I'm trying to get clear why protein folding occurs spontaneously.

$$\ce\Delta G=\Delta H-T\Delta S$$

According to thermodynamics the ΔG should be negative for a process to occur spontaneously. When a protein folds the ΔS (Entropy) is decreasing, because the protein gets more ordered. However I think the forming of the bonds (disulfide and other weak interactions) counterbalance this unfavourable rising entropy by forming an enthalpy (ΔH) which thus would result in a negative ΔG.

However I saw some other explanation online which didn't apply the Gibbs free energy formula to the protein but instead to the water. Because the water is more ordered when the protein is unfolded it would be favourable according to the second law of thermodynamics to fold the protein because after folding the water molecules can move more freely --> higher entropy -->negative ΔG --> spontaneous folding.

So my question is which of these two (or maybe both) are the cause of the spontaneous folding of proteins?

  • $\begingroup$ I have edited the title to avoid confusion with a related question that focuses solely on the entropy change. Hope that is ok. If not revert it. $\endgroup$
    – David
    Sep 7, 2016 at 9:41

1 Answer 1



  • The first explanation is commonly encountered.
  • The second explanation cannot be correct, as it stands, as it ignores the free energy change in the protein.
  • A modification of the second explanation (perhaps what was intended) is that it is necessary to consider the protein folding and change in the water as being coupled, in which case the overall free energy change — the sum of the two considered separately — is the determinant of protein folding. The assertion would then be that a negative free-energy change in the water system is the deciding factor. This view has been persuasively advocated on the basis of experimental measurements.

Free energy change in individual transformations

It is standard practice in biochemistry to consider the Gibbs Free Energy of transformation of the sort A → B in isolation in determining whether it will proceed spontaneously. A chemical reaction for which ΔG is negative may generate heat (i.e. have a negative enthalpy change (ΔH) ) which affects its aqueous surroundings, but it seems justified to consider the reaction in isolation as there is no sense that the change in the vibration of the water molecules is driving or coupled to the reaction.

This approach has been applied to the structural change of protein folding with the conclusion (consistent with the first explanation) that the change in enthalpy (ΔH) is sufficient to produce a negative ΔG and hence drive protein folding (Citation 1, below).

Free energy change in coupled transformations Many biochemical changes involve transformations which individually have a positive free energy change, but are made possible by coupling to another reaction with negative free energy change, of greater magnitude:

A → B , ΔG1 = +x

C → D , ΔG2 = –y

If y>x and these two reactions are coupled (generally through a complex reaction path on an enzyme) , then we have:

A + C → B + D , ΔGoverall = –ve

See also Berg et al.

Although one can reject the second explanation in the question as it stands because it ignores the free energy change in the protein folding, perhaps it was intended to mean that the folding of the protein (A → B) should be considered as coupled to the change in the environment of the water (C → D), and that the negative ΔG for the aqueous environment made a greater contribution to the overall ΔG than that for the protein folding.

Is it valid to consider these two systems as coupled? In the original version of my answer I argued against this point of view, but am no longer convinced by my own arguments. The water environment is clearly essential for the hydrophobic effect — the burying of the hydrophobic residues in the centre of the protein away from the water. This is evident if one considers the same protein in a hydrophobic environment such as a cell membrane — it would not fold. In membrane proteins it is hydrophobic residues that are exposed to the lipid bilayer and it is their interiors that sometimes have hydrophilic channels.

So in this coupled system, what is the determinant of the negative free energy change? Minikel (Citation 2, below) asserts that there is no net enthalpy change for the protein folding, and it is the entropy effect on the ΔG for the aqueous environment that drives the folding. He indicates that this view is supported by differential scanning colorimetry and, although he doesn’t cite references, there is a recent (if rather complex) review of this topic by Christopher M. Johnson.

Citation 1: Assertion of role of ΔH of protein

The following explanation, taken from Essential Biochemistry, treats the protein folding in isolation and asserts that change in enthalpy is sufficient to produce a negative free energy change:

The folding of a protein also provides an example of the "ΔH" and "–TΔS" terms competing with one another to determine the ΔG of the folding process. As described above, the change in entropy of the protein as it folds is negative, so the "–TΔS" term is positive. However, in addition to entropic effects there are enthalpic contributions to protein folding. These include hydrogen bonding, ionic salt bridges, and Van der Waals forces. An input of thermal (heat) energy is required to disrupt these forces, and conversely when these interactions form during protein folding they release heat (the ΔH is negative). When all of these entropic and enthalpic contributions are weighed, the enthalpy term wins out over the entropy term. Therefore the free energy of protein folding is negative, and protein folding is a spontaneous process.

Citation 2: Rebuttal of role of ΔH of protein and assertion of role of water

The following explanation, taken from on-line lecture notes of of Eric V. Minikel of Harvard University, rebutting the point of view above:

An incorrect and simplistic view of protein folding is as follows. An unfolded protein has high configurational entropy but also high enthalpy because it has few stabilizing interactions. A folded protein has far less entropy, but also far less enthalpy. There is a tradeoff between H and S here. Note that because ΔG = ΔH - TΔS, increased temperature weights the S term more heavily, meaning that higher temperature favors unfolding.

That entire explanation only considers the energy of the protein and not that of the solvent. In fact, hydrophobic domains of a protein constrain the possible configurations of surrounding water (see explanation above), and so their burial upon folding increases the water’s entropy. Moreover, it turns out that the hydrogen bonding of polar residues and the backbone is satisfied both in an unfolded state (by water) and in a folded state (by each other). Therefore enthalpy is “zero sum,” and protein folding is driven almost entirely by entropy.

  • $\begingroup$ thankyou great explanation! I understand your point "it would seem very diffucultto specify how many water molecules were affected".However thermodynamics usually calculates the free energy of the total system, so this confuses me a little bit. @david $\endgroup$
    – KingBoomie
    Sep 6, 2016 at 6:21
  • $\begingroup$ I think you are right to be confused and even sceptical about my answer. The warning signal is the personalization in “In my opinion…” and “I find…”. Why should you or anyone else care about my opinion? And the fact that I was writing in that way suggests that I was trying to use my authority in lieu of a convincing argument. I have now modified my answer so that it no longer rejects coupling of the changes. Apologies to anyone who has already up-voted my answer — do feel free to withdraw your vote. $\endgroup$
    – David
    Sep 6, 2016 at 12:37
  • $\begingroup$ I found an explanation about considering the solvent(water) in protien folding thermodynamics (cureffi.org/2014/09/05/molecular-biology-02) it's at the bottom of the page,maybe you are interested @david. It looks like the solvent is important. $\endgroup$
    – KingBoomie
    Sep 6, 2016 at 12:45
  • $\begingroup$ @RickBeeloo — Pretty convincing, despite having to take his word for "it turns out". Need to revise my answer again, but it may have to wait until tomorrow. $\endgroup$
    – David
    Sep 6, 2016 at 17:23
  • $\begingroup$ I have revised my answer once more, including the source cited by @RickBeeloo. $\endgroup$
    – David
    Sep 7, 2016 at 9:37

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