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Transcript
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