SEQanswers Cuffdiff confidence intervals - relative quantitation
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 03-31-2016, 06:30 AM #1 human_urostor Junior Member   Location: York, UK Join Date: Mar 2016 Posts: 1 Cuffdiff confidence intervals - relative quantitation Hi all, I have not found any answers to this kind of question anywhere so I have to ask. I've done RNA-seq and analyzed it in tophat/cuffdiff (then cummeRbund but this is irrelevant here). I also did RT-qPCR on some genes to cross-validate my findings. My reference gene for RT-qPCR was one that was moderately, evenly expressed in all my samples (judging from FPKM). The expression level of all of my other genes was therefore noted as a fraction of the expression level of my reference gene in the same sample (RQ). To make a figure comparing the results from those two methods, I want to present the expression levels for genes of interest from RNA-seq as fractions of expression level of my reference gene too. My problem is: how to calculate the uncertainty in the relative FPKM value? From the research I have done, if FDR is left at 0.05, cuffdiff reports 95% confidence intervals calculated as in this paper: http://bioinformatics.oxfordjournals.../1026.abstract I have found two methods of calculating uncertainty when dividing two values. Method one: http://www.math-mate.com/chapter34_4.shtml (subsection "Division with two numbers with large errors – long method") Method two (let R be the relative expression): https://en.wikipedia.org/wiki/Propag...ce_measurement The problem with the first one is that (using my mathematical common sense) I would get a 95%*95%=90% CI, the resulting uncertainties seem too huge, and it seems not to be supported by any literature. The problem with the second method is that I need to use standard deviation, and the 95% CI is not exactly the SD. I would appreciate any help.

 Tags confidence interval, cuffdiff, cufflinks, rt-qpcr, standard deviation