Assume that the genes w⁺ and cv⁺ are located 20 cM apart on the X chromosome of Drosophila melanogaster. Mutations in w⁺ and cv⁺ give rise to white eyes and crossveinless phenotypes, respectively, which are recessive to the wild-type phenotype. A homozygous wild-type female was crossed to a white-eyed, crossveinless male. The F₁ progeny was sib-mated. What percentage of the progeny will be white-eyed and crossveinless?

Correct option is A

​Problem Breakdown:

• Genes w+ (wild type) and cv+ (wild type) are located 20 cM apart on the X chromosome of Drosophila melanogaster.
• Mutations in w and cv give rise to white eyes and crossveinless phenotypes, respectively. These mutations are recessive to the wild-type phenotypes.
• A homozygous wild-type female (w+ w+ cv+ cv+) was crossed to a white-eyed, crossveinless male (w cv).
• The F1 progeny were sib-mated.
• We need to find the percentage of white-eyed, crossveinless progeny in the F2 generation.

Step-by-Step Solution:

1. Genotypes of the Parents:
• Female: w+ w+ cv+ cv+ (wild type, homozygous)
• Male: w cv (white-eyed, crossveinless; X-linked recessive mutations)
2. F1 Progeny:
• The F1 generation will inherit one X chromosome from each parent. Since the male has the mutations w and cv, the F1 females will inherit w cv from their father and w+ cv+ from their mother. The F1 males will inherit w cv from their father and w+ cv+ from their mother.

The F1 progeny will have the following genotypes:

• F1 females: w+ cv+ / w cv (wild-type eyes, wild-type crossveins; heterozygous for both traits)
• F1 males: w+ cv+ / w cv (wild-type eyes, wild-type crossveins; hemizygous for both traits)
1. Sib-Mating (F1 Progeny):
• When the F1 generation is sib-mated, we cross F1 females (w+ cv+ / w cv) with F1 males (w+ cv+ / w cv).
• The F2 generation will have the following potential gametes from each parent:
• Female (w+ cv+ / w cv): Can produce gametes w+ cv+, w cv+, w+ cv, w cv.
• Male (w+ cv+ / w cv): Can produce gametes w+ cv+, w cv+, w+ cv, w cv.
2. F2 Generation Cross: We now consider the genotypes of the F2 progeny:
• w+ w+ cv+ cv+ (wild type)
• w w+ cv+ cv (white-eyed, wild-type crossveins)
• w+ w cv+ cv+ (wild-type eyes, crossveinless)
• w w cv cv (white-eyed and crossveinless)

F2 Phenotypes:

• White-eyed and crossveinless: These progeny must inherit the w mutation for white eyes and the cv mutation for crossveinless. From the cross, the probability of getting w w cv cv is calculated by considering the probability of both the white-eye and crossveinless traits being inherited together. These genes are 20 cM apart, so there is a 20% chance of recombination between them, meaning an 80% chance that they will be inherited together (as non-recombinant alleles).

Therefore, the percentage of progeny with white eyes and crossveinless will be 20% (recombinant frequency for these two traits).

Correct Answer:

Ans. (1) 20