01 / 20

Hardy-Weinberg
Equilibrium

Master the null model of evolution: predict allele and genotype frequencies in non-evolving populations using the equations \( p + q = 1 \) and \( p^2 + 2pq + q^2 = 1 \).

AP Biology

What Is a Gene Pool?

A gene pool is the total collection of alleles for every gene across all individuals in a population. Population genetics tracks how the frequencies of these alleles change (or stay the same) over time.

Allele
A variant form of a gene. For a two-allele system we label them A (dominant) and a (recessive).
Allele Frequency
The proportion of a specific allele in the gene pool: \( p \) = frequency of A, \( q \) = frequency of a.
Genotype Frequency
The proportion of individuals with a particular genotype: AA, Aa, or aa.

The Null Model of Evolution

In 1908, mathematician G. H. Hardy and physician Wilhelm Weinberg independently showed that allele frequencies in a population will not change from generation to generation as long as certain conditions are met. This is the Hardy-Weinberg Principle: a theoretical baseline that describes a population in genetic equilibrium.

Five Conditions for Equilibrium

1
No mutations. No new alleles are created or existing alleles altered.
2
Random mating. Individuals pair without regard to genotype.
3
No natural selection. All genotypes have equal fitness.
4
Very large population size. No genetic drift (random sampling effects).
5
No gene flow. No immigration or emigration of alleles.

The Two Equations

Allele Frequency
$$ p + q = 1 $$
\( p \) = frequency of dominant allele A
\( q \) = frequency of recessive allele a
Genotype Frequency
$$ p^2 + 2pq + q^2 = 1 $$
\( p^2 \) = freq. of AA
\( 2pq \) = freq. of Aa
\( q^2 \) = freq. of aa

Visualizing Allele Combinations

Random Mating: Allele Punnett Square
\( A\ (p) \)
\( a\ (q) \)
\( A\ (p) \)
\( a\ (q) \)
 AA
\( p^2 \)
 Aa
\( pq \)
 Aa
\( pq \)
 aa
\( q^2 \)
Heterozygote (Aa) appears twice: \( pq + pq = 2pq \)

Allele vs Genotype Frequencies

Allele Frequency Genotype Frequency
What it measures Proportion of a single allele in the gene pool Proportion of individuals with a specific genotype
Variables \( p \) and \( q \) \( p^2 \), \( 2pq \), \( q^2 \)
Sum \( p + q = 1 \) \( p^2 + 2pq + q^2 = 1 \)
Observable? Not directly (calculated) Phenotype of aa is directly countable
Key use Track evolutionary change over time Predict expected distribution of genotypes

How to Solve H-W Problems

1
Identify what you know. Problems usually give the frequency of the recessive phenotype (aa individuals). That frequency equals \( q^2 \).
2
Find \( q \). Take the square root: \( q = \sqrt{q^2} \).
3
Find \( p \). Subtract: \( p = 1 - q \).
4
Calculate genotype frequencies. Compute \( p^2 \) (AA), \( 2pq \) (Aa), and \( q^2 \) (aa).
5
Verify. Confirm that \( p^2 + 2pq + q^2 = 1 \). If not, recheck your arithmetic.

Example 1: Cystic Fibrosis

In a population of 10,000 people, 400 are born with cystic fibrosis (an autosomal recessive condition). Assuming Hardy-Weinberg equilibrium, find all allele and genotype frequencies.

Cystic Fibrosis: Frequency Distribution

0 0.2 0.4 0.6 Freq. 0.64
AA (\( p^2 \))
 0.32
Aa (\( 2pq \))
 0.04
aa (\( q^2 \))
 Genotype Frequency Distribution

Example 2: PTC Tasting

About 70% of a population can taste PTC (a bitter chemical). Non-tasting is recessive (tt). Find the frequency of heterozygous tasters (Tt).

Example 3: Testing for Equilibrium

A biologist samples 500 wildflowers: 320 red (RR), 160 pink (Rr), 20 white (rr). Flower color shows incomplete dominance. Is this population in Hardy-Weinberg equilibrium?

What Disrupts Equilibrium?

Natural Selection
Some genotypes have higher fitness: differential survival and reproduction change allele frequencies.
Genetic Drift
Random fluctuations in small populations: bottleneck and founder effects.
Gene Flow
Immigration or emigration introduces or removes alleles from the gene pool.
Mutation
Creates new alleles. Slow effect alone, but provides raw material for other forces.
Non-random Mating
Assortative mating or inbreeding changes genotype frequencies (but not allele frequencies alone).

Common Mistakes

✕ Forgetting to Square Root
You are given \( q^2 \), not \( q \). Always take \( q = \sqrt{q^2} \) before finding \( p \).
✕ Confusing Phenotype with Genotype
With complete dominance, AA and Aa look the same. The recessive phenotype frequency is \( q^2 \), not \( q \).
✕ Using Dominant Phenotype as \( p^2 \)
The dominant phenotype includes both \( p^2 \) (AA) and \( 2pq \) (Aa). You cannot set it equal to \( p^2 \) alone.

Summary at a Glance

Symbol Meaning How to Find
\( p \) Dominant allele (A) frequency \( 1 - q \)
\( q \) Recessive allele (a) frequency \( \sqrt{q^2} \)
\( p^2 \) Homozygous dominant (AA) frequency Square \( p \)
\( 2pq \) Heterozygous (Aa) frequency \( 2 \times p \times q \)
\( q^2 \) Homozygous recessive (aa) frequency Count recessive phenotype

Question 1 of 5

In a population of 1,000 individuals, 90 have a recessive phenotype (aa). Assuming Hardy-Weinberg equilibrium, what is the frequency of the dominant allele (\( p \))?

Question 2 of 5

Which of the following is NOT a condition required for Hardy-Weinberg equilibrium?

Question 3 of 5

If \( p = 0.6 \) and \( q = 0.4 \) in a population at Hardy-Weinberg equilibrium, what is the expected frequency of heterozygotes?

Question 4 of 5

A researcher compares observed genotype frequencies to expected Hardy-Weinberg frequencies and finds a significant difference. What can the researcher conclude?

Question 5 of 5

In a population, 49% of individuals show the recessive phenotype. Approximately what percentage of the population are carriers (heterozygous)?