A female rat homozygous for a recessive X-linked mutation is mated to a male with wild type phenotype. The phonotypes of the F1 progeny will be

Options

A. all wild type
B. 50% mutants irrespective of sex
C. all females wild type and all males mutant
D. all males wild type and all females mutant

Answer

To solve the problem, we need to analyze the genetic cross between a female rat homozygous for a recessive X-linked mutation and a male rat with a wild type phenotype.

### Step-by-Step Solution:

1. **Identify the Genotypes:**
- The female rat is homozygous for a recessive X-linked mutation. We can denote this as \( X^m X^m \), where \( X^m \) represents the X chromosome with the recessive mutation.
- The male rat has a wild type phenotype, which means he has a normal X chromosome and a Y chromosome. We denote this as \( X^+ Y \), where \( X^+ \) represents the normal X chromosome.

2. **Determine the Gametes:**
- The female can only produce gametes with the \( X^m \) chromosome, so her gametes are \( X^m \).
- The male can produce two types of gametes: \( X^+ \) and \( Y \).

3. **Set Up the Punnett Square:**
- We will create a Punnett square to visualize the possible combinations of gametes from the parents.
- The gametes from the female (both \( X^m \)) will be placed on one side, and the male's gametes (\( X^+ \) and \( Y \)) will be placed on the other side.

| | \( X^+ \) | \( Y \) |
|---------|-----------|---------|
| \( X^m \) | \( X^m X^+ \) | \( X^m Y \) |
| \( X^m \) | \( X^m X^+ \) | \( X^m Y \) |

4. **Analyze the Offspring:**
- The offspring genotypes from the Punnett square are:
- \( X^m X^+ \) (female) - This female is a carrier for the mutation but will express the wild type phenotype because the wild type allele is dominant.
- \( X^m Y \) (male) - This male will express the mutant phenotype because he has the recessive mutation on his only X chromosome.

5. **Determine the Phenotypes:**
- All female offspring (\( X^m X^+ \)) will have a wild type phenotype.
- All male offspring (\( X^m Y \)) will have a mutant phenotype.

### Conclusion:
The phenotypes of the F1 progeny will be:
- All females are wild type.
- All males are mutant.

Thus, the correct answer is **option C: all females wild type and all males mutant.**