Thanks. But sorry for my superficial understanding of methylation sequencing experiment, what puzzles me actually is how the PCR process prevents a new C synthesis, because there is still a G base in the other chains.
I knew that using bisulfite can turn unmethylated C to U, but I still have a problem about this technology:
Supposed a non-methylated C in the forward chain now becomes U, but then during PCR, with a G in reverse chain, will it re-synthesize the C-base in the new chain (may be something happen just like in the pic) ?

Thanks.
1 answer
Unmethylated Cytosin (C) will be converted upon bisulfite treatment into Uracil (U). During PCR U will be replaced with Thymin (T) while methylated C will stay C. Therefore methylated C will be sequenced as C and non-methyl. C will come out as T. If you want details, I suggest a search engine of choice to find extensive articles on that.
Note that most methylation is symmetric, so while you only included one C in your picture there should be a CG in the top strand and a GC in the bottom strand. Therefore, after bisulfite conversion both strands would have a U instead of a C and there would be a signal of this non-methylated position on both strand after PCR amplification.
Yes, you are right. My picture just shows a un-methylation C site and its new chains after PCR. And my confusion is that many blogs say methylation C will keep C and un-methylation C will become a T, but during the PCR , it seems there will generate the un-methylation C.
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