Hi!
I am analysing 16S data and generated OTU tables (part of the GenAp - ampliconSeq pipeline from Mugqic). In my study, participants gave a stool sample and were then given probiotics and sampled again. I want to subtract the values of the initial sampling day from the second OTU table, to have the changes in abundance. I would like to do:
OTU_Table(treatment) - OTU_Table(initial) = OTU_Table(difference)
but am not able to find any software that does this! I have been looking into metagenomeSeq (for its fitZIG function), but I am having format conversion bugs. I was also looking into LefSe (https://bitbucket.org/biobakery/biobakery/wiki/lefse), but am not sure if I can compare only two tables and output one table.
My next step is probably to write my own code to do this substraction task, but I am sure something exists out there that can do this!
Thanks!
Jérémie Auger
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Subtracting is a poor measure of abundance changes here. At the bare minimum, you should look at proportion differences: OTU_Table(treatment) / OTU_Table(initial). But you also have to normalize for unequal sequencing depth, see this review for some issues and strategies:
Normalization and microbial differential abundance strategies depend upon data characteristics
Thanks for the answer. I am working on rarefied data, do you think that accounts for uneven sequencing depth? My problem with differential abundance softwares is that they require groups to make the calculations i.e. a contrast matrix. However, I am still blinded (double blinded study) and don't know who is in which group (placebo or probiotics). I want to be able to make groups and see if they fit with the real groups. I know that simply subtracting is not a good option, especially for the further statistical analyses. I am not looking to compute p-values, only to have a set of tables that represent delta(OTUs) after the treatment and see if I can cluster them appropriately. If i decide to generate these tables manually, I will probably go for a foldchange value, something like [ Fc = log2(b/a) ] or [ (B-A)/A ].