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:

1. Understanding the Inheritance Pattern:
Sickle cell anemia is an autosomal recessive genetic disorder. This means that a person must inherit two copies of the mutated gene (one from each parent) to be diseased (affected).

Let us denote the alleles as:
- ${\mathrm{Hb}}^A$: Normal allele (dominant)
- ${\mathrm{Hb}}^S$: Sickle cell allele (recessive)

2. Genotypes of the Parents:
Both parents are heterozygous for the sickle cell gene. Therefore, their genotypes are:
- Father: ${\mathrm{Hb}}^A{\mathrm{Hb}}^S$
- Mother: ${\mathrm{Hb}}^A{\mathrm{Hb}}^S$

3. Punnett Square Analysis:
When we cross these two heterozygous parents (${\mathrm{Hb}}^A{\mathrm{Hb}}^S\times {\mathrm{Hb}}^A{\mathrm{Hb}}^S$), the possible combinations of alleles in the progeny (offspring) are:

- ${\mathrm{Hb}}^A{\mathrm{Hb}}^A$: Normal, non-carrier (1 out of 4, or 25%)
- ${\mathrm{Hb}}^A{\mathrm{Hb}}^S$: Carrier, unaffected (2 out of 4, or 50%)
- ${\mathrm{Hb}}^S{\mathrm{Hb}}^S$: Diseased (affected) (1 out of 4, or 25%)

4. Conclusion:
Only the individuals with the homozygous recessive genotype (${\mathrm{Hb}}^S{\mathrm{Hb}}^S$) will suffer from sickle cell anemia.
Thus, the percentage of diseased progeny is:

$\frac{1}{4}\times 100=25\%$