Organic Chemistry · The Organic Chemistry of Metabolic Pathways
Catabolism of Carbohydrates: Glycolysis
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Glycolysis Ten-step cytosolic pathway splitting glucose into two pyruvate Full entry → is the ten-enzyme pathway that splits the six-carbon sugar glucose into two three-carbon molecules of Pyruvate Three-carbon end product of glycolysis Full entry →. It is the oldest and most universal energy-harvesting pathway in biology: nearly every organism runs a version of it, and it needs no oxygen. The name comes from the Greek glykys ("sweet") and lysis ("splitting").
The pathway has three jobs: convert glucose into an oxidizable form, harvest some of its stored energy as ATP and NADH Reduced form of NAD⁺ carrying two electrons and a proton Full entry →, and supply three-carbon building blocks for the citric acid cycle, fermentation, and biosynthesis. The net reaction is:
Glucose + 2 NAD+ + 2 ADP + 2 Pi → 2 Pyruvate + 2 NADH + 2 H+ + 2 ATP + 2 H2O
Read it carefully: two ATP are produced net, but the pathway actually spends two ATP early and makes four later. That two-part structure — an energy-investment phase followed by an energy-payoff phase — is the most useful organizing idea for the whole pathway.
Why this matters
- The gateway to glucose metabolism. Glycolysis is the only route by which glucose enters energy metabolism, and red blood cells (which lack mitochondria) depend on it completely.
- Clinical relevance. Enzyme defects cause inherited hemolytic anemias (e.g., pyruvate kinase deficiency); cancer cells rely heavily on aerobic glycolysis (the Warburg effect); lactate accumulation explains muscle fatigue after intense exercise.
- Connects to the rest of the chapter. Glycolysis feeds pyruvate to the pyruvate dehydrogenase complex (Topic 6) and the citric acid cycle (Topic 7), and gluconeogenesis (Topic 8) reverses it.
The college version
Core Concepts
The investment phase (steps 1–5): glucose → fructose-1,6-bisphosphate
The pathway begins by trapping glucose inside the cell and building a molecule that can be split. Step 1: hexokinase transfers a phosphate from ATP to glucose, giving glucose-6-phosphate — the negative charge traps the sugar in the cell. Step 2: phosphoglucose isomerase rearranges the ring to fructose-6-phosphate. Step 3: phosphofructokinase-1 (PFK-1) adds a second phosphate, giving fructose-1,6-bisphosphate (F-1,6-BP). This is the Committed step The irreversible reaction that commits a substrate to the pathway (PFK-1 step) Full entry → — the molecule now has nowhere to go but down the glycolytic path — and PFK-1 is the pathway's main regulatory enzyme.
Step 4: aldolase cleaves F-1,6-BP into two three-carbon isomers, dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P). Step 5: triose phosphate isomerase interconverts them. Because only G3P continues and the isomerase keeps the two in equilibrium, both fragments are metabolized — and from here on, every step happens twice per glucose.
The payoff phase (steps 6–10): energy harvest
Step 6: glyceraldehyde-3-phosphate dehydrogenase oxidizes G3P and adds a phosphate, forming 1,3-bisphosphoglycerate (1,3-BPG) while reducing NAD⁺ to NADH — the pathway's only oxidation step and its first "high-energy" intermediate. Step 7: phosphoglycerate kinase transfers 1,3-BPG's phosphate to ADP, making ATP — the first Substrate-level phosphorylation ATP made directly by transferring phosphate from a high-energy intermediate to ADP Full entry →. Step 8: phosphoglycerate mutase moves the phosphate to carbon 2. Step 9: enolase removes water to form phosphoenolpyruvate (PEP), the second high-energy intermediate. Step 10: pyruvate kinase transfers PEP's phosphate to ADP, making a second ATP and leaving pyruvate.
The ATP balance: 2 spent, 4 made, 2 net
Two ATP are invested in steps 1 and 3; four are made in steps 7 and 10 (two of each, because the payoff phase runs twice). Net: 2 ATP per glucose, plus 2 NADH. Anaerobically, NADH is recycled by lactate dehydrogenase (pyruvate → lactate); aerobically, its electrons go to the electron transport chain for extra ATP.
Regulation: PFK-1 is the master switch
PFK-1 is activated by AMP and Fructose-2,6-bisphosphate Regulatory signal made from F-6-P by a separate enzyme Full entry → (a signaling molecule made from F-6-P) and inhibited by ATP and citrate. High ATP means the cell is energetically comfortable, so the pathway slows; high AMP means energy is scarce, so it speeds up. Hexokinase is inhibited by its product glucose-6-phosphate, and pyruvate kinase by ATP and acetyl CoA. Glycolysis runs fast when energy is needed and slows when it is plentiful — no central command required.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Net 2 ATP | Total ATP made | 2 are invested early and 4 are made later; net = 4 − 2 = 2. Exam questions often ask for gross vs net. |
| Fructose-6-phosphate | Fructose-1,6-bisphosphate | F-6-P is one phosphate on carbon 6; F-1,6-BP has phosphates on carbons 1 and 6 — it is the product of the committed PFK-1 step. |
| PFK-1 | Hexokinase | PFK-1 is the main regulated enzyme (committed step); hexokinase is the first enzyme but is secondary in regulation. Both consume ATP. |
| Substrate-level phosphorylation | Oxidative phosphorylation | Glycolysis makes ATP directly from high-energy intermediates (steps 7, 10); oxidative phosphorylation uses NADH/FADH₂ and oxygen in mitochondria. |
| NADH produced in glycolysis | NAD⁺ consumed | Step 6 reduces NAD⁺ to NADH; the NAD⁺ must be regenerated (by lactate formation or electron transport) or glycolysis stalls. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine glucose is a large bill that is too big to spend. Glycolysis spends two small coins to break it into two medium bills (investment), then each medium bill becomes change that gives the cell two coins of energy twice (payoff). You end with two medium bills (pyruvate) plus two coins of profit.
Worked example
Example 1: ATP accounting with a substitution check
Problem. A muscle cell consumes 3.0 mmol of glucose via glycolysis under anaerobic conditions. How many mmol of ATP and NADH are produced net?
Step 1 — State the ratio from the balanced pathway. Per 1.0 mmol glucose, the net yield is 2.0 mmol ATP and 2.0 mmol NADH.
Step 2 — Set up the conversion with units.
n(ATP) = 3.0 mmol glucose × 2.0 mmol ATP1.0 mmol glucose = 6.0 mmol ATP
n(NADH) = 3.0 mmol glucose × 2.0 mmol NADH1.0 mmol glucose = 6.0 mmol NADH
Answer. 6.0 mmol ATP and 6.0 mmol NADH. The units cancel correctly (mmol glucose × mmol ATP/mmol glucose = mmol ATP), which confirms the setup.
Example 2: Tracing carbons through the pathway
Problem. Glucose is labeled with ¹⁴C at carbon 1. After one pass through glycolysis, where do the labeled atoms end up, and why?
Step 1 — Recall the split. Aldolase cleaves F-1,6-BP between carbons 3 and 4, giving DHAP (carbons 1–3) and G3P (carbons 4–6); carbon 1 is in DHAP.
Step 2 — Apply the isomerase. Triose phosphate isomerase converts DHAP to G3P, moving the label to the carbonyl carbon of that triose.
Step 3 — Trace to the product. Each G3P becomes a pyruvate; the labeled carbon becomes the carboxylate carbon, lost as CO₂ later in the citric acid cycle (not in glycolysis itself).
Answer. After one pass, the label is in one of the two pyruvate molecules, specifically its carboxyl carbon; the other pyruvate is unlabeled. Glycolysis itself releases no CO₂ — the two carbons that become CO₂ are lost in later steps (Topics 6–7).
Example 3: Predicting regulation from energy state
Problem. A cell suddenly has high ATP and high citrate. What happens to PFK-1 activity and to the rate of glycolysis, and what is the biological logic?
Step 1 — Apply the effector rules. ATP and citrate both inhibit PFK-1.
Step 2 — Consequence. PFK-1 activity falls and glycolytic flux drops.
Step 3 — The logic. High ATP signals a full energy charge; high citrate signals a well-supplied citric acid cycle. Slowing glycolysis prevents waste when the cell is already energetically comfortable.
Answer. PFK-1 is inhibited, glycolysis slows, and the cell spares glucose for times of need — classic negative feedback.
Key takeaways
- Glycolysis: glucose (6C) → 2 pyruvate (3C each), 10 enzyme steps, all in the cytosol, no O₂ required.
- Investment: 2 ATP (hexokinase, PFK-1). Payoff: 4 ATP (2 × phosphoglycerate kinase, 2 × pyruvate kinase). Net: 2 ATP + 2 NADH per glucose.
- PFK-1 catalyzes the committed step (F-6-P → F-1,6-BP) and is the main regulatory enzyme: activated by AMP and F-2,6-BP; inhibited by ATP and citrate.
- The only oxidation step is step 6 (G3P dehydrogenase), which makes 1,3-BPG and NADH.
- Substrate-level phosphorylation (direct ADP + high-energy phosphate → ATP) happens at steps 7 and 10 — no oxygen or electron transport needed.
- Anaerobic fate of pyruvate: lactate (animals, bacteria) or ethanol + CO₂ (yeast); aerobic fate: acetyl CoA (Topic 6).
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Write the net equation for glycolysis, including the correct stoichiometry of ATP, NADH, and pyruvate.
Show answer
Glucose + 2 NAD⁺ + 2 ADP + 2 Pᵢ → 2 pyruvate + 2 NADH + 2 H⁺ + 2 ATP + 2 H₂O.
Which step is the committed step, and which enzyme catalyzes it? Name one activator and one inhibitor.
Show answer
Fructose-6-phosphate → fructose-1,6-bisphosphate, catalyzed by phosphofructokinase-1 (PFK-1). Activated by AMP and fructose-2,6-bisphosphate; inhibited by ATP and citrate.
How many ATP are invested and how many are produced gross per glucose? Show the arithmetic.
Show answer
Invested: 2 ATP (hexokinase, PFK-1). Produced gross: 4 ATP (2 × phosphoglycerate kinase, 2 × pyruvate kinase). Net = 4 − 2 = 2 ATP.
Why can red blood cells survive on glycolysis alone?
Show answer
Red blood cells lack mitochondria, so they cannot run oxidative phosphorylation; glycolysis provides their ATP, and pyruvate is reduced to lactate to recycle NAD⁺.
A bacterial cell runs glycolysis anaerobically on 5.0 mmol of glucose. How many mmol of ATP are produced net?
Show answer
5.0 mmol glucose × 2.0 mmol ATP/mmol glucose = 10 mmol ATP net.
Where in the cell does glycolysis occur, and why does the product of step 1 trap glucose inside the cell?
Show answer
In the cytosol. Glucose-6-phosphate is negatively charged (phosphate group), so it cannot cross the hydrophobic cell membrane and is trapped for metabolism.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Glycolysis
- Ten-step cytosolic pathway splitting glucose into two pyruvate
- Substrate-level phosphorylation
- ATP made directly by transferring phosphate from a high-energy intermediate to ADP
- Committed step
- The irreversible reaction that commits a substrate to the pathway (PFK-1 step)
- NADH
- Reduced form of NAD⁺ carrying two electrons and a proton
- Fructose-2,6-bisphosphate
- Regulatory signal made from F-6-P by a separate enzyme
- Pyruvate
- Three-carbon end product of glycolysis
Sources & references
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