A chemist at a pharmaceutical company is measuring equilibrium constants for rea

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 1.76×106 M . Drug A is introduced into the protein solution at an initial concentration of 2.00×106M. Drug B is introduced into a separate, identical protein solution at an initial concentration of 2.00×106M. At equilibrium, the drug A-protein solution has an A-protein complex concentration of 1.00×106M, and the drug B solution has a B-protein complex concentration of 1.40×106M.

Part A Calculate the Kc value for the A-protein binding reaction.

Part B Calculate the Kc value for the B-protein binding reaction.

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

Explanation / Answer

I think, you trying to calculate for Kc and then using the values in the problem in the equation. If yes, that might be your problem.

The equation for question 1 is roughly Protein(aq) + Drug A -> Protein-Drug A

The concentration that you're given, 1.00 x10^-6M, is at equilibrium. This will be the numerator in your equation.

As for the denominator, you have to take the initial molarity minus the change in molarity to find the molarity at equilibrium for both Protein and Drug A.

The initial molarity for Protein-Drug A is 0, and the equilibrium molarity is 1.00x 10^-6M. This will give us the change in molarity, which is +1.00x10^-6M.

Using this value, plug into the equation: Equilibrium concentration = initial - change

Eq protein = 1.76x10^-6M - 1.00x10^-6M = 7.6x10^-7M

Eq Drug A = 2.00x10^-6M - 1.00x10^-6M = 1.00x10^-6M

NOW plug into the Kc equation!

Kc = [Protein-DrugA] / [Protein][Drug A]

Kc = 1.00x10^-6M / (7.6x10^-7M)(1.00x10^-6M)

Kc = 1.31x10^6M This is for drug A

Now calculate same for drug B

Eq protein = 1.76x10^-6M - 1.40x10^-6M = 3.6x10^-7M

Eq Drug B = 2.00x10^-6M - 1.40x10^-6M = 0.6 x10^-6M
Kc = [Protein-DrugB] / [Protein][Drug B]

Kc = 0.6 x10^-6M / (3.6x10^-7M)(0.6 x10^-6M)

According to these results drug B would have a higher affinity for the protein.

Hope you will be agree with my explanation. Write me for any query.

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