A husband and wife have normal vision, although both of their fathers are red - green colour blind , which is inherited as an x - linked recessive trait .
What is the probability that their first child will be ?
i. A normal son
ii. A carrier daughter
iii. A colour - blind son
iv. A colour - blind daughter

Options

A. `i=1/4,ii=1/4,iii=1/4,iv=1/4`
B. `i=1/4,ii=1/4,iii=1/4,iv=0/4`
C. `i=0/4,ii=1/2,iii=1/4,iv=1/4`
D. `i=1/4,ii=1/2,iii=1/4,iv=0`

Answer

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

### Step 1: Determine the Genotypes of the Parents
- The husband has normal vision. Since his father is colorblind (X-linked recessive), he must have inherited a normal X chromosome from his mother. Therefore, his genotype is **XY** (normal vision).
- The wife also has normal vision, but her father is colorblind. This means she must have inherited one X chromosome that carries the colorblind allele (Xᶜ) from her father. Therefore, her genotype is **XXᶜ** (normal vision but a carrier).

### Step 2: Identify Possible Gametes
- The husband can produce two types of gametes: **X** (normal) and **Y** (male).
- The wife can produce two types of gametes: **X** (normal) and **Xᶜ** (colorblind allele).

### Step 3: Create a Punnett Square
Now we can set up a Punnett square to visualize the possible combinations of gametes:

| | X (from husband) | Y (from husband) |
|-------|------------------|------------------|
| X (from wife) | XX (normal daughter) | XY (normal son) |
| Xᶜ (from wife) | XXᶜ (carrier daughter) | XᶜY (colorblind son) |

### Step 4: Analyze the Outcomes
From the Punnett square, we can see the following possible genotypes for the children:
1. **XX** - Normal daughter (1 out of 4)
2. **XY** - Normal son (1 out of 4)
3. **XXᶜ** - Carrier daughter (1 out of 4)
4. **XᶜY** - Colorblind son (1 out of 4)

### Step 5: Calculate the Probabilities
Now we can determine the probabilities for each of the requested outcomes:
- **i. Probability of a normal son (XY)**: 1 out of 4 (1/4)
- **ii. Probability of a carrier daughter (XXᶜ)**: 1 out of 4 (1/4)
- **iii. Probability of a colorblind son (XᶜY)**: 1 out of 4 (1/4)
- **iv. Probability of a colorblind daughter**: 0 out of 4 (0/4) - daughters cannot be colorblind because they inherit one normal X from their mother.

### Final Summary of Probabilities
- Normal son: **1/4**
- Carrier daughter: **1/4**
- Colorblind son: **1/4**
- Colorblind daughter: **0/4**