Expected phenotypic ratios in Mendelian dihybrid crosses If an F1 individual is heterozygous for two independently assorting genes and is selfed (or mated inter se with a like F1), what phenotypic ratio is expected in the F2?

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:

  • One F1 is heterozygous at both loci (AaBb).
  • Cross is an F1 self (AaBb × AaBb) or inter se with another AaBb.
  • Genes assort independently; dominance is complete at each locus.


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:

  • 1:1:1:1: testcross outcome, not F1 self.
  • Other ratios do not follow independent assortment with complete dominance.


Common Pitfalls:
Confusing F1 self with testcross; always check the mating design.



Final Answer:
9:3:3:1

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion