In a population that is in Hardy weinberg equilibrium, the frequency of a recessive allele for a certain heareditary trait is `0.20.` What percentage of the individual in the next generation would be expected to show the dominant trait :-

Options

A. `16%`
B. `32%`
C. `64%`
D. `96%`

Answer

To solve the question regarding the frequency of the dominant trait in a population at Hardy-Weinberg equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Frequency of the Recessive Allele (q)**:
- The problem states that the frequency of the recessive allele is 0.20. Thus, \( q = 0.20 \).

2. **Calculate the Frequency of the Dominant Allele (p)**:
- According to the Hardy-Weinberg principle, the sum of the frequencies of the alleles (p + q) is equal to 1.
- Therefore, we can calculate \( p \) as follows:
\[ p = 1 - q = 1 - 0.20 = 0.80 \]

3. **Use the Hardy-Weinberg Equation**:
- The Hardy-Weinberg equation is given by:
\[ p^2 + 2pq + q^2 = 1 \]
- Here, \( p^2 \) represents the frequency of homozygous dominant individuals, \( 2pq \) represents the frequency of heterozygous individuals, and \( q^2 \) represents the frequency of homozygous recessive individuals.

4. **Calculate the Frequencies**:
- Calculate \( p^2 \):
\[ p^2 = (0.80)^2 = 0.64 \]
- Calculate \( 2pq \):
\[ 2pq = 2 \times (0.80) \times (0.20) = 0.32 \]
- Now, we can find the total frequency of individuals showing the dominant trait (both homozygous dominant and heterozygous):
\[ \text{Total frequency of dominant trait} = p^2 + 2pq = 0.64 + 0.32 = 0.96 \]

5. **Convert to Percentage**:
- To find the percentage of individuals showing the dominant trait in the next generation:
\[ \text{Percentage} = 0.96 \times 100 = 96\% \]

### Final Answer:
The percentage of individuals in the next generation expected to show the dominant trait is **96%**.

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