Question

A Dihybrid cross is done between two parent pea plants (pure line) who differ in two pairs of contrasting traits: Seed colour and seed shape. In the F$_2$ generation the number of phenotypes and genotypes will be:

• A. phenotypes = 4; genotypes = 16
• B. phenotypes = 9; genotypes = 14
• C. phenotypes = 4; genotypes = 8
• D. phenotypes = 4; genotypes = 9

Hint:

In a dihybrid cross, the F$_2$ phenotypic ratio is 9:3:3:1 (4 phenotypes), and the genotypic ratio is 1:2:1:2:4:2:1:2:1 (9 genotypes), assuming independent assortment.

Answer 1

The Correct Option is D

Step 1: A dihybrid cross involves two traits. Assuming seed color (Yellow YY vs. Green yy) and seed shape (Round RR vs. Wrinkled rr), pure line parents are YYRR and yyrr.
Step 2: The F$_1$ generation from YYRR × yyrr is YyRr (yellow round, heterozygous). In the F$_2$ generation (YyRr × YyRr), the phenotypic ratio is 9:3:3:1, giving 4 distinct phenotypes: yellow round, yellow wrinkled, green round, and green wrinkled.
Step 3: The genotypic ratio is 1:2:1:2:4:2:1:2:1, resulting in 9 distinct genotypes: YYRR, YYRr, YyRR, YyRr, YYrr, Yyrr, yyRR, yyRr, and yyrr.
Step 4: Comparing the options, option (D) correctly states 4 phenotypes and 9 genotypes.
Thus, in the F$_2$ generation, there are 4 phenotypes and 9 genotypes.