4.5 - Hardy-Weinberg and reproductive isolation

Counting alleles in a population

An allele frequency measures the proportion of copies of a gene that are a particular allele in the population's gene pool.

Consider one autosomal gene with two alleles, $A$ and $a$, in diploid organisms. Each individual has two copies: $AA$, $Aa$ or $aa$. Let $p$ be the frequency of $A$ and $q$ the frequency of $a$. These frequencies are dimensionless proportions between 0 and 1. The two alleles exhaust the possibilities, so $p+q=1$, or 100%. A dominant allele is not necessarily the more frequent allele: dominance describes a heterozygote's phenotype.

If every individual's genotype is known, count alleles directly. Each $AA$ contributes two $A$ copies; each $Aa$ contributes one of each; each $aa$ contributes two $a$ copies.

A complete genotype census

The following original illustrative data describe all 500 individuals in one breeding generation of a closed study population.

The genotype count ratio is $180:240:80=9:12:4$. The allele frequencies are $p=600/1000=0.60$ and $q=400/1000=0.40$. Thus 40% of allele copies are $a$, whereas only 16% of individuals are $aa$. Heterozygotes account for the additional $a$ copies.

This counting method does not require Hardy–Weinberg equilibrium. It works even when mating is non-random or selection has changed the genotype proportions. A representative sample gives an estimate of population frequency; a complete census, as stipulated here, gives the frequency for that defined population at that time.

Predicting genotype proportions with Hardy–Weinberg

The Hardy–Weinberg model predicts how alleles combine into genotypes when gametes meet at random under stated conditions.

For this two-allele autosomal model, suppose both parental gamete pools have $P(A)=p$ and $P(a)=q$. Random union makes the two gamete draws independent. Multiply the probabilities for one combination; add the probabilities of mutually exclusive routes to the same genotype.

The two heterozygote routes give $pq+pq=2pq$. They are not two different genotypes.

For example, if $p=0.70$, rearranging $p+q=1$ gives $q=1-p=0.30$. Substitution gives $p^2=0.49$, $2pq=2(0.70)(0.30)=0.42$, and $q^2=0.09$. These add to 1. In 1000 offspring, the expected counts are 490 $AA$, 420 $Aa$ and 90 $aa$. They are probability-based expectations, not a quota that every finite set of offspring must meet exactly.

Conditions behind a stable model

For the simple equilibrium model to maintain allele frequencies across generations, assume a very large population so random genetic drift is negligible; random mating with respect to this gene; no mutation changing one allele into another; no immigration or emigration transferring alleles; and no natural selection favouring a genotype through differential survival or reproduction. Use diploid sexual inheritance, ordinary Mendelian segregation and the same allele frequencies in the two parental gamete pools. Compare the same life stage in each generation: selection between fertilisation and adulthood can alter observed genotype proportions.

Random mating explains the $p^2:2pq:q^2$ combination rule. Absence of forces such as selection, migration and mutation explains why $p$ and $q$ remain unchanged in the ideal model over time. These are related claims, but they are not interchangeable. Non-random mating can change the distribution of genotypes without itself changing the frequencies of the alleles being counted.

The model provides a baseline for investigating real populations; real biology need not meet every ideal condition. Neither dominance nor recessiveness alone makes an allele increase or disappear. Hardy's original analysis, reprinted in 2003, established the random-mating baseline for this reasoning.

Inferring allele frequency from a recessive phenotype

Phenotype observations contain less genetic information than complete genotype counts when a dominant phenotype can represent either $AA$ or $Aa$.

Suppose a fictional plant's pale phenotype is caused only by genotype $aa$, while both $AA$ and $Aa$ are green. Assume two alleles, complete expression of this simple dominance relationship, no environmental mimic of the pale phenotype, a representative population count and Hardy–Weinberg genotype proportions at the stage counted. Only under these conditions can the pale proportion be treated as $q^2$ and used to infer $q$.

Working backwards from $q^2$

There are 100 pale plants among 2500 plants. First convert the phenotype count to a proportion:

Solve for $q$ by taking the positive square root, because an allele frequency cannot be negative:

The inferred heterozygote proportion is then

The expected number of heterozygotes is $0.32\times2500=800$. The predicted $AA$ proportion is $0.80^2=0.64$, giving 1600 plants. The count check is $1600+800+100=2500$.

The pale fraction is 4%; the allele $a$ frequency is 20%. Taking the square root of the percentage number 4 would be incorrect: convert to the proportion 0.040 first. The green proportion is $1-q^2=0.96$; it is not $p$, because it combines $AA$ and $Aa$.

If equilibrium is not justified, the number of green plants alone does not tell us how many are heterozygous. Many different mixtures of $AA$ and $Aa$ could give the same dominant phenotype count and different allele frequencies. Genotyping a representative sample can provide the missing information. Similarly, if the phenotype has several genetic or environmental causes, the simple square-root inference is not valid.

Use these conditions before selecting a calculation, not after obtaining a convenient answer.

Distinguishing model fit from change over time

A genotype census at one time can test agreement with Hardy–Weinberg predictions, whereas repeated measurements can reveal whether allele frequencies change over time.

One population at one time

Consider a separate complete census of 500 organisms: 200 $AA$, 100 $Aa$ and 200 $aa$. Direct counting gives

The Hardy–Weinberg expected counts are $0.25\times500=125$ $AA$, $0.50\times500=250$ $Aa$, and 125 $aa$. Observed counts differ markedly, with fewer heterozygotes and more homozygotes than predicted. This is disequilibrium at this gene at this time. It prompts investigation of assumptions, such as non-random mating or pooling subpopulations with different allele frequencies, as well as selection and data quality. It does not give an earlier allele frequency or identify the cause of the mismatch. A one-time mismatch alone cannot demonstrate allele-frequency change over time.

Following a population through generations

Now return to the study population from the first part. The original illustrative table supplies complete genotype censuses of the same defined population at the same life stage in two successive generations.