Final answer:
The phenotypes of the F1 offspring from a cross between true-breeding 'fabulous' (AABB) and 'plain' (aabb) organisms would all be 'fabulous,' as they would inherit dominant alleles from the fabulous parent, resulting in a 100% fabulous phenotype (AaBb).
Step-by-step explanation:
The student is asking about the phenotypes of the F1 offspring of a cross between true-breeding "fabulous" (homozygous dominant) and "plain" (homozygous recessive) organisms, which we can symbolize as AABB and aabb, respectively. In this scenario, each parent contributes one allele for every gene.
The fabulous parent will contribute an 'A' and a 'B' allele, while the plain parent will contribute an 'a' and a 'b' allele. All the F1 offspring will therefore have the genotype AaBb and exhibit the dominant phenotype. If we assume 'fabulous' is the dominant phenotype and 'plain' is recessive, the F1 generation will all express the dominant 'fabulous' phenotype.
Using a Punnett square to visualize this, every box of the Punnett square would result in offspring with the genotype AaBb, which shows the 'fabulous' phenotype due to the dominance of the 'A' and 'B' alleles. Therefore, in the F1 generation, we have a 100% ('all') chance that the offspring exhibit the phenotype of the 'fabulous' parent, who has the dominant alleles. The cross between these true-breeding parents results in uniformly 'fabulous' F1 offspring, and there would be no plain or smashing phenotypes present in the F1 generation.