Molecular Cell Biology Eighth Edition Test Bank

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2 - 1
2 Chemical Foundations
Test Bank
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 D-tryptophan
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.
b. has a positive 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 dissolving
urea in water at room temperature, it becomes cold. How should the 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 for this reaction is 19 under standard conditions. What is the 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

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3
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
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
Protein Structure
and Function

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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
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

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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
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.

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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
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

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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
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

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