Properties of Proteins and their Measurement FST 201

Transcription

1 Properties of Proteins and their Measurement FST 20 Properties we will consider Determination (The ultimate Extrinsic property - How Much?) Optical properties Size, Mr, deviation from sphericity Amino Acid composition Sequence Isoelectric point, Charge at a particular ph Hydrophobicity Activity Solubility functionality (Nutritional Quality, Digestibility) Determination Chromatographic methods Chromatography SEC IEC Affinity HIC Detection A260, A206 Specific absorbances ( Chromatography ) Enzyme activity postcolumn reactions blots, ninhydrin, Lowry, anything Electrophoresis General theory Native PAGE SDS Gradient IEF Detection Coomassie colloidal silver, Gold Activity stains, zymograms blots Spectroscopy CD, ORD fluorescence transitions, quenching, solvent effects Intrinsic Trp Probes ANS, Isodsf[Noric acid Sequencing IEC and RP determinations of amino acids Edman, C-terminal sequencing (chromatography) Concept of overlaps Sequencing the gene Bioinformatics Denaturation NMR spectroscopy Folding

2 Protein Determination FST 20 Methods for Protein Determination Note: "Protein" usually refers to a mixture of molecules with different molecular weights and different properties. Protein concentration is expressed in weight per volume (or percent by weight), not in moles, unless you have a pure protein or have some way of distinguishing the protein of interest from others in the solution. ) Ultraviolet Spectroscopy 280 nm. Soluble proteins at mg/ml concentration contribute an absorbance at 280 nm of absorbance units to the solution. On the average, you can assume an absorbance of A/(mg*mL - ). This number works very well for protein mixtures, because mixtures tend to be average. For pure proteins, there is much more variation. For instance, the nearlyuniversal standard, BSA has an absorbance of 0.66 A per mg*ml -. Why so much variation? It is only the aromatic AA's, and mostly Trp, that absorb appreciably at 280nm: Absorbance Trp Tyr Phe Wavelength (nm) Obviously proteins that are deficient in aromatic AA's (esp. Trp), such as nucleohistones and gelatin, have such a low extinction coefficient at 280nm that the method is not reliable. Also, b vitamins, nucleic acids, ATP, etc. also absorb in this region, so the potential for interference is high nm. The peptide bond also nm, as do all amino acids, because of π-> π* transitions and n -> π* transitions: R R R O O O CH - C CH - C CH - C.. NH NH NH NH Unfortunately, so does nearly every other carboxylate, amide, ester, phosphate. Even air absorbs light of λ < 200 nm ( so it is called the "vacuum UV"). Absorbance at λ < 235nm is very sensitive, but highly subject to interference. Drawbacks of measurements in the UV: Not very sensitive: need ~mg/ml concentrations, ~ml sample. Samples must be soluble! The solution must not be turbid. Many compounds interfere. Not constant from protein to protein (must establish an extinction coefficient). 2) Kjeldahl Procedure (published 883) Consists of three steps i. Digestion at , usually with a catalyst (Hg, Se, Cu) H 2 SO 4 + Protein CO 2 + H 2 O + (NH 4 ) 2 SO 4 +?? ii. Distillation after neutralization NaOH(xs) + (NH 4 ) 2 SO 4 2NH 3 (g) + Na 2 SO 4 + H 2 O NH 3 is distilled out into trap containing a known XS of boric acid or H 2 SO 4. iii. Determination of ammonia Titrate the remaining H 2 SO 4 with NaOH, or 2

3 Protein Determination FST 20 Measure NH 4 directly using Nessler's reagent (the trapping procedure is a bit different for this method). The method is widely used for food and agricultural products because it is a digestion - sample need not be soluble. Also, it gives fairly uniform measurement of all proteins. However, it is a time-consuming, multi-step procedure requiring a specialized apparatus if you have a large # of samples. Its primary drawback is that it is not very sensitive and it does not actually measure protein - it measures N. It assumes proteins are 6% nitrogen, and no other major source of NH 4 + is present. If you work with proteins known to be higher or lower in N, you can change the calibration. 3) Biuret Biuret is a historical name for imidodicarbonic diamide (NH 2 CONHCONH 2 ) that forms a violet-colored complex of with Cu(II) in alkaline medium. Other compounds with 2 peptide bonds form a blue-colored complex with Cu ++ salts (citrate, tartrate) in alkaline solution that has a similar structure and spectrum. A 540 is proportional to peptide concentration. Since the method actually measures peptide bonds, it is often the method of choice. Unfortunately, it is not very sensitive (measures ~2-4mg protein), and often cannot be used. A standard curve is required. 4) Lowry - 00 times more sensitive than Biuret Reagent contains Cu 2+ ion in alkaline solution (tartrate), which is the same as the Biuret method, but also contains phosphotungstic and phosphomolybdic acid ("Folin-Ciocalteau phenol reagent") which can be used to measure phenols. Cu 2+ ion catalyzes the oxidation of aromatic AA's by phosphotungstic and phosphomolybdic acids which also creates a blue color. Measure absorbance between 500 and 750nm (A 750 is the most sensitive), usually at 650nm, and compare to a standard curve. Lowry's paper is the most often-referenced paper in history! - Advantages: - specific (reducing agents, high [sucrose], some buffers interfere) - very sensitive - only moderately time-consuming - Disadvantages: - 2-steps with incubation periods - interference by some common additives (tris, thiols) - depends on aromatics - color yield not constant (gelatin, for instance, has relatively few aromatics) - generally need a standard curve. 5) Bradford - Dye-Binding (mid 970's) sometimes called "Bio Rad" Single-step assay - Coomassie Brilliant Blue G-250 (not R) stains proteins blue; A 595 is proportional to protein concentration. Absorbance of the dye shifts with a change in the dielectric constant of the medium, which it turn alters its λ max. In water, it is brownish (red cationic form plus green neutral form) with λ max = 465; in organic solvent or bound to proteins, it is blue (anionic form) with λ max = 595. Binding to proteins apparently depends on cationic nature of protein, so the extent of binding, hence color yield, can vary. Advantages: - 0X more sensitive than Lowry - similar color yield for different proteins - few common chemicals interfere Disadvantages: - detergents and organic solvents interfere - lipids and flavonoids can interfere - susceptible to contamination (tends to be noisy) -??? Other dyes work also, and are used occasionally. 6) BCA Method ("Pierce Method", after Pierce Chemical Co.) Principle: Cu ++ oxidizes Trp, Tyr, Cys, Cys 2, in alkaline solution producing Cu + ; detect the Cu + produced by using bicinchoninic acid (BCA). The BCA 2 Cu + complex is highly colored (purple) and can be detected at 562 nm. Advantages: few compounds interfere, fairly uniform color yield for different proteins. Interference: reducing agents, oxidizing agents, chelating agents. 3

4 Protein Determination FST 20 7) Infrared analysis Measure near IR spectrum (largely overtones of vibrations). The spectrum is broad and featureless, and contains contributions from water, carbohydrates and lipids as well as proteins. With a massive number of standards, the amounts of each component can be determined. The method is usually carried out using reflectance, so it lends itself to use on solid or intact samples or to on-line measurement. 8) Turbidity (a last resort) Cold precipitating agents (3-0% trichloroacetic acid or sulfosalicylic acid) cause most proteins to precipitate and make the solution cloudy. Turbidity is measured using a spectrophotometer or nephelometer and the "absorbance" is compared to a standard curve, which is best obtained using the same protein, calibrated using Kjeldahl. The wavelength of choice is usually 540nm; avoid wavelengths at which other components absorb - their extinction coefficients could vary with ph. 4

5 Chromatography FST 20 Fundamentals and Mathematical Description of Chromatography Partitioning and the fundamental equations. Let's begin by reviewing some definitions that relate to column chromatography: partition coefficient K = phase ratio β = capacity ratio k = C C s m V V m s K C β = C s m Vs V m amount of solute = concentration * volume, e.g., As = Cs * Vs. In these equations, Cs and Cm, Vs and Vm are concentrations of solute in and the volumes of the stationary and mobile phases, respectively. The capacity ratio (or partition ratio) is the ratio of the amount of the solute in the stationary and mobile phases. The average velocity (or speed < S >) of a solute as it passes through a column can be calculated using a little intuition. The solute can exist in two states, dissolved in the mobile phase, which has a velocity of Sm, or dissolved in the stationary phase, which has a velocity of Ss (= 0). The fraction of the time a particular molecule spends in the mobile phase relative to the stationary phase must be But C C s m Vs V m = K = β Am (A + A ) m s = CmVm C V + C V = m m s s Cs + C m Vs V k, so the fraction of time the solute spends in the mobile phase is the stationary phase is minus this value, or - + k each phase times the velocity of each phase, summed over both phases: m.. The fraction of time spent in + k. The average velocity of the solute is just the fraction of time spent in < S > = Sm + k + 0 [ - + k ] = S m + k. So, the average velocity of an average solute molecule depends only on the partition ratio, k, and on the solvent velocity, Sm. In some kinds of liquid chromatography, it's not convenient to measure the solvent velocity. So, let's get it out of the S to Sm, which we'll call Rf. equation by measuring the ratio of < > < S > Sm = + k = R f This Rf is the "retardation factor" that is measured from TLC plates and paper chromatograms. It is calculated by taking the ratio of the distance traveled by the solute to that traveled by the solvent front. In these kinds of chromatography, all spots (solutes) and the solvent are allowed to run for the same total length of time. A measurement of the distance traveled is therefore a measure of the relative velocities of the solvent and solute. So the Rf is not such a phenomenological parameter as you might have thought; it's a direct measure of the partition ratio, k. In column chromatography, there is no solvent front (although the solvent velocity can be measured by using a marker that does not interact with the stationary phase and is not retarded at all) and distance measurements are not convenient. In fact, instead of allowing everything to run for the same length of time, column chromatography requires that everything travel the same distance, i.e., from the beginning to the end of the column. Of course, all solute molecules spend the same length of time in the mobile phase; they only move when they are in the mobile phase and always move with a velocity of Sm when they are in the mobile phase. But some molecules spend a greater total time in the column because they spend 5

6 Chromatography FST 20 some of the time in the stationary phase, which doesn't move. This total time spent on the column is known as the retention time tr. But all of the concepts are the same as discussed above for Rf's. Assuming a constant flow rate, the time it takes a solute to reach the end of the column is inversely proportional to its velocity: tr /<S>. Also, at constant flow rate, tr will be proportional to the volume at which the solute emerges ("elutes" or "is eluted") from the column. There is no solvent front, so we'll have to label some of the solvent molecules and watch for them to come off in order to measure the solvent velocity. If the time it takes a given molecule of the solvent, which travels with velocity Sm, to reach the end of the column is called tm (or, equivalently, if the volume of mobile phase that comes out of the column during this time is called Vm), it is simple to write an expression analogous to that for Rf but using times or volumes instead of velocities or distances: t t r m = Sm V KV < S > = + k = m + s. V m Check the algebra to be sure the expression on the right is correct. So, place 0 7 molecules on a column and run them through. They should all come out as a nice sharp spike after exactly Vm + KVs milliliters have come out, right? Nope. On the average, the molecules will emerge from the column as a band with a noticeable width and characteristic shape. The elution volume Vm + KVs and the elution time tr are only averages. What is the lineshape, why is it so shaped, and what does the width mean? Plate theory We re not going to derive the whole thing, but it s useful to see how it is set up and what the assumptions are. The idea is that the lineshape depends on the statistical likelihood that a solute will pass from one stage (plate) of the column to the next stage when tall of the mobile phase in the stage flows on to the next stage. That probability depends on the partition coefficient of the solute between the two phases. If the solute partitions more readily into the mobile phase, then it will more likely be transferred to the next stage when the volume of its current stage moves on to the next stage. But the amount of the solute that happens to be in the stationary phase when the flow occurs will be left behind. So the band broadens. But that solute that was left behind is extracted into the small volume of solvent that comes in from the previous stage, so a portion of it moves along to the next stage when the next small volume flows. The differential equation. Let's think about a chromatography column and how molecules of solute become distributed along it. Let's divide the column into N stages, identified by a number n which runs from to N. Each stage contains both mobile phase (its volume is vm) and stationary phase (its volume is vs). To avoid confusion of these volumes, which refer to the volumes in a single stage, with the volumes in the previous section, which refer to the entire column, we'll use lower case v's for the stage volumes. They are related, of course, because Vm is the sum of all N of the vm's, and likewise for Vs and vs. The concentrations of solute in the stationary and mobile phases are Cs and Cm, respectively. The column looks conceptually like this: n N We'll say that the mobile phase is on the upper half of the column, and the stationary phase, on the bottom. The mobile phase flows in the direction of the arrow. When a very small volume dv flows, how does the amount of solute in a stage change (any stage, chosen at random)? The change in amount is simply the amount of material that enters the stage minus the amount that leaves. The amount that flows is the concentration of the solute times the volume that flows: Amount into stage n - Amount out of stage n = Cn- dv - Cn dv = total amount where Cn is the concentration of the solute in the mobile phase of plate "n" and Cn- is its concentration in the previous stage. The change in total amount can be written d[cnvm + Csvs], so the whole equation can be written as Cn- dv - Cn dv = d[cnvm + Csvs]. But we have already defined a relation between Cs and Cn in the same stage; their ratio is the partition coefficient: M S 6

7 Chromatography FST 20 So the whole expression is K = C C s. n (Cn- - Cn) dv = d[cn (vm + Kvs)]. On the right, only Cn is a variable, so that the rest can be taken out from behind the differential sign as a constant: (vm + Kvs) dcn. The entire expression can be rearranged to the following form: dcn dv + v m Cn + Kv s = Cn v + Kv m s (I) This equation is very general and applies to any stage of the column as long as equilibrium is established as the volume dv flows (otherwise K is not equal to Cs/Cn). Let's save ourselves a little work, however and say that a = won't have to keep writing the fraction over and over. v m + Kv s so we Running a column after loading the first stage with solute You can, or at least the engineers among you, can solve this differential equation (I) for each stage; you have to start at the first stage and move forward because the amount of solute that flows into the nth stage depends on how much is in the (n- )th stage. This process would get tedious, but it turns out that a simple pattern develops, and you can write the concentration of solute in the Nth stage Cn, as a sum that depends on the stages before the nth stage. The exact equation depends on the initial conditions what stages contained what at the beginning. The most interesting and relevant cases in chromatography are ones in which only the first stage of the column (or the first m stages) initially contains any solute, and the solvent or buffer (the 0th) stage contains none. This condition is known as square-plug loading. It is usually very desirable to have your sample loaded as a thin band on the top of the column. Luckily, this set of initial conditions also simplifies the arithmetic. Since C o 0 but all other Ci o = 0 and Co = 0, all of the terms drop out but a few: o n (n i) C i (av) -av Cn = e i= (n i)! But, it can be simplified more, because the only Ci o that was not zero was the first one, C o So it's really Cn = C o (av) (n ) (n )! e -av (II) A plot of this equation after, 5, 0 & 20 volumes of /a have flowed is shown below. 0.4 Concentration Plate number This equation actually describes the concentration of solute in each small stage during chromatography. In an ideal case, it describes a TLC or paper chromatography experiment. At the end of a TLC experiment, you have spots distributed along the TLC plate which essentially indicate which stages of the TLC plate the solutes are found in. But that's not the way that column chromatography is usually done. You don't usually leave the solutes on the column, you run the column until they come off and measure the time or volume it took for them to come off. What we really want to know is the concentration of solute in the last (Nth) stage of the column (actually the N+th stage, where the detector is) as a 7

8 Chromatography FST 20 function of volume or time. In fact, a "chromatogram" or "elution profile" is a plot of the concentrations of solutes versus volume, time or chart distance. Let's calculate a chromatogram. We want the concentration in the last plate and we assume our column has many plates or stages. So we can use the equation we already have (IV) and just consider the Nth plate (the last plate) with N chosen to be a very large number (greater than 00 is large enough for the calculation). If N 00, then N- is the same as N to less than % accuracy, so let's leave out the "-". The final equation (thank goodness) is CN ( av) o C = N! N e -av (III) It's also been generalized a little by dividing through by C o so that the expression gives concentration as a fraction of the initial concentration of solute on the first stage of the column. According to this model for chromatographic behavior, a plot of the concentration of a solute as it leaves a column should look like the one below. This plot was generated with the assumption that vm = vs, that K =, and that there were only 30 plates. These are not real-life conditions. For capillary GC, vm = 0 3 vs or more and that there are sometimes thousands of plates. But, the basic lineshape is correct. A point worth mentioning about this ideal lineshape is that it is not symmetrical and not exactly gaussian. There is a tail at the back edge. Tailing of peaks can be a serious problem and is usually associated with some flaw in the chromatographic system or procedure. However, it should be noted that truly symmetrical peaks are not to be expected, even in the ideal case. The symmetry of the peak improves as the number of plates increases. It starts to look gaussian. 0. Concentration Volume (multiples of /a) The lineshape and its statistical interpretation. This equation, a relation between concentration and volume, is the idealized lineshape for a chromatographic peak. If you look up the lineshape in textbooks, you are usually referred to the Poisson distribution (PD) which is P(n) = n (av) e -av. n! It is the same equation as (V) and the textbooks say that it is a good approximation to the ideal lineshape as long as n is large. The poisson distribution is like the binomial distribution, except that it works best to describe processes with a large number of trials and a low probability of success. The PD gives the probability of having n successes after v trials, where a is the probability of success after each trial. If a "success" means for a molecule to move from one plate to the next, then to get to the end of the column, a molecule has to have N successes, one for each plate. Each "trial" is the flow of a volume element v, the volume of a single plate. The probability of success at each trial is the same as the probability of being in the mobile phase, since the entire volume of mobile phase gets transferred to the next plate with each trial. The probability of being in the mobile phase is "a". And "a" is v m + Kv s, which makes perfect sense. Big deal, so what, and why is this even remotely interesting? Plate theory has two main results, both of which can be obtained directly from equation III. The first is that the very top of the curve, which is what you would call the elution volume, can be found by taking the derivative of III with respect to v, setting it equal to zero and solving for V at that point. You ll get (IV) N ve = a = N (v m + Kvs) = Vm + KVs (IV) which is the same as for my simple average speed theory (whew!). So, now we have confidence that V e = V m + KV s because we can arrive at the same equation in two ways. 8

9 Chromatography FST 20 But the simple theory didn t tell us anything about the width of the peak, and in fact implied that it should be infinitely sharp. Equation III indicates that it s not sharp. but a distribution about a maximum. So, how broad is this distribution? To find out, I simplified equation III a little by letting the ratio of concentrations equal y and the quantity av equal x: y = (e -x ) xn N! (IIIsimp) Then, I drew two tangents through the inflection points on either side of the peak, and I defined the width as the segment between the intersection of these lines with the horizontal axis. This sounds hard, but the inflection points are the points where the second derivative equals zero. I already had the slopes of the tangents, because they are just the derivative evaluated at the inflection points. I dropped perpendiculars to the baseline and found that the flat legs of the triangle (see below) had lengths related to the number of column stages that the solute passed through, N (see below). Since you know the coordinates of the inflection points, it s easy to find the distance between them, which turns out to be 2 N. As shown in the figure, the entire width of the peak at the baseline X is 4 N. But X was a placeholder for a V, so X must equal a V, since a is a constant. Y infl Y infl L L 2 N- N N N+ So, we can say that a V = 4 N. But we also know (from the left side of (IV) and other places) that a V e = N. What if we took ratios of these equations: ave N NN N a V = 4 N = 4 N N = N = 4 which means that knowing the elution volume and the width at the base, we can calculate the number of these fictitious stages or plates in the column. This is the Plate Number or number of theoretical plates. It can be calculated as 6 (elution/width) 2, where elution can be volume, time, distance on a chromatogram, anything as long as you measure width as the width at the base, and in the same units you used for elution (time, volume, distance ) Notice that the larger the value of N, the larger the ratio of elution volume to peak width, or comparing two systems with the same elution volume, the solute experiencing more plates was broadened less. The system with more plates was more efficient (see below). The original concept of these stages was just a way to express the idea that the concentration of the solute in the mobile phase was governed by its partition coefficient and therefore reached equilibrium before flow occurred. So, the number of theoretical plates can be thought of as the number of times the solute participated in an equilibrium event decided whether to go to the mobile or to the stationary phase. Why should we be concerned about the plate number of a system? The ultimate goal of chromatography is usually to separate molecules that have different properties tht give them different elution volumes (or retention times). To obtain separation, the elution volumes have to be different (which you ensure by choosing the right kind of column, the right solvent strength, etc.), and the peaks have to be narrow. If the elution volumes differ by 5 ml, then the peaks are well resolved if they are ml wide, but essentially unresolved if they are 25mL wide. Explicitly, resolution is just the difference in elution times (volumes, chart distances ) of two adjacent peaks divided by the mean of the peak widths: Ve V. 4 or Ve Ve R = W + W Ve = W + W 2 Since plate higher plate number signifies narrower peaks, resolution increases with higher plate number (although only at its square root). 9

10 Chromatography FST 20 An easy way to increase the number of plates experienced by a solute in order to increase resolution would be to increase the length of the column. Doubling the length of the column would be predicted to increase resolution by a factor of 2, or 40%. But that doesn t mean that the longer column is more efficient, because it would take twice as long to run the column (unless you run it faster, which would decrease the number of plates; see below). Another common parameter, which is a better indication of column efficiency, is the plate height, H, which is the length of the column divided by the calculated number of theoretical plates. It s also called the Height Equivalent to a Theoretical Plate, or HETP. This parameter could be interpreted as the distance a solute molecule flows before it undergoes an equilibrium event. If this distance is large, the column is not very efficient. Suppose, for instance, that the column is badly packed, and there were channels through which some molecules drifted without coming into contact with the stationary phase. They would elute earlier than solute molecules that passed through better-packed regions of the column. The resulting peak would therefore be broad and the number of plates calculated from the run would be low (the column is inefficient). The plate length would be long, indicating that the average flow between equilibrium events is long. There are other packing problems that could cause band broadening. But you might wonder, even if a column is packed as well as it could be, how high can the plate number be? And, how can the design of the column be changed to improve the plate number? Your would have to model the chromatographic process to include the paths that various solutes take, the velocity of diffusion, the existence of stagnant regions in the column, eddy currents And, you d have to admit that plate theory was a lie at its very premise. A plate was the volume in which the solute partitioned itself between the mobile and stationary phases according to its partition coefficient. In truth, the column doesn t stop and wait for equilibrium to be reached; equilibrium is never really reached; the model should really involve competing rates. Important Results from Rate Theory (Giddings, particle size and flow rates) The mobile phase is always flowing. How fast? It depends; mobile phase that squirts through between two packing particles moves faster than mobile phase that passes through a large open space between particles (i.e., "stagnant mobile phase", see the illustration below). There may also be eddy currents, so that, even though the solvent is moving, it's going around in circles instead of progressing toward the outlet of the column. You can still define a linear flow rate that describes the average solvent flow, but a particular solute molecule might not reside in average-flowing mobile phase., So, while the mobile phase is moving, the solute has to equilibrate between the stationary and mobile phases. How long does it take to equilibrate? It depends; solute that is in a pocket of stagnant mobile phase must diffuse to the surface of the stationary phase before it can decide whether or not to stick. Likewise, a solute molecule that has diffused deeply into the stationary phase (stagnant stationary phase) must diffuse back to the surface before it can decide whether or not to stick. Instead of making the bogus assumption that equilibrium is reached and using an equilibrium constant K, it would be better to use ratios of binding and dissociation rate constants such as k on /k off and formulate chromatography as competition between the on-off kinetics and flow. Of course, at vanishingly small values of flow rate k on /k off. = K, and rate theory simplifies to plate theory. Unfortunately, rate theory is mathematically much more complex than plate theory. It is made even more complex by the fact that some of the rates are not independent, but coupled to one another. Fortunately, people have already worked it out for us under different conditions (e.g., open tubular columns, spherical packing, etc.). The most important result of rate theory not predicted by plate theory is that plate number varies with linear velocity ("flow rate"), with dimensions of the packing, and with the ease of motion of the solute in both the stationary and mobile phases. This relation, usually expressed in terms of H(ETP) rather than N, is generically referred to as the van Deemter equation. The specific solution arrived at by Giddings is most relevant to liquid chromatography in packed columns: H = + C d C dp2 µ / D e p m m + CD d m C µ + smdp 2 µ Cd s f µ + D D m s α β γ δ ε The parameters are as follows: Dm and Ds, diffusion coefficients in the mobile and stationary phase; Ci, constants used to weight the importance of each term; ds, the thickness of the stationary phase film; dp, the diameter of the packing particle; and µ, the linear velocity of the mobile phase. The first term represents eddy diffusion (α, swirling in solvent currents) and mass transfer in the mobile phase (β). The next term, γ, represents the effect of longitudinal diffusion. Term δ represents the escape of a solute molecule from stagnant mobile phase. The last term, ε, takes care of mass transfer in the stationary phase. If a variable appears in the numerator, an increase in that parameter's value causes an increase in H, and consequently, worse performance. Examine each of the terms and see if you can rationalize why each variable appears in the numerator or in the denominator. For the current discussion, two points are important. The first point is that the linear velocity appears in numerator in all terms except γ. Running the column slower nearly always improves performance. More plates are obtained. The only term that indicates that decreased mobile phase velocity makes things worse is the term for longitudinal diffusion. This term isn't nearly as important for liquid chromatography as it is for GC because the coefficient of diffusion of solutes in solvent is much lower that for molecules in the vapor phase. Obviously, if you run a column very slowly, diffusion will be important. The basic shape of the plot of H vs. µ is this: 0

11 Chromatography FST 20 Longitudinal diffusion-limited HETP optimum solvent velocity Mass transferlimited reasonable working range Solvent velocity The other point of major importance is the position of dp in each term in which it appears. I've put it in bold type so you can find it. The particle diameter appears in the numerators. Increasing the particle size always makes performance worse. It creates larger void spaces between particles where stagnant mobile phase exists and makes the effects of alternate paths more severe. The parameter's value is squared in some terms. Normal gravity-flow or low-pressure columns contain particles which are usually mesh. This wet mesh size corresponds to absolute diameters of 75 down to 50µm. If you try to pack 20µm particles in the same column, it won't flow - at all. You need to apply high pressure, or the linear velocity will be so low that the chromatogram would seem never to end and would be broadened by longitudinal diffusion. High Pressure Liquid Chromatography was invented to keep up the flow rates. There's nothing magical about the pressure. It's just necessary to get things to move. The smaller the particles, the better. 5µm particles used to be considered small. Now, every HPLC company makes 5µm particles, and some make 3µm particles. Pressures up to 5000 PSI may be needed (but generally PSI). The figure below, taken from Snyder and Kirkland s book, shows schematically some of the differences between small and larger packings. Smaller column packing Larger packing Stagnant stationary phase Different flow paths Different solvent velocities Stagnant mobile phase Summary We need one more parameter to give a complete description of optimization of resolution in column chromatography, the separation parameter α. It is just the ratio between k's (remember k?) of two closely-spaced peaks: α = k2 where k 2 > k. k Here s the grand equation: the resolution required to separate two closely-spaced peaks is given by the following Rs = 4 k N ( α ) k +. Sometimes, for GC, you see it with the (α-) term written as (α-)/α, because of a slightly different assumption in the derivation. This parameter is only useful when α, so the difference is small. Your first guess about obtaining greater resolution might be to use a longer column (or smaller particles, to increase n). But doubling the column length gives only a

12 Chromatography FST 20 40% increase because of the square root sign. How about changing α? The parameter α is difficult to change predictably. It can be adjusted by altering selectivity, either of the column (use a different stationary phase) or of the mobile phase (choose a different strong component or eluting ion, a different ph, etc.). But, you have quite a lot of control over k, so we'll spend some time optimizing k. This "k" is often described as the average of the k's of the two peaks being separated. In deriving this equation, things work out most conveniently if you assign this k to the second-eluting component. As in the case of α, the two k values must be similar anyway, so the difference in definition isn't large. A plot of Rs versus k looks like this: Rs 0 0 k --> Notice that changing k a little when k is small causes a large change in resolution. This is the sensitive region of the curve. When k is large, resolution is not very sensitive to k. Also, large k's cause long analysis times and broader peaks. So, we'll want to work with k values that are not too large. On the average, people like to work at k values between and 0. If you only have two peaks to separate, maybe.5 < k < 0 is okay. If you have a bunch of peaks to spread out, the range 2 < k < 20 may be more practical. This equation can be very useful to help you select the perfect solvent composition to give them best possible resolution. But, in many kinds of chromatography, the individual k values differ by a large amount, which means that the α value are also large. So, resolution is not as big an issue as simply waiting for the later peaks to elute. Maybe the solvent is too weak to elute them, or they might elute, after hours, as very low, broad humps that are difficult to distinguish from baseline. If you strengthen the solvent (add more of a strong component, more salt, etc., depending on the kind of chromatography), then you run the risk that two peaks that bind the column weakly and would have separated nicely in the weak solvent will elute together in a solvent strong enough to cause elution of a more tightly-bound solute. There are at least four solutions to this problem:. use another kind of column (i.e., side-step the issue), 2. choose a solvent of intermediate strength (I sort of implied that this isn t always possible), 3. elute the two groups separately ("batch elution") using different solvents, and 4. use a continuous gradient of solvent strengths. All of these are reasonable and have different advantages. Let's omit the first by assuming you have chosen the best column for the separation. The second approach, termed isocratic elution, can give very good resolution (maybe too good to be practical) and does not require that you re-equilibrate the column with starting solvent (see approaches 3 and 4) before making another run. But the solutes might be so different that, in order to get adequate resolution among the weak solutes, you have to accept very long retention times for the strong solutes. Not only is time wasted, but there are other serious results. Tighter binding means that the solute undergoes more plates (N is large). Larger N values mean that the peak is broader; the solute is more dilute as it exits from the column. If you are using the column in a preparative procedure, your compound will appear in a greater volume - maybe too large to handle. If you are trying to quantitate the solute, it is bothersome that the compound is harder to detect - short, fat peaks are harder to detect and quantitate than tall, sharp ones. The batch elution approach, changing to a stronger solvent after the weak solutes are out, can also be very useful. Lines are sharp and resolution is adequate. But it requires that you have quite a lot of information before you make a run (how many and which solvents do you use?) It also requires some programming or, at least, timed changes in the solvents and it isn't very flexible. If you are going to be running the same analysis over and over, it might be the best way to go. The major disadvantage would be that you have to re-equilibrate the column with the weak solvent between runs. Gradient elution, something like temperature programming in GC, is also a very attractive possibility. Gradients are not difficult to make. All compounds can be run with similar apparent K or N values (note the "apparent") and therefore have similar linewidths and detectabilities. You can make the gradient cover a wide range so that you need relatively little information before you make your first run. You can even use the information you get from a gradient run to choose the best isocratic or batch solvent to use. The major drawback 2

13 Chromatography FST 20 is that the column must be re-equilibrated between runs. We'll talk about gradients in relation to ion-exchange chromatography. So, all kinds of chromatography are about the same and follow the same general rules. There s always a stationary phase, a mobile phase, and a solute that partitions between them. What differs among different kinds of chromatography is the nature of the interactions that lead to adsorption or partition of the solute onto/into the stationary phase as well as the interactions that cause the solute to partition into the mobile phase. For chromatography of proteins, these mechanisms depend on the properties of the proteins that are exploited in order to effect separation or, perhaps can be measured using chromatography. We ll start with an odd one. Size Exclusion Chromatography An important variety of chromatography for proteins is size exclusion or gel filtration chromatography. The stationary phase is composed of porous particles, packed into a column A particle A packed column This system behaves like partition chromatography: where, V o = volume of solvent between particles V p = volume of pores K = partition coefficient V e = V o + KV p Solutes that are too large to fit into the pores view the particles as solid spheres and (ideally) do not interact with the packing material at all, so K = 0, and Very small solutes diffuse into the pores readily, K = Ve = Vo V e = V o + V p So far, we can use this kind of column to separate very large molecules from very small ones. If we want to separate proteins from one another, we have to get more detailed in our approach and consider molecules that fit into the pores but are not very much smaller than the pores? First, devise a way to measure K but Vp is hard to measure, so let K = Ve Vo Vp V bed = V o + (V particle + V pore ) = V o + V s,, 3

14 Size Exclusion Chromatography FST 20 Vbed can be measured from the geometry of the column or by dumping out the packing and measuring the volume it occupied. The definition of the stationary phase has changed a little, because it is really the pores, but we lump the pores together with the particles (the "inert phase"). So, let's give our partition coefficient a new name, K av, just to be honest. Then, K av = V V e bed Vo V o (note Vs Vp if particles are uniform) Vo is easily measured by measuring Ve of a molecule larger than the pores (Blue Dextran 2000 is common). So, the partition coefficient or K av can be determined for any protein. Common lore says that for a particular protein, K av log (M r ), as exemplified by the Sephadex G series: Selectivity for globular proteins av K G-75 G-00 G Log Mr So, you can select your gel on the basis of expected M r. But there is no theory to support this assertion, and a real case may look like this: Cytochrome c av K Hemoglobin BSA AdHCatalase 0 Urease Ferritin Fibrinogen Log Mr So, what really determines K av? Consider hemispherical pore of radius r and molecule of radius a. What is the volume of the pore as seen by the solute? The true volume of the pore is 2 3 πr3 but the solute can't fit in the outer 'a' of the pore because it bumps the wall; the apparent radius is (r - a), so the apparent volume 2 is 3 π(r - a)3. 4

15 Size Exclusion Chromatography FST 20 r a Our elution equation contains the term KV p, which is K*( 2 3 πr3 ) summed over all of the pores. How can we incorporate the apparent volume K*( 2 3 π(r - a)3 ) into the equation? And what is K, really? K is really.ö..., because the environment inside and outside the pore is the same, so there's no energetic drive for the solute to be inside or outside the pore. But, because they are multiplied together, K and Vp are mathematically indistinguishable - we can tell that their product varies, but not which one is responsible for the change. In chromatography, it is always held that the volume of the stationary phase is constant and that K varies from solute to solute; in this case K is constant and the apparent volume changes from solute to solute. So the K*V term could be either K apparent *Vp or K*V apparent. Let's accept tradition and say that K changes; to make it quantitative, set K apparent *Vp and K*V apparent equal to each other and solve for K apparent : so that K apparent *( 2 3 r3 ) = *( 2 3 (r - a)3 ), K app = (r a) 3 r So, the cube root of K ought to be proportional to the radius of the solute. The solute may not be spherical, so let's use an effective hydrodynamic radius, the Stokes' radius (more about this parameter in a minute). Here are the data shown above, replotted vs R s. 3 K / R s (Å) Not bad. The fact that fibrinogen is on the line suggests that the model is correct, at least approximately. This is only one of several theories. My treatment is based on a model by Porath. An old but thoughtful review with experimental tests is given by L.M Siegel and K.J. Monty, Biochim. Biophys Acta 2, (966). Other models were proposed by Laurent & Killander and by Ackers (and actually, by several others since then). Their models lead to the following equations, which give rise to the following plots. 5

16 Size Exclusion Chromatography FST 20 Laurent & Killander: (-logkav) /2 = α (β + Rs); /2 (-log K av ) R s These models require that you fit the parameters to slope and intercept terms α and β. You have to run standards in order to evaluate α and β. Ackers also gives an equation that yields linear plots. His method has only one calibration parameter, so you only need one standard: (Å) 00 - R s 2 A K = R s A Rs 3 A Rs A 5. Which theory is correct? Who knows! All of them fit the data pretty well, so we can't really distinguish among them on the basis of the experimental work available. Part of the reason for our inability to decide is that K only varies between 0 and. Physicists like to test theories over a range of several orders of magnitude. We're stuck, because we can't go higher than and the method is not sensitive enough to measure K values of.,.0, as being different. In practice, Laurent and Killander's method is commonly used because Kav is much easier to measure experimentally than K. So, the hydrodynamic radius of a protein or other polymer can be determined by gel filtration; its molecular weight can be estimated by comparison to standards that have the same shape as the unknown. The method is gentle, so subunit structure usually remains intact. Size-Exclusion systems can also be used for denatured proteins. Salt, acid, urea, guanidinium chloride, etc., can be included in the buffer (before the column is packed!) and unfolded proteins can be run. As in SDS gel electrophoresis, reducing agents may be added to ensure complete dissociation and unfolding. Unfolded proteins are more likely than native proteins to have similar shapes. What (good) is the Stokes' radius? The trouble with the simple model we just developed is that it requires that we know the radius of the solute molecules. What if they aren't spherical? What is the radius of a non-sphere? Luckily, we are not the first ones to run into this problem; it has come up before in hydrodynamics. The frictional force on an object moving through a non-vacuum is equal to its velocity times a frictional coefficient: f*v. For a sphere, the frictional coefficient can be calculated easily from the Stokes equation: fo = 6πηr, where r is the radius of the sphere and η is the medium (solution) viscosity. The subscript o indicates that this frictional coefficient is for a sphere. If we don't know whether the molecule is a sphere or not, we could still measure the frictional coefficient and calculate a value of "r" from it. This apparent or effective radius is called the Stokes' radius, Rs. The Stokes equation still holds but the f no longer should be written with the o subscript: f = 6πηRs. Anyway, we now have a well-defined effective radius that we can use in our model for size exclusion chromatography of notnecessarily-spherical molecules. It is also a useful parameter used to describe protein structure. If it is put together with other 6

17 Size Exclusion Chromatography FST 20 information, it can give you a good idea of molecular shape. For instance, suppose you can measure the molecular weight independently (by ultracentrifugation or SDS PAGE). Then, you can make some assumptions about what would happen if the molecule were a sphere. The mass is the volume times the density or the volume divided by partial specific volume, v -. The volume of a sphere is 4 3 πr3. So, Mass = 4 3v πr3 and the radius of this supposed sphere is r = 3vM 3 4π. Since we're assuming this molecule to be a sphere, we can calculate fo, the frictional coefficient for a sphere, from this expression. Nearly all globular proteins have v - = ~0.735, so the only real data we need is the molecular weight or mass. 3vM fo = 6πηr = 6πη 3 4π. Now, if we do a gel filtration experiment and get a Stokes radius, we can calculate the real frictional coefficient from f = 6πηRs. The ratio of these two frictional coefficients is called the frictional ratio. Doing the algebra leads to the following expression for the frictional ratio: f f o = Rs 3vM 4π 3. If our molecule is not a sphere, what shape is it? It makes sense, as a first approximation, to say that the molecule is a symmetrically distorted sphere - an ellipsoid (a three-dimensional version of an ellipse). Ellipsoids come in two types. A prolate ellipsoid is the solid shape you get by stretching a sphere so that it looks like an egg or a football. An oblate ellipsoid is obtained by compressing a sphere so that it looks like a discus. Each kind of ellipsoid has two axes defining the degree of distortion, a short one and a long one. The ratio of the long to the short axis is called the axial ratio. The values of frictional ratios can be looked up in various tables where they are listed according to the axial ratio produces a given frictional ratio. Axial ratios are also important in determining the viscosity of a protein solution. An example of the dependence of frictional ratios on axial ratios is shown below prolate f/fo oblate.2 after Van Holde Axial ratio (a/b) 7

18 Size Exclusion Chromatography FST 20 Another common use of the stokes radius obtained from gel filtration is to compare it to velocity ultracentrifugation data. The data obtained from velocity ultracentrifugation is the sedimentation coefficient, S20,w. If you know both the frictional coefficient and S20,w, you can calculate the molecular weight: M = N f o S20w vρ, where No is Avogadro's number and ρ is the density of the solution. So, if you have the sedimentation coefficient and the Stokes radius, you can obtain both the molecular weight and the axial ratio. Neither piece of data alone would give you an unequivocal answer for either parameter! Choosing a size-exclusion system. Although a size-exclusion experiment alone does not give you a good estimate of molecular weight (unless your standards have the same shape and density as your unknown), the molecular weight is generally a good guide as to which column to choose. Catalogs list their products according to their "selectivity range", which is usually expressed in molecular weight units. Sometimes, two selectivity ranges are reported, one for spherical molecules (e.g., globular proteins) and one for rigid rods (e.g., dextrans). The selectivity range is narrower than you might think. Kav can only range between 0 and. Obviously, molecules larger than the "exclusion limit" all have a K of 0 and can not be separated. Unfortunately, molecules that have Kav's of.990 and.999 are not very different from and are essentially entirely included in the pore. So, the range of K's available is much smaller than for "real" partition chromatography. Also, since Kav can't be grater than, there is a fundamental limit on the α value that can be provided by a size-exclusion column. Also, it is a relatively low plate-number process and size-exclusion columns must be run at rather low linear velocities. If the separation is difficult, it might be best to try another kind of chromatography (i.e., another partitioning principle). The fundamental choices of stationary phase material (i.e., inert phase) will depend largely on the kind of mobile phase the sample is soluble in. Systems exist that are compatible with both aqueous and organic solvent. A partial list of available media will be passed out in class. Suggested Reading: L.M. Siegel and K.J. Monty, "Determination of Molecular Weights and Frictional Ratios...Gel Filtration...", Biochim. Biophys. Acta 2, 346 (966). Compares several models for the mechanism of gel filtration on Sephadex G-200. K. Unger, "High-Performance Size Exclusion Chromatography", Methods Enzymol. 04C, (984). K.E. Van Holde, Physical Biochemistry, st or 2nd edition, Prentice-Hall, Inc., Englewood Cliffs, N.J., 97 or 98. Frictional ratios, axial ratios, velocity ultracentrifugation and equilibrium density ultracentrifugation. Related Topics Guessing the diameter of a protein from its molecular weight Proteins have a partial specific volume of about This is a well-known number, because a buoyancy correction ( - v - ρ) must be made for the calculations involved in sedimentation velocity experiments. The reciprocal of the partial specific volume is the molecular density, which is about.35 g/ml. If a protein has 04 amino acids (mitochondrial cytochrome c), it should have a molecular weight of 04aa*2daltons/aa =,648, or about 2kDa. Using the molecular density, a 2kDa protein should occupy 2,000 g*mol - /.35 ml*g - = 8,888 ml*mol -. Dividing by Avogadro's number to obtain volume/molecule, we get about.5 x 0-20 ml/molecule, and using ml = cc = (0-8 ) 3 Å 3, we get the volume in cubic Å as.5 x 0 4 Å 3. Setting this volume equal to the volume of a sphere, 4 3 π r3 and solving for r, we get about (3500) /3, or 5Å. The diameter is therefore about 30Å. Ion Exchange Chromatography The development of partition chromatography assumes that the partition coefficient K is constant. Obviously, at very high solute concentrations, this assumption breaks down because at least one of the phases becomes saturated with the solute. Chromatography is done at lower solute concentrations, where neither phase is "overloaded". Some kinds of chromatography have the phenomenon of saturation built into them explicitly. These are systems in which the stationary phase is a surface to which the solute binds. The most famous case is adsorption chromatography, sometimes called 8