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, we will use the Hardy-Weinberg principle, which provides a way to calculate the expected frequencies of genotypes in a population that is in equilibrium.

### Step-by-Step Solution:

1. **Identify Allele Frequencies**:
- The frequency of allele 'A' (p) = 0.7
- The frequency of allele 'a' (q) = 0.3

2. **Calculate the Frequency of Heterozygous Individuals**:
- According to the Hardy-Weinberg principle, the frequency of heterozygous individuals (genotype Aa) can be calculated using the formula:
\[ 2pq \]
- Here, \( p \) is the frequency of allele 'A' and \( q \) is the frequency of allele 'a'.
- Substituting the values:
\[ 2pq = 2 \times (0.7) \times (0.3) = 2 \times 0.21 = 0.42 \]

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

### Final Answer:
The expected number of heterozygous individuals in the sample of 250 individuals is **105**.

---