6. You are measuring the thermodynamics of an enzyme that catalyzes the conversi

ID: 908528 • Letter: 6

Question

6. You are measuring the thermodynamics of an enzyme that catalyzes the conversion of leucine into isoleucine. Keq = 0.2 at 298K and pH 7.

DG = DG°’ + RTln[([C][D])/([A][B])]

R = 8.315 x 10-3 kJ/mol K

DG°’ = -2.303 RT log10 (K’eq)
K’eq = 10 -DG°’/2.303RT

a. If the [leucine] at equilibrium is 1 nM, what is the [isoleucine] at equilibrium?

b. What is the standard free energy change for this reaction?

c. Calculate the change in free energy given the initial [leucine] = 600 nM and the initial [isoleucine] = 0.1 nM.

d. Will the reaction occur spontaneously under these conditions described in part c? Explain why or why not.

Keq = [isoleucine]/[leucine]

or, 0.2 = [isoleucine]/[leucine]

a) 0.2 = [isoleucine]/(1*10-9)

or, [isoleucine] = 2*10-10 M = 0.2 nM

b) delta G0 = -2.302*R*T*logKeq = -2.303*8.315*298*log0.2 = 3988.7 J = 3.988 kJ

c) delta G = delta G0 + R*T*ln{[isoleucine]/[leucine]} = 3988.7 + 8.315*298*ln(0.1/600) = -17567.6 J = -12.567 kJ

d) Since, delta G < 0 , hence reaction will occur spontaneously under these conditions.

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