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Chem 352 - Fall 2014
Quiz 3
R = 8.314 J/(mol•K) = 0.08206 (L•atm)/(mol•K)
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1. Myoglobin (Mb) and hemoglobin (Hb) are both oxygen binding proteins.
a. Though both Mb and Hb bind oxygen, they do so for different reasons. For mammals, describe
the role that each protein performs when binding oxygen.
Mb: Myoglobin’s role is to bind and store oxygen in the tissues until needed. It has a heme group
cofactor which gives the protein a red color, which in turn can give muscle tissue a red color as
well.
Hb: Hemoglobin’s role is to bind oxygen in the lung and then transport it through the blood to the
tissues. There it releases it, often to an awaiting myoglobin molecule, which in turn binds and stores
it until needed. For this to work, the hemoglobin must have a lower aﬃnity of oxygen in the tissues
than myoglobin does. Hemoglobin also contains heme groups, which why blood is red
b. On the graphs shown to the right,
sketch and label the oxygen
binding curves, fraction bound (Y)
vs. partial pressure of oxygen in
torr (pO2), for Mb with a P50 of
10 torr and Hb with a P50 of 30 torr.
Be sure to label the axes.
Y
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1.1
1.0
0.9
0.8
Mb
0.7
0.6
0.5
Hb
0.4
Hb at pH 7.2
0.3
0.2
0.1
0.0
0
10
20
30
40
50 60 70
pO2 (torr)
80
90 100 110 120
c. Given the pO2 is 100 torr in the lungs on 26 torr in the tissues, explain how the differences in the
binding behaviors for Mb and Hb shown in the graphs above best suites each oxygen binding
protein to the roles you described in part a. When in the lung, hemoglobin will become nearly
100% saturated with oxygen, but as travels out to the tissues, where the partial pressure of O2 is
lower, hemoglobin’s binding aﬃnity for oxygen falls oﬀ more rapidly than myoglobin’s. This allow
it to more readily pass of the oxygen oﬀ to the awaiting myoglobin molecules.
d. During time of heavy exertion, the pH of the blood out in the tissues may drop by a couple of
tenths of a pH unit. On the graph above, sketch and label a second binding curve for Hb that
illustrates the effect that a drop in the pH from 7.4 to 7.2 will have on the Hb’s oxygen binding
behavior.
e. Explain why this behavior would allow for Hb to deliver more O2 to the tissues? The lowering of the pH leads to protonation of groups on the hemoglobin, which leads to the
formation of salt bridges (charge/charge interactions) that stabilize the deoxy (tense) state for
hemoglobin. This will increase the apparent P50 for Hb and shift its sigmoidal binding curve
further to the right. this will allow the hemoglobin to oﬀ load even more oxygen to the awaiting
myoglobin when it reaches the tissues.
Chem 352
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Quiz 3
Fall, 2014
2. Fumarase is an enzyme that catalyzes one of the
reactions in the citric acid cycle in which fumarate is
converted to L-malate.
Lyase
a. What class of enzyme-catalyzed reaction is this? _______________________________________
2.0
The initial rate (vo) for the fumarase
reaction was measured as a function
of the fumarate concentration. The
fumarase concentration used in these
experiments was 5.0 nM. The results
of this experiment are shown in the
graph to the right.
1.8
1.6
1/v0 (s/µM)
1.4
b. What is the name for this type of
plot? Lineweaver-Burke Plot
__________________________
KM =
1.2
1.0
0.8
0.6
0.4
−1
−1
=
= 0.24 mM
x − intercept −4.0 / mM
Vmax =
0.2
-4
-2
0
2
4
1
1
=
= 3.33 µM/s
y − intercept 0.30 s/µM
6
8
10
12
14
1/[fumarate] (1/mM)
c. How may fumarate molecules does fumarase convert to L-malate per second, when fully saturated
666 molecules per second
with fumarate? (Show calculations)__________________________________________________
This is given by the turnover number, or kcat,
k cat =
Vmax
⎡⎣ E ⎤⎦
total
;
k cat =
3.33 µM/s 3.33 x 10 −6 M/s
=
= 666 / s
5.0 nM
5.0 x 10 −9 M
No
d. Under these condition, is fumarase displaying catalytic perfection? _________________________
What is your evidence for this claim?
This is determined by comparing the catalytic eﬃciency,
k cat
KM
=
k cat
KM
, to its theoretical maximum, 108 1/M•s.
666 / s
666 / s
=
= 2.64 x 106 1/M•s, which is << 1 x 108 1/M•s
0.25 mM 0.25 x 10 −3 M
e. What does it mean for an enzyme to be catalytically perfect.
k
The theoretical maximum value for cat of 108 1/M•s is imposed by the rate that the substrate molecules
K
can diﬀuse into the active site of theMenzyme. Since there is nothing that be done to the enzyme the
enzyme to increase the reaction rate beyond this point, we say that the enzyme as attained catalytic
perfection.
f.
Meso-tartrate is a competitive inhibitor of fumarase. On the graph shown
above, show what effect this inhibitor should have on the kinetics of this
reaction. A competitive inhibitor is one that competes with the substrate
for binding to the active site of the enzyme. It has no eﬀect on Vmax, but
increases the apparent KM.
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