Kinetics and Regulation of Enzymes

Importance of Enzyme Kinetics

Primary function of enzymes: To accelerate the rates, or velocities, of reactions.
Purpose of kinetic description: Essential for understanding how enzymes function and for quantifying kinetic parameters.
Quantifiable kinetic parameters:
- How fast an enzyme can operate.
- How fast it operates at substrate concentrations found in a cell.
- What substrate is most readily operated on by the enzyme.

Kinetics and Reaction Rates

Kinetics: The study of the rates of chemical reactions.
Enzyme kinetics: The study of the rates of enzyme-catalyzed reactions.
Simple reaction (A ightarrow P):
- Reactions involve the disappearance of reactants and the appearance of products.
- Velocity (V): The quantity of reactant (A) that disappears in a specified unit of time (t). It is equal to the velocity of the appearance of product (P) in the same unit of time.
- Formula for velocity: V = -d[A]/dt = d[P]/dt

How to Perform Kinetic Measurements

Mix enzyme + substrate.
Record rate: Measure the rate of substrate disappearance or product formation as a function of time (which represents the velocity of the reaction).
Plot initial velocity (V0): Plot V0 versus substrate concentration ([S]).
Repeat: Change substrate concentration and repeat the process to obtain multiple V_0 values.

Reaction Order

First-order reaction: Reaction velocity is directly proportional to reactant concentration. Its units are \text{s}^{-1}.
Biomolecular or second-order reactions: Many important biochemical reactions are biomolecular. Their units are \text{M}^{-1}\text{s}^{-1}. The rate equations for these reactions are generally:
- V = k[A][B]
- V = k[A]^2

The Michaelis–Menten Model

Initial Velocity (V_0) of Catalysis

Definition: The number of moles of product formed per second shortly after the reaction has begun.
Variation: V_0 varies with the substrate concentration ([S]) when enzyme concentration is constant.
Enzyme concentrations in cells: Relatively low.

Product Formation Through Catalysis

Enzyme reaction: An enzyme (E) catalyzes the conversion of substrate (S) into product (P). E + S \xrightleftharpoons[k{-1}]{k1} ES \xrightleftharpoons[k{-2}]{k2} E + P
- k_1: Rate constant for the formation of the enzyme-substrate (ES) complex.
- k_2: Rate constant for the formation of product P.
- k{-1} and k{-2}: Rate constants for the respective reverse reactions.
Focus on initial rate enzyme kinetics: Most useful for determining the exact concentration of the substrate at the beginning of the reaction, as the reaction rate (slope of the tangent line) decreases over time.
Initial rate of catalysis (V_0): The number of moles of product formed per second when the reaction is just beginning.

Michaelis–Menten Equation

Description: Describes the variation of enzyme activity as a function of substrate concentration.
Formula: V0 = \frac{V{max}[S]}{K_M + [S]}
- V_{max}: Maximal velocity possible.
- K_M: Michaelis constant.
- [S]: Substrate concentration.
Michaelis constant (K_M):
- Unique to each enzyme and independent of enzyme concentration.
- Describes the properties of the enzyme-substrate interaction.
- Varies for enzymes that can use different substrates.
Maximal velocity (V_{max}):
- Attained only when all of the total enzyme ([E]_T) is bound to substrate (S); i.e., the enzyme is saturated.
- Formula: V{max} = k2[E]_T
- Directly dependent on enzyme concentration.
Kinetic behavior at different substrate concentrations:
- Very low [S] (when [S] \ll KM): The velocity (V0) is directly proportional to the substrate concentration (first-order kinetics).
- High [S] (when [S] \gg KM): The velocity is maximal (V{max}) and independent of substrate concentration (zero-order kinetics). The enzyme is saturated.
Relationship between KM and V{max}: When V0 = V{max}/2, the value of K_M is numerically equal to the substrate concentration at which the reaction velocity is half its maximal value.

Michaelis–Menten Enzymes

Characteristics: Most enzymes in the cell are simple, monomeric, and unregulated.
Evolution: These enzymes have evolved to be as fast and efficient as possible.
Kinetics: They conform to simple Michaelis-Menten kinetics; their activity is governed simply by mass action (if substrate is present, they catalyze).

Determining KM and V{max}

Curve-fitting programs: Most commonly achieved with the use of computer curve-fitting programs from rates of catalysis measured at various substrate concentrations.
Lineweaver-Burk equation and plots:
- Equation: The Michaelis-Menten equation can be manipulated into a double-reciprocal equation that yields a straight-line plot.
  \frac{1}{V0} = \frac{KM}{V{max}[S]} + \frac{1}{V{max}}
- Advantages: This plot allows for more accurate determination of V_{max} and can help visually differentiate different reaction mechanisms.
- Intercepts: The x-intercept is -1/KM and the y-intercept is 1/V{max}.

K_M, an Important Enzyme Characteristic

Range of values: The K_M values of most enzymes range between 10^{-1} and 10^{-7} \text{ M}.
Dependence: K_M depends on the particular substrate and environmental conditions (e.g., pH, temperature, ionic strength).
Significance: Provides a measure of the substrate concentration required for significant catalysis to take place.
Examples of K_M values:
- Carbonic anhydrase (CO\small_2): 8000 \text{ \mu M} (microMolar)
- Chymotrypsin (Acetyl-L-tryptophanamide): 5000 \text{ \mu M}
- \beta -Galactosidase (Lactose): 4000 \text{ \mu M}
- Penicillinase (Benzylpenicillin): 50 \text{ \mu M}
- Lysozyme (Hexa-N-acetylglucosamine): 6 \text{ \mu M}

k_{cat}, turnover number

Definition: The number of substrate molecules that an enzyme can convert into product per unit time when the enzyme is fully saturated with substrate.
Calculation: The maximal rate, V{max}, reveals the turnover number of an enzyme if the concentration of active sites ([E]T) is known.
- Formula: k{cat} = V{max}/[E]_T
Examples of turnover numbers (per second):
- Carbonic anhydrase: 600,000
- 3-Ketosteroid isomerase: 280,000
- Acetylcholinesterase: 25,000
- Penicillinase: 2000
- Lactate dehydrogenase: 1000
- Chymotrypsin: 100
- DNA polymerase I: 15
- Tryptophan synthetase: 2
- Lysozyme: 0.5

k{cat}/KM, the specificity constant

Conditions: When the substrate concentration ([S]) is much greater than KM (Note: This might be a typo in the slide text, it's usually much less than KM for this constant to be most relevant for efficiency at low [S]). The slide states: "When the substrate concentration is much greater than KM, the enzymatic velocity depends on the values of k{cat}/KM, [S], and [E]T". The formula given is:
- V0 = \frac{k{cat}}{KM}[E]T[S]
Significance: This rate (k{cat}/KM) is called the specificity constant.
- It measures catalytic efficiency.
- It accounts for both the rate of catalysis with a particular substrate and the nature of the enzyme-substrate interaction.

Determining KM and k{cat} (Examples)

Example 1: Calculating K_M for happyase
- Reaction: SAD
  ightarrow HAPPY
- Given: k{cat} = 600 \text{ s}^{-1}, [E]T = 20 \text{ nM}, [SAD] = 40 \text{ \mu M}, V_0 = 9.6 \text{ \mu M/s}.
- First, calculate V{max}: V{max} = k{cat} \times [E]T = 600 \text{ s}^{-1} \times 20 \times 10^{-9} \text{ M} = 1.2 \times 10^{-5} \text{ M/s} = 12 \text{ \mu M/s}
- Using the Michaelis-Menten equation (V0 = \frac{V{max}[S]}{KM + [S]}): 9.6 \text{ \mu M/s} = \frac{12 \text{ \mu M/s} \times 40 \text{ \mu M}}{KM + 40 \text{ \mu M}}
  0.8 = \frac{40 \text{ \mu M}}{KM + 40 \text{ \mu M}} 0.8(KM + 40 \text{ \mu M}) = 40 \text{ \mu M}
  0.8 KM + 32 \text{ \mu M} = 40 \text{ \mu M} 0.8 KM = 8 \text{ \mu M}
  K_M = 10 \text{ \mu M}
Example 2: Calculating k_{cat} for happyase*
- Given: [E]T = 4 \text{ nM}, V{max} = 1.6 \text{ \mu M/s}.
- Calculate k{cat}: k{cat} = V{max} / [E]T = (1.6 \times 10^{-6} \text{ M/s}) / (4 \times 10^{-9} \text{ M}) = 400 \text{ s}^{-1}
Example 3: Calculating KM for happyase* (using k{cat} from previous part)
- Given: k{cat} = 400 \text{ s}^{-1}, [E]T = 1 \text{ nM}, [HAPPY] = 30 \text{ \mu M}, V_0 = 300 \text{ nM/s}.
- Formula provided (rearranged Michaelis-Menten):
  V0 = k{cat} \frac{[E]T[S]}{KM + [S]} (Note: This is equivalent to (V0 = \frac{V{max}[S]}{KM + [S]}) since (V{max} = k{cat}[E]T))
- Substitute values:
  0.300 \text{ \mu M/s} = 400 \text{ s}^{-1} \frac{0.001 \text{ \mu M} \times 30 \text{ \mu M}}{KM + 30 \text{ \mu M}} 0.300 \text{ \mu M/s} (KM + 30 \text{ \mu M}) = 12 \text{ \mu M}^2\text{/s}
  0.300 KM + 9 \text{ \mu M}^2 = 12 \text{ \mu M}^2 0.300 KM = 3 \text{ \mu M}^2
  K_M = 10 \text{ \mu M}

Multi-substrate Reactions

Common occurrence: Most biochemical reactions are bisubstrate reactions, starting with two substrates (A + B) and yielding two products (P + Q).
Mechanism: Many bisubstrate reactions transfer a functional group from one substrate to the other.
Classes of bisubstrate reactions:
1. Sequential reactions:
  - Mechanism: All substrates must bind to the enzyme before any product is released.
  - Ternary complex: A complex consisting of the enzyme and both substrates forms.
  - Types:
    - Ordered: Substrates bind (and products are released) in a defined sequence.
    - Random: Substrates can bind (and products can be released) in any order.
2. Double-displacement (Ping-Pong) reactions:
  - Mechanism: One or more products are released before all substrates bind the enzyme.
  - Defining feature: The existence of a substituted enzyme intermediate, where the enzyme is temporarily modified.
  - Analogy: Substrates and products appear to