Multiple Choice
Calculate the dissociation rate constant (kd) at equilibrium if [P] = 20 mM, [L] = 10 mM, [PL] = 5 mM, and the association rate constant (ka) = 100 mM-1s-1.
8. Protein Function
Introduction to Protein-Ligand Interactions
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Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
Which of the following rate law expressions represents a protein-ligand interaction at equilibrium?
A
[PL]ka = [P][L]kd.
B
[PL][P]ka = [L]kd.
C
[L][PL]kd = [P]ka.
D
[L][P]ka = [PL]kd.
Verified step by step guidance1
Understand the concept of protein-ligand interactions: At equilibrium, the rate of formation of the protein-ligand complex (PL) is equal to the rate of its dissociation.
Identify the components involved: [P] is the concentration of the protein, [L] is the concentration of the ligand, [PL] is the concentration of the protein-ligand complex, ka is the association rate constant, and kd is the dissociation rate constant.
Write the general rate law for the association: The rate of formation of the complex is given by the product of the concentrations of the reactants and the association rate constant, expressed as [P][L]ka.
Write the general rate law for the dissociation: The rate of dissociation of the complex is given by the concentration of the complex and the dissociation rate constant, expressed as [PL]kd.
Set the two rate expressions equal to each other at equilibrium: At equilibrium, the rate of association equals the rate of dissociation, so [P][L]ka = [PL]kd. This matches the correct answer provided.
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