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?

Explanation:

According to hardy weinberg

p+q=1

p2+2pq+q2=1

p stands for the dominant genotype

q stands for recessive genotype

pq stands for population that are not homogenous i.e not the same

Given that p=0.3

P2=0.09

q=1-0.3= 0.7

q2=0.49

2pq= 2×0.3×0.7

=0.42

Therefore, the frequency of heterozygous population is 0.42

p2+2pq+q2

0.09+0.42+0.49 = 1

gold i'm pretty sure i might be wrong

the answer is a) cell division.

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