Level 4 — ApplicationEnzymes & Bioenergetics Basics

Enzymes & Bioenergetics Basics

⏱ 60 minutes50 marksprintable — key stays hidden on paper

Level 4 Examination — Application

Time limit: 60 minutes Total marks: 50 Instructions: Answer all questions. Show reasoning for all applied/data problems. Use $...$ notation for any energy values where required.

Question 1 (10 marks)

A pharmaceutical researcher studies an enzyme Y that converts substrate S into product P. Two candidate drugs are tested. When kinetic data are plotted, they observe:

Drug A: raises the apparent value of the substrate concentration needed to reach half the maximum rate, but the maximum reaction rate ( $V_{max}$ ) stays the same.
Drug B: lowers the maximum reaction rate, and no amount of extra substrate restores the original maximum.

(a) Identify the type of inhibition caused by Drug A and Drug B, justifying each from the data. (4)

(b) For each drug, state precisely where on the enzyme it most likely binds and explain the molecular consequence. (4)

(c) The researcher wants a drug whose effect can be overcome inside a cell that occasionally floods with high substrate. Which drug should they choose and why? (2)

Question 2 (12 marks)

A metabolic pathway proceeds: $A \rightarrow B \rightarrow C \rightarrow D$ . The standard free-energy changes are:

$A \rightarrow B$ : $\Delta G = +15\ \text{kJ/mol}$
$B \rightarrow C$ : $\Delta G = -30\ \text{kJ/mol}$
$C \rightarrow D$ : $\Delta G = -5\ \text{kJ/mol}$

(a) Classify each individual step as exergonic or endergonic. (3)

(b) Calculate the overall $\Delta G$ for $A \rightarrow D$ and state whether the whole pathway is spontaneous. (3)

(c) Step $A \rightarrow B$ does not occur on its own in the cell. Explain, using coupling, how the cell drives this step, and calculate the net $\Delta G$ if this step is coupled to ATP hydrolysis ( $\Delta G = -30.5\ \text{kJ/mol}$ ). (3)

(d) Explain why a favourable (negative) overall $\Delta G$ does not guarantee the pathway proceeds at a measurable rate, and name what the enzyme addresses. (3)

Question 3 (10 marks)

An enzyme from a hot-spring bacterium and a human digestive enzyme are compared. Activity was measured across temperature and pH:

Enzyme	Optimal Temp	Optimal pH
Bacterial enzyme X	80 °C	7.0
Human enzyme Z	37 °C	2.0

(a) Human enzyme Z shows almost zero activity at pH 7. Suggest where in the human body enzyme Z functions, and explain the pH result in terms of protein structure. (3)

(b) When bacterial enzyme X is placed at 95 °C its activity collapses. Explain the molecular event, and state why this differs from the reversible drop seen when the same enzyme is cooled to 10 °C. (4)

(c) A student claims "higher temperature always means faster enzyme reaction." Refute this using the concept of an optimum. (3)

Question 4 (10 marks)

The end-product E of a branched pathway inhibits the very first enzyme of the pathway by binding to a site other than the active site, changing the enzyme's shape.

(a) Name this regulatory phenomenon and the specific type of inhibition/regulation involved. (2)

(b) Explain the logic (the "why") of why cells evolved to have the end product regulate the first enzyme rather than a middle enzyme. (3)

(d) A researcher removes E entirely from the medium. Predict the effect on flux through the pathway and justify. (2)

Question 5 (8 marks)

A coenzyme-requiring dehydrogenase loses all activity when dialysed (small molecules removed), but regains full activity when NAD⁺ and a trace of $\text{Mg}^{2+}$ are added back.

(a) Distinguish between a cofactor and a coenzyme, classifying NAD⁺ and $\text{Mg}^{2+}$ correctly. (4)

(b) Explain why the dialysed enzyme (the protein alone) had no activity, using correct terminology (apoenzyme / holoenzyme). (2)

Answer keyMark scheme & solutions

Question 1 (10)

(a) (4)

Drug A = competitive inhibition (2): apparent substrate concentration for half-max rate increases (higher apparent $K_m$ ) but $V_{max}$ unchanged — hallmark of competition that can be out-competed by substrate. (1 identify + 1 justify)
Drug B = non-competitive inhibition (2): $V_{max}$ falls and cannot be restored by more substrate — inhibitor not competing for the active site. (1 identify + 1 justify)

(b) (4)

Drug A binds the active site itself, physically blocking substrate binding; because it resembles the substrate it competes for the same pocket. (2)
Drug B binds an allosteric / other site, changing enzyme conformation so the active site becomes less catalytically effective regardless of substrate present. (2)

(c) (2) Choose Drug A (1) — competitive inhibition is overcome by high substrate concentration, so its effect is reversible under substrate flooding, whereas Drug B's inhibition persists (1).

Question 2 (12)

(a) (3)

$A\rightarrow B$ : $+15$ → endergonic (1)
$B\rightarrow C$ : $-30$ → exergonic (1)
$C\rightarrow D$ : $-5$ → exergonic (1)

(b) (3) $\Delta G_{total} = +15 + (-30) + (-5) = -20\ \text{kJ/mol}$ (2). Negative ⇒ spontaneous overall (1).

(c) (3) The endergonic step is coupled to an exergonic reaction (ATP hydrolysis) so the combined free-energy change is negative (1, coupling logic). Net $\Delta G = +15 + (-30.5) = -15.5\ \text{kJ/mol}$ (2) — now spontaneous.

(d) (3) A negative $\Delta G$ shows a reaction is thermodynamically feasible but says nothing about rate/kinetics (1). Reactions still face an activation energy barrier (1). The enzyme lowers the activation energy, speeding the reaction without changing $\Delta G$ (1).

Question 3 (10)

(a) (3) Enzyme Z works in the stomach (1) where gastric acid gives pH ~2. At pH 7 the change in H⁺/OH⁺ disrupts ionic bonds/charges (1), altering the tertiary structure and active-site shape so substrate no longer binds (1). (e.g. pepsin)

(b) (4) At 95 °C excess thermal energy breaks the weak bonds (H-bonds, ionic, hydrophobic interactions) maintaining tertiary structure → denaturation, active site permanently deformed (2). This is irreversible (1). At 10 °C molecular motion/collision frequency simply falls, slowing activity, but the structure is intact so activity returns on warming — reversible (1).

(c) (3) False (1). Rising temperature raises rate only up to the optimum by increasing kinetic energy/collisions (1); beyond the optimum denaturation dominates and activity falls sharply, so "always faster" is wrong (1).

Question 4 (10)

(a) (2) Feedback (end-product) inhibition (1) achieved via allosteric (non-competitive) regulation (1).

(b) (3) Regulating the first enzyme prevents wasteful accumulation of intermediates (1); it saves resources/energy by shutting the whole pathway at the entry point rather than after intermediates are made (1); it gives efficient, responsive control of supply matching demand (1).

(c) (3) Feedback/allosteric inhibitor binds an allosteric site (not the active site) whereas a competitive inhibitor binds the active site (1+1). Feedback regulation is typically reversible and non-competitive (cannot be relieved by more substrate), while competitive inhibition is relieved by high substrate (1).

(d) (2) Removing E lifts the inhibition (1) → the first enzyme stays active → flux through the pathway increases (1).

Question 5 (8)

(a) (4) A cofactor is any non-protein helper (broad); an inorganic metal ion like $\text{Mg}^{2+}$ is a cofactor (1+1). A coenzyme is an organic cofactor (often vitamin-derived); NAD⁺ is a coenzyme (1+1).

(b) (2) The protein alone is the apoenzyme, which is inactive (1); only when combined with its cofactor/coenzyme does it form the active holoenzyme (1).

(c) (2) NAD⁺ is regenerated: it is reduced to NADH during the reaction and later re-oxidised back to NAD⁺ elsewhere (1), so it cycles repeatedly rather than being used up permanently like a substrate (1).

[
  {"claim":"Overall pathway dG A to D = -20 kJ/mol","code":"dG=15+(-30)+(-5); result=(dG==-20)"},
  {"claim":"Pathway spontaneous overall (dG<0)","code":"dG=15-30-5; result=(dG<0)"},
  {"claim":"Coupled step A to B net dG = -15.5 kJ/mol and spontaneous","code":"net=15+(-30.5); result=(abs(net-(-15.5))<1e-9 and net<0)"},
  {"claim":"Step A to B alone is endergonic (dG>0)","code":"result=(15>0)"}
]