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the claim
The probability of an AABbCC offspring from a trihybrid cross is one sixteenth.
the verdict
COMMON KNOWLEDGE
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the weight of evidence
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The probability of a specific genotype like AABbCC from a trihybrid cross involving three independently assorting loci with complete dominance is calculated using the product rule (1/4 * 1/2 * 1/4), yielding one sixteenth (1/16). This is a foundational principle of Mendelian genetics that does not require citation.

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The analysis

The claim asks about the basic Mendelian probability of a specific genotypic combination from a trihybrid cross. For three independently assorting loci where parents are heterozygous for each (e.g., AaBbCc x AaBbCc), the probability for each locus is: AA = 1/4, Bb = 1/2, CC = 1/4. Multiplying these independent probabilities (1/4 * 1/2 * 1/4) yields 1/16. This is a matter of basic arithmetic and textbook Mendelian genetics, constituting common knowledge that requires no empirical citation.

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first checked04 Aug 2026
judged → COMMON KNOWLEDGE · 9504 Aug 2026
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