Hardy-Weinberg Equilibrium: Formula, Assumptions & Carriers

Hardy-Weinberg equilibrium says that in a large, randomly mating population with no mutation, migration or natural selection, allele frequencies stay constant from one generation to the next, and genotype frequencies follow $p^2 + 2pq + q^2 = 1$ (with $p + q = 1$). On Step 1 it is a calculation tool: for a rare autosomal recessive disease, prevalence $= q^2$, so $q = \sqrt{\text{prevalence}}$ and carrier frequency $= 2pq \approx 2q$. A disease that affects 1 in 10,000 people has $q = 1/100$ and about 1 in 50 carriers. For an X-linked recessive disease, affected males $= q$ and affected females $= q^2$.

The two equations, term by term

$p$ and $q$ are allele frequencies; $p^2$, $2pq$ and $q^2$ are genotype frequencies. The genotype equation is the allele equation squared: $(p + q)^2 = p^2 + 2pq + q^2 = 1$.

Why these terms: with random mating, a child gets two alleles drawn independently from the population's pool. Two copies of the disease allele occur with probability $q \times q$. A heterozygote can arise two ways (A from the mother and a from the father, or the reverse), hence $2pq$.

The five assumptions and what breaks each one

The model assumes all five conditions; real populations only approximate them. When observed genotype counts deviate significantly from $p^2 : 2pq : q^2$, at least one assumption is being violated (or the genotyping is wrong).

Facts behind common stems:

• Founder effect: a small group separates from a larger population and carries only a subset of its variation, so a rare allele can end up common in that group. It is a form of genetic drift, and together with social or geographic isolation it is the most likely explanation for most rare recessive disorders that are unusually common in particular communities.
• Consanguinity changes genotype frequencies, not allele frequencies. Related parents are more likely to carry the same rare allele, so more children are homozygous than $q^2$ predicts.
• Heterozygote advantage is selection: the classic example is sickle cell trait in malaria-endemic regions.
• One generation of random mating restores Hardy-Weinberg proportions at an autosomal locus. X-linked loci take several generations, because males carry one copy and females two.

Autosomal recessive: the carrier frequency shortcut

The four-step setup:

1. Affected: $q^2 = \text{prevalence}$
2. Disease allele: $q = \sqrt{\text{prevalence}}$
3. Normal allele: $p = 1 - q \approx 1$ when $q$ is small
4. Carriers: $2pq \approx 2q$

Worked example, prevalence 1 in 10,000:

$q^2 = \frac{1}{10{,}000} \Rightarrow q = \frac{1}{100}, \quad p = \frac{99}{100}$

$2pq = 2 \times 0.99 \times 0.01 = 0.0198 \approx \frac{1}{50}$

About 2% of people carry the allele, while 0.01% are affected.

The last column is the concept behind the math: the rarer the disease, the larger the share of its alleles hiding in healthy carriers. Cystic fibrosis is the classic real-world case. MedlinePlus Genetics puts it at 1 in 2,500 to 3,500 white newborns in the US; applying the $2q$ shortcut to those incidence figures gives an estimated 1 in 25 to 1 in 30 carriers among white Americans (a Hardy-Weinberg estimate, not a measured carrier rate).

Going backward: if the stem gives the carrier frequency, halve it to get $q$, then square it. Carriers 1 in 30 → $q = 1/60$ → prevalence $= 1/3{,}600$.

When not to use $2q$: the shortcut assumes $p \approx 1$. For a common allele, calculate $2pq$ in full: if $q = 0.1$ (prevalence 1 in 100), $2pq = 2 \times 0.9 \times 0.1 = 0.18$, not 0.20.

Two-step questions: risk of an affected child

Multiply each parent's probability of being a carrier, then multiply by 1/4, the chance that two carriers have an affected child:

$P(\text{affected child}) = P(\text{parent 1 carrier}) \times P(\text{parent 2 carrier}) \times \tfrac{1}{4}$

The third row is a built-in check: two random people have an affected child at about the population prevalence. The 2/3 comes from Mendelian genetics, not Hardy-Weinberg: an unaffected child of two carriers is AA or Aa in a 1:2 ratio. The pedigree logic is in autosomal recessive inheritance and the 2/3 rule.

X-linked recessive: males give you q directly

Males have one X chromosome, so every male who inherits the allele is affected: the frequency of affected males equals $q$, with no square root. Females need two copies.

Carrier females are about twice as common as affected males, and affected females are vanishingly rare, which is why X-linked recessive disease is almost always seen in males. The pedigree side is in the X-linked disorders flowchart.

Rare autosomal dominant disease (less often tested): nearly everyone affected is heterozygous, so prevalence $\approx 2q$ and $q \approx$ prevalence/2. No square root here either.

Exam traps

• Prevalence is $q^2$, not $q$ for autosomal recessive disease: take the square root first.
• Affected males are $q$ for X-linked recessive disease: don't take a square root.
• An unaffected sibling of an affected person has a 2/3 carrier risk, not 1/2; parents of an affected child are obligate carriers.
• A random partner's carrier risk is $2q$, not $q$.
• $2q$ is an approximation that works for rare diseases; use the full $2pq$ when $q$ is large.
• Founder effect is drift, heterozygote advantage is selection, consanguinity is nonrandom mating. Match the mechanism to the assumption it breaks.

Practice question

A 26-year-old woman comes for preconception counseling. Her older brother has cystic fibrosis; she and both parents are healthy. Her partner has no family history of cystic fibrosis, and neither has had carrier testing. In their population, which is in Hardy-Weinberg equilibrium, cystic fibrosis affects 1 in 2,500 newborns. What is the probability that their first child will have cystic fibrosis?

• A. 1/100
• B. 1/150
• C. 1/200
• D. 1/300
• E. 1/2,500

Answer: B. She is unaffected with an affected sibling, so her carrier probability is 2/3. For her partner, $q^2 = 1/2{,}500$, so $q = 1/50$ and his carrier probability is $2q = 1/25$. Two carriers have an affected child with probability 1/4:

$\tfrac{2}{3} \times \tfrac{1}{25} \times \tfrac{1}{4} = \tfrac{2}{300} = \tfrac{1}{150}$

Why the others are wrong:

• A (1/100): treats her as an obligate carrier (probability 1). That would be right only if she had already had an affected child.
• C (1/200): uses 1/2 for her carrier risk. Once the affected genotype is excluded, 2 of the 3 remaining equally likely outcomes are carriers.
• D (1/300): uses $q$ (1/50) as the partner's carrier risk instead of $2q$ (1/25).
• E (1/2,500): is the risk for two random people, which ignores her family history.

Key takeaways

• $p + q = 1$ describes alleles; $p^2 + 2pq + q^2 = 1$ describes genotypes. Both require a large population, random mating, no mutation, no migration and no selection.
• The violations to recognize: drift and founder effect, consanguinity, new mutations, gene flow, and selection, including heterozygote advantage.
• Autosomal recessive: prevalence $= q^2$ and carriers $\approx 2q$, which is twice the square root of the prevalence.
• Couple risk = carrier probability × carrier probability × 1/4, using 2/3 for an unaffected sibling and $2q$ for a random partner.
• X-linked recessive: affected males $= q$, carrier females $\approx 2q$, affected females $= q^2$.
• The rarer the disease, the more carriers there are for each affected person (about $2/q$).

Sources

• Population Evolution: Hardy-Weinberg principle of equilibrium — OpenStax Biology 2e (2018)
• Population Genetics: genetic drift, bottleneck and founder effects — OpenStax Biology 2e (2018)
• Hardy-Weinberg equilibrium — Modern Genetics, Biology LibreTexts
• The Hardy-Weinberg Principle — Nature Education Scitable (2010)
• Applications in population genetics — Geneva Foundation for Medical Education and Research (2012)
• Inbreeding: population genetics for forensic analysts — National Institute of Justice
• Inbreeding lecture notes — University of Washington Genome Sciences 453
• Founder Effect — National Human Genome Research Institute glossary
• Cystic fibrosis — MedlinePlus Genetics
• Tay-Sachs disease — MedlinePlus Genetics
• Protective effect of sickle cell trait against malaria — CDC