Final answer:
When crossing a white-eyed male fruit fly (XwY) with a heterozygous red-eyed female (XWXw), the expected F2 phenotypic ratio of offspring is 1 red-eyed female to 1 white-eyed female to 1 red-eyed male to 1 white-eyed male, or a 1:1:1:1 ratio.
Step-by-step explanation:
In a classical set of experiments with fruit flies (Drosophila melanogaster), Thomas Hunt Morgan demonstrated that genes are linked to chromosomes. Specifically, he studied the inheritance of eye color in these flies, discovering that the gene for eye color is located on the X chromosome, making it an X-linked trait. In Drosophila, red eyes are wild-type and dominant (symbolized as XW), while white eyes are recessive (symbolized as Xw).
Now, let's consider the cross between a white-eyed male (XwY) and a heterozygous red-eyed female (XWXw). The male can only pass on a Y chromosome to sons and the Xw chromosome to daughters. The heterozygous female can pass on either the XW or Xw chromosome. Using a Punnett square, we can predict the offspring's genotypes and phenotypes.
The resulting genotypic ratio in the F2 generation will be:
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- 50% red-eyed females (XWXw)
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- 50% white-eyed females (XwXw)
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- 50% red-eyed males (XWY)
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- 50% white-eyed males (XwY)
Therefore, the phenotypic ratio of their offspring will be 1 red-eyed female:1 white-eyed female:1 red-eyed male:1 white-eyed male, or simply 1:1:1:1.