A normal girl, whose mother is hemophilic, marries a male with no ancestral history of hemophilia. 
What will be the possible phenotypes of the off springs?  
A. Haemophilic son and hemophilic daughter  
B. Hemophilic son and carrier daughter  
C. Normal daughter and normal son  
D. Normal son and hemophilic daughter  

Choose the most appropriate answer from the options given below.

(B) and (C) only

Haemophilia is an X-linked recessive disorder. The gene for haemophilia is located on the X chromosome. Females have two X chromosomes (XX), while males have one X and one Y chromosome (XY).

In the given case, the woman has normal vision. Hence, she should possess a normal allele on at least one of her X chromosomes (X). Since her mother is haemophilic ($X^h{X}^h$), her other (maternal) X chromosome must be carrying the mutant allele (${\mathrm{X}}^{\mathrm{h}}$). So, she is a carrier of haemophilia ($X^{h}X$).

Her husband has no ancestral history of haemophilia, so he is most likely a normal male without the allele for haemophilia on his single X chromosome (XY).

Since the woman is a carrier (${\mathrm{X}}^{\mathrm{h}}\mathrm{X}$):
50% of her sons are likely to be haemophilic.
50% of her sons are likely to be normal.
Since the woman is a carrier (${\mathrm{X}}^{\mathrm{h}}\mathrm{X}$) and her husband is normal (XY):
50% of her daughters are likely to be normal (XX).
50% of her daughters are likely to be carriers (${\mathrm{X}}^{\mathrm{h}}\mathrm{X}$).
Based on the possibilities above, the most appropriate answer is:
(b) Haemophilic son and carrier daughter.
(c) Normal daughter and normal son.