Question

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. 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. complete parts a through d below. that a son will inherit the disease? 0 (type an integer or a decimal. do not round.) 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? 0 (type an integer or a decimal. do not round.) 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? 0.5 (type an integer or a decimal. do not round.) 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? (type an integer or a decimal. do not round.)

Explanation:

Step1: Analyze inheritance for sons in part a

A son gets Y from father and X from mother. If no information about mother's X - chromosome defect, assume normal. So probability of getting defective x from mother (for son to have disease) is 0 if mother is non - carrier.
Probability = 0

Step2: Analyze inheritance for daughters in part b

A daughter gets one X from father and one from mother. Father has defective x and mother has XX. A daughter will get a non - defective X from mother, so she won't have the disease.
Probability = 0

Step3: Analyze inheritance for sons in part c

Mother has Xx (one defective x and one normal X) and father has XY. Sons get Y from father and either X or x from mother. Probability of getting x from mother is $\frac{1}{2}=0.5$.

Step4: Analyze inheritance for daughters in part d

Mother has Xx and father has XY. Daughters get one X from father (which is non - defective X) and one from mother. Since they always get a non - defective X from father, probability of having the disease (xx) is 0.
Probability = 0

Answer:

a. 0
b. 0
c. 0.5
d. 0