Firefly luciferase is the enzyme that allows fireflies to illuminate their abdom

Question

Firefly luciferase is the enzyme that allows fireflies to illuminate their abdomens. Because this light generation is an ATP-requiring reaction, firefly luciferase can be used to test for the presence of ATP. In this way, luciferase can test for the presence of life. The coupled reactions are luciferin+O2oxyluciferin+light

ATPAMP+PPi

If the overall G of the coupled reaction is -4.60 kJ/mol , what is the equilibrium constant, K, of the first reactions at 30 C ? The G for the hydrolysis of ATP to AMP is 31.6 kJ/mol. Express your answer numerically.

Explanation / Answer

Solution :-

Overall reactions

Luciferrin + O2 + ATP <---> oxyluciferrin + light + AMP ppi

Delta G overall = Delta G (oxidation of luciferon)+ Delta G of conversion of ATP to AMP

-4.60 kJ /mol = Delta G oxd^n of luciferon + (-31.6 kJ /mol)

Delta G ox^n of luciferon = -4.60 + 31.6 kJ /mol = 27.0 kJ

K = Exp ( - Delta G oxidaotion / RT)

K = Exp [ - 27000 J per mol /8.314 J per mol K * 303 K]

K = 2.21 *10^-5

So the K equilibrium for the first reaction is Keq = 2.21*10^-5

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