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

Both parents are heterozygous for the sickle‑cell allele, so each has the genotype $Aa$ where $A$ represents the normal hemoglobin allele and $a$ the sickle‑cell allele.

To find the possible genotypes of their offspring we construct a Punnett square.

Parent 1 can contribute either $A$ or $a$.
Parent 2 can also contribute either $A$ or $a$.

The four possible combinations are:

$AA$ – child receives a normal allele from both parents (healthy).
$Aa$ – child receives a normal allele from one parent and a sickle allele from the other (carrier, but not diseased).
$aA$ – same as $Aa$ (carrier).
$aa$ – child receives a sickle allele from each parent (diseased).

Thus the genotype frequencies are:

1 out of 4 ($AA$) – healthy.
2 out of 4 ($Aa$ and $aA$) – carriers, healthy.
1 out of 4 ($aa$) – diseased.

Therefore the proportion of progeny that will be diseased is $1/4$, which is 25 %.