I attempt to do some early antibody affinity measurement by ELISA. I found a few publications and made some sense out of them.
I believe that a simple setup should get me close to Kd. I simply set up a set of wells that are coated with same amount of Ag (or receptor) and feed them with antibodies of a gradient of concentrations. By solving a set of equations, I should be able to deduce Kd.
By definition, Kd= [Free Ab]*[Free Ag]/[Ab-Ag]. By designate [Ab-Ag] as X and total Ag as Y, I get
X=0.5 ( (Y+Kd+[total Ab])-square root( (y+Kd+[total Ab])^2-4y*[total Ab]) )
From this connection, I may create a set of data, like
Kd(nM) 1.4 1.4 1.4 1.4 1.4 1.4
[Ag] total (nM) 1.63 1.63 1.63 1.63 1.63 1.63
[Ab] total (nM) 0.5 0.625 0.781 0.977 1.221 1.526
[Ag-Ab] (nM) 0.248 0.304 0.370 0.447 0.536 0.634
OD 0.497 0.608 0.740 0.894 1.071 1.268
Among which, the [total Ab] are known value, the [Ag-Ab] are calculated. While the Kd, total [Ag] absorbed on plate and OD are made up. OD numbers would be acquired by ELISA results. Kd and total [Ag] are to be calculated by OD values.
One reasonable assumption I made is that OD and [Ag-Ab] are proportional. It should be true if I don't deplete the substrate.
So I would end up with a set of equations, which go [Ag-Ab]=OD/a
20.497/a= (y+Kd+0.5) - square root ((y+kd+0.5)^2-4y*0.5)
20.608/a= (y+Kd+0.625) - square root ((y+kd+0.625)^2-4y*0.625)
20.740/a= (y+Kd+0.781) - square root ((y+kd+0.781)^2-4y*0.781)
20.894/a= (y+Kd+0.977) - square root ((y+kd+0.977)^2-4y*0.977)
21.071/a= (y+Kd+1.221) - square root ((y+kd+1.221)^2-4y*1.221)
21.268/a= (y+Kd+1.526) - square root ((y+kd+1.526)^2-4y*1.526)
I'm not positive but it seems by solving the equation set, I would get close estimates of Kd, y (Totol [Ag] on plate) and a (OD/[Ag-Ab])
I vaguely remember that Matlab should be helpful. I wonder if I'm on the right track to solve the equation sets?
I've read more sophisticated method like transfering ab solution between 2 identically coated plate to determine the bound antibody. I'm still struggling with the math behind it.
Thank you very much. Field Tian