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5). (14 Points). (A genetic code dictionary is on the last page of the exam. You will probably need to use it). You want to make a collection of amber mutants of phage PhiX184. You have at your disposal an E. coli strain that lacks any nonsense suppressors (supo) and a variety of strains that carry known amber suppressors supE (Gln), supD(Ser) and supF(Tyr). A). Using a simple genetic assay how could you isolate PhiX174 mutants that contain amber mutations suppressed by supE? (6 Points). [Mutagenize the phage with a mutagen that causes transitions and plate with the supE strain as the indicator. After plaques form, test them for growth on both the supo and supE strain. Amber mutants suppressed by supE will form plaques on E. coli supE but not E. coli supo.] Assume you successfully isolated amber mutants in the experiment above. You find that the amber mutants are suppressed by supE at both 30o and 42o C. You test the ability of other suppressors to suppress the amber mutants. The suppression tests are done at 30o and 42o C and the results shown in the table below are shown. 30° Mutant 1 2 3 supD + + 42° supF + supD + + supF - B). For each of the 3 amber mutants, explain the results with the suppressor-containing strains. (6 Points). [Mutant 1- It is not suppressed by supD or supF at any temperature. Therefore, Ser and Tyr are not acceptable amino acid residues at the position of the nonsense codon in the protein. (Gln is acceptable because it forms plaques on the supE strain). Mutant 2- It is suppressed by supD but not supF at both temperatures. Therefore the protein is active at either temperature with either Gln or Ser at the position of the nonsense codon in the protein. Tyr is not an acceptable amino acid for function.] Mutant 3- It is suppressed by supD at both temperatures so Gln and Ser are acceptable amino acids at both temperatures. The pattern with supF is different. Insertion of Tyr makes the protein temperature sensitive.] As we discussed in class, 5BU causes transitions and HA is specific for the CG to TA transition. You decide to isolate revertants of the original amber mutant #1. If you used 5BU as the mutagen, would you expect to isolate mutagen-induced revertants (assume only 1 mutation per phage chromosome)? If so what amino acid residues would be found at the position of the amber codon in revertant proteins? Would HA give you revertants? (2 Points). [5BU causes bidirectional transitions so the UAG codon would be substituted by CUG (Leu), UGG (Trp) and UAA (Ochre). No. Only UAA codons could be generated by HA].