Question

Hardy Weinberg equation is

• A. $p^2+2pq+q^2=1$
• B. $p+2pq+q=1$
• C. $p+q=4$
• D. $p+q^2+2pq=1$

Answer 1

The correct option is A $p^2+2pq+q^2=1$

• The equation is an expression of the principle known as Hardy-Weinberg equilibrium, which states that the amount of genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors.
• To explore the Hardy-Weinberg equation, we can examine a simple genetic locus at which there are two alleles, A and a.
• The Hardy-Weinberg equation is expressed as:

$p^2$ + 2$pq$ + $q^2$ = 1
where $p$ is the frequency of the "A" allele and $q$ is the frequency of the "a" allele in the population.

• In the equation, $p^2$ represents the frequency of the homozygous genotype AA, $q^2$ represents the frequency of the homozygous genotype aa, and 2$pq$ represents the frequency of the heterozygous genotype Aa.

Thus, option A is correct and other options are wrong.