MCB421 FALL2005 EXAM#3 ANSWERS Page 1 of 12. ANSWER: Both transposon types form small duplications of adjacent host DNA sequences.

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1 Page 1 of 12 (10pts) 1. There are two mechanisms for transposition used by bacterial transposable elements: replicative (Tn3) and non-replicative (Tn5 and Tn10). Compare and contrast the two mechanisms with respect to: (2pts each) a. Host DNA sequences adjacent to the ends of the element: ANSWER: Both transposon types form small duplications of adjacent host DNA sequences. b. Formation of co-integrants: ANSWER: Replicative transposons form co-integrants. Non-replicative transposons do not form co-integrants. c. Resolvase enzyme: ANSWER: Used by replicative transposons but not by non-replicative transposons. d. Requirement for DNA synthesis: ANSWER: Both use DNA polymerase I to fill in gaps caused by transposition. e. Structure of DNA in intermediates of the reaction: ANSWER: Non-replicative transposons form excised loops bound by transposase. Replicative transposons form co-integrant structures. (10pts) 2. Many different mutations have been isolated in the puta gene of Salmonella enterica. A few of these mutations are shown on the following deletion map of the puta structural gene. The promoter is at the left, pointing rightward into the puta gene.

2 Page 2 of 12 Based on the relative location of the mutations and the results in the table below, explain the phenotypes observed for each of the single and double mutants shown. [The puta1992::mudj and puta1993::mudk insertions map in the same deletion interval.] (1pt each +1 freebie) Mutations LacZ expression Explanation - proline + proline puta1991::mudj x transcription puta1992::mudj x transcription puta1993::mudk x transcription and translation, therefore translation puta1991::mudj puta2096::tn10 puta1992::mudj puta2096::tn10 puta1993::mudk puta2096::tn10 puta1991::mudj puta2097::tn10 puta1992::mudj puta2097::tn10 puta1993::mudk puta2097::tn10 <2 <2 polar <2 <2 polar <2 <2 polar non-polar <2 <2 polar <2 <2 polar (30pts) 3. Salmonella Typhimurium switches between two antigenic types of flagella by a process called phase variation. The following figure summarizes the initial evidence that this process occurs by an inversion (Zieg et al Science 196: ). (5pts) a. The hin (H-inversion) gene and promoter for the downstream fljb gene are located between two hix sites where site-specific recombination occurs. Given the following restriction map, what results would you expect if the region between the hix sites was inverted:

3 Page 3 of 12 ANSWER: The distance between the Hix sites will remain the same, but the distance between the PstI and ClaI sites will now be 1.6 Kb. (5pts) b. Given the DNA sequence of this region, how would you test for inversion using a PCR assay? [Draw a diagram on the cartoon below showing where the PCR primers would hybridize and what products you would expect.] ANSWER: Primers sets 1&2 and 3&4 will give PCR products if the hix sites are as shown in the figure. Primer sets 1&3 and 2&4 will give PCR products if an inversion occurs. (5pts) c. The DNA sequence of the Salmonella Typhimurium hin gene is shown below. (Only one strand of the sequence is shown for simplicity.) 5'- AAA ATC ATG GCT ACT ATT GGG TAT ATT CGG GTG TCA ACA ATT GAC CAA AAT ATC GAT TTA CAG CGT AAT GCG CTT ACT AGT GCA AAT TGT GAC CGC ATT TTT GAG GAC CGT ATC AGT GGC AAG ATT GCA AAC CGC CCC GGC CTG AAA CGA GCG TTA AAG TAT GTA AAT AAA GGC GAT ACT CTT GTC GTC TGG AAA TTA GAC AGA CTG GGC CGC AGC GTG AAA AAC CTG GTG GCG TTA ATA TCA GAA TTA CAT GAA CGT GGA GCT CAC TTC CAT TCT TTA ACC GAT AGT ATT GAT ACC AGT AGC GCG ATG GGG CGA TTC TTT TTT CAT GTA ATG TCA

4 Page 4 of 12 GCA CTG GCC GAG ATG GAG CGA GAA TTA ATT GTC GAG CGA ACC CTT GCC GGA CTG GCT GCC GCC AGA GCG CAA GGA CGA CTG GGA GGG CGC CCT CGG GCG ATC AAC AAA CAT GAA CAG GAA CAG ATT AGT CGG CTA TTA GAG AAA GGC CAT CCT CGG CAG CAA CTA GCT ATT ATT TTT GGT ATT GGC GTA TCT ACC TTA TAC AGA TAT TTT CCG GCA AGC CGC ATA AAA AAA CGA ATG AAT TAA-3 Given a clone of this sequence in any of the vectors discussed in class, how could you do site directed mutagenesis of this sequence to disrupt the unique ClaI restriction site (boxed)? (5pts) i. Indicate the sequence of the oligonucleotide(s) you would use (clearly labeling the 5 and 3 ends): ANSWER: 5 -T GAC CAA AAT AAA AAA TTA CAG CGT A-3 -There are several different answers to this. -Must have at least 10bp upstream and downstream of the desired mutation. -Must alter the 6bp ClaI site. (5pts) ii. Draw a diagram showing the enrichment you would use to eliminate non-mutant plasmids: ANSWER: The diagram for Quikchange Mutagenesis (Stratagene) is shown below. The dut/ung method and MutS methods are also accepted.

5 Page 5 of 12 (5pts) iii. Give a brief explanation of how you would test for the desired mutant: ANSWER: Prep the DNA and try cutting with ClaI. The correct mutation will not give any cleavage products. Use the template DNA as a positive control to ensure the enzyme is working. Finally, sequence the mutant to ensure there are no secondary mutations. (5pts) iv. How your mutation would alter the amino acid sequence of the resulting Hin protein? (A copy of the genetic code is on the last page of the exam.)

6 Page 6 of 12 ANSWER: Original protein sequence: KIMATIGYIRVSTIDQNIDLQRNALTSANCDRIFEDRISGKIANRPGLKR ALKYVNKGDTLVVWKLDRLGRSVKNLVALISELHERGAHFHSLTDSIDTS SAMGRFFFHVMSALAEMERELIVERTLAGLAAARAQGRLGGRPRAINKHE QEQISRLLEKGHPRQQLAIIFGIGVSTLYRYFPASRIKKRMN* Mutagenized protein sequence: KIMATIGYIRVSTIDQNKKLQRNALTSANCDRIFEDRISGKIANRPGLKR ALKYVNKGDTLVVWKLDRLGRSVKNLVALISELHERGAHFHSLTDSIDTS SAMGRFFFHVMSALAEMERELIVERTLAGLAAARAQGRLGGRPRAINKHE QEQISRLLEKGHPRQQLAIIFGIGVSTLYRYFPASRIKKRMN* (10pts) 4. Given a puta point mutant and a P22 generalized transducing lysate grown on a random pool of Tn10dCam (chloramphenicol resistant) transposon insertions: (Tn10dcam is a derivative of Tn10 that has the cat gene encoding resistance to chloramphenicol.) a. How could you isolate a Tn10dCam near (but not in) the puta - gene? [Hint: this will require several steps.] Draw a diagram showing how you would do the experiment and indicate the medium you would use for each selection or screen. The phenotypes of the WT and puta mutant are shown in the table below, where + indicates growth and - indicates no growth. Min + Glucose + NH 4 Min + Glucose + Proline puta puta ANSWER: 1) Isolate a Tn10dCam linked to the puta point mutation: Random Pool Tn10 puta + puta - X

7 Page 7 of 12 Select Cam R put + on Min + Glucose + Cam + Proline medium. puta + Tn10 2) Make P22 HT lysate of above strain and transduce into Cam S puta -. Select on Min + Glucose + Cam + NH 4 puta - will grow Replica plate onto Min + Glucose + Cam + Proline puta - will not grow Tn10 puta + X 3) Backcross to recipient strain and select as described above. All should be puta -. b. How could you use a Tn10dCam insertion linked to the puta + gene to isolate point mutations in the puta gene? [Draw a figure indicating the donor and recipient and any selection or screen you would use.] ANSWER: Localized mutagenesis of phage, transduce into Cam R, screen puta Chromosomal duplications of the nadc gene were constructed in S. typhimurium. The resulting merodiploids have one mutant nadc allele in copy #1 and a different mutant nadc allele in copy #2 with a mini-mud at the join point, as shown in the figure below. (2pts) a. How could you select for maintenance of this chromosomal duplication? ANSWER: Maintain the duplication in medium with Amp. (8pts) b. If one of the copies of nadc had a Tn10dCam insertion mutation (i.e. nadc::tn10dcam), two types of segregrants can be obtained. Draw a diagram showing the two classes of recombination events that would result in segregration of

8 Page 8 of 12 the chromosomal duplication. [Clearly label the genes on the diagram, indicate the position of any cross-overs required, and show the two types of segregrants obtained.] ANSWER: leu + nadc proa' Mud(Amp, lac) 'leu nadc ::Tn10 dcam proa + 'leu nadc ::Tn10 dcam proa' Mud Mud leu + nadc leu + nadc proa' 'leu nadc ::Tn10 dcam proa + proa+ 'leu nadc ::Tn10 dcam proa + leu + nadc ::Tn10 dcam proa + leu + nadc + proa + (10pts) 6. The srl operon is required for Salmonella to grow on sorbitol as a sole carbon source. In the absence of sorbitol a srl::mudj(lac, Kan R ) operon fusion expresses very low levels of ß-galactosidase. However, when sorbitol is added, the srl::mudj(lac, Kan R ) operon fusion expresses high levels of ß-galactosidase. (4pts) a. How could you use this srl::mudj(lac, Kan R ) fusion to isolate mutations that affect the regulation of the srl operon? ANSWER:

9 Page 9 of 12 (2pts) b. If you isolated a mutation in the srl::mudj(lac, Kan R ) strain that resulted in high constitutive expression of ß-galactosidase, what could you conclude about the mechanism of regulation by the wild-type srl gene? Briefly explain your answer. ANSWER: Because MudJ produces operon fusions, these results indicate that you have mutated a regulator that controls transcription of the srl operon. (Furthermore, if the mutation is due to a null mutation, it suggests that you have inactivated a repressor.) (4pts) c. How could you determine if the regulatory mutation is within the srl operon or elsewhere on the chromosome? Briefly explain your answer. ANSWER: Backcross the mutant with the wild-type strain as shown in the figure below. Grow P1 lysate srl::mudj srlr - P1( x ) Transduce selecting Kan R srl::mudj srlr - ( x ) srl + srlr + Screen for LacZ expression If regulatory mutation is unlinked, all of the Kan R transductants should retain wild-type srl::mudj phenotype. If r egulatory mutation is linked then the re should be two classes of K an R transductants -- with the wild-type srl::mudj phenotype and some with the mutant phenotype. If the mutation is within the lacz gene then all of the Kan R transductants should have the mutant phenotype.

10 Page 10 of 12 (10pts) 7. It is possible to make P22 lysogens in E. coli (under special circumstances that are not relevant to this question). You have constructed the four P22 lysogens shown below. One lysogen contains a wild type P22 prophage. A second, called Del #1, has the DNA within the bracket deleted from the prophage. A second, called Del #2, has the DNA within the bracket deleted. A third strain has a P22 prophage with an amber mutation in a gene called Orf14. Orf14 lies within the DNA deleted in lysogen Del #2. attl P22 attr Prophages: Del #1 Del #2 Orf14 X You have three phage lysates. One is λ vir, which we discussed in class. The second is wild type λ +. The third is λ imm22. This phage has the immunity region of phage P22 but is otherwise all of the λ chromosome. You decide to perform an experiment shown in the table below. You spot about 100 pfu of the indicated phage on the bacterial indicators and check for the appearance of plaques after incubation at 37 C overnight. The results are shown in the table below. (Turbid means turbid plaques. Clear means clear plaques. None means no plaques observed). Plaques With λvir λ + λimm 22 Prophage Wild Type clear turbid turbid Del #1 clear clear clear Del #2 clear turbid none Orf14am clear turbid none (2pts) a. Why does λvir make plaques on all of the indicators? ANSWER: λvir is not repressed by λ or P22 repressors. (4pts) b. What region of the P22 chromosome is deleted in prophage Del #1? Why? What is a plausible explanation for the behavior of prohage Del #1 with λ + and λimm 22?

11 Page 11 of 12 ANSWER: λ + and λimm 22 make clear plaques with Del #1, which indicates lytic growth. You would expect turbid plaques if the phage can undergo lysogeny. Therefore, the Del #1 prophage makes something that promotes lytic growth of both phages. (4pts) c. What is a plausible explanation for the behavior of prophages Del #2 and Orf 14? Why? ANSWER: Del #2 and Orf14am allow λ + to form lysogens as shown by turbid plaques. THis is consistent with the model above that the region defined by Del #2 and Orf14am promotes lytic growth of λ +. The lack of Orf14am allows λ + to form lysogens. However, λimm 22 cannot form any plaques on Del #2 or Orf14. This suggests that the region defined by Del #1 encodes the P22 repressor. (10pts) 8. A strain of E. coli is simultaneously infected with lambda gal and lambda bio transducing phages at a multiplicity of infection (moi) of approximately 5 for each phage. After the culture lyses, approximately 10% of the phage released are wild-type lambda. No wild-type lambda were found in control single infections. (6pts) a. Explain how wild-type lambda can be generated in the double infection. Use a diagram to explain your model and indicate the cellular and/or phage factors required. ANSWER: λdgal gal att int + xis ci λbio att bio Int Xis IHF (FIS) λ + att (4pts) b. Would you expect another new class of phage from this cross? If so, indicate what the structure of this phage would be.

12 Page 12 of 12 ANSWER: λdgalbio gal att bio THE GENETIC CODE