Question
Trace the complete fate of one glucose molecule through aerobic respiration: glycolysis, the Krebs cycle (TCA cycle), and the Electron Transport Chain (ETC). What are the total ATP, NADH, FADH₂, and CO₂ produced?
Solution — Step by Step
Location: Cytoplasm (cytosol)
Input: 1 glucose (6C) + 2 ATP (investment) + 2 NAD⁺ + 4 ADP + 4 Pi
Process:
- Glucose (6C) is phosphorylated twice (consuming 2 ATP) → Fructose-1,6-bisphosphate
- Split into 2 molecules of Glyceraldehyde-3-phosphate (G3P, 3C each)
- Each G3P is oxidised and converted to pyruvate (3C), producing 2 ATP and 1 NADH per G3P
Output per glucose:
- 2 Pyruvate (3C each)
- Net 2 ATP (4 produced – 2 invested)
- 2 NADH
- 0 CO₂ released
Glycolysis does not require oxygen — it works both aerobically and anaerobically.
Each pyruvate (3C) enters the mitochondria and is converted to Acetyl-CoA (2C) by the pyruvate dehydrogenase complex.
For each pyruvate:
- Pyruvate (3C) → Acetyl-CoA (2C) + 1 CO₂ + 1 NADH
For 2 pyruvates (from 1 glucose):
- 2 Acetyl-CoA
- 2 CO₂
- 2 NADH
Location: Mitochondrial matrix
Each Acetyl-CoA (2C) joins with Oxaloacetate (4C) to form Citrate (6C), then goes through a series of oxidation steps.
Per one turn of the Krebs cycle (one Acetyl-CoA):
- 2 CO₂ released
- 3 NADH
- 1 FADH₂
- 1 GTP (= 1 ATP)
For 2 Acetyl-CoA (from 1 glucose — 2 turns of the cycle):
- 4 CO₂
- 6 NADH
- 2 FADH₂
- 2 GTP (= 2 ATP)
Location: Inner mitochondrial membrane (cristae)
NADH and FADH₂ produced in the earlier stages donate electrons to protein complexes (Complex I, II, III, IV) embedded in the inner mitochondrial membrane.
Process:
- Electrons flow from NADH/FADH₂ through the chain to the final electron acceptor: oxygen (O₂)
- As electrons flow, they pump H⁺ ions from the matrix to the intermembrane space at Complexes I, III, IV
- The H⁺ gradient drives ATP synthase (Complex V) as H⁺ flows back into the matrix — this is chemiosmosis
- O₂ accepts the final electrons + H⁺ to form water (H₂O)
ATP yield from NADH and FADH₂ (theoretical, using P/O ratios):
- Each NADH → ~2.5 ATP
- Each FADH₂ → ~1.5 ATP
Total electrons entering ETC (from 1 glucose):
- NADH: 2 (glycolysis) + 2 (pyruvate oxidation) + 6 (Krebs) = 10 NADH
- FADH₂: 2 FADH₂ (from Krebs)
| Stage | ATP | NADH | FADH₂ | CO₂ |
|---|---|---|---|---|
| Glycolysis | 2 (net) | 2 | 0 | 0 |
| Pyruvate oxidation | 0 | 2 | 0 | 2 |
| Krebs cycle (×2) | 2 | 6 | 2 | 4 |
| Subtotal (substrate-level) | 4 ATP | 10 NADH | 2 FADH₂ | 6 CO₂ |
ETC and oxidative phosphorylation:
- 10 NADH × 2.5 = 25 ATP
- 2 FADH₂ × 1.5 = 3 ATP
- ETC total = 28 ATP
Grand total:
- Substrate-level phosphorylation: 4 ATP
- Oxidative phosphorylation: 28 ATP
- Total ≈ 30–32 ATP per glucose (theoretical maximum ~36–38 ATP by older estimates, revised to ~30–32 by modern P/O ratio measurements)
CO₂ released: 6 (accounts for all 6 carbon atoms of glucose)
Why This Works
The entire process is designed to extract energy efficiently from the C–H bonds of glucose. Glycolysis partially oxidises glucose; the Krebs cycle completes the oxidation; the ETC harvests the energy stored in NADH/FADH₂ as a proton gradient and converts it to ATP.
Oxygen’s role is only in the final step — as the terminal electron acceptor. Without oxygen, the ETC cannot proceed, NADH/FADH₂ cannot be reoxidised, and the cell falls back on fermentation (anaerobic, producing only 2 ATP from glycolysis).
Alternative Method
For quick counting: . All 6 carbons go to CO₂ (confirmed), and 6 O₂ molecules are consumed in the ETC.
Common Mistake
Students often write “38 ATP per glucose” without qualification. The older textbook value of 36–38 ATP is based on P/O ratio of 3 for NADH and 2 for FADH₂. Modern biochemistry uses 2.5 and 1.5 respectively, giving ~30–32 ATP. For NEET and CBSE, both values may be accepted, but mention that the 38-ATP figure is a theoretical maximum based on older estimates.