A plant is of $F_1$ generation with genotype "AABbCC". On selfing of this plant what is the phenotypic ratio…
A plant is of $F_1$ generation with genotype "AABbCC". On selfing of this plant what is the phenotypic ratio in $\mathrm{F}_2$ generation?
$3: 1$
$1: 1$
$9: 3: 3: 1$
$27: 9: 9: 9: 3: 3: 3: 1$
Solution
The genotype of $F_1$ generation is AABbCC
Number of gametes that are obtained by crossing can be obtained by formula $=2^n$
(where $n$ represents the number of heterozygous gametes)
Thus, the number of heterozygous gametes $=2^1=2$
2 gametes formed are: $\mathrm{ABC}$ and $\mathrm{AbC}$.
On self crossing $F_1$ generation, the phenotypes that are formed in $\mathrm{F}_2$ generation are in the ratio of $3: 1$