Uniquely mapped reads from BAM file (BWA)

[a_mt]

Hi,

Thank you very much for the reply.

XT:A:U

tag works fine

[Brian Bushnell]

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.

[a_mt]

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:

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