How many ATP and NADPH molecules are required to make one molecule of glucose through the Calvin pathway ?

The Calvin‑Benson cycle fixes carbon dioxide into carbohydrate using three distinct phases: carbon fixation, reduction, and regeneration of ribulose‑1,5‑bisphosphate (RuBP).

1. **Carbon fixation** – Each CO₂ molecule combines with one RuBP (5‑carbon) to give two molecules of 3‑phosphoglycerate (3‑PGA). This step does not require energy.

2. **Reduction** – Every 3‑PGA is first phosphorylated by ATP to form 1,3‑bisphosphoglycerate, then reduced by NADPH to glyceraldehyde‑3‑phosphate (G3P).
$3‑PGA\; +\; ATP\; \to \; 1,3‑BPG\; +\; ADP$
$1,3‑BPG\; +\; NADPH\; \to \; G3P\; +\; NADP⁺\; +\; Pi$

3. **Regeneration of RuBP** – To keep the cycle going, most of the G3P produced is used to regenerate RuBP. This step also consumes ATP.

For each CO₂ fixed, the net energy cost is:

$3\; ATP\; +\; 2\; NADPH$

To synthesize one molecule of glucose (C₆H₁₂O₆) the cycle must fix six CO₂ molecules, because glucose contains six carbon atoms.

**Total ATP needed**:

$3\; ATP/CO₂\; \times \; 6\; CO₂\; =\; 18\; ATP$

**Total NADPH needed**:

$2\; NADPH/CO₂\; \times \; 6\; CO₂\; =\; 12\; NADPH$

Therefore, the Calvin pathway requires **18 ATP and 12 NADPH** to produce one molecule of glucose.