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
72%
b
17%
c
50%
d
34%
determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate the frequency of the
heterozygous genotype.
a
72%
b
17%
c
50%
d
34%
Answers
GPT-4o mini
To find the frequency of the heterozygous genotype (Tt) for the ability to taste PTC, we can use the Hardy-Weinberg principle, which states that in a large, randomly mating population, the frequencies of genotypes can be represented by the equation:
\[
p^2 + 2pq + q^2 = 1
\]
Where:
- \( p \) = frequency of the dominant allele (T)
- \( q \) = frequency of the recessive allele (t)
- \( p^2 \) = frequency of homozygous dominants (TT)
- \( 2pq \) = frequency of heterozygotes (Tt)
- \( q^2 \) = frequency of homozygous recessives (tt)
**Step 1: Determine the frequencies of genotypes.**
From the problem:
- Total sampled = 215 individuals
- Those who can taste PTC (phenotype TT and Tt) = 150
- Those who cannot taste PTC (phenotype tt) = 65
To find \( q^2 \) (the frequency of the recessive genotype tt), we can calculate it by:
\[
q^2 = \frac{\text{Number of tt individuals}}{\text{Total individuals}} = \frac{65}{215} \approx 0.3023
\]
**Step 2: Calculate \( q \) (frequency of allele t).**
We take the square root of \( q^2 \):
\[
q = \sqrt{q^2} = \sqrt{0.3023} \approx 0.55
\]
**Step 3: Calculate \( p \).**
Since \( p + q = 1 \):
\[
p = 1 - q \approx 1 - 0.55 \approx 0.45
\]
**Step 4: Calculate the frequency of heterozygous genotype (Tt).**
According to the Hardy-Weinberg equation:
\[
2pq = 2 \cdot p \cdot q
\]
Substituting in the values for \( p \) and \( q \):
\[
2pq = 2 \cdot 0.45 \cdot 0.55 \approx 0.495
\]
This is the frequency of heterozygous individuals (Tt) in the population.
To convert this frequency into a percentage:
\[
0.495 \times 100 \approx 49.5\%
\]
Rounding it, the frequency can be approximated to 50%, which corresponds to option **c. 50%**.
\[
p^2 + 2pq + q^2 = 1
\]
Where:
- \( p \) = frequency of the dominant allele (T)
- \( q \) = frequency of the recessive allele (t)
- \( p^2 \) = frequency of homozygous dominants (TT)
- \( 2pq \) = frequency of heterozygotes (Tt)
- \( q^2 \) = frequency of homozygous recessives (tt)
**Step 1: Determine the frequencies of genotypes.**
From the problem:
- Total sampled = 215 individuals
- Those who can taste PTC (phenotype TT and Tt) = 150
- Those who cannot taste PTC (phenotype tt) = 65
To find \( q^2 \) (the frequency of the recessive genotype tt), we can calculate it by:
\[
q^2 = \frac{\text{Number of tt individuals}}{\text{Total individuals}} = \frac{65}{215} \approx 0.3023
\]
**Step 2: Calculate \( q \) (frequency of allele t).**
We take the square root of \( q^2 \):
\[
q = \sqrt{q^2} = \sqrt{0.3023} \approx 0.55
\]
**Step 3: Calculate \( p \).**
Since \( p + q = 1 \):
\[
p = 1 - q \approx 1 - 0.55 \approx 0.45
\]
**Step 4: Calculate the frequency of heterozygous genotype (Tt).**
According to the Hardy-Weinberg equation:
\[
2pq = 2 \cdot p \cdot q
\]
Substituting in the values for \( p \) and \( q \):
\[
2pq = 2 \cdot 0.45 \cdot 0.55 \approx 0.495
\]
This is the frequency of heterozygous individuals (Tt) in the population.
To convert this frequency into a percentage:
\[
0.495 \times 100 \approx 49.5\%
\]
Rounding it, the frequency can be approximated to 50%, which corresponds to option **c. 50%**.