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

The correct option is 25%.

Step-by-step Explanation:

Sickle cell anaemia is an autosomal recessive genetic disorder controlled by a single pair of alleles, Hb^A and Hb^S:
- Hb^A represents the normal allele.
- Hb^S represents the mutated sickle cell allele.

Since it is a recessive trait, only homozygous individuals with the genotype Hb^SHb^S will display the diseased phenotype. Heterozygous individuals (Hb^AHb^S) are healthy carriers of the disease.

Let's analyze the cross between two heterozygous parents:
Parents: Hb^AHb^S (Carrier Father) × Hb^AHb^S (Carrier Mother)

Each parent produces two types of gametes:
- 50% carrying the Hb^A allele
- 50% carrying the Hb^S allele

Combining these gametes yields the following genotypic distribution in the progeny:
1. Hb^AHb^A (Normal/Healthy): 1 out of 4 (25%)
2. Hb^AHb^S (Carrier/Healthy): 2 out of 4 (50%)
3. Hb^SHb^S (Diseased): 1 out of 4 (25%)

Only the homozygous recessive offspring (Hb^SHb^S) will be diseased.
Thus, the percentage of diseased progeny is:
$\frac{1}{4}\times 100=25\%$