DESeq2::rlog - does it make sense to use it when I don't have duplicates?
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@chao-jen-wong-7035
Last seen 17 months ago
USA/Seattle/Fred Hutchinson Cancer Rese…

Hi, I (sadly) only have one RNA-seq sample for each condition and want to use log fold change to determine whether the feature counts is different in one condition relative to the other. Apparently, it is not recommended to use the DESeq function to perform differential analysis for it treats the samples as duplicates when calculating dispersion. Does the rlog function do the same thing? Should I use simple log transformation (log2(count+1) to calculate the fold change between conditions or use rlog to do it? Using simple log  gives a lot of noise, especially the low count features and I have to come out with arbitrary filter based on the counts. It is annoying. The rlog gives me a nice list that are less likely to have false positive.  But should I use rlog in this case?

 

Thanks,

Chao-Jen

deseq2 • 1.9k views
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@ryan-c-thompson-5618
Last seen 11 weeks ago
Icahn School of Medicine at Mount Sinai…

You can use rlog, but you will have to use a design of ~1 to estimate the dispersion trend. This will reduce the large magnitude of the fold changes for low-count genes. The rlog transformation serves a similar purpose to the additive constant C in log(count + C), but with a more principled way to adaptively choose the additive constant (and thus the strength of log fold change shrinkage) based on the data. So anywhere that log(count + C) would make sense, rlog probably makes sense too.

On the other hand, if you do decide to use the log(count + C) approach, make sure use the normTransform function, which will add the constant after normalizing the counts.

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The default rlog is blind = TRUE, so the same as design = ~1 DESeq() will also give shrunken log fold changes, just a slightly different prior estimation step.
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Thanks, Michael. It is good to know that setting blind=TRUE leads to design=~1. I don't recall seeing it on the man page. 

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Thanks. That is very helpful. 

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