Genes R and Y follow independent assortment. If RRYY produce round yellow seeds and rryy produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?

To find the phenotypic ratio of the F₂ generation, we analyze a Mendelian dihybrid cross involving two genes that assort independently:

1. Gene R control seed shape:
- R (dominant allele) for round seeds
- r (recessive allele) for wrinkled seeds
2. Gene Y control seed color:
- Y (dominant allele) for yellow seeds
- y (recessive allele) for green seeds

Parental Cross (P Generation):
We cross a true-breeding round yellow seeded plant ($RRYY$) with a true-breeding wrinkled green seeded plant ($rryy$).
The gamete from the first parent is $RY$ and from the second parent is $ry$.

F₁ Generation:
All offspring in the F₁ generation have the heterozygous genotype $RrYy$ and exhibit the dominant phenotype: round yellow seeds.

F₂ Generation (Selfing of F₁):
When the F₁ plants ($RrYy$) are self-pollinated, each parent produces four types of gametes due to independent assortment: $RY$, $Ry$, $rY$, and $ry$, each with equal probability ($\frac{1}{4}$).

Using a Punnett square to combine these gametes, we obtain 16 genetic combinations. Since the genes assort independently, we can also determine the phenotypic outcomes by multiplying the probabilities of the individual traits:
- Probability of round seeds (dominant phenotype from $Rr\times Rr$) = $\frac{3}{4}$
- Probability of wrinkled seeds (recessive phenotype from $Rr\times Rr$) = $\frac{1}{4}$
- Probability of yellow seeds (dominant phenotype from $Yy\times Yy$) = $\frac{3}{4}$
- Probability of green seeds (recessive phenotype from $Yy\times Yy$) = $\frac{1}{4}$

Now we calculate the probability of each phenotype combination in the F₂ generation:
- Round, Yellow seeds: $\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$
- Round, Green seeds: $\frac{3}{4}\times \frac{1}{4}=\frac{3}{16}$
- Wrinkled, Yellow seeds: $\frac{1}{4}\times \frac{3}{4}=\frac{3}{16}$
- Wrinkled, Green seeds: $\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

Therefore, the phenotypic ratio in the F₂ generation is 9 : 3 : 3 : 1.