6. Separations

8/2/2019 6. Separations

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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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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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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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6. Separations

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