Question

The ability to taste PTC is due to a single dominate allele "T". You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate the frequency of the heterozygous genotype. a 17% b 72% c 34% d 50%

Answer 1

To determine the frequency of the heterozygous genotype (Tt) for the ability to taste PTC, we can use the Hardy-Weinberg principle. Here's how we'll do it step by step.

Let's denote:
- \( p \) = frequency of the dominant allele \( T \)
- \( q \) = frequency of the recessive allele \( t \)
- \( p^2 \) = frequency of homozygous dominant genotype (TT)
- \( 2pq \) = frequency of heterozygous genotype (Tt)
- \( q^2 \) = frequency of homozygous recessive genotype (tt)

1. **Find \( q^2 \)**:
The individuals who cannot taste PTC (tt) are 65 out of a sample of 215 individuals. Therefore, the frequency of the recessive phenotype (tt) is:

\[
q^2 = \frac{65}{215} \approx 0.3023
\]

2. **Calculate \( q \)**:
To find \( q \), we take the square root of \( q^2 \):

\[
q = \sqrt{q^2} = \sqrt{0.3023} \approx 0.55
\]

3. **Calculate \( p \)**:
Since \( p + q = 1 \):

\[
p = 1 - q \approx 1 - 0.55 = 0.45
\]

4. **Calculate the frequency of the heterozygous genotype \( 2pq \)**:
Now we can calculate the frequency of the heterozygous genotype (Tt):

\[
2pq = 2 \times p \times q \approx 2 \times 0.45 \times 0.55 \approx 0.495
\]

5. To express this as a percentage, we multiply by 100:

\[
0.495 \times 100 \approx 49.5\%
\]

This rounds to approximately \( 50\% \).

Therefore, the frequency of the heterozygous genotype (Tt) is approximately 50%. The correct answer is:

**d 50%**.