Lecture /1/8 University of Washington Department of Chemistry Chemistry 45 Winter Quarter 8 A. Analysis of Diffusion Coefficients: Friction Diffusion coefficients can be measured by a variety of methods including dynamic light scattering. Information on molecular shape can be obtained from diffusion coefficients. The radius of a spherical protein can be calculated from the diffusion coefficient (i.e. the Stokes radius) and from the specific volume, i.e. the volume occupied per gram of unhydrated protein. These two numbers often differ because the Stokes radius is a hydrodynamic radius, i.e. the radius of a sphere composed of a protein covered with tightly attached water molecules (waters of hydration). The degree of hydration of the protein (i.e. grams of hydrated waters per gram of protein) may be determined by comparison of the Stokes radius with the radius calculated assuming an unhydrated protein sphere. Example: The diffusion coefficient and specific volume of bovine pancreas ribonuclease (an enzyme that digests RNA) have been measured in dilute buffer at T=9K and are found to be: D=1.1x1-11 m /s V =. 77cm / g. The molecular weight of the protein is 1,69 g/mole. a) Calculate the frictional coefficient f. 16 1 k T c18. x1 g cm s K hb98kg B 8 f = = = 8. 1 g / s 7 D 11. 1 cm / s b) Assuming the protein is an unhydrated sphere, calculate its frictional coefficient and its radius. 4 1 69 77 MV MV b, g / molegc. cm / gh πr = r = = N A 4π N A 4π 6. 1 molecules / mole 7 r = 157. 1 cm = 157. Angstroms and 1 1 7 8 f = 6πηr = 6π c. 1g cm s hc 157. 1 cmh =. 95 1 g / s 8 f r 9. 1 c) = = = 15.. 8 f r 95. 1 Therefore F HG I f r V 1V1 115 f KJ F = = H G I δ. r K J = V b115. gc. 77cm / gh. 77cm / g 15. grams 1 δ 1 = 1. cm / g = water/gram protein
d) How many water molecules hydrate each protein molecule 1 moles HO = 15. g / 18g mole =. 58 moles moles protein = g g mole 1 1 / 1, 69 =. 7moles HO / protein =. 58 /. 7 = 79. 5 8 B. Non-spherical Molecules: In many cases, the shapes of rigid, non-spherical molecules can be modeled in terms of some idealized shape. For example, virus particles can be modeled as rods or ellipsoids of rotation. Frictional coefficient for rod-like particle. o Suppose a rod-like particle has a length a and radius b. Its volume is given by the formula: Vrod = π ab. The axial ratio P is defined as P = a. b o For a rod-like particle, the frictional coefficient is defined as ( ) 1/ / / P f = f. ln( P).1 o The term f in the equation above is defined as f = 6πη R. R is defined as the radius of a sphere which has a volume equal to the volume of the rod with axial ratio P=a/b. R is determined as follows: F ab ab Vsphere = Vrod R = ab R = R = H G 1 / 4π I π K J Frictional coefficient for a prolate ellipsoid (cigar-shaped particle). For a prolate ellipsoid the volume is V = 4 π pro ab. The lengths a and b are called the major and minor semi-axis lengths, respectively.
1 / P P 1 For a prolate ellipsoid f = f where again f = 6πη R ln P P 1 Again, R is the radius of a sphere which has a volume equal to the volume of a prolate ellipsoid for which P=a/b. In this case R ab 1 =c h /. In general, the frictional coefficient ration f/f for a prolate ellipsoid does not vary markedly with aspect ration P. For example, for P ranging from 1 to, f/f only varies from 1. to about.. C. Brownian Motion in Liquids: Microscopic View Having considered briefly the macroscopic diffusion equation, let us consider the nature of the motion of individual molecules in fluids, and thus explore microscopic aspects of diffusion. Let us for now consider the motions of a solute molecule, which is surrounded by solvent and is much larger and more massive than the solvent molecules. If the trajectory (pathway) of an individual solute molecule is traced out it appears to be a random walk The displacement of the solute molecule after a time t is indicated by the dark, bold line. Designate this displacement r. if we follow the progress of this molecule, its path is described by a series of short jumps, each jump punctuated by a change in direction, which appear random. If we average the net displacement over time, the time-averaged net displacement is zero. Call this average x = If we equivalently observe an ensemble of molecules execute a jump we see that molecules will be displaced in other directions. For every solute molecule displaced r as shown, there will be another solute molecule with a displacement in the opposite direction -r. The displacements will average to zero over the ensemble. Call the ensemble average x =. From the point of view of forces, the chaotic path of the particle is the result of two forces which govern the motion of the particle. The first is the force of friction which we have defined as F = fv and results from the frictional drag that the f
particle encounters as it moves through a fluid, the latter viewed as a viscous, continuous medium. The second force is the random force arising from incessant collisions between the particle and the solvent. The random nature of this force F(t) effects the seemingly random direction change, while the short hops are the result of the solvent drag on the particle and the fact that the collision displace only slightly the massive solute particle. One is tempted to try to calculate the path of the particle using equations of motion, i.e. Ff F( t) = Ma (.1) where F(t) is the random force. But this is not a particularly useful approach. The reason that (1.1) is not useful for the problem is that, as a result of the large number of solvent molecules and the complex nature of their interactions, the force F(t) is very complex and its exact form is not known. In fact, F(t) can only be understood through average properties but if we average F(t) over time of over all solvent particles that collide in an instant with the massive solute particle, the averages are zero () F() t F t This type of motion is called Brownian motion. = = (.) D. The Random Walk: The fact that (1.1) cannot be used directly and that only the averages of F(t) can be understood mean that we cannot predict with certainty, where the particle will jump to at a given instant. We can only contemplate the probability that a particle will reach a given point and as such we can only calculate average displacements. We have already concluded that x = but this need not be the case for x. Many processes in nature including the motion of Brownian particles, motions of gas molecules, and the average shape of a linear polymer, can be treated as random walk problems. One dimensional random walk: Suppose N particles are placed in the middle of a linear trough. Each particle can hop forward () or can hop backward (-). After each particle has taken a hop, the number that hop forward is N and the number that hop backwards is N -. The probability for a forward hop is N N P = = (.1) N N N. The probability for hopping backwards is N N P = = (.). N N N Note P P =1 and N - =N-N. There are many possible outcomes. All the particles may jump in the direction (N =N). All the particles may jump in the direction (N - =N). There are clearly many combinations of jumps for N fleas. To obtain the probability that any
combination of and jumps occurs, consider the total probability for N fleas jumping N N N1 N ( N 1) N ( P P ) = P N P P P P... (.)! N! N N P N P N P N...!! Each term in the expansion corresponds to the probability of a particular outcome. For example, P N is the probability that all N fleas jump in the N! direction. In general N N P N P N!! is the probability that N fleas jump and N- jump -. The term N! N! W( N, N ) = = (.4) N! N! N! ( N N )! is the number of possible ways N particles can jump such that N go in the direction and N - go in the direction. E. Net Displacement: The Degeneracy Function W(N,N ) W( N, N ) is called the degeneracy and is basically the same parameter encountered in equilibrium statistical mechanics. It quantifies the number of ways of obtaining a given net displacement. The expression for W(N,N ) can be expressed for large N in terms of the net displacement number m. lnw = const ln N! ln N! ln N! For large N ln N! N ln N N (Stirling s Approximation). lnw const N ln N N N ln N N Suppose we want the probability in terms of a net displacement N m N m N = ; N = excess displacement of m steps to the right. Then using these definitions N m 1 N m N m 1 N m lnw = cons tan t ln 1 ln 1 N N (.5) N m m N m m cons tan t ln 1 ln 1 N N and the approximation 1 m m 1 m ln ( 1± x) ± x x or ln 1 ± ± N N N (.6) m m /N lnw const W e (.7) N
m is a number displacement so the net distance displacement is x=lm x / l N x /l N where l is the unit displacement. W e orw = ce where c is a constant. We can determine c because W, as a probability dependet on x must integrate to 1: x / N 1 dxw ( x) = c dxe = 1 c = (.8) π N Suppose the number of hops per unit time is N. Then the number of hops N=N t. Therefore W = 1 x / l N 1 x /l Nt e = e πn t πn t (.9) Because W(x) is a probability, we can use it to calculate averages: x = xw ( x) dx = (.1) which is the same result obtained with the diffusion equation. But the average squared displacement is x = xw x dx= N ( ) (.11) Conclusion, for a random walk process, the root mean square displacement is the microscopic unit displacement times the square root of the total number of hops N. By recalling the macroscopic mass transport coefficient, the diffusion coefficient D is related to the parameters of the random walk model x = Dt = N (.1) F.. Chain Polymers: Mean End-to-End Distance & Random Coils Many biological polymers (e.g. high molecular weight DNA) may be idealized as a repeating structural unit connected by a series of bonds (see diagram). The shape of the polymer is determined by rotations around the bonds. The rotational state of the bond between the (i-1) and the i unit is designated by the torsional angle φ i. The shape of the polymer chain is determined by the set of
torsion angles {φ i }. Two conformationally-dependent parameters associated with a chain polymer are the end-to-end distance and the radius of gyration. Suppose a polymer chain is composed of n1 repeating units. Imagine the n1 repeating units are connected by n bonds. Associated with each bond is a vector l i, which is parallel to the bond n axis. The end-to-end distance vector r is the sum of the n bond vectors: r = li (see diagram). i= 1 The magnitude of r is ( r r) 1/ where r l l l l nl l l n n n n = i j = i j = i j i= 1 j= 1 i= 1 j= 1 j> i There may be many possible shapes for the polymer. Therefore we can identify a mean-squared end-to-end distance r = r = nl l l i j j> i Random Coil Polymers: A particularly simply model of a chain polymer is called a random coil polymer. Basically at each repeating unit the direct of the bond is random. Therefore the shape of the polymer can be visualized as following a random walk making the structural description of such a polymer an exact analog of the diffusion of a particle through a continuous medium. A consequence of the random nature of bond directions is that in the equation r = r = nl l l l l = r = r = nl i j i j j> i j> i