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  • Uniquely mapped reads from BAM file (BWA)

    Hi all,

    I know this question has been asked many times. But, I am still not able to find the proper answer. So, how do you caluclate the number of uniquely mapped reads from BAM file generated by BWA?

    So far, I have tried the folowing:

    Code:
    samtools flagstat LPCWLN_fsorted.bam
    118956972 + 0 in total (QC-passed reads + QC-failed reads)
    0 + 0 duplicates
    117768703 + 0 mapped (99.00%:-nan%)
    118956972 + 0 paired in sequencing
    59478678 + 0 read1
    59478294 + 0 read2
    116464167 + 0 properly paired (97.90%:-nan%)
    117419717 + 0 with itself and mate mapped
    348986 + 0 singletons (0.29%:-nan%)
    834518 + 0 with mate mapped to a different chr
    755636 + 0 with mate mapped to a different chr (mapQ>=5)
    
    samtools view LPCWLN_fsorted.bam | wc -l
    118956972
    
    samtools view -F 256 LPCWLN_fsorted.bam | wc -l
    118956972
    
    samtools view LPCWLN_fsorted.bam | grep XT:A:U | wc -l
    105334435
    I am totally confused now? Does the 3rd line of flagstat o/p gives the no. of uniqely mapped reads ? or the SAM flag -F 256 gives ? or the other one?

    Appreciate any help.

    Thanks

  • #2
    bwa adds "XT:A:R" to to ambiguously-mapped reads and "XT:A:U" to uniquely-mapped reads, so filtering by that tag should work. The 256 bit (0x100) indicates that an alignment is secondary, which is relevant, but won't help you in this case. Note that if you map with BBMap (which also adds XT:A:U and XT:A:R tags), you can use the flag "ambig=toss" to consider ambiguously-mapping reads as unmapped; only uniquely-mapped reads will be considered mapped.
    Last edited by Brian Bushnell; 04-29-2014, 12:05 PM.

    Comment


    • #3
      Hi,

      Thank you very much for the reply.
      Code:
      XT:A:U
      tag works fine

      Comment

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