Final answer:
The proportion of the progeny expected to phenotypically resemble the AABBCc parent from the cross AABBCc × AabbCc is 3/4. This result is obtained by applying the principles of simple dominance, independent assortment, and using the product rule to calculate the probability of the dominant phenotype being expressed in all three gene loci.
Step-by-step explanation:
When examining the cross between AABBCc (showing a dominant phenotype) and AabbCc, we are considering a trihybrid cross with independent assortment and complete dominance. Since these are assumptions similar to those used by Mendel, we can apply the principles of dominance and independent assortment to predict phenotypic ratios. In this case, we want to determine the proportion of offspring that will phenotypically resemble the AABBCc parent.
To resemble the AABBCc parent phenotypically, offspring must have at least one dominant allele for each of the first two gene loci (A and B) and can have either allele for the C locus. The parent AabbCc will always contribute a dominant A allele, meaning all progeny will express the dominant trait for the A locus. For the B locus, the first parent can only contribute a dominant allele, while the second parent can contribute either a dominant or a recessive allele. Since we assume complete dominance, as long as there is one dominant B allele, the trait will be expressed. Therefore, all offspring will also exhibit the dominant B phenotype. Finally, for the C locus, since both parents have the Cc genotype, there is a 3/4 chance that the offspring will exhibit the dominant trait, because three of the four possible combinations (CC, Cc, and cC) result in the dominant phenotype.
Using the product rule, we multiply the individual probabilities of each independent event. The probability of inheriting the dominant phenotype for A is 1, for B is 1, and for C is 3/4. Hence, the probability of an offspring phenotypically resembling the AABBCc parent is 1 × 1 × 3/4, which equals 3/4.