10.1.4 Using the Hardy-Weinberg Theory

Question 1

using hardy-weinberg theory

Answer 1

• Review: The Hardy-Weinberg equationis used to predict allelic and genotypic frequencies in a nonevolving population.
• Given the frequency of one genotype, you can use the Hardy-Weinberg equation and simple algebra to calculate the frequency of the two other genotypes.
- The equation can be used to determine the frequency of carriers of serious conditions in human populations.
- For example, use the equation to determine the number of carriers for Tay-Sachs disease in a population of 20,000. Two people in the population are known to have the condition.
- To solve, first determine the values of p and q, where
p= the frequency of the dominant allele and q= the frequency of the recessive allele.
- Since the disease results from a homozygous recessive condition, the value q^2 represents the frequency of individuals with Tay-Sachs. Solve for q^2 and determine the values of q and p.
q^2=2/20000
q^2= 0.0001
q = 0.01
- Now substitute the value of q in the equation p + q = 1 to determine the value of p.
p + q = 1
p + 0.01 = 1
p = 0.99
- When the values of p and q are determined, they can be entered into the Hardy-Weinberg equation to determine the frequency of carriers.
- Review: The Hardy-Weinberg equation:
p2 + 2pq + q^2 = 1.
- The illustration on the left shows the Hardy-Weinberg
equation and the genotypes represented by p^2, 2pq, and q^2. In the equation, p^2 represents the frequency of homozygous dominant individuals, 2pq represents the frequency of heterozygous individuals, and q^2
represents the frequency of homozygous recessive individuals.
- In the problem, we are interested in determining the frequency of carriers, which can be obtained by determining the value of 2pq.
2pq = 2(0.99)(0.01) = 0.02
- Two percent of the population are carriers for Tay-Sachs disease. Of the 20,000 individuals of the population, 400 are carriers, or 1 out of 50.

Question 2

Sickle cell anemia is a heritable condition in which red blood cells become misshapen. A survey of 40,000 African Americans was conducted. Sixteen individuals showed symptoms of the disease. What is the frequency of the recessive allele, q, assuming Hardy–Weinberg equilibrium?

Answer 2

• 0.02

Question 3

A study on eye color was performed at an elementary school. The allele for brown eyes is dominant over the allele for blue eyes.

Using the data given, what is the number of students in the 1995 sample that was taken that are heterozygous assuming the group is in Hardy–Weinberg equilibrium?

Answer 3

• 420

Question 4

The Hardy–Weinberg formula is used for what purpose?

Answer 4

• To predict the gene or genotype frequency of a population

Question 5

A sample of a large population showed that 3,200 of 20,000 individuals had a particular recessive phenotype. If the population were in Hardy–Weinberg equilibrium, how many individuals out of 100 would be expected to be homozygous dominant?

Answer 5

• 36

Question 6

On the planet Zoom there is a population of animals called snerkles. In snerkles, white nose hairs are dominant to blue nose hairs. In this population 9% of the individuals have blue nose hairs. Assuming Hardy–Weinberg equilibrium what percentage of individuals are heterozygotes, expressing the white nose hair phenotype?

Answer 6

• 42%

Question 7

In a population of cats that is in Hardy–Weinberg equilibrium, 25% of the individuals have long hair, a recessive trait. What is the frequency of the dominant allele in the population?

Answer 7

• 0.5