Lab 2 Drosophilia cross

Lab 2 Drosophilia cross - wild type. We were able to...

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Exercise 2: Linkage and Chromosome Mapping: Drosophilia melanogaster Zachary Bergman Friday PM Section: 75 TA: Rashmi 1. For cross number 2, we determined that all the genes are X-linked recessive. We found them to be recessive because the parental male genotype is known to be homozygous for all three mutations and the parental female genotype is wild type. By looking at the F1 progeny, all of which were wild type, we see that the parental female line must be homozygous for the dominant gene and also that the mutations are recessive because the offspring would receive the mutation gene from the father and a wild type gene from the mother making them heterozygous
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Unformatted text preview: wild type. We were able to determine the mutations were X-linked by looking at the F2 progeny. All females in this group were wild type and the males had a large dispersal between the different mutations. Because none of the females had any of the mutations, we prove not only that it is X-linked, but also again that the mutations are recessive. The F2 females must receive an X from their fathers, and because there is no recombination in Drosophilia males, this X must be wild type. The F2 males, on the other hand, only receive the one X from their mother which has 8 possible recombinations resulting in 8 different phenotypes. 2....
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