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List of top five numerical problems on sex-linked inheritance.
Q.1. Of what type will be the children with reference to colour blindness, when a man is colour-blind and his wife is normal?
Solution:
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The cause of the colour blindness is the presence of recessive (c) gene on the X chromosome.
Because man is colour-blind (XcY) and his wife is normal (XX), following will be the results while crossing:
Following will be the results after fertilization:
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(i) XXc, i.e., normal but carrier daughter.
(ii) XY, i.e., normal son.
Results:
No child will be colour-blind.
Q.2. When a baemophilic male is mated with a heterozygous baemophilic female, what baemophilic proportion will be resulted in each sex.
Solution:
(Haemophilia is a disease that causes delayed clotting of blood. It is due to a recessive gene ‘b’ located on X chromosome).
Haemophilic gene is represented by ‘h’
Haemophilic male = XhY
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Heterozygous haemophilic female = XhX
Gametes:
XhY → Xh and Y
XhX → Xh and X
Q.3. When a haemophilic male is mated with a homozygous non-baemophilic female—What will be the result?
Solution:
Haemophilic male = XhY
Homozygous non-haemophilic female = XX
Q.4. Of what type will be the children with reference to colour blindness, when a woman is colour-blind and her husband is normal?
Result:
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In such a case one normal and one colour-blind son, and one normal and one carrier daughter would be resulted.
Q.5. When both the parents are colour-blind, can they produce a normal daughter?
Results:
The above results indicate that the colour-blind gene (c) is passed to both the X chromosomes of the daughter and so no normal daughter can be produced.
