06Enz01

Describe the overall mechanism of substrate binding on to the active site of enzymes

Enzyme substrate binding involves the substrate to form a complex Complex forms → forms a transition state species → product of reaction The substrate binds to a small portion of the enzyme called an "active site" The active site is situated in a cleft or crevice of a protein They consist of amino acids essential for enzymatic activity Held together by non-covalent interactions

Apply what you learned about enzymes and inhibition in the treatment of AIDS [6D Biochemical Connections]

In AIDS treatment a combination of drug therapies are used as it is more effective - HIV Protease, Integrase and Reverse Transcriptase Inhibitors play important roles in this situation: Using multiple inhibitors ensures that the level of key viral enzymes remain below toxic levels to the cell. HIV Protease - It is an enzyme essential to the production of new virus particles in infected cells Unique to this virus as it catalyzes the processing of viral proteins in an infected cell. - In the structure of HIV protease, including the active site, were designed by scientists to have synthesized compounds to bind to the active site. Integrase - It is also known as the viral enzyme, this is needed for the virus to copy itself in the host cell. Mechanism of Treatment - The HIV protease would be used in the body for it to replicate. - Drugs with a protease inhibitor would then block the action of the protease enzymes and prevent them from doing their part in allowing the HIV to multiply. - This then interrupts the life cycle of the HIV and ultimately stops the virus from multiplying

Explain the mechanism of competitive inhibition

In a competitive inhibition the Inhibitor binds to the active site and blocks substrate access to it. in a situation where there is an equilibrium between the free enzyme and enzyme-inhibitor complex: The substrate competes with the inhibitor for the active site Therefore no product is made unless an ES complex is formed

Understand the implications when the experimental conditions are such that [S] = KM

In experimental conditions such that [S] = KM If the substrate concentration is less than KM, the enzyme is not activated If the substrate concentration is greater than KM, the enzyme is activated

Account for mixed inhibition types in some enzymes

Mixed inhibition occurs when the inhibitor can bind through a mix of the previous mechanisms. Inhibitors bind to the enzyme whether the enzyme is already bound to the substrate, this however has a greater affinity for one state over the other. Some of the effects of MIXED INHIBITION are: An increase in Vmax A decrease in KM

Understand the derivation of the Michaelis-Menten equation based on substrate concentration levels and the assumption of a 'steady-state' condition

Steady-state theory - when enzyme substrate complex is equal to the rate of breakdown Δ[ES]/Δt = -Δ[ES]/Δt and K1[E][S] = K-1[ES] + K2[ES] Solving for [ES] concentration Should know all concentration of species involved The initial concentration is a known experimental condition Substrate concentration is greater than enzyme concentration The total concentration of enzyme [E]T is known but a large portion is known in the complex. The Free Enzyme [E] can be solved with: [E] = [E]T - [ES] Where [E]T is equal to the total concentration Substituting [E] = [E]T - [ES] to K1[E][S] = K-1[ES] + K2[ES] we get: K1 ([E]T - [ES])[S] = K-1[ES] + K2[ES] Collecting all rate constants (K) we get: (([E]T-[ES])[S]) / ([ES]) = ((K-1+ K2 )/ (K1)) = KM KM is called the Michaelis Constant By reorganizing constants Michaelis and Menten derived an equation that defines the reaction velocity in terms of substrate concentration, KM and Vmax V = (-Vmax [S] / KM + [S])

Explain the significance of the KM value to the efficiency of the enzyme

The KM value is the concentration of substrate at which 50% of enzyme active sites are occupied by substrate A Low value for KM → a high affinity for substrate (more reactive) A high value for KM → a low affinity for substrate (less reactive) If the substrate concentration is less than KM then the enzyme is not activated

Determine whether a given inhibitor is an uncompetitive inhibitor from kinetic data

The Lineweaver-Burk plot has parallel lines in a graph The Vmax is decreased because any amount of inhibitor will tie up the ES complex into the useless EIS form KM is decreased and makes it appear that the enzyme is bound to the substrate better.

Evaluate whether the kinetics of an enzymatic reaction falls under the Michaelis-Menten kinetics assumptions or conditions

Allosteric proteins DO NOT follow Michaelis-Menten kinetics. The reaction velocity vs substrate concentration would not exhibit a hyperbolic plot, which is supposed to be in Michaelis-Menten kinetics, rather it exhibits sigmoidal plot when using the Michaelis-Menten equation.

Rate of Reactions

Can be expressed either in terms of "rate of disappearance of one of the reactants" or "rate of appearance of product" The rate of disappearance is expressed as: -Δ[A]/Δt, where Δ = change, [A] = concentration in mole/liter, while t = time In a reaction of A + B → P the rate of reaction can be expressed as: -Δ[A]/Δt = -Δ[B]/Δt = Δ[P]/Δt

Using the examples in Table 6.2, correlate the significance of the functions of the listed enzymes to their reported KM and Vmax values [Why do some enzymes act fast? Why do some enzymes act slowly?]

Catalase and Carbonic Anhydrase are both very reactive compounds Catalase has one of the highest turnover number from all known enzymes - Its high number alludes to its importance in detoxifying the hydrogen peroxide and formation of CO2 bubbles in blood Chymotrypsin and Acetylcholinesterase is within the range of normal metabolic enzymes Lysozyme is an enzyme which degrades the polysaccharide components of bacterial walls thus often found in body tissues - This explains its low catalytic efficiency, and operates to catalyze polysaccharide degradation.

Correlate rate of reaction with activation energy

ΔG‡ or activation energy is closely related to reaction rate as it determines whether the reaction rate runs slower or faster. When ΔG‡ is higher the reaction rate is slower When ΔG‡ is lower the reaction rate is faster Enzymes speed up reactions by lowering the activation energy; catalysts can also do this but enzymes are more efficient.

Induced-Fit Model

Binding of the substrate induces a change of conformation of the enzyme that results in a complementary fit. The binding site has a 3-dimensional shape before the substrate is bound The Model is more attractive in terms of considering the nature of the transition state and lowered activation energy that occurs with an enzyme catalyzed reaction.

Compare and contrast the properties of ordinary catalysts versus biological catalysts

Biological catalysts are widely known as enzymes, they are protein molecules that act as a catalyst in a body cell. Biological catalysts enhance the rate of reaction by 10^20 Biological catalysts can also be fine-tuned wherein they can either speed up or slow down depending on the metabolic needs of the cell. Ordinary catalysts on the other hand enhance rates of reaction by 10^2 to 10^4 Virtually has no reaction in the cell that occurs without enzymes

Differentiate between the behavior of chymotrypsin and of ATCase in terms of the plots of their initial velocity versus substrate concentration

Chymotrypsin is a non-allosteric enzyme and exhibits hyperbolic kinetics Chymotrypsin catalyzes the selective hydrolysis of peptide bonds where the carboxyl is contributed by Phe and Tyr Depending on the reaction of p-nitrophenyl acetate concentration it is catalyzed by chymotrypsin. ATCase is an allosteric enzyme thus exhibiting sigmoidal kinetics. It catalyzes the reaction and follows the rate of reaction where [carbamoyl phosphate] is constant. Note: Allosteric proteins are ones in which subtle changes at one site affect the structure and function at another site

Determine whether a given inhibitor is a non-competitive inhibitor from kinetic data

In the Lineweaver-Burk plot, a presence of a non-competitive inhibitor is seen when both the slope and y-intercept change for the inhibited reaction. This is of course without changing the x-intercept The Vmax decreases but the KM stays the same because of the inhibitor not interfering with the binding of the substrate to the active site.

Generate a typical kinetic plot of enzymatic reactions (reaction velocity vs. substrate concentration, constant enzyme concentration)

In this figure the graphical determination of Vmax and Km (Michaelis constant) are shown from a plot of reaction velocity (V) vs substrate concentration [S] Vmax is the constant rate reached when the enzyme is completely saturated with substrate When the rate of reaction is half its max value, the substrate concentration is equal to the Michaelis Constant

Explain the mechanism of uncompetitive inhibition

In uncompetitive inhibition the inhibitor can bind to the ES complex but not to the free enzyme. These inhibitors distort the active site which prevents the enzyme from being catalytically active but it does not block the binding of the substrate. Inhibition can't occur with an enzyme that only acts on a single substrate at a time.

Account for the effects of irreversible inhibition or of 'suicide substrates'

Irreversible inhibition happens when an inhibitor binds to the enzyme and permanently inactivates it hence why it is called "irreversible" Inhibitor binds covalently to the enzyme and never lets it go Suicide substrates - molecules created for the specific purpose of binding to the enzyme and inactivating it. Example of suicide substrate: Penicillin which binds to a specific serine residue of a bacterial enzyme rendering the bacteria unable to make cell walls.

Understand the significance of the Michaelis constant, KM

KM is equal to the concentration of substrate at which 50% of enzyme active sites are occupied by substrate The significance of the Michaelis constant is rooted in how high or how low it is A low KM value has high affinity for substrate A high KM value has low affinity for substrate

Be aware of other enzyme mechanisms aside from a simple 'one substrate-one product' system (ordered, random, ping-pong)

Ordered Mechanism - an enzyme mechanism where substrates have to bind to the enzyme in a SPECIFIC order Random Mechanism - an enzyme mechanism where substrates can bind to the enzyme in ANY order Ping-pong - the substrate binds to the enzyme and releases a product before the 2nd substrate binds to the enzyme

Differentiate between the progress of ordinary reactions and of enzymatic reactions using the free energy profile (reactants, transition state, products, progress of reaction, free energy, activation energy (or Ea), effect of temperature)

Ordinary reactions and enzymatic reactions both have the same initial and final points of their reactions. The difference however is in the paths these 2 reactions take. Enzymatic or catalyzed reactions do not need a larger activation energy to reach a transition state whilst ordinary uncatalyzed reactions need a larger amount of free energy to reach the transition state. Thus an enzymatic reaction has a greater reaction rate than ordinary ones due to it requiring less energy.

Orders of Reaction

Overall Order Refers to the sum of all exponents in the rate law equation 1st Order Reaction The rate of reaction is directly proportional to the concentration of reactant 2nd Order Reaction The rate of reaction is proportional to the square of the concentration of reactant. Zero Order Reaction Rate of reaction is independent of the concentration of reactant

Explain the details in the individual steps in the Michaelis-Menten mechanism of converting substrates into products

The Michaelis-Menten mechanism is divided into two steps: In the first step the substrate binds to the active site of the enzyme In the second step the substrate is converted into the products and released from the substrate The enzyme catalyzing the reaction is unchanged throughout the process.

Explain the complications or the low-reliability in determining the value of KM using graphical methods from a Michaelis-Menten Plot

The basis of determining KM graphically is when the rate of reaction is half its max value. Vmax used to be difficult to estimate because of the asymptotic nature of a hyperbola where the value is never reached with any finite concentration. In turn this made it difficult to determine the KM of an enzyme

Correlate the free energy of the system (reactants/transition state/product) as the enzymatic reaction progresses

The enzyme and the substrate are the reactants An E + S complex is then formed which has lower energy than the E + S at the start The bound ES must then attain the conformation of the transition state EX‡ If E and S binding was a perfect fit the ES would be at low energy and the difference between the ES and transition state (EX‡) would be large, which slows down the reaction rate. Thus enzymes help out in lowering the energy of the transition state and increasing the energy of the ES complex

Cite the implications of the initial velocity of reactions

The initial velocity of a reaction is a rate measured immediately after enzyme and substrate are mixed This is done so that we can be certain that the product is not converted to substrate to any appreciable extent. It is important to note that velocity measured IS initial velocity. At infinite substrate concentration the reaction would proceed at maximum velocity

Explain why the formation of the enzyme-substrate complex (ES) is a necessary step in the formation of the product in enzymatic reactions

The presence of an enzyme complex lowers the activation energy of a reaction. Without it being formed prior it will not promote the rapid progression of providing certain ions or chemical groups that are needed to form covalent bonds with molecules.

Rate Law

The rate law is expressed as a proportion or equation where: Rate = K[A]f[B]g K = rate constant F = order of reaction with respect to A (determined experimentally) G = order of reaction with respect to B (not necessarily equal to the coefficients of a balanced equation but frequently they are) The values of exponents are related to the number of molecules involved in detailed steps that constitute the mechanism.

Determine whether a given inhibitor is a competitive inhibitor from kinetic data

The slope and x-intercept changes but the y-intercept does not The Vmax is unchanged but the KM increases Vmax is unchanged because it is the measure of velocity of an infinite substrate; more substrate outcompetes an inhibitor.

Lock and Key Model

The substrate bind to a portion of the enzyme with a complementary shape This assumes a high degree of similarity between the shape of substrate and geometry of the binding site on the enzyme The model has intuitive appeal, however it does not consider an important protein, namely their conformational flexibility

Review the difference between thermodynamics and kinetics (Is the reaction "spontaneous"? Is the reaction "fast"?)

Thermodynamics describes the overall properties, behavior and equilibrium composition of a reaction system. Kinetics on the other hand describes the RATE at which a particular process will occur and the pathway by which the reaction will occur.

Explain the significance of the Vmax value to the efficiency of the enzyme (related term 'turnover number')

Vmax is the turnover number of an enzyme, this constant is equally referred to as Kcat Note that enzyme efficiency is Kcat/KM Kcat - describes how many substrate molecules are transferred into products per unit enzyme. By dividing the Kcat with KM it can create a ratio to test enzyme efficiency wherein: The greater the ratio the higher rate of catalysis The lower the ratio the lower rate of catalysis

mechanism of non-competitive inhibition

When the enzyme is in equilibrium with a substrate it forms an ES complex which can then form the product. However, an enzyme can also bind to an inhibitor but the binding of the enzyme to the active site does not prevent the inhibitor from binding to the substrate itself. There is no product formed in an ESI complex Due to the inhibitor not interfering with the binding of the substrate to the active site the KM is unchanged. Increasing substrate concentration will not overcome non-competitive inhibition