I read from a paper:
"14% of cytosines are methylated in Arabidopsis thaliana, 4% in Mus musculus, 2.3% in Escherichia coli, 0.03% in Drosophila, and virtually none (< 0.0002%) in yeast"
but they did not say how to get the number of methylated cytosines. Anybody could show me how to calculate the percentage precisely?
•
written
by
t2g4free •
Tools such as Bismark and bwa-meth align WGBS reads and then extract the methylation status of every cytosine within each aligned read. They also produce summary statistics, including the genome-wide average methylation state.
For an example of what this looks like, you can see the Bismark documentation and an example summary report which reports the genome-wide methylation level.
Yeah, but there is no result just like the I read from the paper.
I know it must have to calculate, but how is the key point. Anyway, thx
If you want for all Cytosines, not context specific (CpG / CHG / CHH as printed in the report) just add the numbers:
12180621963159331815251593319601( (31593318 + 1525159 + 3319601) / 1218062196 ) * 100 = 2.9914792627 %
Maybe the OP is confused because "% methylated cytosines" could be interpreted as "% of genomic positions with a cytosine with some evidence of methylation". So if there are 10 cytosine positions in the genome and each of them has % methylation > 0, then "% methylated cytosines" = 100%. If this is what the OP (and the paper) means, then it's a bit trickier to calculate as one needs to define a threshold to claim that the C to T conversion at a genomic position is significantly smaller than 100% (i.e. methylation is greater than 0%) given sequencing and conversion errors.
btw, what does “OP” mean?
Thx 4 all your reply, @dariober's answer is what I mean.
I don't know what the paper means and confused.
If you post a link to the paper that would help to clarify the question :)
Log in to answer this question.
Of course, they all use WGBS, after alignment, they got the cytosine methylation state in every reads, but then?