If a mother and father both have one mutant gene for a disease with an autosomal dominant inheritance, what is their risk of having an affected child?

Options

A. 0
B. 0.25
C. 0.5
D. 0.75

Answer

To determine the risk of having an affected child when both parents have one mutant gene for a disease with autosomal dominant inheritance, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Genetic Basis**:
- In autosomal dominant inheritance, only one copy of the mutant gene (dominant allele) is needed for an individual to express the trait.
- We denote the mutant gene as "A" (dominant) and the normal gene as "a" (recessive).

2. **Determine the Genotype of the Parents**:
- Both parents have one mutant gene, so their genotypes are heterozygous: Aa (where "A" is the mutant allele and "a" is the normal allele).

3. **Set Up a Punnett Square**:
- To find the possible genotypes of the offspring, we can create a Punnett square. The alleles from each parent are placed on the top and side of the square:
```
A a
---------
A | AA | Aa |
---------
a | Aa | aa |
---------
```

4. **Analyze the Offspring Genotypes**:
- From the Punnett square, we can see the possible genotypes of the offspring:
- AA (homozygous dominant) - affected
- Aa (heterozygous) - affected
- Aa (heterozygous) - affected
- aa (homozygous recessive) - normal
- Therefore, the possible genotypes are:
- 1 AA (affected)
- 2 Aa (affected)
- 1 aa (normal)

5. **Calculate the Ratios**:
- Out of the four possible genotypes, three (AA and Aa) result in an affected child, while one (aa) results in a normal child.
- This gives us a ratio of 3 affected to 1 normal.

6. **Determine the Probability**:
- The probability of having an affected child is the number of affected genotypes divided by the total number of genotypes:
- Probability of affected child = 3 (affected) / 4 (total) = 0.75 or 75%.

### Conclusion:
The risk of having an affected child when both parents have one mutant gene for an autosomal dominant disease is 75%.