Assume the following values of Eo and AGo\': malate + NAD+ oxaloacetate + NADH +

Question

Assume the following values of Eo and AGo': malate + NAD+ oxaloacetate + NADH + H+ lactate + NAD+ pyruvate + NADH + H+ Half reaction Go'z +29.7 k/mol G",= +25.1 kJ/mol Eo (V) Pyruvate + 2H++ 2e. lactate 0.190 Pyruvate + CO2 + H++ 2e malate 0.330 Intracellular steady state concentrations malate-410 M: NADH = 290 M; pyruvate = 3.22 mM: lactate = 1.1 mM; NAD+-20.0 mM; CO2-15.5 torr a) Consider the oxidation of malate to oxaloacetate by NAD: malate + NAD+ oxaloacetate + NADH + H+ In yeast mitochondria, where the pH 8.1, this reaction is exergonic only at low oxaloacetate concentrations. Assuming a pH 8.1, a temperature of 37 °C, and the steady-state concentrations given above, calculate the maximum concentration of oxaloacetate at which the reaction will still be exergonic (b) Now, consider the decarboxylation of oxaloacetate to pyruvate and CO2: oxaloacetate + H+ pyruvate + CO2 Assuming a pH 8.1, a temperature of 37 C, and the steady-state concentrations given above, is this decarboxylation reaction thermodynamically favorable or not at the oxaloacetate concentration you calculated in part (a)? Show the calculations that support your answer.

Explanation / Answer

a) The oxidation reaction is

malate +   NAD+ ---> oxaloacetate +   NADH + H+     G0=+29.7kJ/mol

G0= -RTlnK

Where

K = equilibrium constant

T= temperature = 370 C = 310 K

R= gas constant = 8.314 J / mol K = 8.314 X 10-3 kJ / mol K

Putting values

G0 = +29.7kJ/mol = -8.314 X 10-3 X 310 ln K

lnK = -11.52

Taking antilog

K = 9.93 X 10-6

Now K = [Product] / [Reactant] = [oxaloacetate] X [NADH] / [NAD+] [ Malate]

9.93 X 10-6 = [Oxaloacetate] X [290 M] / 20 mM x 410 M

[Oxaloacetate] = 0.281 X 10-6 M = 2.81 X 10-4 mM

Below this concentration reaction will be exergonic. And this concentration reaction will be in equilibrium.

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