Question

A chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate molecules bind to a protein involved in cancer. The drug molecules bind the protein in a 1:1 ratio to form a drug-protein complex. The protein concentration in aqueous solution at 25 ∘C is 1.50×10⁻⁶ M . Drug A is introduced into the protein solution at an initial concentration of 2.00×10−6M. Drug B is introduced into a separate, identical protein solution at an initial concentration of 2.00×10−6M. At equilibrium, the drug A-protein solution has an A-protein complex concentration of 1.00×10−6M, and the drug B solution has a B-protein complex concentration of 1.40×10−6M.

Calculate the Kc value for the A-protein binding reaction.

Calculate the Kc value for the B-protein binding reaction.

Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?

	Drug A is the better choice for further research.
	Drug B is the better choice for further research.

Answer 1

The reaction is as follow
Drug A + Protein --> DrugA-Protein
To solve for equilibrium constants, you need to calculate the concentration of each of the species at equilibrium. By definition an equilibrium constant is [products]/[reactants].
K_A = [Protein-Drug A] / ( [Protein] * [Drug A] )
[Protein-DrugA] = 1.00*10^-6 M
[Protein] = 1.50*10^-6 - 1.00*10^-6 = 5*10^-7 M

[Drug A] = 2.00*10^-6 - 1.00*10^-6 = 1.00*10^-6 M

K_A = 2.0 x 10^-6 M for Drug A

For Drug B

K_B = [Protein-Drug B] / ( [Protein] * [Drug B] )
[Protein-DrugA] = 1.00*10^-6 M
[Protein] = 1.40*10^-6 - 1.00*10^-6 = 4*10^-7 M

[Drug B] = 2.00*10^-6 - 1.00*10^-6 = 1.00*10^-6 M

K_B = 2.5 x 10^-6 M for Drug B

According to these results drug B would have a higher affinity for the protein