The best way to solve this problem is by systematically analyzing the possible genotypes for the parent generation and by eliminating the improbable genotypes. To solve for the parents’ genotypes you need to look at the F1 and F2 generations.

As described in the passage the F1 generation only exhibits one phenotype: round and yellow. Round and yellow are dominant alleles because you get four different phenotypes when you test cross (cross them with homozygous recessive individuals) a plant from the F1 generation. The F1 generation could be either homozygous dominant or heterozygous for both traits, since they exhibit the dominant alleles.

If F1 offspring were all homozygous dominant then you would only observe dominant traits in F2 offspring. The F1 offspring must be heterozygous for both traits in order to produce the observed ratios in the F2 generation.

F1 test cross: AaBb x aabb

F2 generation: AaBb, Aabb, aaBb, and aabb

Since all F1 offspring are heterozygous, we only have one possible combination of parents: homozygous dominant and homozygous recessive.

P generation: AABB x aabb

F1 outcome: all AaBb (observed)

The best answer is that both parents are homozygotes.

The passage tells us that the disease gene is X-linked and recessive. If the young girl has the disease, then she must have a double recessive genotype, meaning that both of her parents have at least one recessive gene. Her mother must either be a carrier, or have the disease, but we cannot conclude one over the other. Her father must have the disease, since he only has one copy of the X-chromosome. He received his Y chromosome from his father, so we cannot make any judgments about the father's father. He received his X chromosome from his mother, meaning that she must have been a carrier or had the disease. As far as the unborn brother is concerned, he has a 50% chance of getting the disease if the mother is a carrier, and 100% if she has the disease.

The only thing we can say with 100% certainty is that the father has the disease.

Codominance occurs when each allele for a certain gene isn't completely dominant over the other, so both are expressed. We see both fully blue and fully red flowers on each plant, so codominance fits our situation in this question. Incomplete dominance would be the correct answer if the filial plants had flowers that were a combination of blue and red, so they would be purple. Expressivity is the term for when a trait is expressed to different degrees in a population, like some plants having flowers that are more blue than others. Penetrance is when only a certain percentage of the population expresses the trait at all.

The correct answer is 1/16. A dihybrid cross results in a phenotypic ratio of 9:3:3:1. The 1 in the ratio is indicative of the double recessive phenotype, with a homozygous recessive genotype for both traits. Thus, only 1/16 of the offspring will display both recessive phenotypes.

This problem asks you to use concepts of inheritance and Mendelian genetics. The best approach to this problem is to rule out possiblities rather than to find the actual mode of inheritance, as the latter can be a much more difficult and time-consuming process. First off, we know that Y-linked inheritance could not explain this pattern because we see that in generation 1 (G1), the male is affected. If he is affected, all of his sons (who inherit his Y chromosome) would also be affected. There is one son in G2 who is not. Similarly, dominant X-linked inheritance could not explain this pattern; recall that the daughters inherit two copies of the X chromosome, and one is always inactivated. Were the trait X-linked dominant, then the girls of generation 3 (G3) would be affected, having received a copy of the affected gene from their father. Revisiting all other options, we see that any of the remaining inheritance patterns could possibly explain what we see.

Blood type is inherited as a codominant trait and relies on alleles for blood antibodies as well as Rh (Rhesus) factor. The father's blood type is A–, so he has no Rh factor and must be either AA or AO. The mother must be AB with an Rh factor.

Father possibilities: A^-A^- or A^-O^-

Mother possibilities: A^-B⁺ or A⁺B^- or A⁺B⁺

Based on these possibilities, we cannot conclude if the child will be positive or negative for Rh factor; however, since the mother has no allele for O blood type, we can conclude that the child cannot have O type blood. The child could receive AA, AO, BO, or AB.

The results of blood typing crosses can be calculated in the same way as those of traditional crosses: with Punnett squares. If we put the AB (maternal) genotype along one side of the square and the BB (paternal) genotype along the other, we see that only two genotypes are present in the offspring. 50% will be AB and 50% will be BB (a blood type of B); therefore, A is not one of the blood types present in the offspring, so this is our answer.

All of the other answer choices have a possibility of producing A type offspring.

We can depict the parental genotypes by using to signify the dominant white-eye allele and to signify the recessive red-eye allele.

The mother would have genotype since she has red eyes. The father would have genotype since he has white eyes.

We can see that the possible genotype for offspring will be either for a daughter or for a son. Any daughters must inherit the X-chromosome from both parents, and must therefore inherit a white-eye allele from the father. We can conclude that all daughters will have white eyes.

While this question looks complicated it can be broken down into three individual Punnett squares, and the probability of each of the three individual allele genotypes can be multiplied to give the desired answer.

Starting with P: Parent 1 is homozygous dominant (PP) and parent 2 is homozygous recessive (pp), therefore 100% of their offspring will be heterozygous (Pp), or 1/1.

For Q: Both parents are heterozygous (Qq); therefore 1/4 of their offspring will be QQ, 1/2 will be Qq, and 1/4 will be qq.

For R: Parent 1 is heterozygous (Rr) and parent 2 is homozygous recessive (rr); therefore 1/2 their offspring will be Rr and 1/2 will be rr.

Because the question asks the probability that the offspring is PpQQRr, we can multiply 1/1 * 1/4 * 1/2 = 1/8.

The parent cross (P generation)  will be PP x pp, where P is the dominant allele (purple) and p is the recessive allele (white). All offspring from this cross will be heterozygous, Pp, and will represent the F1 generation.

The F2 generation will result from a cross of the F1 generation, meaning Pp x Pp. This cross will result in 1 PP, 2 Pp, and 1 pp genotypes. Only one out of every four F2 offspring will have the recessive phenotype of white flowers.