Question

You are performing a fruit fly cross similar to those that Morgan performed. Unfortunately, you forgot to write down the parents of your cross. You count the progeny, and find you have 40 red-eyed males, 80 red-eyed females, and 40 white-eyed males. Assuming that all genotypes from this cross should have equal survival rates, what are the genotypes of the parent flies?

Answer 1

Female: X⁺X⁻

Male: X⁺Y

Step-by-step explanation:

• The ruit fly eye-color gene is linked to the X chromosome.

• The allele expressing red color (wild) is dominant over the allele expressing white color (mutant)
• F1 Progeny: 40 red-eyed males, 80 red-eyed females, and 40 white-eyed males

Red-eyed males--> X⁺Y

Red-eyed males--> X⁺X⁺

White-eyed males-->X⁻Y

Being X⁺ the one that expresses red eyes and X⁻ the one that expresses white eyes.

• Individuals with Red eyes: 40 males + 80 females = 120 red-eyed individuals

• Individuals with white eyes: 40 males

This is a proportion 3:1 ----> 120 red: 40 white

There were red and white-eyed males, which Suggesting that the female parental was heterozygous for the color-eye trait, giving one X⁺ to 40 males and X⁻ to the other 40 males.

While the parental male was red-eyed. If it was white-eyed, then white-eyed females would have appeared among the offspring.

Parental) X⁺X⁻ x X⁺Y

Punnet square) X⁺ X⁻

X⁺ X⁺X⁺ X⁺X⁻

Y X⁺Y X⁻ Y

F1) 1/4 X⁺X⁺

1/4 X⁺X⁻

1/4 X⁺Y

1/4 X⁻ Y

2/4 red-eyed females

1/4 red- eyed male

1/4 white-eyed male

3/4 red-eyed individuals

1/4 white- eyed individuals