The X-linked gene white in Drosophila melanogaster, has more than 400 known alleles. We will concern ourselves with only four. The mutant allele white (w) we have talked about in class. It causes white eyes in homozygous females and hemizygous males. Another mutation, white-apricot (w), is dominant to w, and produces a pale yellow eye color called apricot. The last mutant is white-coffee (w^d). When homozygous, it causes a light brown color and is dominant to w. The w^a and w^d f alleles are codominant, and w^d w^a flies have an eye color called dark apricot. The wild type allele, w\', is dominant to all of the others, and produces the normal red eye color. Predict the phenotypic ratios (include sex) that would result from the following crosses. We will use X^+ to indicate an X-chromosome with the wild-type allele, X^+ to represent white-apricot, x^d to represent white-coffee, and X^w for white. Y, of course, is the Y chromosome. X^d X^w times X Y Phenotypic ratio A white-eyed female and an apricot-eyed male Phenotypic ratio The female progeny of cross b crossed with with coffee-eyed male. Phenotypic ratio The male progeny of cross b crossed with with dark apricot females Phenotypic ratio
the question says Xcf mutation is dominant to w , and X* means a wild type allele doing the cross,
a)
therefore, the phenotype ratio would be: 2 red eye female : 1 light brown male : 1 white male
b. white eye female : Xw Xw , apricot eye male : Xa Y
therefore the ratio would be: 2 apricot females : 2 white males
c. female of cross b is XwXa and coffee eyed male would be Xcf Y
therefore the ratio for this cross should be : 1 coffee eyed female : 1: darkapricot eye female : 1 white male : 1 apricot eye male
d. male prigeny of cross b is Xa Y and dark apricot female is Xa Xa
therefore the ratio would be 1 or all apricot eye male and females.
| Xcf | Xw |
| X* | XcfX* | X*Xw |
| Y | XcfY | XwY |
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