Which ligand binds the tightest?

Question

Ligand A, with a Kd of 10^-9M

Ligand B, with a Kd of 10^-3M

Ligand C, with a percent occupancy of 30% at one micromolar

Ligand D, with a percent occupancy of 80% at 10 nm

Answer 1

the relative molecular mass of lead 2 trioxonitrate V is (pb=180,N=14, O=16).

Answer 2

Since the dissociation constant is equal to the inverse of the association constant, it is possible to figure this out. It would be the one with the lowest dissociation contant or the one with the highest association constant. Ligand A has a lower dissociation constant then B, so B is eliminated. Since 30% occupancy requires the ligand must greater than 3 times the number of it's dissociation constant, and the units for the dissociation constant are given in M. After converting everything to M, the equations for ligand C and D will be the following:

1x10^-6=3(Kd)

10x10^-9=8(Kd)

Solving for Kd,

Ligand C's Kd=3.3 x10^-7

Ligand D's Kd=1.2 x10^-9

The answer is Ligand A

Answer 3

The answer is A, however the explanation previously given is incorrect.

The correct explanation is that the association fraction (theta) is given by the expression:

theta = [L]/([L] +Kd)

where [L] = ligand native concentration and Kd = the dissociation constant.

Given this equation, you can then solve for Kd in (C) and (D):

0.3 = 1 x 10^-6/(1 x 10^-6 + Kd)
Kd = 2.3 x 10^-6

0.8 = 1 x 10^-9/(1 x 10^-9 + Kd)
Kd = 2.5 x 10^-9

This allows you to compare each Kd, and conclude that (A) has the lowest dissociation constant, and, therefore, the greatest association constant of the ligand (the ligand binds the tightest).