Question

The removal of a glucose residue from glycogen is catalyzed by glycogen phosphorylase: Glycogen (n residues) + Pi ←→ Glycogen (n-1 residues) + Glucose-1-P The standard state free energy change ( ΔG°´ ) for the forward reaction is +3.1 kJ/mole. R = 8.314 J/mole·K. Under standard conditions at 25°C, what is the [Pi]/[G-1-P] ratio? Round the answer to one decimal place.

Answer 1

The equilibrium constant for the reaction is

K = [G-1-P]/[Pi]

We know that

ΔG^0’ = -R*T*ln K

Since the reaction takes place under standard conditions, we have T = 25°C = (25 + 273) K = 298 K and hence,

(+3.1 kJ/mol) = -(8.314 J/mol.K)*(298 K)*ln [G-1-P][Pi]

===> (+3.1 kJ/mol)*(1000 J/1 kJ) = -(8.314 J/mol.K)*(298 K)*ln [G-1-P]/[Pi]

===> ln [G-1-P][Pi] = -(3100 J/mol)/(8.314 J/mol.K).(298 K) = -1.25122

===> [G-1-P]/[Pi] = exp(-1.25122) = 0.28615

===> [Pi]/[G-1-P] = 1/(0.28615) = 3.49467 ≈ 3.50 (ans).