I don’t understand why this has to be such a dichotomy?
Good question. I intentionally didn’t get into the details in the post, but happy to walk through it in the comments.
Imagine some not-intended-to-be-realistic simplified toy organism with two cell types, A and B. This simplified toy organism has 1001 methyl modification sites which control the cell’s type; each site pushes the cell toward either type A or type B (depending on whether the methyl is present or not), and whatever state the majority push toward, that’s the type the cell will express.
Under that setup, what do our two mental models look like?
The no-methyl-turnover model would say: well, somewhere upstream in development the methyl groups almost-all lock in to the same cell state, let’s say type A. And then the cell (and its descendants, barring some programmed change) stays in that state indefinitely, because the methyl groups stay in that state indefinitely. Maybe aging involves the methyl groups slowly flipping as entropy does its work, until eventually the majority-A is lost, and the cell switches to the “wrong” type.
The dynamic steady state model would instead say: even if a few of the 1001 methyl groups flip to the “wrong” state at some point, they’ll just be flipped back soon after. Something is pushing the methyls back toward the consensus value. That means there’s no mechanism for the memory to slowly “decay”; a state change would only happen if some big shock makes a whole bunch of the methyl groups flip at the same time. So under the dynamic model, there’s no gradual loss of memory; there’s just a(n exponentially small under normal conditions) probability of a sudden discrete transition.
Yes, this part was obvious! What I meant with those bit-flips was the exponentially small probability of a sudden discrete transition. Why could accidental transitions like this not accumulate? Because they are selected against fast enough? I wouldn’t expect those epigenetic marks to have enough redundancy to reliably last to the end of an organisms’ lifetime, because methylated cytosine is prone to deaminate (which is why CG is the least frequent 2-mer at a ~1% frequency, rather than the ~6.25% you’d expect on baseline). I am confused how the equilibrium works here, but it seems like mutational load could explain why organisms who rely on methylating cytosine have less CG’s than would be useful to maintain epigenetic information. Things would be different in organisms that don’t rely so heavily on methylating cytosine.
Good question. I intentionally didn’t get into the details in the post, but happy to walk through it in the comments.
Imagine some not-intended-to-be-realistic simplified toy organism with two cell types, A and B. This simplified toy organism has 1001 methyl modification sites which control the cell’s type; each site pushes the cell toward either type A or type B (depending on whether the methyl is present or not), and whatever state the majority push toward, that’s the type the cell will express.
Under that setup, what do our two mental models look like?
The no-methyl-turnover model would say: well, somewhere upstream in development the methyl groups almost-all lock in to the same cell state, let’s say type A. And then the cell (and its descendants, barring some programmed change) stays in that state indefinitely, because the methyl groups stay in that state indefinitely. Maybe aging involves the methyl groups slowly flipping as entropy does its work, until eventually the majority-A is lost, and the cell switches to the “wrong” type.
The dynamic steady state model would instead say: even if a few of the 1001 methyl groups flip to the “wrong” state at some point, they’ll just be flipped back soon after. Something is pushing the methyls back toward the consensus value. That means there’s no mechanism for the memory to slowly “decay”; a state change would only happen if some big shock makes a whole bunch of the methyl groups flip at the same time. So under the dynamic model, there’s no gradual loss of memory; there’s just a(n exponentially small under normal conditions) probability of a sudden discrete transition.
Make sense?
Yes, this part was obvious! What I meant with those bit-flips was the exponentially small probability of a sudden discrete transition. Why could accidental transitions like this not accumulate? Because they are selected against fast enough? I wouldn’t expect those epigenetic marks to have enough redundancy to reliably last to the end of an organisms’ lifetime, because methylated cytosine is prone to deaminate (which is why CG is the least frequent 2-mer at a ~1% frequency, rather than the ~6.25% you’d expect on baseline). I am confused how the equilibrium works here, but it seems like mutational load could explain why organisms who rely on methylating cytosine have less CG’s than would be useful to maintain epigenetic information. Things would be different in organisms that don’t rely so heavily on methylating cytosine.