Glycolysis and fermentation for the MCAT

MCAT content library / Biology & biochemistry / Glycolysis and fermentation

Follow carbon, ATP investment, and NAD⁺ recycling to explain pathway output and redox bottlenecks.

About 17 minutes, including the worked examples and checks.

Useful foundations

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Keep three accounts: carbon, ATP, and electron carriers
Fermentation regenerates the carrier that glycolysis needs
Regulation changes flux without changing the stoichiometric account
Check your reasoning
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Keep three accounts: carbon, ATP, and electron carriers

Glycolysis converts one six-carbon glucose into two three-carbon pyruvates in the cytosol. Its carbon pathway, ATP balance, and NAD⁺ requirement answer different questions. Keeping those accounts separate lets you predict why a pathway slows after an intervention instead of treating every product increase as evidence of greater ATP production. These are central AAMC topics within glycolysis, fermentation, feeder pathways, and metabolic regulation.¹

The investment is paid once; the payoff runs twice

Starting from free glucose, hexokinase uses one ATP to produce glucose 6-phosphate. Isomerization yields fructose 6-phosphate, and phosphofructokinase-1 uses a second ATP to make fructose 1,6-bisphosphate. Cleavage and rearrangement produce two glyceraldehyde 3-phosphate molecules. Every later reaction therefore occurs twice per original glucose. The six carbons remain in the two pyruvates; glycolysis itself releases no CO₂.²

Glyceraldehyde 3-phosphate dehydrogenase oxidizes each three-carbon substrate while reducing NAD⁺ to NADH and incorporating inorganic phosphate. Phosphoglycerate kinase and pyruvate kinase subsequently transfer phosphate from intermediates to ADP. Those two ATP-producing events occur for each of the two three-carbon molecules: four ATP made minus two invested equals two net ATP per glucose. This is substrate-level phosphorylation, which does not require a membrane proton gradient.³

Account	Input or investment	Output or return	Net result
Carbon skeletons	One 6-carbon glucose	Two 3-carbon pyruvates	Six carbons retained
ATP	Two ATP consumed	Four ATP formed	Two ATP gained
NAD carrier	Two NAD⁺ reduced	Two NADH formed	Two NADH gained before regeneration

Accounting per one glucose completed through glycolysis

The NADH row is not another two ATP. NADH carries reducing equivalents; whether and how those electrons support additional ATP production depends on downstream processes. Likewise, “four ATP produced” describes the gross payoff, whereas “two ATP gained” includes the investment. If a question changes the entry substrate, rebuild the account from the entry point rather than copying the free-glucose total.

A feeder pathway can bypass an investment

Glycogen phosphorylase uses inorganic phosphate to release glucose 1-phosphate from appropriate α(1→4)-linked residues. Phosphoglucomutase converts that product to glucose 6-phosphate. This route reaches glycolysis after the hexokinase reaction, so only the phosphofructokinase ATP investment remains. With all subsequent reactions completed, the net is four minus one, or three ATP per such glycogen-derived unit.⁴

This accounting starts with stored glycogen; it does not include the earlier energy cost of making the polymer. It also does not assign three ATP to every residue under every breakdown route. A branch-point glucose released by hydrolysis is free glucose and needs phosphorylation. Dietary starch is digested to absorbable sugars; free glucose entering a cell follows the two-ATP investment account. Specify the actual substrate, not just the food or storage polymer it came from.⁴⁵

Fermentation regenerates the carrier that glycolysis needs

The dehydrogenase step needs oxidized NAD⁺. A finite NAD pool cannot support continuing glycolysis if its molecules become NADH and never return. Lactate dehydrogenase can transfer the reducing equivalents from NADH to pyruvate: pyruvate + NADH + H⁺ → lactate + NAD⁺. That conversion regenerates NAD⁺ without making an additional ATP molecule or removing a carbon as CO₂.⁶

For an original batch of 120 μmol free glucose converted entirely to lactate, glycolysis forms 240 μmol pyruvate, 240 μmol NADH, and 240 μmol net ATP. Converting all the pyruvate to lactate consumes those 240 μmol NADH and regenerates the same amount of NAD⁺. The combined process therefore produces 240 μmol lactate and net ATP, with no net accumulation of NADH. The carrier molecules are recycled; the starting NAD pool need not contain 240 μmol if it turns over repeatedly.

In ethanol fermentation, pyruvate first loses CO₂ to produce acetaldehyde, which is then reduced to ethanol while NAD⁺ is regenerated. The product carbon count therefore differs from lactate fermentation. Neither conversion should be confused with oxidative phosphorylation, and an organism using an electron transport chain with a nonoxygen acceptor is performing anaerobic respiration rather than this fermentation mechanism.⁶

Worked experiment: remove the NAD⁺ regeneration route

The following data are an original illustrative experiment in an anoxic cell-free preparation. Lactate dehydrogenase is selectively blocked at time zero. No alternative NADH-oxidizing route or pyruvate-consuming reaction is present. Total NAD⁺ plus NADH remains 50 μmol, and the enzymes retain normal activity in separate assays with their required substrates supplied. The flux column is an instantaneous tracer measurement, not a cumulative amount.

Time after block (min)	NADH/NAD⁺	Pyruvate pool (μmol)	Glucose-equivalent glycolytic flux (μmol/min)
0	0.25	5	10
1	1.00	20	5
4	4.00	35	1

Illustrative response after the selective block

At time zero, a ratio of 0.25 means NADH occupies 0.25/(1 + 0.25) = 20% of the pool: 10 μmol NADH and 40 μmol NAD⁺. At four minutes, a ratio of 4 means 80% NADH: 40 μmol NADH and only 10 μmol NAD⁺. The total carrier has not disappeared; its oxidized fraction has become less available. Pyruvate accumulates while measured flux falls, consistent with interruption of recycling.

Do not call the four-minute ratio a fourfold increase in NADH concentration. The ratio rose sixteenfold, from 0.25 to 4, while NADH amount rose fourfold, from 10 to 40 μmol. The supplied fixed total makes that conversion possible. Without the total, a ratio alone would not determine either absolute amount.

The strongest mechanistic follow-up would restore NADH oxidation through an independent, supplied reaction and test whether flux recovers while the lactate enzyme remains blocked. More glucose cannot replace oxidized NAD⁺ at the dehydrogenase step. A finite addition of NAD⁺ may delay the bottleneck, but sustained flux requires a recycling route. The three snapshots also do not justify integrating a straight line between flux points; the time course between measurements was not supplied.

Regulation changes flux without changing the stoichiometric account

Hexokinase, phosphofructokinase-1, and pyruvate kinase are major regulatory sites. For the usual regulated phosphofructokinase, ATP can act both as a required substrate and, at a distinct regulatory site, an inhibitory signal of a high-energy state. ADP/AMP and other regulators can favor flux; regulation varies with enzyme form and cellular context. More ATP therefore does not always mean a faster ATP-consuming step.⁷

Feedback can change molecules processed per minute while the complete-pathway ATP yield per specified input stays the same. Finally, increased lactate output alone does not establish an increase in oxidative ATP generation. To make that claim, measure the relevant respiratory pathway and its coupling. Lactate, ATP concentration, and ATP production rate are different observables, even when they change together.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

A preparation converts 50 μmol of glycogen-derived glucose 6-phosphate completely to lactate. Each input bypasses hexokinase, consumes one ATP at phosphofructokinase, and produces four ATP in the payoff phase. No carbon enters another pathway. What are the net ATP gain and lactate yield?

A. 100 μmol ATP; 100 μmol lactate

Review this choice. This ATP result subtracts two investments as though the input were free glucose. The stated entry point requires only one.

B. 150 μmol ATP; 50 μmol lactate

Review this choice. The ATP calculation is correct, but each six-carbon input produces two three-carbon lactates, not one.

C. 150 μmol ATP; 100 μmol lactate

Correct. Net ATP is 50 × (4 − 1) = 150 μmol. Carbon balance gives 50 × 2 = 100 μmol lactate.

D. 200 μmol ATP; 100 μmol lactate

Review this choice. Two hundred micromoles is the gross payoff. Subtract the 50 μmol ATP invested at phosphofructokinase.

The principle to carry forward

Choose the starting substrate, subtract only the investments it encounters, and conserve carbon independently.

Check 2

An anoxic preparation has a fixed NAD pool and no functional lactate dehydrogenase or other NADH-oxidizing route. NADH accumulates and glycolysis slows. Glucose, ADP, phosphate, and the other required substrates are already sufficient. Which addition most directly supports sustained recovery of glycolysis while the lactate block remains?

A. An ATP-regenerating system that does not react with NADH

Review this choice. This addresses ATP availability but does not regenerate the NAD⁺ required at the blocked redox bottleneck.

B. More glucose with no change to the NAD pool

Review this choice. Glucose is already sufficient. Adding more does not oxidize the accumulated NADH.

C. More NADH with no change to enzyme activity

Review this choice. Adding the reduced carrier does not supply oxidized NAD⁺ and can further increase the reduced fraction.

D. An enzyme plus excess oxidized Q that converts NADH and Q to NAD⁺ and reduced Q

Correct. The supplied reaction regenerates NAD⁺ independently of lactate dehydrogenase. Excess oxidized Q provides the stated electron-accepting capacity; no oxygen is required by this hypothetical route.

The principle to carry forward

A rescue should restore the missing biochemical function, not merely add more material upstream.

Check 3

In matched populations with equal viable-cell counts, steady lactate release rises from 12 to 24 μmol/min. There is no lactate import or other lactate source or sink, and intracellular lactate does not accumulate. Lactate dehydrogenase consumes one NADH per lactate formed. Which conclusion is supported without measuring respiration?

A. ATP production by oxidative phosphorylation has doubled.

Review this choice. Lactate formation is not a measurement of oxidative phosphorylation. Respiratory ATP production could change differently.

B. NAD⁺ regeneration through lactate dehydrogenase has doubled.

Correct. Under the supplied steady-state balance, release equals formation. One regenerated NAD⁺ per lactate gives an increase from 12 to 24 μmol/min through that reaction.

C. NADH production within glycolysis has stopped.

Review this choice. Continued lactate formation consumes NADH; it does not demonstrate that glycolysis stopped making it.

D. Total cellular ATP production has doubled.

Review this choice. The result constrains the lactate reaction but does not specify all ATP-producing or ATP-consuming processes in the cells.

The principle to carry forward

Use the supplied stoichiometry for the measured reaction; do not extend its rate to an unmeasured pathway.