The only answer that properly explains a 2:1 ratio is if the homozygous dominant phenotype is lethal. In our punnett square, this would give a two-thirds chance for a heterozygous offspring and a one-third chance for a recessive offspring. We know the recessive phenotype is not lethal because homozygous recessive offspring were produced.

Parents: Aa x Aa

Offspring genotypes: AA, Aa, Aa, aa

Offspring phenotypes: lethal, dominant, dominant, recessive (only three live offspring produced)

Offspring ratio: 2 dominant to 1 recessive

There is also no evidence that codominance or incomplete dominance is present. If black and white spotted offspring or gray offspring were produced then these theories would have merit.

This problem requires a standard dihybrid cross. The crossed genotypes are AaBb x AaBb. This results in a phenotypic ratio of 9 dominant for both traits, 3 dominant for a single trait, 3 dominant for the other trait, and 1 recessive for both traits. In this cross, it will result in: 9 AxBx, 3 Axbb, 3 aaBx, and 1 aabb.

Since we know that the genes are both capable of making the red coloration we actually need to add together all of the choices that contain at least a single dominant allele. Essentially, AxBx, Axbb, and aaBx all show the exact same phenotype. This leaves us with a 15:1 ratio of red to white flowers. 

We are not given the mother's full genotype in the question; she could reasonably carry two A alleles, or an A allele and a recessive O allele. We know that the father must carry one copy of the A allele and one copy of the B allele.

Two punnett squares can answer this question, corresponding to the two possible maternal genotypes: one crossing AA x AB and the other crossing AO x AB. From the first cross there is a 50% chance of blood type A versus 50% chance of blood type AB (half AA and half AB). The second cross shows that there is a potential chance of 50% for type A, 25% for type AB, and 25% for type B (one AA, one OA, one AB, and one OB).

Based on these possibilities, the child could have blood type A, B, or AB. The child cannot have blood type O.

There are nine distinct genotypes present after a standard dihybrid cross. This question can easily be answered by setting up a Punnett square (AaBb x AaBb) and counting the number of unique genotypes present after doing the cross. The numbers also conveniently work out that however many offspring display the dominant phenotype is equal to the number to of genotypes present (this is true for monohybrid and trihybrid crosses as well). 

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.