In a H.W. population of 1600 individuals,256 are showing recessive trait. Find out the heterozygous individuals in that population .

Options

A. 768
B. 576
C. 1344
D. 256

Answer

To solve the problem of finding the number of heterozygous individuals in a Hardy-Weinberg (H.W.) population of 1600 individuals, where 256 individuals show a recessive trait, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the total population and recessive individuals**:
- Total population (N) = 1600
- Number of individuals showing recessive trait = 256

2. **Calculate the frequency of the recessive phenotype (q²)**:
- The recessive phenotype is expressed only in homozygous recessive individuals (genotype = qq).
- Frequency of recessive phenotype (q²) = Number of recessive individuals / Total population
- q² = 256 / 1600 = 0.16

3. **Calculate the frequency of the recessive allele (q)**:
- To find q, take the square root of q².
- q = √(0.16) = 0.4

4. **Calculate the frequency of the dominant allele (p)**:
- According to the Hardy-Weinberg principle, p + q = 1.
- Therefore, p = 1 - q = 1 - 0.4 = 0.6

5. **Calculate the frequency of heterozygous individuals (2pq)**:
- The frequency of heterozygous individuals (genotype = Pq) is given by the formula 2pq.
- 2pq = 2 * p * q = 2 * 0.6 * 0.4 = 0.48

6. **Calculate the number of heterozygous individuals in the population**:
- To find the number of heterozygous individuals, multiply the frequency of heterozygous individuals by the total population.
- Number of heterozygous individuals = 0.48 * 1600 = 768

### Final Answer:
The number of heterozygous individuals in the population is **768**.

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