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Distinguishing Features Separate the protein from all other contents Sequence and fold give overall properties  Molecular weight  Solubility  Exposed.

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Presentation on theme: "Distinguishing Features Separate the protein from all other contents Sequence and fold give overall properties  Molecular weight  Solubility  Exposed."— Presentation transcript:

1 Distinguishing Features Separate the protein from all other contents Sequence and fold give overall properties  Molecular weight  Solubility  Exposed hydrophobic surface  Ability to bind other molecules, metals  pI- the overall charge of the protein  Sequence!!!

2 Protein Analysis Purification A. Simple solubility characteristics- precipitation B. Chromatography- ion exchange, size, hydrophobic, specific affinity C. Gel electrophoresis- native, denaturing (SDS) Characterization A.Sequencing- degradation, Mass Spectrometry B.Spectroscopy- UV, CD, fluorescence, EPR, NMR C.Antibody binding- specificity D.Structure- X-ray crystallography, NMR

3 Proteins Function By Binding Transport- O 2 /CO 2, cholesterol, metals, sugars Storage- metals, amino acids, Immune response- foreign matter (antigens) Receptors- regulatory proteins, transmitters Structure- other structural proteins Enzymes- substrates, inhibitors, co-factors Toxins- receptors Cell functions- proteins, metals, ions Surface properties: steric access, shape, hydrophobic accessible surface, electrostatic surface

4 Regulation of Protein Function  Allosteric Control  Stimulation/inhibition by control factors  Reversible covalent modification  Proteolytic activation/inactivation Static Structure/Dynamic Biology Structures are static snapshots of highly dynamic molecular systems Biological process occur at femtosec - min. timescale

5 NMR of Proteins Challenges Proteins have hundreds/thousands of signals Resonance assignment first…..who do all these signals belong to? Need to use computer programs to convert from NMR data to structures Applications Folded protein? Measure binding constants Assess structural homology/effect of mutations Three-dimensional structure determination Measure flexibility/dynamics

6 Enzymes: Protein Catalysts  Increase rates of reaction, but not consumed  Enable reactions to occur under mild conditions  High reaction specificity/no side products  Catalytic mechanisms: Bond Strain, Proximity/Orientation, Acid/Base Catalysis, Covalent Catalysis, Metal Ions, Electrostatic, Preferential Binding of the Transition State  Activity of enzymes can be regulated  Availability of substrate or enzyme  Reversible covalent modification  Allosteric control (other proteins or co-factors)  Activity of enzymes can be inhibited  Competitive inhibition  Uncompetitive inhibition  Mixed or non-competitive inhibition  Inactivator

7 Energies and Rates of Reactions  The transition state is the highest point on the reaction coordinate diagram.  The height of the energy barrier is the Free Energy of Activation:  G ‡.  Backwards barrier usually higher than forward. The difference in energy between substrate and product is the Free Energy of Reaction.  If the energy barrier is higher for one step than the other, than the rate of this step will be slower. The step with the highest transition state free energy (the highest point on the reaction coordinate) is the Rate Determining Step of the reaction. KEY POINT: Ref. Is  G ‡

8 The Catalytic Effect of Enzymes  Enzymes lower transition state, reducing  G ‡  The catalytic efficiency (  G ‡ cat ) is  G ‡ catalyzed vs. uncatalyzed  Catalytic efficiency reflected in kinetic parameters: rate enhancement for the reaction.  Lowering of the free energy barrier and increase in rate is equal for the forward and the reverse reactions- transition state energy is lowered!!!  Time required to come to equilibrium is less.  Increase in velocity with which products are produced from reactants and vice versa.  No change in the ratio product:substrate:  G reaction

9 First Order Rate Equations (cont.) Equation for a first order reaction: ln[S] = ln[S] o –kt A plot of ln[S] vs. t is a straight line:  The intercept is the starting concentration [S] o  Slope is the negative of the rate constant (k) Half-life of the reaction (t ½ ) is the time required for half of S to be used up. The slope of the line in the plot never changes and the half-life is the same regardless of the starting concentration. t ½ = ln2/k = 0.693/k

10 Enzyme Kinetics E + S  ES  P + E E- enzyme, S- substrate, ES- enzyme-substrate complex, P-product, k 1,k 2 - forward rates, k- 1 - reverse rate, k -2 - negligible) v = d[P]/dt = k 2 [ES] [ES] is difficult to measure because it is time dependent: d[ES]/dt = k 1 [E][S] - k -1 [ES] - k 2 [ES] Only the total concentration of enzyme [E] T = [E] + [ES] can be measured.

11 Steady State Assumption  At steady state: [S] >> [E], [ES] remains constant because all enzyme active sites filled with S, d[ES]/dt = 0  Recall from the last page: d[ES]/dt = k 1 [E][S] - k -1 [ES] - k 2 [ES] v = d[P]/dt = k 2 [ES] [E] T = [E] + [ES]  So, at steady state: v = d[P]/dt = k 2 [ES] = k 1 [E] [S] - k -1 [ES] k 1 term k 1 [E] [S] = k -1 [ES] + k 2 [ES] = (k -1 + k 2 ) [ES] substitute k 1 ([E] T - [ES]) [S] = (k -1 + k 2 ) [ES]

12 Continuing to Reorganize k 1 ([E] T - [ES]) [S] = (k -1 + k 2 ) [ES] shift k 1 ([E] T - [ES]) [S] / [ES] = (k -1 + k 2 ) / k 1 Define the Michaelis Constant (K M ) = (k -1 + k 2 )/k 1 shift [ES ] ([E] T - [ES]) [S] = K M [ES] expand [E] T [S] - [ES] [S] = K M [ES] swap term [E] T [S] = [ES] [S] + K M [ES] = ([S] + K M ) [ES] reorganize [ES] = [E] T [S] / K M + [S]

13 Express With Measurable Quantities [ES] = [E] T [S] / K M + [S] Using this relationship, and working in the regime where the back reaction for [ES] is negligible, the initial velocity can be expressed in [E T ] and [S], which are measurable quantitites! v o = k 2 [ES]  v o = k 2 [E] T [S]/(K M + [S]) At saturation [E] T = [ES], so: v o = V max = k 2 [E] T This gives Michaelis-Menten Equation of enzyme kinetics: v o = V max [S] / (K M + [S])

14 How to Characterize M-M Kinetics  M-M Equation: v o = V max [S] / (K M + [S]) [S] << K M, v o = (V max / K M ) [S] [S] = K M, v o = V max /2 [S] >> K m, v o = V max  Define k cat, turnover number: k cat = V max / [E] T When [S] << K M, very little ES is formed, so [E] T = [E] v o ~ (k cat / K M ) [E] [S]

15 Implications of M-M Equation (cont.) The k cat /K m term is a measure of the enzyme’s catalytic efficiency: how often a molecule of substrate that is bound reacts to give product The upper limit to k cat /K m is k 1 because decomposition of ES to E + P can occur no more frequently than ES is formed. The most efficient enzymes have k cat /K m values near the diffusion-controlled limit: conditions where a reaction occurs almost every time a substrate is bound.


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