Given the following genotypes for two parents, AABBCc × AabbCc, assume that all traits exhibit simple dominance and independent

Given the conditions referenced in the question, which include independent assortment and simple dominance, crossing these two parent genotypes will yield an expected 75% of the offspring resembling the AABBCc genotype parent. To elaborate, independent assortment is when an organism's alleles for a trait separate independently during meiosis, while simple dominance refers to the effect of dominant and recessive alleles for a trait—with the trait appearing if at least one dominant allele is present. Understanding these principles allows us to solve the problem. For Parent 1, the genotype is AABBCc, and the possible allele combinations produced are ABC and ABc. For Parent 2, with a genotype of AabbCc, the assortments include AbC, Abc, abC, and abc. After using a Punnett square to combine these alleles, the resulting genotypes are AABbCC, AABbCc, AaBbCC, AaBbCc, AABbCc, AABbcc, AaBbCc, and AaBbcc, leading to a genotypic ratio of 1AABbCC: 2AABbCc: 1AABbcc: 1AaBbCC: 2AaBbCc: 1AaBbcc. The phenotypic ratio expected from this cross is 6ABC and 2ABc, thus 75% of the offspring should resemble the first parent, calculated by (6/8) x 100 = 75%.