Q.In pea plants, round seed shape ($R$) is dominant over wrinkled ($r$), and yellow seed colour ($Y$) is dominant over green ($y$). The F1 dihybrid ($RrYy$) is self-pollinated. Using a $4\times4$ Punnett square, work out the phenotypic ratio of the F2 generation.

A dihybrid cross tracks the inheritance of two separate characters, each controlled by its own gene with two alleles, through a single cross - for example seed shape (round R dominant over wrinkled r) together with cotyledon colour (yellow Y dominant over green y). Crossing pure-breeding round-yellow (RRYY) with pure-breeding wrinkled-green (rryy) parents gives a uniformly round-yellow F1 (RrYy); because the two genes lie on separate chromosomes and assort independently during meiosis (the Law of Independent Assortment), this F1 produces four kinds of gametes - RY, Ry, rY, ry - in equal proportions. Random fertilisation among these four gamete types in a self-cross produces sixteen zygote combinations that collapse, by dominance, into Mendel's classic F2 phenotype ratio of 9 round-yellow : 3 round-green : 3 wrinkled-yellow : 1 wrinkled-green - including two new recombinant phenotypes (round-green, wrinkled-yellow) not present in either original parent, the clearest possible evidence that the two genes are inherited independently of each other.