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 for the binding reactions between protein E and proteins S and I are given as Keq for ES = 10 and Keq for EI = 2. Based on these values, statement (A) is true, statement (B) depends on the concentration, statement (C) cannot be inferred solely based on the given information, and statement (D) is not directly addressed.

Step-by-step explanation:

The equilibrium constants for the binding reactions between protein E and proteins S and I are given as Keq for ES = 10 and Keq for EI = 2. Based on these equilibrium constant values, we can draw the following conclusions:

1. (A) E binds I more tightly than S: Since the equilibrium constant for EI is lower than the equilibrium constant for ES, it indicates that the binding of protein I to protein E is stronger compared to the binding of protein S.
2. (B) When S is present in excess, no I molecules will bind to E: This statement cannot be concluded solely based on the equilibrium constant values. It depends on the concentration of the proteins S and I.
3. (C) The binding energy of the ES interaction is greater than that of the EI interaction: The equilibrium constants do not directly represent the binding energy. Therefore, this statement cannot be inferred solely based on the given information.
4. (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: