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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. The resulting F1 plant after self fertilization, gives rise to various genotypes. How many of these offsprings in the F2 population will display ABg phenotype?

Respuesta :

Answer:

9/64

Explanation:

Given, AABBGG X aabbgg = AaBbGg (F1)

Resulting F1 progeny is self fertilized.

AaBbGg X AaBbGg :

If we check individual crosses:

     A     a

A  AA  Aa

a   Aa  aa

3/4 of the progeny will display dominant A phenotype ( AA, Aa ). Similarly,

3/4 of the progeny will display dominant B phenotype ( BB, Bb ).

    G      g

G  GG  Gg

g   Gg   gg

1/4 of the progeny will display recessive g phenotype (gg).

Total offspring that will display ABg phenotype =

3/4 * 3/4 * 1/4 = 9/64

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