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DNA double helix and the second law of thermodynamics

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vissi008:
Good afternoon everyone! I am currently preparing for my biochemistry class this fall and I have a question in relation to the formation of the double helix of DNA and the second law of thermodynamics:

Basically I've understood than in order for any process to occur, the entropy of the universe must be increasing. In the case of the formation of the double helix there is an increase of order which leads to a decreased entropy. Following this logic, does it not mean that this process violates the second law of thermodynamics and yet it still occurs? Can someone explain why, please?

jeffmoonchop:
There are papers discussing this topic of entropy in living cells compared to its environment. heres an example, hopefully the link works, if not google entropy in living cells:

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3712629/

Borek:
Common misunderstanding. Local entropy goes down (DNA strand), but overall entropy of the cell and its surroundings goes up ("the entropy of the universe must be increasing").

Following your line of thinking making a smartphone/car/building would be against second law of thermodynamics. It is not, making of each of these things requires dissipation of huge amounts of energy and entropy of the whole universe goes up :)

Yggdrasil:
The base pairing of DNA actually increases entropy because of the hydrophobic effect:

--- Quote ---To a first approximation, hydrogen bonding between two groups in water is not energetically favorable because roughly equivalent hydrogen bonds to water must be exchanged for one such new bond. Thus, in enthalpic terms, solvation effects will not favor a hydrogen bonded pairing of two nucleobases. The bases G and C must first lose several hydrogen bonds to water in order to form a triply-hydrogen bonded pair. In addition, the bases lose entropy of relative translation and rotation in order to form the complex, a destabilizing effect. However, other entropic effects favor this pairing: The entropy of the freed water molecules is likely to be favorable; moreover, the formation of the second and third H-bond in the base pair comes with little additional translational/rotational entropy penalty. This is also true as multiple pairs are formed between two strands. Thus, the hydrogen bonding in a pair does appear to be energetically favorable in the context of a larger double helix.
--- End quote ---
https://www.annualreviews.org/doi/full/10.1146/annurev.biophys.30.1.1

See the full text of the review article for a more comprehensive discussion.

See also the Wikipedia article on the hydrophobic effect: https://en.wikipedia.org/wiki/Hydrophobic_effect

Enthalpy:
Addition to the previous remarks:

ΔQ/T is useful only for heat and work, that is, at engines, for which thermodynamics was invented.

At chemical reactions, other processes occur than heat exchanges, expansion, compression and so on. Then, G and µ decide, rather than S.

And if you mix deuterium and protium (1H2), the process is very much irreversible despite ΔQ=0 and, with the usual definitions, ΔS=0. Other definitions of S would do that one better.

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