Final answer:
The proportion of colorless F2 progeny from a cross between AaBbCc individuals is 9/64, or approximately 14%, considering that the cc genotype inhibits pigment production and genes A and B are independently assorting.
Step-by-step explanation:
When analyzing the inheritance pattern of genes controlling pigment production in an organism, it is important to understand concepts like independent assortment and epistasis. In the given question, genes A, B, and C control the production of black pigment, with dominant alleles coding for functional enzymes and recessive alleles coding for non-functional enzymes. The genotype AABBCC, expressing all dominant alleles, is crossed with aabbcc, expressing all recessive alleles, which results in F1 individuals with heterozygous genotypes (AaBbCc). When these F1 individuals are cross-fertilized, the F2 progeny will express a range of genotypes and phenotypes due to independent assortment.
In the case highlighted, the C gene acts as an inhibitor of the B gene. If the genotype contains the recessive alleles cc for gene C, the pathway for pigment production is blocked, leading to a colorless phenotype regardless of the A or B alleles. Given that the genotype for black pigment requires at least one dominant allele at each locus and the cc genotype is an inhibitor, we need to determine the proportion of F2 progeny with at least one dominant allele at the A and B loci and two recessive alleles at the C locus.