I'm thinking that this isn't actually possible, but I'd like to check before I write it off. I have RNA-seq data from cells under three different conditions, there are no replicates for any of the conditions. The data has been processed with RSEM, and log2 fold changes have been calculated for each control-test pairing using the normalized expected read counts using EBseq.

If possible, I'd like to also calculate the p-value for each of these fold-changes, however, because there are no replicates I don't think that this is possible. Is there any way to do this?

  • $\begingroup$ There are some helpful hints here! $\endgroup$
    – CKM
    Jun 9, 2017 at 19:57

1 Answer 1


DESeq2 is able to adjust p-values for single-replicate samples by estimating shared dispersion across all samples. DESeq2 will give a warning, but try its best to carry out the analysis. More information on doing that can be found in the DESeq2 vignette and the DESeq2 manual page.

The results will not be anywhere near the sensitivity that can be achieved when replicate information is available, but it might be good enough if only very obvious large-scale changes are of importance.

  • $\begingroup$ Agreed. DESeq2 uses an empirical Bayes model that incorporates knowledge about the variation of all genes to estimate the true expression value of any single gene in any single sample. Unfortunately due to its complexity I find it incredibly hard to figure out what's going on under the hood when I use it. $\endgroup$ Jun 10, 2017 at 23:58
  • 1
    $\begingroup$ I just accept that I can never have the depth of knowledge of analysis tools that the tool authors have, and treat data analysis as similar to a biological experiment -- you don't have to know everything about how things work in order to gain understanding from them. $\endgroup$
    – gringer
    Jun 11, 2017 at 1:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.