Haemolytic jaundice is due to dominant gene but only `20%` of the people develop this disease. A heterozygous man marries a homozygous normal woman. What proportion of the children in population would be expected to have this disorder :-

Options

A. `1/5`
B. `1/20`
C. `1/10`
D. `1/2`

Answer

To solve the problem regarding the inheritance of hemolytic jaundice, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotypes**:
- Hemolytic jaundice is caused by a dominant gene. Let's denote the dominant allele (which causes the disorder) as "A" and the normal allele as "a".
- The heterozygous man has the genotype "Aa" (he carries one allele for the disorder).
- The homozygous normal woman has the genotype "aa" (she carries two alleles for the normal trait).

2. **Set Up the Punnett Square**:
- We will create a Punnett square to determine the possible genotypes of their offspring.
- The man (Aa) can pass on either "A" or "a", while the woman (aa) can only pass on "a".

```
| a | a |
-------------------
A | Aa | Aa |
-------------------
a | aa | aa |
```

3. **Determine the Offspring Genotypes**:
- From the Punnett square, we see the possible genotypes of the offspring:
- 2 offspring with genotype "Aa" (affected by hemolytic jaundice)
- 2 offspring with genotype "aa" (normal)

4. **Calculate the Ratio of Affected Offspring**:
- The ratio of affected (Aa) to total offspring is:
- Affected (Aa): 2
- Normal (aa): 2
- Total: 4
- Therefore, the ratio of affected to total offspring is 2/4 or 1/2.

5. **Adjust for Population Prevalence**:
- The problem states that only 20% of the population actually develops the disease.
- To find the expected proportion of children who would develop the disorder in the population, we multiply the ratio of affected offspring by the prevalence:
- Expected proportion = (1/2) * (20/100)
- Expected proportion = (1/2) * (1/5) = 1/10.

### Final Answer:
The proportion of children in the population expected to have hemolytic jaundice is **1/10**.