A women with normal vision marries a man with normal vision and gives birth to a colourblind son. Her husband dies and she marries a colourblind man. Wht is the probability of heer children having the abnormality :-

Options

A. `50%` colourblins sons + `50%` colourblind daughters
B. All sons colourblind & daughter carrier
C. All daughter colourblind & sons normal
D. `50%` sons colourblind and all daughters normal

Answer

To solve the problem of determining the probability of the children of a woman with normal vision (who is a carrier for color blindness) and a colorblind man having the abnormality, we can follow these steps:

### Step 1: Understand the Genetics of Color Blindness
Color blindness is an X-linked recessive trait. This means that the gene responsible for color blindness is located on the X chromosome. Males have one X and one Y chromosome (XY), while females have two X chromosomes (XX).

### Step 2: Determine the Genotype of the Woman
The woman has normal vision but has given birth to a colorblind son. Since color blindness is recessive, she must be a carrier for the colorblind allele. Therefore, her genotype is X^C X (where X^C represents the normal vision allele and X represents the colorblind allele).

### Step 3: Determine the Genotype of the First Husband
The first husband has normal vision, which means his genotype is X^C Y (normal vision allele on the X chromosome and Y chromosome).

### Step 4: Determine the Genotype of the Colorblind Man
The second husband is colorblind, so his genotype is X Y (he has the colorblind allele on his only X chromosome).

### Step 5: Set Up the Punnett Square for the Second Marriage
Now, we need to set up a Punnett square to determine the possible genotypes of the children from the second marriage between the carrier woman (X^C X) and the colorblind man (X Y).

| | X^C (from mother) | X (from mother) |
|---------|--------------------|------------------|
| **X** | X^C X (normal girl) | X X (colorblind girl) |
| **Y** | X^C Y (normal boy) | X Y (colorblind boy) |

### Step 6: Analyze the Punnett Square Results
From the Punnett square, we can see the following possible outcomes for their children:
- 1 normal girl (X^C X)
- 1 colorblind girl (X X)
- 1 normal boy (X^C Y)
- 1 colorblind boy (X Y)

### Step 7: Calculate the Probability
From the results:
- 50% of the daughters will be colorblind (1 out of 2).
- 50% of the sons will be colorblind (1 out of 2).

Thus, the probability of any child (son or daughter) being colorblind is 50%.

### Final Answer
The probability of her children having the abnormality (color blindness) is **50%**.

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