Question

The protein calcineurin binds to the protein calmodulin with an association rate of 8.9 3 103 M21s21 and an overall dissociation constant, Kd, of 10 nM. Calculate the dissociation rate, kd, including appropriate units.

Answer 1

Answer:

K(dissociation) = 8.93 × 10 ⁻⁵ s⁻¹

Explanation:

Given data :

The association rate, K(association) = 8.93 × 10 ³ M⁻¹. s⁻¹

The value of the overall dissociation constant is, Kd = 10 nM = 10 × 10⁻⁹ M

Now,

The dissociation constant (Kd) is given as:

Kd = K(dissociation)/K(association)

on substituting the values, we get

10 × 10⁻⁹ M = K(dissociation) /8.93 × 10 ³ M⁻¹. s⁻¹

or

K(dissociation) = 8.93 × 10 ⁻⁵ s⁻¹

Answer 2

The dissolution rate is K(dissociation) = 8.93 × 10 ⁻⁵ s⁻¹

What will be the dissolution rate for the protein?

It is given that

$\rm The\ dissolution \ rate, \ k(association)=8.93\times10^{3} M^{-1} s^{-1}$

$\rm The\ value \ of\ overall \ dissolution\ constant\ K_d=10nm=10\times10^{-9}M$

Now,

The dissociation constant (Kd) is given as:

$K_d=\dfrac{k( dissociation)}{k(association)}$

on substituting the values, we get

$10\times 10^{-9}=\dfrac{k(dissociation)}{8.93\times10^{3}}$

or

$K(dissociation )=8.93\times10^{-5}s^{-1$

Thus the dissolution rate is K(dissociation) = 8.93 × 10 ⁻⁵ s⁻¹

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