Download 03-131 Genes, Drugs, and Disease Problem Set

03-131 Genes, Drugs, and Disease
Problem Set 6 - Bonus
1. (10 pts) Some individuals cannot produce sufficient quantities of growth hormone which causes a number of problems
during development. You want to insert the coding information for human growth hormone into a plasmid so that
you can produce the growth hormone in bacteria. The DNA sequence (one strand only) and the protein sequence of
the growth hormone are shown below. Give the sequence of the left and right primers that you would use to
generate a PCR product that could be inserted into the plasmid using restriction enzymes. Assume that the total
length of your primers are 12 bases. Also assume that you would use the same plasmid that we have used for HIV
protease, i.e. the final PCR product should have an EcoR1 site on the left and a BamHI site on the right.
GGTGTGATGTTGGTTGTTGATCGTTACGCTTTTAGCGGTGTGTAATTCAC
MetLeuValValAspArgTyrAlaPheSerGlyVal
The start codon is in bold and the stop codon is underlined. Note that the plasmid already has a start and stop codon,
nd
so it is not necessary to include those in the PCR product. You just need to amplify the 2 codon (Leu) to the last one
(Val).
The amplified region is highlighted in yellow. If you wanted to just amplify this sequence, your PCR
primers would be:
LP: 5’ TTGGTT
RP: 5’CTCACC.
HOWEVER, we need to add restriction sites to the ends so that we can make sticky ends, EcoR1 on the
left and BamH1 on the right. These sequences are added to the 5’ end of our primers, so the complete
primers are:
LP: 5’ GAATTCTTGGTT RP: 5’GGATCCCTCACC.
(you will be given the restriction sites on the exam, no need to memorize them).
2. (5 pts) The recognition sequences for TaqI is T^CGA and for ClaI is AT^CGAT
i) The following DNA was cut (digested) with ClaI. How many pieces (fragments) of DNA would be produced and what
are their sizes (The lengths of the top and bottom strand will be different, so just give the size of the top strand).
TTGGATCGTATGCGGATCGATGTATAAT
AACCTAGCATACGCCTAGCTACATATTA
There is a single ClaI site (highlighted). The enzyme will cleave both strands between the T and the C,
thus two fragments would be produced:
TTGGATCGTATGCGGAT
CGATGTATAAT
AACCTAGCATACGCCTAGC
TACATATTA
ii) Fragments that are produced by one enzyme can always be ligated to fragments produced by the same enzyme.
Can fragments that are produced by TaqI be ligated to fragments produced by ClaI? Why or why not?
If the following DNA was treated with TaqI, the products are shown on the right:
-TCGA-T
CGA-AGCT-AGC
TThese fragments have exactly the same sticky ends (highlighted) as the fragments made by ClaI
digestion. Therefore they could be ligated together.
3. (10 pts) Hemophiliacs have a genetic deficiency such that they cannot make a protein required for blood clotting. This
protein can be purified from cows and administered to hemophiliacs.
i) This treatment works well initially, but eventually the treatment is ineffective. Suggest a possible reason for this.
[Hint: The amino acid sequence of the clotting protein from cows is different than humans].
The immune system of the patient will consider the injected protein as foreign and generate
antibodies against it, which will result in removal/destruction of the clotting factor.
ii) You could solve the above problem by obtaining the clotting protein from donated human blood, but that is very
expensive process and the supply is limited. Consequently you want to express the protein in bacteria. You will
use the same expression plasmid that we discussed in class, but instead of using PCR to obtain the codons from
human DNA you will generate the DNA using chemical synthesis. The amino acid sequence of the clotting factor is:
Met-Arg-Leu-Phe-Trp
You would have to synthesize a DNA sequence that coded for this amino acid sequence. Because there
are more than one codon for many of the amino acids, the DNA sequence will not be unique (however
the protein sequence would be). One possibility is:
ATGCGTCTTTTTTGG
4. (5 pts) Why is it not useful to vaccinate someone against HIV by the injection of HIV virus? The virus is lethal – you
would be giving them AIDs.
03-131 Genes, Drugs, and Disease
Problem Set 6 - Bonus
5. (5 pts) Suggest a way to produce an HIV vaccine using the methods that we have been recently discussing in class. You
can assume that the entire RNA sequence of HIV is known and that the sequences that code for the coat proteins are
known. Coat proteins are on the surface of the virus and would be recognized by antibodies that would be produced
after vaccination.
You could somehow kill the virus and then inject that – however there is no guarantee that you will kill
all of the virus, so you could still infect someone. A better way, which would NOT involve exposing
people to the virus, is to produce the coat protein in bacteria – in the same way we have done for HIV
protease. The only difference would be that you would use different PCR primers that would amplify
the coding information for the coat protein.
6. (5 pts) Assuming you were successful in producing a vaccine against HIV, what property of the viral lifecycle would likely
make the vaccine ineffective?←
Although antibodies would be generated that recognize the coat protein, the virus would develop
mutations in the coat protein and the antibodies would no longer bind.
7. (5 pts) You would like to sequence mutations in the HIV protease gene and you believe that the mutations are located
within the first 20 amino acids. What sequencing primer could you use? Give an example sequence.
The sequencing primer needs to be “upstream”, or towards the 5’ end, of the region that you want to
sequence. The beginning part of the protease sequence is (from lecture 16):
5'-ggagccgatagacaaggaactgtatcctttaacttcCCTCAGATCACTCTTTGGCAA57
ProGlnIleThrLeuTrpGln7
Any sequence in the highlighted region would be fine, i.e. TAACTTC. In this case the sequence data
would begin with the first base of the first codon (C).
8. (5 pts) You order the DNA primer for sequencing, but discover that you have sent the company the sequence of the
complementary strand, can you still use it to sequence your HIV protease mutant? Why or why not?
The primer would not work because it would produce sequence data from the opposite strand in the
other direction:
5'-ggagccgatagacaaggaactgtatcctttaacttcCCTCAGATCACTCTTTGGCAA57
cctcggctatctgttccttgacataggaaattgaacGGAGTCTAGTGAGAAACCGTT
ProGlnIleThrLeuTrpGln7
←
The primer is 5’ caagtta, and DNA polymerase would add bases to the 3’ end, the first base to add
would be an A, and the sequence would be aaggatac……, i.e. away from the region that we wanted to
sequence.
9. (4 pts) You are doing DNA sequencing and have run out of ddTTP. Can you obtain the sequence
using just the other three dideoxybases? If so, how would the data change?
Yes, if you know the location of three of the bases, you can infer the fourth.
Fragments that would have ended in ddT will be absent from your data, since the
polymerase would have put in a dTTP when there was an A in the template. The
change in the data is shown to the right. In the case of the +ddTTP data (top),
the first four peaks are due to the production of the following fragments by DNA
polymerase:
Primer-C3’H
Primer-CT3’H
Primer-CTG3’H
Primer-CTGC3’H
When ddTTP is missing, the second fragment cannot be made, and will be absent
from the data.
03-131 Genes, Drugs, and Disease
Problem Set 6 - Bonus
10. The DNA sequence data for a wild-type protein and a mutant are presented below. Translation of the wild-type
sequence by Expasy is also shown below (a “-“ is a stop codon).
i) What is the correct reading frame? The reading frame without stop codons is likely correct (frame 1) ,
unless you knew that the sequence data was at the end of the coding region.
ii) What is the mutation?
1
123 456 789 012 345 6789
ctg cgc ttc ccg gag cgtagcaccgagatcggtaagctgctgagcagctacttgcagaa
L
R
F
P
E
R
The G at position 3 in the wildtype has been changed to an A, this changes the codon to AAG, coding
for Lysine (Lys, K).
5'3' Frame 1
ctgcgcttcccggagcgtagcaccgagatcggtaagctgctgagcagctacttgcagaa
L R F P E R S T E I G K L L S S Y L Q
5'3' Frame 2
ctgcgcttcccggagcgtagcaccgagatcggtaagctgctgagcagctacttgcagaa
C A S R S V A P R S V S C - A A T C R
5'3' Frame 3
ctgcgcttcccggagcgtagcaccgagatcggtaagctgctgagcagctacttgcagaa
A L P G A - H R D R - A A E Q L L A E