Tertiary Structure and the Hydrophobic Core
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You should be familiar with the four levels of protein structure and the local conformations of \(\alpha\)-helices and \(\beta\)-sheets. Basic knowledge of amino-acid side-chain properties (polar, non-polar, charged) is assumed.
Why this matters
The specific three-dimensional shape of a protein is the direct determinant of its biological function.
While secondary structure provides the local building blocks, tertiary structure defines the global architecture of a single polypeptide chain. This folding is driven primarily by the thermodynamic imperative to minimise the system's free energy in an aqueous environment. Understanding how the hydrophobic core forms allows us to predict stability, explain misfolding diseases, and design stable proteins for therapeutic use.
The Driving Force: Solvation and Entropy
Imagine placing a non-polar molecule, such as an oil droplet, into water. Water molecules are highly ordered when they form a cage-like structure around the non-polar surface to maintain hydrogen bonding with each other. This ordering represents a decrease in entropy. When many non-polar groups come together, the total surface area exposed to water decreases. This releases the ordered water molecules back into the bulk solvent, increasing the entropy of the water. This increase in solvent entropy is the primary thermodynamic driver for the burial of non-polar residues in the protein interior. It is not that the protein "likes" to be hydrophobic; rather, the water "likes" to be free. This phenomenon is known as the hydrophobic effect.
Thermodynamics of Folding
The stability of the tertiary structure is governed by the Gibbs free energy change, \(\Delta G\), of the folding process. For a protein to fold spontaneously, \(\Delta G\) must be negative. The free energy change is a balance between enthalpic contributions (bond formation) and entropic contributions (loss of conformational freedom).
The equation for Gibbs free energy is:
For the polypeptide chain itself, \(\Delta S\) is negative: the disordered chain loses conformational entropy when it settles into one ordered structure, and the \(-T\Delta S\) term opposes folding. That cost is paid partly by favourable enthalpy (van der Waals contacts and hydrogen bonds in the packed core) and, to a large extent, by the entropy gained by water released from non-polar surfaces. The net result is only marginally negative: a typical small protein is stable by just 20–60 kJ mol-1.
Composition of the Hydrophobic Core
The interior of a globular protein is densely packed with non-polar side chains. This core is characterised by high packing density, often approaching that of liquid alkanes. The residues most frequently found in the core include leucine, isoleucine, valine, phenylalanine, and methionine. These residues have large, non-polar side chains that maximise van der Waals interactions when packed tightly together.
The packing efficiency is a critical determinant of stability. A well-packed core minimises voids and maximises the number of favourable intermolecular contacts. If the core is poorly packed, water molecules may penetrate the interior, disrupting the hydrophobic effect and leading to unfolding. The specific arrangement of these residues is dictated by the primary sequence, ensuring that compatible shapes and sizes are brought together in three-dimensional space.
Role of Polar and Charged Residues
While the core is predominantly hydrophobic, it is not exclusively so. Some polar residues, such as serine, threonine, and asparagine, can be found in the interior if they are fully satisfied by internal hydrogen bonds. However, charged residues (aspartate, glutamate, lysine, arginine) are rarely found in the core unless they form specific salt bridges or are part of a catalytic site. The energetic cost of desolvating a charged residue is very high, so it is usually reserved for the protein surface where it can interact favourably with water.
The surface of the protein is enriched in polar and charged residues, which form a hydration shell. This amphipathic distribution—hydrophobic inside, hydrophilic outside—is the hallmark of soluble globular proteins. This arrangement ensures solubility in the aqueous cytoplasm or extracellular fluid while maintaining a stable, compact interior.
The concept
The levels of protein structure
From amino-acid sequence to folded, functional shape.
A common misconception is that the hydrophobic effect is an enthalpic process driven by the attraction between non-polar groups. In reality, it is primarily an entropic process driven by the release of ordered water molecules. The van der Waals interactions within the core are important for stability, but the initial driving force for burial is the increase in solvent entropy. C
Worked example
Problem. For a small globular protein, folding at 298 K has \(\Delta H = -250\ \mathrm{kJ\,mol^{-1}}\) and \(\Delta S = -0.750\ \mathrm{kJ\,K^{-1}\,mol^{-1}}\) (these are the overall values for protein plus solvent). Find \(\Delta G\) of folding at 298 K, and estimate the melting temperature \(T_m\) at which folded and unfolded forms are equally stable. Treat \(\Delta H\) and \(\Delta S\) as constant with temperature.
- \(T\Delta S = 298 \times (-0.750) = -223.5\ \mathrm{kJ\,mol^{-1}}\).
- \(\Delta G = \Delta H - T\Delta S = -250 - (-223.5) = -26.5\ \mathrm{kJ\,mol^{-1}}\): folding is favourable, but only just.
- At \(T_m\), \(\Delta G = 0\), so \(T_m = \Delta H/\Delta S = 250/0.750 = 333\ \mathrm{K}\), about 60 °C.
Units check. K × kJ K-1 mol-1 gives kJ mol-1, and kJ mol-1 ÷ kJ K-1 mol-1 gives K. Real proteins also have a large heat-capacity change on unfolding, which is why they can denature on cooling as well as on heating.
Three quick tests
- Why is the hydrophobic effect considered an entropic process?
Answer
Because it is driven by the increase in entropy of water molecules released from ordered cages around non-polar groups. - Which type of residue is most likely to be found on the surface of a soluble protein?
Answer
Polar and charged residues, as they can form favourable interactions with the aqueous environment. - If a mutation replaces a core leucine with a charged aspartate, what is the likely effect on protein stability?
Answer
Stability would likely decrease due to the high energetic cost of desolvating the charged residue and the disruption of the hydrophobic core packing.
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Every lecture and every concept in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.