Final answer:
The possible phenotype ratio of children from a cross between a heterozygous woman for hemophilia and a normal man would result in all daughters being carriers of the disorder and half of the sons being affected by hemophilia.
Step-by-step explanation:
The possible phenotype ratio of children from a woman who is heterozygous for hemophilia and a normal man would be such that all daughters will be normal (heterozygous) because they will inherit a normal X allele from their father and half their sons will be normal and half will display hemophilia. This is because the mother carries one X chromosome with the hemophilia allele and will pass this on to half of her sons, while the father's Y chromosome doesn't carry the normal allele to compensate.
Hemophilia is an X-linked recessive disorder, meaning that the gene causing the trait or the disorder is located on the X chromosome and it is not dominant. Therefore, males are more commonly affected because they only have one X chromosome (XY) and a single recessive gene on that X chromosome will cause the disease. Females, having two X chromosomes (XX), are typically only carriers unless they have the recessive gene on both X chromosomes.
Using this information, we can infer the following: If a daughter receives one X chromosome carrying the gene for hemophilia from her mother and one normal X chromosome from her father, she will be heterozygous, but asymptomatic—a carrier. A son, on the other hand, has no matching X chromosome to mask the effect of the recessive gene if he receives the X chromosome carrying the hemophilia gene from his mother, as he inherits only a Y chromosome from his father.