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Why do denatured proteins tend to be less soluble than the native protein? In terms of hydrophobic effects, could anyone explain this phenomenon for me?

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    $\begingroup$ Put bluntly, soluble proteins have shielded as many hydrophobic residues in their sequence as possible from the solvent (say cytosol, primarily aqueous). As the solvent then has access primarily to hydrophilic residues, the protein is largely soluble. Denatured proteins with large, exposed regions of hydrophobic residues become insoluble because hydrophobic molecules do not dissolve in water. As this is what amounts to a homework question, however, do keep in mind for subsequent questions that this site requires some personal effort to obtain a good answer! $\endgroup$ – CKM Feb 8 '16 at 22:29
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Background information

A native protein has it's hydrophobic amino acids folded inwards, so it has a hydrophobic core. why? Because it's thermodynamically favoured. When a protein is denatured the hydrophobic amino acids won't be longer inside the core of the protein, but instead will be exposed to the hydrophilic environment (assuming you are talking about a protein inside an organism). But this isn't thermodynamically favoured because hydrophobic parts cannot form H-bonds with water and so aren't much soluble in water:

enter image description here

So when this happens, many protein molecules will aggregate like soap micelles (because hydrophobic interactions are favored). This will lower the total surface area of hydrophobic amino acids in contact with the water, thus lower disturbance in H-bonding in water molecules.

Here's my self-made depiction of effect of organics molecule on H-bonding in water (its quite rough, so don't care about fine details).

H-bonding

Summarized hydrophobic amino acids will be exposed to the hydrophilic environment --> the proteins will aggregate together --> insoluble plaque.

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