AP Biology - Understanding Hardy-Weinberg Assumptions and Calculations

Which of the following is a Hardy-Weinberg assumption?

Random mating

Natural selection is in operation

High rate of mutation

Gene flow between populations

Explanation

Random mating is one of the five Hardy-Weinberg assumptions that help maintain equilibrium. If random mating occurs, in tandem with the other assumptions, we can reasonably assume that there will not be a shift in allele frequencies or distributions.

The other Hardy-Weinberg assumptions are that natural selection does not occur, mutation does not occur, genetic drift (gene flow) does not occur, and that the population size is large.

If four percent of the population is homozygous recessive for the trait that carries dimples (recessive), what is the fractional frequency of the dominant allele?

Explanation

Using the Hardy-Weinberg law to solve for allele frequency in populations, you can solve for the answer using the following two equations.

$p^{2}+2pq+q^{2}=1$

p is the fractional frequency of the dominant allele, q is the fractional frequency of the recessive allele, and q2 is the fraction of the population that is homozygous recessive. q2 is given in the question to be 0.04 (or 4%).

$q^{2}= 0.04 \rightarrow q=0.2$

$0.2 + p=1 \rightarrow p=0.8$

A population of snails is in Hardy-Weinberg equilibrium. The snails come in two different colors: red, the dominant phenotype, and white, the recessive phenotype. The population consists of sixty-four red snails and thirty-six white snails.

Assuming that the population is in Hardy-Weinberg equilibrium, what is the value of $\small q$ ?

Explanation

We can solve this question using the Hardy-Weinberg equations:

$\small q$ is equal to the recessive allele frequency, while $\small q^2$ in the second Hardy-Weinberg equation corresponds to the frequency of the recessive phenotype.

The question tells us the number of dominant red snails and the number of recessive white snails. Using these values, we can find the frequency of the recessive phenotype.

$q^2=\frac{#white}{#total}=\frac{36}{64+36}$

From here, take the square root to find the value of $\small q$ .

$\sqrt{q^2}=\sqrt{0.36}$

A population of snails is originally in Hardy-Weinberg equilibrium. The snails come in two different colors: red, the dominant phenotype, and white, the recessive phenotype. The original population has a dominant allele frequency of $\small 0.6$ and a recessive allele frequency of $\small 0.4$ . A new predator is introduced to the habitat that is particularly fond of the red snails. After a few years the dominant allele frequency has been reduced to $\small 0.2$ .

What is the recessive allele frequency after the introduction of this predator?

Explanation

Most of the information in the question is actually superfluous because we are given the final dominant allele frequency. The dominant allele frequency corresponds to the variable $\small p$ in the Hardy-Weinberg equations.

The question tells us that the dominant allele frequency after introduction of the predator is $\small 0.2$ . Use this value in the first Hardy-Weinberg equation to solve for the recessive allele frequency, $\small q$ .

A population of snails is in Hardy-Weinberg equilibrium. The snails come in two different colors: red, the dominant phenotype, and white, the recessive phenotype. There are sixteen homozygous dominant, forty-eight heterozygous, and thirty-six homozygous recessive snails.

What are the allele frequencies for this population?

$p=0.4\ \text{and}\ q=0.6$

$p=1.0\ \text{and}\ q=0.0$

$p=0.35\ \text{and}\ q=0.65$

$p=0.2\ \text{and}\ q=0.8$

Explanation

We can solve this question using the Hardy-Weinberg equations:

In the second equation, $\small p^2$ corresponds to the frequency of homozygous dominant individuals, $\small 2pq$ corresponds to the heterozygous frequency, and $\small q^2$ corresponds to the frequency of homozygous recessive individuals. We are given enough information to find each of these values from the question.

$p^2=\frac{#homo_D}{#total}=\frac{16}{16+48+36}=0.16$

$2pq=\frac{#hetero}{#total}=\frac{48}{16+48+36}=0.48$

$q^2=\frac{#homo_R}{#total}=\frac{36}{16+48+36}=0.36$

We can find the values of $\small p$ and $\small q$ by taking the square root of their squares.

$\begin{matrix} p^2=0.16 & q^2=0.36\ \sqrt{p^2}=\sqrt{0.16}& \sqrt{q^2}=\sqrt{0.36}\ p=0.4& q=0.6 \end{matrix}$

Which of the following is not a Hardy-Weinberg assumption?

Mutation frequencies are high

The population is large

Random mating occurs

Natural selection is not in operation

Explanation

One of the five main assumptions is that mutations are negligible. This makes sense because, if a population is in Hardy-Weinberg equilibrium, evolution is not occurring. A low rate of mutations would help keep a population in equilibrium.

The five assumptions of Hardy-Weinberg equilibrium are a large population size, no natural selection, no mutation rate, no genetic drift, and random mating.

Which of the following are assumptions made by the Hardy-Weinberg principle?

All of these

Random mating and sexual reproduction

Large population size

No migration, mutation, and selection

Explanation

The Hardy-Weinberg principle is a theory that describes how allele frequencies may change within a population absent of evolutionary mechanisms. The theorem is based on certain assumptions regarding the population in question. These assumptions include random mating, large population size, sexual reproduction, and the absence of migration, mutation and selection. It is exceedingly rare for all the Hardy-Weinberg assumptions to be met in nature, but this theory is a tool used to study allele frequencies within populations.

Imagine that a population is in Hardy-Weinberg equilibrium. A certain gene presents as two different alleles, and 49% of the population is homozygous dominant.

What percentage of the population is homozygous recessive?

51%

42%

Further information is needed to solve the problem

Explanation

When a population is in Hardy-Weinberg equilibrium, we can quantitatively determine how the alleles are distributed in the population. P2 is equal to the proprtion of the population that is homozygous dominant based on the equation p2 + 2pq + q2 = 1. We also know that p + q = 1.

Since P2 = 0.49 in this case, we know that p is equal to 0.7. Since there are only two alleles for this gene, we know that the other allele, q in this case, is 0.3. Since homozygous recessive is referred to as q2 in the equation, we can plug in the value of 0.3 and determine that q2 = 0.09. As a result, we confirm that 9% of the population is homozygous recessive.

In a given population of snails, spiral shells are dominant to round shells. If 36% of the population is homozygous for the spiral shell allele, what percentage of the population is heterozygous?

Explanation

We can use the Hardy-Weinberg equations to solve this problem.

We know that spiral shells are dominant, and that 36% of the population is homozygous for the spiral allele. This tells us that 36% of the population is homozygous dominant. The term corresponds to the homozygous dominant percentage.

$p=\sqrt{0.36}=0.6$

is the dominant allele frequency. Now we can solve for , the recessive allele frequency.

The term will give us the frequency of heterozygotes.

48% of the population is heterozygous.

A population is in Hardy-Weinberg equilibrium. In the population, 1% of individuals show the recessive trait for blue eyes. What is the value of in this situation?

Explanation

For a population in Hardy-Weinberg equilibrium, every trait follows the equations:

In these formulas, represents the frequency of the dominant allele and represents the frequency of the recessive allele. $p^{2}$ represents the frequency of the homozygous dominant genotype, represents the frequency of the heterozygous genotype, and $q^{2}$ represents the frequency of the homozygous recessive genotype.

In this case, the individuals with blue eyes would be represented by the homozygous recessive genotype. Using this data, we can solve for the frequency of the recessive allele.

$q^{2} = 0.01$