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Quiz: Thermodynamics and Enzyme Kinetics

Test your understanding of free energy, ATP, redox reactions, and enzyme structure and regulation with these review questions.

1. What are the direct products of ATP hydrolysis at the terminal phosphoanhydride bond?

  1. AMP and pyrophosphate (PPi)
  2. ADP and glucose
  3. cAMP and phosphate
  4. ADP and inorganic phosphate (Pi)
Show Answer

The correct answer is D. Hydrolysis of ATP's terminal phosphoanhydride bond releases inorganic phosphate (Pi) and yields ADP, a reaction with a standard free energy change of approximately −30.5 kJ/mol. This large negative ΔG°' arises from relief of electrostatic repulsion among the phosphate groups and greater stability of the products, making ATP hydrolysis the cell's most common exergonic driver reaction.

Concept Tested: ATP structure and hydrolysis

2. Which statement correctly describes the second law of thermodynamics as it applies to biological systems?

  1. Energy is created during every metabolic reaction
  2. Every energy conversion increases the total entropy of the universe
  3. Living cells always decrease the entropy of the universe
  4. Energy conversions in cells are 100% efficient
Show Answer

The correct answer is B. The second law states that every energy conversion increases the total entropy of the universe, since no conversion is perfectly efficient and some energy is always dissipated as heat. Living organisms maintain their own internal order (low entropy) only by continuously releasing heat to their surroundings, so the universe's overall entropy still increases even as the organism itself stays highly ordered.

Concept Tested: Second law of thermodynamics and entropy

3. Why is a reaction with ΔG greater than zero considered non-spontaneous (endergonic)?

  1. The products have more free energy than the reactants, so the reaction requires an input of energy to proceed
  2. The reaction has reached equilibrium and no further change is possible
  3. The reaction releases free energy spontaneously without any energy input
  4. The enzyme catalyzing the reaction has been denatured
Show Answer

The correct answer is A. A positive ΔG means the products of the reaction contain more free energy than the reactants. Because reactions spontaneously move toward lower free energy, this reaction cannot proceed on its own and requires an input of free energy, typically supplied by coupling it to an exergonic reaction such as ATP hydrolysis.

Concept Tested: Free energy and reaction spontaneity

4. How do enzymes increase the rate of a chemical reaction?

  1. By increasing the free energy of the reactants
  2. By changing the equilibrium position of the reaction
  3. By lowering the activation energy required to reach the transition state, without changing the free energy difference between reactants and products
  4. By making an endergonic reaction release free energy
Show Answer

The correct answer is C. Enzymes accelerate reactions by stabilizing the transition state and lowering the activation energy barrier, allowing the reaction to reach equilibrium much faster. Critically, enzymes do not alter the thermodynamics of a reaction — ΔG, ΔH, ΔS, and the equilibrium position remain the same with or without the enzyme; only the kinetics change.

Concept Tested: Enzyme catalysis and activation energy

5. How do competitive and noncompetitive inhibitors differ in their effects on Vmax and Km?

  1. Competitive inhibitors decrease Vmax; noncompetitive inhibitors increase apparent Km
  2. Both inhibitor types decrease Vmax and increase apparent Km identically
  3. Competitive inhibitors bind irreversibly; noncompetitive inhibitors bind reversibly
  4. Competitive inhibitors increase apparent Km but leave Vmax unchanged; noncompetitive inhibitors decrease Vmax but leave Km unchanged
Show Answer

The correct answer is D. Competitive inhibitors bind the active site and can be outcompeted by high substrate concentration, so Vmax is unchanged but more substrate is needed to reach half-maximal velocity, raising apparent Km. Noncompetitive inhibitors bind an allosteric site and impair catalysis regardless of substrate binding, lowering Vmax while leaving the enzyme's substrate affinity (Km) unchanged.

Concept Tested: Competitive vs. noncompetitive inhibition

6. The phosphorylation of glucose to glucose-6-phosphate has a standard free energy change of +13.8 kJ/mol. When coupled to ATP hydrolysis (ΔG°' = −30.5 kJ/mol), what is the overall ΔG°' of the coupled reaction, and is it spontaneous?

  1. +44.3 kJ/mol; non-spontaneous
  2. −16.7 kJ/mol; spontaneous
  3. +16.7 kJ/mol; non-spontaneous
  4. −44.3 kJ/mol; spontaneous but unrelated to ATP
Show Answer

The correct answer is B. Adding the two ΔG°' values (+13.8 kJ/mol + (−30.5 kJ/mol)) gives an overall ΔG°' of −16.7 kJ/mol. Because the sum is negative, the coupled reaction is spontaneous overall, even though the phosphorylation step alone is endergonic. This is the general strategy cells use to drive unfavorable reactions: couple them to the strongly exergonic hydrolysis of ATP.

Concept Tested: Coupled reactions and free energy calculation

7. A drug binds to a site on an enzyme distinct from the active site and lowers the enzyme's Vmax without changing its Km. How should this drug be classified?

  1. Competitive inhibitor
  2. Allosteric activator
  3. Noncompetitive inhibitor
  4. Coenzyme
Show Answer

The correct answer is C. A noncompetitive inhibitor binds an allosteric site rather than the active site and reduces catalytic efficiency regardless of whether substrate is bound, which lowers Vmax while leaving the enzyme's binding affinity for substrate (Km) unchanged. A competitive inhibitor would instead occupy the active site and raise apparent Km, while an activator would increase rather than decrease activity.

Concept Tested: Allosteric inhibition classification

8. A human enzyme with an optimal temperature of 37°C is heated from 20°C to 37°C, and then further heated to 60°C. What happens to its activity across this range?

  1. Activity increases up to 37°C as collision frequency rises, then falls sharply at 60°C as heat disrupts the enzyme's noncovalent bonds and causes denaturation
  2. Activity increases continuously from 20°C to 60°C without limit
  3. Activity remains constant across this entire temperature range
  4. Activity decreases steadily as temperature rises from 20°C to 60°C
Show Answer

The correct answer is A. Below the optimal temperature, rising temperature increases molecular motion and collision frequency between enzyme and substrate, raising the reaction rate. Once temperature exceeds the optimum, however, heat disrupts the hydrogen bonds and hydrophobic interactions that maintain the enzyme's three-dimensional shape, denaturing the active site and causing activity to fall sharply, as would occur by 60°C for most human enzymes.

Concept Tested: Temperature effects on enzyme activity

9. In a pathway A → B → C → D, the end product D normally binds allosterically to enzyme E1 (which catalyzes A → B) and inhibits it. A mutation eliminates the allosteric site on E1 so that D can no longer bind. What is the most likely consequence for the cell?

  1. The pathway would shut down completely because E1 would be permanently inactive
  2. E1 would become a competitive inhibitor of its own substrate
  3. The pathway would become more efficient with no downside
  4. The pathway would continue converting A to B regardless of D concentration, potentially overproducing D and wasting cellular resources
Show Answer

The correct answer is D. Feedback inhibition normally allows the end product D to signal E1 to slow production once enough D has accumulated. Without a functional allosteric site, E1 can no longer sense D's concentration, so the pathway would continue running at an uncontrolled rate, leading to overproduction of D and inefficient use of the cell's precursor molecules and energy.

Concept Tested: Feedback inhibition

10. In the electron transport chain, electrons flow from NADH (E°' = −0.32 V) to O2 (E°' = +0.82 V). What does this direction of electron flow indicate?

  1. The reaction is endergonic and requires ATP input to proceed
  2. Electrons move spontaneously toward the molecule with higher (more positive) reduction potential, releasing free energy that can be used to pump protons
  3. NADH has a greater tendency to accept electrons than O2
  4. The reaction violates the first law of thermodynamics
Show Answer

The correct answer is B. Electrons spontaneously flow from carriers with lower (more negative) reduction potential to those with higher (more positive) reduction potential, since this direction releases free energy. The large potential difference between NADH and O2 (ΔE°' = +1.14 V) releases roughly 220 kJ/mol, energy the electron transport chain uses to pump protons across the inner mitochondrial membrane and power ATP synthase.

Concept Tested: Redox reactions and reduction potential