Question

A lady carrier for haemophilia (Hh) marries a normal man (HO). Daughters of such a lady would be
A. 50% normal (HH) and 50% carrier (Hh)
B. 50% normal (HH) and 50% haemophilic (hh)
C. 50% carrier (Hh) and 50% haemophilic (hh)
D. 25% carrier (Hh) and 75% haemophilic (hh)
E. 75% carrier (Hh) nad 25% haemophilic (hh)

Answer 1

Hint: Haemophilia is an inherited bleeding disorder in which the blood doesn’t clot properly. People with haemophilia have lower levels of either factor VIII (8) or factor IX (9).

Complete Answer:
Before we find the answer, let us know a little more about haemophilia. It is an X- linked recessive disorder. It is caused by a mutation in the genes that provide instructions for making the clotting factor proteins that are needed for clotting of blood. The genes are situated in the X chromosome. Males have one X chromosome and one Y chromosome (XY) while females have two X chromosomes (XX).

The two most common type of haemophilia are:
1. Haemophilia A (Classic haemophilia)- Caused by lack or decrease of clotting factor VIII
2. Haemophilia B (Christmas disease)- Caused by lack or decrease of clotting factor IX

Let us analyze the correct answer,
Option A: A carrier female has one defective gene on one of the X chromosomes i.e, ($HH^1$) and a normal male has no defective gene on X chromosome (HH). The 4 types of possible progenies are normal male (HO), carrier female ($HH^1$), haemophilic male ($H^{1}O$) and normal female (HH). Daughters of such a lady can be 50% normal (HH) and 50% carrier (Hh). Therefore, this is the correct option.

	H	$H^{1}$
H	HH	$HH^1$
O	HO	$H^{1}O$

Option B: When a lady carrier (Hh) marries a normal man (HO), there is no chance for daughter with haemophilia (hh) because for the disease to be expressed both the genes should be recessive ($H^{1}H^{1}$). Therefore, this is an incorrect option.

Option C: When a lady carrier (Hh) marries a normal man (HO), there is no chance for daughter with haemophilia (hh) because for the disease to be expressed both the genes should be recessive ($H^{1}H^{1}$). Therefore, this is an incorrect option.

Option D: When a lady carrier (Hh) marries a normal man (HO), there is no 25% chance for a daughter to become a carrier and no 75% chance for a daughter with haemophilia (hh) because for the disease to be expressed both the genes should be recessive ($H^{1}H^{1}$). Therefore, this is an incorrect option.

Option E: When a lady carrier (Hh) marries a normal man (HO), there is no 75% chance for a daughter to be a carrier (Hh) and no 25% chance to become haemophilic because for the disease to be expressed both the genes should be recessive ($H^{1}H^{1}$). Therefore, this is an incorrect option.

Hence the correct answer is option A.

Additional Information:
- The X chromosome contains many genes that are not present in the Y chromosome. Therefore, male have one copy of most of the genes on the X chromosome whereas females have two copies.
- So, males can become haemophilic if they inherit an affected X chromosome with mutation in clotting factor VIII or IX. Females can also become haemophilic but it is rarer. A female is a carrier if she has one affected X chromosome and shows symptoms of haemophilia. She can also pass this affected X chromosome to her children.

Note: Some families have haemophilia even though there is no prior history of haemophilia in the family. Sometimes female carriers can have normal sons, just by chance.