Question

The protein calcineurin binds to the protein calmodulin with an association rate of 8.9 × 103 M-1s-1 and an overall dissociation constant, Kd, of 10 nM. The dissociation rate kd is:_____. Please explain step by step.
A. 8.9 × 10^3 M-1s-1
B. 8.9 × 10^2 s-1
C. 1.1 × 10-10 s-1
D. 8.9 × 10-5 s-1

Answer 1

Answer:

D

Explanation:

From the information given:

Association rate $K_a$ = $8.9 \times 10^3 M^{-1}s^{-1}$

dissociation constant $K_D$ = 10 nM

dissociation rate $K_d$ = ???

Using the following relation from equilibrium dissociation constant to determine the dissociation rate, we have:

$K_D =\dfrac{ K_d}{K_a}$

$K_d = K_D \times K_a$

$K_d =(10*10^{-9} \ M) \times (8.9*10^3 \ M^{-1}{s^{-1})$

$\mathbf{K_d =8.9*10^{-5} \ {s^{-1}}}$