Download Biochemistry 440: Problem Set 7 Answers - Oxidation Numbers and Glycolysis - Prof. Rachel and more Assignments Biochemistry in PDF only on Docsity!
Fall, 2008
Problem Set 7 Answers
1a. Give the formal oxidation number of the carbon in the following compounds.
CH 4 , CH 3 OH, and CO 2. Which is the most reduced form of carbon?
- 4 - 2 +4 CH 4 is the most reduced form of carbon. Note that I am calculating this differently from the book (p. 513). I do not expect you to know the book method, only the one I presented in class.
1b. Assign oxidation numbers to each carbon in glucose.
1c. Assign oxidation numbers to each carbon in pyruvate. Compare the oxidation state of
pyruvate to that of glucose.
Assign oxidation numbers to each carbon. Compare the oxidation state of pyruvate to that of glucose. Pyruvate is oxidized compared to glucose. Overall oxidation state of the carbons in pyruvate is +2, in glucose is 0.
2. In the fourth step of glycolysis, fructose 1,6 bisphosphate is cleaved to glyceraldehyde 3-
phosphate and dihydoxyacetone phosphate. Is this an oxidation? Why or why not? Explain in
terms of the oxidation numbers of the carbons.
No, this is not an oxidation. The sum of the oxidation numbers for the carbons in fructose 1,6 bisphosphate is zero. The sum of the oxidation numbers for carbons in C C CH 3 O O O
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glyceraldehyde 3-phosphate is zero. The sum of the oxidation numbers for carbons in dihydroxyacetone phosphate is also zero. So the sum of the oxidation numbers in the reactants equals the sum of the oxidation numbers in the products. Therefore, this is not an oxidation.
3. The book calls the summed, overall standard free energy change of the first five steps of
glycolysis "endergonic". When these reactions happen in a cell, are any of them endergonic?
Explain.
No, none of the steps are truly endergonic. They might be very close to equilibrium under cellular conditions, though, and have unfavorable thermodynamic potentials under standard conditions, which show in the positive ฮG' o values for steps 2, 3 and 5.
- Step1 & 3: Through ATP coupling (which gives these reaction steps a favorable ฮG' o ) and supported by in vivo - Q values, which are considerably smaller than the K'eq values, these reactions have a strongly negative ฮG, and are in fact very exergonic.
- Step 2: Its ฮG' o is slightly positive, but because the product Fructose 6-phosphate is quickly used up by the following, very favorable step 3, the Q-value for this reaction is small and the actual ฮG in the cell is near equilibrium. Steps 4 and 5: Especially the Aldolase step (#4) is under standard conditions very unfavorable in direction of the products DHAP and Glyceraldehyde 3-P (ฮG' o of +23.8 kJ/mol). Also, the isomerization step #5 by Triose-phosphate isomerase has a slightly unfavorable ฮG' o . But since their product glyceraldehyde 3-phosphate is very quickly used up (see e.g. the extremely small in vivo - Q value for the Aldolase reaction presented in lecture, from table 15-3 in the textbook), this step effectively operates near equilibrium under cellular conditions. The reason for this can be found in the very favorable step #7 (under standard conditions) of the "pay-off phase", which can be thought of as "pulling" step #6 and the two preceding steps towards the resonance-stabilized intermediate 3-phospho glycerate (thereby "using up" its favorable potential almost completely and operating in vivo near equilibrium. The whole sequence is then finally "pulled" towards the product pyruvate (stabilized by tautomerization) through step 10.
4. You feed cells glucose that is labeled on the first carbon with 14 C. Which carbon(s) of
pyruvate will be labeled after glycolysis is completed? Do the same for each carbon of glucose.
See figure 14-6 on page 535.
14 C on C1 or C6 of glucose will label CH 3 of pyruvate. 14 C on C2 or C5 of glucose will label C=O of pyruvate. 14 C on C3 or C4 of glucose will label carboxyl group of pyruvate.
Fall, 2008
5B. 1 net mol ATP/mol glucose is formed in this revised cycle, whereas 2 net mole ATP/mole
glucose are formed in the real scheme. If the only thing that matters is net yield of ATP then the
real scheme is better.
Oxidative Reversible (not oxidative)
6. A. In the above drawing, label the oxidative reactions and reversible reactions of the pentose
phosphate pathway.
B. If NADPH is required by the cell, which of these two sets of reactions MUST be active?
The oxidative steps. Notice that of the 4 possible conditions mentioned in class
3 require NADPH. The only reaction pathway above that produces NADPH is
the oxidative pathway. It follows that if the cell requires NADPH, it will use
this pathway. If the cell does not require NADPH (condition #4), then it will
not utilize this pathway.
C. If NADPH and not ribose are required by the cell, will the reversible reactions also be
activated? If so, in which direction will they occur (to the right or to the left)? Are
gluconeogenic or glycolytic enzymes also involved? Are they depicted in this diagram?
Reversible reactions will occur towards the right. The ribose- 5 - phosphate (C5)
produced by the oxidative reactions is not required by the cell and is converted
into fructose- 6 - phosphate and glyceraldehyde 3-phosphate by the reversible
reactions above. Then the fructose 6-phosphate is converted into glucose 6-
phosphate by phosphohexose isomerase (a glycolytic enzyme, not shown in the
diagram) and the glyceraldehyde 3-phosphate is converted to glucose 6-
phosphate by part of the gluconeogenic pathway (not shown in diagram). The
glucose 6-phosphate may then be used again in the oxidative reactions to make
more NADPH.
C 3 C 3
C 6 C 5 C 6
CO 2 C 2 C 3
C 7
C 6 C 5 C 5 C 4
CO 2 C 6
C 2 C 2
C 6 C 5
CO 2 C 3 C 3
C 3 C 3
C 6 C 5 C 6
CO 2 C 2 C 3
C 7
C 6 C 5 C 5 C 4
CO 2 C 6
C 2 C 2
C 6 C 5
CO 2 C 3 C 3
A
B
C
D E
F
G
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D. If NADPH and ribose are both required by the cell, will the reversible reactions occur? If
so, in which direction?
The reversible reactions will not occur, because the product of the oxidative reactions (Ribose) is required by the cell.
E. If NADPH and other building blocks are required by the cell (such as pyruvate), will the
reversible reactions occur? If so, in which direction? Are gluconeogenic or glycolytic
enzymes also involved?
The reversible reactions will occur from the left to the right. Ribose will be converted to fructose 6-phosphate (C6) and glyceraldehyde 3-phosphate (C3). These molecules will then enter glycolysis to produce pyruvate, which can be converted into amino acids.
F. What if the cell only requires ribose? Will the oxidative reactions occur? In which
direction will the reversible reactions occur? Are gluconeogenic or glycolytic enzymes
also involved
If the cell doesnโt require NADPH, then the oxidative reactions will not occur. Ribose will be formed through the reversible reactions alone. Ribose (3) will be made from fructose- 6 - phosphate (2) and glyceraldehyde- 3 - phosphate (1). Because F6P and G3P are glycolytic intermediates, glycolytic enzymes are required to form them from glucose 6-P.
G. Correlate the diagram above with the actual chemical reactions below.
The conversion of glucose 6-P to ribulose 5-P are the oxidative reactions. The letters below correspond to the steps shown in the diagram above.
CH 2 OH
C O^
TPP
CH 2 OH
C O^
TPP
CH 2 OH
C O^
TPP
CH 2 OH
C O^
TPP
CH 2 OH
C
H
O
OH C
H C OH
H 2 C OPO 32 -
xylulose 5 - P
CH 2 OH
C
H
O
OH C
H C OH
H 2 C OPO 32 -
xylulose 5 - P
CH 2 OH
C
H
O
OH C
H C OH
H 2 C OPO 32 -
xylulose 5 - P
C
H C OH
H 2 C OPO 32 -
O (^) H โ โ โ โ โ โ
CH 2 OH
C O^
TPP glyceraldehyde 3 - P
C
H C OH
H 2 C OPO 32 -
O (^) H โ โ โ โ โ โ
CH 2 OH
C O^
TPP
CH 2 OH
C O^
TPP
CH 2 OH
C O^
TPP glyceraldehyde 3 - P
H C OH
C
OH
O
H C
H C OH
H 2 C OPO 32 -
H
ribulose 5 - P
H C OH
C
OH
O
H C
H C OH
H 2 C OPO 32 -
H
ribulose 5 - P
C
C
OH
OH
H C
H C OH
H 2 C OPO 32 -
H
O H
ribose 5 - P
C
C
OH
OH
H C
H C OH
H 2 C OPO 32 -
H
O H
C
C
OH
OH
H C
H C OH
H 2 C OPO 32 -
H
O H
ribose 5 - P H C OH H C OH H 2 C OPO 32 -
C
O H
erythrose 4 - P
H C OH
H C OH
H 2 C OPO 32 -
C
O H
erythrose 4 - P
H C OH
H C OH
H 2 C OPO 32 -
C
O H
H C OH
H C OH
H 2 C OPO 32 -
C
O H
erythrose 4 - P
C
C
OH
OH
H C
H C OH
H 2 C OPO 32 -
H
HO
C
H
O
CH 2 OH
sedoheptulose
C 5
C 2
C
C
OH
OH
H C
H C OH
H 2 C OPO 32 -
H
HO
C
H
O
CH 2 OH
sedoheptulose
C 5
C 2
C 5
C 2
Fructose 6 - P
CH 2 OH
C
HO C
OH
H
H C
H C OH
H 2 C OPO 32 -
O
Fructose 6 - P
CH 2 OH
C
HO C
OH
H
H C
H C OH
H 2 C OPO 32 -
O
Fructose 6 - P
CH 2 OH
C
HO C
OH
H
H C
H C OH
H 2 C OPO 32 -
O
CH 2 OH
C
HO C
OH
H
H C
H C OH
H 2 C OPO 32 -
O
A&C
D
E
G
F
B
