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Enzyme Function
Active Site Chemistry

Explore how enzymes catalyze biochemical reactions through precise molecular recognition, transition state stabilization, and the chemistry of the active site.

Biochemistry

What Are Enzymes?

Enzymes are biological catalysts: proteins (and some RNA molecules) that dramatically accelerate chemical reactions without being consumed. Nearly every metabolic process in living cells depends on enzymes to proceed at rates fast enough to sustain life.

Catalyst
Speeds up a reaction without being permanently altered
Specificity
Each enzyme typically acts on one substrate or a small group of related substrates
Regulation
Activity can be turned up or down by inhibitors, activators, and cellular signals

The Active Site: A Molecular Pocket

Enzyme Active Site Substrate Asp His Ser Key catalytic residues interact with the substrate

Why Enzymes Matter

Without enzymes, most biochemical reactions would take thousands to millions of years to reach equilibrium at body temperature. Enzymes can accelerate reactions by factors of \( 10^{6} \) to \( 10^{17} \), making life possible.

Uncatalyzed
OMP decarboxylase substrate: half-life of \( 7.8 \times 10^{7} \) years
Catalyzed
Same reaction with enzyme: half-life of \( 18 \) milliseconds

Activation Energy and the Transition State

Reaction Progress Free Energy
\( E_a \)
\( E_a \) (cat)
 Transition State S P Uncatalyzed Enzyme-catalyzed

Lock-and-Key vs Induced Fit

Lock-and-Key S Rigid enzyme Shape is pre-formed Induced Fit S Flexible enzyme Shape changes upon binding

Catalytic Mechanisms at the Active Site

1
Acid-base catalysis: Active site residues donate or accept protons \( (\text{H}^+) \) to stabilize charged intermediates and transition states.
2
Covalent catalysis: A nucleophilic residue (e.g., Ser, Cys) forms a transient covalent bond with the substrate, creating a reactive intermediate.
3
Metal ion catalysis: Metal cofactors (e.g., \( \text{Zn}^{2+} \), \( \text{Mg}^{2+} \)) stabilize negative charges or activate water molecules for nucleophilic attack.
4
Proximity and orientation: The active site holds substrates in the optimal geometry for bond formation or cleavage.

The Michaelis-Menten Equation

$$ v_0 = \frac{V_{\max}[\text{S}]}{K_m + [\text{S}]} $$

\( v_0 \)
Initial reaction velocity: the rate measured before significant product accumulates
\( V_{\max} \)
Maximum velocity when all enzyme active sites are saturated with substrate
\( K_m \)
Michaelis constant: the \( [\text{S}] \) at which \( v_0 = \frac{V_{\max}}{2} \). A measure of substrate affinity

Interpreting the Saturation Curve

[S]
\( v_0 \)
\( V_{\max} \)
\( V_{\max}/2 \)
\( K_m \)
 Saturation region Linear region

The Enzyme-Substrate Reaction Pathway

$$ \text{E} + \text{S} \xrightleftharpoons[k_{-1}]{k_1} \text{ES} \xrightarrow{k_2} \text{E} + \text{P} $$

1
Binding: Free enzyme (E) binds substrate (S) to form the enzyme-substrate complex (ES) at rate \( k_1 \).
2
Catalysis: The ES complex undergoes a chemical transformation through the transition state, converting S to product P at rate \( k_2 \) (also called \( k_{\text{cat}} \)).
3
Release: Product P is released and the free enzyme E is regenerated, ready for another catalytic cycle.

Factors Affecting Enzyme Activity

Temperature
Higher temperature increases kinetic energy and collisions, but excessive heat denatures the protein, destroying the active site
pH
Each enzyme has an optimal pH. Deviations alter ionization states of active site residues and disrupt catalysis
Cofactors
Metal ions or organic coenzymes (e.g., NAD+, FAD) are required by many enzymes for full catalytic function

Competitive vs Non-Competitive Inhibition

Competitive I S blocked Binds active site
\( K_m \) increases, \( V_{\max} \) unchanged
 Non-Competitive S I Binds allosteric site
\( V_{\max} \) decreases, \( K_m \) unchanged

How to Solve Enzyme Kinetics Problems

1
Identify given values: \( V_{\max} \), \( K_m \), \( [\text{S}] \), or \( v_0 \). Determine what is unknown.
2
Write the equation: \( v_0 = \frac{V_{\max}[\text{S}]}{K_m + [\text{S}]} \). Plug in known values.
3
Solve algebraically: Rearrange to isolate the unknown. If finding \( K_m \): \( K_m = [\text{S}]\left(\frac{V_{\max}}{v_0} - 1\right) \).
4
Check special cases: When \( [\text{S}] = K_m \), then \( v_0 = \frac{V_{\max}}{2} \). When \( [\text{S}] \gg K_m \), then \( v_0 \approx V_{\max} \).
5
Interpret the result: A low \( K_m \) means high affinity. A high \( k_{\text{cat}}/K_m \) (catalytic efficiency) means the enzyme is very effective.

Example 1: Finding \( v_0 \)

An enzyme has \( V_{\max} = 100\,\mu\text{mol/min} \) and \( K_m = 5\,\text{mM} \). Calculate \( v_0 \) when \( [\text{S}] = 10\,\text{mM} \).

Example 2: Finding \( K_m \)

An enzyme has \( V_{\max} = 200\,\mu\text{mol/min} \). At \( [\text{S}] = 3\,\text{mM} \), the measured velocity is \( v_0 = 120\,\mu\text{mol/min} \). Find \( K_m \).

Example 3: Effect of a Competitive Inhibitor

With the enzyme from Example 1 (\( V_{\max} = 100 \), \( K_m = 5\,\text{mM} \)), a competitive inhibitor raises the apparent \( K_m \) to \( 15\,\text{mM} \). Find \( v_0 \) at \( [\text{S}] = 10\,\text{mM} \).

Common Mistakes in Enzyme Chemistry

✕ Enzymes "provide energy"
Enzymes do not supply energy. They lower the activation energy barrier, allowing existing thermal energy to drive the reaction more readily.
✕ Low \( K_m \) = low affinity
It is the opposite: a low \( K_m \) means the enzyme reaches half-maximal velocity at a low substrate concentration, indicating high affinity.
✕ Enzymes change equilibrium
Enzymes speed up both the forward and reverse reactions equally. They do not shift the equilibrium position of a reaction.

Summary: Inhibition Types Compared

Competitive Non-Competitive Uncompetitive
Binds to Active site Allosteric site ES complex only
Effect on \( K_m \) Increases No change Decreases
Effect on \( V_{\max} \) No change Decreases Decreases
Overcome by \( \uparrow [\text{S}] \)? Yes No No
Resembles substrate? Yes No No

Question 1 of 5

Enzymes accelerate reactions by:

Question 2 of 5

An enzyme has \( V_{\max} = 80\,\mu\text{mol/min} \) and \( K_m = 4\,\text{mM} \). What is \( v_0 \) when \( [\text{S}] = 4\,\text{mM} \)?

Question 3 of 5

A competitive inhibitor affects enzyme kinetics by:

Question 4 of 5

Which model best describes how most enzymes interact with their substrates?

Question 5 of 5

Enzyme A has \( K_m = 0.5\,\text{mM} \) and Enzyme B has \( K_m = 10\,\text{mM} \) for the same substrate. Which statement is correct?