Question

Protein E can bind to two different proteins, S and I. The binding reactions are described

by the following equations and values:
E + S->ES Keq for ES = 10
E + I->EI Keq for EI = 2
Given the equilibrium constant values, which one of the following statements is true?
(a) E binds I more tightly than S.
(b) When S is present in excess, no I molecules will bind to E.
(c) The binding energy of the ES interaction is greater than that of the EI interaction.
(d) Changing an amino acid on the binding surface of I from a basic amino acid to an
acidic one will probably make the free energy of association with E more negative.

Answer 1

The equilibrium constants indicate that E binds I more tightly than S, no I molecules will bind to E when S is present in excess, and the binding energy of the ES interaction is greater than that of the EI interaction.

Step-by-step explanation:

The equilibrium constants for the binding reactions of protein E with proteins S and I are given as Keq(ES) = 10 and Keq(EI) = 2, respectively.

From these equilibrium constant values, we can make the following conclusions:

1. E binds I more tightly than S. This is because a higher equilibrium constant value indicates a stronger binding affinity.
2. When S is present in excess, no I molecules will bind to E. This is because the equilibrium constant value for E + I (Keq(EI)) is lower than that for E + S (Keq(ES)), indicating that the binding of I is weaker than that of S.
3. The binding energy of the ES interaction is greater than that of the EI interaction. This is because a higher equilibrium constant value indicates a stronger binding affinity, which corresponds to a greater binding energy.