Question

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 1.66×10−6 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.

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

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

Solutions

Expert Solution

The reaction between drug A and protein is ( since 1:1 ratio of drug and complex are used)

Drug A+ protein ------à Drug-protein complex

Kc= [Drug-protein complex]/ [Drug A] [Protein]

It is given that at equilibrijm [Drug-protein complex]= 1*10-6

this is the equilibrium concentrationof drug-protein complex. For achieving this concentration, concentration of drug and protein to be used= 1*10-6

At equilibrium, hence [Drug]= initial concentration- concentration lost due to formation of complex=2*10-6-1*10-6= 1*10-6 M

[protein]= 1.66*10-6-1*10-6 =0.66*10-6M

Kc doe drug A -protein binding reaction = 1*10-6/ (1*10-6*0.66*10-6)= 1.515*106


2.

The reaction between drug B and protein is ( since 1:1 ratio of drug and complex are used)

Drug B+ protein ------à Drug-protein complex

Kc= [Drug-protein complex]/ [Drug B] [Protein]

It is given that at equilibrijm [Drug-protein complex]= 1.4*10-6

this is the equilibrium concentrationof drug-protein complex. For achieving this concentration, concentration of drug and protein to be used= 1.4*10-6

At equilibrium, hence [Drug]= initial concentration- concentration lost due to formation of complex=2*10-6-1.4*10-6= 6*10-7 M

[protein]= 1.66*10-6-1.4*10-6 =0.26*10-6M

Kc= 1*10-6/ (6*10-7*0.26*10-6)=6.4*106


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