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  • Differences between bioanalyzer results and chip-seq results.

    Hi,
    I did chip-seq on a histone modification of treated cells and untreated cells, I expect to receive more fragments at the treated fraction and less at the untreated one. But for the sequencing the same concentration of DNA is used that neutralizes differences between the fractions. For example, in the bioanalyzer one sample was 4 times higher in treated then untreated, but I didn’t saw this different in the results of MACS2. So, how can I overcome it with bioinformatics tools?
    Thanks,

  • #2
    I think to do qPCR on some common peaks that l got by MACS2 and follow the factor between the treated cells and untreated cell to decrease the signal peaks. Then I can compare between them.
    Any ideas? Did anyone do it?
    Last edited by tamars; 03-31-2016, 04:29 AM.

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    • #3
      Your bioanalyzer traces had 4x more DNA for treated but when libraries were made from your samples they must have been loaded in equimolar amounts on the sequencer. That fact should not affect composition of DNA (it should be sampled equally from all samples).

      Have you inspected the alignments to see if data looks reasonable? That you see reads piling up in regions that you expect them to and not where you don't. Perhaps there are other reasons you are not seeing peaks with MACS2.

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      • #4
        I'm not sure I understand what you say. We know that our treatment leads to increase in modification across the whole genome. This is why IP of the treated cells leads to 4 fold more DNA than IP of untreated cells. My concern is that the loading of the "equimolar amounts" on the sequencer actually abolished the differences between the two IPs (of treated vs untreated cells). So how can I quantitatively compare between the two? This is why we thought of finding the difference between the two IPs using qPCR of few specific locations, and then use this factor to correct the signal in the "untreated samples". For example if we find that there is a three fold difference, we can randomly analyze one third of the original fastq reads of the untreated cells, without doing any "library size scaling" during downstream processing.
        Does it make sense? Do I miss something?
        Thanks,

        Comment


        • #5
          To clarify: "equimolar amounts" part was only referring to the libraries getting loaded on the sequencer so that they yield comparable amounts of reads per sample.

          As long as your sampling was correct the two samples (treated and untreated) should be representative of the respective populations. If there was a difference that should show up in the results.

          Have you inspected the alignments to see if the results (in terms of where you expect the reads to align) appear logical?

          Comment


          • #6
            Originally posted by tamars View Post
            I'm not sure I understand what you say. We know that our treatment leads to increase in modification across the whole genome. This is why IP of the treated cells leads to 4 fold more DNA than IP of untreated cells. My concern is that the loading of the "equimolar amounts" on the sequencer actually abolished the differences between the two IPs (of treated vs untreated cells). So how can I quantitatively compare between the two? This is why we thought of finding the difference between the two IPs using qPCR of few specific locations, and then use this factor to correct the signal in the "untreated samples". For example if we find that there is a three fold difference, we can randomly analyze one third of the original fastq reads of the untreated cells, without doing any "library size scaling" during downstream processing.
            Does it make sense? Do I miss something?
            Thanks,
            I see few common sense issues:
            1- Higher yield of IP in comparison to control could be due to other factors.
            2- If treatment increases histone modification as assumed, then there should be more peaks corresponding to modified region regardless of sequencing depth or input DNA amount used for library prep. For instance if there are 100K region in control then 400K region in treatment should show at least 300K more peaks unless the treatment does not affect number of loci but positions.
            3- I assume experiment was set up to test the hypothesis that treatment increases histone modification. Results does not reject null hypothesis.
            4- Was there sufficient number of biological replicates?

            Comment


            • #7
              Hey Tamars,

              If I interpret your question correctly, you're looking for some sort of quantification like you would use for ChIP-qPCR (eg. Percent Input). To my understanding, standard ChIP-seq protocols won't give you anything like this. I've seen some protocol variations that allow for this (2016 paper from Brad Bernstein's group: http://www.cell.com/molecular-cell/a...765(15)00863-1. There's also this spike-in approach: http://genome.cshlp.org/content/earl.../gr.168260.113).

              But then again, what are you trying to show by quantifying it? I've always struggled with interpreting what higher signal, or even what a higher % input from a qPCR actually represents biologically. A higher proportion of the cells in your dish/tissue have the modification at that particular locus? If it were a transcription factor, maybe binding is more stabilized and more likely to be successfully crosslinked? Not so relevant for a histone mod though. Maybe you're working with a system where this makes sense, not sure.

              As others have mentioned above, it's also very possible that there is some sort of technical explanation for this variation in DNA yield, but let's assume it's not for now. If it's a consistent difference, you could always report absolute yield in your results. I would guess that the large difference in yield would be due to the treatment causing the addition of this modification to a bunch of new loci across the genome. I would first do standard peak calling on the two conditions independently, compare the number of peaks between groups, and the locations of the peaks.

              Comment

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