According to Mendel’s second law, in the crossing between homozygous individuals concerning two pairs of nonlinked alleles, AABB x aaBB, what are the genotypical and phenotypical proportions in F1 and F2?
Parental genotypes: AABB, aaBB. Gametes from the parental generation: Ab and aB. Thus F1 will present 100% AaBb gametes (and the phenotypical correspondent form). As F1 are AaBb individuals the gametes from their crossing can be: AB, Ab, aB and ab. The casual combination of these gametes forms the following genotypical forms: one AABB, two AABb, two AaBb, four AaBB, one Aabb, one Aabb, one aaBB, two aaBb and two aabb. The phenotypical proportion then would be: nine A_B_ (double dominant); three A_bb (dominant for the first pair, recessive for the second); three aaB_ (recessive for the first pair, dominant for the second); one aabb (double recessive)
