Metabolism and bioenergetics sample question 1

Muscle glycogen cannot be broken down to glucose 1-phosphate, so little substrate enters glycolysis and less lactate forms.

Metabolism and bioenergetics sample question 2

To regenerate NAD⁺ so that glycolysis can continue

Metabolism and bioenergetics sample question 3

The enzyme does not change ΔG°′, so the reaction proceeds only if coupled to an exergonic reaction such as ATP hydrolysis

Metabolism and bioenergetics sample question 4

PFK-1 is activated and FBPase-1 is inhibited, so glycolysis is favored over gluconeogenesis.

Metabolism and bioenergetics sample question 5

It hydrolyzes ATP, because the catalytic sites lie in F₁, but cannot make ATP without proton flow through F₀ turning the stalk.

MCAT · Biological and Biochemical Foundations · Biochemistry

Metabolism and bioenergetics

Metabolism and bioenergetics is part of Biochemistry in the MCAT Biological and Biochemical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.

Question types you will see here include glycolysis regulation, aTP and reducing-equivalent accounting, electron transport inhibitors and uncouplers and fed-to-fasted metabolic switch.

Common traps: treating hexokinase as the main control in liver; reversing the effect of citrate or AMP on PFK-1; missing tissue differences (liver versus muscle).

Sample questions

Question 1
During ischemic forearm exercise, a patient with deficiency of muscle glycogen phosphorylase shows almost no rise in venous lactate, whereas a healthy person shows a large rise. Which explanation is correct?
- A.
  Muscle glycogen cannot be broken down to glucose 1-phosphate, so little substrate enters glycolysis and less lactate forms.
- B.
  Glycolysis is blocked at phosphofructokinase-1, so glycogen accumulates in the muscle and glucose is not used for lactate formation.
- C.
  Lactate is formed normally but is taken up by the liver faster in these patients.
- D.
  Pyruvate cannot be reduced to lactate, because lactate dehydrogenase requires no NADH in muscle.
Answer: A. Muscle glycogen cannot be broken down to glucose 1-phosphate, so little substrate enters glycolysis and less lactate forms.
Without phosphorylase, muscle glycogen cannot supply glucose 1-phosphate for glycolysis during ischemic exercise, so anaerobic glycolysis and lactate production are low and patients tire quickly (McArdle disease). A PFK-1 block (B) would be a different disorder, rapid hepatic uptake (C) is not the mechanism, and lactate dehydrogenase requires NADH (D).
Question 2
During intense exercise, skeletal muscle converts pyruvate to lactate. What is the main purpose of this reaction?
- A.
  To regenerate NAD⁺ so that glycolysis can continue
- B.
  To convert pyruvate into glucose within the muscle
- C.
  To produce additional ATP directly from pyruvate
- D.
  To make FADH₂ for the electron transport chain
Answer: A. To regenerate NAD⁺ so that glycolysis can continue
Lactate dehydrogenase oxidizes NADH to NAD⁺, which glyceraldehyde-3-phosphate dehydrogenase needs to keep glycolysis running when oxidative capacity is limited. The lactate dehydrogenase reaction itself makes no ATP; the benefit is the continued ATP production by glycolysis.
Question 3
An enzyme catalyzes a reaction with ΔG°′ = +15 kJ/mol. Which statement is correct?
- A.
  The enzyme makes ΔG°′ negative by lowering the activation energy, so no coupling to another exergonic reaction is needed
- B.
  The enzyme shifts the equilibrium toward products, so the reaction proceeds without any coupling in the cell
- C.
  The enzyme makes the reaction spontaneous by adding free energy to the reaction from its active site
- D.
  The enzyme does not change ΔG°′, so the reaction proceeds only if coupled to an exergonic reaction such as ATP hydrolysis
Answer: D. The enzyme does not change ΔG°′, so the reaction proceeds only if coupled to an exergonic reaction such as ATP hydrolysis
D: enzymes change rates and not thermodynamics, so an uphill reaction needs coupling. A confuses activation energy with ΔG. B is wrong because enzymes cannot alter K_eq. C is wrong because an enzyme supplies no free energy.
Question 4
After a meal, insulin raises fructose 2,6-bisphosphate in hepatocytes. What is the effect on phosphofructokinase-1 (PFK-1) and fructose 1,6-bisphosphatase (FBPase-1)?
- A.
  PFK-1 is inhibited and FBPase-1 is activated, so gluconeogenesis is favored.
- B.
  Both enzymes are activated, so the two pathways run together.
- C.
  Both enzymes are inhibited, so the pathways stop.
- D.
  PFK-1 is activated and FBPase-1 is inhibited, so glycolysis is favored over gluconeogenesis.
Answer: D. PFK-1 is activated and FBPase-1 is inhibited, so glycolysis is favored over gluconeogenesis.
D is correct: fructose 2,6-bisphosphate is a potent activator of PFK-1 and an inhibitor of FBPase-1, so it directs flux toward glycolysis in the fed state. A describes the effect of glucagon, which lowers it. B would create a futile cycle. C would stall both pathways, which fructose 2,6-bisphosphate does not do.
Question 5
The F₁ head of ATP synthase can be separated from the membrane-embedded F₀ portion. Which statement about isolated F₁ is correct?
- A.
  It hydrolyzes ATP, because the catalytic sites lie in F₁, but cannot make ATP without proton flow through F₀ turning the stalk.
- B.
  It makes ATP from ADP and Pi, because proton flow is needed only to release ATP and not to form it at the catalytic sites in the first place.
- C.
  It does nothing, because F₁ binds neither ATP nor ADP unless F₀ is attached.
- D.
  It moves protons across the membrane, because the proton channel lies in F₁.
Answer: A. It hydrolyzes ATP, because the catalytic sites lie in F₁, but cannot make ATP without proton flow through F₀ turning the stalk.
Catalytic sites are in the F₁ head; in the cell, protons flowing through the F₀ c-ring turn the central stalk, and this rotation changes the conformations of the F₁ sites to release newly made ATP. Without F₀, F₁ runs in reverse and hydrolyzes ATP. B and C are wrong about F₁'s role, and D wrongly assigns the proton channel to F₁ (it is F₀).

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