Competitive ELISA Standard Curves
Posted 26 May 2009 - 02:55 AM
I have finally optimised my competition ELISA with the aim to quantify my protein.
My standards are pure protein, at a known concentration, diluted TEN FOLD for 5 points, ie 800ng, 80ng, 8ng, 0.8ng, 0.08ng.
I then run unknown samples at the same time, to compare the values against the standards.
My question is, which type of curve do it draw, and do i use the equation for the curve to work out the concentration? Or, do i simply try and read from the curve?
Should the chart be a line or a scatter? Should it be a log trendline or polynomial? On which axis should the OD and concentration be? I have so many different protocols and books, but no one seems to give a definitive answer for a serial diluted set of standards...
Thanks for any help,
Posted 26 May 2009 - 04:38 PM
Plot the standard curve as a scatter plot with OD on the y axis (it is the dependant variable, concentration is independent).
The line/curve you choose will depend on the graph. Some ELISA protocols generate standard curves which can be approximated by fitting a straight line (sometimes in log scale). Most of the others end up S shaped and use a variation of the Hill equation:
StdOD = MaxOD / (1+ ([X] / IC50)^slope)
If you donít have access to fitting software you can get just as good a fit on a spreadsheet by manually adjusting IC50 and slope until the line looks close to your data.
To convert your sample ODs into concentrations, you just need the inverse equation (apologies in advance if I have done this wrong, you need to check my working)
[X] = IC50 * exp( ln(( MaxOD/SmpOD)-1)/slope)
Posted 27 May 2009 - 04:11 AM
hope this helps you. One point, interpret results at the high and low ends of the dose reponse curve with caution as these results will be the least accurate.
Posted 27 May 2009 - 07:00 AM
I've found a 'standard curves analysis' function on sigma plot which I think gives me the sort of thing I am looking for. I'll check out your other suggestions first though and get back to you if I need more help...
Thanks very much again,