S-63S-63
Enzymes
chapter 6
1. Keeping the Sweet Taste of Corn The sweet taste of freshly picked corn (maize) is due to the
high level of sugar in the kernels. Store-bought corn (several days after picking) is not as sweet,
because about 50% of the free sugar is converted to starch within one day of picking. To preserve the
sweetness of fresh corn, the husked ears can be immersed in boiling water for a few minutes
(“blanched”) then cooled in cold water. Corn processed in this way and stored in a freezer maintains
its sweetness. What is the biochemical basis for this procedure?
2. Intracellular Concentration of Enzymes To approximate the actual concentration of enzymes in a
bacterial cell, assume that the cell contains equal concentrations of 1,000 different enzymes in solution
in the cytosol and that each protein has a molecular weight of 100,000. Assume also that the bacterial
cell is a cylinder (diameter 1.0 mm, height 2.0 mm), that the cytosol (specific gravity 1.20) is 20% solu-
ble protein by weight, and that the soluble protein consists entirely of enzymes. Calculate the average
molar concentration of each enzyme in this hypothetical cell.
Answer There are three different ways to approach this problem.
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S-64 Chapter 6 Enzymes
3. Rate Enhancement by Urease The enzyme urease enhances the rate of urea hydrolysis at pH 8.0
and 20 C by a factor of 10
14
. If a given quantity of urease can completely hydrolyze a given quantity of
urea in 5.0 min at 20 C and pH 8.0, how long would it take for this amount of urea to be hydrolyzed
under the same conditions in the absence of urease? Assume that both reactions take place in sterile
systems so that bacteria cannot attack the urea.
Answer
4. Protection of an Enzyme against Denaturation by Heat When enzyme solutions are heated,
there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solution
of the enzyme hexokinase incubated at 45 C lost 50% of its activity in 12 min, but when incubated at
45 C in the presence of a very large concentration of one of its substrates, it lost only 3% of its activ-
ity in 12 min. Suggest why thermal denaturation of hexokinase was retarded in the presence of one of
its substrates.
5. Requirements of Active Sites in Enzymes Carboxypeptidase, which sequentially removes
carboxyl-terminal amino acid residues from its peptide substrates, is a single polypeptide of 307 amino
acids. The two essential catalytic groups in the active site are furnished by Arg
145
and Glu
270
.
(a) If the carboxypeptidase chain were a perfect ahelix, how far apart (in Å) would Arg
145
and
Glu
270
be? (Hint: see Fig. 4–4a.)
(b) Explain how the two amino acid residues can catalyze a reaction occurring in the space of a few
angstroms.
Chapter 6 Enzymes S-65
6. Quantitative Assay for Lactate Dehydrogenase The muscle enzyme lactate dehydrogenase cat-
alyzes the reaction
Pyruvate Lactate
H
NADH
CH
3
C COO
O
NAD
CH
3
COO
H
C
OH
NADH and NAD
are the reduced and oxidized forms, respectively, of the coenzyme NAD. Solutions of
NADH, but not NAD
, absorb light at 340 nm. This property is used to determine the concentration of
NADH in solution by measuring spectrophotometrically the amount of light absorbed at 340 nm by the
solution. Explain how these properties of NADH can be used to design a quantitative assay for lactate
dehydrogenase.
7. Effect of Enzymes on Reactions Which of the following effects would be brought about by any en-
zyme catalyzing the simple reaction?
S
k1
88z
y88
k2
P where K
eq
(a) Decreased K
eq
; (b) Increased k
1
; (c) Increased K
eq
; (d) Increased G
‡
; (e) Decreased G
‡
;
(f) More negative G; (g) Increased k
2
.
8. Relation between Reaction Velocity and Substrate Concentration: Michaelis-Menten Equation
(a) At what substrate concentration would an enzyme with a k
cat
of 30.0 s
1
and a K
m
of 0.0050
M
operate at one-quarter of its maximum rate?
(b) Determine the fraction of V
max
that would be obtained at the following substrate concentrations
[S]:
1
2
K
m
, 2K
m
, and 10K
m
.
[P]
[S]
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S-66 Chapter 6 Enzymes
(c) An enzyme that catalyzes the reaction X z
yY is isolated from two bacterial species. The enzymes
have the same V
max
, but different K
m
values for the substrate X. Enzyme A has a K
m
of 2.0
M
,
while enzyme B has a K
m
of 0.5
M
. The plot below shows the kinetics of reactions carried out
with the same concentration of each enzyme and with [X] 1
M
. Which curve corresponds to
which enzyme?
Answer
9. Applying the Michaelis-Menten Equation I A research group discovers a new version of happyase,
which they call happyase*, that catalyzes the chemical reaction
HAPPY 88z
y88 SAD
The researchers begin to characterize the enzyme.
(a) In the first experiment, with [E
t
] at 4 n
M
, they find that the V
max
is 1.6
M
s
1
. Based on this ex-
periment, what is the k
cat
for happyase*? (Include appropriate units.)
Time
[Y]
Chapter 6 Enzymes S-67
(b) In another experiment, with [E
t
] at 1 n
M
and [HAPPY] at 30
M
, the researchers find that V
0
300 n
M
s
1
. What is the measured K
m
of happyase* for its substrate HAPPY? (Include appropriate
units.)
(c) Further research shows that the purified happyase* used in the first two experiments was actually
contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from
the happyase* preparation, and the two experiments repeated, the measured V
max
in (a) is in-
creased to 4.8
M
s
1
, and the measured K
m
in (b) is now 15
M
. For the inhibitor ANGER, calcu-
late the values of ␣and ␣.
(d) Based on the information given above, what type of inhibitor is ANGER?
10. Applying the Michaelis-Menten Equation II Another enzyme is found that catalyzes the reaction
A88z
y88 B
Researchers find that the K
m
for the substrate A is 4
M
, and the k
cat
is 20 min
1
.
(a) In an experiment, [A] 6 m
M
, and the initial velocity, V
0
was 480 n
M
min
1
. What was the [E
t
] used
in the experiment?
(b) In another experiment, [E
t
] 0.5
M
, and the measured V
0
5
M
min
1
. What was the [A] used
in the experiment?
(c) The compound Z is found to be a very strong competitive inhibitor of the enzyme, with an ␣of
10. In an experiment with the same [E
t
] as in part (a), but a different [A], an amount of Z is
added that reduces the rate V
0
to 240 n
M
min
1
. What is the [A] in this experiment?
(d) Based on the kinetic parameters given above, has this enzyme evolved to achieve catalytic
perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic
perfection.
Answer
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S-68 Chapter 6 Enzymes
11. Estimation of V
max
and K
m
by Inspection Although graphical methods are available for accurate
determination of the V
max
and K
m
of an enzyme-catalyzed reaction (see Box 6–1), sometimes these
quantities can be quickly estimated by inspecting values of V
0
at increasing [S]. Estimate the V
max
and
K
m
of the enzyme-catalyzed reaction for which the following data were obtained.
12. Properties of an Enzyme of Prostaglandin Synthesis Prostaglandins are a class of eicosanoids,
fatty acid derivatives with a variety of extremely potent actions on vertebrate tissues. They are re-
sponsible for producing fever and inflammation and its associated pain. Prostaglandins are derived
from the 20-carbon fatty acid arachidonic acid in a reaction catalyzed by the enzyme prostaglandin en-
doperoxide synthase. This enzyme, a cyclooxygenase, uses oxygen to convert arachidonic acid to
PGG
2
, the immediate precursor of many different prostaglandins (prostaglandin synthesis is described
in Chapter 21).
(a) The kinetic data given below are for the reaction catalyzed by prostaglandin endoperoxide syn-
thase. Focusing here on the first two columns, determine the V
max
and K
m
of the enzyme.
(b) Ibuprofen is an inhibitor of prostaglandin endoperoxide synthase. By inhibiting the synthesis of
prostaglandins, ibuprofen reduces inflammation and pain. Using the data in the first and third
columns of the table, determine the type of inhibition that ibuprofen exerts on prostaglandin
endoperoxide synthase.
[S] (
M
)
V
0
(m
M
/min) [S] (
M
)
V
0
(m
M
/min)
2.5 10
6
28 4 10
5
112
4.0 10
6
40 1 10
4
128
110
5
70 2 10
3
139
210
5
95 1 10
2
140
Rate of formation of PGG
2
[Arachidonic acid] Rate of formation of PGG
2
with 10 mg/mL ibuprofen
(m
M
)(m
M
/min) (m
M
/min)
0.5 23.5 16.67
1.0 32.2 25.25
1.5 36.9 30.49
2.5 41.8 37.04
3.5 44.0 38.91
Chapter 6 Enzymes S-69
Answer
(a) Calculate the reciprocal values for the data, as in parentheses below, and prepare a
double-reciprocal plot to determine the kinetic parameters.
13. Graphical Analysis of V
max
and K
m
The following experimental data were collected during a study
of the catalytic activity of an intestinal peptidase with the substrate glycylglycine:
Glycylglycine H
2
O 88n 2 glycine
Product formed Product formed
[S] (m
M
)(mmol/min) [S] (m
M
)(mmol/min)
1.5 0.21 4.0 0.33
2.0 0.24 8.0 0.40
3.0 0.28 16.0 0.45
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S-70 Chapter 6 Enzymes
Use graphical analysis (see Box 6–1) to determine the K
m
and V
max
for this enzyme preparation
and substrate.
14. The Eadie-Hofstee Equation There are several ways to transform the Michaelis-Menten equation so
as to plot data and derive kinetic parameters, each with different advantages depending on the set being
analyzed. One transformation of the Michaelis-Menten equation is the Lineweaver-Burk, or double-
reciprocal, equation. Multiplying both sides of the Lineweaver-Burk equation by V
max
and rearranging
gives the Eadie-Hofstee equation:
V
0
(K
m
)
[
V
S
0
]
V
max
A plot of V
0
vs. V
0
/[S] for an enzyme-catalyzed reaction is shown below. The curve labeled “Slope –K
m
”
was obtained in the absence of inhibitor. Which of the other curves (A, B, or C) shows the enzyme activity
when a competitive inhibitor is added to the reaction mixture? Hint: See Equation 6–30.
A
B
C
V
max
V
0
Slope K
m
V
0
[S]
V
max
K
m
Chapter 6 Enzymes S-71
15. The Turnover Number of Carbonic Anhydrase Carbonic anhydrase of erythrocytes (M
r
30,000)
has one of the highest turnover numbers known. It catalyzes the reversible hydration of CO
2
:
H
2
O CO
2
88z
y88 H
2
CO
3
This is an important process in the transport of CO
2
from the tissues to the lungs. If 10.0 mg of pure
carbonic anhydrase catalyzes the hydration of 0.30 g of CO
2
in 1 min at 37 C at V
max
, what is the
turnover number (k
cat
) of carbonic anhydrase (in units of min
1
)?
Answer The turnover number of an enzyme is the number of substrate molecules trans-
formed per unit time by a single enzyme molecule (or a single catalytic site) when the enzyme
is saturated with substrate:
16. Deriving a Rate Equation for Competitive Inhibition The rate equation for an enzyme subject
to competitive inhibition is
V
0
a
V
K
m
m
ax
[S
[
]
S]
Beginning with a new definition of total enzyme as
[E
t
] [E] [ES] [EI]
and the definitions of aand K
I
provided in the text, derive the rate equation above. Use the derivation
of the Michaelis-Menten equation as a guide.
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S-72 Chapter 6 Enzymes
17. Irreversible Inhibition of an Enzyme Many enzymes are inhibited irreversibly by heavy metal ions
such as Hg
2
, Cu
2
, or Ag
, which can react with essential sulfhydryl groups to form mercaptides:
EnzXSH Ag
888n EnzXSXAg H
The affinity of Ag
for sulfhydryl groups is so great that Ag
can be used to titrate OSH groups
quantitatively. To 10.0 mL of a solution containing 1.0 mg/mL of a pure enzyme, an investigator
added just enough AgNO
3
to completely inactivate the enzyme. A total of 0.342 mmol of AgNO
3
was
required. Calculate the minimum molecular weight of the enzyme. Why does the value obtained in
this way give only the minimum molecular weight?
Chapter 6 Enzymes S-73
18. Clinical Application of Differential Enzyme Inhibition Human blood serum contains a class of
enzymes known as acid phosphatases, which hydrolyze biological phosphate esters under slightly
acidic conditions (pH 5.0):
RXOXPO
2
3
H
2
O 888n RXOH HOXPO
2
3
Acid phosphatases are produced by erythrocytes, the liver, kidney, spleen, and prostate gland. The
enzyme of the prostate gland is clinically important because its increased activity in the blood can be
an indication of prostate cancer. The phosphatase from the prostate gland is strongly inhibited by tar-
trate ion, but acid phosphatases from other tissues are not. How can this information be used to
develop a specific procedure for measuring the activity of the acid phosphatase of the prostate gland
in human blood serum?
19. Inhibition of Carbonic Anhydrase by Acetazolamide Carbonic anhydrase is strongly inhibited by
the drug acetazolamide, which is used as a diuretic (i.e., to increase the production of urine) and to lower
excessively high pressure in the eye (due to accumulation of intraocular fluid) in glaucoma. Carbonic an-
hydrase plays an important role in these and other secretory processes because it participates in regulat-
ing the pH and bicarbonate content of several body fluids. The experimental curve of initial reaction ve-
locity (as percentage of V
max
) versus [S] for the carbonic anhydrase reaction is illustrated below (upper
curve). When the experiment is repeated in the presence of acetazolamide, the lower curve is obtained.
From an inspection of the curves and your knowledge of the kinetic properties of competitive and mixed
enzyme inhibitors, determine the nature of the inhibition by acetazolamide. Explain your reasoning.
V (% of V
max
)
No inhibitor
Acetazolamide
[S] (mM)
0.2
100
50
0.4 0.6 0.8 1
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S-74 Chapter 6 Enzymes
20. The Effects of Reversible Inhibitors Derive the expression for the effect of a reversible inhibitor
on observed K
m
(apparent K
m
aK
m
/a). Start with Equation 6–30 and the statement that
apparent K
m
is equivalent to the [S] at which V
0
V
max
/2a.
21. pH Optimum of Lysozyme The active site of lysozyme contains two amino acid residues essential for
catalysis: Glu
35
and Asp
52
. The pK
a
values of the carboxyl side chains of these residues are 5.9 and 4.5,
respectively. What is the ionization state (protonated or deprotonated) of each residue at pH 5.2, the
pH optimum of lysozyme? How can the ionization states of these residues explain the pH-activity pro-
file of lysozyme shown below?
22. Working with Kinetics Go to the Living Graphs for Chapter 6.
(a) Using the Living Graph for Equation 6–9, create a Vversus [S] plot. Use V
max
100 m
M
s
1
, and
K
m
10 m
M
. How much does V
0
increase when [S] is doubled, from 0.2 to 0.4 m
M
? What is V
0
when [S] 10 m
M
? How much does the V
0
increase when [S] increases from 100 to 200 m
M
?
Observe how the graph changes when the values for V
max
or K
m
are halved or doubled.
100
50
0246810
pH
Activity (% of maximal)
Chapter 6 Enzymes S-75
(b) Using the Living Graph for Equation 6–30 and the kinetic parameters in (a), create a plot in
which both aand aare 1.0. Now observe how the plot changes when a2.0; when a3.0;
and when a2.0 and a3.0.
(c) Using the Living Graphs for Equation 6–30 and the Lineweaver-Burk equation in Box 6–1, create
Lineweaver-Burk (double-reciprocal) plots for all the cases in (a) and (b). When a2.0, does
the xintercept move to the right or to the left? If a2.0 and a3.0, does the xintercept
move to the right or to the left?
Answer
Data Analysis Problem
23. Exploring and Engineering Lactate Dehydrogenase Examining the structure of an enzyme
results in hypotheses about the relationship between different amino acids in the protein’s structure
and the protein’s function. One way to test these hypotheses is to use recombinant DNA technology to
generate mutant versions of the enzyme and then examine the structure and function of these altered
forms. The technology used to do this is described in Chapter 9.
One example of this kind of analysis is the work of Clarke and colleagues on the enzyme lactate
dehydrogenase, published in 1989. Lactate dehydrogenase (LDH) catalyzes the reduction of pyruvate
with NADH to form lactate (see Section 14.3). A schematic of the enzyme’s active site is shown below;
the pyruvate is in the center:
NH2
H3C
O
CH
H
C
CH
3
O(NADH)
Pyruvate
O
C
O
N
C
H
H3CCH
2
CH3
Ile250
Lactate dehydrogenase
Gln102
OH
H
C
Thr246
–
H
C
O
O
Asp168
His195 +
–
H
N
N
C
NH
H
HN
H
NH
Arg109
+
C
H
NH
NH
H
HN
Arg171
+
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S-76 Chapter 6 Enzymes
The reaction mechanism is similar to many NADH reductions (Fig. 13–24); it is approximately the
reverse of steps 2 and 3 of Figure 14–8. The transition state involves a strongly polarized carbonyl
group of the pyruvate molecule as shown below:
(a) A mutant form of LDH in which Arg
109
is replaced with Gln shows only 5% of the pyruvate bind-
ing and 0.07% of the activity of wild-type enzyme. Provide a plausible explanation for the effects
of this mutation.
(b) A mutant form of LDH in which Arg
171
is replaced with Lys shows only 0.05% of the wild-type
level of substrate binding. Why is this dramatic effect surprising?
(c) In the crystal structure of LDH, the guanidinium group of Arg
171
and the carboxyl group of pyru-
vate are aligned as shown in a co-planar “forked” configuration. Based on this, provide a plausible
explanation for the dramatic effect of substituting Arg
171
with Lys.
(d) A mutant form of LDH in which Ile
250
is replaced with Gln shows reduced binding of NADH.
Provide a plausible explanation for this result.
Clarke and colleagues also set out to engineer a mutant version of LDH that would bind and
reduce oxaloacetate rather than pyruvate. They made a single substitution, replacing Gln
102
with Arg;
the resulting enzyme would reduce oxaloacetate to malate and would no longer reduce pyruvate to
lactate. They had therefore converted LDH to malate dehydrogenase.
(e) Sketch the active site of this mutant LDH with oxaloacetate bound.
(f) Provide a plausible explanation for why this mutant enzyme now “prefers” oxaloacetate instead
of pyruvate.
(g) The authors were surprised that substituting a larger amino acid in the active site allowed a
larger substrate to bind. Provide a plausible explanation for this result.
Answer
OOC
A
CH3
A
C
OG
G
O
–
Chapter 6 Enzymes S-77
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