Complementary Gene Interaction and F2 Phenotypic Ratio Explained

[translate:Green, yellow and white phenotypes] in this cross follow a classic complementary gene interaction where both dominant alleles [translate:A and B] are required for full conversion to the green product. Plants with at least one dominant [translate:A] and one dominant [translate:B] (de>A_B_) are green, those with only de>A_ but de>bb are yellow, and those with de>aa (irrespective of B) remain white. The expected F2 phenotypic ratio after selfing de>AaBb is Green 9 : Yellow 3 : White 4, so the correct option is (1).

Question Recap and Pathway Logic

The pathway shown is:
[translate:X (white substrate) –A–> Y (yellow intermediate) –B–> Z (green product)].
Mutant alleles [translate:a and b] encode non‑functional enzymes, so:
de>A_ = functional enzyme A ([translate:X → Y] occurs, yellow made).
de>aa = no A activity ([translate:X cannot be converted, plant remains white regardless of B]).
de>B_ = functional enzyme B ([translate:Y → Z] occurs, green made), but only if Y is present.
de>bb = no B activity ([translate:Y accumulates as yellow, provided A_ is present]).

Thus:
de>A_B_ → green (both steps work).
de>A_bb → yellow (first step works, second blocked).
Any genotype with de>aa (de>aaB_ or de>aabb) → white (first step blocked, X stays white).

Dihybrid Self Cross of AaBb

Selfing de>AaBb involves a standard dihybrid cross with independent assortment. The F2 genotypes form a 16‑cell Punnett square in the usual 9:3:3:1 pattern.
de>A_ occurs in 3/4 of F2 (AA, Aa).
de>B_ occurs in 3/4 of F2 (BB, Bb).
Therefore de>A_B_ occurs in 3/4 × 3/4 = 9/16 of F2 → green.
de>A_bb occurs in 3/4 × 1/4 = 3/16 → yellow.
de>aaB_ occurs in 1/4 × 3/4 = 3/16 and de>aabb in 1/4 × 1/4 = 1/16; together 4/16 → white.

So phenotypic ratio:
Green = 9
Yellow = 3
White = 4

Correct Answer

Green (9) : White (4) : Yellow (3) = option (1).

Option-wise Explanation

• Option (1): Matches the complementary gene interaction where gene A is epistatic at the first step and gene B acts at the second step. 9 de>A_B_ green, 3 de>A_bb yellow, 3 de>aaB_ + 1 de>aabb white = 9:3:4, expressed as Green 9 : White 4 : Yellow 3, so this is correct.
• Option (2): Green (9) : Yellow (4) : White (3) would require 4 yellow and 3 white, but only one class (de>A_bb) is yellow (3), so this does not match the logic.
• Option (3): Green (9) : Yellow (6) : White (1) ratio usually arises from duplicate gene action, not from this linear pathway. Here, de>aaB_ cannot be yellow because without A, Y is not produced.
• Option (4): Green (9) : White (7) ratio corresponds to complementary genes producing a single phenotype, but the pathway clearly shows a distinct yellow intermediate phenotype, so combining yellow and white into one class is incorrect.

Introduction

In many CSIR NET and postgraduate genetics exams, questions on biochemical pathways controlled by two independently assorting genes test understanding of gene interaction and phenotypic ratios. Here, a white substrate (X) converts to a yellow intermediate (Y) and then to a green product (Z) by enzymes from genes A and B. Selfing an de>AaBb plant produces green, yellow, and white progeny with a characteristic ratio.

Summary of Pathway and Genotype-Phenotype Mapping

Gene A encodes the enzyme converting X (white) to Y (yellow); mutant de>a blocks this step.
Gene B encodes the enzyme converting Y (yellow) to Z (green); mutant de>b blocks this step.
Phenotypes:
de>A_B_ → green (both steps active)
de>A_bb → yellow (first step active, second blocked)
de>aaB_, de>aabb → white (first step blocked)

Cross Details and Phenotypic Ratios

Independent assortment of alleles in self-cross of de>AaBb produces:
9/16 green, 3/16 yellow, and 4/16 white progeny, consistent with complementary gene interaction and recessive epistasis.