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 1.58×10⁻⁶ M . Drug A is introduced into the protein solution at an initial concentration of 2.00×10^-6 M. Drug B is introduced into a separate, identical protein solution at an initial concentration of 2.00×10^-6 M. At equilibrium, the drug A-protein solution has an A-protein complex concentration of 1.00×10^-6 M, and the drug B solution has a B-protein complex concentration of 1.40×10^-6 M.

Part A: Calculate the K_c value for the A-protein biding reaction.

Part B: Calculate the K_c value for the B-protein biding reaction.

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

Answer 1

PART A

Reaction for drud A

Protein + drug A    Protein - drug A comlex

K_c = [ Protein - drug comlex] / [Protein] [drug A ]

initial concentration of protein = 1.58×10^-6M

initial concentration of drug A = 2.00×10^-6M

final concentration of Protein - drug A comlex = 1.00×10^-6M

final concentration of protein = 1.58×10^-6 - 1.00×10^-6 = 5.8×10^-7M

final concentration of drug A = 2.00×10^-6 - 1.00×10^-6 = 1.00×10^-6M

K_c = [ Protein - drug comlex] / [Protein] [drug A ]

substitute value of final (equilibrium) concentrration

K_c = [1.00×10^-6] / [5.8×10^-7] [1.00×10^-6]

K_c = 1.72×10⁶M

PART B

Reaction for drud B

Protein + drug B    Protein - drug B comlex

K_c = [ Protein - drug B comlex] / [Protein] [drug B ]

initial concentration of protein = 1.58×10^-6M

initial concentration of drug B = 2.00×10^-6M

final concentration of Protein - drug B comlex = 1.40×10^-6M

final concentration of protein = 1.58×10^-6 - 1.40×10^-6 = 1.8×10^-7M

final concentration of drug B = 2.00×10^-6 - 1.40×10^-6 = 6.00×10^-7M

K_c = [ Protein - drug comlex] / [Protein] [drug A ]

substitute value of final (equilibrium) concentrration

K_c = [1.40×10^-6] / [1.8×10^-7] [6.00×10^-7]

K_c = 1.29×10⁷M

Part C

K_c for drug A = 1.72×10⁶M and K_c for drug B =  1.29×10⁷M  K_c for drug B is higher than  K_c for drug A

Hence dug B reaction is more favorable bind more strongly thus, drug B is better choice.