A woman (whose father is colour blind but mother is normal) marries a haemophiliac man with hypertichosis. What percentage of progeny will show genotypically any two of the traits out of the three mentioned above at a given time?

Options

A. 0
B. 0.25
C. 50
D. 0.75

Answer

To solve the problem, we need to analyze the genetic traits involved and how they are inherited. The traits in question are color blindness, hemophilia, and hypertrichosis. We'll break down the inheritance patterns for each trait and then calculate the percentage of progeny that will show any two of these traits.

### Step 1: Determine the Genotype of the Woman
The woman has a colorblind father (X^cY) and a normal mother (XX). Since color blindness is an X-linked recessive trait, the woman must be a carrier for color blindness. Therefore, her genotype is X^cX (where X^c represents the X chromosome with the colorblind allele).

### Step 2: Determine the Genotype of the Man
The man is hemophiliac (X^hY) and has hypertrichosis (which is an autosomal trait). His genotype is X^hY, where X^h represents the X chromosome with the hemophilia allele.

### Step 3: Set Up the Punnett Square
To find the possible genotypes of their offspring, we can set up a Punnett square for the X-linked traits (color blindness and hemophilia). The woman can contribute either X^c or X, and the man can contribute either X^h or Y.

| | X^h (from father) | Y (from father) |
|----------|--------------------|------------------|
| X^c (from mother) | X^cX^h (carrier for color blindness and hemophilia) | X^cY (colorblind and hemophiliac) |
| X (from mother) | XX^h (normal but carrier for hemophilia) | XY (normal) |

### Step 4: Analyze the Offspring Genotypes
From the Punnett square, we have the following offspring genotypes:
1. X^cX^h (carrier for color blindness and hemophilia)
2. X^cY (colorblind and hemophiliac)
3. XX^h (normal but carrier for hemophilia)
4. XY (normal)

### Step 5: Identify Progeny with Any Two Traits
Now, we need to identify which of these genotypes express any two of the traits:
- X^cX^h: Carrier for color blindness and hemophilia (shows two traits)
- X^cY: Colorblind and hemophiliac (shows two traits)
- XX^h: Normal but carrier for hemophilia (shows one trait)
- XY: Normal (shows no traits)

The genotypes that show any two traits are:
- X^cX^h
- X^cY

### Step 6: Calculate the Percentage
Out of the four possible genotypes, two show any two traits. Therefore, the percentage of progeny that will show genotypically any two of the traits is:
\[ \frac{2}{4} \times 100 = 50\% \]

### Final Answer
The percentage of progeny that will show genotypically any two of the traits at a given time is **50%**.

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