Hardy Weinbery Equillibrium

March 13, 2021 | 1 Minute Read

Hardy Weinberg Equillibrium(HWE): Allele frequency stays constant

Assumptions:

1. Random mating
2. No selection
3. Large and diploid population
4. Discrete generation

Proof that allele frequency will stay constant in diploid individuals, 2 alleles.

Say in generation $1$, given allele frequency $P(A)=p$, $P(B)=q$, where $p+q=1$

Diploid individuals: $P(AA)=P(A)\times P(A) = p^{2}$ $P(Aa)=2*P(A)\times P(a) = 2pq$ $P(aa)=P(a)\times P(a) = q^{2}$

In next generation, the probability of getting allele $A$ is finding a parent of AA or a parent of Aa and choose the A with a $1/2$ chance: $P(A) = p^{2} + 1/2 \times 2pq = p(p+q) = p$.

Proof: Extend to multiple allele

Now we have multiple allele $1,2….H$ with frequency $\theta_{1}, \theta_{2}…\theta_{H}$

Homozygous: $P(ii)= \theta_{i}\theta_{i}$ Heterozygour: $P(ij)= 2\theta_{i}\theta_{j}$

Next generation: $P(i) = P(ii) + \sum_{j \neq i}{1/2 \times P(ij)}$

$= \theta_{i}((\sum_{j \neq i} {\theta_{j}})+ \theta_{i} ) = \theta_{i}$

# Why do we care about Hardy Weinbery Equillibrium

Because violation of HWE indicate violation of assumption, which is interesting biological phenomenon.

## Example 1:

1/10000 in the population has a recessive disease, what fraction of the population carry the mutation?

$q^{2} = 1/10000$

$q = 1/100$

carrier population = $P(Aa) = 2pq = 2 \times 1/100 \times 99/100 \approx 0.02$

Example 2: Sickle cell disease

$S$ is the disease allele. $AA:AS:SS = 9365:2993:29$ HWE is violated, therefore there might be selection

Example 3: Wahlund effect: heterozygous population is less than expected because population is not randomly mating

If we sample from New York City, we will find heterozygous population less than expected.

This is because the population is not randomly mating. Say New York City is consisted of two equal sized community, A and B. 90% community A carry mutation P while 10% of community B carry mutaiton P, on average, New York City as a whole might have 50% population carrying P. So we expect 20.50.5 population is heterozygote. But in fact intra-community marriage is more often, therefore there is less heterozygote population.