Question

BIOCHEM; The intramitochondrial concentration of malate under some conditions is 0.22 mM and the NAD+/ NADH ratio is 20:1. Calculate the maximum intramitochondrial concentration of oxaloacetate under these conditions. (Assume 25oC)

Answer 1

Given [Malate] = 0.22 mM = 2.2x10^-4 M

The conversion of malate to oxaloacetate occurs inside mitochondrial matrix during TCA cycle. The complete reaction is

Malate + NAD⁺ <------ > Oxaloacetate + NADH + H+ 2e- ; G⁰ = 29700 J/mol

If we consider the pH of mitochondrial matrix as 7.0, [H+] = 10^-7.0 M

Under equilibrium condition,

G⁰ = 29700 J/mol = - 2.303RTlogK

log[Oxaloacetate]x[NADH]x [H+] / [Malate]x[NAD⁺] = - 29700 J/mol / 2.303x8.314 JK-1mol-1x310K

log([Oxaloacetate]x10^-7 M / 20x2.2x10^-4 xM = - 5.0037

[Oxaloacetate]x10^-7 M / 20x2.2x10^-4 xM = 9.915x10^-6

=> [Oxaloacetate] = 0.436 M