Question

A woman heterozygous for haemophilia marries a haemophilic man. Choose the correct options with the ratios of carrier daughter, haemophilic daughter, normal son and haemophilic son.

• A. 1: 2: 2:1
• B. 1:1:1:1
• C. 2:1:1:2
• D. 2:1:1:1

Answer 1

The correct option is B 1:1:1:1

Haemophilia is an X-linked recessive disorder which is expressed in homozygous condition in females and in hemizygous condition in males. In females, if the allele is present in heterozygous condition, the female is said to possess the trait but shows no typical phenotypic expression. She leads a normal life just like an unaffected individual because the normal allele present on the other chromosome makes up for the defective allele. A woman heterozygous for haemophilia would have a genotype $X^h$X, and a haemophilic man would have a genotype X${}^h$Y. Given below are all the possible outcomes of this cross.

From the attempted cross we can see that the cross yields a carrier daughter, haemophilic daughter, normal son, and haemophilic son in the ratio 1:1:1:1.
Hence option b is correct.