Difficulty: Easy
Correct Answer: 9:3:3:1
Explanation:
Introduction / Context:The hallmark ratio for a Mendelian dihybrid cross under independent assortment is 9:3:3:1 in the F2 generation. This assumes complete dominance at each locus, no epistasis, and random segregation.
Given Data / Assumptions:
Concept / Approach:Each locus segregates 3:1 for the dominant phenotype. The joint probability across two independent loci yields the product distribution: 9 double-dominant phenotypes, 3 dominant at A only, 3 dominant at B only, and 1 double-recessive phenotype.
Step-by-Step Solution:
Compute A locus: 3 dominant : 1 recessive.Compute B locus: 3 dominant : 1 recessive.Combine: (33)=9 both dominant; (31)=3 A dominant only; (13)=3 B dominant only; (11)=1 both recessive → 9:3:3:1.Verification / Alternative check:Punnett square or probability tree confirms the same ratio. A testcross (AaBb × aabb) would instead yield 1:1:1:1, highlighting the importance of the specified mating.
Why Other Options Are Wrong:
Common Pitfalls:Confusing F1 self with testcross; always check the mating design.
Final Answer:9:3:3:1
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