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- . An Hfr+ strain is used to map three genes: ala, leu, and thr in an interrupted mating experiment. Hfr/ala+, leu+, thr+, camS x F-/ala-, leu-, thr-, camR The cross is initiated at time = 0, and at various times, the mating mixture is plated on three types of medium. Each plate contains minimal medium with chloramphenicol and the amino acids as listed in the table. The data represents the number of colonies growing on each plate. Time of interruption Nutrients added 5 min 10 min 15 min 20 min alanine, leucine 8 21 58 85 leucine, threonine 0 3 19 46 alanine, threonine 0 0 7 16 Based on these data, what is the order of gene transfer relative to one another and to the F-factor.The map below was constructed using information from an interrupted mating experiment. Str# refers to the strain numnber from which the data were obtained. The strains that are referenced by the Str#'s on the map are considered to be F+ strains. The units for the map distance values on this map are minutes Assume that four genes were transferred by each strain in the interrupted mating experiment. Based on this map, list in order of transfer (first to last) the genes transferred by Str4. Type your answer by typing in the letters that represent the genes without any spaces so that Moodle can grade it correctly. K Str2 10 9 Strl W B Str5 8. H. R Str3 Str6. 11 A Str4 12You also obtained the MIC for a variety of common antifungals by testing them against a single strain of Aspergillus fumigatis. Six replicates of each were tested. MIC (ug/mL) Amphotericin B 1 Miconazole 0.06 Ciclopirox 0.01 Nystatin 0.625 New product 0.1 Two-way ANOVA Amp v. New 3.70E-04 Nys v. New 4.16E-05 Mico v. New 0.04 Cic vs. New 1.50E-05 Additionally, Candida aurus is a fungus that is rapidly becoming resistant to multiple antifungals. Because strains now exist that are completely resistant to three or more families of antifungals, there is room in the market for an antifungal that is effective against C. aurus. If the data above is representative for the efficacy against both A. fumigatus, a fungus that is plentiful and sensitive to most antifungals, and C. aurus, determine whether this efficacy data supports room for this product on the existing marketplace. If you think more data is required in order to make a final decision about this…
- 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?. 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?You are studying a microorganism that contains a chloramphenicol acetyltransferase (CAT) enzyme, and looking for a clone of this microorganism that no longer contains the gene encoding CAT. Imagine you had an LB plate containing chloramphenicol. You streak an isolated colony of the Parent Strain and Clone A onto the same plate. Which of the following statements are true about the growth pattern after 24 hours in the incubator? Select all that apply? L-hlareeheticol LB chloremphenicel Parent Clune A Parent Clone A LD chlerephenicol 18 chloremphenitol Paremt Clone A Paront Clone A Plate A is the expected growth pattern as clone A should grow in the LB plate with chloramphenicol Plate B is the expected growth pattern as LB plate with chloramphenicol parent strain should grow in the Plate B is the expected growth pattern as clone A should not grow in the LB plate with chloramphenicol Plate A is the expected growth pattern as the parent strain should not grow in the LB plate with…
- 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.Make a calculation.In a transformation experiment involving a recipientbacterial strain of genotype a- b-, the following resultswere obtained. What can you conclude about the locationof the a and b genes relative to each other? Transformants (%)Transforming DNA a+ b- a-b+ a+b+a+b- 3.1 1.2 0.04a+b- and a-b+ 2.4 1.4 0.03Recombinant 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.…
- Linkage maps in an Hfr bacterial strain are calculated in units of minutes (the number of minutes between genes indicates the length of time that it takes for the second gene to follow the first in conjugation). In making such maps, microbial geneticists assume that the bacterial chromosome is transferred from Hfr to F − at a constant rate. Thus, two genes separated by 10 minutes near the origin end are assumed to be the same physical distance apart as two genes separated by 10 minutes near the F −attachment end. Suggest a critical experiment to test the validity of this assumption.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 cotransformation, and the following results are obtained: Pair of genes Cotransformation Pair of genes Cotransformatic a* and b* No b* and d* No a* and c* No b* and e* Yes a* and d* Yes c* and d* No a* and e* Yes c* and e* Yes b* and c* Yes d* and e* No On the basis of these results, what is the order of the genes on the bacterial chromosome?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…