ANSWERS TO Exam Questions from Exam 2 Mutations Bacterial Genetics and Bacterial Gene Regulation 1 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 tRNA 2 You are studying the regulation of a bacterial gene called nytT which is expressed 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 T light B A T light A B T 3 After you perform the experiments from Question 2 you decide to continue studying the regulation of the bacterial gene nytT which is expressed only when the bacterial strain is grown in the dark a malA galA thrA sucA araA alaA ileA alaA nytB1 valA trpA Tn Kanr trpA b 30 c Tn Kanr B d Tn B A e 20 A lacA leuA 4 You are studying a strain of E coli whose total genome size is 4 639 kilobase pairs kbp Type of cell An F bacterial cell An F bacterial cell An Hfr cell named Hfr A resulting from recombination between IS 4 and IS 3 A cell resulting from recombination between IS 2 and IS 1 in Hfr A What is the What is the size of size of the the circular E coli extrachromosomal chromosome circle of DNA in the cell in the cell Can hisF be transferred efficiently inefficiently OR never Never Can trpD be transferred efficiently inefficiently OR never Never 4 639 kbp 0 kbp there isn t one 4639 95 Never Never 4734 0 efficiently Inefficiently 4620 114 Efficiently Efficiently 5 You have isolated a mutation in the Lac I gene this mutation causes constitutive LacZYA gene expression Propose two different explanations for why the amber suppressing mutant allele of the gene encoding tRNAtrp fails to suppress this particular amber mutation Answer Either the efficiency of the tRNA suppressor only creates low levels of the wildtype form of the protein and such low levels are not sufficient for the protein to function properly OR Inserting a tryptophan residue at that position in the protein creates a nonfunctional protein 6 You have isolated an E coli mutant which you call Lac1 This mutant cannot grow on the sugar lactose as the only carbon source a 80 b Tn Kanr 1 2 c Z S d P or I 7 You have identified a new strain of E coli that can grow on starch a b c d e f operator because constitutive cis dominant repressor because constitutive trans recessive activator because trans uninducible recessive D is earlier than B D is earlier than C amylase starch D C B 8 Below is a diagram of the F factor showing the direction of the origin of transfer ori T and an IS2 element carried on this F plasmid a b TrpA PyrD c The F plasmid B C d the mutation in PurB gives a dominant phenotype 9 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 same result c Tn Kanr 1 2 Tn 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 transcription 10 Raffinose is a sugar that requires the lactose permease the LacY gene product to enter an E coli cell d a I or OC or I b I d c OC or I d on the chromosome 11 You are studying the regulation of ubiquinone synthesis in bacteria a repressor because constitutive recessive trans b ubiquinone A B Ubi1 ubiquinone B A Ubi1 c because the double mutant phenotype would be the same for the two models since the two mutations cause the same phenotype d ubiquinone B A Ubi1 e BS activates the activity expression of A regardless of the presence of ubiquinone 12 You have isolated a Tn5 insertion in an otherwise wild type E coli strain that you think may be linked to the Lac operon a 70 b Tn Kanr 2 1 c the second cross d KanR 2 1 would give constitutive expression 13 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 amber 14 The region of the E coli chromosome near the Lac operon is diagrammed below a PhoA b no LacI LacPOZYA ProB c look for constitutive expression of LacZ in strains grown without lactose d six e PhoA l l LacI LacPOCZ YA l ProB LacPO Z YA ProB LacI f an Hfr 15 For each of the two following subparts one for the lac operon and one for the mal operon predict the number of units of enzyme activity that will be displayed by a strain of the given genotype grown under the given conditions a galactosidase activity IPTG IPTG Lac O Z F Lac Oc Z 1 100 Lac I O Z Y F Lac I O Z Y 2 200 Lac I Oc Z F Lac I O Z 101 200 Lac I d Oc Z F Lac Is P O Z 100 100 b maltase activity maltose maltose MalT Q F MalT Q 1 100 MalTC Q F MalT Q 100 100 MalTC Q F MalT Q 100 100 16 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 that encodes a form of B that represses even when methanol is present e methanol B A Mox methanol A B Mox f because B2 causes a phenotype that is distinguishable from the phenotype caused by the A mutation g methanol A B Mox 17 Phage T4 expresses an enzyme lysozyme which enables the phage to lyse infected bacterial cells a 1 m u b 120 000 base pairs in length c one possibility CGA arginine 5 CGA 3 3 GCT 5 the bottom strand is used as a template in transcription the other possibility UGG tryptophan 5 TGG 3 3 ACC 5 the bottom strand is used as a template in transcription 18 You have isolated a Tn5 insertion in an otherwise wild type E coli strain this transposon is near to but not within the group of lac genes on the E coli chromosome a 60 b Number of transuctants galactosidase permease Genotype 578 uninducible regulated Z Y I 400 constitutive constitutive Z Y I 20 uninducible constitutive Z Y I regulated regulated Z Y I 2 c 58 d Tn Kanr Z I 19 An enzyme that you are interested in from E coli is regulated by the following scheme a inducer A B …
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