According to Hardy-Weinberg's principle, if allele one is denoted as 'A' and allele two as 'a' and their frequencies are denoted by p and q, and if random mating occurs. The frequency of heterozygous individual would be `:`

Options

A. 2pq
B. `q^(2)`
C. pq
D. `p^(2)`

Answer

To solve the question regarding the frequency of heterozygous individuals according to Hardy-Weinberg's principle, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Hardy-Weinberg Principle**: The Hardy-Weinberg principle states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.

2. **Define Allele Frequencies**: In this case, we denote allele 'A' as the dominant allele with frequency 'p' and allele 'a' as the recessive allele with frequency 'q'. According to the principle, the sum of the frequencies of the alleles must equal 1:
\[ p + q = 1 \]

3. **Identify Genotypic Frequencies**: The genotypic frequencies can be expressed using the formula:
\[ p^2 + 2pq + q^2 = 1 \]
- \(p^2\) represents the frequency of homozygous dominant individuals (AA).
- \(2pq\) represents the frequency of heterozygous individuals (Aa).
- \(q^2\) represents the frequency of homozygous recessive individuals (aa).

4. **Focus on Heterozygous Frequency**: The question specifically asks for the frequency of heterozygous individuals. From the formula, we see that the frequency of heterozygous individuals is given by:
\[ 2pq \]

5. **Conclusion**: Therefore, the frequency of heterozygous individuals (Aa) in the population is:
\[ \text{Frequency of heterozygous individuals} = 2pq \]

### Final Answer:
The frequency of heterozygous individuals would be \(2pq\).

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