A monohybrid cross between two homozygous plants would involve a parental generation that looked like this: SS (tall) x ss (short). The F1 generation would produce only heterozygous tall plants (Ss). The F2 generation would produce offspring from the following cross: Ss x Ss. A punnett square would reveal that the F2 generation would have 25% homozygous tall (SS), 50% heterozygous tall (Ss), and 25% homozygous short plants. Note that we do not need information regarding which trait is dominant in this case, and we would still get the correct answer if we took short as the dominant phenotype.

The crossing of two heterozygous parents will yield 1 homozyougs dominant offspring, 1 homozygous recessive offspring and 2 heterozygous offspring. Since white blossoms is the dominant allele, the heterozygous offspring will be white leading to a phenotypic ratio of 3 white : 1 pink blossom.

Since we had pure-breeding parents (also known as homozygotes for their respective colors), we can safely say the F1 offspring are heterozygotes and have a red allele and a white allele. When crossing these offspring with one another, we will expect to get a 3:1 ratio of red to white flowers. Not all flowers will be red, but 75% of the flowers will be.

Shown below is the punnett square that reflects this conclusion. (Note that "A" represents a red allele and "a" represents a white allele):

                           A                      a

               A       AA (red)         Aa (red)

               a     Aa (red)           aa (white)

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.

Solve for the recessive allele frequency.

To solve this question, we must use the Hardy-Weinberg equations:

There are two ways to solve this problem. The easier way is to use the second Hardy-Weinberg equation. We are told that the population is in Hardy-Weinberg equilibrium, so the observed genotype frequencies equal the expected genotype frequencies. Each term in the second Hardy-Weinberg equation can be used in coordination with the given phenotypic frequencies.

The dominant allele frequency is 0.6 and the recessive allele frequency is 0.4.

The second method involves using our population and the total number of alleles. Since our population totals to 100, we have a total of 200 alleles (two alleles per member in the population. Next, to get the frequencies we simply have to divide the total number of a single allele by the total number of alleles in the population. For the dominant allele frequency it would look like this:

We can then get the recessive allele frequency from the first Hardy-Weinberg equation:

Both methods result in the same answer. 

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.

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.

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.

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.

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.

Use the two Hardy-Weinberg equations:

Above,  is the frequency of the dominant allele, and  is the frequency of the recessive allele in the isolated population. 

Since there are 20 people in total on the island, that means that there are 40 alleles for eye color. 2 of the 40 are for the blue allele:

We are looking for the blue eye phenotype, which can only result from two recessive alleles.