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 \(25^{\circ} \mathrm{C}\) is \(1.50 \times 10^{-6} \mathrm{M}\) . Drug \(\mathrm{A}\) is introduced into the protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\) Drug \(\mathrm{B}\) is introduced into a separate, identical protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\) . At equilibrium, the drug \(\mathrm{A}-\) protein solution has an \(\mathrm{A}\) -protein complex concentration of \(1.00 \times 10^{-6} \mathrm{M},\) and the drug \(\mathrm{B}\) solution has a B-protein complex concentration of \(1.40 \times 10^{-6} \mathrm{M}\) Calculate the \(K_{c}\) value for the A-protein binding reaction and for the B-protein binding reaction. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research?

Short answer

The equilibrium constants for the A-protein binding reaction (\(K_{cA}\)) and the B-protein binding reaction (\(K_{cB}\)) are 2 and 23.33, respectively. Since \(K_{cB}\) is greater than \(K_{cA}\), drug B binds more strongly to the protein, making it the better choice for further research.

Step-by-step solution

Calculate the change in concentrations for reactants and products

For drug A, the initial concentration of A-protein complex is given to be 0 M, and at equilibrium, it is \(1.00 \times 10^{-6} \mathrm{M}\). Therefore, the change in concentration for A-protein complex is \(+1.00 \times 10^{-6} \mathrm{M}\). Since the protein and drug A react in a 1:1 ratio, the change in concentration for both protein and drug A is -\(+1.00 \times 10^{-6} \mathrm{M}\) each. For drug B, the initial concentration of B-protein complex is given to be 0 M, and at equilibrium, it is \(1.40 \times 10^{-6} \mathrm{M}\). Therefore, the change in concentration for B-protein complex is \(+1.40 \times 10^{-6} \mathrm{M}\). Since the protein and drug B react in a 1:1 ratio, the change in concentration for both protein and drug B is -\(+1.40 \times 10^{-6} \mathrm{M}\) each.

Calculate the equilibrium concentrations for all species

For drug A, the initial concentrations are: [protein] = \(1.50 \times 10^{-6} \mathrm{M}\) [drug A] = \(2.00 \times 10^{-6} \mathrm{M}\) [A-protein complex] = 0 M The equilibrium concentrations for drug A can be calculated by adding the change in concentrations: [protein] = \(1.50 \times 10^{-6} - 1.00 \times 10^{-6} = 0.50 \times 10^{-6} \mathrm{M}\) [drug A] = \(2.00 \times 10^{-6} - 1.00 \times 10^{-6} = 1.00 \times 10^{-6} \mathrm{M}\) [A-protein complex] = 0 M + \(1.00 \times 10^{-6} \mathrm{M}\) = \(1.00 \times 10^{-6} \mathrm{M}\) For drug B, the initial concentrations are: [protein] = \(1.50 \times 10^{-6} \mathrm{M}\) [drug B] = \(2.00 \times 10^{-6} \mathrm{M}\) [B-protein complex] = 0 M The equilibrium concentrations for drug B can be calculated by adding the change in concentrations: [protein] = \(1.50 \times 10^{-6} - 1.40 \times 10^{-6} = 0.10 \times 10^{-6} \mathrm{M}\) [drug B] = \(2.00 \times 10^{-6} - 1.40 \times 10^{-6} = 0.60 \times 10^{-6} \mathrm{M}\) [B-protein complex] = 0 M + \(1.40 \times 10^{-6} \mathrm{M}\) = \(1.40 \times 10^{-6} \mathrm{M}\)

Calculate the equilibrium constant (\(K_c\))

For both reactions, the equilibrium constant (\(K_c\)) can be calculated as: \(K_c = \dfrac{[\text{protein-drug complex}]}{[\text{protein}][\text{drug}]}\) For drug A-protein binding reaction: \(K_{cA} = \dfrac{[A\text{-protein complex}]}{[\text{protein}][\text{drug A}]}\) \(K_{cA} = \dfrac{1.00 \times 10^{-6} \mathrm{M}}{(0.50 \times 10^{-6} \mathrm{M})(1.00 \times 10^{-6} \mathrm{M})} = 2\) For drug B-protein binding reaction: \(K_{cB} = \dfrac{[B\text{-protein complex}]}{[\text{protein}][\text{drug B}]}\) \(K_{cB} = \dfrac{1.40 \times 10^{-6} \mathrm{M}}{(0.10 \times 10^{-6} \mathrm{M})(0.60 \times 10^{-6} \mathrm{M})} = 23.33\)

Compare the equilibrium constants to determine the stronger binding drug

We have calculated the equilibrium constants for both drug A and drug B: \(K_{cA} = 2\) \(K_{cB} = 23.33\) Since \(K_{cB} > K_{cA}\), drug B binds more strongly to the protein than drug A. Therefore, drug B is the better choice for further research.

Key concepts

Binding Reactions in Pharmacology

Understanding how drugs interact with proteins in the body is a cornerstone of pharmacology. Binding reactions involve the formation of a temporary association between a drug molecule (ligand) and a target protein, typically an enzyme or receptor. The strength and duration of this interaction are key to the drug's efficacy. In essence, when a drug binds to a protein, it can either activate or inhibit its function, which impacts the biological processes it controls.

It's helpful to imagine the binding process like a lock and key mechanism. The drug works as the key, and the protein serves as the lock. A high-affinity binding reaction means the key fits perfectly, with strong bonds between the drug and protein. Pharmacologists will often refer to the 'affinity' of a drug for its protein target, which describes how tightly the drug binds. Measuring the equilibrium constants for these reactions helps researchers determine which drug candidate would be most effective. This binding can be reversible or irreversible, and the speed at which a drug binds and then detaches from its target is genuinely important for its pharmacological profile.

Drug-Protein Complex

A drug-protein complex forms when a drug molecule and a protein molecule interact with each other. This complex is a non-covalent association, meaning the drug is not permanently linked to the protein. This reversible nature allows the drug to detach after a certain period, which is crucial for the regulation of its effects. For instance, a drug meant to inhibit an enzyme will form a complex with that enzyme, blocking its active site temporarily and preventing it from catalyzing a reaction.

Researchers need to measure the concentrations of these complexes to understand the drug's pharmacodynamics. This involves calculating the ratio of drug to protein at equilibrium, which gives insight into the likelihood and strength of the drug's effect. The formation of the complex is influenced by factors like molecular shape, charge, and the presence of other competing molecules. The aim is to design a drug that forms the optimal complex with its target protein to elicit the desired response with minimal side effects.

Chemical Equilibrium

At the heart of drug-protein interactions, like the one described in the exercise, is the concept of chemical equilibrium. Chemical equilibrium is achieved when the rate of the forward reaction forming the drug-protein complex and the backward reaction, where the complex dissociates, are equal. In other words, it's the point at which the concentrations of reactants (drug and protein) and products (drug-protein complex) no longer change with time while in a closed system.

The equilibrium constant, denoted as Km, provides a quantitative measure of the relative concentrations of reactants and products at this point. It's a critical value that indicates the potency of the interaction; a higher Km signifies a stronger binding affinity. By calculating the Km value for different drugs, as demonstrated in the provided exercise, chemists can predict which drug will be more effective based on its ability to remain bound to the protein. This concept is vital for the development of new pharmaceuticals and for clinicians to determine the right drug dosage for patients.