1. Types of Centrifuges used to measure the S-value
Analytical Ultracentrifuge (monitor the distribution of material by absorption or dispersion) as a function of time
Method of choice, but requires specialized equipment
Beckman “Optima” centrifuge
small sample, but must be pure - optical detection
used to determine sedimentation velocity  S
(frontal analysis (moving boundary method) - not zonal method)
Preparative Ultracentrifuge
common instrumentation
sedimentation coefficient obtained by a “zonal method”
 requires a density gradient to stabilize against turbulence / convection
 obtaining S usually requires comparison to a set of standards of known S value

2. Measuring the sedimentation coefficient with an analytical ultracentrifuge
The sedimentation coefficient is usually measured in an analytical ultracentrifuge. Currently the Beckman “Optima” is the only commercial machine available. This is a relatively specialized piece of equipment.
In this centrifuge, the sample is located in a centrifuge tube that allows the absorbance to be measured as a function of location in the tube. One starts with the protein distributed evenly through the tube and as the centrifuge spins, the material sediments towards the bottom of the tube. The speed with which the boundary moves is measured by monitoring the midpoint of the absorbance changes. The velocity of this moving boundary yields the sedimentation coefficient of the protein.

3. ln rb(t) = ln rb(to) + w2 S (t-to)
Measuring S using an analytical ultracentrifuge
Moving boundary
t = 0
[C]
t t2
velocity =
drb dt
rb = boundary position
top r bottom
drb rb(t)
rb dt rb(to)
• = S ln =  2 S (t-to)
ln rb(t) = ln rb(to) + w2 S (t-to)
plot ln rb(t) vs time
slope S  for known w
time
ln (rb)

4. The sedimentation coefficient of DNA and a protein (bovine serum albumin) as a function of concentration.
Note the very large concentration dependence of S for DNA, which makes this a very impractical method for studying DNA.

5. and  is not constant (f = 6Rs)
Using the Preparative Ultracentrifuge
Finding the S-value
typically a sucrose gradient is used to prevent turbulence due to convection
But the gradient makes computation of the S-value very difficult because in the sucrose density gradient,
r is not constant
and  is not constant (f = 6Rs)
A set of “standards” of known S-value is used to get S for the unknown.
M (1-V2) Nf
S =

6. Using the Sedimentation Coefficient to get molecular parameters
M (1 - V2) Nf
Once you measure it
there are 3 unknowns : S =
1 measure, calculate or estimate V2 (inverse density)
for proteins: V2  0.75 mL/g
-can compute from amino acid composition
2 f can be obtained from another experiment: gel filtration, diffusion (Stokes
radius  f)
combination of sedimentation and gel filtration can yield M.
Practical strategy to get M
f = 6Rs

7. Combining Sedimentation and Diffusion
The value of the specific volume of the anhydrous particle, V2, can be a sticking point when the sedimenting species is not a simple protein, but is a complex of components with different densities. Typical difficult systems would be membrane proteins with bound lipids or detergent molecules, or glycoproteins with sugars
attached to the protein. In this case, one cannot just make an estimate of the value of V2, which will be a weighted average of the densities of the components of the complex particle.
kT kT
f Rs
D = =
Any method yielding Rs can be used.
This is now most frequently obtained by gel filtration chromatography
M (1 - V2) Nf
S =
S M(1-V2 )
D RT
( ) =
The frictional coefficient drops out of
the equation

8. This used to be one of the standard methods to obtain protein molecular weight
infinite dilution
So20,w
20oC
in water (correct for viscosity and temperature from conditions of actual measurement)
To obtain the molecular weight of the native form of an enzyme (subunit association), the combination of sedimentation - gel filtration (RS) is very useful
(from Tanford)

9. Examples of Combining Sedimentation plus Stokes radius to get M.
The following is illustrative of how molecular weight (M), size and shape combine determine the values of the sedimentation coefficient and Stokes radius.
Classically, the value of Rs would come from the translational diffusion coefficient, but this would be rare today. Instead, gel filtration or molecular sieve chromatography would be used to obtain Rs.
Let’s compare the effect of lowering the pH of solutions of two proteins, hemoglobin and serum albumin. In the former case, the protein dissociates from a tetramer to monomers (or perhaps a mixture containing dimers), whereas in the latter case, the protein expands.

10. 4 x S M(1-V2 ) = D RT Example 1: Hemoglobin Dissociation at low pH D
(Rs) S
M  Rs
S  D
4 x
The protein dissociates from a tetramer to monomers (or perhaps a mixture containing dimers).
The effect of lower pH on hemoglobin: the sedimentation coefficient decreases and the diffusion coefficient goes up (Stokes radius goes down).
Whereas the diffusion constant varies as the inverse of Rs, the sedimentation coefficient varies as the ratio of M/Rs. The ratio of S/D depends on molecular weight, M, only if V2 is constant. kT
6Rs
D =
M (1 - V2)
N 6Rs
S =
S M(1-V2 )
D RT =

11. S M(1-V2 ) = D RT Example 2: Expansion of BSA at low pH kT 6Rs D =
In contrast to hemoglobin, the protein bovine serum albumin expands during acidification.
Example 2: Expansion of BSA at low pH
D 
(Rs ) S
M  Rs
S constant D kT
6Rs
D =
M (1 - V2)
N 6Rs
S =
S M(1-V2 )
D RT =

12. Conclusions from the two proteins
Note that in both the examples of hemoglobin and serum albumin, the sedimentation coefficient gets smaller at acid pH, i.e., the protein moves more slowly in the centrifuge tube.
But the causes are totally different, and cannot be distinguished without the additional information from another experiment to yield the Stokes radius independently.
In the case of serum albumin, the increase in radius is not accompanied by any change in the molecular weight, so the ratio of S/D remains constant as a function of pH.
Also note that this information is not available from X-ray crystallography but requires these solution techniques.

13. Using set of “Standards” of known S-value to
estimate molecular weight of an unknown from S values
1 The best method to obtain M from a measured S-value is to independently get V2 and f, and then obtain M in the equation below:
S =
2 Can convert S  M if you compare with known standards.
However: M must be the only independent parameter
Example stands set of “spherical” proteins
 same V2
increasing M
f  Rs  M1/ In general (for spherical compact proteins)
S  M2/ increasing
But: if unknown is ellipsoid f > fsphere (then f and M are not correlated)
S > Ssphere, get wrong M
M (1 - V2) Nf
f = 6Rs
unknown

14. Using standards to get M from the S-value
The key to using a comparison to standards to obtain M is that molecular weight be the only independent parameter. (Also true for electrophoresis and for gel filtration chromatography)
S =
If all standards and the unknown are same shape and have the same V2 , then S will be a smooth and predictable function of Molecular weight.
S  
You will get the wrong value of M if the shape of the unknown is not the same as the shape of the standards to which it is compared.
You will get the wrong value of M if the value of V2 is incorrect. Glycoproteins, proteins with lipids and/or detergents bound to them are examples where the anhydrous volume per gram of the particle is not a trivial matter to obtain.
M (1 - V2 ) Nf
f = 6Rs
M M
f f (M)

15. relative distance moved
Example: determination of the S-value of GlpF
the glycerol transport facilitator protein from E. coli
M (1-V2) Nf
S =
Native
tetramer
Empirical correlation of the
S-value of soluble protein
standards using
sedimentation velocity in a 5-20%
sucrose density (neither  nor f are
constant!)
relative distance moved
PNAS (2001) 98,

16. Using the Preparative Ultracentrifuge to obtain an S-value
Most biochemical laboratories are equipped with a preparative ultracentrifuge. These instruments do not have optics needed to measure the distribution of protein or any other substance in a tube. Instead, if one wants to determine the distribution of the material after a centrifuge run, it is necessary to remove the tube after a specified time and take out the sample drop by drop from top to bottom (or vice versa). Bio-assays can then be used to determine the contents of each fraction. It is necessary to use a zonal method with a preparative ultracentrifuge rather than the boundary method. In the zonal method, the material is layered in a zone at the top of the tube at the beginning of the experiment, and then this zone of material migrates down the tube:

17. Because of the size of the tubes (several ml), one of the major problems to obtain quantitative information on sedimentation velocity from a preparative ultracentrifuge is the disturbance of the material distribution by turbulence from physical vibrations or due to convection from thermal differences in different parts of the tube. This is partly overcome by the use of a sucrose gradient, in which sucrose is added to the sample with a dense solution on the bottom. If there is a local place in the tube that is slightly warmer than the surrounding sample, this would normally become less dense and rise in the solution, thus causing mixing. The sucrose density prevents this mixing.
However, other problems are introduced that complicate determination of the S-value. Two of the factors that must be known to determine the S-value are now different in each level of the tube, and they are not constants: ρ (the solution density) and f (6πηRs), which depends on the solution viscosity (η).
In practice this makes it very difficult to convert a distance migrated down the tube into a meaningful S-value. In order to do this, it is necessary to use a set of known samples where the S-value is known and make the measurements along with that of the unknown. This is not the best way to determine an S-value.

18. Separations in the Preparative Centrifuge: Differential Sedimentation
The three parameters that determine the sedimentation coefficient are all important in allowing the separation of molecules, complexes, organelles or other particles from each other. This technique is called differential sedimentation. In the extreme case, particles with a large S-value will migrate rapidly to the bottom of the centrifuge tube and form a pellet, and those with smaller S-values will remain in the supernatant. Separating on the basis of zonal migration, as is the case with DNA species of varying supercoil density, requires a stabilizing sucrose gradient.

19. Using the Preparative Ultracentrifuge
Preparative Applications: Differential Sedimentation
1) Larger size particles (larger M/f): sediment faster (pellet vs supernatent).
2) Higher density particles (smaller V2): sediment faster
3) More compact particles (smaller f): sediment faster
M (1-V2) Nf
S =

20. Using the Preparative Ultracentrifuge
Examples: lipoproteins ( separate on the basis of density)
2.protein - nucleic acid complexes (separated from protein and DNA by density differences)
3. RNA: usually must denature to eliminate 3o structure to separate on the basis of size
4 .DNA forms - differ in shape/size from each other
closed circular plasmid linear concatented supercoiled

21. more supercoils makes the DNA more compact
DNA Supercoiling:
induced by binding intercalating molecules (such as ethidium bromide)
that insert between basepairs and unwind the double helix
more supercoils makes the DNA more compact
change in shape results in a change in S-value

22. Discovery of DNA Supercoiling by Analytical Centrifugation (1968)
Titrate DNA with ethidium bromide
Closed circular
DNA
Nicked DNA
M (1-V2) Nf
S =
Negative supercoils:
Compact/high S value
No supercoils:
Low S value
Positive supercoils:
Compact/high S value

23. Some particles cannot be separated based on differences in S value
The technique of sedimentation equilibrium is used to separate species on the basis of density, thus extending the versatility of centrifugation in biomolecular separations.

24. Sedimentation Equilibrium
 = 0 + RT ln c M (1-V2)
2r2 2
 = radial velocity
 = partial molar free energy C
top r bottom
dm dc
dr c dr r a
= 0 at equilibrium 0 = RT 2rM(1-Vp)
 =
dC 2rM (1-V2)
c dr RT
at the top of the
sample (meniscus, r = a),
C = Co
integrate: ln C(r) = (r2 - a2)
2M (1-V2)
2RT
ln C
meniscus
get M from the slope
-must know  and V2 and 
r2 - a2
Note: f is not in the equation

25. Sedimentation equilibrium
Best method is to use an analytical centrifuge
measure C(r) vs r optically
Need pure material to do this

26. Sedimentation equilibrium using an analytical ultracentrifuge
Alkyl hydroxyperoxide reductase from S. typhimurium
Recombinant heterodimer generated to study electron
transfer properties
Two subunits are co-expressed in E. coli:
subunit size: 34,000 and 55,000 (AhpF and AhpC)
Question: what is the molecular weight of the isolated complex
with enzymatic activity?
Biochemistry (2001) 40,

27. Sedimention equilibrium of AhpF/AhpC complex
Beckman Optima analytical ultracentrifuge: monitor A280 in the
centrifuge cell
Conditions:
several protein concentrations from 3 to 34 µM
115 µl per sample
calculated V = cm3/g
measured solvent density,  = g/cm3
Temperature = 20o C
Rotor speed varied from 8000, 9500 and rpm
collect data at 8, 10 and 12 h for each speed

28. Sedimentation equilibrium data of
Alkyl hydroxyperoxide reductase
M = d ln C/dr2(2 R T/(2(1 - V)))
8000 rpm
14000 rpm
9500 rpm
Conclusion: Enzyme contains
one copy of each subunit
Result: M = 86,200 ± 200

29. Sedimentation Equilibrium without an analytical ultracentrifuge
Alternatives to the analytical ultracentrifuge
1) the airfuge
2) the preparative ultracentrifuge

30. Preparative Ultracentrifuge
Airfuge
enables you to measure the distribution of material in a centrifuge tube qusing biochemical techniques
- small, high speed (desktop)
- sample solution for biochemical assays
- get ln c(r) vs r2 from impure material
- enzyme assays
- binding assays
- antibody detection
1. Can use like the airfuge and remove sample to measure material distribution
2. Can use a density gradient centrifugation to separate particles on the basis of V2
take samples vs
depth
Preparative Ultracentrifuge

31. Sedimentation equilibrium in a density gradient
- use sucrose, CsCl, KBr, etc to establish a density gradient in the centrifuge tube
-either make the gradient prior to centrifugation or
establish the gradient during centrifugation
This creates a 3-component system
1) solute - 2) water - 3) density forming co-solvent (eg, sucrose)
must use the partial specific volume of hydrodynamic particle S=
when (r) =
S = 0
Vol/gram (particle)
Solution density (varies with position in the tube, r)
M [1 - Vh (r)]
N f 1 Vh
 < 1 / Vh particle density greater than density of solution
 = 1 / Vh particle density equal to density of solution
 < 1 / Vh particle density less than density of solution
Sedimentation
Equilibrium
Flotation
S > 0
S = 0
S < 0

32. = For a linear density gradient: (r) = o + (r - ro) ( )
where ro is where  =
at equilibrium d dr
dc 2 M [1 - Vh (r)]
c dr RT
c(r) = c(ro) e
Gaussian Distribution
about r = ro
Note: Width of the distribution depends on Molecular weight
At ro: = Vh = d dr 1 Vh
substitute
= 0 = d dr
- 2 ro M Vh( ) (r - ro)2 / 2 R T ro
gradient r 1 o
V2 + H2O
1 +  H2O
at the peak (this does not take into account any salt or ions)

33. Applications of sedimentation equilibrium in a density gradient for
analysis and for preparative purposes
1 Different membranes or organelles can be separated on the basis of density
E.coli: inner + outer membranes
eukaryotic: rough + smooth ER
2 Serum lipoproteins: LDL, HDL etc
3 DNA with different GC content
 = = • fGC Vh
4 Single vs double-strand DNA
5 15N vs 14N containing DNA (or DNA with Br-U instead of T)
6 Supercoils with different amounts of bound ethidium bromide - have different Vh values

34. Some particles cannot be separated based on differences in S value
Differential Sedimentation
Density gradient Sedimentation Equilibrium
pellet

35. Case Study: Actin-binding Nebulin Fragments
Nebulin binds to the actin/tropomyosin/troponin thin filament
of skeletal muscle
Multiples of repeated sequences arranged in modules
Fragments containing modules retain actin binding
function
Two fragments: NA3 and NA4
Question: Are NA3 and NA4 aggregated in solution?

36. Model of Nebulin binding to actin
fragment

37. Sedimentation equilibrium of nebulin fragments
Using the Beckman Airfuge
Conditions:
0.1 mg/ml protein
1 mM Ca++, pH 7 buffer
20% sucrose to stabilize the protein distribution
during rotor deceleration
Spin solution for 80 h at 54,000 rpm ( = 5700 s-1)

38. Using a syringe, remove liquid from the top of the
tube and measure protein concentration
for each fraction

39. M = d ln C/dr2(2 R T/(2(1 - V)))
R (gas constant) = 8.3 x 107 g cm2 s-2mol-1K-1
 = 5700 s-1
 = 1.08 g cm-3 density of 20% sucrose
V = partial specific volume of nebulin
fragments in 20% sucrose.
Calculated from amino acid sequence
0.739 for NA3
0.745 for NA4
T = 300 oK
Conclusion:
Nebulin fragments
Are not aggregated
NA3: M = 37 kDa (monomer = 31kDa)
NA4: M = 35 kDa (monomer = 25 kDa)

40. What is the shape of the nebulin fragment?
Sedimentation velocity in a Beckman L5-50 preparative
ultracentrifuge: SW41 swinging bucket rotor
Conditions:
0.05 mg/ml protein concentration
1 mM Ca++, pH 7 buffer
20% sucrose
Remove sample from the top
and measure protein to
determine the profile of the
trailing boundary.
Spin for 18 h at 35,000 rpm at 20o C

41. Sedimentation Velocity of Nebulin Fragments
trailing boundary
top of
Centrifuge tube
t = 64,800 sec
 = 3665 s-1
rb = boundary position
rm = meniscus position
So = (ln rb - ln rm)/2t

42. Sedimentation velocity of nebulin fragments
Must correct for the fact that the
measurement is done in 20% sucrose:
1. Density is 1.08 (+sucrose) instead of 1.00 g/cm3
2. Viscosity is P (+ sucrose) instead of 0.01 P
3. Partial specific volumes are different
V = (+ sucrose) instead of cm3/g for NA3
V = (+ sucrose) instead of cm3/g for NA4
Results:
Sw,20 = 1.1 S for
both NA3 and NA4
Sw,20 = So (1 - Vww)
w(1 - V)
Measured in 20% sucrose

43. Interpretation of the Sedimentation of nebulin fragments
1. We can now calculate the Diffusion coefficients:
Dw,20 = Sw,20 RT/(M(1 - V)
2. From this, calculate f/fmin:
f/fmin = (kT/Dw,20)(6(3MV/4No)1/3)
(Note: use values of ,  and V that apply to the protein in water)
From sedimentation equilibrium
Results: NA3 Dw,20 = 3.2 x 10-7 cm2s f/fmin = 3.27
NA4 Dw,20 = 3.6 x 10-7 cm2s f/fmin = 2.96

44. f/fmin experimental RS = (kT/Dw,20)/6 Anhydrous sphere
kT kT
f RS
since f = 6Rs
RS = (kT/Dw,20)/6
experimental
vol = [V2 • ]
= R3min M N 4 3
Anhydrous
sphere
Rmin
Rmin = (3V2M/4N) 1/3
Define: fmin = 6Rmin
Rs f
Rmin fmin :
measured = =
(kT/Dw,20)/6
(3V2M/4N)1/3
anhydrous sphere

45. Nebulin fragments are highly asymmetric
NA3 f/fmin = 3.27
NA4 f/fmin = 2.96
If anhydrous prolate ellipsoids: a/b = 50/1 to 60/1
about 600 Å long