Which of the following statements is correct?
- Female carrier for haemophilia may transmit the disease to sons.
- Thalassemia is a qualitative problem.
- Change in whole set of chromosomes is called aneuploidy
- Sickle cell anaemia is a quantitative problem.
The Correct Option is A
Approach Solution - 1
Haemophilia is an X-linked recessive disorder. A female carrier (with one normal X chromosome and one X chromosome carrying the defective gene) has a 50% chance of transmitting the defective X chromosome to her sons. Since sons inherit the Y chromosome from their fathers, they can only inherit the X chromosome from their mother. Therefore, a female carrier can pass on the disease to her sons.
The correct answer is (A) : Female carrier for haemophilia may transmit the disease to sons.
Approach Solution -2
The correct answer is: (A) Female carrier for haemophilia may transmit the disease to sons.
Haemophilia is a genetic disorder that impairs the blood's ability to clot properly. It is an X-linked recessive disease, which means the defective gene causing the disorder is located on the X chromosome. Since males have only one X chromosome (XY), a single defective allele on the X chromosome will result in haemophilia. In contrast, females have two X chromosomes (XX), so a single defective allele typically does not cause the disease in females unless both X chromosomes carry the mutation.
A female who carries the haemophilia gene (heterozygous carrier) may pass on the X chromosome with the defective gene to her children. If she passes this X chromosome with the mutation to a son (who inherits only one X chromosome from his mother), the son will have haemophilia. Therefore, a female carrier of haemophilia has a 50% chance of transmitting the disease-causing allele to her sons.
Thus, the correct statement is that a female carrier for haemophilia may transmit the disease to her sons, as stated in option (A).
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