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Population Genetics &
Hardy–Weinberg Equilibrium

Master the allele frequency equations, understand the five conditions for equilibrium, and learn how to identify evolutionary forces disrupting a gene pool.

MCAT Biology Foundational Concept 1C

What Is a Gene Pool?

A gene pool is the complete set of all alleles for every gene in a population. Population genetics studies how allele frequencies change (or stay the same) across generations.

Allele Frequency
How common a specific allele is in the population, expressed as a proportion from 0 to 1.
Genotype Frequency
The proportion of individuals in a population with a particular genotype (e.g., AA, Aa, or aa).

Allele vs Genotype Frequencies

Population: 100 Individuals AA 36 individuals frequency = 0.36 Aa 48 individuals frequency = 0.48 aa 16 individuals frequency = 0.16
Allele A: \( \frac{2 \times 36 + 48}{200} = \frac{120}{200} = 0.60 \)
Allele a: \( \frac{2 \times 16 + 48}{200} = \frac{80}{200} = 0.40 \)
\( p = 0.60, \quad q = 0.40, \quad p+q=1 \)

Why Hardy–Weinberg Matters

The Hardy–Weinberg principle is the null hypothesis of population genetics. It predicts what genotype frequencies should look like if no evolution is occurring. Any deviation from predicted frequencies signals that an evolutionary force is at work.

Null Model
Like a "control" experiment: it tells us what happens with no evolutionary change.
Diagnostic Tool
Compare observed genotype frequencies to HW predictions to detect evolution.
Clinical Use
Estimate carrier frequencies for autosomal recessive diseases (e.g., cystic fibrosis).

The Hardy–Weinberg Equations

ALLELE FREQUENCIES
\( p + q = 1 \)
 Two alleles must sum to 1 GENOTYPE FREQUENCIES
\( p^2 + 2pq + q^2 = 1 \)
 Expansion of \( (p + q)^2 \)
\( p^2 \)
 = frequency of homozygous dominant (AA)
\( 2pq \)
 = frequency of heterozygotes (Aa)
\( q^2 \)
 = frequency of homozygous recessive (aa)

Five Conditions for Equilibrium

Hardy–Weinberg equilibrium holds only when all five conditions are met simultaneously. In reality, they are almost never all true: this is why populations evolve.

1
No mutations. No new alleles are created or converted.
2
Random mating. No sexual selection or assortative mating.
3
No natural selection. All genotypes have equal fitness.
4
Large population size. No genetic drift (random changes).
5
No gene flow. No migration of individuals in or out.

Forces That Disrupt Equilibrium

Force Primary Genetic Effect Random or Directed?
Natural Selection Increases adaptive alleles Directed
Genetic Drift Random fluctuation in allele frequencies Random
Gene Flow Equalizes allele frequencies Directed
Mutation Introduces new alleles Random
Non-random Mating Changes genotype frequencies Directed

Genetic Drift: Bottleneck vs Founder

Bottleneck Effect
A sudden environmental event drastically reduces population size. Surviving alleles may not represent the original gene pool. Example: cheetah population crash.
Founder Effect
A small group colonizes a new area with a non-representative sample of alleles. Example: Amish populations with high frequency of Ellis-van Creveld syndrome.

How to Apply Hardy–Weinberg

1
Identify what you know. Typically the frequency of the recessive phenotype (\( q^2 \)) or a given allele frequency.
2
Find \( q \). If given \( q^2 \), take the square root: \( q = \sqrt{q^2} \).
3
Find \( p \). Use \( p = 1 - q \).
4
Calculate genotype frequencies. Plug into \( p^2 \), \( 2pq \), and \( q^2 \).
5
Answer the question. Often they ask for carrier frequency (\( 2pq \)) or the number of affected individuals.

Worked Example: Cystic Fibrosis

Cystic fibrosis (CF) is autosomal recessive. About 1 in 2,500 Caucasians is affected. What fraction of the population are carriers?

1
Affected individuals are aa, so \( q^2 = \frac{1}{2500} = 0.0004 \).
2
\( q = \sqrt{0.0004} = 0.02 \)
3
\( p = 1 - 0.02 = 0.98 \)
4
Carrier frequency \( = 2pq = 2(0.98)(0.02) = 0.0392 \)

Worked Example: Sickle Cell Allele

In a West African population, the frequency of the sickle cell allele (\( \text{Hb}^S \)) is 0.12. What percentage of individuals are heterozygous carriers (\( \text{Hb}^A\text{Hb}^S \))?

1
\( q = 0.12 \), so \( p = 1 - 0.12 = 0.88 \).
2
Carrier frequency \( = 2pq = 2(0.88)(0.12) = 0.2112 \).
3
About 21.1% of the population are carriers.

Worked Example: PTC Tasting

In a class of 200 students, 32 cannot taste PTC (non-tasters are tt, autosomal recessive). Calculate all genotype and allele frequencies.

Step 1:
\( q^2 = \frac{32}{200} = 0.16 \)
 Step 2:
\( q = \sqrt{0.16} = 0.40 \)
 Step 3:
\( p = 1 - 0.40 = 0.60 \)
 \(p^2\) (TT)
\( = (0.60)^2 = 0.36 \rightarrow 72 \text{ students} \)
 \(2pq\) (Tt)
\( = 2(0.60)(0.40) = 0.48 \rightarrow 96 \text{ students} \)
 \(q^2\) (tt)
\( = (0.40)^2 = 0.16 \rightarrow 32 \text{ students} \)

Types of Natural Selection

Directional
Favors one extreme phenotype. Shifts the mean of the trait distribution in one direction. Example: antibiotic resistance.
Stabilizing
Favors the intermediate phenotype. Reduces variation. Example: human birth weight.
Disruptive
Favors both extremes over the intermediate. Increases variation. Example: beak sizes in finches.

Fitness and Selection Coefficient

On the MCAT, you may see problems referencing fitness (\( w \)) and the selection coefficient (\( s \)). These describe how strongly selection acts against a genotype.

Fitness (\( w \))
Reproductive success relative to the most fit genotype. Ranges from 0 (lethal) to 1 (most fit).
Selection Coefficient (\( s \))
\( s = 1 - w \). Measures the reduction in fitness. If \( w = 0.7 \), then \( s = 0.3 \) (30% disadvantage).

Common Mistakes on HW Problems

✕ Confusing \( q \) with \( q^2 \)
The recessive phenotype frequency is \( q^2 \), not \( q \). You must take the square root before finding \( p \).
✕ Forgetting to Square Root
Given "1 in 10,000 affected," students use \( q = 0.0001 \) instead of \( q = \sqrt{0.0001} = 0.01 \).
✕ Assuming HW Always Applies
HW is a model. If the question describes selection, drift, or migration, frequencies will deviate from predictions.
✕ Mixing Phenotype and Genotype
Dominant phenotype = \( p^2 + 2pq \), not just \( p^2 \). Heterozygotes show the dominant trait too.

Summary Reference Table

Symbol Meaning Example (CF)
\( p \) Dominant allele frequency 0.98
\( q \) Recessive allele frequency 0.02
\( p^2 \) Homozygous dominant (AA) 0.9604
\( 2pq \) Heterozygous carriers (Aa) 0.0392
\( q^2 \) Homozygous recessive (aa) 0.0004
\( w \) Fitness (0 to 1) Relative reproductive success
\( s \) Selection coefficient (\( 1 - w \)) Measure of disadvantage

Question 1 of 5

In a population in Hardy–Weinberg equilibrium, the frequency of the recessive allele is \( q = 0.3 \). What is the expected frequency of heterozygous individuals?

Question 2 of 5

Phenylketonuria (PKU) is an autosomal recessive disorder affecting 1 in 10,000 newborns. What is the approximate carrier frequency?

Question 3 of 5

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

Question 4 of 5

A devastating hurricane reduces an island bird population from 10,000 to 50 individuals. The surviving birds have different allele frequencies than the original population. This is an example of:

Question 5 of 5

In a population where \( p = 0.6 \) and \( q = 0.4 \), what proportion of individuals showing the dominant phenotype are actually heterozygous?