Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Mon

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

Test Bank for All Chapters (Chapter 2-24)

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2-1

Chemical Foundations

Section 2.1 1. Covalent bonds between which of the following pairs of atoms are nonpolar? a. C–C b. C–H c. O–H d. a and b Ans: d Question Type: Multiple Choice Chapter: 2.1 Blooms: Remembering Difficulty: Easy 2. Which of the following is a noncovalent interaction? a. hydrophobic effect b. ionic interactions c. van der Waals interactions d. all of the above Ans: d Question Type: Multiple Choice Chapter: 2 Blooms: Remembering Difficulty: Easy 3. Which of the following is the strongest interaction? a. hydrogen bond b. ionic bond c. phosphoanhydride bond d. van der Waals interaction Ans: c Question Type: Multiple Choice Chapter: 2 Blooms: Understanding Difficulty: Easy 4. Which of the following is the weakest interaction? a. hydrogen bond b. ionic bond c. phosphoanhydride bond d. van der Waals interaction Ans: d Question Type: Multiple Choice

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Chapter: 2 Blooms: Understanding Difficulty: Easy 5. When two atoms differing in electronegativity are joined in a covalent bond, then the: a. electrons are shared equally between the atoms. b. bond is nonpolar. c. resulting compound is devoid of any dipole moment. d. atom with the greater electronegativity attracts the bonded electrons more strongly.

Ans: d Question Type: Multiple Choice Chapter: 2 Blooms: Understanding Difficulty: Moderate 6. The interactions between two proteins such as an antibody and an antigen or a hormone and its receptor are quite strong despite the fact that these interactions consist of relatively weak noncovalent bonds. How can this be?

Ans: Two proteins can bind tightly because of molecular complementarity, in which multiple noncovalent bonds participate.

Although each individual bond is weak, the cumulative effect of many noncovalent bonds is a relatively strong and highly specific interaction.

Question Type: Essay Chapter: 2 Blooms: Analyzing Difficulty: Moderate 7. What produces the dipole of a water molecule?

Ans: The dipole of a water molecule is caused by the difference in electronegativity between O and H. The oxygen atom has a greater electronegativity than the hydrogen atom. As a result, oxygen attracts the electrons in the O–H bond more strongly, and the oxygen side of the bond has a slight net negative charge. This results in a dipole moment.

Question Type: Essay Chapter: 2.1 Blooms: Understanding Difficulty: Moderate Section 2.2 8. Which of the following is a negatively charged amino acid? a. alanine b. aspartate c. glutamine d. histidine Ans: b Question Type: Multiple Choice Chapter: 2 Blooms: Remembering Difficulty: Easy 9. Which of the following is/are a hydrophilic amino acid? a. aspartate b. serine c. tryptophan d. aspartate and serine

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Ans: d Question Type: Multiple Choice Chapter: 2 Blooms: Remembering Difficulty: Easy 10. Adenosine is a: a. component of RNA. b. nucleoside. c. pyrimidine. d. a and b Ans: d Question Type: Multiple Choice Chapter: 2 Blooms: Remembering Difficulty: Moderate 11. Which of the following is a monosaccharide? a. fructose b. galactose c. glucose d. all of the above Ans: d Question Type: Multiple Choice Chapter: 2 Blooms: Remembering Difficulty: Easy 12. Based on what you know about hydrophobic interactions, which of the following is/are composed of a bilayer? a. a cell’s membrane b. spontaneously aggregated phospholipids surrounding an aqueous interior c. lipid vesicles that have budded off the cell’s membrane d. all of the above Ans: d Question Type: Multiple Choice Chapter: 2 Blooms: Remembering Difficulty: Easy 13. You discover that you suffer from a deficiency in the amino acid tryptophan. At the pharmacy, you find both Dtryptophan and L-tryptophan supplements. Which do you purchase? Why? Ans: You should choose L-tryptophan. All amino acids can exist as one of two stereoisomers ( D or L) because of asymmetry around the α carbon. Proteins consist of the L form of amino acids, and as these stereoisomers possess distinct biological properties and are not readily interconverted, you should choose the form that is normally utilized by cells.

Question Type: Essay Chapter: 2 Application Difficulty: Moderate 14. Cysteine often plays an important role in stabilizing protein structure. Explain how this works.

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Ans: Two adjacent sulfhydryl (SH) groups can oxidize to form a covalent disulfide (S–S) bond. Disulfide bonds can stabilize the structure of folded peptides or sometimes link two separate peptide chains together.

Question Type: Essay Chapter: 2 Blooms: Understanding Difficulty: Easy 15. Triacylglycerol and cholesterol esters are nonpolar; in contrast, phospholipids are amphipathic molecules.

Biomembranes are based on phospholipids rather than on triacylglycerols. Why? Ans: Biomembranes are based on phospholipids rather than on triacylglycerols because phospholipids as amphipathic molecules can form planar lipid bilayers, whereas the nonamphipathic nonpolar triacylglycerols cannot. Their amphipathic property, the presence of a polar and nonpolar domain at opposite ends of the same molecule, allows phospholipids to form hydrophilic associations with water at the same time as forming hydrophobic associations with each other through their hydrophobic tails. Triacylglycerols are strictly hydrophobic in nature and hence in an aqueous environment tend to associate with one another to form lipid droplets. This minimizes the contact of triacylglycerol with water. Recall the old adage: oil and water do not mix.

Question Type: Essay Chapter: 2 Blooms: Evaluating Difficulty: Difficult 16. A nucleotide can vary in a. the base b. the sugar c. the phosphate group d. the sugar and the base

.

Ans: d Question Type: Multiple Choice Chapter: 2 Blooms: Understanding Difficulty: Moderate

17. Which of the following is NOT one of the ways RNA differs from DNA? a. Ribonucleotides have a hydroxyl group on the 2 carbon of their sugar subunit. b. Ribonucleotides can have enzymatic activity. c. Ribonucleotides contain a phosphate group. d. Ribonucleotides can contain the base uracil.

Ans: c Question Type: Multiple Choice Chapter: 2 Blooms: Remembering Difficulty: Moderate

18. What is the major structural difference between starch and cellulose? a. the types of monosaccharide subunits in the molecules b. the amount of branching that occurs in the molecule c. that humans can only ingest starch d. the type of glycosidic linkages in the molecule Ans: d Question Type: Multiple Choice

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Chapter: 2 Blooms: Understanding Difficulty: Moderate

19. How do phospholipids interact with water molecules? a. The polar heads interact with water; the nonpolar tails do not. b. Phospholipids don't interact with water because water is polar and lipids are nonpolar. c. The polar heads avoid water; the nonpolar tails attract water (because water is polar and opposites attract). d. Phospholipids dissolve in water.

Ans: a Question Type: Multiple Choice Chapter: 2 Blooms: Remembering Difficulty: Easy 20. Which of the following is the BEST explanation for why vegetable oil is a liquid at room temperature while animal fats are solid? a. Vegetable oil has fewer double bonds than animal fats. b. Animal fats have no amphipathic character. c. Vegetable oil has longer fatty-acid tails than do animal fats. d. Vegetable oil has more double bonds than do animal fats.

Ans: a Question Type: Multiple Choice Chapter: 2 Blooms: Understanding Difficulty: Easy Section 2.3 21. A 1-mL solution of 0.05 M H2SO4 is diluted to 100 mL at 25°C. What is the pH of the resulting solution? a. 1 b. 2 c. 3 d. 4 Ans: c Question Type: Multiple Choice Chapter: 2 Blooms: Applying Difficulty: Moderate 22. An Archaea cell living in an abandoned mine is found to contain a very high concentration of protons. It is likely that this cell: a. has a high ph and is acidic. b. has a high ph and is alkaline. c. has low ph and is acidic. d. has a low pH and is alkaline.

Ans: c Question Type: Multiple Choice Chapter: 2 Blooms: Understanding Difficulty: Easy

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23. A 1-mL solution of 0.1 M NaOH is diluted to 1 L at 25°C. What is the pH of the resulting solution? a. 1 b. 7 c. 10 d. 13 Ans: c Question Type: Multiple Choice Chapter: 2 Blooms: Applying Difficulty: Moderate 24. The pKa of the weak base NH3 is 9.25. When present in lysosomes, a subcellular organelle—ammonia—is almost totally protonated. Which of the pH values listed below is most likely to be that of the lysosome lumen? a. 1 b. 5 c. 8 d. 14 Ans: b Question Type: Multiple Choice Chapter: 2 Analysis Difficulty: Difficult 25. If the equilibrium constant for the reaction A → B is 0.5 and the initial concentration of A is 25 mM and of B is 12.5 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.

Ans: d Question Type: Multiple Choice Chapter: 2 Analysis Difficulty: Difficult 26. For the binding reaction A + B → AB, the dissociation constant is equal to: a. ( + ) b. ([A] + [B])/[AB]. c. Keq d. The first and third answers are correct.

Ans: b Question Type: Multiple Choice Chapter: 2 Blooms: Applying Difficulty: Easy 27. What is the effect of an enzyme on the end equilibrium concentration of reactants and products?

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Ans: An enzyme has no effect on the end equilibrium concentration of reactants and products.

Question Type: Essay Chapter: 2 Blooms: Analyzing Difficulty: Easy 28. The enzyme alcohol dehydrogenase is capable of catalyzing the oxidation of a number of different substances, including ethanol, ethylene glycol, and methanol, to an aldehyde. The metabolic products of both ethylene glycol and methanol are highly toxic to humans. A standard medical treatment for prevention of ethylene glycol or methanol poisoning is the administration of a dose of ethanol. Why is this treatment effective?

Ans: The ethanol-like ethylene glycol and methanol are capable of binding to the enzyme, alcohol dehydrogenase, and competing with its other substrates. A sufficient dosage of ethanol can out-compete the other substrates, and hence the ethylene glycol and methanol are not metabolized to toxic products. Gradually the ethylene glycol or methanol will be excreted from the body.

Question Type: Essay Chapter: 2 Blooms: Evaluating Difficulty: Moderate 29. How do cells maintain a relatively constant pH despite the fact that many metabolic processes produce acids?

Ans: All cells contain buffers such as phosphate ions that can absorb or release protons or hydroxyl ions to stabilize pH changes near neutral pH. Question Type: Essay Chapter: 2 Blooms: Evaluating Difficulty: Difficult Section 2.4 30. In a biochemical reaction in which H < 0 and S > 0: a. the reaction is spontaneous. b. the reaction is endothermic. c. the reaction is endergonic. d. G is positive.

Ans: a Question Type: Multiple Choice Chapter: 2 Blooms: Applying Difficulty: Easy 31. In the reaction NAD+ + H+ + 2e− → NADH, NAD+ becomes: a. dehydrated. b. hydrolyzed. c. oxidized. d. reduced.

Ans: d Question Type: Multiple Choice Chapter: 2 Blooms: Understanding Difficulty: Easy 32. The ultimate source of chemical energy for all cells is:

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a. electricity. b. heat. c. light. d. magnetism.

Ans: c Question Type: Multiple Choice Chapter: 2 Blooms: Applying Difficulty: Moderate 33. Hydrolysis of ATP: a. is endothermic.

G value. c. must be coupled to an energetically favorable reaction. d. none of the above Ans: d Question Type: Multiple Choice Chapter: 2 Blooms: Understanding Difficulty: Easy 34. What is [P]/[R] when G =G°´? a. –1 b. 0 c. 1 d. 2.3 Ans: c Question Type: Multiple Choice Chapter: 2 Blooms: Applying Difficulty: Moderate 35. A reaction with a positive G value can be made energetically favorable by increasing the: a. G. b. starting concentration of products. c. starting concentration of reactants. d. The first two answers are correct.

Ans: c Question Type: Multiple Choice Chapter: 2 Blooms: Analyzing Difficulty: Moderate 36. Photosynthesis by plants and certain microbes traps the energy in light and uses it to: a. reduce glucose into carbon dioxide. b. synthesize ATP from ADP and inorganic phosphate. c. generate ATP from the oxidation of reduced inorganic compounds. d. none of the above Ans: b Question Type: Multiple Choice Chapter: 2

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Blooms: Understanding Difficulty: Moderate 37. NAD+ and FAD are often referred to as: a. redox proteins. b. polymers. c. reduced dinucleotides. d. electron-carrying coenzymes.

Ans: d Question Type: Multiple Choice Chapter: 2 Blooms: Remembering Difficulty: Easy 38. A solution of 8 M urea is sometimes used in the isolation of protein molecules. When the solution is prepared by G for this process change if you tried to dissolve urea in the cold room, rather than at room temperature?

Ans: Urea will be less soluble at cold temperatures than at room temperature because the decrease in temperature will decrease the term TS, increasing the value of G, because G = H − TS. The values of H and S are relatively independent of temperature.

Question Type: Essay Chapter: 2 Blooms: Analyzing Difficulty: Moderate 39. Phosphoglucomutase converts glucose 1-phosphate, the product of the reaction catalyzed by glycogen phosphorylase, into glucose 6-phosphate. The Keq G°´ for the reaction?

Ans: −1.741 kcal/mol, G°´ = −2.3RT log Keq, G°´ = −2.3 (1.987) (298) log Keq Question Type: Essay Chapter: 2 Application Difficulty: Moderate 40. Under what conditions is the G for a reaction different from the G°´?

Ans: G°´ is the Gibbs free energy of a reaction under standard conditions: pH 7.0, 1 M initial concentration of all reactants and products except protons and water, 1 atm pressure, 298°K (25°C). Variation of any of these parameters from standard conditions, depending on the reaction, can produce a different G value.

Question Type: Essay Chapter: 2 Blooms: Understanding Difficulty: Easy

41. The amount of free energy released when bonds are broken during a reaction is higher when the molecule has more electronegative atoms.

Ans: False Question Type: True/False Chapter: 2 Blooms: Applying Difficulty: Moderate

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42. Which of the following is true about an observed change in free energy (G)? a. Free energy was created when the Big Bang occurred. b. It can be calculated from the total change in energy, temperature, and change in entropy. c. If a reaction's free energy is greater than zero, it is likely to happen spontaneously. d. Free energy is comparable to unusable energy.

Ans: b Question Type: Multiple Choice Chapter: 2 Blooms: Understanding Difficulty: Easy

3

Protein Structure and Function

Section 3.1 1. Which of the following is defined as the tertiary structure of a protein? a. the primary amino acid sequence b. structural domains such as a DNA-binding domain c. folded structures such as an α helix d. structural features such as a turn Ans: b Question Type: Multiple Choice Chapter: 3 Blooms: Understanding Difficulty: Easy 2. Monomeric proteins do not contain a: a. primary structure. b. secondary structure. c. tertiary structure. d. quaternary structure.

Ans: d Question Type: Multiple Choice Chapter: 3 Blooms: Understanding Difficulty: Easy

3. Which of the following is NOT part of a zinc-finger motif? a. zinc ion b. proline residue c. cysteine residue

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d. histidine residue Ans: b Question Type: Multiple Choice Chapter: 3 Blooms: Remembering Difficulty: Easy 4. Describe the types of bonds/interactions that hold together or stabilize the primary, secondary, tertiary, and quaternary structures of proteins.

Ans: The primary structure of a protein is linked by covalent peptide bonds. The secondary structure is stabilized by hydrogen bonds between atoms of the peptide backbone. The tertiary structure is stabilized by hydrophobic interactions between the nonpolar side groups and hydrogen bonds between polar side groups. The quaternary structure is held together by noncovalent bonds between protein subunits.

Question Type: Essay Chapter: 3 Blooms: Understanding Difficulty: Easy

5. Many proteins contain one or more motifs built from particular combinations of secondary structure. Describe the three common structural motifs discussed in this chapter.

Ans: The three structural motifs described in this chapter include the coiled coil motif, the helix-loop-helix motif, and the zinc finger motif. The coiled-coil motif consists of two or more  helices wrapped around one another. The helix-loop-helix motif consists of two helices connected by a loop that contains certain hydrophilic residues at invariant positions in the loop.

The zinc-finger motif consists of an  helix and two  strands held together by a zinc ion in a fingerlike bundle.

Question Type: Essay Chapter: 3 Application Difficulty: Moderate 6. What types of bonds are apt to be more common in the nonaqueous, interior environment of a protein than in the aqueous, surface environment of a protein?

Ans: Proteins are arranged so that hydrophilic amino acids are on the surface of the protein and hydrophobic amino acids are in the interior. Hence, hydrogen bonding and ionic interactions with water are particularly common at the protein surface; hydrophobic interactions are more common in the protein interior.

Question Type: Essay Chapter: 3 Blooms: Applying Difficulty: Moderate 7. There are many important roles for the dynamic nature of proteins in a cell. Which of the following is NOT likely to describe one such reason? a. A protein’s structure determines its function. b. Other molecules could be needed to allow proteins to fold into their active (ordered) conformation. c. Quaternary structures are usually very transient (occur for short periods of time). d. Proteins are crucial for many cell functions.

Ans: c Question Type: Multiple Choice Chapter: 3 Blooms: Applying Difficulty: Moderate

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9. You are studying an oligopeptide composed of eight amino acids. The four amino acids nearest the C terminus are nonpolar. The two amino acids nearest the N terminus are charged. The middle two amino acids are polar. Which amino acid is likely to be labeled as number 2? a. threonine b. phenylalanine c. glutamine d. lysine Ans: d Question Type: Multiple Choice Chapter: 3 Blooms: Applying Difficulty: Moderate 10. A protein containing several proline residues is: a. not likely to form quarternary structures. b. likely to be an integral membrane protein. c. not likely to form alpha helices. d. likely to be found in beta turns.

Ans: c Question Type: Multiple Choice Chapter: 3 Blooms: Remembering Difficulty: Moderate 11. Which of the following is true about protein folding? a. Amino acids cluster in the primary sequence so that all the hydrophobic amino acids are near each other to facilitate folding into the hydrophobic core of the tertiary structure. b. All known proteins have well-ordered conformations. c. Amino acids with hydrophobic, nonpolar side chains stabilize the tertiary structure through hydrogen bonding with water molecules surrounding the proteins. d. Elements from the secondary structure are maintained in the tertiary structure.

Ans: d Question Type: Multiple Choice Chapter: 3 Blooms: Understanding Difficulty: Moderate 12. When comparing domains and structural motifs, which of the following is NOT true? a. Motifs are found in secondary structures, while domains are found in tertiary structures. b. Helices are observed in motifs and domains. c. Structural domains appear in different proteins with similar functions, while structural motifs have been less conserved over evolution. d. A domain may be repeated in the same protein, but multiple copies of the same motif are rare.

Ans: c Question Type: Multiple Choice Chapter: 3 Blooms: Understanding Difficulty: Moderate 13. You disrupt all hydrogen bonds in a protein. What level of structure will be preserved? a. secondary structure

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b. primary structure c. tertiary structure d. quaternary structure Ans: b Question Type: Multiple Choice Chapter: 3 Blooms: Understanding Difficulty: Easy 14. Two proteins that have a similar function: a. will share similar amino acid sequences if they are homologs. b. must have similar amino acid sequences. c. will have identical primary structures. d. belong in families together.

Ans: a Question Type: Multiple Choice Chapter: 3 Blooms: Remembering Difficulty: Moderate Section 3.2 15. All the following statements about molecular chaperones are true EXCEPT: a. they play a role in the proper folding of proteins. b. they are located in every cellular compartment. c. they are found only in mammals. d. they bind a wide range of proteins.

Ans: c Question Type: Multiple Choice Chapter 3.2 Blooms: Understanding Difficulty: Easy 16. Hsp90 family members are present in all organisms EXCEPT: a. archaea. b. bacteria. c. fungi. d. plants.

Ans: a Question Type: Multiple Choice Chapter 3.2 Blooms: Remembering Difficulty: Easy 17. Describe the mechanism by which the bacterial chaperonin GroEL promotes protein folding.

Ans: The bacterial chaperonin GroEL forms a barrel-shaped complex of 14 identical subunits. A partially folded or misfolded polypeptide is inserted into the GroEL barrel, where it binds to the inner wall and folds into its native conformation. In an ATP-dependent step, the GroEL barrel expands to a more open state, which results in release of the folded protein.

Question Type: Essay Chapter 3.2

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Blooms: Understanding Difficulty: Moderate 18. What role does aberrant protein folding play in the development of a disease such as Alzheimer’s disease?

Ans: Misfolding of a protein marks it for degradation by proteolytic cleavage. In Alzheimer’s disease, misfolding and subsequent proteolytic degradation of the amyloid precursor protein generates a short fragment called -amyloid protein, which changes from an -helical to a -sheet conformation. This aberrant structure aggregates into highly stable filaments called amyloid plaques that accumulate in the brains of Alzheimer’s patients.

Question Type: Essay Chapter 3.2 Blooms: Understanding Difficulty: Easy 19. Which of the following does NOT impose limits on protein folding? a. ability of side chains to form hydrogen and ionic bonds b. backbone sequence of the polypeptide c. rotations of the planes around the peptide bonds d. size of side chains Ans: b Question Type: Multiple Choice Chapter 3.2 Blooms: Remembering Difficulty: Easy 20. Eggs are protein-rich foods. An uncooked egg can catalyze a reaction that breaks down bacterial cell walls. After cooking, this activity is almost abolished. This is likely because: a. the enzyme became denatured. b. bacteria can grow on cooked eggs. c. the cell membranes were liquefied. d. cooking sped up chemical reactions.

Ans: a Question Type: Multiple Choice Section 3.2 Blooms: Understanding Difficulty: Moderate 21. The correct order for molecular chaperone–mediated protein folding is: I – exchange of ATP for ADP on chaperone II – chaperone undergoes conformational change, which affects protein folding III – chaperone binds to exposed hydrophobic residues on unfolded protein IV – folded protein is released a. I, II, III, IV b. III, II, I, IV c. III, I, II, IV d. II, III, I, IV Ans: b Question Type: Multiple Choice Section 3.2 Blooms: Understanding Difficulty: Easy Section 3.3

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22. All the following statements about enzymes are true EXCEPT: a. they function in an aqueous environment. b. they lower the activation energy of a reaction. c. they increase the rate of a reaction. d. a single enzyme typically reacts with many different substrates.

Ans: d Question Type: Multiple Choice Chapter 3.3 Blooms: Understanding Difficulty: Easy 23. The Km for an enzyme-catalyzed reaction: a. determines the shape of the kinetics curve. b. determines the Vmax for the reaction. c. is a measure of the affinity of the substrate for the enzyme. d. is a measure of the rate of the reaction.

Ans: c Question Type: Multiple Choice Chapter 3.3 Blooms: Understanding Difficulty: Moderate

24. For an enzyme-catalyzed reaction, doubling the concentration of enzyme will: a. double the Vmax. b. halve the Vmax. c. double the Km. d. halve the Km. Ans: a Question Type: Multiple Choice Chapter 3.3 Application Difficulty: Easy

25. A small molecule that binds directly to the active site of an enzyme and disrupts its catalytic reaction is called: a. an allosteric inhibitor. b. a competitive inhibitor. c. a noncompetitive inhibitor. d. RNAi. Ans: b Question Type: Multiple Choice Chapter 3.3 Blooms: Remembering Difficulty: Easy 26. Changes in the conformational shape of an enzyme that diminish the size of its ligand-binding pocket are likely to affect an enzyme’s: a. specificity. b. affinity. c. epitope.

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d. specificity and affinity.

Ans: d Question Type: Multiple Choice Section 3.3 Blooms: Remembering Difficulty: Easy 27. Which of the following is true about enzymes? a. The catalytic site is responsible for substrate specificity. b. Reactions catalyzed by enzymes give the products more free energy than reactions that occur spontaneously. c. The rate of an enzymatic reaction is always proportional to the concentration of the substrate . d. Enzymes increase reaction rates by lowering the activation energy needed to reach the transition state.

Ans: d Question Type: Multiple Choice Section 3.3 Blooms: Understanding Difficulty: Moderate Section 3.4 28. Which of the following plays a role in the degradation of proteins? a. RNAi b. ubiquitin c. proteasome d. b and c Ans: d Multiple select Chapter 3.4 Blooms: Understanding Difficulty: Easy 29. Which of the following modifications marks a protein for degradation in proteasomes? a. phosphorylation b. ubiquitinylation c. acetylation d. glycosylation Ans: b Question Type: Multiple Choice Chapter 3.4 Blooms: Remembering Difficulty: Easy 30. Protein self-splicing: a. is autocatalytic. b. occurs in all eukaryotes. c. is an ATP-dependent process. d. is autocatalytic and occurs in all eukaryotes.

Ans: a Question Type: Multiple Choice Chapter 3.4 Blooms: Remembering

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Difficulty: Easy 31. Proteases that attack selected peptide bonds within a polypeptide chain are synthesized and secreted as inactive forms called: a. carboxypeptidases. b. aminopeptidases. c. zymogens. d. none of the above Ans: c Question Type: Multiple Choice Chapter 3.4 Blooms: Remembering Difficulty: Easy 32. Which of the following is a mechanism for regulating protein activity? a. proteolytic processing b. phosphorylation/dephosphorylation c. ligand binding d. all of the above Ans: d Question Type: Multiple Choice Chapter 3.4 Blooms: Understanding Difficulty: Easy 33. Protein kinase A is converted from an inactive state to an active state by binding: a. ATP. b. calcium. c. cAMP. d. ATP and cAMP. Ans: c Question Type: Multiple Choice Chapter 3.4 Blooms: Remembering Difficulty: Easy 34. Kinases, which are responsible for the activation or inactivation of a number of proteins, add phosphate groups onto: a. tryptophan residues. b. serine residues. c. cysteine residues. d. tryptophan and cysteine residues.

Ans: b Question Type: Multiple Choice Chapter 3.4 Blooms: Remembering Difficulty: Easy 35. The conversion of inactive chymotrypsinogen to active chymotrypsin is an example of: a. proteolytic activation. b. positive cooperativity. c. allostery. d. ligand-induced activation.

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Ans: a Question Type: Multiple Choice Chapter 3.4 Blooms: Understanding Difficulty: Easy 36. Describe the general mechanism by which a multisubunit protein can be activated by binding an allosteric effector molecule.

Ans: A multisubunit protein often contains both regulatory and catalytic subunits. In the absence of the allosteric effector molecule, the active site of the enzyme is masked by the regulatory subunit. Upon binding the allosteric effector molecule, a conformational change occurs, which relieves the suppression by the regulatory subunit on the catalytic subunit.

Question Type: Essay Chapter 3.4 Blooms: Understanding Difficulty: Difficult

37. What is positive cooperativity?

Ans: The activity of a protein can be modulated by binding a ligand. Cooperativity describes a phenomenon in which the binding of one ligand molecule affects the binding of subsequent ligand molecules. This allows a protein molecule to respond more efficiently to small changes in ligand concentration. In positive cooperativity, the binding of one ligand molecule enhances the binding of subsequent ligand molecules.

Question Type: Essay Chapter 3.4 Blooms: Understanding Difficulty: Moderate 38. A misfolded protein targeted to the proteasome will undergo: a. unfolding using energy released by ATPases. b. entry into the proteasome through the narrow channel in the beta channel. c. death by a thousand cuts in the alpha subunit rings. d. complete cleavage into its amino acid monomers.

Ans: a Question Type: Multiple Choice Section 3.4 Blooms: Remembering Difficulty: Easy 39. Modification of proteins by ubiquitin and ubiquitin-like E3 ligases can stimulate all of the following EXCEPT: a. recognition of intracellular viruses. b. regulation of the cell cycle. c. mRNA stability. d. nuclear import.

Ans: c Question Type: Multiple Choice Section 3.4 Blooms: Remembering Difficulty: Moderate

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40. GTPases serve in many signal transduction pathways and the presence of GTP or GDP dictates whether the pathway is on or off, respectively. Which of the following statements is true regarding guanine nucleotide exchange factors (GEF) and their role in these signaling pathways? a. They hydrolyze GTP into GDP and Pi. b. They decrease the GTPase activity of the G-protein. c. They catalyze the dissociation of GDP on the G-protein to therefore promote the replacement of GTP. d. none of the above Ans: c Question Type: Multiple Choice Section 3.4 Blooms: Remembering Difficulty: Moderate Section 3.5 41. Which of the following methods can separate proteins based on their mass? a. centrifugation b. ion exchange chromatography c. SDS polyacrylamide gel electrophoresis d. centrifugation and SDS polyacrylamide gel electrophoresis Ans: d Question Type: Multiple Choice Chapter 3.5 Blooms: Understanding Difficulty: Moderate 42. In two-dimensional gel electrophoresis, proteins are first resolved by a. IEF; SDS-PAGE b. SDS-PAGE; affinity chromatography c. SDS-PAGE; ion exchange d. IEF; gel filtration

and then by

.

Ans: a Question Type: Multiple Choice Chapter 3.5 Blooms: Remembering Difficulty: Easy 43. Gel filtration chromatography separates proteins on the basis of their: a. charge. b. mass. c. affinity for a ligand. d. mass and charge.

Ans: b Question Type: Multiple Choice Chapter 3.5 Blooms: Remembering Difficulty: Easy 44. Starting with 1 mCi (milliCurie) of a phosphorus-32-labeled compound, how long would it take until only 0.125 mCi remains? a. 14.3 days b. 28.6 days

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c. 42.9 days d. 57.2 days Ans: c Question Type: Multiple Choice Chapter 3.5 Application Difficulty: Moderate 45. Western blotting is a method for detecting: a. DNA. b. RNA. c. protein. d. carbohydrate.

Ans: c Question Type: Multiple Choice Chapter 3.5 Blooms: Remembering Difficulty: Easy 46. What is the basis for separation of proteins by two-dimensional gel electrophoresis? Why is this better for resolving a mixture of proteins?

Ans: In the first dimension, proteins are separated by isoelectric focusing, which separates proteins on the basis of their charge. In the second dimension, the proteins that have been separated by charge are further separated by their molecular weight (mass). The advantage of the two-dimensional technique is its ability to separate proteins more effectively. For example, two proteins with the same molecular weight could not be separated by one-dimensional SDS polyacrylamide gel electrophoresis. However, if these proteins differed in charge, then the two-dimensional gel would be able to separate these proteins into unique spots.

Question Type: Essay Chapter 3.5 Application Difficulty: Moderate

47. How can gel filtration chromatography separate proteins based on their mass?

Ans: In gel filtration chromatography, a column of porous beads made from acrylamide, dextran, or agarose is poured into a column. Proteins flow around the spherical beads. Because the surface of the beads contains large depressions, smaller proteins will penetrate into the depressions more easily than larger proteins and thus will travel more slowly through the column than larger proteins.

Question Type: Essay Chapter 3.5 Blooms: Understanding Difficulty: Easy

48. What is Western blotting? How can this technique be used to detect proteins?

Ans: Western blotting or immunoblotting is a method for identifying proteins separated on a gel using a specific antibody.

The proteins are first separated by molecular weight using polyacrylamide gel electrophoresis and then transferred from the gel to a membrane. The membrane is incubated with a primary antibody specific for the desired protein. After unbound antibody is washed away, the presence of the bound primary antibody is detected using a secondary enzyme-linked antibody.

The presence of the antibody-enzyme complex can then be detected using a chromogenic substrate.

Question Type: Essay

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Chapter 3.5 Blooms: Understanding Difficulty: Easy 49. A chunk of tissue is treated so that each cell’s membrane is broken open to release the contents inside, and then subjected to differential centrifugation. Which of the following is true at the end of the centrifugation? a. Depending on the speed of the centrifugation, the proteasome is more likely to be in the supernatant than a chaperone. b. Proteins of similar density will be found in the same fraction (either pellet or supernatant). c. The pellet contains the least dense material. d. The pellet will contain a purified protein for further analysis.

Ans: b Question Type: Multiple Choice Section 3.5 Application Difficulty: Moderate Section 3.6 50. Medical researchers are developing new clinical tests that detect and analyze the expression of multiple proteins and protein complexes in the hope that they might improve diagnosis of diseases such as early stage cancers. What techniques might researchers use in these studies?

Ans: They might use protein separation techniques such as two-dimensional gel electrophoresis and high-throughput LCMS/MS (liquid chromatography/mass spectroscopy) to separate and identify proteins and protein fragments on a global scale.

Question Type: Essay Chapter 3.6 Blooms: Understanding Difficulty: Easy 51. Proteomics allows researchers to: a. examine where a protein is located within the cytosol of a cell. b. compare thousands of samples from different people in the same experiment. c. examine which proteins differ in abundance between a normal sample and a disease sample. d. use antibodies to label specific proteins on an SDS-PAGE gel.

Ans: c Section 3.6 Application Difficulty: Moderate 52. Mass spectrometry techniques are used in proteomics for all of the following purposes EXCEPT: a. identification of thousands of proteins’ amino acid sequences within a single cell. b. characterization of proteolytically digested pieces of proteins from the sample. c. identification of all the protein complexes present in certain yeast species. d. identification of proteins within each organelle in liver tissue.

Ans: a Question Type: Multiple Choice Section 3.6 Blooms: Understanding Difficulty: Easy

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22

Culturing and Visualizing Cells

Section 4.1 1. A myeloma cell is best described as: a. a precursor cell that gives rise to gametes. b. an immortal immune cell that cannot produce antibodies. c. a self-renewing stem cell. d. The first and third answers are correct.

Ans: b Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Easy 2. What factors necessary for growth of animal cells in culture are provided by serum? a. amino acids b. precursors of DNA synthesis c. growth factors d. vitamins Ans: c Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Easy 3. Which one of the following is the best technique/approach to allow you to localize catalase in peroxisomes? a. a catalase monoclonal antibody and transmission electron microscopy b. platinum or gold and scanning electron microscopy c. FRAP and FRET d. all of the above Ans: a Question Type: Multiple Choice Chapter: 4 Application Difficulty: Moderate 4. All of the following are produced by animal cells in culture and help the cells adhere to the culture dish EXCEPT: a. glycoproteins. b. collagen. c. phospholipase A.

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d. hyaluronic acid.

Ans: c Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Moderate 5. Characteristics of transformed cells can include all of the following EXCEPT: a. aneuploidy. b. ability to differentiate into different cell types. c. tight junctions. d. presence of integrated viral genes.

Ans: c Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Easy 6. Separation of most blood cells is difficult, if not impossible, to achieve because they have similar properties and/or densities. What procedure is used to separate T-cells of the immune system from the many other different types of white blood cells or spleen cells? What feature of the T-cell facilitates the isolation protocol?

Ans: Flow cytometry and fluorescence-activated cell sorting is used to select and isolate T cells from numerous other cell types. Briefly, T cells, unlike other cells, express CD3 and Thy1.2 proteins on their cell surface. Antibodies specific to these markers are linked to a fluorescent dye and incubated with the pool of cells. The antibodies bind the cell-surface markers on the T-cell surface, and when all cells are placed in the FACS machine, a laser is used to excite the dye, causing it to fluoresce.

Fluorescing T cells selectively sorted from the non-fluorescing cells can be cultured in vitro.

Question Type: Essay Chapter: 4 Application Difficulty: Moderate 7. The purpose of treating tissue from an embryonic chick with trypsin and EDTA when generating a primary cell culture is to: a. break down proteins present in the serum so the cells can use the amino acids for energy. b. cleave any cell surface proteins so you can grow the chick cells as a suspension culture for Difficulty: Easy use in an experiment. c. break the protein–protein interactions that hold cells together in a tissue. d. prevent cell senescence.

Ans: c Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Moderate 8. Primary cultures contain (this cell type) which quickly predominate over the other cell types. Given the proper space and nutrients needed to grow, these cells can divide about 50 times, such that after 50 doublings, an original culture of 100 cells would become cells. a. fibroblasts; 2500 b. fibroblasts; 1.1 × 1017 c. epithelial cells; 2500

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d. epithelial cells; 4.5 × 1015 Ans: b Question Type: Multiple Choice Chapter: 4 Blooms: Analyzing Difficulty: Moderate 9. In your cell biology laboratory class, you are given a sample of blood. Which of the following results are NOT possible from your analysis of this blood using flow cytometry? a. isolation of the largest and most dense cells from the smaller blood cells b. measurement of the amount of DNA in white blood cells c. relative quantitation of the sizes and shapes of the cells in the blood sample d. after the addition of fluorescent antibodies specific to T cells, the analysis of T cell abundance and size relative to other cells in the sample Ans: a Question Type: Multiple Choice Chapter: 4 Application Difficulty: Moderate 10. Epithelial cells have distinct surfaces. Which of these surfaces tends to be involved in secreting proteins out of the cell and into the bloodstream? a. apical b. basal c. lateral d. all of the above Ans: a Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Easy 11. When studying a new protein, it is useful to generate an antibody against it to tag it in experiments designed to study its function and localization. Which of the following is NOT a step needed to generate this new antibody? a. Mouse spleen cells are fused with myeloma cells. b. Cells are selected for their ability to divide and grow in the absence of purines in the medium. c. Mice are injected with the antibody that will tag the new protein. d. The spleen is isolated from a laboratory mouse.

Ans: c Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Easy 12. Why are cells cultured in the lab such a useful model system for studying human disease? a. Cells only function when they are not organized into tissues. b. Humans are a varied population and extrapolating from physiological studies yields mixed results. c. Cells cultured in the lab behave exactly the same as cells in the human body. d. Cells are very inexpensive to grow.

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Ans: b Question Type: Multiple Choice Chapter: 4 Application Difficulty: Difficult 13. Drugs used to inhibit which molecules would reduce the amount of antibody produced by hybridomas? a. proteasome inhibitors b. protease inhibitors c. transcription inhibitors d. antibiotics Ans: c Question Type: Multiple Choice Chapter: 4 Application Difficulty: Difficult Section 4.2 14. The phenomenon in which a chemical absorbs light at one wavelength and emits it at a specific and longer wavelength is called: a. differential interference contrast. b. fluorescence. c. deconvolution. d. shadowing.

Ans: b Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Easy 15. Which of the following could be used to visualize subcellular structure in living cells? a. transmission electron microscopy b. scanning electron microscopy c. bright-field microscopy d. differential interference light microscopy Ans: d Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Easy

16. To visualize cells by immunofluorescence microscopy, the cells must be: a. placed in a vacuum. b. living. c. sectioned. d. permeabilized.

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Ans: d Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Moderate 17. The fluorescent properties of dyes such as SNARF-1 can provide information on the: a. location of specific proteins. b. concentration of H+ ions in specific regions of the cell. c. the amount of RNA in a cell. d. volume of a cell.

Ans: b Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Easy 18. How does the wavelength of the light used to illuminate a specimen affect the ability to resolve objects within the specimen?

Ans: Because the limit of resolution is given by D = (0.61/(N sin ), shorter wavelength light (e.g., blue) will provide better resolution than longer wavelength light (e.g., red).

Question Type: Essay Chapter: 4 Blooms: Understanding Difficulty: Moderate 19. Fixatives such as formaldehyde are routinely used in certain types of electron microscopy and light microscopy.

However, fixatives may introduce complications in the analysis of the resulting images. What problems may result from using fixatives?

Ans: Most fixatives are cross-linking agents that immobilize proteins and other cellular molecules. As part of the fixation process, samples are often also dehydrated. Such chemical changes may alter the normal structure and spatial relationship of cellular components, and this must be taken into account when interpreting images of fixed specimens.

Question Type: Essay Chapter: 4 Blooms: Analyzing Difficulty: Moderate 20. Which of the following allows one to circumvent the theoretical resolution of the microscope? a. total internal reflection fluorescence microcopy b. photo-activated localization microscopy c. indirect immunofluorescence microscopy d. double-label fluorescence microscopy Ans: b Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Moderate 21. A small tumor is excised from a patient’s body. The pathologist wants to examine the number, size, and arrangement of cells within the tumor. The best technique to use would be:

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a. DIC microscopy. b. phase contrast microscopy. c. bright-field microscopy after fixation, sectioning, and staining. d. fluorescence microscopy.

Ans: c Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Moderate 22. Fluorescence microscopy of cells that are labeled with a green fluorescent antibody against microtubules (a cytoskeletal component) differs from cells expressing green fluorescent protein (GFP) tagged to beta tubulin (a component of microtubules) in all the following ways EXCEPT: a. the antibody labeled cells need to be permeabilized before visualization, while the GFP cells do not. b. the microtubules could change their localization over time in the GFP cells, but not in the antibody labeled cells. c. the GFP cells do not need to be treated with fixatives, but the antibody labeled cells do. d. the antibody labeled cells will identify the microtubule structures, while the GFP will be observed in other places in the cell because it is not as specific as an antibody.

Ans: d Question Type: Multiple Choice Chapter: 4 Application Difficulty: Difficult 23. Fluorescence microscopy of intact organisms or large cells results in the generation of blurred images. Which of the following is NOT a technique that reduces the out-of-focus signal that causes blurring? a. two-photon excitation microscopy b. confocal microscopy c. fluorescence recovery after photobleaching (FRAP) d. deconvolution microscopy Ans: c Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Easy Section 4.3 24. Describe how a Förster resonance energy-transfer (FRET) biosensor like cameleon, which consists of CFP linked to YFP by the protein calmodulin, can detect local changes in calcium ion concentration. For your information, CFP excites at 440 nm and emits at 480 nm, whereas YFP emits at 535 nm.

Ans: Basically, if calcium is absent and the cameleon construct is excited by UV light at 440 nm, there is emission of energy at 480 nm, but because the distance between CFP and YFP is greater than 10 nm, the energy cannot be transferred to YFP and hence there is no detection of light at 535 nm. In the presence of calcium ions, however, calmodulin undergoes a conformational change, which brings CFP in close proximity to YFP. When CFP is excited with UV light at 440 nm, the energy is transferred to YFP, allowing it to fluoresce and be detected at 535 nm.

Question Type: Essay Chapter: 4 Blooms: Analyzing Hard

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25. Osmium tetroxide is commonly used to: a. stain specimens for light microscopy. b. coat specimens for metal shadowing electron microscopy. c. stain specimens for transmission electron microscopy. d. measure the Ca2+ concentration inside living cells.

Ans: c Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Easy 26. You are studying lamins and use an antibody method to follow their expression and subcellular localization at the electron microscope level. What process would you use to find these proteins in the bacteria E. coli?

Ans: Normally, antibody labeling can be used in tandem with electron microscopy to visualize proteins in thin sections.

Cells are lightly fixed to avoid denaturing the epitopes on the desired protein and, after freezing, the tissue is sectioned at very low temperature. The tissue is thawed and a specific antibody to the desired protein is applied as in immunofluorescence microscopy. To detect the antibody, however, the tissue is incubated with electron-dense gold particles attached to protein A, a bacterial protein that binds the Fc segment of all antibodies. These gold particles are seen at the level of the electron microscope; however, you would not be able to see lamins in E. coli because bacteria do not contain membrane-bounded organelles and lamins are a component of the nuclear envelope.

Question Type: Essay Chapter: 4 Application Hard 27. The best 3-D images of cellular organelles come from: a. cryoelectron microscopy. b. immunoelectron microscopy. c. thin-section electron microscopy. d. low-angle rotary shadowing with TEM. Ans: a Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Moderate

Section 4.4 28. If a cellular homogenate were subjected to differential centrifugation, which of the following would be expected to pellet first? a. the endoplasmic reticulum b. mitochondria c. the cytosol d. nuclei Ans: d Question Type: Multiple Choice Chapter: 4

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Application Difficulty: Moderate 29. The disruption of a cell is necessary to release its organelles and contents for subsequent isolation. One method, called , uses ultrahigh-frequency sound to disrupt the cell plasma membrane. a. tomography b. epitope tagging c. sonication d. centrifugation Ans: c Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Easy 30. Although many types of vesicles are similar in size and density, it is possible to isolate specific types of vesicles through the use of: a. a fluorescent-activated cell sorting machine. b. antibodies attached to bacterial carriers and low speed centrifugation. c. ultracentrifugation. d. light microscopy.

Ans: b Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Easy 31. Ultracentrifuges allow cell biologists to isolate mitochondria from lysosomes based on organelle differences in: a. isoelectric point. b. ionic composition. c. equilibrium density. d. size.

Ans: c Question Type: Multiple Choice Chapter: 4 Application Difficulty: Moderate 32. Rough endoplasmic reticulum can be separated from smooth endoplasmic reticulum by differential centrifugation. What is the basis for this fractionation?

Ans: During cell disruption, the endoplasmic reticulum fragments and reseals into small vesicle-like compartments termed microsomes. Microsomes derived from the rough endoplasmic reticulum contain ribosomes and these ribosomes provide additional mass, which allows separation from microsomes derived from the smooth endoplasmic reticulum.

Question Type: Essay Chapter: 4 Blooms: Understanding Difficulty: Moderate 33. The enzymatic functions of specific organelles can best be examined:

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a. in intact cells where the organelles are working as they normally would. b. in cells that have been prepared for SEM (scanning electron microscopy). c. in fractions taken from cell homogenates that have been separated by differential centrifugation. d. in fractions taken from differential centrifugation of cellular homogenates that are further incubated with antibodies specific to the organelle under study, which can be used to precipitate the organelles away from contaminates.

Ans: d Question Type: Multiple Choice Chapter: 4 Blooms: Understanding Difficulty: Moderate 34. Centrifugation of homogenized tissue from a mouse liver at 15,000 × g for 5 minutes will pellet: a. mitochondria b. nuclei c. plasma membrane d. mitochondria and nuclei Ans: d Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Easy 35. Rank the following organelles in terms of most dense to least dense: I. mitochondria II. peroxisomes III. lysosomes a. I, II, III b. II, I, III c. III, II, I d. III, I, II Ans: b Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Moderate 36. Proteomic studies allow for the identification of all proteins within an organelle, assuming the organelle can be purified sufficiently well. Proteomic analyses of mitochondria from different cell types revealed what interesting finding? a. Mitochondrial proteins have all been identified over the years. b. Mitochondria contain less than 40 different proteins. c. Mitochondria in different cell types can contain different proteins. d. Mitochondrial proteins cannot be examined using proteomic approaches.

Ans: c Question Type: Multiple Choice Chapter: 4 Blooms: Remembering Difficulty: Moderate

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5 Fundamental Molecular Genetic Mechanisms PART A: Linking Concepts and Facts 5.1 Structure of Nucleic Acids 1. In biology the central dogma refers to: a. protein to DNA to RNA. b. RNA to DNA to protein. c. DNA to RNA to protein. d. RNA to protein to DNA. Ans: c Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy 2. There are four basic molecular processes in a cell, three of which lead to the production of protein. Which of the following shows the correct order? a. translation, RNA processing, transcription b. RNA processing, translation, transcription c. transcription, RNA processing, translation d. transcription, translation, RNA processing Ans: c Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy 3. Nucleotide bases within a single strand of DNA are held together by which of the following bonds? a. phosphodiester b. pentose-phosphate c. polynucleotide d. phosphoester Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 4. In DNA, which of the following base pairing normally occurs between two strands of DNA? a. A to C

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6 - 32 b. T to A c. G to T d. C to T Ans: b Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy 5. In the laboratory, which of the following two factors are most useful in manipulating DNA? a. magnesium ions and sugar concentrations b. UV light and calcium ions c. formaldehyde and iron ions d. temperature and pH Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 6. To study the melting temperature (Tm) of your DNA, you have plotted the absorbance of UV light at 260 nm against temperature and have come up with the the following graph. Based on the evidence, the Tm of your double-stranded DNA is approximately:

a. 75 degrees b. 80 degrees c. 85 degrees d. none of the above Ans: b Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Difficult

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7. Which of the following is not a constituent of deoxyribonucleotides? a. phosphate moieties b. deoxyribose c. ribose d. organic bases Ans: c Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy 8. Which base pair(s) typically occur(s) in double-stranded DNA? a. G·C b. G·T c. G·A d. G·G Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy 9. When the circular double-stranded DNA of certain viruses is underwound, the DNA can twist back on itself and form: a. supercoils b. pseudoknots c. relaxed coils d. hairpins Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 10. The ability of DNA to denature is important for which process? a. DNA synthesis b. nucleic acid hybridization experiments c. RNA synthesis d. all of the above Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy

5.2 Transcription of Protein-Coding Genes and Formation of Functional mRNA 11. The most common arrangement of protein-coding genes in bacteria are those where the genes encoding proteins that work together are arranged in a contiguous array of DNA called: a. an operon. b. a terminator. c. a promoter. d. an operator.

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Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 12. Which of the following are removed from mRNAs during processing? a. exons b. noncoding sequences c. RNA cap structure d. poly(A) tail Ans: b Question Type: Multiple Choice Chapter: 5 Blooms: Difficulty: 13. The base in the wobble position of a codon a. is the 5´ (first) base. b. is the 3´ (third) base. c. is the second base. d. often contains adenine.

Ans: b Question Type: Multiple Choice Chapter: 5 Blooms: Difficulty:

14. Approximately what percent of all human genes are expressed as alternatively spliced mRNAs? a. 2 b. 25 c. 50 d. 90 Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Difficulty:

5.3 The Decoding of mRNAs by tRNAs 15. The genetic code is said to be ―degenerate‖ because: a. amino acid sequences are always read in the 3’→5’ direction of the mRNA. b. amino acids are encoded only by DNA sequences found in introns. c. a particular amino acid can be specified by more than one codon. d. tryptophan is the first amino acid in all polypeptide chains.

Ans: c Question Type: Multiple Choice

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6 - 35 Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy 16. In eukaryotic cells, protein synthesis in the cytoplasm utilizes three types of RNA molecules. Which of the following contains a three nucleotide sequence called the anticodon? a. mRNA b. tRNA c. rRNA d. all of the above Ans. b Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy

5.4 Stepwise Synthesis of Proteins on Ribosomes 17. In humans, the ribosome responsible for translating proteins consists of which of following two subunits? a. 60S and 30S b. 40S and 50S c. 18S and 40S d. 60S and 40S Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Difficult 18. Which of the following techniques would you need to use in order to determine the size of the large and small ribosomal subunits in E. coli? a. X-ray crystallography b. plaque assay c. centrifugation d. hybridization Ans: c Question Type: Multiple Choice Chapter: 5 Blooms: Applying, Analyzing Difficulty: Moderate 19. The first stage of protein translation occurs when Met-tRNA Met base pairs with the mRNA being translated in the i ribosomal: a. P site. b. A site. c. E site. d. T site.

Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding

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6 - 36 Difficulty: Difficult 20. Which of the following is seen when a single mRNA is simultaneously translated to form multiple copies of a polypeptide chain? a. phagosome b. multisome c. nucleosome d. polysome Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 21. You are studying polysomes and want to see if their numbers are actually increased in cells translating protein from an expressed retroviral oncogene. Which of the following methods are used to visualize these structures? a. sedimentation analyses b. electron microscopy c. PCR d. sedimentation analyses and electron microscopy Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Applying, Analyzing Difficulty: Moderate 22. Which of the following is not a recognized stage of protein synthesis in either prokaryotes or eukaryotes? a. elongation b. initiation c. translation d. termination Ans: c Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 23. Which of the following factors recognizes the UAG, UAA, and UGA codons? a. RNA polymerase b. DNA polymerase c. termination factors d. elongation factors Ans: c Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy 24. Which codon serves as the start codon in mRNA for translation? a. AGU b. AUG c. UGA

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

6 - 37 d. UGG Ans: b Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: : Easy 25. Which of the following is a protein that is involved in translation? a. topoisomerase b. ribosomal RNA c. RNA polymerase d. aminoacyl-tRNA synthetase Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 26. Which of the following structures interacts with ribosomes? a. tRNA b. mRNA c. rRNA d. all of the above Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy 27. Cellular protein synthesis proceeds in which direction? a. carboxyl to amino terminus b. amino to carboxyl terminus c. 3´' to 5´ d. 5´ to 3´ Ans: b Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 28. In eukaryotes, the circular structure of mRNA molecules undergoing translation serves to a. increase translation efficiency. b. target the mRNA for destruction. c. prevent the formation of polysomes. d. promote the termination step in protein synthesis.

Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Difficult

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

6 - 38 29. In which of these polymers are the monomers added one at a time? a. DNA b. rRNA c. protein d. all of the above Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate

5.5 DNA Replication 30. Okazaki fragments are small fragments of DNA that eventually are ligated to form which of the following strands of DNA? a. lagging b. parent c. leading d. joining Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Difficult 31. DNA replication begins at sequences called a. promoters. b. initiators. c. origins. d. Okazaki fragments.

Ans: c Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 32. An enzyme that breaks DNA, dispels the tension, and reseals the strand ahead of a DNA replication growing fork is called a(n) a. topoisomerase. b. DNA polymerase. c. phosphodiesterase. d. aminoacyl-tRNA synthetase.

Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate

4.6 DNA Repair and Recombination

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

6 - 39 33. Silent mutations that occur under a variety of conditions in DNA results in which of the following? a. a stop codon b. a change of an amino acid to another c. the formation of an anticodon d. no change in the amino acid Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 34. Which of the following lead(s) to a point mutation? a. deamination of a cytosine base into a uracil base b. benzo(a)pyrene conversion of guanine to a thymine base c. deamination of 5-methyl cytosine into thymine d. all of the above Ans: d Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Difficult 35. A Holliday structure is a(n) a. intermediate in genetic recombination. b. double-stranded DNA break. c. collapsed replication fork. d. thymine-thymine dimer.

Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 36. Which of the following are enzymes that play a key role in the base excision repair of nucleotide mismatches and damaged bases? a. glycosamines b. glycosidases c. glycosylases d. none of the above Ans: c Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Difficult 37. Thymine–thymine dimers are chemical adducts that develop in the DNA as a result of damage caused by which of the following? a. UV light b. X-rays c. RNA polymerase reading errors d. gamma radiation

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

6 - 40 Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate

5.7 Viruses: Parasites of the Cellular Genetic System 38. Viruses that use the lytic cycle of growth generally proceed through five general stages. Which one of the following shows the correct order of three of these stages? a. replication, assembly, adsorption b. penetration, adsorption, release c. replication, assembly, release d. assembly, replication, penetration Ans: c Question Type: Multiple Choice Chapter: 5 Blooms: Remembering, Understanding Difficulty: Difficult 39. You have received a saliva sample from a group of veterinarians who think one of the dogs they are treating has rabies.

Which of the following techniques would you use to determine not only if this sample contains viral particles, but how many viral particles are in the sample? a. centrifugation b. nucleic acid hybridization c. plaque assay d. all of the above Ans: c Question Type: Multiple Choice Chapter: 5 Blooms: Applying, Analyzing Difficulty: Difficult 40. This type of assay can be used to count the number of viral particles in a sample. a. plaque assay b. pap smear c. lytic cycle d. transformation assay Ans: a Question Type: Multiple Choice Chapter: 5 Blooms: Difficulty:

PART B: Testing on the Concepts 5.1 Structure of Nucleic Acids 41.

A double-stranded piece of DNA containing the sequence GCATGGCCACTACCG has a higher Tm than one containing the sequence GAATGGTAACAACTG. Describe the properties of DNA that make this true.

Ans: The first sequence contains 10/15 G·C base pairs. G·C base pairs are more stable than A·T base pairs because G and C

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

6 - 41 form three hydrogen bonds, whereas A and T form only two. The greater stability conferred by the additional hydrogen bond in G·C base pairs means that this sequence requires more energy (and thus a higher temperature) for denaturation than does the second sequence, which contains 6/15 G·C base pairs.

Question Type: Essay Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 42. What is the difference between a nucleoside and a nucleotide?

Ans: Nucleosides contain a base and a sugar without a phosphate. Nucleotides are nucleosides that have one, two, or three phosphate groups esterified at the 5´ hydroxyl.

Question Type: Essay Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy 43. To incorporate radiolabeled nucleotides into newly synthesized DNA, researchers use -phosphorus32-labeled nucleotides, in a DNA synthesis reaction, where the  denotes the position of the radioactive phosphate moiety. Explain why the  position and not the  or  position is the best position for the radioactive group in these experiments.

Ans: When nucleotides are added to the DNA polymer by DNA polymerase, the  phosphate of the incoming nucleotide attaches to the 3´ hydroxyl of the deoxyribose of the preceding residue to form a phosphodiester bond, releasing a pyrophosphate. If the radioactive label were contained on any position other than the  position, then the radioactive group would be present in the released pyrophosphate and not in the DNA chain.

Question Type: Essay Chapter: 5 Blooms: Evaluating, Creating Difficulty: Difficult

5.2 Transcription of Protein-Coding Genes and Formation of Functional mRNA 44. Which parts of the newly synthesized mRNA molecule do not transmit information for the synthesis of protein?

Ans: The 5´ 7-methylguanylate cap provides stability only and is not used to generate information. Also, the 5´UTR, 3´UTR, and intron sequences in the body of the mRNA molecule do not code for protein sequence either. The poly(A) tail, which is added to the 3´UTR after mRNA synthesis, is also noncoding.

Question Type: Essay Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate

5.3 The Decoding of mRNAs by tRNAs 45. If perfect Watson-Crick base pairing were demanded between codons and anticodons, cells would need 61 different tRNAs. If there are only 20 amino acids used in protein synthesis, how would you explain this excess number of tRNAs compared to amino acids? Conversely, how would you explain the fact that some cells contain fewer than 61 tRNAs? Ans: The number of tRNAs in most cells is more than the 20 amino acids used in protein synthesis because many amino acids have more than one tRNA to which they can attach. The fact that a single tRNA anticodon is able to recognize more than one, but not necessarily every, codon corresponding to an amino acid would explain why some cells have fewer than the expected 61 tRNAs. This broader recognition occurs because of the nonstandard pairing between bases in the wobble position corresponding to the 3´ base in the mRNA and the complementary 5´ base in the tRNA anticodon.

Question Type: Essay

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

6 - 42 Chapter: 5 Blooms: Applying, Analyzing Difficulty: Moderate

5.4 Stepwise Synthesis of Proteins on Ribosomes 46. When the CAU anticodon of a tRNAMet was modified to UAC, the anticodon for tRNAVal, valine aminoacyl-tRNA synthetase recognized the altered tRNAMet and added valine rather than methionine to it. When the converse modification was made, the altered tRNAVal containing a CAU anticodon (rather than UAC) was recognized and activated by methionine aminoacyl-tRNA synthetase. What do these data suggest about the mechanism by which aminoacyl-tRNA synthetases recognize their cognate tRNAs? Ans: These data suggest that the anticodon region of a tRNA is recognized by the corresponding aminoacyl-tRNA synthetase.

However, some tRNAs apparently contain other identity elements that are of primary importance in recognition of these species by the appropriate aminoacyl-tRNA synthetase.

Question Type: Essay Chapter: 5 Blooms: Evaluating, Creating Difficulty: Difficult 47. Many antibiotics, most of which have been isolated from fungi, act by inhibiting initiation, elongation, or termination of peptides during protein synthesis. These compounds have been very useful in the study of protein synthesis, mainly because an individual antibiotic usually interferes with protein synthesis at a single well-defined step in the complex process of translation. Assume that you have isolated a novel antibiotic from the fungus Pilobolus. This antibiotic inhibits bacterial growth and bacterial protein synthesis but does not affect growth or protein synthesis in eukaryotic cells. How could you identify the component of the translational machinery that is affected by this novel antibiotic?

Ans: Incubation of the radiolabeled antibiotic with cell extracts followed by chromatographic or electrophoretic separation could be used to identify the cell component(s) that binds the antibiotic. In addition, a genetic approach could be taken. The translational components directly involved in the step inhibited by the antibiotic can be identified by growing bacteria in the presence of lethal concentrations of the antibiotic and obtaining mutants that are resistant to its effects. Comparison of the translational components (tRNA, rRNA, initiation and elongation factors, etc.) in the resistant and sensitive cells would likely reveal which component of the translational machinery is altered in the resistant mutant.

Question Type: Essay Chapter: 5 Blooms: Evaluating, Creating Difficulty: Difficult

5.5 DNA Replication 48. The replication of DNA is quite complicated and requires the participation of many different enzymatic activities. These include an RNA polymerase (DNA primase) that synthesizes short segments of RNA, called primers, which are base paired to the DNA template. The 3´-OH ends of these RNA primers serve as initiation sites for the actual DNA polymerase activity.

Obviously, DNA primase can synthesize polynucleotides without the benefit of a 3´-OH primer; indeed, it can catalyze the hydrolysis and subsequent linkage of two nucleoside triphosphates without the need for a complementary strand whatsoever.

The short segments of RNA must be eliminated and replaced with DNA before replication can be completed. Why do you think that RNA, rather than DNA, primers are employed in the DNA replication process?

Ans: Because DNA is the molecule of inheritance, replication errors must be scrupulously avoided. Base pairing without a 3´-OH primer, as necessarily performed by any RNA polymerase, is very error prone. If a DNA polymerase performed this function, errors in the DNA sequence (mutations) would be introduced during replication and then transmitted to future generations. An error rate of 1 base in 105, which is not unusual for RNA polymerases, would result in an enormous increase in the mutation rate. However, the RNA primers made during DNA replication are erased and replaced with high-fidelity DNA copies; any mismatched bases will be replaced before being passed on to the next generation.

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

6 - 43 Question Type: Essay Chapter: 5 Blooms: Applying, Analyzing Difficulty: Moderate

5.7 Viruses: Parasites of the Cellular Genetic System 49. What is the difference between lytic and lysogenic bacteriophages?

Ans: Lytic bacteriophages infect cells, replicate, and lyse the cell as they exit to infect another group of cells. Temperate bacteriophages can either enter the lytic cycle or integrate their genomes into the chromosome of the host bacterium. Once integrated, the phage chromosome replicates as if it were a bacterial gene. With some frequency, the integrated phage chromosome is excised from the bacterial chromosome. At this point, it carries out a lytic cycle to infect additional cells.

Question Type: Essay Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate 50. Why is the enzyme reverse transcriptase found within retroviral virions?

Ans: Reverse transcriptase is found within a retrovirus particle because this enzyme is not expressed by the host-cell genome.

A retrovirus requires this enzyme to replicate; therefore, it must bring it into the cell along with its genome.

Question Type: Essay Chapter: 5 Blooms: Remembering, Understanding Difficulty: Easy 51. Polio virus affects only intestinal cells and motor neurons. By analogy to HIV, explain why only these particular cell types are susceptible to polio infection.

Ans: Prior to entering the cell, the polio virus binds to specific receptor proteins on the host cell surface. Once the viral particle binds, the virus enters the cell through endocytosis. The only cells that are susceptible to the polio virus are those that have the specific polio virus receptor protein on their surface.

Question Type: Essay Chapter: 5 Blooms: Remembering, Understanding Difficulty: Moderate

6 Molecular Genetic Techniques Section 6.1 1. Organisms that are considered polyploid have:

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

15 - 44 a. more than one genome. b. more than two copies of each chromosome. c. only one copy of each chromosome. d. multiple copies of a specific chromosome.

Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 2. The chemical mutagen ethylmethane sulfonate (EMS) was used to treat cells, and after 24 hours of incubation, DNA was isolated from these cells and sent for sequencing. Based on the DNA sequence from untreated cells, those treated with EMS showed several of which of the following conversions? a. G–C to A–T b. G–A to C–T c. G–T to A–C d. G–G to T–G Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Moderate 3. One common difference between meiosis and mitosis is that in meiosis, the end result is: a. four cells each having two chromosomes. b. two cells each having one chromosome. c. four cells each having one chromosome. d. two cells each having two chromosomes.

Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 4. Which of the following tests is used to determine if different recessive mutations are in the same gene? a. centimorgan linkage test b. conditional test c. cohabitation test d. complementation test Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 5. A mutation that changes a cysteine codon to a tryptophan codon is called a a. missense mutation. b. nonsense mutation. c. silent mutation. d. none of the above.

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

15 - 45 Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 6. Crossing of a homozygous wild type with a mutant that is heterozygous for a dominant mutation will result in F1 progeny of which a. all show the mutant phenotype. b. half show the wild-type phenotype and half show the mutant phenotype. c. three-fourths show the wild-type phenotype and one-fourth show the mutant phenotype. d. all show the wild-type phenotype.

Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Moderate 7. A mutation in one gene that counteracts the effects of a mutation in another gene is known as a a. temperature-sensitive mutation. b. recessive mutation. c. conditional mutation. d. suppressor mutation.

Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 8. Describe how genetic complementation can be used in yeast to determine whether two different recessive mutations are in the same or different genes.

Ans: Yeasts are normally a haploid organism but can be made to form a diploid organism by mating. Genetic complementation is a phenomenon whereby the wild-type phenotype can be restored after mating two recessive mutants. If two recessive mutations are in the same gene then a diploid organism containing both mutations will show a mutant phenotype because neither allele provides a functional copy of the gene. If two recessive mutations are in different genes then the diploid yeast will show a wild-type phenotype because a wild-type allele of each gene will be present.

Question Type: Essay Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult

9. Describe the properties and utility of temperature-sensitive mutations.

Ans: A temperature-sensitive mutation is a mutation that expresses a wild-type phenotype at one temperature and a mutant phenotype at another temperature. For example, the protein may exist in a wild-type conformation at a permissive temperature of 23C; however, at a nonpermissive temperature of 36C, the protein undergoes a conformational change to a mutant form. Temperature-sensitive or conditional mutations are especially useful for isolation of mutations in essential genes. In this case, cells with the mutation can be propagated normally at the permissive temperature, and the effect of the mutation can be studied at the nonpermissive temperature.

Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

15 - 46 Difficulty: Difficult Section 6.2 10. A recombinant DNA vector: a. cannot replicate in a host. b. does not contain bacterial DNA sequences. c. contains DNA sequences from only one source. d. contains DNA sequences that can be cleaved by restriction enzymes.

Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 11. A polylinker increases the versatility of a DNA plasmid because it contains: a. a DNA recognition sequences for several different restriction enzymes. b. a DNA sequence to allow the vector to replicate. c. a DNA sequence that confers resistance to ampicillin. d. b and c.

Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 12. You have an E. coli plasmid containing a mammalian cDNA and want to isolate the cDNA to use in another experiment.

The plasmid map reveals that you can isolate the complete cDNA by digesting the plasmid with EcoRI. How many bands would there be after a successful EcoRI restriction enzyme digest? a. one b. two c. three d. four Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Moderate 13. Construction of a cDNA library involves several steps, the first usually involves: a. digesting noncoding regions of DNA. b. cloning complementary double-strand DNA into a plasmid. c. isolating total RNA from the organism or cell type of interest. d. reverse transcribing RNA into cDNA. Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

15 - 47 14. Sequencing methods were used to test the integrity of the cDNA library and when this was done you found many of the sequences were those found in introns. After troubleshooting you determined that the reason these intronic sequences were present was because during the construction of this cDNA library: a. genomic DNA was present. b. retroviral DNA was a contaminant. c. endonucleases should have been added. d. RNA polymerase was present.

Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 15. In addition to sequencing, intronic sequences can be identified in a genomic DNA library using which of the following methods? a. Southern blotting b. in situ hybridization c. PCR d. Southern blotting and PCR Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 16. Which of the following series of steps does the polymerase chain reaction follow in order to amplify DNA in a test tube? a. DNA denaturation, primer elongation, primer annealing b. primer annealing, DNA ligation, primer elongation c. primer elongation, primer annealing, DNA ligation d. DNA denaturation, primer annealing, primer elongation Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 17. You want to amplify a region of yeast DNA using PCR so that this fragment can be cloned into a plasmid. Which of the following is needed for the PCR? a. RNA polymerase b. DNA ligase c. DNA helicase d. Taq polymerase Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 18. To see if your polymerase chain reaction was successful and it amplified the right sequence of interest, you electrophorese the products from the reaction on an agarose gel. After staining the gel you find there are two bands of different sizes. Which one of the following is a possible explanation for these results? a. The PCR induced a frame shift mutation into the DNA sequence.

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

15 - 48 b. RNA polymerase copied the DNA into two copies of RNA. c. The primers annealed to two different templates. d. a and b.

Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 19. Which of the following polymerase chain reaction techniques is widely used to study the amount of a specific mRNA within a cell or tissue? a. qualitative regional transcribed-PCR b. quantitative reverse targeted-PCR c. quantitative reverse transcriptase-PCR d. qualitative regional transposase-PCR Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy

20. Which of the following enzymes will produce a blunt end (the cut site is indicated by the * in the recognition sequence)? a. TaqI (T*CGA) b. EagI (C*GGCCG) c. EcoRV (GAT*ATC) d. NsiI (ATGCA*T) Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 21. DNA ligase a. synthesizes DNA from an RNA template. b. forms a phosphodiester bond. c. joins Okazaki fragments. d. b and c Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 22. Which of the following is a functional element of a plasmid? a. origin of replication b. drug-resistance gene c. polylinker sequence d. all of the above Ans: d Question Type: Multiple Choice

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

15 - 49 Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate

23. Next generation sequencing is much more efficient than the Sanger method because a. it uses an RNA template instead of DNA. b. it uses gel electrophoresis to resolve end-labeled strands of DNA. c. it uses PCR amplification. d. it uses gel electrophoresis to resolve end-labeled strands of DNA, and it uses PCR amplification.

Ans. c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 45. Describe some typical features of a restriction enzyme recognition sequence.

Ans: Restriction enzyme recognition sequences are typically 4–8 base pairs in length and are often palindromic, which means that the sequence read in the 5´ to 3´ direction is the same on each DNA strand. For example, the recognition sequence for EcoRI is 5´ GAATTC 3´. The complementary sequence is 3´ CTTAAG 5´, which is the same sequence when read in the 5´ to 3´ direction.

Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 24. Describe the essential features of a yeast shuttle vector.

Ans: A yeast shuttle vector is a plasmid that can replicate in both bacteria (E. coli) and yeast. For replication in bacteria, the plasmid needs a bacterial origin of replication and a gene for selection in transformed E. coli (e.g., for resistance to ampicillin). For replication in yeast, the plasmid needs a yeast origin of replication (autonomously replicating sequence, ARS), a gene for selection in transformed yeast (e.g., ura3), and a centromere. In addition, a polylinker sequence for the efficient cloning of foreign DNA is necessary.

Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult Section 6.3 25. A specific nucleic acid fragment is used in all but one of the following techniques? a. Southern blotting b. Western blotting c. Northern blotting d. in situ hybridization Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 26. Which of the following is used to compare gene expression in cells under different conditions? a. Southern blotting

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

15 - 50 b. P-element insertion c. stable transfection d. microarray analysis Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 27. The process by which genes cloned into specialized eukaryotic vectors are introduced into cultured animal cells for transient expression analysis is called: a. ligation. b. hybridization. c. transfection. d. transformation.

Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 28. You are growing a population of cells that appear to be resistant to G418, a chemical that is normally toxic to cells. Upon further analysis of the cells’ genome you find that it contains a genomic sequence that encodes the protein neomycin phosphotransferase. What would you do to test if this protein confers resistance to the G418 chemical? a. determine if the cells are also making β-galactosidase b. tag the protein with green fluorescent protein to see if it is degraded by G418 c. label a fragment of the gene to see where it is expressed d. clone the gene into a G418-sensitive cell line, then treat these cells with G418 Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 29. Modifying the DNA sequence of a gene of interest by appending to it a DNA sequence that encodes a stretch of amino acids recognized by a known monoclonal antibody is called: a. epitope tagging. b. in situ hybridization. c. polymerase chain reaction. d. next-generation sequencing.

Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 30. Northern blotting is a technique used to detect which one of the following in cells and tissues? a. DNA. b. RNA. c. protein. d. carbohydrate.

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

15 - 51 Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 31. In the large-scale production of a particular human protein in E. coli cells, the cDNA corresponding to the protein was modified so that the expressed protein would have six histidine residues at the C-terminus. The purpose of this modification was to a. facilitate transfer of the cDNA into the E. coli cells. b. provide a promoter for the transcription of the cDNA in E. coli. c. facilitate purification of the expressed protein though binding to an affinity column containing chelated nickel atoms. d. prevent degradation of the expressed protein by E. coli proteases.

Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 32. In situ hybridization is a powerful tool used in gene expression studies because it provides the investigator with information pertaining to a. the cellular and tissue-specific localization of the mRNA encoded by a particular gene. b. the activity of the protein translated from a particular mRNA. c. the size of the mRNA transcript. d. all of the above Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Moderate 33. What is epitope tagging? What is its application?

Ans: Epitope tagging is a method by which a cloned cDNA sequence is modified by the addition of a short DNA sequence that encodes a peptide recognized by a known monoclonal antibody. The short encoded peptide is called the epitope. This modified cDNA can be introduced and expressed in cells, and the location of the modified protein can be determined immunologically. Because the modified protein now expresses the epitope, the monoclonal antibody to the epitope can be used to detect the presence of the epitope-tagged protein. Epitope tagging allows the use of a single antibody for detection of any epitope-modified protein and eliminates the need to generate an antibody to each protein of interest.

Question Type: Essay Chapter: 6 Blooms: Applying, Creating Difficulty: Easy 34. Compare the advantages and limitations of microarrays and Northern blots for analyzing gene expression.

Ans: Microarrays allow a more global analysis of gene expression by analyzing thousands of genes simultaneously. Using the cluster-analysis program, groups of known and unknown genes that are regulated in a coordinated fashion can be revealed. Northern blots allow the analysis of only a few genes at a time; however, a Northern blot can reveal more information about the RNAs than a microarray. For example, a Northern blot can reveal the presence of multiple mRNAs that may be differentially expressed. The presence of multiple mRNAs could be missed by microarray analysis.

Question Type: Essay Chapter: 6 Blooms: Analyzing, Evaluating

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

15 - 52 Difficulty: Moderate Section 6.4 35. Which human disease is the result of an inherited X-liked recessive mutation? a. Tay-Sachs disease b. Duchenne muscular dystrophy c. cystic fibrosis d. none of the above Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 36. Microsatellites, which often vary in DNA sequence between individuals, are also called: a. single-nucleotide polymorphisms. b. nucleotide tandem replacements. c. short tandem repeats. d. autosomal dominant polymorphism.

Ans: c Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 37. The identification of disease genes can be made more complicated because of a. genetic heterogeneity. b. location of the gene on the X chromosome. c. polygenic traits. d. genetic heterogeneity and polygenic traits.

Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 38. Linkage studies can map disease genes with a resolution of about one centimorgan. Typically, the physical distance of a DNA region this size is approximately: a. 7.5 × 102 base pairs b. 7.5 × 103 base pairs c. 7.5 × 104 base pairs d. 7.5 × 105 base pairs Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 39. A haplotype is a set of closely linked genetic markers on a particular chromosome that tend to be inherited together. The genetic technique that looks at inheritance patterns and uses haplotypes in determining gene locations is: a. linkage mapping.

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15 - 53 b. linkage disequilibrium mapping. c. candidate gene approach. d. all of the above Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 40. How can linkage analysis determine the position of genes on a chromosome?

Ans: Linkage analysis examines the frequency of genetic recombination between two genes or markers on a chromosome. A genetic map or linkage map contains many markers mapped to a chromosome. Linkage analysis between the unknown gene and one of the known markers allows the rough localization of the gene on the chromosome. The basis for recombinational analysis is that two genes that are far apart on a chromosome will have a higher frequency of recombination than two genes that are close together. Thus, if recombination between the gene of interest and a marker is very low, then the gene is likely located near that marker gene.

Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 41. How does genetic heterogeneity or polygenic traits make the identification of a disease gene more difficult?

Ans: Some inherited diseases result from a mutation in a single gene; for example, a mutation in the -globin chain of hemoglobin causes sickle-cell anemia. However, some diseases result from mutations in not just one but multiple genes in an individual (a polygenic trait), whereas others result from mutations in any one of multiple different genes in different individuals, a phenomenon known as genetic heterogeneity. When more than one gene is involved in a single individual or among individuals, then mapping of the disease gene to a marker gene on a linkage map becomes much more complex.

Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult Section 6.5 42. Although small hairpin RNAs are used in RNAi-mediated destruction of a target mRNA, in order for them to be effective, the double-stranded RNA must be cleaved by which of one of the following? a. RISC b. RNA polymerase c. exonuclease d. dicer Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Moderate 43. CRISPR, one of the latest technologies to allow investigators to edit DNA sequences within a genome, can inactivate gene function if it produces a frameshift mutation leading to which one of the following sequences? a. TAG b. TTT c. TGT d. TTA

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Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 44. A step in the generation of knockout mutations in mice includes selection of embryonic stem (ES) cells that are a. resistant to G418 and resistant to ganciclovir. b. sensitive to G418 and resistant to ganciclovir. c. resistant to G418 and sensitive to ganciclovir. d. sensitive to G418 and sensitive to ganciclovir.

Ans: a Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult 45. As an investigator you want to functionally inactivate a gene without altering its sequence. Which of the following would you use to accomplish this? a. gene knockout b. RNA interference c. dominant negative mutation d. RNA interference and dominant negative mutation Ans: d Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 46. In RNA interference studies, the double-stranded RNA a. disrupts the target DNA sequence. b. results in the destruction of the target mRNA. c. destroys the target protein. d. all of the above Ans: b Question Type: Multiple Choice Chapter: 6 Blooms: Remembering, Understanding Difficulty: Easy 47. To study the function of the essential cytosolic Hsc70 genes in yeast, researchers constructed a shuttle vector in which a copy of the Hsc70 gene was ligated to the GAL1 promoter. The vector was then introduced into haploid yeast cells in which all four copies of the Hsc70 genes had been disrupted. Following introduction of the vector, you would expect that a. the yeast cells would grow on both glucose and galactose media. b. the yeast cells would grow on glucose but not galactose medium. c. the yeast cells would grow on galactose but not glucose medium. d. on transfer to either glucose or galactose medium, the vector-carrying cells would eventually stop growing because of insufficient Hsc70 activity.

Ans: c Question Type: Multiple Choice Chapter: 6

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15 - 55 Blooms: Applying, Analyzing Difficulty: Difficult 48. To inactivate the function of a wild-type small GTPase one could introduce a. a dominant negative allele of a GTPase gene that binds to and inactivates a guanine nucleotide exchange factor. b. an RNAi to silence the expression of the guanine nucleotide exchange factor. c. Cre recombinase. d. a and b.

Ans. d Question Type: Multiple Choice Chapter: 6 Blooms: Applying, Analyzing Difficulty: Difficult 49. RNAi is used to functionally inactivate genes in cells and whole organisms like C. elegans. Describe the basics of how you would knock down the expression of a gene required for muscle formation in C. elegans and what method could you use to confirm that your results were specifically attributed to the RNAi? Ans: The cDNA coding sequence of the muscle target gene is cloned, in both the sense and antisense orientations, adjacent to a strong promoter into plasmids. In vitro transcription of both constructs using RNA polymerase and rNTPs yields RNA copies complementary to the coding and antisense sequences. These copies are allowed to anneal and are then microinjected into the gonad of C. elegans. In the developing embryos, Dicer cleaves the double-stranded RNA into siRNAs, which bind to the corresponding endogenous mRNA target. Binding leads either to the degradation of the message or it serves to prevent the translation of the mRNA into a functional protein. In situ hybridization, Northern blot, or RT-PCR analyses are used to detect changes in the expression of the mRNA transcript, whereas immunocytochemistry or Western blot analysis are used to detect changes at the protein level.

Question Type: Essay Chapter: 6 Blooms: Applying, Analyzing, Evaluating, Creating Difficulty: Difficult 50. How can the function of an essential gene required for embryonic development be studied in an adult knockout mouse?

Ans: A standard knockout mouse cannot be created to study an essential embryonic gene because the loss of both copies of the gene would lead to embryonic death. The loxP-Cre system, however, can be employed to generate a conditional knockout mouse. In this case, the essential embryonic gene would be flanked by loxP sites. A transgenic mouse homozygous for the loxP modified embryonic gene would be mated with a mouse heterozygous for the loxP modified embryonic gene, which also expresses the Cre protein. The Cre protein would be expressed from a development-specific promoter that turns on Cre expression in a neonatal or adult mouse, leading to loss of the target gene only in post-embryonic mice.

Question Type: Essay Chapter: 6 Blooms: Remembering, Understanding Difficulty: Difficult

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7

Biomembrane Structure

Section 7.1 1. The plasma membrane around a eukaryotic cell is composed of: a. hydrophilic fatty acyl side chains. b. hydrophobic phospholipids. c. a phospholipid bilayer. d. none of the above Ans: c Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 2. Which of the cellular organelles are enclosed by two cellular membranes? a. Golgi b. lysosome c. nucleus d. endoplasmic reticulum Ans: c Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 3. Sphingolipids are considered amphipathic glycolipids if: a. their polar head groups are sugars that are not linked to the tails via a phosphate group. b. they are phospholipids with a one-chain hydrophobic tail. c. have one fatty acyl chain attached to carbon 2 of glycerol. d. none of the above Ans: a Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Hard 4. An investigator wants to use FRAP to quantify the lateral movement of a specific plasma membrane protein, but to do so this investigator must first: a. focus a laser on one region of the cell surface. b. measure the intensity of the fluorescence of the cell surface. c. label the cell with a fluorescent reagent that binds specifically to the cell surface. d. add detergents to make the cell surface more fluid.

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Ans: c Question Type: Multiple Choice Chapter: 7 Blooms: Applying, Analyzing Difficulty: Moderate 5. Which two organelles are responsible for producing phospholipids and sphingolipids? a. mitochondria and proteasome b. endoplasmic reticulum and mitochondria c. Golgi complex and persoxisome d. endoplasmic reticulum and Golgi complex Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 6. When using immunofluorescence microscopy with labeled Annexin V antibodies you found that there is a change in the distribution of the fluorescence when the cells were treated with tumor necrosis factor, an apoptosis-inducing agent. Your data show that this increase in Annexin V fluorescence increases specifically in the: a. mitochondria. b. exoplasmic face of the plasma membrane. c. Golgi. d. endoplasmic reticulum.

Ans: b Question Type: Multiple Choice Chapter: 7 Blooms: Applying, Analyzing Difficulty: Difficult 7. Cholesterol mixes with phospholipids in a biomembrane because cholesterol molecules are a. amphipathic. b. steroid derivatives. c. entirely hydrophobic. d. phospholipid derivatives.

Ans: a Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 8. All the following statements describe biomembranes except: a. Different biomembranes may contain different proportions of the same phospholipids. b. The two leaflets of a biomembrane may contain different phospholipids. c. Some biomembranes have free edges. d. Some phospholipids and cholesterol may cluster to form lipid rafts.

Ans: c Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Moderate

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9. Which of the following classes of lipids is (are) present in biomembranes? a. phosphoglycerides b. sphingolipids c. sterols d. all of the above Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 10. Phospholipids with short or unsaturated fatty acyl chains a. decrease membrane fluidity. b. increase membrane fluidity. c. cause biomembranes to become thicker. d. allow hydrophilic molecules to diffuse across the lipid bilayer.

Ans: b Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 11. Lipid droplets arise from the a. endoplasmic reticulum. b. plasma membrane. c. cytosol. d. exoplasm.

Ans: a Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 12. How do animal cells maintain membrane fluidity, and hence membrane function, in response to decreased temperature?

Ans: Membrane fluidity normally decreases with decreasing temperature and the bilayer becomes more gel-like. To maintain sufficient fluidity and hence function, animal cells could increase the ratio of unsaturated to saturated phospholipids in the membrane.

Question Type: Essay Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 13. What experimental evidence supports the fluid mosaic model of biomembranes?

Ans: Results from fluorescence recovery after photobleaching (FRAP) experiments have demonstrated the two-dimensional movement of membrane components and allow quantitative measurement of the extent of membrane fluidity.

Question Type: Essay Chapter: 7 Blooms: Applying, Analyzing Difficulty: Moderate

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15 - 59 14. Describe how you would prepare liposomes from a biological membrane.

Ans: The first step would be to treat the biological membrane with an organic solvent such as a mixture of chloroform and methanol. This would dissolve phospholipids and cholesterol, but not proteins or carbohydrates. Once the dissolved lipids were separated from the proteins and carbohydrates, you would evaporate the solvent. The final step would be to disperse the residue (which consists of phospholipids and cholesterol) in water to form liposomes.

Question Type: Essay Chapter: 7 Blooms: Applying, Creating Difficulty: Moderate Section 7.2 15. What type of simple-pass membrane protein would contain a hydrophobic membranea. transmembrane protein b. integral membrane protein c. peripheral protein d. transmembrane protein and integral membrane protein

-helix?

Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Difficult 16. Which of the following is TRUE regarding lipid-anchored membrane proteins? a. They are only present on the exoplasmic leaflet of the cell. b. They are only present on the endoplasmic leaflet of the cell. c. They are only present on the cytosolic leaflet of the cell. d. They can be present on either the cytosolic or exoplasmic leaflet of the cell.

Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Moderate 17. Peripheral proteins bound to the exoplasmic face of the plasma membrane can also bind to: a. cytoskeletal proteins. b. proteins of the ECM. c. the hydrophobic core of the phospholipid bilayer. d. proteins of the outer mitochondrial membrane.

Ans: b Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Moderate 18. Proteins can be attached to the phospholipid bilayer plasma membrane by covalently linked hydrocarbon groups. Which of the following mechanisms are employed in this anchoring? a. GPI b. prenylation c. acylation d. all of the above

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15 - 60 Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 19. In immunofluorescence studies you always see the small G-protein Ras in a specific region of the cell and, given its protein sequence along with many other proteins localized to this area, you hypothesize that their C-terminal Cys-Ala-Ala-X (X can be any amino acid) motifs may be responsible for targeting these proteins to this subcellular region. To test this hypothesis, you genetically engineer the sequence encoding this motif onto a cDNA encoding green fluorescent protein.

When immunofluorescence is used, where specifically would you expect to see GFP localized in the cell? a. in the nucleus b. in the cytoplasm c. in the Golgi d. at the plasma membrane Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Applying, Analyzing Difficulty: Moderate 20. Human alkaline phosphatase protein is attached to the plasma membrane by a GPI anchor. In an experiment you treat cells with each of the following enzymes and find one of them inhibits the activity of the protein because it releases it into the extracellular matrix. Which specific enzyme led you to conclude that this anchor on alkaline phosphatase was both necessary and sufficient for binding the protein to the membrane where it can perform its normal activity? a. glycosyltransferases b. phospholipase C c. flippases d. none of the above Ans: b Question Type: Multiple Choice Chapter: 7 Blooms: Applying, Analyzing Difficulty: Difficult 21. Peripheral membrane proteins a. contain many hydrophobic amino acid residues. b. contain membrane spanning domains. c. have covalently attached lipid or fatty acid anchors. d. may noncovalently interact with phospholipid heads.

Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Moderate 22. The a. lipid-anchored membrane proteins b. integral membrane proteins c. peripheral membrane proteins d. extracellular matrix proteins

are transmembrane proteins.

Ans: b

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

15 - 61 Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 23. Porins a. are peripheral membrane proteins. b. contain no hydrophobic amino acid residues. c. have many hydrophobic -helical regions. d. allow small hydrophilic molecules to pass through a membrane.

Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Moderate 24. Glycolipids and glycoproteins are especially abundant in the a. nucleus. b. mitochondrial inner membrane. c. cytosol. d. plasma membrane.

Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 25. Movement of phospholipids from one leaflet to the other a. occurs routinely. b. requires cholesterol. c. requires flippases. d. is impossible.

Ans: c Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 26. What are the primary functions of the plasma membrane in all cells?

Ans: In both bacteria and higher eukaryotic cells, the plasma membrane provides similar functions. These include regulation of nutrient transport into the cell and release of metabolic waste to the extracellular environment. By allowing certain material to pass in and out of the cell, and preventing other material from passing in and out of the cell, the plasma membrane acts to set up a molecular environment inside the cell that is different from the extracellular environment.

Question Type: Essay Chapter: 7 Blooms: Remembering, Understanding Difficulty: Moderate 27. When examined by fluorescence recovery after photobleaching (FRAP), certain integral membrane proteins are significantly less mobile than others. What accounts for this reduced mobility?

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15 - 62 Ans: The decreased mobility of certain integral proteins is due to interactions with the cytoskeleton or the extracellular matrix.

Question Type: Essay Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 28. Describe how charged amino acids guide the assembly of membrane complexes such as the T-cell receptor.

Ans: The T-cell receptor is composed of four dimers. Each contains specific charged residues as part of its transmembrane segments. The charged residues of neighboring segments interact with each other electrostatically in such a way that the overall charge of the transmembrane portion of the complex is minimized. These interactions serve to anchor dimers together to form the complex.

Question Type: Essay Chapter: 7 Blooms: Remembering, Understanding Difficulty: Hard Section 7.3 29. Free and unesterified fatty acids bind to one of a number of chaperone proteins in order to be transported in the aqueous environment of the cytosol. Which of the following facilitates this intracellular transport? a. fatty acid-binding proteins b. aquaporins c. flippases d. fatty acid synthase Ans: a Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 30. The major site of lipid synthesis in eukaryotic cells is the a. nucleus. b. endoplasmic reticulum (ER). c. peroxisome. d. mitochondria.

Ans: b Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 31. Phosphate-containing lipids include all of the following except a. plasmalogen. b. phospholipids. c. sphingolipid. d. triglyceride.

Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy

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32. Cholesterol, bile acids, ergosterol, and stigmasterol share all the following common structural features except a. a four-ring structure. b. a hydroxyl group on first ring. c. a carboxylic acid group. d. a multiple carbon chain extending off the ring structure.

Ans: c Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Moderate 33. The enzyme in cholesterol biosynthesis subject to feedback inhibition is a. ABCB4. b. desaturase. c. fatty acid synthase. d. HMG CoA reductase.

Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy 34. Cholesterol and phospholipids are transported between organelles by a. Golgi-dependent mechanisms. b. incompletely characterized vesicle populations. c. direct contact between membranes and, to some extent, small, soluble lipid-transfer proteins. d. b and c Ans: d Question Type: Multiple Choice Chapter: 7 Blooms: Remembering, Understanding Difficulty: Hard 35. Describe how unesterified fatty acids unlinked to CoA are able to move through the aqueous environment of the cytoplasm.

Ans: Fatty acid-binding proteins contain a hydrophobic pocket lined by  sheets. The ability of unesterified fatty acids to fit inside this pocket and interact non-covalently with the surrounding protein facilitates their intracellular movement.

Question Type: Essay Chapter: 7 Blooms: Remembering, Understanding Difficulty: Moderate 36. There are several enzymes involved in cholesterol biosynthetic pathway. Which of these is subject to feedback regulation? How does this enzyme sense cholesterol levels?

Ans: HMG-CoA reductase catalyzes the key rate-controlling step in cholesterol biosynthesis, the conversion of HMG CoA to mevalonate. The activity of this enzyme is tightly regulated. The enzyme is an endoplasmic reticulum (ER) membrane protein and has a number of transmembrane segments. Five of the transmembrane segments compose the sterol-binding domain. The sterol-binding domain senses the level of cholesterol in the ER membrane.

Question Type: Essay Chapter: 7

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15 - 64 Blooms: Remembering, Understanding Difficulty: Moderate 37. How are cholesterol and phospholipids transported between organelles?

Ans: The final steps in the synthesis of cholesterol and phospholipids take place primarily in the ER. Transport from there to other organelles is by poorly understood mechanisms. The transport is Golgi-independent. It is proposed to be either by membrane-limited vesicles or other protein-lipid complexes, direct contact between membranes, or by transfer via small, soluble lipid-transfer proteins. Some combination of these processes is likely important.

Question Type: Essay Chapter: 7 Blooms: Remembering, Understanding Difficulty: Easy

8

Genes, Genomics, and Chromosomes

Section 8.1 1. Which one of the following regarding pseudogenes is NOT true? a. They are present in the eukaryotic genome. b. They encode miRNAs. c. They mark the region of gene duplications. d. They always encode functional products.

Ans: b Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 2. Which of the following is a typical feature of prokaryotic genes? a. polycistronic messenger RNAs b. complex transcription units c. introns d. a and c Ans: a Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 3. The chicken lysozyme gene is considered to be a solitary gene because a. it contains no introns. b. it is not present on a chromosome. c. it is represented only once in the haploid genome. d. none of the above

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Ans. c Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy

4. All the following statements about complex transcription units are true except: a. They can have multiple poly(A) sites. b. They can generate multiple mRNAs. c. They can generate multiple polypeptides. d. They are common in bacteria.

Ans: d Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 5. In eukaryotes, tandemly repeated genes encode a. rRNAs. b. cytoskeletal proteins. c. -globin. d. all of the above Ans. a Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 6. Short micro RNAs (miRNAs) a. code for proteins. b. are common in bacteria but not eukaryotes. c. are involved in regulation of gene expression. d. have no known function.

Ans: c Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 7. Give a functional definition of a gene.

Ans: A gene consists of the entire DNA sequence required for synthesis of a functional protein or RNA molecule. In addition to the coding regions or exons, a gene includes transcription control regions, such as enhancers, and other critical noncoding regions such as poly(A) sites and splice sites. Sometimes essential control regions can even be located in introns.

Question Type: Essay Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 8. What is the advantage of complex transcription units over simple transcription units?

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15 - 66 Ans: A complex transcription unit can encode mRNAs processed in multiple ways to generate different proteins. A simple transcription unit can code for only one RNA and one protein. Complex transcription units allow for a greater diversity of proteins from the same number of genes.

Question Type: Essay Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate Section 8.2 9. After a diagnostic sequencing analysis of an individual’s DNA, you find that this person has a number of microsatellite triplet repeats within a region of their huntingtin gene. Specifically, these CAG repeats code for long stretches of: a. glycines. b. prolines. c. glutamines. d. stop codons.

Ans: c Question Type: Multiple choice Chapter: 8 Blooms: Applying, Analyzing Difficulty: Moderate 10. Blood samples were retrieved from a crime scene, and three suspects were arrested on suspicion of committing the crime.

Which of the following techniques could be used to identify the suspect(s) responsible for crime? a. DNA fingerprinting b. polymerase chain reaction c. in situ hybridization d. DNA fingerprinting and polymerase chain reaction Ans: d Question Type: Multiple choice Chapter: 8 Blooms: Applying, Analyzing Difficulty: Moderate 11. Which of the following organisms has the greatest amount of DNA per cell? a. chicken b. fruit fly c. tulip d. human Ans: c Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 12. All the following statements about microsatellite DNA are true except: a. It consists of a repeat length of 1–13 base pairs. b. It can cause neurological diseases such as myotonic dystrophy. c. It can occur within transcription units. d. all of the above Ans: d

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15 - 67 Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Difficult 13. Which of the following classes of repetitive DNA is most abundant in the human genome? a. simple-sequence DNA b. non-LTR transposons c. LTR transposons d. DNA transposons Ans: b Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 14. Describe the general organization of protein coding genes in the yeast and human genomes.

Ans: In yeast, the protein coding regions are closely spaced along the DNA sequence. In contrast, in the human genome, only a small fraction of the DNA encodes for protein. Thus, the density of protein coding genes per length of DNA is higher in yeast than it is in humans. Put another way, the human genome contains a much higher proportion of noncoding to coding sequences than does the yeast genome.

Question Type: Essay Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 15. Describe the proposed mechanism discussed in this chapter for the origin of gene families.

Ans: A gene family consists of a set of duplicated genes that encode proteins with similar but not identical amino acid sequences. An example of a gene family is the genes encoding the -like globins. The different genes in the gene family probably arose by duplication of an ancestral gene, most likely as a result of an unequal crossover during meiotic recombination. Over time, these duplicated genes accumulated random mutations. In some cases, a protein with a slightly different function emerged; in other cases, the mutations led to a nonfunctional gene known as a pseudogene.

Question Type: Essay Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 16. What is the underlying mechanism behind why gene mutations that lead to Huntington’s disease act as dominant mutations?

Ans: The mutations that lead to Huntington’s disease are examples of expanded microsatellite repeats. In the case of the gene responsible for Huntington’s disease, there is a triplet CAG repeat in the first exon. Expansion of this repeat results in synthesis of long stretches of polyglutamine. Over time, the protein products that contain long stretches of polyglutamine aggregate. Protein aggregation leads to neuronal cell death, which in turn gives rise to the symptoms of Huntington’s disease.

These microsatellite mutations are dominant because the presence of aggregated proteins causes symptoms, even though some normal proteins are produced from the normal allele.

Question Type: Essay Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate Section 8.3

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17. Drosophila is considered a model system because it is quite easy to create transgenic lines harboring a variety of different genes. As a Drosophila geneticist, which one of the following is a DNA transposon that you would exploit to create a transgenic line? a. copia element b. N element c. P element d. Ty element Ans: c Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 18. All the following steps are performed by the enzyme transposase during transposition of bacterial insertion sequences except a. excision of the IS element from the donor DNA molecule. b. introduction of staggered cuts into the target DNA molecule. c. ligation of the IS element to the target DNA. d. synthesis of DNA to fill in the single-stranded gaps.

Ans: d Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Difficult 19. Which of the following is not a mobile DNA element? a. transposon b. long terminal repeats (LTR) c. long interspersed elements (LINES) d. insertion sequence (IS) elements Ans: b Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 20. Which of the following mobile elements is a retrotransposon? a. yeast Ty element b. bacterial IS sequence c. Drosophila P element d. maize activator (Ac) element Ans: a Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 21. SINES (short interspersed elements) a. are approximately 300 base pairs long. b. are LTRs containing retrotransposons.

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15 - 69 c. are present in over 1 million copies in the human genome. d. a and c Ans: d Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 22. Mobile DNA elements likely contributed to the evolution of higher organisms by the a. generation of gene families by gene duplication. b. creation of new genes by exon shuffling. c. formation of more complex regulatory regions. d. all of the above Ans: d Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Difficult Section 8.4 23. Which of the following is an algorithm designed to compare the sequence of a newly identified protein with sequences already stored in the GenBank database? a. LINES b. BLAST c. HATs d. 3C Ans: b Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 24. The DNA and protein sequences of the -tubulin genes in humans and in fish are similar, and because each arose due to speciation, these genes would be considered: a. homologous. b. orthologous. c. paralogous. d. autologous.

Ans: b Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 25. Describe the two major pathways for transposition of mobile elements.

Ans: Mobile elements fall into two major classes. Insertion sequences and transposons move via a DNA intermediate, whereas retrotransposons transpose via an RNA intermediate. DNA elements encode a transposase enzyme, which catalyzes the transposition event. A retrotransposon is first transcribed into RNA, which is then used as a template for synthesis of

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15 - 70 double-stranded DNA by the action of the retrotransposon-encoded enzyme, reverse transcriptase. The resulting doublestranded DNA is then integrated into the host genome.

Question Type: Essay Chapter: 8 Blooms: Remembering, Understanding Difficulty: Difficult Section 8.5 26. To examine the folding and compaction of chromatin during mitosis, you will need to isolate and stain chromosomes at a particular stage using a special spreading preparation technique. For the best analysis, the chromosomes must be at which one of the following stages? a. metaphase b. interphase c. telophase d. anaphase Ans: a Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 27. There are five major types of histone proteins, but only four of them are considered as core histones. Which one of the following is NOT considered a core histone protein? a. H1 b. H2B c. H3 d. H4 Ans: a Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 28. Histone modifications play integral roles in chromatin condensation and function. Which of the following is NOT considered to be a histone modification? a. acetylation b. methylation c. phosphorylation d. prenylation Ans: d Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 29. You are studying the regulation of a group of genes and have determined that the full activation of transcription of these genes occurs when histone acetyl transferases have made post-translational modifications specifically to which one of the following amino acids? a. glycine b. glutamine c. lysine d. proline

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Ans: c Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 30. “3C” or chromosome conformation capture methods, used to determine the three-dimensional spatial organization of chromatin within nuclei of interphase cells, rely on a series of steps where the end result is the sequence analysis of purified DNA fragments. Which one of the following presents the correct order of steps you as an investigator need to follow in a 3C method strategy? a. shear DNA to 200–600 bp; cross-link proteins and DNA with formaldehyde b. ligate linkers marked with biotin onto DNA fragments; dilute and ligate the fragments c. cross-link streptavidin to DNA; purify and shear biotin-labeled fragments d. none of the above Ans: b Question Type: Multiple choice Chapter: 8 Blooms: Applying, Analyzing Difficulty: Difficult 31. Which of the following pairs of proteins are considered to be paralogous? a. yeast -tubulin and yeast -tubulin b. yeast -tubulin and worm -tubulin c. fly -tubulin and human -tubulin d. worm -tubulin and human -tubulin Ans: a Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 32. How many genes are estimated to be in the human genome? a. 21,000 b. 35,000 c. 75,000 d. 100,000 Ans: a Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 33. Open reading frame (ORF) analysis is not effective in identifying genes in higher eukaryotes because of the presence of a. promoters. b. enhancers. c. introns. d. repetitious DNA. Ans: c Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding

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15 - 72 Difficulty: Moderate 34. Which of the following lines of evidence is indicative of the presence of a gene in an unknown DNA sequence? a. alignment to a partial cDNA sequence b. sequence similarity to genes of other organisms c. ORF consistent with the rules for exon and intron sequences d. all of the above Ans: d Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate Section 8.6 35. Which of the following terms describes the phenomenon of genes occurring in the same order on a chromosome in two different species? a. heterochrony b. neoteny c. synteny d. phylogeny Ans: c Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 36. Which of the following terms describes when a chromosome is replicated everywhere except the telomeres and centromere, but the daughter chromosomes do not separate? a. hybridization b. polytenization c. polymerization d. heterochromatization Ans: b Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 37. Which of the following is NOT a functional element required for any eukaryotic chromosome to replicate and segregate correctly? a. replication origin b. centromere c. kinetochore d. telomeres Ans: c Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 38. All the following statements are true about a nucleosome except:

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15 - 73 a. It contains an octamer core of histones. b It is about 10 nm in diameter. c. It is the ―string‖ of the ―beads-on-a-string‖ appearance. d. It contains approximately 150 base pairs of DNA. Ans: c Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 39. DNA that is transcriptionally active a. is more susceptible to DNase I digestion. b. is tightly packed into a solenoid arrangement. c. contains nonacetylated histones. d. is more condensed than nontranscribed DNA. Ans: a Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 40. All of the following can be found in chromatin except a. DNA. b. histones. c. RNA. d. transcription factors.

Ans: c Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 41. Which of the following statement(s) is (are) true of a eukaryotic chromosome? a. It is a linear structure. b. It consists of a single DNA molecule. c. It can contain greater than a billion base pairs of DNA. d. all of the above Ans: d Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 42. In mammals, X-chromosome inactivation a. occurs in half the diploid cells of the adult female. b. results from the ionization of the X-chromosome. c. is considered an epigenetic event. d. b and c Ans. c Question Type: Multiple choice Chapter: 8

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15 - 74 Blooms: Remembering, Understanding Difficulty: Easy 43. Describe how modification of histone tails can control chromatin condensation.

Ans: The amino termini of histones, which are known as histone tails, extend from the structure of the nucleosome. Positively charged lysine side chains present in the histone tails may interact with linker DNA or other nucleosomes. Acetylation of the lysine side chains neutralizes the positive charges, thereby eliminating the potential interaction with the negatively charged DNA phosphate groups. Thus, acetylation of histones makes the chromatin less likely to form a condensed structure.

Deacetylation of the histones once again allows the positively charged lysines to interact with the DNA phosphate groups, leading to chromatin condensation.

Question Type: Essay Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate Section 8.7 44. The karyotype for any particular species is characterized by a. the number of metaphase chromosomes. b. the size and shape of the metaphase chromosomes. c. the banding pattern of the metaphase chromosomes. d. all of the above Ans: d Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 45. Chromosome painting involves a. staining chromosomes with Giemsa reagent. b. hybridizing fluorescent probes to chromosomes. c. hybridizing radioactive probes to chromosomes. d. a and b Ans: b Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy 46. All the following statements about heterochromatin except: a. It is a dark-staining area of a chromosome. b. It is usually transcriptionally active. c. It is often simple sequences of DNA. d. It is a region of condensed chromatin.

Ans: b Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy

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15 - 75 47. Telomeres a. consist of repetitive sequences with high G content. b. are a few hundred base-pairs long in vertebrates. c. have specific proteins bound at the DNA ends. d. a and c Ans: d Question Type: Multiple choice Chapter: 8 Blooms: Remembering, Understanding Difficulty: Moderate 48. Why is there a need for a specialized structure at the ends of eukaryotic chromosomes and for the enzyme telomerase?

Ans: Because all known DNA polymerases elongate DNA in the 5´ to 3´ direction, all require a RNA or DNA primer to initiate synthesis. As the replication fork approaches the end of the chromosome, DNA synthesis on the leading strand continues to the end of the chromosome without a problem. However, because the lagging strand is synthesized discontinuously, it cannot be replicated in its entirety. When the RNA primer is removed, a short segment of DNA remains single-stranded with no way to make this region double-stranded. If there were no specialized mechanism for replicating DNA at the ends, then the chromosome would shorten with each round of replication. Telomerase is the enzyme that completes DNA synthesis at the telomeres.

Question Type: Essay Chapter: 8 Blooms: Remembering, Understanding Difficulty: Easy

9

Transcriptional Control of Gene Expression

Section 9.1 1. In bacteria, an operon: a. is a region of DNA that is transcribed as a single mRNA encoding several proteins. b. encodes for miRNAs. c. contains a promoter unique for each individual gene in the operator. d. none of the above Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 2. Sigma factors present in bacteria are proteins required to: a. allow translation to proceed. b. terminate DNA replication. c. initiate transcription. d. terminate DNA replication and initiate transcription.

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15 - 76 Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 3. Operator constitutive mutants of the lac operon would a. express the lac repressor constitutively. b. block the binding of RNA polymerase to the promoter. c. express -galactosidase constitutively. d. prevent the inducer from binding to the repressor.

Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 4. How does binding of the lac repressor to the lac operator block transcription initiation? a. lac repressor binding blocks RNA polymerase from interacting with DNA at the start site. b. lac repressor binding induces a DNase that cleaves the DNA at the transcription start site. c. lac repressor binding causes a conformational change in RNA polymerase. d. lac repressor binding induces a protease that degrades the sigma subunit of RNA polymerase.

Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 5. Which of the following proteins does not ―footprint‖ the lac operon control region? a. lac repressor b. -galactosidase c. RNA polymerase d. cAMP-CAP Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: : Remembering, Understanding Difficulty: Moderate Section 9.2 6. You are studying the expression of the gene that appears to be under the control of three different transcription-control regions during mouse embryonic development. Which one of the following is the BEST method to use to determine when each of these regions are active in the developing mouse embryo? a. DNA affinity chromatography b. polymerase chain reaction c. reporter gene assay d. DNAse 1 footprinting Ans: c Question Type: Multiple Choice Chapter: 9

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15 - 77 Blooms: Applying, Analyzing Difficulty: Moderate 7. Reporter genes employ fragments of DNA encoding proteins that when translated do not have any obvious effects in the cells and tissues. Which one of the following is NOT a reporter protein? a. luciferase b. green fluorescent protein c. β-galactosidase d. RNA polymerase 1 Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 8. You are studying the effects of α-amanitin, a poisonous cyclic octapeptide, on eukaryotic cells and have noticed that following treatment, there is no miRNA transcription. Based on this evidence you conclude that α-amanitin must be inhibiting: a. RNA polymerase I. b. RNA polymerase II. c. RNA polymerase III. d. RNA polymerase I and RNA polymerase III. Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Applying, Analyzing Difficulty: Difficult 9. Which of the following is a fundamental difference between gene regulation in bacteria compared with eukaryotes? a. In bacteria, but not eukaryotes, there is a specific sequence that specifies where RNA polymerase binds and initiates transcription. b. In eukaryotes, but not bacteria, transcription can be influenced by how effectively the DNA sequence of a promoter region interacts with histone octamers. c. Transcription regulation is the most widespread form of control of gene expression in bacteria but not in eukaryotes. d. Gene regulation is readily reversible in eukaryotes but not bacteria.

Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 10. All of the following statements about the essential carboxy terminal domain (CTD) of RNA polymerase are true except: a. The CTD is present in RNA polymerase I, II, and III. b. The CTD can become phosphorylated. c. The CTD is critical for viability. d. The CTD of mammals contains more than 50 repeats of a heptapeptide.

Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate

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15 - 78 11. Define the terms cis-acting DNA sequences and trans-acting proteins.

Ans: Cis-acting DNA elements affect only the expression of genes on the same DNA molecule that are linked to the DNA element. In contrast, trans-acting proteins can diffuse through the cell to bind to their target DNA sequence.

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 12. Describe the similarities and differences between prokaryotic and eukaryotic RNA polymerases.

Ans: In prokaryotes, there is only one RNA polymerase, which consists of five subunits. In eukaryotes, there are three RNA polymerases, which are more complex than the bacterial RNA polymerase. RNA polymerase I synthesizes ribosomal RNA; RNA polymerase II synthesizes messenger RNA; and RNA polymerase III synthesizes tRNA and other small RNAs. All three contain two large subunits and 12–15 smaller subunits, which contain some sequence homology to the E. coli RNA polymerase subunits (, , and ´).

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 13. Describe the structure and function of the carboxy terminal domain (CTD) of RNA polymerase II. Ans: The carboxy terminal domain (CTD) of RNA polymerase II consists of a heptapeptide repeat, with a consensus sequence of Tyr-Ser-Pro-Thr-Ser-Pro-Ser. Yeast RNA polymerase II contains 26 or more repeats of this sequence, while the mammalian RNA polymerase II contains 52 repeats. The CTD is critical for viability, and at least 10 copies of the repeat must be present for survival. During formation of the transcription initiation complex, the CTD is unphosphorylated. When the RNA polymerase transcribes downstream of the promoter, the CTD is phosphorylated at serine and threonine residues.

One hypothesis is that phosphorylation of the CTD causes the release of RNA polymerase from the transcription initiation complex.

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult Section 9.3 14. You have identified a transcription factor and hypothesize that it binds to the promoter region of a gene that encodes a protein that causes cells to stop dividing. In order to test the interaction between the transcription factor and the DNA you will need to do a specific assay. Which one of the following would you use to test your hypothesis? a. fluorescent in situ hybridization b. chromatin immunoprecipitation c. immunocytochemistry d. high-throughput DNA sequencing Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Applying, Analyzing Difficulty: Moderate 15. The human genome encodes transcription factors that contain an acidic activation domain that is phosphorylated in response to increased levels of the second messenger cAMP. Which one of the following contains one of these activation domains? a. CBP

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15 - 79 b. CDK9 c. CREB d. CTD Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 16. Enhancers are considered transcription-control elements that regulate the expression of eukaryotic genes. Which one of the following is true regarding these elements? a. They are only found upstream of the transcription start site. b. They are never found more than one kilobase away of the transcription start site. c. They are only found in introns. d. They generally range in length from about 50–200 base pairs.

Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 17. Which one of the following terms is used to describe the protein:DNA complex containing several transcription factors bound to a single enhancer? a. nucleosome b. chromosome c. enhanceosome d. proteasome Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 18. The TATA box a. serves as a promoter sequence for genes transcribed by RNA polymerase III. b. is located approximately 100 base pairs upstream of the start site for mRNAs. c. is present in all eukaryotic genes. d. acts to position RNA polymerase II for transcription initiation.

Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 19. All the following elements can function as eukaryotic promoters except a. a TATA box. b. an initiator element. c. CpG islands. d. an enhancer.

Ans: d Question Type: Multiple Choice

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15 - 80 Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 20. Which of the following is the correct order of binding of general transcription factors to initiate transcription at RNA polymerase II promoters? a. TFIID, TFIIB, Pol II, TFIIH b. PolII, TFIID, TFIIB, TFIIH c. TFIIB, PolII, TFIIH, TFIID d. TFIID, TFIIH, TFIIB, PolII Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 21. What is the function of TFIIH in the transcription initiation complex? a. binding to the TATA box b. unwinding the DNA duplex c. catalyzing the synthesis of RNA d. all of the above Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 22. This serves as the promoter for 70% of eukaryotic genes and typically serves as a control region for genes that are transcribed at relatively low rates. a. TATA box b. enhancers c. CpG islands d. UAS (upstream activating sequences) Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 23. Describe the structure of the RNA polymerase II transcription initiation complex.

Ans: The RNA polymerase II transcription initiation complex is a multiprotein complex. This complex consists of a DNA promoter element to which general transcription factors (i.e., TFIIA, TFIIB, TFIID, TFIIE, TFIIH) bind along with RNA polymerase II. This multisubunit nucleoprotein complex consists of 60−70 polypeptides with a mass of approximately 3 MDa and is nearly as large as a eukaryotic ribosome.

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 24. Describe the functional properties of TFIID and TFIIH.

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15 - 81 Ans: TFIID is a large, multisubunit complex of approximately 750 kDa. TFIID consists of a 38 kDa TATA box-binding protein (TBP) and 11 TBP-associated factors (TAFs). TBP is the first protein to bind to a TATA box-containing promoter.

TFIIH is the last protein to bind to the initiation complex. TFIIH contains helicase activity, which unwinds the DNA duplex at the start site. As the polymerase transcribes from the promoter, a subunit of TFIIH phosphorylates the carboxy terminal domain (CTD) of RNA polymerase II. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult Section 9.4 25. An enhancer a. is a DNA element that stimulates transcription of eukaryotic promoters. b. binds to RNA polymerase and stimulates transcription. c. acts as a binding site for RNA polymerase. d. interacts with repressor proteins to enhance transcriptional repression.

Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 26. Which of the following is not used in the electrophoretic mobility shift assay (EMSA)? a. a radiolabeled DNA fragment b. a polyacrylamide gel c. a DNA binding protein d. DNase I Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 27. A leucine-zipper motif contains a. a stretch of five leucine residues in a row. b. a leucine residue at every seventh position. c. a leucine residue complexed with a zinc ion. d. an alternating leucine-alanine-proline structure.

Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 28. Which of the following is not a structural motif found in a DNA-binding domain? a. homeodomain b. zinc-finger c. helix-loop-helix d. random-coil acidic domain

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15 - 82 Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 29. What is the functional difference between enhancers and promoter proximal elements?

Ans: Enhancers can stimulate transcription from a promoter tens of thousands of base pairs away. In contrast, promoterproximal elements are located 100 to 200 base pairs upstream of the start site and usually lose their ability to stimulate transcription from a promoter when moved only several tens of base pairs away.

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 30. Describe how the electrophoretic mobility shift assay (EMSA) and the DNase I footprinting techniques are used to identify DNA-protein interactions.

Ans: In the electrophoretic mobility shift assay (EMSA), DNA-protein interactions are detected by changes in the mobility of a DNA fragment bound to a protein. A DNA fragment containing a putative protein binding site is first radiolabeled and then incubated in the presence of sequence-specific DNA binding proteins. The DNA fragment containing a bound protein migrates slower in a gel, causing a shift in the location of the radiolabeled DNA detected by autoradiography. In the DNase I footprinting technique, a DNA fragment is first labeled at only one end with 32P. The radiolabeled DNA fragment is incubated with a DNA binding protein and then digested with a limiting concentration of DNase I. The DNase I concentration is set such that on average each DNA molecule is cut only once. The resulting DNA fragments are separated by denaturing gel electrophoresis and visualized by autoradiography. In the absence of a DNA binding protein, a ladder of DNA bands is detected on the autoradiogram. Binding of a protein to the DNA prevents DNase I from digesting the radiolabeled DNA at the site of the DNA-protein interaction, resulting in a blank area (or ―footprint‖) in the DNA ladder.

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 31. What is an enhanceosome?

Ans: An enhanceosome is a large nucleoprotein complex bound to an enhancer element. This complex is formed by the cooperative assembly of transcription factors to their multiple binding sites in an enhancer.

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 32. How can transcription factors be purified using sequence-specific DNA-affinity chromatography?

Ans: Sequence-specific DNA-affinity chromatography is a technique that takes advantage of the binding specificity of a protein to a specific DNA sequence. Once the DNA sequence to which a transcription factor binds is identified, this DNA sequence can be coupled to a bead in a column. A protein mix containing the transcription factor is applied to this column.

Proteins that do not bind to the DNA fragment are washed off the column. The bound transcription factor can then be eluted from the column in the presence of a high concentration of a salt.

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate

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33. Describe the structure and function of a zinc-finger motif.

Ans: A zinc finger is a structural motif found in DNA binding domains; it consists of a short length of the polypeptide chain folded around a Zn2+ ion. The two basic classes of zinc finger domains are the C2H2 and C4 structures. The C2H2 zinc finger domain consists of two cysteine (C) and two histidine (H) residues bound to one Zn 2+ ion. The C4 zinc finger contains four cysteines bound to one Zn2+ ion. The three-dimensional structure of the zinc finger forms a compact domain, which can insert its  helix into the major groove of DNA. Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult Section 9.5 34. Bromodomains are found in chromosome-associated proteins that contribute to transcriptional activation. To facilitate this activation, the bromodomains bind to histones, specifically their lysine residues that have been post-translationally modified by: a. methylation. b. acetylation. c. phosphorylation. d. ubiquitination.

Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 35. Which of the following is NOT true regarding the chromatin-remodeling SWI/SNF complex? a. It acts as tumor suppressor. b. It serves as a co-activator of transcription. c. It has homology to DNA helicases. d. It can stabilize DNA-histone interactions.

Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 36. All the following statements about heterochromatin are true except: a. DNA dyes stain heterochromatin more darkly than euchromatin. b. The DNA of heterochromatin is more highly condensed than that of euchromatin. c. Heterochromatin is associated with inactive genes. d. Heterochromatin is more susceptible to DNaseI than is euchromatin.

Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 37. All of the following events play a role in yeast-mating type switching except a. methylation of the silent-mating-type locus.

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15 - 84 b. transcription of the gene at the MAT locus. c. chromatin condensation at the silent mating type locus. d. a recombination event known as gene conversion.

Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult 38. The mediator complex a. can form a molecular bridge between activators of transcription and DNA replication machinery. b. can function to maintain a promoter in a hypoacetylated state. c. has histone acetylase activity. d. none of the above Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 39. Transcriptionally inactive genes a. are always located within euchromatin. b. are not located within nucleosomes. c. often are methylated. d. are not resistant to DNase I. Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 40. Describe the role of histone deacetylation and hyperacetylation in yeast transcriptional control.

Ans: Histone deacetylation/hyperacetylation is one mechanism for regulating transcriptional control in yeast. Repressor proteins can cause deacetylation of lysine residues in histone N-termini in nucleosomes. Unacetylated histones contain positive charges due to the N-terminal lysines and interact strongly with DNA phosphates. These strong interactions may restrict access of general transcription factors, thus leading to transcriptional repression. In contrast, histones with hyperacetylated lysines in their N-termini are neutral in charge, eliminating the strong electrostatic interactions with the DNA phosphates. This more open chromatin configuration facilitates access of general transcription factors and induces transcriptional activation.

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult Section 9.6 41. The nuclear-receptor superfamily consists of several proteins that bind to consensus sequences of DNA response elements. Which of the following is NOT considered a member of this superfamily? a. retinoic acid receptor b. acetylcholine receptor c. glucocorticoid receptor

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15 - 85 d. progesterone receptor Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 42. Which one of the following techniques would be best suited to follow how a thyroxine-bound receptor translocates from the cytoplasm to its DNA response element? a. in situ hybridization and a radioactive fragment of the DNA response element b. fluorescence microscopy and a GFP-tagged receptor fusion protein c. pulse-chase radiolabeling d. none of the above Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Applying, Analyzing Difficulty: Difficult 43. Lipid soluble hormones activate transcription by a. binding to specific cell-surface receptors. b. phosphorylating a protein kinase. c. binding to a nuclear receptor. d. inhibiting a histone deacetylase.

Ans: c Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 44. Which protein domains are found in nuclear-receptor family members? a. variable region, DNA-binding domain, ligand-binding domain b. acetylase domain, DNA-binding domain, ligand-binding domain c. variable region, acetylase domain, ligand-binding domain d. variable region, DNA-binding domain, acetylase domain Ans: a Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 45. Regulation of transcription by steroid hormones a. involves hormone receptors only found in the nucleus. b. involves cytoplasmic hormone receptors that can move to the nucleus. c. involves two ligase domains. d. always activates transcription.

Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding

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15 - 86 Difficulty: Moderate 46. Describe how lipid soluble hormones, glucocorticoid for example, regulate gene transcription acting through nuclear hormone receptors.

Ans: Glucocorticoid is a lipid soluble hormone that binds to a member of the nuclear hormone receptor family, the glucocorticoid receptor (GR), which regulates gene transcription. In the absence of glucocorticoid, the GR in the cytoplasm is bound to the protein HSP90. When glucocorticoid diffuses through the cell membrane, it binds to the GR ligand-binding domain and causes a conformational change in the GR, releasing HSP90. The GR bound to glucocorticoid is then translocated into the nucleus, where it interacts with glucocorticoid response elements (GRE) and regulates transcription of responsive genes.

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 47. Describe the mechanism of transcriptional control for the heat shock genes. What advantage does this type of control impart to the cell?

Ans. During transcription of the heat-shock genes, RNA Pol II pauses after transcribing ≈25 nucleotides. Under stress conditions, where intracellular proteins are denatured or may become denatured, heat shock transcription factor (HSTF) is activated. In this state, HSTF binds to specific regions in the promoter of the heat shock genes, stimulating RNA Pol II to continue chain elongation. Binding also facilitates the rapid re-initiation by other RNA Pol II molecules, leading to a significant up-regulation in heat-shock-gene-expression. Thus, the mechanism of stalling the RNA Pol II and having partially completed transcripts ready to finish elongation and undergo translation when the need arises is a safeguard, protecting cells against unexpected, stressful conditions.

Question Type: Essay Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult Section 9.7 48. Epigenetics marks refer to modifications to DNA and proteins that in turn regulate gene expression. Which statement is true regarding these specific types of modifications? a. They have the ability to both silence and activate genes. b. They can involve the methylation of cytosine bases. c. They are linked to the acetylation of histones. d. All of the above Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate 49. What family of proteins plays an essential role in the repression of genes that help to direct the formation of specific tissues and organs in a developing embryo? a. Retinoblastoma b. Trithorax c. Polycomb d. Pax Ans: c Question Type: Multiple Choice

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15 - 87 Chapter: 9 Blooms: Remembering, Understanding Difficulty: Easy 50. X chromosome inactivation in mammals is mediated by a. micro RNAs (miRNA). b. long non-coding RNAs (ncRNA). c. messenger RNA (mRNA). d. short RNA-directed methylation of histones and DNA. Ans: b Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Moderate Section 9.8 51. Which of the following statement(s) regarding transcription initiation and RNA Pol III is (are) true? a. ATP hydrolysis is not required for initiation. b. Pol III is responsible for synthesizing tRNAs and 5S-rRNA. c. The promoter elements of tRNA genes lie entirely within the transcribed sequence. d. all of the above Ans: d Question Type: Multiple Choice Chapter: 9 Blooms: Remembering, Understanding Difficulty: Difficult

10

Post-transcriptional Gene Control

PART A: Linking Concepts and Facts 10.1

Processing of Eukaryotic Pre-mRNA

1. pre-mRNPs that are capped, spliced, and cleaved must be a. elongated b. phosphorylated c. polyadenylated d. transcribed

before they are called nuclear mRNPs.

Ans: c Question Type: Multiple choice

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15 - 88 Chapter: 10 Blooms: Remembering, Understanding Difficulty: Easy 2. The RNA recognition motif is the most common RNA-binding domain in hnRNP proteins. Another name for this motif is: a. RNA-binding domain. b. RNA motif. c. KH motif. d. all of the above Ans: a Question Type: Multiple choice Chapter: 10 Blooms: Remembering, Understanding Difficulty: Easy 3. In an experiment you have used recombinant DNA technology to create hnRNP C protein tagged with green fluorescent protein and hnRNP A1 protein tagged with red fluorescent protein. What would you expect to see when the proteins are expressed and visualized in Xenopus cells? a. green fluorescence only in the cytoplasm b. red fluorescence only in the nucleus c. red fluorescence only in the cytoplasm d. green and red fluorescence in the nucleus Ans: d Question Type: Multiple choice Chapter: 10 Blooms: Applying, Analyzing Difficulty: Difficult 4. Mutations that affect the binding of an SR protein to an exonic splicing enhancer can cause exon skipping in some genes, producing mRNAs that when translated yield nonfunctional proteins. In addition to an RNA-binding domain, SR proteins also contain an RS domain that is involved in binding: a. DNA. b. RNA. c. protein. d. none of the above Ans: c Question Type: Multiple choice Chapter: 10 Blooms: Remembering, Understanding Difficulty: Moderate 5. Sequencing of small RNAs isolated from metazoan cells revealed low levels of short, capped RNAs transcribed from both the sense and antisense strands of DNA. What is the term used to describe the fact that the majority of the metazoan genome is transcribed? a. permissive transcription. b. persuasive transcription. c. pervasive transcription. d. progressive transcription.

Ans: c Question Type: Multiple choice Chapter: 10 Blooms: Remembering, Understanding

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15 - 89 Difficulty: Moderate

6. The consensus sequence for poly(A) addition is a. the site of poly(A) tail addition. b. AAUAAA. c. downstream of the cleavage site. d. none of the above Ans: b Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Easy 7. Histone mRNAs lack a. poly(A) tails. b. introns. c. a 3´UTR. d. all of the above Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Easy 8. Which process involves two transesterification reactions? a. splicing b. RNA editing c. capping d. nuclear transport Ans: a Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 9. Splice sites in pre-mRNA are marked by two universally conserved sequences located a. in the middle of introns. b. at the ends of exons. c. at the ends of introns. d. none of the above Ans: c Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 10. Splicing joins a. two intron sequences. b. two polypeptides. c. two DNA molecules. d. two exon sequences.

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Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Easy 11. The branch-point A residue involved in lariat formation is part of the a. intron. b. exon. c. 5´'UTR. d. 3´UTR. Ans: a Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 12. Indicate the order in which the following steps occur in the production of a mature mRNA. a. initiation of transcription, splicing, addition of 5´ cap, addition of poly(A) tail, transport to cytoplasm b. initiation of transcription, addition of 5´ cap, splicing, addition of poly(A) tail, transport to cytoplasm c. initiation of transcription, addition of poly(A) tail, addition of 5´ cap, splicing, transport to cytoplasm d. initiation of transcription, addition of 5´ cap, addition of poly(A) tail, splicing, transport to cytoplasm Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Difficult 13. Components of the spliceosome include a. a pre-mRNA b. proteins that react immunologically with the sera of patients with systemic lupus erythematosus. c. U1 snRNA, which interacts with the 5´ splice site in pre-mRNA. d. all of the above Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Difficult 14. This type of RNA functions in the removal of introns from pre-RNAs. a. snRNA (small nuclear RNA) b. snoRNA (small nucleolar RNA) c. siRNA (small interfering RNA) d. miRNA (micro RNA) Ans: a Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate

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15 - 91 10.2

Regulation of Pre-mRNA Processing

15. In birds, a gene encoding a Ca2+-activated K+ channel is expressed in auditory hair cells as multiple mRNAs, which encode for proteins that open at different Ca2+ concentrations. The Ca2+ concentration at which the channel opens allows these cells to respond to different sound frequencies. Which one of the following would explain the appearance of the various isoforms of this particular channel? a. the presence of chain-terminating mutations b. exon skipping c. the absence of compensating mutations d. alternative splicing Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 16. Sex-lethal protein in Drosophila can best be described as a(n) a. splicing regulatory factor. b. RNA editing factor. c. transcription factor. d. all of the above Ans: a Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 17. Which of the following does not require enzymes? a. RNA editing b. excision of group II introns c. trans-splicing d. excision of group III introns Ans: b Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 18. RNA editing is a. post-transcriptional alteration of mRNA sequences. b. pretranscriptional alteration of RNA sequences. c. post-transcriptional joining of two RNA molecules. d. none of the above Ans: a Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Easy

10.3

Transport of mRNA Across the Nuclear Envelope

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15 - 92 19. The export of mRNAs outside the nucleus requires several proteins that are post-translationally modified by: a. acetylation. b. methylation. c. phosphorylation. d. ubiquitination.

Ans: c Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 20. Some have evolved a constitutive transport element within their genome, which allows for the export of unspliced RNAs into the cytoplasm. a. nematodes b. mammals c. retroviruses d. trypanosomes Ans: c Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Easy

21. Which of these events does not occur within the nucleus? a. RNA editing in mammals b. RNA capping c. polyadenylation d. RNA editing in protozoans Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 22. Which type of RNA participates in nuclear export of mRNA? a. snRNA b. hnRNA c. tRNA d. rRNA Ans: b Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Easy 23. Transport of unspliced HIV mRNA from the nucleus to the cytoplasm of host cells is promoted by a virus-encoded protein named a. Tat. b. Rev. c. nucleoplasmin.

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15 - 93 d. Ran. Ans: b Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate

10.4

Cytoplasmic Mechanisms of Post-transcriptional Control

24. The RISC complex contains which one of the following proteins? a. Argonaute b. DGCR8 c. Dicer d. Drosha Ans: a Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 25. Knocking out the dicer gene in mammals would lead to a loss of: a. mRNAs. b. miRNAs. c. shRNAs. d. snRNAs. Ans: b Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 26. You are using a variety of techniques to study how the RISC complex differs between siRNAs and miRNAs and have found that what distinguishes an RISC complex containing an siRNA from one containing an miRNA is that: a. the miRNA base-pairs perfectly with its target mRNA. b. the siRNA base-pairs perfectly with its target mRNA. c. the miRNA-RISC complex inhibits transcription. d. the siRNA-RISC complex blocks translation.

Ans: b Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Difficult 27. The mammalian target of rapamycin (mTOR) complex 1 plays an integral role in all but one of these processes. Which one is mTOR not directly involved in? a. Pol III transcription b. protein degradation c. ribosome biogenesis d. promoting cell growth Ans: b

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15 - 94 Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 28. The cell contains numerous organelles, all with specific functions, but the one that plays a key role in the digestion of ribosomes, mitochondria, and other organelles is the: a. autophagosome. b. endosome. c. nucleosome. d. proteasome.

Ans: a Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Easy 29. Which of the following regarding the P-body is FALSE? a. it contains decapping enzymes b. it is a deadenylase complex c. it has exoribonuclease activity d. it is the site where polysomes form Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 30. Cells use several ways to avoid the translation of improperly processed mRNA molecules. mRNA surveillance mechanisms. a. Nonsense-mediated decay b. Non-stop decay c. No-go decay d. all of the above

is(are) considered

Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Difficult 31. microRNAs play a key role in which of the following? a. translational repression b. viral RNA degradation c. RNA interference d. all of the above Ans: a Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Easy

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33. Which of the following does not take part in the degradation process of eukaryotic mRNAs? a. capping b. endonucleolytic cleavage c. exonucleolytic decay d. poly(A) shortening Ans: a Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate

10.5 Processing of rRNA and tRNA 34. Which of the following are NOT found within the nucleus? a. Cajal bodies b. histone locus bodies c. nucleoli d. P-bodies Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 35. Synthesis of pre-rRNA occurs in the a. nucleolus. b. endoplasmic reticulum. c. extranucleolar area of the nucleus. d. cytosol.

Ans: a Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Easy 36. The 45S pre-rRNA molecule a. can organize a nucleolus when present in a single copy. b. is encoded by tandemly arranged genes. c. is methylated on specific bases. d. all of the above Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 37. A ribozyme is an RNA sequence a. that uses Mg2+ ions as a cofactor. b. with catalytic ability to cleave RNA. c. that acts in the spliceosome.

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15 - 96 d. all of the above Ans: d Question Type: Multiple choice Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate

PART B: Testing on the Concepts 10.1

Processing of Eukaryotic Pre-mRNA

38. How is the 5´ cap added to nascent RNAs? Ans: A capping enzyme removes the -phosphate from the 5´ end of the nascent RNA emerging from the surface of a RNA polymerase II complex. A separate subunit of the capping enzyme then transfers a GMP moiety from a GTP donor to the 5´ diphosphate of the nascent transcript, creating a 5´-5´ triphosphate structure. Separate enzymes transfer a methyl group from an S-adenosinemethionine donor to the N7 position of the guanine and the 2´ oxygen of riboses at the 5´ end of the nascent RNA. Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 39. In animal cells, nearly all cytoplasmic mRNAs have a 3´ poly(A) tail, which is added to the pre-mRNA before splicing.

What proteins are involved in polyadenylation? Indicate their order of association with pre-mRNA and their functions.

Ans: (1) Poly(A) signal, which often is an AAUAAA sequence and binds the cleavage and polyadenylation specificity factor (CPSF); (2) poly(A) site, at which cleavage occurs and addition of A residues begins; and (3) G/U-rich region, which binds cleavage stimulatory factor (CStF). Polyadenylation of pre-mRNA begins with binding of CPSF, which is composed of several proteins, to the poly(A) signal. Then, at least three other proteins, including CStF, bind to the CPSF-RNA complex; interaction of CStF with the downstream GU-rich sequence stabilizes the entire complex. Binding of poly(A) polymerase to the complex then stimulates cleavage of the RNA at the poly(A) site and subsequent addition of A residues. Polymerization of A residues initially occurs slowly, but its rate is enhanced by binding of multiple copies of a protein called PABII. The mechanism by which the length of the poly(A) tail is restricted to about 200 nucleotides is not known.

Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Difficult 40. What are hnRNP proteins? How were they identified?

Ans: hnRNP proteins are the major protein components of heterogeneous nuclear RNA particles, which consist of unspliced nuclear mRNA and other nuclear RNAs. To identify hnRNP proteins, investigators exposed cells to UV irradiation, which causes covalent cross-links to form between RNA and closely associated proteins. Chromatography of nuclear extracts from irradiated cells on an oligo-dT cellulose column binds the poly(A) tails of unspliced mRNAs and can be used to recover proteins that have become cross-linked to these RNAs. Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 41. snRNP-dependent splicing of pre-mRNA is thought to have evolved from the self-splicing properties inherent in the sequence of either group I or II introns. Alternative splicing of pre-mRNAs processed by spliceosomes has been

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15 - 97 demonstrated, whereas this phenomenon does not occur in RNA transcripts that undergo self-splicing. Explain this difference.

Ans: In the case where splicing is self-mediated in response to sequence features, splicing is an intrinsic property of the molecule. This is the case with group I and II introns. In snRNP-mediated splicing, the splicing process, although responsive to the pre-mRNA sequence, is not dictated by the sequence of the RNA being spliced. For this reason, splicing of the molecule may be regulated and alternative RNA splicing may occur.

Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Difficult

42. The finding that the short consensus sequence at the 5´ end of introns is complementary to a sequence near the 5´ end of U1 snRNA suggested that this snRNA must interact with pre-mRNA for splicing to occur. Describe three types of experimental evidence that indicate U1 snRNA is required for splicing.

Ans: Addition of antiserum specific for U1 snRNP prevents in vitro splicing. A synthetic oligonucleotide of the same sequence as the 5´ end of U1 snRNA competes for the normal U1 snRNA and prevents splicing. Mutations in either the 5´ splice site of pre-mRNA or U1 snRNA prevent splicing; however, if a compensatory mutation that restores base pairing is present in the second component, then splicing occurs.

Question Type: Essay Chapter 10 Blooms: Applying, Analyzing Difficulty: Moderate

43. The spliceosomal splicing cycle involves ordered interactions among a pre-mRNA and several U snRNPs. According to the current model of spliceosomal splicing, which intermediate(s) in the splicing of a pre-mRNA containing one intron should be immunoprecipitated by anti-U2 snRNP? Which additional intermediate(s) should be immunoprecipitated by antiU2AF? Ans: Four different potential intermediates should be immunoprecipitated by anti-U2 snRNP: (1) a structure involved in the process of joining the two exons together but still containing the intron; (2) a structure that contains the excised intron in lariat form; (3) the pre-mRNA with U2 snRNP bound to the 5´ end of the intron; and (4) a structure consisting of the premRNA, U1 snRNP, and U2 snRNP bound to the branch site. Because U2AF assists U2 snRNP in binding the pre-mRNA, antibodies against this protein will immunoprecipitate the same complexes.

Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Difficult 44. In yeast, U2 snRNA base-pairs to a short sequence near branch-point A in introns. In higher eukaryotes, this branch-point sequence is not highly conserved, and a protein called U2AF promotes binding of U2 snRNA to pre-mRNA. You have produced mice with a knockout mutation in the U2AF gene. Would you expect mice heterozygous for the U2AF knockout mutation to be viable? Would you expect mice homozygous for the U2AF knockout mutation to be viable?

Ans: Association of the U2 snRNP with pre-mRNA is a necessary step in splicing. In higher eukaryotes, viability depends on proper splicing of pre-mRNA. However, assuming that U2AF normally is produced in excess, heterozygous knockout mice most likely would have sufficient U2AF to support splicing. Thus, little, if any, effect on the viability of these mice would be observed. Because proper splicing of pre-mRNA is a necessity for the viability of higher eukaryotes, a homozygous knockout mutation for UA2AF would be expected to be lethal in mice. Nonlethality would indicate the existence of redundancy in the pre-mRNA splicing mechanism. Because biological systems often exhibit redundancy to protect the organism, the homozygous knockout mice might survive.

Question Type: Essay Chapter 10

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15 - 98 Blooms: Remembering, Understanding Difficulty: Difficult

10.2

Regulation of Pre-mRNA Processing

45. Describe how the Sex-lethal (Sxl) protein is regulated during the development of Drosophila females.

Ans: Early in development, females utilize the Pe promoter to synthesize sxl mRNA containing exons 1 and 2, which is spliced normally, resulting in the production of early Sxl protein. Later in development, the Pl promoter located upstream is utilized, producing exons 1 through 4. The Sxl protein made earlier in development binds to this sxl mRNA, preventing splicing of exons 2 and 3. The resulting mRNA, containing only exons 1, 2, and 4, is translated into functional late Sxl protein, which also binds to the late sxl pre-mRNA, ensuring its continued production.

Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Difficult

10.3

Transport of mRNA Across the Nuclear Envelope

46. How does HIV bypass the normal restriction that prevents unspliced mRNAs from being transported from the nucleus to the cytoplasm?

Ans: The HIV genome codes for a viral protein called Rev. Rev binds to a sequence in the viral RNA called the Rev response element (RRE). Once the Rev protein builds up to a sufficient concentration in the host cell, it binds to the RRE and allows viral RNA to be exported from the host cell nucleus.

Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate

10.4

Cytoplasmic Mechanisms of Post-Transcriptional Control

47. The oocytes of multicellular animals contain stored mRNAs that encode numerous proteins required for early embryonic development. These proteins, however, are not translated until after fertilization and, therefore, a mechanism must be in place to ensure the stored mRNAs remain intact and untranslated before they are needed. Discuss the mechanism that keeps these stored mRNAs from being translated in the oocyte.

Ans: Stored mRNAs in oocytes have short poly(A) tails, consisting of ~20–40 residues. These short tails can bind only a few molecules of cytoplasmic poly(A)-binding protein (PABPI), which is not enough to interact with the initiation factor eIF4G. Following fertilization, the poly(A) tail increases in length with the addition of ≈150 A residues. This facilitates the binding of several PABPI molecules, allowing them to interact with eIF4G in a multimeric complex with eIF4E, other initiation factors, and the cap at the 5´ end of the mRNA. The stable conformation that forms is required for the initiation of translation. Thus, stored mRNAs are not translated efficiently because their short poly(A) tails do not provide enough binding sites to allow PABPI to implement the stability the complex requires for translation initiation.

Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate 48. How do researchers visualize the cellular locations of specific RNA molecules?

Ans: In one method, researchers manipulate the RNA sequence to include high-affinity binding sites for RNA binding proteins. They fuse the RNA binding proteins with other proteins that fluoresce as different colors (such as the green

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15 - 99 fluorescent protein or the red fluorescent protein). Using these kinds of probes, the subcellular locations of RNA molecules can be determined by fluorescence microscopy.

Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Moderate

10.5 Processing of rRNA and tRNA 49. The tissue-specific expression of antisense RNA is one experimental approach for selectively shutting down production of a protein. For example, some researchers have proposed that this approach could be used to regulate the production of pollen in tobacco, oilseed rape, and maize. The controlled production of sterile male plants, for example, would eliminate the problem of self-fertilization in the production of hybrid maize seeds. In this approach, expression of antisense RNA would be controlled by coupling it to a promoter that is specific to anthers, the part of flowers where pollen is produced. Alternatively, the RNase activity inherent in self-splicing RNA might provide a sequence-specific means to regulate pollen production.

Discuss how a catalytic RNA (i.e., a ribozyme) might be designed to prevent the expression of proteins needed for pollen production.

Ans: Some RNAs are capable of both sequence-specific base pairing and catalytic activity as an RNase. For example, when the 400-nucleotide-long intron sequence from Tetrahymena rRNA, a group I self-splicing RNA, is synthesized in a test tube, it folds and can bind two substrates, a guanine nucleotide and a substrate RNA chain. This synthetic intron then catalyzes the covalent attachment of the G to the substrate RNA, thereby cleaving the substrate RNA at a specific site. The release of the two RNA fragments frees the catalytic RNA for repeated rounds of catalysis. In principle, through the inclusion of the appropriate sequence for base pairing, a catalytic RNA can be designed that will bind to any substrate RNA and sever it at a specific site. Engineering a DNA sequence encoding a properly designed catalytic RNA under control of a tissue-specific set of promoter/enhancer elements and incorporating it into the germ line of plants could result in the tissue-specific synthesis of a ribozyme capable of selectively destroying a differentiation-specific mRNA required for pollen production.

Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Difficult

51. What are the main features of splicing pre-tRNAs that distinguish it from splicing pre-mRNAs? Ans: Splicing of pre-tRNA does not involve spliceosomes. In the first step, an endonuclease-catalyzed reaction excises the intron, which is released as a linear fragment, and a 2,3-cyclic monophosphate ester forms on the cleaved end of the 5 exon.

A multistep reaction that requires the energy derived from hydrolysis of one GTP and one ATP then joins the two exons. In contrast, pre-mRNA splicing occurs in spliceosomes, involves two transesterification reactions, releases the intron as a lariat structure, and does not require GTP. Although these transesterification reactions do not require ATP hydrolysis, it probably is necessary for the rearrangements that occur in the spliceosome.

Question Type: Essay Chapter 10 Blooms: Remembering, Understanding Difficulty: Difficult

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11

Transmembrane Transport of Ions and Small Molecules

Sections 11.1 1. The partition coefficient K, the equilibrium constant for partitioning between oil and water, for butyric acid is about 10 –2 , and for 1,4-butanediol it is about 10–4. You add liposomes containing only water to a solution with an initial concentration of 1 mM butyric acid and 100 mM 1,4-butanediol outside the liposomes. What is the relative rate of diffusion of the two substances into the liposome interior? a. Butyric acid diffuses into the liposome 100-fold faster than butanediol. b. The two diffuse into the liposome at equal rates. c. Butanediol diffuses into the liposome 100-fold faster than butanediol. d. Not enough information is provided to permit an answer.

Ans: b Question Type: Multiple Choice Chapter: 11 Blooms: Analyzing Difficulty: Difficult 2. Classes of membrane transport proteins include all of the following EXCEPT: a. ATP-powered pumps. b. ion channels. c. protein translocons. d. transporters.

Ans: c Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Easy 3. The glucose uniporter GLUT1 has a Km of 1.5 mM for D-glucose and 30 mM for D-galactose. At a concentration of 5 mM for each, what is the rate of glucose transport relative to galactose transport? The Vmax may be assumed to be the same for both. a. 3.7-fold slower b. equal c. 5.5-fold faster d. 20-fold faster Ans: c Question Type: Multiple Choice Chapter: 11 Blooms: Analyzing Difficulty: Moderate 4. Both ethanol and glycine are small molecules of approximately equal molecular weight. However, the membrane is much more permeable to ethanol than glycine. What accounts for the large difference in membrane permeability between ethanol and glycine?

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15 - 101 Ans: Ethanol is a small alcohol; glycine is a small amino acid. Glycine, like all amino acids, is a zwitterion. At neutral pH, the amino group of glycine is positively charged and the carboxyl group is negatively charged. Charged groups are impermeable to lipid bilayers.

Question Type: Essay Chapter: 11 Application Difficulty: Moderate 5. In a laboratory activity, red-labeled glucose, blue-labeled water, and green-labeled ethanol are added to a solution placed over an artificial, pure phospholipid membrane. Which colors will be observed on the other side of the membrane after 10 minutes? a. red b. blue c. green d. b and c Ans: d Multiple select Chapter: 11 Blooms: Understanding Difficulty: Easy 6. In which of the following cases is energy NOT needed for transmembrane transport? a. Lysine moves into the cell against its concentration gradient via the Na+/lysine symporter. b. Potassium ions (K+) move out of the cell down the K+ concentration gradient via potassium channels. c. Glucose moves into the cell down its concentration gradient via a glucose uniporter. d. The second and third answers are correct. e. all of the above Ans: d Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Moderate Section 11.2 7. How does uniport transport compare with simple diffusion? a. Similar to simple diffusion, uniport transport is nonspecific. b. Uniport transport is slower but more specific than simple diffusion. c. Uniport transport is much faster and more specific than simple diffusion. d. Simple diffusion is reversible but uniport transport is not.

Ans: c Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Moderate 8. Aquaporins are: a. -barrel proteins in the outer membrane of bacteria. b. ABC proteins. c. symporters. d. water channels.

Ans: d

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15 - 102 Question Type: Multiple Choice Chapter: 11 Blooms: Remembering Difficulty: Easy 9. When computing the osmotic pressure that must be placed across the membrane to stop the flow of water, what is the glucose osmotic equivalent of 1 M CaCl2? a. 1 M b. 2 M c. 3 M d. 4 M Ans: c Question Type: Multiple Choice Chapter: 11 Application Difficulty: Moderate 10. Describe how aquaporins facilitate the movement of water across the plasma membrane.

Ans: Aquaporins are water-channel proteins that specifically increase the permeability of biomembranes to water. The level of aquaporin 2 is rate-limiting for water transport by the kidney and is essential for resorption of water in the kidney.

Question Type: Essay Chapter: 11 Blooms: Understanding Difficulty: Easy 11. What is the approach that plants use to respond to differences in osmotic pressure between the inside and outside of the cell?

Ans: Plants have rigid cell walls. The normal concentration of solutes is higher inside the plant vacuole than in the cytosol.

Likewise, the solute concentration is higher in the cytosol than in the extracellular space. Hence, the plant cell exerts pressure against the cell wall and is retained within the cell wall.

Question Type: Essay Chapter: 11 Blooms: Understanding Difficulty: Moderate 12. Amino acid entry into cells can occur via uniporters or symporters. If the rate of leucine entry into the cell increases when the pH decreases, this suggests:

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15 - 103

a. leucine is being transported by a uniporter. b. leucine is being actively pumped across the membrane against its concentration gradient. c. leucine is being transported by a H+ symporter. d. leucine is crossing the membrane via simple diffusion.

Ans: c Question Type: Multiple Choice Chapter: 11 Blooms: Analyzing Difficulty: Moderate 13. Glucose enters erythrocytes via a GLUT-1 uniporter. As the levels of glucose in the bloodstream decrease between meals, what happens to the glucose in the cells? a. Glucose leaves the cell through the GLUT-1 uniporter, traveling down the new concentration gradient. b. Glucose remains in the cell because uniporters can only transport in one direction. c. Glucose remains in the cell because the GLUT-1 uniporters are gated and the gates close at low glucose concentrations. d. Glucose remains in the cell because it has been phosphorylated and no longer has affinity for the GLUT-1 uniporter.

Ans: d Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Moderate 14. Which of the following is true about glucose transporters? a. GLUT-2 and GLUT-4 are expressed on all cell types. b. All the members of the GLUT family have the same affinity for glucose. c. Cells that express GLUT-1 and GLUT-3 display more glucose uptake than cells expressing GLUT-2 at physiological blood concentrations (5 mM). d. GLUT proteins are specific for glucose molecules and will not transport other sugars.

Ans: c Question Type: Multiple Choice Chapter: 11 Application Difficulty: Moderate 15. The osmotic potential of a hypotonic solution should cause an animal cell to effect by . a. burst; not expressing aquaporins b. shrink; not expressing aquaporins c. burst; expressing more aquaporins d. shrink; expressing more aquaporins

, but with frog oocytes prevent this

Ans: a Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Moderate Section 11.3 16. Which of the four classes of ATP-powered pumps share overall similarity: several subunits, the same general organization, and a similar function as H+ transporters? a. ABC superfamily and P-class pumps

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15 - 104 b. ABC superfamily and P-class pumps c. F-class pumps and P-class pumps d. F-class pumps and V-class pumps Ans: d Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Moderate 17. The major ATP-powered pump responsible for maintaining ion gradients across the plasma membrane of mammalian cells is: a. the calmodulin-activated plasma membrane Ca2+ ATPase. b. the sarcoplasmic reticulum Ca2+ ATPase. c. the vacuolar F-class proton pump. d. the plasma-membrane Na+/K+ ATPase.

Ans: d Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Easy 18. ABC superfamily proteins are thought to act as ATP-dependent flippases in transporting: a. lipophilic drugs out of mammalian cells. b. Ca2+ out of mammalian cells. c. H+ out of mammalian cells. d. Na+ out of mammalian cells, K+ into mammalian cells.

Ans: a Question Type: Multiple Choice Chapter: 11 Blooms: Remembering Difficulty: Easy 19. Which statement describes the mode of action of the ABCB1 transporter (the first eukaryotic ABC transporter to be recognized)? a. During transport, the ligand binding site is alternately exposed to the exoplasmic and the cytoplasmic side of the membrane. b. During transport, a conserved aspartate residue is phosphorylated. c. This class of pumps transports only H+ ions. d. This transporter acts as a chloride channel.

Ans: a Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Easy 20. Which of the following statement(s) is (are) true of V-class proton pumps? a. They are ATPases. b. They are present in membranes of plant vacuoles. c. They serve to decrease the pH inside a lysosome. d. all of the above Ans: d

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15 - 105 Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Moderate 21. Describe in general terms how the muscle Ca2+ ATPase pumps Ca2+ ions from the cytosol into the sarcoplasmic reticulum (SR).

Ans: A P-class Ca2+ ATPase located in the SR membrane of the skeletal muscle pumps Ca 2+ from the cytosol into the lumen of the SR. This pump is the major integral membrane protein in SR membranes and hence can be readily purified. Ca2+ pumping from the cytosol to the SR involves a series of ordered steps and two conformational states of the pump, termed E1 and E2. In the E1 state, the pump binds two cytosolic Ca2+ ions and ATP. The ATP is cleaved to ADP with phosphorylation of an aspartic acid with a high-energy acyl phosphate. Next, a reduction in the energy state of the aspartate acyl phosphate produces a conformational change in the protein from the E1 to E2 state. In the E2 state, the affinity of the pump for Ca 2+ is 1000-fold less and both calcium ions are released into the SR lumen. With dephosphorylation of the pump, there is a second conformational change and the pump reverts again to the E1 state.

Question Type: Essay Chapter: 11 Blooms: Understanding Difficulty: Difficult 22. What evidence suggests that all of the P-class ion pumps evolved from a common ancestor even though they now transport different ions?

Ans: The catalytic subunits of all P-class ion pumps share a similar sequence, including a conserved aspartate residue that is phosphorylated during transport. The 3-D structures of the catalytic subunits are similar for those P-class ion pump structures that are known.

Question Type: Essay Chapter: 11 Blooms: Understanding Difficulty: Moderate 23. What is the basic structural organization of an ABC superfamily transport protein?

Ans: All members of the ABC superfamily of proteins contain two transmembrane domains and two cytosolic ATP-binding domains, which couple ATP-hydrolysis to solute movement. The transmembrane domains associate with each other. A transmembrane domain and associated cytosolic domain together form what may be thought of as a monomer. In nature, these two domains may be present in one polypeptide and the overall ABC transport protein then consists of two polypeptides. Alternatively, the four domains may be present as separate subunits or in some cases fused as a single protein.

Question Type: Essay Chapter: 11 Application Difficulty: Difficult 24. Explain why ATP-powered proton pumps cannot by themselves acidify the lumen of the lysosome.

Ans: The V-class proton pump responsible for acidifying the lumen of the lysosome, where the net movement of electric charge occurs during transport, is electrogenic. As each proton is pumped into the lumen, it leaves behind a negatively charged ion in the cytosol. Positive and negative ions attract each other across the membrane of the lysosome, which causes an electric potential. Soon, pumping leads to a buildup of protons in the lumen, which repels other protons. At this point, a significant transmembrane proton concentration gradient cannot be established. Thus, to generate an acidic environment necessary to breakdown components imported into the lysosome, proton transport must be accompanied by an equal number of anions, in this case in the form of Cl–, into the lumen.

Question Type: Essay Chapter: 11 Blooms: Analyzing

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15 - 106 Difficulty: Difficult 25. The many pumps present in the membrane have interconnected and important functions. Which of the following functions is NOT correct? a. A P-class pump is involved in muscle relaxation. b. A V-class pump is involved in maintaining the low pH of lysosomes. c. A P-class pump is involved in keeping the cytosolic levels of calcium low. d. A P-class pump is involved in keeping the cytosolic levels of potassium low.

Ans: d Question Type: Multiple Choice Chapter: 11 Blooms: Remembering Difficulty: Moderate 26. Which of the following is true about cystic fibrosis? a. The normal CFTR channel protein is a chloride uniporter, but CF patients have a mutation that makes it require ATP. b. CFTR mutations result in a mutated protein, which pumps too many chloride ions out of the cell. c. There are no treatments for patients with CF that are specifically targeted to the CFTR receptor. d. Patients with the most common mutation make a CFTR protein that is not present in the membrane.

Ans: d Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Moderate Sections 11.4 27. The resting membrane potential in animal cells depends largely on nongated a. Ca2+ b. H+ c. K+ d. Na+

channels.

Ans: c Question Type: Multiple Choice Chapter: 11 Blooms: Remembering Difficulty: Easy 28. The magnitude of the membrane electrical potential is calculated by: a. the Nernst equation. b. the Michaelis–Menten equation. c. the Faraday equation. d. the Bose–Einstein equation.

Ans: a Question Type: Multiple Choice Chapter: 11 Blooms: Remembering Difficulty: Easy 29. Ion channels achieve selectivity in transport through all of the following mechanisms EXCEPT: a. evolution from distinct and different parent proteins. b. divergent evolution from a single type of channel protein.

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15 - 107 c. low activation energy for selective passage of the dehydrated ion. d. use of P segments to form an ion selectivity filter.

Ans: a Question Type: Multiple Choice Chapter: 11 Blooms: Remembering Difficulty: Moderate 30. Why are nongated channels important in the generation of a negative electric potential (voltage) of 50–70 mV inside the cell with respect to the outside?

Ans: If there is no ion movement across the membrane, there is no membrane potential. This is true even if there is a difference in ion concentrations on either side of the membrane. The presence of K+ channels that are usually open allows for ion movement from the inside of the cell to the outside and the creation of a negative membrane potential inside the cell.

Question Type: Essay Chapter: 11 Blooms: Understanding Difficulty: Moderate 31. Calculate the G for the movement of Na+ from inside a typical mammalian cell to outside.

Ans: As summarized in Table 11-2 on text page 448, the cellular concentration of Na + is 12 mM and the blood concentration is 145 mM. The inside of the cell has a negative electric potential of –70 mV. Therefore, for K+ the G for Na+ export will have a positive value for the Gc term (Na+ is being moved against a concentration gradient) and a positive value for the Gm term (Na+ is being moved against the negative electric potential inside the cell). The sum of these two terms is +3.06 kcal/mol. The Na+/K+ ATPase uses energy from ATP cleavage to export 3 Na + for every 2 K+ imported.

Question Type: Essay Chapter: 11 Application Difficulty: Difficult 32. Describe the patch-clamping technique. investigations?

Why do investigators often use frog oocytes in their patch-clamping

Ans: By pressing a special electrode against a ―patch‖ of plasma membrane and forming a tight seal, an investigator can ―clamp‖ the voltage (or current) at a constant value and study the opening, closing, regulation, and ion conductance of a single ion channel. The technique can be used on whole cells or isolated membrane patches and there are advantages to using one over the other. Frog oocytes are often used in patch-clamping experiments because they are readily available and large enough to microinject in vitro transcribed mRNAs encoding channel proteins, which will be expressed on the cell surface.

Since frog oocytes normally do not express any channel proteins of their own, only the channel protein expressed from the microinjected mRNA will be present, allowing it to be studied in isolation.

Question Type: Essay Chapter: 11 Application Difficulty: Difficult

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15 - 108 33. A mutant channel protein is expressed in an oocyte. Compared to patch clamping experiments in oocytes expressing the normal channel, the length of downward deviations is diminished by half. This indicates: a. the mutant channel doesn’t make a functional channel. b. the mutant channel doesn’t stay open as long. c. the mutant channel opens more frequently. d. the mutant channel behaves the same as the wild-type channel.

Ans: b Question Type: Multiple Choice Chapter: 11 Blooms: Analyzing Difficulty: Easy Section 11.5 34. The G calculation for the two-Na+/one-glucose symporter includes which of the following terms? a. RT ln [glucose in]/[glucoseout] b. 2RT ln [Na+ ]/[Na+ ] in

out

c. 2FE d. all of the above Ans: d Question Type: Multiple Choice Chapter: 11 Application Difficulty: Moderate 35. How does inhibition of the Na+/K+ ATPase increase the force of heart muscle contraction? a. It increases cytosolic Na+ and therefore decreases Ca2+ export. b. It increases cytosolic K+ and therefore decreases Ca2+ export. c. It decreases cytosolic Na+ and therefore decreases Ca2+ export. d. It decreases cytosolic K+ and therefore decreases Ca2+ export.

Ans: a Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Moderate 36. What is the expected effect on salt accumulation in the plant vacuole of a mutation in the plant vacuolar ATPase that decreases the H+ concentration in the vacuole? a. increased Na+ accumulation b. decreased Na+ accumulation c. no effect d. complete seed germination failure Ans: b Question Type: Multiple Choice Chapter: 11 Application Difficulty: Moderate 37. Propose a rationale for why the import of amino acids or sugars into cells is typically coupled to Na+ ion import.

Ans: The Na+/K+ ATPase establishes and maintains large differences in Na+ and K+ distributions across the plasma membrane. The concentration of Na+ is low inside cells and high outside. The membrane potential is negative. The import

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15 - 109 of Na+ is both concentration and membrane-potential favorable and can therefore drive the import of amino acids or sugars.

The exact opposite is true for K+. Coupling to Na+ movement is accomplished by symporters.

Question Type: Essay Chapter: 11 Blooms: Analyzing Difficulty: Difficult 38. Propose a rationale for why the import of sucrose into the plant vacuole is coupled to the export of H + ion into the plant cytosol.

Ans: The plant vacuole is acidified by proton pumps. Moreover, the proton pumps generate a positive membrane potential in the vacuole relative to the cytosol. The export of H+ into the cytosol is both concentration and membrane-potential favorable and can therefore drive the import of sugars (i.e., sucrose) into the vacuole. Coupling of sucrose import to H+ export is accomplished by antiporters.

Question Type: Essay Chapter: 11 Blooms: Analyzing Difficulty: Difficult 38. The movement of dietary glucose and sodium from the intestine into the bloodstream relies on which of the following transport mechanisms? a. sodium glucose symporter b. transcellular transport c. sodium/potassium ATPase d. all of the above Ans: d Question Type: Multiple Choice Chapter: 11 Blooms: Remembering Difficulty: Moderate Section 11.6 39. Transepithelial glucose transport uses a symporter to transport glucose up a concentration gradient by: a. coupling glucose transport to proton movement. b. coupling glucose transport to Na+ movement. c. coupling glucose transport to Ca2+ movement. d. coupling glucose transport to Cl– movement.

Ans: b Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Easy 40. Parietal cells acidify the stomach contents while maintaining a neutral cytosolic pH by: a. exporting ―excess‖ cytosolic OH– as HCO 3–. b. exchanging HCO3– for Cl–. c. preserving electroneutrality by accompanying the movement of each Cl– ion into the stomach lumen by a K+. d. all of the above.

Ans: d Question Type: Multiple Choice Chapter: 11 Blooms: Understanding

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15 - 110 Difficulty: Moderate 41. Transepithelial transport requires a a. polarized b. sealed c. polarized and sealed d. permeable

cell layer.

Ans: c Question Type: Multiple Choice Chapter: 11 Blooms: Understanding Difficulty: Easy 42. How can a uniporter, GLUT2, be sufficient for entry of glucose from intestinal epithelial cells into the bloodstream?

Ans: Glucose is actually present in a higher concentration inside the intestinal epithelial cell than in the bloodstream. Hence, a uniporter on the basolateral surface of the intestinal epithelial cell can be effective in facilitating the entry of glucose into the bloodstream. The two-Na+/one-glucose symporter located on the apical surface, facing the intestinal lumen, is an example of secondary active transport generating the high glucose concentration inside the epithelial cell.

Question Type: Essay Chapter: 11 Blooms: Understanding Difficulty: Difficult 43. The H+/K+ ATPase on the apical surface of parietal cells exports H+ and imports K+. How is the buildup of excess K+ ions in the parietal cell cytosol prevented?

Ans: A K+ channel on the apical surface of the parietal cell removes the excess K +. The outcome of transepithelial transport is the summation of processes mediated by a number of transport proteins.

Question Type: Essay Chapter: 11 Application Difficulty: Moderate

12

Cellular Energetics

Section 12.1 1. is the carbon-containing compound produced by glycolysis. a. Pyruvate b. Glucose c. CO2 d. Acetyl CoA Ans: a Question Type: Multiple choice Section 12.1 Blooms: Remembering

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15 - 111 Difficulty: Easy 2. In the glycolytic pathway, which of the following are allosterically controlled enzymes? a. triose phosphate isomerase b. phosphofructokinase-1 c. enolase d. none of the above Ans: b.

Question Type: Multiple choice Section 12.1 Blooms: Understanding Difficulty: Easy 3. Yeasts use fermentation to synthesize ATP: a. in the absence of oxygen. b. because fermentation produces more ATP than oxidative phosphorylation. c. because yeasts lack mitochondria. d. to produce lactic acid.

Ans: a Question Type: Multiple choice Section 12.1 Blooms: Understanding Difficulty: Moderate 4. What is the role of substrate-level phosphorylation in glycolysis?

Ans: During glycolysis, substrate level phosphorylation is used to synthesize ATP. This process occurs twice during glycolysis and involves the transfer of a high-energy phosphate group (from 1,3-bisphosphate or phosphoenolpyruvate) to ADP. Question Type: Essay Section 12.1 Applying Difficulty: Moderate 5. During prolonged exercise, oxygen is scarce in muscle tissue. Under these conditions, muscle cells convert pyruvate to two molecules of lactic acid. What happens to the lactic acid that is generated in this way?

Ans: Lactic acid is secreted from muscle cells into the bloodstream. Some is taken up from the bloodstream by the liver, where it is either reoxidized to pyruvate then metabolized to generate energy and CO2 or it is converted back to glucose and stored as glycogen in the liver. Some of the lactic acid is metabolized to CO 2 in the heart.

Question Type: Essay Section 12.1 Applying Difficulty: Difficult

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15 - 112 6. In the overall reaction for cellular respiration, glucose is: a. oxidized to CO2. b. reduced to CO2. c. oxidized to O2. d. reduced to O2.

Ans: a Question Type: Multiple choice Section 12.1 Blooms: Remembering Difficulty: Easy 7. Phosphofructokinase is active in the glycolytic pathway when the levels of ATP are high in the cell because a. less; ATP is an allosteric inhibitor b. less; ATP is a competitive inhibitor c: more; ATP stimulates phosphofructokinase through allosteric interactions d. more; lower ADP levels activate phosphofructokinase

.

Ans: a Question Type: Multiple choice Section 12.1 Blooms: Remembering Difficulty: Easy 8. In the absence of oxygen, NAD+ is recovered by each glucose molecule broken down. a. fermentation; two b. fermentation; four c. aerobic respiration; two d. anabolism; four

, which leads to a net production of

ATP molecules for

Ans: a Question Type: Multiple choice Section 12.1 Blooms: Remembering Difficulty: Easy Section 12.2 9. Which of the following is FALSE regarding mitochondrial structure? a. The inner mitochondrial membrane contains more surface area than the outer membrane and contains many of the proteins required for making ATP. b. The outer mitochondrial membrane contacts the cellular cytosol and the mitochondrial matrix. c. Mitochondria contain DNA in the matrix. d. The curved morphology of cristae are due to MICOS protein complexes.

Ans: b Question Type: Multiple choice Section 12.2 Blooms: Understanding Difficulty: Moderate 10. Which of the following health-related conditions are NOT related to mitochondrial function? a. Tay-Sachs disease b. Parkinson’s disease c. Kearns-Sayre syndrome

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15 - 113 d. aging Ans: a Question Type: Multiple choice Section 12.2 Blooms: Remembering Difficulty: Moderate Section 12.3 11. Electron transport from NADH and FADH2 to O2 occurs in the: a. mitochondrial matrix. b. cytosol. c. mitochondrial inner membrane. d. mitochondrial outer membrane.

Ans: c Question Type: Multiple choice Section 12.3 Blooms: Remembering Difficulty: Easy 12. Compared with glucose, oxidation of which of the following is more important in humans as a source of ATP? a. cellulose b. sucrose c. proteins d. fats Ans: d Question Type: Multiple choice Section 12.3 Blooms: Understanding Difficulty: Easy 13. The four stages of glucose oxidation are listed below. Place them in the correct order.

I. pyruvate oxidation to CO2 in the mitochondrion via a 2-carbon acetyl CoA intermediate (citric acid cycle) II. electron transport to generate a proton motive force III. conversion in the cytosol of one 6-carbon glucose molecule to two 3-carbon pyruvate molecules (pyruvate) (glycolysis) IV. ATP synthesis in the mitochondrion (oxidative phosphorylation) a. I, II, III, IV

b. II, III, I, IV

c. III, II, IV, I

d. III, I, II, IV

Ans: d Question Type: Multiple choice Section 12.3 Applying Difficulty: Moderate 14. What is the function of the malate–aspartate shuttle?

Ans: The malate–aspartate shuttle functions to deliver the electrons produced during glycolysis and carried by cytosolic NADH across the mitochondrial inner membrane to the matrix, where the electrons serve to reduce NAD+ to NADH in the matrix. This NADH in the matrix can then donate electrons to the electron transport chain.

Question Type: Essay Section 12.3 Applying Difficulty: Difficult

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15. The molecule that immediately enters the citric acid cycle is formed by which process? a. chemiosmosis b. glycolysis c. the light reactions d. pyruvate conversion Ans: d Question Type: Multiple choice Section 12.3 Blooms: Remembering Difficulty: Easy 16. The first step in the citric acid cycle occurs when acetyl CoA reacts with oxaloacetate to form: a. succinate. b. citrate. c. pyruvate d. cytochrome c Ans: b Question Type: Multiple choice Section 12.3 Blooms: Remembering Difficulty: Easy 17. Glucose is not the only energy-containing molecule that can enter cellular respiration pathways. Which food source is most likely to enter the citric acid cycle as fatty acids? a. fats b. carbohydrates c. amino acids d. DNA Ans: a Question Type: Multiple choice Section 12.3 Blooms: Understanding Difficulty: Easy Section 12.4 18. NADH-CoQ reductase and CoQH2–cytochrome c reductase each use the energy derived from electron transfer to transport the mitochondrial matrix. a. four protons into b. two protons into c. four protons out of d. two protons out of Ans: a Question Type: Multiple choice Section 12.4 Blooms: Remembering Difficulty: Easy 19. is a lipid soluble molecule that acts to shuttle electrons within the mitochondrial inner membrane. a. Cytochrome c b. NADH

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15 - 115 c. CoQ d. FADH2 Ans: c Question Type: Multiple choice Section 12.4 Blooms: Remembering Difficulty: Easy 20. In mitochondria, the proton-motive force is due largely to: a. a voltage gradient across the outer membrane. b. a voltage gradient across the inner membrane. c. a pH gradient across the outer membrane. d. a pH gradient across the inner membrane.

Ans: b Question Type: Multiple choice Section 12.4 Blooms: Understanding Difficulty: Moderate 21. Each cytochrome in the electron transport chain has a different reduction potential. What is the importance of these differences for electron transport?

Ans: The different reduction potential (or tendency to accept an electron) of the cytochromes in the electron transport chain allows these molecules to establish a unidirectional electron flow along the chain.

Question Type: Essay Section 12.4 Blooms: Understanding Difficulty: Moderate 22. During cellular respiration, the electron transport chain utilizes the energy produced from passing electrons from one molecule to the next to create: a. a proton gradient. b. NADH. c. CO2. d. an electron gradient.

Ans: b Question Type: Multiple choice Section 12.4 Blooms: Understanding Difficulty: Easy Section 12.5 23. Which of the following statement(s) regarding the origin of the mitochondria is(are) TRUE? a. A bacterium invaded and established a symbiotic relationship with a eukaryotic host cell. b. The outer mitochondrial membrane is derived from the bacterial plasma membrane. c. The globular F1 domain points toward the mitochondria’s intermembrane space. d. all of the above Ans: a Question Type: Multiple choice Section 12.5 Blooms: Understanding Difficulty: Moderate

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24. During ATP synthesis, protons move ―down‖ their electrochemical gradient through: a. the F0 complex of ATP synthase. b. the F1 complex of ATP synthase. c. a proton channel protein. d. CoQH2–cytochrome c reductase.

Ans: a Question Type: Multiple choice Section 12.5 Blooms: Remembering Difficulty: Easy 25. Transport of pyruvate into the mitochondrial matrix depends on energy provided by: a. ATP hydrolysis. b. ATP synthesis. c. a Na+ gradient. d. the proton-motive force.

Ans: d Question Type: Multiple choice Section 12.5 Blooms: Remembering Difficulty: Easy 26. Brown-fat mitochondria uncouple oxidative phosphorylation to produce: a. ADP. b. oxygen. c. heat. d. fat.

Ans: c Question Type: Multiple choice Section 12.5 Blooms: Remembering Difficulty: Easy 27. A major source of reactive oxygen species (ROS) in animal cells is: a. glycolysis. b. electron transport in the mitochondria. c. the reactions catalyzed by catalase and glutathione peroxidase. d. vitamin E and  lipoic acid.

Ans: b Question Type: Multiple choice Section 12.5 Blooms: Understanding Difficulty: Easy 28. ATP synthase is composed of two oligomeric proteins, F0 and F1. What is the function of each protein complex and where is each found in mitochondria?

Ans: F0 is a proton-channel protein and the F1 complex is an ATPase running in reverse. F0 is found in the mitochondrial inner membrane and F1 is associated with F0 on the matrix face of the inner membrane.

Question Type: Essay Section 12.5

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15 - 117 Applying Difficulty: Moderate Section 12.6 29. In chloroplasts, light absorption, electron transport, and ATP synthesis all occur: a. in the stroma. b. in the thylakoid lumen. c. in or on the thylakoid membrane. d. in or on the inner membrane.

Ans: c Question Type: Multiple choice Section 12.6 Blooms: Understanding Difficulty: Moderate 30. The principal pigment involved in photosynthesis is: a. carotenoid. b. chlorophyll a. c. chlorophyll b. d. heme.

Ans: b Question Type: Multiple choice Section 12.6 Blooms: Remembering Difficulty: Easy

31. In photosynthesis, all of the following reactions are directly dependent on light, except: a. carbon fixation. b. synthesis of ATP. c. electron transport. d. removal of electrons from H2O. Ans: a Question Type: Multiple choice Section 12.6 Applying Difficulty: Moderate 32. Which stages of photosynthesis can occur only in the light and which can also occur in the dark?

Ans: Of the four stages of photosynthesis, the first three (light absorption, electron transport, and ATP synthesis) can occur only when light is available. The last stage, carbon fixation, can occur whether or not light is available. Although carbon fixation can take place in the dark, the reactions involved are often turned off in the dark to conserve ATP for other cellular processes.

Question Type: Essay Section 12.6 Blooms: Understanding Difficulty: Moderate 33. What is the role of quinone in generating the charge separation needed to remove electrons from H2O for use in electron transport?

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15 - 118 Ans: Quinone Q is a strong reducing agent and accepts an electron from reaction-center chlorophyll when this chlorophyll is in a photon-induced excited state. This leaves the reaction-center chlorophyll in a strong oxidizing state that is powerful enough to remove electrons from H2O. Question Type: Essay Section 12.6 Blooms: Understanding Difficulty: Difficult

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15 - 119 34. Based on what you know about the action spectra of photosynthesis, irradiating a leaf with which of the following light types would result in the release of the greatest quantities of oxygen? a. red and orange light b. red and blue light c. green and blue light d. violet and yellow light Ans: b Question Type: Multiple choice Section 12.6 Applying Difficulty: Easy 35. The "tail" of chlorophyll is hydrophobic, which is important for: a. absorbing blue light. b. anchoring it in the thylakoid membrane. c. transferring electrons. d. giving plants their green color.

Ans: b Question Type: Multiple choice Section 12.6 Blooms: Understanding Difficulty: Moderate 36. Plastoquinone, like ubiquinone, can move freely within the membrane. This is important for its function as: a. an electron and proton shuttle. b. an electron carrier. c. a reaction center. d. a pump.

Ans: a Question Type: Multiple choice Section 12.6 Blooms: Understanding Difficulty: Moderate 37. What is the role of water in the light reactions of photosynthesis? a. provides energy b. accepts electrons c. accepts protons d. source of electrons Ans: d Question Type: Multiple choice Section 12.6 Blooms: Understanding Difficulty: Easy 38. Early investigators thought the oxygen produced by photosynthetic plants came from carbon dioxide. In fact, it comes from: a. water. b. air. c. electrons from NADPH. d. glucose.

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15 - 120 Ans: a Question Type: Multiple choice Secion 12.6 Blooms: Understanding Difficulty: Easy Section 12.7 39. Cyclic electron flow in the thylakoid membrane generates: a. oxygen. b. a proton-motive force. c. sulfur. d. NADPH. Ans: b Question Type: Multiple choice Section 12.7 Blooms: Understanding Difficulty: Easy 40. During photosynthesis, O2 is produced: a. on the stromal face of the thylakoid membrane. b. on the luminal face of the thylakoid membrane. c. throughout the stromal space. d. throughout the entire chloroplast.

Ans: b Question Type: Multiple choice Section 12.7 Blooms: Understanding Difficulty: Easy 41. During cyclic electron flow, electron transport: a. involves neither PSI nor PSII. b. takes place only in PSI. c. takes place only in PSII. d. cycles electrons back and forth between PSI and PSII. Ans: b Question Type: Multiple choice Section 12.7 Blooms: Understanding Difficulty: Moderate 42. All of the following statements describe the process of photorespiration, except: a. photorespiration consumes O2. b. photorespiration generates CO2. c. photorespiration generates substantial amounts of ATP. d. photorespiration competes with photosynthesis.

Ans: c Question Type: Multiple choice Section 12.7 Applying Difficulty: Moderate

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15 - 121 43. What molecule acts as an electron donor during photosynthesis in chloroplasts? What alternative is used by some photosynthetic bacteria (e.g., purple bacteria)?

Ans: H2O is the electron donor during photosynthesis in chloroplasts, whereas some photosynthetic bacteria (e.g., purple bacteria) can use H2S or H2 as an electron donor.

Question Type: Essay Section 12.7 Blooms: Understanding Difficulty: Moderate 44. Which components of PSII are responsible for producing the proton-motive force?

Ans: Several components of PSII contribute to the proton-motive force, albeit by different mechanisms. First, the removal of electrons from water by P680 generates protons in the thylakoid space. Next, the delivery of electrons by quinone Q to cytochrome bf is accompanied by the transfer of two protons from the stroma to the thylakoid lumen. Finally, protons may be transported from the stroma to the thylakoid space by cytochrome bf functioning in a Q cycle.

Question Type: Essay Section 12.7 Blooms: Understanding Difficulty: Difficult Section 12.8 45. The enzymes that catalyze the Calvin cycle are found in the: a. thylakoid lumen. b. phloem. c. cytosol. d. stromal space.

Ans: d Question Type: Multiple choice Section 12.8 Blooms: Remembering Difficulty: Easy 46. The fixation of CO2 into carbohydrates is catalyzed by: a. ribulose 1,5-bisphosphate carboxylase. b. thioredoxin. c. rubisco activase. d. 3-phosphoglycerate.

Ans: a Question Type: Multiple choice Section 12.8 Blooms: Remembering Difficulty: Easy 47. Plants use a. the xylem b. the phloem c. mesophyll cells d. root cells

to transport sucrose to all regions of the organism.

Ans: b Question Type: Multiple choice Section 12.8 Blooms: Remembering

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15 - 122 Difficulty: Easy 48. What is the source of the energy for carbon fixation?

Ans: The reactions that fix CO2 are powered by energy released by ATP hydrolysis and by the reducing agent NADPH. The ATP and NADPH were previously generated from the energy of absorbed photons of light.

Question Type: Essay Section 12.8 Blooms: Understanding Difficulty: Moderate 49. What is photorespiration? How is it related to photosynthesis?

Ans: Photorespiration competes with the process of photosynthesis. During photorespiration, O2 and ATP are consumed and CO2 is generated. Rubisco, which acts to fix CO2, also catalyzes photorespiration. Photorespiration is favored when stomata close to prevent moisture loss, and CO2 levels inside the leaf fall below the Km of rubisco for CO2.

Question Type: Essay Section 12.8 Blooms: Understanding Difficulty: Moderate 50. Explain how the rates of photosynthesis in plants like corn and sugarcane can be two to three times faster than the rates of photosynthesis in plants like wheat or rice.

Ans: Sugarcane and corn are C4 plants, whereas wheat and rice are C3 plants. C4 plants have evolved a two-step system that reduces the rate of photorespiration and enhances the rate of photosynthesis. This system works via a CO 2 shuttle that involves binding CO2 in mesophyll cells. The carbon dioxide is stored in the oxaloacetate. Oxaloacetate is a four-carbon molecule that gives the C4 plants their name. Oxaloacetate is converted to malate, which is transferred to bundle cells. Malate reacts to release CO2 in the bundle cells. This increases the CO2 concentration in bundle cells and, as a result, increases the rate of photosynthesis.

Question Type: Essay Section 12.8 Applying Difficulty: Difficult

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51. Why doesn't the Calvin cycle end after a three-carbon sugar is produced? a. because electrons are still excited in the photosystems b. because O2 must be produced to act as a final electron acceptor c. because RuBP must be generated for the cycle to continue d. because there are not enough carbon building blocks in the cell Ans: c Question Type: Multiple choice Section 12.8 Blooms: Understanding Difficulty: Easy 52. The Calvin cycle: a. converts glucose into energy in the form of NADH and ATP. b. uses glucose and electrons from NADPH to make ATP. c. uses ATP and glucose to make NADPH. d. uses electrons from NADPH and ATP to produce glucose.

Ans: d Question Type: Multiple choice Section 12.8 Blooms: Understanding Difficulty: Easy

13

Moving Proteins into Membranes and Organelles

Section 13.1 1. Which of the following proteins involved in cotranslational translocation of proteins into the ER membrane is NOT a GTPhydrolyzing protein? a.  subunit of the SRP receptor b. elongation factors in ribosome-mediated mRNA translation c. P54 subunit of SRP d. Sec61 translocon Ans: d Question Type: Multiple choice Section 13.1 Blooms: Analyzing Difficulty: Moderate 2. Protein insertion into the mammalian ER membrane is typically: a. cotranslational. b. post-translational. c. pretranslational.

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15 - 124 d. quasitranslational.

Ans: a Question Type: Multiple choice Section 13.1 Blooms: Understanding Difficulty: Moderate 3. Post-translational translocation of some secretory proteins in yeast is powered by: a. ATP hydrolysis by BiP. b. cAMP hydrolysis by cAMP phosphodiesterase. c. GTP hydrolysis EF-Tu. d. phospholipid hydrolysis by phospholipase C. Ans: a Question Type: Multiple choice Section 13.1 Blooms: Remembering Difficulty: Moderate 4. In the absence of targeting information, what is the default location of proteins synthesized on cytosolic ribosomes?

Ans: Proteins synthesized on cytosolic ribosomes that contain no information for targeting to organelles diffuse throughout the cytosol.

Question Type: Essay Section 13.1 Blooms: Understanding Difficulty: Easy 5. What are the general features of an N-terminal signal sequence that targets secretory proteins to the ER? Ans: N-terminal signal sequences targeting proteins to the ER are 16 to 30 amino acids in length and have a hydrophobic core of 6 to 12 amino acids. Preceding the core is one or more positively charged amino acids. Otherwise, N-terminal signal sequences have little in common.

Question Type: Essay Section 13.1 Blooms: Applying Difficulty: Moderate 6. In a cell-free protein synthesis system utilizing microsomes from fragmented ER, under which condition could you determine if the new protein was imported into the microsome? a. Ribosomes and mRNA are incubated with microsomes, then a protease is added and the results are analyzed. b. Ribosomes and mRNA are incubated with microsomes, then a protease and detergent are added and the results are analyzed. c. Ribosomes and mRNA are incubated with protease, then microsomes are added and the results are analyzed. d. Ribosomes and mRNA are incubated with microsomes, then detergent is added and the results are analyzed.

Ans: a Question Type: Multiple choice Section 13.1 Blooms: Analyzing Difficulty: Moderate 7. Which of the following is true about ER import?

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15 - 125 a. N-terminal signal sequences have been determined to be necessary for ER import because if they are added to sequences which are not normally targeted to the ER, they will end up in the ER. b. Signal recognition particles (SRPs) are needed during co-translational import. c. Ribosomal translation of a 6-12 amino acid sequence on the N-terminus will initiate the process through interactions with SRP. d. GTP hydrolysis by the Sec61 translocon causes ribosomal translation to begin again after it is halted by SRP binding.

Ans: c Question Type: Multiple choice Section 13.1 Blooms: Understanding Difficulty: Difficult Section 13.2 8. Type I membrane proteins have all of the following properties, except: a. cleavable signal sequence. b. internal signal-anchor sequence. c. internal stop-transfer sequence. d. N-out, C-in topology.

Ans: b Question Type: Multiple choice Section 13.2 Blooms: Understanding Difficulty: Moderate 9. GPI-anchoring serves a special function, especially in polarized epithelial cells, because this modification serves to target proteins to the: a. RER. b. Golgi. c. plasma membrane. d. nucleus.

Ans: c Question Type: Multiple choice Section 13.2 Blooms: Understanding Difficulty: Easy 10. The topology of membrane proteins can often be predicted by computer programs that identify a. glycosylation-specific b. hydrophilic c. hydrophobic d. cleavable signal sequence

topogenic segments.

Ans: c Question Type: Multiple choice Section 13.2 Blooms: Understanding Difficulty: Easy 11. In multipass membrane proteins synthesized in association with membrane-bounded ribosomes of the rough ER, signalanchor and stop-transfer anchor sequences alternate. What do these sequences do?

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15 - 126 Ans: Signal-anchor sequences direct insertion of internal segments of the protein into the ER membrane; stop-transfer sequences stop the transfer of the protein across the membrane. The alternation of the two produces a protein that loops in and out of the membrane multiple times.

Question Type: Essay Section 13.2 Blooms: Understanding Difficulty: Easy 12. Having misfolded soluble or secretory proteins in the RER contributes to what investigators call the ―traffic jam,‖ a scenario associated with a number of human diseases where the normal transport of proteins is blocked by these abnormal proteins and the inability of protein complexes to arrive at their correct site and function properly. Briefly describe how the cell overcomes this particular traffic jam by exporting the misfolded proteins out of the RER into the cytosol, where they are degraded by the proteasome.

Ans: Essentially, the misfolded proteins have N-linked carbohydrate chains that are trimmed by the enzyme α-mannosidase.

Once trimmed, these proteins are recognized by the lectin-like protein EDEM and/or OS-9, which targets the protein to an ER-associated degradation or ERAD complex, which serves as a type of channel needed to export the protein into the cytosol. Once in the cytosol, these proteins are subjected to enzymes that eventually target them to the proteasome for degradation.

Question Type: Essay Section 13.2 Blooms: Applying Difficulty: Difficult 13. A transmembrane receptor that functions at the cell membrane has an exoplasmic N-terminal sequence, a signal-anchor sequence, and a stop-transfer-anchor sequence. This protein was first inserted into the membrane where? a. at the plasma membrane b. in the cis-Golgi c. in the late endosome d. in the ER Ans: d Question Type: Multiple choice Section 13.2 Blooms: Understanding Difficulty: Moderate Section 13.3 14. Glycosylation, a post-translational modification of proteins, occurs in the: a. Golgi. b. proteasome. c. mitochondria. d. none of the above Ans: a Question Type: Multiple choice Section 13.3 Blooms: Remembering Difficulty: Easy 15. All the following proteins interact with exposed amino acids during protein folding in the ER, except: a. BiP. b. calnexin. c. PDI. d. prolyl isomerase.

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Ans: b Question Type: Multiple choice Section 13.3 Blooms: Remembering Difficulty: Moderate 16. Unassembled or misfolded proteins in the RER can be damaging to the physiology of a cell and therefore are transported to the cytosol where they are degraded. This transport process is referred to as: a. polyubiquitination. b. disulfide isomerization. c. dislocation. d. O-linked glycosylation.

Ans: c Question Type: Multiple choice Section 13.3 Blooms: Remembering Difficulty: Easy 17. What is the meaning of ―quality control in the ER?‖ Ans: ―Quality control within the ER‖ refers to the need for proteins to be properly modified and folded before they can exit from the ER and travel to the Golgi apparatus. Improperly modified and folded proteins are typically translocated into the cytosol for degradation.

Question Type: Essay Section 13.3 Blooms: Understanding Difficulty: Easy 18. Why are bacteria often a poor choice for the production of proteins for therapeutic purposes?

Ans: Typically, proteins used for therapeutic purposes are secreted proteins in animals; disulfide bonds stabilize their structures. Disulfide-bond formation occurs spontaneously in the lumen of the ER but not within bacteria. With this realization, animal cells became the preferred choice for the production of such proteins.

Question Type: Essay Section 13.3 Blooms: Applying Difficulty: Moderate Section 13.4 19. Sorting of proteins to mitochondria and chloroplasts is: a. cotranslational. b. post-translational. c. pretranslational. d. quasitranslational.

Ans: b Question Type: Multiple choice Section 13.4 Blooms: Remembering Difficulty: Moderate 20. Tom/Tim and Toc/Tic protein complexes are involved in: a. post-receptor recognition events in the cytosolic folding of proteins prior to import into mitochondria or chloroplasts.

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15 - 128 b. pre-proteasomal steps in tagging aged proteins for degradation. c. protein translocation into mitochondria and chloroplasts, respectively. d. resetting biological clocks following rounds of intense protein synthesis.

Ans: c Question Type: Multiple choice Section 13.4 Blooms: Applying Difficulty: Moderate 21. Protein sequences for targeting to mitochondria or chloroplasts are located at: a. the C-terminus of the precursor protein. b. amino acid position 173 in most mitochondrial and chloroplast proteins. c. the N-terminus of the precursor protein. d. the second and third answers are correct Ans: c Question Type: Multiple choice Section 13.4 Blooms: Remembering Difficulty: Moderate 22. Protein import into the mitochondrial matrix is supported by energy input from: a. ATP hydrolysis by chaperone proteins in the cytosol. b. ATP hydrolysis by chaperone proteins in the mitochondrial matrix. c. the proton-motive force across the inner mitochondrial membrane. d. all of the above Ans: d Question Type: Multiple choice Section 13.4 Blooms: Understanding Difficulty: Moderate 23. During in vitro translation of mitochondrially targeted proteins, when must mitochondria be added for import of proteins synthesized on cytosolic ribosomes?

Ans: Proteins are imported into mitochondria post-translationally. Therefore, although mitochondria can be added during the translation process for import to occur, cotranslational presence is not a requirement as it is for import into the ER. The mitochondria can be added post-translationally.

Question Type: Essay Section 13.4 Blooms: Applying Difficulty: Moderate 24. How are proteins imported into the thylakoids of chloroplasts?

Ans: For cytosolically synthesized proteins targeted to chloroplast thylakoids, multiple N-terminal uptake-targeting sequences are required. These act sequentially with the N-terminus, most targeting sequences being removed in the chloroplast stroma to expose the next targeting sequence. Four different pathways are known for the import of proteins from the chloroplast stroma into thylakoids. Three are for proteins imported from the cytosol and one is for proteins made in the chloroplast stroma. All pathways are variations of those used for export of proteins by bacteria. Examples of proteins homologous between bacteria and chloroplasts have been identified.

Question Type: Essay Section 13.4 Blooms: Applying

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15 - 129 Difficulty: Difficult 25. A polypeptide chain contains an amphipathic helix, with arginine and lysine residues on one side and hydrophobic residues on the other. It will likely enter: a. the peroxisome. b. the ER. c. the lysosome. d. the mitochondria.

Ans: d Question Type: Multiple choice Section 13.4 Blooms: Remembering Difficulty: Moderate 26. In a cell that lacks cytosolic Hsc70: a. import of proteins into the mitochondrial matrix would be diminished. b. proteins in the ER would not fold properly. c. the peroxisome would contain more catalase. d. generation of ATP from the electron transport chain would happen at the plasma membrane.

Ans: a Question Type: Multiple choice Section 13.4 Blooms: Applying Difficulty: Moderate 27. Which of the following components of mitochondrial import are NOT required for a sequence containing a matrixtargeting sequence and an intermembrane-space-targeting sequence? a. Tom b. Tim c. Oxa1 d. membrane-bound protease Ans: c Question Type: Multiple choice Section 13.4 Blooms: Remembering Difficulty: Moderate Section 13.5 28. Many peroxisomal matrix proteins are imported as: a. folded proteins. b. nascent chains in the process of completing their elongation. c. protein fragments that are spliced together within the peroxisome. d. unfolded proteins.

Ans: a Question Type: Multiple choice Section 13.5 Blooms: Remembering Difficulty: Easy 29. PTS1- and PTS2-bearing matrix proteins are targeted to: a. a common cytosolic receptor.

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15 - 130 b. a common import receptor and translocation machinery on the peroxisomal membrane. c. a common receptor on the nuclear pore that catalyzes entry into the nucleus via pore targeting sequences. d. a common receptor protein within the peroxisomal matrix that activates protein processing for PTS1- and PTS2-bearing proteins.

Ans: b Question Type: Multiple choice Section 13.5 Blooms: Remembering Difficulty: Moderate 30. Unlike mitochondria and chloroplasts, peroxisomes can arise preexisting organelles. a. as condensate b. as dispersions c. de novo d. as vaporware

from precursor membranes, as well as by division of

Ans: c Question Type: Multiple choice Section 13.5 Blooms: Understanding Difficulty: Easy 31. To what extent do peroxisomal matrix protein import and peroxisomal membrane protein import share the same machinery?

Ans: The fact that mutated cells giving rise to Zellweger syndrome, a defect in peroxisomal matrix protein import, still form peroxisomal membranes (peroxisomal ghosts) with the normal composition of peroxisomal membrane proteins strongly indicates that the import machinery for membrane proteins is very different from that for matrix proteins. This situation is different from that for other organelles such as the ER, mitochondria, and chloroplasts.

Question Type: Essay Section 13.5 Blooms: Applying Difficulty: Moderate 32. What is meant by de novo formation of peroxisomes?

Ans: This is the concept that peroxisomes can arise from nonperoxisomal membranes. The peroxisomal proteins Pex19, Pex3, and Pex16 are involved. The nature of the precursor membrane is unclear.

Question Type: Essay Section 13.5 Blooms: Understanding Difficulty: Moderate 33. Many peroxisomal matrix proteins are imported as: a. folded proteins. b. nascent chains in the process of completing their elongation. c. protein fragments that are spliced together within the peroxisome. d. unfolded proteins.

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Ans: a Question Type: Multiple choice Section 13.5 Blooms: Understanding Difficulty: Moderate Section 13.6 34. The nuclear pore complex allows for: a. passive diffusion of smaller molecules. b. import of proteins. c. active transport of very large molecules. d. all of the above Ans: d Question Type: Multiple choice Section 13.6 Blooms: Understanding Difficulty: Easy 35. During the import of proteins into the nucleus, the  importin subunit binds directly to: a. FG nucleoporins. b. Ran·GDP c. basic nuclear localization signals in cargo proteins. d. all of the above Ans: c Question Type: Multiple choice Section 13.6 Blooms: Understanding Difficulty: Moderate 36. Which type of RNA participates in nuclear export of mRNA? a. snRNA b. hnRNA c. tRNA d. rRNA Ans: b Question Type: Multiple choice Section 13.6 Remebering Difficulty: Moderate 37. Which of the following is present in the nuclear export sequence of PKI (an inhibitor of protein kinase A)? a. a proline-rich sequence b. a leucine-rich sequence c. a lysine-rich sequence d. all of the above Ans: b Question Type: Multiple choice Section 13.6 Remebering Difficulty: Moderate

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38. Transport of unspliced HIV mRNA from the nucleus to the cytoplasm of host cells is promoted by a virus-encoded protein named: a. Tat. b. Rev. c. nucleoplasmin. d. Ran. Ans: b Question Type: Multiple choice Section 13.6 Remebering Difficulty: Easy 39. How does Ran·GTP participate in the nuclear export of the HIV Rev protein?

Ans: In the nucleus, Ran·GTP binds to the nuclear export receptor exportin 1 and then to the leucine-rich nuclear export sequence (NES) in Rev. Exportin 1, in this trimolecular cargo complex, interacts transiently with FG repeats in FGnucleoporins, allowing it to traverse the nuclear pore complex (NPC). In the NPC, the cargo complex encounters Ran·GAP, stimulating Ran to hydrolyze GTP, which reduces its affinity for exportin 1. Exportin 1 subsequently loses its affinity for the NES, releasing Rev to the cytoplasm.

Question Type: Essay Section 13.6 Blooms: Applying Difficulty: Difficult 40. The nuclear pore complex (NPC) contains structures that form a gel-like matrix that allow small molecules to diffuse through, but require larger proteins to enter via importin or other nuclear chaperones. a. nuclear localization signals (NLS) b. nuclear lamina c. structural nucleoporin d. FG-nucleoporin Ans: d Question Type: Multiple choice Section 13.6 Blooms: Remembering Difficulty: Easy 41. During the process of nuclear import, a GEF works in the: a. cytoplasm to exchange GTP for GDP bound to Ran. b. cytoplasm to use GTP to release Ran from importin. c. nucleus to exchange GTP for GDP bound to Ran. d. nucleus to activate the intrinsic GTPase activity of Ran. Ans: c Question Type: Multiple choice Section 13.6 Blooms: Understanding Difficulty: Moderate

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14

Vesicular Traffic, Secretion, and Endocytosis

Section 14.1 1. The discovery of green fluorescent protein (GFP) has greatly facilitated living cell experiments because: a. GFP is green. b. GFP requires a jellyfish-specific cofactor. c. GFP sequences may be readily fused to those of other proteins. d. wild-type GFP folding is adapted to normal seawater temperatures, 15–25 °C. Ans: c Question Type: Multiple choice Section 14.1 Blooms: Applying Difficulty: Moderate 2. Endoglycosidase D is a useful reagent because it allows scientists to distinguish glycosylated proteins that: a. remain in the cis-Golgi. b. remain in the trans-Golgi. c. remain in the ER. d. get secreted.

Ans: c.

Question Type: Multiple choice Section 14.1 Blooms: Understanding Difficulty: Moderate 3. Yeast sec mutations: a. provide little evidence regarding the mechanism, necessitating other assays or information. b. invariably affect nonessential genes. c. affect protein transport into mitochondria but not chloroplasts. d. all fall into the same complementation class.

Ans: a Question Type: Multiple choice Section 14.1 Blooms: Analyzing Difficulty: Moderate 4. Cell-free transport assays: a. complement genetic approaches to the secretory pathway. b. often probe for changes in the glycosylation state of transported proteins. c. provide a means to test the effect of added purified proteins. d. all of the above Ans: d

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15 - 134 Question Type: Multiple choice Section 14.1 Blooms: Understanding Difficulty: Difficult 5. Why is VSV G protein one of the more useful tools in analyzing membrane trafficking?

Ans: The usefulness of VSV G protein resides mainly in its folding properties or, more specifically, those of the mutant VSV G protein, tsO45 G. When expressed in cells, this protein is temperature sensitive in its folding properties. At 39.5 °C, the protein folds abnormally in the endoplasmic reticulum (ER) and fails to exit from the ER. At 32 °C, the protein folds normally and exits the ER. Most important, misfolded VSV G will fold normally when the culture temperature is shifted to 32 °C. This permits a pulse-chase situation in which tsO45 G accumulates in the ER and then may be chased from the ER by a temperature shift in the presence of a protein-synthesis inhibitor to prevent the synthesis of new VSG G. The protein has two N-linked oligosaccharide chains. Therefore, its subcellular distribution may be inferred from its glycosylation state. As a protein, it may be fused to GFP. As a viral protein, high levels of VSV G expression are tolerated by cells.

Question Type: Essay Section 14.1 Blooms: Applying Difficulty: Difficult 6. How can the direction in which vesicles move in a VSV G–based, cell-free system for transport between Golgi compartments be distinguished?

Ans: Certainly, the results from such a cell-free system provide strong evidence for transfer between Golgi compartments and the role of various proteins in this transfer. In a cisternal progression context, the results are interpreted to mean that the Golgi enzyme is transported to the compartment containing the cargo protein, VSV G. In other words, a retrograde vesicle is formed by budding from the wild type Golgi, and this retrograde vesicle transports the resident Golgi enzyme Nacetylglucosaminyltransferase to the mutant Golgi containing VSV G protein. Using this assay system, vesicles can be isolated, and the composition of the transport vesicles can be characterized. The data indicate that the vesicles are enriched in Golgi enzymes, not VSV G protein. Such data are sufficient to establish that the vesicles are retrograde carriers.

Question Type: Essay Section 14.1 Blooms: Analyzing Difficulty: Difficult 7. The first step in the secretory pathway that should be inhibited by a non-functional mutant of NSF is: a. ER to Golgi transport. b. intra-Golgi transport. c. trans-Golgi network (TGN) transport to the plasma membrane. d. trans-Golgi network (TGN) transport to endosomes.

Ans: a Question Type: Multiple choice Section 14.1 Blooms: Applying Difficulty: Moderate 8. Given the wild type (normal) yeast on the left and the mutant yeast on the right, identify the defective phenotype for the mutant. a. entry into the ER b. fusion of transport vesicles from the ER to the Golgi c. budding from the Golgi to secretory vesicles d. fusion of secretory vesicles with the plasma membrane Ans: b

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15 - 135 Question Type: Multiple choice Section 14.1 Blooms: Understanding Difficulty: Moderate Section 14.2 9. Vesicle budding recruits proteins that are needed for subsequent: a. invagination of the vesicle into the ER lumen. b. selective vesicle targeting and fusion. c. shedding of integral membrane proteins into the cytosol. d. assembly of chromosome folding machinery.

Ans: b Question Type: Multiple choice Section 14.2 Blooms: Applying Difficulty: Moderate 10. The presence of clathrin mediates vesicular transport: a. from ER to cis-Golgi. b. trans-Golgi to endosome. c. nuclear membrane to endosome. d. plasma membrane to trans-Golgi.

Ans: b Question Type: Multiple choice Section 14.2 Blooms: Remembering Difficulty: Easy 11. Which of the following small GTPases are NOT involved in vesicle budding or docking? a. ARF b. rab1 c. ras d. Sar1p Ans: c Question Type: Multiple choice Section 14.2 Blooms: Remembering Difficulty: Moderate 12. How are different coat proteins recruited to different sites within the cell?

Ans: There are three known classes of coat proteins: clathrin, COPI, and COPII. Small GTPases recruit each to membranes.

Sar1 recruits COPII. Sar1 itself is activated by a guanine nucleotide exchange factor, Sec12, an ER integral membrane protein. ARF recruits both clathrin and COPII. How ARF, or different isoforms of ARF, and guanine nucleotide exchange factor(s) recruit different coat protein/adapters to different sites is not yet known.

Question Type: Essay Section 14.2 Blooms: Applying Difficulty: Difficult 13. How are SNARE proteins thought to bring about specific membrane fusion?

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15 - 136 Ans: SNARE proteins are thought to bring about specific membrane fusion by a pairing process. During vesicle budding, vSNARE proteins are recruited into the budding vesicle membrane. The cytosolic surface of the target membrane expresses exposed t-SNARE proteins. v-SNARE and t-SNARE proteins pair to form coiled-coil complexes, drawing the vesicle and target membranes very close together. Membrane fusion then occurs by a process that is not well understood.

Question Type: Essay Section 14.2 Blooms: Applying Difficulty: Moderate 14. If Sar1 is inserted into the membrane: a. it is bound to GTP and recruits COPII coat proteins. b. it is bound to GDP and recruits COPII coat proteins. c. it is bound to GTP and recruits cargo. d. it is not bound to GTP or GDP. Ans: a Question Type: Multiple choice Section 14.2 Blooms: Understanding Difficulty: Easy 15. Proteins that function in the ER will encounter which of the following? a. enzymatic modification of the ER resident soluble protein to add the KDEL sequence b. release from KDEL receptor binding in the Golgi due to a pH change c. anterograde transport in COPII coated vesicles d. sequestration in the ER by KDEL receptors Ans: c Question Type: Multiple choice Section 14.2 Blooms: Understanding Difficulty: Moderate 16. What phenotype would be observed in a cell containing a nonhydrolyzable form of ATP with respect to the vesicles of the secretory pathway? a. Secretion of peptides from the cell would be increased. b. Vesicles would be delivered to incorrect target membranes. c. Uncoated vesicles would accumulate. d. Coated vesicles would accumulate.

Ans: d Question Type: Multiple choice Section 14.2 Blooms: Analyzing Difficulty: Moderate Section 14.3 17. COPI coat proteins mediate a. anterograde b. enterograde c. retrograde d. siderograde

transport between the Golgi apparatus and other organelles.

Ans: c

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15 - 137 Question Type: Multiple choice Section 14.3 Blooms: Remembering Difficulty: Easy 18. Which of the following is a forward transport sorting signal acting at the ER? a. diacidic amino acid motif within the cytosolic domain b. KDEL (Lys-Asp-Glu-Leu) C-terminal sequence c. KKXX (Lys-Lys-X-X) within the cytosolic domain d. NPXY (Asn-Pro-X-Tyr) within the cytosolic domain Ans: a Question Type: Multiple choice Section 14.3 Blooms: Remembering Difficulty: Easy 19. Soluble and membrane proteins advance through the Golgi complex by: a. cisternal progression. b. stable continuities between Golgi cisterna. c. transient continuities between Golgi cisterna. d. vesicular transport.

Ans: a Question Type: Multiple choice Section 14.3 Blooms: Understanding Difficulty: Moderate 20. What is the role of vesicles in the early stages of the secretory pathway such as ER to Golgi trafficking?

Ans: There are two classes of coat proteins involved in the early stages of the secretory pathway. COPII coat proteins are involved in vesicle budding at the ER. The newly formed COPII-coated vesicles act as anterograde (forward) carriers. COPI coat proteins are involved in vesicle budding at the cis-Golgi network and within the Golgi. COPI vesicles mediate retrograde (backward) traffic that recycles soluble and membrane proteins to their site of residence.

Question Type: Essay Section 14.3 Blooms: Applying Difficulty: Easy 21. How are soluble, luminal ER proteins that ―leak‖ out of the ER retrieved to the ER? Ans: Soluble proteins of the ER have a Lys-Asp-Glu-Leu (KDEL) amino acid sequence in their C-terminus. This sequence is a retrieval signal. ER-resident proteins that are missorted to the cis-Golgi network bind to a KDEL receptor located in the cis-Golgi network. The missorted ER-resident protein KDEL receptor complex is recognized by COPI coat proteins and carried back to the ER via retrograde COPI vesicles.

Question Type: Essay Section 14.3 Blooms: Understanding Difficulty: Moderate Section 14.4 22. Protein sorting of anterograde cargo to different destinations within the Golgi complex occurs in the: a. cis-Golgi. b. medial-Golgi. c. trans-Golgi.

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15 - 138 d. trans-Golgi network.

Ans: d Question Type: Multiple choice Section 14.4 Blooms: Understanding Difficulty: Easy 23. In hepatocytes, the process by which apically destined proteins travel from the basolateral region across the cell before fusing with the apical membrane is called: a. exocytosis. b. transcytosis. c. endocytosis. d. none of the above Ans: b Question Type: Multiple choice Section 14.4 Blooms: Remembering Difficulty: Easy 24. The mannose 6-phosphate residue is important, as it is required to target soluble enzymes to the lysosome. The two enzymes responsible for attaching this residue onto these soluble enzymes reside in the: a. RER. b. cis-Golgi. c. medial-Golgi. d. trans-Golgi.

Ans: b Question Type: Multiple choice Section 14.4 Blooms: Remembering Difficulty: Moderate 25. In MDCK cells, which of the following is a sorting signal that allows proteins to be targeted to the apical membrane? a. GPI b. HA c. KDEL d. all of the above Ans: a Question Type: Multiple choice Section 14.4 Blooms: Remembering Difficulty: Easy 26. How are clathrin-coated vesicles pinched off?

Ans: Clathrin-coated vesicles are pinched off in a dynamin-mediated process. Dynamin is a cytosolic GTPase that forms a collar around the necks of clathrin-coated buds. It forces the neck membranes close together and membrane fusion (i.e., pinching off) occurs. The GTPase activity is necessary for this.

Question Type: Essay Section 14.4 Blooms: Understanding Difficulty: Moderate

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15 - 139 27. Professor George Palade’s elegant experiment to follow protein synthesis and trafficking, published nearly 60 years ago, provided us with a great deal of information and has been used as a tool by several investigators. If you had access to all the reagents needed to repeat the in vitro experiment, describe what you would need to do to see the progression of newly synthesized proteins and their transport in the cell.

Ans: Expose pancreatic tissue slices to radioactive leucine for a short period of time (referred to as a pulse). The radioactive pulse of leucine will then be replaced with nonradioactive leucine (the chase), and at varying time points the slices will be fixed and processed for electron microscopy and autoradiography. Autoradiographs of tissue sections will be compared to electron microscopy images to visualize the location of radioactive granules in the cell. At time points shortly after the chase begins, granules should be visualized in the ER, but at later ―post-chase‖ time points the granules would be in the Golgi or near the plasma membrane.

Question Type: Essay Section 14.4 Creating Difficulty: Moderate Section 14.5 28. The LDL receptor is a receptor for: a. apolipoprotein B. b. receptor-mediated endocytosis of LDL. c. cell signaling via activation of adenylate cyclase. d. apolipoprotein B and receptor-mediated endocytosis of LDL. Ans: d Question Type: Multiple choice Section 14.5 Blooms: Understanding Difficulty: Easy 29. Lipoproteins are effective in transporting lipid molecules in an aqueous environment because their surface layer is: a. hydrophobic. b. glycosidic. c. amphipathic. d. hydrophobic and glycosidic.

Ans: c Question Type: Multiple choice Section 14.5 Blooms: Understanding Difficulty: Moderate 30. Acidification of endosomes is important in dissociating: a. cholesterol from LDL. b. iron from transferrin. c. transferrin from the transferrin receptor. d. all of the above Ans: b Question Type: Multiple choice Section 14.5 Blooms: Understanding Difficulty: Moderate 31. Formation of the late endosome/multivesicular endosome occurs by mechanisms similar to those of:

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15 - 140 a. exocytosis of insulin in response to glucose levels in the blood. b. GCA protein–mediated budding at the trans-Golgi network. c. retrovirus budding from the plasma membrane. d. none of the above Ans: c Question Type: Multiple choice Section 14.5 Blooms: Understanding Difficulty: Moderate 32. Describe the types of mutations in the LDL receptor that would cause familial hypercholesterolemia.

Ans: A mutation that prevents the synthesis of the LDL receptor protein would block LDL uptake. Likewise, mutations that prevent the proper folding of the receptor in the endoplasmic reticulum would lead to its premature degradation. Mutations that interfere or reduce the receptor’s ability to bind LDL would also contribute to the disease. Finally, a mutation in the NPXY-sorting signal, which has no apparent effect on receptor-ligand interaction, would block the incorporation of the LDLLDL receptor into coated pits, thereby preventing internalization.

Question Type: Essay Section 14.5 Blooms: Applying Difficulty: Difficult 33. Which portion of a clathrin coat recognizes internalization signals such as Leu-Leu, Asn-Pro-X-Tyr, and in the cytosolic domain of cell-surface endocytic receptors?

Tyr-X-X-

Ans: Adapter proteins of the AP2 class recognize internalization signals located within the cytosolic domain of cell-surface endocytic proteins. Adapter proteins associate directly with the plasma membrane. The clathrin triskelion then associates with the adapter proteins. Overall, the complex forms a clathrin/AP2-coated vesicle.

Question Type: Essay Section 14.5 Blooms: Understanding Difficulty: Moderate 34. What is autophagy?

Ans: Autophagy (―eating oneself‖) is the delivery of bulk amounts of cytosol or entire organelles to lysosomes with subsequent degradation. Autophagy is often a regulated process and is typically induced in cells placed under conditions of starvation or other types of stress. The autophagic process begins with the formation of a cup-shaped membrane structure that envelops a portion of the cytosol or an organelle. The source of this membrane is not clear.

Question Type: Essay Section 14.5 Blooms: Understanding Difficulty: Moderate 35. An important molecule for generating fatty acids in the cell enters via receptor-mediated endocytosis. The complex formed between the receptor on the plasma membrane and the important molecule is stable only at neutral pH. Based on this knowledge, you would predict: a. a COPII-coated vesicle will be required for import. b. the important molecule enters the cell via a protein channel. c. both the molecule and the receptor are degraded to release the molecule from the receptor. d. the molecule is released from the receptor in the endosome.

Ans: c Question Type: Multiple choice Section 14.5

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15 - 141 Blooms: Applying Difficulty: Moderate 36. The topogenic sequence of transferrin must include: a. a signal sequence and an internal signal-anchor sequence. b. an internal signal-anchor sequence with positive charges N-terminal to the SA. c. an internal signal-anchor sequence with positive charges C-terminal to the SA. d. alternating signal-anchor sequences with stop-transfer sequences.

Ans: b Question Type: Multiple choice Section 14.5 Blooms: Applying Difficulty: Difficult 37. Receptor-mediated endocytosis of iron-carrying transferrin results in: a. release of iron and proteolysis of the transferrin protein for recycling inside the cell. b. transferrin/transferrin receptor complexes in the cytosol of the cell. c. release of iron from the endosome to the cytosol. d. vesicle formation at the plasma membrane mediated by COP proteins.

Ans: d Question Type: Multiple choice Section 14.5 Blooms: Understanding Difficulty: Moderate Section 14.6 38. The mannose 6-phosphate residue is important, as it is required to target soluble enzymes to the enzymes responsible for attaching this residue onto these soluble enzymes reside in the a. peroxisome; RER b. lysosome; cis-Golgi c. nucleus; medial-Golgi d. endosome; trans-Golgi

. The two .

Ans: b Question Type: Multiple choice Section 14.6 Blooms: Remembering Difficulty: Moderate 39. Which of the following is TRUE about lysosomes? a. They contain enzymes only capable of breaking down nucleic acids. b. They are bound by a single membrane but can engulf organelles containing double membranes. c. Proteins targeted to the lysosome are glycosylated in the ER and a specific mannose is phosphorylated. d. The final, functional state of lysosomal enzymes contains mannose-6-phosphate.

Ans: b Question Type: Multiple choice Section 14.6 Blooms: Understanding Difficulty: Moderate

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15 - 142

15 Signal Transduction Receptors

and G-Protein Coupled

Sections 15.1 1. In paracrine signaling, the signaling molecule: a. acts on cells in close proximity to the secreting cell. b. acts on target cells far away from the secreting cell. c. acts on the same cells that secreted the signaling molecule. d. is carried to the target cells via the circulation.

Ans: a Question Type: Multiple choice Section 15.1 Blooms: Understanding Difficulty: Easy 2. GTPases serve in many signal transduction pathways and the presence of GTP or GDP dictates whether the pathway is on or off, respectively. Which of the following statements is TRUE regarding guanine nucleotide exchange factors (GEF) and the role in these signaling pathways? a. They hydrolyze GTP into GDP and Pi. b. They decrease the GTPase activity of the G-protein. c. They catalyze the dissociation of GDP on the G-protein and promote the replacement of GTP. d. none of the above Ans: c Question Type: Multiple choice Section 15.1 Blooms: Understanding Difficulty: Moderate 3. Explain the differences between endocrine, paracrine, and autocrine signaling.

Ans: In endocrine signaling, signaling molecules are synthesized by one organ and act on distant target cells. In animals, the signaling molecule is carried to target cells via the circulation or other extracellular fluids. In paracrine signaling, the signaling molecules are released and affect only target cells in close proximity. In autocrine signaling, the cell that releases the signaling molecule is also affected by the released signaling molecule.

Question Type: Essay Section 15.1 Blooms: Understanding Difficulty: Easy

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16 - 143 Sections 15.2 4. Cell sensitivity to an external signal is determined by: a. kon. b. koff. c. Kd. d. the number of surface receptors.

Ans: d Question Type: Multiple choice Section 15.2 Blooms: Remembering Difficulty: Moderate 5. If [R] = the free receptor concentration and [L] = the free ligand concentration, Kd is: a. [R]/[L]. b. [L]/[R]. c. [R][L]/[RL]. d. [RL]/[R] [L].

Ans: c Question Type: Multiple choice Section 15.2 Blooms: Analyzing Difficulty: Moderate 6. Describe the differences between an agonist and an antagonist.

Ans: Agonists are molecules that mimic the function of a natural ligand by binding to the receptor and inducing the normal response. In contrast, an antagonist binds to the receptor but does not induce a response. An antagonist can block the binding of the natural ligand, thus reducing the normal physiological response to the ligand.

Question Type: Essay Section 15.2 Blooms: Applying Difficulty: Moderate 7. A particular antagonist for an epinephrine-receptor protein is under consideration as a new drug. What values would you use to measure how tightly the drug binds to the target protein compared with epinephrine binding? What technique would you use to measure drug binding?

Ans: The dissociation constant, Kd, is a measure of how tightly a ligand binds to its receptor. If the Kd for the drug is less than that for epinephrine, that would mean the antagonist binds more tightly to the receptor protein than does epinephrine. You could use a competitive binding assay to measure drug binding.

Question Type: Essay Section 15.2 Blooms: Analyzing Difficulty: Difficult 8. What experimental approach was used to identify functional domains of G protein–coupled receptors?

Ans: The functional domains of G protein–coupled receptors were determined by experiments using recombinant chimeric receptor proteins containing parts of the 2 and 2 adrenergic receptors. These chimeric receptors were tested for their ligand-binding specificity and their ability to activate or inhibit adenylyl cyclase. The results of these studies demonstrated that the cytosolic loop (C3 loop) between  helices 5 and 6 interacts with G proteins.

Question Type: Essay Section 15.2

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16 - 144 Blooms: Applying Difficulty: Difficult 9. Pull-down assays can be used to detect when small G proteins like Rac1 have been activated by various growth factors. If you were given two lysates, one having been treated with platelet-derived growth factor (PDGF) and the other treated with PDGF that had been heat-inactivated, briefly describe the pull-down assay and what you would expect to see following the assay.

Ans: Active Rac1 (i.e., in the GTP-bound form) binds specifically to the binding domain (PBD) of p21-activated protein kinase (PAK1). This PAK1-PBD bound to agarose beads can therefore be used as a tool to pull down active Rac GTP from protein lysates by centrifugation. To assay for levels of Rac GTP, PAK1-PDB-agarose is added to each of the lysates described above. The tubes are mixed and centrifuged, the proteins re-suspended and analyzed by SDS-polyacrylamide gel electrophoresis, and then transferred to a membrane and blotted against anti-Rac and anti-actin antibodies. The actin antibody is used to ensure equal amounts of protein were loaded in each lane. The blot probed with the anti-Rac antibody should only show a signal from the lysate of cells that were treated with PDGF, because PAK1-PDD agarose beads will recognize and pull down the GTP-bound, active form of Rac during centrifugation. Rac is not activated by the heatinactivated PDGF and therefore little to no signal should be seen in these lysates.

Question Type: Essay Section 15.2 Blooms: Analyzing Difficulty: Difficult 10. Pull-down assays using proteins that ONLY bind to the GTP-bound version of the small G-protein can be used to detect when small G proteins like Rac1 have been activated by various growth factors. You were given two lysates, one having been treated with platelet-derived growth factor (PDGF) and the other not treated with PDGF. This method is similar to performing co-IPs, so based on that knowledge, which of the following steps is erroneous? a. To assay for levels of active Rac1, PAK1-beads (the protein that only binds GTP-bound Rac1-conjugated to beads similar to Nickel beads) are added to each of the lysates described above and incubated on a rocking platform (on ice). b. The tubes are vortexed well, loading buffer is added, and the samples are heated to aid in protein denaturation. c. The samples are loaded and analyzed by SDS-polyacrylamide gel electrophoresis, then the proteins are transferred to a membrane. d. The membrane (a.k.a. blot) will be probed with the anti-Rac antibody.

Ans: b Question Type: Multiple choice Section 15.2 Blooms: Analyzing Difficulty: Moderate Sections 15.3 11. Of the components of a heterotrimeric G protein, which subunit(s) is(are) able to activate downstream responses? a. only the alpha subunit b. only the beta/gamma subunits c. only the delta subunit d. both the alpha subunit and the beta/gamma subunits Ans: d Question Type: Multiple choice Section 15.3 Blooms: Remembering Difficulty: Easy 12. In trimeric G proteins, GTP binds to: a. the  subunit. b. the  subunit.

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16 - 145 c. the  subunit. d. the activated trimer.

Ans: a Question Type: Multiple choice Section 15.3 Blooms: Remembering Difficulty: Easy 13. Which of the following is NOT a common intracellular second messenger? a. inositol 1,4,5-trisphosphate (IP3) b. 1,2 diacylglycerol (DAG) c. adenosine triphosphate (ATP) d. 3´–5´ cyclic guanine monophosphate (cGMP) Ans: c Question Type: Multiple choice Section 15.3 Blooms: Understanding Difficulty: Moderate 14. Phospholipase C is activated by: a. Gs. b. Gi. c. Gq. d. none of the above Ans: c Question Type: Multiple choice Section 15.3 Blooms: Remembering Difficulty: Easy 15. Which of the following general statement(s) about a G protein–coupled receptor is (are) TRUE? a. It contains 12 transmembrane domains. b. It is positioned with the N-terminus on the cytoplasmic face of the membrane. c. It is positioned with the C-terminus on the cytoplasmic face of the membrane. d. all of the above Ans: c Question Type: Multiple choice Section 15.3 Blooms: Applying Difficulty: Easy 16. All the following statement(s) about cholera toxin are TRUE, except: a. it chemically modifies the Gs protein. b. it is a G protein-coupled receptor. c. it prevents hydrolysis of bound GTP to GDP. d. it leads to continuous activation of adenylyl cyclase.

Ans: b Question Type: Multiple choice Section 15.3 Blooms: Remembering Difficulty: Moderate

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17. Phosphatidylinositol 4,5-bisphosphate (PIP2) is cleaved by phospholipase C into: a. 1,2-diacylglycerol (DAG). b. phosphatidylinositol (PI). c. inositol 1,4,5-trisphosphate (IP3). d. 1,2-diacylglycerol (DAG) and inositol 1,4,5-trisphosphate (IP3).

Ans: d Question Type: Multiple choice Section 15.3 Blooms: Remembering Difficulty: Easy 18. 1,2-diacylglycerol (DAG) and inositol 1,4,5-trisphosphate (IP3) are cleaved from phosphatidylinositol 4,5-bisphosphate (PIP2) by the enzyme: a. adenylyl cyclase. b. phosphodiesterase. c. phospholipase C. d. protein kinase C. Ans: c Question Type: Multiple choice Section 15.3 Blooms: Remembering Difficulty: Easy Sections 15.4 19. Summarize the steps in the cycling of GTPase switch proteins from active to inactive states.

Ans: GTPase switch proteins cycle from inactive to active states depending upon whether GTP (active) or GDP (inactive) is bound. Signal-induced receptor activation leads to the conversion of the GTPase switch protein from an inactive to an active state mediated by a guanine nucleotide exchange factor (GEF). GDP is released and replaced by GTP, resulting in a conformational change that allows the protein to activate downstream effector proteins. The intrinsic GTPase activity causes hydrolysis of GTP to GDP and the conversion back to the inactive state.

Question Type: Essay Section 15.3 Blooms: Applying Difficulty: Moderate 20. cGMP phosphodiesterase catalyzes the conversion of: a. cAMP to cGMP. b. cGMP to 5´-GMP. c. GTP to cGMP. d. cGMP to GDP. Ans: b Question Type: Multiple choice Section 15.4 Blooms: Applying Difficulty: Easy 21. Which enzyme plays a role in regulating rhodopsin-induced closing of cation channels? a. guanylyl cyclase b. adenylyl cyclase

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16 - 147 c. phosphodiesterase d guanylyl cyclase and phosphodiesterase Ans: d Question Type: Multiple choice Section 15.4 Blooms: Remembering Difficulty: Easy 22. Which of the following is a common step in the opening/closing of ion channels by acetylcholine and rhodopsin binding to their receptors? a. The G·GTP subunit dissociates from the G complex. b. The released G·GTP subunit interacts with the ion channel. c. The released G complex interacts with the ion channel. d. The released G·GTP subunit activates cGMP phosphodiesterase.

Ans: a Question Type: Multiple choice Section 15.4 Blooms: Applying Difficulty: Moderate 23. Rhodopsin, a light-sensitive GPCR whose role in vision is dependent on its phosphorylation status, is influenced in part by the protein arrestin. Which of the following is TRUE with regard to rhodopsin and vision? a. Rhodopsin phosphatase increases the degree of rhodopsin phosphorylation. b. Arrestin binds to the completely phosphorylated opsin to inhibit signaling. c. Rhodopsin activation promotes the opening of a cGMP-gated ion channel. d. none of the above Ans: b Question Type: Multiple choice Section 15.4 Blooms: Applying Difficulty: Moderate 24. Describe the proposed mechanism for the opening of K+ channels by cardiac muscarinic acetylcholine receptors.

Ans: Cardiac muscarinic acetylcholine receptors are linked to K+ channels by a trimeric G protein. Binding of acetylcholine to its receptor triggers the activation and release of the G i-GTP subunit from the G subunit. The released G subunit, rather than the Gi subunit, then binds to and opens the K+ channel. Activation is terminated by the hydrolysis of GTP to GDP and reformation of the ternary complex.

Question Type: Essay Section 15.4 Blooms: Applying Difficulty: Moderate 25. The G protein stimulated by light via the Rhodopsin receptor is a. Gαi; sodium channels b. Gαo; adenylyl cyclase c. Gαt; PDE d. Gαq; guanylate cyclase

, while the effector is

.

Ans: c Question Type: Multiple choice Section 15.4

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16 - 148 Blooms: Remembering Difficulty: Easy Sections 15.5 26. Which of the following statements about adenylyl cyclase stimulation/inhibition in adipose cells is TRUE? a. Prostaglandin E1 stimulates adenylyl cyclase. b. Glucagon inhibits adenylyl cyclase. c. Epinephrine stimulates adenylyl cyclase. d. Glucagon inhibits adenylyl cyclase and epinephrine stimulates adenylyl cyclase.

Ans: c Question Type: Multiple choice Section 15.5 Blooms: Understanding Difficulty: Moderate 27. Which of the following events occur during the epinephrine-stimulated conversion of glycogen to glucose-1-phosphate? a. activation of PKA by cAMP b. inhibition of glycogen synthase c. activation of glycogen phosphorylase d. all of the above Ans: d Question Type: Multiple choice Section 15.5 Blooms: Applying Difficulty: Moderate 28. The activity of -adrenergic receptors is regulated by: a. -adrenergic receptor kinase (BARK). b. calmodulin. c. -tubulin. d. all of the above Ans: a Question Type: Multiple choice Section 15.5 Blooms: Understanding Difficulty: Easy 29. Which of the following mechanisms can terminate an intracellular signaling pathway once the concentration of an external signal decreases? a. degradation of the second messenger b. desensitization of receptors c. deactivation of a signal transduction protein d. all of the above Ans: d Question Type: Multiple choice Section 15.5 Blooms: Understanding Difficulty: Easy 30. Describe experimental evidence supporting that intrinsic GTPase activity of the G subunit is important for terminating effector activation.

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16 - 149 Ans: The role of the intrinsic GTPase activity of the G subunit has been investigated using GTP analogs that can bind to the G subunit but cannot be hydrolyzed by its intrinsic GTPase activity. In these nonhydrolyzable compounds, the terminal phosphodiester (P–O–P) bond in GTP is replaced with P–CH2–P or P–NH–P bonds. In the presence of these nonhydrolyzable GTPs, the response of the effector protein is prolonged upon ligand-induced activation of the receptor. This is because the displacement of GDP with the modified GTP results in continuous activation of the effector protein. Because the bound GTP analog cannot be hydrolyzed to GDP, the effector remains in an active state permanently.

Question Type: Essay Section 15.5 Blooms: Applying Difficulty: Moderate 31. How does the body respond to decreases in blood glucose levels below about 5 mM? Ans: Reduced blood glucose levels induce  cells in the pancreas to release glucagon into the blood. Glucagon binds to the glucagon receptor in liver cell membranes. The glucagon receptor is coupled to a G s protein. Stimulation of the glucagon signal pathway in liver cells activates PKA. The overall effect is to inhibit glycogen synthesis and increase glycogenolysis.

Glycogen breakdown yields glucose 1-phosphate, which liver cells convert to glucose. Glucose is released into the blood, bringing the blood glucose level back up to normal.

Question Type: Essay Section 15.5 Blooms: Applying Difficulty: Moderate 32. When the regulatory domain of PKA is bound to cAMP: a. the released catalytic subunit travels to the nucleus where it can phosphorylate CREB. b. the catalytic subunit acts on CRE to stimulate gene transcription. c. the AKAP adaptor protein sequesters the catalytic subunit. d. PDE is inhibited to cause cAMP levels to rise.

Ans: a Question Type: Multiple choice Section 15.5 Blooms: Understanding Difficulty: Moderate 33. Muscarinic acetylcholine receptors are GPCRs that slow the rate of heart muscle contraction upon ligand binding/activation. Activation of this receptor leads to opening of potassium channels triggered by decreases in cAMP levels.

The muscarinic acetylcholine receptor likely couples to: a. Gαq. b. Gαs. c. Gαi. d. Gαo.

Ans: c Question Type: Multiple choice Section 15.5 Blooms: Applying Difficulty: Easy 34. Cholera toxin can make you sick through overactivation of a Gs pathway even if your body stops making the ligand associated with activation of the pathway. Knowing this, which of the following CANNOT be true about cholera toxin? a. It chemically modifies the Gs protein. b. Cholera toxin activates a ligand-activated G protein-coupled receptor. c. It prevents hydrolysis of bound GTP to GDP. d. It leads to continuous activation of adenylyl cyclase.

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Ans: b Question Type: Multiple choice Section 15.5 Blooms: Analyzing Difficulty: Moderate 35. Signal amplification is an important part of GPCR-mediated signaling. Which of the following steps do NOT directly amplify the signal? a. effector activation b. binding of second messenger to target protein/ion channel c. kinase acting on substrates d. second messenger generation Ans: b Question Type: Multiple choice Section 15.5 Blooms: Understanding Difficulty: Moderate 36. One form of receptor desensitization can be mediated by negative feedback involving phosphorylation of the receptor itself by a kinase activated downstream of a second messenger. The phosphorylated receptor likely: a. would be resensitized by a kinase. b. would activate heterotrimeric G proteins at a faster rate. c. would be unable to bind ligand. d. could be resistant to desensitization if Ser/Thr phosphatases were acting on the same receptor, Ans: d Question Type: Multiple choice Section 15.5 Blooms: Applying Difficulty: Moderate 37. Which type of experimental evidence shows that the intrinsic GTPase activity of the G subunit is important for terminating effector activation? a. A nonhydrolyzable GTP analog that can bind to the G subunit but cannot be hydrolyzed by the intrinsic GTPase, thereby activating the effector protein longer upon ligand-induced activation of the receptor. b. A nonhydrolyzable GTP analog causes displacement of GDP with the modified GTP resulting in continuous activation of the receptor because the bound GTP analog cannot be hydrolyzed to GDP. c. A dominant active (constant activity) GEF causes stimulation of the effector protein for longer upon ligand-induced activation of the receptor. d. A dominant negative (no activity) GEF causes stimulation of the effector protein for longer upon ligand-induced activation of the receptor.

Ans: a Question Type: Multiple choice Section 15.5 Blooms: Analyzing Difficulty: Difficult Sections 15.6 38. A mutation renders a Gαq subunit constitutively active (also known as dominant active). Which of the following effects might you observe in a cell with this mutation? a. PIP2 levels in the cell membrane diminish b. PKC activity decreases

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16 - 151 c. IP3 is sequestered at the membrane d. beta/gamma subunits are less active Ans: a Question Type: Multiple choice Section 15.6 Blooms: Analyzing Difficulty: Moderate 39. Diacylglycerol (DAG) formation aids in cell signaling by: a. acting as a substrate for PKC. b. acting as a soluble second messenger that can directly activate kinases near the nucleus. c. acting as a specific docking site to activate a downstream kinase. d. none of the above Ans: c Question Type: Multiple choice Section 15.6 Blooms: Understanding Difficulty: Easy 40. β-blockers inhibit an adrenergic receptor found on many cell types in the body, including heart cells. Patients who have arrhythmias (irregular heartbeats) most likely benefit from what downstream action of these drugs? a. increased frequency of IP3 release b. activation of PLC c. increased adenylyl cyclase activity d. decreasing cAMP-mediated effects on contraction rate Ans: d Question Type: Multiple choice Section 15.6 Blooms: Analyzing Difficulty: Moderate 41. Calmodulin: a. is a ubiquitous protein in eukaryotic cells. b. binds Ca2+ in a cooperative fashion. c. is a membrane-bound protein. d. is a ubiquitous protein in eukaryotic cells that binds Ca2+ in a cooperative fashion.

Ans: d Question Type: Multiple choice Section 15.6 Blooms: Understanding Difficulty: Moderate 42. DAG activates: a. PKC. b. PKA. c. PLC. d. calmodulin.

Ans: a Question Type: Multiple choice Section 15.6 Blooms: Remembering

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16 - 152 Difficulty: Easy 43. In muscle, glycogen phosphorylase kinase can be activated by nerve stimulation even in the absence of hormonal signals.

Nerve stimulation alone results in activation of glycogen phosphorylase kinase as a result of: a. elevated cytosolic Ca2+. b. phosphorylation by cAMP dependent PKA. c. both of the above d. none of the above Ans: a Question Type: Multiple choice Section 15.6 Blooms: Understanding Difficulty: Moderate 44. After a meal, when blood glucose rises, circulating insulin binds to insulin receptors on various cell types and reduces blood glucose levels by: a. fusion of intracellular vesicles containing GLUT4 glucose transporters with the plasma membrane. b. stimulation of the conversion of glucose to glycogen. c. inhibition of glucose synthesis from smaller molecules. d. all of the above Ans: d Question Type: Multiple choice Section 15.6 Blooms: Understanding Difficulty: Moderate 45. Describe the steps in the synthesis of 1,2-diacylglycerol (DAG) and inositol 1,4,5-trisphosphate (IP3) from phosphatidylinositol (PI).

Ans: Phosphatidylinositol (PI) is phosphorylated to form PI 4-phosphate (PIP), which is in turn phosphorylated to form PI 4,5-bisphosphate (PIP2). PIP2 is a substrate for phospholipase C, which cleaves PIP 2 into 1,2-diacylglycerol (DAG) and inositol 1,4,5-trisphosphate (IP3).

Question Type: Essay Section 15.5 Blooms: Applying Difficulty: Easy

16 Signaling Pathways That Control Gene Activity Section 16.1

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17 - 153 1. Latent TGF is converted to mature TGF by: a. dephosphorylation. b. phosphorylation. c. proteolysis. d. translocation. e. none of the above Ans: c Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy 2. A loss-of-function mutation in which of the following would inhibit TGF signaling? a. I-Smad b. R-Smad c. Ski d. SnoN e. all of the above Ans: b Question Type: Multiple choice Chapter: 16 Blooms: Analyzing Difficulty: Moderate 3. A gain-of-function mutation in which of the following would promote malignancy in cells whose proliferation is inhibited

by TGF? a. co-Smad b. I-Smad c. R-Smad d. all of the above e. none of the above Ans: b Question Type: Multiple choice Chapter: 16 Blooms: Analyzing Difficulty: Moderate 4. Which of the following receptors binds TGF? a. type I b. type II c. type III d. types I and II e. types II and III Ans: e Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Moderate 5. What feature allows TGF signaling molecules to be quickly mobilized?

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17 - 154 Ans: TGF signaling molecules are secreted and stored in the extracellular matrix in an inactive form. When needed, they can be rapidly activated by proteolytic digestion.

Question Type: Essay Chapter: 16 Blooms: Understanding Difficulty: Moderate 6. Ski and SnoN were originally identified as oncoproteins. Explain how constitutive expression of these proteins promotes cancer.

Ans: Ski and SnoN both inhibit TGF signaling by binding to nuclear Smad complexes and preventing these complexes from modulating expression of target genes. Because TGF signaling inhibits the proliferation of many cells, constitutive expression of Ski and SnoN would render cells insensitive to this antiproliferative signal, leading to unregulated cell division.

Question Type: Essay Chapter: 16 Blooms: Applying Difficulty: Moderate 7. TGF-β normally promotes: a. cell division. b. cell motility. c. tissue organization. d. all of the above Ans: c Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy Section 16.2 8. Which of the following contain(s) an SH2 domain? a. SHP1 b. SOCS c. STAT5 d. all of the above e. none of the above Ans: d Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy 9. Binding of erythropoietin to its extracellular receptor engages which of the following signaling pathways? a. JAK/STAT b. Ras/MAP kinase c. PI-3 kinase d. all of the above e. none of the above Ans: d Question Type: Multiple choice Chapter: 16 Blooms: Remembering

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17 - 155 Difficulty: Moderate 10. Compare and contrast the mechanisms by which SHP1 and SOCS proteins modulate erythropoietin signaling.

Ans: SHP1 and SOCS are both induced by negative feedback loops to down-regulate erythropoietin signaling. SHP1 is a phosphatase that contains SH2 domains that bind to activated erythropoietin receptors (EpoR). SHP1 bound to EpoR induces short-term down-regulation by dephosphorylating and inhibiting JAKs. SOCS proteins also contain SH2 domains that bind EpoR to prevent the binding of signaling molecules, bind the activation lip of JAKs to inhibit their activity, and recruit ubiquitin ligases to target JAKs for degradation. Degradation of JAKs leads to long-term down-regulation.

Question Type: Essay Chapter: 16 Blooms: Applying Difficulty: Difficult 11. Which of the following mechanisms is NOT used to terminate cytokine signaling and the JAK/STAT pathway? a. transcription of SOCS proteins b. dephosphorylation of phosphotyrosine residues c. heterodimerization of cytokine receptors d. receptor internalization Ans: c Question Type: Multiple choice Chapter: 16 Blooms: Understanding Difficulty: Easy Section 16.3 12. The EGF receptor is a(n): a. adapter protein. b. guanine nucleotide exchange protein. c. kinase. d. protease. e. none of the above Ans: c Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy 13. Binding of hormone to a receptor tyrosine kinase causes all of the following, except: a. dimerization of the receptor. b. autophosphorylation of the receptor. c. activation of Ras through an interaction with GRB2 and Sos. d. hydrolysis of GTP bound to Ras. Ans: d Question Type: Multiple choice Chapter: 16 Blooms: Understanding Difficulty: Moderate 14. Which of the following is a protein kinase? a. EGF receptor b. erythropoietin receptor

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17 - 156 c. STAT5 d. EGF receptor and erythropoietin receptor e. all of the above Ans: a Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Moderate Section 16.4 15. Ras is a(n): a. adapter protein. b. guanine nucleotide exchange factor. c. kinase. d. protease. e. none of the above Ans: e Question Type: Multiple choice Chapter: 16 Blooms: Understanding Difficulty: Easy 16. In which of the following double-mutant flies will R7 photoreceptors develop normally? a. constitutively active Raf and inactive MAPK b. constitutively active MAPK and inactive Ras c. constitutively active Ras and inactive Raf d. all of the above e. none of the above Ans: b Question Type: Multiple choice Chapter: 16 Blooms: Analyzing Difficulty: Moderate 17. Serum response factor (SRF) is phosphorylated by: a. MEK. b. MAP kinase. c. p90RSK. d. Raf. e. Ras. Ans: c Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy 18. Describe the role of adapter proteins in the activation of Ras by receptor tyrosine kinases.

Ans: Following dimerization and autophosphorylation of receptor tyrosine kinases, GRB2 and Sos proteins couple the receptor to the inactive Ras·GDP complex. Sos promotes dissociation of GDP from Ras, allowing GTP to bind. Sos acts as a

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17 - 157 guanine nucleotide exchange factor (GEF), which helps convert inactive Ras·GDP to active Ras·GTP. The Ras·GTP complex can then activate downstream effector molecules.

Question Type: Essay Chapter: 16 Blooms: Understanding Difficulty: Moderate 19. Describe the mechanism by which Ras is cycled from its active to inactive form.

Ans: Ras is cycled from its active to inactive form by the action of two proteins: guanine nucleotide exchange factor (GEF) and GTPase activating protein (GAP). When bound to GDP, Ras is in its inactive form. Binding of GEF to the Ras·GDP complex causes release of GDP and binding of GTP. Ras bound to GTP is in its active form. The active Ras·GTP complex possesses low intrinsic GTPase activity, which is elevated when GAP binds, causing conversion to inactive Ras·GDP. Question Type: Essay Chapter: 16 Blooms: Understanding Difficulty: Moderate 20. Describe the experimental approach used to determine the order of events in a signaling pathway.

Ans: The order of events in a signaling pathway can be determined by the analysis of mutants. For example, quiescent cells can be induced to proliferate in the absence of growth factors if they contain a constitutively active mutant Raf protein. In addition, cells that express a mutant defective Raf protein cannot be stimulated to proliferate uncontrollably by in the presence of a constitutively active RasD protein. These results show that Ras is upstream of Raf in the signaling pathway.

Question Type: Essay Chapter: 16 Blooms: Applying Difficulty: Moderate 21. How can multiple MAP kinase pathways be segregated when they share a common component?

Ans: Despite different MAP kinase pathways sharing common components, they are able to avoid cross talk between pathways. This is accomplished by formation of pathway-specific complexes assembled on molecular scaffolds. In this way, a specific component can be present in more than one MAP kinase pathway but will only result in one specific signaling pathway because it is complexed with a set of pathway-specific components.

Question Type: Essay Chapter: 16 Blooms: Understanding Difficulty: Moderate 22. Explain why, upon ligand binding, cell-surface receptors are often subjected to receptor-mediated endocytosis.

Ans: Removal of transmembrane receptors by receptor-mediated endocytosis is one mechanism to down-regulate receptor signaling so that the signal is not too robust and does not persist long after the signaling molecule is removed. Receptors that are endocytosed may eventually be recycled back to the cell surface, but this takes time. They may even be degraded in lysosomes, an even longer-term mechanism for down-regulation of signaling.

Question Type: Essay Chapter: 16 Blooms: Applying Difficulty: Moderate 23. Transcriptional activation downstream of the MAPK pathway involves: a. MEK. b. CREB phosphorylation. c. binding of transcription factors to the SRE of c-Fos. d. recruitment of Raf’s N-terminal regulatory domain.

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17 - 158 e. Ras activation upon Sos binding.

Ans: c Question Type: Multiple choice Chapter: 16 Blooms: Understanding Difficulty: Moderate 24. Before ligand binding, receptor tyrosine kinases: a. contain binding sites for SH2-domain containing proteins. b. contain activation lip tyrosines within the kinase active site. c. bind to GRB2 within the cytosolic domains. d. have excellent tyrosine kinase activity.

Ans: b Question Type: Multiple choice Chapter: 16 Blooms: Understanding Difficulty: Easy 25. Many PTB-containing proteins act as docking sites for multiple proteins. If these proteins are involved in the RTK signal transduction pathway, they most likely: a. contain GRB2 domains. b. are involved in autophosphorylation of the RTK. c. are phosphorylating the receptor and the docking protein. d. are adopting a different conformation (shape) to expand the pathways affected by ligand binding.

Ans: d Question Type: Multiple choice Chapter: 16 Blooms: Applying Difficulty: Easy 26. An SH2-containing protein contains a mutation that changes its binding pocket such that tyrosine and phosphotyrosine bind with equal affinity. As a result, MEK activity: a. decreases due to changes in Raf activation. b. decreases due to allosteric inhibition of SH2-domain binding. c. increases with ligand binding-induced dimerization. d. does not change with receptor dimerization and transautophosphorylation.

Ans: d Question Type: Multiple choice Chapter: 16 Blooms: Analyzing Difficulty: Moderate 27. Which of the following is NOT true about the role of adapter proteins in the activation of Ras by receptor tyrosine kinases? a. Following dimerization and autophosphorylation of receptor tyrosine kinases, GRB2 and Sos proteins couple to the receptor. b. GRB2 and Sos proteins couple the receptor to the inactive Ras·GDP complex. c. GRB2 interacts with Sos via its SH3 domain. d. Sos acts as a GAP, which helps convert Ras·GDP to active Ras·GTP. Ans: d Question Type: Multiple choice

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17 - 159 Chapter: 16 Blooms: Understanding Difficulty: Moderate 28. The order of events in a signaling pathway can be determined by the analysis of mutants. Cells that express a mutant defective Raf protein cannot be stimulated to proliferate uncontrollably by constitutively active RasD (dominant active). This indicates: a. quiescent cells can be induced to proliferate in the absence of growth factors if they contain a constitutively inactive mutant Raf protein. b. Raf is upstream of Ras in the signaling pathway. c. Ras is upstream of Raf in the signaling pathway. d. Ras mutants are recessive to Raf mutants.

Ans: c Question Type: Multiple choice Chapter: 16 Blooms: Analyzing Difficulty: Moderate 29. How can multiple MAP kinase pathways be segregated when they share a common component, like a downstream kinase? a. formation of pathway-specific complexes assembled on molecular scaffolds b. activation of different small G proteins c. differential activation of activation lip tyrosines d. phosphorylation of PTB-containing proteins Ans: a Question Type: Multiple choice Chapter: 16 Blooms: Understanding Difficulty: Easy 30. Which of the following explains why Ras is activated quickly by RTKs? a. RTKs phosphorylate Ras. b. Ras changes conformation upon ligand binding to prepare for activation. c. Ras binds phosphotyrosine residues on RTKs. d. Ras is maintained at the plasma membrane through a lipid-mediated attachment.

Ans: d Question Type: Multiple choice Chapter: 16 Blooms: Analyzing Difficulty: Moderate 31. Which protein stabilizes an intermediate in the Ras GTP hydrolysis reactions? a. Sos b. GRB2 c. RTK d. Raf Ans: a Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy

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17 - 160 Section 16.5 32. By what mechanism does PI-3 phosphate promote activation of protein kinase B (PKB)? a. recruiting PKB to the plasma membrane b. recruiting the activating kinase PDK1 to the plasma membrane c. releasing inhibition of the catalytic site by the PH domain d. the first and second answers are correct e. all of the above Ans: e Question Type: Multiple choice Chapter: 16 Blooms: Applying Difficulty: Moderate 33. Which of the following are enzyme pairs that catalyze opposite reactions? a. MEK and MAP kinase b. NF-B and I-B c. PI-3 kinase and PTEN phosphatase d. JAKs and STATs e. none of the above Ans: c Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Moderate 34. Which of the following signaling pathways can be activated by cytokines? a. JAK/STAT b. PI-3 kinase c. Ras/MAP kinase d. JAK/STAT and PI-3 kinase e. all of the above Ans: e Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy 35. How does activation of protein kinase B promote cell survival?

Ans: Protein kinase B phosphorylates and inhibits Bad and other pro-apoptotic proteins. Protein kinase B also phosphorylates the transcription factor Forkhead-1, which results in Forkhead-1 binding to 14-3-3 and becoming sequestered in the cytosol where it cannot activate pro-apoptotic genes.

Question Type: Essay Chapter: 16 Blooms: Understanding Difficulty: Moderate 36. Many kinases, including MAP kinase, protein kinase B, receptor tyrosine kinases, and JAKs, possess activation lips.

What is an activation lip? What modification of the activation lip is required for activation of these kinases?

Ans: The activation lip is a region in the protein kinase containing one or more amino acids that must be modified by phosphorylation to induce a conformational change that activates the kinase.

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17 - 161 Question Type: Essay Chapter: 16 Blooms: Understanding Difficulty: Easy Section 16.6 37. All genes regulated by PKA contain a cis-acting DNA sequence that binds to the phosphorylated form of a transcription factor called: a. -catenin. b. CREB. c. c-Jun. d. TCF. Ans: b Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy 38. Predict the consequences of a temperature-sensitive mutation in which phosphorylation of the -catenin protein is blocked. Above the permissive temperature: a. -catenin levels will increase and -catenin will be constitutively active. b. -catenin levels will increase and -catenin will be inactive. c. -catenin levels will decrease. d. -catenin will remain constant but cells will be unresponsive to Wnt signaling.

Ans: a Question Type: Multiple choice Chapter: 16 Blooms: Analyzing Difficulty: Easy 39. In the NF-B signaling pathway, which of the following molecules is downstream of I-B kinase? a. NF-B b. TAK1 c. TNF- d. NF-B and TAK1 e. all of the above Ans: a Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy 40. Using a luciferase reporter system in tissue culture cells, researchers from the University of California, San Francisco, found that ethanol stimulates transcription of genes in brain cells possibly involved in the adaptive responses to alcohol. This process was found to depend on PKA activation. By analogy to other PKA-dependent transcription activation pathways, describe a possible pathway for this transcription induction. What other proteins would be involved?

Ans: All genes regulated by PKA contain a cis-acting DNA sequence called the CRE (cAMP response element.) Activated PKA phosphorylates the CREB protein. Together with the co-activator, CBP/300, phosphorylated CREB protein can bind to CRE sequences of target genes to activate transcription.

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17 - 162 Reference for the research cited in this question: Asher, O., et al. 2002. Ethanol stimulates cAMP-responsive element (CRE)mediated transcription via CRE-binding protein and cAMP-dependent protein kinase, J. Pharmacol Exp Ther. 301(1): 66–70. http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&list_uids=11907158&dopt=Abstract Question Type: Multiple choice Chapter: 16 Blooms: Analyzing Difficulty: Difficult Section 16.7 41. Which of the following is cleaved by regulated intramembrane proteolysis (RIP)? a. IB b. Notch c. TGF d. IB and TGF e. Notch and TGF Ans: b Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy 42. NF-B signaling is important in: a. development. b. immunity. c. inflammation. d. immunity and inflammation. e. all of the above Ans: c Question Type: Multiple choice Chapter: 16 Blooms: Remembering Difficulty: Easy 43. What feature distinguishes the ligand Delta from other ligands such as EGF, TGF, and erythropoietin that bind transmembrane receptors?

Ans: The other ligands are diffusible molecules that can travel some distance from the cell that secretes the ligand to the cell that expresses the receptor. In contrast, Delta is a transmembrane protein and therefore must be expressed on a cell that is in direct contact with the target cell expressing the receptor Notch.

Question Type: Essay Chapter: 16 Blooms: Applying Difficulty: Moderate

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Cell Organization and Movement I: Microfilaments Section 17.1 1. The plasma membrane of eukaryotic cells is supported by: a. actin filaments. b. microtubules. c. lamins. d. intermediate filaments.

Ans: a Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 2. Three major groups of filament systems comprising the cytoskeleton are all composed of polymers of assembled subunits, which vary in thickness when assembled. Which is the correct order, from smallest to largest, of the filament systems? a. microtubules, microfilaments, intermediate filaments b. intermediate filaments, microtubules, microfilaments, c. microfilaments, microtubules, intermediate filaments d. none of the above Ans: d Question Type: Multiple choice Chapter: 17 Blooms: Applying Difficulty: Moderate Section 17.2 3. Actin-binding proteins that generate actin filament bundles: a. are long and flexible. b. bind only to the ends of actin filaments. c. can also bundle microtubules. d. are short and inflexible.

Ans: d Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Moderate 4. Decoration of actin filaments with myosin S1 is commonly used to: a. attach actin filaments to cell membranes. b. disassemble actin filaments. c. reveal the polarity of actin filaments. d. reveal the polarity of myosin filaments.

Ans: c Question Type: Multiple choice Chapter: 17 Blooms: Understanding

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18 - 164 Difficulty: Easy 5. Many actin cross-linking proteins contain: a. an ATP-binding cleft. b. a head domain. c. a CH domain. d. only one actin-binding site.

Ans: c Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Moderate 6. Within an actin filament, each actin subunit is surrounded by a. one b. two c. four d. eight

neighboring actin subunits.

Ans: c Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Moderate 7. How do actin filaments appear when viewed by negative stain electron microscopy?

Ans: When viewed by negative stain electron microscopy, an actin filament appears as a twisted string of beads with a diameter of 7–9 nm.

Question Type: Essay Chapter: 17 Blooms: Remembering Difficulty: Easy 8. What cellular components cause some actin filaments to form bundles and others to form networks?

Ans: Different actin-binding proteins (ABPs) cause actin filaments to cluster together in different ways. While all ABPs have two actin-binding sites, those ABPs that promote bundle formation tend to be short and inflexible, whereas the ABPs that promote network formation tend to be longer and more flexible.

Question Type: Essay Chapter: 17 Blooms: Remembering Difficulty: Moderate 9. At ATP-G-actin concentrations that are intermediate between the Cc for the (+) end and the Cc for the (–) end, be observed. The Cc needed for elongation at the (+) end is than that at the (–) end of actin microfilaments. a. growth; lower b. shrinking, higher c. treadmilling, lower d. none of the above

will

Ans: c Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Moderate

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Section 17.3 10. All of the following statements about actin assembly are correct EXCEPT: a. ATP-actin can assemble into filaments. b. actin subunits can treadmill through an actin filament. c. actin assembly can produce force for movement. d. actin (−) ends assemble more rapidly than actin (+) ends.

Ans: d Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Moderate 11. Gelsolin is activated by: a. ATP binding. b. phosphorylation. c. Ca2+ binding. d. dephosphorylation.

Ans: c Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 12. During treadmilling, actin subunits add: a. predominantly to filament (+) ends. b. predominantly to filament (−) ends. c. equally to both filament ends. d. along the length of filaments.

Ans: a Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Moderate 13. Which of the following proteins promotes actin assembly and is involved in signaling pathways controlling actin assembly at the plasma membrane? a. myosin b. profilin c. thymosin 4 d. filamin Ans: b Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 14. Listeria is a bacterial parasite that has evolved unique ways to enter animal cells and then use actin polymerization for its intracellular movement. Which of the following statements is true in regards to Listeria and actin polymerization? a. Listeria ActA protein binds and inactivates the Arp2/3 complex.

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

18 - 166 b. Listeria ActA binds VASP, which enhances ATP-actin assembly. c. Cofilin is necessary to accelerate the assembly of the (−) end of the filament. d. CapZ promotes the elongation of the filament.

Ans: b Question Type: Multiple choice Chapter: 17 Blooms: Applying Difficulty: Moderate 15. What actin-binding protein mediates gel-sol transitions, and how is this protein’s activity regulated?

Ans: Gel-sol transitions are mediated by gelsolin, an actin-binding protein that severs actin filaments and caps the (+) ends of the resulting fragments. Gelsolin is activated by Ca2+ binding.

Question Type: Essay Chapter: 17 Blooms: Understanding Difficulty: Easy 16. What is the function of thymosin 4?

Ans: Thymosin 4 is a G-actin sequestering protein that functions to maintain a relatively high G-actin concentration in cells.

Presumably, the cell can access this G-actin pool when new filament assembly is needed.

Question Type: Essay Chapter: 17 Blooms: Applying Difficulty: Moderate 17. A well-coordinated mechanism involving cell-surface receptors and the actin cytoskeleton allows leukocytes to bind and engulf invading bacteria. Briefly describe how bacteria are phagocytosed.

Ans: In a process termed opsonization, antibodies circulating in the blood recognize surface proteins on the bacteria. The Fc region of the bound antibodies is then recognized by specific receptors on the leukocyte cell surface, causing the cell to begin engulfing the bacterium. Binding also leads to the assembly of an actin network at the engulfment site where, together with myosin, force is generated to draw the bacteria into the cell. Eventually, the bacterium is completely engulfed, forming a phagosome in the leukocyte. The phagosome fuses to lysosomes, where acidic enzymes kill and degrade the bacterium.

Question Type: Essay Chapter: 17 Blooms: Understanding Difficulty: Moderate 18. Which of the following is a G-actin sequestering protein that allows cells to maintain a relatively high G-actin concentration in cells? a. thymosin 4 b. FH2 domain of formin c. profilin d. cofilin Ans: a Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Easy 19. Cofilin: a. cleaves actin by binding the ATP-actin in the filament.

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

18 - 167 b. cleaves actin by twisting adjacent F-actin monomers in the filament. c. recruits G-actin monomers to the elongating F-actin microfilament. d. promotes exchange of ADP for ATP on G actin.

Ans: b Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Moderate 20. If the activity of thymosin β4 was inhibited in fibroblasts, the overall effect would be: a. a decrease in cell locomotion because less actin would be recruited to the leading edge. b. an increase in cell locomotion because there would be a higher concentration of actin available. c. no change in cell locomotion because thymosin β4 doesn’t interact with Rho, Rac, or Cdc42. d. none of the above Ans: b Question Type: Multiple choice Chapter: 17 Blooms: Applying Difficulty: Moderate 21. The two proteins that play the most important role in actin microfilament elongation are: a. thymosin β4 and cofilin. b. cofilin and profilin. c. profilin and thymosin β4. d. CapZ and cofilin.

Ans: c Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Easy 22. What would occur if CapZ was inhibited in cells? a. increased organization of actin in stereocilia due to stabilized actin dynamics b. increased rates of polymerization due to reduced availability for ADP-G-actin c. increased steady-state treadmilling of F-actin d. faster movement of cells toward a chemical signal Ans: c Question Type: Multiple choice Chapter: 17 Blooms: Analyzing Difficulty: Moderate

Section 17.4 23. Which of the following proteins is involved in formation of actin bundles in microvilli by providing crosslinks between actin filaments? a  actinin b. cofilin c. fimbrin d. profilin

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Ans: c Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 24. Human erythrocytes depend on microfilament networks to provide strength and flexibility. Describe this network and explain how microfilaments are attached to the plasma membrane in human erythrocytes.

Ans: The microfilament network that underlies the plasma membrane in human erythrocytes is based on short actin filaments of about 28 subunits in length. These are held together in ―hubs‖ by six spectrin molecules. Overall, this creates a fishnet structure. The network is attached to the plasma membrane in two ways. The protein ankyrin links the microfilament network to the bicarbonate transporter in the plasma membrane, and band 4.1 protein links the network to glycophorin C. Question Type: Essay Chapter: 17 Blooms: Understanding Difficulty: Moderate 25. How do the different types of actin-binding proteins relate to the ability of actin to form bundles and networks? a. Proteins promoting bundle formation have one actin-binding site, whereas proteins promoting network formation have two actin-binding sites. b. Different types of actin bind different actin-binding proteins to form bundles or networks. c. Proteins that promote bundle formation bind ADP-actin, whereas proteins that promote network formation bind ATP-actin. d. Proteins promoting bundle formation tend to be short and inflexible, whereas proteins that promote network formation tend to be longer and more flexible.

Ans: d Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Moderate Section 17.5 26. Which region of myosin interacts with actin filaments? a. the head domain b. the rod domain c. the light chains d. the tail domain Ans: a Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 27. Which of the following properties is not shared by all myosins? a. the ability to bind ATP b. the ability to form dimers c. the ability to bind actin d. the presence of a head domain Ans: b Question Type: Multiple choice Chapter: 17

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

18 - 169 Blooms: Remembering Difficulty: Easy 28. All myosins move toward the (+) end of actin filaments EXCEPT: a. myosin I. b. myosin II. c. myosin V. d. myosin VI. Ans: d Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 29. Describe the functional properties of the head, neck, and tail domains of myosin.

Ans: The head domain of myosin is a specialized ATPase that couples the hydrolysis of ATP with motion. The activity of the myosin-ATPase is activated by binding to actin. The neck is an -helical region that is critical for converting small conformational changes in the head domain into large movements of the molecule and for regulating the activity of the head domain. The tail domain contains the binding sites for other molecules such as the thick filament in muscle.

Question Type: Essay Chapter: 17 Blooms: Understanding Difficulty: Moderate Section 17.6 30. In the operational model for movement of myosin along an actin filament, the power stroke occurs during: a. binding of ATP. b. hydrolysis of ATP. c. release of phosphate (Pi). d. release of ADP. Ans: c Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 31. What is the function of CapZ and tropomodulin in the sarcomere? a. to center myosin thick filaments b. to attach actin thin filaments to the Z disk c. to maintain a constant actin thin filament length d. to make contraction sensitive to Ca2+ Ans: c Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Moderate 32. In the budding yeast S. cerevisiae, which of the following is NOT transported into the bud by myosin V? a. peroxisome b. vacuole

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

18 - 170 c. mRNA d. nucleus Ans: d Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 33. Multinucleated cells may result from a defect in: a. myosin V. b. myosin I. c. stress fiber formation. d. myosin II. Ans: d Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Moderate 34. What are the functions of the myosin head domain and tail domain, respectively?

Ans: The myosin head domain is the portion of the motor protein that binds actin filaments and uses the energy from ATP hydrolysis to generate movement along the filament. All types of myosins have similar head domains. The myosin tail domain acts to recognize and bind cellular cargo. Each type of myosin has a different tail domain and thus can potentially move a different kind of cargo (although myosin II forms a bipolar filament).

Question Type: Essay Chapter: 17 Blooms: Understanding Difficulty: Moderate 35. What contractile structure is essential for cell division?

Ans: The contractile ring is a transient contractile bundle formed toward the end of mitosis. It is laterally attached to the cell membrane, and as the ring contracts, the plasma membrane is brought together between the spindle poles until the membrane fuses and two new cells are created.

Question Type: Essay Chapter: 17 Blooms: Remembering Difficulty: Moderate 45. Which of the following is NOT a function of myosin-powered movements? a. skeletal muscle contraction b. cytoplasmic streaming c. cytokinesis d. flagellum-mediated cell motility Ans: d Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy Section 17.7 36. Membrane extension during cell locomotion is driven by: a. myosin II.

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

18 - 171 b. actin depolymerization. c. contraction. d. actin polymerization.

Ans: d Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 37. Lamellipodia are located: a. at a moving cell’s trailing edge. b. at a moving cell’s leading edge. c. around the entire periphery of a nonmotile cell. d. throughout the cytosol of a moving cell.

Ans: b Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 38. The elastic Brownian ratchet model has been proposed to explain: a. membrane extension. b. focal adhesion formation. c. cell-body translocation. d. gel-sol transitions.

Ans: a Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Easy 39. Small G proteins, including Rho, Rac, and Cdc42, contribute to the coordinated movement and overall polarity of a migrating cell. Assuming that the cell is migrating in a left-to-right fashion, which of the following is correct? a. active Cdc42 at the leading edge of the cell and active Rho at the back of the cell b. active Rac at the back of the cell and active Cdc42 at the front of the cell c. active Cdc42 at the back of the cell and active Rac at the leading edge of the cell d. active Rac at the leading edge of the cell and active Rho at the back of the cell Ans: a Question Type: Multiple choice Chapter: 17 Blooms: Analyzing Difficulty: Moderate 40. How do cells grip the substrate to generate locomotion?

Ans: Locomoting cells use focal adhesions to get a grip on substrate material (e.g., the extracellular matrix). Focal adhesions are complexes of many different proteins, which form at the leading edge of moving cells just after membrane extension and dissolve (or are left behind) at the rear end of the moving cell.

Question Type: Essay Chapter: 17 Blooms: Understanding

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

18 - 172 Difficulty: Easy 41. How are actin-binding proteins involved in gel-sol transitions?

Ans: The transformation between sol and gel states results from the disassembly and reassembly of actin networks in the cytosol. Proteins such as cofilin sever actin filaments to form the sol state, whereas profilin promotes actin assembly, and actinin and filamin form the networks needed for the transition to the gel state.

Question Type: Essay Chapter: 17 Blooms: Understanding Difficulty: Easy 42. Like Listeria, other bacterial pathogens have also evolved to take advantage of actin-based cell motility systems in their hosts. For example, some pathogenic strains of E. coli make a cytotoxic factor (CNF1) that converts a specific glutamine residue on RhoGTPases to glutamate. This change blocks both the intrinsic and GAP-stimulated GTP hydrolysis activity of the Rho protein. Predict the effects of CNF1 on human epithelial cells in culture.

Ans: If GTP hydrolysis were blocked, Rho would remain constitutively active. Similar to the effects of dominant active Rho mutants, this would result in the formation of stress fibers in the tissue culture cells. (Stress fiber formation would allow the cultured epithelial cells to spread.)

Reference for the data cited in this question: Fiorentini, C., et al. 1997. Escherichia coli Cytotoxic Necrotizing Factor I (CNF1), a Toxin That Activates the Rho GTPase. J. Biol. Chem. 272(31):19532–19537.

Question Type: Essay Chapter: 17 Blooms: Analyzing Difficulty: Moderate 43. The correct order of events in cell locomotion is: a. rear focal adhesion, membrane protrusion, front focal adhesion, cell body translocation, de-adhesion. b. membrane protrusion, front focal adhesion, cell body translocation, de-adhesion, rear focal adhesion. c. front focal adhesion, cell body translocation, de-adhesion, rear focal adhesion, membrane protrusion. d. cell body translocation, de-adhesion, rear focal adhesion, membrane protrusion, front focal adhesion.

Ans: a Question Type: Multiple choice Chapter: 17 Blooms: Understanding Difficulty: Easy 44. In response to a chemotactic signal, a cell forms structures to aid in locomotion. In lamellipodia, active Rac will stimulate F-actin polymerization at the leading edge via , whereas actin that will form stress fibers will be recruited by formin downstream of activation of this small GTPase . a. Arp 2/3; Rho b. Arp 2/3; Cdc42 c. WASP; Rho d. Rho kinase; Cdc42 Ans: a Question Type: Multiple choice Chapter: 17 Blooms: Remembering Difficulty: Moderate 45. In a scratch wound assay, cells are treated with inhibitors of Rho kinase. What would be observed? a. The cells completely close the scratch, and only lamellipodia and filapodia are seen with actin staining. b. The distance the cells migrate into the scratch is small, and actin staining reveals stress fibers, lamellipodia, and filapodia. c. The cells are stimulated to undergo mitosis. d. The cells migrate into the scratch, but no stress fibers can be observed with actin staining.

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Ans: b Question Type: Multiple choice Chapter: 17 Blooms: Analyzing Difficulty: Difficult

18 Cell Organization and Movement II: Microtubules and Intermediate Filaments Section 18.1 1. A microtubule protofilament is formed by the: a. lateral association of only -tubulin subunits. b. head-to-tail association of only -tubulin subunits. c. lateral association of tubulin dimers. d. head-to-tail association of tubulin dimers.

Ans: d Question Type: Multiple choice Chapter 18.1 Blooms: Understanding Difficulty: Moderate 2. In cells, the -tubulin ring complex is found: a. in the hollow core of the microtubule. b. at the microtubule (−) end. c. at the microtubule (+) end. d. along the outer wall of the microtubule.

Ans: b Question Type: Multiple choice Chapter 18.1 Blooms: Understanding Difficulty: Moderate 3. In most cells, where do all microtubules originate?

Ans: In most cells, microtubules originate from the MTOC, and the minus ends of the microtubules nucleated by this structure usually remain associated with the MTOC. Question Type: Essay Chapter 18.1 Blooms: Remembering

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19 - 174 Difficulty: Easy 4. What is a microtubule protofilament?

Ans: A microtubule protofilament is a head-to-tail assembly of tubulin dimers. Thirteen such structures compose the cylindrical wall of the microtubule.

Question Type: Essay Chapter 18.1 Blooms: Understanding Difficulty: Easy 5. The alpha and beta tubulin proteins can bind: a. to ATP or ADP. b. to GTP or GDP. c. only to GDP. d. none of the above Ans: d Question Type: Multiple choice Chapter 18.1 Blooms: Remembering Difficulty: Easy 6. Where are microtubules observed to be present in different polarities? a. axons of nerve cells b. animal cells in interphase c. animal cells in mitosis d. dendrites of nerve cells Ans: d Question Type: Multiple choice Chapter 18.1 Blooms: Remembering Difficulty: Easy Section 18.2 7. At MTOCs, microtubule nucleation is facilitated by: a. centrioles. b. -tubulin. c. GDP-tubulin dimers. d. basal bodies.

Ans: b Question Type: Multiple choice Chapter 18.2 Blooms: Remembering Difficulty: Easy 8. Growing microtubule ends are normally stabilized by: a. a GDP cap. b. a GTP cap. c. phosphorylation of tubulin subunits. d. -tubulin.

Ans: b

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

19 - 175 Question Type: Multiple choice Chapter 18.2 Blooms: Understanding Difficulty: Easy 9. The drug taxol acts to: a. block microtubule assembly. b. stabilize microtubules against depolymerization. c. promote cell division. d. sever microtubules.

Ans: b Question Type: Multiple choice Chapter 18.2 Blooms: Remembering Difficulty: Moderate 10. What happens to a microtubule that loses its GTP cap?

Ans: Loss of the GTP cap will cause a microtubule to convert from growth to shortening. Shortening will continue until the microtubule disappears or a new cap forms.

Question Type: Essay Chapter 18.2 Blooms: Blooms: Applying Difficulty: Moderate 11. What are the effects of colchicine and taxol on cells?

Ans: Colchicine blocks assembly of the cell’s microtubules; taxol stabilizes the cell’s tubulin in polymer form. In either case, the normal microtubule dynamics will be disrupted, and the cells will be unable to divide.

Question Type: Essay Chapter 18.2 Blooms: Blooms: Applying Difficulty: Moderate 12. A drug that prevents microtubules from depolymerizing could be used to: a. stimulate cytokinesis. b. inhibit mitosis. c. promote cell division. d. treat Alzheimer’s disease.

Ans: b Question Type: Multiple choice Chapter 18.2 Blooms: Blooms: Applying Difficulty: Easy Section 18.3 13. Microtubule assembly requires: a. microtubule-associated proteins. b. incubation at 4C. c. ATP. d. a tubulin concentration in excess of the Cc. Ans: d

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

19 - 176 Question Type: Multiple choice Chapter 18.3 Blooms: Understanding Difficulty: Moderate 14. MAP2 and Tau are examples of microtubule: a. destabilizing proteins. b. motor proteins. c. nucleating proteins. d. stabilizing proteins.

Ans: d Question Type: Multiple choice Chapter 18.3 Blooms: Remembering Difficulty: Easy

15. The EB1 protein has several functions. Which of the following is/are true regarding EB1? a. It promotes microtubule growth by enhancing polymerization at the (+) end. b. Other microtubule plus-end tracking proteins use EB1 to “hitchhike” onto the growing microtubule. c. It can bind further back from the blunt end of microtubules. d. all of the above Ans: d Question Type: Multiple choice Chapter 18.3 Blooms: Understanding Difficulty: Difficult 16. Which of the following is NOT a way in which a microtubule switches from growing to shortening? a. loss of GTP cap b. treatment with colchicine c. binding of MAP2 d. binding of stathmin Ans: b Question Type: Multiple choice Chapter 18.3 Blooms: Understanding Difficulty: Easy Section 18.4 17. The region of a motor protein that interacts with the motor’s cellular cargo is the: a. head domain. b. tail domain. c. rod domain. d. light chains.

Ans: b Question Type: Multiple choice Chapter 18.4 Blooms: Remembering Difficulty: Easy

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19 - 177 18. All of the following statements describe kinesin-I EXCEPT: a. kinesin-I is a (−) end-directed motor. b. kinesin-I transports vesicles along microtubules. c. kinesin-I binds and hydrolyzes ATP to produce movement. d. kinesin-I is composed of two heavy chains and two light chains.

Ans: a Question Type: Multiple choice Chapter 18.4 Blooms: Blooms: Applying Difficulty: Moderate 19. The serves as a template for the unusual structure of axoneme microtubules. a. -tubulin ring complex b. pericentriolar material c. centrosome d. basal body Ans: d Question Type: Multiple choice Chapter 18.4 Blooms: Remembering Difficulty: Easy 20. For kinesin motors, the direction of movement along a microtubule is specified by the motor’s: a. motor (head) domain. b. neck region. c. stalk domain. d. tail domain.

Ans: b Question Type: Multiple choice Chapter 18.4 Blooms: Remembering Difficulty: Easy 21. Which of the following is true regarding the transport of cargo by cytoplasmic dynein? a. Dynactin is the protein that links dynein to microtubules. b. Transport is toward the (+) end of the microtubules. c. GTP binds to the head region of dynein. d. LIS1 associates with the head region of dynein to facilitate transport.

Ans: d Question Type: Multiple choice Chapter 18.4 Blooms: Understanding Difficulty: Moderate 22. Why would neuronal vesicles probably contain both kinesin and cytosolic dynein?

Ans: Neuronal vesicles may contain both motors to allow the vesicle to travel in either direction along a microtubule. Kinesin can carry the vesicle from the cell body to the axon terminus, while dynein can carry the vesicle back to the cell body.

Question Type: Essay Chapter 18.4 Blooms: Analyzing Difficulty: Moderate

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23. What effect will addition of AMP-PNP have on axonal transport?

Ans: AMP-PNP induces tight binding of kinesin (and most other microtubule motors) to microtubules; therefore, vesicle transport would cease because the motors would be locked onto the microtubules.

Question Type: Essay Chapter 18.4 Blooms: Analyzing Difficulty: Difficult Section 18.5 24. The force for axoneme bending is derived from the: a. sliding movement of central pair microtubules. b. contraction of central pair microtubules. c. sliding movement of outer doublet microtubules. d. contraction of outer doublet microtubules.

Ans: c Question Type: Multiple choice Chapter 18.5 Blooms: Understanding Difficulty: Easy 25. The primary cilium: a. is nonmotile because it lacks the ―central pair‖ of microtubules. b. is relatively susceptible to microtubule destabilizing drugs like colchicine. c. has no known role in humans. d. none of the above Ans: a Question Type: Multiple choice Chapter 18.5 Blooms: Understanding Difficulty: Moderate 26. What motor protein generates the force to cause axoneme bending?

Ans: Outer arm axonemal dynein generates the sliding of outer doublets relative to each other, and it is this movement that ultimately produces axoneme bending.

Question Type: Essay Chapter 18.5 Blooms: Remembering Difficulty: Moderate 27. What is the role of the basal body in generating axoneme structure?

Ans: The basal body is a bundle of nine triplet microtubules. Two tubules of each triplet (the A and the B tubules) act as a template for tubulin assembly and the generation of the A and B tubules of the axoneme.

Question Type: Essay Chapter 18.5 Blooms: Understanding Difficulty: Easy 28. Which of the following is NOT true about cilia? a. All cilia arise from nine sets of outer triplet microtubules (similar to centrioles). b. Cilia contain the same transport system as flagella, called IFT.

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19 - 179 c. All cells with cilia are motile because of the axonemal dynein motor. d. Primary cilia contain receptors that increase the cell’s responsiveness to its environment.

Ans: c Question Type: Multiple choice Chapter 18.5 Blooms: Understanding Difficulty: Moderate Section 18.6 29. Separation of spindle poles during spindle formation and anaphase B most likely depends on which of the following? a. (+) end-directed microtubule motors at the cell cortex b. (+) end-directed microtubule motors at the kinetochore c. (−) end-directed microtubule motors in the microtubule overlap zone d. (+) end-directed microtubule motors in the microtubule overlap zone Ans: d Question Type: Multiple choice Chapter 18.6 Blooms: Understanding Difficulty: Easy 30. Which of the following occurs during anaphase A? a. The spindle elongates. b. Kinetochores remain attached to shortening kinetochore microtubules. c. Chromosomes move to the spindle equator. d. The spindle poles move closer together.

Ans: b Question Type: Multiple choice Chapter 18.6 Blooms: Remembering Difficulty: Moderate 31. In the mitotic spindle, astral microtubules function to: a. connect the spindle poles. b. attach chromosomes to the spindle. c. carry out cytokinesis. d. anchor the spindle poles to the plasma membrane.

Ans: d Question Type: Multiple choice Chapter 18.6 Blooms: Understanding Difficulty: Moderate 32. Treadmilling through kinetochore microtubules can be observed by: a. simple fluorescence microscopy. b. optical trap microscopy. c. fluorescence speckle microscopy. d. electron microscopy.

Ans: c Question Type: Multiple choice Chapter 18.6

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

19 - 180 Blooms: Remembering Difficulty: Easy 33. Kinetochores assemble at the: a. centrosome. b. spindle pole. c. telomere. d. centromere.

Ans: d Question Type: Multiple choice Chapter 18.6 Blooms: Remembering Difficulty: Easy 34. Capture of microtubule (+) ends by chromosomes occurs during: a. metaphase. b. prometaphase. c. anaphase. d. telophase.

Ans: b Question Type: Multiple choice Chapter 18.6 Blooms: Remembering Difficulty: Easy 35. Poleward movement of chromosomes during anaphase A requires: a. microtubule polymerization. b. ATP. c. kinetochore motor proteins. d. BimC. Ans: c Question Type: Multiple choice Chapter 18.6 Blooms: Understanding Difficulty: Easy 36. What function do kinetochore-bound motor proteins perform during anaphase A? Ans: During anaphase A, kinetochore-bound motor proteins are thought to maintain kinetochore attachment to shortening kinetochore microtubules.

Question Type: Essay Chapter 18.6 Blooms: Understanding Difficulty: Easy 37. What role do astral microtubules play in spindle elongation?

Ans: Astral microtubules may serve as ―ropes‖ by which (−) end-directed microtubule motors at opposite sides of the cell cortex reel in the spindle poles and thereby produce spindle elongation.

Question Type: Essay Chapter 18.6 Blooms: Understanding Difficulty: Easy

Test Bank for Molecular Cell Biology 8th Edition by Harvey Lodish, Arnold Berk, Chris A. Kaiser, Monty Krieger, Anthony Bretscher, Hidde Ploegh, Angelika Amon, Kelsey C. Marti

19 - 181 38. During late anaphase and telophase, the microfilament-based contractile ring facilitates pinching the cell into two, but the ring must first be positioned equidistant between the two spindle poles before this occurs. Describe the mechanism whereby signals from the spindle direct the positioning of the contractile ring.

Ans: The chromosome passenger complex (CPC) is part of the signal responsible for regulating microtubule attachment to kinetochores during pro-metaphase. At anaphase the CPC leaves the centromeres and becomes associated with polar microtubules at the spindle center. Here, the CPC recruits the centralspindlin protein complex which includes a kinesin motor protein that concentrates at the middle of the spindle. Later, during anaphase B, centralspindlin recruits a RhoA exchange factor, which exchanges GDP for GTP on RhoA. RhoA, in this GTP-bound state, then activates a formin protein to drive the nucleation of actin into the microfilaments that make up the ring. Thus, the spindle position directly serves to delineate where the contractile ring must form to properly complete cytokinesis.

Question Type: Essay Chapter 18.6 Blooms: Blooms: Applying Difficulty: Difficult Section 18.7 39. Which of the following does NOT belong to the intermediate filament protein family? a. vimentin b. keratin c. laminin d. desmin Ans: c Question Type: Multiple choice Chapter 18.7 Blooms: Remembering Difficulty: Easy 40. During mitosis, the breakdown of the nuclear envelope depends on the disassembly of lamin filaments that form a meshwork supporting the membrane. How is that breakdown accomplished?

Ans: Phosphorylation of nuclear lamins by a cyclin-dependent kinase active in early mitosis induces their disassembly and prevents their reassembly until the phosphates are removed later in mitosis by specific phosphatases.

Question Type: Essay Chapter 18.7 Blooms: Understanding Difficulty: Moderate 41. The important role that intermediate filaments play in the epithelial cells of the skin is evident in which of the following? a. Individual keratin filaments span the cell membrane and directly connect one cell to the next. b. Knockout mice for a keratin gene like K14 exhibit stronger skin that can withstand abrasion. c. The polarity of keratin filaments allows them to withstand shear stress. d. Patients with mutations in keratin genes exhibit skin problems.

Ans: d Question Type: Multiple choice Chapter 18.7 Blooms: Understanding Difficulty: Moderate 42. Which of the following is true about intermediate filaments? a. They are named based on their location in cells between actin microfilaments and microtubules. b. All cells express the same class II cytoplasmic intermediate filament proteins. c. Staggered, antiparallel tetramers give intermediate filaments strength.

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19 - 182 d. Acidic and basic keratins provide dynamic paths on which organelles may travel.

Ans: c Question Type: Multiple choice Chapter 18.7 Blooms: Understanding Difficulty: Moderate Section 18.8 43. In studies of wound healing, it was noticed that when the cells at the edge are induced to polarize and move to fill the wound, the Golgi complex moves to the front of the nucleus toward the cell front. What is the purpose of this reorientation and how is it accomplished?

Ans: Cdc42 activation at the front of the cell binds to the polarity factor Par6. This initiates a series of interactions that recruits the dynein-dynactin complex to the front of the cell. The dynein-dynactin complex pulls microtubules to orient the centrosome toward the front of the cell. This microtubule reorientation serves to polarize the secretory pathway toward the front of the cell to allow for the delivery of adhesion molecules.

Question Type: Essay Chapter 18.8 Blooms: Blooms: Applying Difficulty: Difficult

19 Regulating the Eukaryotic Cell Cycle 1. During which stage of the cell cycle is the chromosome content of a mammalian liver cell 1n? a. G1 b. S c. G2 d none of the above Ans: d Question Type: Multiple choice Chapter 19.1 Blooms: Understanding Difficulty: Easy 2. What is the difference between a kinetochore and a centromere?

Ans: The kinetochore is a multiprotein complex present at the centromeric region of each sister chromatid. Microtubules from each spindle pole attach at their respective kinetochore. A centromere is actually part of the chromosome, where two sister chromatids are physically attached.

Question Type: Essay Chapter 19.1 Blooms: Applying

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Difficulty: Moderate 3. Human genes/proteins that regulate the cell cycle are most easily isolated by: a. biochemical purification. b. complementation of Xenopus mutants. c. complementation of yeast mutants. d. positional cloning.

Ans: c Question Type: Multiple choice Chapter 19.2 Blooms: Applying Difficulty: Moderate 4. Injection of an immature Xenopus oocyte with MPF induces: a. meiosis. b. oocyte maturation. c. synthesis of progesterone. d. meiosis and oocyte maturation.

Ans: d Question Type: Multiple choice Chapter 19.2 Blooms: Analyzing Difficulty: Easy 5. All the following statements about destruction of cyclins are true, except: a. Destruction is carried out by proteasomes. b. Destruction is preceded by polyubiquitination. c. Ubiquitination occurs at specific sites on target molecules. d. Ubiquitin targets Cdks.

Ans: d Question Type: Multiple choice Chapter 19.2 Blooms: Understanding Difficulty: Moderate

6. Mitotic cyclin levels in early Xenopus embryos are regulated by: a. phosphorylation. b. synthesis and degradation of their mRNAs. c. degradation by APC. d. degradation by MPF. Ans: c Question Type: Multiple choice Chapter 19.2 Blooms: Understanding Difficulty: Moderate 7. What is a cdc mutant? What protein is encoded by cdc2 in fission yeast?

Ans: The abbreviation ―cdc‖ stands for ―cell-division cycle.‖ Yeast cdc mutants are typically organisms with a temperaturesensitive mutation in a gene required for progression through a particular phase of the cell cycle. At the nonpermissive temperature, cdc mutants arrest at a particular place in the cell cycle. The cdc2 gene encodes the cyclin-dependent kinase

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(CDK) protein in the fission yeast Schizosaccharomyces pombe. In addition to encoding CDKs, cdc genes may encode cyclin subunits or activators of CDKs such as the phosphatase Cdc25.

Question Type: Essay Chapter 19.2 Blooms: Understanding Difficulty: Moderate 8. Which of the following is a target of the APC? a. Cdc25 b. cyclin B c. separase d. Cdc28 Ans: b Question Type: Multiple choice Chapter 19.3 Blooms: Remembering Difficulty: Moderate

9. Site-directed mutagenesis of Tyr-15 to Phe in cdc2: a. results in the wee phenotype. b. results in a phenotype similar to that caused by mutation in cdc25. c. increases the possible number of phosphate groups on Cdc2. d. prevents binding of Cdc13.

Ans: a Question Type: Multiple choice Chapter 19.3 Blooms: Applying Difficulty: Difficult 10. Which of the following S. pombe mutants is(are) larger than normal? a. cdc2D b. cdc25− c. wee1− d. cdc25− and wee1− Ans: b Question Type: Multiple choice Chapter 19.3 Blooms: Remembering Difficulty: Moderate 11. Exit from mitosis depends upon the degradation of: a. cohesion. b. condensin. c. cyclin B. d. securin.

Ans: c Question Type: Multiple choice Chapter 19.3 Blooms: Understanding Difficulty: Moderate

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12. Name three activities of G1 cyclin-CDKs that ensure progression through the cell cycle.

Ans: G1 cyclin-CDKs inactivate the APC, activate expression of S phase cyclins, and phosphorylate the S phase inhibitor, which targets it for degradation by the SCF. Question Type: Essay Chapter 19.3 Blooms: Applying Difficulty: Moderate

13. Investigators found that the cyclin B concentration rises and falls in synchrony with mitotic events and MPF activity.

However, correlation is not causality. How did they show that cyclin B was responsible for these cell-cycle events?

Ans: To show that cyclin B was responsible for the rise and fall in MPF activity synchronous with mitotic events, investigators treated Xenopus extracts with RNase, which abolished these events. Upon addition of only cyclin B mRNA, these events resumed.

Question Type: Essay Chapter 19.3 Blooms: Understanding Difficulty: Difficult 14. What are the biochemical activities of Wee1 and Cdc25 proteins? How do these activities regulate MPF? Ans: Wee1 is a kinase that phosphorylates Cdc2 on Tyr-15 and inhibits its activity. Cdc25 is a phosphatase that removes this phosphate and re-establishes activity in the presence of a phosphate group on Thr-161 in Cdc2.

Question Type: Essay Chapter 19.3 Blooms: Understanding Difficulty: Moderate 15. The S. cerevisiae homolog of Xenopus MPF is: a. Cdc13-Cdc2. b. Cdc25-Cdc2. c. Clb1-Cdc28. d. Cln3-Cdc28.

Ans: c Question Type: Multiple choice Chapter 19.4 Blooms: Remembering Difficulty: Moderate 16. Which of the following is a substrate of Cln3-Cdc28? a. SBF b. SPF c. MPF d. MBF Ans: a Question Type: Multiple choice Chapter 19.4 Blooms: Remembering Difficulty: Moderate 17. Which of the following is targeted by the SCF complex?

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a. Cdh1 b. cyclin B c. securin d. Sic1 Ans: d Question Type: Multiple choice Chapter 19.4 Blooms: Remembering Difficulty: Moderate 18. During G1, prereplication complexes: a. are phosphorylated by B-type cyclin-CDKs. b. are phosphorylated by DDK. c. bind to ORC. d. all of the above Ans: c Question Type: Multiple choice Chapter 19.4 Blooms: Understanding Difficulty: Moderate 19. Cyclin D/Cdk4 functions during: a. G1. b. S. c. G2. d. M. Ans: a Question Type: Multiple choice Chapter 19.4 Blooms: Remembering Difficulty: Easy 20. What is START? How does nutritional status determine whether S. cerevisiae cells will pass START? Ans: START defines the point in the cell cycle at which yeast cells become irreversibly committed to cell-cycle progression, even if nutrients are withdrawn. In S. cerevisiae, START occurs late in G1 and regulates entry into S phase. Entry into S phase depends upon synthesis of the cyclin Cln3. Translation of CLN3 mRNA is regulated by a short upstream open reading frame (ORF) that inhibits translation of the CLN3 ORF. When nutrients are abundant, there are sufficient translation initiation factors that allow the CLN3 ORF to be translated.

Question Type: Essay Chapter 19.4 Blooms: Applying Difficulty: Moderate 21. What stimulus is required for quiescent cells to re-enter the cell cycle? What events occur after stimulation?

Ans: Quiescent (G0) cells re-enter the cell cycle after stimulation by growth factors. Early response genes are transcribed first; these include genes encoding transcription factors such as fos and jun. These transcription factors induce expression of delayed-response genes, including the genes encoding cyclins and CDKs. In particular, expression of cyclin D is required for cyclin D-CDK4/6 phosphorylation of Rb, which leads to inactivation of Rb as a repressor of E2F. E2F is the transcription factor required for expression of S phase genes, including the S phase cyclins E and A. Question Type: Essay Chapter 19.4

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Blooms: Applying Difficulty: Difficult

22. After addition of serum to G0-arrested mammalian cells, two classes of genes are expressed: early-response genes and delayed-response genes. Explain why expression of delayed-response genes is blocked by inhibitors of protein synthesis but expression of early-response genes is not.

Ans: Transcription of early-response genes after addition of growth factors results in mobilization of signal-transduction cascades that activate preexisting transcription factors in the cytosol or nucleus. Many of the early-response genes encode transcription factors, such as c-Fos and c-Jun, which stimulate transcription of the delayed-response genes. Thus, activation of delayed response genes requires protein synthesis.

Question Type: Essay Chapter 19.4 Blooms: Applying Difficulty: Moderate 23. Breakdown of the nuclear envelope during mitosis is accomplished by CDK-dependent phosphorylation of lamins: a. A and C, but not B. b. A and B, but not C. c. B and C but not A. d. A, B, and C. Ans: d Question Type: Multiple choice Chapter 19.5 Blooms: Remembering Difficulty: Easy 24. The stable attachment of sister kinetochores to microtubules emanating from opposite spindle poles is called: a. merotelic attachment. b. syntelic attachment. c. amphitelic attachment. d. monotelic attachment.

Ans: c Question Type: Multiple choice Chapter 19.5 Blooms: Understanding Difficulty: Moderate 25. Describe the experiment that demonstrated that phosphorylation of nuclear lamins is required for nuclear envelope breakdown during mitosis.

Ans: Site-directed mutagenesis was used to generate nuclear lamin A that could not be phosphorylated by MPF by converting the serines that are normally phosphorylated to alanines. Expression vectors encoding wild-type or nonphosphorylatable nuclear lamin A were transfected into mammalian cells in culture. In the cells expressing wild-type lamin A, immunofluorescence revealed that lamin A was localized to the nuclear membrane during interphase and then diffusely throughout the cytoplasm during mitosis. Nonphosphorylatable nuclear lamin A remained localized to an intact nuclear envelope, even when cells were in mitosis, as determined by chromosome condensation.

Question Type: Essay Chapter 19.5 Blooms: Analyzing Difficulty: Moderate

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26. Separase initiates sister chromatid segregation at anaphase by cleaving: a. APC. b. Cdc20. c. cyclin B. d. Scc1.

Ans: d Question Type: Multiple choice Chapter 19.6 Blooms: Remembering Difficulty: Easy 27. Describe how Cdc14 triggers the exit from mitosis in the budding yeast, S. cerevisiae.

Ans: Cdc14 is a protein phosphatase, which at anaphase is activated by the mitotic exit network. Once activated, Cdc14 dephosphorylates APC/CCdh1 and Sic1, which promotes mitotic cyclin degradation and mitotic CDK inactivation, respectively.

Question Type: Essay Chapter 19.6 Blooms: Understanding Difficulty: Moderate 28. Which of the following inhibit(s) cyclin A/Cdk2 activity? a. INK4 b. p21CIP c. Rb d. INK4 and Rb Ans: b Question Type: Multiple choice Chapter 19.7 Blooms: Remembering Difficulty: Moderate 29. Failure of which cell-cycle checkpoint is most likely to result in nondysjunction? a. chromosome-segregation checkpoint b. DNA-damage checkpoint c. unreplicated DNA checkpoint d. dysreplicated DNA checkpoint Ans: a Question Type: Multiple choice Chapter 19.7 Blooms: Understanding Difficulty: Moderate 30. Chk1is activated by: a. ATR. b. Mad1. c. p53. d. Mad1 and p53.

Ans: a Question Type: Multiple choice Chapter 19.7

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Blooms: Remembering Difficulty: Easy 31. How does p53 function as a tumor-suppressor protein?

Ans: When p53 becomes stabilized in response to damaged DNA, it arrests the cell cycle at G 1, S, or G2 by inducing the transcription of p21, a stoichiometric inhibitor of all cyclin-CDK complexes. In some circumstances, p53 will induce the expression of genes that promote cell death by apoptosis. By either arresting the cell cycle or triggering cell death, p53 prevents propagation of mutations into the next cell generation. The formation of a tumor depends upon the accumulation of multiple somatic cell mutations.

Question Type: Essay Chapter 19.7 Blooms: Applying Difficulty: Moderate 32. How does the presence of unreplicated DNA prevent entry into mitosis?

Ans: Inter–S phase checkpoint control involves ATR and Chk1. The association of ATR with replication forks is thought to activate its protein kinase activity. Active ATR phosphorylates and activates the Chk1 kinase. Active Chk1 kinase phosphorylates and inactivates the Cdc25 phosphatase, which otherwise removes the inhibitory phosphate from mitotic CDKs. This prevents initiation of mitosis. ATR continues to initiate this protein-kinase cascade until all replication forks complete DNA replication and disassemble.

Question Type: Essay Chapter 19.7 Blooms: Applying Difficulty: Difficult 33. Centromeric Rec8 becomes cleaved during: a. meiosis I. b. meiosis II. c. mitosis. d. none of the above Ans: b Question Type: Multiple choice Chapter 19.8 Blooms: Remembering Difficulty: Moderate 34. How do chromosome movements differ between meiosis I and meiosis II? Is spindle composition the basis for this distinction?

Ans: During meiosis I, synapsed homologous chromosomes are segregated into the two daughter cells. During meiosis II, sister chromatids are segregated as in a mitotic division. Factors associated with the chromosomes rather than the spindles provide the basis for this distinction because a meiosis I spindle is capable of segregating meiosis II chromosomes properly, and a meiosis II spindle will segregate meiosis I chromosomes properly.

Question Type: Essay Chapter 19.8 Blooms: Applying Difficulty: Difficult 35. Compare cohesin function during mitosis and meiosis.

Ans: During mitosis, sister chromatids are initially associated with cohesin complexes along the full length of the chromatids. Cohesin complexes become restricted to the region of the centromere at metaphase. Separase cleaves the Scc1

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cohesin subunit, which allows sister chromatids to separate. In metaphase of meiosis I, crossing over between maternal and paternal chromatids produces synapsis of homologous parental chromosomes. The chromatids of each replicated chromosome are cross-linked by cohesin complexes along their entire length. Rec8, a meiosis-specific homolog of Scc1, is cleaved in chromosome arms but not in the centromere, allowing homologous chromosome pairs to segregate into daughter cells. Centromeric Rec8 is cleaved during meiosis II, allowing individual chromatids to segregate into daughter cells.

Question Type: Essay Chapter 19.8 Blooms: Analyzing Difficulty: Difficult

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New Questions 36. During mitosis, the breakdown of the nuclear envelope depends on the: a. disassembly following proteasomal degradation of intermediate filaments. b. disassembly of lamin filaments following phosphorylation by mitotic cyclin/cdk complexes. c. ubiquitination of mitotic cyclin proteins. d. lamin filament dephosphorylation by cdc14.

Ans: b Question Type: Multiple choice Chapter 19.5 Blooms: Understanding Difficulty: Moderate 37. Which of the following is NOT a way G1/S cyclin/cdks ensure progression through the cell cycle? a. inactivate APC via Cdh1 phosphorylation b. phosphorylate the S phase inhibitor (CKI) c. activate SCF E3 ligase d. activate expression of S-phase cyclins Ans: c Question Type: Multiple choice Chapter 19.3 Blooms: Understanding Difficulty: Moderate 38. DNA replication at each origin occurs only once during the cell cycle because of: a. specificity of the origin-recognition complex (ORC). b. S phase Cdk phosphorylating MCM helicase. c. MCM helicase loading by M phase cyclins. d. G1 cyclin/cdk activation of E2F. Ans: b Question Type: Multiple choice Chapter 19.4 Blooms: Understanding Difficulty: Easy 39. Separation of spindle poles during spindle formation and anaphase B most likely depends on which of the following? a. (+) end-directed microtubule motors at the cell cortex b. (+) end-directed microtubule motors at the kinetochore c. (−) end-directed microtubule motors in the microtubule overlap zone d. (+) end-directed microtubule motors in the microtubule overlap zone Ans: d Question Type: Multiple choice Chapter 19.6 Blooms: Understanding Difficulty: Moderate 40. Which of the following occurs during anaphase A? a. The spindle elongates. b. Kinetochores remain attached to shortening kinetochore microtubules. c. Chromosomes move to the spindle equator. d. The spindle poles move closer together.

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Ans: b Question Type: Multiple choice Chapter 19.6 Blooms: Understanding Difficulty: Easy 41. You perform an experiment where you prevent Cdc20 from binding to APC in a cell line that usually completes the cell cycle in 60 minutes. Your experiment prevents Cdc20 from binding to APC for 120 minutes, and then you fix and stain the DNA, securin, separase and cohesin proteins. Which of the following would NOT be found in the observed cell? a. The cell would contain condensed chromosomes aligned at the metaphase plate. b. Cohesins will be localized around the sister chromatids. c. Securin would be bound to separase (appear to co-localize). d. The DNA will be decondensed because the cell is in interphase.

Ans: d Question Type: Multiple choice Chapter 19.6 Analysis Difficulty: Moderate

20 Integrating Cells into Tissues Section 20.1 1. The functions of the extracellular matrix include: a. supporting differentiation. b. inducing morphogenesis. c. binding growth hormones. d. all of the above Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 2. The major families of cell surface adhesion molecules include: a. cadherins and selectins. b. integrins. c. the Ig-superfamily. d. all of the above

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21 - 193 Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 3. In a classic experiment, H. V. Wilson studied aggregation of mechanically dissociated individual sponge cells from two different species. He found that the cells of each species would adhere to one another but not to cells of the other species.

Describe the factors involved in this species-specific aggregation.

Ans: The sponge cells aggregate as a result of calcium-dependent species-specific interactions that involve homotypic proteoglycan aggregation factors in the extracellular matrix that bind to cells via surface receptors.

Question Type: Essay Chapter: 20 Blooms: Applying Difficulty: Moderate 4. Cadherin cellular adhesion molecules promote: a. collagen binding. b. multiadhesive matrix protein binding. c. cell-specific homophilic interactions. d. all of the above Ans: c Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Easy Section 20.2 5. Desmosomes: a. associate with actin filaments on the cytoplasmic side. b. contain the transmembrane proteins desmoglein and desmocollin. c. inhibit transfer of membrane proteins from the basolateral to the apical domain. d. contain integrins.

Ans: b Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 6. Which of the following statements best describes the difference between low-affinity integrins and high-affinity integrins? a. Many integrins can exist in two conformations: a low-affinity (bent) conformation and a high-affinity (straight) conformation. b. Dissociation of the  heterodimer converts many integrins from the low-affinity to the high-affinity state. c. Association of the  heterodimer converts many integrins from the low-affinity to the high-affinity state. d. Proteolytic cleavage of the C-terminal tails of the two subunits converts many integrins from the low-affinity to the highaffinity state.

Ans: a Question Type: Multiple choice Chapter: 20 Blooms: Applying

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21 - 194 Difficulty: Moderate 7. Vertebrate gap junctions are composed of: a. adherins. b. collagens. c. connexins. d. integrins.

Ans: c Question Type: Multiple choice Chapter: 20 Blooms: Remembering Difficulty: Easy 8. A tight junction is made up of which of the following proteins? a. tricellulin b. occludin c. claudin d. all of the above Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Remembering Difficulty: Moderate 9. EDTA is a divalent cation chelator. Propose an explanation for how EDTA promotes the dissociation of animal cell tissue?

2+

Ans: Much of cell-cell adhesion is mediated by cadherins. Cadherin interactions between cells are Ca 2+ chelates divalent cations, including Ca , leading to dissociation of cadherin cell-cell interactions.

Question Type: Essay Chapter: 20 Blooms: Applying Difficulty: Moderate

dependent. EDTA

10. What is the function of a gap junction?

Ans: Gap junctions are communicating junctions between cells that allow for the ready exchange of small molecules up to about 2000 daltons. Gap junctions are composed of connexin protein molecules, which interact between adjacent cells to form small amino acid–lined pores between the cells.

Question Type: Essay Chapter: 20 Blooms: Understanding Difficulty: Moderate 11. What are the cytoskeletal proteins associated with hemidesmosomes?

Ans: Hemidesmosomes are macromolecular complexes that are linked to intermediate filaments composed of keratin.

Question Type: Essay Chapter: 20 Blooms: Understanding Difficulty: Easy

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21 - 195 12. What is a polarized epithelial cell?

Ans: A polarized epithelial cell is one whose membrane is differentiated into two domains, the apical domain and the basolateral domain. These characteristics can be maintained even in tissue culture. For example, Madin-Darby canine kidney cells in culture demonstrate microvilli on their apical surfaces and tight junctions and desmosomes that separate the apical and basolateral domains.

Question Type: Essay Chapter: 20 Blooms: Understanding Difficulty: Moderate 13. Cellular responses to adhesion receptor signaling do NOT include: a. cell proliferation. b. cytoskeletal organization. c. gene transcription. d. all of the above Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Easy Section 20.3 14. Which of the following is the term used to describe a thin, sheet-like meshwork of extracellular matrix components that can be found in epithelial cells? a. basal lamina b. basement membrane c. gap junction d. cell wall Ans: a Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Easy 15. Basal lamina include all of the following, except: a. type I collagen. b. type IV collagen. c. laminin. d. nidogen.

Ans: a Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Easy 16. Which one of the following is NOT a property of perlecan? a. It contains laminin-like LG domains. b. It is a proteoglycan. c. It is only found in the basal lamina. d. It is a glycoprotein.

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Ans: c Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Easy 17. Multiadhesive matrix proteins form adhesive bridges between what components or structures?

Ans: Multiadhesive matrix proteins link cell-surface components such as integrins and ECM, extracellular matrix components such as collagen, and the glycosaminoglycan (GAG) portion of proteoglycans. Both laminin and fibronectin interact with integrins, collagen, and proteoglycan GAGs. Question Type: Essay Chapter: 20 Blooms: Understanding Difficulty: Moderate Section 20.4 18. The collagen triple-helix domain is NOT: a. rich in glycine. b. an  helix. c. rich in proline. d. rich in hydroxyproline.

Ans: b Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 19. Proteoglycans are: a. located exclusively at the cell surface. b. located exclusively in the extracellular matrix. c. highly positively charged. d. glycoproteins that contain glycosaminoglycans.

Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 20. Syndecans are cell-surface proteoglycans that: a. bind to collagens. b. bind to multi-adhesive matrix proteins. c. anchor cells to the extracellular matrix. d. all of the above Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate

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21 - 197 21. Biological roles of proteoglycans and hyaluronan include all of the following, except: a. maintenance of porosity for the diffusion of small molecules between cells and tissues. b. presentation of growth factors to cells. c. resistance to compression. d. storage sites for extracellular energy reserves.

Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 22. Which extracellular matrix component is expressed in a cell-specific manner and binds to the tripeptide sequence ArgGly-Asp? a. integrins b. collagen c. proteoglycans d. fibronectins Ans: a Question Type: Multiple choice Chapter: 20 Blooms: Remembering Difficulty: Easy 23. Polymerization of collagen into large collagen fibers occurs: a. in the endoplasmic reticulum. b. in the Golgi complex. c. in secretory vesicles. d. extracellularly.

Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 24. Hydroxyproline is important in the formation of staple collagen polymers within cells. The formation of hydroxyproline requires: a. vitamin A. b. vitamin B. c. vitamin C. d. vitamin D. Ans: c Question Type: Multiple choice Chapter: 20 Blooms: Remembering Difficulty: Easy 25. Which of the following is NOT true regarding matrix metalloproteases? a. Their activity depends on zinc ions. b. They are responsible for cleaving -catenin from E-cadherin. c. They are inhibited by TIMPs. d. Some of the members are called gelatinases.

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21 - 198 Ans: b Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 26. How are multiple forms of fibronectin generated?

Ans: Multiple molecular forms of fibronectin are generated by differentially splicing the precursor RNA products derived from one gene.

Question Type: Essay Chapter: 20 Blooms: Understanding Difficulty: Moderate 27. What are the three unusually abundant amino acids in collagen?

Ans: The three unusually abundant amino acids in collagen are glycine, proline, and hydroxyproline. hydroxyproline is the hydroxylated form of proline and is not one of the standard 20 amino acids.

Question Type: Essay Chapter: 20 Blooms: Remembering Difficulty: Moderate

Note that

28. In mice engineered to overexpress the syndecan-1 gene in the hypothalamic region of the brain and other tissues, normal control of feeding by antisatiety peptides is disrupted, and the animals overeat and become obese. Other studies have shown that syndecan-3 may also be involved in the regulation of feeding behavior. Given the following information, which of these outcomes might you expect in experimentally manipulated mice?

Hypothalmic expression of syndecan-3 is increased by fasting.

Syndecan-3 knockout mice are resistant to diet-induced obesity.

Heparinase can cleave syndecan-3 to release heparin sulfate GAG chains that may suppress the activity of antisatiety peptides and feeding behavior.

A. Overexpression of heparinase in mice with normal levels of syndecan-3 would result in resistance to diet-induced obesity.

B. In mice with normal levels of syndecan-3, fasting followed by free access to a high-fat diet would lead to resistance to diet-induced obesity.

C. Lack of heparinase in mice with normal levels of syndecan-3 would result in animals that overeat and become obese.

D. Overexpression of heparinase in syndecan-3 knockout mice would reverse the resistance to diet-induced obesity.

E. There are two plausible hypotheses: 1) Overexpression of heparinase in mice with normal levels of syndecan-3 would result in resistance to diet-induced obesity; and 2) lack of heparinase in mice with normal levels of syndecan-3 would result in animals that overeat and become obese.

Ans: E Question Type: Essay Chapter: 20 Blooms: Evaluating Difficulty: Difficult

29. What kinds of polymerized structures do collagen types form?

Ans: Collagens form three types of polymerized structures: fibrillar collagens, fibril-associated collagens, and sheet-forming collagens.

Question Type: Essay Chapter: 20 Blooms: Remembering

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21 - 199 Difficulty: Moderate 30. Heparan sulfate, an example of a(n) a. GAG; FGFR b. proteoglycan; FGF c. multi-adhesive protein; FGFR d. integrin; cell motility

, aids in the activation of

.

Ans: a Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Easy Section 20.5 31. NCAMs, a group of cell-adhesion proteins belonging to the Ig superfamily: a. are more heavily sialylated in embryonic tissues than in adult tissues. b. bind to proteoglycans.

2+

c. mediate Ca -dependent cell-to-cell binding. d. all of the above Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate

32. Dystroglycan is a large glycoprotein that binds to dystrophin in muscle cells. It is also present in other cells and can also bind to: a. laminin. b. the virus that causes Lassa fever. c. the bacterium that causes leprosy. d. all of the above Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 2+

33. What is the role of Ca in NCAM-mediated homophilic cell-cell adhesion?

2+

2+

Ans: Ca has no role in NCAM-mediated cell-cell adhesion. NCAMs are not Ca -dependent.

Question Type: Essay Chapter: 20 Blooms: Remembering Difficulty: Easy 34. What is the oligomeric structure of integrins?

Ans: Integrins are heterodimers composed of  and  subunits.

There are three different known  subunits and approximately 13 different known  subunits. Hence, much of the difference in integrins is attributed to differences in the 

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21 - 200 subunit.

Question Type: Essay Chapter: 20 Blooms: Understanding Difficulty: Moderate 35. Describe the importance of extravasation and how leukocytes use this mechanism to fight infection/inflammation.

Ans: In response to signals in the areas of inflammation and infection, endothelial cells move vesicle-sequestered selectins to the cell surface where they mediate weak binding to circulating leukocytes. Leukocytes roll along the endothelial surface, where they encounter platelet-activating factor (PAF) and ICAM-1, which also appeared in response to inflammatory signals.

PAF, other secreted activators, and chemokines induce morphological changes in the leukocyte, as well as activation of L2 integrins on the surface. Integrin binding to CAMs on the endothelium holds the leukocyte, thereby allowing it to alter its shape and then move (or transmigrate) between endothelial cells to the site of inflammation and/or infection in the underlying tissue.

Question Type: Essay Chapter: 20 Blooms: Applying Difficulty: Moderate 36. Which of the following best describes the structure of integrins? a. Integrins cluster in cis until coming into contact with another cell, where a trans network forms. b. Integrins are homodimers of subunits expressed in a cell-type specific manner, c. A single integrin subunit forms heterophilic interactions with fibronectin and laminin. d. Integrins are heterodimers composed of  and  subunits.

Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 37. What signal is released by endothelial cells to recruit circulating leukocytes? a. Vesicle-sequestered P-selectins are moved to the cell surface. b. Vesicle-sequestered αLβ2 molecules are moved to the cell surface. c. Integrins are cleaved from the plasma membrane so that the extracellular domain can bind leukocytes. d. Secreted glycoproteins trigger leukocytes to initiate ―rolling‖ motility.

Ans: a

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21 - 201 Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate 38. People with a defect in the synthesis of the integrin β2 subunit exhibit reduced: a. leukocyte attachment. b. leukocyte rolling. c. endothelial cell activation. d. extravasation.

Ans: d Question Type: Multiple choice Chapter: 20 Blooms: Applying Difficulty: Moderate 39. What effect might an injection of RGD peptide have on tumor cells moving through the blood/lymph system? a. block cell attachment to new tissue and therefore prevent metastasis b. encourage clumping of cancer cells for removal by the immune system c. promote tumor cell binding to platelets and potentially lead to a fatal clot d. block integrin-mediated cell cycle progression Ans: a Question Type: Multiple choice Chapter: 20 Blooms: Applying Difficulty: Moderate 40. As fibroblasts form a three-dimensional structure in contact with the ECM, the cells bound to ECM: a. exhibit less mobility. b. make the same adhesion structures as fibroblasts grown in two-dimensional culture. c. proliferate at a faster rate. d. de-differentiate.

Ans: c Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Easy Section 20.6 41. Pectins and hyaluronic acid are both: a. secreted by animal cells. b. very negatively charged. c. proteins. d. intracellular substances.

Ans: b Question Type: Multiple choice Chapter: 20 Blooms: Understanding Difficulty: Moderate

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21 - 202 42. Which of these is NOT a component of plant cell walls? a. collagen b. pectin c. cellulose d. hemicellulose Ans: a Question Type: Multiple choice Chapter: 20 Blooms: Remembering Difficulty: Easy 43. How are cellulose fibrils laid down in parallel arrays?

Ans: Cellulose fibrils are synthesized by the cell-surface enzyme cellulose synthase. Cellulose synthase movement is guided by underlying parallel arrays of microtubules.

Question Type: Essay Chapter: 20 Blooms: Applying Difficulty: Easy 44. What is the function of plasmodesmata in plants?

Ans: The function of plasmodesmata is to connect plant cells. They form the functional equivalent of gap junctions between plant cells. However, as cytoplasmic continuities between cells, larger molecules including various proteins can be exchanged between cells.

Question Type: Essay Chapter: 20 Blooms: Understanding Difficulty: Easy

21 Cell Birth, Lineage, and Death Section 21.1 1. In mammalian development, which of the following is the correct chronological order of events? a. blastocyst, zygote, four-cell, compacted morula, b. zygote, four-cell, compacted morula, blastocyst c. compacted morula, blastocyst, zygote, four-cell d. four-cell, compacted morula, blastocyst, zygote Ans: b Question Type: Multiple choice Chapter: 21

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23 - 203 Blooms: Understanding Difficulty: Easy 2. Describe gamete fusion that occurs during mammalian fertilization.

Ans: The mouse sperm must penetrate a layer of cumulus cells before it encounters the amorphous zona pellucida (ZP) layer surrounding the oocyte. It is here where the sperm surface protein GaIT encounters and binds ZP3 glycoprotein in the ZP, which triggers the acrosomal reaction. The acrosomal vesicle, containing hydrolytic and proteolytic enzymes and present within the tip of the sperm head, is exocytosed as a result of the acrosomal reaction. These enzymes serve to degrade the ZP, facilitating sperm-egg plasma membrane interactions. Sperm penetration triggers a Ca 2+ release from within the egg, triggering the exocytosis of egg cortical granules. Enzymes within these granules alter the properties of the ZP to prevent the binding of additional sperm (polyspermy).

Question Type: Essay Chapter: 21 Blooms: Applying Difficulty: Difficult 3. How can somatic cell nuclear transfer be used to clone a mouse?

Ans: Although a relatively inefficient method of cloning, as evidenced from the data that emerged following cloning Dolly the sheep, the technique relies on removing the nucleus from a fully differentiated cell and transplanting it into an enucleated egg of a donor mouse. In one example, the nucleus from an olfactory sensory neuron of a mouse that expresses green fluorescent protein (GFP) was transplanted into an enucleated egg and then differentiated in culture until the inner cell mass stage. GFP-expressing ES cells were isolated and injected into a tetraploid blastocyst host. When this chimeric blastocyst was transplanted into the uterus of a pseudopregnant female, the tetraploid cells formed the placenta and the GFP-expressing cells the embryo proper. If all cells express GFP when the pups were born, this would indicate that their nuclei had been derived from that originally isolated from the GFP-expressing olfactory neuron.

Question Type: Essay Chapter: 21 Blooms: Understanding Difficulty: Difficult 4. The first differentiation event following fertilization occurs when: a. the morula separates from the blastocyst. b. the trophoectoderm separates from the inner cell mass. c. the four-cell stage undergoes cleavage to generate the eight-cell stage. d. the zygote divides to form two cells.

Ans: b Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Easy Section 21.2 5. In the mouse embryo, pluripotent stem cells are found in the: a. inner cell mass. b. trophectoderm. c. zona pellucida. d. blastocoel.

Ans: a Question Type: Multiple choice Chapter: 21 Blooms: Remembering

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23 - 204 Difficulty: Easy 6. Daughter cells that form as a result of asymmetric cell division may differ in: a. size. b. shape. c. protein composition. d. all of the above Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Easy 7. Which of the following genes are expressed exclusively in pluripotent embryonic stem cells? a. Nanog b. Sox2 c. Oct4 d. Nanog and Oct4 Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Moderate 8. Fully differentiated mouse fibroblasts can be reprogrammed or induced to form pluripotent stem cells when transfected with retroviral vectors that express: a. KLF4. b. c-MYC. c. SOX2. d. all of the above Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Moderate 9. Dolly, the famous sheep, was cloned using: a. induced pluripotent stem cells. b. homologous recombination. c. somatic cell nuclear transfer. d. Cre recombinase.

Ans: c Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Easy 10. Briefly describe how Yamanaka and colleagues induced differentiated cells to form stem cells.

Ans: Research had shown that certain transcription factors (KLF4, SOX2, OCT4, and c-MYC) were expressed in embryonic stem cells, and down-regulation of their expression coincided with differentiation into one of the three germ layer lineages.

Recognizing this, Yamanaka infected mouse fibroblasts with retroviruses expressing these transcription factors and found

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23 - 205 that the cells lost their fibroblast-like properties and adopted characteristics of stem cells. Later on, these experiments were repeated using keratinocytes (skin cells) that were repeatedly transfected with specific mRNAs encoding the four canonical transcription factors described above. Over time, these cells adopted a stem cell fate, and there was no trace of the introduced mRNAs, indicating that they had in fact been reprogrammed.

Question Type: Essay Chapter: 21 Blooms: Applying Difficulty: Difficult 11. The use of cell-based models to understand or treat human disease processes is aided by stem cells in all the following ways, except: a. growth of neurons in cell culture containing mutations associated with ALS for drug screening. b. culture of patient-specific cells to replace those lost to disease, such as for type 2 diabetes. c. differentiation of neurons from iPS cells to replace those damaged by neurodegenerative diseases. d. provides proof that ES cells are pluripotent and can divide to make a normal animal.

Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Moderate Section 21.3 12. In the Drosophila ovary, which of the following proteins is secreted from the cap cell and is responsible for creating and maintaining the niche for germ-line stem cells? a. Hedgehog b. FGF c. Arm d. Ptc Ans: a Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Easy 13. Which of the following is FALSE regarding the stem cells located in the epithelial lining of the small intestine? a. They express -catenin. b. Lgr5 encodes an R-spondin receptor. c. The Lgr5 gene is induced by Wnt signaling. d. They are located in pits called crypts.

Ans: b Question Type: Multiple choice Chapter: 21 Blooms: Applying Difficulty: Moderate 14. Which of the following is(are) in the cell lineage that gives rise to T cells? a. GM-CFC b. myeloid stem cell c. pluripotent hematopoietic stem cell d. myeloid stem cell and pluripotent hematopoietic stem cell Ans: c

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23 - 206 Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Moderate 15. Floral meristems give rise to: a. leaves. b. petals. c. stamens. d. petals and stamens.

Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Moderate 16. Describe how the stem cell niche contributes to the maintenance of germ-line stem cells in Drosophila.

Ans: In the fly ovary, there is a niche where oocyte precursors form and begin to differentiate. Cap cells in the tip of the germarium secrete two TGF--like proteins (Dpp and Gbb) as well as Hh. Binding of these ligands to TGF- receptor I and II, and Ptc, respectively, on the surface of a germ-line stem cell, signals two transcription factors Mad and Med and a corepressor Schnurri, to repress expression of the bag of marbles (bam) gene, which encodes a key differentiation factor. The maintenance of the niche is dependent on homotypic E-cadherin inteactions between cap cells and germ-line stem cells, and armadillo, the fly homolog of -catenin, which tethers E-cadherin to the actin cytoskeleton within each cell type. When a germ-line stem cell divides, the daughter cell that maintains the E-cadherin-association with the cap cell continues its life as a stem cell. The other daughter cell, however, is displaced from the Dpp, Gbb, and Hh signals, allowing the Bam gene to be expressed. Translation of Bam in this cell induces it to differentiate into a cystoblast, which after four rounds of division, produces 16 interconnected cells, one of which becomes the oocyte.

Question Type: Essay Chapter: 21 Blooms: Understanding Difficulty: Difficult 17. Describe how Cre recombinase is used in lineage-tracing studies to follow the fate of Lgr5 +-expressing intestinal stem cells.

Ans: Investigators have genetically engineered a strain of mice in which Cre recombinase fused to the estrogen-binding domain of the estrogen receptor is placed under the control of the Lgr5 promoter. The Cre recombinase-ER chimera would therefore only be expressed in intestinal stem cells of these mice, and the protein would be localized to the cytoplasm.

Treating these cells, however, with the estrogen analog tamoxifen causes this fusion protein containing the domain of the estrogen receptor to translocate to the nucleus. Another strain of mice was engineered to express the bacterial galactosidase (-gal) gene preceded by two loxP sites flanking a blocking segment of DNA. Thus, in the absence of Cre recombinase, there would be no -gal expression. Mice from the two strains were mated, and offspring containing both marker transgenes were identified. Only mice treated with tamoxifen showed -gal expression in cells of the intestinal crypt.

In this manner, tamoxifen had facilitated the translocation of the Cre-ER chimeric protein into the nucleus, where Cre had interacted with the loxP sites to excise the blocking segment of DNA. The -gal gene was then capable of being expressed, and when the protein was translated, it was detected using a histochemical stain in which a substrate in contact with -gal forms a blue precipitate. In the lineage-tracing experiment, mice exposed to tamoxifen for 1 day showed blue staining in cells deep in the intestinal crypt. Those exposed to tamoxifen for longer periods showed blue-stained cells higher up in the villus, indicating that they had migrated away from their ―birthplace.‖ Question Type: Essay Chapter: 21 Blooms: Applying

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23 - 207 Difficulty: Difficult 18. Define stem cells. Which of the following cells are stem cells: (a) fertilized egg, (b) intestinal crypt cell, (c) granulocytemacrophage colony-forming cell (GM-CFC)?

Ans: Stem cells are cells that can both give rise to differentiated cell types or their precursors and self-renew. Stem cells are found in both embryonic and adult tissues. (a) The fertilized egg is not really a stem cell. It gives rise to all cell types in an animal but does not self-renew. (b) Within the intestinal crypt, there is a population of true stem cells that give rise to all intestinal epithelial cells and can self-renew. (c) GM-CFCs are progenitor cells, not stem cells. They can give rise to multiple differentiated blood cells but cannot self-renew. Myeloid stem cells are the stem cells for this lineage.

Question Type: Essay Chapter: 21 Blooms: Applying Difficulty: Difficult 19. Which of the following provides evidence for the role of Paneth cells in supporting the intestinal stem cell niche? a. Mutant mice with reduced numbers of Paneth cells have fewer intestinal stem cells. b. Co-culture of Paneth cells with intestinal stem cells leads to formation of villus-like structures. c. Paneth cells produce Wnt. d. all of the above Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Easy Section 21.4 20. The mating projection or shmoo in budding yeast cells relies on having: a. Cdc42 recruited and localized to the region near the highest concentration of mating pheromone. b. formin proteins to nucleate the assembly of actin filaments. c. myosin V to move secretory vesicles to the plus (+) ends of microfilaments. d. all of the above Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Applying Difficulty: Moderate 21. Even before the first cell division, the one-cell C. elegans zygote has an asymmetric distribution of: a. Par3 and aPKC in the cortex of the posterior half of the cell. b. Par6 in the cortex of the posterior half of the cell. c. Par2 in the cortex of the posterior half of the cell. d. all of the above Ans: c Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Moderate 22. In C. elegans, which of the following is located in the lineage that gives rise to germ cells? a. P1 b. P2

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23 - 208 c. P3 d. all of the above Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Moderate 23. The apical-basolateral polarity of epithelial cells depends on which one of the following? a. the Scribble complex in the apical domain of the cell b. the Crumbs complex in the basal domain of the cell c. the interaction between nectin and JAM-A d. all of the above Ans: c Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Moderate 24. Which is the correct combination of proteins in polarized cells involved in planar-cell polarity? a. Flamingo-Strabismus on one side and Dishevelled-Frizzled on the other b. Frizzled-Strabismus on one side and Flamingo-Dishevelled on the other c. Strabismus-Dishevelled on one side and Flamingo-Frizzled on the other d. none of the above Ans: a Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Moderate 25. Shmoo formation allows budding yeast to grow toward a concentration of mating factor. Describe how asymmetric changes in the haploid yeast cell promote the formation of this mating projection.

Ans: Haploid yeast exist in either an a or  state, where a cells secreting a factor prefer to mate with  cells secreting  factor. Each cell type expresses a surface G-protein coupled receptor (GPCR) for the other type’s mating factor, and when the factors are detected, each cell synchronizes their cell cycle and arrest at G o. GPCR activation results in the accumulation and activation of the small G-protein Cdc42 closest to the highest concentration of the mating factor. Cdc42, now in the active GTP-bound state, activates formin proteins that promote the assembly and elongation of microfilaments that have their (+) ends directed toward the cell cortex. Secretory vesicles, carried by the myosin-V motor protein toward the (+) end of these microfilaments, promote the localized growth of the projection (shmoo). Cdc42 that has moved away from the growing tip as a result of this increase in outward growth gets endocytosed and transported back to the shmoo tip, where it can continue to participate in formin activation and microfilament assembly. When shmooing cells of opposite mating types eventually touch, they fuse at the shmoo tips, and the haploid nuclei combine to restore the diploid state.

Question Type: Essay Chapter: 21 Blooms: Understanding Difficulty: Difficult 26. Describe two ways that stem cells can be induced to divide asymmetrically.

Ans: In the first method, a stem cell responds to an external cue by becoming polarized, causing fate determinants to localize to one region of the cell. The daughter cell inheriting these determinants remains a stem cell, whereas the other daughter cell differentiates. The other method involves a niche, which causes the stem cell to orient its mitotic spindle. Following

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23 - 209 division, the daughter cell in association with the niche remains as a stem cell, whereas the one distant to the niche differentiates.

Question Type: Essay Chapter: 21 Blooms: Applying Difficulty: Moderate 27. Epithelial cell tight junctions recruit: a. Crumbs complexes. b. Scribble complexes. c. apical PAR complexes. d. all of the above Ans: c Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Easy 28. When a cell divides asymmetrically: a. PAR proteins are spread homogenously throughout the cell. b. the spindle orients the new daughter cell toward the niche cell. c. exogenous factors bind cell surface receptors to induce actin cytoskeletal rearrangements. d. the resulting two cells are identical to each other.

Ans: c Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Easy Section 21.5 29. Which of the following is a feature that defines a cell dying in response to tissue damage (necrosis)? a. The nucleus condenses and then fragments. b. Small membrane bodies are released and then engulfed by other cells. c. The cell shrinks. d. none of the above Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Moderate 30. Which of the following genes encodes a caspase in C. elegans? a. ced-3 b. ced-4 c. ced-9 d. ced-8 Ans: a Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Easy

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23 - 210

31. Which of the following apoptotic proteins is homologous to human Bcl-2? a. CED-3 b. CED-4 c. CED-9 d. none of the above Ans: c Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Moderate 32. Which of the following forms a dimer that gets displaced from the surface of the mitochondria by EGL-1? a. CED-3 b. CED-4 c. CED-9 d. CED-8 Ans: b Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Moderate 33. Which of the following triggers apoptosis? a. Fas ligand b. NGF c. TNFα d. Fas ligand and TNFα Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Moderate 34. Explain the role of mitochondria in apoptosis.

Ans: When mitochondria are triggered to release cytochrome c into the cytosol, cytochrome c binds to and activates Apaf-1, which leads to activation of the caspases. Release of cytochrome c release occurs when Bax homodimers form channels that permit ion influx through the mitochondrial membrane. Bax can also promote apoptosis in a caspase-independent manner that involves mitochondrial depolarization.

Question Type: Essay Chapter: 21 Blooms: Understanding Difficulty: Moderate 35. Apoptosis can be induced by a mitochondrial protein being released into the cytosol. This protein binds directly to what in the cytosol? a. caspase 3 b. Bax c. Bcl-2 d. Apaf-1

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23 - 211 Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Remembering Difficulty: Easy 36. Mitochondrial permeabilization can be induced by compounds that poke small holes in the outer membrane. This would cause: a. cytochrome c to be released into the cytosol. b. increased ATP synthesis through oxidative phosphorylation. c. increased Bad-Bcl2 interactions. d. upregulation of caspase expression.

Ans: a Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Moderate 37. Examination of a cell’s structure reveals condensed chromatin, a shrunken cytoplasm, but unfragmented DNA and a lack of blebbing. If this cell has been triggered to undergo apoptosis by an extrinsic factor, which of the following events has NOT likely happened yet? a. activation of a TNF-α receptor b. recruitment of TRADD c. recruitment of FADD d. activation of caspases Ans: d Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Easy 38. XIAP proteins: a. inhibit initiator and effector caspases. b. are released from the mitochondria through Bad/Bax oliogomers. c. are activated by SMAC/DIABLO proteins. d. cleave the cytoskeleton during apoptosis.

Ans: a Question Type: Multiple choice Chapter: 21 Blooms: Understanding Difficulty: Easy 39. In cells that have been signaled to die via TNF-a receptor signaling but lack caspase-8 activity: a. apoptosis occurs via the other caspases. b. necroptosis occurs downstream of RIP protein activation of MLKL. c. the cell remains alive and viable. d. none of the above Ans: b Question Type: Multiple choice Chapter: 21

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23 - 212 Blooms: Understanding Difficulty: Easy

Nerve Cells

22 Section 22.1 1. In a human brain, the number of glial cells is: a. about 1/10 the number of neurons. b. about the same as the number of neurons. c. about 10 times the number of neurons. d. about 100–200 times the number of neurons.

Ans: c Question Type: Multiple choice Chapter 22.1 Blooms: Blooms: Remembering Difficulty: Moderate 2. The resting potential of a typical neuron is: a. −60 mV. b. 0 mV. c. 20 mV. d. 50 mV. Ans: a Question Type: Multiple choice Chapter 22.1 Blooms: Blooms: Remembering Difficulty: Easy 3. Action potentials are: a. pulses of ion flow along axons. b. brief local voltage changes within a neuron from inside-negative to inside-positive. c. transmembrane potentials that result from the action of the Na+/K+ ion pump. d. electrical signals used by glial cells to transmit information.

Ans: a Question Type: Multiple choice Chapter 22.1 Blooms: Understanding Difficulty: Moderate 4. Which cells produce myelin sheaths? a. Schwann cells and oligodendrocytes b. astrocytes c. afferent neurons

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23 - 213 d. interneurons Ans: a Question Type: Multiple choice Chaper 22.1 Blooms: Blooms: Remembering Difficulty: Easy 5. What starts an action potential in a post-synaptic neuron?

Ans: When an action potential reaches the terminus of a presynaptic cell, it stimulates exocytosis of synaptic vesicles and the release of neurotransmitter molecules into the synapse. Neurotransmitter molecules diffuse across the synapse and bind to receptors on the dendrite of the post-synaptic cell. This triggers opening or closing a specific ion channels embedded in the plasma membrane of the post-synaptic cell dendrite. In most cases, this depolarizes the localized area of the post-synaptic cell. If this depolarization is large enough, it triggers an action potential in the axon of the post-synaptic cell.

Question Type: Essay Chapter 22.1 Blooms: Understanding Difficulty: Difficult 6. Which of the following represents a correct structure-function relationship for neurons? a. cell body – sends signal to another cell b. axon – contains the nucleus of the cell c. dendrite – forms synapses with other neurons d. axon termini – receive incoming signals Ans: c Question Type: Multiple choice Chapter 22.1 Blooms: Understanding Difficulty: Easy 7. Action potentials: a. vary in intensity b. vary in frequency and timing c. result in a decrease in resting potential d. all of the above Ans: b Question Type: Multiple choice Chapter 22.1 Blooms: Blooms: Remembering Difficulty: Easy 8. During a knee jerk reflex, which of the following steps occurs first? a. stretch receptor is activated b. interneuron sends an inhibitory signal to the hamstring muscle c. motor neuron stimulates the quadriceps muscle d. action potential travels along a sensory neuron Ans: a Question Type: Multiple choice Chapter 22.1 Blooms: Blooms: Remembering Difficulty: Easy

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23 - 214 9. Which of the following is NOT a function of astrocytes? a. stimulate blood vessels in the brain to maintain tight junctions b. provide growth factors to neurons c. produce extracellular matrix d. excite muscle cells Ans: d Question Type: Multiple choice Chapter 22.1 Blooms: Blooms: Remembering Difficulty: Easy 10. The birth of new neurons in the adult brain occurs in: a. the dentate gyrus. b. the brainstem. c. olfactory bulb. d. neural tube.

Ans: a Question Type: Multiple choice Chapter 22.1 Blooms: Blooms: Remembering Difficulty: Easy Section 22.2 11. The resting membrane potential in animal cells depends largely on nongated a. Ca2+ b. H+ c. K+ d. Na+

channels.

Ans: c Question Type: Multiple choice Chapter 22.2 Blooms: Blooms: Remembering Difficulty: Moderate 12. During an action potential, which happens first? a. opening of voltage-gated Na+ channels b. closing of voltage-gated Na+ channels c. opening of voltage-gated K+ channels d. closing of voltage gated K+ channels Ans: a Question Type: Multiple choice Chapter 22.2 Blooms: Understanding Difficulty: Moderate 13. Repolarization during the refractory period is largely due to the: a. opening of non-voltage-gated K+ channels. b. opening of voltage-gated K+ channels. c. opening of voltage-gated Na+ channels. d. action of the Na+/K+ pump.

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23 - 215 Ans: b Question Type: Multiple choice Chapter 22.2 Blooms: Understanding Difficulty: Moderate 14. Fruit flies carrying the shaker mutation have motor neurons with an abnormally long action potential because of: a. a defect in the Na+/K+ pump. b. a defect in voltage-gated Na+ channels. c. a defect in non-voltage-gated K+ channels. d. a defect in voltage-gated K+ channels.

Ans: d Question Type: Multiple choice Chapter 22.2 Blooms: Understanding Difficulty: Moderate 15. How are most voltage-gated channels inactivated? a. by ligand binding b. by changes in membrane potential c. through interaction with other membrane proteins d. by closing spontaneously soon after opening Ans: d Question Type: Multiple choice Chapter 22.2 Blooms: Understanding Difficulty: Easy 16. Where are voltage-gated Na+ channels concentrated? a. spaced uniformly along the axon membrane b. clustered at the nodes of Ranvier c. clustered at the axon terminus d. embedded within the myelin sheath Ans: b Question Type: Multiple choice Chapter 22.2 Blooms: Blooms: Remembering Difficulty: Moderate 17. Which cells produce myelin? a. oligodendrocytes b. Schwann cells c. astrocytes d. Schwann cells and oligodendrocytes e. all of the above Ans: d Question Type: Multiple choice Chapter 22.2 Blooms: Blooms: Remembering Difficulty: Easy

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23 - 216 18. Why are nongated channels important in the generation of an inside-negative electric potential (voltage) of 50–70 mV across the plasma membrane of cells?

Ans: If there is no ion movement across the membrane, there is no membrane potential. This is true even if there is a difference in ion concentrations on either side of the membrane. The presence of K+ channels that are usually open allows for ion movement from the inside of the cell to the outside and the creation of a negative membrane potential inside the cell.

Question Type: Essay Chapter 22.2 Blooms: Understanding Difficulty: Difficult 19. How do researchers measure ion movements through single channels?

Ans: Patch clamping is a technique that measures the electric current caused by the movement of ions across a small patch of the plasma membrane. The membrane is electrically depolarized or hyperpolarized and maintained (clamped) at that potential by an electronic feedback device. The inward or outward movement of ions across a patch of membrane can be quantified from the amount of electric current needed to maintain the membrane potential at the designated “clamped” value.

Question Type: Essay Chapter 22.2 Blooms: Understanding Difficulty: Difficult 20. Action potentials are propagated in only one direction, down the axon. Explain how the absolute refractory period of the voltage-gated Na+ channels and the brief hyperpolarization resulting from K + efflux produces this outcome.

Ans: The Na+ influx due to opening of the voltage-gated Na+ channel is self-limiting and is closed off by movement of the channel-inactivating segment into the open channel. As long as the membrane remains depolarized, the channel-inactivating segment remains in the channel opening, and influx is prevented. This is the refractory period. Only after the inside-negative resting potential is restored does the channel-inactivating segment swing away from the pore and the Na + channel return to the closed resting state. Only now can it be opened in response to membrane depolarization. The brief hyperpolarization is a consequence of the closing of the Na + channel by the channel-inactivating segment. The refractory period has the consequence that the upstream, recently activated, but refractory Na + channels cannot be activated in response to local diffusion of Na+. Only downstream channels can be activated in response to local diffusion of Na +. Therefore, the propagation occurs in only one direction.

Question Type: Essay Chapter 22.2 Blooms: Applying Difficulty: Difficult 21. How does opening and closing of voltage-gated cation channels occur?

Ans: All voltage-gated ion channels have similar structures. Therefore, there is a general pattern to their opening and closing.

Each must sense changes in membrane potential, transduce detection into opening of the channel, and finally inactivate the channel shortly after its opening. Overall, individual steps must be accomplished in a cyclic manner. Voltage-gated channels consist of four domains, arranged either in one polypeptide or spread over four homologous subunits. The association of the four domains within the membrane forms the channel. If one polypeptide is present, there is a single channel-inactivating segment. If four polypeptides are present, there are four channel-inactivating segments. All have evolved from a single polypeptide that contained six transmembrane  helices. Voltage sensing is achieved by protein-bound positive charges carried on the S4, voltage-sensing or gating, helices. These positively charged lysine or arginine residues move when the membrane is depolarized, and therefore the protein conformation changes. The gate itself is composed of the cytosolic facing N-termini of the four S5 helices and the C-termini of the four S6 helices. Movement of the channel-inactivating segment into the open pore blocks ion flow. Upon repolarization, this ball-and-chain segment no longer blocks the channel, and the channel is gated and may be reopened in response to depolarization.

Question Type: Essay Chapter 22.2 Blooms: Applying Difficulty: Difficult 22. Myelinated axons have the conduction velocity of much larger axons (unmyelinated of course) because of all the following reasons, except:

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23 - 217 a. voltage channels are only present at the nodes of Ranvier. b. saltatory conduction. c. more ion channels are present in myelinated axons. d. insulation.

Ans: c Question Type: Multiple choice Chapter 22.2 Blooms: Understanding Difficulty: Easy Section 22.3 23. Two signaling proteins, agrin and neuregulin, have been identified in studies of neuromuscular junction development. In the absence of neuregulin: a. Schwann cells die. b. postsynaptic structures do not form in muscle cells. c. the acetylcholine receptor protein is not synthesized. d. all of the above Ans: a Question Type: Multiple choice Chapter 22.3 Blooms: Understanding Difficulty: Moderate 24. During the development of specialized postsynaptic structures at a neuromuscular junction, preexisting AChR proteins on the surface of the uninnervated muscle are induced to aggregate by: a. agrin, a signal released from the neuron. b. the muscle membrane protein, MuSK. c. the cytosolic protein, rapsyn. d. all of the above Ans: d Question Type: Multiple choice Chapter 22.3 Blooms: Understanding Difficulty: Moderate 25. Botulinum toxin inhibits nerve transmission by targeting: a. the acetylcholine receptor. b. VAMP. c. dynamin. d. synaptotagmin.

Ans: b Question Type: Multiple choice Chatper 22.3 Blooms: Understanding Difficulty: Moderate 26. Neurotransmitters are stored in: a. the cell body. b. the dendrite. c. the axon hillock.

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23 - 218 d. synaptic vesicles.

Ans: d Question Type: Multiple choice Chapter 22.3 Blooms: Blooms: Remembering Difficulty: Easy 27. Which of the following proteins is NOT part of a four-helix complex that mediates fusion of synaptic vesicles with the plasma membrane? a. NSF b. SNAP-25 c. syntaxin d. VAMP Ans: a Question Type: Multiple choice Chapter 22.3 Blooms: Blooms: Remembering Difficulty: Easy 28. Evidence that synaptotagmin is the Ca2+ sensor for exocytosis of neurotransmitters includes: a. Ca2+ binding by synaptotagmin. b. partial loss-of-function mutations of synaptotagmin in Drosophila and C. elegans that result in neurons that are defective in Ca2+-stimulated vesicle exocytosis. c. uncoordinated embryonic muscle contractions in Drosophila and C. elegans mutants that lack synaptotagmin. d. all of the above Ans: d Question Type: Multiple choice Chapter 22.3 Blooms: Analyzing Difficulty: Difficult 29. Signaling at synapses is usually terminated by: a. calcium influx. b. potassium influx. c. inhibitory neurotransmitters. d. degradation or reuptake of neurotransmitters.

Ans: d Question Type: Multiple choice Chapter 22.3 Blooms: Understanding Difficulty: Moderate 30. Acetylcholine receptor loss is observed in people with: a. schizophrenia. b. drug addiction. c. myasthenia gravis. d. all of the above Ans: d Question Type: Multiple choice Chapter 22.3 Blooms: Blooms: Remembering

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23 - 219 Difficulty: Easy 31. Electrical synapses depend on: a. neurotransmitters and receptor proteins. b. calcium channels. c. gap junction channels. d. all of the above Ans: c Question Type: Multiple choice Chapter 22.3 Blooms: Understanding Difficulty: Moderate 32. Botulinum toxin acts on: a. SNARES to promote vesicle fusion. b. AMPA receptors to inhibit LTP. c. calcium receptors to increase acetylcholine release. d. VAMP to prevent fusion with SNARE complexes.

Ans: d Question Type: Multiple choice Chapter 22.3 Blooms: Understanding Difficulty: Moderate Section 22.4 33. Explain how the expression of the H+/acetylcholine antiporter and the choline acetyltransferase enzyme are coordinated.

Ans: Expression of the H+/acetylcholine antiporter and the choline acetyltransferase enzyme are coordinated by an evolutionarily conserved mechanism that involves the location of the antiporter gene within the first intron of the gene encoding choline acetyltransferase.

Question Type: Essay Chapter 22.4 Blooms: Understanding Difficulty: Moderate 34. How does the nicotinic acetylcholine receptor function as a ligand-gated ion channel at nerve-muscle synapses?

Ans: The nicotinic acetylcholine receptor is a pentameric protein with five related polypeptides. The receptor has fivefold symmetry; the actual cation channel is lined by homologous segments from each of the five subunits. The channel opens when two acetylcholine molecules bind to sites located at interfaces between subunits. Opening is within microseconds of acetylcholine binding. The pore is large enough to accommodate hydrated Na + and K+. The exact conformational changes are not well understood.

Question Type: Essay Chapter 22.4 Blooms: Applying Difficulty: Difficult 35. How are neurotransmitters packaged at synaptic endings for quantal release?

Ans: Neurotransmitters are packaged into synaptic vesicles in the presynaptic nerve ending in an antiport-mediated process.

Filled vesicles are docked at the synaptic ending awaiting a signal. The vesicles are then released at the synaptic ending in response to a Ca2+ influx. This produces release of neurotransmitters as vesicle quanta rather than one neurotransmitter at a time.

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23 - 220 Question Type: Essay Chapter 22.4 Blooms: Understanding Difficulty: Moderate 36. Describe the common features of the GABA, norepinephrine, dopamine, and serotonin transporter proteins.

Ans: Transporter proteins are present in presynaptic membranes and are responsible for reuptake of neurotransmitters from the synapse following neurotransmitter release. The GABA, norepinephrine, dopamine, and serotonin transporters are all Na+ symporters. They all have 12 transmembrane  helices, and they have 60–70% identical amino acid sequence homology.

Question Type: Essay Chapter 22.4 Blooms: Applying Difficulty: Difficult 37. Reception of taste causes: a. an action potential that travels along the axon of a taste cell. b. release of neurotransmitters that are detected by adjacent neurons. c. an increase in cGMP that inactivates ligand gated Na+/Ca+2 channels. d. activation of TRPV1, which is a ligand gated Na+/Ca+2 channel.

Ans: b Question Type: Multiple choice Chapter 22.4 Blooms: Applying Difficulty: Moderate 38. Taste receptors are either channel proteins or: a. sodium ion pumps. b. potassium ion pumps c. calcium ion pumps d. G protein-coupled receptors.

Ans: d Question Type: Multiple choice Chapter 22.4 Blooms: Understanding Difficulty: Easy 39. Touch receptors are: a. non-gated cation channels. b. gated cation channels. c. calcium ion pumps. d. G protein-coupled receptors.

Ans: b Question Type: Multiple choice Chapter 22.4 Blooms: Blooms: Remembering Difficulty: Easy 40. Why do chili peppers seem hot?

Ans: Chili peppers contain capsaicin, which binds to a pain receptor called TRPV1. This is the same pain receptor that responds to temperatures greater than 43 °C. Question Type: Essay Chaper 22.4

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23 - 221 Blooms: Blooms: Remembering Difficulty: Easy Section 22.5 41. Long-lasting effects of habituation involve: a. stimulation of sensitizing interactions. b. decreasing numbers of connections between sensory and motor neurons. c. movement of the Golgi apparatus out into the dendrites. d. rewiring of neurons in the spinal cord.

Ans: b Question Type: Multiple choice Chapter 22.5 Blooms: Understanding Difficulty: Moderate 42. Silencing of the gene encoding FMRP: a. decreases translation of mRNAs near the dendrites. b. is associated with Huntington’s disease. c. results in less spatiotemporal regulation of neuron-specific protein expression. d. causes mice to solve mazes more quickly.

Ans: c Question Type: Multiple choice Chapter 22.5 Blooms: Understanding Difficulty: Moderate 43. Long-lasting memories: a. are formed in the subventricular zone. b. trigger a loss of CAMKII alpha activity. c. do not require protein expression to be potentiated. d. trigger increases in AMPA receptors postsynaptically.

Ans: c Question Type: Multiple choice Chapter 22.5 Blooms: Understanding Difficulty: Moderate

23 Immunology Section 23.1

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1. Where do most of the interactions between the cells and molecules required for the immune response occur? a. in the bloodstream b. in endothelial tissues c. in lymph nodes d. in the thymus Ans: c Question Type: Multiple choice Chapter 23.1 Blooms: Understanding Difficulty: Easy 2. Which complement components show chemoattractant activity and help recruit neutrophils to sites of complement activation? a. C1q b. C1rC1s complexes c. C2C4 complexes d. small fragments released from C3 Ans: d Question Type: Multiple choice Chapter 23.1 Blooms: Remembering Difficulty: Moderate 3. Fluid and immune system cells can escape from blood vessels to provide nutrients and defensive proteins to tissues. How are the cells and fluid replaced to maintain homeostasis in the bloodstream?

Ans: Fluid and cells that leave the circulation return in the form of lymph. Lymphatic vessels collect interstitial fluid and drain into lymph nodes. Immune system cells also congregate in lymph nodes, where they interact to activate immune responses. Fluid and activated immune system cells re-enter the bloodstream via lymphatic vessels that drain into the circulation.

Question Type: Essay Chapter 23.1 Blooms: Understanding Difficulty: Moderate 4. What roles do natural killer cells play in the innate immune response?

Ans: Natural killer cells seek out viral infected host cells and kill them. Natural killer cells also secret large quantities of interferon-, a cytokine essential for orchestrating many aspects of antiviral defenses.

Question Type: Essay Chapter 23.1 Blooms: Applying Difficulty: Moderate

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24 - 223 5. Which of the following is NOT an example of a mechanical or chemical barrier to pathogens? a. lysozymes in tears b. acids secreted in the stomach c. phagocytes d. tight junctions between epithelial cells Ans: c Question Type: Multiple choice Chapter 23.1 Blooms: Understanding Difficulty: Easy 6. Which cell type is responsible for providing an innate immune response to viruses? a. macrophages b. natural killer cells c. dendritic cells d. epithelial cells Ans: b Question Type: Multiple choice Chapter 23.1 Blooms: Remembering Difficulty: Easy Section 23.2 7. Treatment of IgG antibodies with the protease pepsin yields: a. fragments that retain antigen-binding capacity but are monovalent. b. separated heavy and light chains. c. single chain antibodies similar to those made by camelids. d. bivalent fragments called F(ab)s.

Ans: d Question Type: Multiple choice Chapter 23.2 Blooms: Understanding Difficulty: Moderate 8. Which of the following is NOT a heavy chain isotype? a.  b.  c.  d.  Ans: c Question Type: Multiple choice Chapter 23.2 Blooms: Remembering Difficulty: Easy 9. Which immunoglobulin is secreted as a pentamer? a. IgA b. IgE c. IgG d. IgM

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Ans: d Question Type: Multiple choice Chapter 23.2 Blooms: Remembering Difficulty: Easy 10. Tear fluids and other secretions are rich in: a. IgA. b. IgE. c. IgG. d. IgM. Ans: a Question Type: Multiple choice Chapter 23.2 Blooms: Remembering Difficulty: Easy 11. Describe how the heavy chains and light chains are arranged in IgA or IgG antibody molecules.

Ans: IgA and IgG immunoglobulins are each composed of two identical heavy chains covalently attached to two identical light chains. Both the light chains and the heavy chains have variable and constant regions, with the variable regions located at their N-terminal ends. Both the variable and constant regions are folded into compact domains composed exclusively of  sheets.

Question Type: Essay Chapter 23.2 Blooms: Understanding Difficulty: Moderate 12. Somatic recombination is catalyzed by RAG1 and RAG2 recombinases. How does the RAG1-RAG2 complex recognize the cleavage site at the exact boundary of the coding and signal sequences in the variable regions of immunoglobulin genes?

Ans: At the 3´ end of each light-chain V gene, there is a conserved sequence composed of a heptamer and a nonamer separated by a 23-bp spacer. At the 5´ end of each J element, there is a similar recognition sequence composed of the antiparallel and complementary heptamer and nonamer, in this case separated by a 12-bp spacer. The 12-bp and 23-bp spacers separate the conserved recognition sequences by one or two full turns of the DNA helix. The RAG1-RAG2 recombinase complex makes a single-stranded cut only when the two complementary heptamer-nonamer sequences with different length spacers are juxtaposed. Similar recognition sequences are present in the heavy-chain V, D, and J regions.

Question Type: Essay Chapter 23.2 Blooms: Applying Difficulty: Moderate 13. Explain why only one in three attempts at somatic recombination results in a productive VJ or VDJ combination.

Ans: The final steps in somatic recombination involve ligation of the coded regions. This step is imprecise. Whenever V > J, D > J, or V > DJ elements recombine, the resulting reading frame is random. Only one in three attempts will result in a reading frame that is compatible with either heavy-chain or light-chain synthesis.

Question Type: Essay Chapter 23.2 Blooms: Understanding Difficulty: Difficult 14. During B-cell development, antibody production switches from membrane-bound IgM to secreted antibody production.

How does this switch occur?

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Ans: Somatic recombination of the immunoglobulin heavy-gene locus first creates a rearranged gene that can express a membrane-bound  chain. Antigen binding is required before the B cell can switch to secreted antibody production. The choice between membrane-bound versus secreted IgM depends on differential use of polyadenylation sites in the  transcript.

The  transcript has two potential polyadenylation sites. If the downstream site is chosen, the resulting protein includes a Cterminal membrane anchor, which yields a membrane-bound form of . If the upstream polyadenylation site is used, the membrane anchor is bypassed, and the transcript is processed to yield the secreted version of the  chain.

Question Type: Essay Chapter 23.2 Blooms: Understanding Difficulty: Difficult 15. Describe the role of Fc receptor proteins in antibody-dependent cell-mediated cytotoxicity.

Ans: Cells such as monocytes and natural killer cells have receptors for the Fc portion of IgG molecules that allow the natural killer cells to kill cells that display viral or other antigens to which antibodies are attached. Fc receptors interact with the constant regions of the antibodies bound at the cell surface. This stimulates the monocytes or natural killer cells to release toxic small molecules such as oxygen radicals or release the contents of cytotoxic granules. Proteins contained in the released cytotoxic granules attach to the surface of the target cells, inflict membrane damage, and kill the target cell.

Question Type: Essay Chapter 23.2 Blooms: Applying Difficulty: Moderate

16. When comparing transcytosis of IgA in tears and IgG in neonates, which of the following is NOT true? a. Both processes involve transport into and back out of an epithelial cell. b. Both processes require recognition of the immunoglobulin by a cell surface receptor. c. Both processes result in cleavage by proteolysis to release the bound immunoglobulin. d. Both processes rely on vesicle trafficking within a cell.

Ans: c Question Type: Multiple choice Chapter 23.2 Blooms: Understanding Difficulty: Moderate 17. Which of the following would occur from B-cell class switching? a. rearrangement of the light chain b. new VDJ recombination that still retains antigen specificity c. production of IgA secreted into the bloodstream with the same antigen specificity d. production of IgGγ2 with the same antigen specificity Ans: d Question Type: Multiple choice Chapter 23.2 Blooms: Understanding Difficulty: Moderate Section 23.3 18. Which class of immunoglobulins is made first? a. IgA b. IgE c. IgG

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24 - 226 d. IgM Ans: d Question Type: Multiple choice Chapter 23.3 Blooms: Remembering Difficulty: Easy 19. During the early stages of B cell differentiation, heavy chains are complexed with: a.  light chains. b.  light chains. c. surrogate light chains. d. none of the above Ans: c Question Type: Multiple choice Chapter 23.3 Blooms: Remembering Difficulty: Easy 20. ITAMs are part of: a. Ig and Ig b. heavy chains. c. light chains. d. surrogate light chains.

Ans: a Question Type: Multiple choice Chapter 23.3 Blooms: Remembering Difficulty: Moderate 21. Patients with an X-linked inherited disease called Hyper IgM (XHIGM) syndrome have a deficiency in CD40 ligand, a protein found on the surface of T lymphocytes. Patients with XHIGM are unable to make the class switch between IgM and either IgA or IgG. From this information, you could conclude that: a. B-cell production of IgM does not depend on the CD40 ligand. b. the class switch between IgM and either IgA or IgG requires T cells. c. the switch between membrane-bound IgM and secreted IgM requires T cells. d. B-cell production of IgM does not depend on the CD40 ligand, and the class switch between IgM and either IgA or IgG requires T cells.

Ans: c Question Type: Multiple choice Chapter 23.3 Blooms: Applying Difficulty: Moderate 22. Class switch recombination generates antibodies with: a. different light-chain variable regions. b. different light-chain constant regions. c. different heavy-chain variable regions. d. different heavy-chain constant regions.

Ans: d Question Type: Multiple choice Chapter 23.3

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24 - 227 Blooms: Remembering Difficulty: Easy 23. Monoclonal antibodies are produced by fusing mouse spleen cells with: a. primary B cells. b. immature B cells. c. HGPRT-deficient myeloma cells. d. T-cells.

Ans: c Question Type: Multiple choice Chapter 23.3 Blooms: Remembering Difficulty: Easy 24. Explain why Bence-Jones proteins are different for different B-cell tumors from different patients but identical when isolated from a single tumor.

Ans: B-cell tumors are clonal expansions of individual lymphocytes. Each tumor cell secretes large quantities of the same immunoglobulin. Some of the light chains of the tumor-derived immunoglobulins are secreted in the urine of tumor patients.

These are called Bence-Jones proteins. Bence-Jones proteins are different when isolated from different patients because immunoglobulin production by activated B cells involves random gene rearrangement of the genes for the immunoglobulin proteins, thus producing variable light chains. Bence-Jones proteins are the same when isolated from an individual tumor because the tumor cells are all derived from a single activated B cell.

Question Type: Essay Chapter 23.3 Blooms: Applying Difficulty: Difficult Section 23.4 25. Autoimmune diseases are associated with particular alleles of genes for: a. cytokines. b. immunoglobulins. c. MHC proteins. d. T-cell receptors.

Ans: c Question Type: Multiple choice Chapter 23.4 Blooms: Remembering Difficulty: Easy 26. Compare the class I and class II pathways of antigen processing and presentation. What kinds of antigens are involved?

Which types of T cells recognize the antigen–MHC protein complexes?

Ans: In both pathways, host cells acquire antigens, process them, and then display them in a form that can be recognized by T lymphocytes. The class I pathway involves class I MHC proteins and, except for cross loading, primarily applies to presentation of proteins synthesized by the class I positive cell itself. T-cell receptor proteins on CD8+ T cells recognize antigens complexed with class I MHC molecules.

The class II pathway involves class II MHC proteins, which are present only on professional antigen-presenting cells (APCs).

In this case, the antigenic materials come from sources outside the APCs themselves (such as invading bacteria.) APCs absorb the antigenic materials and process them into small peptides. APCs present the small peptides from the processed antigens as a complex bound to class II MHC proteins on their surface. CD4+ T cells recognize antigens complexed with class II MHC molecules.

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24 - 228 Question Type: Essay Chapter 23.4 Blooms: Applying Difficulty: Difficult

27. Experiments that first identified a role for the MHC gene locus in tissue identification were successful in mice because: a. mice can yield large litters, and great numbers of offspring were needed for genetic analyses. b. mice have very small genomes, making it easier to link genetic changes to functional changes. c. mice with identical genetic backgrounds were able to accept tissue transplants, but other strains of mice rejected the tissue. d. mice without intact immune systems are routinely used to study the effects of MHC molecules.

Ans: c Question Type: Multiple choice Chapter 23.4 Blooms: Applying Difficulty: Easy 28. Which of the following is NOT a way that MHC proteins play a role in the immune response? a. presentation of antigen by class II MHCs on T-cells b. restrict cytotoxic T-cell killing of virally-infected cells c. allow lymphocytes to distinguish between ―self‖ and ―non-self‖ d. presentation of antigen by class II MHCs on B-cells Ans: a Question Type: Multiple choice Chapter 23.4 Blooms: Understanding Difficulty: Moderate Section 23.5 29. In the thymus, T cells can be deleted only if: a. the appropriate self-antigen is represented adequately in the thymus as MHC-peptide complexes. b. the appropriate self-antigen is not present in the thymus as a MHC-complex. c. the T cell cannot engage self MHC proteins. d. the appropriate self-antigen is represented adequately in the thymus as MHC-peptide complexes, and the T cell cannot engage self MHC proteins.

Ans: d Question Type: Multiple choice Chapter 23.5 Blooms: Applying Difficulty: Moderate 30. In cells with mutations in either TAP1 or TAP2: a. ubiquitin conjugation would be disrupted and antigen proteins would not be tagged for destruction. b. ubiquitin-conjugated proteins would not be destroyed by proteasomal proteolysis. c. peptides generated by proteasomal proteolysis would not be complexed with class I MHC molecules. d. class I MHC–antigen complexes would be unable to cross the ER membrane.

Ans: d Question Type: Multiple choice Chapter 23.5 Blooms: Applying

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24 - 229 Difficulty: Moderate 31. Which receptor proteins are displayed by dendritic cells and specialize in the recognition of microbial and viral products? a. Toll-like receptors (TLRs) b. class I MHC proteins c. class II MHC proteins d. T-cell receptor proteins Ans: a Question Type: Multiple choice Chapter 23.5 Blooms: Remembering Difficulty: Moderate 32. Activated CD4 T cells recognize an antigen-experienced B cell by means of: a. the class I MHC–peptide complexes displayed by the B cell. b. the class II MHC–peptide complexes displayed by the B cell. c. both the class I and class II MHC–peptide complexes displayed by the B cells. d. none of the above Ans: b Question Type: Multiple choice Chapter 23.5 Blooms: Understanding Difficulty: Moderate 33. Which cells can activate macrophages and stimulate an inflammatory response through the production of IFN a. CD4 Th1 T cells b. CD8 T cells c. regulatory T cells d. plasma cells Ans: c Question Type: Multiple choice Chapter 23.5 Blooms: Remembering Difficulty: Moderate 34. Irradiation prevents mice from generating their own T cells. When the immune systems of irradiated mice are reconstituted with unfractionated CD4 cells, the recipients show no signs of autoimmune disease. However, if the T cells are depleted of the CD4+CD25+ subset prior to transfer, recipients show multi-organ autoimmunity. This result demonstrates the role of in suppressing autoimmunity. a. cytotoxic T cells b. helper T cells c. regulatory T cells d. all of the above Ans: c Question Type: Multiple choice Chapter 23.5 Blooms: Analyzing Difficulty: Easy 35. Self-antigens in the retina and central nervous system are the beneficiaries of a form of tolerance called: a. anergy. b. central tolerance.

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24 - 230 c. clonal exhaustion. d. immune privilege.

Ans: d Question Type: Multiple choice Chapter 23.5 Blooms: Understanding Difficulty: Easy 36. When tissue damage occurs, what attracts neutrophils to the site of damage?

Ans: Resident fibroblasts produce a chemokine, IL-8, that attracts neutrophils.

Question Type: Essay Chapter 23.5 Blooms: Remembering Difficulty: Easy 37. Patients who have received an organ transplant take an immunosuppressant called cyclosporine to: a. encourage the patient’s T cells to recognize the new organ cells as ―self‖. b. upregulate T-regulatory cells to prevent any immune response against the new organ. c. prevent clonal expansion through inhibition of IL-2 transcription. d. inhibit calcineurin which results in less NFAT phosphorylation.

Ans: c Question Type: Multiple choice Chapter 23.5 Blooms: Applying Difficulty: Difficult Section 23.6 38. Which of the following peaks earliest in the time course of a viral infection? a. type I interferons b. NK cells c. virus-specific CTLs d. antibody titers Ans: a Question Type: Multiple choice Chapter 23.6 Blooms: Understanding Difficulty: Easy 37. sense the presence of cytoplasmic pathogen-derived nucleic acids. a. Inflammasomes b. Antigen-presenting cells c. Toll-like receptors d. High-affinity antibodies Ans: a Question Type: Multiple choice Chapter 23.6 Blooms: Remembering Difficulty: Moderate

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38. Some Toll-like receptor proteins recognize and destroy CpG-containing bacterial DNA. differentiate between host DNA and DNA from invading bacteria?

How do these receptors

Ans: Mammalian genomes show an underrepresentation of CpG dinucleotides. In addition, many of the CpG dinucleotides in mammalian genomes are methylated. Thus, DNA containing unmethylated CpG is characteristic of microbes.

Question Type: Essay Chapter 23.6 Blooms: Applying Difficulty: Easy 39. Why has the live-attenuated polio vaccine been recently discontinued?

Ans: The risk of re-emergence of more virulent strains of the poliovirus currently outweighs the benefits of the liveattenuated poliovirus vaccine over the killed poliovirus vaccine.

Question Type: Essay Chapter 23.6 Blooms: Analyzing Difficulty: Moderate 40. Cancer cells should be recognized by the immune system and destroyed. Which of the following cell and receptors combinations are often dysregulated in tumors? a. colon cancer epithelial cells – CTLA-4 b. macrophages – class I MHC c. Tregs – CD40 d. T cells – PD-1 Ans: d Question Type: Multiple choice Chapter 23.6 Blooms: Understanding Difficulty: Difficult

24 Cancer Section 24.1 1. Which of the following is a characteristic of malignant tumors? a. localized to tissue of origin

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24 - 232 b. metastatic c. well differentiated d. sense signals that restrict cell division Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate 2. Gain-of-function mutations in which of the following genes typify colorectal carcinomas? a. APC b. p53 c. K-ras d. all of the above Ans: c Question Type: Multiple choice Chapter: 24 Blooms: Remembering Difficulty: Easy 3. During metastasis, cells may undergo an epithelial-to-mesenchymal transition whereby there is a(n): a. increase in cell polarity. b. loss of cell-cell-adhesion. c. down-regulation in the expression of the Snail and Twist transcription factors. d. all of the above Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate 4. Which of the following promotes angiogenesis? a. EGF b. PDGF c. TGF d. VEGF Ans: d Question Type: Multiple choice Chapter: 24 Blooms: Remembering Difficulty: Easy 5. In females, all the cells in a tumor have the same inactive X chromosome. The reason for this is that: a. all of the cells in a tumor are derived from a single progenitor cell. b. all late-stage tumors have chromosome abnormalities. c. multiple mutations are required for tumor induction. d. many tumor-suppressor genes are on the X chromosome.

Ans: a Question Type: Multiple choice Chapter: 24 Blooms: Applying

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24 - 233 Difficulty: Easy 6. Name the six fundamental properties of malignant tumors. Which of these properties are amenable to study in a cell culture model of cancer?

Ans: Malignant tumors exhibit (1) self-sufficiency in growth signals; (2) insensitivity to antigrowth signals; (3) evasion of apoptosis; (4) limitless replicative potential; (5) sustained angiogenesis; and (6) tissue invasion and metastasis. The first four properties are amenable to study in a cell-culture model because they are autonomous properties of cancer cells. In contrast, sustained angiogenesis and tissue invasion and metastasis involve the interaction of cancer cells with other tissues and are better studied in vivo.

Question Type: Essay Chapter: 24 Blooms: Applying Difficulty: Difficult 7. Leukemias are different from most other classes of cancers. Which of the fundamental properties of cancer cells are likely to be lacking in leukemias and why?

Ans: Leukemias are derived from blood cells or their precursors and proliferate as individual cells in the blood rather than as solid tumors. Because they already exist in the circulation, leukemias do not need to promote angiogenesis and likewise do not need to escape their tissue of origin and metastasize, although some leukemias may do so.

Question Type: Essay Chapter: 24 Blooms: Understanding Difficulty: Moderate 8. What is the multi-hit model of cancer? What data support this model?

Ans: The multi-hit model proposes that multiple, successive mutations are required to produce a cancer cell. Many lines of evidence support this model, including the fact that tumor cells are clonally derived and possess the same genetic alterations.

The increased incidence of cancer with age also supports the model. Finally, the cooperative effect of oncogenes in producing tumors in mice supports the model.

Question Type: Essay Chapter: 24 Blooms: Analyzing Difficulty: Difficult 9. Describe the Warburg effect and how it applies to cancer cells.

Ans: Otto Warburg described how cancer cells, regardless of the level of oxygen in their environment, produce large amounts of lactate as a result of using aerobic glycolysis to produce energy.

Question Type: Essay Chapter: 24 Blooms: Understanding Difficulty: Easy 10. Which of the following is NOT a microscopic characteristic of tumor cells? a. high nuclear to cytoplasmic ratio b. greater percentage of mitotic cells c. larger size d. few specialized structures Ans: c Question Type: Multiple choice Chapter: 24 Blooms: Remembering

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24 - 234 Difficulty: Easy 11. 2-hydroxyglutarate is produced in cancer cells: a. because cancer cells exhibit increased aerobic glycolysis. b. that have mutations in isocitrate dehydrogenase. c. to counteract all the lactate produced by aerobic glycolysis. d. all of the above Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate 12. Which of the following is NOT a role for the tumor microenvironment? a. tumor cells contain mutations b. interaction with immune cells c. neighboring cells relay information to cancer cells d. HCV-mediated inflammation Ans: a Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate 13. Which of the following proteins involved in angiogenesis is paired correctly with its function? a. HIF – tyrosine kinase b. VEGF – receives a secreted signal to induce blood vessel growth c. oxygen sensor – transcription factor d. VEGF receptor – tyrosine kinase Ans: d Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate 14. Which of the following interventions would NOT be likely to prevent tumor cells from metastasizing? a. inhibitors of adhesion molecules b. inhibitors of enzymes that degrade the basement membrane c. inhibitors of cell motility d. inhibitors of EGF Ans: a Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate 15. A bone tumor sample appears to be heterozygous for a p53 mutation. The most likely explanation for this finding is: a. Mutated p53 is an oncogene. b. The p53 mutation is resulting in a dominant negative p53 protein. c. The p53 mutation is resulting in a premature stop codon, and no functional protein is made from that allele. d. There was a mistake in labeling the tissue sample—all tumors musts lack p53.

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24 - 235 Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate Section 24.2 16. Direct-acting carcinogens are those that: a. chemically react with nitrogen and phosphorous atoms in DNA. b. work in conjunction with cytochrome P-450 in the endoplasmic reticulum. c. are mainly reactive electrophiles. d. all of the above Ans: c Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate 17. Telomerase: a. induces apoptosis. b. contains reverse transcriptase. c. is inactivated by Bcl-2. d. is active in all normal adult cells.

Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate 18. Exposure to benzo(a)pyrene is most frequently associated with which type of cancer? a. lung cancer b. Burkitt’s lymphoma c. colorectal cancer d. breast cancer Ans: a Question Type: Multiple choice Chapter: 24 Blooms: Remembering Difficulty: Easy 19. The first line of defense against point mutations is: a. BRCA1. b. cytochrome P-450. c. DNA polymerase. d. MSH2.

Ans: c Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate

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24 - 236 20. Describe the differences between direct-acting and indirect-acting carcinogens.

Ans: Direct-acting carcinogens have reactive electrophiles that react with nitrogen and oxygen atoms in DNA. These modified nucleotides distort the normal pattern of base pairing. Indirect-acting carcinogens are generally unreactive, waterinsoluble compounds. These compounds must first be converted to ultimate carcinogens by the introduction of electrophilic centers, usually by cytochrome P-450 enzymes in the liver.

Question Type: Essay Chapter: 24 Blooms: Understanding Difficulty: Difficult 21. Which of the following is NOT evidence that smoking causes lung cancer? a. Epidemiological rates of lung cancer dramatically increased after more people started to smoke. b. Exposure of lung cells to the active carcinogen found in cigarettes causes mutations in p53. c. Mutations in p53 are found in the same codons in patients with lung cancer who smoke as in cultured cells treated with benzo(a)pyrene. d. Epidemiological rates of lung cancer in women increased years after the rates in men increased.

Ans: d Question Type: Multiple choice Chapter: 24 Blooms: Applying Difficulty: Moderate 22. If colon cancers could be prevented by intervening in the earliest molecular changes during multi-hit progression, the molecule you would target for intervention would be: a. p53. b. Ras. c. APC. d. VEGF. Ans: c Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Easy 23. Transgenic mice that express an activated oncogene like rasD under the control of the TetOFF system are likely to undergo which of the following events when tetracycline is given? a. growth of tumors b. shrinkage of tumors c. no effect on tumors d. tumors become metastatic Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Applying Difficulty: Moderate Section 24.3 24. Which of the following is(are) a tumor suppressor gene? a. APC b. ras

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24 - 237 c. Rb d. APC and Rb Ans: d Question Type: Multiple choice Chapter: 24 Blooms: Remembering Difficulty: Easy 25. Which of the following is a proto-oncogene? a. APC b. myc c. ptc1 d. Rb Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Remembering Difficulty: Easy 26. Germ-line mutations in which of the following has(have) been implicated in hereditary cancers? a. HPV b. BRCA1 c. Rb d. BRCA1 and Rb Ans: d Question Type: Multiple choice Chapter: 24 Blooms: Remembering Difficulty: Easy 27. Hereditary cancers typically possess loss-of-heterozygosity in: a. proto-oncogenes. b. tumor-suppressor genes. c. both proto-oncogenes and tumor-suppressor genes. d. neither proto-oncogenes nor tumor-suppressor genes.

Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Applying Difficulty: Moderate 28. Nearly all malignant tumors possess a loss-of-function mutation in one or more cell cycle checkpoints. However, a loss of checkpoints is not a requisite characteristic of cancer cells per se. Explain this paradox.

Ans: Because multiple mutations are required to produce the cancer phenotype, a loss of checkpoint function will greatly increase the likelihood that cancer-promoting mutations will develop. It is possible that these mutations could occur by chance, even with all checkpoint pathways intact. However, cancer would be a much less frequent event if mutations in checkpoint genes never occurred.

Question Type: Essay Chapter: 24 Blooms: Analyzing Difficulty: Difficult

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29. What are the differences and similarities between the transforming genes of retroviruses and those of DNA tumor viruses?

Ans: The oncoproteins encoded by transducing retroviruses are derived from normal cell proteins, whereas those encoded by DNA viruses are required for viral replication. Indeed, the presence of oncogenes in transducing retroviruses renders these viruses defective (unable to replicate). Both types of viruses express oncoproteins from integrated genomes.

Question Type: Essay Chapter: 24 Blooms: Applying Difficulty: Difficult 30. Describe gain-of-function and loss-of-function mutations with respect to cancer.

Ans: A gain-of-function mutation converts a proto-oncogene into an oncogene. Such mutations are dominant. A loss-offunction mutation occurs in tumor-suppressor genes, abrogating their function in preventing cancer. Such mutations are recessive and require alteration of both alleles.

Question Type: Essay Chapter: 24 Blooms: Understanding Difficulty: Moderate 31. How can DNA microarrays be used to detect gene amplifications in cancer cells?

Ans: Using genomic DNA from cancer cells as probes and DNA microarrays containing fragments of normal genomic DNA, it is possible to tell the difference between amplified DNA and normal DNA. Amplified genes give stronger signals than do normal genes.

Question Type: Essay Chapter: 24 Blooms: Applying Difficulty: Moderate 32. A patient with B cell leukemia donates some cancer cells for analysis. The cells do not contain the usual chromosomal translocations associated with leukemia. Instead, a sporadic mutation seems to have arisen in a certain gene, followed by loss of heterozygosity in the second copy. This gene product is most likely: a. Ras. b. Rb. c. VEGF. d. telomerase.

Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Applying Difficulty: Moderate 33. Cancer genome analysis has been utilized for all the following applications, except: a. to identify which patients are good candidates for molecularly targeted drugs. b. to identify cancerous cells in a biopsy based on morphological characteristics. c. to identify driver mutations. d. to identify the link between chromothripsis and aggressive neuroblastoma.

Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Applying

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24 - 239 Difficulty: Moderate Section 24.4 34. Ras is a: a. growth factor. b. kinase. c. phosphatase. d. none of the above Ans: d Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Easy 35. Src is a: a. growth factor. b. kinase. c. phosphatase. d. transcription factor.

Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Easy 36. A loss-of-function mutation in a. Crk b. Csk c. Myc d. NF1

would have the same effect as a gain-of-function mutation in Ras.

Ans: d Question Type: Multiple choice Chapter: 24 Blooms: Applying Difficulty: Moderate 37. Burkitt’s lymphoma results from the overproduction of: a. Fos. b. Myc. c. Ras. d. Src. Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Remembering Difficulty: Easy 38. What renders erbB and her2 oncogenic? What are their normal proto-oncogene counterparts?

Ans: ErbB and her2 both encode constitutively active receptor tyrosine kinases. That is, they encode proteins that transduce a growth-promoting signal even in the absence of ligand binding. ErbB is a mutant form of the EGF receptor, and Neu is a

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24 - 240 mutant form of the Her2 receptor. ErbB possesses a deletion in the extracellular domain; Neu possesses a point mutation.

Both mutant receptors dimerize and become active as kinases in the absence of ligand.

Question Type: Essay Chapter: 24 Blooms: Understanding Difficulty: Moderate 39. Explain how the chemotherapeutic drug Gleevec works in the treatment of myelogenous leukemia.

Ans: Gleevec specifically targets the fusion protein Abl kinase that is present in CML cells. It inhibits Abl kinase activity and is highly lethal to CML cells but does not inhibit normal tyrosine kinase activity.

Question Type: Essay Chapter: 24 Blooms: Understanding Difficulty: Moderate Section 24.5 40. Which of the following proteins belong to the control system that regulates cells past a certain point in late G1 (START)? a. CDK4 b. cyclin B c. Rb d. CDK4 and Rb Ans: d Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate 41. A gain-of-function mutation in a. cyclin D b. p16 c. Rb d. cyclin D and p16

will bypass restriction point controls.

Ans: a Question Type: Multiple choice Chapter: 24 Blooms: Applying Difficulty: Moderate 42. The human papillomavirus E7 protein inhibits the function of: a. p53. b. Rb. c. PDGF receptor. d. all of the above Ans: b Question Type: Multiple choice Chapter: 24 Blooms: Remembering Difficulty: Easy 43. MicroRNAs are a new class of oncogenic factors because they function: a. as tumor suppressors.

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24 - 241 b. as oncogenes. c. to induce errors during DNA replication. d. as tumor suppressors and as oncogenes.

Ans: d Question Type: Multiple choice Chapter: 24 Blooms: Understanding Difficulty: Moderate 44. Mutation of p53 is described as a dominant-negative mutation. Describe the mechanism by which this mutation causes the dominant-negative phenotype.

Ans: Mutation of p53 is a dominant mutation because the active protein is a tetramer of identical subunits; the presence of even one defective subunit in the complex abrogates its function. It is a negative mutation because the normal function of the protein is lost.

Question Type: Essay Chapter: 24 Blooms: Analyzing Difficulty: Moderate

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