Whilst eventually my email seems to be unscrambled, my math hasn't as I couldn't read my own scribble and turned my 10^20 into 10^26(thanks mario). The excess is not that high as 10nM are actually 6* 10^12 molecules which makes it 6*10^6 per cell. That leaves a 120 fold excess on and average cell with 50000 epitopes to be covered. Still a lot of molecules but makes it look less funny but more realistic. regards Gerhard -----Original Message----- From: gerhard nebe-von-caron [SMTP:Gerhard.Nebe-von-Caron@unilever.com] Sent: Tuesday, February 29, 2000 1:29 PM To: Cytometry Mailing List Subject: RE: dilution of antibody Considering the concentration of 1.5ug antibody per ml to reachreasonable staining saturation within less than 30 minutes as a ballparkfigure, this is equivalent to 10nM of antibody. This means that in 1mlthere are 10^18 molecules hopping around, or 10^12 antibodies hittingeach cell if you have a million cells in there. One can calculatefurther how many antibodies are within a certain diffusion distance ofeach cell to get even more precise, but the number of molecular involvedmight already help to illustrate the funny effects of on and off rates,depletion and mass transport limitations. They are also an importantfactor to consider when working with particle based assays. In general lower concentrations mean longer incubation times, so dolower temperatures as they reduce the 'hit-rates'. If cells are unfixedthat in turn can have effects on cell stimulation, change binding sitedensities.... so it is not at all easy to decide what to do. Regards Gerhard -----Original Message----- From: Mario Roederer[SMTP:Roederer@drmr.com] Sent: Friday, February 25, 2000 10:09 PM To: Cytometry Mailing List Subject: RE: dilution of antibody Joost: >Or am I mistaken? You are mistaken. Aaron Kantor and I went into this topic in some depth in our chapter in Handbook of Experimental Immunology (5th ed) "FACS Analysis of Leukocytes". I recommend that you look into it for a thorough description. Albert was correct in his analysis--for nearly all antibodies, the number of cells is irrelevant (until you get to numbers of 10's of millions that bind the antibody); the concentration of the antibody, however, is paramount. (Incidentally, so is time and temperature--make sure that you titrate your antibodies at the same temperature and for the same amount of time as your experiments). I think the reason for the origin of this error is that for a long time, manufacturers gave their effective titres as "ug per million cells" rather than "ug per ml". mr (PS: cell number can become relevant for very high affinity antibodies. This is because with high affinity antibodies, the amount of antibody supplied can be much lower, thus, in much lower excess over antigen. However, the vast majority of antibodies have an affinity low enough such that a significant excess of antigen (in typically staining experiments) is required to achieve near-saturation.). >Dear Albert, > >Sorry, but I don't agree with you. When you should look at a marker 80% of >your cells is positive for. And you determine the amount of antibody >(= your >concentration) for 300.000 cells. It is not possible to get even a >peak far >from your negative peak when you stain 5 million cells. This means >that you >throw away at least 5 times the amount of antibody when you stain300.000 >cells. Because you determined (titrated) your antibody for 5 million >cells. >And the definition of concentration is the amount of Ab in a certain >volume. >So when you got 100.000 cells and need 0.01 µg of Ab to stain them you can >add this in 50 µl or in 5 ml. The only difference is thetimeperiod >it will >take to get your Ab on the cells. Or am I mistaken? > >Best regards, > >Joost Schuitemaker
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