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

D

Step-by-step explanation:

From the information given:

Association rate
`K_a` =
`8.9 * 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 =( K_d)/(K_a)`

`K_d = K_D * K_a`

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

`\mathbf{K_d =8.9*10^(-5) \ {s^(-1)}}`