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Introduction to the Hardy-Weinberg Equation

Hardy-Weinberg Equilibrium | B.Sc. Zoology E-Content
B.Sc. Zoology · Population Genetics · UGC Four Quadrant E-Content

Introduction to the
Hardy-Weinberg Equation

A complete interactive learning module on genetic equilibrium, allele frequencies, and evolutionary forces in populations.

Dr. Bhabesh Nath Assistant Professor, Department of Zoology
B. N. College, Dhubri, Date of Creation 26 February, 2024
zoologys.co.in Unit 1.1
H-W Equation
+2pq+=1 | p + q = 1

p = frequency of dominant allele A; q = frequency of recessive allele a

e-Text: Study Material

The Hardy-Weinberg principle (also called Hardy-Weinberg equilibrium or HWE) is the foundational null model of population genetics. It was independently formulated in 1908 by the British mathematician Godfrey Harold Hardy and the German physician Wilhelm Weinberg. It states that in a large, randomly mating population free from evolutionary forces, allele and genotype frequencies will remain constant from generation to generation.

Key Insight: Hardy-Weinberg equilibrium does not describe how populations actually evolve — it describes what would happen if they did not evolve. Any deviation from HWE signals that at least one evolutionary force is acting on the population.

1.1 The Hardy-Weinberg Equation

Consider a gene locus with two alleles: A (dominant) with frequency p, and a (recessive) with frequency q. Since these are the only two alleles at the locus:

p + q = 1
For a locus with exactly two alleles, the sum of their frequencies must equal 1 (100% of all alleles at that locus).

With random mating, the genotype frequencies in the next generation are predicted by the binomial expansion of (p + q)²:

p² — Homozygous Dominant

Expected frequency of genotype AA. These individuals carry two copies of the dominant allele.

2pq — Heterozygous

Expected frequency of genotype Aa. These individuals are carriers of the recessive allele but phenotypically dominant.

q² — Homozygous Recessive

Expected frequency of genotype aa. These individuals express the recessive phenotype and are detectable in the population.

1.2 The Five Conditions for Hardy-Weinberg Equilibrium

Hardy-Weinberg equilibrium is maintained only when all five conditions below are met simultaneously. In natural populations, these conditions are rarely — if ever — fully satisfied, making HWE a theoretical baseline rather than a description of reality.

  • 1
    Large Population Size: Genetic drift (random fluctuations in allele frequency) is negligible only in very large populations. In small populations, chance events can significantly alter allele frequencies.
  • 2
    Random Mating (Panmixia): Every individual must have an equal probability of mating with any other individual. Non-random mating, such as assortative mating or inbreeding, distorts genotype frequencies.
  • 3
    No Mutation: The rate of mutation must be negligible so that no new alleles are introduced into the gene pool and existing alleles are not converted into other forms.
  • 4
    No Gene Flow (Migration): There must be no immigration or emigration that would bring in or remove alleles from the population.
  • 5
    No Natural Selection: All genotypes must have equal fitness — that is, equal survival and reproductive success. Selection for or against any genotype will shift allele frequencies over time.

1.3 Significance of Hardy-Weinberg Equilibrium

The HWE principle is fundamental to population genetics and evolutionary biology for several reasons. First, it provides the null hypothesis for population studies — if observed genotype frequencies deviate significantly from expected HWE frequencies (tested by chi-square analysis), researchers infer that one or more evolutionary forces are acting. Second, it allows estimation of carrier frequencies for recessive genetic disorders in human populations — a critical tool in genetic counselling. Third, it confirms that Mendelian inheritance in itself does not change allele frequencies, refuting early misconceptions that dominant alleles would automatically increase in frequency over generations.

Application — Estimating Carrier Frequency: If the frequency of a recessive disorder (e.g., cystic fibrosis) in a population is 1 in 2500 (q² = 0.0004), then q = 0.02 and p = 0.98. The expected carrier frequency (2pq) ≈ 2 × 0.98 × 0.02 ≈ 0.039, or roughly 1 in 25 individuals.

1.4 Population Stability vs. Genetic Equilibrium

Population stability refers to maintenance of relatively constant population size and genetic diversity. It is governed by birth rates, death rates, immigration, emigration, mutation, and selection. Genetic equilibrium (HWE) is a more specific concept — constancy of allele and genotype frequencies — and requires the five conditions listed above. A population can be numerically stable but genetically evolving, or vice versa.

Deviations from HWE may indicate genetic drift (random allele frequency changes), gene flow (migration of individuals), mutation pressure, natural selection, or non-random mating. Each deviation is itself a potential evolutionary mechanism worth investigating.

📚 Key Terms Glossary

Click any term to reveal its definition.

Allele Frequency
The relative proportion of a particular allele among all alleles at a given locus in the gene pool of a population. Expressed as a value between 0 and 1.
Genotype Frequency
The proportion of individuals in a population possessing a specific genotype (e.g., AA, Aa, or aa). HWE predicts these as p², 2pq, and q² respectively.
Genetic Drift
Random fluctuations in allele frequencies due to chance events, especially pronounced in small populations. A violation of the large-population condition of HWE.
Gene Flow
The transfer of alleles between populations via migration of individuals or gametes. Introduces or removes alleles, disrupting HWE.
Panmixia
Random mating within a population, where every individual has an equal probability of mating with any other individual regardless of genotype or phenotype.
Null Hypothesis
In population genetics, HWE serves as the null hypothesis — the baseline of no evolution. Significant deviation from HWE (tested by χ² analysis) suggests evolutionary forces are at work.
Simulations & Visual Learning

🔢 Allele Frequency Calculator & Genotype Bar Chart

Set allele frequencies p and q (must sum to 1.0) to see predicted HWE genotype frequencies.

Frequency of allele A (p) 0.60
Frequency of allele a (q) 0.40
⚠ p + q must equal 1.00 — q has been auto-adjusted.
AA
p² = 0.360
Aa
2pq = 0.480
aa
q² = 0.160
p² + 2pq + q² = 0.360 + 0.480 + 0.160 = 1.000 — Population is in Hardy-Weinberg Equilibrium.
📊 When Does H-W Equilibrium Break Down?

The table below summarises the five evolutionary forces, the H-W condition each violates, and the direction of their effect on allele frequencies.

Evolutionary Force H-W Condition Violated Effect on Allele Frequencies Example
Natural Selection No selection; equal fitness Increases frequency of favoured allele; decreases deleterious allele Sickle-cell allele maintained in malaria-endemic regions
Genetic Drift Large population size Random increase or decrease; fixation or loss possible Founder effect in isolated human populations
Gene Flow (Migration) No migration Introduces new alleles; homogenises frequencies between populations Colour morphs in bird populations across continents
Mutation No mutation Introduces new alleles; generally very slow rate per generation De novo mutations in germ cells
Non-random Mating Random mating (panmixia) Alters genotype frequencies without changing allele frequencies Inbreeding — increases homozygosity
🧮 Worked Problem — Calculating Carrier Frequency

In a population, albinism (autosomal recessive) affects 1 in 10,000 individuals. Assuming Hardy-Weinberg equilibrium, calculate: (a) the frequency of the recessive allele q, (b) the frequency of the dominant allele p, and (c) the frequency of carriers.

1

Identify the frequency of homozygous recessives (q²) from the problem statement.

q² = 1/10,000 = 0.0001
2

Calculate q by taking the square root of q².

q = √0.0001 = 0.01
3

Calculate p using the fundamental relationship p + q = 1.

p = 1 − q = 1 − 0.01 = 0.99
4

Calculate the carrier frequency (heterozygotes, Aa) using the HWE formula.

2pq = 2 × 0.99 × 0.01 = 0.0198 ≈ 1 in 50
Approximately 1 in 50 individuals is a carrier — far more common than the 1 in 10,000 affected individuals, highlighting the importance of HWE in genetic counselling.
Self-Assessment

Test your understanding of the Hardy-Weinberg principle. Answer each question to see immediate feedback with explanation.

Question 01 / 07
The Hardy-Weinberg principle was independently formulated in:
The principle was published in 1908 by G. H. Hardy (mathematician) and Wilhelm Weinberg (physician) independently, establishing the mathematical foundation of population genetics.
Question 02 / 07
In a population at Hardy-Weinberg equilibrium, if the frequency of the dominant allele (p) is 0.7, what is the expected frequency of heterozygotes (2pq)?
With p = 0.7, q = 1 − 0.7 = 0.3. Therefore 2pq = 2 × 0.7 × 0.3 = 0.42. The value 0.49 is p², and 0.09 is q².
Question 03 / 07
Which of the following is NOT a condition required for Hardy-Weinberg equilibrium?
HWE requires NO mutation (or negligible mutation rate). A high mutation rate would continuously introduce new alleles, disrupting the equilibrium. The other three options (large population, random mating, no selection) are correct conditions.
Question 04 / 07
In a population of 10,000 individuals, 900 have the genotype aa (homozygous recessive). What is the frequency of the recessive allele q?
q² = 900/10,000 = 0.09. Therefore q = √0.09 = 0.3. Note: 0.09 is q², not q itself — a common error! q = 0.3, and p = 1 − 0.3 = 0.7.
Question 05 / 07
Inbreeding in a population primarily causes:
Inbreeding (non-random mating) alters genotype frequencies — specifically increasing homozygosity — without necessarily changing allele frequencies themselves. This violates the random mating condition of HWE.
Question 06 / 07
The Hardy-Weinberg principle is best described as a:
HWE is a null model — it predicts what would happen in the absence of evolutionary forces. Any deviation from HWE suggests evolution is occurring. It does not describe the mechanisms of evolutionary change itself.
Question 07 / 07
If a recessive disorder affects 1 in 10,000 people, using HWE, what is the approximate frequency of heterozygous carriers in the population?
q² = 0.0001, so q = 0.01 and p = 0.99. Carrier frequency 2pq = 2 × 0.99 × 0.01 ≈ 0.02 = 1 in 50. Carriers are 200 times more common than affected individuals!
0/7
Complete the quiz to see your score
✏️ Fill in the Blanks

Complete each statement using your knowledge of the Hardy-Weinberg principle.

1. In a population at HWE, if p = 0.6, the frequency of homozygous dominant genotype (AA) is .

2. The sum of allele frequencies at a two-allele locus is always .

3. The evolutionary force that involves random changes in allele frequency in small populations is called .

4. Non-random mating violates the HWE condition of .

5. The genotype frequency of heterozygotes (Aa) in a HWE population is given by the expression .

Interaction & Application
🧩 Activity 1: Sort the Conditions

Drag each item into the correct category: conditions required for HWE, or factors that disrupt HWE.

Random mating
Natural selection
Large population
Genetic drift
No mutation
Gene flow
No natural selection
Inbreeding
✅ Required for HWE
❌ Disrupts HWE
🔬 Activity 2: Population Dynamics Simulator

Simulate how allele frequencies change over generations when HWE conditions are violated. Adjust the settings and observe how frequency of allele a (q) evolves.

Frequency of q (allele a)
Frequency of p (allele A)
📝 Activity 3: Application Problems

Solve these problems and click to reveal the answer and explanation.

Problem 1: In a butterfly population, brown wing colour (B) is dominant over yellow (b). In a sample of 500 butterflies, 180 are yellow. Calculate: (a) frequency of b, (b) frequency of B, (c) number of heterozygous brown butterflies expected.
▶ Show Solution

Step 1: q² = 180/500 = 0.36

Step 2: q = √0.36 = 0.6 (frequency of b)

Step 3: p = 1 − 0.6 = 0.4 (frequency of B)

Step 4: 2pq = 2 × 0.4 × 0.6 = 0.48 → 0.48 × 500 = 240 heterozygous butterflies

Problem 2: A wildlife survey of 1000 deer finds that 490 are homozygous dominant (AA), 420 are heterozygous (Aa), and 90 are homozygous recessive (aa). Is this population in Hardy-Weinberg equilibrium? Show your calculation.
▶ Show Solution

Observed: AA = 490/1000 = 0.49; Aa = 420/1000 = 0.42; aa = 90/1000 = 0.09

Calculate allele frequencies: p = (2×490 + 420)/(2×1000) = 1400/2000 = 0.7; q = 0.3

Expected under HWE: p² = 0.49; 2pq = 0.42; q² = 0.09

Conclusion: Observed = Expected ✅ → Population IS in HWE.

🎯 Learning Outcome Reflection
After completing this module, you should be able to: (1) state and explain the Hardy-Weinberg equation; (2) list and explain all five conditions of HWE; (3) calculate allele and genotype frequencies given observed data; (4) identify and explain the evolutionary forces that disrupt HWE; and (5) apply HWE to estimate carrier frequencies in human genetics problems.

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