Question

In the situation of Malate Dehydrogenase generating Oxaloacetate, the ∆G^o = + 29.7 kJ/mol. What is the equilibrium ratio of Malate versus Oxaloacetate in this reaction?

	a.	Cannot determine
	b.	2.63 x 10-6
	c.	6.22 x 10-6
	d.	9.0 x 10-6
	e.	1.3 x 10-6

Citrate Synthase uses Oxaloacetate and Acetyl-CoA to synthesize Citrate, with a highly exergonic ∆G^o = - 31.5 kJ/mol. Why would this be necessary for the TCA Cycle? HINT: What was asked in Question 7 and which answer is MOST correct given these data?

	a.	Oxaloacetate is a required intermediate for the TCA Cycle.
	b.	Highly favorable citrate synthesis drives the unfavorable Malate Dehydrogenase reaction.
	c.	Oxaloacetate is a required intermediate for Gluconeogenesis.
	d.	Malate conversion to Oxaloacetate is highly unfavorable.
	e.	Citrate synthesis is a rate-determining step in the TCA Cycle.

Answer 1

Ans-e. 1.3 x 10-6

Explanation- G° = +29.7kJ/mol

By formula, G° = -RTlnKeq

29.7 = -8.31*298*lnKeq      R= gas constant = 8.31 J mol-1 K-1

-29.7/8.31/298 = 2.303 log keq       T = temperature = 25°C (for biochemical reaction)=298K

log Keq = - 0.0052    Keq = [products]/[reactants] = [oxaloacetate]/[malate]

Keq = 0.98

[malate]/[oxaloacetate] = 1/Keq = 1.3 x 10-6

Ans2.

b.	Highly favorable citrate synthesis drives the unfavorable Malate Dehydrogenase reaction.

Explanation- Since,  the reaction for conversion of malate to oxaloacetate is endergonic because G° is positive. hence, this energy can be acquired from energy released by citrate synthesis.