In a cross between a male and female, both heterozygous for sickle cell anaemia gene, what percentage of the progeny will be diseased

Options

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

Explanation

To solve the problem of determining the percentage of progeny that will be diseased in a cross between two heterozygous individuals for sickle cell anemia, we can follow these steps:

### Step 1: Understand the Genetics of Sickle Cell Anemia
Sickle cell anemia is an autosomal recessive disorder. This means that an individual must inherit two copies of the sickle cell allele (HbS) to express the disease. The normal allele is represented as HbA.

### Step 2: Determine the Genotypes of the Parents
Both parents are heterozygous for the sickle cell anemia gene. Therefore, their genotypes can be represented as:
- Male: HbA HbS
- Female: HbA HbS

### Step 3: Set Up a Punnett Square
To find the possible genotypes of the progeny, we can set up a Punnett square. The gametes produced by each parent will be:
- From the male: HbA and HbS
- From the female: HbA and HbS

The Punnett square will look like this:

| | HbA | HbS |
|---------|-----------|-----------|
| **HbA** | HbA HbA | HbA HbS |
| **HbS** | HbA HbS | HbS HbS |

### Step 4: Analyze the Results
From the Punnett square, we can see the possible genotypes of the progeny:
1. HbA HbA (Normal)
2. HbA HbS (Carrier)
3. HbA HbS (Carrier)
4. HbS HbS (Diseased)

### Step 5: Calculate the Percentage of Diseased Progeny
Out of the four possible genotypes:
- 1 genotype (HbA HbA) is normal.
- 2 genotypes (HbA HbS) are carriers.
- 1 genotype (HbS HbS) is diseased.

Thus, the percentage of progeny that will be diseased (HbS HbS) is:
- Diseased = 1 out of 4 = 25%

### Final Answer
Therefore, the percentage of the progeny that will be diseased is **25%**.

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