Consider the following interaction between a protein and its ligand: [P] + [L] $\harr$ [PL]. If [L] = 5 mM, [P] = 3 mM, and [PL] = 2 mM, what is the Ka for the interaction?
A) 7.5 mM
B) 15 mM
C) 0.133 mM
D) 0.4 mM

Question

Consider the following interaction between a protein and its ligand: [P] + [L]   $\harr$ [PL]. If [L] = 5 mM, [P] = 3 mM, and [PL] = 2 mM, what is the Ka for the interaction?
A) 7.5 mM
B) 15 mM
C) 0.133 mM
D) 0.4 mM

Quizplus · Accepted Answer

The equation for the dissociation constant (Kd) for this interaction is Kd = ([P][L])/[PL]. 
Using the given values: 
Kd = ([P][L])/[PL] = (3 mM)(5 mM)/(2 mM) = 7.5 mM 
Therefore, the equilibrium constant (Ka) for the interaction is the inverse of Kd: 
Ka = 1/Kd = 1/7.5 mM = 0.133 mM. 
Thus, the correct answer is C.

Consider the Following Interaction Between a Protein and Its Ligand ↔\harr↔

Consider the Following Interaction Between a Protein and Its Ligand $\harr$