Part C Firefly luciferase is the enzyme that allows fireflies to illuminate thei

Question

Part C 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 + O2 ATP oxyluciferin + light AMP + PP, If the overall G of the coupled reaction is 5 00 k mo what is the equilibrium constant K of the first reactions at 23 °C ? The G for the hydrolysis of ATP to AMP is-31.6 kJ/mol Express your answer numerically. Submit Hints My Answers Give Up Review part

Explanation / Answer

Part C. dG = -5 kJ/mol

dG for the first reaction would be = -5 + 31.6 = 26.6 kJ/mol

dG = -RTlnKeq

T = 23 oC = 23 + 273 = 296 K

Feed value,

26.6 = -8.314 x 296 lnKeq

Keq = 0.99

Part A. For a spontaneous reaction dG = -ve

dG = dH - TdS

0 = 180.5 - T x 0.0248

T = 7278.226 K

So above temperature = 7278.226 K the reaction would be spontaneous.

Part B. dG = dH - TdS

Taking vaues from above,

dG = 180.5 - (298 x 0.0248) = 173.11 kJ

173.11 = -8.314 x 298 lnKeq

Keq = 0.93

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