ANSWERS TO Problem set questions from Exam 2 Unit – Mutations, Bacterial Genetics,and Bacterial Gene RegulationCentral Dogma, Mutagens and Mutations1. The three stop codons in the genetic code are 5’UAG3’, 5’UAA3’, and 5’UGA3’.(a) gln-tRNA(b) gln-tRNA: 5’-CAA-3’ this strand is used as a template3’-GTT-5’(c) mutant is: 5’-TAA-3’ this strand is used as a template3’-ATT-5’2. You are studying an E. coli gene that encodes an enzyme of interest to you.(a) The +1 frameshift sets the amber mutation out of frame so that the stop is not read,and the –1 frameshift after the amber mutation returns the reading frame back to theoriginal reading frame, so that all amino acids will be as they were originally intendedfollowing the second frameshift.(b) There must be an out-of-frame stop codon that lies in between where the –1frameshift is placed and where the +1 frameshift is placed. This out-of-frame stopcodon must be put into its reading frame by the –1 frameshift.(c) if AT and GC equal, p(no stop for 50 codons) = (61/64)^50 = 0.0907if AT is 40%, p(no stop for 50 codons) = (0.968)^50 = 0.1793. One way to isolate nonsense suppressor mutations in tRNA genes is to select forthe simultaneous suppression of the mutant phenotype of a strain carrying nonsensemutations in two different genes.Explain why it would be a bad idea to start with an original strain that has an ambermutation (TAG) in the his1 gene and an ochre mutation (TAA) in the his2 gene.You cannot isolate a nonsense suppressor mutation that will recognize both TAG andTAA, and you would need such a nonsense suppressor mutation to see a His+phenotype from a strain that has a TAG mutation in his1 and a TAA mutation in his2.The only way you could ever see His+ is if you got mutations in two different tRNAgenes in the same strain, such that one tRNA gene became a TAG nonsensesuppressor and the other tRNA gene became a TAA nonsense suppressor. Such adouble mutational event is highly unlikely.4. Consider a phage gene that encodes the enzyme lysozyme.(a) 1: missense, 2: frameshift or nonsense(b) 12/1000 would be wild-type progeny(c) There must be an out-of-frame stop codon that lies in between where the –2frameshift is placed and where the +2 frameshift is placed. This out-of-frame stopcodon must be put into its reading frame by the +2 frameshift (but not by the +1frameshift).(d) p (no stop for 15 codons) = (61/64)^46 = 0.115. You are trying to isolate mutations in the gene encoding tRNAtrp that will producemutant forms of this tRNA that will recognize a stop codon, as opposed to the normal trpcodon.(a) 5’-CCA-3’(b) 5’-CCA-3’ 3’-GGT-5’(c) 5’-TCA-3’ 3’-AGT-5’5’-CTA-3’3’-GAT-5’5’-TTA-3’3’-AAT-5’(d) 5’-UCA-3’ 5’-CUA-3’ 5’-UUA-3’(e) UGA and UAGTransposons and Cotransduction Mapping in bacteria (Moving DNA between bacterialcells by transduction [using phage])1. You have isolated two E. coli mutants in the PyrF gene, called PyrF-1 and PyrF-2.(a) it must be near to PyrF (within 100 kb)(b) 30%(c) Tn5 F1 F22. The E. coli ser1 gene is required for synthesis of the amino acid serine, and strainsharboring mutations in this gene will not grow unless serine is provided in the growthmedium.(a) There will still be a stop codon within ser1, which will still stop translation of the Ser1protein, even if a second mutation is acquired within the ser1 gene. The only kind ofmutation that could override a stop codon is a frameshift that puts the stop codon out offrame, but a frameshift also changes the frame of the entire rest of the protein after thestop codon.(b) 40%(c) it is extragenic because it is unlinked(d) it could be intragenic(e) yes3. In a transduction experiment, phage P1 is grown on a bacterial host of genotype A+B+ C+ and the resulting lysate is used to infect a recipient strain of genotype A– B– C–.(a) true(b) true(c) true(d) false4. You have used mutagenesis with the chemical EMS to isolate four different E. colimutants that will not grow unless the amino acid histidine is provided in the growthmedium.(a) That colony arose from a single cell that received a piece of bacterial DNA from a P1phage that contained a Tn5 insertion that was linked to the his1 locus.(b) 80%(c) they are not linked, so they must be alleles of different genes(d) make sure they are both recessive and then do a complementation test(e) 1 and 4 are linked to each other because they both are linked to the Tn5 insertion(f) Tn5 4 1(g) you can’t conclude either way just because they are linked5. Wild-type E. coli have flagella that allow them to swim towards nutrient sources.(a) because you cannot perform a selection for cells that receive mot DNA – the motilityphenotype (or the inability to be mobile) cannot be selected for(b) 60%(c) PhoS and mot1 must be on different sides of the Tn5 insertion, such that both arelinked to the insertion, but neither are linked (within 100 kb) of each other(d) Tn5 PhoS 1 2F plasmids, Hfrs, F’ plasmids (Moving DNA between bacterial cells by conjugation [usingmating])1. The region of the E. coli chromosome surrounding the Lac operon contains themarkers PhoA — Lac I – LacZ,Y,A — ProB, in that order.(a) the F’ plasmid:(b) early(c) recombination event between the chromosome and the F’ plasmid:(d) complementation tests, cis/trans testsLac I+ O+ Z--- Y+ A+Lac I+ O+ Z--- Y+ A+Lac I+ OC Z+ Y+ A+PhoAProB2. The diagram below shows the F factor plasmid, and a portion of the E. colichromosome that contains three different insertion sequences (IS) of the same type asthat which is carried on the F plasmid.(a) 1st Hfr: A B C D2nd Hfr: A B C D3rd Hfr: A B C D(b) 1st Hfr transfers C then D early2nd Hfr transfers only A early3rd Hfr transfers C then B then A early(c) 1st Hfr forms this F’: 2nd Hfr forms this F’: 3rd Hfr forms this F’:CBBCBC(d) 1st Hfr’s F’ transfers B and C2nd Hfr’s F’ transfers B
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