§ B-06 Mendelian Genetics & Hardy–Weinberg Equilibrium: Genotype Frequency Dynamics (Grade 9–11 · Mendelian & Population Genetics)
Punnett Square, 3:1 & 9:3:3:1 Segregation Ratios & Population Allele Equilibrium p² + 2pq + q² = 1. “In a large random-mating population free from selection, mutation, and migration, allele frequencies p and q remain constant across generations, and genotype frequencies AA, Aa, and aa follow the binomial expansion (p + q)² = p² + 2pq + q² = 1.”
Theory Summary & Scientific Mechanism
Gregor Mendel’s laws of segregation (1:2:1 genotypic and 3:1 phenotypic ratios in a monohybrid cross Aa × Aa) and independent assortment (9:3:3:1) arise from meiotic allele separation and random gamete union visualized by the Punnett square.
At the population level, the Hardy–Weinberg principle (1908) states that in a large, random-mating diploid population free from selection, mutation, migration, and drift, allele frequencies p = f(A) and q = f(a) (p + q = 1) remain constant across generations.
Genotype frequencies obey the binomial expansion (p + q)² = p² + 2pq + q² = 1, where p² is the homozygous dominant frequency (AA), 2pq is the heterozygous carrier frequency (Aa, peaking at 0.50 when p = q = 0.5), and q² is the homozygous recessive frequency (aa). Under selection s against aa, mean fitness is w = 1 - s q² and the next-generation allele frequency shifts to p' = (p² + pq)/(1 - s q²).
Everyday Analogy
If a bag holds 60% blue marbles (p=0.6) and 40% yellow marbles (q=0.4), drawing two at random gives two blue with probability p² = 0.36, two yellow with q² = 0.16, and a mixed pair in two ways with 2pq = 0.48! That binomial law is Hardy–Weinberg equilibrium.
Interactive Question & Derivation
In a population at Hardy–Weinberg equilibrium, the frequency of the recessive phenotype (homozygous aa) is q² = 0.16 (16%). What are the frequencies of the recessive allele q, the dominant allele p = 1 - q, and the heterozygous carrier genotype 2pq (Aa)?
- A: Recessive allele q = 0.16, dominant allele p = 0.84, and heterozygotes 2pq = 0.16
- B: Recessive allele q = √(0.16) = 0.40, dominant allele p = 1 - 0.40 = 0.60, and heterozygous carriers Aa make up 2pq = 2 · 0.60 · 0.40 = 0.48 (48%)
- C: All individuals showing the dominant phenotype (84%) are homozygous AA
- undefined: Heterozygote frequency is always 50% regardless of p and q
Study Hint: From q² = 0.16 we get q = √(0.16) = 0.40, so p = 1 - 0.40 = 0.60 and heterozygous carriers 2pq = 2(0.60)(0.40) = 0.48 (48%).
Proof & Explanation: Under Hardy–Weinberg equilibrium, alternative allele frequencies sum to unity (p + q = 1) and genotype frequencies are f(AA) = p², f(Aa) = 2pq, and f(aa) = q². Since the recessive phenotype occurs only in aa homozygotes, q² = 0.16 yields q = √(0.16) = 0.40. The dominant allele frequency is p = 1 - q = 0.60. Homozygous dominants AA represent p² = 0.36 (36%), while heterozygous carriers Aa represent 2pq = 2 · 0.60 · 0.40 = 0.48 (48%). When p = q = 0.5 (monohybrid cross Aa × Aa), we recover Mendel’s classic 1:2:1 genotype ratio (25% : 50% : 25%) and 3:1 dominant-to-recessive phenotype ratio.
Key Equations & 3D Simulation Parameters
f(AA) = p², f(Aa) = 2pq, f(aa) = q², f_dom = p² + 2pq = 1 - q²
- Dominant allele frequency p (q = 1 − p; p = 0.5 → Mendel 3:1): 0.1 … 0.9 (default: 0.6)
- Inheritance mode (1=Complete dominance 3:1, 0.5=Incomplete dominance 1:2:1): 0.5 … 1 (default: 1)
- Natural selection coefficient s against recessive homozygotes aa: 0 … 0.8 (default: 0)
- Sampled offspring population size Nᵢnd: 50 … 500 ind. (default: 200 ind.)