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10 December, 06:34

In a small population of Brazilian natives, the frequency of gene p, responsible for this disease, is 0.3. What must be the frequency of people who are heterozygous for this disease? (p + q = 1, p2 + 2pq + q2 = 1)

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  1. 10 December, 09:31
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    Hello,

    If:

    p - the frequency of dominant allele P,

    q - the frequency of recessive allele p,

    the frequencies of the genotypes are:

    p² - for PP genotype (dominant homozygote without the disease),

    2pq - for Pp genotype (heterozygote for disease),

    q² - for pp genotype (recessive homozygote with the disease).

    It is given:

    p = 0.3

    2pq = ?

    Since p + q = 1 ⇒ q = 1 - p

    q = 1 - 0.3 = 0.7

    Knowing p and q, we can calculate the frequency of people heterozygous for this disease (2pq):

    2pq = 2 · 0.3 · 0.7 = 0.42

    Therefore, the frequency of people heterozygous for this disease is 0.42
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