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Answer:
B quality is bugs controls the quantity of eyes. The predominant allele, B, is answerable for 2 eyes and the latent allele, b, is dependable dor 3 eyes.
At first, the populace is absolutely homozygous, homozygous prevailing (BB) and homozygous latent (bb). All bugs will have either 2 (BB) or 3 (bb) eyes.
In any case, after the test, the populace ends up being homozygous latent (bb) and heterozygous predominant (Bb). The genotype changes, however the phenotype continues as before, either 2 eyes (Bb) or 3 eyes (bb).
If a population is in Hardy-Weinberg equilibrium, its allelic and genotypic frequencies will remain the same through many generations. The expected phenotypic frequencies are BB + Bb = 0.51 / bb = 0.49.
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The Hardy-Weinberg equilibrium theory states that the allelic frequencies in a locus are represented as p and q. Assuming a diallelic gene,
- p is the frequency of the dominant allele,
- q is the frequency of the recessive allele.
The genotypic frequencies after one generation are
- p² (H0m0zyg0us dominant genotypic frequency),
- 2pq (Heter0zyg0us genotypic frequency),
- q² (H0m0zyg0us recessive genotypic frequency).
If a population is in H-W equilibrium, it gets the same allelic and genotypic frequencies generation after generation.
The addition of the allelic frequencies equals 1
p + q = 1
The sum of genotypic frequencies equals 1
p² + 2pq + q² = 1
We know that the starting population has 20 individuals.
- 10 of them are BB ⇒ frequency = 0.5
- the other 10 are bb ⇒ frequency = 0.5
These are the genotypic frequencies at the starting population.
Now, we need to get the allelic frequencies.
To get the allelic frequency - f(b)-, we will use the recessive genotypic frequency, F(bb).
F(bb) = q² = 0.5
f(b) = q = √0.5 = 0.707 ≅ 0.7
So, q = 0.7
We will get the dominant allelic frequency, f(B) = p, by crealing the following equation,
p + q = 1
p + 0.7 = 1
p = 1 - 0.7
p = 0.3
So, up to here we have
- p = 0.3
- q = 0.7
Let us calsulate genotypic frequency after one generation
- F(BB) = p² = 0.3² = 0.09
- F(Bb) = 2pq= 2 x 0.3 x 0.7 = 0.42
- F(bb) = q² = 0.7² = 0.49
The expected phenotypic frequencies are
- BB + Bb = 0.09 + 0.42 = 0.51
- bb = 0.49
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