​Two genes a and b are located at a distance of 10cM. Individuals of the genotype AaBb are sib-mated. The two genes are linked in trans. What percentage of the progeny is expected to have the genotype aabb?​

Correct option is A

Explanation-

Understand "linked in trans":
If genes a and b are linked in trans, the AaBb genotype has : One chromosome Ab and other chromosome  aB.
So AaBb = Ab/aB
Parental vs Recombinant Gametes:
Parental types (no recombination): Ab and aB
→ These occur with 90% probability (because recombination = 10%)
Recombinant types (with recombination): AB and ab
→ These occur with 10% probability
→ So each recombinant type = 10% ÷ 2 = 5% each
Cross: AaBb × AaBb = (Ab/aB) × (Ab/aB)
Gamete possibilities with their probabilities:
Gamete Probability
Ab 45%
aB 45%
ab 5%
AB 5%
Probability of aabb progeny:
To get aabb:
        Gamete 1: ab
        Gamete 2: ab
So:
         P(aabb) = P(ab) × P(ab) = 0.05 × 0.05 = 0.0025 = 0.25%

Final Answer: Option a: 0.25%