Micro 411/Genome Sci411 Exam Three

Transcription

1 Name Question 1. 20pts Diagram the initiation of recombination by RecBCD starting with double stranded DNA. Make sure to label all enzymes, the location of the Chi site and where RecA is required. A B. Diagram RecBCD action at the site B in first diagram. C. Diagram RecBCD action at site C. D. Diagram D-loop formation with another double-stranded DNA molecule.

2 Question 2. 20pts. Four metabolites (A, B, C, D) are known to be involved in the biosynthesis of amino acid X. Structural analogs of each of these metabolites (A*, B*, C*, D*) are lethal to cells because they mimic the appropriate molecule and are used to form an aberrant form of X (= X*) that causes spontaneous termination of translation whenever it is incorporated into a polypeptide. Minimal glucose (MinGlu) medium containing each of these analogs individually was used to isolate conditional mutants that could grow in the presence of a given analog at 42 C, but not at 30 C. These mutants were then screened individually for growth at 42 C in the presence of each of the remaining three analogs. As shown in the table below, four mutant classes were identified by this analysis. (The double mutant is described in part D of this question.) Mutant class Growth (+/-) at 42 C on MinGlu with: A* B* C* D* No analog Number of isolates Double mutant n/a A) Assuming that all genetic loci involved are of equal mutational target size, how many individual gene products were probably mutated in this analysis? B) Assuming that all the mutations identified are conditional (ts) knockouts, are these coldsensitive or heat-sensitive conditional mutations? Briefly explain your logic ( 20 words). C) Again assuming that all mutations are ts knockouts, diagram the expected pathway for X biosynthesis based on these data. (Your diagram should account for all of the genes identified in part A.) Indicate on your diagram the specific catalytic steps affected for each of the four mutant classes. D) The specific mutation present in a Class 3 (D*-resistant) mutant was transduced into a Class 1 (C*-resistant) mutant, creating a recombinant cell carrying both mutations. As shown in the last row of the table, this double-mutant did not grow at 42 C under any of the conditions tested. Briefly explain this result ( 20 words).

3 Question 3. 15pts Lambda phage is a temperate phage and the plaque morphology is an indicator of host and phage gene interplay. You want to understand how lysogeny works and have built the following F plasmids with the genes shown in the table. The F plasmids have a high level of expression for each of the individual genes listed. You infect individual strains of E.coli carrying each of the F plasmids with a lambda phage encoding penicillin resistance. What effect would these F plasmids have on the ability to recover penicillin-resistant colonies relative to having only F+ alone? (Answer in terms of increased, decreased, or the same frequency relative to an F+ cell). F plasmid Results F -cii F -N protein F -ci F -int F -xis Question 4. 14pts. Below is a set of three genes that are on the E.coli chromosome. MogA is a molybdenum cofactor biosynthesis protein, yaah allows cells to transport acetate to use as a carbon source and htga regulates the ability to withstand heat shock. The Tn10 insertion into moga encodes tetracycline resistance, and you have bioassays for the other two genes, so each marker can easily be selected and/or screened. You use P1 phage to map the genes. The results of 3 transduction crosses are shown below. Based on these data, draw the map order of these genes. Donor Recipient Selected Phenotype Recombinants moga::tn10 htga- moga+ htga+ Tet r htgahtga+ yaah+ htga- yaah- htga+ YaaH+ htgahtga+ moga::tn10 yaah+ moga+ yaah- Tet r yaahyaah+ Number Obtained You also have another gene, yaai::pen (penicillin resistant) that co-transduces with htga 99% of the time. How could you use this information to confirm your map experimentally?

4 Question pts You fortuitously discover a previously unknown outer membrane protein in E. coli and name it "OmpV" because it appears to be the receptor for the lytic phage Phage V. You also determine that it is a nonspecific channel allowing the passage of hydrophilic substances into and out of the cell, and is thus similar in function to OmpC and OmpF. You decide to further characterize the gene for OmpV 1. How might you experimentally identify mutants for OmpV expression following a transposon mutagenesis? 2. The results from your screen could yield two kinds of mutants. What are they? 3. You want to map the chromosomal location of the ompv mutant. You transduce the ompv mutation into each of the Hfr mapping strains shown below. You mate each of the Hfr ompv strains into wild type E.coli and select for phage resistance. The results of these mating are shown below. Where does the ompv mutation map? thra (0 min) (90min) Hfr3 met A cysg Hfr2 Hfr1 Hfr 6 Hfr 5 Hfr4 (73 min) cysa (50 min) pyrc (23 min) Hfr Strain Number of Phage resistant colonies Hfr1 20 Hfr2 210 Hfr3 80 Hfr4 2 Hfr 5 8 Hfr6 300

5 Question 6. 12pts You have performed chemical mutagenesis on a plasmid that has a lacz gene with a non-inducible (constitutive) promoter and screened individual mutants for an altered lacz phenotype. The plasmid replicates using the RNA I, RNA II, and RopA system for replication. The bacterial host cell is chromsomally deleted for the lac operon, so the amount of betagalactosidase activity in cells containing the lacz plasmid is directly proportional to the copy number of the plasmid. You have determined that none of the mutations are in the lacz gene or in the promoter for lacz. This means that any mutations reside elsewhere on the plasmid. You get the following results: Wild type 50 units Mutant 1 10 units Mutant 4 0 units Mutant units Name the potential mutations that would result in the lacz phenotype of each mutant shown. Mutant 1 Mutant 4 Mutant 15