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  • Tefina
    Member
    • Jun 2010
    • 13

    #1

    Allowed maximum number of differences in BWA

    The paper says the following:

    Given a read of length m, BWA only tolerates a hit with at most k differences (mismatches or gaps), where k is chosen such that <4% of m-long reads with 2% uniform base error rate may contain differences more than k. With this configuration, for 15–37 bp reads, k equals 2; for 38–63 bp, k=3; for 64–92 bp, k=4; for 93–123 bp, k=5; and for 124–156 bp reads, k=6.

    Can somebody explain me how the exact calculation is done?

    best,
    tefina
  • dp05yk
    Member
    • Dec 2010
    • 66

    #2
    This is just the default -n value. You can override this by specifying your own -n value, but remember, the more errors you allow, the longer BWA will take to run.

    Comment

    • Tefina
      Member
      • Jun 2010
      • 13

      #3
      Thanks for your answer.
      I am aware that this is just the default value and can be changes easily. I am just wondering how they calculate these thesholds (from the given 2% error rate and the specified 4% of maximum allowed differences).

      Comment

      • dp05yk
        Member
        • Dec 2010
        • 66

        #4
        There's a function in the file "bwtaln.c", called "bwa_cal_max_diff" that does this.



        The first paramater, l, is the length of the sequence.
        The second parameter, err, is the average error specified by the paper (a constant set to 0.02)
        The third parameter, thres, is the error threshold, which I am assuming corresponds to the 4% (0.04) listed in the paper.

        Comment

        • Tefina
          Member
          • Jun 2010
          • 13

          #5
          Thanks a lot for the info!
          Now I just have to make sense out of this

          Comment

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