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 1.60×10−6 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. c. assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?

a. kc = 1,67x10⁶

b. kc = 1,17x10⁷

c. Drug B is the better choice.

Explanation:

The bind of drug-protein is described as:

Protein + Drug ⇄ Drug-protein

Where kc is:

kc = [Drug-protein] / [Protein] [Drug] (1)

a. For A, the equilibrium concentration of each specie is:

[Protein]: 1,60x10⁻⁶M - x

[Drug]: 2,00x10⁻⁶M - x

[Drug-protein]: x = 1,00x10⁻⁶M

-Where x is reaction coordinate-

Thus:

[Protein]: 1,60x10⁻⁶M - 1,00x10⁻⁶M = 0,60x10⁻⁶M

[Drug]: 2,00x10⁻⁶M - 1,00x10⁻⁶M = 1,00x10⁻⁶M

Replacing in (1):

kc = [1,00x10⁻⁶] / [1,00x10⁻⁶] [0,60x10⁻⁶]

kc = 1,67x10⁶

b. For B:

[Protein]: 1,60x10⁻⁶M - x

[Drug]: 2,00x10⁻⁶M - x

[Drug-protein]: x = 1,40x10⁻⁶M

-Where x is reaction coordinate-

Thus:

[Protein]: 1,60x10⁻⁶M - 1,40x10⁻⁶M = 0,20x10⁻⁶M

[Drug]: 2,00x10⁻⁶M - 1,40x10⁻⁶M = 0,60x10⁻⁶M

Replacing in (1):

kc = [1,40x10⁻⁶] / [0,20x10⁻⁶] [0,60x10⁻⁶]

kc = 1,17x10⁷

c. The drug with the bigger kc will be the more effective because will be the drug that binds more strongly with the protein. Thus, drug B is the better choice.

I hope it helps!

away from the curb

explanation:

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