The mistake: methyl groups usually do not stick around long-term; they turn over regularly. Here are two studies which measured the turnover. Turnover timescale varies by location on the DNA strand, but turnover every few days is typical.
I don’t know if someone else was making this particular mistake. I certainly find it quite tricky to describe these things with language unless I am extremely careful. I am still confused, but I do find your suggested pathway via transposons more plausible, and the story for why transposons fits into my head. I don’t think a correct story for why not epigenetic “bit-flips” currently fits properly into my head, and that correct story would have to be created in order to convince other people. Ideally, some testable predictions[1].
But if the methyl groups are instead part of a dynamic equilibrium, especially with high redundancy (and therefore stability), then that’s a whole different situation.
I don’t understand why this has to be such a dichotomy? What I hear you say is that these dynamic equilibria are too stable to be a root cause of aging. I think this is true in the case of X-inactivation[2], which is redundant across most of the X-chromosome. But as far as I can tell, the number of redundant parts that make up one “unit” that can bit-flip lies on a spectrum.
Example: one methylated C is probably too unstable to encode something. But what if I have a local cluster of five of them where each one reinforces each other (because enzymes that change the methylation often check the methylation status of close-by C’s)?
That’s the point of CpG islands, right? So some of these CpG islands are probably more stable than others, and some could be just stable/unstable enough to be a root cause of aging, no?
For example, my guess would be that the transposon story and the accumulating epigenetic chaos story make different predictions about which methylations would serve as an aging clock and how well. They would also predict different things if we compare aging in different species, because transposons and mitochondria are (quite?) universal in eukaryotes, while epigenetics between eukaryotes is quite different.
My understanding is X-inactivation almost never bit-flips in adults. If it were unstable on the timescale of years, you’d sometimes see one white hair in a spot where a cat has black hair, for example. Maybe it is more common in cells with faster turnover?
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.
I don’t know if someone else was making this particular mistake. I certainly find it quite tricky to describe these things with language unless I am extremely careful. I am still confused, but I do find your suggested pathway via transposons more plausible, and the story for why transposons fits into my head. I don’t think a correct story for why not epigenetic “bit-flips” currently fits properly into my head, and that correct story would have to be created in order to convince other people. Ideally, some testable predictions[1].
I don’t understand why this has to be such a dichotomy? What I hear you say is that these dynamic equilibria are too stable to be a root cause of aging. I think this is true in the case of X-inactivation[2], which is redundant across most of the X-chromosome. But as far as I can tell, the number of redundant parts that make up one “unit” that can bit-flip lies on a spectrum. Example: one methylated C is probably too unstable to encode something. But what if I have a local cluster of five of them where each one reinforces each other (because enzymes that change the methylation often check the methylation status of close-by C’s)? That’s the point of CpG islands, right? So some of these CpG islands are probably more stable than others, and some could be just stable/unstable enough to be a root cause of aging, no?
For example, my guess would be that the transposon story and the accumulating epigenetic chaos story make different predictions about which methylations would serve as an aging clock and how well. They would also predict different things if we compare aging in different species, because transposons and mitochondria are (quite?) universal in eukaryotes, while epigenetics between eukaryotes is quite different.
My understanding is X-inactivation almost never bit-flips in adults. If it were unstable on the timescale of years, you’d sometimes see one white hair in a spot where a cat has black hair, for example. Maybe it is more common in cells with faster turnover?
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.