Dihybrid cross: probability of a double recessive
Two plants heterozygous for two unlinked genes (AaBb × AaBb) are crossed. What fraction of the offspring will show both recessive phenotypes?
Answer A: explanation for every choice
A
Correct. P(aa) = ¼ and P(bb) = ¼. For independent genes, multiply: ¼ × ¼ = 1/16.
B
3/16 is each class with one dominant and one recessive phenotype.
C
1/4 is the probability of the recessive phenotype for one gene alone.
D
9/16 is the fraction showing both dominant phenotypes.
TakeawayFor independent genes, multiply the probabilities.
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Reviewed 2026-11-05 by a medical student. Spot an error? Report it with question ID bb54.