Transcript Chapter 6 Problem Set

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.
Chap. 6. Problem 4.
The data are consistent with the possibility that the hexokinasesubstrate 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.
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 ∆G’0 for a reaction, are
not affected by an enzyme, (a), (c), and (f) are not observed.
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 = Vmax[S]/(Km +[S]) or 0.25Vmax = Vmax[S]/(Km + [S])
After canceling Vmax terms and rearranging the equation it becomes
0.25Km + 0.25[S] = [S] or 0.25Km = 0.75[S]
On solving for [S] we get
[S] = 0.33Km = (0.33)(0.005M) = 1.7 x 10-3 M.
Part (b). See next slide.
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
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).
Chap. 6. Problem 12 (cont.).
1/[S] (mM-1)
1/V0 (min/mM)
1/V0 (min/mM)
without ibuprofen 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
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
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).
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
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.
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 sidechain 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.