Maximum Heterozygote Frequency

In H.W. equilibrium, genotypic frequency of heterozygote will be highest when gene frequency of one gene is-

Correct Answer: 0.50

In Hardy-Weinberg equilibrium, the frequency of heterozygotes is 2pq. This expression reaches its maximum value when p = q = 0.5, giving 2pq = 2 × 0.5 × 0.5 = 0.50 (50%).

Mathematical Proof

• Heterozygote frequency: H = 2pq; where p + q = 1, so q = 1 − p
• H = 2p(1−p) = 2p − 2p²
• To maximise: dH/dp = 2 − 4p = 0 → p = 0.5
• Maximum H = 2 × 0.5 × 0.5 = 0.50

Key Notes on Heterozygote Frequency

• Frequency of heterozygotes (2pq) cannot exceed 50%
• Maximum is achieved when p = q = 0.5
• H-W equation: p² + 2pq + q² = 1
• p + q = 1 (sum of all allele frequencies = 1)
• P + H + Q = 1 (sum of all genotypic frequencies = 1)

Why Other Options Are Wrong

• 0.60 → 2 × 0.6 × 0.4 = 0.48 < 0.50; not the maximum
• 0.40 → 2 × 0.4 × 0.6 = 0.48 < 0.50; not the maximum
• 0.9 → 2 × 0.9 × 0.1 = 0.18; very low heterozygote frequency