The two-state denaturation of a protein: N <> D has an equilibrium constant: Kd
ID: 866234 • Letter: T
Question
The two-state denaturation of a protein:
N <> D
has an equilibrium constant: Kd = [D] / [N]
where N is the native state and D is the denatured state. The native state specifically
binds a ligand, A, with association constant, a:
a = [NA] / [N] [A]
a. Derive an expression for the equilibrium constant for denaturation, KdA in the
presence of ligand A in terms of [A], a and Kd. Please show all work.
b. What happens to the equilibrium constant for denaturation as the concentration of A is
increased? i.e. Does ligand binding stabilize or destabilize the native state? Please show all work.
Explanation / Answer
If N and A are free, so:
A+N-(K1)->NA. K1: KINETIC ASSOCIATION CONSTANT
NA-(K-1)->N+A. K-1: kINETIC DISSOCIATION CONSTANT
IF
(K-1)[NA]=(K1)[A][N]
THEN
(K-1)/(K1)=[A][N]/[NA]=Kd
Kd: DISSOCIATION CONSTANT DIFFERENT TO K1 AND K-1
DENATURATION
N+A->NA->D+A
Kd=[D][A]/[NA]
as A increases Kd is bigger. Which means that the more ligand the more it denaturates.
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