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03-131 Genes, Drugs, and Diseases
Problem set 4
Skills/Concepts that you should have obtained from this problem set:
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September 29, 2015
Distinguish between a competitive inhibitor and allosteric inhibitor, based on structure
Competitive inhibitors of HIV reverse transcriptase also cause chain termination,
Determine whether a compound is DNA or RNA based on its structure
Convert DNA sequences to RNA sequences
Write duplex DNA given a single strand.
Recognize that polymerases incorporate the correct base using both the number of
hydrogen bonds and the size – a purine is always paired with a pyrimidine.
HIV reverse transcriptase is error prone because it lacks a 3’-5’ exonuclease
proofreading activity, therefore mutations arise in drug targets (protease &
transcriptase), leading to drug resistant viruses.
Drugs can be redesigned to effectively inhibit the altered (mutated) drug targets. The
redesigned drugs optimize complementary interactions between the drug and the mutant
enzyme.
1. (6 pts, 10 min) Shown to the right is one of the substrates for reverse transcriptase and two drugs,
compound I and compound II, that are currently used to inhibit the enzyme.
O
i) Briefly describe the role of this enzyme in the
O
H
CH3
HS
N
N
H
CH3
HIV lifecycle and indicate why it would be a
N
O
N
particularly good drug target.
O
N
CH2OH
O
N
O
CH2OH
It copies the viral RNA to DNA.
S
O
O
H3C
O
It is a good drug target because:
N
HO
a) this step is essential for the life
N
Compound II
Compound I
Substrate
N
cycle of the virus
b) This activity is absent in humans, so inhibitors are unlikely to have side-effects.
ii) Based on the structure of the drugs, which drug is most likely to be an allosteric inhibitor and which is
most likely to be a competitive inhibitor. Be sure to justify your answer.
Compound II is the competitive inhibitor because it looks like the substrate.
Compound I is the allosteric inhibitor, because it does not look like the substrate.
2. (5 pts, 10 min) Both compound I and compound II can inhibit HIV reverse transcriptase. Which will be the
better inhibitor if the substrate concentrations are high in the cell? Briefly justify your answer.
Competitive inhibitors cannot inhibit at high substrate, since they are displaced from the active
site by the substrate. Therefore an allosteric inhibitor would be better.
3. (5 pts, 5 min) Draw dAMP and label the 5’ and 3’ carbons. How does
dATP differ from dAMP?
Structure of dAMP shown on right. dATP would have three
phosphates.
5'
3'
4. (4 pts, 2 min) Write the corresponding RNA sequence for the following
DNA sequence. ATGCGTGACTAG. Please label the 5’ end.
Simply replace the T bases with U: 5’-AUGCGUGACUAG
4. (4 pts, 2 min) Write the duplex DNA representation of this DNA sequence. ATGCGTGACTAG. Please label
the 5’ end.
The top strand is in the 5’-3’ direction, by convention. The complementary (bottom) strand is
written in the opposite direction, to reflect the actual structure of the DNA duplex:
5’-ATGCGTGACTAG-3
3’-TACGCACTGATC-5’
O
5. (9 pts, 15 min) The diagram to the right shows three different basepairs, T-T, A-T
O
H
N
(normal basepair) and A-G.
N
N
i) Identify hydrogen bonds between each pair of bases. How many
N
H
O
ribose
ribose
hydrogen bonds do you see for each basepair?
O
T
T
Each basepair can form two hydrogen bonds, as indicated.
ii) What is the relative size of each basepair, what is the distance between the
H
N H
O
riboses (e.g. large, medium, or small)? The T-T basepair is the smallest,
N
N
the AT larger, and the AG the largest.
H N
N
N
ribose
N
ribose
iii) When DNA is replicated by polymerases, the correct pairing of bases is
O
A with T; other pairings, e.g. T-T, A-G are incorrect and would lead to
A
T
mutations. Based on your answer to part i) and part ii), what other criteria do
O
H
N
polymerases use to insert the correct base, besides numbers of hydrogen bonds?
N H
N
N
N
In addition to matching up hydrogen bonds, only a purine-pyrimidine
ribose
N
N
N
match fits properly in the helix. Thus polymerases use both
H N
ribose
N
H
the number of hydrogen bonds and the size to make sure they
G
A
are adding the right base.
Jmol Questions – You will need to view two Jmol pages for this problem set
7. Visit the first Jmol page and answer the following questions.
i) Is the label “A” at the 5’ or 3’ end of the molecule? Why?
The first carbon that is encountered as you proceed down the chain is the 5’ carbon.
Remember that the base is attached to the 1’ carbon, just count the carbons around the
ring.
ii) Is this DNA or RNA, justify your answer. Several possibilities:
the 2’ carbon has an –OH group, so it is RNA.
The helical structure is like that shown in the notes for RNA, in particular the bases are not
on the helix axis.
8. (16 pts, 20 min) The second Jmol page associated with this problem
set shows wild-type and a mutant HIV protease in complex with a
number of different HIV drugs. One of these drugs is the same as
the one presented in class. This particular drug contains a
cyclohexane ring at one end and it binds to the wild-type enzyme
with high affinity.
Four different drugs, with alteration in the cyclohexane ring, have
been developed for the purpose of inactivating a mutant HIV
protease. Reciprocal plots are provided that give the activity of the
wild-type protein without and with the cyclohexane drug. A
separate plot is given for the mutant protein, without drug, with the
cyclohexane drug, and with four other drugs. Please answer the
following questions.
i) What feature of the HIV life-cycle leads to a high level of
mutations in the HIV genetic material (2 pts)?
The enzyme that copies the viral RNA to DNA, HIV
reverse transcriptase, has a high error rate because it
lacks a proofreading function (3’-5’ exonuclease).
ii) Which residue is altered in the mutant HIV protease? How has it been changed? (1 pt)
The amino acid at position 82 has been changed from Valine to Aspartic Acid.
iii) Does this residue contribute mostly to catalysis or specificity? Justify your answer. (2 pts)
Specificity of substrate binding, the two catalytic aspartic acid residues are unchanged.
iv) Explain, with reference to the change in KI (which you can obtain from the slopes of the reciprocal plots)
and the structure of the two enzymes, why the affinity to the original cyclohexane drug has been
affected by this mutation (3 pts) In calculating KI values, assume an inhibitor concentration of 1 nM.
The KI values for the cyclohexane containing drug are obtained from the slope of the lines,
first calculate α from the ratio of the slopes and the KI=[I]/(α-1).
Enzyme-Drug Combination
α
KI
Wt-cyclohexane
1.00/0.2 = 5
1nM/4 = 0.25 nM
Mut-cyclohexane
0.36/0.2 = 1.8
1nM/0.8 = 1.25 nM
The KI is higher for the mutant enzyme, indicating weaker binding of the cyclohexane
containing inhibitor to the mutant enzyme. In the case of the wild-type enzyme there is a
favorable van der Waals contact with Val82 and the six carbon ring (cyclohexane) on the
drug. In addition, both groups are non-polar so the hydrophobic effect would also stabilize
the interaction. Replacement of Val with Asp in the mutant decreases both of these
interactions, van der Waals because of the poorer contact, and the hydrophobic effect
because Asp is polar, therefore the KI increases.
Enzyme-Drug Combination
α
KI
Mut-Drug 1
1.00/0.2 = 5
1nM/4 = 0.25 nM
Mut-Drug 2
0.50/0.2 = 2.5
1nM/1.5 = 0.66 nM
Mut-Drug 3
0.25/0.2 = 1.25
1nM/0.25 = 4 nM
v) Which of the three drugs would be the worst inhibitor of the mutant protease? Justify your answer with
reference to the reciprocal plot, as well as the interaction between the drug and the mutant enzyme. A
simple sketch of the interaction between the drug and the inhibitor would be useful (4 pts).
Drug 3 is the worst, because it has the largest KI value, or the smallest slope. The
cyclohexane ring on the drug has been replaced by a large negatively charged COOH
functional group. This would lead to poor binding because both the drug and Asp82 have
negative charges and the atoms from both groups overlap with each other.
vi) Which of the three drugs would be the best inhibitor of the mutant protease? Justify your answer with
reference to the reciprocal plot, as well as the interaction between the drug and the mutant enzyme. A
simple sketch of the interaction between the drug and the enzyme would be useful (4 pts).
Drug 1 is the best drug because it has the smallest KI value, or largest slope. The
cyclohexane ring has been replaced by an amino functional group that shows good van der
Waals contact with Asp82. In addition, their charges are opposite, leading to good
electrostatic interactions.
OH
H
N
H
O
H
N
O
N
H
O
O
O
O
O
O
OH
H
N
+
NH3
H
O
NH2
Drug 3
O
O
+
N
H
O
+
O
O
H
N
NH2
NH3
O
Asp82
(HIV Protease)
Worst Drug, High Ki
Asp82
Best Drug, low Ki
Drug 1
NH3
O
J-mol page instructions:
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The “unmodified” button will load the “wild-type” or non-mutant enzyme.
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The “simple view” button will show the backbone of the protein, the sidechain of key residues, and the bound drug.
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The check boxes will add surfaces to the indicated features, to orient yourself with respect to the molecule.
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The “Mut+cyclohexane” button will load the mutant HIV protease and the original, non-modified drug. This drug has
a cyclohexane group that contacts the enzyme and is the same as “drug B” from lecture.
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The buttons labeled “Drug1”, “Drug2”, etc. will load the structure of the mutant HIV protease with a different drug
bound in the active site.
Wild-Type (Val82)
Protease
12
Mutant Protease
No Inh([S] alone)
No Inh
[S] alone
Cyclohex drug
Drug 1
[S]+Cyclo
hex
10
Drug 2
Drug 3
y = x + 0.2
12
1
10
y = x + 0.2
6
8
1/rate
1/rate
8
4
y = 0.2x + 0.2
6
2
y = 0.50x + 0.2
4
2
y = 0.36x + 0.2
y = 0.25x + 0.2
3
2
y = 0.20x + 0.2
0
0
5
1/[S]
10
0
0
5
1/[S]
10