Name: __________________________________
MCB 111/211
G = - RT ln K
Midterm 2
k = (kT/h) e-G=/RT
2005
R = 1.987 cal/deg-mole
1. Find the letter below that best matches the following statements. Use a letter only once. (10 pts.)
A.
B.
C.
D.
E.
F.
G.
H.
I.
J.
K.
L.
M.
N.
O.
P.
Q.
i)
ii)
iii)
iv)
v)
vi)
vii)
viii)
ix)
x)
Dead-End Elimination
ribosome display
PHD
m value
often largely an effect on the entropy change of a reaction
110 M
1 mM
protein stability curve
Tm
Cm
pKa changes between folded and unfolded states
hydrophobic effect
assumption in Chou-Fasman method for predicting secondary structure
assumption in the 3D-1D profile (threading) method for predicting structure
free energy of transfer from water to octanol
heat required to change the temperature by 1 °K
Polar interactions
__L___
__P___
__H___
__E___
__A___
__N___
__J___
__O___
__M___
__K___
accounts for much of the temperature dependence of protein stability
Cp
small difference between large, nearly balancing effects
effective concentration
finds the global energy minimum sequence in computational protein design
the protein folds
midpoint of solvent-induced denaturation curve
measure of the hydrophobic effect
interactions with residues nearby in the sequence determine structure
measure of the contribution of electrostatic interactions to protein stability
2. HIV and RSV are related viruses that each encode a protease that cleaves their respective Gag
polyproteins. The sizes of the proteases are 98 (HIV) and 124 residues (RSV). BLAST reveals that
they contain a segment of 43 residues with 19 identities. The E-value for the alignment is 6.5.
2a. What is the definition of the E-value? (4 pts.)
The number of hits of that quality expected using a random sequence to search the database.
1/5
Name: __________________________________
2b. Do you conclude that the HIV and RSV proteases are homologous or not? Why or why not? (5
pts.)
No. The sequence match is likely to be random. The e-value is >0.001.
Or
Yes. HIV and RSV are related.
2c. Suggest a way to test your answer to 2b. Using that method, what results would expect if the folds
are similar? (6 pts.)
Threading. E-value <0.001 or Z-score >7.
Solve the structures. C rmsd <2.5 Å.
3. Dahiyat and Mayo (Science 278, 82 (1997)) designed a sequence (called FSD-1) that adopted a
zinc-finger fold without binding zinc. The zinc-binding residues (Cys2His2) in the natural protein (Zif268)
were replaced by Ala, Lys, Phe and Phe. More hydrophobic surface area was buried in the folded
FSD-1 structure than in the wild-type protein (which had zinc in the core).
3a. Although the design calculations examined ~2 x1027 sequences, over 1062 structural variants were
ranked computationally. What accounts for the much larger number of structural variants than
sequence variants? (6 pts.)
Most residues were tried in multiple rotamers.
3b. Assume that unfolding the natural zinc finger causes release of the Zn2+ ion. How do you expect
the replacement of the Zn2+ site in the natural protein by hydrophobic residues in FSD-1 to influence the
Cp of unfolding? Justify your answer. (6 pts.)
Cp goes up because the core of FSD-1 is more hydrophobic. The hydrophobic effect has a
large, positive Cp.
2/5
Name: __________________________________
3c. Briefly describe two ways you would measure Cp of unfolding. (6 pts.)
DSC. Cp corresponds to the offset of the folded and unfolded baselines.
Measure H at different temperatures.
H (T) = H (T0) + Cp (T-T0)
Cp = (H (T) - H (T0)) / (T-T0)
3d. (9 pts.) Considering only the effects of replacing the Zn2+ site in the natural protein by hydrophobic
residues in FSD-1, do you expect Zif268 or FSD-1 to be more destabilized (or less stabilized) in:
i) 20% ethanol (vs. no ethanol)
FSD-1
ii) 1 M ammonium sulfate (vs. low salt)
Zif268
iii) 0 °C (low temperature vs. room temperature)
FSD-1
3e. FSD-1 and the natural zinc finger are both 28 amino acids long. Assuming each residue gains 12
degrees of freedom upon unfolding, estimate the Gconf for unfolding at 37 °C. (8 pts.)
S = R lnWu = R ln1228 = R x 28 ln12 cal/mol-deg = 138.25 cal/mol-deg
Gconf = -TSconf = -310 OK x 138.25 cal/deg-mole = -42.858 kcal/mol
3f. At 20 °C, the stability (Gu) of FSD-1 was 1.2 kcal/mole. Assuming a two-state model, what is the
ratio of unfolded protein to folded protein at this temperature? (8 pts.)
Gu = -RT ln K, where K = [U]/[F]
K = e -Gu/RT = e -1200/(1.987 x 293) = 0.127
3/5
Name: __________________________________
NOT COVERED IN 2006
4. Bolon and Mayo (Proc Natl Acad Sci
USA 98, 14274 (2001)) used the same
design method to make a mutant of the
protein thioredoxin that catalyzed
hydrolysis of an ester substrate. They
created the design by calculating the
location in the protein that would accept
a model for the transition state of the
reaction in which a histidine attacked the
ester bond (Fig. 1). The best designed
mutant contained the changes
Phe12Ala, Leu17His and Tyr70Ala.
Figure 1. A. Mechanism of ester
hydrolysis. Nu = nucleophile. B.
Transition state analog involving a
histidine nucleophile.
4a. What is the logic of using a model of the transition state in this way to create an enzyme? (6 pts.)
1. The enzyme binds the transition state tighter than the ground states.
2. The enzyme changes the chemical mechanism of the the reaction (His nucleophile vs. H2O).
4b. The rate of the catalyzed reaction was 180-fold faster than the uncatalyzed reaction at 20 °C. By
how much did the mutant thioredoxin lower the activation energy for the reaction? (8 pts.)
kcat/kuncat = 180 = exp ((-G≠cat - -G≠uncat)/RT) = exp ((G≠uncat - G≠cat)/RT)
RT ln 180 = (G≠uncat - G≠cat) = 3.023 kcal/mole
5. A 23-residue fragment of the transcriptional activator CREB folds upon binding to the coactivator
CBP. Ser133 of CREB is phosphorylated. At 25° C, the pKa for protonation of the di-anionic
phosphoryl group is 5.4 in free CREB and 4.3 in the CREB/CBP complex.
5a. What is the free energy of the electrostatic interaction of CBP with the Ser133-phosphoryl group?
(Mestas & Lumb, Nature Structural Biology 6, 613 (1999)). (6 pts.)
pKa = -log Ka = 2.303 ln Ka, 10-pKa = Ka
G
= -RT (ln Ka(free) – ln Ka(bound))
= -RT ln (Ka(free)/ Ka(bound))
= -RT ln (10-5.4/10-4.3) = -1.5 kcal/mole
4/5
Name: __________________________________
5b. Does phosphorylation of CREB favor or oppose CBP binding? (6 pts.)
Favor
5c. Suggest two possible mechanisms by which CREB phosphorylation could regulate CBP binding.
(6 pts.) NOT COVERED IN 2006
1. CBP directly contacts the PO4 in the complex.
2. Long range electrostatic intereactions with CBP
3. PO4 promotes the folding of CREB to favor CBP binding (allosteric interaction)
5/5