Q.105. In Drosophila, the red eye phenotype ( W ) is dominant over the recessive white eye mutant (w). In a mixed
population of red and white eye flies of 10,000 individuals, 3,600 flies were white eyed. The percentage of
the heterozygous red eye flies in this population is ____ .

In Drosophila melanogaster, the red-eye allele (W) is dominant over the
white-eye allele (w). In a population of 10,000 flies,
there are 3,600 white-eyed individuals.
Using the Hardy–Weinberg principle, we can calculate the percentage of
heterozygous red-eye flies.

Final Answer: 48% of the population consists of heterozygous red-eye flies.

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Hardy–Weinberg Calculation

White-eyed flies are homozygous recessive (ww), so:

q² = 3600 / 10000 = 0.36

Taking the square root:

q = √0.36 = 0.6

Since p + q = 1:

p = 1 − 0.6 = 0.4

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Frequency of Heterozygotes

The frequency of heterozygous individuals is:

2pq = 2 × 0.4 × 0.6 = 0.48

Thus:

Heterozygous red-eye flies = 48%

Out of 10,000 flies:

0.48 × 10,000 = 4,800 heterozygous red-eye flies

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Cross-Check Using Phenotypes

• Total red-eyed flies = 1 − q² = 1 − 0.36 = 0.64 (6,400 flies)
• Homozygous dominant (WW) = p² = 0.16 (1,600 flies)
• Heterozygous (Ww) = 6,400 − 1,600 = 4,800 flies

4,800 / 10,000 × 100 = 48%

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Autosomal vs Sex-Linked Clarification

Although eye color in Drosophila is often taught as X-linked,
this problem uses W / w notation and provides no sex-specific data.
Therefore, the trait is treated as autosomal and solved using
standard Hardy–Weinberg equilibrium, as expected in exam questions.

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Common Answer Choices Explained

• 36%: Incorrect — equals q², the frequency of white-eyed flies.
• 48%: Correct — equals 2pq, the heterozygous frequency.
• 64%: Incorrect — total red-eyed flies (p² + 2pq).
• 16%: Incorrect — homozygous dominant (p²) only.

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Final Takeaway

Using Hardy–Weinberg equilibrium:

The percentage of heterozygous red-eye flies in the population is 48%.