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 colorless?


A B C
Colorless ---> Colorless ---> Colorless ---> Red

Answer 1

In the F2 progeny of the cross between AABBCC and aabbcc, the proportion of colorless individuals is ⅖³ or 1/64, as each gene has a 1/2 chance of contributing a recessive allele required for the colorless phenotype.

Step-by-step explanation:

When considering a cross between an individual of genotype AABBCC (which produces black pigment) with one of genotype aabbcc (colorless), the F1 progeny will all have the genotype AaBbCc, being heterozygous at all three loci.

Since all three genes assort independently and contribute pigment production, for an F2 progeny to be colorless, it must receive the recessive allele from both parents for each gene. The chance of receiving a recessive allele from a heterozygous parent is 1/2 for each gene.

The probability an F2 progeny is colorless is therefore (1/2)^6, or 1/64, because each gene has two alleles and a colorless individual must have a homozygous recessive combination (aa, bb, cc). This fraction is derived from applying the probability for a single gene across all three genes.