Question

The following reactions can be coupled to give alanine and oxaloacetate:

glutamate + pyruvate ↔ ketoglutarate + alanine ∆Go ’rxn,303 = − 1004 J/mol

glutamate + oxaloacetate ↔ ketoglutarate + aspartate ∆Go ’rxn,303 = − 4812 J/mol

(a) Write the form of the equilibrium constant for pyruvate + aspartate ↔ alanine + oxaloacetate and calculate the numerical value of the equilibrium constant at 30o C.

(b) In the cytoplasm of a certain cell, the components are at the following concentrations: pyruvate = 10-2 M, aspartate = 10-2 M, alanine = 10-4 M, and oxaloacetate = 10-5 M. Calculate the Gibbs free energy change for the reaction of part (a) under these conditions. What conclusion can you reach about the direction of this reaction under cytoplasmic conditions?

Answer 1

(a) Equilibrium constant expression for the given reaction would be,

K = [alanine][oxaloacetate]/[pyruvate][aspartate]

dGo = -1004 + 4812 = 3808 J/mol

dGo = -RTlnK

3808 = -8.314 x 298 lnK

K = 0.215

(b) With the given concentrations,

dGo = -8.314 x 298 ln(10^-4 x 10^-5/10^-2 x 10^-2)

        = 28.524 kJ/mol

The reaction is non-spontaneous in the forward direction and would proceed towards the reactant side under cytoplasmic conditions