​In Drosophila melanogaster, a cross was performed, and the resulting progeny are indicated below.​

​The $$F_1$$  progeny were sib-mated and the  $$F_2$$ progeny were analysed. The following statements were made based on the above crosses and analysis of the progeny:

A. The mutation leading to sepia eye color is located on an autosome.
B. Yellow body is a dominant phenotype.

C. One fourth of the  $$F_2$$  progeny will be males with yellow body color.

​D. In this dihybrid cross, as the  $$F_2$$​ progeny do not show a 9:3:3:1 typical Mendelian ratio, the two genes can be assumed to be linked.

Which one of the following options correctly identifies each statement as True (T) or False (F) from A to D, respectively?

Correct option is A

Parental (P):

• Yellow body, sepia eyes (mutant)

• Gray body, red eyes (wild type)

F1 Generation:

• Gray body, red eyes

• Yellow body, red eyes

→ Indicates that gray body and red eyes are dominant.
→ Male progeny inherit X from mother → express yellow (X-linked), suggesting both genes are on the X chromosome.

Now Evaluate Each Statement:

A. The mutation leading to sepia eye color is located on an autosome - True

 Although yellow body color is X-linked, sepia eye is autosomal recessive (gene se on chromosome 3 in Drosophila).

B. Yellow body is a dominant phenotype - False

•  F1 females (from yellow × gray ) are gray, showing gray is dominant.

C. One fourth of the F2 progeny will be males with yellow body color. -True

•  In X-linked recessive inheritance:

  • From F1: X⁺Xʸ × X⁺Y

  • F2: 1/4 of total (males with XʸY) will be yellow

D. In this dihybrid cross, as the F2 progeny do not show a 9:3:3:1 typical Mendelian ratio, the two genes can be assumed to be linked. - False

•  One gene is autosomal (sepia), the other X-linked (yellow). Since they’re on different chromosomes, they are unlinked.