Concept explainers
A study was conducted in an attempt to determine which functional regions of a particular conjugative transfer gene (tral) are involved in the transfer of plasmid R27 in Salmonella enterica. The R27 plasmid is of significant clinical interest because it is capable of encoding multiple-antibiotic resistance to typhoid fever. To identify functional regions responsible for conjugal transfer, an analysis by Lawley et al. [(2002). J. Bacteriol. 184:2173-2180] was conducted in which particular regions of the tral gene were mutated and tested for their impact on conjugation. Shown here is a map of the regions tested and believed to be involved in conjugative transfer of the plasmid. Similar coloring indicates related function. Numbers correspond to each functional region subjected to mutation analysis.
Accompanying the map is a table showing the effects of these mutations on R27 conjugation.
Effects of Mutations in Functional Regions of Transfer Region 1 (tral) on R27 Conjugation
- (a) Given the data, do all functional regions appear to influence conjugative transfer?
- (b) Which regions appear to have the most impact on conjugation?
- (c) Which regions appear to have a limited impact on conjugation?
- (d) What general conclusions might one draw from these data?
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Concepts of Genetics (12th Edition)
- By conducting conjugation experiments between Hfr and recipientstrains, Wollman and Jacob mapped the order of many bacterialgenes. Throughout the course of their studies, they identified severaldifferent Hfr strains in which the F-factor DNA had been integratedat different places along the bacterial chromosome. A sample of theirexperimental results is shown in the following table:What information do you know based on the question and your understanding of the topic?arrow_forwardWe have two specific strains of E. coli that have shown horizontal gene transfer (HGT) when mixed. To experimentally determine the method of HGT that is happening, the following conditions are set up in different tubes of culture media: A) Donor and recipient strain mixed together (control - no treatment). B) Donor and recipient strains mixed together, DNase added (can digest DNA in solution, not within cells).C) Special tube containing a membrane filter (with pores that allow DNA and viruses to pass through, but not bacterial cells) that separates two compartments. Donor strain is added on one side, the recipient strain on the other (they are separated by the filter).D) Donor and recipient strains mixed together, with chemical that inactivates viruses (chemical affects bacteriophages in solution so they are unable to attach to cells). The results: Tubes A, B, and D: HGT was observed. Tube C: HGT was NOT observed. Based on this, which type of HGT was occurring? Conjugation,…arrow_forwardBacteriophage P22 was used in generalised transduction experiments to infect the Salmonella typhimurium donor strains described in the table below. The resulting phage lysates were then used to infect the recipient strains of S. typhimurium recipient strains listed in the table. In each cross, a phenotype was selected for one of the selected for one of the three genetic markers studied (str, aceA, thrA), and were made to select the recombinants corresponding to the other two markers. markers. The results are given in the following table: Strain I donor str thrA aceA thrA str aceA+ Strain recipient strs thrA+ aceA thrA str aceA Phenotype selected Str Ace+ Str recombinants selected ThrA ThrA ThrA ThrA Ace Ace Number 60 40 95 5 10 90 str: gene involved in streptomycin resistance, aceA: gene involved in the use of acetate as a carbon source, thrA: gene involved in threonine biosynthesis. 1) What are the selective media used in these three transduction experiments? to obtain the selected…arrow_forward
- By conducting conjugation experiments between Hfr and recipientstrains, Wollman and Jacob mapped the order of many bacterialgenes. Throughout the course of their studies, they identified severaldifferent Hfr strains in which the F-factor DNA had been integratedat different places along the bacterial chromosome. A sample of theirexperimental results is shown in the following table:Explain how these results are consistent with the idea that thebacterial chromosome is circular?arrow_forwardBacteriophage P22 was used in generalized transduction experiments to infect the Salmonella typhimurium donor strains described in the table below. The resulting phage lysates were then used to infect the S. typhimurium recipient strains listed in the table. In each cross, a phenotype was selected for one of the three genetic markers studied (str, aceA, thrA), and then replicates were performed to select the corresponding recombinants for the other two markers. The results are given in the following table: Recipient strain Selected phenotype Selected recombinants Donor strain str thrA aceA+ thrA str aceA+ strs thrA+ aceA thrA+ str aceA Str Ace+ Str ThrA ThrA+ ThrA ThrA+ Ace Ace str: gene involved in streptomycin resistance, aceA gene involved in the use of acetate as a carbon source, thrA: gene involved in the biosynthesis of threonine. Number 60 40 95 5 10 90 Determine the order of the genes and draw a genetic map showing this orderarrow_forwardBy conducting conjugation experiments between Hfr and recipientstrains, Wollman and Jacob mapped the order of many bacterialgenes. Throughout the course of their studies, they identified severaldifferent Hfr strains in which the F-factor DNA had been integratedat different places along the bacterial chromosome. A sample of theirexperimental results is shown in the following table:Analyze data. Compare and contrast. Make a drawing.arrow_forward
- The strain of λ phage t is cI857. That tells you that the cI DNA segment is disabled by a specific mutation. What is the exact genetic change in cI857? What specific property of the cI gene product does this mutation change, and how does this help titering for a plaque assay?arrow_forwardIn 1944, Avery, Macleod, and McCarty provided strong evidence that DNA is the hereditary material in Streptococcus pneumoniae by Group of answer choices showing that avirulent cells could become virulent by the process of transduction none of these is true. showing that virulent cells could become avirulent if the DNA was destroyed after transformation showing that avirulent cells could not gain the ability to become virulent cells if conjugation was interrupted. showing that avirulent cells could not gain the ability to become virulent if DNA was destroyed after transformation.arrow_forwardDNA sequencing of the entire H. influenzae genomewas completed in 1995. When DNA from the nonpathogenic strain H. influenzae Rd was compared tothat of the pathogenic b strain, eight genes of the fimbrial gene cluster (located between the purE andpepN genes) involved in adhesion of bacteria to hostcells were completely missing from the nonpathogenic strain. What effect would this deletion have oncotransformation of purE and pepN genes using DNAisolated from the nonpathogenic versus the pathogenic strain?arrow_forward
- In a study, bacterial culture of E. coli was infected with bacteriophage. How- ever, these cultures were protected from phage infection. Interestingly, it was found that the resistance is due to endonudeases present in E. coli, which deaved the phage DNA. How is the E. coli genomic DNA protected from the action of these endonuclease enzymes?arrow_forwardIn E. coli, the gene bioD+ encodes an enzyme involved in biotin synthesis, and galK+ encodes an enzyme involved in galactose utilization. An E. coli strain that contained wild-type versions of both genes was infected with P1 phage, and then a P1 lysate was obtained. This lysate was used totransduce (infect) a strain that was bioD− and galK−. The cellswere plated on a medium containing galactose as the sole carbonsource for growth to select for transduction of the galK+ gene.This medium also was supplemented with biotin. The resultingcolonies were then restreaked on a medium that lacked biotin tosee if the bioD+ gene had been cotransduced. The following resultswere obtained:What information do you know based onthe question and your understanding of the topic?arrow_forwardIn E. coli, the gene bioD+ encodes an enzyme involved in biotin synthesis, and galK+ encodes an enzyme involved in galactose utilization. An E. coli strain that contained wild-type versions of both genes was infected with P1 phage, and then a P1 lysate was obtained. This lysate was used totransduce (infect) a strain that was bioD− and galK−. The cellswere plated on a medium containing galactose as the sole carbonsource for growth to select for transduction of the galK+ gene.This medium also was supplemented with biotin. The resultingcolonies were then restreaked on a medium that lacked biotin tosee if the bioD+ gene had been cotransduced. The following resultswere obtained:What topic in genetics does this question address?arrow_forward
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