Using the product rule to check your answer
Because the three genes assort independently, any offspring probability is the product of three single-gene probabilities. For AaBbCc × AaBbCc:
- P(dominant for all three) = ¾ × ¾ × ¾ = 27/64
- P(AaBbCc) = ½ × ½ × ½ = 8/64
- P(recessive for all three) = ¼ × ¼ × ¼ = 1/64
Two genes are easier to visualise — try the dihybrid cross calculator.
FAQ
How many boxes are in a trihybrid Punnett square?
64. Each fully heterozygous parent (AaBbCc) makes 2³ = 8 different gametes, and 8 × 8 = 64 combinations.
What is the trihybrid phenotype ratio?
For AaBbCc × AaBbCc it is 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1 — eight phenotype classes. That is (3:1)³ expanded. The calculator lists them in a fixed trait order, so the 9s and 3s may appear interleaved.
Is there a shortcut instead of drawing 64 boxes?
Yes — the product rule. Treat each gene as its own monohybrid cross and multiply probabilities. For example, P(aabbcc) = ¼ × ¼ × ¼ = 1/64.