You sampled 1000 individuals from a population of mosquitoes to test whether they are in Hardy-Weinberg Equilibrium for a gene D, where a recessive allele (d) is responsible for conferring pesticide resistance. After genotyping your 1000 sampled mosquitoes, go obtain the following number of individuals for the three possible genotypes: a. Genotype dd: 22 individuals b. Genotype Dd: 188 individuals c. Genotype DD: 790 individuals

The population is not following the Hardy-Weinberg equilibrium.

Explanation:

Hardy-Weinberg  proposed that the population will remain in the equilibrium if they are not affected with the external natural force like genetic drift, mutation and many others. The gene pool of a population will remain in equilibrium state for generations.

The mathematical expression for the Hardy-Weinberg are

Genotypic frequency is p²+q²+2pq=1

Allele frequency is p+q=1

where p-dominant allele,

q-recessive allele

Therefore,calculating expected frequency of allele q,

q² (recessive genotype)-  q²+2pq/1000

= 22+188/1000

=210/1000

= 0.21

q= 0.45

Calculating expected frequency of allele p,

Using p+q=1

p=1-q

p= 0.55 (expected)

Calculating observed value of frequency of p,

p²= p²+2pq/1000

= 790+188/1000

= 978/1000

= 0.978

p = 0.988 (observed)

Since the expected frequency does not match the observed frequency therefore the population is not under the Hardy-Weinberg equilibrium and the gene pool is disturbed.

d is the answer

sorry if its wrong!