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.

When crossing organisms that are heterozygous for two traits, the result is a dihybrid cross.

AaBb x AaBb

Possible offspring: 1 AABB, 1 aaBB, 1 AAbb, 1 aabb, 2 Aabb, 2 aaBb, 2 AaBB, 2 AABb, 4 AaBb

The phenotypic ratio of a heterozygous dihybrid cross for autosomal (none-sex-linked) traits is always 9:3:3:1. Nine offspring will show both dominant traits (AABB, AaBB, AABb, AaBb). Three offspring will show dominance for one trait and recessiveness for the other (Aabb, AAbb) and three offspring will show the reciprocal (aaBb, aaBB). Only one offspring will be recessive for both traits (aabb). For this cross, nine mice will have black fur and brown eyes, three will have black fur and blue eyes, three will have brown fur and brown eyes, and one will have brown fur and blue eyes.

When crossing organisms that are heterozygous for two traits, the result is a dihybrid cross.

AaBb x AaBb

Possible offspring: 1 AABB, 1 aaBB, 1 AAbb, 1 aabb, 2 Aabb, 2 aaBb, 2 AaBB, 2 AABb, 4 AaBb

The phenotypic ratio of a heterozygous dihybrid cross for autosomal (none-sex-linked) traits is always 9:3:3:1. The genotype in this question, Aabb, corresponds to black fur and blue eyes. Two mice out of every sixteen will have this genotype.

When crossing organisms that are heterozygous for two traits, the result is a dihybrid cross.

AaBb x AaBb

Possible offspring: 1 AABB, 1 aaBB, 1 AAbb, 1 aabb, 2 Aabb, 2 aaBb, 2 AaBB, 2 AABb, 4 AaBb

The phenotypic ratio of a heterozygous dihybrid cross for autosomal (none-sex-linked) traits is always 9:3:3:1. The genotype in this question, AaBb, corresponds to black fur and brown eyes. Four mice out of every sixteen will have this genotype.