Question

A plant of F${}_1$generation with genotype "AABbCC", on selfing of this plant, what is the phenotypic ratio in F${}_2$ generation?

• A. 3:1
• B. 1:1
• C. 9:3:3:1
• D. 27:9:9:9:3:3:3:1

Answer 1

The correct option is A 3:1

	ABC	AbC
ABC	AABBCC	AABbCC
AbC	AABbCC	AAbbCC

The genotype of the $F_1$ generation plants is AABbCC. The number of gametes involved in the given cross can be calculated by $2^n$ where n= 1 ( n is the number of heterozygotes). Hence there will two gametes ABC and AbC. On selfing these plants the phenotype of the $F_2$ generation will be 3:1 as shown in the Punnett square. Thus the correct answer is option A.