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03-131 Genes, Drugs, and Disease
Problem Set 5
Due October 14, 2015
What you should have learned from this problem set:
 Only triphosphates can be added by polymerases
 Antiviral drugs that are competitive inhibitors of polymerases can become terminators after the
addition of three phosphates.
 How to determine the effect of DNA mutations on HIV protease – reading a sequencing gel,
determining the reading frame, using the codon table.
 How to counteract changes at Val82 in the protease by introducing complementary interactions in a
new drug.
 PCR can be used to amplify tandem repeats, provided the primers are outside the repeated region.
 If the tandem repeats differ between two samples, the samples must have come from different
individuals.
 If the tandem repeats are the same in two samples (e.g. blood from a crime scene and blood taken
from a possible suspect) they do not necessarily have to come from the same person, many
individuals will have the same tandem repeats. Thus, this technique can only be used to exclude
identity matches, not prove them.
1. (10 pts, 10 min) AZT is a competitive inhibitor of HIV reverse
transcriptase. Its structure is shown on the right, along with
one of the normal substrates for the enzyme, dTTP.
i) Why is it not possible for AZT to be incorporated into the
growing polymer?
It
is missing three phosphate groups. Only
Substrate
triphosphates can be added to the 3’OH by
(dTTP)
polymerases.
ii) What would have to be added to the AZT such that it could be added to the primer by a polymerase?
(AZT)
Three phosphates, to make it a triphosphate.
iii) If AZT were used by a polymerase, would elongation (addition of the next base) be possible? Why or why not?
No, it has no 3’ OH so it is a chain terminator, just like the dideoxy bases used in DNA sequencing. This
would prevent the entire HIV vRNA to be converted to DNA.
2. (10 pts, 5 min) Acyclovir is an antiviral agent. The structure of acyclovir and
guanosine (deoxy) are shown on the right.
i) Use the web to find out what kind of viral infections are typically treated with
acyclovir.
Herpes simplex.
ii) How do you think acyclovir works to inhibit replication of the virus?
Same as AZT. The drug itself would be a competitive inhibitor of
replication, when phosphorylated to the triphosphate, it would
also terminate growing chains because it lacks a 3’OH.
3. (15 pts, 25 min) Two sequencing reactions are shown below. The top sequence
is of a mutant HIV protease gene and the bottom is from a wild-type virus.
Your ultimate goal is to determine how a mutation may have affected binding
of the drug to the HIV protease. You will need to download the problem set to
get the data in color. Sequence files are also downloadable, if you want to use
bioinformatics tools (recommended). You can translate the DNA sequence at:
http://web.expasy.org/translate/. (This will give the one letter code for the amino acids, a conversion chart is given on
the last page of this problem set.) Alternatively you can use Blastn to align the mutant sequence data to the wildtype
sequence.
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03-131 Genes, Drugs, and Disease
Problem Set 5
Due October 14, 2015
i) Determine the change in the DNA sequence and the resultant change in protein sequence. You will need to consult
the known DNA/protein sequence of HIV protease (see lecture notes) to determine the correct reading frame.
ii) Does this mutation affect the substrate specificity of HIV protease (the types of substrates it likes to act on), or the
ability of the enzyme to cleave peptide bonds? Why?
iii) The anti-viral drug that inhibits HIV protease is shown on the right.
OH
O
H
H
+
How might you modify the drug such that it would bind, and
H3C
N
N
NH3
N
effectively inhibit, the activity of the virus containing the mutant
H
O
O
O
protease? To aid in the visualization of the enzyme-inhibitor
complex, you can utilize the HIV jmol page for this problem set. This
O
NH2
Jmol page will allow you to load the structure of HIV protease with a
number of different mutations – you need to determine the correct mutation (part i) to select the correct structure to
view.
i) The mutation is an Asp at residue 82, There are three ways you could have done this:
a) “Manually” find the sequence difference – it occurs at position 44 in the sequence data. Starting
reading from 41:
WT: CTGTCAACA
Mut:CTGACAACA
Now you have to find this in the HIV protease sequence:
238
GGACATAAAGCTATAGGTACAGTATTAGTAGGACCTACACCTGTCAACATAATTGGAAGA297
68GlyHisLysAlaIleGlyThrValLeuValGlyProThrProValAsnIleIleGlyArg87
We see that the third reading frame is correct for how I read the sequences:
WT: CT GTC AAC A
Mut:CT GAC AAC A
So Valine82 (GTC) has been changed to Asp(GAC).
b) you can translate the DNA sequence using Expasy and find the mutation that way.
c) you can align the mutant sequence to the wild-type using Blastn
Methods b and c were done in class, see lecture 16.
The mutation (Val to Asp) would affects substrate specificity because it changes the specificity pocket.
An effective drug would replace the cyclohexane ring with a + charged amino group (the identical
change was on the last problem set, see Q8, problem set 4).
4. (10 pts, 15 min) Solve the following forensic case.
Tandem repeats(3)
DNA was isolated from the victim, collected from
the crime scene, and from three suspects. The DNA
Chromosome
was subjected to PCR using primers that amplify a
region of the chromosome that is known to contain
tandem repeats, which are sequences that are
PCR product
repeated multiple times on the same chromosome.
The number of tandem repeats differs from one individual to another, causing the length of the PCR product to differ.
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03-131 Genes, Drugs, and Disease
Problem Set 5
Due October 14, 2015
For example one chromosome could look like this, with three tandem repeat (see above), while a chromosome might
have four, giving a larger PCR product. Note that since we have two copies of each chromosome there are two possible
PCR products, one from each chromosome. It is possible to have the same number of tandem repeats on each
chromosome.
After PCR the DNA was separated by size and stained to visualize the
DNA fragments. In the image on the right shows the sizes of DNA
fragments for each of the following samples:
 Lane 1 – Victim’s blood
 Lane 2 – Blood at crime scene (victim+criminal)
 Lane 3 - Suspect #1
 Lane 4 - Suspect #2
 Lane 5 - Suspect #3
Bands towards the bottom of the diagram are smaller. Bands at the
same position have the same size. The circled band will be used for
problem 5.
i) Who is no longer a suspect? Please justify your answer.
ii) On the basis of the above data, who is still a suspect for the crime?
Please justify your answer.
iii) Is the individual identified in part ii) definitely guilty?
Each person will give rise to two PCR products, one from the
chromosome they inherited from their mother, and one they inherited from their father.
The size of the PCR products will be different if there a different number of repeats. They will be the
same if the number of repeats is the same. If the number of repeats is the same, only one band will
be seen after electrophoresis.
Lane #2: Contains PCR products from two people, the victim and the culprit. Only three bands are seen
because both the victim and the culprit have the same number of repeats on one of their
chromosomes. So it is not possible to determine unambiguously which of the lower bands come from
the culprit. The top band must come from the culprit.
i) Lane 4, or suspect 2, cannot be a suspect since the DNA from this person has different repeats that
the culprit.
ii) Lanes 3 and 5 remain as suspects since the number of repeats they have could be the same as the
culprit. They both have the upper product, and their other band matches the other two bands in lane
2.
iii) No, they are not necessarily guilty because many other people in the population will have the same
repeats as lane 3 or lane 5.
5. (10 pts, 15 min) The smallest PCR product (circled in the above image) was sequenced and its sequence was:
GCTGTAGCCATTGGGCCCTTTAAAGGGCCCTTTAAAGGGCCCTTTAAACCGCGTGAGTCA
i) What is the sequence of the tandem repeat?
ii) How many tandem repeats are present in this DNA?
iii) Give the sequence of the left and right PCR primers that would have been used to produce this PCR product. Assume
that the PCR primers are 8 bases long.
i) GGGCCCTTTAAA
ii ) three
iii) The left primer is the underlined sequence on the left., the right primer is the complement of the
underlined region on the right. The exact primer isn’t important as long as they flank the repeats.
Left primer 5’ CTGTAGCCATT
Right primer 5’ TGACTCACGC
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