaths in the United States. 4. Which statement concerning the difference between cancer cells and normal cells is false? a. Normal cells display density-dependent inhibition of cell proliferation. b. Cancer cells have reduced requirements for extracellular growth factors. c. Malignant cells generally secrete proteases that digest extracellular matrix components. d. Cancer cells undergo normal differentiation, but excessive proliferation. Answer: d Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. Feedback A: Incorrect. This is true. Feedback B: Incorrect. This is true. Feedback C: Incorrect. This is true. Feedback D: Correct! This is false. Cancer cells do not differentiate normally. In fact, many therapies are designed to push cancer cells toward terminal differentiation where they would be likely to stop proliferating.

5. Angiogenesis contributes to cancer development by a. providing nutrients and oxygen to tumors and by facilitating metastasis. b. initiating a mutation in a gene that causes uncontrolled cell growth. c. inhibiting apoptosis (programmed cell death). d. preventing the normal inhibition of growth that occurs when cells make contact. Answer: a Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Summarize the properties of cancer cells. Feedback A: Correct! Tumors stimulate angiogenesis, or new blood vessel formation, which brings them nutrients and oxygen. In addition, the blood vessels provide a path of entry into the circulatory system, from which tumors can spread to other tissues. Feedback B: Incorrect. Angiogenesis does not occur until after uncontrolled cell growth has been initiated. Feedback C: Incorrect. Inhibition of apoptosis contributes to the development of cancer, but it is not related to angiogenesis. Feedback D: Incorrect. A loss of contact inhibition is characteristic of cancerous cells, but it does not involve angiogenesis. 6. Infection with which of the following viruses is associated with development of liver cancer in humans? a. Simian virus 40 (SV40) b. Hepatitis B viruses c. Epstein-Barr virus d. Papillomaviruses Answer: b Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 1. Remembering Learning Objective: Compare cancer induction by chemicals and viruses. Feedback A: Incorrect. SV40 is associated with Merkel cell carcinoma in humans. Feedback B: Correct! Chronic infection with hepatitis B along with hepatitis C is associated with a hundredfold increase in the risk of developing liver cancer. Feedback C: Incorrect. The Epstein-Barr virus is associated with Burkitt’s lymphoma and nasopharyngeal carcinoma. Feedback D: Incorrect. Papillomaviruses are associated with cervical cancers. 7. The more than 40 oncogenic retroviruses that have been identified share expression of all of the following genes except a. src. b. pol. c. env. d. gag. Answer: a Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Explain how retroviral oncogenes were identified.

Feedback A: Correct! src is found in some retroviruses, but not all. It encodes a tyrosine kinase. Feedback B: Incorrect. pol encodes a viral reverse transcriptase. Feedback C: Incorrect. env encodes viral envelope glycoproteins. Feedback D: Incorrect. gag encodes a viral protease. 8. Which statement does not describe a way in which oncogenes incorporated into viral genomes can differ from their normal cellular counterparts (proto-oncogenes)? a. They can contain fusions to viral sequences, resulting in structural changes that deregulate the protein. b. They can contain point mutations in regulatory domains, resulting in a loss of protein regulation. c. They can be present in many tandem copies, as opposed to the single copy present in the cell. d. They can be expressed from much stronger promoters than the normal cellular promoter. Answer: c Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Contrast oncogenes and proto-oncogenes. Feedback A: Incorrect. This is common among viral oncogenes. Feedback B: Incorrect. Point mutations that deregulate protein activity are common among viral oncogenes, such as the ras oncogenes. Feedback C: Correct! Viruses have very compact genomes and do not possess tandem copies of genes. Feedback D: Incorrect. Viral promoters are often very strong, and incorporation of a gene into a virus often brings it under the control of one of these promoters. 9. Burkitt's lymphoma is caused almost exclusively by a. point mutations in the ras proto-oncogene. b. translocation of the abl proto-oncogene. c. translocation of the c-myc proto-oncogene. d. amplification of the N-myc gene. Answer: c Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Contrast oncogenes and proto-oncogenes. Feedback A: Incorrect. Ras mutations are common in colon and lung cancers. Feedback B: Incorrect. Translocation and fusion of abl to bcr are common in chronic myelogenous leukemia. Feedback C: Correct! Burkitt’s lymphoma is caused by a translocation of c-myc to one of the immunoglobulin gene loci, causing its unregulated expression. Feedback D: Incorrect. This occurs in neuroblastomas. 10. The PDGF receptor is a proto-oncogene (a normal cellular gene that becomes oncogenic when mutated). Which statement about the PDGF receptor is true?

a. The oncogenic mutation occurs at the intracellular amino terminal end of the receptors. b. The oncogenic mutation constitutively represses the intracellular kinase activity of the protein. c. The oncogenic mutation generates a receptor that is no longer dependent on ligand for activation. d. The PDGF protein monomer activates cell proliferation via tyrosine kinase activity. Answer: c Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Contrast oncogenes and proto-oncogenes. Feedback A: Incorrect. The mutation occurs at the extracellular amino terminal. Feedback B: Incorrect. The mutation activates intracellular kinase activity. Feedback C: Correct! Once the Tel/PDGFR fusion protein dimerizes, activity of the intracellular kinase domain produces unregulated proliferation even in the absence of growth factor. Feedback D: Incorrect. PDGFR fused with Tel create a fusion protein dimer. 11. Oncogenic conversion of the two dimerizing proto-oncogene components of the AP-1 transcription factor can lead to abnormal cell proliferation. Which two gene products form the active AP-1 transcription factor complex? a. Fos and Jun b. Fos and Myc c. Myc and Jun d. Raf and Myc Answer: a Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Describe the ways in which oncogenes are formed in human cancers. Feedback A: Correct! Feedback B: Incorrect. Feedback C: Incorrect. Feedback D: Incorrect. 12. The translocation that places the transcription factor Tel at the amino terminus of the PDGF receptor converts PDGFR to an oncogene by a. constitutive repression of the fusion protein. b. forming Tel dimers that produce unregulated signals for proliferation. c. activating the tyrosine kinase domain in monomeric form. d. increasing the transcriptional activation capability of Tel. Answer: b Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Describe the ways in which oncogenes are formed in human cancers. Feedback A: Incorrect. Expression level of the fusion protein does not significantly alter tyrosine kinase activity. Feedback B: Correct! The Tel portion of the Tel/PDGFR fusion protein forms dimers in

the absence of growth factor binding, leading to activation of tyrosine kinase domains in the cytoplasmic domain and unregulated proliferative signals from the oncogene protein. Feedback C: Incorrect. Tyrosine kinases must form dimers and phosphorylate each other to be activated. Feedback D: Incorrect. Tel is not an oncogene, and the fusion protein does not travel to the nucleus. 13. The viral raf gene is oncogenic because a. it is more highly expressed under control of the viral env promoter. b. it encodes a constitutively active protein kinase that is missing a regulatory N terminal domain. c. it carries a loss of function point mutation. d. it carries a gain of function point mutation. Answer: b Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Describe the ways in which oncogenes are formed in human cancers. Feedback A: Incorrect. Expression levels would not change kinase activity. Feedback B: Correct! Loss of the N terminal regulatory domain allows Raf to constitutively activate the growth-promoting ERK signaling pathway. Feedback C: Incorrect. Viral raf is not altered by a point mutation. Feedback D: Incorrect. Viral raf is not altered by a point mutation. 14. The majority of oncogene proteins are a. metabolic enzymes. b. structural proteins, such as nuclear lamins. c. proteins involved in cell sorting. d. components of signaling pathways that regulate cell proliferation. Answer: d Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Summarize the functions of oncogene proteins. Feedback A: Incorrect. Metabolic enzymes are not likely to cause cancer when mutated. Feedback B: Incorrect. Defective structural proteins do not cause cancer. Feedback C: Incorrect. Defective cell sorting may cause chaos in the cell, but it is unlikely to cause cancer. Feedback D: Correct! Defective versions of proteins involved in signaling cell proliferation are the most common causative agents of cancer. 15. The protein encoded by the PTEN gene acts as a tumor suppressor by a. phosphorylating p53 and increasing its proapoptotic activity. b. antagonizing CDK4, 6/cyclin D activity to activate the G1 cell cycle restriction point. c. dephosphorylating phosphatidylinositol 3,4,5-bisphosphate, thus antagonizing PI 3 kinase/Akt activity. d. increasing expression of BRCA1 to protect cells against DNA damage. Answer: c

Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. Feedback A: Incorrect. PTEN is not a kinase and does not target p53. Feedback B: Incorrect. This is the mechanism of CDK inhibitors, such as p16 and p21. Feedback C: Correct! PTEN is a phosphatase that dephosphorylates PIP3, rendering Akt unable to stimulate cell proliferation. Feedback D: Incorrect. PTEN does not influence BRCA1 or BRCA2 expression. 16. Which statement about the tumor suppressor gene p53 is false? a. It plays a role in up to 50% of human cancers. b. Its loss causes retinoblastoma, a rare childhood cancer of the eye. c. It blocks cell cycle progression in response to DNA damage. d. It is required for apoptosis in response to DNA damage. Answer: b Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. Feedback A: Incorrect. This is true. p53 is mutated in a wide variety of malignancies. Feedback B: Correct! This is false. A gene called Rb is the cause of retinoblastoma. Feedback C: Incorrect. This is true. p53 induces the synthesis of p21, which inhibits the Cdk/cyclin complex as well as PCNA, thus blocking progression of the cell cycle until after the damage has been repaired. Feedback D: Incorrect. This is true. Cells with unrepaired DNA normally die by undergoing apoptosis. However, cells lacking p53 fail to undergo apoptosis, and thus can survive and become malignant. 17. PTEN is a tumor suppressor gene in the Akt signaling pathway. Which statement about PTEN/Akt is false? a. PIP2 can be oncogenically mutated to induce cell survival. b. PTEN antagonizes (has the opposite effect) of PI 3-kinase. c. Akt can be oncogenically mutated to induce cell survival. d. It is a lipid phosphatase that dephosphorylates PIP 3 to PIP2. Answer: a Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Give examples of the functions of tumor suppressor gene products in signal transduction, cell cycle progression, and cell survival. Feedback A: Correct! This is false. PIP2 is not a protein and cannot be mutated. Feedback B: Incorrect. This is true. Feedback C: Incorrect. This is true. Feedback D: Incorrect. This is true. 18. Which of the following genes has been associated with colon cancer?

a. APC b. bcl-2 c. Rb d. erbB-2 Answer: a Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 1. Remembering Learning Objective: Explain the major types of genetic alterations in human cancers. Feedback A: Correct! APC, as well as other genes, has been linked to colon cancer. Feedback B: Incorrect. bcl-2 is associated with chronic lymphocytic leukemia. Feedback C: Incorrect. Rb is associated with retinal blastoma. Feedback D:

Essay 1. What is a possible reason that carcinomas, which are tumors derived from epithelial cells, are the most common types of cancer in humans? Answer: Most cells in adults have ceased to proliferate and have entered the G0, or quiescent, stage of the cell cycle. In contrast, epithelial cells are continually turned over and thus continue to divide. Because many of the signals that stimulate cell division are still active, they may be more susceptible to uncontrolled cell proliferation. Epithelial cells come into frequent contact with cancer-causing substances from the environment because they line both the outer (skin) and inner (e.g., the intestinal and respiratory tracts) surfaces of the body. Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. 2. Why are cancer cells less dependent on growth factors than their normal cellular counterparts? Answer: 1) Cancer cells frequently secrete their own growth factors in an autocrine fashion, making them less dependent on growth factors from normal signaling pathways. 2)Oncogenic mutations frequently occur in the signaling pathways themselves, making the pathways constitutively active and less dependent on any growth factors that would normally trigger the pathways (in effect, they bypass the initiating event of the cascade altogether). Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. 3. What is the likely reason that antiangiogenic therapies hold so much promise for treatment of tumors? Answer: Without angiogenesis, tumors would be limited in size to approximately a million cells. After that, angiogenesis is required for the tumor to have adequate blood flow for survival. Blood vessels grow in response to growth factors secreted by the

tumors, which stimulate the endothelial cells to proliferate in the walls of capillaries in surrounding tissues. This results in the outgrowth of new capillaries into the tumor and also facilitates metastasis. Thus, drugs that block this angiogenesis could have a dramatic effect on tumor growth and metastasis. These drugs are currently in clinical trials and development in many pharmaceutical companies and laboratories. Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 3. Applying Learning Objective: Summarize the properties of cancer cells. 4. Describe how growth of transformed cells in a tissue culture dish differs from growth of untransformed cells. Answer: Transformed cells display three properties that untransformed cells do not: altered morphology, loss of contact inhibition, and loss of density-dependent inhibition of growth. In a background of untransformed cells, single transformed cells will form a colony of morphologically altered cells that overgrow the background of normal cells. These foci of transformed cells can be detected from within a few days to two weeks. Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. 5. When observed under a microscope, cancer cells often have a more rounded morphology than untransformed cells. What accounts for this difference in morphology? Answer: Cancer cells tend to be less dependent on cell−cell and cell−matrix interactions for survival. They are often less adhesive to neighbors and/or extracellular matrix, due to reduced expression of cell surface adhesion molecules, such as E-cadherin. This reduced adhesiveness of cancer cells results in morphological and cytoskeletal alterations, such as rounding. Textbook Reference: The Development and Causes of Cancer Bloom’s Category: 2. Understanding Learning Objective: Summarize the properties of cancer cells. 6. Describe how the first oncogene was discovered. Answer: The src oncogene was discovered in a study of mutant versions of the Rous sarcoma virus (RSV), which causes tumors in birds. Some of the mutants were able to replicate normally in birds but did not cause tumors, indicating that (1) viral replication itself did not cause tumors, and (2) the gene that was mutated was involved in tumor development. The gene was eventually identified and found to transform avian cells by itself, and thus was the first oncogene to be discovered. Equally important, studies of src led to the discovery that retroviral oncogenes are derived from normal genes (protooncogenes) picked up from host cells. Textbook Reference: Oncogenes Bloom’s Category: 1. Remembering Learning Objective: Explain how retroviral oncogenes were identified. 7. What is the most common mechanism by which -catenin is converted from a protooncogene to an oncogene?

Answer: -catenin acts together with Tcf to activate genes such as c-myc and cyclin D1, whose products promote survival and cell proliferation. It is responsive to the Wnt family of protein paracrine hormones. In the absence of Wnt signal, -catenin is phosphorylated, ubiquitinated, and rapidly degraded by the proteasome. In the presence of a Wnt signal, -catenin is not phosphorylated, allowing it to accumulate in the cytoplasm, translocate to the nucleus, and interact with Tcf to initiate transcription of target genes. Point mutations in -catenin that prevent its phosphorylation stabilize it in the absence of Wnts, making cell proliferation Wnt independent. Textbook Reference: Oncogenes Bloom’s Category: 2. Understanding Learning Objective: Describe the ways in which oncogenes are formed in human cancers. 8. Explain why the disease retinoblastoma is inherited in a dominant fashion, despite the fact that it is caused by a recessive (loss-of-function) mutation in the tumor suppressor gene Rb. Answer: Statistically, 50% of the progeny of an individual with retinoblastoma (heterozygous for Rb) will inherit the mutant version of the gene. However, because human cells are diploid, and loss of Rb function is recessive, not all cells in these individuals will become cancerous. Loss of the second allele of Rb (the normal copy from the unaffected parent) by mutation in a single cell, however, results in that cell becoming cancerous and the subsequent development of disease. Somatic mutations of this allele are common enough that almost all children who receive one mutant copy of the allele will lose the normal allele in at least one retinal cell. Inheritance of the disease therefore follows the pattern for inheritance of a dominant allele. In other diseases caused by recessive mutations, mutation of both alleles of the disease gene in just few a cells is, on the whole, insignificant. In the case of Rb, it can be lethal. Textbook Reference: Tumor Suppressor Genes Bloom’s Category: 2. Understanding Learning Objective: Explain the major types of genetic alterations in human cancers. 9. Why is cancer more often cured when detected early? Answer: Cancer develops in a stepwise fashion: a cell first begins to proliferate uncontrollably. This is followed by an increased invasiveness of one of the cell’s progeny, followed by an increased metastatic potential of one of that cell’s progeny. Thus, the gradual accumulation of mutations leads to a malignant tumor with metastatic potential. If the tumor is caught early on, at the stage where it is still just a benign growth, it can be treated locally by surgery. In addition, the earlier a malignant tumor is treated, the more likely it will be to respond to radiation and chemotherapy; unlike normal cells, cells in late-stage tumors often do not cease cell proliferation in response to these treatments. Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 2. Understanding Learning Objective: Explain the importance of early diagnosis. 10. What is the therapeutic strategy for using Herceptin to treat breast cancer? Answer: In about 30% of breast cancers, ErbB-2 (a receptor protein tyrosine kinase) is overexpressed as a result of amplification of the erbB-2 gene. In vitro, cells that

overexpress ErbB-2 exhibit elevated proliferation. Herceptin is a monoclonal antibody that binds the extracellular domain of ErbB-2 and blocks proliferation in the ErbB-2 overexpressing cell lines. In clinical trials, Herceptin was shown to reduce tumor size and increase human survival, and it became the first antibody to gain FDA approval. It is now widely used in treating ErbB-2-overexpressing breast tumors. Textbook Reference: Molecular Approaches to Cancer Treatment Bloom’s Category: 2. Understanding Learning Objective: Describe the basis for selectivity of oncogene-targeted drugs.