The hydrophobic interaction is thought to be driven by an entropic force. If it is, shouldn't hydrophobic interactions be stronger at higher temperatures, where states of higher entropy are favoured? Why, then, are protein hydrophobic cores denatured by heat?
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$\begingroup$ I'll just guess a bit... Is it maybe that the hydrophobic effect requires both entropic stabilization as well as van der waals interactions between non-polar moieties? Maybe with the van der waals forces overcome, the entropy of the hydrophobic moieties themselves becomes more stabilizing than water expulsion from hydrophobic surface area? $\endgroup$– JoryAug 5, 2016 at 3:21
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3 Answers
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You are correct that the hydrophobic interactions are stabilized at higher temperatures. However, I believe the addition of heat mainly disrupts protein structure by breaking hydrogen bonds through increased thermal motion.
Once the secondary structure starts to go, any tertiary interactions will get scrambled, likely resulting in the separation of hydrophobic residues. The refolding of the polypeptide backbone must be stronger than the enhanced hydrophobic interactions.
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$\begingroup$ I like this answer. I guess the energy of the hydrophobic interaction alone is not sufficient to conformationally restrict core residues in an efficiently packed structure. So just to be clear, if you heat oil in water (or any hydrophobic solute in water), you would, in fact, be favouring coalescence of the hydrophobic contents, right? $\endgroup$– JoryAug 7, 2016 at 15:25
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1$\begingroup$ Yep, exactly, but there are exceptions. When an organic solute has a polar group (e.g. the alcohol group in octanol) increased temperatures can also increase solubility, for much the same reason hydrophobic cores of proteins can be denatured by heat. $\endgroup$ Aug 8, 2016 at 0:06
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To add to the answer of Dan Burden: the interactions might be stronger, but you're still making it easier for the protein to cross the energy barriers towards unfolding if you heat the protein.
Also, one thing that helped me understand protein physics is the following: an interaction might be worth 2 kcal/mole in a vacuum, but if you break this interaction you probably don't go to 0 kcal/mole. It's likely the protein forms another hydrogen bond worth only 1.8 kcal/mole with water, or forms a similar hydrophobic interaction with a non-fitting patch of the protein that's almost as favorable.
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$\begingroup$ The energy barrier is a good point. After all denaturation involves transition to a kinetically trapped state. In terms of such replacement interactions, I'm familiar with that idea in terms of electrostatic interactions such as when secondary structural h bonds are replaced by ones with water with little energy difference. Where do the replacement interaction partners come from for the hydrophobic moieties in the core though? Are you thinking it's more just intramolecular interactions with other core residues, but in a way less tightly packed and organized manner? Or are you thinking more th $\endgroup$– JoryAug 7, 2016 at 15:31
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1$\begingroup$ I think in the end the hydrophobic residues will also interact with the residues of other proteins that are starting to unfold and aggregate. $\endgroup$– VonBecheAug 10, 2016 at 16:34
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As has been pointed out by @VonBeche adding energy (heating) is eventually going to unfold/denature a protein. Intermolecular (or other) forces cannot ultimately withstand thermal energy.
However, the use of 'entropic forces' while common is misleading since no forces are involved due to entropy. Entropy has units energy/temperature and so $T\Delta S$ units of energy not force, which is energy/distance. I try to explain, below, why it appears as a force.
When placing a protein into water, a 'cavity' in the water has to be made which means breaking some hydrogen bonds (expensive in energy terms) and thus also some loss of entropy and the water has to bridge the cavity, also expensive. If the protein is unfolded and in water it has very many different configurations, possibly all of about the same high energy, relative to its folded state.
The water will try to hydrogen bond to polar residues but for non-polar residues hydrogen bonds will have to be broken and then remade as water tries to bridge around the residue (Remaking bonds recoups some of the energy lost and bridging recoups some of the entropy). In thermally induced, random diffusion of the protein chain, at some instant residues will come close enough to one another for relatively long range intermolecular forces can act between them. If water can be pushed out from between residues the intermolecular forces can increase, first, because there is no longer a polar molecule in the way (high energy interaction) and second because the residues can become closer.
In this manner in a globular protein the hydrophobic core can form, with polar residues on the outside hydrogen bonded to the water. So its not that entropy is a driving force but that by expelling water internal energy (enthalpy) can be lowered and water can now join with other water molecules making hydrogen bonds (enthalpy lowered) and increasing entropy in the water, thus lowering $\Delta G$ overall. The reduced entropy of forming a hydrophobic core is more than offset by removing water and allowing it to enter the bulk.
There is a decrease in free energy by forming a large cluster of particles out of several smaller (hydrophobic) ones provided that the combined particle is large enough; calculation indicate a volume/surface area ratio > 1 nm. The decrease in free energy is largely enthalpic, whereas for smaller particles free energy change is largely entropic.
(Some other points about globular proteins, hydrophilic residues do hydrogen bond with water but there is still an entropic cost since now these water have restricted rotational & translational entropy vs. bulk water. When a hydrophobic residue is at the surface, waters have to bridge round it at large entropic cost that is roughly proportional to surface area. If a polar group is in the hydrophobic core (for functional reasons) its energy must be controlled by neutralising its charge or dipole with other residues.)
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$\begingroup$ Right and all that is exactly why I thought heat would favour the interaction. If it is driven by the entropy difference of water expulsion, the entropy term should be an even more negative contribution to the Delta G at higher temperature. $\endgroup$– JoryAug 7, 2016 at 15:22