Types of Centrifuges used to measure the S-value

1. Types of Centrifuges used to measure the S-value • AnalyticalUltracentrifuge (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 analyticalultracentrifuge 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 thiscentrifuge, the sample is located in a centrifuge tube that allows the absorbance to be measured as a function oflocation 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 measuredby monitoring the midpoint of the absorbance changes. The velocity of this moving boundary yields thesedimentation coefficient of the protein.

3. ln rb(t) = ln rb(to) + w2 S (t-to) plot ln rb(t) vs time slope S  for known w rb = boundary position drb dt velocity = ln (rb) time Measuring S using an analytical ultracentrifuge Moving boundary t = 0 [C] t1 t2 0 top r bottom 1 drb rb(t) rb2 dt rb(to) • = Sln = 2 S (t-to)

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

5. M (1-V2) Nf S = 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.

6. M (1 - V2) Nf Using the Sedimentation Coefficient to get molecular parameters 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. S M(1-V2) D RT Combining Sedimentation and Diffusion The value of the specific volume of the anhydrous particle, V2, can be a sticking point when the sedimentingspecies is not a simple protein, but is a complex of components with different densities. Typical difficult systemswould 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 aweighted average of the densities of the components of the complex particle. kT kT f 6Rs D = = Any method yielding Rs can be used. This is now most frequently obtained by gel filtration chromatography M (1 - V2) Nf S = ( ) = The frictional coefficient drops out of the equation

8. infinite dilution So20,w 20oC in water (correct for viscosity and temperature from conditions of actual measurement) This used to be one of the standard methods to obtain protein molecular weight 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 getM. The following is illustrative of how molecular weight (M), size and shape combine determinethe values of the sedimentation coefficient and Stokes radius. Classically, the value of Rs would come from thetranslational diffusion coefficient, but this would be rare today. Instead, gel filtration or molecular sievechromatography 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 theformer case, the protein dissociates from a tetramer to monomers (or perhaps a mixture containing dimers),whereas in the latter case, the protein expands.

10. Example 1: Hemoglobin Dissociation at low pH kT 6Rs D = M (1 - V2) N 6Rs S = S M(1-V2) D RT = 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.

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

12. Conclusionsfrom 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 M1/3 In general (for spherical compact proteins) S  M2/3 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. M (1-V2) Nf S = Native tetramer Example: determination of the S-value of GlpF the glycerol transport facilitator protein from E. coli 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, 2888-2893

16. Using the Preparative Ultracentrifugeto 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. M (1-V2) Nf S = 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

20. Using the Preparative Ultracentrifuge Examples: 1. 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. 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. M (1-V2) Nf S = Discovery of DNA Supercoiling by Analytical Centrifugation (1968) Titrate DNA with ethidium bromide Closed circular DNA Nicked DNA 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. 2r2 2  = 0 + RT ln c - M (1-V2) r a 1 dC 2rM (1-V2) c dr RT 2M (1-V2) 2RT integrate: ln C(r) = (r2 - a2)  = Sedimentation Equilibrium  = radial velocity  = partial molar free energy C top r bottom dm 1 dc dr c dr = 0 at equilibrium 0 = RT - 2rM(1-Vp) at the top of the sample (meniscus, r = a), C = Co 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, 3912-3919

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 = 0.743 cm3/g measured solvent density,  = 1.0058 g/cm3 Temperature = 20o C Rotor speed varied from 8000, 9500 and 14000 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. 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 1 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  = 1 = 1.66 + 0.09 • 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 2 - 15 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 0.0195 P (+ sucrose) instead of 0.01 P 3. Partial specific volumes are different V = 0.739 (+ sucrose) instead of 0.728 cm3/g for NA3 V = 0.745 (+ sucrose) instead of 0.733 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-1 f/fmin = 3.27 NA4 Dw,20 = 3.6 x 10-7 cm2s-1 f/fmin = 2.96

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