huge explanation
In Drosophila, the sex of the individual is determined by the presence of X and Y chromosomes, with females having two X chromosomes (XX) and males having one X and one Y chromosome (XY). The genes located on the X chromosome are said to be X-linked and follow different inheritance patterns in males and females.
In the cross involving two X-linked genes, we can use the following notation:
- w+ = wild-type allele for gene 1
- w = mutant allele for gene 1
- y+ = wild-type allele for gene 2
- y = mutant allele for gene 2
- X = X chromosome
- Y = Y chromosome
Assuming that the genes are located on the same X chromosome, we can predict the genotypes of the F1 females by crossing a w+ y/X male with a w y+/X female. The Punnett square for this cross would be:
| w+ y | w y+
--- | --- | ---
X/Y | w+ y/X | w y+/X
X/X | w+ y/X | w y+/X
From the Punnett square, we can see that all F1 females would be heterozygous for both genes, with the genotype w+ y/X. This is because the X chromosome from the male carries the w+ allele for gene 1 and the Y chromosome carries the y allele for gene 2, while the X chromosome from the female carries the w allele for gene 1 and the y+ allele for gene 2. Therefore, all F1 females would inherit one wild-type allele and one mutant allele for both genes.
In the F2 generation, we would expect to see a 1:1 ratio of w+ y/Y males to w y+/Y males, as the Y chromosome determines maleness and does not carry the genes of interest. The fact that almost 98% of the males in the F2 generation are either w+ y/Y or w y+/Y suggests that the genes are closely linked on the X chromosome, with very few recombinant offspring. This further supports our assumption that the genes are located on the same X chromosome and that the F1 females have the genotype w+ y/X.