Protein separation methods

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Protein separation methods
Ultracentrifugation
Mixture of proteins
Up to 100,000 x g
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Ultracentrifuge
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centrifugal force m(omega)2r
Causing sedimentation
m mass omega angular velocity r
distance from the center of rotation
Opposing sedimentation friction foV.
Constant velocity is soon reached centrifugal
force frictional force
So m(omega)2r   foV fo frictional
coefficient (depends on shape)
And  V m(omega)2r/fo,
V proportional to mass (MW) V inversely
proportional to fo (shape) V inversely
proportional to non-sphericity (Spherical shape
moves fastest)
Or  V (omega)2r x m / fo
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Glass plates
Large, high positive charge
Large, low positive charge
Small, High positive charge
Small, Low positive charge
Molecules shown after several hours of
electrophoresis
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Glass plates
Winner Small, High positive charge
Loser Large, low positive charge
Intermediate Large, high positive charge
Intermediate Small, Low positive charge
Molecules shown after several hours of
electrophoresis
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Upper resevoir
Cut out for contactof buffer with gel
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Clamped glass sandwich
Electrode connection( 150 V)
Reservoir for buffer
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Power supply
Happy post-doc
Tracking dyes
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SDS PAGE SDS polyacrylamide gel electrophoresis

• sodium dodecyl sulfate, SDS (or SLS)
  CH3-(CH2)11- SO4-
• CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-SO
  4-

SDS
All the polypeptides are denatured and behave as
random coils All the polypeptides have the same
charge per unit length All are subject to the
same electromotive force in the electric
field Separation based on the sieving effect of
the polyacrylamide gel Separation is by molecular
weight only SDS does not break covalent bonds
(i.e., disulfides) (but can treat with
mercatoethanol for that) (and perhaps boil
for a bit for good measure)
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P.A.G.E.
12 18 48 80 110 130
160
140
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Molecular sieve chromatography (gel filtration,
Sephadex chromatography)
Sephadex bead
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Molecular sieve chromatography
Sephadex bead
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Molecular sieve chromatography
Sephadex bead
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Molecular sieve chromatography
Sephadex bead
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Molecular sieve chromatography
Sephadex bead
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Larger molecules get to the bottom faster, and
. Non-spherical molecules get to the bottom
fatser
infrequent orientation
Non-spherical molecules get to the bottom faster
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Handout 4-3 protein separations
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Largest and most spherical
Lowest MW
Winners
Largest and least spherical
Similar to handout, but Winners Native PAGE
added
Most chargedand smallest
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Enzymes protein catalysts
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Each arrow an ENZYME
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Chemical reaction between 2 reactants
Spontaneous reaction Energy released Goes
to the right H-I is more stable than H-H or I-I
here I.e., the H-I bond is stronger, takes more
nergy to break it Thats why it goes to the
right, i.e., it will end up with more products
than reactants i.e., less tendency to go to the
left, since the products are more stable
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Change in Energy (Free Energy)
H2 I2

-3 kcal/mole
2 HI
Reaction goes spontaneously to the right
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Different ways of writing chemical reactions
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2H 2I
say, 100 kcal/mole
say, 103 kcal/mole
Change in Energy (Free Energy)
H2 I2

-3 kcal/mole
2 HI
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2H 2I
100 kcal/mole
Change in Energy
H-H I-I (TS)
Say, 20 kcal/mole
H2 I2
Activation Energy

-3 kcal/mole
2 HI
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HHII (TS)
Allows it to happen
Energy needed to bring molecules together to
form a TS complex
Change in Energy (new scale)
determines speed VELOCITY rate of a
reaction
Activation energy
H2 I2

3 kcal/mole
2 HI
Net energy change Which way it will end up
2 separate concepts
DIRECTION of the reaction, independent of the rate
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Biosynthesis of a fatty acid
3 glucose
18-carbon fatty acid
Free energy change 300 kcal per mole of
glucose is REQUIRED
3 glucose
18-carbon fatty acid
So getting a reaction to go in the direction you
want is a major problem (to be discussed next
time)
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Concerns about the cells chemical reactions

• Direction
• We need it to go in the direction we want
• Speed
• We need it to go fast enough to have the cell
  double in one generation
• Catalysts deal with this second problem, which we
  will now consider

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The velocity problem is solved by catalysts
The catalyzed reaction
The catalyst takes part in the reaction, but it
itself emerges unchanged
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HHII (TS)
Activation energy without catalyst
TS complex with catalyst
Change in Energy
Activation energy WITH the catalyst
H2 I2
2 HI
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Reactants in an enzyme-catalyzed reaction
substrates
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Reactants (substrates)
Active site or substrate binding site (not
exactly synonymous, could be part of the active
site)
Not a substrate
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Unlike inorganic catalysts, enzymes are specific
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Small molecules bind with great specificity to
pockets on protein surfaces
Too far
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Unlike inorganic catalysts, enzymes are specific

•                                      succinic
  dehydrogenase
• HOOC-HCCH-COOH lt--------------------------------gt
  HOOC-CH2-CH2-COOH
  
  
  2H
• fumaric acid                                  
                     succinic acid
• NOT a substrate for the enzyme
• 1-hydroxy-butenoate 
    HO-CHCH-COOH
• (simple OH instead of one of
  the carboxyl's)
• Maleic acid
• 
  
• 
  
  maleic
  acid
• Platinum will work with all of these,
  indiscriminantly

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• Enzymes work as catalysts for two reasons
• They bind the substrates putting them in close
  proximity.
• They participate in the reaction, weakening the
  covalent bonds
• of a substrate by its interaction with the
  enzymes amino acid residue side groups (e.g.,
  by stretching).

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Chemical kinetics

• Substrate ? Product (reactants in enzyme
  catalyzed reactions are called substrates)
• S ? P
• Velocity V ?P/ ? t
• So V also -?S/ ?t (disappearance)
• From the laws of mass action
• ?P/ ?t - ?S/ ?t k1S k2P
• For the INITIAL reaction, P is small and can be
  neglected
• ?P/ ?t - ?S/ ?t k1S
• So the INITIAL velocity Vo k1S

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Vo ?P/ ? t
P vs. t Slope Vo
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Effect of different initial substrate
concentrations
0.6
S4
S3
0.4
P
S2
0.2
S1
0.0
t
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Vo the slope in each case
Effect of different initial substrate
concentrations
0.6
S4
S3
0.4
P
S2
0.2
Vo k1S
S1
Slope k1
0.0
t
Considering Vo as a function of S (which wil be
our usual useful consideration)
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Now, with an enzyme
We can ignore the rate of the non-catalyzed
reaction
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Enzyme kinetics (as opposed to simple chemical
kinetics)
Vo independent of S
Vo proportional to S
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Michaelis and Menten mechanism for the action of
enzymes (1913)
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Michaelis-Menten mechanism
X

• Assumption 1. E S lt--gt ES this is how enzymes
  work, via a complex
• Assumption 2. Reaction 4 is negligible, when
  considering INITIAL velocities (Vo, not V).
• Assumption 3. The ES complex is in a
  STEADY-STATE, with its concentration unchanged
  with time during this period of initial rates. 
• (Steady state is not an equilibrium condition, it
  means that a compound is being added at the same
  rate as it is being lost, so that its
  concentration remains constant.)

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E S
ES
E P
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Michaelis-Menten Equation(s)
See handout at your leisure for the derivation
(algebra)
k3EoS
Vo
(k2k3)/k1 S
If we let Km (k2k3)/k1, just gathering 3
constants into one, then
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All the ks are constants for a particular enzyme
At high S (compared to Km), Rate is constant Vo
k3Eo
At low S (compared to Km), rate is proportional
to S Vo k3EoS/Km
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At high S, Vo here k3Eo, Vmax
So the Michaelis-Menten equation can be written
Vmax S
Vo
Simplest form
Km S
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• Now, Vmax k3Eo
• So k3 Vmax/Eo
• the maximum (dP/dt)/Eo, the maximum
  (-dS/dt)/Eo
• k3 the TURNOVER NUMBER
• the maximum number of moles of substrate
  converted to product per mole of enzyme per
  second
• max. molecules of substrate converted to product
  per molecule of enzyme per second
• Turnover number then is a measure of  the
  enzyme's catalytic power.

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Some turnover numbers

• Succinic dehydrogenase 19 (below average)
• Most enzymes 100 -1000
• The winner
• Carbonic anhydrase (CO2 H20 H2CO3)
• 600,000
• Thats 600,000 molecules of substrate, per
  molecule of enzyme, per second.
• Picture it!
• You cant.

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Got this far
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Km ?
Vmax/2 is achieved at a S turns out to be
numerically equal to Km
So Km is numerically equal to the concentration
of substrate required to drive the reaction at ½
the maximal velocity Try it Set Vo ½ Vmax and
solve for S.
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Another view of Km
Consider the reverse of this reaction (the
DISsociation of the ES complex)
The equilibrium constant for this dissociation
reaction is
Kd ES/ES k2/k1
(Its the forward rate constant divided by the
backward rate constant. See the Web lecture if
you want to see this relationship derived)
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Consider in reverse
Kd k2/k1
Km (k2k3)/k1 (by definition)
IF k3 ltlt k2, then Km k2/k1 But Kd k2/k1
(from last graphic) so Km Kd for the
dissociation reaction
(and 1/Km the association constant)
So the lower the Km, the more poorly it
dissociates. That is, the more TIGHTLY it is held
by the enzyme
And the greater the Km, the more readily the
substrate dissociates, so the enzyme is binding
it poorly
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Km ranges

• 10-6M is good
• 10-4M is mediocre
• 10-3M is fairly poor

So Km and k3 quantitatively characterize how an
enzyme does the job as a catalyst