The probability of an AABbCC offspring from a trihybrid cross is one sixteenth.
Is it true that the probability of an AABbCC offspring from a trihybrid cross is one sixteenth?
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.
Verdict: COMMON KNOWLEDGE · 0 sources for, 0 against · Last checked
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.
What does the evidence say?
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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