a.
To draw:
The map that shows relative positions of markers and origins of transfer in HfrA and HfrB strains and also the distance between these markers.
Introduction:
Genetic markers are the genes that are identifiable through
b.
To draw:
The order of the genes based on the given cotransduction data.
Introduction:
Cotransduction refers to the transfer of different bacterial genes together in one phage by transduction process.
c.
To determine:
The way an individual increases the chances of mapping gly gene relative to at least some of the other markers and the composition of the medium that would be used to map gly.
Introduction:
Tranductants refer to the cells resulting from the gene transfer. Transduction process is mainly mediated by bacteriophages.
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Genetics: From Genes to Genomes, 5th edition
- If an E. coli auxotroph X could grow only on a medium containing leucine, and an auxotroph Y could grow only on a medium containing threonine. (i) Write the genotype of both E. coli strains. (ii) How would you test whether DNA from X could transform Y?arrow_forwardIn Escherichia coli, four different Hfr strains, derived from the same F* strains, were mated with F strains auxotrophic for a number of nutritional requirements (Arg Bio Cys Trp Gal His Lac Mal Xyl Leu Met). Matings were interrupted at various intervals and cells were plated on minimal medium supplemented with particular nutrients to test for gene transfer. The following results show the time of entry for each of the genes in four different Hfr strains. Table 1: Time-of-Entry Mapping Data* Hfr strains Genes lac" his" arg' bio 9. Cys gal" trp' 5 11.5 2.5 mal" xyl" leu met Hfr 1 Hfr 2 Hfr 3 Hfr 4 6.5 3.5 11 15 15 4 6. 17.5 5 14.5 3 20 * The numbers denote the number of minutes elapsed before a gene enters the F cells. Draw a circular map of E. coli chromosome.arrow_forwardWild-type bacteria were used as donors to transduce the following recipient genotypes with the results indicated as the number of wild type colonies: Recipient Recipient leu- phe+ his- met+ tyr+ leu- phe+ his+ met- tyr+ leu- phe+ his+ met+ tyr- leu- phe- his+ met+ tyr+ leu+ phe+ his-met- tyr+ 127 leu+ phe-his- met+ tyr+ #wild type 37 138 132 1 leu+ phe+ his- met+ tyr- leu+ phe-his+ met- tyr+ leu+ phe-hist met+ tyr- leu- phe+ his+ met+ tyr+ 22 6 leu+ phe+ his+ met- tyr- 148 31 304 leu+ phe+ his+ met+ tyr- What is the order of the genes? tyr-leu-his-phe-met O leu-met-phe-his-tyr leu-his-met-tyr-phe his-tyr-leu-phe-met #wild type 356 10arrow_forward
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- What is the simplest explanation for why patients have been identified with only one copy of the phosphofructokinase-1 gene (heterozygous), but no patients have been identified that lack both copies of the phosphofructokinase-1 gene (homozygous)? Patients lacking both copies of the phosphofructokinase-1 genes will be found once DNA sequencing technology can sequence whole genomes. Phosphofructokinase-1 is needed for nitrogen metabolism and there are no enzymes to replace this function, so cells die from ammonia toxicity. Phosphofructokinase-1 is a required enzyme for carbohydrate metabolism in all living cells, complete loss of this enzyme would be lethal. There are 6 phosphofructokinase-1 paralogous genes in humans and it is impossible to lack both type 1 copies when there are also types 4, 5, and 6.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_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
- What order should the steps be in for this culture method?arrow_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:What information do you know based on the question and your understanding of the topic?arrow_forwardA wild-type mouse that is heterozygous for two immunoglobulin heavy chain alleles (IgHa/b) generates the population of B cells shown on the left of the figure below. A mouse strain, also IgHa/b, carries an inactivating mutation in the VpreB gene. In addition to producing fewer mature B cells than the wild-type mice, the VpreB-deficient mice generate B cells as shown on the right. What is the explanation of the difference seen between the wild-type and the VpreB-mutant B cells?arrow_forward
- Biology: The Dynamic Science (MindTap Course List)BiologyISBN:9781305389892Author:Peter J. Russell, Paul E. Hertz, Beverly McMillanPublisher:Cengage Learning