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

Answer:

Female: X⁺X⁻  

Male: X⁺Y

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