1. 2 pq 2. pq 3. q² 4. p²

Answer: 1. 2 pq The Hardy-Weinberg equation is p² + 2pq + q² = 1 . p² and q² are the homozygous genotypes, and 2pq is the frequency of heterozygotes.

In Hardy-Weinberg equation, the frequency of heterozygous individual is represented by

1. 2\(pq\)

2. \(pq\)

3. \(q^2\)

4. \(p^2\)

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 2 Practice test neet Save WhatsApp Report

Best answer

Answer: 1. 2\(pq\)

The Hardy-Weinberg equation is \(p^2 + 2pq + q^2 = 1\).

\(p^2\) and \(q^2\) are the homozygous genotypes, and 2pq is the frequency of heterozygotes.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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