In a genetic cross between two true-breeding lines—one with sepia-eyed flies and one with wild-type eyes—if sepia is autosomal recessive, what percent of the F2 generation is expected to have sepia eyes?

In a genetic cross between two true-breeding lines—one with sepia-eyed flies and one with wild-type eyes—if sepia is autosomal recessive, what percent of the F2 generation is expected to have sepia eyes?

• A. 50%
• B. 25%
• C. 0%
• D. 75%

Answer: (B)

Autosomal recessive traits require two copies of the recessive allele to show the phenotype. Crossing true-breeding lines where one parent is sepia (homozygous recessive, ss) and the other is wild-type (homozygous dominant, SS) gives F1 offspring that are all heterozygous (Ss) and display the wild-type phenotype. When F1 individuals mate (Ss × Ss), the F2 genotypes segregate in a 1:2:1 ratio (SS : Ss : ss), which translates to a 3:1 phenotypic ratio of wild-type to sepia. So 1 out of 4, or 25%, of the F2 generation are expected to have sepia eyes. Being autosomal means this pattern is the same in males and females.

Autosomal recessive traits require two copies of the recessive allele to show the phenotype. Crossing true-breeding lines where one parent is sepia (homozygous recessive, ss) and the other is wild-type (homozygous dominant, SS) gives F1 offspring that are all heterozygous (Ss) and display the wild-type phenotype. When F1 individuals mate (Ss × Ss), the F2 genotypes segregate in a 1:2:1 ratio (SS : Ss : ss), which translates to a 3:1 phenotypic ratio of wild-type to sepia. So 1 out of 4, or 25%, of the F2 generation are expected to have sepia eyes. Being autosomal means this pattern is the same in males and females.