GRE Subject Test: Biochemistry, Cell, and Molecular Biology - Help with Hardy-Weinberg

Which of the following is not an assumption of the Hardy-Weinberg equilibrium?

Non-random mating within a population

No natural selection within a population

No genetic drift within a population

No gene flow between populations

No mutations within a population

Explanation

Non-random mating is not an assumption of the Hardy-Weinberg equilibrium, in fact, in order to make predictions about the next generation, random mating must be assumed. Additionally, no new mutations, no gene flow, no genetic drift, and no natural selection must also occur. If any of these phenomenon are present in a population, we can not estimate allele frequencies in subsequent generations due to chance, rather selective pressures may favor one allele over another allele.

Which of the following choices are likely to change the allele frequencies of the indicated populations?

I. A geographic barrier isolating a small subset of a larger population

II. The introduction of a predator that only preys upon the homozygous dominant members of the population

III. A population that displays completely random mating

I and II

I, II, and III

III only

II only

Explanation

Allele frequencies are the measure of an allele in relation to the total number of alleles in the given population. Introducing a predator that only preys upon homozygous dominant members will cause the number of dominant alleles to drop significantly and will, therefore, change allele frequencies. This would be an example of the bottleneck effect. Isolating a small subset of a population is going to change allele frequencies because that small subset is not likely to accurately represent the original population. This is an example of the founder effect.

Random mating is actually a factor that helps maintain allele frequencies, and is a requirement for Hardy-Weinberg equilibrium.

A population of insects exists in which black coloration is dominant to white. If there are 64 black insects and 36 white insects in the population, what is the recessive allele frequency?

Explanation

We can use the white insect population to figure out our recessive allele frequency if we use the Hardy-Weinberg equations:

From this equation we know that is actually the frequency of our homozygous recessive genotype. We can determine the value of this genotypic frequency based on the information in the question. Any white insects must be homozygous recessive.

$\frac{36}{36+64}=0.36$

Solve for the recessive allele frequency.

$q=\sqrt{0.36}=0.6$

Within his rat population, a scientist is trying to generate twice as many recessive homozygotes as heterozygotes. What allelic frequency would accomplish this?

Explanation

Use the Hardy-Weinberg equations:

The equation he will need to set up is the following:

Solve for and substitute the first equation into the equation above.

Simplify.

$q=\frac{4}{5}=0.8$

Lastly, find .

Which of the following is not a tenet of Hardy-Weinberg equilibrium?

Genetic drift

Large population

Randomized mating

No natural selection

No migration

Explanation

The Hardy-Weinberg equilibrium does not account for genetic drift. The Hardy-Weinberg law states that genetic frequencies will remain constant in a population from generation to generation in the absence of evolutionary influences. Therefore, there is no migration, natural selection, nonrandom mating, or small populations in a Hardy-Weinberg population.

What are the phenotypic ratios for a given population for which the proportion of the dominant allele is 0.55 and that of the recessive allele is 0.45?

Homozygous dominant: 0.30

Heterozygous: 0.50

Homozygous recessive: 0.20

Homozygous dominant: 0.20

Heterozygous: 0.50

Homozygous recessive: 0.30

Homozygous dominant: 0.25

Heterozygous: 0.50

Homozygous recessive: 0.25

Homozygous dominant: 0.50

Heterozygous: 0.30

Homozygous recessive: 0.20

Homozygous dominant: 0.30

Heterozygous: 0.20

Homozygous recessive: 0.50

Explanation

To solve this problem, assume Hardy-Weinberg equilibrium and use the associated equations to solve:

is dominant allele and is recessive allele

To find the phenotype ratios:

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

$p^{2} =$ homozygous dominant

${p^{2 ^}} = 0.55^{2 } = 0.30$

heterozygous

$q^{2} =$ homozygous recessive

$q^{2} = 0.45^{2 } = 0.2$

Which of the following conditions are not necessary for a population to be in Hardy-Weinberg equilibrium?

Natural selection affects the alleles under consideration

Population size must be large

Mating must happen at random

There are no mutations

No migration between populations occurs

Explanation

The Hardy-Weinberg equilibrium states that the frequency of alleles at a locus remains constant from generation to generation. In order for this to be the case, natural selection cannot affect the alleles under consideration. All other answer choices describe conditions that do need to be met for Hardy-Weinberg equilibrium to be displayed. Note that the conditions for Hardy-Weinberg equilibrium are not met in nature.

A population of beetles exists in which black coloration is dominant to white. The allele frequencies of the population were originally o.4 for the dominant allele and 0.6 for the recessive allele. A predator was introduced that selectively ate the white beetles. The new population consists of 36 homozygous dominant black beetles, 48 heterozygous beetles, and 16 white beetles. What are the new allele frequencies?

$p = 0.6,\ q = 0.4$

$p = 0.75,\ q = 0.25$

$p = 0.4,\ q = 0.6$

$p = 0.25,\ q = 0.75$

Explanation

The original allele frequencies are actually superfluous information that we will not need to use in our calculation. We are given the populations of each genotype, so we can use the Hardy-Weinberg equations to solve for the allele frequencies.

In the second equation, gives the frequency of the homozygous dominant phenotype and gives the frequency of the homozygous recessive phenotype. Using the population ratios from the question we can solve for these values to find the allele frequencies.

$p^2=\frac{36}{36+48+16}=0.36$

$p=\sqrt{36}=0.6$

$q^2=\frac{16}{36+48+16}=0.16$

$q=\sqrt{0.16}=0.4$

The problem could also be solved by summing the total number of alleles, and dividing the total of each individual allele by this number. Keep in mind that each individual carries two alleles.

$p=\frac{2(36)+48}{2(36+48+16)}=0.6$

$q=\frac{2(16)+48}{2(36+48+16)}=0.4$

Which of the following is not a condition for Hardy-Weinberg equilibrium?

Natural selection is operating on the population

Completely random mating

Negligible mutation frequencies

Large population size

Explanation

Of the choices, the only one that is not a Hardy-Weinberg assumption is that natural selection is occurring on the population. In fact, the exact opposite is a Hardy-Weinberg assumption. If natural selection is occurring on a population, over a large period of time, it is likely to have an effect on allele frequencies within the population.

All other answers are requirements in order for Hardy-Weinberg equilibrium to be in effect: large population size, random mating, and negligible mutation frequencies.

An isolated population consists of 10 males and 10 females. Two individuals are carriers of the recessive blue eye allele. Assuming all Hardy-Weinberg conditions are met. What is the frequency of the blue eye phenotype in the population?