Let's say that I've done a RNA-Seq with three replicates for Mock and three replicates for Treated.
After read counting, Gene 1 has a count value of 100 (average of the three replicates) in Mock.
Gene 2 has a count value of 1000 (average of the three replicates) also in Mock.
Is it possible to say that Gene 2 is 10 fold more expressed in Mock compare to Gene 1?
As soon as I posted my answer, I noticed I misread your question. No, you can't compare counts directly because gene lengths differ, and longer genes are expected to have more counts. You can use TPM (transcripts per million) to normalize for gene length.
original answer:
Not directly as you described, because RNAseq counts are compositional, that is, they convey relative information (proportions), not absolute values. This happens because RNAseq libraries are amplified (thus the original number of RNA molecules is lost), different samples are normalized to equivalent cDNA concentrations, and sequenced at an arbitrary depth. All this means without a good set of spike-ins, one has no idea at all of how many molecules were present at the beginning of the process.
Differential abundance analysis packages (such as DESeq2 or edgeR, for example) use a range of assumptions and statistical techniques to circumvent this limitation. So, you can infer if a particular gene is more or less expressed in Mock vs Treated if you test all the genes, then search for the genes you have interest in these results.
Thanks h.mon. So now if I have a TPM value of 100 for Gene1 and a TPM value of 1000 for Gene2, then I may say that Gene 2 is 10 fold more expressed in Mock compare to Gene 1?
Thanks h.mon. So now if I have a TPM value of 100 for Gene1 and a TPM value of 1000 for Gene2, then I may say that Gene 2 is 10 fold more expressed in Mock compare to Gene 1?