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urgent! calculating concentration of compound with different cell amounts

calculation concentration dose

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3 replies to this topic

#1 irisgzz

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Posted 16 April 2012 - 01:00 AM

Hi guys!

This might seem a stupid question, but just to be sure!

I want to test a couple of inhibitors in vitro to check expression of genes related with immune response in my cells. I know the IC50 value of these compounds from an assay with 2x10^5 cells/mL, and I have some data about previous titration of this inhibitors (using different concentrations of the inhibitors on 2x10^5 cells/mL to find what is good to inhibit my target enzymes). So, lets say that 1µM is a concentration that is good to inhibit my enzymes without finding cytotoxic effects on my cells.

Now, the thing is that I have to stimulate cells to check for immune response (with LPS, TNF) and see if my inhibitors have some effect. The protocol that I have for stimulation is for 1x10^6 cells/mL. If I would like to add my inhibitors directly to these cells on 1x10^6 cells/mL together with my stimuli, would it be valid to say:

- 1µM of inhibitor for 2x10^5 cells/mL --> for 1x10^6 cells/mL (5x more cells), then 5µM of inhibitor?

Thanks a lot in advance!

#2 pito

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Posted 16 April 2012 - 01:13 AM

1µM for 2x10^5 cells/ml


1µM for 2x10^5 cells/ml
II
II : 2x10^5
II X 1x10^6
V
? µM for 1x10^6 cells/ml




Use simple math: for 1x10^6 cells/ml it is 1µM divided by 2x10^5 and multiplied by 1x10^6 cells/ml, which gives you: 5 µM , or indeed as you allready noticed, its 5 times more cells.. so indeed multiply by 5.

Edited by pito, 16 April 2012 - 01:15 AM.

If you don't know it, then ask it! Better to ask and look foolish to some than not ask and stay stupid.


#3 irisgzz

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Posted 16 April 2012 - 02:35 AM

1µM for 2x10^5 cells/ml


1µM for 2x10^5 cells/ml
II
II : 2x10^5
II X 1x10^6
V
? µM for 1x10^6 cells/ml

Thanks a lot! = )


Use simple math: for 1x10^6 cells/ml it is 1µM divided by 2x10^5 and multiplied by 1x10^6 cells/ml, which gives you: 5 µM , or indeed as you allready noticed, its 5 times more cells.. so indeed multiply by 5.



#4 bob1

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Posted 16 April 2012 - 05:32 PM

I would say that the amount of compound present in the solution at 1 uM/L is likely to be enough to work for any number of cells, so you shouldn't need to adjust it.





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