Question

let A and a represent dominant and recessive alleles whos respective frequencies are p and q in a given interbreeding population at equilibrium (with p+q=1). A) if 16 percent of the individuals in the population have recessive phenotypes, what percentage of the total number of recessive genes exist in the heterozygous condition? B) if 1.0 percent of the individuals were homozygous recessive, what percentage of the recessive genes would occur in heterozygotes?

Answer 1

Answer:

A) $48$ %

B) $18$ %

Explanation:

A) Given

Frequency of individuals in the population having recessive phenotypes $= 16$%

Frequency of individuals in the population having recessive phenotypes is represented by $q^2$

$q^ 2 = 0.16\\q= \sqrt{0.16}\\ q = 0.4$

Thus frequency of recessive genes is equal to $40%$

frequency of dominant genes is equal to

$p = 1-q\\= 1-0.4\\= 0.6$

Now,

$p^2 + q^2+2pq = 1\\2pq = 1-0.16-0.36\\2pq = 0.48$

$48$ % of genes exist in the heterozygous condition

B.

Frequency of individuals in the population having recessive phenotypes $= 1$%

Frequency of individuals in the population having recessive phenotypes is represented by $q^2$

$q^ 2 = 0.01\\q= \sqrt{0.01}\\ q = 0.1$

Thus frequency of recessive genes is equal to $10%$

frequency of dominant genes is equal to

$p = 1-q\\= 1-0.1\\= 0.9$

Now,

$p^2 + q^2+2pq = 1\\2pq = 1-0.01-0.81\\2pq = 0.18$

$18$ % of genes exist in the heterozygous condition