Answer:
The correct option is;
D. 45%
Explanation:
From the Hardy- Weinberg law, we have;
p² + 2·p·q + q² = 1
p + q = 1
Where:
p = Dom inant allele frequency in the population
q = Recessive allele frequency in the population
p² = The percentage of individuals in the population that are hom ozygous dominant
q² = The percentage of individuals in the population that are homo zygous recessive
2×p×q = The percentage of hete rozyous individuals in the population
The number of individuals that express the recessive phenotype = 86
The number of individuals in the population = 200
The percentage of individuals that express the recessive phenotype, q² = 86/200 = 0.43
Therefore;
q = √0.43 = 0.656
p + q = 1
p = 1 - q = 1 - 0.656= 0.344
∴ The frequency of individuals that express the do minant phe notype, p = 0.344
The percentage of heterozyous individuals in the population = 2×p×q × 100 = 2 × 0.656 × 0.344 × 100 = 45.15% ≈ 45%
Answer:
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