Question

Find out the phenotypic ratio of offspring found in F2 generation, when homozygous yellow and round seed of pea plant cross with homozygous green and wrinkled seed plant. With the help of checker board.

Hint:

A 9:3:3:1 ratio is characteristic of a dihybrid cross, where two traits are inherited independently.

Answer 1

Step 1: Understanding the cross.
The problem involves a dihybrid cross between two homozygous pea plants:
- Parent 1: Homozygous yellow (YY) and round (RR) seeds.
- Parent 2: Homozygous green (yy) and wrinkled (rr) seeds.
The F1 generation will be heterozygous for both traits: \( YyRr \).

Step 2: Punnett Square Analysis.
To find the phenotypic ratio in the F2 generation, we set up a Punnett square for the F1 cross \( YyRr \times YyRr \). The alleles for seed color (Y for yellow, y for green) and shape (R for round, r for wrinkled) segregate independently, following Mendel's law of independent assortment. 

\[ \begin{array}{c|c c c c} & YR & Yr & yR & yr \\ \hline YR & YYRR & YYRr & YyRR & YyRr \\ Yr & YYRr & YYrr & YyRr & Yyrr \\ yR & YyRR & YyRr & yyRR & yyRr \\ yr & YyRr & Yyrr & yyRr & yyrr  \end{array} \]

Step 3: Conclusion.
The phenotypic ratio from the F2 generation is as follows:
- **Yellow and round (dominant traits)**: 9/16
- **Yellow and wrinkled**: 3/16
- **Green and round**: 3/16
- **Green and wrinkled**: 1/16
Thus, the phenotypic ratio is **9:3:3:1**.