The frequency of two alleles in a gene pool is 0.19 (A) and 0.8 (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 provides a mathematical model to study genetic variation in a population at equilibrium.

### Step-by-Step Solution:

**Step 1: Identify Allele Frequencies**
- We are given the frequencies of two alleles:
- Frequency of allele A (p) = 0.19
- Frequency of allele a (q) = 0.8

**Step 2: Verify the Allele Frequencies**
- According to Hardy-Weinberg equilibrium, the sum of the allele frequencies must equal 1.
- p + q = 0.19 + 0.8 = 0.99
- Since the sum is not equal to 1, we need to adjust the frequencies. Let's assume the frequency of allele a is actually 0.81 (which is a common mistake in transcription). Thus:
- p = 0.19
- q = 0.81

**Step 3: Calculate the Percentage of Heterozygous Individuals**
- The formula for the frequency of heterozygous individuals (2pq) is:
- 2pq = 2 * p * q
- Substituting the values:
- 2 * 0.19 * 0.81 = 0.3078
- To convert this to a percentage:
- 0.3078 * 100 = 30.78%

**Step 4: Calculate the Percentage of Homozygous Recessive Individuals**
- The formula for the frequency of homozygous recessive individuals (q²) is:
- q² = (0.81)²
- Calculating this:
- (0.81)² = 0.6561
- To convert this to a percentage:
- 0.6561 * 100 = 65.61%

### Final Answers:
(a) Percentage of heterozygous individuals = 30.78%
(b) Percentage of homozygous recessive individuals = 65.61%