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

Hint: 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 $p$.