In Exercises 25–32, find the probability and answer the questions.

X-Linked Genetic Disease Men have XY (or YX) chromosomes and women have XX chromosomes. X-linked recessive genetic diseases (such as juvenile retinoschisis) occur when there is a defective X chromosome that occurs without a paired X chromosome that is not defective. In the following, represent a defective X chromosome with lowercase x, so a child with the xY or Yx pair of chromosomes will have the disease and a child with XX or XY or YX or xX or Xx will not have the disease. Each parent contributes one of the chromosomes to the child.

a. If a father has the defective x chromosome and the mother has good XX chromosomes, what is the probability that a son will inherit the disease?

b. If a father has the defective x chromosome and the mother has good XX chromosomes, what is the probability that a daughter will inherit the disease? c. If a mother has one defective x chromosome and one good X chromosome and the father has good XY chromosomes, what is the probability that a son will inherit the disease?

d. If a mother has one defective x chromosome and one good X chromosome and the father has good XY chromosomes, what is the probability that a daughter will inherit the disease?

• It is given that males have XY or YX pair of chromosomes, and females have XX pair of chromosomes.
• A genetic disease is given to be X-linked if it occurs without a paired non-defective X chromosome.
• A defective chromosome is represented by x, while a healthy chromosome is represented by X.
• Also, a chromosome from each parent is inherited by the child.

Theprobability of an eventis the chance of an event happening. For an arbitrary event A,

$P\left(A\right)=\frac{\text{Number}\text{of}\text{favorable}\text{outcomes}\text{of}\text{A}}{\text{Total}\text{number}\text{of}\text{outcomes}}$

The collection of all possible outcomes of an experiment is called thesample space.

It is known that one X and one Y chromosome result in a male while two X chromosomes result in a female.

The following combinations of chromosomes will result in the disease:

• xY
• Yx

The following combinations of chromosomes will not result in the disease:

• XY
• YX
• xX
• Xx
• XX

One chromosome is carried from each of the two parents to the child.

a.

Let the father’s chromosome combination be xY (Yx).

Let the mother’s chromosome combination be XX.

The possible outcomes for the genotype of the child can be listed as follows:

• If the x chromosome is inherited from the father, and the X chromosome is inherited from the mother, the child will have the xX chromosome and be a healthy daughter.
• If the X chromosome is inherited from the mother, and the x chromosome is inherited from the father, the child will have the Xx chromosome and be a healthy daughter.
• If the Y chromosome is inherited from the father, and the X chromosome is inherited from the mother, the child will have the YX chromosome and be a healthy son.
• If the X chromosome is inherited from the mother, and the Y chromosome is inherited from the father, the child will have the XY chromosome and be a healthy son.

Therefore, the sample space becomes (xX, Xx, YX, XY)

Son would have one X-type chromosome (either defective or not) and one Y chromosome.

The total number of chromosome combinations for the son is two.

The number of combinations that result in a diseased son is 0.

The probability of a diseased son, say event E, is calculated as follows:

$$\begin{gathered} P\left(E\right)=\frac{0}{2} \\ =0 \end{gathered}$$

Therefore, the probability that a son will inherit the disease if the father has a defective chromosome is equal to 0.

b.

Here, the father’s chromosome combination be xY (Yx), and the mother’s chromosome combination be XX.

Therefore, the sample space becomes (xX, Xx, YX, XY).

The daughter would have two X chromosomes(either defective or not).

The total number of chromosome combinations for daughter = 2.

The number of combinations that result in a diseased daughter = 0.

Thus, the probability that a daughter would inherit the disease; event F is computed as follows:

$$\begin{gathered} P\left(F\right)=\frac{0}{2} \\ =0 \end{gathered}$$

Therefore, the probability that a daughter will inherit the disease if the father has a defective chromosome is equal to 0.

c.

Step 1:

Let the father’s chromosome combination be XY (or YX).

Let the mother’s chromosome combination be xX/Xx.

Step 2:

The following possible outcomes for the genotype of the child can be listed:

• At first, the X chromosome is inherited from the father, and then the x chromosome is inherited from the mother; the child will have the Xx chromosome and be a healthy daughter.
• At first, the x chromosome is inherited from the mother, and then the X chromosome is inherited from the father; the child will have the xX chromosome and be a healthy daughter.
• At first, the Y chromosome is inherited from the father, and then the x chromosome is inherited from the mother; the child will have the Yx chromosome and be a diseased son.
• At first, the x chromosome is inherited from the mother, and then the Y chromosome is inherited from the father; the child will have the xY chromosome and be a diseased son.
• At first, the X chromosome is inherited from the father, and then the X chromosome is inherited from the mother; the child will have the “XX” chromosome and be a healthy daughter.
• At first, the X chromosome is inherited from the mother, and then the X chromosome is inherited from the father; the child will have the XX chromosome and be a healthy daughter.
• At first, the Y chromosome is inherited from the father, and then the X chromosome is inherited from the mother; the child will have the YX chromosome and be a healthy son.
• At first, the X chromosome is inherited from the mother, and then the Y chromosome is inherited from the father; the child will have the XY chromosome and be a healthy son.

Therefore,the sample space becomes (xX, Xx, Yx, xY, XX, YX, XY).

The total number of chromosome combinations for son= 4.

The number of combinations that result in a diseased son = 2.

The probability of a diseased son, event G, is calculated as follows:

$$\begin{gathered} P\left(G\right)=\frac{2}{4} \\ =0.5 \end{gathered}$$

Therefore, the probability that a son will inherit the disease if the mother has a defective chromosome is equal to 0.

(d)

For defective chromosomes in the mother,the sample space becomes (xX, Xx, Yx, xY, XX, YX, XY).

The total number of chromosome combinations for the daughter is 3.

The number of combinations that result in a diseased daughter is 0.

Thus, the probability that the daughter will inherit the disease, event H, is computed as follows:

$$\begin{gathered} P\left(H\right)=\frac{0}{3} \\ =0 \end{gathered}$$

Therefore, the probability that a daughter will inherit the disease if the mother has a defective chromosome is equal to 0.