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 red-green color blindness, which is an X-linked recessive trait. We will determine the probabilities of different outcomes for the first child of a couple where both parents have normal vision, but both of their fathers are color blind.

### Step 1: Determine the Genotypes of the Parents
- Since both fathers are color blind (X-linked recessive), we can infer that:
- The husband's genotype (normal vision) must be X^N Y (where X^N represents the normal vision allele).
- The wife's genotype (normal vision but with a color blind father) must be X^N X^c (where X^c represents the color blind allele, making her a carrier).

### Step 2: Gamete Formation
- The husband can produce two types of gametes:
- X^N (normal vision)
- Y (male)

- The wife can produce two types of gametes:
- X^N (normal vision)
- X^c (carrier)

### Step 3: Punnett Square Analysis
We can create a Punnett square to visualize the possible combinations of gametes from both parents:

| | X^N (from wife) | X^c (from wife) |
|----------|------------------|------------------|
| **X^N (from husband)** | X^N X^N (normal daughter) | X^N X^c (carrier daughter) |
| **Y (from husband)** | X^N Y (normal son) | X^c Y (color blind son) |

### Step 4: Determine the Probabilities
From the Punnett square, we can summarize the possible genotypes and their corresponding probabilities:
1. **Normal Daughter (X^N X^N)**: 1 out of 4 (25%)
2. **Carrier Daughter (X^N X^c)**: 1 out of 4 (25%)
3. **Normal Son (X^N Y)**: 1 out of 4 (25%)
4. **Color Blind Son (X^c Y)**: 1 out of 4 (25%)

### Step 5: Determine the Probability for Each Outcome
- **i. Probability of a normal son**: 1 out of 4 (25%)
- **ii. Probability of a carrier daughter**: 1 out of 4 (25%)
- **iii. Probability of a color-blind son**: 1 out of 4 (25%)
- **iv. Probability of a color-blind daughter**: 0 out of 4 (0%)

### Final Summary of Probabilities
- Normal Son: 1/4
- Carrier Daughter: 1/4
- Color Blind Son: 1/4
- Color Blind Daughter: 0/4