The G°’ value for the malate DH reaction is +7.10 kcal/mol, and the G°’ value fo

Question

The G°’ value for the malate DH reaction is +7.10 kcal/mol, and the G°’ value for the citrate synthase reaction is -7.53 kcal/mol. Determine Keq for the combined, coupled reaction (e.g., L-malate <--> citrate, plus ancillary cofactors, reactants, and side-products), and compare this value to Keq for the malate DH reaction alone. By what factor do you have to multiply the Keq of the malate DH reaction to equal the Keq for the coupled reaction. Assume a temperature of 37 °C and report your answer to the nearest ones.

Explanation / Answer

Given data:

I) For malate DH, (G0’)m = + 7.10 kcal/mol.

Let us calculate Keq for this reaction, say (Keq)m,

We have T = 37 0C = 37 + 273 = 210 K, R = 1.987 cal.K-1.mol-1.

General formula,

(G0) = -RT ln(Keq)

For malat DH reaction, with all known values,

+7.10 = - 1.987 x 310 x ln(Keq)m.

ln(Keq)m. = (7.10) / (-1.987x310)

ln(Keq)m = -1.15 x 10-2

(Keq)m = e-1.15 x 10-2

(Keq)m = 0.9885…………… ( 1)

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II) For citrate synthesis, (G0’)c = -7.53 kcal/mol.

Overall free-energy change (G0’) for a chemically coupled series of reactions is equal to the sum of the free-energy changes to the individual steps.

Hence for a combined coupled reaction, free energy change be (G0’)

(G0’) = (G0’)c + (G0’)m

(G0’) = -7.53 + (+7.10)

(G0’) = - 0.43 kcal/mol.

For combined coupled reaction free-energy change (G0’) is -0.43 kcal/mol.

Let us calculate Keq for combined coupled reaction,

(G0’) = -RTln(Keq)

-0.43 = -1.987 x 310 x ln(Keq)

ln(Keq) = (-0.43)/(-1.987x310)

ln(keq) = 6.98 x 10-4.

Keq = e6.98x10-4.

Keq = 1.0007

Keq for combined coupled reaction is 1.0006.

Now,

(Keq) / (Keq)m = 1.0007/0.9885 = 1.01.

Keq / (Keq)m = 1.01

Keq = 1.01 x (Keq)m

Keq for malate DH reaction needed to be multiply by 1.01 to equal Keq for combined coupled reaction.

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