The Hardy-Weinberg equation is a means by which the frequency of two alternate alleles can be predicted within a population

For two alleles of a given genetic characteristic, three genotypes are possible (assuming Mendelian inheritance): AA, Aa and aa

- Dominant allele is A, with a frequency of p
- Recessive allele is a, with a frequency of q

**The Hardy-Weinberg Equation**

• The total frequency of both alleles will be 100% – in other words: ** p + q = 1**

• Because genotypes consist of two alleles, this equation must be squared: **( p + q ) **^{2} = 1

• This gives the expanded Hardy-Weinberg equation: p^{2} + 2pq + q^{2}** = 1 **

*(whereby p*

^{2}=**AA**; 2pq =**Aa**; q^{2}=**aa**)__Graphical Representation of the Hardy-Weinberg Principle__

Hardy-Weinberg Conditions

For the Hardy-Weinberg equation to be accurate, certain population conditions are assumed:

- The population is large with random mating
- There is no mutation or gene flow
- There is no natural selection or allele-specific mortality

**Worked Example**

Suppose we had a population of 500 people, in which 9% were albino (albinism is a recessive characteristic)

- How many individuals in this population are heterozygous?

Using the equations: **p + q = 1** __and__ p^{2} + 2pq + q^{2}** = 1**

- If q
^{2}= 0.09 ⇒ q = 0.3*(√ 0.09)* - If q = 0.3 ⇒ p = 0.7
*(p + 0.3 = 1)* - If p = 0.7 ⇒ p
^{2}= 0.49*(0.7*^{2}) - 2pq = 0.42 ⇒
*(0.49 + 2pq + 0.09 = 1)*

Substituting these numbers for frequencies and applying them to the original population shows that:

- 49% of people are homozygous dominant (AA), which is
**245**individuals (0.49 × 500) - 9% of people are homozygous recessive (aa), which is
**45**individuals (0.09 × 500) - 42% of people are heterozygous (Aa), which is
**210**individuals (0.42 × 500)

**Practice Question**