Final answer:
The expected phenotypic ratio for the offspring is 1 normal daughter to 1 son with hypophosphatemia when an unaffected woman and an affected man mate, with the disorder being X-linked dominant.
Step-by-step explanation:
The expected phenotypic ratio of their offspring when an unaffected woman mates with a male with hypophosphatemia, which is an X-linked dominant disorder, would be 1 normal daughter: 1 son with hypophosphatemia. This can be derived from understanding that the father, who is affected, will pass the X chromosome with the disease gene to all of his daughters, making them affected since it is a dominant trait. However, he will pass his Y chromosome to his sons, resulting in unaffected boys because they do not receive the X chromosome carrying the disease allele.
In this case, the unaffected woman is homozygous for the normal gene and the male has hypophosphatemia (vitamin D-resistant rickets) which is inherited as an X-linked dominant disorder. When an X-linked dominant disorder is passed from a male to his offspring, all of his daughters will be affected and none of his sons will be affected. Therefore, the expected phenotypic ratio of their offspring would be 1 normal daughter: 1 daughter with hypophosphatemia.