7. (10pts.) The diagram below shows the immunity region of phage l. The genes are:

Transcription

1 7. (10pts.) The diagram below shows the immunity region of phage l. The genes are: ci codes for ci repressor. The ci857 allele is temperature sensitive. cro codes for Cro protein. N codes for the transcription anti-terminator. cii codes for ci activator. ciii codes for the ciii protease inhibitor. OL 1,2, and 3 and OR 1,2, and 3 are operators. P L and P R are the rightward and leftward promoters. P RE is the promoter for repressor establishment. P RM is the promoter for repressor maintenance. P L P R cii N O L1 O L2 O L3 ci O R1 O R2 O R3 cro ciii P RM P RE Calef et al. isolated a mutated lysogen that deleted most lambda DNA on the right and left sides of the immunity region as shown below. The mutant also has the ci857 TS allele (regions not shown below are deleted): P L P R O L1 O L2 O L3 ci 857 O R1 O R2 O R3 cro The lysogen had always been maintained at 30 C previously to the experiment described below. They grew the lysogen at 30 C until cells reached mid-log growth. They split the culture in half and grew one at 30 C and the other at 42 C for several hours. They then used the cells from each culture as indicator bacteria for the ability of wt l and limm434 to form plaques to 30 C. They obtained the results shown in the Table: Indicator Cells Infected with wt l Infected with limm434 Grown at 30 C no plaques normal, turbid plaques Grown at 42 C clear plaques normal, turbid plaques

2 7 cont. The indicator cells grown at 42 C were returned to 30 C and grown for several generations and checked again for plaquing by wt l and limm434. They obtained the same results as with the cells grown at 42 and not shifted back to 30 (i.e. they still obtained clear plaques with wt l and turbid plaques with limm434). a. Why doesn't wt l form plaques on the indicator lysogen grown at 30 C? ANSWER: At 30 C, the ci857 repressor in the lysogen is active and will bind to the right and left operators of the infecting phage. This will prevent transcription of any phage proteins and thus prevent lysis. b. Why can limm434 (as opposed to wt l, above) form plaques on the indicator lysogen grown at 30? Why are the plaques turbid? ANSWER: limm434 has the immunity region from phage 434, so the l repressor is unable to bind its operators and repress transcription. So, the infecting phage may lyse the indicator. The plaques are turbid because the limm434 will eventually start making its own ci-like repressor (probably once the MOI becomes high) and form lysogens. c. Why does wt l form clear plaques on the indicator strain if it is grown at 42 C? ANSWER: l is able to form plaques on the lysogen at 42 because the ci857 repressor is inactive at 42 and so does not prevent lysis by the infecting phage. The lack of ci in the recipient lysogen leads to accumulation of cro protein. Cro protein prevents transcription of ci, both in the lysogen and in the incoming phage. This ensures that incoming phage will only be able to grow lytically, and so the plaques are clear. d. Why doesn't growth of the indicator lysogen at 30 C for several hours restore the original phenotype (no plaques with wt l) of indicator cells once they have been grown at 42? Once ci is deactivated at 42, transcription of cro protein starts. Once cro protein is present, it prevents transcription of ci, even when the temperature is shifted back down to 30 C. Thus, the lysogen loses immunity even when shifted back to 30.

3 1. (10pts) Draw a diagram showing how could you use a Tn10 Tet R insertion 80% linked to the S. typhimurium trpa + gene to isolate point mutations in trpa. [Your diagram should describe the donor and recipient strains, any selections and screens required, and the media used.] ANSWER: You should start with a trpa + strain closely linked to Tn10 as stated in the question. Grow a P22HT lysate on the strain and mutagenize in vitro (or mutagenize the strain before making the lysate). Infect a wt Salmonella strain with the lysate, and select for Tet-resistance. Such colonies are likely to have received the donor trpa allele. Screen them for Trp- phenotype by replica-plating onto minimal media with and without his. If they grow on the former but not on the latter, they are likely the desired trpa mutants. If your diagram showed crossovers, it must show a double crossover, with one on either side of the trpa/tn10 area.

4 2. (10 pts.) You have an E. coli B strain, an E. coli K strain, and a strain of E. coli B, JG880, carrying a plasmid containing a large segment of DNA from E. coli K. You grow phage l on each of these strains and then assay the capability of the resulting phage to grow lytically on each strain. The results are shown below, where a (+) means phage were capable of forming plaques on a strain and a ( ) means no plaques could be formed. Strain infected l grown on B l grown on K l grown on JG880 K B JG Explain the above results in terms of the restriction and modification systems present in each strain. Be concise, but be sure to explain each result. What genes from E. coli K must be present on the plasmid in JG880? How can you tell? [Hint: remember that JG880 is a B type strain.] ANSWER: DNA synthesized in a cell is immediately modified by methylation enzymes at specific sequences. The methylation prevents cutting by restriction enzymes at the same sites, so that the cell does not digest its own DNA. Different strains have different modification/restriction systems that methylate or cut at different sites. l grown on a B strain is able to infect a B strain because it has been modified by the B modification system and thus is not restricted by the B restriction enzymes. It is unable to infect a K strain, because it has not been modified at K modification sites, and so is cut by K restriction enzymes. The same logic applies for why l grown on K can infect K but not B strains. l grown on JG880 can infect either K or B strains, so it must have K's modification system on its plasmid. JG880 cannot be infected by l grown on B, even though it is a B strain. Therefore, it must also have K's restriction system on its plasmid.

5 3. (20pts.) Lysogens of phage l loop out of the chromosome upon induction and begin lytic growth. a. What is the signal for induction of lysogens? ANSWER: Host DNA damage, specifically ssdna. b. What phage-encoded protein performs the recombination event that loops out phage l? ANSWER: Xis (partial credit for Int) c. What type of recombination event is this an example of? ANSWER: Site-specific recombination. d. Name another example of this type of recombination, not involving phage. ANSWER: Resolution of co-integrative intermediates during replicative transposition. e. Draw the integration and excision of phage l, being sure to label the attb, attp, attl and attr sites (if present) in each step. Also draw an aberrant excision event that could lead to the formation of lbio specialized transducing phage, being sure to include the expected frequency of this event. Use the back of the page if necessary. INTEGRATION: l attp attb Int attl l attr Normal excision bio bio Aberrant excision leading to lbio specialized transducing phage

6 4. (20 pts.) You have isolated a new gene, yssa, from Salmonella. After obtaining the DNA sequence of the gene and performing a GenBank search you find that the gene has homology to known Type III topoisomerases from a variety of bacteria. It has a tyrosine residue that is part of the active site of all known Type III topoisomerases. In described topoisomerases, this residue is required for activity. Assume that you have a method to isolate YssA protein in small quantities and that you have detected topoisomerase activity in YssA isolates. Assume that you have a simple method to assign a positive or negative phenotype for isolated YssA protein (i.e. you can say whether a given version of the YssA protein is functional or not). a. Explain how you would use the Quickchange PCR method of site-directed mutagenesis to test whether the conserved tyrosine residue is required for YssA function. Draw a diagram showing where each primer would anneal on a plasmid containing yssa. Show the location of the mutagenic nucleotide on each primer in your diagram. ANSWER: Your diagram should show a plasmid containing yssa and two anticomplementary primers annealing on either side of the plasmid within the yssa gene. The mutagenic nucleotide should be in the middle of each primer. b. What would you want to do with the Quickchange PCR reaction after PCR and before transformation? Why? ANSWER: digest with DpnI to eliminate residual template DNA.

7 4c. If the relevant Tyr codon is UAU, suggest a one-nucleotide change you could make to change the amino acid using the above protocol. What new amino acid would this result in? Do you think this change in amino acid would definitively test the importance of the Tyr residue? Why or why not? [Use the genetic code dictionary and table of AA structures on the back of the exam to answer this question.] ANSWER: The codon can be changed to Phe, Ser, Cys, His, Asn, or Asp. Regardless of the amino acid, it cannot be a definitive test. Each of these amino acids has something in common with or similar to Tyrosine. Phe has an aromatic group, Ser has a hydroxyl, Cys has a thiol (sometimes similar to hydroxyl), His, Asn, and Asp can sometimes act as proton donors or acceptors, as can Tyr. 3 points for a codon change, 2 points for talking in some sensible way about the new amino acid, 1 point for correctly saying that it would not be a definitive test. d. Describe a genetic experiment that could help you purify a large amount of the protein for biochemical studies. Put the gene on an overexpression vector, perhaps one with a T7 promoter system. A His tag could also be attached if you so desire. Full credit for an overexpression vector, 2 points for His tag. e. What is one possible problem with the experiment described in part d? Several possible answers overexpression could be toxic, His-tag could affect activity, etc.

8 5. (10 pts.) You have isolated a novel operon, the yuk operon, that contains two genes required for virulence in Salmonella. You have constructed operon fusions with the galk gene (encoding galactose kinase) and gene fusions with the lacz gene (encoding ß- galactosidase) for each of the two genes, yukf and yukg. The expression of each fusion was assayed with or without exposure to superoxide (superoxide is an oxygen radical that pathogens can encounter in hosts). The results are shown in the following table. Gene galk operon fusion Galactose kinase activity lacz gene fusion ß-galactosidase activity - superoxide + superoxide - superoxide + superoxide yukf yukg What do the results indicate about the regulation of the yukf and yukg genes? Briefly explain your answer. ANSWER: yukf is not transcriptionally regulated by superoxide because we see no increase in reporter activity in response to superoxide in the operon fusion. yukf is regulated translationally in response to superoxide because we see a marked increase in reporter gene acitivity in the gene fusion in response to superoxide. Since there was no increase in the operon fusion, we know that the entire difference must be due to translational regulation. yukg is regulated transcriptionally in response to superoxide because we see an increase in reporter gene activity in the operon fusion. It is not regulated translationally because, although we see a difference in the gene fusion, the ratio is the same as in the operon fusion, so we know that the effect can be entirely accounted for by transcriptional regulation.

9 6. (20 pts.) You have constructed a pir-dependent plasmid that contains a tetracyline resistance gene and the E. coli thra promoter region and the first half of the coding sequence of the thra gene. (Draw a plasmid with that structure). The plasmid is introduced into a pir- strain that also contains a mutation in the thra promoter region. The remainder of the thra gene is wild type. Tetracycline resistant colonies are selected. a. Draw the two classes of integrants (relative to the chromosomal thra region) that could be isolated. This time, be sure to also draw a recombination event between the suicide plasmid and chromosome that would lead to each class of integrant, being sure to make clear the location of the recombination event relative to the two mutations. Finally, indicate the likely Thr phenotype of each class of integrant (i.e. Thr+ or Thr-). Briefly explain your reasoning. If you need more space, use the back of the page. ANSWER: NOTE THAT ONLY A SINGLE CROSSOVER EVENT IS REQUIRED HERE! If the crossover occurs before the promoter mutation: thra

10 The result is an integrated plasmid and 1 and one-half copies of the thra gene. The half copy has the good promoter, and the whole copy has the bad promoter. The cell would be Thr-.

11 If the crossover occurs after the promoter mutation: The result is an integrated plasmid and One and one-half copies of the thra gene. The half copy is with the defective promoter, and the full copy is with the good promoter. The phenotype of the cell would be Thr+.

12 b. If you wanted to isolate segregants that lost the plasmid, what feature(s) would you have included on the original plasmid? ANSWER: a marker for reverse selection, like sacb.

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