Question

In a random mating population frequency of disease causing recessive allele is 80%. What would be the frequency of carrier individual in population and how?
(1) 64%
(2) 32%
(3) 16%
(4) 100%

Answer 1

Answer on Question#54555 – Biology – Genetics

Question:

In a random mating population frequency of disease causing recessive allele is 80%. What would be the frequency of carrier individual in population and how?

(1) 64%

(2) 32%

(3) 16%

(4) 100%

Solution:

According to the Hardy-Weinberg principle:

$$(p + q)^2 = p^2 + 2pq + q^2 = 1;$$

P – the frequency of «A» in the population;

Q – the frequency of «a» in the population;

AA – healthy individuals;

Aa – carriers;

aa – diseased;

q = 0.8

p = 1 – 0.8 = 0.2

$$(p + q)^2 = p^2 + 2pq + q^2 = 1;$$ $$0.2 * 0.2 + 2 * 0.8 * 0.2 + 0.8 * 0.8 = 0.04 + 0.32 + 0.64 = 1;$$ $$2pq = 2 * 0.8 * 0.2 = 0.32$$ $$2pq = 0.32 \text{ (Aa - carriers)}$$

Answer: (2) 32%

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