In a population of 200, 86 individuals express the recessive phenotype. What percent of the population would you predict would be heterozygous?
A. 65%
B. 55%
C. 35%
D. 45%

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%

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