Medium MCAT Glycolysis Practice Questions

Concept Explanation

Glycolysis is a metabolic pathway occurring in the cytosol that breaks down one molecule of glucose into two molecules of pyruvate, producing a net gain of 2 ATP and 2 NADH. This fundamental anaerobic process is essential for energy production in nearly all living organisms and serves as the primary entry point for carbohydrate metabolism. The pathway consists of ten enzymatic steps divided into two main phases: the energy investment phase and the energy payoff phase.

During the investment phase, two ATP molecules are consumed to phosphorylate glucose and its derivatives, effectively trapping the sugar in the cell and preparing it for cleavage. Key regulatory enzymes in this phase include hexokinase (or glucokinase in the liver) and phosphofructokinase-1 (PFK-1), which is the rate-limiting enzyme of glycolysis. PFK-1 is allosterically inhibited by ATP and citrate, while being activated by AMP and fructose-2,6-bisphosphate.

The payoff phase involves the oxidation of glyceraldehyde-3-phosphate, resulting in the reduction of $\text{NAD}^+$ to $\text{NADH}$ and the generation of 4 ATP molecules through substrate-level phosphorylation. Since 2 ATP were initially spent, the net yield is 2 ATP. In aerobic conditions, pyruvate enters the mitochondria for the citric acid cycle; however, in anaerobic conditions, pyruvate is reduced to lactate to regenerate $\text{NAD}^+$, allowing glycolysis to continue. Understanding these mechanisms is crucial for mastering general chemistry principles applied to biological systems.

Solved Examples

1. Problem: Calculate the net ATP yield of glycolysis if a cell starts with one molecule of glucose-6-phosphate instead of glucose.
   Solution:
   1. Identify the normal investment: Usually, 1 ATP is used by hexokinase to convert glucose to glucose-6-phosphate (G6P), and 1 ATP is used by PFK-1 to convert fructose-6-phosphate to fructose-1,6-bisphosphate.
   2. Adjust for the starting material: If the pathway begins at G6P, the hexokinase step is bypassed, meaning only 1 ATP is invested.
   3. Calculate the payoff: The payoff phase remains the same, producing 4 ATP.
   4. Final Calculation: $4 \text{ ATP (produced)} - 1 \text{ ATP (invested)} = 3 \text{ ATP net}$.
2. Problem: Explain how high levels of Citrate affect the rate of glycolysis.
   Solution:
   1. Identify the target enzyme: Citrate is an intermediate of the Krebs cycle and acts as an allosteric regulator of PFK-1.
   2. Determine the signal: High citrate indicates that the energy needs of the cell are being met and the citric acid cycle is saturated.
   3. Mechanism: Citrate binds to PFK-1, increasing the enzyme's $K_m$ for its substrate, thereby inhibiting its activity.
   4. Result: The glycolytic flux decreases.
3. Problem: A researcher inhibits Glyceraldehyde-3-phosphate dehydrogenase (GAPDH). What is the immediate effect on the production of ATP and NADH?
   Solution:
   1. Locate the enzyme: GAPDH catalyzes the conversion of glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate.
   2. Identify the products of this step: This step produces 2 NADH (one per triose sugar).
   3. Assess downstream effects: Since 1,3-bisphosphoglycerate is the substrate for the first substrate-level phosphorylation step (phosphoglycerate kinase), inhibiting GAPDH prevents the subsequent production of 4 ATP.
   4. Conclusion: Both ATP and NADH production from glycolysis will drop to zero (net -2 ATP due to the investment phase).

Practice Questions

1. Which of the following enzymes catalyzes an irreversible step in glycolysis and is inhibited by high levels of ATP?

1. Phosphoglycerate kinase
2. Phosphofructokinase-1
3. Glyceraldehyde-3-phosphate dehydrogenase
4. Enolase

2. In an anaerobic environment, the conversion of pyruvate to lactate is vital because it:

1. Produces an additional 2 ATP per glucose molecule.
2. Reduces $\text{NAD}^+$ to $\text{NADH}$ for the electron transport chain.
3. Oxidizes $\text{NADH}$ to $\text{NAD}^+$ to sustain the payoff phase.
4. Transports pyruvate into the mitochondria.

3. Arsenate is a chemical that can substitute for inorganic phosphate in the reaction catalyzed by Glyceraldehyde-3-phosphate dehydrogenase, resulting in a product that spontaneously hydrolyzes to 3-phosphoglycerate without producing ATP. How many net ATP are produced from one glucose molecule in the presence of arsenate?

1. -2
2. 0
3. 2
4. 4

4. Which of the following molecules is a potent allosteric activator of PFK-1 in the liver, even when ATP levels are high?

1. Glucose-6-phosphate
2. Fructose-2,6-bisphosphate
3. Citrate
4. Acetyl-CoA

5. If a patient has a deficiency in Pyruvate Kinase, which of the following would most likely be observed in their red blood cells?

1. Increased concentrations of ATP
2. Decreased concentrations of 2,3-bisphosphoglycerate (2,3-BPG)
3. Increased concentrations of Phosphoenolpyruvate (PEP)
4. Increased rates of aerobic respiration

6. Glucokinase differs from Hexokinase in that Glucokinase:

1. Has a lower $K_m$ for glucose.
2. Is found in most peripheral tissues.
3. Is inhibited by its product, glucose-6-phosphate.
4. Is induced by insulin in the liver.

7. During the conversion of Fructose-1,6-bisphosphate to Glyceraldehyde-3-phosphate and Dihydroxyacetone phosphate, which type of enzyme is utilized?

1. Isomerase
2. Lyase
3. Transferase
4. Hydrolase

8. What is the net reaction of glycolysis?

1. $\text{Glucose} + 2 \text{ADP} + 2 \text{P}_i + 2 \text{NAD}^+ \rightarrow 2 \text{Pyruvate} + 2 \text{ATP} + 2 \text{NADH} + 2 \text{H}^+ + 2 \text{H}_2 \text{O}$
2. $\text{Glucose} + 4 \text{ADP} + 4 \text{P}_i + 2 \text{NAD}^+ \rightarrow 2 \text{Pyruvate} + 4 \text{ATP} + 2 \text{NADH}$
3. $\text{Glucose} + 2 \text{ATP} + 2 \text{NAD}^+ \rightarrow 2 \text{Pyruvate} + 4 \text{ATP} + 2 \text{NADH}$
4. $\text{Glucose} + 2 \text{ADP} + 2 \text{P}_i + 2 \text{NADH} \rightarrow 2 \text{Lactate} + 2 \text{ATP} + 2 \text{NAD}^+$

Answers & Explanations

1. Answer: B. PFK-1 is one of the three irreversible enzymes in glycolysis (alongside hexokinase and pyruvate kinase). It is the primary control point and is inhibited by high ATP, which signals that the cell has sufficient energy. While pyruvate kinase is also inhibited by ATP, it is not listed in a way that makes it a better answer than the rate-limiting step.

2. Answer: C. In the absence of oxygen, the electron transport chain cannot oxidize NADH back to $\text{NAD}^+$. Without $\text{NAD}^+$, the GAPDH step of glycolysis stops. Lactate dehydrogenase reduces pyruvate to lactate while simultaneously oxidizing NADH to $\text{NAD}^+$, allowing glycolysis to continue. For more on electron movement, see medium MCAT redox practice questions.

3. Answer: B. Normally, glycolysis produces 4 ATP and uses 2 ATP (Net 2). Arsenate bypasses the phosphoglycerate kinase step where 2 ATP are normally produced. Therefore, the total ATP produced is only 2 (from the pyruvate kinase step). Since 2 ATP were invested at the start, the net yield is $2 - 2 = 0$.

4. Answer: B. Fructose-2,6-bisphosphate (F2,6BP) is produced by PFK-2 in response to insulin. It is the most potent activator of PFK-1 and can override the inhibitory effects of ATP, ensuring glycolysis continues in the liver when glucose levels are high.

5. Answer: C. Pyruvate kinase converts PEP to pyruvate. If this enzyme is deficient, PEP and all upstream intermediates (like 2,3-BPG and 3-phosphoglycerate) will accumulate. Red blood cells lack mitochondria, so they cannot perform aerobic respiration.

6. Answer: D. Glucokinase is the liver/pancreatic isoform. It has a high $K_m$ (low affinity), is not inhibited by G6P, and is upregulated by insulin. Hexokinase is found in peripheral tissues and has a low $K_m$.

7. Answer: B. Aldolase cleaves Fructose-1,6-bisphosphate into two three-carbon sugars. Enzymes that break bonds without using water or redox are classified as lyases. This is a common topic in reaction mechanism studies.

8. Answer: A. This represents the balanced stoichiometry of the pathway. Note the net gain of 2 ATP and 2 NADH. Option D describes fermentation, not the general glycolytic pathway.

1. Which enzyme is responsible for the "trapping" of glucose within the cell by converting it to glucose-6-phosphate?

• APhosphofructokinase-1
• BHexokinase
• CPyruvate Kinase
• DAldolase

Frequently Asked Questions

What is the rate-limiting enzyme of glycolysis?

Phosphofructokinase-1 (PFK-1) is the rate-limiting enzyme because it catalyzes the slowest and most highly regulated step of the pathway. It is controlled by various allosteric effectors like ATP, AMP, and Fructose-2,6-bisphosphate to match the cell's energy needs.

Why does glycolysis occur in the cytosol?

Glycolysis occurs in the cytosol because all the necessary enzymes are located there, and it allows the pathway to function independently of the mitochondria. This is essential for cells that lack mitochondria, such as mature red blood cells, which rely entirely on glycolysis for energy.

How many ATP are produced in the energy payoff phase?

The energy payoff phase produces a total of 4 ATP molecules per glucose molecule. Specifically, two ATP are produced by phosphoglycerate kinase and two are produced by pyruvate kinase as the two three-carbon molecules are processed.

What happens to pyruvate in the absence of oxygen?

In the absence of oxygen, pyruvate undergoes fermentation to become lactate (in humans) or ethanol (in yeast). This process does not produce extra ATP but is vital for regenerating $\text{NAD}^+$ from $\text{NADH}$, which is required to keep glycolysis running.

What are the three irreversible steps of glycolysis?

The three irreversible steps are catalyzed by hexokinase (or glucokinase), phosphofructokinase-1 (PFK-1), and pyruvate kinase. These steps have large negative Gibbs free energy changes and serve as the primary sites of metabolic regulation. For more details on energetic favorability, consult Nature Scitable's guide to metabolism.