If a haemophilic man marries a carrier woman than which of the following holds true for their progenies ?

Options

A. 50 % daughters are carrier and 50 % are haemophilic
B. All the daughters are haemophilic
C. All sons are haemophilic and all daughters are normal
D. All sons normal, all daughters carriers

Explanation

To solve the problem of determining the genetic outcomes for the progeny of a hemophilic man and a carrier woman, we can follow these steps:

### Step 1: Understand the Genetic Background
- Hemophilia is an X-linked recessive disorder. This means that the gene responsible for hemophilia is located on the X chromosome.
- Males have one X and one Y chromosome (XY), while females have two X chromosomes (XX).

### Step 2: Identify the Genotypes
- The hemophilic man has the genotype X^hY, where X^h represents the X chromosome carrying the hemophilia gene.
- The carrier woman has the genotype X^hX, where one X chromosome carries the hemophilia gene (X^h) and the other is normal (X).

### Step 3: Determine Gametes
- The man can produce two types of gametes: X^h (from his X chromosome) and Y (from his Y chromosome).
- The woman can produce two types of gametes: X^h (from her hemophilic X) and X (from her normal X).

### Step 4: Create a Punnett Square
- Set up a Punnett square to visualize the possible combinations of gametes from the parents.

| | X^h (from mother) | X (from mother) |
|---------|--------------------|------------------|
| X^h (from father) | X^hX^h (daughter, hemophilic) | X^hX (daughter, carrier) |
| Y (from father) | X^hY (son, hemophilic) | XY (son, normal) |

### Step 5: Analyze the Results
- From the Punnett square, we can see the possible genotypes of the offspring:
- 1 daughter with genotype X^hX^h (hemophilic)
- 1 daughter with genotype X^hX (carrier)
- 1 son with genotype X^hY (hemophilic)
- 1 son with genotype XY (normal)

### Step 6: Calculate the Ratios
- The results show:
- 50% of the daughters are carriers (X^hX)
- 50% of the daughters are hemophilic (X^hX^h)
- 50% of the sons are hemophilic (X^hY)
- 50% of the sons are normal (XY)

### Conclusion
The correct statement regarding their progeny is that 50% of the daughters are carriers and 50% are hemophilic.