Hemophilia is an X-linked recessive disease. A phenotypically normal woman whose father was hemophiliac is married to a man who has normal blood-clotting. They have a son and a daughter. The daughter is hemophiliac and has Turner syndrome. The nondisjunction occurred in.

a. Meiosis I in the father

b. Meiosis II in the father

c. Meiosis I in the mother

d. Meiosis II in the mother

e. Either A or B

The mother is phenotypically normal but her father was hemophiliac, so she inherited a Xʰ chromosome from her father and the normal Xᴴ from her mother. Her genotype is heterozygous XᴴXʰ.

Her husband has normal blood-clotting, so his genotype is XᴴY.

If meiosis in both parents were normal, the possible daughers they could have would have the genotypes: XᴴXᴴ (normal girl) or XᴴXʰ (carrier girl).

They had a daughter with hemophilia and Turner Syndrome (only one X chromosome). Because the only allele that her father can pass on to them is the normal Xᴴ, making them phenotypically normal, we know that the fathers' meiosis is the one that failed, and the daughter received only the Xʰ from her mother.

During meiosis I homologous chromosomes separate (in the case of sex chromosomes in the father, it's the X and Y). If they failed to separate, both would migrate to the same daughter cell, and as a result after meiosis II four daughter cells would be produced: two with both an X and a Y chromosome and two with neither. The latter would be the gamete that combined with the mother with the Xʰ allele and caused hemophilia and Turner syndrome.

During meiosis II, sister chromatids separate. The result of nondisjunction of the X chromosome in meiosis II would cause two normal gametes with a single Y chromosome, a gamete with two X chromosomes and a gamete with no X chromosomes. The latter would cause the daughter's diseases.

The answer is e. either A or B.