Question

In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa and the remaining 160 to aa. Based on this data, the frequency of allele a in the population is:
A. $0.4$
B. $0.5$
C. $0.6$
D. $0.7$

Answer 1

Hint: An allele is one of the two or more versions of a gene (DNA is formed of various segments of genes). Genotypes refer to the genetic makeup of an organism.

Complete answer: Hardy-Weisberg principle is a principle which states that the genetic variation in a population will remain constant from one generation to the next in the absence of any external factors. The Hardy-Weisberg principle relies on a number of assumptions like random mating, the absence of natural selection, a very large population size, no gene flow or migration etc.
According to Hardy-Weinberg equilibrium:
$\begin{align} & {{p}^{2}}+2pq+{{q}^{2}}=1\\ &\\ \end{align}$
${{p}^{2}}$ Is the frequency of homozygous genotype AA.
${{q}^{2}}$ Is the frequency of genotype aa.
$pq$ Is the frequency of genotype Aa.
There are 1000 individuals and 360 among them belong to genotype AA.
$p=\sqrt{0\cdot 36}$
$p=0\cdot 6$
Hence, the correct answer is option C.

Additional information: There are around seven assumptions underlying Hardy–Weinberg equilibrium which are stated as follows:
i) organisms are diploid i.e. they have two sets of chromosomes
ii) Mating is random
iii) Only sexual reproduction occurs
iv) Generations are non-overlapping
v)population size is infinitely large
vi) No migration, gene flow, mutation or selection

Note: If the allele frequencies are the same for both generations, then the population in Hardy-Weinberg is equilibrium. For a graphical representation of the equation, a triangular plot is used. It represents the distribution of the three genotype frequencies in relation to each other.