DNA obtained from the indicated donor type recombinants are scored. The number of wild-types for each cross is given in the chart below. What is the order of the genes? Donor Recipient wid type colonies a-b-C+ a+ b+C 273 a- b+ c- a+ b- c+ 462 a- b+ c+ ab-c- 2 ab- C+ be ce bec cannot be determined
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Genetic Recombination
Recombination is crucial to this process because it allows genes to be reassorted into diverse combinations. Genetic recombination is the process of combining genetic components from two different origins into a single unit. In prokaryotes, genetic recombination takes place by the unilateral transfer of deoxyribonucleic acid. It includes transduction, transformation, and conjugation. The genetic exchange occurring between homologous deoxyribonucleic acid sequences (DNA) from two different sources is termed general recombination. For this to happen, an identical sequence of the two recombining molecules is required. The process of genetic exchange which occurs in eukaryotes during sexual reproduction such as meiosis is an example of this type of genetic recombination.
Microbial Genetics
Genes are the functional units of heredity. They transfer characteristic information from parents to the offspring.
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- DNA obtained from the indicated donor strains is used to TRANSFORM the indicated recipient strains. The resulting progeny are plated on minimal medium so that only wild-type recombinants are scored. The number of wild-types for each cross is given in the chart below. donor donor arg+ val- ser- val+ ser- arg- val- ser+ val- ser+ arg+ val- ser- O valine recipient recipient none of these. O arginine arg- val+ ser+ arg+ val- ser+ Which locus is closest to serine? arg+ val+ ser- wild-types wild-types 70 98 0 2 arg- arg-E. coli strain B (permissive) were co-infected with rlla and rllb mutant phage T4. Progeny were plated in a dilution series with the following results: Lawn Dilution Number of plaques Strain B 10-6 10 Strain K 10-4 What is the recombination frequency? Remember that the readout is viability and you cannot see the inviable progeny. Answer: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 Cotransformation 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?
- Draw a diagram/figure to explain the conjugation process (e.g. use PowerPoint or draw one by hand and include a photo of it). You should include in the diagram the F- recipient, Hfr Donor and the transconjugant/recombinant recipient. Make sure to include the genes encoding for Leucine, Threonine, Thiamine and Streptomycin resistance in your diagram. How does an Hfr strain of coli transfers chromosomal DNA to an F- strain? What determines how much of the chromosomal DNA is transferred?DNA from a strain with genotype a+b+c+d+ is used to transform a strain with a genotype a-b-c-d-. Pair of genes are checked for cotransformation and the data was collected. Pair of genes Cotransformation Pair of genes Cotransformation a+b+ Yes c+e+ Yes b+c+ No d+e+ No a+e+ Yes b+e+ No d+c+ Yes b+d+ No d+a+ No a+c+ No Using the data in #7, determine which locus is closest to a. c e b can’t determine from the data What are the methods of genetic material transfer in bacteria? Transformation Transduction Conjugation None of these All of these Along with mutation, which of the following is another way that bacteria acquire antibiotic resistance through horizontal gene transfer? Transformation Transduction Conjugation None of these All of theseA cross is made between two E. coli strains: Hfr arg + bio + leu+ × F− arg − bio − leu−. Interrupted mating studies show that arg+ enters the recipient last, and so arg+ recombinants are selected on a medium containing bio and leu only. These recombinants are tested for the presence of bio + and leu+. The following numbers of individuals are found for each genotype: arg+ bio+ leu+ 320 arg+ bio- leu+ 0 arg+ bio+ leu- 8 arg+ bio- leu- 48a. What is the gene order? b. What are the map distances in recombination percentages?
- cterium of genotype a+b+c+ is the donor in a cotransformation mapping experiment. The recipient is a-b-c-. percentages of co-transformation are: a-c = 7; b-c = 20: a-b =2 nich two markers are furthest apart? O d) cannot be determined O b-c a-b a-cA bacterium of genotype a+ b+ c+ d+ is the donor in a cotransformation mapping. The recipient is a-b-c-d-. Data from the transformed cells is shown below. What is the order of the genes? a+ and b+ 5 a+ and c+ 2 a+ and d+ 0 b+ and c+ 0 b+ and d+ 0 c+ and d+ 5 cbad cadb bcda acbd bacdThe following recombinants are recovered when conjugation occurs between an a*d*g+ donor and an adg recipient. at d+ g+ = 84% a d g+ = 6% at d g+ = 10% a dt g+ = less than 1% What is the map distance between the a and d genes? 10 map units 74 map units less than 1 map unit 84 map units 6 map units
- 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 is the map distance (in minutes) between these two genes?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?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.