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counting reads over repeats

To identify the expression levels of repeats in my RNA-Seq samples, I'm trying to count reads over repeats. The RNA-seq data (strand specific, dUTP ...) has been mapped and non-uniquely mapped reads (NH:i:X>1) were separated for further analysis. But which program should I use to count reads over gtf file (RepeatMasker) ?

Htseq-count (-s reverse) returns me only zero values and report all as __alignment_not_unique: (reads with more than one reported alignment)

rna-seq
• 4,097 views • link
updated by Ram Ram • written by R • R

1 answer

If you are using default settings with HTseq-count, then it will ignore all the reads with mapping quality of less than 10. Thus, it will ignore non-unique reads (reads aligned to more than one position) as Tophat scores these reads with mapping quality of 3 (2 locations),2 (3 locations), 1 (4-9 locations) and 0 ( 10 or more locations). Use -a <minaqual> as - a 0 so that HTSeq-count will counts these reads towards expression counts.

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updated by Ram Ram • written by Ashutosh Pandey Ashutosh Pandey

But I used default, which is 0

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written by R • R

The default for the latest version is 10. So it seems you are using an older version. I looked more into it and here is what I found. From version 0.4.6 onwards, HTSeq count skips reads that are non-uniquely mapped according to the 'NH' field. Source: http://www-huber.embl.de/users/anders/HTSeq/doc/history.html#version-0-5-4

I just grep'd for "NH" in count.py script that comes with HTSeq and found the following lines:

grep "NH" HTSeq-0.6.1/HTSeq/scripts/count.py
               if r.optional_field( "NH" ) > 1:
               if ( r[0] is not None and r[0].optional_field( "NH" ) > 1 ) or \
                     ( r[1] is not None and r[1].optional_field( "NH" ) > 1 ):

You need to modify "1" to some number suitable for your analysis and HTSeq-count should work for you. Though highly unlikely but please make sure that there are no restrictions with modifying the code.

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updated by Ram Ram • written by Ashutosh Pandey Ashutosh Pandey

Thank you so much :)

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updated by Ram Ram • written by R • R

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