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MBD-seq (or any enrichment-seq) read number <-> biology?
Let's assume we have the same cell type in conditions A and B.
A non-sequencing experiment shows that there is overall more DNA methylation in condition A than in condition B.
Then we perform MBD-seq on conditions A and B to see where this difference in methylation lies in the genome. (To secure that we put exactly the same amount of DNA starting material in A and B, we carefully quantified it using Qubit etc). We align using BWA. Immediately we see that in all biological replicates, in condition A there are more uniquely mapped reads (up to 25 %) than in condition B. (In some cases there were also more raw sequencing reads in condition A). We are talking about 16 - 25 million uniquelly mapped reads / sample, single end, single sample seq on GAII. Inputs perfectly same. A first question is: does this higher number of uniqely mapped reads in A reflect the biology of A and B (more overall methylation in A)?
However, after loading wig files into a browser, we can't see any differentially methylated regions - A and B look like perfect replicates of eachother
. The differential methylation analysis is currently being run (calling peaks with BALM and using MeDIPS to quantify methylation), however preliminary results show there is no much difference between A and B.
Does anyone have an explanation for this, assuming that the non-sequencing experiment was valid, and that there indeed is quantitative overall difference in methylation between A and B?
The only things that come to my mind for now are:
1. differential methylation between A and B occurs in repetitive DNA sequences, which are exluded from the analysis by BWA?
2. maybe the MBD protein used for enrichment recognizes other citosine modifications, not only methylation, so the difference in methylation between A and B could be in the state of another modification in B, but still recognised by MBD?
3. Is there a normalisation step in the algorithms used that would divide peak hight (= quantity) by total number of reads (which is higher in A). If more reads in A reflect biological presence of more methylation, would dividing each peak quantity by this higher total number of reads diminish the difference between A and B?
I'm not a computational person, but in my understanding this normalisation step wouldn't affect the quantitative analysis and identification of differentially methylated regions (DMRs) only if we assume that DMRs will behave like a microarray experiment: most of the regions don't change, and only a few do. But is it possible that the change in methylation is more uniformly distributed across the genome so this normalisation is affecting quantitative analysis?
We are quite confused with this experiment, as two different experiments that both worked perfectly don't agree: one says there is more methylation overall in A than in B, but then MBD-seq shows that A and B are identical, like they are replicates of each other. The DNA used in both experiments is exactly the same, so no possibility of inter-replicate effect variability.
Many thanks!
A non-sequencing experiment shows that there is overall more DNA methylation in condition A than in condition B.
Then we perform MBD-seq on conditions A and B to see where this difference in methylation lies in the genome. (To secure that we put exactly the same amount of DNA starting material in A and B, we carefully quantified it using Qubit etc). We align using BWA. Immediately we see that in all biological replicates, in condition A there are more uniquely mapped reads (up to 25 %) than in condition B. (In some cases there were also more raw sequencing reads in condition A). We are talking about 16 - 25 million uniquelly mapped reads / sample, single end, single sample seq on GAII. Inputs perfectly same. A first question is: does this higher number of uniqely mapped reads in A reflect the biology of A and B (more overall methylation in A)?
However, after loading wig files into a browser, we can't see any differentially methylated regions - A and B look like perfect replicates of eachother
. The differential methylation analysis is currently being run (calling peaks with BALM and using MeDIPS to quantify methylation), however preliminary results show there is no much difference between A and B.Does anyone have an explanation for this, assuming that the non-sequencing experiment was valid, and that there indeed is quantitative overall difference in methylation between A and B?
The only things that come to my mind for now are:
1. differential methylation between A and B occurs in repetitive DNA sequences, which are exluded from the analysis by BWA?
2. maybe the MBD protein used for enrichment recognizes other citosine modifications, not only methylation, so the difference in methylation between A and B could be in the state of another modification in B, but still recognised by MBD?
3. Is there a normalisation step in the algorithms used that would divide peak hight (= quantity) by total number of reads (which is higher in A). If more reads in A reflect biological presence of more methylation, would dividing each peak quantity by this higher total number of reads diminish the difference between A and B?
I'm not a computational person, but in my understanding this normalisation step wouldn't affect the quantitative analysis and identification of differentially methylated regions (DMRs) only if we assume that DMRs will behave like a microarray experiment: most of the regions don't change, and only a few do. But is it possible that the change in methylation is more uniformly distributed across the genome so this normalisation is affecting quantitative analysis?
We are quite confused with this experiment, as two different experiments that both worked perfectly don't agree: one says there is more methylation overall in A than in B, but then MBD-seq shows that A and B are identical, like they are replicates of each other. The DNA used in both experiments is exactly the same, so no possibility of inter-replicate effect variability.
Many thanks!
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