Both father and mother have allele for PKU but are phenotypically normal. What is the probability that their child will have the disease?

Options

A. 0.25
B. 0.5
C. 0.75
D. 1

Answer

To solve the question regarding the probability of their child having Phenylketonuria (PKU) when both parents are phenotypically normal but carry the allele for PKU, we can follow these steps:

### Step 1: Understand the Genetics of PKU
PKU is an autosomal recessive disorder. This means that a child must inherit two copies of the mutated allele (one from each parent) to express the disease. If an individual has one normal allele and one mutated allele, they are a carrier but do not show symptoms.

### Step 2: Define the Genotypes of the Parents
Since both parents are phenotypically normal but carry the allele for PKU, we can denote their genotypes as follows:
- Mother: Cc (where C = normal allele, c = PKU allele)
- Father: Cc (same as the mother)

### Step 3: Set Up a Punnett Square
To determine the probability of their child inheriting PKU, we can set up a Punnett square. The possible gametes from each parent are:
- Mother: C, c
- Father: C, c

The Punnett square will look like this:

| | C | c |
|-------|-------|-------|
| **C** | CC | Cc |
| **c** | Cc | cc |

### Step 4: Analyze the Outcomes
From the Punnett square, we can see the possible genotypes for the offspring:
- CC (normal)
- Cc (carrier, normal phenotype)
- Cc (carrier, normal phenotype)
- cc (affected by PKU)

### Step 5: Calculate the Probability
Now, we count the outcomes:
- 1 CC (normal)
- 2 Cc (carriers, normal)
- 1 cc (affected)

Thus, the probability of having a child with PKU (cc) is 1 out of 4 possible outcomes.

### Step 6: Final Probability
The probability that their child will have PKU is:
\[ \text{Probability} = \frac{\text{Number of affected outcomes}}{\text{Total outcomes}} = \frac{1}{4} = 0.25 \]

### Conclusion
Therefore, the probability that their child will have the disease is **0.25**.

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