Second Exam Biochem 104b
Question 1
You work in a protein engineering lab and you want to redesign a protein to introduce a
new hydrogen bond to a solvent exposed carbonyl oxygen. Using computer graphics you
identify an alanine residue that, when replaced by a tyrosine, would allow the formation
of that new hydrogen bond. However, there are two possible conformations for the new
tyrosine side-chain.
In conformation 1 the tyrosine can form the hydrogen bond to the desired solventexposed backbone carbonyl, but the aromatic ring remains solvent exposed on all sides.
In conformation 2 the tyrosine side chain stacks on top of another aromatic side
chain. It is not clear how closely the two side-chains will pack, but it is clear that in
conformation 2 the interface between the two rings will be shielded completely from
solvent.
a) What interactions, if any, favor conformation 1 and 2.
Answer:
Conformation 1 would be favored by the hydrogen bond formed between the
tyrosine residue and the carbonyl. However, formation of this hydrogen bond
requires the disruption of other hydrogen bonds these two groups are forming
with solvent. Therefore forming this hydrogen bond should have essentially no
stabilizing effect.
In conformation 2 we burry two hydrophobic surfaces. That of the tyrosine and
that of the other aromatic side chain. The net effect is equivalent to burying an
aromatic ring in a hydrophobic environment.
b) Among those interactions there is one, for which the exact strength is hard to
estimate. What is that interaction and why is it hard to estimate?
Answer: If the packing of the two aromatic chains is very tight, we might also get
substantial VDW interactions, but it would be hard to predict the magnitude of
this interaction. The reason is the steep (1/r6 ) distance dependence of the energy
from VDW interactions, a small structural difference can make a large difference
in the interaction energy.
c) Using the interactions you listed, calculate the expected difference in free energy
between the two conformations.
Answer: The main energy difference between the two conformations is the burial
of a hydrophobic surface corresponding to 6 methyl groups (i.e. 6 time 0.8 kcal/mol =
4.8 kcal/mol) stability of the two conformations will be dominated by
d) Determine the equilibrium constant between the two conformations.
Answer: Keq=[conf2]/[conf1]=exp(-Gconf1-conf2/RT)= 2980. At any given time
only one in ~3000 molecules is expected to adopt the conformation with the
hydrogen bond.
Question 2
You have a protein structure that contains a narrow cavity near the surface that’s so
narrow that it is not filled with solvent. This cavity is a little bit longer than a phenylalanine sidechain, but just a little bit shorter than a tryptophan residue. The
thermodynamic data for the unfolding reaction of the WT protein are:
∆S0 0.275 kcal mol-1 K-1 ∆H0 94 kcal mol-1 giving a ∆G0 (at 25 deg C) of 12 kcal mol-1
For the phenylalanine mutation the values are
∆S0 0.258 kcal mol-1 K-1 ∆H0 95.2 kcal mol-1 giving a ∆G0 (at 25 deg C) of 18 kcal
mol-1
Now you make the tryptophan mutation and get the following melting curve.
Temp (C)
55
60
65
70
75
80
Fluorescent signal
0.0065
0.053
0.31
0.77
0.96
1
a) For the unfolding reaction of the mutant determine the Tm, ∆S0, ∆H0 and ∆G0 (at
25 deg C) of unfolding.
Answer: ∆S0 0.27588 kcal mol-1 K-1 ∆H0 93.8 kcal mol-1 Tm = 67 deg C and a
∆G0 (at 25 deg C) of unfolding of 11.6 kcal/mol.
b) What values would you have expected to find, if your design had worked and you
had perfectly filled the cavity?
Answer: A perfectly buried tryptophan would have removed the equivalent of 8
methyl groups from an aqueous environment. At room temp this corresponds to
an increase in the mutant’s ∆G0 of unfolding of 0.8 kcal/mol * 8 = 6.4 kcal/mol.
The hydrophobic effect is entropic in nature and the corresponding change in
entropy is 6.4 kcal/mol / 298 deg. K) = 0.0215 kcal/mol*K so that the ∆S0 of
unfolding of the mutant would be 0.2535 kcal mol-1 K-1. There might also be a
small contribution from VDW interactions between the trypotophan and the
pocket walls that would be expressed in a 2-3 kcal increase in the enthalpy of
unfolding so that the ∆H0 of unfolding of the tryptophan mutant might be 96
kcal/mol. Then Tm=∆H0/∆S0 = 379 K or 106 deg C and ∆G0 at 25 deg. C = ∆H0298 K * ∆S0 = 20.5 kcal/mol
c) Are your results consistent with what you would have expected if the mutant had
filled the cavity? What could have gone wrong? Propose a hypothesis that would
explain the results you got for the tryptophan mutant.
Answer: The results for the tryptophan mutant look very similar to those of the
wild-type protein, certainly we do not see anywhere near the stabilization we
would expect if we had successfully filled the cavity. One scenario that would
explain the observations is that the tryptophan would not fit into the pocket
without causing very bad VDW interactions. The tryptophan may therefore adopt
a conformation, in which the aromatic ring simply sticks out into the solvent. As a
result the aromatic sidechain is solvent exposed in both the unfolded and folded
conformation so that its presence has no stabilizing or destabilizing effect.
Question 3
In most protein structures a very large fraction of the residues are found either in alpha
helices or in beta-sheets.
a) List the key structural features of an alpha helix (hydrogen bonding pattern,
direction, in which side chains point, etc. etc.)
Answer: Please see the list in the solutions of problem 3a in problem set 3.
b) Large aromatic side-chains prefer to reside in beta-sheets rather than alphahelices. Why?
Answer: The environment in a helix is far more crowded than in a betasheet.
Where the side chains point perpendicularly away from the backbone. Also, in
betasheets sidechains from neighboring strands are nicely positioned to interact
with one another via hydrophobic and VDW interactions.
c) For phenylalanine the relative propensities are 1.16 for alpha helices and 1.33 for
beta sheets. Estimate the average energetic advantage for a phenylalanine residue
to be in a betasheet. Based on this result, do you think that this propensity for
betasheets will play a large role in determining the conformation a specific
phenylalanine adopts in the context of a protein?
Answer: We can think of the ratio of relative propensities as an equilibrium
constant, which, in turn, allows us to calculate the average difference in free
energy for a phenylalanine being in a helix and in a beta sheet. With Keq helix->sheet
= propensitysheet / propensityhelix = 1.33 / 1.16 we can calculate ∆G0helix->sheet= -RT
ln Keq helix->sheet = - 0.08 kcal/mol. This energy difference is so small relative to
typical interaction energies derived from hydrophobic interaction, hydrogen
bonds or electrostatics, that one would not expect the slightly higher propensity
helixes to seriously effect the conformation in a specific case.
d) Based on the fact that so many residues in proteins reside in alpha-helices one
may expect that isolated peptides corresponding to individual alpha helices would
also be very stable. Is this true? If not, what then stabilized alpha-helices and why
are they so prevalent.
Answer: No, small peptides corresponding to sections of proteins that adopt
helices in the context of the fully folded proteins will rarely form stable helices on
their own. The reason is the high entropic price of forming a helix, or any well
defined structure for that matter. The formation of hydrogen bonds during the
formation of a helix typically requires the disruption of an equal number of
hydrogen bonds that the backbone groups can form with solvent and therefore
provide no net contribution to helix formation. Helices are usually only stable,
when they are able to pack against other helices, which allows the burial of
hydrophobic residues. A different way of looking at it is that helices are a great
way to satisfy the hydrogen bonding valences of the peptide backbone in the
hydrophobic interior of a protein.