Final answer:
In the cross between a white-eyed male (XY) and a heterozygous red-eyed female (XWXw) in fruit flies, the expected outcome is a 1:1:1:1 ratio of red-eyed females, white-eyed females, red-eyed males, and white-eyed males, leading to equal numbers of each phenotype and genotype among the offspring.
Step-by-step explanation:
Genotypes in X-Linked Crosses
The question pertains to the genotypic outcome of a cross involving an X-linked trait, specifically the eye color in Drosophila melanogaster (fruit flies). When considering a cross between a white-eyed male (XY) and a female that is heterozygous for red eye color (XWXw), we can predict the offspring ratio using a Punnett square. Since eye color is an X-linked trait in fruit flies, males have only one X-chromosome and therefore display the phenotype of whatever allele is present on it. Females, having two X-chromosomes, can be homozygous or heterozygous for the trait.
In this cross, the male's genotype is XY, and the heterozygous female's genotype is XWXw. The possible gametes for the male are X and Y, while the female can produce XW and Xw gametes. This results in the following offspring genotypes:
- 50% of females will be red-eyed (XWX) - receiving XW from the female and X from the male.
- 50% of females will be white-eyed (XwX) - receiving Xw from the female and X from the male.
- 50% of males will be red-eyed (XWY) - receiving XW from the female and Y from the male.
- 50% of males will be white-eyed (XY) - receiving Xw from the female and Y from the male.
This results in a 1:1:1:1 ratio of red-eyed females, white-eyed females, red-eyed males, and white-eyed males among the offspring. Therefore, the expected genotypic ratio from the cross of a white-eyed male and a heterozygous red-eyed female is equal numbers of each phenotype and genotype.