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Neutralization assay calculation

calculation antibody

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#1 monroe

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Posted 08 July 2013 - 11:21 AM

Hi All,

I am new to this assay and I had a very basic calculation question. In neutralization assay I incubate 50ul of 100ug/ml ab with 50ul virus for 1H at 37 and then add 100ul of cells. My question is what is the final concentration of the Ab that I am testing. Is it 50ug/ml (1+1 = Ab+Virus) or 25ug/ml (considering the dilution factor after adding cells (1+1+2: Ab+Virus:cells).
I appreciate you response.

Thanks.

#2 bob1

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Posted 08 July 2013 - 12:52 PM

The way to calculate this is using C1V1=C2V2 you have initial concentration (100 ug/ml) and volume (50 ul) and you have final volume (50 + 50 + 100)...

#3 monroe

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Posted 09 July 2013 - 08:59 AM

Thanks Bob,
I think I was not able to clarify. As an example, suppose I get 50% neutralization say at a particular Ab concentration. In that case the calculated inhibitory concentrations will refer to the antibody concentrations in the preincubation mixture (virus+ Ab: 50+50) which will be 50ug/ml of Ab or after adding the cells (50+50+100: Ab+Virus:cells) which will then be 25ug/ml. So when you report the neutralization activity of the ab in this particular case what will you say "50% neutralization was seen at 50ug/ml or 25ug/ml of Ab.

#4 bob1

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Posted 09 July 2013 - 12:34 PM

Ah, I see. I think, but am not sure, that it would be the virus and antibody only, as this is where the neutralization occurs, the cells are just the read and are added after the neutralization.

#5 monroe

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Posted 11 July 2013 - 06:20 AM

Thanks, Bob. that is what I thought that actually virus neutralization is occuring in the preincubation mixture (virus+ Ab: 50+50). I got confused as one of my colleague mentioned the other way around. Thanks Bob.





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