To solve this problem, we need to determine the percentage of progeny that will be diseased in a cross between two individuals heterozygous for the sickle cell anaemia gene.$\newline$Let's denote the normal allele as $S$ and the sickle cell allele as $s. \newline$Both parents are heterozygous, so their genotypes are $Ss. \newline$We can set up a Punnett square to find the genotypes of the progeny:$\newline \bullet$The possible gametes from each parent are $S$ and $s. \newline \bullet$The Punnett square is as follows:$\newline$ - Parent 1 (top): $S, s \newline$ - Parent 2 (side): $S, s \newline$ - Resulting genotypes:$\newline$ - $SS$ (normal)$\newline$ - $Ss$ (carrier, not diseased)$\newline$ - $Ss$ (carrier, not diseased)$\newline$ - $ss$ (diseased)$\newline$The genotypic ratio is:$\newline \bullet$1 $SS : 2$ $Ss : 1$ $ss \newline$The percentage of progeny that are diseased ($ss$) is:$\newline \bullet \frac{1}{4} \times 100$Therefore, the percentage of the progeny that will be diseased is 25%.$\newline$This corresponds to Option 3.