abruptness directly follows from the strong temperature dependence ofDG,
which is largely due toDHbeing very large (see Figure 7.5). Unfolding due
to a decrease in pH must be due to a change in ionization of side groups. For
a simple ion, however, a change in ionization from 0.1 to 0.9 occurs over a
pH range of 2 units (see Figure 2.9a), whereas Figure 7.4c shows a transition
over only 0.5 pH units. The explanation must be that the protein involved
has some buried His groups in its native conformation and if the molecule
unfolds, all of these can become ionized. This then happens at pH 3.8 rather
than at the pKof 6.4 (Table 7.1), where the driving force for unfolding
presumably is too small.
Several workers have tried to calculate the variouscontributions to the
free energyinvolved in stabilizing the native conformation. The subject is,
however, still somewhat controversial, at least partly because of differences
in terminology. We will now consider what terms are involved in the
hypothetical stability equation and how large they may be for a small
molecule or domain of about 100 residues. For most terms, reliable
estimates cannot be made.
The first four terms are negative, hencepromote the native conforma-
tion:
- Hydrogen bonds. It should be realized that all of the groups
potentially involved can also make H-bonds with water, and the latter
mostly are stronger. This would mean that no internal H-bonds are formed.
However, the predominantly apolar interior of a globular protein molecule
may have a dielectric constant as low as 5, as compared to 80 in water. That
would cause the internal H-bonds to be much stronger, having a net bond
energy of about 16 kJ?mol^1. Assuming 50 H-bonds per molecule, this
would lead to a term inDU?NGof roughly800 kJ?mol^1. - Hydrophobic interactions. It has long been accepted that these
provide a major stabilizing force, but at present many workers assume them
to play a fairly small, although not a negligible, part. However, since
changes in enthalpy as well as entropy are involved, and since it is difficult to
separate hydrophobic interaction from other forces, the author feels that a
definite answer has not yet been given. Anyway, the presence of
hydrophobic residues is essential for obtaining the native conformation,
since they allow H-bonds to form in an apolar environment; in that sense,
the 800 kJ?mol^1 mentioned in item 1 does include hydrophobic
interaction. - Van der Waals attraction. Of course, this also occurs between
groups and water molecules, but the net effect is likely to be stabilizing. Van
der Waals attraction cannot be fully separated from hydrophobic
interaction. Moreover, some dipole–dipole interaction may be involved. It
concerns an enthalpic contribution.