Question

Two different proteins X and Y are dissolved in aqueous solution at 37 ∘C. The proteins bind in a 1:1 ratio to form XY. A solution that is initially 1.00 mM in each protein is allowed to reach equilibrium. At equilibrium, 0.24 mM of free X and 0.24 mM of free Y remain.

what is the mM?

Answer 1

The reaction for for the given proteins are as follows:

X + Y $\rightarrow$ XY

Then the equilibrium constant can be given as follows:

Kc = [XY] /[X][Y]

At equilibrium , we have 0.24 mM of X and 0.24 mM of Y

[X] = [Y] = 0.24 mM = 2.4 x 10-4 M

Calculate the concentration of XY at equilibrium:

[XY] = 1.00 - 0.24 = 0.76 mM = 7.6 x 10-4 M

Then

Kc = 7.6 x 10-4 M /( 2.4 x 10-4 M)2

Kc = 1.32 x 104 mM

Therefore, at equilibirium [XY] = 7.6 x 10-4 M and Kc = 1.32 x 104 M.

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