31. A plant of the genotype AaBb is selfed. The two genes are linked and are 50 map units apart. What proportion of the progeny will haw the genotype aabb?
(1) 1/2 (2) 1/4
(3) 1/8 (4) 1/16
Key concept
Distance 50 map units means 50% recombination frequency, which is the maximum; such loci behave as if unlinked because parental and recombinant gametes occur in equal frequencies [web:59][web:60].
So an AaBb individual produces the four gametes AB, Ab, aB, ab each with probability 1/4 (independent assortment).
Selfing AaBb × AaBb with independent assortment
For any dihybrid self (AaBb × AaBb):
- Each locus segregates as Aa × Aa → genotypes 1/4 AA, 1/2 Aa, 1/4 aa
- Probability of aa = 1/4
- Probability of bb = 1/4
Loci assort independently, so:
de>P(aabb) = P(aa) × P(bb) = 1/4 × 1/4 = 1/16
Thus, 1/16 of the progeny are expected to be aabb [web:33].
Option-wise explanation
- 1/2: would require half of all offspring to be double recessive, which is impossible
- 1/4: equals the probability of being recessive at a single locus (aa or bb), not both
- 1/8: does not match any simple product of 1/2 or 1/4 Mendelian probabilities
- 1/16: correct – matches the product of independent probabilities for aa and bb
