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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Ответ:
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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