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+O2ATP⇌⇌oxyluciferin+lightAMP+PPi If the overall ΔG of the coupled reaction is -7.70 kJ/mol , what is the equilibrium constant, K, of the first reactions at 17 ∘C ? The ΔG for the hydrolysis of ATP to AMP is −31.6 kJ/mol.

Answer 1

The overall reaction is

luciferin + O2 + ATP <----> oxyluciferin + light + AMP + PPi , G1 = - 7.70 KJ/mol ------ (1)

The hydrolysis reaction of ATP to form AMP and PPi is

ATP < ----- > AMP + PPi , G2 = - 31.6 KJ/mol

The reverse of the above reaction can be written as

AMP + PPi <-----> ATP, G3 = - (- 31.6 KJ/mol) = + 31.6 KJ/mol ------ (2)

Now adding equation (1) and (2) we get the required reaction

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luciferin + O2 + ATP + AMP + PPi <------> oxyluciferin + light + AMP + PPi + ATP,

Now ATP, AMP, and PPi will cancel out from both sides. Hence the overall reaction becomes

luciferin + O2 <------> oxyluciferin + light , G = G1 + G3 = - 7.70 KJ/mol+31.6 KJ/mol = 23.9 KJ/mol

Now equilibrium constant, K can be calculated from the following relation

G = - RTxlnK

T = 17 DegC = 17 + 273 = 290 K

=> 23.9 KJ/mol = 23900 J/mol = - 8.314 JK^-1mol^-1x 290 K x lnK

=> lnK = - 23900 /290x8.314

=> K = exp( - 23900 /290x8.314) = 4.95 x 10^-5 (answer)