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  • PCR duplicates

    Hi all. Sorry to bother you. After reading several threads about PCR duplicates, I'm totally confused.

    The last step of library preparation in illumina RNA-Seq is PCR which means a fragment will have many copies in the library. Are these copies the so-called PCR duplicates? People suggest to remove PCR duplicates during RNA/ChIP-Seq analysis. I'm confused that why the PCR step is added to the library preparation if the info of these copies will be finally removed in the analysis. I think I must misunderstand what is PCR duplicate. Could anyone help me out?

    Thanks very much! I will appreciate it.

    Best

  • #2
    PCR is typically done to increase yield to more tractable levels (ie visible on gel, bioanalyzer, nanodrop, etc)...and does indeed generate exact copies of fragments (thus paired end reads will both have the exact same start positions). This would manifest as bias in any quantitation/read-counting experiment as different fragments may have different amplification efficiencies.

    If one quantitates their libraries by qPCR, has a fairly complex library (in relation to the number of reads collected), and is confident in the size distribution, one can force the levels of PCR duplicates very very low (<1%).

    Sounds like you understand it perfectly. =]

    Comment


    • #3
      Blithely removing "PCR duplicates" from a data set is bogus.

      There is no way to determine whether a group of sequence reads come from a single fragment of DNA/RNA or result from over-amplification of a single initial target strand in a typical data set.

      I think this is especially the case for ChIP seq or RNAseq as these would tend to have the highest initial start-site bias of commonly constructed libraries.

      --
      Phillip

      Comment


      • #4
        Thanks for your reply.

        According to Illumina manual, 15 cycle PCR is performed. So there are about 30,000 copies for each fragment. How could the PCR duplicate level be so low? I guess only a small part of PCR product is added to the hybridization cell, isn't it? I'm lack of this information.


        Originally posted by ECO View Post
        PCR is typically done to increase yield to more tractable levels (ie visible on gel, bioanalyzer, nanodrop, etc)...and does indeed generate exact copies of fragments (thus paired end reads will both have the exact same start positions). This would manifest as bias in any quantitation/read-counting experiment as different fragments may have different amplification efficiencies.

        If one quantitates their libraries by qPCR, has a fairly complex library (in relation to the number of reads collected), and is confident in the size distribution, one can force the levels of PCR duplicates very very low (<1%).

        Sounds like you understand it perfectly. =]

        Comment


        • #5
          Thanks for your reply. I agree with you. Especially our data are single reading, it is difficult to mark PCR duplicates.


          Originally posted by pmiguel View Post
          Blithely removing "PCR duplicates" from a data set is bogus.

          There is no way to determine whether a group of sequence reads come from a single fragment of DNA/RNA or result from over-amplification of a single initial target strand in a typical data set.

          I think this is especially the case for ChIP seq or RNAseq as these would tend to have the highest initial start-site bias of commonly constructed libraries.

          --
          Phillip

          Comment


          • #6
            Originally posted by aquleaf View Post
            Thanks for your reply.

            According to Illumina manual, 15 cycle PCR is performed. So there are about 30,000 copies for each fragment. How could the PCR duplicate level be so low? I guess only a small part of PCR product is added to the hybridization cell, isn't it? I'm lack of this information.
            Well if you cluster at 10 pM, that amounts to roughly 6 million amplicons/ul. So you can do the math. 600 ul is added the MiSeq whereas 120 ul/lane is added to the HiSeq by the cBot.

            I think 10 pM on the MiSeq would give you about 12-16 million clusters. Whereas you would be at around 100 million in a HiSeq lane. (These are pretty rough figures.)

            --
            Phillip

            Comment

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