If one prepares three biological replicates of RNAseq libraries for control and treated samples each, with average 150 bp insert size for control and 270 bp for treated samples, what would be the best approach to identify differentially expressed genes between control and treated samples. Is there a need for insert sizes to be taking into account?
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Differential gene expression between samples with non-equal inserts size
Last edited by nucacidhunter; 05-13-2014, 01:45 AM.Tags: None
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I would say that's a bad design. Identifying differential expression accurately is hard enough with identical library prep for different conditions.
But, you could generate some synthetic data with 150bp and 270bp insert sizes, and a flat distribution (no differential expression), run the tests, and see how much difference it makes. You could even measure the apparent differential expression (if any) and use that to normalize the real data.
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Thanks for your comments Brian. We have not done this experiment and just were exploring the possibility. One argument is that it should not make any difference in results. For simplicity assume we have only one transcript in our RNA which is 1 kb long. If it is fragmented to 100 bp in one library it will result in 10 fragment and 5 fragments in another library if fragmented to 200 bp. If it is amplified and sequenced and we pick 1M reads the FPKM in both libraries would be equal.
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The issue is that the different fragment sizes may result in differences in mappability between the samples. If this occurs then you have a batch effect non-uniformly affecting genes, which is a situation to be avoided whenever possible. This is assuming that you're using paired-end sequencing, of course (if not, it won't really matter). If that's correct, you'd be best served either not doing things that way or generating the synthetic reads as Brian suggested and then seeing how mappability might be affected.
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