27. Genes A, B and C. control three phenotypes which assort independently; A plant with the genotype Aa Bb Cc is selfed. What is the probability for progeny which show the dominant phenotype for AT LEAST ONE of the phenotypes controlled by genes A, B and C?
(1) 1/64 (2) 27/64
(3) 63/64 (4) Cannot be predicted
Step-by-step solution
Genes A, B, C assort independently and control three separate phenotypes.
Parent genotype: Aa Bb Cc (triple heterozygote) selfed.
For each single gene on selfing
Aa × Aa → genotypes: AA, Aa, aa in ratio 1:2:1.
Phenotypes: 3/4 dominant, 1/4 recessive.
So for any given gene:
de>P(recessive phenotype) = 1/4
de>P(dominant phenotype) = 3/4
Probability calculation
Because genes assort independently, probabilities across loci multiply.
Find probability of NO dominant phenotype at all three loci (aabbcc):
de>P(aabbcc) = (1/4)A × (1/4)B × (1/4)C = 1/64
Use complement rule for “at least one dominant”:
de>P(at least one dominant) = 1 – 1/64 = 63/64
So the required probability is 63/64 [web:27][web:30].
Why other options are wrong
- 1/64: This is the probability of the opposite event (all three traits recessive: aabbcc).
- 27/64: This would be (3/4)3, i.e., probability of dominant at all three loci simultaneously.
- Cannot be predicted: The situation follows straightforward Mendelian segregation with independent assortment.
Therefore, the probability that a progeny shows the dominant phenotype for at least one of the three traits is 63/64 (option 3).
