Mon have XY (or YX) chromosomes and women have XX chromosomes. X-linked recessiv

Question

Mon 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 good. 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?

Explanation / Answer

Solution :

(a) If father has one defective x chromosome then total number pairs possible when each parent transfers one of his/ her chromosome –

xX , xX, YX, YX then events in which boy is born are YX and YX

Thus required probability is zero as both are the cases when disease doesn't occur.

(b) Events in wwhich a girl is born are xX and xX , again required probability comes zero.

(c) If mother has one defective x chromosome then total number pairs possible when each parent transfers one of his/ her chromosome –

xX , xY, XX, XY then favourable events are xY .

Thus required probability = 1/2 ( as boy occurs in two ways xY and XY )

(d) If mother has one defective x chromosome then total number pairs possible when each parent transfers one of his/ her chromosome –

xX , xY, XX, XY then favourable events are no one as both the cases in which a girl is born ( xX and XX ) , disease doesn't occur.

Thus required probability = 0

Answer

TY!

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