For immunoassays the four parameter logistic function is a general curve fit function proposed by David Rodbard. The form of the equation is:
y = d + (a - d) / (1 + (x/c)^b)
c describes the inflection point of the curve. At this point half of the maximal signal a is reached and thus the concentration of unbound antibody and bound antibody is equal.
Can we draw the conclusion that c is equal the equilibration constant K for this assay?
That equation is just the standard dose-response curve or the 4-parameter logistic curve. It is used to get an estimate of the potency (EC50) of a given ligand interacting with a receptor.
The parameter c is the concentration at which half maximal saturation of antibody is reached (that is, EC50 or effective concentration at 50%). One would be tempted to consider the parameter c as the dissociation constant (Kd) for that interaction, but, although EC50 and Kd are related, that identification is not true.
In fact, EC50 = Kd + [Ab]/2. In order to understand this equation, please, check this link within RG:
https://www.researchgate.net/post/Does_IC-50_values_of_ligand_or_inhibitor_depend_on_the_concentration_of_protein_receptors_or_proteins
Kd is useful to rank affinities, but EC50 is not appropriate for that purpose. Imagine you have two antigens, one with Kd = 0.1 nM and the other with Kd = 1 nM. If you perform an immunoassay with 1 microM antibody, the EC50 for both antigens will be 0.5 microM, no matter their affinity. The reason for that is that the concentration of antigen required to reach 50% saturation of antibody will be at least half the concentration of the antibody).
Conclusion: when Kd > [Ab], EC50 = Kd
That equation is just the standard dose-response curve or the 4-parameter logistic curve. It is used to get an estimate of the potency (EC50) of a given ligand interacting with a receptor.
The parameter c is the concentration at which half maximal saturation of antibody is reached (that is, EC50 or effective concentration at 50%). One would be tempted to consider the parameter c as the dissociation constant (Kd) for that interaction, but, although EC50 and Kd are related, that identification is not true.
In fact, EC50 = Kd + [Ab]/2. In order to understand this equation, please, check this link within RG:
https://www.researchgate.net/post/Does_IC-50_values_of_ligand_or_inhibitor_depend_on_the_concentration_of_protein_receptors_or_proteins
Kd is useful to rank affinities, but EC50 is not appropriate for that purpose. Imagine you have two antigens, one with Kd = 0.1 nM and the other with Kd = 1 nM. If you perform an immunoassay with 1 microM antibody, the EC50 for both antigens will be 0.5 microM, no matter their affinity. The reason for that is that the concentration of antigen required to reach 50% saturation of antibody will be at least half the concentration of the antibody).
Conclusion: when Kd > [Ab], EC50 = Kd
Hi there,
The only way to get an accurate determination of Kd is to work at equilibrium with a large excess of antigen towards antibody. The following article may help:
https://www.ncbi.nlm.nih.gov/pubmed/3981007
as well as the following video:
https://www.youtube.com/watch?v=in0ykyFRLCw
The explanation of Adrian is absolutely correct. For practical purposes, based on the equation EC50 = Kd + [Ab]/2, we usually estimate Kd by diluting the antibody step by step to very low concentrations until there is no change of the EC50 anymore. Then EC50 approaches Kd. An example is shown in Fresenius J Anal Chem (1997) 358: 614–622.
This approach is equivalent to the conclusion of above: Kd >> [Ab], EC50 = Kd.
http://link.springer.com/article/10.1007/s002160050479
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