In a random mating population of 1 lac individuals -75,000 individuals were seen to have black eyes (Dominant phenotype), the rest had blue eyes. How many black eyed individuals are hybrid?

Options

A. 0
B. 25000
C. 50000
D. 75000

Answer

To solve the question of how many black-eyed individuals are hybrid in a population of 100,000 individuals where 75,000 have black eyes (dominant phenotype) and 25,000 have blue eyes (recessive phenotype), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Phenotypes and Their Frequencies:**
- Total population = 100,000
- Black-eyed individuals (dominant phenotype) = 75,000
- Blue-eyed individuals (recessive phenotype) = 25,000

The frequency of the blue-eyed phenotype (recessive) can be calculated as:
\[ q^2 = \frac{25,000}{100,000} = 0.25 \]

2. **Calculate the Frequency of the Recessive Allele (q):**
- To find the frequency of the recessive allele (q), take the square root of \(q^2\):
\[ q = \sqrt{0.25} = 0.5 \]

3. **Calculate the Frequency of the Dominant Allele (p):**
- Since \(p + q = 1\), we can find the frequency of the dominant allele (p):
\[ p = 1 - q = 1 - 0.5 = 0.5 \]

4. **Determine the Genotype Frequencies:**
- The genotype frequencies can be represented as follows:
- Homozygous dominant (BB) = \(p^2\)
- Heterozygous (Bb) = \(2pq\)
- Homozygous recessive (bb) = \(q^2\)

Using the values of p and q:
- \(p^2 = (0.5)^2 = 0.25\) (Homozygous dominant)
- \(2pq = 2 \times 0.5 \times 0.5 = 0.5\) (Heterozygous)
- \(q^2 = (0.5)^2 = 0.25\) (Homozygous recessive)

5. **Calculate the Number of Hybrid Individuals (Bb):**
- The frequency of heterozygous individuals (Bb) is 0.5. To find the number of hybrid individuals in the population:
\[ \text{Number of Bb individuals} = 0.5 \times 100,000 = 50,000 \]

### Final Answer:
Thus, the number of black-eyed individuals that are hybrid (heterozygous) is **50,000**.

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