Enzymes and General Principles of Metabolism: Catalysis, Kinetics, Inhibition, Regulation, Coenzymes and ATP

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

1. Which statement correctly describes the effect of an enzyme on the reaction it catalyses?

   1. It makes ΔG°′ more negative, so that the equilibrium lies further towards the products
   2. It raises the equilibrium constant by binding the product more tightly than the substrate
   3. It lowers ΔG‡, speeding the forward and reverse reactions equally without changing K′eq
   4. It supplies free energy to the substrate so that an unfavourable reaction proceeds by itself
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   Answer: C — It lowers ΔG‡, speeding the forward and reverse reactions equally without changing K′eq

   A catalyst lowers the activation free energy by stabilising the transition state; because the same transition state lies on the forward and reverse paths, both rates rise by the same factor and K′eq (fixed by ΔG°′) is unchanged. An enzyme cannot change ΔG°′ or supply free energy; making an unfavourable reaction go requires coupling to a favourable one, such as ATP hydrolysis, which is a different matter from catalysis.

2. An enzyme that uses pyridoxal phosphate catalyses L-glutamate + pyruvate ⇌ α-ketoglutarate + L-alanine. To which EC class does it belong?

   1. Oxidoreductases, because glutamate is converted into an α-keto acid in the reaction
   2. Transferases, because an amino group is moved from one molecule to another
   3. Ligases, because a new carbon–nitrogen bond is formed in the alanine product
   4. Lyases, because a group is removed from glutamate without the use of water
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   Answer: B — Transferases, because an amino group is moved from one molecule to another

   This is an aminotransferase (transaminase), EC class 2: the amino group passes from glutamate to the enzyme's pyridoxal phosphate and then to pyruvate, with no external electron acceptor such as NAD⁺. The oxidoreductase answer is tempting because glutamate becomes a keto acid, but the electrons stay within the paired substrates. Ligases need ATP, and lyases do not transfer a group to a second substrate.

3. Glycolysis and gluconeogenesis share most of their enzymes but use different enzymes at three steps. What is the main reason these three glycolytic steps are not simply run in reverse?

   1. The three steps occur in different cell compartments, so a single enzyme could never reach both of its substrates
   2. Gluconeogenesis uses NADPH rather than NADH, so it needs enzymes with a different coenzyme site
   3. The glycolytic enzymes are degraded during fasting and must be replaced by different proteins
   4. Those steps have large negative ΔG in cells, so separate irreversible reactions are needed and allow separate control
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   Answer: D — Those steps have large negative ΔG in cells, so separate irreversible reactions are needed and allow separate control

   Hexokinase, phosphofructokinase-1 and pyruvate kinase operate far from equilibrium with large negative ΔG, so their reverse is strongly unfavourable; gluconeogenesis bypasses them with different, also exergonic reactions (pyruvate carboxylase with PEP carboxykinase, fructose 1,6-bisphosphatase, glucose 6-phosphatase). Having distinct enzymes lets the two pathways be regulated reciprocally and avoids futile cycling. The bypass is not explained by NADPH, which gluconeogenesis does not use, nor by enzyme degradation.

4. An enzyme obeys Michaelis–Menten kinetics. With the same enzyme concentration, the initial rate is 30 µmol min⁻¹ at [S] = 2.0 mM and 50 µmol min⁻¹ at [S] = 6.0 mM. What is Vmax, in µmol min⁻¹?

   Numerical answer — type the value.

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   Answer: 75

   Vmax = v₀(Km + [S])/[S] must be the same at both points: 30(Km + 2)/2 = 50(Km + 6)/6, so 15Km + 30 = 8.333Km + 50, giving Km = 3.0 mM. Then Vmax = 15 × (3 + 2) = 75 µmol min⁻¹. Check: 75 × 6/9 = 50. Assuming the higher rate is already Vmax (50) is the usual slip.

5. A Lineweaver–Burk plot (1/v₀ in min µM⁻¹ against 1/[S] in µM⁻¹) for an enzyme is a straight line with slope 0.040 min and y-intercept 0.0125 min µM⁻¹. What is Km in µM? Give the answer to one decimal place.

   Numerical answer — type the value.

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   Answer: 3.2

   y-intercept = 1/Vmax, so Vmax = 1/0.0125 = 80 µM min⁻¹. Slope = Km/Vmax, so Km = 0.040 × 80 = 3.2 µM. Equivalently, the x-intercept is −0.0125/0.040 = −0.3125 µM⁻¹ = −1/Km, giving Km = 3.2 µM.

6. A purified enzyme at a total active-site concentration of 10 nM gives Vmax = 5.0 µM s⁻¹ and Km = 25 µM. What is its specificity constant kcat/Km, expressed in units of 10⁶ M⁻¹ s⁻¹?

   Numerical answer — type the value.

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   Answer: 20

   kcat = Vmax/[E]T = (5.0 × 10⁻⁶ M s⁻¹)/(10 × 10⁻⁹ M) = 500 s⁻¹. kcat/Km = 500/(25 × 10⁻⁶ M) = 2.0 × 10⁷ M⁻¹ s⁻¹ = 20 × 10⁶ M⁻¹ s⁻¹, about a tenth of the diffusion limit. Forgetting to convert µM and nM to M gives answers off by powers of ten.

7. For an enzyme catalysing A + B → P + Q, initial rates are measured against [A] at several fixed concentrations of B. Which of the following statements are correct?

   1. In a ping-pong mechanism, plots of 1/v₀ against 1/[A] at the different fixed [B] are parallel lines
   2. Lines in such plots that intersect to the left of the 1/v₀ axis show that no ternary complex EAB forms
   3. Aminotransferases are the standard example of an ordered sequential mechanism with a ternary complex
   4. In a ping-pong mechanism the first product leaves before the second substrate binds, leaving a substituted enzyme
   Show answer

   Answer: A — In a ping-pong mechanism, plots of 1/v₀ against 1/[A] at the different fixed [B] are parallel lines; D — In a ping-pong mechanism the first product leaves before the second substrate binds, leaving a substituted enzyme

   In ping-pong kinetics [B] changes only the intercept of the 1/v₀ against 1/[A] plot, so the lines are parallel, and the mechanism passes through a substituted enzyme after the first product leaves. Intersecting lines are the signature of sequential mechanisms, which do form a ternary complex, so that statement is reversed. Aminotransferases are the textbook ping-pong case, carrying the amino group as pyridoxamine phosphate.

8. In the presence of a drug, the Lineweaver–Burk plot of an enzyme is a line parallel to the control line but shifted upward. What kind of inhibitor is the drug, and how does it change the kinetic constants?

   1. Uncompetitive; it binds only the ES complex and lowers apparent Vmax and Km by the same factor
   2. Competitive; it binds free enzyme at the active site and raises Km while Vmax stays unchanged
   3. Pure noncompetitive; it binds E and ES equally and lowers Vmax while Km stays unchanged
   4. Mixed; it binds E more tightly than ES, so it raises apparent Km and lowers apparent Vmax
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   Answer: A — Uncompetitive; it binds only the ES complex and lowers apparent Vmax and Km by the same factor

   An uncompetitive inhibitor divides both Vmax and Km by α′, so the slope Km/Vmax is unchanged and only the intercept 1/Vmax rises: parallel lines. Competitive inhibition changes the slope but not the y-intercept (lines meet on the 1/v₀ axis); pure noncompetitive inhibition changes the slope and y-intercept but not the x-intercept (lines meet on the 1/[S] axis); mixed inhibition gives lines that cross left of the 1/v₀ axis.

9. At 8.0 µM, a competitive inhibitor raises the apparent Km of an enzyme from 1.5 mM to 4.5 mM and leaves Vmax unchanged. What is the Ki of the inhibitor, in µM?

   Numerical answer — type the value.

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   Answer: 4

   For competitive inhibition, apparent Km = αKm with α = 1 + [I]/Ki. α = 4.5/1.5 = 3, so [I]/Ki = 2 and Ki = 8.0/2 = 4 µM. Taking Ki = [I]/α (2.7 µM) is the common slip.

10. Compound Z lowers the Vmax of an enzyme and leaves its Km unchanged. Which single experiment would best show whether Z is a reversible pure noncompetitive inhibitor or an irreversible inactivator?

    1. Measure initial rates at much higher substrate concentrations to see whether the inhibition is overcome
    2. Dialyse or gel-filter the enzyme–Z mixture and assay again; activity returns only if Z binds reversibly
    3. Draw a Lineweaver–Burk plot and check whether the lines with and without Z meet on the 1/[S] axis
    4. Measure Km at several concentrations of Z and check whether it stays constant as Z increases
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    Answer: B — Dialyse or gel-filter the enzyme–Z mixture and assay again; activity returns only if Z binds reversibly

    Both kinds of inhibitor remove a fraction of the enzyme from action without changing the affinity of the rest, so both lower Vmax, leave Km unchanged, are not overcome by substrate and give lines meeting on the 1/[S] axis. Only removal of free Z separates them: a reversible inhibitor dissociates during dialysis or gel filtration and activity returns, whereas a covalently bound inactivator stays attached.

11. Which of the following statements about drugs that act by inhibiting enzymes are correct?

    1. Statins are uncompetitive inhibitors that bind only the enzyme–substrate complex of HMG-CoA reductase
    2. Aspirin irreversibly inactivates cyclooxygenase by acetylating a serine residue
    3. Allopurinol is converted by xanthine oxidase into oxypurinol, which stays tightly bound and inactivates the enzyme
    4. Methotrexate stops DNA synthesis by covalently alkylating the active site of thymidylate synthase
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    Answer: B — Aspirin irreversibly inactivates cyclooxygenase by acetylating a serine residue; C — Allopurinol is converted by xanthine oxidase into oxypurinol, which stays tightly bound and inactivates the enzyme

    Aspirin transfers its acetyl group to a serine of cyclooxygenase, an irreversible covalent change, and allopurinol is a mechanism-based (suicide) inactivator because xanthine oxidase itself makes the tightly bound oxypurinol. Statins resemble HMG-CoA and compete with it at the active site; they are competitive, not uncompetitive. Methotrexate is a folate analogue that binds dihydrofolate reductase extremely tightly and non-covalently; thymidylate synthase is affected only indirectly, through loss of tetrahydrofolate.

12. Aspartate transcarbamoylase of E. coli shows sigmoidal kinetics, is inhibited by CTP and is activated by ATP. If its catalytic trimers are separated from the regulatory chains and assayed alone, what is observed?

    1. No activity at all, because each active site is built at the interface between a catalytic and a regulatory chain
    2. Sigmoidal kinetics that is inhibited even more strongly by CTP, because the trimers are freed from restraint
    3. Catalytic activity with hyperbolic kinetics that is no longer affected by either CTP or ATP
    4. Sigmoidal kinetics unaffected by ATP but still inhibited by CTP, which binds the catalytic sites
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    Answer: C — Catalytic activity with hyperbolic kinetics that is no longer affected by either CTP or ATP

    The active sites lie within the catalytic trimers (between adjacent catalytic chains), so isolated trimers are active. CTP and ATP bind the regulatory chains, so without them there is no feedback control, and the cooperative T⇌R transition of the whole c₆r₆ assembly is lost, giving hyperbolic kinetics. This experiment showed that allosteric and catalytic sites are on different subunits; the claim that CTP acts at the catalytic site is therefore wrong.

13. In the liver of a fasting person, glucagon raises cAMP and activates protein kinase A. Which consequence helps switch the liver from glycolysis to gluconeogenesis?

    1. PKA phosphorylates the bifunctional PFK-2/FBPase-2, lowering fructose 2,6-bisphosphate
    2. PKA phosphorylates and activates glycogen synthase, so glucose 6-phosphate is stored as glycogen
    3. PKA moves GLUT4 transporters into the hepatocyte membrane so glucose leaves the liver faster
    4. PKA raises fructose 2,6-bisphosphate, which inhibits PFK-1 and diverts fructose 6-phosphate
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    Answer: A — PKA phosphorylates the bifunctional PFK-2/FBPase-2, lowering fructose 2,6-bisphosphate

    Phosphorylation of the liver bifunctional enzyme inhibits its kinase (PFK-2) and activates its phosphatase (FBPase-2), so fructose 2,6-bisphosphate falls. Since this compound activates PFK-1 and inhibits fructose 1,6-bisphosphatase, its fall slows glycolysis and speeds gluconeogenesis. Fructose 2,6-bisphosphate activates rather than inhibits PFK-1, so raising it would do the opposite; phosphorylation inactivates glycogen synthase; GLUT4 is an insulin-regulated transporter of muscle and fat, not liver.

14. Which of the following vitamin – coenzyme – reaction pairings are correct?

    1. Thiamine – thiamine pyrophosphate – carrying one-carbon units for thymidylate synthesis
    2. Vitamin K – reduced vitamin K – hydroxylation of proline residues in collagen
    3. Biotin – enzyme-bound biotin – ATP-dependent carboxylation of pyruvate to oxaloacetate
    4. Pyridoxine – pyridoxal phosphate – transamination of amino acids
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    Answer: C — Biotin – enzyme-bound biotin – ATP-dependent carboxylation of pyruvate to oxaloacetate; D — Pyridoxine – pyridoxal phosphate – transamination of amino acids

    Pyruvate carboxylase carries CO₂ on biotin, and aminotransferases use pyridoxal phosphate. One-carbon units for thymidylate are carried by tetrahydrofolate (from folate), not by thiamine pyrophosphate, which serves α-keto acid decarboxylation and transketolase. Prolyl hydroxylation in collagen needs ascorbate (vitamin C); vitamin K serves γ-carboxylation of glutamate in clotting factors.

15. A man with chronic alcoholism presents with confusion, an unsteady gait and weakness of eye movements. His blood pyruvate and lactate are raised and his erythrocyte transketolase activity is low. Deficiency of which coenzyme best explains these findings?

    1. FAD, which is needed by succinate dehydrogenase and by acyl-CoA dehydrogenase
    2. Pyridoxal phosphate, which is needed for transamination and decarboxylation
    3. Tetrahydrofolate, which carries one-carbon units for purine and thymidylate synthesis
    4. Thiamine pyrophosphate, which is needed by pyruvate dehydrogenase and transketolase
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    Answer: D — Thiamine pyrophosphate, which is needed by pyruvate dehydrogenase and transketolase

    The picture is Wernicke's encephalopathy from thiamine deficiency, common in alcoholism. Without thiamine pyrophosphate, pyruvate dehydrogenase fails, so pyruvate accumulates and is reduced to lactate, and transketolase activity (the standard test of thiamine status) falls; the brain, which relies on glucose oxidation, is hit first. Pyridoxal phosphate deficiency would not lower transketolase or block pyruvate oxidation, and folate deficiency presents as megaloblastic anaemia.

16. In a cell at 310 K, [ATP] = 3.0 mM, [ADP] = 0.30 mM and [Pi] = 3.0 mM. Take ΔG°′ for ATP + H₂O → ADP + Pi as −30.5 kJ mol⁻¹ and R = 8.314 J mol⁻¹ K⁻¹, and ignore activity coefficients and Mg²⁺ binding. What is the actual ΔG of ATP hydrolysis in this cell, in kJ mol⁻¹? Give the answer with its sign, to one decimal place.

    Numerical answer — type the value.

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    Answer: -51.4

    Mass-action ratio = [ADP][Pi]/[ATP] = (0.30 × 10⁻³)(3.0 × 10⁻³)/(3.0 × 10⁻³) = 3.0 × 10⁻⁴ (concentrations in M); ln(3.0 × 10⁻⁴) = −8.112. RT = 8.314 × 310 = 2577 J mol⁻¹ = 2.577 kJ mol⁻¹. RT ln Q = −20.91 kJ mol⁻¹. ΔG = −30.5 − 20.91 = −51.4 kJ mol⁻¹. Writing the ratio upside down gives −9.6, a sign that the products and reactants were swapped.

17. Standard free energies of hydrolysis (ΔG°′, kJ mol⁻¹) are: phosphoenolpyruvate −61.9, phosphocreatine −43.0, ATP to ADP + Pi −30.5, glucose 6-phosphate −13.8 and glycerol 3-phosphate −9.2. Which phosphoryl transfer is favourable under standard conditions?

    1. Glucose 6-phosphate + ADP → glucose + ATP
    2. Phosphoenolpyruvate + ADP → pyruvate + ATP
    3. Creatine + ATP → phosphocreatine + ADP
    4. Glycerol 3-phosphate + ADP → glycerol + ATP
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    Answer: B — Phosphoenolpyruvate + ADP → pyruvate + ATP

    For donor-P + ADP → donor + ATP, ΔG°′ = ΔG°′(donor hydrolysis) + 30.5. PEP: −61.9 + 30.5 = −31.4 kJ mol⁻¹, favourable (the pyruvate kinase step). Glucose 6-phosphate: +16.7; glycerol 3-phosphate: +21.3. Making phosphocreatine from ATP: −30.5 + 43.0 = +12.5 kJ mol⁻¹, unfavourable under standard conditions; it proceeds in resting muscle only because of the cellular concentrations.

18. In working muscle, [ATP] falls by only about 10% while [AMP] rises several-fold, and AMP rather than ATP acts as the main signal that activates phosphofructokinase-1, glycogen phosphorylase b and AMP-activated protein kinase. Why is AMP the more sensitive signal?

    1. Myosin ATPase splits ATP directly into AMP and pyrophosphate during contraction, so AMP reports work done
    2. AMP is the only adenine nucleotide that is able to cross the plasma membrane, so it reports the energy state of the whole body
    3. Adenylate kinase keeps 2 ADP ⇌ ATP + AMP near equilibrium, so with [ATP] ≫ [AMP] a small fall in ATP is a large rise in AMP
    4. AMP binds every ATP site on these enzymes with far higher affinity, so a few AMP molecules displace all the ATP
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    Answer: C — Adenylate kinase keeps 2 ADP ⇌ ATP + AMP near equilibrium, so with [ATP] ≫ [AMP] a small fall in ATP is a large rise in AMP

    Adenylate kinase equilibrates the adenylates, so [AMP] ≈ K[ADP]²/[ATP]. When ATP is used, ADP rises and AMP rises roughly with the square of ADP; because AMP starts at a small concentration, its fractional change is many times that of ATP. This amplification makes AMP the alarm signal. Myosin produces ADP + Pi, not AMP; AMP's action is intracellular, not a blood-borne report; and AMP acts at its own allosteric sites rather than displacing ATP from every site.