Biochemistry
Final exam
Block 3, 2008
Name
Answer all of the following questions (all are worth 8 pts). Be sure to include molecular details in your answers
that support the cause and effect relationships that exist between the chemical mechanisms and physiological results.
You may use notes and the textbook for this exam. On-line references are not allowed.
1.
6-Phosphogluconate dehydrogenase catalyses the oxidative decarboxylation of 6-phosphogluconate to form
ribulose-5-phosphate. The reactions are outlined in figure 14-21 of your text. Suggest a mechanism for this
reaction showing the movement of all electrons and protons, as well as the structure of the intermediate that
is formed. (Hint: as with any organic mechanism, don’t try to do everything at once.)
2.
In the non-oxidative phase of the pentose phosphate pathway, transketolase and transaldolase are used to
transfer carbon chains among the various pathway intermediates. Transketolase uses thiamine pyrophosphate
(TPP) as its cofactor whereas transaldolase does not require a cofactor. Chemically explain why these
enzymes have different cofactor requirements. (See the Figures 23-25 in Chpt 14).
Both enzymes lyse a carbon-carbon bond and require the correct positioning of a carbonyl to help
stabilize the carbanion that forms. In the case of transaldolase, the lysed bond is beta to the existing
carbonyl and can be cleaved simply by forming the imine of the carbonyl. The imine is conjugated to
the beta carbon and stabilizes the forming anion.
The transketolase reaction requires the TPP cofactor because the lysed bond is alpha to the existing
carbonyl and cannot be stabilized through resonance conjugation. However, the TPP thiazolium
ring carries its own imine, equivalent to that found in the transaldolase mechanism. Nucleophilic
attack of the thiazolium anion on the existing carbonyl in the substrate places the imine beta to the
bond that needs cleaved allowing the forming anion to be stabilized.
In summary, the enzymes have different cofactor requirements due to the location of the bond being
cleaved relative to the location of the carbonyl in the substrate and its ability to stabilize the anion
formed during the reaction.
Exam_final_ansrs
B. Nowak-Thompson
11/26/2008
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3.
Describe the fate of pyruvate, formed by glycolysis in animal skeletal muscle, under two conditions: (a) at
rest, and (b) during an all-out sprint. Show enough detail in your answer to explain why pyruvate metabolism
is different in these two cases.
(a) At rest, plenty of O2 is being delivered to the muscle, and pyruvate formed during glycolysis
is oxidized to acetyl-CoA by the pyruvate dehydrogenase complex. Acetyl groups then enter the
citric acid cycle and are oxidized to CO2.
(b) Under the conditions of all-out exertion, skeletal muscle cannot be supplied with enough O2
to keep metabolism completely aerobic; under these conditions, muscle tissue must function
anaerobically. Pyruvate is reduced to lactate to recycle NADH, formed by glycolysis, to NAD+,
so that glycolysis can continue.
4.
A recently discovered bacterium carries out ATP synthesis coupled to the flow of electrons through a chain
of carriers to some electron acceptor. The components of its electron transfer chain differ from those found in
mitochondria; they are listed below with their standard reduction potentials.
Electron carriers in the newly discovered bacterium:
—————————————————————————————————————
E'°
Oxidant
Reductant
(V)
—————————————————————————————————————
+
NAD
NADH
–0.32
flavoprotein b (FPb)
flavoprotein b
–0.62
(oxidized)
(reduced)
3+
2+
cytochrome c (Fe )
cytochrome c (Fe )
+0.22
Fe-S protein
Fe-S protein
+0.89
(oxidized)
(reduced)
flavoprotein a (FPa)
flavoprotein a
+0.77
(oxidized)
(reduced)
—————————————————————————————————————
a.
Place the electron carriers in the order in which they are most likely to act in carrying electrons.
FPb→ NAD+ → cyt c → FPa → Fe-S
b.
Is it likely that O2 (for which E'° = 0.82 V) is the final electron acceptor in this organism? Why or why
not?
No; Fe-S has a larger E'°, so will probably be the terminal acceptor.
c.
Using inhibitors and artificial electron donors, you find that the pathway is inhibited by CN- ion but
inhibition is overcome by the addition of NADH. What is the actual sequence of electron transport in
this system?
FPb→ cyt c → NAD+ → FPa → Fe-S
d.
What is the major factor explaining why your prediction in (a) turned out to be incorrect?
intracellular concentrations greatly affect the actual E values and can alter the order of
electron transfer.
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B. Nowak-Thompson
11/26/2008
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5.
Compound X inhibits ATP production in mitochondrial membrane preparations. It was observed that when
compound X was added to a membrane prep, the NAD+/NADH ratio decreased. Would you expect cells
treated with compound X to be more or less susceptible to oxidative damage? Explain your reasoning.
It is an inhibitor of electron transfer; its addition lowers the NAD+/NADH ratio because
NADH produced by oxidative reactions in mitochondria can no longer be reoxidized by
electron flow to O2. As a result, lifetimes of reactive oxygen species on the electron transport
are increased allowing them to react with other cellular components, damaging the cell.
6.
If propionate labeled with 14C at C1 is added to a liver homogenate, radiolabeled glucose can be detected.
Briefly describe the path the label takes (chemical structures are not necessary) and identify which carbon(s)
of glucose would be labeled? (use the standard numbering system for glucose or draw a picture).
C3 and C4 of glucose are labeled. Briefly, the propionate is convereted into oxaloacetate which
would be labeled a both carboxylate carbons due to the scrambling of the label via succinate. OAA is
then converted to PEP via loss of one carboxylate and converted to glucose via gluconeogenesis. If you
follow the caroxylate of PEP through gluconeogenesis, you find that it labels the carbons of DHAP and
GAP that correspond to C3 and C4 of glucose.
b.
The glucose in this experiment was not labeled to a high degree. In fact, very little of the label was
actually retained within the system. Where did the label go? (it did not decay; the half-life of 14C is about
12,000 years.)
The propionate gets transformed into succinyl CoA that enters the TCA cycle. Because the label is at
C1 (carboxylate) of propionate, much of it ends up being lost as CO2 in the first pass through the
cycle.
7.
Cardiac muscle is apparently not capable of synthesizing fatty acids, but nevertheless expresses an isoform of
acetyl-CoA carboxylase (ACC-280). ACC-280 synthesizes the precursor to fatty acid biosynthesis, malonylCoA. Why is it necessary for this enzyme to be expressed in cardiac muscle even though cardiac muscle does
not use malonyl-CoA synthetically?
The ability to produce malonyl-CoA is necessary to allow transport of fatty acids into the mitochondria
for b-oxidation. The carnitine transferase that provides a route for longer chain fatty acids to cross the
inner mitochondrial membrane, is inhibited by malonyl-CoA. Thus, regulation of b-oxidation in relation
to other metabolic processes such as glycolysis requires malonyl-CoA production.
Exam_final_ansrs
B. Nowak-Thompson
11/26/2008
3 of 3