In humans,the genes for red-green colour blindness (R = normal,r = colour-blind)and haemophilia A (H = normal,h = haemophilia)are both X-linked and only 3 map units apart.Suppose a woman has four sons,and two are colour-blind but have normal blood clotting and two have haemophilia but normal colour vision.What is the probable genotype of the woman?
A) HR/hr
B) Hr/hr
C) hr/hR
D) Hr/hR
E) HR/Hr

Question

Quizplus · Accepted Answer

Since the woman has four sons, each son has a 50% chance of inheriting either X chromosome from the mother. Therefore, the two colour-blind sons inherited the mother's X chromosome with the R allele, and the two sons with haemophilia inherited the mother's X chromosome with the h allele. 
Since the two genes are only 3 map units apart, they are likely to be inherited together most of the time, unless a crossover event occurred between them during meiosis. 
To produce the observed pattern of inheritance, the woman must be a carrier of both R and H alleles on one of her X chromosomes and h and normal allele on the other X chromosome. 
Therefore, the probable genotype of the woman is Hr/hR.

In Humans,the Genes for Red-Green Colour Blindness (R = Normal,r