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Chapter 6 Enzymes 60

Chapter 6 Enzymes

Multiple Choice Questions 1. An introduction to enzymes Pages: 183-184 Difficulty: 1 Ans: A One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis. Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the: A) B) C) D) E)

apoenzyme. coenzyme. holoenzyme. prosthetic group. substrate.

2. An introduction to enzymes Page: 185 Difficulty: 1 Ans: D Which one of the following is not among the six internationally accepted classes of enzymes? A) B) C) D) E)

Hydrolases Ligases Oxidoreductases Polymerases Transferases

3. How enzymes work Pages: 186-187 Difficulty: 2 Ans: E Enzymes are potent catalysts because they: A) B) C) D) E)

are consumed in the reactions they catalyze. are very specific and can prevent the conversion of products back to substrates. drive reactions to completion while other catalysts drive reactions to equilibrium. increase the equilibrium constants for the reactions they catalyze. lower the activation energy for the reactions they catalyze.

4. How enzymes work Pages: 186-187 Difficulty: 1 Ans: D The role of an enzyme in an enzyme-catalyzed reaction is to: A) B) C) D) E)

bind a transition state intermediate, such that it cannot be converted back to substrate. ensure that all of the substrate is converted to product. ensure that the product is more stable than the substrate. increase the rate at which substrate is converted into product. make the free-energy change for the reaction more favorable.

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5. How enzymes work Pages: 186-188 Difficulty: 2 Ans:D Which one of the following statements is true of enzyme catalysts? A) B) C) D) E)

Their catalytic activity is independent of pH. They are generally equally active on D and L isomers of a given substrate. They can increase the equilibrium constant for a given reaction by a thousand fold or more. They can increase the reaction rate for a given reaction by a thousand fold or more. To be effective, they must be present at the same concentration as their substrate.

6. How enzymes work Pages: 186-188 Difficulty: 2 Ans: D Which one of the following statements is true of enzyme catalysts? A) They bind to substrates, but are never covalently attached to substrate or product. B) They increase the equilibrium constant for a reaction, thus favoring product formation. C) They increase the stability of the product of a desired reaction by allowing ionizations, resonance, and isomerizations not normally available to substrates. D) They lower the activation energy for the conversion of substrate to product. E) To be effective they must be present at the same concentration as their substrates. 7. How enzymes work Pages: 186-188 Difficulty: 1 Ans: C Which of the following statements is false? A) B) C) D) E)

A reaction may not occur at a detectable rate even though it has a favorable equilibrium. After a reaction, the enzyme involved becomes available to catalyze the reaction again. For S  P, a catalyst shifts the reaction equilibrium to the right. Lowering the temperature of a reaction will lower the reaction rate. Substrate binds to an enzyme's active site.

8. How enzymes work Pages: 189-190 Difficulty: 1 Ans: B Enzymes differ from other catalysts in that only enzymes: A) B) C) D) E)

are not consumed in the reaction. display specificity toward a single reactant. fail to influence the equilibrium point of the reaction. form an activated complex with the reactants. lower the activation energy of the reaction catalyzed.

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9. How enzymes work Page: 190 Difficulty: 2 Ans: A Compare the two reaction coordinate diagrams below and select the answer that correctly describes their relationship. In each case, the single intermediate is the ES complex.

A) (a) describes a strict “lock and key” model, whereas (b) describes a transition-state complementarity model. B) The activation energy for the catalyzed reaction is #5 in (a) and is #7 in (b). C) The activation energy for the uncatalyzed reaction is given by #5 + #6 in (a) and by #7 + #4 in (b). D) The contribution of binding energy is given by #5 in (a) and by #7 in (b). E) The ES complex is given by #2 in (a) and #3 in (b). 10. How enzymes work Pages: 190-191 Difficulty: 2 Ans: B Which of the following is true of the binding energy derived from enzyme-substrate interactions? A) B) C) D) E)

It cannot provide enough energy to explain the large rate accelerations brought about by enzymes. It is sometimes used to hold two substrates in the optimal orientation for reaction. It is the result of covalent bonds formed between enzyme and substrate. Most of it is derived from covalent bonds between enzyme and substrate. Most of it is used up simply binding the substrate to the enzyme.

11. Enzyme kinetics as an approach to understanding mechanism Page: 192 Difficulty: 1 Ans: D The concept of “induced fit” refers to the fact that: A) B) C) D)

enzyme specificity is induced by enzyme-substrate binding. enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction. enzyme-substrate binding induces movement along the reaction coordinate to the transition state. substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation. E) when a substrate binds to an enzyme, the enzyme induces a loss of water (desolvation) from the substrate.

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12. Enzyme kinetics as an approach to understanding mechanism Pages: 192-193 Difficulty: 2 Ans: A In the following diagram of the first step in the reaction catalyzed by the protease chymotrypsin, the process of general base catalysis is illustrated by the number ________, and the process of covalent catalysis is illustrated by the number _________.

A) B) C) D) E)

1; 2 1; 3 2; 3 2; 3 3; 2

13. Enzyme kinetics as an approach to understanding mechanism Page: 194 Difficulty: 1 Ans: B The benefit of measuring the initial rate of a reaction V0 is that at the beginning of a reaction: A) B) C) D) E)

[ES] can be measured accurately. changes in [S] are negligible, so [S] can be treated as a constant. changes in Km are negligible, so Km can be treated as a constant. V0 = Vmax. varying [S] has no effect on V0.

14. Enzyme kinetics as an approach to understanding mechanism Pages: 195-199 Difficulty: 3 Ans: B Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows MichaelisMenten kinetics is false? A) B) C) D) E)

As [S] increases, the initial velocity of reaction V0 also increases. At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km. Km is the [S] at which V0 = 1/2 Vmax. The shape of the curve is a hyperbola. The y-axis is a rate term with units of m/min.

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15. Enzyme kinetics as an approach to understanding mechanism Page: 196 Difficulty: 2 Ans: D Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed reaction could be written as k1 k2 E+S ES  E + P k-1 Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by the expression: A) B) C) D) E)

k1 ([Et]  [ES]). k1 ([Et]  [ES])[S]. k2 [ES]. k-1 [ES] + k2 [ES]. k-1 [ES].

16. Enzyme kinetics as an approach to understanding mechanism Page: 196 Difficulty: 2 Ans: C The steady state assumption, as applied to enzyme kinetics, implies: A) B) C) D) E)

Km = Ks. the enzyme is regulated. the ES complex is formed and broken down at equivalent rates. the Km is equivalent to the cellular substrate concentration. the maximum velocity occurs when the enzyme is saturated.

17. Enzyme kinetics as an approach to understanding mechanism Pages: 196-199 Difficulty: 3 Ans: C An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 mol. If, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 mol) of product to be formed? A) B) C) D) E)

1.5 min 13.5 min 27 min 3 min 6 min

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18. Enzyme kinetics as an approach to understanding mechanism Pages: 196-201 Difficulty: 3 Ans: D Which of these statements about enzyme-catalyzed reactions is false? A) At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration. B) If enough substrate is added, the normal Vmax of a reaction can be attained even in the presence of a competitive inhibitor. C) The rate of a reaction decreases steadily with time as substrate is depleted. D) The activation energy for the catalyzed reaction is the same as for the uncatalyzed reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed reaction. E) The Michaelis-Menten constant Km equals the [S] at which V = 1/2 Vmax. 19. Enzyme kinetics as an approach to understanding mechanism Page: 197 Difficulty: 2 Ans: C The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: V0 Substrate added (mol/min) (mmol/L) ————————————— 217 0.8 325 2 433 4 488 6 647 1,000 ————————————— The Km for this enzyme is approximately: A) B) C) D) E)

1 mM. 1,000 mM. 2 mM. 4 mM. 6 mM.

20. Enzyme kinetics as an approach to understanding mechanism Page: 196-197 Difficulty: 2 Ans: C For enzymes in which the slowest (rate-limiting) step is the reaction k2 ES  P Km becomes equivalent to: A) B) C) D) E)

kcat. the [S] where V0 = Vmax. the dissociation constant, Kd, for the ES complex. the maximal velocity. the turnover number.

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21. Enzyme kinetics as an approach to understanding mechanism Page: 197 Difficulty: 2 Ans: D The Lineweaver-Burk plot is used to: A) B) C) D) E)

determine the equilibrium constant for an enzymatic reaction. extrapolate for the value of reaction rate at infinite enzyme concentration. illustrate the effect of temperature on an enzymatic reaction. solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration. solve, graphically, for the ratio of products to reactants for any starting substrate concentration.

22. Enzyme kinetics as an approach to understanding mechanism Page: 197 Difficulty: 3 Ans: A The double-reciprocal transformation of the Michaelis-Menten equation, also called the LineweaverBurk plot, is given by 1/V0 = Km /(Vmax[S]) + 1/Vmax. To determine Km from a double-reciprocal plot, you would: A) B) C) D) E)

multiply the reciprocal of the x-axis intercept by 1. multiply the reciprocal of the y-axis intercept by 1. take the reciprocal of the x-axis intercept. take the reciprocal of the y-axis intercept. take the x-axis intercept where V0 = 1/2 Vmax.

23. Enzyme kinetics as an approach to understanding mechanism Page: 198 Difficulty: 2 Ans: E To calculate the turnover number of an enzyme, you need to know: A) B) C) D) E)

the enzyme concentration. the initial velocity of the catalyzed reaction at [S] >> Km. the initial velocity of the catalyzed reaction at low [S]. the Km for the substrate. both A and B.

24. Enzyme kinetics as an approach to understanding mechanism Page: 198 Difficulty: 1 Ans: E The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the: A) B) C) D) E)

dissociation constant. half-saturation constant. maximum velocity. Michaelis-Menten number. turnover number.

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25. Enzyme kinetics as an approach to understanding mechanism Pages: 201-202 Difficulty: 3 Ans: B In a plot of l/V against 1/[S] for an enzyme-catalyzed reaction, the presence of a competitive inhibitor will alter the: A) B) C) D) E)

curvature of the plot. intercept on the l/[S] axis. intercept on the l/V axis. pK of the plot. Vmax.

26. Enzyme kinetics as an approach to understanding mechanism Pages: 201-202 Difficulty: 1 Ans: D In competitive inhibition, an inhibitor: A) B) C) D) E)

binds at several different sites on an enzyme. binds covalently to the enzyme. binds only to the ES complex. binds reversibly at the active site. lowers the characteristic Vmax of the enzyme.

27. Enzyme kinetics as an approach to understanding mechanism Pages: 201-205 Difficulty: 2 Ans: D Vmax for an enzyme-catalyzed reaction: A) B) C) D) E)

generally increases when pH increases. increases in the presence of a competitive inhibitor. is limited only by the amount of substrate supplied. is twice the rate observed when the concentration of substrate is equal to the Km. is unchanged in the presence of a uncompetitive inhibitor.

28. Enzyme kinetics as an approach to understanding mechanism Page: 204 Difficulty: 2 Ans: B Enzyme X exhibits maximum activity at pH = 6.9. X shows a fairly sharp decrease in its activity when the pH goes much lower than 6.4. One likely interpretation of this pH activity is that: A) B) C) D) E)

a Glu residue on the enzyme is involved in the reaction. a His residue on the enzyme is involved in the reaction. the enzyme has a metallic cofactor. the enzyme is found in gastric secretions. the reaction relies on specific acid-base catalysis.

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29. Examples of enzymatic reactions Pages: 203-204 Difficulty: 3 Ans: A Phenyl-methane-sulfonyl-fluoride (PMSF) inactivates serine proteases by binding covalently to the catalytic serine residue at the active site; this enzyme-inhibitor bond is not cleaved by the enzyme. This is an example of what kind of inhibition? A) B) C) D) E)

irreversible competitive non-competitive mixed pH inhibition

30. Examples of enzymatic reactions Page: 212 Difficulty: 2 Ans: B Both water and glucose share an —OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water. The best explanation is that: F) glucose has more —OH groups per molecule than does water. G) the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis. H) the —OH group of water is attached to an inhibitory H atom, while the glucose —OH group is attached to C. I) water and the second substrate, ATP, compete for the active site resulting in a competitive inhibition of the enzyme. J) water normally will not reach the active site because it is hydrophobic. 31. Examples of enzymatic reactions Pages: 210-211 Difficulty: 2 Ans: B A good transition-state analog: A) B) C) D) E)

binds covalently to the enzyme. binds to the enzyme more tightly than the substrate. binds very weakly to the enzyme. is too unstable to isolate. must be almost identical to the substrate.

32. Examples of enzymatic reactions Pages: 210-211 Difficulty: 1 Ans: C A transition-state analog: A) B) C) D) E)

is less stable when binding to an enzyme than the normal substrate. resembles the active site of general acid-base enzymes. resembles the transition-state structure of the normal enzyme-substrate complex. stabilizes the transition state for the normal enzyme-substrate complex. typically reacts more rapidly with an enzyme than the normal substrate.

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33. Examples of enzymatic reactions Page: 213 Difficulty: 2 Ans: D The role of the metal ion (Mg2+) in catalysis by enolase is to A) B) C) D) E)

act as a general acid catalyst act as a general base catalyst facilitate general acid catalysis facilitate general base catalysis stabilize protein conformation

34. Examples of enzymatic reactions Page: 216-217 Difficulty: 2 Ans: C Penicillin and related drugs inhibit the enzyme A) B) C) D) E)

; this enzyme is produced by

.

-lacamase; bacteria transpeptidase; human cells transpeptidase; bacteria lysozyme; human cells aldolase; bacteria

35. Regulatory enzymes Pages: 220-221 Difficulty: 2 Ans: E Which of the following statements about allosteric control of enzymatic activity is false? A) B) C) D) E)

Allosteric effectors give rise to sigmoidal V0 vs. [S] kinetic plots. Allosteric proteins are generally composed of several subunits. An effector may either inhibit or activate an enzyme. Binding of the effector changes the conformation of the enzyme molecule. Heterotropic allosteric effectors compete with substrate for binding sites.

36. Regulatory enzymes Pages: 220-221 Difficulty: 1 Ans: A A small molecule that decreases the activity of an enzyme by binding to a site other than the catalytic site is termed a(n): A) B) C) D) E)

allosteric inhibitor. alternative inhibitor. competitive inhibitor. stereospecific agent. transition-state analog.

37. Regulatory enzymes Pages: 220-221 Difficulty: 1 Allosteric enzymes: A) B) C) D) E)

Ans: C

are regulated primarily by covalent modification. usually catalyze several different reactions within a metabolic pathway. usually have more than one polypeptide chain. usually have only one active site. usually show strict Michaelis-Menten kinetics.

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38. Regulatory enzymes Pages: 221-222 Difficulty: 2 Ans: C A metabolic pathway proceeds according to the scheme, R  S  T  U  V  W. A regulatory enzyme, X, catalyzes the first reaction in the pathway. Which of the following is most likely correct for this pathway? A) Either metabolite U or V is likely to be a positive modulator, increasing the activity of X. B) The first product S, is probably the primary negative modulator of X, leading to feedback inhibition. C) The last product, W, is likely to be a negative modulator of X, leading to feedback inhibition. D) The last product, W, is likely to be a positive modulator, increasing the activity of X. E) The last reaction will be catalyzed by a second regulatory enzyme. 39. Regulatory enzymes Pages: 223-226 Difficulty: 3 Ans: A Which of the following has not been shown to play a role in determining the specificity of protein kinases? A) B) C) D) E)

Disulfide bonds near the phosphorylation site Primary sequence at phosphorylation site Protein quaternary structure Protein tertiary structure Residues near the phosphorylation site

40. Regulatory enzymes Page: 226 Difficulty: 1 Ans: C How is trypsinogen converted to trypsin? A) B) C) D)

A protein kinase-catalyzed phosphorylation converts trypsinogen to trypsin. An increase in Ca2+ concentration promotes the conversion. Proteolysis of trypsinogen forms trypsin. Trypsinogen dimers bind an allosteric modulator, cAMP, causing dissociation into active trypsin monomers. E) Two inactive trypsinogen dimers pair to form an active trypsin tetramer.

Short Answer Questions 41. An introduction to enzymes Page: 184 Difficulty: 1 Define the terms “cofactor” and “coenzyme.” Ans: A cofactor is any chemical component required for enzyme activity; it includes both organic molecules, called “coenzymes,” and inorganic ions. 42. How enzymes work Page: 187 Difficulty: 2 Draw and label a reaction coordinate diagram for an uncatalyzed reaction, S  P, and the same reaction catalyzed by an enzyme, E.

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Ans: See Fig. 6-3, ...