Problem Set 7

 

1)                  Is the PFK reaction in muscle more or less exergonic than under standard conditions, given the following concentrations:

Fructose-6-phosphate

1.0 mM

Fructose-1,6-bisphosphate

10.0 mM

AMP

0.10 mM

ADP

0.50 mM

ATP

5.0 mM

Pi

10.0 mM

 

 

2)                  You make wine that is 10% w/v EtOH. What molar concentration of glucose (or equivalent) would be required for this to occur?

 

MW of EtOH is about 46 g/mol, so:

This is an unlikely concentration. However, there are other carbohydrate sources in barley including starches.

 

 

3)                  Which carbons of glucose, catabolized through glycolysis and the Krebs cycle, will be lost most rapidly as CO2?

 

The bond between carbons 3 and 4 of glucose is the one that is cleaved in the aldolase reaction. As a result, C3 becomes C1 of DHAP (the non-phosphorylated terminal carbon) and C4 becomes the aldehyde carbon of glyceraldehyde-3-phosphate. Triose phosphate isomerase converts the non-phosphorylated terminal carbon of DHAP (originally C3 of glucose) to the aldehyde carbon of another glyceraldehyde-3-phosphate. The aldehyde carbon of glyceraldehyde-3-phosphate eventually becomes the carboxyl carbon of pyruvate, which is the carbon that gets cleaved off by the PDH complex.

 

 

4)                  Calculate DGo' for oxidation of malate by malate dehydrogenase:

Malate

Oxaloacetate + 2e- + 2H+

+0.17 V

NAD+ + 2e- + 2 H+

NADH + H+

-0.32 V

Malate + NAD+

Oxaloacetate + NADH + H+

-0.15 V

 

5)                  What would DGo' be for an enzyme that oxidizes succinate with NAD+ instead of FAD? If [succinate]/[fumarate] = 10, what is the minimum [NAD+]/[NADH] ratio to make the reaction exergonic?

Succinate

Fumarate + 2e- + 2H+

-0.03 V

NAD+ + 2e- + 2 H+

NADH + H+

-0.32 V

Succinate + NAD+

Fumarate + NADH + H+

-0.35 V

 

In order for the reaction to be exergonic, DG must be < 0.

(This differs a little from the book value)

 

6)                  .

For the reaction:

GSSG + 2e- + 2H+

2GSH

-0.23 V

NADPH + H+

NADP+ + 2e- + 2 H+

+0.32 V

GSSG + NADPH + H+

2GSH + NADP+

+0.09 V

 

a)      Calculate DGo

b)      What would be the effect on DGo of using NADH instead of NADPH?

DGo would be unchanged, since NADPH and NADH have the same standard reduction potential.

c)      What about under real circumstances?

In real tissues, the [NAD+]/[NADH] ratio tends to be high, compared to the [NADP+]/[NADPH] ratio. Therefore DG would be driven lower for an NADH-linked enzyme and the reaction would be driven more to the left.