Chap. 6. Problem 1.

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Chap. 6. Problem 1.
After harvesting, the simple sugars present in
the kernels of corn continue to be enzymatically
converted into starch. This sugar has a very
sweet taste, whereas starch does not. Thus,
within a day or so of harvest, untreated corn no
longer tastes as sweet. To preserve sweetness,
freshly picked corn can be heated (blanched) for
a few minutes to denature the enzymes that
synthesize starch. This preserves the sweet
flavor of fresh corn. During freezing, any
remaining starch synthesis enzyme activity is
minimal due to the low temperature of storage.
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Chap. 6. Problem 4.
The data are consistent with the possibility that
the hexokinase-substrate complex is actually
thermodynamically more stable than free
hexokinase. The additional interactions formed
between the enzyme and its substrate must
contribute to the stabilization of hexokinase.
Because the energy of the hexokinase-substrate
complex is lower than free hexokinase, the
activation energy barrier separating the folded
and unfolded states of the enzyme is increased in
the presence of substrate. This reduces the
extent of denaturation that occurs when the
substrate is present in solution with the enzyme.
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Chap. 6. Problem 7.
Enzymes are catalysts that change the rates of
reactions without affecting reaction equilibria.
Thus, (b), (e), and (g) are correct as all of
these parameters are connected to the reaction
rate. (d) is not correct as the energy of the
transition state actually is decreased, not
increased, by the enzyme due to the
complementarity of its structure to the
transition state. Because reaction equilibria,
and the related term ?G0 for a reaction, are not
affected by an enzyme, (a), (c), and (f) are not
observed.
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Chap. 6. Problem 8a.
Part (a). In this problem, we are given the
values of V0 (0.25 Vmax) and Km. On substitution
of these parameters into the MM equation, we can
solve for S. Namely, V0 VmaxS/(Km S) or
0.25Vmax VmaxS/(Km S) After canceling
Vmax terms and rearranging the equation it
becomes 0.25Km 0.25S S or 0.25Km
0.75S On solving for S we get S 0.33Km
(0.33)(0.005M) 1.7 x 10-3 M. Part (b). See
next slide.
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Chap. 6. Problem 8b.
Part (b). To determine the fraction of Vmax
(V0/Vmax) that is obtained at any S relative to
the Km, first rearrange the MM equation to the
form V0/Vmax S/(Km S) And then substitute
in S values expressed as a function of Km. 1)
For S 0.5Km V0/Vmax 0.5Km/(Km 0.5Km)
0.5/1.5 0.33 2) For S 2Km V0/Vmax 2Km/(Km
2Km) 2/3 0.67 3) For S 10Km V0/Vmax
10Km/(Km 10Km) 10/11 0.91
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Chap. 6. Problem 12.
The Vmax and Km for prostaglandin endoperoxide
synthase, and the type of inhibition that occurs
in the presence of ibuprofen, can easily be
determined from double-reciprocal plots of the
kinetic data. To construct these graphs, we first
have to calculate 1/S and 1/V0 from the data in
the table (next slide).
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Chap. 6. Problem 12 (cont.).
1/S (mM-1) 1/V0 (min/mM) without ibuprofen 1/V0 (min/mM) with ibuprofen
2.0 0.043 0.0600
1.0 0.031 0.0396
0.67 0.027 0.0328
0.40 0.024 0.0270
0.28 0.023 0.0257
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Chap. 6. Problem 12 (cont.).
Part (a). The double-reciprocal plots obtained
with the transformed kinetic data are
The y-intercept is equivalent to 1/Vmax the
x-intercept is equivalent to -1/Km. In the
absence of inhibitor, Vmax and Km therefore
are 1/Vmax 0.0190 min/mM and Vmax 52.6
mM/min -1/Km -1.7 mM-1 and Km 0.59 mM
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Chap. 6. Problem 12 (cont.).
Part (b). The plot obtained with ibuprofen
intersects the y-axis at the same value of 1/Vmax
as obtained without the inhibitor. Thus the Vmax
for the reaction is unchanged. The inhibitor plot
intersects the x-axis at a smaller value of -1/Km
(that is a greater value of Km). The plot in the
presence of the inhibitor therefore reveals that
ibuprofen is a competitive inhibitor of
prostaglandin endoperoxide synthase. (See Fig. 1
in Box 6-2 of the textbook).
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Chap. 6. Problem 15.
The turnover number for an enzyme is the number
of substrate molecules converted to product per
unit time by a single enzyme molecule when the
enzyme is saturated with substrate. The turnover
number, kcat is calculated from the relationship
kcat Vmax/Et, where Et is the total moles of
enzyme.
The turnover number in units of min-1 can be
obtained if we first convert the weights of the
enzyme and substrate to molar amounts Vmax
(0.30 g/min)/(44 g/mol) 6.8 x 10-3 mol/min Et
(10 ?g)(1 g/106 ?g)/30,000 g/mol 3.3 x 10-10
mol The turnover number now can be obtained by
dividing the moles of CO2/min by the moles of
enzyme present, or Kcat (6.8 x 10-3
mol/min)/(3.3 x 10-10 mol) 2.0 x 107 min-1
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Chap. 6. Problem 18.
To determine acid phosphatase activity in the
blood that is derived from the prostate gland,
one would need to measure activity in the
presence and absence of tartrate. If for example
100 U of activity were measured in the absence of
tartrate, and 10 U were measured in its presence,
then the blood would contain 90 U of activity
originating from the prostate. In other words,
the difference between the two activities
represents the activity of acid phosphatase
originating from the prostate gland.
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Chap. 6. Problem 21.
The pH optimum of lysozyme (pH 5.2) is
approximately 1 unit below and above the
respective pKas of Glu35 and Asp52 in the active
site. Thus, about 90 of the Glu35 side-chain
would be present in its conjugate acid
(protonated) form, and 90 of the Asp52
side-chain would be present in its conjugate base
(ionized) form at the pH optimum. Because
activity declines above pH 5.2, it suggests that
the Glu35 side-chain must be in its protonated
form for catalysis. Likewise, because activity
declines below pH 5.2, Asp52 must be in its
ionized form for catalysis.