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  • ArtificialBreeze
    Junior Member
    • Jun 2015
    • 2

    #1

    de novo assembly with de Bruijn colored graphs

    Hi all!
    I am currently working on an undergrad de novo assembly project to call errors in reads. In his thesis, D. Zerbino mentions that Cortex / Velvet stores variants / errors / alternative samples under new colors. My question is how would we efficiently identify those? I don't really see an optimal implementation for this. Like de Bruijn graphs are fine to construct and traverse if the reads are ideal, but whenever you have a small difference in base pairs, it starts being more complex.
    Any thoughts welcome , thanks!
    Last edited by ArtificialBreeze; 06-15-2015, 12:15 PM.
  • Brian Bushnell
    Super Moderator
    • Jan 2014
    • 2709

    #2
    Just store kmers and counts, and look for branches. When you have a kmer X, there are 4 possible next kmers. Assuming Y is the last K-1 bases of X. Then the 4 possible next kmers are:
    YA
    YC
    YG
    YT

    So, starting from X, just look up the 4 possible next kmers and get their counts. You'll end up with something like this:

    YA 0
    YC 100
    YG 0
    YT 1

    In that case, you have a branch - the next base should be C or T. But T only has a count of 1, so it's probably C; with 101x coverage of a region, it's not unlikely to have 100 correct copies and 1 error.

    However, it also might not be an error (though it probably is). It could be that there are 100 times as many YC kmers in the genome as compared to YT kmers, and both are correct. How you decide whether or not this is an error is up to your heuristics. But regardless, the optimal implementation is generally to store kmer counts in a way they can be quickly looked up (generally by hashing), and only store them for kmers that actually occur.

    If you decide YT is a single substitution error, then for reads containing YT (which in this case would be only 1 read), you would replace the T with a C.
    Last edited by Brian Bushnell; 06-15-2015, 12:20 PM.

    Comment

    • ArtificialBreeze
      Junior Member
      • Jun 2015
      • 2

      #3
      Oh, I didn't think of that, it is true that this works since we have left and right k-1 mers, which I completely forgot.
      Thanks very much !

      Comment

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