Question

100 POINTS!!! hardy weinberg problem

Scenario #2 In 2001, a population of 2,500 poison dart frogs lived in the Amazon rain forest. Due to increased deforestation, the population dwindled to 25 frogs in 2019. New government regulations were enacted in 2022, successfully putting an end to the deforestation of the Amazon rain forest. Once deforestation was stopped, the poison dart frog population was able to recover. By 2050, the population reached 8,000 frogs. Of that population, 20 are homozygous recessive for being spotted (ss genotype).

Solve for the frequency's below

q2

q

p

p2

2pq

Answer 1

Sure, here are the frequencies:

- q^2 = 0.0025 (frequency of homozygous recessive individuals)

- q = 0.05 (frequency of the recessive allele)

- p = 0.95 (frequency of the dominant allele)

- p^2 = 0.9025 (frequency of homozygous dominant individuals)

- 2pq = 0.095 (frequency of heterozygous individuals)

These frequencies were calculated using the Hardy-Weinberg equilibrium equation, which relates the frequencies of alleles and genotypes in a population that is not evolving. In this case, we knew the frequency of one genotype (ss) and used the equation to solve for the frequencies of the other genotypes and alleles.

Answer 2

q^2 = 0.0025

q ≈ 0.05

p ≈ 0.95

p^2 ≈ 0.9025

2pq ≈ 0.095

Step-by-step explanation:

To solve for the frequencies, we can use the Hardy-Weinberg equation, which relates allele frequencies to genotype frequencies in a population. In this case, we have the following information:

The population reached 8,000 frogs by 2050.

The frequency of the homozygous recessive genotype (ss) is 20.

Let's calculate the frequencies:

q^2 represents the frequency of the homozygous recessive genotype (ss). In this case, q^2 = 20/8000 = 0.0025.

q represents the frequency of the recessive allele. To find q, we can take the square root of q^2: q = √(0.0025) ≈ 0.05.

p represents the frequency of the dominant allele. Since p + q = 1, p ≈ 1 - q ≈ 1 - 0.05 = 0.95.

p^2 represents the frequency of the homozygous dominant genotype (SS). p^2 ≈ (0.95)^2 ≈ 0.9025.

2pq represents the frequency of the heterozygous genotype (Ss). 2pq ≈ 2(0.95)(0.05) ≈ 0.095.