The MOLECULES
of LIFE
Physical and Chemical Principles
Solutions Manual
Prepared by James Fraser and Samuel Leachman
Chapter 16
Principles of Enzyme
Catalysis
Problems
True/False and Multiple Choice
1. The initial reaction velocity for an enzyme reaction
reaches a maximum at high substrate concentration
because the free enzyme can no longer regenerate at
the end of each reaction cycle.
True/False
6. An enzyme inhibitor is observed to alter the KM but not
the Vmax of a reaction. This inhibitor is most likely:
a.
b.
c.
d.
e.
A noncompetitive inhibitor.
A competitive inhibitor.
An allosteric inhibitor.
A substrate-dependent noncompetitive inhibitor.
A covalent inhibitor.
2. The turnover number for an enzyme obeying Michaelis–
Menten kinetics is:
a. k2.
b. kcat/KM.
c. k1/k–1.
d. (k1 + k2).
e. ΔG‡.
7. Due to its extremely slow dissociation kinetics, the
protein bovine pancreatic trypsin inhibitor (BPTI) has
broad specificity and inhibits more proteases than
protease inhibitors that are small molecules.
3. Catalytic antibodies are generally less efficient than
natural enzymes that catalyze the same reactions.
Fill in the Blank
True/False
4. A metabolic enzyme generates the amino acid
methionine. For a given substrate concentration, an
experiment conducted in the presence of high initial
concentrations of methionine generates less new
methionine than an experiment conducted with no
initial methionine present. This is likely an example of:
a. A ping-pong mechanism of substrate binding.
b. A proximity effect.
c. Substrate strain.
d. Product inhibition.
e. A reaction intermediate.
5. Which of the following is not a commonly observed
feature of proteases?
a. The catalytic triad in the active site.
b. Exclusively hydrophobic residues in the active
site.
c. A cysteine residue in the active site.
d. Metal ions coordinated in the active site.
e. A pair of acidic residues in the active site.
True/False
8. In the schemes for the catalyzed reactions considered
in this chapter, S, E, and P refer to _______________,
______________, and ______________, respectively.
Answer: substrate, enyzme, product
9. The specificity constant or catalytic efficiency is the
ratio between __________ and _________.
Answer: kcat, KM
10. In a plot of initial velocity versus substrate
concentration, an allosteric enzyme displays a
____________ curve, whereas a non-allosteric enzyme
that obeys Michaelis–Menten kinetics displays a
____________ curve.
Answer: sigmoidal, hyperbolic
11. The geometry of competitive inhibitors commonly
mimics the _____________ of the reaction that the
enzyme normally catalyzes.
Answer: transition state
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
Figure Q16A
2
Chapter 16: Principles of Enzyme Catalysis
12. Proteins are not the only polymers that act as catalysts.
Catalytic ______ molecules are also essential for cells,
including playing an essential role in protein synthesis.
a.
Answer: RNA/ribozyme
14. Show that the equations plotted in Lineweaver–Burk
and Eadie–Hofstee plots are equivalent.
Answer:
Start with the Lineweaver–Burk equation:
1/v0 = 1/Vmax + KM/Vmax[S]
Multiply all by Vmax,
Vmax/v0 = 1 + KM/[S]
Multiply all by v0,
Vmax = v0 + v0KM/[S]
Rearrange for v0,
v0 = –v0KM/[S] + Vmax
(which is the Eadie–Hofstee equation).


→ E•⋅ S → E + P , calculate
15. In a reaction, E + S ←

k
k2
−1
the value of KM if the forward rate constant (k1) for
E•S formation is 4.3 × 106 sec–1•M–1, the reverse rate
constant (k–1) for E•S dissociation is 2.4 × 102 sec–1, and
the turnover number (k2) is 1.2 × 103 sec–1.
Answer:
KM = (k–1 + k2)/k1 = (2.4 × 102 sec–1 + 1.2 × 103
sec–1)/4.3 × 106 sec–1•M–1
= 3.3 × 10–4 M
16. Presented below are Lineweaver–Burk plots for
enzymatic reactions with (red) and without (blue)
inhibitor. What type of inhibition is occurring in each
case?
1.0
Figure Q16B0
0
0.2
0.4
0.6
0.8
–1
1/[S] (M )
1.0
1.2
0
0.2
0.4
0.6
0.8
–1
1/[S] (M )
1.0
1.2
0
0.2
0.4
1.0
1.2
b.
0.7
0.6
1/v (M–1•sec)
Answer:
o‡cat ‒ ΔGo‡uncat)/RT)
kcat/kuncat = e(‒(ΔG
o‡cat
‒ ΔGo‡uncat)/8.314 × 298)
105 = e(‒(ΔG
5
o‡
ln(10 ) = (‒(ΔG cat ‒ ΔGo‡uncat)/2.5)
ΔGo‡cat ‒ ΔGo‡uncat = –28.5 kJ•mol–1
1.5
0.5
0.5
0.4
0.3
0.2
0.1
Figure Q16C
0
c.
4.5
4.0
3.5
1/v (M–1•sec)
13. At 25°C, an enzyme accelerates a reaction by a factor
of 105 over the uncatalyzed reaction in water. If the
effect of the enzyme is solely to reduce the energy of
the transition state, by what amount does it reduce the
energy of the transition state (EA)?
1/v (M–1•sec)
2.0
Quantitative/Essay
k1
2.5
3.0
2.5
2.0
1.5
1.0
0.5
0
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
0.6
0.8
1/[S] (M–1)
PROBLEMS and solutions
Answer:
a. Competitive inhibition.
b. Substrate-dependent noncompetitive inhibition.
c. Noncompetitive inhibition.
Substitute slope into:
1/v = slope × 1/[S] + 1/Vmax
0.5 = 0.8 × 0.5 + 1/Vmax
0.1 = 1/Vmax
Vmax = 10 mM•sec–1
KM/Vmax = 0.8, K*
M = 8 mM
17. The table below lists initial velocities measured for an
enzymatic reaction at different substrate concentrations
in the presence and absence of an inhibitor. The enzyme
concentration is identical in both reactions.
v, no inhibitor (mM•sec–1)
v, with inhibitor (mM•sec–1)
1
2.50
1.11
2
4.00
2.00
5
6.25
3.85
10
7.69
5.56
20
8.70
7.14
[S] (mM)
a. Graph a Lineweaver–Burk plot.
b. What are the apparent values of Vmax and KM for
each experiment?
c. What is the inhibition mechanism?
d. If the concentration of inhibitor is 0.5 mM, what is
the value of KI?
Answer:
Q16.17a
a. Red
line is with inhibitor, blue line is with no
inhibitor.
1/v (mM–1•sec)
1.0
no inhibitor
with inhibitor
0.8
0.6
3
c. The slope changes but the y intercept does not.
This is reflected in the apparent KM increase upon
addition of inhibitor and the constant Vmax between the
two reactions. These observations are indicative of a
competitive inhibitor.
d. For a competitive inhibitor, the apparent KM with
inhibitor is related to the KM without inhibitor:
K*
M = KM(1 + [I]/KI)
8 mM = 3 mM (1 + 0.5 mM/KI)
8 mM = 3 mM + 1.5 mM/KI
5 mM = 1.5 mM/KI
KI = 1.5 mM/5 mM
KI = 0.3 mM
18. The table below lists initial velocities measured for an
enzymatic reaction at different substrate concentrations
in the presence and absence of an inhibitor. The enzyme
concentration is identical in both reactions.
[S] (mM)
v, no inhibitor
(mM•sec–1)
v, with inhibitor (mM•sec–1)
1
1.000
0.923
5
1.154
1.053
10
1.176
1.071
50
1.195
1.087
100
1.198
1.089
a. Graph a Lineweaver–Burk plot for each set of data.
b. What are the apparent values of Vmax and KM for
each experiment?
c. What is the inhibition mechanism?
d. If the concentration of inhibitor is 10 nM, what is
the value of KI?
0.4
0.2
0.0
0.0
0.2
0.4
1/[S]
0.6
0.8
–1
(mM )
1.0
1.2
Answer:
a. Red
line is with inhibitor, blue line is with no
Q16.18a
inhibitor.
1.2
1.0
1/v (mM–1•sec)
b. First calculate the slope of the lines—the data are
nearly perfectly linear, so we will just use the first two
data points to calculate the relevant parameters.
For no inhibitor:
Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (0.4 – 0.25)/
(1 – 0.5) = 0.3
Substitute slope into:
1/v = slope × 1/[S] + 1/Vmax
0.25 = 0.3 × 0.5 + 1/Vmax
0.1 = 1/Vmax
Vmax = 10 mM•sec–1
KM/Vmax = 0.3, KM = 3 mM
For with inhibitor:
Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (0.9 – 0.5)/(1 – 0.5)
= 0.8
0.8
0.6
0.4
no inhibitor
with inhibitor
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1/[S] (mM–1)
1.0
1.2
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
4
Chapter 16: Principles of Enzyme Catalysis
c. The y intercept changes, but the slope does not.
This is reflected in the apparent KM decrease upon
addition of inhibitorand the apparent Vmax decrease,
but the maintenance of a constant ratio between the
two constants. These observations are indicative of a
substrate-dependent noncompetitive inhibitor.
d. 10 nM is 0.00001 mM, K*M = KM/(1 + [I]/KI)
K*M = KM/(1 + [I]/KI)
0.182 mM = 0.2 mM/(1 + 0.00001 mM/KI)
0.00001 mM/KI = 0.2 mM /0.182 mM – 1
KI = 0.1 μM or 100 nM
19. The table below lists initial velocities measured for an
enzymatic reaction at different substrate concentrations
in the presence and absence of an inhibitor. The enzyme
concentration is identical in both reactions.
[S] (µM)
v, no inhibitor (µM•sec–1)
v, with inhibitor (µM•sec–1)
10
8.93
6.94
20
10.42
8.93
30
11.03
9.87
100
12.02
11.57
200
12.25
12.02
a. Graph a Lineweaver–Burk plot for each set of data.
b. What are the values of Vmax and KM for each
experiment?
c. What is the inhibition mechanism ?
d. If the concentration of inhibitor is 100 nM, what is
the value of KI?
Q16.19a
Answer:
a.
0.16
0.14
1/v (µM–1•sec)
b. For no inhibitor:
Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (1 – 0.85)/(1 – 0.1)
= 0.167
Substitute slope into:
1/v = slope × 1/[S] + 1/Vmax
1 = 0.167 × 1 + 1/Vmax
Vmax = 1.2 mM•sec–1
KM/Vmax = 0.167, KM = 0.167 × 1.2 = 0.2 mM
With inhibitor:
Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (0.95 – 0.92)/(0.2 –
0.02) = 0.167
Substitute slope into:
1/v = slope × 1/[S] + 1/Vmax
0.95 = 0.167 × 0.2 + 1/Vmax
Vmax = 1.09 mM•sec–1
KM/Vmax = 0.167, KM* = 0.167 × 1.09 = 0.182 mM
0.12
0.10
0.08
0.06
0.04
no inhibitor
with inhibitor
0.02
000
0.0
0.02
0.04
0.06
1/[S]
0.08
–1
(µ M )
0.10
0.12
b. First calculate the slope of the lines—the data are
nearly perfectly linear, so we will just use the first two
data points to calculate the relevant parameters.
For no inhibitor:
Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (0.112 – 0.096)/
(0.1 – 0.05) = 0.32
Substitute slope into:
1/v = slope × 1/[S] + 1/Vmax
0.112 = 0.32 × 0.1 + 1/Vmax
0.08 = 1/Vmax
Vmax = 12.5 μM•sec–1
KM/Vmax = 0.32, KM = 4 μM
For no inhibitor:
Slope = KM/Vmax = (y1 – y2)/(x1 – x2) = (0.144 – 0.112)/
(0.1 – 0.05) = 0.64
Substitute slope into:
1/v = slope × 1/[S] + 1/Vmax
0.112 = 0.64 × 0.05 + 1/Vmax
0.08 = 1/Vmax
Vmax = 12.5
KM/Vmax = 0.64, K*
M = 8 μM
c. The slope changes but the y intercept does not.
This is reflected in the apparent KM increase upon
addition of inhibitor and the constant Vmax between the
two reactions. These observations are indicative of a
competitive inhibitor.
d. For a competitive inhibitor, the apparent KM with
inhibitor is related to the KM without inhibitor:
K*
M = KM (1 + [I]/KI)
8 = 4(1 + 0.1/KI)
8 = 4 + 0.4/KI
4 = 0.4/KI
KI = 0.4/4
KI = 0.1 μM or 100 nM
20. An experiment with 10 nM of an enzyme obeying
Michaelis-Menten kinetics yields a Vmax of 7 × 10–3
M•sec–1.
a. What is the turnover number (k2)?
b. The experiment is repeated in the presence of a
noncompetitive inhibitor and the Vmax is reduced to
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
PROBLEMS and solutions
5 × 10–4 M•sec–1. What fraction of the enzyme is bound
to the inhibitor?
Answer:
Recognize v0 = Vmax/(1 + KM/[S]), so a concentration
near KM will show half the velocity of the highest
velocity observed, assuming it is close to Vmax.
a. ~3–5 × 10–3 M
b. ~4 × 10–9 M
c. ~200 × 10–3 M
Answer:
a. Vmax = k2[E]
7 × 10–3 M•sec–1 = (k2)(1 × 10–8 M)
k2 = 7 × 105 sec–1
b. Vmax = k2(1 – f)[E]
(1 – f) = 5 × 10–4 M–1•sec–1/(7 × 105 sec–1 × 1 × 10–8 M)
1 – f = 0.07
f = 0.93
Therefore 93% of the enzymes are bound to the
inhibitor.
21. Given the following three data tables of substrate
concentrations and initial velocities for enzymes that
obey Michaelis–Menten kinetics, estimate KM for each
enzyme in molar units.
22. When the bi-substrate analog PALA is added to the
enzyme ATCase at low concentration it increases the
rate of reaction of aspartate and carbamylphosphate.
However, at higher concentrations it decreases the
reaction rate. How can PALA act as both an activator
and inhibitor of ATCase?
Answer:
PALA mimics the two substrates of ATCase. Its
affinity is greater than the substrate, and therefore
outcompetes substrate for the active site. At moderate
concentrations (where not every site on ATCase is
occupied by PALA), the equilibrium of ATCase is shifted
to favor the active conformation. Since the active
conformation has a higher affinity for substrate, binding
sites that are not occupied by PALA are able to increase
the reaction velocity. At high concentrations of PALA
all binding sites become occupied by inhibitor and
turnover decreases.
a.
[S] (mM)
v0 (mM•sec–1)
1
266.7
3
553.8
5
705.9
50
1121.5
500
1191.7
5000
1199.2
b.
[S] (nM)
v0 (mM•min–1)
4
123.5
5
137.4
6
148.5
10
177.3
100
240.2
1000
249.0
[S] (mM)
v0 (M•hour–1)
c.
1
0.00
10
0.01
100
0.07
200
0.10
1000
0.17
5000
0.19
5
23. In the search for the catalytic mechanism of an enzyme,
three mutations of charged residues to alanine (which
is uncharged) are made and compared with the wild
type (WT) enzyme. At otherwise identical conditions
and concentrations of enzymes, the following initial
velocities (µM•sec–1) are measured as a function of pH.
pH
WT
Arg55Ala
Glu63Ala
Lys113Ala
4
1.3 × 103
1.3 × 103
1.3 × 102
1.5 × 103
5
8.7 × 105
8.7 × 105
1.3 × 102
9.5 × 105
6
8.1 × 104
8.1 × 104
1.3 × 102
8.1 × 104
7
5.3 × 103
5.3 × 103
1.3 × 102
5.2 × 103
a. Explain which residue likely acts as a general acid/
base during catalysis?
b. What is a possible mechanism for the slightly
increased reaction velocities observed in the Lys113Ala
mutants at lower pH?
Answer:
a. Glu63 likely participates directly in catalysis. All
other mutations have much higher catalytic efficiency
(comparable to wild type). Additionally, Glu63Ala no
longer displays any pH dependency when mutated to an
uncharged residue.
b. Having a Lys near the catalytic Glu will shift the pKa
down and make it less likely to donate a hydrogen to
the reaction. Removing this charge with the Lys113Ala
mutant shifts the pKa slightly and leads to increased
reactivity and increased reaction velocities.
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science
6
Chapter 16: Principles of Enzyme Catalysis
24. Why is triose phosphate isomerase considered to be an
example of a “perfect enzyme”?
Answer:
Triose phosphate isomerase is considered a “perfect”
enzyme because it catalyzes its reaction so fast that it
is diffusion controlled.
The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science