1.  A man with hemophilia (a recessive, sex-linked blood clotting disorder) has a daughter of normal phenotype.  She marries a man who is normal for the trait.
            X^H - normal
            X^h - hemophilia
    Since this is a recessive disorder, one X^H allele will give a normal phenotype.  When the offspring do not have a X^H and only have the X^h allele/alleles they will have the hemophilia condition.
            Man's genotype is X^hY.
            Daughter's genotype is X^HX^h  (She got the X^h from her father.  She has a normal phenotype, so we know her other allele is X^H.
            Daughter's Husband is  X^HY  (Man is normal for the hemophilia trait)

    A) What is the probability that a daughter of this mating with be a hemophiliac?

 

	XH	Xh
XH	XH XH	XHXh
Y	XHY	XhY

    B) What is the probability that a son will be a hemophiliac?
        The probability that a son will be a hemophiliac is 1/2 or 50%.

    C) If the couple has four sons, what is the probability that all four will be born with hemophilia.
        Each occurrence has a 1/2 chance of happening.
        1/2 x 1/2 x 1/2 x 1/2 = 1/16

2.  Pseudohypertrophic muscular  dystrophy is a disorder that causes gradual deterioration of the muscles.  It is seen only in boys born to apparently normal parents and usually results in death in the early teens.

    A) Is this disorder caused by a dominant or a recessive allele?
        Since the parents have a normal phenotype, this disorder must be caused by a recessive allele.

    B) Is its inheritance sex-linked or autosomal?  How do you know?
        Since this disorder is seen only in boys, It must be sex-linked.

    C) Explain why this disorder is seen only in boys and never in girls.
       A female won't be able to get a second  recessive allele from an affected male because he won't survive to breeding age.

3.  Red-green color blindness is caused by a sex-linked recessive allele.  A color-blind man marries a woman with normal vision whose father was color-blind. note: I changed the wording of this question.
        Man X^bY
        Woman X^BX^b.  We know this because she has normal vision (X^B).  Her father was color-blind and gave her her X^b allele.

    A) What is the probability that their daughter will be color-blind? note: Because they are telling you in the problem that there is a daughter, you do not have to figure in the probability that the child will be a girl.
 

 

	XB	Xb
Xb	XBXb	XbXb
Y	XBY	XbY

    B) What is the probability that their son will be color-blind? note: Because they are telling you in the problem that there is a son, you do not have to figure in the probability that the child will be a boy.
       The chance that a son will be color-blind is also 1/2.

6.  An aneuploid person is obvious female, but her cells have two Barr bodies.  What is the probable complement of sex chromosomes in this individual?
        The person is aneuploid, therefore she has an abnormal number of chromosomes.   Each normal, female, nucleated, somatic cell contains one Barr body (an inactivated X chromosome).  Each cell needs an active X chromosome, so if two are inactive, she wouls still have one active X.  Therefore she must be XXX.