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.

50% occupancy

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.

  • 1
    $\begingroup$ Welcome to Biology.SE. Unfortunately, general chemistry questions without a strong biological component are off-topic. Just mentioning enzymes or antibodies as reactants isn't sufficient. Please delete your question here and ask it at Chemistry (cross-posting is frowned upon), or flag it and request that it be migrated. $\endgroup$
    – MattDMo
    Jul 8, 2022 at 12:37
  • 4
    $\begingroup$ Are you sure? Protein-binding, which is definitely the most relevant tag, is not even an option in the chemistry stackexchange. $\endgroup$
    – Justin
    Jul 8, 2022 at 13:09

1 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:

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


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