Enzyme Action and Factors Affecting Enzyme Activity
Enzyme-Substrate Interaction and the Active Site
An Enzyme interacts with its Substrate through a specialised region called the Active Site. The active site is made up of two definite regions: the Binding Site and the Catalytic Site. The binding site recognises and attaches to the correct substrate, forming an Enzyme-Substrate Complex. This interaction activates the catalytic site, which then converts the substrate into product(s). After the reaction, the enzyme detaches from the products unchanged and is free to catalyse another reaction. Enzymes require an aqueous medium for their activity.
Binding Site: Recognises and binds the correct substrate molecule through complementary shape and charge
Catalytic Site: Performs the chemical transformation of the substrate into product(s) once activated by substrate binding
Enzyme-Substrate Complex: A transient intermediate formed when substrate occupies the active site
Enzyme Reuse: After releasing products, the enzyme remains unchanged and can participate in further reactions
Most enzymes are not freely floating in the cytoplasm. They are attached to membrane systems inside the cell in specific and orderly arrangements. Mitochondria and Chloroplasts are notable examples where enzymes are organised on internal membranes. This compartmentalisation allows for efficient regulation of metabolic pathways and prevents unwanted side reactions.
Membrane Attachment: Enzymes are organised on cellular membranes rather than diffusing freely in the cytoplasm
Compartmentalisation: Specific membranes like those in mitochondria and chloroplasts house organised enzyme systems
Spatial Organisation: The orderly arrangement of enzymes on membranes ensures efficient substrate channeling
Models of Enzyme Action
The Lock and Key Model was proposed by Emil Fischer in 1890 to explain enzyme-substrate specificity. According to this model, the Active Site is a rigid, pre-formed structure — just as only one specific key fits into a particular lock, only a specific Substrate can bind to a particular enzyme's active site. The enzyme acts merely as a template without changing shape. While this model explains specificity, later research showed it does not account for all enzymatic reactions.
Rigid Active Site: The lock and key model treats the active site as an inflexible, fixed structure
One-to-One Specificity: Each enzyme catalyses only one specific substrate, like a key-lock pair
Proposed by Fischer: Emil Fischer introduced this concept in 1890
Limitation: Later evidence showed that some enzymes change shape upon substrate binding, contradicting the rigid-site assumption
Koshland (1959) proposed the Induced Fit Model as a modification of Fischer's model. This model states that when a Substrate binds to the Active Site, it induces a conformational change in the enzyme's structure. This change enables the enzyme to perform its catalytic activity more effectively. Unlike the lock and key model, the active site is not rigid — it adjusts its shape to accommodate the substrate, resulting in a closer fit during catalysis.
Flexible Active Site: The induced fit model proposes that the active site moulds itself around the substrate
Conformational Change: Substrate binding causes a structural change in the enzyme that enhances catalysis
Proposed by Koshland: Daniel Koshland introduced this model in 1959 based on new experimental evidence
Improved Fit: The enzyme-substrate complex has a tighter, more precise arrangement than the lock and key model predicts
Effect of Temperature on Enzyme Activity
The rate of an enzyme-controlled reaction increases with temperature up to a certain limit. The specific temperature at which an enzyme works at its maximum rate is called the Optimum Temperature. For human body enzymes, the optimum temperature is 37°C. Heat provides Activation Energy and kinetic energy to the reacting molecules, increasing their movement and the frequency of effective collisions between enzyme and substrate.
$$Q_{10} = \frac{\text{Rate at } (T+10°\text{C})}{\text{Rate at } T°\text{C}} \approx 2$$
The Q10 temperature coefficient expresses how much the reaction rate increases for a 10°C rise in temperature
$Q_{10}$=Temperature coefficient(dimensionless)
$T$=Temperature in degrees Celsius(°C)
$T = 27°\text{C}$
→Rate at 37°C is approximately double the rate at 27°C
Rate Increase: Reaction rate roughly doubles for every 10°C rise in temperature (up to the optimum)
Optimum Temperature: The temperature at which the enzyme achieves maximum catalytic activity (37°C for human enzymes)
Kinetic Energy: Higher temperature increases molecular motion, leading to more frequent enzyme-substrate collisions
Activation Energy: Heat supplies the energy needed for molecules to reach the transition state
Beyond the Optimum Temperature, further increase in heat energy causes violent vibrations of the atoms that make up the enzyme molecule. If these vibrations become too intense, the globular structure essential for enzyme activity is lost. The enzyme is said to be Denaturation — its active site is permanently altered and it can no longer catalyse the reaction. This loss of structure is irreversible under normal conditions.
Excessive Vibration: Very high temperatures cause violent atomic vibrations within the enzyme
Globular Structure Loss: The three-dimensional shape of the enzyme is destroyed
Denaturation: Permanent loss of enzyme structure and function due to extreme heat
Irreversibility: Denatured enzymes cannot regain their original shape and catalytic ability
Effect of pH on Enzyme Activity
Every enzyme functions most effectively over a narrow range of pH known as the Optimum pH. A slight change in pH can alter the ionisation of amino acids at the Active Site, affecting the enzyme's ability to bind the substrate. pH changes may also affect the ionisation of the substrate itself. Under these altered conditions, enzyme activity is either reduced or blocked entirely. Extreme pH changes cause bonds within the enzyme to break, resulting in Denaturation.
Narrow pH Range: Each enzyme has a specific optimum pH at which it works most efficiently
Ionisation Effects: pH changes alter the charge of amino acid residues at the active site
Substrate Ionisation: Changes in pH can also affect the charge state of the substrate, reducing binding affinity
Extreme pH: Causes bond breakage within the enzyme, leading to denaturation
Optimum pH Values for Selected Enzymes
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Pepsin: 2.00
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Sucrase: 4.50
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Enterokinase: 5.50
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Salivary amylase: 6.80
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Catalase: 7.60
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Chymotrypsin: 7.00–8.00
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Pancreatic lipase: 9.00
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Arginase: 9.70
Effect of Enzyme and Substrate Concentration
When substrate concentration is unlimited, the reaction rate is directly proportional to the amount of Enzyme present. Doubling the enzyme concentration doubles the rate of reaction because more active sites are available to convert substrate into product. However, beyond a certain limiting concentration, the reaction rate no longer depends on further increases in enzyme amount — the substrate becomes the limiting factor.
Direct Proportionality: Reaction rate increases linearly with enzyme concentration (at unlimited substrate)
More Active Sites: Increasing enzyme molecules increases the number of active sites available for catalysis
Limiting Concentration: After a threshold, additional enzyme does not increase rate — substrate becomes the limiting factor
At low Substrate concentration, the reaction rate is directly proportional to the amount of substrate available. As substrate concentration increases while enzyme concentration remains constant, the rate rises until a point is reached where all active sites are occupied. Beyond this point, the enzyme is said to be saturated — further increase in substrate does not increase the reaction rate. This plateau represents the maximum velocity ($V_{\max}$) for the given enzyme concentration.
$$\text{Rate} \propto [S] \text{ (at low } [S]\text{)}; \quad \text{Rate} \to V_{\max} \text{ (at high } [S]\text{)}$$
The relationship between reaction rate and substrate concentration shows direct proportionality at low substrate levels and a plateau at maximum velocity when the enzyme is saturated
$[S]$=Substrate concentration(mol/L or M)
$V_{\max}$=Maximum reaction rate when enzyme is fully saturated(mol/L/s)
All active sites occupied
→Rate equals $V_{\max}$ and no longer increases with more substrate
Low Substrate: Rate is directly proportional to substrate concentration — many active sites are free
Increasing Substrate: Rate rises as more active sites become occupied
Saturation: All active sites are occupied — the enzyme is saturated and cannot process substrate any faster
Maximum Velocity: The plateau rate ($V_{\max}$) is reached when every active site is occupied at all times
Enzyme Inhibitors
An Enzyme Inhibitor is a chemical substance that reacts with the enzyme in place of the substrate but is not transformed into product(s). Inhibitors block the Active Site temporarily or permanently. Examples include poisons (such as cyanide), antibiotics, anti-metabolites, and certain drugs. Inhibitors are classified into two main categories: Irreversible Inhibitor and Reversible Inhibitor.
Definition: Inhibitors bind to enzymes and prevent normal catalytic activity without being converted to products
Examples: Cyanide (poison), antibiotics, anti-metabolites, and some drugs act as enzyme inhibitors
Temporary or Permanent: Inhibition can be reversible or irreversible depending on the nature of binding
An Irreversible Inhibitor checks the reaction rate by occupying the active sites or destroying the globular structure of the enzyme. They form strong Covalent Bond interactions with amino acid residues at the active site, or they may physically block the active site. Because the inhibition cannot be reversed by simply increasing substrate concentration, these inhibitors cause permanent loss of enzyme activity. Cyanide is a classic example, which irreversibly inhibits cytochrome c oxidase in the electron transport chain.
Covalent Bonding: Irreversible inhibitors form permanent covalent bonds with amino acid residues in the active site
Structural Destruction: Some irreversible inhibitors destroy the globular structure of the enzyme entirely
Permanent Inactivation: The effect cannot be reversed by increasing substrate concentration
Cyanide Example: Binds irreversibly to cytochrome c oxidase, blocking cellular respiration
Reversible Inhibition: Competitive and Non-competitive
Reversible Inhibitor molecules form weak linkages with the enzyme, and their effect can be neutralised by increasing the concentration of substrate. They are divided into two types: Competitive Inhibitor and Non-competitive Inhibitor. A Competitive Inhibitor has structural similarity with the substrate, which allows it to be selected by the Binding Site. However, it cannot activate the Catalytic Site, so no product is formed. It competes with the substrate for the active site. Since both the substrate and the inhibitor compete for the same site, increasing substrate concentration can overcome competitive inhibition. A classic example is malonic acid, which competitively inhibits succinic dehydrogenase by competing with succinic acid for the active site.
Structural Similarity: The competitive inhibitor closely resembles the substrate in shape and structure
Competes for Active Site: Both substrate and inhibitor compete for binding to the same active site
No Product Formation: Although the inhibitor binds the active site, it cannot activate the catalytic site
Overcome by Substrate: Increasing substrate concentration displaces the inhibitor and restores reaction rate
Malonic Acid Example: Competitively inhibits succinic dehydrogenase (the natural substrate is succinic acid)
A Non-competitive Inhibitor binds to the enzyme at a location other than the Active Site — an allosteric site. This binding alters the enzyme's three-dimensional structure in such a way that even if the genuine substrate binds the active site, catalysis fails to take place. Because the inhibitor does not compete for the active site, increasing substrate concentration does NOT overcome non-competitive inhibition. The enzyme's globular structure is changed, rendering the catalytic site ineffective regardless of substrate availability.
Allosteric Binding: Non-competitive inhibitors bind to a site distinct from the active site
Structural Alteration: Binding changes the overall shape of the enzyme, distorting the active site
Catalytic Failure: Even when the substrate binds, the distorted active site cannot catalyse the reaction
Not Overcome by Substrate: Increasing substrate concentration does not reverse non-competitive inhibition because the inhibitor does not compete for the active site
Application Strategy
Worked Examples
Problem Statement
If an enzyme-catalysed reaction proceeds at a rate of 20 units/min at 20°C, what is the approximate rate at 40°C (assuming the temperature is still below the optimum)?
Q10 temperature coefficient
Strategic Toolkit
$$Q_{10} = \frac{\text{Rate at } (T+10°\text{C})}{\text{Rate at } T°\text{C}} \approx 2$$
Step 1
Identify the temperature increase
The temperature increases from 20°C to 40°C, which is a rise of 20°C — equivalent to two intervals of 10°C each.
$$\Delta T = 40 - 20 = 20°\text{C} \implies 2 \text{ intervals of } 10°\text{C}$$
Step 2
Apply the Q10 coefficient for the first interval
The rate doubles for each 10°C rise. At 30°C (first interval), the rate doubles from 20 to 40 units/min.
$$\text{Rate at } 30°\text{C} = 20 \times 2 = 40 \text{ units/min}$$
Step 3
Apply Q10 for the second interval
At 40°C (second interval), the rate doubles again from 40 to 80 units/min.
$$\text{Rate at } 40°\text{C} = 40 \times 2 = 80 \text{ units/min}$$
Step 4
Final answer
The approximate rate at 40°C is 80 units/min, which is four times the original rate at 20°C.
$$\text{Rate at } 40°\text{C} = 20 \times 2^2 = 80 \text{ units/min}$$
Problem Statement
An enzyme-catalysed reaction produces 30 μmol of product per minute with 5 mg of enzyme at unlimited substrate concentration. How much product is formed per minute if the enzyme amount is increased to 15 mg?
Enzyme concentration and rate proportionality
Strategic Toolkit
$$\text{Rate} \propto [E] \text{ (at unlimited substrate)}$$
Step 1
Determine the concentration factor
The enzyme amount is increased from 5 mg to 15 mg, which is a threefold increase.
$$\text{Factor} = \frac{15}{5} = 3$$
Step 2
Apply direct proportionality
Since substrate is unlimited, the rate is directly proportional to enzyme concentration. A threefold increase in enzyme triples the reaction rate.
$$\text{New rate} = 30 \times 3 = 90 \text{ μmol/min}$$
Step 3
Final answer
With 15 mg of enzyme at unlimited substrate, the reaction produces 90 μmol of product per minute.
$$\text{Product rate} = 90 \text{ μmol/min}$$
Problem Statement
Malonic acid inhibits succinic dehydrogenase, which normally converts succinic acid to fumaric acid. When the concentration of succinic acid is increased, the inhibition is reduced. What type of inhibition is this, and why?
Competitive inhibition mechanism
Step 1
Analyse the structural relationship
Malonic acid has a structure similar to succinic acid, the natural substrate of succinic dehydrogenase. This structural similarity allows malonic acid to be recognised by the binding site of the enzyme.
Step 2
Identify the binding location
Malonic acid competes with succinic acid for the same active site on succinic dehydrogenase. Once bound, it cannot activate the catalytic site, so no fumaric acid is produced.
Step 3
Evaluate the effect of increasing substrate
The fact that increasing succinic acid concentration reduces inhibition confirms that the inhibitor competes for the active site. This is the defining characteristic of competitive inhibition.
Step 4
Conclusion
Malonic acid is a competitive inhibitor of succinic dehydrogenase because it has structural similarity to the substrate, binds the active site, and its effect is overcome by increasing substrate concentration.
Problem Statement
An enzyme has a maximum reaction rate (Vmax) of 120 μmol/min. At a substrate concentration of 2 mM, the observed rate is 60 μmol/min. At 8 mM, the rate reaches 120 μmol/min. Explain what happens at 8 mM substrate concentration.
Substrate saturation and Vmax
Strategic Toolkit
$$\text{Rate} \to V_{\max} \text{ when active sites are fully occupied}$$
Step 1
Analyse the rate at 2 mM substrate
At 2 mM, the rate is 60 μmol/min, which is half of Vmax. This indicates that about half the active sites are occupied at this concentration.
$$\frac{60}{120} = 0.5 \implies \text{approximately 50\% active sites occupied}$$
Step 2
Analyse the rate at 8 mM substrate
At 8 mM, the rate equals Vmax (120 μmol/min), meaning every active site on every enzyme molecule is occupied at all times.
$$\text{Rate} = V_{\max} = 120 \text{ μmol/min}$$
Step 3
Explain saturation
At 8 mM the enzyme is fully saturated. Adding more substrate beyond this concentration will not increase the reaction rate because there are no free active sites available. The rate has plateaued.
Memorize formulas
Q10 Temperature Coefficient
Reaction rate approximately doubles for every 10°C rise in temperature below the optimum. Used to estimate rate changes with temperature for enzyme-catalysed reactions.
Formula
$$Q_{10} = \frac{\text{Rate at } (T+10°\text{C})}{\text{Rate at } T°\text{C}} \approx 2$$
Rate vs Substrate Concentration
At low substrate concentration, rate is proportional to [S]. At high [S], rate plateaus at Vmax when all active sites are saturated. Units: rate in mol/L/s, [S] in mol/L (M).
Formula
$$\text{Rate} \propto [S] \text{ (low } [S]\text{)}; \quad \text{Rate} \to V_{\max} \text{ (high } [S]\text{)}$$
Review
1
The region of an enzyme where the substrate binds and the chemical reaction takes place is called the:
Ans
Active site
Review
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Identify the functional region
The active site is the combined region where substrate recognition and catalysis both occur.
Active site = binding site + catalytic site
2
According to the lock and key model of enzyme action, which statement is correct?
Ans
The substrate has a complementary shape to the enzyme's active site
Review
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Lock and key model principle
Proposed by Emil Fischer; both enzyme active site and substrate maintain rigid, pre-existing shapes that are complementary to each other.
Rigid complementarity between enzyme and substrate
3
The induced fit model differs from the lock and key model because in the induced fit model:
Ans
The active site undergoes a conformational change upon substrate binding
Review
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Compare the two models
Lock and key = rigid active site; Induced fit = flexible active site that reshapes upon substrate binding.
Key difference = conformational change in the enzyme
4
When an enzyme binds to its substrate, the temporary intermediate formed is known as the:
Ans
Enzyme-substrate complex
Review
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Reaction pathway
Enzyme + Substrate → Enzyme-Substrate Complex → Enzyme-Product Complex → Enzyme + Product
ES complex = enzyme-substrate complex (first intermediate)
5
Human enzymes typically have an optimum temperature of approximately:
Ans
37 °C
Review
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Recall optimum temperature for human enzymes
Human body temperature is approximately 37 °C, and enzymes have evolved to work best at this temperature.
$$T_{\text{opt}} \approx 37°\text{C}$$
6
An enzyme-catalyzed reaction proceeds at a rate of 40 units/min at 20 °C. If the $Q_{10}$ coefficient is 2, what will be the approximate rate at 40 °C?
Ans
160 units/min
Review
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Identify the temperature interval
From 20 °C to 40 °C is a 20 °C increase, which equals 2 intervals of 10 °C.
$$\Delta T = 20°\text{C}, \quad n = \frac{20}{10} = 2$$
Apply Q10 formula
Rate at new temperature = Rate at initial temperature × (Q10)^(n)
$$\text{Rate} = 40 \times 2^2 = 40 \times 4 = 160 \text{ units/min}$$
7
At temperatures above the optimum, enzyme activity declines rapidly because:
Ans
The enzyme's tertiary structure is disrupted, altering the active site shape
Review
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Mechanism of high-temperature enzyme inactivation
Excessive thermal energy disrupts weak bonds maintaining the enzyme's 3D fold.
Denaturation → loss of active site geometry → loss of function
8
Pepsin, a digestive enzyme in the stomach, has an optimum pH of approximately 2. What happens when pepsin is transferred to the small intestine (pH ≈ 8)?
Ans
Its substrate specificity changes from proteins to carbohydrates
Review
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Analyze pH compatibility
Pepsin's optimum pH is ~2. The small intestine pH is ~8, which is far outside its functional range.
$$\text{pH}_{\text{small intestine}} = 8 \gg \text{pH}_{\text{opt}} \text{ of pepsin} = 2$$
Determine the consequence
Drastic pH change alters ionization of active site residues, causing denaturation.
Pepsin denatures at pH 8 and loses all catalytic activity
9
If the concentration of enzyme is doubled while substrate concentration remains constant and non-limiting, the reaction rate will:
Ans
Double
Review
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Identify the limiting factor
Since substrate is non-limiting (in excess), enzyme concentration is the rate-determining variable.
$$\text{Rate} \propto [E] \text{ when } [S] \text{ is in excess}$$
Calculate the new rate
Doubling E doubles the rate.
$$\text{Rate}_{\text{new}} = 2 \times \text{Rate}_{\text{original}}$$
10
As substrate concentration increases while enzyme concentration is held constant, the reaction rate:
Ans
Initially increases proportionally, then plateaus at $V_{\max}$
Review
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Low substrate region
At low S, rate increases proportionally because active sites are mostly unoccupied.
$$\text{Rate} \propto [S] \text{ (low } [S]\text{)}$$
High substrate region
At high S, all enzyme active sites are saturated — no free sites remain for additional substrate.
$$\text{Rate} \to V_{\max} \text{ (plateau)}$$
11
A competitive inhibitor reduces enzyme activity by:
Ans
Denaturing the enzyme at high concentrations
Review
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Mechanism of competitive inhibition
The inhibitor resembles the substrate structurally and binds reversibly to the active site, competing directly with the substrate.
Competitive inhibitor blocks the active site — substrate cannot bind
Overcoming competitive inhibition
Since inhibition is reversible and occurs at the active site, excess substrate can outcompete the inhibitor.
Increasing S restores reaction rate toward $V_{\max}$
12
A non-competitive inhibitor binds to an enzyme at a site distinct from the active site. What effect does this have on $V_{\max}$ and the enzyme's ability to reach this maximum?
Ans
$V_{\max}$ is reduced because the inhibitor decreases the total number of functional enzyme molecules
Review
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Binding location
Non-competitive inhibitor binds to an allosteric site, separate from the active site.
Allosteric binding → conformational change in enzyme
Effect on kinetic parameters
The conformational change reduces catalytic efficiency of the affected enzyme molecules, decreasing the maximum rate achievable.
$$V_{\max \text{ (with inhibitor)}} < V_{\max \text{ (without inhibitor)}}$$
13
Which of the following is a characteristic of irreversible enzyme inhibition?
Ans
The inhibitor forms a covalent bond with the enzyme
Review
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Identify the bond type
Irreversible inhibitors form covalent bonds with the enzyme, permanently modifying its structure.
Covalent bond → permanent inactivation
14
An experiment shows that adding more substrate partially restores enzyme activity that was reduced by an inhibitor. This observation is consistent with which type of inhibition?
Ans
Competitive inhibition
Review
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Analyze the experimental observation
More substrate restoring activity implies the inhibitor and substrate compete for the same binding location.
Competitive binding at the active site
Confirm with kinetic reasoning
At high S, substrate molecules outcompete the inhibitor for active sites, partially restoring the rate toward $V_{\max}$.
Consistent with competitive inhibition
15
Enzymes increase the rate of biochemical reactions primarily by:
Ans
Increasing the concentration of reactants
Review
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Define the catalytic mechanism
Enzymes stabilize the transition state, reducing the energy barrier that reactants must overcome.
$$\text{Enzyme lowers } E_a \text{ (activation energy)}$$
Clarify what is NOT changed
The enzyme does not alter the overall ΔG of the reaction, temperature, or reactant concentrations.
Only the energy barrier is lowered — the thermodynamic outcome is unchanged
Question 1 of 15
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1
The region of an enzyme where the substrate binds and the chemical reaction takes place is called the: