[BioC] The proper way of analysisng FACS-data; Question off topis?

[Nolwenn LeMeur]

Hi Ulrik,

I'm far from being an expert in analysing FACS-data but I can suggest 
you some tools implemented in rflowcyt (see the vignette):
- you should use visualization tools (e.g. ECDF plot, density plot)  to 
assess the quality of  your replicates and verify the reproducibility of 
your experiment
-  rflowcyt also proposes some statistical methods (WLR.flowcytest, 
KS.flowcytest...) to test for the difference between two univariate 
distributions, e.g. treated and untreated with or without a time parameter.

You can also try to post your message to Purdue University Cytometry 
Laboratories'  e-mail archive
http://www.cyto.purdue.edu/hmarchiv/cytomail.htm
It is the main cytometry e-mail archive.

Hope it helps,
Nolwenn

Ulrik Stervbo wrote:

>Hello All,
>
>I have a question regarding the proper way of analysing FACS-data. I
>realise that the question may be off topic and I apologise. If it is
>too far off, please advise me where else to go.
>
>I am relatively new to FACS measurements (though not new to
>programming). In the laboartory where I just started, they treat cell
>cultures in triplicates and measure for some parameter on the FACS,
>e.g. DNA fragmentation or exposure of phosphatidylserine. After FACS
>measurement, the percentage of cells with for instance fragmented DNA
>or exposed PS is found, and the mean form the triplicates is taken.
>Later bar-graphs with standard variations are created.
>
>As we collect at least 3 x 10.000 events, I somehow feel that
>collapsing them all onto basically just three measurements, and then
>ask if those three measurements are different from three other
>experiments may not be the best way to go.
>
>I would like to know if there is a significant difference after
>treatment with some compound, and also compare treatment times. Is
>there a better way than taking the means of triplicates? Any solution
>should produce bar-graphs with quantitative differences.
>
>Thanks in advance.
>Ulrik
>
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