Question

In a certain population which is in Hardy-Weinberg equilibrium, there are only two eye colours: brown (dominant) and blue (recessive). 16 % have blue eyes in the population. What percentage of the population are heterozygous for brown eyes?

• A. 24%
• B. 48%
• C. 60%
• D. 64%

Answer 1

The correct option is B 48%

Hardy-Weinberg law states that the gene pool of a population remains constant under certain conditions. This is called genetic equilibrium or Hardy-Weinberg equilibrium.

The Hardy-Weinberg equation is an algebraic equation which is an expression of the Hardy-Weinberg law. The frequencies of alleles and the genotypic ratios in a population can be calculated with the help of this equation.

According to Hardy Weinberg equation, if there are two alleles for a particular gene where A represents the dominant allele and a represents the recessive allele, then the frequency of A is p and a is q.
Combined frequencies of all alleles for a gene in a population is equal 1.
Therefore, p+q=1.
The frequency of homozygous dominant individuals (AA) in the population would be equal to pxp or $p^2$
The frequency of homozygous recessive individuals (aa) in the population would be equal to qxq or $q^2$
Since there are two ways of forming the heterozygote Aa, (allele A from the father and a from mother and vice versa) the frequency of heterozygous dominant individuals (Aa) in the population would be 2pq.
The sum of all three genotypic frequencies is $p^2+2pq+q^2=1$ which is a binomial expansion of $(p+q{)}^2$

Therefore,
p = frequency of the dominant allele (A)
q = frequency of the recessive allele (a)
$p^2$ = frequency of homozygous dominant individuals (AA)
$q^2$ = frequency of homozygous recessive individuals (aa)
2pq = frequency of heterozygous dominant individuals (Aa)

As per the question, the population has two variants in the eye colour, brown and blue. Brown eye colour (BB/Bb) is dominant over blue (bb).

Let us denote p as the frequency of the dominant allele, B and q as the frequency of the recessive allele, b.
Now, it is given in the question that the percentage of the population with blue eyes (bb) is 16%
Hence, the frequency of homozygous recessive individuals $\left(q^2\right)$ = 0.16; therefore, $q=\sqrt{0.16}=0.4$
Since this is a Hardy-Weinberg population, we can assume that p + q = 1, so p = 1–0.4 = 0.6.
The frequency of heterozygous dominant individuals with brown eyes (Bb) is therefore 2pq = 2(0.6)(0.4) = 0.48=48%.

So, 48% of the population is heterozygous for brown eyes.