Question

Aspartate transaminase catalyzes the following reaction: Aspartate + α-ketoglutarate ↔ oxaloacetate + glutamate. It has a cofactor called PLP and the deltaG of Aspartate transaminase to PLP is -8.2 kJ/mol under nominal biological conditions. 1. What is the Keq 2. If the deltaG is -6 kJ/mol how is Keq changed?

Answer 1

Aspartate transaminase catalyzes the following reaction: Aspartate + a-ketoglutarate oxaloacetate + glutamate transaminase AG = -8.2 kJ mol-1 = -8.2 x 10-3 Jmol-1 The reaction is occurring normal biological condition so, T = 298 K, R = 8.314 Imol-1K-1 Now, calculating the value of Keq as follows: AG = -RT In keq -Inkeq = 4G -Inkeq=RT -8.2 x 10-3 Jmol-1 -Inkeq = 8.314 Imol-1K-1 x 298 K keq = 0.033x10-3 keq = 1 If the value of AG is changes to -6 kJmol-7, there will not be any change. It is calculated as follows. -6 X 10-3 Jmol-1 -Inkeq = 8.314 Imol-1K-1 x 298 K keq = 0.024x10-3 keq = 1