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 1.78×10−6 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.

Answer 1

For Drug A and protein complex      

         DrugA + protein           -->   DrugA -protein

Initial    a    b               0

Final    a-x      b-x x   

Kc = equilibrium constant = x /(a-x)(b-x)

Kc = 1.00×10−6 / 1.00×10−6 X 0.78 X 10^-6

Kc = 1.28 X 10^6

For Drug A and protein complex   

         Drug B + protein --> DrugB -protein

Initial    a    b               0

Final    a-x      b-x x   

Kc = equilibrium constant = x /(a-x)(b-x)

x = 1.4×10−6

a = 2.00×10−6M

b = 1.78×10−6

a-x = 0.6×10−6

b-x = 0.38 X 10^-6

Kc = 1.4×10−6 / 0.6×10−6 X 0.38 X 10^-6

Kc = 6.14 X 10^6