Question

In an attempt to create a biological battery for energy storage, you examine the possibility of incorporating a rare enzyme that facilitates the phosphorylation of 3-phosphoglycerate (3PG) in the presence of H+ to form 1,3 bisphosphoglycerate (1,3-BPG). The delta G for this enzymatically driven process is +59.6 kj/mol.

A.) Draw the reaction scheme for 3PG to (1,3-BPG).

B.) is this a spontaneous or nonspontaneous reaction?

C.) What is the Keq under standard biological conditions, based on the delta G value above?

D.) What effect should raising the pH habe on the free energy of the reaction?

Answer 1

A)

B) Since deltaG is positive for this reaction, it is a  nonspontaneous reaction.

C) Given G = + 59.6 KJ/mol = +59600 J/mol

DeltaG and Keq are related through the following formulae

G = - RTln(Keq)

=> ln(Keq) = -G /RT = + 59.6 KJ/mol = +59600 J/mol

If we consider the temperature as body temperature then T = 37 C = 310 K

=> ln(Keq) = - 59600 Jmol-1 / (8.314 JK-1mol-1 x 310K) = -23.1246

=> Keq = 9.06x10^-11 (answer)

(d) G and G⁰ are related as

G = G⁰ + RTln[1,3-biPG] / ([H+]x[3-PGh])

when pH = - log[H+] increases the value of [H+] decreases. Hence the value of RTln[1,3-biPG] / ([H+]x[3-PGh]) increases and hence the value of G also increases.

Hence free energy increases (answer)