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.
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?
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?
(type an integer or a decimal. do not round.)

Explanation:

Step1: Analyze son's inheritance from parents in part a

A son gets Y - chromosome from father and X - chromosome from mother. Father has x and mother has XX. Son will get Y from father and X from mother, so chromosome pair is XY.

Step2: Determine probability for son in part a

Since son's chromosome pair is XY (non - defective), probability of son having disease is 0.

Step3: Analyze daughter's inheritance from parents in part b

A daughter gets one X from father and one X from mother. Father has x and mother has XX. Daughter's possible chromosome pairs are Xx or XX, both non - defective.

Step4: Determine probability for daughter in part b

Since daughter's chromosome pairs are non - defective, probability of daughter having disease is 0.

Step5: Analyze son's inheritance from parents in part c

A son gets Y from father (father has good XY) and X from mother. Mother has Xx. Probability of getting x from mother is $\frac{1}{2}$.

Answer:

a. 0
b. 0
c. 0.5