MCB421 FALL2005 EXAM#3 ANSWERSPage 1 of 12

(10pts) 1. There are two mechanisms for transposition used by bacterial transposableelements: replicative (Tn3) and non-replicative (Tn5 and Tn10). Compare and contrastthe 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 DNAsequences.

b. Formation of co-integrants:

ANSWER: Replicative transposons form co-integrants. Non-replicativetransposons 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 Salmonellaenterica. A few of these mutations are shown on the following deletion map of the putAstructural gene. The promoter is at the left, pointing rightward into the putA gene.

MCB421 FALL2005 EXAM#3 ANSWERSPage 2 of 12

Based on the relative location of the mutations and the results in the table below, explainthe phenotypes observed for each of the single and double mutants shown. [TheputA1992::MudJ and putA1993::MudK insertions map in the same deletion interval.](1pt each +1 freebie)

Mutations LacZ expression Explanation- proline + proline

putA1991::MudJ 5 100 20x transcription

putA1992::MudJ 10 200 20x transcription

putA1993::MudK 7 70 10x transcription and translation,therefore ↓ translation

putA1991::MudJputA2096::Tn10

<2 <2 polar

putA1992::MudJputA2096::Tn10

<2 <2 polar

putA1993::MudKputA2096::Tn10

<2 <2 polar

putA1991::MudJputA2097::Tn10

5 100 non-polar

putA1992::MudJputA2097::Tn10

<2 <2 polar

putA1993::MudKputA2097::Tn10

<2 <2 polar

(30pts) 3. Salmonella Typhimurium switches between two antigenic types of flagella by aprocess called phase variation. The following figure summarizes the initial evidence thatthis process occurs by an inversion (Zieg et al. 1977. Science 196: 170-172).

(5pts) a. The hin (H-inversion) gene and promoter for the downstream fljB gene arelocated between two hix sites where site-specific recombination occurs. Given thefollowing restriction map, what results would you expect if the region between the hixsites was inverted:

MCB421 FALL2005 EXAM#3 ANSWERSPage 3 of 12

ANSWER: The distance between the Hix sites will remain the same, but the distancebetween 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 aPCR assay? [Draw a diagram on the cartoon below showing where the PCR primerswould hybridize and what products you would expect.]

ANSWER: Primers sets 1&2 and 3&4 will give PCR products if the hix sites areas shown in the figure. Primer sets 1&3 and 2&4 will give PCR products if aninversion occurs.

(5pts) c. The DNA sequence of the Salmonella Typhimurium hin gene is shownbelow. (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

1 2 3 4

MCB421 FALL2005 EXAM#3 ANSWERSPage 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 dosite 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 labelingthe 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-mutantplasmids:

ANSWER: The diagram for Quikchange Mutagenesis® (Stratagene) is shown below.The dut/ung method and MutS methods are also accepted.

MCB421 FALL2005 EXAM#3 ANSWERSPage 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 notgive any cleavage products. Use the template DNA as a positive control to ensure theenzyme is working. Finally, sequence the mutant to ensure there are no secondarymutations.

(5pts) iv. How your mutation would alter the amino acid sequence of the resulting Hinprotein? (A copy of the genetic code is on the last page of the exam.)

MCB421 FALL2005 EXAM#3 ANSWERSPage 6 of 12

ANSWER: Original protein sequence:KIMATIGYIRVSTIDQNIDLQRNALTSANCDRIFEDRISGKIANRPGLKRALKYVNKGDTLVVWKLDRLGRSVKNLVALISELHERGAHFHSLTDSIDTSSAMGRFFFHVMSALAEMERELIVERTLAGLAAARAQGRLGGRPRAINKHEQEQISRLLEKGHPRQQLAIIFGIGVSTLYRYFPASRIKKRMN*

Mutagenized protein sequence:KIMATIGYIRVSTIDQNKKLQRNALTSANCDRIFEDRISGKIANRPGLKRALKYVNKGDTLVVWKLDRLGRSVKNLVALISELHERGAHFHSLTDSIDTSSAMGRFFFHVMSALAEMERELIVERTLAGLAAARAQGRLGGRPRAINKHEQEQISRLLEKGHPRQQLAIIFGIGVSTLYRYFPASRIKKRMN*

(10pts) 4. Given a putA point mutant and a P22 generalized transducing lysate grown on arandom pool of Tn10dCam (chloramphenicol resistant) transposon insertions:(Tn10dcam is a derivative of Tn10 that has the cat gene encoding resistance tochloramphenicol.)

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 dothe experiment and indicate the medium you would use for each selection orscreen. The phenotypes of the WT and putA mutant are shown in the tablebelow, where “+” indicates growth and “-“ indicates no growth.

Min + Glucose + NH4 Min + Glucose + ProlineputA+ + +putA- + -

ANSWER:1) Isolate a Tn10dCam linked to the putA point mutation:

Random Pool

putA+

putA-

Tn10

X

MCB421 FALL2005 EXAM#3 ANSWERSPage 7 of 12

Select CamR put+ on Min + Glucose + Cam + Proline medium.

2) Make P22 HT lysate of above strain and transduce into CamS putA-.Select on Min + Glucose + Cam + NH4 putA- will growReplica plate onto Min + Glucose + Cam + Proline putA- will not grow

3) Backcross to recipient strain and select as described above. All should beputA-.

b. How could you use a Tn10dCam insertion linked to the putA+ gene to isolatepoint mutations in the putA gene? [Draw a figure indicating the donor andrecipient and any selection or screen you would use.]

ANSWER: Localized mutagenesis of phage, transduce into CamR, screenputA-.

5. Chromosomal duplications of the nadC gene were constructed in S. typhimurium. Theresulting merodiploids have one mutant nadC allele in copy #1 and a different mutantnadC 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 diagramshowing the two classes of recombination events that would result in segregration of

putA+

Tn10

putA+ XTn10

MCB421 FALL2005 EXAM#3 ANSWERSPage 8 of 12

the chromosomal duplication. [Clearly label the genes on the diagram, indicate theposition of any cross-overs required, and show the two types of segregrants obtained.]

ANSWER:

..

leu+ nadC proA'

'leu nadC ::Tn10 dCam proA+Mud(Amp, lac)

'leu nadC ::Tn10 dCam proA+

leu+ nadC

proA'

Mud

'leu nadC ::Tn10 dCam proA+

Mud

leu+ nadC ::Tn10 dCam proA+ leu+ nadC+ proA+

leu+ nadC proA'

'leu

proA+

nadC ::Tn10 dCam

(10pts) 6. The srl operon is required for Salmonella to grow on sorbitol as a sole carbonsource. In the absence of sorbitol a srl::MudJ(lac, KanR) operon fusion expresses very lowlevels of ß-galactosidase. However, when sorbitol is added, the srl::MudJ(lac, KanR) operonfusion expresses high levels of ß-galactosidase.

(4pts) a. How could you use this srl::MudJ(lac, KanR) fusion to isolate mutationsthat affect the regulation of the srl operon?

ANSWER:

MCB421 FALL2005 EXAM#3 ANSWERSPage 9 of 12

(2pts) b. If you isolated a mutation in the srl::MudJ(lac, KanR) strain that resultedin high constitutive expression of ß-galactosidase, what could you conclude aboutthe mechanism of regulation by the wild-type srl gene? Briefly explain youranswer.ANSWER: Because MudJ produces operon fusions, these results indicatethat you have mutated a regulator that controls transcription of the srloperon. (Furthermore, if the mutation is due to a null mutation, it suggeststhat you have inactivated a repressor.)

(4pts) c. How could you determine if the regulatory mutation is within the srloperon or elsewhere on the chromosome? Briefly explain your answer.

ANSWER: Backcross the mutant with the wild-type strain as shown in thefigure below.

Transduce selecting KanR

Screen for LacZ expression

Grow P1 lysate

( )

x( )

srl+

P1

srl::MudJsrlR-x

srl::MudJsrlR-

srlR+

If regulatory mutation is unlinked, all of the KanR transductantsshould retain wild-type srl::MudJ phenotype. If r egulatory mutationis linked then the re should be two classes of K anR transductants --with the wi ld-type srl::MudJ phenotype and some with the mutantphenotype. If the mutation is within the lacZ gene then all of theKanR transductants should have the mutant phenotype.

MCB421 FALL2005 EXAM#3 ANSWERSPage 10 of 12

(10pts) 7. It is possible to make P22 lysogens in E. coli (under special circumstances thatare not relevant to this question). You have constructed the four P22 lysogens shownbelow. One lysogen contains a wild type P22 prophage. A second, called Del #1, has theDNA within the bracket deleted from the prophage. A second, called Del #2, has theDNA within the bracket deleted. A third strain has a P22 prophage with an ambermutation in a gene called Orf14. Orf14 lies within the DNA deleted in lysogen Del #2.

You have three phage lysates. One is λ vir, which we discussed in class. The second iswild type λ +. The third is λ imm22. This phage has the immunity region of phage P22but is otherwise all of the λ chromosome.

You decide to perform an experiment shown in the table below. You spot about 100 pfuof the indicated phage on the bacterial indicators and check for the appearance of plaquesafter incubation at 37°C overnight. The results are shown in the table below. (Turbidmeans turbid plaques. Clear means clear plaques. None means no plaques observed).

Plaques Withλvir λ+ λimm22

ProphageWild Type clear turbid turbid

Del #1 clear clear clearDel #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+ andλimm22?

attL attRP22

Prophages:

Del #1

Del #2

Orf14 X

MCB421 FALL2005 EXAM#3 ANSWERSPage 11 of 12

ANSWER: λ+ and λimm22 make clear plaques with Del #1, which indicates lyticgrowth. You would expect turbid plaques if the phage can undergo lysogeny.Therefore, the Del #1 prophage makes something that promotes lytic growth ofboth phages.

(4pts) c. What is a plausible explanation for the behavior of prophages Del #2 and Orf14? Why?

ANSWER: Del #2 and Orf14am allow λ+ to form lysogens as shown by turbidplaques. 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+ toform lysogens.

However, λimm22 cannot form any plaques on Del #2 or Orf14. This suggeststhat 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 biotransducing 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 adiagram to explain your model and indicate the cellular and/or phage factors required.

ANSWER:

(4pts) b. Would you expect another new class of phage from this cross? If so, indicatewhat the structure of this phage would be.

λdgal

gal attint+ xis cI

λbio

bioatt

IntXisIHF(FIS)

λ+

att

MCB421 FALL2005 EXAM#3 ANSWERSPage 12 of 12

ANSWER:

THE GENETIC CODE

λdgalbio

bioattgal