Final answer:
The number of possible offspring combinations from a cross between two individuals with genotype AaBbCc is 64. This result is calculated as 3 combinations for each gene times 8 for the gamete combinations, totaling 64 unique genotypes.
Step-by-step explanation:
The question asked how many possible offspring combinations can be formed by a cross between two individuals with the genotype AaBbCc. To find the number of different genotypes that can be produced by this trihybrid cross, we must consider the possible combinations for each gene separately and then calculate the total number of combinations. Each gene has three possible combinations: homozygous dominant (AA or BB or CC), heterozygous (Aa, Bb, or Cc), and homozygous recessive (aa, bb, or cc). The number of combinations can be found using the principle of independent assortment, which states that the alleles of different genes segregate independently of one another.
So, for each gene, we have three possible genotypes, and since we have three genes, the total number of combinations is 3 (for A) × 3 (for B) × 3 (for C), which equals 27. However, each combination can occur in two forms, either from the male parent or the female parent (e.g., Aa can come as A from father and a from mother, or vice versa), making the final number of individual genotypes in the offspring 2³ × 27 = 64 unique genotypes. This is evident from the Punnett square representation of a trihybrid cross, which shows 64 cells for all the possible combinations.
To answer the original question, the number of possible offspring combinations in a cross between two individuals with the genotype AaBbCc is 64.