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Sedbober [7]
3 years ago
12

1. In a 1000 of African people, 4% of the population is born with sickle cell anemia. What percentage of the population possess

the heterozygous advantage
Biology
1 answer:
Bingel [31]3 years ago
7 0

The percentage of the population that possesses the heterozygous advantage is 0.32 (2pq). It is a principle known as the Hardy Weinberg equilibrium.

<h3>Hardy Weinberg equilibrium</h3>

The Hardy-Weinberg equilibrium is a principle in population genetics that states that allele and genotype frequencies in a population will remain constant across generations in the absence of different evolutionary forces.

The equation of the Hardy Weinberg equilibrium is p² + 2pq + q² = 1, where p² is the dominant homo-zygous frequency, q² is the recessive homo-zygous frequency, and 2pq is equal to the heterozygous frequency.

In this case,

  • q² = 0.04 >> q = √0.04 = 0.2
  • p + q = 1 >> p + 0.2 = 1 >> p = 0.8

According to the Hardy Weinberg equilibrium, the number of heterozygous is equal to 2pq.

In this case,

  • 2pq = 2 x (0.2 x 0.8) = 0.32

Learn more about the Hardy Weinberg equilibrium here:

brainly.com/question/3406634

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