Question

1. A man with hemophilia (a recessive, sex-linked condition) and a woman that carries the hemophilia allele have a daughter with normal phenotype.
   1. Define allele symbols for the hemophilia and normal alleles.
   2. Draw a pedigree and indicate the genotype of each individual in the pedigree.
   3. The daughter marries a man who is normal for the trait. What is the probability that a daughter of this pair will be a hemophiliac? What is the probability that a son will be a hemophiliac?
   4. Challenge! If this couple (in part “c”) has four sons, what is the probability that all four will have hemophilia?
   5. Challenge! What is the probability that if the couple (in part “c”) has four children, they will all be sons with hemophilia?

Answer 1

a).

Normal allele = H; Hemophilic allele = h

b).

c).

Daughter (XHXh) x (XHY) normal man ---Parents

	XH	Y
XH	XHXH (normal daughter)	XHY (normal son)
Xh	XHXh (normal but carrier daughter)	XhY (hemophilic son)

Normal = ¾

Hemophilia = ¼

The probability that a daughter of this pair will be a hemophiliac = 0 (zero)

The probability that a son will be a hemophiliac= ¼ or 25%

d).

If this couple (in part “c”) has four sons, the probability that all four will have hemophilia = ¼ * ¼ * ¼ * ¼ = 1/256

e).

Probability of a sone = ½ (as only two geneders and each gender has an equal proportion, ½)

The probability that if the couple (in part “c”) has four children, they will all be sons with hemophilia = (¼* ½) * (¼* ½) *(¼* ½) *(¼* ½) = 1/4096