Difficulty: Intermediate | Prerequisites: Basic probability, understanding of dominant and recessive alleles.
Hardy-Weinberg equilibrium is the bridge between genetics and statistics. It gives you a mathematical model for predicting genotype frequencies in a population when nothing interesting is happening evolutionarily (no selection, no mutation, no migration, random mating, large population). The model matters because departures from it tell you that something is happening, whether that is natural selection, genetic drift, or non-random mating. If you are comfortable with allele notation (uppercase = dominant, lowercase = recessive) and squaring probabilities, you have the prerequisites.
If you know the frequency of two alleles in a population, Hardy-Weinberg lets you predict how many individuals will be homozygous dominant, heterozygous, or homozygous recessive. The three genotype frequencies must sum to 1, and so must the two allele frequencies. It is the genetic equivalent of a null model.
Allele frequency
The proportion of a specific allele in a population's gene pool. Conventionally, p represents the dominant allele frequency and q represents the recessive allele frequency, where p + q = 1.
Genotype frequency
The proportion of a population carrying a particular combination of alleles (e.g. BB, Bb, or bb). Under Hardy-Weinberg equilibrium, these are determined entirely by allele frequencies.
Hardy-Weinberg equilibrium (HWE)
The principle that allele and genotype frequencies in a population remain constant from generation to generation in the absence of evolutionary influences. Think of it as the "nothing is changing" baseline for population genetics.
Homozygous dominant (e.g. BB)
An individual carrying two copies of the dominant allele. Genotype frequency under HWE = p².
Heterozygous (e.g. Bb)
An individual carrying one dominant and one recessive allele. Genotype frequency under HWE = 2pq.
Homozygous recessive (e.g. bb)
An individual carrying two copies of the recessive allele. Genotype frequency under HWE = q².
The entire model rests on two equations:
Allele frequencies: p + q = 1
Genotype frequencies: p² + 2pq + q² = 1
These follow directly from expanding (p + q)². Each term maps to a genotype:
p² → homozygous dominant (BB)
2pq → heterozygous (Bb)
q² → homozygous recessive (bb)
Given: B (dominant allele) has frequency p = 0.8, and b (recessive allele) has frequency q = 0.2, in a population of 100 individuals.
BB = p² = (0.8)² = 0.64 → 64 individuals
Bb = 2pq = 2(0.8)(0.2) = 0.32 → 32 individuals
bb = q² = (0.2)² = 0.04 → 4 individuals
Check: 0.64 + 0.32 + 0.04 = 1.00 ✓
Note: the value for bb is 0.04, not 0.4. This is a common arithmetic slip, since (0.2)² = 0.04.
The model holds only when all of the following are true:
No natural selection (no genotype has a survival or reproduction advantage)
No mutation (alleles are not changing)
No migration (no alleles entering or leaving the population)
Random mating (individuals do not choose partners based on genotype)
Large population size (no genetic drift)
Violating any assumption means observed genotype frequencies will deviate from Hardy-Weinberg predictions, and that deviation itself becomes evidence of an evolutionary force at work.
Allele frequencies:
p + q = 1
Genotype frequencies (Hardy-Weinberg equation):
p² + 2pq + q² = 1
where:
p = frequency of the dominant allele
q = frequency of the recessive allele
p² = frequency of homozygous dominant genotype
2pq = frequency of heterozygous genotype
q² = frequency of homozygous recessive genotype
To find number of individuals: multiply each genotype frequency by the total population size N.
Genetic counsellors use Hardy-Weinberg to estimate carrier frequencies for recessive disorders. If 1 in 2,500 people has cystic fibrosis (q² = 1/2,500), then q = 1/50 and the carrier frequency is 2pq ≈ 2(49/50)(1/50) ≈ 1 in 25. That figure guides screening recommendations.
Conservation biologists test whether small, isolated populations have drifted away from Hardy-Weinberg equilibrium as a measure of genetic health.
"(0.2)² = 0.4." It is 0.04. Squaring a number less than 1 makes it smaller, not larger. This arithmetic error appeared in the homework and is easy to make under time pressure.
"Hardy-Weinberg tells you what is happening in a population." It tells you what would happen if nothing evolutionary were going on. Its value is as a null model; real populations almost always deviate to some degree.
"The heterozygous frequency is pq." It is 2pq. The factor of 2 accounts for two possible arrangements: inheriting B from the mother and b from the father, or vice versa.
"If a population is not in Hardy-Weinberg equilibrium, it must be undergoing natural selection." Any of the five assumptions could be violated. Migration, drift, non-random mating, or mutation could equally be responsible.
⚠️ Be able to go from allele frequencies to genotype frequencies and back. This is a near-guaranteed calculation question.
⚠️ Know all five assumptions. Exams often present a scenario and ask which assumption is being violated.
⚠️ Watch the arithmetic on q². Squaring 0.2 gives 0.04, not 0.4. Write out the multiplication explicitly if you are prone to this error.
⚠️ Remember the factor of 2 in the heterozygote term (2pq). Forgetting it is one of the most common mistakes.
Fill in the blank: If p = 0.7, then q = __________.
True or false: Under Hardy-Weinberg equilibrium, genotype frequencies change every generation.
Fill in the blank: The expected frequency of heterozygotes is calculated as __________.
True or false: (0.3)² = 0.9.
Fill in the blank: Hardy-Weinberg equilibrium assumes the population is __________ (small / large).
Answers: 1. 0.3. 2. False. 3. 2pq. 4. False (it is 0.09). 5. Large.
Q: In a population of 200, the frequency of the dominant allele (A) is 0.6 and the recessive allele (a) is 0.4. How many individuals do you expect to be heterozygous under Hardy-Weinberg equilibrium?
A: 2pq = 2(0.6)(0.4) = 0.48. In a population of 200, that is 0.48 × 200 = 96 heterozygous individuals.
Q: You observe that 1 in 10,000 individuals in a population is homozygous recessive for a trait. Assuming Hardy-Weinberg equilibrium, what is the frequency of the recessive allele?
A: q² = 1/10,000 = 0.0001, so q = √0.0001 = 0.01.
Q: A researcher finds that the observed genotype frequencies in a population differ significantly from Hardy-Weinberg expectations. Name two possible explanations.
A: Any two of: natural selection favouring or disfavouring a genotype, non-random mating (e.g. assortative mating), genetic drift in a small population, migration introducing new alleles, or mutation altering allele frequencies.
Q: Why does the heterozygote frequency formula include a factor of 2?
A: Because a heterozygote (Bb) can form in two ways: B from the mother and b from the father, or b from the mother and B from the father. Each has probability pq, so the combined probability is 2pq.
Hardy-Weinberg equilibrium connects to the chi-squared goodness-of-fit test, which you use to determine whether observed genotype counts differ significantly from expected Hardy-Weinberg frequencies.
It also links to the broader theme of null models in statistics: just as a null hypothesis in a t-test assumes "no difference," Hardy-Weinberg assumes "no evolution." In both cases, you learn the most from the departures.
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