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How many turns of Calvin cycle are required for the formation of three molecules of glucose?
1 Glucose = 6 Calvin Turns; 3 Glucose = 3 × 6 = 18 Turns!
The Calvin cycle consists of three primary stages: Carboxylation (catalyzed by RuBisCO), Reduction, and Regeneration of RuBP. Each single cycle incorporates and fixes 1 molecule of inorganic carbon dioxide (CO₂).
Because each turn brings in only 1 carbon atom, 6 separate turns of the cycle are mandatory to fix 6 carbon atoms and yield 1 net molecule of glucose.
For 3 complete molecules of glucose, the number of required turns scales proportionately:
Option (1) is 6 (turns for 1 glucose); Option (2) is 3 (number of glucose molecules requested); Option (3) is 1; Option (4) is 18. The correct choice is Option (4).
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Incorrect. 6 turns produce only 1 molecule of glucose, not 3.
Incorrect. 3 turns fix only 3 CO₂ molecules, yielding half a triose phosphate (1/2 glucose).
Incorrect. 1 turn fixes only 1 CO₂.
100% mathematically and physiologically accurate: 3 × 6 = 18 turns.
A notorious NTA trick is embedding a multiplier ("three molecules") while students have memorized "6 turns for glucose" on reflex, causing them to hurriedly circle 6 (Option 1).
"Fixation of each CO₂ requires one cycle turn. Synthesis of one hexose requires 6 turns; synthesis of n hexoses requires 6n turns."
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