Problem 1 (Source Material — Classic)

Given: 16% of population shows recessive phenotype. Find: Dominant allele frequency and carrier frequency.

$q^{2}$ = 0.16
q = √0.16 = 0.4
p = 1 - 0.4 = 0.6

Dominant allele frequency = p = 0.6 (ANSWER)
Carrier frequency = 2pq = 2 × 0.6 × 0.4 = 0.48 = 48%

Problem 2 (Standard NEET Format)

Given: In a population of 10,000, 900 individuals have albinism (aa). Find: Number of carriers.

$q^{2}$ = 900/10,000 = 0.09
q = √0.09 = 0.3
p = 1 - 0.3 = 0.7
Carrier frequency = 2pq = 2 × 0.7 × 0.3 = 0.42
Number of carriers = 10,000 × 0.42 = 4,200

Problem 3 (Given Dominant Allele Frequency)

Given: Frequency of allele A = 0.8. Find: All genotype frequencies.

p = 0.8, q = 1 - 0.8 = 0.2
AA = $p^{2}$ = 0.64 = 64%
Aa = 2pq = 2 × 0.8 × 0.2 = 0.32 = 32%
aa = $q^{2}$ = 0.04 = 4%
Verify: 0.64 + 0.32 + 0.04 = 1.00 ✓

Problem 4 (Rare Disease — Approximation)

Given: Carrier frequency for a rare disease = 2%. Find: Disease frequency.

2pq = 0.02
Since disease is rare, p ≈ 1
∴ 2q ≈ 0.02 → q ≈ 0.01
Disease frequency = $q^{2}$ = (0.01)^{2} = 0.0001 = 0.01%
(1 in 10,000 individuals affected)

Key Relationships to Memorize

1. Carrier frequency is always LARGER than disease frequency for rare recessives.
2. When p = q = 0.5 → maximum heterozygosity (2pq = 0.5).
3. When q is small → disease is rare → most recessive alleles hidden in carriers.
4. Verify all answers: $p^{2}$ + 2pq + $q^{2}$ must = 1.