Question

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

RXN 1: glutamate + pyruvate --> ketoglutarate + alanine
RXN 2 glutamate + oxaloacetate --> ketoglutarate + aspartate

(a) Write the form of the equilibrium constant for the reaction: Pyruvate + aspartate --> alanine + oxaloacetate and calculate the numerical value of the equilibrium constant at 30 C.

(b) In the cytoplasm of a certain cell, the components are at the following concentrations:
[aspartate] = 1.00 x 10-2 M
[pyruvate] = 1.00 x 10-2 M
[alanine] = 1.00 x 10-4 M
[oxaloacetate] = 1.00 x 10-5 M
Calculate the change in free energy (G’) for the reaction in part (a) under these conditions (i.e., 303K). What conclusion can you reach about the direction of the reaction under cytoplasmic conditions?

Answer 1

Reaction 1 the equilibrium constant if Kequ(1)= [alanine]* [keoglutarate]/[glutamate]*[pyruvate] . Similarly, for reaction 2 the equilibrium constant if Kequ ( 2)= [aspartate]*[keoglutarate]/[glutamate]* [oxaloacetate].Now for the given reaction the equilibrium constant if K3 then it can be represented in the following way K3= [alanine]*[oxaloacetate]/[pyruvate]*[aspartate]= Kequ(1)/Kequ(2). From the given data K3= 10^-4*10^-5/10^-2*10^-2 =1* 10^-5 . At 303 K , ∆G = - RTlnK= -8.314* 303* 2.303 log (1*10^-5)= 29007.92 joule, as ∆G is positive so it indicates that the reaction is non- spontaneous in forward direction under cytoplasmic condition.