Respuesta :
Generation after generation, populations in H-W equilibrium have the same allelic and genotypic frequencies.
f(Sb) = p = 0.626. f(Sb+) = q = 0.374. F(SbSb+) = 2pq = 0.468
What is Hardy-Weinberg Equilibrium?
According to the Hardy-Weinberg equilibrium theory, assuming a diallelic gene.
The dominant allele p(X) has a frequency of p.
The recessive allele p(x) has a frequency of q.
After one generation, the genotypic frequencies are
p² (Homozygous dominant genotypic frequency),
2pq (Heterozygous genotypic frequency),
q² (Homozygous recessive genotypic frequency).
In a population that is in H-W equilibrium, genotypic and allelic frequencies don't change over generations.
The allelic frequencies added together equal 1.
p + q = 1.
The sum of genotypic frequencies equals 1
p² + 2pq + q² = 1
In the exposed:
Sb is dominant
Sb+ is recessive ⇒ wild type
Population size ⇒ 137 + 836 = 973
wildtype bristles ⇒ 137 ⇒ Sb + Sb+
stubble bristles ⇒ 836 ⇒ SbSb and SbSb+
Phenotypic frequency:
F(wildtype) = q² = 137/973 = 0.14
F(stubble) = p² + 2pq = 836/973 = 0.86
To determine the genotypic and allelic frequencies, we will now use the frequency of the recessive trait.
F(Sb+Sb+) = q² = 137/973 = 0.14 ⇒ Recessive Genotypic frequency
f(Sb+) = q = √0.14 = 0.374 ⇒ Recessive Allelic frequency
We will now utilize q (recessive allelic frequency) to determine the dominant genotypic frequency and the dominant allelic frequency since this population is in H-W equilibrium. We will solve the following equation to achieve this.
p + q = 1
p + 0.374 = 1
p = 1 - 0.374
p = 0.626 ⇒ Dominant Allelic frequency
F( SbSb) = p² = 0.626² = 0.392 ⇒ Dominant Genotypic frequency
The heterozygous frequency will be the last thing we learn. There are 2 possibilities, The dominant genotypic frequency is subtracted from the dominant phenotypic frequency.
F(SbSb + SbSb+) - F(SbSb) = 0.86 - 0.392 = 0.468
p² + 2pq + q² = 1
0.392 + 2pq + 0.14 = 1
2pq = 1 - 0.392 - 0.14
2pq = 0.468
Answers:
- f(Sb) = p = 0.626 ⇒ Dominant Allelic frequency
- f(Sb+) = q = 0.374 ⇒ Recessive Allelic frequency
- F(SbSb+) = 2pq = 0.468 ⇒ Heterozygous frequency
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