if a plant with genotype AaBbcc is self crossed the no. of genotype and phenotype will be?

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Solution

ABc Abc aBc abc

ABc AABBcc AABbcc AaBBcc AaBbcc

Abc AABbcc AAbbcc AaBbcc Aabbcc

aBc AaBBcc AaBbcc aaBBcc aaBbcc

abc AaBbcc Aabbcc aaBbcc aabbcc

Parents: AaBbcc X AaBbcc
Gametes : ABc, aBc, abc, Abc
Punnett square:

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F_{2}

On selfing a plant of $F_{1}$ generation with genotype AABbCC, the phenotypic ratio in $F_{2}$ generation will be:

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