Question

Consider the following ligand binding equilibrium at 25 °C:

                                                  P+LPL            ∆G° = –10 kJ/mol

a) What is the K_a for the binding of P to L?

Use the relationship between free energy and equilibrium constant: ∆G°=-RT ln Keq. The answer is 56 /M.

b) What is the K_d for the dissociation of PL to P and L?

Kd=1/Ka, or ∆G° = +10 kJ/mol

c) The rate constant for binding of P to L is 10⁶ M^-1s^-1.  What is the rate constant for dissociation of L from PL?

For this problem, recognize that the forward and reverse rate constants can be related to equilibrium constant as follows;

kf [P][L]=kr[PL]; [P][L]/[PL]=kr/kf = Kd

0.017*10⁶ = 17x10³

^{***Answers are shown in bold, I just need to know how to get there***}

Answer 1