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(p^{2}) |
Aa(pq) |

a(q) | Aa(pq) | aa(q^{2}) |

** **

So, **Frequency of individuals with genotype AA in a population = p ^{2}**

^{ }, 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., p

^{2}

**Frequency of individuals with genotype aa in a population = q ^{2}**

**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:

p^{2} + 2pq + q^{2} =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) = N
_{AA} - Number of individuals that are homozygous for the a allele (aa) = N
_{aa} - Number of individuals that are heterozygous (Aa) = N
_{Aa} - N
_{AA}+ N_{Aa}+ N_{aa}= 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 N_{AA }and 1 time in N_{Aa }, 2N_{AA}+ N_{Aa } - Total number of copies of both alleles present in the population is 2N, because each individual is diploid.

- So, Frequency of allele
*A*= p = (2N_{AA}+ N_{Aa })/2N

Frequency of allele *a *= q = (2N_{aa}+ N_{Aa })/2N

Frequency of genotype AA = N_{AA}/ N

Frequency of genotype Aa = N_{Aa} / N

Frequency of genotype aa = N_{aa }/ N

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