The ratio between carrier, disease free & diseased individual on marriage between two carriers of sickle cell anaemia will be -

Options

A. `0 : 2 : 1`
B. `0 : 1 : 1`
C. `1 : 2 : 1`
D. `2 : 1 : 1`

Answer

To solve the question regarding the ratio of carrier, disease-free, and diseased individuals resulting from a marriage between two carriers of sickle cell anemia, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Genetics of Sickle Cell Anemia**:
- Sickle cell anemia is an autosomal recessive disorder. This means that an individual must have two copies of the sickle cell gene (HBS) to express the disease. The normal gene is represented as HBA.

2. **Identify the Genotypes of the Parents**:
- Since both parents are carriers, their genotypes are HBA/HBS.

3. **Set Up a Punnett Square**:
- Create a Punnett square to determine the possible genotypes of the offspring. The alleles from one parent (HBA and HBS) will be placed on one side, and the alleles from the other parent (HBA and HBS) will be placed on the other side.

4. **Fill in the Punnett Square**:
- The combinations from the Punnett square will be:
- HBA from parent 1 and HBA from parent 2 = HBA/HBA (normal)
- HBA from parent 1 and HBS from parent 2 = HBA/HBS (carrier)
- HBS from parent 1 and HBA from parent 2 = HBA/HBS (carrier)
- HBS from parent 1 and HBS from parent 2 = HBS/HBS (diseased)

5. **Count the Offspring Genotypes**:
- From the Punnett square, we have:
- 1 HBA/HBA (normal, disease-free)
- 2 HBA/HBS (carriers)
- 1 HBS/HBS (diseased)

6. **Determine the Ratio**:
- The ratio of carrier (HBA/HBS) to disease-free (HBA/HBA) to diseased (HBS/HBS) individuals is:
- Carriers: 2
- Disease-free: 1
- Diseased: 1
- Therefore, the ratio is 2:1:1.

### Final Answer:
The ratio between carrier, disease-free, and diseased individuals when two carriers of sickle cell anemia marry is **2:1:1**.

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