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:
- Random mating (individuals are not allowed to choose a mate)
- Extremely large population size / No genetic drift
- No natural selection
- No gene flow (No genes are transferred to or from other sources (no immigration or emigration takes place)
- 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
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