Principle of Inheritance and VariationMediumBloom L3
Question
In the F2-generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are
Options
A.phenotypes-4, genotypes-16
B.phenotypes-9, genotypes-4
C.Phenotypes-4, genotypes-8
D.Phenotypes-4, genotypes-9
Solution
Sol.(d) [Key Concept: Dihybrid F2 — Phenotypes and Genotypes]
In F2 of a dihybrid cross (AaBb × AaBb): Number of PHENOTYPES = 4 (A_B_, A_bb, aaB_, aabb) in 9:3:3:1 ratio. Number of GENOTYPES = 9 (AABB, AABb, AaBB, AaBb, AAbb, Aabb, aaBB, aaBb, aabb) in ratio 1:2:2:4:1:2:1:2:1. For a dihybrid cross with n independent gene pairs: phenotypes = 2^n (here 2^2=4), genotypes = 3^n (here 3^2=9). Answer: 4 phenotypes, 9 genotypes = option (d).
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