If a carrier haemophilic female (XHXh) is married to a haemophilic male (XhY). What will be the ratio of the presence of haemophilia in the children? Select the best answer from the given options.

Explanation

As per the cross given above, there are four possible genotypes and hence, four possible phenotypes. These are; carrier daughter, hemophiliac daughter, normal son, and hemophiliac son, and all four have an equal probability or likelihood of occurring, which is 25 percent. Option C describes this scenario correctly so, it is the correct option. Option A describes a scenario in which the mother is a hemophiliac (i.e. has a genotype of XhXh) and the father is also a hemophiliac (genotype XhY), which results in all the children being hemophiliacs. Option B describes an impossible scenario, as in order for all the males to be hemophiliac, the mother must have a genotype of XhXh, but in this scenario, the daughters will either be all hemophiliacs or all carriers (father's genotype being either XhY or XHY, respectively). Hence, for all males to be hemophiliacs, it is impossible for the females to have a 50% chance of being hemophiliacs. Option D also describes the given scenario, because if you look at the figure above, males will either be normal or hemophiliac, and females will either be carriers or hemophiliac, as such, each gender has a 50% chance of being hemophiliacs. However, option C better describes this scenario as it specifically lists the probabilities of all four possible phenotypes. (NOTE: The question asks the candidate to select the BEST possible answer from the given options).