What is the Hardy-Weinberg principle?

The Hardy-Weinberg principle involves a mathematical equation that helps calculate the frequencies of alleles for a particular gene within a population. Allele frequency refers to how often an allele appears in a population.

In this context, a population is defined as a group of organisms of the same species living in a specific area at a certain time, with the potential to interbreed.

The assumptions of the Hardy-Weinberg principle

The Hardy-Weinberg principle states that allele frequencies in a population will remain constant across generations if certain conditions are fulfilled.

The key assumptions are:

• No mutations occur.
• There is no migration into or out of the population.
• Mating is random.
• The population size is large.
• There are no natural selection pressures.

In essence, the gene pool, which comprises all alleles of all genes in all individuals of a population at a given time, stays consistent over time. This concept of genetic equilibrium is crucial for analysing changes in gene frequencies over time.

The Hardy-Weinberg equation

The Hardy-Weinberg equation models the relationship between allele frequencies.

The variables it includes are as follows:

• p - The frequency of the dominant allele.
• q - The frequency of the recessive allele.
• p² - The frequency of homozygous dominant individuals (derived as if the genotype were pp).
• 2pq - The frequency of heterozygous individuals (derived as the genotype could be either pq or qp).
• q² - The frequency of homozygous recessive individuals (derived as if the genotype were qq).

The total of all possible allele combinations must equal 1:

$p^2+2pq+q^2=1$

This equation links genotype frequencies directly to allele frequencies.

The sum of all alleles for this gene in the population must also equal 1:

$p+q=1$

These two equations can be rearranged to calculate allele frequencies.

Worked example - Calculating allele frequencies in a population

Imagine a scenario where a recessive allele a causes a rare genetic disorder in 1 out of 40,000 people in a large population. Calculate the frequency of the dominant allele A and the heterozygous genotype Aa.

Step 1: Equations

$p^2+2pq+q^2=1$

$p+q=1$

$q^2=\text{ frequency of homozygous recessives (aa)}$

Step 2: Identify q² and calculate q

as 1 in 40,000 individuals are homozygous recessive (aa), we can calculate q²

$q^2=\frac{1}{\mathrm{40,000}}=0.000025$

$q=\text{ frequency of recessive allele (a) }=\sqrt{q^2}=\sqrt{0.000025}=0.005$

Step 3: Rearrangement to find p

$p=1-q$

$p=\text{ frequency of dominant allele (A) }=1-0.005=0.995$

Step 4: Substitution and correct evaluation

$2pq=\text{ frequency of heterozygotes (Aa)}$

$2pq=2\times 0.995\times 0.005=0.00995$

therefore, the frequency of the heterozygous genotype Aa is 0.00995, or 0.995%, within the population