The frequencies of alleles 'A' and 'a' in a population at Hardy-Weinberg equilibrium are 0.7 and 0.3, respectively. In a random sample of 250 individuals taken from the population, how many are expected to be heterozygous?

Options

A. 112
B. 81
C. 105
D. 145

Answer

To solve the problem of how many individuals in a sample of 250 are expected to be heterozygous for the alleles 'A' and 'a', we can follow these steps:

### Step-by-Step Solution:

1. **Identify Allele Frequencies:**
- The frequency of allele 'A' (dominant) is given as \( p = 0.7 \).
- The frequency of allele 'a' (recessive) is given as \( q = 0.3 \).

2. **Verify Hardy-Weinberg Equilibrium:**
- According to the Hardy-Weinberg principle, the sum of the frequencies of the alleles must equal 1:
\[ p + q = 1 \]
- Check: \( 0.7 + 0.3 = 1 \) (This confirms that the frequencies are correct).

3. **Calculate the Heterozygous Frequency:**
- The frequency of heterozygous individuals (genotype Aa) is given by the formula:
\[ 2pq \]
- Substitute the values of \( p \) and \( q \):
\[ 2pq = 2 \times 0.7 \times 0.3 = 0.42 \]

4. **Calculate the Expected Number of Heterozygous Individuals:**
- To find the expected number of heterozygous individuals in a sample of 250:
\[ \text{Expected heterozygous individuals} = 0.42 \times 250 \]
- Calculate:
\[ 0.42 \times 250 = 105 \]

5. **Conclusion:**
- Therefore, the expected number of heterozygous individuals in the sample of 250 is **105**.

### Final Answer:
The expected number of heterozygous individuals is **105**.