Packaging of P22 DNA requires a pac site while packaging of lambda DNA requires a cos site. Briefly describe:

Transcription

1 1). (12 Points) Packaging of P22 DNA requires a pac site while packaging of lambda DNA requires a cos site. Briefly describe: 1. The mechanisms used by P22 and lambda to package DNA. P22 uses a headfull mechanism, [The first cut is made at a pac site and a headfull of DNA is packaged. The next chromosome is then packaged starting at the end of the first chromosome until another headfull is packaged, etc.] Lambda DNA is packaged by cleavage at a specific site called cos [followed by packaging until the next cos site is encountered and cleaved to end the chromosome). 2. The consequences of the packaging mechanisms with respect to the termini of the DNA in lambda and P22 phage heads. (P22 chromosomes have terminal redundancies at the ends while lambda chromosomes have complementary sticky end cos site). 3. The mechanisms that the two phage use to circularize their genomes after infection. (P22 circularizes by homologous recombination. Lambda circularizes by pairing of the complementary cos ends and ligation by host DNA ligase). 4. Why does P22, but not lambda, perform generalized transduction? (P22 performs generalized transduction by initiating packaging at chromosomal pseudopac sites. Lambda DNA cleavage enzyme apparently does not show relaxed specificity for cleavage of chromosomal DNA).

2 2). (10 Points). Consider a region of chromosomal DNA that contains 1000bp repeated sequences. In one case, the sequences are orientated as direct repeats as shown in Figure 1 below. Figure 1: In the second case, they are orientated as inverted repeats as shown in Figure 2 below. Figure 2: For each construct, draw the products of a homologous recombination event between the segments of homology. Be sure your diagram is very explicit in showing the recombinants. The top figure will give a deletion of the B DNA between the ends of the 2 repeated sequences. A C The bottom figure will give an inversion where the order BC between the repeated elements is reversed to CD. See Kuzminov lecture for details. A C B D

3 3). (12 Points). To determine the gene order of the add, man, and uida genes, threefactor crosses were done. Given the results shown below, what is the order of these three genes? Draw a diagram of each cross to show your rationale for this conclusion Donor: add man uida Selected Phenotype Man Add UidA Recipient: add man uida Recombinants Number Obtained Add UidA Add UidA Add UidA ANSWER: Draw a diagram of each cross Note that you can infer the middle marker from the rare class of recombinants (the Man Add UidA recombinants). This result suggests that the uida allele is the middle marker in the recipient because four crossovers are required to produce this class of recombinants. The question asked you to draw a diagram of each of the crosses, but only the crossovers yielding the rarest class of recombinants is shown in the figure below add uida man add uida man Correct gene order: 4pts Each correct cross drawing: 2pts.

4 4). (10 Points). Below is a deletion map of the thra gene of E. coli. The thra gene is required for synthesis of threonine. Phage P1 was grown on a revertible thra mutant strain and the lysate was used to transduce the different deletion strains shown below. (The boxes indicated DNA that is deleted. The lines represent DNA that is still present in the genome of the strain). The results of the transductions are shown under the Growth column. A). What medium would you plate the transductions on? Minimal medium 2pts B). Indicate the smallest region of the thra gene where the donor mutation maps. thra Growth Begins at the end of Box 11 and ends at the start of Box 7, 8pts.

5 5). (15 Points). Phages Phi 21, and Phi 80 were tested for the ability to infect two strains of E. coli. As a control, phage lambda was grown on E. coli K12 r m was tested. The results are shown in the table below. Clear indicates that the plaques were clear while Turbid indicates the plaques were turbid. () indicates that plaques were not made. Recipient bacteria E. coli B E. coli K12 Phage r m r m r m r m Phi 21 clear clear clear lambda Phi 80 A). (2 Points). What can you infer about the lifestyle of phage Phi 21 based on the type of plaques formed? (Phi 21 appears to be a lytic phage because it makes clear plaques). B). (5 Points). Suggest a likely explanation for the growth of each phage (Phi 21, Phi 80 and lambda) on the E. coli B and K12 strains. (Phi 21 comes from E. coli B (1pt) because it is not restricted by B but is restricted by K12 (1pt). Lambda comes from K12 (1pt) because it is restricted by B but not by K12 (1pt). Phi 80 does not grow on either B or K12 even when they are r (1pt) so it is likely that it cannot absorb to B or K 12 or else it can t undergo morphogenesis in B and K12). C). (2 Points). Suggest two different reasons why Phi 80 can t grow on E. coli B or E. coli K. (The E. coli strains might lack a receptor for Phi 80 or lack a gene product needed for Phi 80 morphogenesis). 1pt for each. D), (6 Points). Draw the predicted results from a onestep growth curve for each of the two possibilities in the following figure. Also include the results expected for infection of a sensitive cell by Phi80. (The Y axis should be PFU / ml and the X axis should be TIME (in minutes).

6 (See handout on lytic growth of phage). 2pts for each. 6). (15 Points). The transposon Tn10 has inverted repeats at the end of the transposon. The same inverted repeats are found at the end of every Tn10 insertion. A). (3 Points). Briefly describe the important features of these inverted repeats and their function. They encode transposase (1pt) and the ends of the inverted repeats (1pt) provide binding sites for transposase and are essential for transposition (1pt). B). (3 Points). Short direct repeats are observed flanking each Tn10 insertion. The sequences of these direct repeats are different for each Tn10 insertion. Briefly explain how these direct repeats are formed and why they are different for each Tn10 insertion.

7 They are host DNA (1pt) from the transposition target site that are generated by staggered cuts which are subsequently filled in by host DNA replication (2pt). C), (3 Points). Some transposable elements make very small amounts of transposase while others make a large amount. Give an example of each type and explain why it makes low or high amounts of transposase. Most transposons and IS elements make low amounts of transposase because if they made large amounts of transposase the elements would transpose at high frequencies and the host would accumulate insertion mutations which would be harmful. On the other hand, Phage Mu must transpose by replicative transposition to replicate its DNA. Thus it needs a large amount of transposase to accomplish this. D). (3 Points). Why do some transposable elements that use replicative transposition contain a RES site and a resolvase protein? (Transposons like Tn1000 that transpose by replicative transposition form cointegrate intermediates that occur as direct repeats. These must be resolved by resolvase recombination at the RES sites to produce the donor and product replicons to complete the transposition event. (Or resolution could occur by HR).) E). (3 Points). Transposon Tn5 encodes resistance to Kanamycin. It encodes its own transposase. Briefly describe a way that you could deliver Tn5 to recipient cells that would select for transposition events. (You could infect with a defective phage that carries a copy of Tn5 that is unable to replicate or lysogenize the host and select for KanR. You could use a suicide plasmid that cannot replicate in the host and select for KanR. You could electroporate in EZTn and select for KanR.

8 7). (10 Points) You have two strains of Salmonella typhimurium (shown below). As part of a series of experiments you need to perform, you need a strain that has a deletion of the sera and serb genes. You have the two strains shown below and P22 HT at your disposal to make the indicated deletion. arga mutants require arginine and sera and serb mutants require serine. Tn10 encodes resistance to tetracycline. A). (5 Points). How could you construct the desired strain? (Be sure to draw carefully and legibly the events required). (See diagram below). B). (2 Points). What media would you use? (Minimal Ser Tet) C). (3 Points). How would you show that the sera and serb genes are deleted by a simple genetic test? (Look for Ser revertants. There should be none. Or make a transducing lysate of a ser mutant. The lysate should not transduce the deletion to Ser).

9 8) (15 points) Slauch and Mann [(1997) Genetics 146:447456] used phage P22 HT to attempt to transduce the plasmid psc101 and some of its derivatives. They used the following three plasmids in their experiment: pbr322 This plasmid (4.3 Kb) carries a gene that confers resistance to ampicillin (Amp R ). It has no homology with the Salmonella chromosome. psc101 This plasmid 5.4 Kb) carries a gene that confers resistance to ampicillin (Amp R ). It has no homology with the Salmonella chromosome. pjs225 This plasmid (10.5 Kb) is the same as psc101 except that it carries a 5 kb fragment of Salmonella chromosomal DNA carrying the region encoding the pura gene shown below. pura Chromosomal DNA in psc101 to make pjs225 All three plasmids were introduced into Salmonella strain JS107 which is wild type except that it contains a kan R cassette in the Salmonella chromosome. The site of the insertion is near pura cloned into psj225, (It is in the middle of 5 Kb fragment in pjs225 except it is in the Salmonella chromosome). pura Kan R pura region of JS107 chromosome with Kan R insertion The recipient strain used in the transductions below was JS110 which is a pyrc auxotroph but otherwise wild type. The pura and pyrc genes are unlinked in the Salmonella chromosome (i. e. They are never cotransduced). Next they performed the following experiment:

10 They grew P22 HT on JS107 / pbr322 and on JS107 / psc101 and made lysates after lytic growth of the phage. They then infected JS110 with one of the lysates. Equal volumes of the transduction mixtures were plated on LB plates containing ampicillin to select Amp R transductants and on glucose minimal medium to select for Pyr transductants. The plates were incubated for colony formation and they determined the efficiency of plasmid transduction (EOPT) by the following formula: EOPT = Number of Amp R transductants / Number of Pyr transductants. The results are shown in the Table below. Donor Plasmid Size (Kb) EPOT (%) psc pjs pbr A). (3 Points). Why was it necessary to perform the transduction to Pyr in each transduction? (ie. Why not just plate for Amp R and count the colonies?). This control corrects for differences in the amount of transducing phages in different lysates. (Each lysate has a different titer from another. However the ratio of phage that carry a plasmid or pyr should be the same). B). (3 Points). Compare the results for transduction of pbr322 relative to psc101. What do the results so far indicate about the mechanism of transduction of the 2 plasmids? Why? [The data show that pbr322 is transduced at 300% the frequency of the chromosomal Pyr marker while psc101 is transduced at 8 % the frequency of the chromosomal Pyr marker, This is a huge difference that likely means that the 2 plasmids are transduced by different mechanisms. We know from lecture that P22HT transduces pbr322 by inducing rolling circle replication to form long concatomers which are packaged into phage heads. The DNA is injected into the recipient and circularized by recombination. Apparently the psc101 plasmid is not induced to perform rolling circle replication so it can t be packaged into phage heads. ] They also did an experiment where they transduced pjs225 in a similar experiment. The results are also shown in the Table.

11 C). (3 Points). Compare the results for transduction of pjs225 with the other 2 plasmids. What can you conclude about the mechanism of transduction of pjs225? (Hint, it is not byrolling circle replication). [The frequency of transduction of pjs225 has increased markedly relative to its psc101 parent. The only difference between psc101 and pjs225 is the 5 Kb chromosomal insert in pjs225. This means that the 5 kb insert must be doing something that increases the transduction frequency. One (unlikely) possibility is that the insert allows rolling circle replication and packaging of concatomer DNA. This is not likely because there is no reason to think that the chromosomal insert carries a gene or site that promotes rolling circle replication. A second (more likely) possibility is that the psj225 plasmid can integrate into the host chromosome by HR between the homologous chromosomal DNA sequences. The DNA is then packaged from the host chromosome into phage heads. When injected into the JS110 recipient, the DNA can be circularized by HR to reform the plasmid. SINCE psc101 HAS NO CHROMOSOMAL HOMOLOGY IT CAN T INTEGRATE AND BE PACKAGED so it is not transduced efficiently. D). (3 Points). Make a detailed diagram that shown the mechanism of your proposed answer to (C). (A diagram of rolling circle replication is not an acceptable answer). See Figure 1 of Mann and Slauch (The plasmid integrates by HR into the region of homology. During lytic growth of P22 HT, the DNA containing the plasmid can be packaged into phage heads and transduced into a new host and integrated into the chromosome.) In another experiment, Mann and Slauch plated the cells from the transductions with the donor psc101 (JS 107) lysate grown and JS 110 recipient and plated on LB Amp Kan plates and got no colonies. When they did the same experiment with the donor pjs 225 (JS 107) lystae they got several hundred Amp R Kan R colonies. E). (3 Points). Can you explain the mechanism of transduction of the Kan R marker in this experiment? Draw out the genetic events that allow transduction of the Kan R.

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