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1. (16 points total) The double-stranded DNA sequence below includes the first 13 codons of wild type human â-globin. Nucleotide numbering is shown above the upper strand., with dots below the lower strand corresponding to the nucleotide number. 10 * 20 30 ̂ 40 50 5' AACAGACACCATGGTGCATCTGACTCCTGAGGAGAAGTCTGCCGTTACTGC 3' 3' TTGTCTGTGGTACCACGTAGACTGAGGACTCCTCTTCAGACGGCAATGACG 5' . . . . . a) Identify the portion of the sequence (from nucleotide number _ to nucleotide number _) containing the coding region for this peptide. (Hint: look for potential start codons.) (4 points) b) Which strand of the DNA (upper or lower) is the coding strand, i.e., carries the information for the peptide (and therefore is copied in the mRNA), and what is the direction of reading the sequence on this page (i.e., left-to-right or right-to-left)? Will the mRNA sequence encoding the peptide differ from the coding strand of the DNA sequence? Why? (3 points) c) Using the Genetic Code Table below, derive the sequence of the peptide encoded by this DNA. Type the amino acid abbreviations above a copy of the DNA sequence. (3 points) d) Beta-thalassemia is a disease caused by mutations in the gene for the hemoglobin â (beta) chain. Would deletion of the T:A base pair under the * in the sequence above (nucleotide 15), along with insertion of an A:T base pair under the ^ symbol (i.e., A inserted between A33 and G34 on the upper strand and T inserted between C34 and T33 on the lower strand), cause Beta-thalassemia? Would â- globin be made in this mutant? Why? (6 points) 2. (16 points) Genes encoding many amber suppressor tRNAs have been made in vitro on DNA synthesizers, by utilizing the known sequences of individual tRNA genes and replacing the naturally occurring anticodon triplet sequences with CTA. When individually introduced into E. coli cells, these synthetic suppressor genes are expressed, and found to be capable of inserting any one of 14 naturally occurring amino acids, respectively, into protein in response to amber codons. It has not been possible to design a gene for a suppressor tRNA capable of inserting into protein any of the other 6 naturally occurring amino acids. a) Suggest an explanation for this observation. (10 points) b) Describe what happens in cells when a synthetic gene derived from a tRNA gene for one of the six remaining amino acids is introduced. Will amber codons be suppressed? If yes, by any particular amino acid? Why? (6 points) 3. (16 points) Puromycin is an analog of the 3' end of aminoacyl-tRNA, so it can enter the A site of the ribosome just like a charged tRNA. If there is an amino acid or peptide attached to a tRNA in the P site, this can be transferred to puromycin, which is then released from the ribosome as peptidyl-puromycin. In a purified system, aminoacylated tRNA can associate with both the A and P sites on the ribosome, while non-aminoacylated (uncharged) tRNA can associate with all three (A, P, and E) sites on the ribosome. An in vitro experiment was performed with purified components in which puromycin was premixed with ribosomes, followed by addition of aminoacylated tRNA. The products of the reaction were analyzed in two ways: (1) formation of peptidyl puromycin was measured and (2) filter binding assays were used to quantitate the amount of tRNA bound per ribosome. It was found that peptidyl puromycin was formed and that one tRNA was bound per ribosome. A second experiment was performed in which puromycin was premixed with ribosomes, but this time non-aminoacylated tRNA was added. Predict the results of this experiment for both assays (peptidyl puromycin formation and filter binding to quantitate amount of tRNA bound). Compare the results of this experiment with the first experiment, and justify your prediction based on the properties of the A, P, and E ribosomal sites. 4. (16 points total) An important method of recombination in E. coli involves mating between male and female cells. When the donor male cell is in the form of an Hfr (High frequency of recombination), mating proceeds from the position of the integrated F factor, with sequential copying and transfer of the genes in the donor cell to the recipient cell, over a time course of up to 100 minutes at 37 degrees, at which point the entire donor chromosome will have been copied and transferred to the recipient cell. This is the reason why genes are mapped as "minutes" on the E. coli chromosome. One of the genes required for biosynthesis of histidine (His) is hisC, which maps at about 44 minutes on the E. coli chromosome. A temperature-sensitive mutation in hisC was found such that the cells can grow at 30 degrees but not 37 degrees (in the absence of exogenous His). Therefore, the cell is a His auxotroph at 37 degrees, but not at 30 degrees. E. coli grow about twice as fast at 37 degrees compared to 30 degrees. An F– strain carrying this mutation is mated with a wild type Hfr strain under varying conditions. For each condition below, will the F– progeny cells be able to grow at 37 degrees without histidine? Explain. (2 points for each for a - e) a) Mating lasted 25 minutes at 37 degrees, with exogenous His in the medium. b) Mating lasted 25 minutes at 37 degrees, without exogenous His in the medium. c) Mating lasted 50 minutes at 37 degrees, with exogenous His in the medium. d) Mating lasted 50 minutes at 37 degrees, without exogenous His in the medium. e) Mating lasted 90 minutes at 30 degrees, without exogenous His in the medium. --- f) Prophage DNA of bacteriophage lambda is integrated in a lysogen between the gal and bio genes, at about 17 minutes on the chromosome. If the prophage was induced with UV light to a lytic cycle, would the progeny phage be expected to transfer (via specialized transduction) the wild type hisC gene to recipient cells carrying the temperature-sensitive hisC allele? Explain. (3 points) g) Wild type E. coli were infected with bacteriophage P1, which does not integrate its DNA into the chromosome. Would the lysate from this infection be able to convert a strain (via generalized transduction) carrying the temperature-sensitive hisC mutation to grow at 37 degrees without His? Explain. (3 points) 5. (18 points) Suppose globin were synthesized starting at the carboxy terminus, proceeding linearly towards the amino terminus to make the complete molecule. What patterns of labeling in tryptic peptides would Dintzis have found when he examined short and long radioactive pulses in both soluble and nascent globin peptides? Explain your reasoning. (Although not required, it would be helpful if you graphed your results for each condition (i.e., short term soluble, long term nascent, etc.) in the form of Y-axis for amount of radioactivity and X-axis for position in the protein chain, from amino terminal on the left to carboxy terminal on the right.) 6. (18 points) An E. coli strain bears amber mutations in both the galactose and leucine operons. (The galactosose operon encodes enzymes that allow the cells to metabolize the sugar galactose. The amber mutation prevents the cells from growing on galactose as a sole carbon source. The leucine operon encodes enzymes that allow the cells to biosynthesize leucine. The amber mutation prevents the cells from growing in the absence of added leucine.) Introduction into the cells of a gene which codes for an isoleucine amber suppressor tRNA (i.e., inserts isoleucine in response to the UAG codon) leads to a phenotype in which the cells no longer require leucine added to the medium for growth, but are still unable to metabolize galactose. By contrast, introduction into the cells of a gene which codes for a tyrosine amber suppressor tRNA leads to a phenotype in which the cells also no longer require added leucine, but now can metabolize galactose as well. How would you explain the different responses to the different amber suppressors?