1. [2 pts] For the fruit fly Drosophila melanogaster, females have two X chromos
ID: 140115 • Letter: 1
Question
1. [2 pts] For the fruit fly Drosophila melanogaster, females have two X chromosomes, and males have only a single X chromosome. The w mutation results in white eyes, and the ry mutation results in rosy eyes. The symbols for the wildtype alleles are w* and, ry respectively. You observe the following crosses: Cross 1: A wild-type male mates with an rosy female. All of the progeny have normal eyes. All of the progeny have normal eyes All of the female progeny have normal eyes. Cross 3: A wild-type male mates with a white eyed female All of the male progeny have white eyes. Answer the following questions. (a) Which alleles are recessive? Which alleles are dominant? Explain your reasoning. (b) Which mutations, if any, must be on the Xchromosome? Which mutations, if any, must be on an autosome? Explain your reasoning.Explanation / Answer
Ans.
w+ and ry+ are dominant alleles and w and ry are the recessive alleles.
this can be explained by representing the crosses given above by punett squares.
Cross 1- wild type male with rosy female
All the progeny have normal eyes.
cross 2-
All of the progeny have normal eyes.
in both the cases, the phenotype of the progeny is resembling the phenotype of one of their parents. so the allele present in parents must be dominant allele.
b)
The mutation resulting in white eye must be in x chromosome as the reciprocal cross i.e., cross 3 ,in which wild type male mates with a white eyed female results in different progeny phenotypes as compared to cross 2.
this is because, male only has one x chromosome wheras female has two.
cross 3-
the mutation resulting in rosy eye must be present in autosome.
this is because, if we had the rosy eye mutation in x chromosome then only half of the progeny would have had normal eyes, similar to cross 3. but, as all of the progeny are normal, the mutation must be autosomal.
ry ry ry+ ry+ ry ry+ ry ry+ ry+ ry ry+ ryRelated Questions
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