Carrier female marries a normal visioned male . How manay daughters would be colorblind carriers?

Options

A. Zero
B. 0.25
C. 0.5
D. 1

Answer

To solve the problem of how many daughters would be colorblind carriers when a carrier female marries a normal visioned male, we can follow these steps:

### Step 1: Understand the Genetics of Colorblindness
Colorblindness is a sex-linked recessive trait carried on the X chromosome. A female has two X chromosomes (XX), while a male has one X and one Y chromosome (XY).

- A normal vision female has two normal X chromosomes (XX).
- A carrier female has one normal X chromosome and one X chromosome with the colorblind gene (XcX).
- A normal vision male has one normal X chromosome and one Y chromosome (XY).

### Step 2: Set Up the Parental Genotypes
In this case:
- The carrier female's genotype is XcX (where Xc represents the X chromosome with the colorblind gene).
- The normal vision male's genotype is XY.

### Step 3: Create a Punnett Square
To determine the possible genotypes of their offspring, we can set up a Punnett square:

| | X (normal) | Y |
|--------|------------|----------|
| Xc | XcX (carrier daughter) | XcY (colorblind son) |
| X | XX (normal daughter) | XY (normal son) |

### Step 4: Analyze the Offspring
From the Punnett square, we can see the potential offspring:
- Daughters:
- XcX (carrier female)
- XX (normal female)

- Sons:
- XcY (colorblind male)
- XY (normal male)

### Step 5: Calculate the Proportions
From the results:
- 50% of the daughters (1 out of 2) will be carriers (XcX).
- 50% of the daughters (1 out of 2) will have normal vision (XX).

### Conclusion
Therefore, if the couple has two daughters, one of them would be a colorblind carrier.

**Final Answer:** 50% of the daughters would be colorblind carriers.

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