13. If plant with genotype AaBb is self-pollinated Where the A and B are not linked, then the probability of getting AABB genotype will be
(1) 1/4         (2) 1/8
(3) 1/16       (4) 1/2

Step-by-step explanation

Genotype of plant: AaBb

with A and B on different (unlinked) chromosomes.

Consider each gene separately in a self-cross Aa × Aa and Bb × Bb.

For gene A: Aa × Aa

→ genotypes AA, Aa, aa in ratio 1:2:1.

Probability of AA = 1/4.

For gene B: Bb × Bb

→ genotypes BB, Bb, bb in ratio 1:2:1.

Probability of BB = 1/4.

Because A and B are unlinked, their inheritance is independent.

The probability of getting AABB is:

de>P(AABB) = P(AA) × P(BB) = 1/4 × 1/4 = 1/16

So the correct probability is 1/16, i.e. option (3).

Why other options are wrong

• 1/4: would be correct for a single locus (AA from Aa × Aa), but not for AA and BB together.
• 1/8: typical for some two-locus genotypes like AaBB from AABB×aabb, but not AABB from AaBb×AaBb.
• 1/2: far too high; that would imply one gene fixed or not segregating.