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  • vivi7
    Member
    • Mar 2014
    • 10

    #1

    Stampy - Mapping multiple individuals to genome

    Stampy - Mapping multiple individuals to genome
    Hi everybody,

    I am very very new to this world, so apologize if the following questions are too basic...
    I am trying to use Stampy to map my RAD-seq reads to a reference genome.
    I have double digested, paired end data from multiple individuals of 6 species and the genome of one of those species.

    I created the genome file and the hash table. According to the manual when mapping paired end data to a genome the script should look something like that:

    ./stampy.py -g hg18 -h hg18 -M solexareads_1.fastq solexareads_2.fastq

    My questions are:

    1) is it possible to map multiple individuals to the genome in one go or should I map each individual singularly to it? Could I use for example something like:
    ./stampy.py -g hg18 -h hg18 -M ind1_1.fastq ind1_2.fastq ind2_1.fastq ind2_2.fastq ind3_1.fastq ind3_2.fastq ?

    2) which option can be added to specify the number of mismatches allowed between reads and genome? I didn't understand it from the help...

    Any help would be greatly appreciated

    Many many thanks
    Vivi
    Last edited by vivi7; 03-18-2014, 12:39 AM.
  • vivi7
    Member
    • Mar 2014
    • 10

    #2
    Hi again,

    I am posting the reply I got from Gerton Lunter,
    so that if someone new to the field like me will have same problems will find an answer hopefully

    '1: no, you should map one sample in one go. If you want, you can merge the BAM files afterwards; if you want to do that, you should 'tag' each set of reads with a unique 'read group' identifier; the --help will tell you how to do this.

    2. This is not possible, but also not necessary. Stampy will map reads if it finds a location in the genome that is more similar to the read than would be expected purely by chance for the best-matching locus in a fully random genome of this size -- up to some threshold that is implicitly set by the algorithm, and corresponds to ~10-15% divergence.

    Hope this helps!
    Best wishes

    Gerton'

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