The dissociation constant, Kd for a complex between protein A and protein B is 4.1 μM. If the two proteins are mixed together at initial concentrations of [A]= 0.025 μM and [B] = 4.7 μM, calculate (a) the equilibrium concentrations of A, B, and AB (the dimer formed by A and B) (b) the percentage of A bound to B

Biochemistry
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Chapter1: Biochemistry: An Evolving Science
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The dissociation constant, Kd for a complex between protein A and protein B is 4.1 μM. If the two proteins are mixed together at initial concentrations of [A]= 0.025 μM and [B] = 4.7 μM, calculate (a) the equilibrium concentrations of A, B, and AB (the dimer formed by A and B) (b) the percentage of A bound to B

Expert Solution
Equilibrium dissociation constant

The equilibrium association between proteins A & B to form AB and the dissociation of AB to give A & B can be represented as shown below.

A + B KdKa AB

Here Ka is the equilibrium association constant and Kd is the equilibrium dissociation constant.

Ka =[AB]eq[A]eq [B]eqKd =[A]eq [B]eq[AB]eq

Here we introduce 0.025μM of A and 4.7μM of B. These two begin to bind with each other and proceed towards reaching equilibrium. Hence the concentration of A and B starts to decrease , while the concentration of AB increases. Once equilibrium is reached, the rate of association of A & B to form AB  becomes equal to the rate of dissociation of AB to A & B. Hence once equilibrium is reached, the concentrations of A , B and AB does not change any further. The concentrations of A , B and AB at equilibrium are [A]eq , [B]eq and [AB]eq respectively.

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