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📝 Activation energy and enzymes (15 MCQs)

📖 From Principles of Biochemistry • 1. The Foundations of Biochemistry • 15 questions available

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What is Activation energy and enzymes?

Definition:
Activation energy ( $E_a$ ) is the minimum energy required for a chemical reaction to occur, representing the energy barrier that reactants must overcome to form products, and enzymes lower this activation energy by providing an alternative reaction pathway that stabilizes the transition state, thereby increasing the reaction rate without altering the overall free energy change ( $\Delta G$ ), and this lowering of $E_a$ is the key to their catalytic efficiency.

Working:
Enzymes work by binding substrates and orienting them in a way that facilitates bond breaking and formation, and they stabilize the transition state through non-covalent interactions, reducing the energy input needed, and the relationship between rate constant ( $k$ ) and activation energy is given by the Arrhenius equation $k = A e^{-E_a/(RT)}$ , where lowering $E_a$ exponentially increases the rate; enzymes do not change the equilibrium constant but speed up the attainment of equilibrium, making them powerful biological catalysts.

Example:
A simple example is the breakdown of sucrose (table sugar) to glucose and fructose, which has a high $E_a$ without a catalyst, but the enzyme sucrase lowers the $E_a$ significantly, allowing the reaction to occur at body temperature, making sugar digestion fast enough to provide energy for metabolism.

Reason:
Understanding activation energy and how enzymes lower it is fundamental to biochemistry because it explains how reactions occur at physiological temperatures, and it underpins enzyme kinetics, drug design, and metabolic regulation, making it a key concept in life sciences.

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📝 All Activation energy and enzymes MCQs

Q1. A reaction has a large positive $\Delta G^\ddagger$ but a negative overall $\Delta G$ . Which conclusion best explains why the reaction can still be thermodynamically favorable yet proceed slowly?

A.The negative

\Delta G

makes the activation barrier disappear

B.The reaction is favorable overall, but reactant molecules still need sufficient energy to reach the transition state ✅

C.A negative

\Delta G

means the reaction must be fast

D.The enzyme must increase the overall

\Delta G

before the reaction can occur

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Overall $\Delta G$ indicates the thermodynamic driving force, whereas $\Delta G^\ddagger$ represents the barrier to reaching the transition state. A reaction can therefore be favorable but kinetically slow when its activation barrier is large.

Q2. Two reactions have identical reactants and products. Reaction X has a smaller $\Delta G^\ddagger$ than reaction Y. Under identical conditions, what is the most defensible prediction?

A.Reaction X generally proceeds faster because more molecules can reach the transition state per unit time ✅

B.Reaction X must have a more negative equilibrium constant

C.Reaction Y must release more free energy

D.Reaction X must produce different products

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A lower activation free energy increases the rate at which reactant molecules can cross the transition-state barrier. It does not necessarily change the reaction's equilibrium position or the identities of its products.

Q3. An enzyme accelerates a reaction by stabilizing the transition state. A student argues that this means the enzyme makes the products more stable than the reactants. What is the main flaw in the reasoning?

A.Transition-state stabilization primarily lowers the kinetic barrier rather than necessarily changing the relative free energies of reactants and products ✅

B.Enzymes always destabilize products

C.Product stability has no relationship to free energy

D.Transition states exist only after products form

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Enzymatic catalysis mainly lowers the activation barrier by preferentially stabilizing the transition state. The relative free energies of reactants and products determine overall thermodynamic favorability and are not necessarily altered by catalysis.

Q4. A laboratory compares an uncatalyzed reaction with an enzyme-catalyzed version. Both reach the same final equilibrium composition, but the enzyme-catalyzed reaction reaches equilibrium much sooner. Which explanation best fits the observation?

A.The enzyme changes the equilibrium constant

B.The enzyme lowers the activation barrier for the forward reaction without necessarily changing the equilibrium constant ✅

C.The enzyme increases the products' free energy

D.The enzyme makes the reverse reaction impossible

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: An enzyme can accelerate the approach to equilibrium by lowering activation barriers. Because it generally lowers the barriers for relevant forward and reverse processes, it does not need to change the equilibrium constant or final equilibrium composition.

Q5. A mutation causes an enzyme to bind the substrate extremely tightly but greatly decreases its catalytic rate. Which interpretation is most reasonable?

A.Strong substrate binding always guarantees rapid catalysis

B.The mutation may stabilize the substrate state more than the transition state, making transition-state formation less favorable ✅

C.The mutation must make

\Delta G

more positive

D.The mutation eliminates the need for activation energy

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Effective catalysis requires preferential stabilization of the transition state relative to the reactant state. Extremely strong substrate binding can sometimes trap the enzyme-substrate complex without providing the structural or energetic changes needed to reach the transition state efficiently.

Q6. An enzyme-catalyzed reaction is measured at two temperatures. The rate increases substantially with temperature before eventually decreasing at very high temperature. Which combined explanation is most appropriate?

A.Temperature initially increases molecular crossing of the activation barrier, while excessive heat can disrupt enzyme structure ✅

B.Temperature always decreases

\Delta G^\ddagger

and therefore rate must increase indefinitely

C.High temperature changes every reaction into an equilibrium-controlled process

D.The initial increase proves that the enzyme becomes permanently more stable

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Increasing temperature generally increases the fraction of molecules able to overcome activation barriers, raising reaction rate. At sufficiently high temperatures, however, protein structure may be disrupted, reducing catalytic activity.

Q7. A researcher compares four catalysts. Their measured activation free energies are 80, 65, 50, and 35 kJ/mol, respectively. Assuming all other conditions are comparable, which catalyst should produce the fastest reaction?

A.The catalyst with 80 kJ/mol

B.The catalyst with 65 kJ/mol

C.The catalyst with 50 kJ/mol

D.The catalyst with 35 kJ/mol ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: A smaller $\Delta G^\ddagger$ means a smaller energetic barrier to the transition state. Therefore, when other variables are comparable, the catalyst associated with 35 kJ/mol should give the greatest reaction rate.

Q8. A student observes that adding an enzyme increases product formation during the first five minutes and concludes that the enzyme has increased the reaction's thermodynamic favorability. Which additional observation would most strongly challenge that conclusion?

A.The uncatalyzed reaction eventually reaches the same equilibrium composition as the catalyzed reaction ✅

B.The enzyme has a specific active site

C.The enzyme is composed of amino acids

D.The reaction requires substrate molecules

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If catalyzed and uncatalyzed reactions reach the same equilibrium composition, the enzyme has changed the rate rather than the equilibrium position. This directly challenges the claim that catalysis altered thermodynamic favorability.

Q9. A reaction-coordinate graph shows two pathways connecting the same reactants and products. Pathway A has a peak much higher than pathway B, while both endpoints are identical. Which statement is best supported by the graph?

A.Pathway B has a lower activation barrier and should generally have a higher rate ✅

B.Pathway B must have a different equilibrium constant

C.Pathway A must produce more energy

D.The endpoints prove pathway A is more spontaneous

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The height of the transition-state peak relative to the reactants represents the activation barrier. Since pathway B has the lower peak and identical endpoints, it should generally permit faster conversion without changing the overall thermodynamic difference.

Q10. A graph of reaction progress shows an uncatalyzed curve peaking at 90 kJ/mol above the reactants and an enzyme-catalyzed curve peaking at 55 kJ/mol, with identical reactant and product levels. What does the graph most directly demonstrate?

A.The enzyme decreases the activation free energy by about 35 kJ/mol while leaving the overall free-energy change unchanged ✅

B.The enzyme increases product stability by 35 kJ/mol

C.The enzyme changes the equilibrium constant by 35 kJ/mol

D.The enzyme supplies 35 kJ/mol of energy to the products

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The activation barrier is reduced from 90 to 55 kJ/mol, a difference of 35 kJ/mol. Because the reactant and product energy levels are unchanged, the overall free-energy change remains the same.

Q11. A student says: 'If an enzyme lowers $\Delta G^\ddagger$ , it must make the reaction more exergonic.' Which reasoning error is present?

A.The student confuses kinetic control of reaction rate with thermodynamic control of reaction favorability ✅

B.The student assumes enzymes are consumed during reactions

C.The student confuses substrates with enzymes

D.The student assumes all reactions have zero activation energy

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Activation free energy controls how readily molecules reach the transition state and therefore strongly influences rate. Overall $\Delta G$ , not $\Delta G^\ddagger$ , determines thermodynamic favorability under specified conditions.

Q12. An inhibitor binds specifically to an enzyme's active region and prevents the substrate from adopting the geometry needed for the transition state. Which effect is most directly expected?

A.An increase in the effective activation barrier for the catalyzed reaction ✅

B.A guaranteed decrease in the reaction's equilibrium constant

C.A guaranteed increase in the products' free energy

D.Complete elimination of the reaction's overall

\Delta G

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: If an inhibitor prevents the enzyme-substrate complex from achieving transition-state geometry, fewer molecules can progress efficiently toward the transition state. The effective activation barrier therefore increases, reducing the reaction rate.

Q13. Two enzymes catalyze the same reaction. Enzyme A lowers $\Delta G^\ddagger$ by 20 kJ/mol, while enzyme B lowers it by 35 kJ/mol. If substrate concentration and temperature are identical, what is the strongest prediction?

A.Enzyme B should generally provide the greater rate enhancement because it creates the larger reduction in the activation barrier ✅

B.Enzyme A must produce more product at equilibrium

C.Enzyme B must make the reaction more thermodynamically favorable

D.Both enzymes must have identical rates because the products are identical

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A larger reduction in $\Delta G^\ddagger$ generally corresponds to a stronger catalytic rate enhancement, assuming comparable enzyme concentrations and mechanisms. Neither catalyst necessarily changes the equilibrium position or overall $\Delta G$ .

Q14. An enzyme follows one pathway in which the transition state is strongly stabilized and another pathway in which only the substrate is strongly stabilized. Which pathway is more likely to produce efficient catalysis, and why?

A.The transition-state-stabilizing pathway, because catalysis depends on preferentially lowering the barrier relative to the starting state ✅

B.The substrate-stabilizing pathway, because tighter binding always means faster chemistry

C.Both pathways must have identical rates because they use the same substrate

D.The substrate-stabilizing pathway, because activation barriers are independent of molecular interactions

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Catalysis is most effective when interactions preferentially stabilize the transition state relative to the reactant state. Stabilizing the substrate alone can lower its energy without proportionally lowering the barrier, potentially reducing catalytic efficiency.

Q15. A hypothetical catalyst lowers the activation barrier for the forward reaction but leaves the reverse barrier unchanged. Why would this observation require careful interpretation before concluding that the catalyst behaves like a typical enzyme?

A.Changing only one barrier could alter the relative forward and reverse rates and potentially affect the equilibrium relationship, unlike ordinary catalytic acceleration of both directions ✅

B.Any reduction in activation energy necessarily destroys the enzyme

C.The catalyst must be consumed during the reaction

D.A lower barrier automatically proves the reaction has become more endergonic

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For a catalyst that simply accelerates a reversible reaction without changing equilibrium, the kinetic barriers for the relevant directions must be affected consistently with the unchanged thermodynamic relationship. Lowering only one barrier would require additional mechanistic explanation.

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