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MCAT Glycolysis, Gluconeogenesis and Fermentation Practice Questions

By StudyStore · Published 2026-10-20 · 8 original questions · about 12 min

Eight free MCAT-style practice questions on glycolysis yield and regulation, gluconeogenic bypass steps, fermentation, the Cori cycle, the pentose phosphate pathway and glucose handling, with explanations.

Quick concept primer

Glycolysis turns one glucose into two pyruvate, with a net gain of 2 ATP and 2 NADH. Three steps are essentially irreversible and are the regulated ones: hexokinase or glucokinase, phosphofructokinase-1 (PFK-1, the main control point) and pyruvate kinase. Gluconeogenesis is not simply glycolysis in reverse. It uses the same reversible steps but bypasses those three irreversible reactions with four enzymes: pyruvate carboxylase, PEP carboxykinase, fructose-1,6-bisphosphatase and glucose-6-phosphatase. The two pathways are regulated reciprocally so they do not run at full speed together.

Without oxygen (or without mitochondria), pyruvate is reduced to lactate or ethanol so that NAD+ is regenerated and glycolysis can continue. The pentose phosphate pathway is a side branch off glucose-6-phosphate that makes NADPH and ribose-5-phosphate without producing or consuming ATP.

Traps: fermentation does not make extra ATP, it only regenerates NAD+; and muscle has no glucose-6-phosphatase, so it cannot release free glucose to the blood.

Free notes for this topic: The Henderson-Hasselbalch Equation for the MCAT

8 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

What is the net yield of the payoff and investment phases of glycolysis for one molecule of glucose converted to pyruvate under anaerobic conditions?

  1. 2 pyruvate, 2 ATP, 2 NADH
  2. 2 pyruvate, 2 ATP, 0 NADH
  3. 2 pyruvate, 4 ATP, 2 NADH
  4. 1 pyruvate, 2 ATP, 1 NADH
Show answer and explanation

Answer: A. 2 pyruvate, 2 ATP, 2 NADH

Two ATP are spent in the investment phase (hexokinase and PFK-1) and four are made by substrate-level phosphorylation in the payoff phase, for a net of 2. Glyceraldehyde-3-phosphate dehydrogenase generates one NADH per triose, so two per glucose. Under anaerobic conditions the NADH is then reoxidized by lactate dehydrogenase, but it is still produced in glycolysis, so choice B is wrong.

Question 2

In liver, which change would most directly increase PFK-1 activity?

  1. A rise in glucagon-driven cAMP signaling
  2. A rise in citrate concentration
  3. A rise in fructose-2,6-bisphosphate
  4. A rise in ATP concentration
Show answer and explanation

Answer: C. A rise in fructose-2,6-bisphosphate

Fructose-2,6-bisphosphate is the most potent allosteric activator of PFK-1 in liver, and insulin raises its level. ATP and citrate inhibit PFK-1, signaling that energy and biosynthetic precursors are plentiful. Glucagon lowers fructose-2,6-bisphosphate through PKA-dependent phosphorylation of the bifunctional enzyme, which slows glycolysis and favors gluconeogenesis.

Question 3

All of the following enzymes are needed for gluconeogenesis from pyruvate EXCEPT

  1. fructose-1,6-bisphosphatase
  2. pyruvate kinase
  3. pyruvate carboxylase
  4. phosphoenolpyruvate carboxykinase
Show answer and explanation

Answer: B. pyruvate kinase

Pyruvate kinase catalyzes an irreversible glycolytic step, PEP to pyruvate. Gluconeogenesis bypasses it with pyruvate carboxylase (pyruvate to oxaloacetate) and PEP carboxykinase (oxaloacetate to PEP). Fructose-1,6-bisphosphatase bypasses PFK-1, and glucose-6-phosphatase bypasses hexokinase and glucokinase.

Question 4

A sprinter's muscle fibers convert pyruvate to lactate during a maximal effort. What is the primary metabolic purpose of this reaction?

  1. To produce NADPH for biosynthesis
  2. To reduce the amount of pyruvate carboxylated in gluconeogenesis
  3. To regenerate NAD+ so that glyceraldehyde-3-phosphate dehydrogenase can keep running
  4. To generate additional ATP beyond that made in glycolysis
Show answer and explanation

Answer: C. To regenerate NAD+ so that glyceraldehyde-3-phosphate dehydrogenase can keep running

Glycolysis consumes NAD+ at the glyceraldehyde-3-phosphate dehydrogenase step. When oxygen delivery or mitochondrial capacity cannot keep up, NADH accumulates and glycolysis would stall. Lactate dehydrogenase uses NADH to reduce pyruvate, restoring NAD+. The reaction itself yields no ATP, and the NADH it uses is the cytosolic NADH from glycolysis, not NADPH.

Question 5

Lactate released by working muscle is converted to glucose in the liver and returned to muscle (the Cori cycle). Using standard accounting, what is the net energy cost of one full turn of the cycle (2 lactate to 1 glucose in liver, 1 glucose to 2 lactate in muscle)?

  1. Net gain of 2 ATP
  2. Net cost of 8 ATP-equivalents
  3. No net cost, because lactate to glucose is a reversal
  4. Net cost equal to 4 ATP-equivalents (6 consumed in liver, 2 made in muscle)
Show answer and explanation

Answer: D. Net cost equal to 4 ATP-equivalents (6 consumed in liver, 2 made in muscle)

Gluconeogenesis from 2 pyruvate consumes 2 ATP (pyruvate carboxylase), 2 GTP (PEP carboxykinase) and 2 ATP (phosphoglycerate kinase running backward), a total of 6 high-energy phosphate bonds. Muscle glycolysis nets only 2 ATP. The 4 ATP-equivalent difference is paid by the liver, which is why the cycle shifts the metabolic burden of anaerobic work from muscle to liver.

Question 6

After eating fava beans, a patient with glucose-6-phosphate dehydrogenase deficiency develops hemolytic anemia. Which explanation is most accurate?

  1. The red cells cannot make ATP by glycolysis
  2. The red cells cannot make enough NADPH to keep glutathione reduced, so oxidative damage lyses them
  3. The red cells cannot make ribose for DNA synthesis
  4. The red cells overproduce lactate and acidify the blood
Show answer and explanation

Answer: B. The red cells cannot make enough NADPH to keep glutathione reduced, so oxidative damage lyses them

Glucose-6-phosphate dehydrogenase is the rate-limiting enzyme of the oxidative pentose phosphate pathway, the only NADPH source in mature red blood cells. NADPH keeps glutathione in its reduced form, which neutralizes peroxides. With the enzyme deficient, an oxidative stress such as certain foods or drugs overwhelms the defense and membrane damage causes hemolysis. Mature red cells have no nucleus, so ribose for DNA is not their issue, and glycolysis is unaffected.

Question 7

Insulin stimulates glucose uptake into skeletal muscle and adipose tissue mainly by

  1. activating hexokinase directly
  2. converting glucose to fructose, which crosses the membrane more easily
  3. increasing the number of GLUT4 transporters in the plasma membrane by vesicle translocation
  4. opening an ATP-gated glucose channel
Show answer and explanation

Answer: C. increasing the number of GLUT4 transporters in the plasma membrane by vesicle translocation

GLUT4 is stored in intracellular vesicles. Insulin signaling triggers fusion of those vesicles with the plasma membrane, so more transporters are available for facilitated diffusion of glucose. GLUT1 handles basal uptake in many tissues, and GLUT2 serves liver and pancreatic beta cells. Hexokinase acts after the glucose has entered the cell.

Question 8

After a meal the portal blood glucose concentration rises well above the fasting level, and the liver takes up and stores a large fraction of it. Which property of glucokinase supports this behavior?

  1. It has a lower Km than hexokinase, so it works best at low glucose
  2. It is strongly inhibited by its product, glucose-6-phosphate
  3. Its Km for glucose is much higher than that of hexokinase, so its rate keeps rising across the physiological range
  4. It is found only in muscle
Show answer and explanation

Answer: C. Its Km for glucose is much higher than that of hexokinase, so its rate keeps rising across the physiological range

Glucokinase, the liver (and beta cell) enzyme, has a high Km, around 10 mM, and is not inhibited by glucose-6-phosphate. It therefore tracks the blood glucose concentration and sets a high capacity for uptake when glucose is abundant. Hexokinase, with a low Km and product inhibition, saturates at low glucose and keeps working in tissues that need steady supply regardless of the fed state.

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.