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

a

explanation:

many plants naturally reproduce this way, but it can also be induced artificially. horticulturalists have developed asexual propagation techniques that use vegetative plant parts to replicate plants. success rates and difficulty of propagation vary greatly. monocotyledons typically lack a vascular cambium and therefore are harder to propagate.