Final answer:
Sons of a mother who is a carrier for a recessive X-linked disorder have a 50% chance of being affected, while daughters have a 50% chance of being carriers.
Step-by-step explanation:
The question being asked relates to a son's chances of inheriting a recessive X-linked disorder based on his mother's genotype. In the case of X-linked disorders, if the mother is a carrier (heterozygous) for a recessive trait and the father has normal (dominant) alleles, their sons have a 50% chance of being affected by the disorder. This is because the son will inherit an X chromosome from his mother and if it carries the recessive allele, he will express the disorder, while a daughter inheriting one X chromosome with the recessive allele remains a carrier, just like her mother, due to her second X chromosome providing a dominant allele for the trait.
Additionally, when discussing dihybrid crosses involving two traits, such as seed texture and color in plants, the principle of independent assortment allows us to predict phenotypic ratios. Dihybrid crosses demonstrate a 9:3:3:1 ratio, which can be broken down into two separate 3:1 monohybrid ratios using the product rule. If considering only one trait at a time, 3/4 of the offspring would display the dominant phenotype and 1/4 would display the recessive phenotype.