NEET 2025 Zoology Question 145 Solution

Mendel crossed a pea plant homozygous for round yellow seeds with one having wrinkled green seeds:

• • P generation: $RRYY$ (Round, Yellow) $\times$ $rryy$ (Wrinkled, Green).
• • Gametes: $RY$ and $ry$.
• • F1 generation: All $RrYy$ individuals displaying the dominant phenotype (Round, Yellow seeds).

Because genes R and Y assort independently:

• • The F1 dihybrid ($RrYy$) produces 4 types of pollen grains and 4 types of eggs in equal proportions (25% each): $RY$, $Ry$, $rY$, $ry$.
• • Random union of these gametes yields a $4 \times 4 = 16$-box Punnett square in the F2 generation.

Combining monohybrid independent probabilities:

• • Round Yellow ($R\_Y\_$): $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$.
• • Round Green ($R\_yy$): $\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$.
• • Wrinkled Yellow ($rrY\_$): $\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$.
• • Wrinkled Green ($rryy$): $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$.

The F2 phenotypic ratio is 9 : 3 : 3 : 1, matching Option (3).