Hardy-Weinberg Law | Zoology Optional Notes for UPSC PDF Download

Conditions of the Hardy-Weinberg population

• The population must be large.
• Mating must be random, with no selection for certain genotypes.
• There must be no emigration or immigration.
• There may be no mutations.
• There may be no incestuous mating.

The Equations

• Suppose in a population of plant species, a gene has two allele, ‘A’ n ‘a’ having frequency ‘p’ n ‘q’ respectively.
• The frequencies of their genotypes AA, 2Aa and aa are p,2pq and q.
• According to this law, the frequency of p and q in a population remains constant generation after generation unless there’s any factor’s influence.

Mathematically,
If there is a gene with two alleles, A and a.
p q = 1....    (Eq. 1)

• p = frequency of A allele
• q = frequency of a allele

p² 2pq q² = 1....    (Eq. 2)

• p² = frequency of AA genotype
• 2pq = frequency of Aa genotype
• q² = frequency of aa genotype

Inference

As shown by Hardy and Weinberg, alleles segregating in a population tend to establish equilibrium with reference to each other. Thus, if two alleles should occur in equal proportion in a large, isolated breeding population and neither had a selective or mutational advantage over the other, they would be expected to remain in equal proportion generation after generation.

Applications

The Hardy–Weinberg principle may be applied in two ways, either a population is assumed to be in Hardy–Weinberg proportions, in which the genotype frequencies can be calculated, or if the genotype frequencies of all three genotypes are known, they can be tested for deviations that are statistically significant.

Sample Problem Involving Hardy-Weinberg Equilibrium

Question: Sickle cell anemia, an autosomal recessive disease, occurs in 1 of every 400 African-American births. What is the frequency of carriers of the sickle cell gene in this population?
Ans:
Carriers (or heterozygotes) are represented by 2pq, so we need to solve for p and q.
We are told that q ² is equal to 1/400 or 0.0025.
Thus, q = 0.05
We know that p q = 1, so p = 1 – q. In this case,
p = 1 – 0.05 = 0.95.
We then solve for 2pq, which is equal to 2(0.95) (0.05) or 0.095.
So 95 of every 1000 African-Americans carry the sickle cell gene.