Physics Asked on August 18, 2021
Living polymers are well described by equilibrium statistical physics. Now I would like to consider a case were living polymers undergo fragmentation due to chaperones. I can think of a kinetic description, but can I still use equilibrium statistical mechanics?
Edit :
I’d like to make the question a bit more general. Here are my interrogations:
When there is no chaperone activity, i.e. polymers polymerize and de polymerize freely, I am comfortable with equilibrium statistical mechanics because I can define a free energy, accounting for polymer binding energy, and entropy. See for example the Flory-Huggins theory.
When it turns to chaperone activity, it seems to me obvious that non equilibrium physics is adapted to this case : it looks like a dynamical system, and I don’t see how to define an energy in the context of equilibrium stat mech.
In an other hand, for the former case (no chaperone), I can also think of a kinetic model, with association and dissociation rates, leading, at equilibrium, to the same result as equilibrium physics.
So, what is fundamentally different between the two cases? Why one can be described by equilibrium physics, while the other can’t (assuming it can’t, that was actually my first question).
In the case with chaperone activity, I see that energy is spent to break each polymer. In the dynamic description, this is not a problem, as long as we can define reaction rates, but if one wants to use equilibrium description, I think the crucial (and difficult) point is to consider this energy used by chaperones.
Thanks for any comment that could make me understand a bit more the boundaries between equilibrium and non equilibrium systems
Edit 2
After some reflections, I think I got what’s make the difference. With chaperone activities, at time +infinite, you expect an equilibrium state, with a polymer size distribution. But this equilibrium is intrinsically dynamic : energy is continuously provided by the chaperones, breaking the polymers. This input of energy is clearly not described by equilibrium stat mech (see micro-canonical, canonical, grand canonical ensembles)
Any comment, reading suggestion is more than welcomed.
I'm not sure whether I understood your actual question correctly. But in my opinion chaperones, in the context of statistical mechanics, can act in two ways. 1) They act as a catalyst, i.e. they merely assure that the minimum in free energy is found fast enough.
2) They can keep the polymer in a steady state. The polymer system without the chaperones is not closed. The interaction with the chaperone system may favour one of the possible states of polymer. If this interaction is strong enough, it can keep the polymer in or near this state even if this is not the equilibrium state.
I don't know much about chaperones but I would expect most of their action to be of the first type.
Answered by taupunkt on August 18, 2021
Life is inherently a non-equilibrium phenomenon, characterized by constant increase of entropy in the outside world, while it decreases inside the living organism, e.g., via polymer synthesis.
In particular, on the molecular level, life is a chain of chemical reactions, where the initially unstable composition of chemicals relaxes towards equilibrium via along chain of chemical reactions. Chaperones are among the molecules that direct these chemical reactions along a certain path, which would not be taken under equilibrium conditions.
Polymerization in living systems is also a non-equilibrium process (see here and here), which is being catalized by various polymerazes. This is however not how it is done in a lab, where the reagents are essentially given time to come to an equilibrium state, where the rate at which molecules are being synthesized equals the rate at which already existing molecules desintegrate. It seems that this is what the OP refers to in Edit 2: equilibrium chemical reaction does not mean that nothing is happening, but that the system has comed to a state characterzised by the detailed balance, where the rate of the forward processes equals to teh rate of the reverse processes.
Answered by Roger Vadim on August 18, 2021
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