Men have XYâ (or YX) chromosomes and women have XX chromosomes.â X-linked recessive genetic diseasesâ (such as juvenileâ 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 lowercaseâ 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. Required:
a. If a father has the defective x chromosome and the mother hasgood XX chromosomes, what is the probability that a son willinherit the disease?
b. If a father has the defective x chromosome and the mother asgood XX chromosomes, what is the probability that a daughter willinherit the disease?
c. If a mother has one defective x chromosome and one good Xchromosome, and the father has good XY chromosomes, what is theprobability that a daughter will inherit the disease?
d. If a mother has one defective x chromosome and one good Xchromosome, and the father has good XY chromosomes, what is theprobability that a daughter will inherit the disease?

A) 0

B) 0

C) 0

D) 0.5

Step-by-step explanation:

We are told that Men have XY (or YX) chromosomes and women have XX chromosomes

A) If a father has the defective x chromosome and the mother has good XX chromosomes, the four possible outcomes will be;

{xX, YX, YX, xX}

The outcome for the son is;

{YX, YX}

The son will have the disease only if it's xY. Thus, probability of son having the disease is; 0/2 = 0

B) If a father has the defective x chromosome and the mother has good XX chromosomes, the four possible outcomes will be;

{xX, YX, YX, xX}

The outcome for the daughter is;

{xX, xX}

The daughter will have the disease only if it's xx. Thus, probability of daughter having the disease is; 0/2 = 0

C) If a mother has one defective x chromosome and one good Xchromosome, and the father has good XY chromosomes, the possible outcomes are;

{xX, xY, XX, XY}

Outcome for daughter is {xX, XX}

The daughter will have the disease it's xx. Thus, probability of daughter having the disease is; 0/2 = 0

D) If a mother has one defective x chromosome and one good Xchromosome, and the father has good XY chromosomes, the possible outcomes are;

{xX, xY, XX, XY}

The male outcome is;

{XY, xY}

The son will have the disease only if it's xY. Thus, probability of son having the disease is; 1/2 = 0.5

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step-by-step explanation:

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step-by-step explanation: