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niya11057
15.10.2020 •
Mathematics
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?
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Ответ:
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
Ответ:
1 participant slept 5 hours.