Question

In the citric acid cycle, malate is dehydrogenated to oxaloacetate in a highly endergonic reaction with a ΔG’o of +30 kJ mol-1: L‐malate + NAD+ ⇌ oxaloacetate + NADH + H+

A. Calculate the equilibrium constant K’eq of this reaction. What is the implication of this result?

B. Let us suppose that you have gathered two sets of data concerning the malate dehydrogenase reaction: (a) From one set of data obtained from three different model systems (rat, mouse and pig) you have determined that the concentration of L-malate in the liver is about 0.25 mM when the [NAD+]/[NADH] ratio is approximately 10 under standard conditions (pH 7 and 25oC); (b) From another set of data obtained from mitochondria isolated from mice you have determined that oxaloacetate concentration ranges from 3-16 nM under the same standard conditions (pH 7 and 25oC) regardless of whether the mice are maintained on high or low glucose diets.

Based on the first set of data, (a), calculate the concentration of oxaloacetate at equilibrium. Compare your result with the data in (b). What is the likely significance of the data in (a) and (b)?

Answer 1

Given,   ΔG0 = 30 kJ mol-1

(A) We know, ΔG0 = - RT ln keq

or, log keq = -  $1585459970946_blob.png$

or, keq =

10-46°/2.303RT

or, keq =

10-30000/2.303 8.314x298

Therefore, keq = 5.524 × 10-6

Since, the value of keq is less than one, it implies that the reaction mixture contains mostly reactants, i.e., at equilibrium [L-malate] > [oxaloacetate].

(B) (a) Given, [L-malate] = 0.25 mM = 0.00025 M

                 $1585459971712_blob.png$ = 10

                                  pH = 7

              Therefore, [H+] = 10-7 M

Now, keq =

[oxaloacetate] INADH][H+1 [L-malats][NAD+)

or, [oxaloacetate] =

5.524 x 10-6x10x0.00025 10-7

Therefore, [oxaloacetate] = 13.81 × 10-2 M

(b) Given, [oxaloacetate] = 3-16 nM = 3-16 × 10-19 M

Therefore, [oxaloacetate] obtained from data (a) is greater than that obtained from data (b).

Data (a) has been obtained from rat, mouse and pig. On the other hand, data (b) was obtained from only mouse. Thus, for higher animals like rat and pig, in the dehydrogenation reaction the rate of forward reaction is greater than that in case of mouse.