Analytical ultracentrifugation
Directed movement in solution induced by an external force
(gravitation, centrifugation, electric field)
short period (10-9 sec) to reach constant velocity
Resisting force fv
from viscous drag
Driving force
Fx
Time after force is applied
− f ⋅v+ Fx = 0, or Fx = fv
⇒ if we measure the velocity of motion produced by a known force
€ can determine the friction coefficient and diffusion coefficient
we
Determination of sedimentation- and diffusion
coefficients by analytical ultracentrifugation (AUC)
absorbance
measurements
Fc: centrifugal force
Fb
Ff
Ff: frictional force
Fc
Fb: buoyant force
Determination of molecular weight and hydrodynamic shape (sedimentation-,
diffusions- and friction coefficient) of biological macromolecules in solution
Idealized sedimentation velocity profile in the absence of diffusion
t0
t1
t2
t3
t0
t1
t2
t3
radial position r
Sedimentation velocity analysis
t0
t1
t0
t1
t2
t3
t2
radial position r
t3
- velocity of band movement -> sedimentation coefficient
- spreading of band boundary -> diffusion coefficient
- sedimentation and diffusion coefficient -> molecular weight
Parameters that describe the hydrodynamic
properties of macromolecules in solution
• diffusion coefficient D
kT
D=
f
• frictional coefficient f
dr dt M ⋅(1-vρ )
s= 2 =
N A ft
ϖ r
• sedimentation coefficient s
v
• partial specific volume v bar ( )
protein: 0.73 ml g-1, DNA: 0.55 ml g-1
€
∂v
v=
∂m
• mass M of the molecules
• density ρ of buffer, 0.9982 g ml-1 for water at 20 ˚C
• viscosity η of buffer, 1.002 mPa second for water at 20 ˚C
Relations that involve the frictional coefficient
€
4
Vsphere = ⋅π ⋅r 3
3
A sphere has radius r and volume V.
M ⋅v
Vsphere =
NA
The volume of a spherical particel can be calculated
from its molecular mass, the partial specific volume v
bar and Avogadros number NA.
€
ft = 6⋅π ⋅η ⋅r
€
For the translation the resulting frictional
coefficient ff in a medium with viscosity η can
be calculated according to Stokes law.
Determination of sedimentation- and diffusion
coefficients by analytical ultracentrifugation (AUC)
Fc: centrifugal force
absorbance
measurements
Fc = ω 2 ⋅r ⋅m
proportional to the mass m,
distance r from the center and
€ angular velocity ω (2π/60 x rpm)
Fb
Ff
Ff: frictional force
Fc
Ff = – f ⋅v
€
Fb: buoyant force
proportional to the mass of the displaced solvent m0,
which can be calculated from the density of the solvent ρ
and the partial specific volume v bar of the molecule
Fb = – ω 2 ⋅r ⋅mo
with
€
m0 = m ⋅ ν ⋅ ρ
Determination of the sedimentation coefficient
Fc
€
+
Fb
+ Ff = 0
At constant velocity the sum of all forces is zero.
ω 2 ⋅r ⋅m – ω 2 ⋅r ⋅mo – f ⋅v = 0
M⋅(1-ν ⋅ρ )
v
= 2 =s
NA ⋅ f
ω ⋅r
kT
D=
f
Rearranging and using the molecular weight M of
avogadro number NA (=1 mol) particles.
This defines the sedimentation coefficient s in
Svedberg (10-13 sec =1 S) as the ratio of velocity to
field strength.
With the molecular weight M the diffusion
coefficient D can be calculated from s or f.
Measuring the velocity with which the boundary moves
Experimental determination of the
sedimentation coefficient
M ⋅(1-ν ⋅ρ )
v
= 2 =s
NA⋅f
ω ⋅r
Definition of the sedimentation coefficient s in
Svedberg (10-13 sec =1 S) as derived before
€
drb
v=
= rb ⋅ω 2 ⋅s
dt
€
€
rb (t)
ln
= ω 2 ⋅s⋅(t – t 0 )
rb (t 0 )
The speed v is determined from the
movement of the boundary
after integration
Logarithmic plot of boundary versus time
How to included information on spreading of boundary
(= diffusion) during sedimentation velocity experiment?
Concentration
Determination of the diffusion coefficient from a distribution
of apparent sedimentation coefficients g(s*)
t0
“true” s value
without diffusion
r
s* = 2 ⋅ln 
ω ⋅t  rm 
1
higher apparent s due
to forward diffusion
t
lower apparent s due
to forward diffusion
The mobility of a particle at a certain time
€ corresponds to a sedimentation coefficient
radial position r
converting the distance traveled by the particle after time t into a sedimentation
coefficient, yields a distribution of apparent sedimenation coefficient g(s*) or c(s*)
g(s*) (A260 Svedberg-1)
The distribution of apparent sedimentation coefficients
for two short DNA duplexes of 32 and 59 base pairs
0.3
DNA 59
bp
0.2
0.1
0
DNA 32
bp
1
2
3
4
s* (Svedberg)
5
Lamm equation to describe the temporal changes of the
concentration distribution of a molecule during sedimentation
s: sedimentation coefficient

∂c 1 ∂   ∂c
2
=
 x D − sω xc

∂t x ∂x   ∂x
D: diffusion coefficient
x: distance from the center of the rotor
ω: angular velocity
or
2


∂c
∂ c 1 ∂c 
2  ∂c
= D 2 +
 − sω  x + 2c
 ∂x

∂t
 ∂x x ∂x 
no analytical solution but can be solved for specific cases or by using numerical
methods to derive s and D from the change of the concentration gradient over time.
Determination of the molecular weight from s and D
M⋅(1-ν ⋅ρ )
v
= 2 =s
NA ⋅ f
ω ⋅r
k T RT
D=
=
f
NA f
The sedimentation coefficient s in Svedberg (10-13 sec =1 S)
D can be determined directly from the shape of
the sedimentation band.
€
s M(1−v⋅ρ) N A f M(1−v⋅ρ)
=
=
D
RT N A f
RT
This is the Svedberg equation according to which
the molecular weight M can be calculated from s
and D.
Frictional coefficients of spheres
4
Vsphere = ⋅π ⋅r 3
3
€
€
€
M ⋅v
Vsphere =
NA
f = 6⋅π ⋅η ⋅r
M⋅(1-ν ⋅ρ)
s=
NA ⋅ f
A sphere has radius r and volume V.
The volume of a spherical particel can be calculated
from its molecular mass, the partial specific volume v
bar and Avogadros number NA.
For the translation the resulting frictional
coefficient f in a medium with viscosity η can
be calculated according to Stokes law.
The sedimentation coefficient s
Frictional coefficients of spheres
1
3
 3 Mv
4
M
⋅v
M
⋅v
3
Vsphere = ⋅π ⋅r 3 =
⇔r 3
⋅ ⇒r =

3
NA
N A 4π
 4 π NA 
M(1−v⋅ρ )
s=
N A ⋅6 π η r
€
€
s=
substitute r from equation given above
M(1−v⋅ρ)
1
3
=
 3 Mv
N A ⋅6 π η 

 4 π NA 
€
s = 0.012
2
M 3 (1−v⋅ρ)
6π η
2
M 3 (1−v⋅ρ)
1
v3
2
NA 3 
1
3 1
 v3
3
4 π 
after substituting all the
constants and water viscosity
€
for s spherical
molecule s is proportional to M2/3
1000
sedimentation coefficient s
20｡C, w
(S)
Calculated sedimentation coefficients for spherical proteins
forbidden region
theoretical for unhydrated spheres
100
globular, hydrated
3.9S 6.3S 9.9S 15.8S
10
hydrated globular
proteins
1
real proteins
(nonglobular, hydrated)
1
10
50 100 200 400 1000
molecular weight (kDa)
104
Hydration of nucleosome core particle
Protein hydration
Svergun, D.I. et al 1998, Proc. Natl. Acad. Sci. USA. 95:2267-72.
First 3Å hydration layer around lysozyme ~10% denser than bulk
water
About 0.3 to 0.4 g H2O per g of protein
The amount of protein hydration can be calculated
from the amino acid composition
ionic
polar
Nonpolar
Amino acid
Hydration
Asp-
6
Glu-
7
Arg+
3
Lys+
4
Asn, Gln, Ser, Thr,Trp
2
Pro, Tyr
3
Ala, Gly, Val,IIe, Leu,
Met
Phe
2
0
Kuntz, I.D., Jr., and W. Kauzmann. 1974. Hydration of proteins and
polypeptides. Adv Protein Chem. 28:239-345.
Preferred hydration sites of DNA bases
DNA “Spine” of hydration
in the minor groove of
DNA
because of its high negative charge
density DNA is strongly hydrated
about 0.8-1.0 g H2O per g of DNA
per base pair about 22-24
molecules water in direct contact
with the DNA
another 16-18 H2O are thought to
be also in the primary hydration
shell
Molecular weight determination by
sedimentation equilibrium centrifugation
0.01
0
-0.01
ES-2 DNA
(59 bp)
NtrC + ES-2
ES-1 DNA
(32 bp)
NtrC protein
NtrC + ES-1
NtrC-P + ES-2
Analysis of sedimentation equilibrium centrifugation determination of M independent of D or f
 -Ej 
exp 
Pj
 kT 
=
-E i 
Pi
exp 
 kT 
 -E i 
Pi ∝ gi ⋅exp 
 kT 
€
1

E j - Ei = M ⋅(1- ν ⋅ ρ )⋅ω 2 ⋅ ⋅ rj2 −ri2 

2
Fz = Meff ⋅ω 2 ⋅r = M -(1- v⋅ρ)⋅ω 2 ⋅r
€
Wz = -Meff ⋅ω 2 ⋅r ⋅(rj - ri )
rj
Wz = ∫ -M eff ⋅ω 2 ⋅r⋅dr
ri
€
rj
1 2
= -M eff ⋅ω ⋅ r
2 r
i
1

= -M eff ⋅ω 2 ⋅ ⋅ rj2 −ri2 

2
(2)
(1)
€
Meff = M - M0 = M - M ⋅v ⋅ρ = M ⋅(1-v
€ ⋅ρ)
2
 E j −E i 
= exp−

ci
 RT 
cj
€
 M ⋅(1-ν ⋅ρ )⋅ω 2 ⋅ r 2 −r 2
cj
j
i
= exp

2⋅RT
ci

(
 M ⋅(1-ν ⋅ρ )⋅ω 2 ⋅ r 2 −r 2
0
Ar = A0 ⋅exp

2⋅RT

(
) 


)  + E


(3)
Making hydrodynamic models
Frictional coefficients for
ellipsoids of revolution
Ellipsoid Volume V
a, b half length of
two axes
€
4
V = ⋅π ⋅a ⋅b2
3
a
p=
axial ratio
b
ft
Re: radius of sphere
Ft =
6 π η Re with the same Volume
€
Friction coefficients for different shapes
2b
M ⋅(1-vρ )
s=
N A ft
M
∝ 1/2 ∝ M 1/2
M
b
L
p=
2b
L
contour length
L = b·N
1/2
€
€
cylinder
f is proportional to ≈ L1/2
random coil of N segments with
length b, Rg radius of gyration
f proportional to ≈ N1/2 or L1/2
Frictional coefficients for oligomers and polymers
Assume a polymer with N segment and frictional coefficient f1 per segment is fixed and
the fluid is moving
Without hydrodynamic interactions the frictional coeffcient would be fN = N · f1
In the presence of hydrodynamic interactions fN < N · f1 This is because, on the average,
each segment decreases the fluid velocity near it, and thus each experiences a smaller
frictional force
Kirkwood approximation to calculate the
friction coefficients for complex shapes

n
n
f1
1
fn = n⋅ f1 ⋅ 1+
∑ ∑

6⋅
π
⋅
η
⋅n
R
i=1 j≠i

ij 
−1
For a complex of n spheres of identical
size and frictional coeffizient f1 the total
frictional coefficient can be estimated
according to the Kirkwood approximation.
DNA-protein complex of NtrC with its
enhancer binding sites
Rippe, K., N. Mücke, and A. Schulz. 1998. Association
states of the transcription activator protein NtrC from E. coli
determined by analytical ultracentrifugation. J. Mol. Biol.
278:915-933.
Kirkwood approximation to calculate the
friction coefficients for complex shapes

n
n
f1
1
fn = n⋅ f1 ⋅ 1+
∑ ∑

6⋅
π
⋅
η
⋅n
R
i=1
j≠i

ij 
−1
For a complex of n segments with frictional coeffizient f1 the total frictional
coefficient can be estimated according to the Kirkwood approximation. In this
equation Rij is the distance between segments and every distance is counted
twice according to the summation. For an object that consists of identical spheres
with radius r according to Stokes law we obtain
−1
n n 1 

r
fn
= n 1+ ∑ ∑ 
f1
 n i=1 j≠i Rij 
Kirkwood approximation for a dimer
−1
n n 1 

r
fn
= n 1+ ∑ ∑ 
f1
 n i=1 j≠i Rij 
For direct contact R12 = R21 = 2 r
€
−1
−1


 1 
r 1 1
fn
= 2⋅1+  +  = 2⋅1+  = 2⋅0.66 =1.33
f1
 2 
 2  2r 2r 
€
fn = 2⋅ f1 ⋅0.66
i. e. 66 % of two spheres
Kirkwood approximation for a
dimer with a long linker

r
fn
= 2⋅1+
f1
 2
−1
 1 1 
 + 
 R12 R21 
for a long friction free linker R12 and R21 is very large so that
€
r
2
 1 1 
 + ≈ 0
 R12 R21 
f = 2⋅ f1
€N
€
i. e. that of two separate spheres
Kirkwood approximation to calculate the
sedimentation coefficient s for bead models
n n
1
sn
r
=1+ ∑ ∑
s1
n i=1 j≠i Rij
see van Holde p. 205
€
Dependence of s and D on molecular mass
dr dt M ⋅(1-vρ )
M
s= 2 =
∝ 1/2
1/3
N
f
ϖ r
M or M
A t
DNA
protein:
D" M
#
protein (sphere)
1
3
double mass M => 0.8 fold lower D
!
sedimentation coefficient s
increases with mass
DNA:
D" M
#
1
2
double mass M => 0.7 fold lower D
!
Experimental strategies
What do we know and want to know
Calculate from the sequence
• mass M of monomer (units)
• extinction coefficient
• partial specific volume v bar
• density ρ of buffer at exp. Temp
• viscosity η of buffer at exp. Temp
Sedimentation velocity (3 samples, 5-6 h)
• species present in the mixture
• sedimentation coefficient s
• diffusion/frictional coefficient D or f
Sedimentation equilibrium (9 samples, 24 h)
• mass M of the complex
• Equilibrium dissociation constant (if in ~µM
range)
AUC - Sample Cells & Rotor
• Samples are loaded into
cells with clear windows
(quartz or sapphire)
(sample vs. reference)
• Cells are placed in a rotor
with vertical holes
• N.B. - Balance is critical.
Low throughput
Absorbance optical
system of analytical
ultracentrifuge
Screw Ring
(301922)
Screw Ring
(301922)
Screw-Ring
Washer
(362328)
Screw-Ring
Washer
(362328)
Window Holder
(305037)
Window Holder
(305037)
Window Gasket
(327071)
Window Gasket
(327071)
Window Liner
(362329)
Window Liner
(362329)
quartz
Window (301730)
sapphire
(307177)
quartz
(301730)
Window
sapphire
(307177)
Gasket (aluminum
centerpiece only)
(330446)
Keyway
Keyway
Centerpiece
(366755)
Centerpiece
(see Table 1)
filling holes (6)
filling holes (2)
Gasket (aluminum
centerpiece only)
(330446)
quartz
(301730)
Window
sapphire
(307177)
quartz
(301730)
Window
sapphire
(307177)
Window Liner
(362329)
Window Liner
(362329)
Window Gasket
(327071)
Window Gasket
(327071)
Window Holder
(305037)
Window Holder
(305037)
Cell Housing
(368115, includes
plugs and plug
gaskets)
Equilibrium
External-Fill
Cell Housing
(334602, includes
plugs and plug
gaskets)
Housing Plug (2)
(362327)
Plug Gasket (2)
(327022)
Housing Plugs (6)
(362327)
Plug Gaskets (6)
(327022)
Double-Sector
Cell assembly for sedimentation
equilibrium (left) and velocity (right) runs
with 12 mm centerpieces