DNA containing the genes gin+ phe+ lys+ was used to transform bacterial cells of genotype gln- phe- lys-. The following cell types were obtained with the number of each cell type shown below: Genotype Number Igin+phe+lys+ 25 gln+|phe-lys- gin phe-lys+ 4. gin-phe+ lys- 14 gin-phe+lys+ gin-phe iys+ gin-phe-tys- 19 Which gene is in the center? Which two genes are the closest together? 2. 00
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- Streptococcus pneumoniae cells of genotype str s mtl - are transformed by donor DNA of genotype strr mtl+ and (in a separate experiment) by a mixture of two DNAs with genotypes strr mtl - and str s mtl+. The accompanying table shows the results.a. What does the first row of the table tell you? Why? b. What does the second row of the table tell you? Why?As mentioned in question 2 of More Genetic TIPS, origins oftransfer can be located in many different places on a bacterial chromosome,and their direction of transfer can be clockwise or counterclockwise.Let’s suppose a researcher conjugated six differentHfr strains that were thr + leu + tons str r azis lac + gal + pro + met +to an F − strain that was thr − leu − tonr str s azir lac − gal − pro −met −, and obtained the following results: Draw a map of the circular E. coli chromosome that shows thelocations and orientations of the origins of transfer in these six Hfrstrains.A student cloned Claudin 1 (877 bp) gene into pTRE3G-BI-ZsGreen1 (total size is 8905 bp) vector using Not1 and EcoRV restriction sites located at 215bp and 779bp respectively. What is the size of the recombinant plasmid? A) 9782 bp B) 9218 bp C) 10776 bp E) 8028 bp F) None of the answers given
- DNA from a strain of Bacillus subtilis with genotype a+ b+ c+ d+ e+ is used to transform a strain with genotype a– b– c– d– e–. Pairs of genes are checked for co-transformation and results shown below are obtained. Based on these results, what is the order of the genes on the bacterial chromosome? please show how you worked out the order by diagraming, thank you! Pair of genes Co-transformation a+ and b+ no a+ and c+ no a+ and d+ yes a+ and e+ yes b+ and c+ yes b+ and d+ no b+ and e+ no c+ and d+ no c+ and e+ yes d+ and e+ noWhen the interrupted mating technique was used withfive different strains of Hfr bacteria, the following orders ofgene entry and recombination were observed. On the basisof these data, draw a map of the bacterial chromosome.Do the data support the concept of circularity?HfrStrain Order1 T C H R O2 H R O M B3 M O R H C4 M B A K T5 C T K A BBy 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.
- 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 geneBy 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?Four E. coli strains of genotype a+ b- are labeled 1, 2, 3, and 4. Four strains of genotype a- b+ are labeled 5, 6, 7, and 8. The two genotypes are mixed in all possible combinations and (after incubation) are plated to determine the frequency of a+ b+ recombinants. The following results are obtained, where M = many recombinants, L = low numbers of recombinants, and 0 = no recombinants:On the basis of these results, assign a sex type (either Hfr, F+, or F-) to each strain.
- 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+?Let’s suppose a new strain of P1 phage has been identified thatpackages larger pieces of the E. coli chromosome. This P1 strainpackages pieces of the E. coli chromosome that are 5 minuteslong. If two genes are 0.7 minute apart along the E. coli chromosome,what would be the cotransduction frequency using a normalstrain of P1 and using the new strain of P1 that packages largerpieces? What would be the experimental advantage of using thisnew P1 strain?Recombinant protein production by a genetically modified Escherichia coli strain is proportional to cell growth. Ammonia is used as a nitrogen source for aerobic glucose respiration. The recombinant protein has the general formula CH1,55O0,31N0,25, while that of the cellular biomass is CH1,77O0,49N0,24. The biomass yield from glucose equals 0.50 g/g, while the recombinant protein yield from glucose corresponds to 20% of the cell yield from substrate.a) How much ammonia is required? What is the oxygen demand? (b) If the biomass yield remains the same, what are the ammonia and oxygen requirements for a wild-type strain of E. coli, with cell biomass of the same elemental composition, but unable to synthesize the recombinant protein? (c) On an industrial scale, cultivation takes place in a continuous fermenter at 28°C and the desired recombinant protein production rate is 7 g/h. Since the viscosity of the culture broth is considerable, the energy input due to agitation cannot be neglected.…