Chapter 6 Enzymes
Multiple Choice Questions
1. An introduction to enzymes
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) apoenzyme.
B) coenzyme.
C) holoenzyme.
D) prosthetic group.
E) substrate.
2. An introduction to enzymes
Which one of the following is not among the six internationally accepted classes of enzymes?
A) Hydrolases
B) Ligases
C) Oxidoreductases
D) Polymerases
E) Transferases
3. How enzymes work
Enzymes are potent catalysts because they:
A) are consumed in the reactions they catalyze.
B) are very specific and can prevent the conversion of products back to substrates.
C) drive reactions to completion while other catalysts drive reactions to equilibrium.
D) increase the equilibrium constants for the reactions they catalyze.
E) lower the activation energy for the reactions they catalyze.
4. How enzymes work
The role of an enzyme in an enzyme-catalyzed reaction is to:
A) bind a transition state intermediate, such that it cannot be converted back to substrate.
B) ensure that all of the substrate is converted to product.
C) ensure that the product is more stable than the substrate.
D) increase the rate at which substrate is converted into product.
E) make the free-energy change for the reaction more favorable.
Chapter 6 Enzymes
61
Which one of the following statements is true of enzyme catalysts?
A) Their catalytic activity is independent of pH.
B) They are generally equally active on D and L isomers of a given substrate.
C) They can increase the equilibrium constant for a given reaction by a thousand fold or more.
D) They can increase the reaction rate for a given reaction by a thousand-fold or more.
E) To be effective, they must be present at the same concentration as their substrate.
6. How enzymes work
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
Which of the following statements is false?
A) A reaction may not occur at a detectable rate even though it has a favorable equilibrium.
B) After a reaction, the enzyme involved becomes available to catalyze the reaction again.
C) For S → P, a catalyst shifts the reaction equilibrium to the right.
D) Lowering the temperature of a reaction will lower the reaction rate.
E) Substrate binds to an enzyme’s active site.
8. How enzymes work
Enzymes differ from other catalysts in that only enzymes:
A) are not consumed in the reaction.
B) display specificity toward a single reactant.
C) fail to influence the equilibrium point of the reaction.
D) form an activated complex with the reactants.
E) lower the activation energy of the reaction catalyzed.
Chapter 6 Enzymes
62
9. How enzymes work
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
Which of the following is true of the binding energy derived from enzyme-substrate interactions?
A) It cannot provide enough energy to explain the large rate accelerations brought about by enzymes.
B) It is sometimes used to hold two substrates in the optimal orientation for reaction.
C) It is the result of covalent bonds formed between enzyme and substrate.
D) Most of it is derived from covalent bonds between enzyme and substrate.
E) Most of it is used up simply binding the substrate to the enzyme.
11. Enzyme kinetics as an approach to understanding mechanism
The concept of “induced fit” refers to the fact that:
A) enzyme specificity is induced by enzyme-substrate binding.
B) enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction.
C) enzyme-substrate binding induces movement along the reaction coordinate to the transition state.
D) 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.
Chapter 6 Enzymes
63
12. Enzyme kinetics as an approach to understanding mechanism
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) 1; 2
B) 1; 3
C) 2; 3
D) 2; 3
E) 3; 2
13. Enzyme kinetics as an approach to understanding mechanism
The benefit of measuring the initial rate of a reaction V0 is that at the beginning of a reaction:
A) [ES] can be measured accurately.
B) changes in [S] are negligible, so [S] can be treated as a constant.
C) changes in Km are negligible, so Km can be treated as a constant.
D) V0 = Vmax.
E) varying [S] has no effect on V0.
14. Enzyme kinetics as an approach to understanding mechanism
Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows Michaelis-
Menten kinetics is false?
A) As [S] increases, the initial velocity of reaction V0 also increases.
B) At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km.
C) Km is the [S] at which V0 = 1/2 Vmax.
D) The shape of the curve is a hyperbola.
E) The y-axis is a rate term with units of m/min.
Chapter 6 Enzymes
64
15. Enzyme kinetics as an approach to understanding mechanism
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) k1 ([Et] − [ES]).
B) k1 ([Et] − [ES])[S].
C) k2 [ES].
D) k-1 [ES] + k2 [ES].
E) k-1 [ES].
16. Enzyme kinetics as an approach to understanding mechanism
The steady state assumption, as applied to enzyme kinetics, implies:
A) Km = Ks.
B) the enzyme is regulated.
C) the ES complex is formed and broken down at equivalent rates.
D) the Km is equivalent to the cellular substrate concentration.
E) the maximum velocity occurs when the enzyme is saturated.
17. Enzyme kinetics as an approach to understanding mechanism
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) 1.5 min
B) 13.5 min
C) 27 min
D) 3 min
E) 6 min
18. Enzyme kinetics as an approach to understanding mechanism
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.
Chapter 6 Enzymes
65
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
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) 1 mM.
B) 1000 mM.
C) 2 mM.
D) 4 mM.
E) 6 mM.
20. Enzyme kinetics as an approach to understanding mechanism
For enzymes in which the slowest (rate-limiting) step is the reaction
k2
ES → P
Km becomes equivalent to:
A) kcat.
B) the [S], where V0 = Vmax.
C) the dissociation constant, Kd, for the ES complex.
D) the maximal velocity.
E) the turnover number.
21. Enzyme kinetics as an approach to understanding mechanism
For the simplified representation of an enzyme-catalyzed reaction shown below, the
statement “ES is in steady-state” means that:
k1 k2
E + S ES E + P
k-1 k-2
A) k2 is very slow.
Chapter 6 Enzymes
66
B) k1= k2.
C) k1= k-1.
D) k1[E][S] = k-1[ES] + k2[ES].
E) k1[E][S] = k-1[ES].
22. Enzyme kinetics as an approach to understanding mechanism
The Lineweaver-Burk plot is used to:
A) determine the equilibrium constant for an enzymatic reaction.
B) extrapolate for the value of reaction rate at infinite enzyme concentration.
C) illustrate the effect of temperature on an enzymatic reaction.
D) solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration.
E) solve, graphically, for the ratio of products to reactants for any starting substrate concentration.
23. Enzyme kinetics as an approach to understanding mechanism
The double-reciprocal transformation of the Michaelis-Menten equation, also called the Lineweaver-
Burk plot, is given by
1/V0 = Km /(Vmax[S]) + 1/Vmax
To determine Km from a double-reciprocal plot, you would:
A) multiply the reciprocal of the x-axis intercept by −1.
B) multiply the reciprocal of the y-axis intercept by −1.
C) take the reciprocal of the x-axis intercept.
D) take the reciprocal of the y-axis intercept.
E) take the x-axis intercept, where V0 = 1/2 Vmax.
24. Enzyme kinetics as an approach to understanding mechanism
To calculate the turnover number of an enzyme, you need to know:
A) the enzyme concentration.
B) the initial velocity of the catalyzed reaction at [S] >> Km.
C) the initial velocity of the catalyzed reaction at low [S].
D) the Km for the substrate.
E) both A and B.
25. Enzyme kinetics as an approach to understanding mechanism
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) dissociation constant.
B) half-saturation constant.
C) maximum velocity.
D) Michaelis-Menten number.
E) turnover number.
Chapter 6 Enzymes
67
26. Enzyme kinetics as an approach to understanding mechanism
In a plot of l/V against 1/[S] for an enzyme-catalyzed reaction, the presence of a competitive inhibitor
will alter the:
A) curvature of the plot.
B) intercept on the l/[S] axis.
C) intercept on the l/V axis.
D) pK of the plot.
E) Vmax.
27. Enzyme kinetics as an approach to understanding mechanism
In competitive inhibition, an inhibitor:
A) binds at several different sites on an enzyme.
B) binds covalently to the enzyme.
C) binds only to the ES complex.
D) binds reversibly at the active site.
E) lowers the characteristic Vmax of the enzyme.
28. Enzyme kinetics as an approach to understanding mechanism
Vmax for an enzyme-catalyzed reaction:
A) generally increases when pH increases.
B) increases in the presence of a competitive inhibitor.
C) is limited only by the amount of substrate supplied.
D) is twice the rate observed when the concentration of substrate is equal to the Km.
E) is unchanged in the presence of a uncompetitive inhibitor.
29. Enzyme kinetics as an approach to understanding mechanism
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) a Glu residue on the enzyme is involved in the reaction.
B) a His residue on the enzyme is involved in the reaction.
C) the enzyme has a metallic cofactor.
D) the enzyme is found in gastric secretions.
E) the reaction relies on specific acid-base catalysis.
30. Examples of enzymatic reactions
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) Irreversible
Chapter 6 Enzymes
68
B) Competitive
C) Non-competitive
D) Mixed
E) pH inhibition
31. Examples of enzymatic reactions
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:
A) glucose has more —OH groups per molecule than does water.
B) 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.
C) the —OH group of water is attached to an inhibitory H atom, while the glucose —OH group is
attached to C.
D) water and the second substrate, ATP, compete for the active site resulting in a competitive
inhibition of the enzyme.
E) water normally will not reach the active site because it is hydrophobic.
32. Examples of enzymatic reactions
A good transition-state analog:
A) binds covalently to the enzyme.
B) binds to the enzyme more tightly than the substrate.
C) binds very weakly to the enzyme.
D) is too unstable to isolate.
E) must be almost identical to the substrate.
33. Examples of enzymatic reactions
A transition-state analog:
A) is less stable when binding to an enzyme than the normal substrate.
B) resembles the active site of general acid-base enzymes.
C) resembles the transition-state structure of the normal enzyme-substrate complex.
D) stabilizes the transition state for the normal enzyme-substrate complex.
E) typically reacts more rapidly with an enzyme than the normal substrate.
34. Examples of enzymatic reactions
The role of the metal ion (Mg2+) in catalysis by enolase is to:
A) act as a general acid catalyst.
B) act as a general base catalyst.
C) facilitate general acid catalysis.
D) facilitate general base catalysis.
E) stabilize protein conformation.
35. Examples of enzymatic reactions
Chapter 6 Enzymes
69
Penicillin and related drugs inhibit the enzyme ; this enzyme is produced by .
A) -lacamase; bacteria
B) transpeptidase; human cells
C) transpeptidase; bacteria
D) lysozyme; human cells
E) aldolase; bacteria
36. Regulatory enzymes
Which of the following statements about allosteric control of enzymatic activity is false?
A) Allosteric effectors give rise to sigmoidal V0 vs. [S] kinetic plots.
B) Allosteric proteins are generally composed of several subunits.
C) An effector may either inhibit or activate an enzyme.
D) Binding of the effector changes the conformation of the enzyme molecule.
E) Heterotropic allosteric effectors compete with substrate for binding sites.
37. Regulatory enzymes
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) allosteric inhibitor.
B) alternative inhibitor.
C) competitive inhibitor.
D) stereospecific agent.
E) transition-state analog.
38. Regulatory enzymes
Allosteric enzymes:
A) are regulated primarily by covalent modification.
B) usually catalyze several different reactions within a metabolic pathway.
C) usually have more than one polypeptide chain.
D) usually have only one active site.
E) usually show strict Michaelis-Menten kinetics.
39. Regulatory enzymes
Which of the following has not been shown to play a role in determining the specificity of protein
kinases?
A) Disulfide bonds near the phosphorylation site
B) Primary sequence at phosphorylation site
C) Protein quaternary structure
D) Protein tertiary structure
E) Residues near the phosphorylation site
Chapter 6 Enzymes
70
40. Regulatory enzymes
How is trypsinogen converted to trypsin?
A) A protein kinase-catalyzed phosphorylation converts trypsinogen to trypsin.
B) An increase in Ca2+ concentration promotes the conversion.
C) Proteolysis of trypsinogen forms trypsin.
D) 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.
41. Regulatory enzymes
The allosteric enzyme ATCase is regulated by CTP, which binds to the T-state of ATCase. CTP is a:
A) positive regulator.
B) negative regulator.
C) co-factor.
D) competitive inhibitor.
E) coenzyme.
42. Regulatory enzymes
Blood coagulation involves:
A) a kinase cascade.
B) zymogen activation.
C) serine proteases.
D) A and B.
E) B and C.
Short Answer Questions
43. An introduction to enzymes
Page: 184 Difficulty: 1
Define the terms “cofactor” and “coenzyme.”
44. 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.
Chapter 6 Enzymes
71
45. How enzymes work
Page: 187 Difficulty: 1
The difference in (standard) free energy content, G’°, between substrate S and product P may vary
considerably among different reactions. What is the significance of these differences?
46. How enzymes work
Page: 187 Difficulty: 2
For a reaction that can take place with or without catalysis by an enzyme, what would be the effect of
the enzyme on the:
(a) standard free energy change of the reaction?
(b) activation energy of the reaction?
(c) initial velocity of the reaction?
(d) equilibrium constant of the reaction?
47. How enzymes work
Pages: 187−188 Difficulty: 2
Sometimes the difference in (standard) free-energy content, G’°, between a substrate S and a product
P is very large, yet the rate of chemical conversion, S → P, is quite slow. Why?
48. How enzymes work
Page: 188 Difficulty: 2
Write an equilibrium expression for the reaction S → P and briefly explain the relationship between
the value of the equilibrium constant and free energy.
49. Enzyme kinetics as an approach to understanding mechanism
Pages: 192−193 Difficulty: 2
What is the difference between general acid-base catalysis and specific acid-base catalysis? (Assume
that the solvent is water.)
Chapter 6 Enzymes
72
50. Enzyme kinetics as an approach to understanding mechanism
Pages: 194−195 Difficulty: 2
Michaelis-Menten kinetics is sometimes referred to as “saturation” kinetics. Why?
51. Enzyme kinetics as an approach to understanding mechanism
Pages: 195−197 Difficulty: 3
Two different enzymes are able to catalyze the same reaction, A → B. They both have the same Vmax,
but differ their Km the substrate A. For enzyme 1, the Km is 1.0 mM; for enzyme 2, the Km is 10 mM.
When enzyme 1 was incubated with 0.1 mM A, it was observed that B was produced at a rate of
0.0020 mmoles/minute. a) What is the value of the Vmax of the enzymes? b) What will be the rate of
production of B when enzyme 2 is incubated with 0.1 mM A? c) What will be the rate of production
of B when enzyme 1 is incubated with 1 M (i.e., 1000 mM) A?
52. Enzyme kinetics as an approach to understanding mechanism
Pages: 196−197 Difficulty: 3
An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was found to
be 2.0 mM, and the Km, for S2 was found to be 20 mM. A student determined that the Vmax was the
same for the two substrates. Unfortunately, he lost the page of his notebook and needed to know the
value of Vmax. He carried out two reactions: one with 0.1 mM S1, the other with 0.1 mM S2.
Unfortunately, he forgot to label which reaction tube contained which substrate. Determine the value
of Vmax from the results he obtained:
Tube number Rate of formation of product
1 0.5
2 4.8
53. Enzyme kinetics as an approach to understanding mechanism
Pages: 196−198 Difficulty: 3
Write out the equation that describes the mechanism for enzyme action used as a model by Michaelis
and Menten. List the important assumptions used by Michaelis and Menten to derive a rate equation
for this reaction.
Chapter 6 Enzymes
73
54. Enzyme kinetics as an approach to understanding mechanism
Pages: 196−198 Difficulty: 2
For the reaction E + S → ES → P the Michaelis-Menten constant, Km, is actually a summary of three
terms. What are they? How is Km determined graphically?
55. Enzyme kinetics as an approach to understanding mechanism
Pages: 196−198 Difficulty: 3
An enzyme catalyzes a reaction at a velocity of 20 mol/min when the concentration of substrate (S)
is 0.01 M. The Km for this substrate is 1 10-5 M. Assuming that Michaelis-Menten kinetics are
followed, what will the reaction velocity be when the concentration of S is (a) 1 10-5 M and (b) 1
10-6 M?
56. Enzyme kinetics as an approach to understanding mechanism
Pages: 198, 222 Difficulty: 2
Give the Michaelis-Menten equation and define each term in it. Does this equation apply to all
enzymes? If not, to which kind does it not apply?
57. Enzyme kinetics as an approach to understanding mechanism
Pages: 196−197 Difficulty: 2
A biochemist obtains the following set of data for an enzyme that is known to follow Michaelis–
Menten kinetics.
Substrate Initial
concentration velocity
(M) (mol/min)
—————————————
1 49
2 96
8 349
Chapter 6 Enzymes
74
50 621
100 676
1,000 698
5,000 699
—————————————
(a) Vmax for the enzyme is __________. Explain in one sentence how you determined Vmax.
(b) Km for the enzyme is _________. Explain in one sentence how you determined Km.
58. Enzyme kinetics as an approach to understanding mechanism
Page: 197 Difficulty: 2
Why is the Lineweaver-Burk (double reciprocal) plot (see Box 6, p. 206) more useful than the
standard V vs. [S] plot in determining kinetic constants for an enzyme? (Your answer should
probably show typical plots.)
59. Enzyme kinetics as an approach to understanding mechanism
Pages: 196−197 Difficulty: 3
An enzyme catalyzes the reaction A → B. The initial rate of the reaction was measured as a function
of the concentration of A. The following data were obtained:
[A], micromolar V0, nmoles/min
0.05 0.08
0.1 0.16
0.5 0.79
1 1.6
5 7.3
10 13
50 40
100 53
500 73
1,000 76
5,000 79
10,000 80
20,000 80
a) What is the Km of the enzyme for the substrate A?
b) What is the value of V0 when [A] = 43?
The above data were plotted as 1/ V0 vs. 1/[A], and a straight line was obtained.
Chapter 6 Enzymes
75
c) What is the value of the y-intercept of the line?
d) What is the value of the x-intercept of the line?
60. Enzyme kinetics as an approach to understanding mechanism
Pages: 196−197 Difficulty: 3
The turnover number for an enzyme is known to be 5000 min-1. From the following set of data,
calculate the Km and the total amount of enzyme present in these experiments.
Substrate Initial
concentration velocity
(mM) (mol/min)
1 167
2 250
4 334
6 376
100 498
1,000 499
(a) Km = __________. (b) Total enzyme = __________ mol.
61. Enzyme kinetics as an approach to understanding mechanism
Page: 198 Difficulty: 3
When 10 g of an enzyme of Mr 50,000 is added to a solution containing its substrate at a
concentration one hundred times the Km, it catalyzes the conversion of 75 mol of substrate into
product in 3 min. What is the enzyme’s turnover number?
62. Enzyme kinetics as an approach to understanding mechanism
Page: 198 Difficulty: 3
Fifteen g of an enzyme of Mr 30,000 working at Vmax catalyzes the conversion of 60 mol of
substrate into product in 3 min. What is the enzyme’s turnover number?
63. Enzyme kinetics as an approach to understanding mechanism
Chapter 6 Enzymes
76
Page: 198 Difficulty: 2
How does the total enzyme concentration affect turnover number and Vmax?
64. Enzyme kinetics as an approach to understanding mechanism
Pages: 198−199 Difficulty: 3
Enzymes with a kcat / Km ratio of about 108 M-1s-1 are considered to show optimal catalytic efficiency.
Fumarase, which catalyzes the reversible-dehydration reaction
fumarate + H2O malate
has a ratio of turnover number to the Michaelis-Menten constant, (kcat / Km) of 1.6 108 for the
substrate fumarate and 3.6 107 for the substrate malate. Because the turnover number for both
substrates is nearly identical, what factors might be involved that explain the different ratio for the
two substrates?
65. Enzyme kinetics as an approach to understanding mechanism
Pages: 202−203 Difficulty: 3
Methanol (wood alcohol) is highly toxic because it is converted to formaldehyde in a reaction
catalyzed by the enzyme alcohol dehydrogenase:
NAD+ + methanol → NADH + H+ + formaldehyde
Part of the medical treatment for methanol poisoning is to administer ethanol (ethyl alcohol) in
amounts large enough to cause intoxication under normal circumstances. Explain this in terms of
what you know about examples of enzymatic reactions.
66. Enzyme kinetics as an approach to understanding mechanism
Pages: 202−203 Difficulty: 3
You measure the initial rate of an enzyme reaction as a function of substrate concentration in the
presence and absence of an inhibitor. The following data are obtained:
[S] V0
−Inhibitor +Inhibitor
0.0001 33 17
0.0002 50 29
Chapter 6 Enzymes
77
0.0005 71 50
0.001 83 67
0.002 91 80
0.005 96 91
0.01 98 95
0.02 99 98
0.05 100 99
0.1 100 100
0.2 100 100
a) What is the Vmax in the absence of inhibitor?
b) What is the Km in the absence of inhibitor?
c) When [S] = 0.0004, what will V0 be in the absence of inhibitor?
d) When [S] =0.0004, what will V0 be in the presence of inhibitor?
e) What kind of inhibitor is it likely to be?
67. Enzyme kinetics as an approach to understanding mechanism
Page: 210 Difficulty: 2
Define the term “suicide inhibitor.”
68. Enzyme kinetics as an approach to understanding mechanism
Pages: 211−212 Difficulty: 3
Cells can develop a resistance to drugs by increasing the cellular concentration of the enzyme that
that drug inhibits. If a cell increases its concentration of a given enzyme to 10 times the normal
amount, which parameters listed below will be increased ten-fold?
Km
KI
[S]
Vmax
v0 when [S] = Km
kcat
Catalytic efficiency
69. Enzyme kinetics as an approach to understanding mechanism
Pages: 202−203 Difficulty: 3
An enzyme follows Michaelis-Menten kinetics. Indicate (with an “x”) which of the kinetic
parameters at the left would be altered by the following factors. Give only one answer for each.
Chapter 6 Enzymes
78
70. Examples of enzymatic reactions
Pages: 207, 213−216 Difficulty: 3
The enzymatic activity of lysozyme is optimal at pH 5.2 and decreases above and below this pH
value. Lysozyme contains two amino acid residues in the active site essential for catalysis: Glu35 and
Asp52. The pK value for the carboxyl side chains of these two residues are 5.9 and 4.5, respectively.
What is the ionization state of each residue at the pH optimum of lysozyme? How can the ionization
states of these two amino acid residues explain the pH-activity profile of lysozyme?
71. Examples of enzymatic reactions
Page: 207 Difficulty: 2
Why does pH affect the activity of an enzyme?
Pages: 208−209 Difficulty: 3
Chymotrypsin belongs to a group of proteolytic enzymes called the “serine proteases,” many of which
have an Asp, His, and Ser residue that are crucial to the catalytic mechanism. The serine hydroxyl
functions as a nucleophile. What do the other two amino acids do to support this nucleophilic
reaction?
73. Examples of enzymatic reactions
Pages: 208−209 Difficulty: 3
For serine to work effectively as a nucleophile in covalent catalysis in chymotrypsin a nearby amino
acid, histidine, must serve as general base catalyst. Briefly describe, in words, how these two amino
acids work together.
Chapter 6 Enzymes
79
74. Examples of enzymatic reactions
Pages: 216−217 Difficulty: 2
Penicillin and related antibiotics contain a 4-membered -lactam ring. Explain why this feature is
important to the mechanism of action of these drugs.
75. Regulatory enzymes
Page: 212 Difficulty: 3
On the enzyme hexokinase, ATP reacts with glucose to produce glucose 6-phosphate and ADP five
orders of magnitude faster than ATP reacts with H2O to form phosphate and ADP. The intrinsic
chemical reactivity of the —OH group in water is about the same as that of the glucose molecule, and
water can certainly fit into the active site. Explain this rate differential in two sentences or less.
76. Examples of enzymatic reactions
Pages: 210−211 Difficulty: 3
Why is a transition-state analog not necessarily the same as a competitive inhibitor?
77. Regulatory enzymes
Pages: 220−221 Difficulty: 2
Explain how a biochemist might discover that a certain enzyme is allosterically regulated.
78. Regulatory enzymes
Pages: 226−227 Difficulty: 2
What is a zymogen (proenzyme)? Explain briefly with an example.