An attached-X female fly, XXY (see the Insights and Solutions box), expresses the recessive X-linked white-eye phenotype. It is crossed to a male fly that expresses the X-linked recessive miniature wing phenotype. Determine the outcome of this cross in terms of sex, eye color, and wing size of the offspring.

The problem states that we have an attached-X female fly with genotype XXY, showing the X-linked recessive white-eye phenotype. This means the genotype of this female fly is (X^w X^w Y), with X^w representing the allele for white eyes. We also have a male fly expressing the X-linked recessive miniature wing phenotype, which means its genotype is (X^m Y), where X^m represents the allele for miniature wings.

Now, let's consider the possible gametes each parent fly can produce. The female fly (X^w X^w Y) can produce two types of gametes: X^w and Y. The male fly (X^m Y) can also produce two types of gametes: X^m and Y.

To predict the genotypes of the offspring, we will use a Punnett square. In this case, we have a 2x2 Punnett square with female gametes on the top and male gametes on the side: X^w Y ____________ X^m | | X^w X^m | X^m Y | ____________ Y | | X^w Y | Y Y | ____________

We can now analyze the genotypes and phenotypes of the offspring. The four possible genotypes are: 1. X^w X^m: Female with one dominant eye-color allele and one recessive miniature wing allele. (Wild-type eyes, Miniature wings) 2. X^w Y: Male with a recessive eye color allele (White eyes, Wild-type wings) 3. X^m Y: Male with a recessive wing-size allele (Wild-type eyes, Miniature wings) 4. Y Y: This genotype is not viable in flies – no offspring. Overall, the offspring will have the following genotypes and phenotypes: Females: - 50% with Wild-type eyes and Miniature wings (genotype X^w X^m). Males: - 50% with White eyes and Wild-type wings (genotype X^w Y). - 50% with Wild-type eyes and Miniature wings (genotype X^m Y).