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