Final answer:
In a cross involving homozygous axial (AA) and terminal (aa) pea plants, the F1 generation is heterozygous (Aa) with axial flowers. The F2 generation produced by crossing two F1 individuals follows Mendel's 9:3:3:1 phenotypic ratio.
Step-by-step explanation:
The gene for flower position in pea plants has two alleles, axial (A) and terminal (a), with axial being dominant to terminal. When crossing two parents that are homozygous for each trait, in this case, one parent with axial flowers (AA) and another with terminal flowers (aa), the F₁ generation will all be heterozygous (Aa) and display the dominant axial flower phenotype. A Punnett square can be used to predict that the F₂ generation will have a 9:3:3:1 ratio of the phenotypes when crossing two Aa heterozygotes, which corresponds to the possible genotypes and phenotypes of axial (AA or Aa) and terminal (aa) flower positions.
As for the case mentioned where the parents are AABB (axial) and aabb (terminal), the expected F₁ genotype is AaBb, showing the axial phenotype. The F₂ generation would indeed exhibit a 9:3:3:1 phenotypic ratio consisting of axial and terminal flowers, if we consider a different gene along with axial and terminal for flower position (like flower color or shape).