Download The MOLECULES of LIFE

The MOLECULES
of LIFE
Physical and Chemical Principles
Solutions Manual
Prepared by James Fraser and Samuel Leachman
Chapter 5
Evolutionary Variation
in Proteins
Problems and Solutions
True/False and Multiple Choice
1. The BLOSUM scoring matrix gives a measure of how
conservative a mutation is. For substitutions of aspartic
acid (Asp, D), which of the following orderings correctly
places the amino acids from most conservative to least
conservative?
a.
b.
c.
d.
K,L,A,C,E,S
E,S,K,A,C,L
L,C,A,K,S,E
A,C,E,K,L,S
2. An environment profile in the 3D-1D profile method
compares:
I. the stability of the amino acid in varying solvents
II. the burial of each amino acid in the structure
III. the hydrophobicity surrounding each amino acid
IV. the type of secondary structure element containing
each amino acid
a.
b.
c.
d.
I, II, III, and IV
I and IV
II, III, and IV
II and IV
7. The core of a protein generally contains residues from
the ________ class of amino acids.
Answer: hydrophobic/nonpolar
8. Globin proteins bind the iron-containing ________
cofactor.
Answer: heme
9. A disulfide bond links two _______ residues.
Answer: cysteine
10. According to the BLOSUM substitution matrix, the most
conservative mutation from tryptophan (W), other than
to itself, is to ______, which has a score of ______.
Answer: tyrosine, 2
11. Many soluble human proteins can be expressed in the
E. coli bacteria or using an in vitro translation system.
How can these proteins fold without the cellular
machinery present in human cells?
Answer: The “thermodynamic hypothesis” states
that proteins adopt native structures that optimize
thermodynamic properties. Since sequence determines
structure and most proteins do not require posttranslational modifications, external templates, or
specific molecular chaperones to fold, many proteins
can fold after translation by a different organism or even
a cell-free (in vitro) translation system.
3. Protein domains can be assembled together in many
different ways, because surface sidechains can be
mutated easily without losing protein stability.
True/False
4. In contrast to ribonuclease, some proteins cannot fold
without the assistance of proteins known as molecular
chaperones. This means the thermodynamic hypothesis
of protein folding does not apply to these proteins.
True/False
5. Two proteins that share more than 50% sequence
identity over a 100-residue stretch are likely to have the
same three-dimensional fold.
True/False
Fill in the Blank
6. Two common chemical denaturants of proteins are
guanidinium and _______.
Answer: urea
12. What level of activity (1, 10, or 100%) is predicted
for ribonuclease-A when it is subject to each of the
following stepwise procedures?
a. i.
denatured, then
ii.
reduced, then
iii.
exposed to oxygen, then
iv.
refolded by removing urea
b. i.
denatured, then
ii.
reduced, then
iii.
refolded by removing urea while exposed to oxygen
c. i.
denatured, then
ii.
refolded by removing urea
Rationalize the predictions based on the effect
of denaturation and reduction–oxidation of the
ribonuclease-A cysteine residues. Assume that all
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Chapter 5: Evolutionary Variation in Proteins
However, we must correct for the vertical ordering: it
does not matter which lysine–glutamate pair is picked
first, second, etc., just that the correct pairs are picked.
This means we need to divide the total number by 6!
(6 pairs to choose first × 5 pairs to chose second...etc).
518,400/(6 × 5 × 4 × 3 × 2 × 1) = 720.
Off the 720 random structures populated during crosslinking in experiment (b) only one will be correct. Thus
the activity expected is 1/720 = 0.0014, or 0.14%.
possible unfolded conformations are equally likely and
that folding is faster than oxidation.
Answer:
a. Low activity (~1%). (i) The protein is first denatured,
forming mostly random structures; however, all disulfide
bonds remain intact. (ii) Reduction causes all disulfide
bonds to be broken. (iii) Exposure to oxygen causes the
disulfide bonds to reform. They reform with random
pairings, since the protein is unfolded and populating
many random structures. (iv) The protein refolds, but
only those molecules in which the disulfides formed
correctly are active.
b. High activity (~100%). (i) The protein is first
denatured forming mostly random structures; however,
all disulfide bonds remain intact. (ii) Reduction causes
all disulfide bonds to be broken. (iii) Exposure to oxygen
causes the disulfide bonds to reform, but since folding
is faster than oxidation, the disulfide bonds reform
correctly.
c. High activity (~100%). (i) The protein is first
denatured forming mostly random structures; however,
all disulfide bonds remain intact. (ii) The protein refolds
and since the disulfide bonds are already intact, full
activity is restored.
13. A folded protein structure contains six ion pairs
between lysine and glutamate. There are no other
possible ion pairs in the protein. A chemical crosslinker forms a covalent bond between ion-paired lysine
and glutamate sidechains. By analogy to the Anfinsen
experiment, the following experiments are done:
a. folded protein → unfold with urea → remove urea
to refold → add cross-linker → remove excess crosslinker → measure activity
b. folded protein → unfold with urea → add crosslinker → remove excess cross-linker → remove urea to
refold → measure activity
The cross-linker does not by itself alter the activity
of the protein when the correct ion pairs are formed.
A protein with incorrect ion pairs cross-linked would
be inactive. The activity measured at the beginning
and end of experiment (a) is 100%. What percentage
of the activity do you expect to observe at the end of
experiment (b)? Assume that all unfolded conformations
are equally likely.
Answer:
There are six ways of picking the first lysine and six
ways of picking the first glutamate (6 × 6 = 36).
There are five ways of picking the second lysine and five
ways of picking the second glutamate (5 × 5 = 25).
There are four ways of picking the third lysine and four
ways of picking the third glutamate (4 × 4 = 16).
There are three ways of picking the fourth lysine and
three ways of picking the fourth glutamate (3 × 3 = 9).
There are two ways of picking the fifth lysine and two
ways of picking the fifth glutamate (2 × 2 = 4).
There is one way of picking the last lysine and one way
of picking the last glutamate (1 × 1 = 1).
The total number of ways to pick is: 36 × 25 × 16 × 9 × 4
× 1 = 518,400.
14. Use the BLOSUM substitution matrix (Figure 5.11) to
compute the sum of the substitution scores (Sij) and
the overall likelihood ratio (L) of the following short
alignments:
a. PADKTN
PEEKSA
b. KFLASV
ATWDPE
Answer:
a. Sequence:
P
A
D
K
T
N
Sequence:
P
E
E
K
S
A
Score:
(7)+(–1)+(2)+(5)+(1)+(2)
Sum of Scores = 12
Likelihood = 2(Score/2) = 2(12/2) = 64
b. Sequence:
K
F
L
A
S
V
Sequence:
A
T
W
D
P
E
Score:
(–1)+(–2)+(–2)+(–2)+(–1)+(–2)
Sum of Scores = –10
Likelihood = 2(Score/2) = 2(–10/2) = 1/32 = 0.3125
15. Based on the BLOSUM matrix, how much more likely is
it that:
a. tryptophan is substituted by a tyrosine than a
tryptophan is substituted by a cysteine?
b. sequence (i) DPKRFL is related to sequence (ii)
EPKRFI than sequence (i) is related to sequence (iii)
KGKRYA?
To answer this question, you must calculate the ratio
of the likelihood ratios for each case. Explain the
significance of higher likelihood.
Answer:
a. The score for W → Y substitution = 2; The
likelihood = 2(Score/2) = 2(2/2) = 2. The score for a
W → C substitution = (–2); The likelihood = 2(Score/2) =
2(–2/2) = 0.5.
Lij/Lik = 2/0.5 = 4 times higher likelihood.
This means that a W → Y substitution is more
conservative than a W → C subsitution.
b. The sum of scores for (i) → (ii) = 27; The likelihood
= 11,585.2. The sum of scores for (i) → (iii) = 9; The
likelihood = 22.6.
Lij/Lik = 11,585.2/22.6 = 512 times higher likelihood.
This means that (i) is more likely to be related to (ii) than
to (iii).
16. Proteins known as cyclophilins catalyze proline cis-trans
isomerization. A catalytic arginine residue is invariant
in all cyclophilins. All other positions change residue
identity in different cyclophilins despite the fact the
variant proteins have the same overall fold and general
catalytic activity. Explain, given the relationship between
protein sequence and structure, how catalytic activity is
PROBLEMS and solutions
retained even though most residues can change.
Answer:
The relationship between sequence and structure
is asymmetric. Many proteins can have the same
structure despite different (degenerate) sequences.
However, sequences fold into only one structure
(following the thermodynamic hypothesis). Only
residues that perform specific functions, such as the
catalytic arginine in cyclophilins and the histidine in
helix F in globins are invariant.
17. Why is the sequence similarity generally higher when
comparing two globins from mammals than when
comparing a globin from a mammal and a globin from a
plant?
Answer:
It is likely that the both the mammalian and plant
globins derived from a single ancestral globin and have
expanded through gene duplication. The mammalian
globins share a common ancestor with each other that
is more recent than the mammalian globins share with
plant globins.
18. How might the tolerated variation in the hydrophobic
core of the lambda repressor change if the hydrophobic
core of wild type lambda repressor were more tightly
packed?
Answer:
The protein would likely be less tolerant of
mutations. Proteins undergo fluctuations, even in the
hydrophobic core. They are not perfectly packed as
interlocking pieces of a puzzle. This property allows for
accommodation of differently shaped sidechains and for
toleration of mutation.
3
Explain why the distribution of protein sizes has the
periodicity that is seen in the diagram and estimate a
value for x.
Answer:
A reasonable value for x is approximately 100–150.
This is the average size of a protein domain (which are
normally 50–200 residues). The periodicity is observed
because proteins are modular and are expanded by
addition of different domains. The peaks at 1x, 2x, and
3x derive from proteins containing 1, 2, or 3 domains.
21. The number of distinct protein folds is limited. Why
might this be so? Approximately how many folds are
there (hundreds, thousands, millions, or billions)?
Answer:
There are likely to be only thousands of folds because
natural selection will favor protein sequences that will
fold into stable structures. In order to fold into a stable
structure, it is necessary to be built up of secondary
structure elements and have a hydrophobic core. It is
probable that there are a finite number of orientations
and combinations of secondary structure elements
that will produce a hydrophobic core of acceptable
geometry to enable protein folding.
22. How many folds are represented below? Describe what
CATH class each fold belongs to and how the secondary
structure is arranged in each fold.
Q5.12 (seq_struct_70_v1)
19. What characteristics define a protein domain?
Answer:
Domains have a distinct topology and a self-contained
hydrophobic core. They generally contain 50–200
residues.
A
B
C
D
E
F
20. The diagram below shows the size distribution for
globular proteins produced by the bacterium E. coli.
frequency of occurrence
Q5.10 (seq_struct_71_v1)
Answer:
There are four folds. A and E have the same Rossmannlike mixed α/β fold. B has a unique mixed α/β fold. C has
an all β sheet structure arranged in a barrel. D and F
have a four helix bundle.
1x
2x
3x
number of residues in the protein
23. A threading program is used and it gives two possible
predicted folds for a particular sequence. They differ
mainly in the placement of a single helix. In predicted
fold (A), the helix is entirely within the hydrophobic
core. In predicted fold (B), the residues of the helix are
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Chapter 5: Evolutionary Variation in Proteins
mostly exposed to solvent. The sequence of the helix is
LIVFLAIL. Explain how the 3D-1D profile method could
be used to distinguish between the two possible folds.
Answer:
All of the residues in the helix are hydrophobic and have
positive scores as α helical buried positions. In contrast,
these residues have negative scores for exposed
α helical positions. Thus, fold A correctly predicts that
the α helix should be part of the hydrophobic core of
the protein.
24. What structural features of the Rossmann domain
enable it to bind nucleotides?
Answer:
The negatively charged phosphate groups of the
nucleotides interact with the P loop at the positive pole
of the helix dipole generated by the first α helix of the
Rossmann fold.
25. Both thioredoxin reductase and glutathione reductase
use fused FAD- and NADPH-binding domains and
dimerization to accomplish their cellular functions.
Which structural feature likely evolved first, the fused
domains or dimerization?
Answer:
The fused FAD and NADPH domain structure is
conserved between both structures. However, the
relationship between the individual subunits of the
dimers is different for each protein (see Figure 5.43).
This implies that domain structure evolved before the
split of individual thioredoxin and glutathione reductase
lineages and that dimerization evolved later.