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Lab 3 Tetrad Analysis - This value was normalized when...

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Exercise 3: Tetrad Analysis and Crossing Over Zachary Bergman Friday PM Section: 75 TA: Rashmi 1. Both of these formulas map the same distance between the genes because in both cases you are taking ½ of the recomibinants divided by the total number of ascospores or asci. The ½ is used in both cases because in all of the recombinants, 4 of the 8 ascospores are recombinant and the other 4 are parental. This holds true because we only recorded the ascospores that followed this trend and ignored those who had multiple crossovers. 2. These values are similar because the entire class examined the same strain and the map distance between the gene and the centromere should be the same and give the same values for everyone. The 1-2 cM difference is due to the fact that, by chance, I observed slightly more recombinants than parental phenotypes.
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Unformatted text preview: This value was normalized when compared to the rest of the class's data. 3. In the first crossover, the 2 nd and 4 th genes cross giving a tan, bl, bl, tan genotype. These then replicate to give the tan, tan, bl, bl, bl, bl, tan, tan final phenotype. In the second cross the 1 st and 3 rd genes cross to give a bl, tan, tan, bl genotype and again they replicate to produce the final bl, bl, tan, tan, tan, tan, bl, bl phenotype. 4. 66.7% recombination frequency is highest possible, because assuming that there is an equal distribution between the 6 possible phenotypes, 4 are recombinant and 2 are parental. This gives a 4/6 ratio of recombination. By definition, if there were more of the 4 recombinant phenotypes than the 2 parental, the recombinants would have to be considered the parental genotypes....
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