. In an interrupted-conjugation experiment in E. coli, thepro gene enters after the thi gene. A pro+ thi+ Hfr iscrossed with a pro- thi- F- strain, and exconjugants areplated on medium containing thiamine but no proline. Atotal of 360 colonies are observed, and they are isolatedand cultured on fully supplemented medium. These cultures are then tested for their ability to grow on mediumcontaining no proline or thiamine (minimal medium),and 320 of the cultures are found to be able to grow butthe remainder cannot.a. Deduce the genotypes of the two types of cultures.b. Draw the crossover events required to produce thesegenotypes.c. Calculate the distance between the pro and thi genesin recombination units.

. In an interrupted-conjugation experiment in E. coli, thepro gene enters after the thi gene. A pro+ thi+ Hfr iscrossed with a pro- thi- F- strain, and exconjugants areplated on medium containing thiamine but no proline. Atotal of 360 colonies are observed, and they are isolatedand cultured on fully supplemented medium. These cultures are then tested for their ability to grow on mediumcontaining no proline or thiamine (minimal medium),and 320 of the cultures are found to be able to grow butthe remainder cannot.a. Deduce the genotypes of the two types of cultures.b. Draw the crossover events required to produce thesegenotypes.c. Calculate the distance between the pro and thi genesin recombination units.

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter17: Bacterial And Viral Genetics

Section: Chapter Questions

Problem 1ITD

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An Hfr strain that is leuA+ and thiL+ was mixed with a strain thatis leuA− and thiL−. In the data points shown in the following graph,the conjugation was interrupted at different time points, and thepercentage of recombinants for each gene was determined bystreaking on a medium that lacked either leucine or thiamine.What information do you know based on the question and your understanding of the topic?
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 oftheir experimental results is shown in the following table: A. E xplain how these results are consistent with the idea that thebacterial chromosome is circular.B. Draw a map of the bacterial chromosome that shows the orderof genes and the locations of the origins of transfer among thesedifferent Hfr strains.
You found five T4 rII− mutants that will not grow onE. coli K(λ). You mixed together all possible combinations of two mutants (as indicated in the followingchart), added the mixtures to E. coli K(λ), and scoredfor the ability of the mixtures to grow and makeplaques (indicated as a + in the chart).1 2 3 4 51 − + + − +2 − − + −3 − + −4 − +5 −a. How many genes were identified by this analysis?b. Which mutants belong to the same complementation groups?
In 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?
In 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?
. a. You want to perform an interrupted-mating mappingwith an E. coli Hfr strain that is Pyr+, Met+, Xyl+,Tyr+, Arg+, His+, Mal+, and Strs. Describe anappropriate bacterial strain to be used as theother partner in this mating.b. In an Hfr × F− cross, the pyrE gene enters therecipient in 5 minutes, but at this time point thereare no exconjugants that are Met+, Xyl+, Tyr+,Arg+, His+, or Mal+. The mating is now allowed toproceed for 30 minutes and Pyr+ exconjugants areselected. Of the Pyr+ cells, 32% are Met+, 94% areXyl+, 7% are Tyr+, 59% are Arg+, 0% are His+, and71% are Mal+. What can you conclude about theorder of the genes?
In your laboratory, you have an F− strain of E. coli that is resistantto streptomycin and is unable to metabolize lactose, but it can metabolizeglucose. Therefore, this strain can grow on a medium thatcontains glucose and streptomycin, but it cannot grow on a mediumcontaining lactose. A researcher has sent you two E. colistrains in two separate tubes. One strain, let’s call it strain A, hasan F′ factor (an F prime factor) that carries the genes that are requiredfor lactose metabolism. On its chromosome, it also has thegenes that are required for glucose metabolism. However, it is sensitiveto streptomycin. This strain can grow on a medium containinglactose or glucose, but it cannot grow if streptomycin is addedto the medium. The second strain, let’s call it strain B, is an F−strain. On its chromosome, it has the genes that are required forlactose and glucose metabolism. Strain B is also sensitive to streptomycin.Unfortunately, when strains A and B were sent to you, thelabels had fallen…
You co-culture the following bacterial strains: an Hfr prototroph and an F- auxotroph for the genes mal, met, mtl, and xyl. You interrupt conjugation at various time points and place the mixtures on media plates lacking each of the nutrients. Based on the results shown on the right, what is the order of these four genes along the bacterial chromosome? Explain your reasoning.
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 table:Draw a map that shows the order of genes and the locations ofthe origins of transfer among these different Hfr strains?
In E. coli, four Hfr strains donate the following markers,shown in the order donated:Strain 1: M Z X W CStrain 2: L A N C WStrain 3: A L B R UStrain 4: Z M U R BAll these Hfr strains are derived from the same F+ strain.What is the order of these markers on the circularchromosome of the original F+?
In a generalized-transduction experiment, phages arecollected from an E. coli donor strain of genotype cys+leu+ thr+ and used to transduce a recipient of genotypecys- leu- thr-. Initially, the treated recipient populationis plated on a minimal medium supplemented with leucine and threonine. Many colonies are obtained.a. What are the possible genotypes of these colonies?b. These colonies are then replica plated onto threedifferent media: (1) minimal plus threonine only, (2)minimal plus leucine only, and (3) minimal. Whatgenotypes could, in theory, grow on these three media?c. Of the original colonies, 56 percent are observed togrow on medium 1, 5 percent on medium 2, and nocolonies on medium 3. What are the actual genotypes ofthe colonies on media 1, 2, and 3?d. Draw a map showing the order of the three genes andwhich of the two outer genes is closer to the middle gene
Production of a recombinant protein by E. coli is proportional to cell growth. Ammonia is used as a source of N and glucose as a source of C, under aerobic conditions. The recombinant protein has a general formula CH1.55N0.31O0.25 and the cell CH1.77N0.4900.24. The yield of biomass from glucose is 0.48 gcel/gglic, and the yield of recombinant protein from glucose is about 20% of that for biomass.a) How much ammonia is needed to produce 50 g of cells producing the recombinant protein?b) What is the oxygen demand in this process?c) For the cultivation of a wild E. coli not producing the recombinant protein, how different would the ammonia and oxygen demand be if the biomass yield remained at 0.48 gcel/gglyc ?