A parent having autosomal dominant disease then what will be the probability of diseased offspring irrespective of sex of the child?

Options

A. `90%`
B. `10%`
C. `50%`
D. `100%`

Answer

To solve the problem of determining the probability of offspring affected by an autosomal dominant disease when one parent has the disease, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Autosomal Dominant Inheritance:**
- In autosomal dominant diseases, only one copy of the mutated gene (dominant allele) is sufficient to express the disease. This means that if a parent has the disease, they can pass on the affected allele to their offspring.

2. **Identify the Genotypes:**
- Let’s denote the dominant allele (disease-causing) as "A" and the normal allele as "a".
- The affected parent can have one of two genotypes: AA (homozygous dominant) or Aa (heterozygous). However, for the purpose of this question, we will consider the more common scenario where the affected parent is heterozygous (Aa).

3. **Determine Possible Gametes:**
- The affected parent (Aa) can produce two types of gametes: A (disease allele) and a (normal allele).
- The unaffected parent (aa) can only produce one type of gamete: a (normal allele).

4. **Set Up a Punnett Square:**
- Create a Punnett square to visualize the potential offspring genotypes:
```
A | a
-----------------
a | Aa | aa
-----------------
a | Aa | aa
```
- From the Punnett square, we can see the possible genotypes of the offspring:
- 50% (2 out of 4) will be Aa (affected)
- 50% (2 out of 4) will be aa (unaffected)

5. **Calculate the Probability:**
- Therefore, the probability of having an affected offspring (Aa) is 50%.

### Final Answer:
The probability of having a diseased offspring, irrespective of the sex of the child, is **50%**.