A normal woman whose father was haemophilic `(X^(h)Y)` marries a normal man (XY). The offspring shall be

Options

A. all normal
B. all sons haemophilic
C. all daughters haemophilic
D. some sons are haemophilic

Answer

To solve the problem, we need to analyze the genetic inheritance of hemophilia, which is an X-linked recessive condition. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Genotypes
- The normal woman has a father who is hemophilic. Since hemophilia is X-linked, the father’s genotype is \(X^hY\) (where \(X^h\) represents the X chromosome carrying the hemophilia allele).
- The woman must have inherited one X chromosome from her father, which means she has the genotype \(X^hX\) (one normal X from her mother and one hemophilic X from her father).
- The normal man has the genotype \(XY\).

### Step 2: Set Up the Punnett Square
- We will set up a Punnett square to determine the possible genotypes of their offspring.
- The woman’s alleles are \(X^h\) and \(X\), and the man’s alleles are \(X\) and \(Y\).

### Step 3: Fill in the Punnett Square
- The Punnett square will look like this:

| | X | Y |
|-------|---------|---------|
| **X** | \(XX\) | \(XY\) |
| **X^h** | \(X^hX\) | \(X^hY\) |

### Step 4: Analyze the Results
- From the Punnett square, we can see the possible genotypes of the offspring:
- \(XX\) (normal female)
- \(XY\) (normal male)
- \(X^hX\) (carrier female, normal phenotype)
- \(X^hY\) (hemophilic male)

### Step 5: Determine the Phenotypes
- The phenotypes of the offspring are:
- 1 normal female (\(XX\))
- 1 carrier female (\(X^hX\))
- 1 normal male (\(XY\))
- 1 hemophilic male (\(X^hY\))

### Step 6: Calculate the Ratios
- The ratio of the offspring is:
- 1 normal female: 1 carrier female: 1 normal male: 1 hemophilic male
- This means that there is a 25% chance for a hemophilic son.

### Conclusion
- The offspring will include some sons who are hemophilic. Therefore, the correct answer is option D: "some sons are hemophilic."

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