The Principle
Hardy-Weinberg — the baseline for detecting evolution
The Hardy-Weinberg principle (proposed independently by Godfrey Hardy and Wilhelm Weinberg in 1908) states that allele frequencies and genotype frequencies in a population remain constant from generation to generation in the absence of evolutionary forces. This 'equilibrium' is the genetic equivalent of Newton's first law — a population at rest stays at rest unless acted upon by a force.
The principle is not describing what real populations do — they are almost never in perfect equilibrium. Rather, it provides a mathematical null hypothesis: if none of the five evolutionary forces are operating, allele frequencies will not change. Any deviation from Hardy-Weinberg equilibrium signals that at least one evolutionary force is acting on that locus.
💡 Why Hardy-Weinberg Matters Clinically
Hardy-Weinberg calculations are used in clinical genetics to estimate the frequency of carriers of recessive genetic diseases in a population — which cannot be directly measured from disease incidence alone.
Example — cystic fibrosis: CF affects approximately 1 in 2,500 European-Americans (q² = 1/2,500 = 0.0004). Therefore q = √0.0004 = 0.02. p = 1 - 0.02 = 0.98. Carrier frequency = 2pq = 2(0.98)(0.02) ≈ 0.039 — approximately 1 in 25 European-Americans is a CF carrier. This is crucial for genetic counseling: two carrier parents each have a 1/4 chance of having an affected child.
Example — PKU: Phenylketonuria affects 1/10,000 newborns (q² = 0.0001). q = 0.01. Carrier frequency = 2pq ≈ 2(0.99)(0.01) ≈ 1/50. Hardy-Weinberg calculations reveal that for every child born with PKU, approximately 50 unaffected carriers are born — a ratio that has major implications for screening program design.
Eq
The Hardy-Weinberg equations
For a gene with two alleles (A and a), let p = frequency of allele A and q = frequency of allele a. Since there are only two alleles: p + q = 1.
The expected genotype frequencies under Hardy-Weinberg equilibrium are:
• AA (homozygous dominant): p²
• Aa (heterozygous): 2pq
• aa (homozygous recessive): q²
And: p² + 2pq + q² = 1
This is simply the binomial expansion of (p + q)² = 1. The 2pq term represents heterozygotes — there are two ways to be heterozygous (receive A from mom and a from dad, or vice versa), hence the factor of 2.
Key insight: under Hardy-Weinberg equilibrium, allele frequencies (p and q) do NOT change between generations, even though genotype frequencies are reshuffled each generation by random mating. Alleles are not destroyed or created — they are just recombined.
Memory trick: p + q = 1 (allele frequencies). p² + 2pq + q² = 1 (genotype frequencies). The 2pq is always the heterozygote frequency. If you know q², take its square root to get q, then p = 1 - q.
Calc
Using the equation — solving Hardy-Weinberg problems
Hardy-Weinberg problems always give you one piece of information and ask you to calculate the rest. The most common starting point is the frequency of the homozygous recessive genotype (aa = q²), because it is the only genotype whose frequency is immediately obvious from phenotype counts (only aa individuals show the recessive phenotype).
Standard approach:
1. Find q²: the proportion of individuals showing the recessive phenotype.
2. Find q: q = √q²
3. Find p: p = 1 - q
4. Find heterozygote frequency: 2pq
5. Find homozygous dominant frequency: p²
Example: In a population of 10,000, 900 individuals have albinism (aa). q² = 900/10,000 = 0.09. q = √0.09 = 0.3. p = 1 - 0.3 = 0.7. Carriers (Aa) = 2pq = 2(0.7)(0.3) = 0.42 = 4,200 individuals. AA = p² = (0.7)² = 0.49 = 4,900 individuals.
Memory trick: Always start with q² (the recessive homozygote frequency) because it's the only one you can observe directly from phenotype. Then q = √q², then p = 1 - q, then calculate 2pq and p².
Cond
Five conditions for Hardy-Weinberg equilibrium
Hardy-Weinberg equilibrium requires ALL five conditions to be met simultaneously. Violation of any one causes allele frequencies to change — i.e., evolution occurs:
1. No mutation: No new alleles introduced by mutation. (Mutation is always occurring at low rates — this condition is never perfectly met.)
2. No gene flow: No immigration or emigration of individuals with different allele frequencies. Isolated populations.
3. No genetic drift: Population must be infinitely large (or at least very large). In small populations, allele frequencies change randomly by chance — genetic drift.
4. Random mating: All individuals equally likely to mate with any other. Non-random mating (assortative mating — choosing mates based on genotype) changes genotype frequencies without changing allele frequencies.
5. No natural selection: All genotypes have equal fitness — no differential survival or reproduction.
Memory trick: 'My Gorilla Gave Random Smiles' = Mutation, Gene flow, Genetic drift, Random mating, Selection. Violate ANY of these = population is evolving.
🔬 Applied Scenario — Detecting Evolutionary Forces with Hardy-Weinberg
Deviations from Hardy-Weinberg equilibrium are diagnostic for specific evolutionary forces:
A
Excess heterozygotes → balancing selection or inbreeding avoidance. If a population shows more heterozygotes than 2pq predicts, this suggests balancing selection (heterozygote advantage) is maintaining diversity. MHC (major histocompatibility complex) genes in humans consistently show excess heterozygosity — individuals with more diverse MHC alleles can recognize a wider range of pathogens, suggesting strong balancing selection at this locus.
B
Deficient heterozygotes → inbreeding or population subdivision. If a population shows fewer heterozygotes than 2pq predicts (excess homozygotes), this suggests inbreeding (mating between relatives) or the Wahlund effect (the population is actually composed of several subpopulations with different allele frequencies that are being treated as one). Conservation genetics uses this to detect inbreeding in small populations of endangered species.
C
Changing allele frequencies over time → selection or drift. If p and q are changing significantly between generations, natural selection or genetic drift is operating. Comparing allele frequencies across generations at a candidate locus vs neutral loci allows biologists to determine whether selection (which acts specifically on the candidate locus) or drift (which acts on all loci equally) is responsible.
D
Newborn screening programs. Hardy-Weinberg calculations underpin the design of newborn screening programs. For PKU, CF, sickle cell disease, and dozens of other recessive conditions, the calculated carrier frequency (2pq) tells public health officials how common carriers are in the population — informing decisions about population-wide carrier screening programs, prenatal testing recommendations, and resource allocation for genetic counseling.
📌 Exam Application
Hardy-Weinberg is heavily tested — master the calculation and the conditions:
1. The equation: p + q = 1. p² + 2pq + q² = 1. Know what each term represents.
2. Solving problems: Always start with q² (recessive phenotype frequency). q = √q². p = 1 - q. Heterozygotes = 2pq.
3. Five conditions: No mutation, No gene flow, No genetic drift (large population), Random mating, No selection. Violation of ANY condition = evolution occurring.
4. Clinical application: Calculate carrier frequency for recessive diseases. CF: 1/2,500 affected → 1/25 carriers. PKU: 1/10,000 affected → 1/50 carriers.
5. What deviations mean: Excess heterozygotes = balancing selection. Deficient heterozygotes = inbreeding or population subdivision. Changing allele frequencies = selection or drift.
⚠️ The Most Common Hardy-Weinberg Mistakes
p² is NOT the carrier frequency — 2pq is. Students frequently confuse p² (homozygous dominant, AA) with 2pq (heterozygous carrier, Aa). The carrier frequency is always 2pq. In problems asking 'what proportion of the population are carriers of a recessive disease,' the answer is 2pq — not p² or q².
Always start with q², not q. You can observe q² (frequency of recessive phenotype) from the population data. You cannot directly observe q. The path is always: q² → q → p → 2pq → p². Students who try to start with other values make calculation errors.
Hardy-Weinberg equilibrium does NOT mean no evolution is possible. It means no evolution IS occurring at that moment. A population can be in Hardy-Weinberg equilibrium for one locus while evolving rapidly at another. And a population that deviates from Hardy-Weinberg is not necessarily going extinct — it is simply evolving. Hardy-Weinberg is a mathematical null hypothesis, not a statement about the health or stability of a population.
✓ Quick Self-Test
1. State the Hardy-Weinberg equation for genotype frequencies.
2. What are the five conditions required for Hardy-Weinberg equilibrium?
3. In a population where 1 in 400 people have phenylketonuria (PKU, autosomal recessive), what is the carrier frequency?
4. What does excess homozygosity relative to Hardy-Weinberg predictions suggest about a population?
5. Why is the heterozygote frequency 2pq and not pq?
Answers:
1. p² + 2pq + q² = 1, where p² = frequency of homozygous dominant (AA), 2pq = frequency of heterozygotes (Aa), q² = frequency of homozygous recessive (aa). Also: p + q = 1.
2. No mutation, no gene flow (no immigration/emigration), no genetic drift (infinitely large population), random mating, no natural selection. ALL five must be met simultaneously for the population to remain in equilibrium.
3. q² = 1/400 = 0.0025. q = √0.0025 = 0.05. p = 1 - 0.05 = 0.95. Carrier frequency = 2pq = 2(0.95)(0.05) = 0.095 ≈ 1 in 10 individuals are carriers.
4. Excess homozygosity (fewer heterozygotes than 2pq predicts) suggests inbreeding — mating between relatives increases the probability that both alleles at a locus are identical by descent. It can also indicate population subdivision (the Wahlund effect): if the 'population' actually consists of separate subpopulations with different allele frequencies that are not interbreeding, treating them as one population will show apparent excess homozygosity.
5. There are two ways to be heterozygous: receive allele A from the mother and a from the father (probability p × q) OR receive a from the mother and A from the father (probability q × p). Both combinations produce the Aa genotype, so the total frequency is pq + qp = 2pq.