Question

Consider the malate dehydrogenase reaction from the citric acid cycle. Given the following concentrations, calculate the free energy change for this reaction at 37.0 °C (310 K). ΔG°\' for the reaction is 29.7 kJ/mol. Assume that the reaction occurs at pH 7. [malate] = 1.53 mM [oxaloacetate] = 0.260 mM [NAD ] = 250 mM [NADH] = 1.0 × 102 mM

Answer 1

We know that free energy change

ΔG = ΔG°+RTInK

Calculation of K, equilibrium constant:

Given that pH = 7

Then, [H+] = 10^-pH = 10^-7 M = 10^-10 mM

  [malate] = 1.53 mM

[oxaloacetate] = 0.260 mM

[NAD ] = 250 mM

[NADH] = 1.0 × 10² mM

malate dehydrogenase + NAD+ <=> Oxaloacetate + NADH + H+

Hence,

Equilibrium constant

K = [oxaloacetate] [NADH] [H+] /   [malate] [NAD ]

= [0.260 mM x 1 x 10² mM x 10^-10 mM] / [1.53 mM x 250 mM]

= 6.8 x 10^-12

K = 6.8 x 10^-12

ΔG

Given that T = 37.0 °C = 37 + 273 K = 310 K

ΔG° = 29.7 kJ/mol = 29700 J/mol

ΔG = ΔG°+RTInK

= 29700 J/mol + (8.314 J/K/mol) ( 310 K) In (6.8 x 10^-12)

= -36574 J/mol

Therefore,

free energy change = -36574 J/mol