## Hardy-Weinberg’s Principle Hey All AIPMT Biology Aspirants, learn about Hardy-Weinberg’s principle, which is one of the four major evolutionary process, apart from Gene Flow,Genetic drift and Natural selection, important for AIPMT Biology.

Hardy-Weinberg’s principle says that allele frequencies in a population are stable and is constant from generation to generation.

• For applying algebraic equations of Hardy–Weinberg principle, Five important assumptions must be met. This principle is based on following five assumptions:
1. Random mating (individuals are not allowed to choose a mate)
2. Extremely large population size / No genetic drift
3. No natural selection
4. No gene flow (No genes are transferred to or from other sources (no immigration or emigration takes place)
5. No mutation
• Individual frequencies are represented by p and q, in diploid population.
• Because there are only two alleles, two frequencies must add up to 1, p + q = 1. Allele frequency range from 0 to 1.

Frequency of allele A = p

Frequency of allele a = q

 A(p) a(q) A(p) AA(p2) Aa(pq) a(q) Aa(pq) aa(q2)

So, Frequency of individuals with genotype AA in a population = p2  , or we can say probability that an allele A with a frequency of p appear on both the chromosomes of a diploid individual is simply the product of the probabilities, i.e., p2

Frequency of individuals with genotype aa in a population  =  q2

Frequency of individuals with genotype Aa in a population =  pq+pq= 2pq

Genotype frequencies in the offspring generation must add up to 1, which means that:

p2 + 2pq + q2 =1

• In any population, where total number of individuals in the population is N
• Number of individuals that are homozygous for the A allele (AA) = NAA
• Number of individuals that are homozygous for the a allele (aa) = Naa
• Number of individuals that are heterozygous (Aa) = NAa
• NAA + NAa + Naa = N
• Each AA individual has two copies of the A allele, and each Aa individual has one copy of the A Therefore, the total number of A alleles in the population is 2 times in NAA and 1 time in NAa , 2NAA+ NAa
• Total number of copies of both alleles present in the population is 2N, because each individual is diploid.
• So, Frequency of allele A = p =  (2NAA+ NAa )/2N

Frequency of allele a =  q =  (2Naa+ NAa )/2N
Frequency of genotype AA = NAA/ N

Frequency of genotype Aa = NAa / N

Frequency of genotype aa = Naa / N