Question

Consider the final reaction in the citric acid cycle, catalyzed by malate dehydrogenase:

Malate + NAD+  ⇄ oxaloacetate + NADH + H+

Equilibrium concentrations of the reaction components were measured as follows:

[malate] = 5.28 mM

[NAD+] = 75 uM

[oxaloacetate] = 14.5 uM

[NADH] = 14.5 uM

pH = 8.83

a) From these data calculate the equilibrium constant, Keq, for this reaction.

b) Calculate the standard free-energy change, ΔG°’, for this reaction (hint: remember that

the prime means that the standard [H+] is 1.0 × 10-7 M, not 1 M)

c) Calculate the actual free-energy, ΔG, under the following conditions:

 [malate] = 6.0 mM

[NAD+] = 100 uM

[oxaloacetate] = 0.001 uM

[NADH] = 0.0001 uM

pH = 8.83

Answer 1

a) Malate + NAD+  ⇄ oxaloacetate + NADH + H+

Equilibrium concentrations of the reaction components were measured as follows:

[malate] = 5.28 mM

[NAD+] = 75 uM = 0.075 mM

[oxaloacetate] = 14.5 uM = 0.0145 mM

[NADH] = 14.5 uM = 0.0145 mM

pH = 8.83

[H+] = 1.0 × 10-7 M = 1.0 × 10^-5 mM

the equilibrium constant, Keq

Keq=[oxaloacetate ] [NADH] [ H+] / [ Malate ] [NAD+]

=(0.0145*0.0145*1*10^-5)/(5.28*0.075)

Keq =5.3093*10^-09

b) the standard free-energy change, ΔG°

ΔG° = -RT ln Keq

= -RT 2.303 Log₁₀ Keq

Note that at 25 to 37°C the value for RT 2.303 is approximately 1.4kcal

ΔG°= -1.4   Log₁₀ Keq

  = -1.4   Log₁₀ 5.3093*10^-9

=(-1.4 ) (-8.2750)

ΔG° =11.5849 kcal/mol

c) the actual free-energy, ΔG

[malate] = 6.0 mM

[NAD+] = 100 uM = 0.1 mM

[oxaloacetate] = 0.001 uM = 0.000001  mM

[NADH] = 0.0001 uM = 0.0000001  mM

pH = 8.83

the equilibrium constant, Keq

Keq=[oxaloacetate ] [NADH] [ H+] / [ Malate ] [NAD+]

= (0.000001 * 0.0000001 * 1.0 *10^-5) / (6.0 * 0.1)

= (1 * 10^-18) / (0.6)

Keq = 1.6667 * 10^-18

the standard free-energy change, ΔG°

ΔG° = -RT ln Keq

= -RT 2.303 Log₁₀ Keq

Note that at 25 to 37°C the value for RT 2.303 is approximately 1.4kcal

ΔG°= -1.4   Log₁₀ Keq

  = -1.4   Log₁₀ 1.6667 * 10^-18

=(-1.4 ) (-17.7781)

   ΔG° =24.8894 kcal/mol

actual free-energy, ΔG is

ΔG = ΔG°+ RT ln Keq

   = 24.8894 +( (-1.4 ) (-17.7781))

= 24.8894+24.8894 =49.7788 kcal/mol