29. Assuming that the A, B, C and D genes are not linked. The probability of a progeny being AaBBccDd from a corss between AABbccDd and aaBBccDD parents will be
(1) 4/32 (2) 3/16
(3) 1/4 (4) 3/32
Step-by-step solution
Cross: AABbccDd × aaBBccDD
Target progeny genotype: Aa BB cc Dd
Consider each gene independently (genes assort independently) [web:47][web:31].
Gene A
Parent 1: AA → gamete always A
Parent 2: aa → gamete always a
Offspring: always Aa → de>P(Aa) = 1
Gene B
Parent 1: Bb → gametes B or b (½ each)
Parent 2: BB → gamete always B
de>P(BB) = P(B from P1) × P(B from P2) = 1/2 × 1 = 1/2
Gene C
Parent 1: cc → gamete always c
Parent 2: cc → gamete always c
Offspring: always cc → de>P(cc) = 1
Gene D
Parent 1: Dd → gametes D or d (½ each)
Parent 2: DD → gamete always D
de>P(Dd) = P(d from P1) × P(D from P2) = 1/2 × 1 = 1/2
Combine probabilities
Since loci assort independently:
de>P(AaBBccDd) = P(Aa) × P(BB) × P(cc) × P(Dd) = 1 × 1/2 × 1 × 1/2 = 1/4
So the probability is 1/4, matching option (3) [web:30].
Why other options are incorrect
- 4/32 = 1/8: too high, would require only one variable locus
- 3/16 ≈ 0.1875: does not arise from simple 1/2 probabilities
- 3/32: would require mixed 3/4 and 1/2 terms, not present here
