The ability of certain people to taste the chemical phenylthiocarbamide (PTC) is governed by a dominant allele T, and the inability to taste PTC by its recessive allele t. If 24% of a population is homozygous taster and 40% is heterozygous taster, what is the frequency of t?

Options

A. (a) 0.3
B. (b) 0.9
C. (c) 0.4
D. (d) 0.6

Answer

To solve the problem, we will use the Hardy-Weinberg principle, which provides a mathematical model to study genetic variation in a population. The principle states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.

### Step-by-Step Solution:

1. **Identify the given information**:
- Homozygous tasters (TT) = 24% of the population = 0.24
- Heterozygous tasters (Tt) = 40% of the population = 0.40

2. **Set up the Hardy-Weinberg equation**:
The Hardy-Weinberg equation is given by:
\[ p^2 + 2pq + q^2 = 1 \]
where:
- \( p^2 \) = frequency of homozygous dominant (TT)
- \( 2pq \) = frequency of heterozygous (Tt)
- \( q^2 \) = frequency of homozygous recessive (tt)

3. **Assign values to the variables**:
- From the information provided:
- \( p^2 = 0.24 \) (homozygous tasters)
- \( 2pq = 0.40 \) (heterozygous tasters)

4. **Calculate \( p \)**:
To find \( p \) (the frequency of the dominant allele T), we take the square root of \( p^2 \):
\[ p = \sqrt{0.24} \approx 0.49 \]

5. **Use the value of \( p \) to find \( q \)**:
Since \( p + q = 1 \), we can find \( q \):
\[ q = 1 - p = 1 - 0.49 = 0.51 \]

6. **Calculate the frequency of the recessive allele t**:
The frequency of the recessive allele \( t \) is simply \( q \):
\[ q \approx 0.51 \]

7. **Final answer**:
The frequency of the allele \( t \) is approximately 0.51.