The correct genetic pathway is: gene A suppresses gene B, and gene B promotes flowering transition (Option B).

Understanding the data in the question

From the table:

• Wild type: 30 days to flower; A and B transcripts normal.

• a mutant: 15 days (earlier flowering); A transcript nil, B transcript increased.

• b mutant: 60 days (late flowering); A transcript normal, B transcript nil.

• ab double mutant: 60 days (late flowering); A and B transcripts nil.

Key deductions:

1. Loss of B (b mutant and ab double mutant) always causes late flowering (60 days), showing that B promotes flowering; when B is absent, the plant cannot flower on time.

2. In a mutant, A is absent but B is increased, and flowering is earlier (15 days), so when A is lost, B goes up and flowering is strongly promoted.

3. Therefore, gene A normally suppresses B; when A is removed, B is overexpressed and flowering is accelerated.

This logic corresponds exactly to option (B) A gene suppresses B, which promotes flowering transition.

Why option B is correct

• In wild type, A keeps B at a normal level, resulting in intermediate flowering time (30 days).

• When A is mutated (a mutant), suppression on B is relieved, B transcript becomes high, and flowering becomes very early (15 days).

• When B is mutated (b mutant), flowering is late (60 days) even though A is normal, so B must be a positive regulator of flowering.

• In the ab double mutant, B is still nil, so flowering remains late (60 days); the presence or absence of A does not matter without B, placing B downstream of A in the pathway.

Thus, the linear pathway that fits all observations is:

                                                                                      A  ⊣  B  →  Flowering transition

where ⊣ means suppresses and → means promotes.

Why the other options are wrong

Option (A): “A gene activates B, which suppresses flowering transition”

• If A activated B and B suppressed flowering, then:

  • In an a mutant, B should be low (no activation), so flowering should be earlier, which partially matches early flowering but contradicts the observation that B is increased in the a mutant.

  • In b mutant, loss of B would remove a supposed suppressor, so flowering should be earlier, but the table shows late flowering (60 days).

• Therefore, this model is inconsistent with both the transcript pattern and the flowering times.

Option (C): “B gene activates A, which suppresses flowering transition”

• If B activated A, and A suppressed flowering, then:

  • In b mutant, A should be nil (no activation), and flowering should be earlier because the suppressor A is gone; instead, the b mutant flowers late.

  • In a mutant, losing A should not necessarily increase B, but the table shows increased B.

• Late flowering in b mutant directly contradicts the idea that A is a suppressor downstream of B.

Option (D): “B gene suppresses A, which promotes flowering transition”

• Here A is a promoter of flowering, B a suppressor of A.

  • In b mutant, loss of B would release A from suppression, so A activity (or transcript) should increase and flowering should be earlier, but the b mutant shows late flowering with normal A transcript.

  • In a mutant, loss of A would eliminate a promoter of flowering; flowering should be late, but it is actually earlier (15 days).

• Hence this pathway cannot explain any of the mutant phenotypes.

Introduction

Understanding the genetic pathway of flowering time with gene A and gene B is essential for mastering gene interaction and epistasis in plant developmental biology and for cracking competitive exams like CSIR NET. This classic problem uses flowering time (days to flower) and transcript levels in wild type, single mutants, and a double mutant to deduce the regulatory relationship between two genes controlling the flowering transition. By carefully analyzing phenotypes and mRNA expression, one can determine which gene promotes flowering, which gene suppresses it, and which option best represents the underlying pathway.