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a plant homozygous for the unlinked dominant alleles a, b and g is crossed with another plant that is homozygous for the recessive alleles a, b and g. how many different genotypes in the f2 population will give rise to the abg phenotype?

Respuesta :

A plant that is homozygous for the unconnected dominant alleles a, b, & g crossed with some other plant that is homozygous again for recessive alleles a, b, and g will result in the abg phenotype in the f2 population is 9÷64 genotypes.

Given, AABBGG X aabbgg = AaBbGg (F1)

The resultant self-fertilized F1 offspring.

AaBbGg X AaBbGg :

If we look at certain crosses:

    A     a

A  AA  Aa

a   Aa  aa

3÷4 will exhibit the dominant A phenotype in the offspring ( AA, Aa ). Similarly,

3÷4 will exhibit the dominant B trait in the progeny ( BB, Bb ).

   G      g

G  GG  Gg

g   Gg   gg

1÷4 if the offspring will have a recessive g trait (gg).

The total number of children who will exhibit the ABg phenotype:

3÷4 × 3÷4 × 1÷4 = 9÷64

Learn more about dominant alleles at

https://brainly.com/question/14053639?referrer=searchResults

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