A researcher has three strains of mice: black, brown, and albino. She crosses the true-breeding albino...

A researcher has three strains of mice: black, brown, and albino. She crosses the true-breeding albino mice with the true-breeding black mice and finds that the F1 progeny are all brown. She then crosses the F1 generation with each other, producing 17 brown mice, 7 black mice and 8 albino mice in the F2 generation. Based on this observation, what is the most consistent mode(s) of inheritance would you propose to explain this results?

A. Double recessive epistasis

B. Recessive epistasis

C. Dominant epistasis

D. More than one mode of inheritance is possible

Expert Solution

Let gene C control the black body color of mice and gene A control the brown body color and is hypostatic ti recessive allele c. The dominant allele C in the presence of 'a' gives black mice. In the presence of dominant allele C, A gives rise to brown mice. So, CCaa gives black mice and ccAA gives albino. When true breeding black mice (CCaa) is crossed with albino (ccAA), brown mice (CcAa) appear in F1. cc masks the effect of AA and is therefore epistatic. Consequently, ccAA is albino and the combination ccaa is also albino due to the absence of both the dominant alleles.

Parents: CCaa (black) ccAA (albino)

gametes: Ca cA

F1: CcAa (brown)

When F1 are selfed, the cross is as follows:

Parents: CcAa (brown) CcAa (brown)

gametes: CA Ca cA ca CA Ca cA ca

F2: CCAA (brown), CCAa (brown), CcAA (brown), CcAa (brown), CCAa (brown), CCaa (black), CcAa (brown), Ccaa (black), CcAA (brown), CcAa (brown), ccAA 9albino), ccAa (albino), CcAa (brown), Ccaa (black), ccAa (albino), ccaa (albino).

The phenotypic ratio of F2 generation is 9 (brown) : 3 (black) : 4 (albino) which is close to the ratio of the number of brown, black and albino mice in F2 (17:7:8). So the correct explanation of this mode of inheritance is Recessive epistasis. Option (B) is the correct one.

gladiator answered 2 years ago

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