Enzymes

Question 1

1. Explain how enzymes speed up chemical reactions

Answer 1

🔑 Enzymes Bind the Transition State Best — Summary
Enzymes bind substrates, but their active sites are most complementary to the transition state.

The transition state is a high-energy, unstable intermediate where bonds are partially broken/forming.

Stabilizing the transition state lowers the activation energy, speeding up the reaction.

If enzymes bound the substrate too tightly in its stable form, it would slow the reaction.

Question 2

Explain relationship between enzymes and binding energy

Answer 2

What is binding energy in this context?
In enzyme catalysis, binding energy refers to:

The energy released when the enzyme binds to the substrate or more importantly, the transition state.

This energy is used to lower the activation energy and help stabilize the transition state.

✅ What enzymes actually do:
Enzymes use binding energy to their advantage.

In fact, the stronger the interactions with the transition state, the more the enzyme can lower the activation energy.

Question 3

2. Define the following terms: enzyme, active site, activation energy, transition state.

Answer 3

🧪 Key Definitions
🔹 Enzyme
A biological catalyst (usually a protein) that speeds up reactions by lowering the activation energy, without being consumed, and with high specificity for its substrate.
- Enzymes do not affect:
o The free energy change, delta G for the S <—->P reaction
- The maximum interactions between S and E occur when the free energy (binding free energy, released by the S-E interactions partly overcomes the energy needed to get to the “top of the hill”

🔹 Active Site
The region on the enzyme where the substrate binds and the reaction is catalyzed. Lined with functional groups (often amino acids) that interact specifically with the substrate.

🔹 Activation Energy (Ea)
The energy barrier that must be overcome for a chemical reaction to proceed.
The difference in free energy of S at the ground state and S at the transition state
📉 Enzymes lower this energy to accelerate the reaction.

🔹 Transition State
A high-energy, unstable intermediate between reactants and products.
⏳ Enzymes stabilize this state to lower the activation energy and speed up the reaction.

Question 4

3. Discuss with examples why enzymes are important in biology, medicine & industry.

Answer 4

• Living organisms must be able to catalyse the conversion of carbon fuel sources into cellular energy (ATP) in an appropriate timescale
• Some diseases are caused by excessive enzymatic activity, whilst others are caused by a deficiency in enzymatic activity; e.g. Phenylketonuria is caused by a deficiency in the enzyme Phenylalanine Hydroxylase
• Many drugs target enzymes either by inhibiting or activating the enzyme target, e.g. RELENZA inhibits the enzyme NEURAMINIDASE from the flu virus

Question 5

4. Describe using examples the importance of cofactors and coenzymes.

Answer 5

• Enzymes: cofactors and coenzymes
  Not all enzymes act alone
• Some enzymes require one or more additional chemical components for activity
• Such ‘chemical components’ can be:
  o Small inorganic molecules called ‘cofactors’, e.g. Mg2+, K+
  o More complex molecules called ‘coenzymes’ transiently carry functional groups during the catalysis of a reaction

Question 6

5. Describe different types of enzymes using examples

Answer 6

1. Kinases - add phosphate using ATP, ADP biproduct (hexokinase)
2. Phosphorylases - Catalyses the covalent addition of inorganic phosphate (Pi) to a molecule (glycogen phosphorylase)
3. Phosphatases - cleavage of phosphate to yield the dephosphorylated product and Pi (glucose 6 phosphatase)
4. Dehydrogenases - Catalyses an oxidation/reduction reaction commonly using NADH/NAD+, NADPH/NADP+ or FADH2/FAD as cofactors; e.g. Glyceraldehyde-3-phosphate dehydrogenase (transfer e) (Glyceraldehyde-3-phosphate dehydrogenase)
5. Mutases - Catalyses the shift of a phosphoryl group from one atom to another within the same molecule (Phosphoglycerate mutase)
6. Isomerases - Catalyses the conversion of one isomer to another (Triose Phosphate Isomerase)
7. Hydratases - Catalyses the addition/removal of water (enolase)
8. Synthases - Catalyses the synthesis of a product (citrate synthase)

Question 7

6. Explain the difference between a ‘reaction intermediate’ and a ‘transition state’

Answer 7

• Most reactions occur in multiple steps via species called “reaction intermediates” such as ES and EP in the example below
• Reaction intermediates lower the free energy (available to do work) of the transition state
• Intermediate are “stable” states - minima on a free energy plot
• Transition states are transient species (maxima on a free energy plot)

Question 8

7. Describe how enzyme concentration affects reaction rate

Answer 8

• Enzymes reduce the activation energy and accelerate rates of reaction by:
  o Binding substrates in the correct orientation relative to the active groups
  o Providing catalytically active groups (side chains, acids, bases, metal ions)
  o Polarising bonds, stabilising charged species (usually unstable)
  o Stabilising the transition state
  o Weak binding interactions between the enzyme and the substrate provide a substantial driving force for enzymatic catalysis
• Rate of reaction depends on (substrate)
  o The rate decreases with time due to a number of possible reasons (or a combination):
   (main reason) S is depleted by conversion to product
   The reaction is reversible, so as the product conc increases, the rate of the reverse reaction increases
   The enzyme may be unstable under the reaction conditions

Question 9

8. Define the following terms: binding energy, initial rate, Km, and Vmax.

Question 10

9. List the assumptions associated with the Michaelis-Menten steady-state description of enzyme kinetics

Answer 10

Assumption 1: ES conversion to E+P is irreversible

Assumption 2: Steady state conditions (ES) is constant means the rate of formation of ES is equal to the rate of breakdown of ES; the conc of the species remains constant

Assumption 3: S > Et - You have a lot of substrate and very little enzyme (total enzyme conc)

Assumption 4: S > P (initial conditions) - We are looking at initial rates – the initial conditions of a reaction where the rate is constant; hence, the conc of substrate must be much higher than conc of product (at least 10-fold higher)

 When we combine these assumptions with this kinetic model, we can derive an expression describing the initial rate (V0) in this form:

Question 11

10. Know and use the Michaelis-Menten equation

Question 12

11. Generate and analyze plots describing enzyme kinetics

Question 13

12. Describe how Km is related to Kd

Question 14

13. Explain what Km means in terms of substrate affinity

Question 15

14. Describe what kcat or turnover number means

Answer 15

• K2 is the rate constant for rate-limiting step
• K2 is also Kcat or turnover number
  > Turnover number is defined as:
  
  • n.o molecules of substrate converted to product (S to P) per unit to time per enzyme molecule saturated with substrate (when [ES] = [Et])

Question 16

15. Explain what the specificity constant is and what it means

Answer 16

• the specificity constant Kcat/Km is defined as the rate constant for the conversion of E+S to E+P
• When [S] < Km, V0 is proportional to Kcat/Km
• Kcat/Km reflects both substrate affinity and catalytic efficiency
• a higher value for Kcat/Km indicates more efficient use of the substrate

Question 17

16. Describe the difference between irreversible and reversible inhibitors

Answer 17

> Irreversible inhibitors: bind covalently to the active site, destroy a functional group essential for enzyme activity, or form a stable noncovalent complex with the enzyme. “Suicide inhibitors”

> Reversible inhibitors: bind reversibly to enzymes and inhibit the enzyme either by competitive, uncompetitive or mixed modes of inhibition

Question 18

17. List the apparent changes to Vmax and Km caused by competitive, uncompetitive and mixed inhibitors

Answer 18

> Competitive inhibitors:
- The inhibitor binds at the active site blocking access to substrate
- Vmax of the enzyme remains the same, a change in the Km is seen by some factor (alpha)
- Note: alpha is a constant, a positive number (>1)
- We say the enzyme has a different “apparent Km” – it appears to behave differently compared to the Km where the enzyme had no inhibitors present

> Uncompetitive inhibitors:
- The inhibitor affects both Vmax and Km; Vmax and Km are both reduced compared to the absence of the inhibitor

 Mixed inhibitors:
- Has components of both competitive (alpha) and uncompetitive inhibitors (alpha prime) present
- Vmax is reduced, and Km is affected by both alpha and alpha prime

Question 19

18. Describe the mechanisms of a competitive, uncompetitive and mixed inhibitors

Answer 19

COMPETITIVE ENZYME
- binds to active site of the enzyme
- reversible if substrate conc overcomes inhibitor conc - so Vmax is unchanged
- Affinity for substrate, however, is reduced by binding to inhibitor - so Km increases

UNCOMPETITIVE
- Inhibitor binds to ES complex nd locks substrate in place
- Increases affinity (lower Km) but decreases Vmax
- unlike competitive inhibition the inhibition cant simply be overcome by increasing substrate conc cuz that just produces more ES complex that inhibitor can bind to - hence Vmax decreases
- However, still a reversible inhibitor because non-covalent weak interaction between inhibitor and ES complex can be overcome
- binds at allosteric site

MIXED
- The inhibitor binds to either enzyme (high km) or ES complex (low km)
- the Vmax is low overall
- Mixed inhibitor produces 2 curves both intersecting on a point not on either axis
- binds at allosteric site

Question 20

19. List the properties of the factor α and α’

Answer 20

✅ 𝛼 and 𝛼′ are constants that tell us how strongly the inhibitor binds to:

α: the free enzyme (E) → this affects the competitive component
α′: the enzyme-substrate complex (ES) → this affects the uncompetitive component

Question 21

20. Describe how α and α’ change the apparent Vmax and apparent Km

Question 22

21. Generate and analyse Michaelis-Menten and double reciprocal plots that show the effect of competitive, uncompetitive and mixed inhibitors

Question 23

22. Be able to calculate α and α’ from kinetic data

Question 24

Why doesnt Michaelis-Menten show cooperativity?

Answer 24

Michaelis-Menten cannot show cooperativity because it assumes non-interacting, single-site enzymes with no conformational change.
To capture cooperativity, you need allosteric models that incorporate T and R states or use the Hill coefficient as a simplification.

Question 25

23. Describe the features of an allosteric enzyme

Answer 25

Scheme of how an allosteric enzyme works: - C – catalytic subunit - R – regulatory subunit - M, the positive modulator, binds to the regulatory site and causes a change in conformation at the active site - S can now bind with high affinity - Allosteric enzymes often regulate metabolic pathways by changing activity in response to changes in the concentration of molecules around them - Allosteric enzymes are regulated by compounds called “allosteric modulators” or “allosteric effectors” - “Positive modulators” activate and “negative modulators” inhibit allosteric enzymes - Modulators bind reversibly and non-covalently to the enzyme – they can “come and go”

Question 26

24. Explain how positive and negative allosteric modulators control enzyme activity

Answer 26

- In the diagram above, you can see a modulator that binds to the regulatory site and changes the conformation of the active site - The modulator binds to a regulatory site usually on a separate regulatory subunit and this causes a conformational change to the active site usually in the catalytic subunit - If the modulator is a positive modulator, then the catalytic subunit can bind substrate with a higher affinity - It behaves as if its got a smaller Km - The reverse is true for a negative modulator, a negative allosteric modulator would bind to a regulatory site which would cause a conformational change in the catalytic subunit but that conformational change results in the catalytic subunit having a higher Km binding substrate less strongly

Question 27

25. Describe how different modulators stabilize different conformational states

Answer 27

- The plot above shows that substrate binding shifts the conformation of an allosteric enzyme from the low activity T state, to the high activity R state - Substrate binding stabilises the R state over the T state, as we increase the amount of substrate concentration, the amount of R state increases, depicted as a sigmoidal curve - The second curve shows how negative modulators (allosteric inhibitors) increase the T state and increase its population - Positive modulators (allosteric activators) increase the R state and increase its population - Note: we can’t use Km to describe the midpoint because these curves don’t show Michaelis-Menten kinetics, so this K0.5 is used instead (it’s the substrate concentration at which we get half-maximal velocity for an allosteric sigmoidal curve) Allosteric enzymes: Show a sigmoidal (S-shaped) curve due to cooperative binding (like hemoglobin with oxygen). Binding of one substrate molecule increases the enzyme's affinity for the next (positive cooperativity). This means there's no single fixed Km — the enzyme’s affinity changes as more substrate binds. 🧩 So instead, we use: K₀.₅ = The substrate concentration at half-maximal velocity (like Km but adapted for sigmoidal behavior). This lets us compare how cooperative enzymes respond to increasing substrate levels, even if their binding affinity isn’t constant.

Question 28

Compare ligand binding curve to michaelis menten

Answer 28

MICHAELIS MENTEN - reaction velocity - Km constant - Hyperbolic (non-allosteric) or Sigmoidal (allosteric) - Km = [S] at 0.5 Vmax (for M-M); K₀.₅ used for allosteric enzymes LIGAND BINDING CURVE - Fraction of binding sites occupied (θ or Y) - Kd (dissociation constant) - Hyperbolic (non-cooperative) or sigmoidal (cooperative) - Present in multimeric proteins like hemoglobin (sigmoidal curve) - Kd = [L] at 50% saturation (Y = 0.5)

Question 29

Compare simple enzyme Michalis kinetics and regulatory enzyme (multisubunit) kinetics

Answer 29

- Hyperbolic - Michaelis - Sigmoidal - Regulatory - Michaelis-Menten Kinetics describes how the rate of an enzyme-catalysed reaction depends on substrate conc, producing a hyperbolic curve where rate increases with substrate until it levels off @ Vmax (max velocity)

Question 30

26. Analyze plots of kinetic data for allosteric enzymes to identify the action of allosteric activators and inhibitors

Question 31

27. Use the example of the ATCase from E. coli as an example of allosteric regulation

Answer 31

ALLOSERIC REGULATION - ATCase converts carbomoyl phosphate + aspartate into an intermediate N-carbomoyl aspartate and later CTP pyrimidine base for use in DNA - When CTP conc high, it acts as a negative inhibitor by binding to allosteric site on ATCase and decreases its affinity for the intermediate N-carbomoyl aspartate COOPERATIVITY: - ATCase has a complex quaternary structure of 12 subunits: o 6 catalytic subunits, arranged as 2 x trimeric complexes. The catalytic subunits function cooperatively o 6 regulatory subunits, arranged as 3 x dimeric complexes - ATP is the activator and CTP is the inhibitor - As ATP increases, the V0 vs [S] plot becomes more hyperbola-like - As CTP increases, the V0 vs [S] plot shifts to the right – the substrate concentration at ½ V0 (K0.5) increases

Question 32

28. Integrate concepts of primary, secondary, tertiary and quaternary structure to describe the structure of α-chymotrypsin

Question 33

29. Rationalize the optimal pH for α-chymotrypsin activity with the protonation/deprotonation of His and the N-ter of the B chain