Question

Two different ligands (A and B) bind to the same receptor on the cell surface. The Kd for ligand A is 10⁻⁷ M; for ligand B it is 10⁻⁹ M. Which ligand has the higher affinity for the receptor?

1) Ligand A
2) Ligand B

Answer 1

Ligand B has a higher affinity for the receptor than ligand A, as indicated by its smaller equilibrium dissociation constant (Kd) of 10⁻⁹ M compared to the 10⁻⁷ M of ligand A.

Step-by-step explanation:

In this scenario, the binding affinity of ligands to a receptor is represented by the equilibrium dissociation constant (Kd). This constant provides an indication of how tightly a ligand binds to its receptor, with a smaller Kd value signifying a stronger affinity. Ligand A has a Kd of 10⁻⁷ M, while ligand B has a Kd of 10⁻⁹ M. Because ligand B has a smaller Kd value compared to ligand A, it means that ligand B has a higher affinity for the receptor; it takes a smaller concentration of ligand B to have half the receptors occupied compared to ligand A.