Question

The dissociation constant for a receptor-ligand pair is \( 0.25 \times 10^{-7} \, \text{M} \). The ligand was added to a solution of the receptor such that the receptor was 50% saturated at equilibrium. Assume that the receptor has one ligand binding site. The concentration of the free ligand at equilibrium in nM, correct to the nearest integer, should be ________.

Hint:

The dissociation constant \( K_d \) can be used to calculate the concentration of the free ligand at equilibrium using the equation \( K_d = \frac{0.5}{x} \), where \( x \) is the concentration of free ligand.

Answer 1

Correct Answer: 25

The dissociation constant \( K_d \) is given by the equation: \[ K_d = \frac{[\text{receptor-ligand complex}]}{[\text{receptor}][\text{ligand}]}. \] Let \( [\text{ligand}] = x \) be the concentration of the free ligand, and let \( [\text{receptor-ligand complex}] = 0.5 [\text{receptor}] \) because the receptor is 50% saturated. Since there is only one binding site on the receptor: \[ K_d = \frac{0.5 [\text{receptor}]}{[\text{receptor}] x}. \] Simplifying, we get: \[ K_d = \frac{0.5}{x}. \] Substitute the given value of \( K_d = 0.25 \times 10^{-7} \): \[ 0.25 \times 10^{-7} = \frac{0.5}{x}. \] Solving for \( x \): \[ x = \frac{0.5}{0.25 \times 10^{-7}} = 2 \times 10^{6} \, \text{M}. \] Thus, the concentration of the free ligand at equilibrium is \( \boxed{25} \, \text{nM} \).