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 A is introduced into the protein solution at an initial concentration of \(2.00 \times 10^{-6} \mathrm{M}\). Drug 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 A-protein solution has an 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 constant for drug A (\(K_{cA}\)) is found to be 2, while the equilibrium constant for drug B (\(K_{cB}\)) is found to be 23.333. Since drug B has a higher \(K_c\) value, it is more effective at binding to the protein. Therefore, drug B is the better choice for further research.

Step-by-step solution

Write down the reaction for drug A and drug B binding to the protein

For both drugs A and B, the reaction is the same: Protein + Drug \(\rightleftharpoons\) Drug-Protein Complex Now, we will focus on solving for Drug A first.

Write down the expression for the equilibrium constant, \(K_c\), for drug A binding to the protein

The equilibrium constant for drug A (\(K_{cA}\)) can be expressed as: \(K_{cA} = \dfrac{[\text{A-Protein Complex}]}{[\text{Protein}][\text{Drug A}]}\)

Calculate the equilibrium concentrations of protein and drug A

In order to use the equilibrium expression, we need the equilibrium concentrations of the protein and drug A: Initial concentration of drug A, [Drug A]\(_0 = 2.00 \times 10^{-6}\) M Equilibrium concentration of A-protein complex, [A-Protein Complex]\(_e = 1.00 \times 10^{-6}\) M Now, we know that for every A-protein complex that is formed, one molecule of drug A and one molecule of protein are consumed. Hence, we can use stoichiometry to find the equilibrium concentrations: Equilibrium concentration of drug A, [Drug A]\(_e= 2.00 \times 10^{-6} - 1.00 \times 10^{-6} = 1.00 \times 10^{-6}\) M Given the initial concentration of protein: Initial concentration of protein, [Protein]\(_0 = 1.50 \times 10^{-6}\) M Similarly, calculating the equilibrium concentration of the protein: Equilibrium concentration of protein, [Protein]\(_e= 1.50 \times 10^{-6} - 1.00 \times 10^{-6} = 0.50 \times 10^{-6}\) M

Calculate the equilibrium constant, \(K_{cA}\), for drug A

Now, we can use the calculated equilibrium concentrations to find the value of \(K_{cA}\): \(K_{cA} = \dfrac{1.00 \times 10^{-6}}{(1.00 \times 10^{-6})(0.50 \times 10^{-6})} = 2\) Next, we'll follow the same steps for drug B.

Write down the expression for the equilibrium constant, \(K_c\), for drug B binding to the protein

The equilibrium constant for drug B (\(K_{cB}\)) can be expressed as: \(K_{cB} = \dfrac{[\text{B-Protein Complex}]}{[\text{Protein}][\text{Drug B}]}\)

Calculate the equilibrium concentrations of protein and drug B

Just like we calculated for drug A, we need the equilibrium concentrations of the protein and drug B: Initial concentration of drug B, [Drug B]\(_0 = 2.00 \times 10^{-6}\) M Equilibrium concentration of B-protein complex, [B-Protein Complex]\(_e = 1.40 \times 10^{-6}\) M Calculation for equilibrium concentration of drug B: Equilibrium concentration of drug B, [Drug B]\(_e= 2.00 \times 10^{-6} - 1.40 \times 10^{-6} = 0.60 \times 10^{-6}\) M Calculating the equilibrium concentration of the protein: Equilibrium concentration of protein, [Protein]\(_e= 1.50 \times 10^{-6} - 1.40 \times 10^{-6} = 0.10 \times 10^{-6}\) M

Calculate the equilibrium constant, \(K_{cB}\), for drug B

Now, we can use the calculated equilibrium concentrations to find the value of \(K_{cB}\): \(K_{cB} = \dfrac{1.40 \times 10^{-6}}{(0.60 \times 10^{-6})(0.10 \times 10^{-6})} = 23.333\)

Compare \(K_{cA}\) and \(K_{cB}\) to determine the more effective drug

The higher the equilibrium constant, the more effective the drug is at binding to the protein. Therefore, we can compare the calculated \(K_c\) values for both drugs: \(K_{cA} = 2\) \(K_{cB} = 23.333\) Since \(K_{cB} > K_{cA}\), drug B is more effective at binding to the protein and should be chosen for further research.

Key concepts

Protein Binding

Protein binding is a critical process in the field of biochemistry and pharmacology. It refers to the reversible interaction between a protein and another molecule, in this case, a drug. This interaction is often measured by the formation of a drug-protein complex. In our scenario, we see that both drugs A and B can bind to a protein involved in cancer treatment.
The ability of a drug to bind to a protein can affect its distribution, effectiveness, and the duration of its action in the body. Binding is typically quantified in terms of the equilibrium constant (\(K_c\)), which provides insight into how strongly a drug binds to its target protein. A high \(K_c\) value indicates stronger binding affinity, suggesting more interaction between the drug and the protein. This interaction is essential for understanding how drugs can be designed and optimized for maximum effectiveness.

Drug Effectiveness

Drug effectiveness is not only about how well a drug can eliminate a disease but also how efficiently it interacts with target molecules within the body, such as proteins. In this context, a more effective drug binds more strongly to its specific protein target, as demonstrated by a higher equilibrium constant (\(K_c\)).
When comparing drugs A and B, drug B exhibited a higher \(K_c\) value of 23.333 compared to 2 for drug A. This implies that drug B has a stronger affinity for the protein, making it potentially more effective at carrying out its intended biological effect. Higher binding affinity can lead to better therapeutic outcomes, as the drug is more likely to reach and remain bound to its target in the body for a longer duration.

Chemical Equilibrium

Chemical equilibrium is a state in which the concentrations of all reactants and products remain constant over time. It occurs in reversible reactions, such as when a drug binds to a protein to form a drug-protein complex. In the given exercise, we see the equilibrium between free protein, free drug, and the drug-protein complex.
The equilibrium constant (\(K_c\)) plays a significant role in understanding these processes. It is derived from the concentrations of the reactants and products at equilibrium and gives a mathematical representation of the extent of the reaction. \(K_c\) is calculated using the formula:

• \(K_c = \frac{[ ext{Drug-Protein Complex}]}{[ ext{Protein}][ ext{Drug}]}\)This relationship means that at equilibrium, the product of the concentrations of the reactants (free protein and free drug) is proportional to the concentration of the product (drug-protein complex). Understanding chemical equilibrium is essential for predicting how changes in conditions can affect the concentrations of these components and, therefore, the overall efficacy of the drug.