Persons carrying the dominant gene(s) can taste a particular chemical and are called tasters. The others are nontasters. In a large random-mating human population at the hardy-weinberg equilibrium. There are 36% non-tasters and 64% tasters. The frequency of heterozygotes in the populations is

Options

A. (a) 0.06
B. (b) 0.08
C. (c) 0.32
D. (d) 0.48

Answer

To solve the problem regarding the frequency of heterozygotes in a population where tasters and non-tasters are present, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Phenotypes and Genotypes**:
- Tasters (dominant phenotype) can be either homozygous dominant (TT) or heterozygous (Tt).
- Non-tasters (recessive phenotype) must be homozygous recessive (tt).

2. **Identify Given Frequencies**:
- The frequency of non-tasters (tt) is given as 36%, which can be expressed as a decimal:
\[ q^2 = 0.36 \]
- The frequency of tasters (TT and Tt combined) is 64%, which can be expressed as:
\[ p^2 + 2pq = 0.64 \]

3. **Calculate q**:
- Since \( q^2 = 0.36 \), we can find \( q \) by taking the square root:
\[ q = \sqrt{0.36} = 0.6 \]

4. **Calculate p**:
- Using the relationship \( p + q = 1 \):
\[ p = 1 - q = 1 - 0.6 = 0.4 \]

5. **Calculate the Frequency of Heterozygotes (2pq)**:
- Now that we have \( p \) and \( q \), we can calculate the frequency of heterozygotes using the formula \( 2pq \):
\[ 2pq = 2 \times p \times q = 2 \times 0.4 \times 0.6 \]
- Performing the multiplication:
\[ 2pq = 2 \times 0.4 \times 0.6 = 0.48 \]

6. **Conclusion**:
- The frequency of heterozygotes in the population is:
\[ \text{Frequency of heterozygotes} = 0.48 \]

### Final Answer:
The frequency of heterozygotes in the population is **0.48**.

---