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Enzymes. CH339K. Transition State. On the way from reactants to products, the reaction goes through a high-energy intermediate structure Amount of energy to reach transition state controls rate Picture is from an O-Chem text – sorry - ugggghhhhhhhh. Reaction Pathways.
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Enzymes CH339K
Transition State • On the way from reactants to products, the reaction goes through a high-energy intermediate structure • Amount of energy to reach transition state controls rate • Picture is from an O-Chem text – sorry - ugggghhhhhhhh
Reaction Pathways • DGo’ is the difference in Free Energy between reactants and products • DGo’determines Keq • Rate depends on EA or DG‡how much energy is required to reach transition state.
What are enzymes? • There are two basic ways to increase the rate of a reaction • Increase the energy of the reactants (heat it up) • Lower the activation energy (catalysis) • Enzymes are protein catalysts. • Virtually every biochemical reaction is mediated by an enzyme.
For the Visual Thinkers: Two Options Raise the temperature of the reactants Use a catalyst Which is: Faster? Less likely to result in being eaten?
Activation Energy (you knew it would get to math) • Let’s look at a first-order reaction where substance X is converted to substance Y, going through the transition state X’: X ⇌ X’ ⇌ Y • The rate can be expressed as a rate constant k times the concentration of the reactant X: [1] v = k * [X]
Activation Energy (cont.) But if we think about it, Y is actually produced from the transition state, so the reaction velocity should really be governed by the concentration of transition state: [2] v = k’ * [X’] Now let’s assume we have a situation where X’ is far more likely to fall back to X than to proceed to form the product Y. Then X’ will essentially be in equilibrium with X, and the equilibrium will be governed by the same thermodynamic rules we have seen before: [3] X ⇌ X’ (forget Y for a second) [4] or [5]
Activation Energy (cont.) • so combining [2] and [5] [6] • Lowering the activation energy from EA1 to EA2 thus changes the rate by a factor of [7] i.e. lowering EA from 20 kJ/mol to 8 kJ/mol increases the rate by over 100x at 37o C!!!
Enzymes tend to be really good catalysts 21,000,000-fold rate increase
How do enzymes work? • Enzymes bind substrates in an active site, where the reaction takes place • Lock and key vs. induced fit (distortion of both enzyme and substrate plays a role in catalysis)
Active site • The active site frequently forms a cleft in the molecule • Substrate binding typically includes van der Waals contacts, H-bonds, and salt links, but can include covalent links.
Mechanisms • Increase effective concentrations • Orient the substrates • Stabilize the transition state • The energy of binding can subsidize conformational strain in the substrate • Acids and bases can participate in catalysis • Covalent or redox participation by the enzyme • Use of enzyme cofactors
Acid-base catalysis:Triose Phosphate Isomerase • Triose phosphate isomerase interconverts the two three-carbon sugars formed by the action of aldolase on fructose-1,6-bisphosphate in the glycolytic pathway
Acid-base catalysis:Triose Phosphate Isomerase • Glu165 acts as a base, extracting a proton from the substrate • His95 acts as an acid, donating a proton. G3P DHAP
Acid-Base + Conformational ChangeLysozyme • In 1922, Alexander Fleming plated bacterial cultures along with samples of his own snot. • Bacteria near his nasal mucus dissolved away. • The active ingredient, lysozyme, cleaves bacterial cell wall polysaccharides. • There is an extended substrate binding cleft that bonds a stretch of 6 sugars. • Lysozyme cleaves its substrate between the fourth and fifth residues in a hexasaccharide
Repeating Structure of Cell Wall Lysozyme cleaves here
Acid-Base + Conformational ChangeLysozyme (from egg white, not snot) Active Site Cleft
Acid-Base + Conformational ChangeLysozyme • Glu35 acts as an acid (has abnormally high pKa). • Asp52 stabilizes the charge on the oxycarbonium transition state. • Binding of 6 sugars subsidizes the torsion of the target sugar bond into a half-chair conformation. • This mimics the conformation of the intermediate, decreasing DG‡ for reaching the transition state.
Transition State Analogs Raising Monoclonal Antibodies
Transition State Analogs • One can create catalytic antibodies by rearing antibodies against transition state analogs Hydrolysis of Aryl Carbonates using p-nitrophenyl-4- carboxybutanephosphonate as antigen – rate acceleration > 104. Patten, P.A. et al, (1996) Science271: 1086-1091.
Covalent Participation - Chymotrypsin • Three key catalytic side chains - Far apart in sequence but adjacent in active site • Ser195 • His57 • Asp102
Covalent Participation - Chymotrypsin • Chymotrypsin is a serine protease. • Serine, Histidine, and Aspartic Acid form a charge relay system.
Subtilisin • From Bacillus subtilis • Same catalytic mechanism • Totally different protein; no evolutionary connection. • Triad: Ser221, His64, Asp32
Example of use of a cofactor • Histidine Decarboxylase
Example of a redox cofactor Glycolate Oxidase Uses Flavin Mononucleotide (FMN) FMN (oxidized) FMN (reduced) + H+ Glyoxylic acid (oxidized) Glycolic acid (reduced)
Enzyme Kinetics • Let's make a simple model of an enzyme-catalyzed reaction that converts one molecule of substrate (S) to one molecule of product: • Let's also make a few assumptions: • The reaction has just started, so [P] = 0 and k4 can be ignored. • [S] >>> [E], so substrate is in no way limiting. • As a result of (2), [S] isn't going to change appreciably during our observation of the reaction and we can assume [ES] is approximately constant.
Enzyme Kinetics • We will define [Etotal] as the total concentration of the enzyme. [Etotal] = [E] + [ES] • We will define v as the reaction velocity for formation of product v = k3[ES]
Enzyme Kinetics • Since [ES] is approximately constant, its rate of formation is equal to its rate of destruction: k1 [E][S] = (k2 + k3)[ES] [E][S] = ((k2 + k3)/k1)[ES] • Let's give a name to the combined constants. Let's call it Km (for the Michaelis constant): [E][S] = Km[ES] ([Etotal] - [ES])[S] = Km[ES] (since [Etotal] = [E] + [ES]) [Etotal][S] - [ES][S] = Km[ES]
Enzyme Kinetics [Etotal][S] - [ES][S] = Km[ES] • Rearranging: Km[ES] + [ ES][S] = [Etotal][S] [ES] = [Etotal][S] / (Km + [S]) • Remember v = k3[ ES], so v = k3[Etotal][S] / (Km + [S]) • The maximum velocity for the rxn is when every enzyme molecule is part of an ES complex: Vmax = k3[Etotal]
Enzyme Kinetics • The maximum velocity for the rxn is when every enzyme molecule is part of an ES complex: Vmax = k3[Etotal] and v = k3[Etotal][S] / (Km + [S]) • Simplifying the above: v = Vmax[S] / (Km + [S]) • This is the Michaelis - Menten equation, which does a pretty good job of describing the overall kinetics of many enzyme catalyzed reactions. There’s that blue arrow again!
kcat • Vmax will change with changing enzyme concentration. It would be nice if we could define a term equivalent to Vmax which was independent of enzyme concentration. • In addition, there are many enzyme catalyzed reactions that have several intermediate steps in the pathway from reactants to products. For example: where EI1, EI2, EI3 are complexes between the enzyme and successive intermediates. • What we can normally measure is not the set of individual reaction rates but rather an overall 'k3apparent' or kcat. • kcat is enzyme-adjusted measured Vmax; that is, • kcat = Vmax/[Etotal] (units of sec-1)
Km • Km is th Substrate concentration at which the reaction is occurring at one-half its maximal rate. It is thus a measure of how much substrate is required for reasonable enzyme activity. Km is often looked on as a dissociation constant for the Enzyme -Substrate complex. Since Km = (k2 + k3)/k1, • this will only be true when k3 << k2! This is frequently the case, but not necessarily!
kcat/Km • At low substrate concentrations, the Michaelis-menten equation reduces to v = (kcat/Km)*[Etotal][S] • i.e. at low [S], [E] ~ [Etotal] and Km + [S] ~ Km. • kcat/Km is thus a rate constant and as such is a measure of catalytic efficiency. • Theoretical maximum for the reaction rate is in the range of 108 - 109 M-1sec-1. • Many enzymes approach this limit and are thus said to have achieved catalytic perfection.
Measuring Km and Vmax (if you only have a pencil and a ruler) Rearrange Michaelis-Menten Equation: • Lineweaver-Burke • Eadie-Hofstee
Control by Inhibition • Inhibitors alter enzyme activity (Km and kcat) • Often used to control enzyme activity • Often used as toxins • Two basic flavors • Irreversible • Reversible • Which are pretty much like they sound
Nerve Gases Sarin Tabun Soman VX Symptoms: Contraction of pupils, profuse salivation, convulsions, involuntary urination and defecation and eventual death by asphyxiation as control is lost over respiratory muscles.
