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Sagot :
To solve this problem, let's consider the Hardy-Weinberg principle which states that in a genetic equilibrium, the genotype frequencies in a population can be expressed as:
[tex]\[ p^2 + 2pq + q^2 = 1 \][/tex]
where:
- [tex]\( p \)[/tex] is the frequency of the dominant allele (B)
- [tex]\( q \)[/tex] is the frequency of the recessive allele (b)
- [tex]\( p^2 \)[/tex] is the frequency of the homozygous dominant genotype (BB)
- [tex]\( 2pq \)[/tex] is the frequency of the heterozygous genotype (Bb)
- [tex]\( q^2 \)[/tex] is the frequency of the homozygous recessive genotype (bb)
Given 128 students with brown eyes (BB or Bb) and 72 students with blue eyes (bb), let's determine the correct genotype frequencies.
1. Calculate the total number of students:
[tex]\[ 128 + 72 = 200 \][/tex]
2. Determine the frequency of the recessive phenotype (bb), which is [tex]\( q^2 \)[/tex]:
[tex]\[ q^2 = \frac{\text{number of students with blue eyes}}{\text{total number of students}} \][/tex]
[tex]\[ q^2 = \frac{72}{200} \][/tex]
[tex]\[ q^2 = 0.36 \][/tex]
3. Find [tex]\( q \)[/tex] (frequency of recessive allele):
[tex]\[ q = \sqrt{q^2} \][/tex]
[tex]\[ q = \sqrt{0.36} \][/tex]
[tex]\[ q = 0.6 \][/tex]
4. Calculate [tex]\( p \)[/tex] (frequency of the dominant allele):
[tex]\[ p = 1 - q \][/tex]
[tex]\[ p = 1 - 0.6 \][/tex]
[tex]\[ p = 0.4 \][/tex]
5. Determine the frequencies of the genotypes [tex]\( p^2 \)[/tex] (BB), [tex]\( 2pq \)[/tex] (Bb), and [tex]\( q^2 \)[/tex] (bb):
[tex]\[ p^2 = (0.4)^2 \][/tex]
[tex]\[ p^2 = 0.16 \][/tex]
[tex]\[ 2pq = 2 \cdot 0.4 \cdot 0.6 \][/tex]
[tex]\[ 2pq = 0.48 \][/tex]
[tex]\[ q^2 = 0.36 \][/tex] (already calculated)
Now, we compare these frequencies with the given options:
A. [tex]\( p^2 = 0.16, 2pq = 0.48, q^2 = 0.36 \)[/tex]
B. [tex]\( p^2 = 0.65, 2pq = 0.302, q^2 = 0.048 \)[/tex]
C. [tex]\( p^2 = 0.24, 2pq = 0.34, q^2 = 0.42 \)[/tex]
Given our calculated values, the correct frequencies are:
[tex]\[ \boxed{(0.16, 0.48, 0.36)} \][/tex]
Thus, the correct answer is:
A. [tex]\( p^2 = 0.16, 2pq = 0.48, q^2 = 0.36 \)[/tex]
[tex]\[ p^2 + 2pq + q^2 = 1 \][/tex]
where:
- [tex]\( p \)[/tex] is the frequency of the dominant allele (B)
- [tex]\( q \)[/tex] is the frequency of the recessive allele (b)
- [tex]\( p^2 \)[/tex] is the frequency of the homozygous dominant genotype (BB)
- [tex]\( 2pq \)[/tex] is the frequency of the heterozygous genotype (Bb)
- [tex]\( q^2 \)[/tex] is the frequency of the homozygous recessive genotype (bb)
Given 128 students with brown eyes (BB or Bb) and 72 students with blue eyes (bb), let's determine the correct genotype frequencies.
1. Calculate the total number of students:
[tex]\[ 128 + 72 = 200 \][/tex]
2. Determine the frequency of the recessive phenotype (bb), which is [tex]\( q^2 \)[/tex]:
[tex]\[ q^2 = \frac{\text{number of students with blue eyes}}{\text{total number of students}} \][/tex]
[tex]\[ q^2 = \frac{72}{200} \][/tex]
[tex]\[ q^2 = 0.36 \][/tex]
3. Find [tex]\( q \)[/tex] (frequency of recessive allele):
[tex]\[ q = \sqrt{q^2} \][/tex]
[tex]\[ q = \sqrt{0.36} \][/tex]
[tex]\[ q = 0.6 \][/tex]
4. Calculate [tex]\( p \)[/tex] (frequency of the dominant allele):
[tex]\[ p = 1 - q \][/tex]
[tex]\[ p = 1 - 0.6 \][/tex]
[tex]\[ p = 0.4 \][/tex]
5. Determine the frequencies of the genotypes [tex]\( p^2 \)[/tex] (BB), [tex]\( 2pq \)[/tex] (Bb), and [tex]\( q^2 \)[/tex] (bb):
[tex]\[ p^2 = (0.4)^2 \][/tex]
[tex]\[ p^2 = 0.16 \][/tex]
[tex]\[ 2pq = 2 \cdot 0.4 \cdot 0.6 \][/tex]
[tex]\[ 2pq = 0.48 \][/tex]
[tex]\[ q^2 = 0.36 \][/tex] (already calculated)
Now, we compare these frequencies with the given options:
A. [tex]\( p^2 = 0.16, 2pq = 0.48, q^2 = 0.36 \)[/tex]
B. [tex]\( p^2 = 0.65, 2pq = 0.302, q^2 = 0.048 \)[/tex]
C. [tex]\( p^2 = 0.24, 2pq = 0.34, q^2 = 0.42 \)[/tex]
Given our calculated values, the correct frequencies are:
[tex]\[ \boxed{(0.16, 0.48, 0.36)} \][/tex]
Thus, the correct answer is:
A. [tex]\( p^2 = 0.16, 2pq = 0.48, q^2 = 0.36 \)[/tex]
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