The frequency of two alleles in a gene pool is 0.19 ( A) and 0.81 ( a) . Assume that the population is in Hardy-Weinberg equilibrium and
(a) Calculate the percentage of heterozygous individuals in the population.
(b) Calculate the percentage of homozygous recessive in the population .

Options

A. `60% and 40%`
B. `66% and 31%`
C. `40% and 60%`
D. `31% and 66%`

Answer

To solve the problem, we will use the Hardy-Weinberg principle, which states that in a population at equilibrium, the allele frequencies can be used to predict the genotype frequencies. The equations we will use are:

1. \( p + q = 1 \) (where \( p \) is the frequency of the dominant allele and \( q \) is the frequency of the recessive allele)
2. \( 2pq \) (for heterozygous individuals)
3. \( q^2 \) (for homozygous recessive individuals)

Given:
- Frequency of allele A (p) = 0.19
- Frequency of allele a (q) = 0.81

### Step-by-step Solution:

**(a) Calculate the percentage of heterozygous individuals in the population.**

1. **Identify the values of p and q:**
- \( p = 0.19 \)
- \( q = 0.81 \)

2. **Use the formula for heterozygous individuals (2pq):**
\[ 2pq = 2 \times (0.19) \times (0.81) \]

3. **Calculate the value:**
\[ 2pq = 2 \times 0.19 \times 0.81 = 0.3078 \]

4. **Convert the value to a percentage:**
\[ \text{Percentage of heterozygous individuals} = 0.3078 \times 100 = 30.78\% \]
Rounding this gives approximately 31%.

**(b) Calculate the percentage of homozygous recessive individuals in the population.**

1. **Use the formula for homozygous recessive individuals (q²):**
\[ q^2 = (0.81)^2 \]

2. **Calculate the value:**
\[ q^2 = 0.81 \times 0.81 = 0.6561 \]

3. **Convert the value to a percentage:**
\[ \text{Percentage of homozygous recessive individuals} = 0.6561 \times 100 = 65.61\% \]
Rounding this gives approximately 66%.

### Final Answers:
- Percentage of heterozygous individuals: **31%**
- Percentage of homozygous recessive individuals: **66%**