nerettilab / RepEnrich2

RepEnrich2 is an updated method to estimate repetitive element enrichment using high-throughput sequencing data.
36 stars 9 forks source link

What results use for further analysis #8

Closed iramai closed 5 years ago

iramai commented 5 years ago

Hi ! Maybe this is not the best place to write my dobts, because what I am writing is not an issue with the program. I followed all the steps, and finally obtained my DE repetitive elements list. Now my question is about interpreting the results list. Do you think that I can work with all differentially expressed repetitive elements (<0.05 pvalue); or do I need to filter results > 2FoldChange and <0.5FoldChange as usually it is done when working with genes?

nskvir commented 5 years ago

Hi there, Thank you for your interest in our software - the answer to this question is ultimately something that is dependent on experimental design and sample sizes. While I would definitely recommend utilizing standard p-value and FDR cutoffs, fold change cutoffs are a little less clear cut. If the samples have particularly low counts it likely makes more sense to set a higher threshold for fold change, similarly to how you perform differential expression with normal genes.

iramai commented 5 years ago

Thanks for the answerd! I will take it into account. While following the edgeR pipeline you add as the final step of the protocol, I have a doubt with the defininf lib.size step. You use the total mapping reads, calculated using the bowtie logs, but when looking to the edgeR protocol there is an option to estimate lib.size on its own. " lib.size for the library size or sequencing depth for each sample. If not specified by the user, the library sizes will be computed from the column sums of the counts". And obiously the final numbers for the library size is not the same. Which one do I need to use for the DEG analysis with edgeR? Thanks in advance

nskvir commented 5 years ago

I would normalize based on the true library size as shown in the example, as opposed to the sum of the counts you get for just the repetitive elements

Best, Nick