ANSWERS TO Exam Questions from Exam 2 – Mutations, Bacterial Genetics, andBacterial Gene Regulation1. Drawn below is part of a wild-type gene.(a)…GCTAAGTATTGCTCAAGATTAGGATGATAAATAACTGG–3’…CGATTCATAACGAGTTCTAATCCTACTATTTATTGACC–5’change TA basepair to CG (for example)(b)…GCTAAGTATTGCTCAAGATTAGGATGATAAATAACTGG–3’…CGATTCATAACGAGTTCTAATCCTACTATTTATTGACC–5’add in a TA basepair (for example)(c) gln-tRNA, glu-tRNA, lys-tRNA, ser-tRNA, leu-tRNA, tyr-tRNA2. You are studying the regulation of a bacterial gene called nytT, which isexpressed only when the bacterial strain is grown in the dark.(a) yes, uninducible, strain 2(b) yes, recessive, strain 4(c) yes, trans, the P1 transduction experiment(d) light --] A B Tlight --] B A Tlight --] A/B TTn-KanralaA+trpA+malA+araA+sucA+ileA+leuA+lacA+trpA+valA+nytB1–alaA+galA+thrA+Tn-Kanr3. After you perform the experiments from Question #2, you decide to continuestudying the regulation of the bacterial gene nytT, which is expressed only when thebacterial strain is grown in the dark.(a)(b) 30%(c) B A(d) Tn B– A–(e) 20%4. You are studying a strain of E. coli whose total genome size is 4,639 kilobasepairs (kbp).Type of cellWhat is thesize of thecircular E. colichromosomein the cell?What is the size oftheextrachromosomalcircle of DNAin the cell?Can hisFbe transferredefficiently,inefficiently,OR never?Can trpDbe transferredefficiently,inefficiently,OR never?An F– bacterial cell4,639 kbp0 kbp(there isn’tone)NeverNeverAn F+ bacterialcell463995NeverNeverAn Hfr cell(named “Hfr A”)resulting fromrecombinationbetweenIS#4 and IS#347340efficientlyInefficientlyA cell resultingfromrecombinationbetweenIS#2 and IS#1 in “Hfr A”4620114EfficientlyEfficiently5. You have isolated a mutation in the Lac I gene; this mutation causes constitutiveLacZYA gene expression.Propose two different explanations for why the amber-suppressing mutant allele of thegene encoding tRNAtrp fails to suppress this particular amber mutation.Answer: Either the efficiency of the tRNA suppressor only creates low levels of the wild-type form of the protein, and such low levels are not sufficient for the protein to functionproperly OR Inserting a tryptophan residue at that position in the protein creates a non-functional protein.Tn-Kanr6. You have isolated an E. coli mutant which you call Lac1–. This mutant cannot growon the sugar lactose as the only carbon source.(a) 80%(b) 1 2(c) Z–(d) P– or IS7. You have identified a new strain of E. coli that can grow on starch.(a) operator, because constitutive cis dominant(b) repressor, because constitutive trans recessive(c) activator, because trans uninducible, recessive(d) D is earlier than B(e) D is earlier than C(f)starch D --] C B8. Below is a diagram of the F factor showing the direction of the origin of transfer (oriT) and an IS2 element carried on this F plasmid.(a)amylaseTn-Kanr(b) TrpA+ PyrD+(c) The F’ plasmid: B+ C+(d) the mutation in PurB gives a dominant phenotype9. The Mot genes of E. coli are required for motility (swimming) of these bacteria.(a) 70%(b) no – the two loci could be unlinked or could be tightly linked and you’d get the sameresult(c) 1 2Tn-Kanr(d) start mot2 mot1 stop(e) an amber nonsense mutation(f) 5’-CTA-3’ 3’-GAT-5’the bottom strand is used as a template for transcription10. Raffinose is a sugar that requires the lactose permease (the LacY gene product)to enter an E. coli cell.(a) I– or OC or I-d(b) I–(c) OC or I-d(d) on the chromosome11. You are studying the regulation of ubiquinone synthesis in bacteria.(a) repressor, because constitutive recessive trans(b) ubiquinone A B --] Ubi1ubiquinone B A --] Ubi1(c) because the double mutant phenotype would be the same for the two models, sincethe two mutations cause the same phenotype(d) ubiquinone B A --] Ubi1(e) BS activates the activity/expression of A regardless of the presence of ubiquinoneTn-Kanr12. You have isolated a Tn5 insertion in an otherwise wild-type E. coli strain that youthink may be linked to the Lac operon.(a) 70%(b) 2 1(c) the second cross(d) KanR 2– 1– would give constitutive expression13. The codon for tryptophan is 5’UGG3’.(a) 5’-CCA-3’(b) 5’-TGG-3’ 3’-ACC-5’(c) 5’-CUA-3’(d) yes(e) amber14. The region of the E. coli chromosome near the Lac operon is diagrammed below:(a) PhoA LacI LacPOZYA ProB | | | |(b) no(c) look for constitutive expression of LacZ in strains grown without lactose(d) six(e) l l l LacI+ LacPOCZ+YA ProB+ PhoA+ LacI+ LacPO+ Z–YA ProB– | | | |(f) an Hfr15. For each of the two following subparts (one for the lac operon and one for themal operon), predict the number of units of enzyme activity that will be displayed by astrain of the given genotype, grown under the given conditions.(a) ß-galactosidase activity–IPTG +IPTGLac O+ Z+ / F' Lac Oc Z– 1 100Lac I+ O+ Z+ Y– / F' Lac I– O+ Z+ Y+ 2 200Lac I+ Oc Z+ / F' Lac I– O+ Z+ 101 200Lac I–d Oc Z+ / F' Lac Is P– O+ Z+ 100 100(b) maltase activity–maltose +maltoseMalT– Q+ / F' MalT+ Q– 1 100MalTC Q+ / F' MalT+ Q– 100 100MalTC Q– / F' MalT– Q+ 100 10016. You are studying the regulation of methanol utilization in bacteria.(a) repressor, because trans, recessive, constitutive(b) because the B2 mutation gives a dominant phenotype(c) repressor(d) B1 causes a loss of function of the B gene. B2 is a superrepressor allele thatencodes a form of B that represses even when methanol is present(e) methanol --] B A --] Moxmethanol --] A B --] Mox(f) because B2 causes a phenotype that is
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