STD 12 - 6. evolution — NEET STD 11 Science — Question

Hardy-Weinberg Principle

It was proposed by GH Hardy an English mathematician and W Weinberg a German physician independently in $1908$

$(i)$ It describes a theoretical situation in which a population is undergoing no evolutionary change. This is called genetic or Hardy-Weinberg equilibrium

$(ii)$ It can be expressed as $p^{2}+2 p q+q^{2}=1$ or $|p+q|^{2}=1$

$(iii)$ Evolution occurs when the genetic equilibrium is up set (evolution is a departure from Hardy-Weinberg equilibrium principle)

The sum of total of Allelic frequency $\langle p+q|$ is $=1$

$p^{2}+2 p q+q^{2} \text { or }|p+q|^{2}$

Where, $p^{2}=\%$ homozygous dominant individuals

$p=i$ frequency of dominant allele

$q^{2}=\%$ homozygous recessive individuals

$q=i$ frequency of recessive allele

$2 p q=\%$ heterozygous individuals

Realize that $|p+q|^{2}=1$ (three are only $2$ alleles)

$p^{2}+2 p q+q^{2}=1$ (these are the only genotypes)

Example An investigator has determined by the inspection that $16\, \%$ of a human population has a recessive trait. Using this information, we can calculate all the genotypes and allele frequencies for the population, provided the conditions for Hardy-Weinberg equilibrium are met

Given $q^{2}=16 \,\%=0.16$ are homozygous recessive individuals

Therefore,

$q=\sqrt{0.16}=0.4=i$ frequency of recessive allele

$p=1.0-0.4=0.6=i$ frequency of dominant allele

$p^{2}=0.6 \times 0.6=0.36$ or $36 \%$ are homozygous dominant individuals

$2 p q=2 \times 0.6 \times 0.4=0.48=48 \%$ are heterozygous individuals

$i\, 1.00-0.52$

$i \,0.48$

Thus, $84 \%$ ( $36+48$ ) have the dominant phenotype