Hardy-Weinberg Equilibrium: Study Notes

Hardy-Weinberg Equilibrium: Concept and Implications

Core idea: In a large population with random mating and no evolutionary forces, allele frequencies stay constant across generations.
If there are two alleles at a locus, say A and a, and their frequencies are p and q respectively, then under random mating the genotype frequencies in the next generation follow:
- P(AA) = p^2\,,\quad P(Aa) = 2pq\,,\quad P(aa) = q^2
The sum of allele frequencies is 1: p + q = 1\,, so q = 1 - p\$.
If you mix alleles at random (random union of gametes), you would expect those genotype proportions, and that scenario is described by Hardy-Weinberg equilibrium.
When observed genotype frequencies match these expected frequencies, the population is said to be in Hardy-Weinberg equilibrium; if they differ, evolution or other factors may be at play.
The transcript’s framing: the idea of "no selection" or no evolution leads to random sampling of alleles and the resulting genotype distribution following the p^2, 2pq, q^2 pattern.

Very large population size (to avoid genetic drift).
Random mating (no mating bias or sexual selection).
No mutation introducing new alleles.
No migration (no gene flow between populations).
No natural selection (genotypes have equal viability and fertility).
Usually non-overlapping generations (for simple modeling), though the core results can apply more broadly.

Define:
- p = \text{frequency of the A allele}
- q = \text{frequency of the a allele}
Relationship:
- p + q = 1\,, hence q = 1 - p\,.
Under random mating, genotype frequencies are:
- P(AA) = p^2\,
- P(Aa) = 2pq\,
- P(aa) = q^2\,
Example with numbers: if p = 0.6, q = 0.4 then
- P(AA) = 0.36\,
- P(Aa) = 0.48\,
- P(aa) = 0.16\,

From the allele frequencies in gametes, the formation of zygotes is like drawing two alleles independently:
- Probability of AA = p\times p = p^2
- Probability of Aa = p\times q + q\times p = 2pq
- Probability of aa = q\times q = q^2
This is equivalent to expanding the binomial model of two alleles: (p + q)^2 = p^2 + 2pq + q^2 , and using p + q = 1.
The key takeaway: genotype frequencies in equilibrium are determined solely by current allele frequencies; they do not depend on initial genotype frequencies, provided assumptions hold.

You can test whether a population is in HWE by comparing observed genotype counts to expected counts under HWE.
Steps:
1. Collect observed counts: O{AA}, O{Aa}, O{aa} for a sample size N = O{AA} + O{Aa} + O{aa}.
2. Estimate allele frequencies from data (from observed counts):
- \hat{p} = \frac{2\,O{AA} + O{Aa}}{2N}
- \hat{q} = 1 - \hat{p}
1. Compute expected counts under HWE:
- E_{AA} = N \hat{p}^2\,
- E_{Aa} = N \cdot 2\hat{p}\hat{q}\,
- E_{aa} = N \hat{q}^2\,
1. Use a chi-squared test to compare observed vs expected:
- \chi^2 = \sum{i \in {AA,Aa,aa}} \frac{(Oi - Ei)^2}{Ei}
1. Degrees of freedom (df): typically 1 for a biallelic locus when p is estimated from data; df = 2 if p and q are fixed a priori.
2. Decision: compare \chi^2 to the critical value at your chosen significance level (e.g., 0.05). If \chi^2 is larger, reject HWE; if not, fail to reject HWE.
Important caveats:
- Deviations can indicate selection, drift, mutation, migration, nonrandom mating, or sampling/genotyping errors.
- Small sample sizes or population structure (Wahlund effect) can masquerade as deviation from HWE.

Concept: compare observed vs expected frequencies to determine if evolutionary forces are acting.
If observed frequencies match the HWE expectations, the population is in Hardy-Weinberg equilibrium for that locus (no evolution affecting that locus under the model).
The transcript describes this as: “these two numbers are the same, so this is Hardy-Weinberg.”
Example scaffold (numbers can vary):
- Suppose observed counts: O{AA} = 36,\; O{Aa} = 48,\; O_{aa} = 16 in a sample of N = 100 individuals.
- Estimate allele frequencies: \hat{p} = \frac{2\cdot 36 + 48}{200} = \frac{144}{200} = 0.72\,, which would imply \hat{q} = 0.28\,.
- Expected counts: E{AA} = 100 \cdot 0.72^2,\; E{Aa} = 100 \cdot 2 \cdot 0.72 \cdot 0.28,\; E_{aa} = 100 \cdot 0.28^2.
- Compare with observed, compute \chi^2, and decide if in HWE.
Practical note: In many classroom or real-world datasets, you first estimate p from the data, then compute expected counts, then apply the chi-square test as above.

Relationship to Mendelian genetics: HWE provides a bridge from allele-level frequencies to genotype-level expectations in populations.
Foundational principle: Under no-evolutionary-forces assumptions, allele frequencies do not change across generations, and genotype frequencies stabilize as a function of p and q.
Real-world relevance:
- Used in population genetics, evolutionary biology, and forensic genetics.
- In human genetics, HWE checks are a standard quality-control step in genotyping data; deviations can flag errors or population structure.
- Helps infer whether part of the population is experiencing selection, inbreeding, or substructure.

If deviations from HWE are detected, consider possible causes:
- Natural selection: some genotypes confer higher viability/fertility.
- Nonrandom mating: assortative mating or inbreeding increases homozygosity.
- Population structure/genetic subdivision: Wahlund effect can reduce heterozygosity.
- Mutation or migration introducing new alleles.
- Genotyping errors or sampling biases.
Ethical/philosophical/real-world angle:
- In human studies, deviations can reflect social or historical population structures; careful interpretation is needed to avoid misattributing social factors to biology.
- The model assumes idealized conditions; real populations often violate one or more assumptions, so HWE serves as a diagnostic baseline rather than a universal law.

Allele frequencies and sum rule:
- p + q = 1\,.
Genotype frequencies under random mating:
- P(AA) = p^2\, , \quad P(Aa) = 2pq\, , \quad P(aa) = q^2\,
Observed and estimated allele frequency from data:
- \hat{p} = \dfrac{2\,O{AA} + O{Aa}}{2N}\,, \quad \hat{q} = 1 - \hat{p}\,
Expected genotype counts under HWE:
- E{AA} = N \hat{p}^2\, , \quad E{Aa} = N \cdot 2\hat{p}\hat{q}\, , \quad E_{aa} = N \hat{q}^2\,
Chi-squared test statistic:
- \chi^2 = \dfrac{(O{AA} - E{AA})^2}{E{AA}} + \dfrac{(O{Aa} - E{Aa})^2}{E{Aa}} + \dfrac{(O{aa} - E{aa})^2}{E_{aa}}\,
Degrees of freedom: typically df = 1 for a single biallelic locus with allele frequencies estimated from data.

Hardy-Weinberg equilibrium provides a baseline expectation for genotype frequencies given allele frequencies, assuming no evolutionary forces.
The key test involves comparing observed genotype counts to HW-predicted counts using the chi-square test and interpreting deviations in the context of possible evolutionary forces or data issues.
The fundamental formulas are p+q=1\, ,\quad P(AA)=p^2,\; P(Aa)=2pq,\; P(aa)=q^2\,,$$ with practical calculation of p from observed data and chi-square testing to assess equilibrium.