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In: Chemistry

A chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate...

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 Kc value for the A-protein biding reaction.

Part B: Calculate the Kc 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?

Solutions

Expert Solution

PART A

Reaction for drud A

Protein + drug A    Protein - drug A comlex

Kc = [ 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

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

substitute value of final (equilibrium) concentrration

Kc = [1.00×10-6] / [5.8×10-7] [1.00×10-6]

Kc = 1.72×106M

PART B

Reaction for drud B

Protein + drug B    Protein - drug B comlex

Kc = [ 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

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

substitute value of final (equilibrium) concentrration

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

Kc = 1.29×107M

Part C

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

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


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