A man sufferin g rom recessive Xlinked disease marries a normal woman .In the progeny

Options

A. All sons are normal
B. All daughter are diseased
C. All sons are diseased
D. None opf the above

Answer

To solve the problem of a man suffering from a recessive X-linked disease marrying a normal woman, we need to analyze the inheritance pattern of X-linked traits. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Genetic Makeup
- The man has the X-linked recessive disease, which means his genotype is X^dY (where X^d represents the X chromosome carrying the disease allele).
- The normal woman has two normal X chromosomes, so her genotype is XX (both X chromosomes are normal).

### Step 2: Determine Possible Gametes
- The man can produce two types of gametes: X^d (carrying the disease allele) and Y (normal male chromosome).
- The woman can produce two types of gametes: X (normal).

### Step 3: Create a Punnett Square
- Set up a Punnett square to visualize the combinations of gametes from the man and the woman.

| | X (from mother) | X (from mother) |
|--------|------------------|------------------|
| X^d (from father) | X^dX (carrier daughter) | X^dX (carrier daughter) |
| Y (from father) | XY (normal son) | XY (normal son) |

### Step 4: Analyze the Offspring
- From the Punnett square, we can see the possible offspring:
- Daughters: X^dX (both daughters will be carriers of the disease but will not express it).
- Sons: XY (both sons will be normal).

### Step 5: Conclusion
- Therefore, in the progeny:
- All daughters will be carriers of the disease (X^dX).
- All sons will be normal (XY).

### Final Answer
- All daughters are carriers, and all sons are normal.

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