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  • nucacidhunter
    replied
    I know that there are some open source software that gives the number and size of the in silico digest fragments. Hopefully someone in the forum can give the names. I found the following thesis interesting in this regard:
    http://www.masterbioinformatica.com/...Z_FERNANDO.pdf

    Leave a comment:


  • AP38
    replied
    Thanks nucacidhunter,

    You are right about the calculation method. It is not the best because of the point you made but should still give a rough estimate. The empirical method is definitely the most accurate way but I would prefer to have a precise in silico estimate prior to jumping into lab work.

    Would you have a good bioinformatic method to propose?

    Thanks!

    Leave a comment:


  • nucacidhunter
    replied
    Calculation method that you have described is incorrect because it assumes that nucleotide sequences in genomes are random and ignores the repetitive content. Two correct methods:

    1- Bioinformatically if there is a reference as you have mentioned
    2- Empirical (most accurate)

    Leave a comment:


  • AP38
    started a topic Estimating the number of ddRAD fragments

    Estimating the number of ddRAD fragments

    Hi all,

    I am wondering if there is an easy way to estimate the number of fragments resulting from a double digest reaction (in my case ddRAD). For a single digest, one can do the following:

    ((proportion of cut sites)^(enzyme cut frequency)) x (genome size) = number of RAD fragments

    For instance, with ecoRI (6 cutter) and a proportion of 0.25 across a genome of 2.5Gbs:

    (0.25)^6 x 2,500,000,000 = 610,351.6 RAD fragments


    Is there a similar way for calculate this for a double digest reaction? One could use simRAD in R or even use grep or cat commands on the reference fasta file (cat ref.fasta | tr -d "\n" | grep -o -E “enz1sequence.{200,400}enzyme2sequence” | wc -l).

    However, I would like to know if there is an easier solution to this such as the one described above for single digest.

    Thanks!

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