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1
SolubilityChromatographyElectrophoresis
Mr
is the sum of the molecular weights of amino acids(-H2O) plus weights of other components.
AAs have different Mr; 113/residue is a good averageVolume can be estimated from the partial specificvolume of the protein, which is between 0.72 and 0.75mL/g for most proteins
A protein of 200 residues is about 22,600 Da, or 3.8 x10-20 g/molecule
Its volume is about 2.8 x 10-20 cm3 or 2.8 x 104 3Its radius is therefore about 19
Compound Molecular Weight
Glycine 33. Da
Lysozyme 14. kDa
Trypsin 24
Pepsin 34
Horserdish peroxidase 40.7
-Amylase 50
LDH 140
Catalase 250
Glutamate dehydrogenase 1,000
Semipermeable membrane Has a molecular weight cut-offLeads to selective dilution:
Small molecules/ions diffusefreely
They are diluted to the totalvolume (5:1005, in the
example)
Large molecules are trappedinside
(unless the small molecules are
1.
counterions to or2. bound by the macromolecule)
1L
5 mL
Depends on interaction with solvent waterCharge-dipole (depends on pH)Dipole-dipoleH-bonding
Proteins with >30% hydrophobic residues cannotcover them with polar groups, and mustAssociate with other proteinsAssociate with membrane
Charge-charge attraction between proteins candecrease solubility (depends on pH)
or
KA=
[H+
][A]
[HA]
KA= [H
+
][A
]
[HA]
Take logs
logKA = log[H+
]+ log[A]
[HA]
rearrange
log[H+]= logKA+ log
[A]
[HA]
HA H+ + A-Equilibrium
pH= pKA + log[A]
[HA]
Henderson-Hasselbalch equation
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Charge vs pH
Imagine a polymer Rwith attachedacidic group AH
R-AH R-A- + H+
Net charge on thepolymer followsthe HH equation
-1
-0.8
-0.6
-0.4
-0.2
0
1 2 3 4 5 6
pH
Ch
X
pH= pKA+ log
[A]
[HA]
pH= pKA + log[A]
[HA]
H3N+-R-COOH H3N+-R-COO-H2N-R-COO-
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1 3 5 7 9 11
pH
Charge
-25
-20
-15
-10
-5
0
5
10
15
1 2 3 4 5 6 7 8 9 1 0 11 12
pH
Charge
Isoelectric point (pH)
Positively
charged
Negatively
charged
Example: 22 Acidic Groups and 12 Basic Groups uniform pKas
Guessing the Isoelectric Point
More Acidic groups - so, pI will be acidic
Which means all basic groups will be protonated (+)
You must deprotonate enough acidic groups to balance
them:
12+ charges (Lys + Arg sidechains, N-terminus)
you need 12 - charges (-COO-)
leaving 10 protonated -COOH
22 Acidic Groups and 12 Basic Groups
pI= pH= pKa + log12
10
= 4 + .08 4.1
Better: use Excel with real pKas
Isoelectric PointProteins are LEAST soluble at their isoelectric point
No net charge - minimal repulsionPerhaps +/- local attractive charge interactions
Proteins with different isoelectric points MAY beseparable by adjusting the pH
Is NaCl equivalent to (NH4)2SO4?
First, define concentration of salt solution, Ionic Strength
where mi is the molal concentration of component iand Zi is the charge on component i
I=1
2m
iZ
i
2
I=11
2+1 (1)2
2
=1M
I=21
2+1 (2)2
2
= 3M
1M NaCl ( 1m) 1M Na+ plus 1M Cl-
1M (NH4)2SO4 2M NH4+ plus 1M SO4
2-
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Complication for weak acids or bases
Ionic Strength depends on pH:
H2CO3 H+ + HCO3
-
H+ + CO32-
Z = 0, doesnt matter
Z = -1
Z = -2
[H+] is usually negligible
Use the H-H equation to
Calculate [HCO3-] and
[CO32-]
[H2CO
3] + [HCO
3
-] + [CO3
2-] = 0.20 M
At pH 10, [H2CO3] 0, so [HCO3-] + [CO3
2-] = 0.20 M
pH= pKa+ log
[CO32]
[HCO3]
10 =10.3+ log[CO3
2]
[HCO3]
[CO3
2]
[HCO3
]= 0.5
0.2M = [HCO3-] + [CO3
2-] = [HCO3-] + 0.5 [HCO3
-]
[HCO3-] = 0.13 M and [CO3
2-] = 0.07 M
So, per liter, 0.13 moles ofNaHCO3 and 0.07 moles ofNa2CO3
I=1
2{[Na
+
](+1)2+[HCO
3
](1)
2+ [CO
3
2](2)
2}
log
(solubility)
Enzyme 2
Enzyme 1
Ammonium sulfate (% of saturation)
Enzyme 3
40%20% 60% 80%
Proteins may associate by charge-charge interactions
Less interaction with solvent means lower solubility
The charge-charge force is described as
in a vacuum, or
in a solvent of dialectric constant D
For Water, D >80
For EtOH, D = 35
Mixtures, in between proteins may precipitate
Use any water-miscible solvent
Forcevacuum
=Z
+
Z
r2
Forcesolvent
=Z
+
Z
Dr2
A stationary phase A mobile phase Solutes that partition
between them
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Systems
Mobile Stat MethodGas Solid GC (fract., distil.)Gas Liquid G(L)PC (capillary)Liquid Solid LSC (silica, alumina)Liquid Liquid Partition LC (RP)Liquid Liquid countercurrent
distribution
General Theory- Definitions
Partition Coefficient
Phase Ratio
Capacity Ratio
Amount of solute As = CsVs
K =CS
CM
=VM
VS
k =K
=
CS
CM
VS
VM
General Theory- Ave.Speed
Ave. Speed of Solute
Fraction of time in mobile phase
So,
S = SM fM + SS fS = SM fM
AM
AM+ A
S
=
CMVM
CMVM+C
SVS
=
1
1+CS
CM
VS
VM
=
1
1+ k
S = SM
1
1+ k
General Theory- Ave.Speed
Rearrange:
If solvent & solute travelfor the same length oftime, speed distance.
So, measure:
S
SM
=
1
1+ k= R
f
R f =Spotdistance
Solventdistance
TLC or Paper
Column Chromatography
Solvent and solute must travel the samedistance (in different times)
So, invert everything, measure retention
time, tr, or elution volume, Vetr
tm
=
Ve
VM
=
SM
S=1+ k
(At constant flow rate t V chart distance)
Play with the Equation
Substitute for k:
Multiply by VM
Ve
VM
=1+CSVS
CMVM
Ve=V
M+
CSVS
CM
=VM+KV
S
Ve
VM
=1+ k Always works
Only works if youcan measure (or
define) VSi.e., partition
chromatography
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Solve for K or k
k=V
e
VM
1 duh( )
K =V
eV
M
VS
Different for every peak
Only if Vs is defined -
Not adsorption chromatog
Problem:predicts verysharp peaks
Try another approach:
model the process as a sequential extraction
1. Put solvent (stationary phase) in tubes2. Put solute in first tube only3. Put immiscible solvent in first tube4. Shake, solute distributes itself according to K
CM
CS
Sequential Extraction
Transfer only the upper phase to the next tube
Fill first tube with fresh upper solvent
Shake, repeat
Sequential ExtractionTransfer only the upper phase to the next tube
Solute that prefers the mobile phase travels faster
Model the process as material balance about the nthtube as small volume dv is transferred
Amtn = Amt in - Amt out = Cn-1dv-Cndv
EquationsCi = conc of solute in the mobile phaseof tube iAmtn = Amt in - Amt out = Cn-1dv - CndvBut the change in total amount = d[CnVm+ CsVs]or,Cn-1dv - Cndv = d[CnVm+ CsVs]
Substitute,for Cs using K = Cs/Cn
Cn-1dv - Cndv = d[Cn(Vs + KVs)] = (Vs + KVs)Cn
Rearrange:
dCn
dV+
Cn
VM+KV
S
=
Cn1
VM+KV
S
Cn as function of volume flowing
For each Cn, you have to know Cn-1, so youhave to calculate ALL previous C i
Luckily, a pattern develops:
Also, all except for
Cn=
Ci
0(aV)ni
(n 1)!eaV
i=1
n
Ci
0= 0
C1
0
Cn=C
i
0 (aV)n i
(n 1)!eaV
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6
Plot Cn vs volume that flowsC
n=C
i
0 (aV)ni
(n 1)!eaV
Cn vs tube # as solvent flows
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0 5 10 15 20 25 30 35 40
Tube number
Concentrationi
nt
ube
n
V = 1
V = 5
V = 10
V = 20
Interpretation
Make a simpler variable a = 1/(VM + KVS)Volume is in units of 1/aOn a TLC plate, as solvent flows
Solute moves along the plateSpots broadenIf different solutes have different K, they move
at different rates (distances)
Column Chromatography
Dont care how solutes are distributed onthe column, but when (at what volume) theycome OUT
We need to know the solute concentrationin the LAST tube, and how that changeswith the volume that flowsachromatogram
Chromatogram
Let the number of tubes N be a very largenumber (>100), so that N N-1
Calculate CN
orCN =Ci
0 (aV)N
(N)!eaV
CN
Ci
0=
(aV)N
(N)!eaV
ChromatogramCN
Ci0=
(aV)N
(N)!eaV
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
50 70 90 110 130 150
V (multiples of 1/a)
Ci
n
lasttube
Who cares? We do
What is the elution volume? What controls the width?
For Ve, take first derivative, set tozero, solve for V
I simplified to
And used d(uw) = udw + wdu = 0I got X = N, or aV = N or
Where the Vs are for each tubemultiply by the # of tubes.
Y = (eX)X
N
N!
Ve =
N
a=N(VM +KVS)
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Who cares? Ve is the same as for the simple treatment What else do we learn? Width?
Take 2nd derivative & set to zero Draw tangents thru the inflection points
Xinflection
= N N
Yinflection
Yinflection
N + N = Xinflection
L1 L2
N
Width at Base?
Xinflection
= N N
Yinflection
Yinflection
N + N = Xinflection
L1 L2
N
Get the Ls from the slope of tangent(rise over run = derivative)
Add the four segments:
Width = aV= L1+ (N-Xinf) + (Xinf-N) + L2
Get
So, width = aV =
We also know that aVe = N, so
or
L1= N 1
L2= N +1
4 N
aVe
aV=
N
4 N=
N
4
N =16Ve
V
2
Calculate the number of Plates
From the elution volume and width at base(in the same units), vol, time, chartdistance
More plates means lessbroadening per unitflow, higher column efficiency
Theoretical Plates or Plate Number
N =16Ve
V
2
Who cares?We doResolution depends on having different Ve,
but also narrow lines
W1 W2
Ve1
Ve2R =
2(Ve)
W1 +W2
N =16Ve
W
2
N is important
What determines N?
Unfortunately, Plate Theory is a lie. The solute never reaches equilibrium There is continuous flow It is the competition between partitioning
and flow that matters Use Rate Theory - Giddings Too hard for me
The Giddings Equation
L/H is the height equivalent to atheoretical plate, H or HEPT
H =1
1
Cedp
+1
Cmdp2 / D
m
+
CdD
m
+
Csmdp2
Dm
+
Cs
df
Ds
, linear solvent velocity
dp, particle diameter
ds, thickness of stat phase filmDi, diffusion coeffs, s and m
phases
. eddy currents
. mass transfer in m phase
. longitudinal diffusion. stagnant m phase escape
. mass xfer in stat phase
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Lessons from Giddings
Biphasic behaviorThere is an optimum gas velocityProbably always above opt velocity for liquids, so usually you
can increase resolution by decreasing the flow rate
Longitudinaldiffusion-limited
Mass transfer-limited
optimum solvent velocity
reasonable working range
HETP
Solvent velocity
Lessons from Giddings - 2
Smaller particles are always betterRemove particles altogether for best
performance (capillaries)
Different flow paths
Stagnant mobile phase
Different solvent
velocities
Stagnant stationaryphase
Sm aller colum n pack ing Larg er p ac king
A stationary phase A mobile phase Solutes that partition
between them
Pour a
column
Remove buffer
and layer thesample on top
Let sample enter
collect or measure eluatelayer buffer on top
Attach to
resevoir
For all kinds of chromatography, retention time:
tr= t
o+ k t
o
tr
to
=1+ k
Ve=V
o+K
dV
s
[solute]
Ve1
Ve2
Ve3
time or volume
or
If the volume of stationary phase is defined:
Ve
Vo
=1+ kor
0
50
100
150
200
250
0 50 100 150 200 250
Volume (mL)
[Protein](mg/mL)
Activity(U/mL)
Protein
Activity
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Column is packed with porous beadsVbed is the volume of packing
Vs is the volume of the beads
Vo is between beads (void volume)
Bed height
Vbed= r2 (bed height)
Vo
Vs
Column is packed with porous beadsSmall molecules can enter pores, large molecules cannot
Enters pores freely
Ve = Vo + 1Vs
Enters partially
Ve = Vo + KVs
Does not enter
Ve = Vo
Vo = volume between beads
Vs = volume of the beads, including pores
Vbed= Vo + Vs so Vs = Vbed - Vo
Measure Vbed physically, or calculate it geometricallyMeasure Vo as Ve of a very large molecule (blue dextran)
Kav
=
VeV
o
VS
=
VeV
o
Vbed
Vo
Kav
Log (Mr)
approximate
correlation
Ve=V
0+K
avV
s
1
0
av
K
Log Mr
Cytochrome c
Hemoglobin
BSAAdH
CatalaseUrease
FerritinFibrinogen
6543 7
K depends on apparent
molecular size, not Mr
Assume hemispherical
pores, with volume of
Large molecule of radius
a cant fit as easily. The
pore appears to have a
volume of
vol =2
3r
3
r
a
volapp =2
3(r a)
3
Ve=V
0+KV
S r
a
K
app V
pore =K
true V
app
K is actually the same for all solutes, K = 1
Different sized molecules see pores differently
Lets pretend K changes
Kapp =
Ktrue
Vapp
Vpore=1 Ktrue
Vapp
Vpore=
23 (r a)3
2
3(r)
3=(r a)3
(r)3Solve:
This a is the Stokes radius, Rs from
f = 6Rs
Kapp( )
1/3
vs a should give a straight line1
0 20 40 60 8 0
R s ()
1
100
K(1/3)
DesaltingRemoval of substrate or inhibitorChange bufferSeparate Proteins of different sizeEstimate molecular weight
Kav Log Mr (if standard and unknown have thesame shape)
Estimate Stokes radius (apparent hydrodynamic radius)Several equations exist
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11
Commonly use sepharose or Sephadex
Activated with CNBr
H2N(CH2)6NH2 and H2N(CH2)6CO2H Are common spacer arms
OH
OH
N
C
BrOH
CO
N
CO
O
NHC
O
O
N
H2N R X
+
R X
OH
O
C
O
NH R X
+
Bead-N(CH2)6NH2 is a nucleophile and can be
attached to many biochemicals (e.g., 8-Br-AMP) or to
proteins by making amide bonds to Asp & Glu
sidechains (using carbodiimide reagents)
Bead-N(CH2)6CO2H can be attached to amines,
including proteins, by making amide bonds (to Lys
sidechain, using carbodiimide reagents)
Several other spacer arms are common [Protein]
[Activity]
2. Wash with loading buffer to remove un-bound
and nonspecifically-bound protein
washload
1. Load the column, low salt, etc.
Most proteins do not bind Stop loading if activity starts to pass through
3. Elute with competitor (second chance at specificity)
elute
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We understand why an enzyme might bind immobilized AMP
But some enzymes bind immobilized aromatic dyes (e.g., Affigel Blue)
even though they do not resemble substrate or inhibitorotherwise, behave just like affinity ligands
Examples (on sepharose, agarose or cellulose)
Cibacron Blue 3GA Reactive Green-19Reactive Red-120
Buy a set of immobilized dyes and screen for binding, elution
Attach marcomolecules as a stationary phaseAntibodiesLectins
e.g., Concanavalin AAvidin/Steptavidin Inhibitor/antagonist
e.g., egg ovomucoid trypsin inhibitorOr immobilize metal ions (e.g., Ni2+) by chelation
chromatograph His-tagged proteins
Began as control experiments for Affinity Chromatog. Commonly used stationary phases
Butyl sepharose or agaroseOctyl sepharose or agarosePhenyl sepharose or agarose
Use Ammonium Sulfate to decrease solubility of the proteins Elute with gradient of amm. sulf., & gradient of eth. glycol
Hydrophobic hairs on beadprotrude into hydrophobic pockets
polarsurfaces interact, alsoBead
t
u
b
e
s
l
a
b
horizo
ntal
CH
2CH C
O
NH2n + m
acrylamide N,N-methylene-bisacrylamide
Casting gels (free radical polymerization)CH
2CH C
N
O
H
CH 2
C
O
C H C H2
N
H
C
OH2N C
O NH
CO
NH2
CH2
NH
C
O
CNH2
O
COH2N
Free-radical polymerization leads to a cross-linked gelAcrylamide concentration determines overall density of gelBis- methyleneacrylamide determines the extent of crosslinkingGels are characterized by %T and %C%T = [grams of (acrylamide + bis) / volume (mL)] x 100%C = [grams of bis / grams of (acrylamide + bis)] x 100
Total acrylamide (approximately % by wt)
5% < T < 20%
Determines pore size
Crosslinker (% of acrylamide that is bis, by wt)
2% < C < 5%
Determines tensile properties
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1. Polymerize gel in tube or between glass plates2. Put samples in wells of gel, include tracking dye3. Suspend gel between two buffer reservoirs 4. Apply constant voltage (100V)5. Turn off electricity when tracking dye approaches the
bottom of gel
6. Remove gel from tube or plates7. Fix and stain (methanol/acetic acid/water, Coomassie Blue)
fixing - methanol/acetic acid precipitates the proteins Coomassie Blue stains everything
8. Destain (methanol/acetic acid/water)
Ion with Z charges of e units
in an electric field E
feels a force and accelerates
FE= ZeE
a =F
mass
Particle with frictional coefficient f
moving with velocity V
feels a frictional force (drag)
Ff = f V
velocity
unit field strength
charge on protein
friction (size)
Acceleration stops and the
particle moves at constant speed when FE= Ff
ZeE=Vfor
V
E=
Ze
f
Which can be re-written as
Separation depends on chargeCan vary the charge by changing the pH
Intercept
depends
on charge
(pH)
Slope depends
on size largestleast charged
Separation depends on sizeCant vary size, but you can Vary frictional drag by increasing %TAnalogy to Stokes eqn.:f= 6Rs, (%T )
Log(mobility)
0 5 10 15 20
%T
At constant pH, can vary %T and plot mobility vs %T
Remove uncertainty about Shape, by denaturationCharge, by coating protein with ionic material
O S O-
O
O
Sodium dodecyl sulfate (and heat)Unfolds the protein
randomizes the shape, sof sizeCoats hydrophobic portions
provides negative chargelarger proteins bind more, soacceleration = FE/mass is the same for all proteins
Mobility depends only on friction, which depends on M rInclude 2-mercaptoethanol to reduce disulfide bondsBeware that glycoproteins bind less detergent
Lower acceleration, thereforeAppear larger
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Mobility depends only on size (not shape or charge)Plot the mobility vs Log (Mr) (Ferguson Plot)
Mobility(cmor%o
ftrackingdye)
Log (Mr)
Unknown
Standards
Mobility depends only on size (not shape or charge)
Same experiment, but pour the gels with a gradient maker
High%T
Low%T
Pour gel containing
ampholytes,
a mixture of
amphoteric amines/
acids
pH 2-3
pH 11
Use acidic buffer at
bottom, alkaline at top
Negative ampholytes
move toward anode
until they becomeprotonated, stop at pI
+
Proteins behave like
ampholytes
Distribute according to
their pI
+
Run IEF gel in one dimension
Treat with SDS
Place treated IEF gel on top of SDS gel
Electrophorese into SDS Gel to get a 2-Dimensional Separation
MonomerM
r
pI
1. Coomassie Brilliant Blue R in methanol/acetic acid water destain in similar solution without dye scan with densitometer
2. Colloidal gold or silver 10x more sensitive
3. Activity stains - very selective Infuse substrates into gel after electrophoresis
colorimetric (UV/Vis) indicator Zymogram
e.g., look for clarification of starch gel4. Antibody stain (Western Blot)
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Recommended