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+O2ATPoxyluciferin+lightAMP+PPi If the overall G of the coupled reaction is -8.30 kJ/mol , what is the equilibrium constant, K, of the first reactions at 25 C ? The G for the hydrolysis of ATP to AMP is 31.6 kJ/mol. Express your answer numerically.

Explanation / Answer

In the first step, luciferin combines with the energy-giving molecule adenosine triphosphate (ATP) on the surface of luciferase to form luciferyl adenylate and pyrophosphate (PPi):

luciferin + ATP luciferyl adenylate + PPi

In the second step, luciferyl adenylate combines with oxygen (O2) to give the products oxyluciferin, adenosine monophosphate (AMP), and light. The oxyluciferin and AMP are then released from the enzyme’s surface.

luciferyl adenylate + O2 oxyluciferin + AMP + light

The light given off by the reactions contains wavelengths between 510 and 670 nanometers, which results in a pale yellow light. The special abdominal cells that make the light also have uric acid crystals in them that help to reflect the light away from the abdomen.

This equation should be used to find K.

K = e^(-standard G/(RT))

R is 8.314J/(molK) and T is 298K. The standard G is found by subtracting the standard G for the hydrolysis of ATP to AMP from the standard G of the coupled reaction.

standard G of first reaction = G of the coupled reaction - G for the hydrolysis of ATP to AMP
standard G of first reaction = -8.30kJ/mol - (31.6 kJ/mol)
standard G of first reaction = -8.30kJ/mol + 31.6 kJ/mol
standard G of first reaction = 23.3kJ/mol
standard G of first reaction = 23300J/mol

Plug in the known values into the first equation to find K.

K = e^(-(23300J/mol)/(8.314J/(molK)*298K))
K = e^(-(23300J/mol)/(2477.572J/mol))
K = e^(-9.40436847)
K = 8.2363*10-5 -------------------    Answer

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