Question

Genes A, B, and C are independently assorting and control production of a black pigment. The dominant alleles A, B and C code for functional enzymes whereas the recessive alleles a, b, and c code for nonfunctional enzymes. An individual of genotype AABBCC is crossed with one of genotype aabbcc, and the resulting F1 individuals are cross-fertilized. For the following pathway state what proportion of the F2 progeny is brown?

Answer 1

Answer: For the pathway, the proportion of the F2 progeny that is brown is 1 : 64 (1/64).

Explanation: Since ABC codes for functional enzyme and abc codes for non-functional enzyme, an individual with genotype AABBCC or AaBbCc can be said to have black pigment while an individual with genotype aabbcc can be said to have brown pigment.

The result of F1 crossing between AABBCC and aabbcc will yield progeny that are heterozygous for the trait (AaBbCc). The number of gametes that can be produced from the F1 progeny (AaBbCc) when selfed (AaBbCc x AaBbCc) is 64 (i.e 2^3 x 2^3 = 64) since there are 3 heterozygous loci in each of the genotype.

When these gametes are crossed, only one offspring will have brown pigment out of the 64 as shown in the attached image.