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Distinguishing Features

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Distinguishing Features. Separate the protein from all other ... Surface properties: steric access, shape, hydrophobic accessible surface, electrostatic surface ... – PowerPoint PPT presentation

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Title: Distinguishing Features


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
  • Sequencing- degradation, Mass Spectrometry
  • Spectroscopy- UV, CD, fluorescence, EPR, NMR
  • Antibody binding- specificity
  • Structure- X-ray crystallography, NMR

3
Proteins Function By Binding
  • Transport- O2/CO2, 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 DG.
  • ? 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 DG

8
The Catalytic Effect of Enzymes
  • ? Enzymes lower transition state, reducing DG
  • ? The catalytic efficiency (DDGcat) is DDG
    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 productsubstrate
    DGreaction

9
First Order Rate Equations (cont.)
  • Equation for a first order reaction lnS
    lnSo kt
  • A plot of lnS vs. t is a straight line
  • The intercept is the starting concentration So
  • Slope is the negative of the rate constant (k)
  • Half-life of the reaction (t½) 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.
  • t½ ln2/k 0.693/k

10
Enzyme Kinetics
  • E S ? ES ? P E
  • E- enzyme, S- substrate, ES- enzyme-substrate
    complex, P-product, k1,k2 - forward rates, k-1 -
    reverse rate, k-2 - negligible)
  • v dP/dt k2 ES
  • ES is difficult to measure because it is time
    dependent
  • dES/dt k1 ES - k-1ES - k2ES
  • Only the total concentration of enzyme ET E
    ES can be measured.

11
Steady State Assumption
? At steady state S gtgt E, ES remains
constant because all enzyme active sites filled
with S, dES/dt 0 ? Recall from the last
page dES/dt k1 ES - k-1ES - k2ES v
dP/dt k2 ES ET E ES ? So, at
steady state v dP/dt k2 ES k1E S -
k-1ES k1 term k1 E S k-1 ES k2 ES
(k-1 k2) ES substitute k1 (ET - ES)
S (k-1 k2) ES
12
Continuing to Reorganize
k1(ET - ES) S (k-1 k2) ES shift
k1 (ET - ES) S / ES (k-1 k2) / k1
Define the Michaelis Constant (KM) (k-1
k2)/k1 shift ES (ET - ES) S KM
ES expand ET S - ES S KM ES swap
term ET S ES S KM ES (S KM)
ES reorganize ES ET S / KM S
13
Express With Measurable Quantities
ES ET S / KM S Using this
relationship, and working in the regime where the
back reaction for ES is negligible, the initial
velocity can be expressed in ET and S, which
are measurable quantitites! vo k2 ES ?
vo k2 ET S/(KM S) At saturation ET
ES, so vo Vmax k2 ET This gives
Michaelis-Menten Equation of enzyme
kinetics vo Vmax S / (KM S)
14
How to Characterize M-M Kinetics
? M-M Equation vo Vmax S / (KM
S) S ltlt KM, vo (Vmax
/ KM ) S S KM,
vo Vmax /2 S gtgt Km,
vo Vmax ? Define kcat, turnover
number kcat Vmax / ET When S ltlt
KM, very little ES is formed, so ET E vo
(kcat / KM) E S
15
Implications of M-M Equation (cont.)
  • The kcat/Km term is a measure of the enzymes
    catalytic efficiency how often a molecule of
    substrate that is bound reacts to give product
  • The upper limit to kcat/Km is k1 because
    decomposition of ES to E P can occur no more
    frequently than ES is formed.
  • The most efficient enzymes have kcat/Km values
    near the diffusion-controlled limit conditions
    where a reaction occurs almost every time a
    substrate is bound.
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