How does one derive a KD from an equilibrium titration experiment?

Question

If I have an antibody A and a target B, and experimentally titrate the antibody against a single concentration of B, and then measure the % of B that is bound after the solutions reach equilibrium, I should be able to determine the KD of the interaction.

Source below (and others) shows that determining KD is as simple as determining the EC50 of such a curve. Source.

How does that square with the following equation for KD?

$KD = \frac{([A]*[B])}{[AB]}$ where A is unbound antibody, $B$ is unbound antigen, and $AB$ is the bound complex.

Let's say I have 10 pM of target antigen, 50% of which is bound when incubated to equilibrium with 10 pM of antibody. That would mean the antibody and antigen are each 50% bound and 50% unbound.

$KD = 5pM * 5pM / 5pM$ $KD = 5 pM$

So I have calculated a 5 pM KD, not the 10 pM that I would think would be similar to the graph below (assuming they had pM units on the X axis) I'm assuming these types of example graphs refer to the input concentration, right?

Even more confusing, what if the antibody is a really tight binder, such that 10pM of antibody is able to essentially bind 100%, such that 10 pM antibody is able to bind 50% of 20pM of antigen. (1:1 binding assumed)

$KD = ~0.001pM * ~10pM / ~10pM$

Now you have a KD of 0.001pM, but the graph would suggest that 10 pM of antibody is necessary to bind to 50% of the target antigen.

Answer 1

The models you are using are based on the Langmuir Isotherm. The classical use of this model is for gas adsorption to a solid. In these experiments, the pressure of the gas is measured against the fraction of sites bound.

This model is easily transferred over to aqueous systems, but you have to be careful about making the correct analogies. The pressure of the gas is related to free concentration of the gas, not the total concentration. So the X-axis of your plot must be the free concentration of your ligand.

Most experiments are done where the ligand is in major excess of the receptor. Therefore, the free ligand concentration is roughly equivalent to the total ligand concentration. However, when this isn't the case it is not hard to adjust based on a known ratio of ligand to receptor.

At 50% binding we know that:

$K_d = [L_f]$, where $[L_f]$ is the free concentration of ligand

From conservation of mass we know:

$[L_f] = [L_t] - [LR]$, where $[L_t]$ is the total ligand in the solution and $[LR]$ is the concentration of bound ligand

Since we are at 50% bidning we know:

$[LR] = \frac{1}{2}[R_t]$, where $[Rt]$ is the total receptor concentration.

This lets us solve:

$K_d = [L_t] - \frac{1}{2}[R_t]$