The disease is recessive and found on the X-chromosome. Remember that males only have one X-chromosome, while females have two. As a result, males only require one mutant allele to express a recessive phenotype, while females need to have two mutant alleles in order to have hemophilia.

In this scenario, both parents are healthy, but the female is a carrier for the disease. This means that one of her alleles codes for hemophilia.

Parents: XX^h x XY

Since daughters will need two copies of the mutant allele, they will all be healthy; by default they will receive the healthy X-chromosome from the father.

Daughters: XX or XX^h; all healthy phenotype

Sons, however, have a 50% chance of receiving the mutant allele from the mother, because the father will always contribute the Y-chromosome to sons.

Sons: XY or X^hY; half healthy, half hemophilic

As a result, 50% of the sons will have hemophilia.

Since the disease is found on the X-chromosome, we need to find the scenario in which both sons and daughters have an equal 50% probability of getting the disease. Regardless of gender, mothers will always donate one X-chromosome to the offspring. If the mother is heterozygous for the disease, she has a 50% chance of giving an offspring the diseased allele. As a result, a heterozygous mother will have children that display the disease in the observed ratio.

Parents: XX^R x XY

Offspring: XX, XX^R, XY, X^RY

Note that this ratio of expression is only possible when the allele for the disorder is dominant; otherwise the heterozygous female would be a carrier, and not express the disorder.

First, we have to look at the genotype of the mother and father. The mother is XX, because she is a woman, while the father is XY. The mother is a carrier for the gene, which means she has the traits on one chromosome, but since it is recessive she does not express it physically (she is not colorblind). Therefore, we can denote her genotype to be X Xc (with c being the color blind trait). When we cross the mother and the father (XY), we receive the following possibilities for their children: XX , XcX, XY, XcY. The trait shows up twice, they could have a girl that is a carrier, but no female will express colorblindness from this couple. If they have a boy, there is a 50% chance that he will be color blind (remember, males only have 1 chromosome so they will inherit genes directly from their mother). But, the question is asking what percent of their children will be colorblind, regardless of it being a girl or boy, therefore the answer is 25%.

Since Phil is male, he has one x-chromosome and one y-chromosome. Muscular dystrophy is x-linked recessive, which means his x-chromosome carries the disease allele. He must have inherited this x-chromosome from his mother because his father passed his y-chromosome. Since we know that Phil's parents are not affected, we know that his mother is a carrier of the disease. Phil's sister has two x-chromosomes: one from their mother and one from their father. In order to express a recessive trait, all copies of that gene must be the disease allele. Phil's father can only pass on a healthy x-chromosome to his daughter, and Phil's mother has a 50% chance of passing on her healthy x-chromosome. Thus, there is no chance that Phil's sister will be affected with the disease; there is a 50% chance that she is a carrier, though.

Since Bill is has the disease and is a male, his chromosomes will appear as: XY. Since Emily does not have the disease and there is no family history of the disease, her chromosomes will appear as: XX. The chances of Bill and Emily having a baby girl is 0.5. This is because they can either have a boy (0.5) or a girl (0.5). These events are independent of each other.

0.5 (baby girl) * 1 (X-chromosome from Bill because he will not donate a Y-chromosome to a female) * 1 (X-chromosome from Emily-does not matter which X-chromosome is given since both are normal) * 0 (No matter what, the baby girl cannot have Fabry because it is an X-linked recessive disease. To get the disease, you would need XX. This is not genotypically possible with the genotypes of the parents provided).

The problem can be solved using the following Punnett square. Since short is recessive, the only genotype that will result in a short appearance is tt. tt occurs one time on the Punnett square, out of four possible combinations; therefore, there is a 1-in-4 chance of giving birth to a short child, or 25%.

The genotypes for the two plants in question can be written as PP for the homozygous dominant plant, and Pp for the heterozygous plant. Remember that a white flower can only be created if it receives two recessive alleles, one from each parent. The homozygous dominant plant can only contribute a dominant allele, so every offspring will have a dominant allele. As a result, every flower will be purple when these two plants are crossed.

PP x Pp

Offspring: Half PP and half Pp. All carry a dominant allele, and will display the dominant phenotype.

When dealing with two traits, it helps to approach each trait separately. The question asks for the fraction of plants in the second generation that have purple flowers and wrinkled peas.

First, we will look at the first generation. The parents are both pure breeding, meaning they are homozygous for each trait. One is purple (dominant) and wrinkled (recessive), while the other is white (recessive) and round (dominant). The result will be dihybrid offspring.

Parent cross: PPrr x ppRR

Offspring: All offspring will be PpRr and exhibit both dominant phenotypes (purple and round).

Now we will look at the first generation self-cross for each trait.

First generation: PpRr x PpRr

Offspring for color: 1 PP, 2 Pp, 1 pp; 3 purple and 1 white.

Offspring for seeds: 1 RR, 2 Rr, 1 rr; 3 round and 1 wrinkled.

We can see that three-fourths of the offspring will be purple, since they carry the dominant allele, and one-fourth of the offspring will be wrinkled, since only one of four will carry two recessive alleles.

Since we are looking for plants that have both of these traits, we multiply these two probabilities together.

In other words,  of the second generation will have purple flowers and wrinkled peas.

When dealing with two traits, it helps to approach each trait separately. The question asks for the fraction of plants in the second generation that are heterozygous for both traits.

First, we will look at the first generation. The parents are both pure breeding, meaning they are homozygous for each trait. One is purple (dominant) and wrinkled (recessive), while the other is white (recessive) and round (dominant). The result will be dihybrid offspring.

Parent cross: PPrr x ppRR

Offspring: All offspring will be PpRr and exhibit both dominant phenotypes (purple and round).

Now we will look at the first generation self-cross for each trait.

First generation: PpRr x PpRr

Offspring for color: 1 PP, 2 Pp, 1 pp; 3 purple and 1 white.

Offspring for seeds: 1 RR, 2 Rr, 1 rr; 3 round and 1 wrinkled.

We can see that half of the offspring will be heterozygous for color, and half of the offspring will be heterozygous for seed shape.

Since we are looking for plants that have both of these traits, we multiply these two probabilities together.

Since the parents are pure breeding, we can designate the purple flowers as PP and the white flowers as pp. Upon breeding, there will only be one possible plant created, which has a genotype of Pp.

PP x pp

Offspring: all offspring are Pp and will display the dominant phenotype (purple)

Upon self fertilization, there will be three genotypes in the second generation: PP, Pp, and pp. A punnet square will show that 50% of the second generation will be homozygous and 50% of the second generation will be heterozygous for the color trait.

Pp x Pp

Offspring: one PP, two Pp, one pp. PP and pp are homozygous, and the two Pp are heterozygous.