Question

The allele for brown eyes is dominant over blue eyes . In a population, 36% of individuals have blue eyes. Assuming Hardy-Weinberg equilibrium, calculate:
Frequency of allele B
Percentage of heterozygous individuals
(iii) Percentage of individuals that are homozygous dominant
(Working to be shown)

Hint:

In any Hardy-Weinberg problem, always begin by writing down the recessive phenotype value ($q^2$). Taking its square root gives you $q$ immediately. Never start with the dominant trait percentage, because it contains a hidden mix of both $p^2$ and $2pq$ individuals.

Answer 1

Step 1: Understanding the Question:
We are given a population in Hardy-Weinberg equilibrium where brown eyes ($B$) are dominant over blue eyes ($b$). The recessive blue-eyed phenotype ($bb$) makes up $36\%$ of the population. We need to find the frequency of allele $B$ and the percentages of the other genotypes. Since these calculations are mathematically dependent on one another, they are presented together.

Step 2: Key Formula or Approach:
The Hardy-Weinberg equilibrium expressions are:
1. Allele frequencies: $p + q = 1$
2. Genotype frequencies: $p^2 + 2pq + q^2 = 1$
Where:
$p = $ frequency of the dominant allele ($B$)
$q = $ frequency of the recessive allele ($b$)
$p^2 = $ frequency of homozygous dominant individuals ($BB$)
$2pq = $ frequency of heterozygous individuals ($Bb$)
$q^2 = $ frequency of homozygous recessive individuals ($bb$)

Step 3: Detailed Explanation:
Step 3A: Calculate allele frequencies ($q$ and $p$) The percentage of blue-eyed individuals ($bb$) is given as $36\%$. Expressed as a decimal frequency:
$$q^2 = 36\% = 0.36$$ Take the square root of both sides to find the frequency of the recessive allele $b$:
$$q = \sqrt{0.36} = 0.6$$ Now, using $p + q = 1$, find the frequency of the dominant allele $B$:
$$p = 1 - q = 1 - 0.6 = 0.4$$ Step 3B: Calculate the percentage of heterozygous individuals ($Bb$) The frequency of heterozygotes is given by the term $2pq$:
$$\text{Frequency} = 2pq = 2 \times 0.4 \times 0.6 = 0.48$$ Convert this frequency into a percentage:
$$\text{Percentage} = 0.48 \times 100\% = 48\%$$ Step 3C: Calculate the percentage of homozygous dominant individuals ($BB$) The frequency of homozygous dominant individuals is given by the term $p^2$:
$$\text{Frequency} = p^2 = (0.4)^2 = 0.16$$ Convert this frequency into a percentage:
$$\text{Percentage} = 0.16 \times 100\% = 16\%$$ Verification check: $16\% \text{ (BB)} + 48\% \text{ (Bb)} + 36\% \text{ (bb)} = 100\%$. The mathematics is completely consistent.

Step 4: Final Answer:
(i) The frequency of allele $B$ is $0.4$. (ii) The percentage of heterozygous individuals is $48\%$. (iii) The percentage of homozygous dominant individuals is $16\%$.