The starting substrate and active site of a Type I topoisomerase is shown below. During this reaction, a small molecule is introduced that removes free hydroxyl groups from DNA (but not protein). Please draw the resulting product under these conditions, including the arrow pushing mechanisms that lead to the product(s). Tyr- s' CH₂ DNA Base O 01P=0 H Base
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- Step 1 Lys Lys Ligase-AMP NH2 NH2 NH2 `NH2 PP АТР он он OH Step 2 Lys Lys NH2 NH2 NH2 NH2 DNA- OH OH Adenylate OH OH HO. Step 3 NH2 NH2 NH2 NH2 AMP Phospho- diester OH OH OH OH OH Describe the mechanism shown above for DNA Ligase. Describe the chemistry of each step How the enzyme appears or might facilitate the chemistry How the enzyme increases the reaction rate.Ultraviolet light can cause covalent linkages between consecutive pyrimidine bases in DNA (up to 100 per second in a single cell in sunlight!). These bulky lesions (i.e. cyclobutane pyrimidine dimers and 6-4 photoproducts), which inhibit DNA and RNA polymerases, are mostly reversed by CPD photolyase when light >300 nm is available to power the reaction. In the dark, however, which DNA repair system is best able to correct these errors? a) non-homologous end-joining b) mismatch repairc) nucleotide-excision repaird) base-excision repair e) homology-directed repair. Propose a mechanism by which a type II topoisomerase could use the energy of ATP hydrolysis to scan a large DNA molecule and, thereby, to direct that the enzyme will catalyze largely “disentan- gling" reactions (decatenation and unknotting).
- Original sequence: Consider the following coding 71 nucleotide DNA template sequence (It does not contain a translational start): 5’-GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3’ Question: 4) In a mutant you discovered that the underlined nucleotide has been deleted. What would the resulting peptide sequence be? What type of mutation is this? 5’-GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3In the Inhibition of telomerase activity. Explain: (a) What is the process affected? (b) What is the Effect on the process? (c) Does it affect prokaryotes, eukaryotes or both?Suggest a reasonable strategy for the specific phosphorylation of the5’ –OH group of a nucleoside.
- Explain why the active site of poly(A) polymerase is much narrower than that of DNA and RNA polymerases.Indicate the biochemical activities for the enzymes listed below. Use the lettered list of activities to mark the activities of the enzymes involved in DNA replication and transcription. E. coli DNA polymerase I Human telomerase E. coli RNA polymerase holoenzyme yeast RNA polymerase || E. coli primase A. 5' to 3' RNA-dependent DNA polymerase B. 5' to 3' DNA-dependent DNA polymerase C. 3' to 5' RNA-dependent DNA polymerase D. 3' to 5' DNA-dependent DNA polymerase E. 5' to 3' DNA-dependent RNA polymerase F. 3' to 5' DNA-dependent RNA polymerase G. 5' to 3' exonuclease H. 3' to 5' exonucleaseRestriction sites of Lambda (A) DNA - In base pairs (bp) The sites at which each of the 3 different enzymes will cut the same strand of lambda DNA are shown in the maps (see figure 3 B-D), each vertical line on the map is where the respective enzymes will cut. A DNA A (bp) 48502 10 000 20 000 30 000 40 000 9162 17 198 B Sal I 7059 14 885 28 338 35 603 42 900 (bp) Hae III 11 826 21 935 29 341 38 016 (bp) 11648 29,624 Eco R1 (bp) 10 592 16 246 28 915 41 864 Figure 3: Restrictrion site map showing the following A) inear DNA that is not cut as reference B) DNA CLt with Sal L C) DNA cut with Hae , D) DNA cut with Eco RI 1. Calculate the size of the resulting fragments as they will occur after digestion and write the sizes on the maps below. Note that linear DNA has a total size of 48 502 bp (see figure 3A). Page 3 of 7 9162 17 198 Sal i (bp) 7059 14 885 28 338 35 603 42 900 Hae I (bp) 11 826 21 935 29 341 38 016 11648 29,624 Eco R1 (bp) 10 592 16 246 28 915 41 864
- Telomerase supplies its own template RNA molecule as shown in Figure 3 below: B AAUCCCAAU TTAGGGTTAGGGTTAGGGTTAGGGTTAGGGTTAGGG-W' JAATCCCAATCCCAATCCCAA-X' Figure 3 (i) Label the ends (5' and 3') on the DNA and RNA strand at position X, Y and Z, in the Figure 3. (ii) Draw and explain the two loop structures at the end of telomere.gene. If the JM109 strain is transformed by the PBKSK plasmid, the strain will produce the B-galactosidase (from the lac gene) and will hydrolyze X-gal to produce the blue compound. Therefore, colonies that were transformed and contain the pBSKS wil you appear blue. IPTG & X-Gal & NO colonies Amp E. coli JM109 E. coli JM109 50 mM calcium chloride-15% glycerol lac lac lac IPTG & I Recovery X-Gal solution at -702C PBSKS White colonies E. coli JM109 E. coli JM109 ampR amp I amp lac lac Heat Shock Non-transformed 42°C E. coli JM109 E. coli JM109 amps amps lac lac IPTG & X-Gal lac I Recovery lac PBSKS BLUE colonies PBSKS ampRI (amp Transformed IPTG & X-Gal & BLUE colonies Amp Hypotheses: Circle the correct answer 1. If PBSKS is transformed into JM109 cells, colonies will be (able/not able) to grow in the presence of ampicillin. a. Why? _ 2. If PBSKS is transformed into JM109 cells, colonies in media with IPTG (will/will not) induce the production the B- galactosidase enzyme. a. Why?_ 3. If…Preparing plasmid DNA (double stranded, circular) for Sanger sequencing involves annealing a complementary, single-stranded oligonucleotide DNA primer to one strand of the plasmid template. This is routinely accomplished by heating the plasmid DNA and primer to 90°C and then slowly bringing the temperature down to 25°C. Why does this protocol work? What enzyme is used and what other components are required in the sequencing reaction? How does the Sanger method determine the sequence?