Question

A series of Equilibrium Dialysis Experiments were set up where a fixed concentration and volume of the protein is contained within dialysis tubing and "dialyzed" until an equilibrium is reached against a solution containing a certain concentration of the ligand were set up with varying concentrations of ligand. This allows the "free" ligand to reach an equilibrium between the inside and outside regions of the experiment, while of course the "bound" ligand is contained entirely inside the dialysis tubing the following data was obtained:

Protein Concentration in Experiment: 0.918mg/mL

Polypeptide Molecular Weight of Protein: 55,800

Molecular Weight by Gel Filtration: 224,000

Data Obtained:

Tube #	Total Ligand Concentration Inside Tube	Total Ligand Concentration Outside Tube
1	12.4840mM	8.904mM
2	7.5000	4.208
3	5.3192	2.4832
4	3.9184	1.5064
5	3.2548	1.0948
6	2.3896	0.7016
7	1.6064	0.4264
8	1.0916	0.2596
9	0.5256	0.1136

Calculate the dissociation constant for the ligand binding to the protein.

How many binding sites per native molecule are there?

Answer 1

Consider the equation

P + L ------ [PL]

Keq = [PL]/[P][L]

Kd = [P][L]/[PL]

where P = protein

L = free ligand

PL = protein-ligand complex

Keq = equilibrium constant

Kd = dissociation constant

from the two equations we can write:

Y = [L]/[Kd] +[L]

where Y = [PL]/[P]total

Y/[L] =[ -1/Kd]Y + 1/[Kd]

A plot of Y/L vs Y gives a staright line with slope = -1/Kd

In the data given:

1) [P] = 0.918 mg/ml; mol wt = 55800 g/mol

[P] = 0.0165 mM

2) [L] = free ligand concentrations i.e total concentration out side the testube

3) [PL] = [ligand inside testube]-[ligand outside]

4) [P]total = [P] + [PL]

The plot is depicted as follows

Fit equation: y = -245.64x + 244.63

slope = 1/kd ; kd = 1/slope = 1/245.64 = 4.07 * 10^-3