ATP and NADPH in Reduction Phase — NEET Biology
✅ Asked in NEET 2022When one CO₂ molecule is fixed as one molecule of triose phosphate, which of the following photochemically made, high-energy chemical intermediates are used in the reduction phase? (NEET 2022 Phase 2)
When one CO₂ molecule is fixed as one molecule of triose phosphate, which of the following photochemically made, high-energy chemical intermediates are used in the reduction phase? (NEET 2022 Phase 2)
Answer & NCERT explanation
Correct answer: A — 2 ATP + 2 NADPH
For fixing one CO₂ molecule and producing one triose phosphate molecule, the reduction phase of Calvin cycle requires 2 ATP molecules (for converting 3-PGA to triose phosphate) and 2 NADPH molecules (as reducing power). This is the stoichiometric requirement per CO₂ fixed.
Read more NCERT concept on the PYQ
Reduction – These are a series of reactions that lead to the formation of glucose. The steps involve utilisation of 2 molecules of ATP for phosphorylation and two of NADPH for reduction per CO2 molecule fixed. The fixation of six molecules of CO2 and 6 turns of the cycle are required for the formation of one molecule of glucose from the pathway.
The reduction phase of the Calvin cycle requires 2 ATP molecules and 2 NADPH molecules for every single CO2 molecule that is fixed. To produce one glucose molecule, 6 CO2 molecules must be fixed, requiring a total of 12 ATP and 12 NADPH. Students often confuse the per-CO2 requirement with the total requirement, or mix up the exact numbers (3:2 ratio). The key is remembering that reduction uses these energy carriers to convert 3-PG into G3P. Since NEET repeatedly tests the stoichiometry of photosynthesis, this ATP-NADPH ratio remains crucial for both MCQs and numerical problems about Calvin cycle efficiency.
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