Practice question
Question
A dihybrid cross AaBb × AaBb produces how many phenotypic classes in F2?
Explanation
A dihybrid individual AaBb with unlinked genes can form four gamete types AB, Ab, aB, ab through independent assortment during meiosis. Selfing AaBb x AaBb produces 4x4=16 genotypic combinations collapsed by complete dominance at each locus. Phenotypically, genotype classes group into A-B- displaying both dominant traits, A-bb showing first dominant alone, aaB- second dominant alone, aabb both recessive. Thus four distinct phenotypic classes appear despite nine genotypic classes. Distinction highlights phenotype versus genotype numbers and forms basis for modified ratios when epistasis operates.