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March 30, 2015

March 30, 2015

Posted by **Patrick** on Thursday, January 31, 2013 at 3:45pm.

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

- Chemistry -
**Anonymous**, Thursday, January 31, 2013 at 4:05pmthe relative molecular mass of lead 2 trioxonitrate V is (pb=180,N=14, O=16).

- Chemistry/Biochemistry -
**Devron**, Friday, February 1, 2013 at 1:35amSince 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

- Chemistry -
**Anonymous**, Friday, October 3, 2014 at 2:58pmThe 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).

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