Q.96 What would be the number of genotypes and phenotypes, respectively, from a cross
between genotypes AaBBCcDd and AaBBCcDd ? Assume independent assortment
and simple dominant–recessive relationship in each gene pair.
(A) 8 and 4
(B) 12 and 4
(C) 27 and 8
(D) 14 and 8

The correct answer is (C) 27 and 8. This cross between AaBBCcDd × AaBBCcDd produces 27 distinct genotypes and 8 phenotypes under independent assortment and complete dominance.​

Genotype Analysis

Each parent AaBBCcDd has three heterozygous gene pairs (A/a, C/c, D/d) and one homozygous dominant pair (B/B), yielding 8 gamete types per parent (2³ = 8). The offspring genotypes follow a (3:1) ratio per heterozygous gene: AA/Aa/aa for A, CC/Cc/cc for C, and DD/Dd/dd for D, while all offspring are BB. Thus, total genotypes = 3 × 1 × 3 × 3 = 27.​

Phenotype Analysis

Phenotypes depend on dominance: 2 possibilities per heterozygous gene (dominant or recessive) across A, C, D, with B always dominant. Total phenotypes = 2 × 1 × 2 × 2 = 8, matching the 27:9:9:9:3:3:3:1 trihybrid ratio collapsed for fixed B.​

Option Breakdown

• (A) 8 and 4: Incorrect; confuses gametes (8) with offspring and undercounts phenotypes (treats as dihybrid).​

• (B) 12 and 4: Incorrect; no basis for 12 genotypes (possibly misreading two heterozygous genes).​

• (C) 27 and 8: Correct, as calculated for three heterozygous loci.​

• (D) 14 and 8: Incorrect; 14 lacks genetic justification (not 3ⁿ or 2ⁿ).​

Introduction to Genotypes and Phenotypes in Multi-Gene Crosses

In genetics, determining the number of genotypes and phenotypes from crosses like AaBBCcDd × AaBBCcDd relies on Mendel’s laws of segregation and independent assortment. This trihybrid-style cross (three heterozygous loci) is common in CSIR NET Life Sciences exams, testing calculation of genotypic (3ⁿ) and phenotypic (2ⁿ) diversity where n = heterozygous pairs.​

Step-by-Step Solution for AaBBCcDd Cross

Identify heterozygous loci: Aa (A), Cc (C), Dd (D) → n = 3; BB fixed.​
Gametes per parent: 2³ = 8 (ABCd, ABCD, etc.).​
Genotypes: Each heterozygous locus yields 3 outcomes (e.g., AA/Aa/aa), so 3³ × 1 = 27 total.​
Phenotypes: Each gives 2 outcomes (dominant/recessive), so 2³ × 1 = 8 total.​

Why CSIR NET Students Must Master This

For competitive exams, recognize fixed loci reduce complexity from full tetrahybrid (81 genotypes, 16 phenotypes) to effective trihybrid. Practice confirms option (C).​

Common Mistakes in Such Questions

• Ignoring homozygous BB (treats as 4 loci → 81/16).

• Gamete count confusion (8 gametes ≠ 8 genotypes).

• Forgetting simple dominance (phenotypes ≠ genotypes).​