Here is the detailed solution and explanation for the given genetics problem involving Drosophila melanogaster, alongside an SEO-friendly article format including title, slug, keyphrase, and meta description.

Introduction:

This article explains the genetic cross involving two X-linked genes, w+ and cv+, located 20 centiMorgans apart in Drosophila melanogaster. It focuses on finding the percentage of progeny showing the recessive white-eyed and crossveinless phenotypes after sib-mating the F1 generation.

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Detailed Solution and Explanation:

Genetic Background and Cross Setup:

• Genes w+ (wild-type eye color) and cv+ (wild-type crossvein) are located 20 cM apart on the X chromosome.

• Mutant alleles w (white eyes) and cv (crossveinless) are recessive.

• A homozygous wild-type female (w+ cv+ / w+ cv+) is crossed with a white-eyed, crossveinless male (w cv / Y).

• F1 progeny are sib-mated.

Step 1: Genotypes of F1 Progeny

• Female parent X chromosomes: w+ cv+ and w+ cv+ (homozygous wild-type).

• Male parent X chromosome: w cv.

• F1 females receive one X from each parent: w+ cv+ / w cv (heterozygous).

• F1 males receive the X from the female parent only: w+ cv+ / Y (wild-type).

Step 2: Gametes Produced by F1 Females (w+ cv+ / w cv)

• Genes are linked 20 cM apart; recombination frequency = 20%.

• Parental (non-recombinant) gametes = 80% total, split as:

  • w+ cv+ (40%)

  • w cv (40%)

• Recombinant gametes = 20% total, split as:

  • w+ cv (10%)

  • w cv+ (10%)

Step 3: Gametes Produced by F1 Males (w+ cv+ / Y)

• Males produce gametes with either X (w+ cv+) or Y.

• Female progeny get X from male and X from female.

• Male progeny get X from female and Y from male.

Step 4: Progeny Phenotypes from F1 Sib-Mating

• Female progeny genotypes can be:

  • w+ cv+ / w+ cv+ (wild-type)

  • w+ cv+ / w cv (heterozygous)

  • w cv / w cv (white-eyed, crossveinless; recessive homozygote)

  • And recombinant combinations with crossover phenotypes.

• Male progeny phenotypes depend on their single X chromosome inherited from the F1 female.

Step 5: Calculating Percentage of White-Eyed, Crossveinless Progeny

• White-eyed, crossveinless phenotype requires both recessive alleles (w cv):

  • In females: homozygous recessive w cv / w cv.

  • In males: w cv / Y.

• Using recombination frequency and the genotype ratios, calculation shows:

  • 40% of male progeny will be white-eyed and crossveinless.

  • Males constitute about 50% of total progeny.

  • Therefore, total white-eyed crossveinless progeny = 50% males × 40% affected = 20%.

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Explanation of Options:

• (1) 20% — Correct answer based on the recombination frequency and genotype calculations.

• (2) 40% — Incorrect; this would suggest all males plus females affected.

• (3) 10% — Incorrect; underestimates recombinant contribution and genotype proportions.

• (4) 5% — Incorrect; too low considering recombination and inheritance patterns.

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This detailed genetic cross analysis clarifies how linkage and recombination distances affect inheritance of linked X chromosome traits in Drosophila melanogaster, leading to the conclusion that 20% of the progeny will show the white-eyed and crossveinless phenotype after sib-mating F1 progeny.

This analysis is essential for understanding classical genetics and chromosome mapping in model organisms like Drosophila melanogaster.