ANSWERS TO Problem set questions from Exam 2 Unit Mutations Bacterial Genetics and Bacterial Gene Regulation Central Dogma Mutagens and Mutations 1 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 template 3 GTT 5 c mutant is 5 TAA 3 this strand is used as a template 3 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 the original reading frame so that all amino acids will be as they were originally intended following the second frameshift b There must be an out of frame stop codon that lies in between where the 1 frameshift is placed and where the 1 frameshift is placed This out of frame stop codon 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 0907 if AT is 40 p no stop for 50 codons 0 968 50 0 179 3 One way to isolate nonsense suppressor mutations in tRNA genes is to select for the simultaneous suppression of the mutant phenotype of a strain carrying nonsense mutations in two different genes Explain why it would be a bad idea to start with an original strain that has an amber mutation 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 and TAA 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 tRNA genes in the same strain such that one tRNA gene became a TAG nonsense suppressor and the other tRNA gene became a TAA nonsense suppressor Such a double 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 2 frameshift is placed and where the 2 frameshift is placed This out of frame stop codon must be put into its reading frame by the 2 frameshift but not by the 1 frameshift d p no stop for 15 codons 61 64 46 0 11 5 You are trying to isolate mutations in the gene encoding tRNAtrp that will produce mutant forms of this tRNA that will recognize a stop codon as opposed to the normal trp codon 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 UAG Transposons and Cotransduction Mapping in bacteria Moving DNA between bacterial cells 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 F2 2 The E coli ser1 gene is required for synthesis of the amino acid serine and strains harboring mutations in this gene will not grow unless serine is provided in the growth medium a There will still be a stop codon within ser1 which will still stop translation of the Ser1 protein even if a second mutation is acquired within the ser1 gene The only kind of mutation that could override a stop codon is a frameshift that puts the stop codon out of frame but a frameshift also changes the frame of the entire rest of the protein after the stop codon b 40 c it is extragenic because it is unlinked d it could be intragenic e yes 3 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 b c d true true true false 4 You have used mutagenesis with the chemical EMS to isolate four different E coli mutants that will not grow unless the amino acid histidine is provided in the growth medium a That colony arose from a single cell that received a piece of bacterial DNA from a P1 phage 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 linked 5 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 motility phenotype 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 are linked to the insertion but neither are linked within 100 kb of each other d Tn5 PhoS 1 2 F plasmids Hfrs F plasmids Moving DNA between bacterial cells by conjugation using mating 1 The region of the E coli chromosome surrounding the Lac operon contains the markers PhoA Lac I LacZ Y A ProB in that order a the F plasmid Lac I O Z Y A b early c recombination event between the chromosome and the F plasmid PhoA Lac I OC Z Y A ProB Lac I O Z Y A d complementation tests cis trans tests 2 The diagram below shows the F factor plasmid and a portion of the E coli chromosome that contains three different insertion sequences IS of the same type as that which is carried on the F plasmid a 1st Hfr A B C D 2nd Hfr A B C D 3rd Hfr A B C D b 1st Hfr transfers C then D early 2nd Hfr transfers only A early 3rd Hfr transfers C then B then A early c 1st Hfr forms this F 2nd Hfr forms this F 3rd Hfr forms this F B C B C B C d 1st Hfr s F transfers B and C 2nd Hfr s F transfers B and C 3rd Hfr s F transfers B and C 3 Transposons are not only useful as portable genetic markers they can also serve as portable regions of homology for recombination a The F recombining with the chromosome due to homology between the Tn5 insertions Tn5 S Tn5 Mot1 …
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