Question

Calculate the frequency of ${L^N}$ alleles:

Blood group	Genotype	Number of Individuals
M	${L^M}$ ${L^M}$	1787
MN	${L^M}$ ${L^N}$	3089
N	${L^N}$ ${L^N}$	1303

A. $54\%$
B. $64\%$
C. $46\%$
D. $36\%$

Answer 1

In the year 1908, two scientists G.H Hardy and Wilhelm Weinberg independently described the field of the population which was later called as Hardy-Weinberg equation. It is derived from the Hardy-Weinberg equilibrium which states that the genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors.
The Hardy-Weinberg equation is
$$$$$${p^2} + {q^2} + 2pq = 1$$
So, suppose there are two alleles A and B.

In the equation, $p$ will represent the frequency of A allele while $q$ will represent the frequency of B allele
${p^2}$ will represent the frequency of homozygous allele AA while ${q^2}$ will represent the frequency of homozygous alleles BB. the $pq$ represents the frequency of heterozygous AB.

Also, they further stated that the sum of allele (A and B) frequencies presented at the particular locus or position at a chromosome is also equal to 1. Or we can say that the population to be in genetic equilibrium the sum of both alleles should be equal to 1.
$p + q = 1$
So, through these two equations, we can find the frequency of allele and vice versa.

Complete answer:
let’s consider the LM be p and LN be q
In this question the frequency of homozygous alleles ${L^M}$ ${L^M}$and ${L^N}$ ${L^N}$ are 1787 and 1303 respectively. While the frequency of heterozygous allele ${L^M}$ ${L^N}$ is 3089.
As we know that
${p^2} + {q^2} + 2pq = 1$

$${p^2} = {L^M}{L^M} \\\ {q^2} = {L^N}{L^N} \\\ 2pq = {L^M}{L^N} \\\$$

Putting these values in this equation.

Frequencies are as follows-
${L^M}{L^M} = 1787 \\\ {L^M}{L^N} = 3089 \\\ {L^N}{L^N} = 1303 \\\ {L^M} = 3089 + (1303 \times 2) = 5695 \\\ {L^N} = 3089 + (1787 \times 2) = 6663 \\\$
We get the frequency of ${L^N}$ by dividing the frequency of ${L^N}$ by total frequency-
$\dfrac{{5695}}{{6663 + 5695}} \times 100 \\\ \dfrac{{5695}}{{12362}} \times 100 \\\ 0.46 \times 100 \\\ 46\% \\\$

Hence, the correct answer is option (C).

Note: The frequencies of homozygous alleles individuals contain twice the number of that particular allele. For example, ${L^M}$ ${L^M}$individuals representing ${p^2}$will contain twice the number of alleles ${L^M}$or allele $p$.