Chem 406 Biophysical Chemistry Lecture 1 Transport Processes, Sedimentation & Diffusion

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1 Chem 406 Biophysical Chemistry Lecture 1 Transport Processes, Sedimentation & Diusion I. Introduction A. There are a group o biophysical techniques that are based on transport processes. 1. Transport processes involve observing how macromolecules move (are transported) when they experience orces acting upon them. 2. These techniques include a. Sedimentation b. Electrophoresis c. Diusion measurements. 3. Each involves a dierent orces a. Sedimentation - centriugal orce b. Electrophoresis - electromotive orce (electrical potential) c. Diusion - Thermal energy 4. In addition to their use or isolating macromolecules, these techniques provide inormation about the ollowing physical properties o a macromolecule various combinations: a. Mass - sedimentation b. Size - sedimentation, electrophoresis, diusion c. Shape - sedimentation, electrophoresis, diusion d. Charge - electrophoresis e. Degree o hydration - sedimentation, electrophoresis, diusion 5. In addition to using these techniques to characterize macromolecules, they are also used to isolate and puriy macromolecules. B. We will ocus on sedimentation as a representative example o a technique that is based on transport processes. 1. We will examine how sedimentation provides inormation about a macromolecule's mass, size, shape and degree o hydration. II. Theory o Sedimentation A. We will ocus irst on an individual macromolecule and then later expand to consider molar quantities o macromolecules. 1. We will consider a macromolecule that is dissolved in an aqueous solvent. a. This solution may also contain buer salts and other small molecular weight components; we will consider these as part o the solvent. b. The solvent will be treated as a continuum. B. Centriugal orce, Fc. 1. In sedimentation, the orce that causes a macromolecule to move is a centriugal orce.

2 2. According to Newton's laws o motion, orce is equal to the product o mass (m) times acceleration (a). F = ma a. When a orce acts upon a molecule with mass m, it accelerates: a = F m b. When a macromolecule revolves around an axis, as it does in a centriuge, it experiences a centriugal acceleration, ac : a c = ω 2 r i. Centriugal acceleration is equal to the square o the angular velocity, ω 2, times the distance rom the axis o rotation, r. 1. ω has units o radians/s. c. Making this substitution the centriugal orce is F c = mω 2 r. C. Frictional drag, Fd 1. As a macromolecule in solution moves under the inluence o a centriugal orce, it experiences an opposing orce. a. The solvent lowing past it causes this orce. b. This orce is proportional to the velocity, v, o the macromolecule: F d = v i. Where is the proportionality constant. ii. is called the rictional coeicient. 1. As we will see, it is a property o the macromolecule and is determined by the macromolecule's size and shape. 2. For a sphere, the rictional coeicient is give by Stoke's equation: o = 6πηR o a. Where R is the radius o the sphere and η is the viscosity o the solvent. D. Buoyant orce, Fb. 1. As the macromolecule move it displaces the solvent. a. The displaced solvent must move in a direction opposite to the macromolecule. 2

3 b. The displaced solvent will also experience a centriugal orce and because the displaced solven moves in a direction opposite to the macromolecule, this orce will oppose the movement o the macromolecule. Mass o displaced volume, m o m o = Vρ F b = ( Vρ)ω 2 r i. Where V is the volume o the macromolecule and r is the density o the solvent. ii. The volume o a macromolecule can be determined rom its mass, m, and its partial speciic volume, ν, V = mν 1. ν, can be determined experimentally. 2. For proteins, ν can be also be predicted rom the amino acid composition. iii. Making this substitution or V: F b = mν ρω 2 r E. Sedimentation velocity 1. The centriugal and buoyant orces are independent o the macromolecule's velocity, however, the drag orce is proportional to the velocity and opposes the movement o the macromolecule. (See Fig 5.3, p196, diagram o sector cell) a. As the macromolecule accelerates under the combined inluences o the centriugal and buoyant orces, the drag orce increases. b. When the orces balance one another, the macromolecule will experience no net orce and will stop accelerating. i. It will then move at a constant velocity known as the steady-state velocity. ii. For sedimentation experiment, this velocity is also called the sedimentation velocity. c. It takes only about a nanosecond or a macromolecule to attain its steady-state velocity. (See Fig 5.1, p193, velocity vs time) 2. At steady-state, the sum o the three orces acting on a macromolecule is 0: Fc - Fb - Fd = 0 a. The signs o the buoyant and drag orces are negative because the oppose the movement o the macromolecule. 3

4 3. Substitution o the expressions derived above or the these three orces gives: mω 2 r mνρω 2 r v = 0 ( ) v = 0 mω 2 r 1 νρ 4. Solving or the sedimentation velocity, v: v = ω 2 rm( 1 ν ρ) F. Sedimentation coeicient 1. The sedimentation velocity is a quantity that can be measured a. It depends not only the properties o the macromolecule and the solvent (m, ν, ρ and ), but also on how ast the centriuge is run (ω) and how ar rom the axis o rotation the solution is placed (r). 2. I the sedimentation velocity, v, is divided by the centriugal acceleration, ω 2 r, a parameter is obtained which is independent o the instrument settings: v ω 2 r = m( 1 ν ρ) 3. This parameter is represented by the letter s and is called the sedimentation coeicient: s = v ω 2 r s = m( 1 ν ρ) a. The sedimentation coeicient has units o seconds and or macromolecules has a magnitude o around seconds. b. Sedimentation coeicients are thereore typically given in units o Svedbergs, S. i. One Svedberg is equal to second. ii. The unit is named in honor o Theodor Svedberg, a Swedish chemist who won the Nobel Prize in 1926 or inventing the analytical ultricentriuge. 4. So ar we have been considering only a single macromolecule, i we wish to consider mol o macromolecules, he numerator and denominator o the 4

5 equation or the sedimentation coeicient is multiplied by Avogadro's number, R : s = Rm( 1 ν ρ) R = M ( 1 ν ρ ) R c. Where M, represents the molecular weight o the macromolecule. III. Moving Boundary Sedimentation A. Determining the sedimentation coeicient 1. In an analytical ultracentriuge the solution containing the macromolecule is placed in a sector shaped sample cell with windows that allow the solution to be optically analyzed during the centriugation. (See Fig 5.5, p198, Schematic o centriuge) 2. As the centriugation proceeds the region o the sample cell closest to the axis o rotation becomes void o macromolecules. a. This produces a solvent/solution boundary. b. This boundary moves with a velocity equal to the sedimentation velocity o the macromolecule. 3. There are various optical methods or monitoring the movement o this boundary with time. a. One o these is a UV scanner. i. Since both proteins and nucleic acids absorb ultraviolet (UV) light, the location o the boundary can be ound by measuring the absorption o UV light as a unction o the radial distance rom the axis o rotation. (See Fig 5.4, p197) 4. The velocity o the boundary is given by derivative equation, drb/dt, which can be set equal to the sedimentation velocity at the rb s = v ω 2 r b v = r b ω 2 s dr b dt = r b ω 2 s 5

6 b. Rearranging this equation gives the ollowing dierential equation dr b r b = ω 2 sdt c. Integrating both sides o this equation gives dr b = ω 2 s dt r b dr b = ω 2 s dt r b ( ) = ω 2 st + constant ln r b d. This equation has the orm o a straight line equation when ln(rb ) is plotted against time (t). i. The slope o the line is equal to ω 2 s and the y-intercept is equal to the constant. IV. Corrections and Adjustments to the Sedimentation Coeicient A. Solvent density and viscosity. 1. The sedimentation coeicient is dependent on properties o the solvent. 2. To see how it depends on solvent properties, we substitute Stoke's equation or the rictional coeicient: s = M( 1 ν ρ) R ( ) = M 1 ν ρ R6πηR e a. The only quantities in this equation that depend on the solvent are the density, ρ and the visconsity, η. 3. To produce values or the sedimentation coeicient that are comparable rom one determination to the another, the values are typically converted to a values expected i the solvent has the viscocity and density o water at 20 C (See Table 5.2, p200 or densities and viscocities or water) 6

7 ( ) ( ) 1 νρ 20,w s 20,w = s T,b 1 νρ T,b η T,b η T,w η T,w η 20,w b. The viscosity is actored into two terms because η T,b η T,w, which is called the relative viscosity, is easily determined by measure the time it takes the buer to low through a capillary tube relative to the time that it takes water to low through the same capillary tube. B. Eect o diusion on the boundary 1. As sedimentation proceeds the boundary will spread. a. This is due to diusion, which is also a transport process. 2. In diusion the macromolecules are moving as a consequence o thermal energy. a. The movement o an individual macromolecule is random in direction and occurs with equal probability in all directions. 3. Like sedimentation, the movement o the macromolecule is opposed by rictional drag. 4. Einstein derived an equation the related diusion to the thermal energy, RT, rictional coeicient, : D = RT R a. Where R, is the ideal gas law constant, and T is the absolute temperature. b. The parameter D is called the diusion coeicient, and like the sedimentation coeicient, it is inversely proportional to a macromolecule's rictional coeicient. i. It can be determined by measuring the rate o boundary spreading. (We will no discuss the details or determining D) ii. Another method o measuring a diusion coeicient is dynamic light scatter, in which the time dependent luctuations in scattered light intensity rom a solution can be used to determine the diusion coeicient o the solute molecules in the solution. (The luctuating scattered light intensity is due to the diusion o the solute molecules.) c. Einstein s equation can be combined with the sedimentation equation to obtain and equation in which the rictional coeicient is actored out: 7

8 s = M( 1 νρ) R R = RT D s = solving or the molecular weight : M = s D MD( 1 νρ) RT RT ( 1 νρ) iii. The sedimentation and diusion coeicients or a macromolecule can be combined to determine the molecular weight o a macromolecule. C. Eect o concentration 1. At inite concentrations, interactions between molecules can cause changes in the determined value o the sedimentation coeicient. a. To eliminate this eect, the determination o the sedimentation coeicient is done at several concentrations, plotted against the concentration, and extrapolated to zero concentration, where interactions can no longer take place. i. The extrapolated, zero concentration value is represented by a o subscript. o ii. s w,20 is a sedimentation coeicient expected at zero concentration at 20 C in water. D. Analysis o the rictional coeicient 1. The rictional coeicient contains inormation about the size and shape o the macromolecule. 2. The rictional coeicient can be obtained rom either the sedimentation or diusion coeicients: 8

9 ( ) M 1 νρ = = or o Rs w,20 RT o RD w,20 3. Using Stokes equation, the radius o a sphere having the same rictional coeicient as the macromolecule can be determined: = 6πηR S R S = 6πη a. This radius, RS, is called the Stoke s radius. 4. I the mass o a molecule is molded into a perect sphere would usually be less than the Stoke s radius a. This radius, Ro, can be determined rom the molecular weight and partial speciic volumes o the macromolecule. V o = 4π 3 ( R o) 3 R o = = 3 3 3V o 4π 3 o Mν 4π R b. Ro is usually less than Rs i. This can be due to deviations rom a spherical shape ii. I can also be due to the binding o solvent molecules to the macromolecule s surace (hydration), which are being dragged along with the macromolecule and increasing its eective volume. iii. The rictional coeicient can be actored into two terms which account or each o these eects: 9

10 iv. o = sp sp o Where 1. is the observed rictional coeicient 2. o is the rictional coeicient or the unhydraded sphere. It can be calculated rom Ro: o = 6πηR o = 6πη 3 3 o Mν 4π R 3. sp is the rictional coeicient or a sphere having the same degree o hydration as the macromolecule v. The ratio sp o is called the hydration actor and can be used to calculate the volume o solvent bound to the macromolecule per unit volume o the macromolecule: sp o = 1+ δ ( ) Where δ represents the degree o hydration in ml o solvent per ml o macromolecule. vi. The ratio sp is called the shape actor and can be used to estimate the deviation o the shape o the macromolecule rom spherical symmetry 1. This usually done be modeling the macromolecule as either a prolate or oblate ellipsoid o revolution: 10

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