Question

For the association of a ligand, L, to a membrane receptor, R: L + R 1 LR

A. The equilibrium constant for dissociation is Kd = 3.4 x 10-4 M. Intuitively, how much association do you expect, a little or a lot? Justify answer.

B. If [L] = 3 x 10-5 M, [R] = 2 x 10-6 M, calculate [LR].

C. For the association reaction at equilibrium calculate ∆Go at 25 o C (as written).

D. Which reaction is favored, dissociation or association?

Please provide explanations as well. Thank you

Answer 1

A) the dissociation constant Kd = 3.4 x 10-4 M

So the assoication must be high = 1/Kd

B) Kd = [L] [ R] / [RL]

3.4 x 10-4 M = 3 x 10-5 X 2 x 10-6 / [LR]

[LR] = 1.76 X 10^-7 M

C) For association Kass = 1/3.4 x 10-4 = 2941.17

We know that

∆Go = -RTlnK

∆Go = - 8.314 X 298 X ln(2941.17 ) = -8.314 X 298 X 7.99 =- 19795 Joules = 19.795 KJ / mole

D) The negative value for association reveals that the association reaciton will be spontaneous.

So this will be favoured.