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)

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

Probably a behavioral adaptation