Suppose the three non-linked autosomal genes A , B and C control coat colour in an animal and the dominants alleles A , B and C are responsible for dark colour and the recessive alleles a, b and care responsible for light colour. If a cross between a male of AABBCC genotype and a female of aabbcc genotype produces 640 offsprings in the F2- generation, how many of them are likely to be of the parental genotype?

Question

Suppose the three non-linked autosomal genes A , B and C control coat colour in an animal and the dominants alleles A , B and C are responsible for dark colour and the recessive alleles a, b and care responsible for light colour. If a cross between a male of AABBCC genotype and a female of aabbcc genotype produces 640 offsprings in the F2- generation, how many of them are likely to be of the parental genotype? (A) 10 (B) 20 (C) 160 (D) 640

Tardigrade Admin · Accepted Answer

In a Mendelian trihybrid cross between A A B B C C and aabbcc parents the F 2 progenies produced are 640 . ∴ The number of parental combinations =(2/64) × 640=20

10

20

160

640

In a Mendelian trihybrid cross between AABBCC and aabbcc parents the F2​ progenies produced are 640. ∴ The number of parental combinations =642​×640=20

In a Mendelian trihybrid cross between $AA BBCC$ and aabbcc parents the $F_{2}$ progenies produced are $640.$
$∴$ The number of parental combinations
$= \frac{2}{64} \times 640 = 20$