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?

Step by step solution

01

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.

02

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: \(K_{cA} = \dfrac{[\text{A-Protein Complex}]}{[\text{Protein}][\text{Drug A}]}\)

03

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

04

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.

05

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: \(K_{cB} = \dfrac{[\text{B-Protein Complex}]}{[\text{Protein}][\text{Drug B}]}\)

06

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

07

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

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

08

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!