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- Now that you have counted the non-recombinant and recombinant asci we want to determine the recombination frequency between the tan gene and the centromere. To calculate the recombination frequency we need an accurate count of the number of non-recombinant recombinant individuals. In the asci the individuals are the ascospores. and 1118 WWW 4 non- recombinant ascospores 4 non- recombinant ascospores 1838 W M M M 500 2 non- recombinant ascospores 2 recombinant ascospores 2 recombinant ascospores 2 non- recombinant ascospores Figure 1.6. The number of ascospores in non-recombinant and recombinant asci. In the non-recombinant asci all 8 ascospores are non-recombinant (see Figure 1.6 above). In the non-recombinant asci 4 of the ascospores are recombinant and 4 of the ascospores are non-recombinant (see Figure 1.6 above). To determine how many recombinant and non-recombinant ascospores where in the two pictures above we need to do the math below. You need to use the total number of…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.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?
- 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?For each genotype in the table below, determine whether or not functional B-gal will be produced in the presence or absence of the inducer. Write a plus (+) if B-gal is produced or a minus (-) if it is not. ma Chromosome F' lac (plasmid) - Inducer + Inducer ---- --- --- I*O*Z I*O°Z+ I*O*Z* I* = wildtype repressor |- - no functional repressor produced O* - wildtype operator OC = operator mutation prevents repressor from binding %3DFrom a cross between e+ f+ g+ and e− f − g− strains ofNeurospora, recombination between these linkedgenes resulted in a few octads containing the followingordered set of spores:e+ f+ g+e+ f+ g+e+ f − g+e+ f − g+e− f − g−e− f − g−e− f − g−e− f − g−a. Where was recombination initiated?b. Did crossing-over occur between genes e and g?Explain.c. Why do you end up with 2 f+ : 6 f − but 4 e+: 4 e−and 4g+: 4g−?d. Could you characterize these unusual octads as MIor MII for any of the three genes involved?Explain.
- 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?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 hisE+ and pheA+ was mixed with a strain thatis hisE− and pheA−. The conjugation was interrupted and the percentageof recombinants for each gene was determined by streakingon a medium that lacked either histidine or phenylalanine. Thefollowing results were obtained:Determine the map distance (in minutes) between these twogenes.
- You and your lab partner performed a complementation test for five recessive histidine auxotrophs. Your plate looks like this: 1 2 3 4 5 1 2 3 4 5 growth no growth What do the gray patches represent ✓ [Select ] What do the tan patches represent? diploid histidine auxotrophs diploid prototrophs haploid histidine auxotrophs haploid prototrophs What is the composition of the medium for this test? [Select] The data contains a conflicting result due to a false positive (that is, it showed growth where it should not have grown) because your partner inoculated a patch too heavily. Which cross most likely led to this result? [Select] How many complementation groups are described by this data, taking into account that a cross yielded a false positive result? [Select]Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. After crossing the F1 generation of the cross between mutant strains 1 and 3, you count and determine the phenotypes of 1,000 colonies (here a colony is equivalent to an individual): 563 colonies that can grow on minimal medium alone; 437 colonies that require adenine…Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other? B. What phenomenon is occurring in the cross between mutant strains 1 and 3? After crossing the F1 generation of the cross between mutant strains 1…