SPRING 2011 Phys 450 Solution set 2

Transcription

1 SPRING 011 Phys 450 Solution set Problem 1 (a) Estimate the diameter of a ater molecule from its self-diffusion coefficient, and using the Stokes-Einstein relation, assuming that it is a spherical molecule. Look up the diffusion coefficient in some reference book and identify your source. Self-diffusion constant of ater (D ). m /s (Source: F. Franks, Water, The Royal Society of Chemistry, London, 198). From the Stokes-Einstein relation D = k BT / 6πηa e can estimate the diameter (d) of the ater molecule as d = a = k J Therefore, d = m Å π Pa. s. m / s B T / πηd This value of the diameter is an underestimate. Note that the van der Waals radius of an oxygen atom is ~1.4 Å, so the diameter of the ater molecule should be at least tice this radius. This underestimate is most likely because the Stokes-Einstein relation, in hich the medium is treated as a continuum, breaks don hen describing the selfdiffusion of ater. (b) A particle undergoing Bronian motion is observed in a microscope, hich gives only x-y information about the position. From a series of measurements, < x + y > is found to gro linearly in time, ith a slope of. 8 cm /s. What is the diffusion constant D for the particle? (c) The Svedberg constant for the ribosome (the large complex in the cell responsible for protein synthesis) is 70S. Estimate the size of the ribosome. You may assume that the density of the ribosome is about 0% larger than the density of ater. Svedberg Constant = v a R = R ρ = = 70S = 70 9η s = 1nm 1

2 Problem Consider a globular (spherical) protein of diameter ~6 nm, amd molecular mass ~0 kda. (a) What is the Reynolds number for this protein as it moves through aqueous solution? In problem set 1, you estimated the thermal velocity of this protein to be 8.5 m/s. Therefore, Reynolds number = kg / m 6 m 8.5m / s ρ Lv / η = 0.05 Pa. s (b) What is the average magnitude of the instantaneous drag force on this protein? F drag = k v = 6πηav = 6π Pa. s m 8.5m / s 480 pn drag (c) The drag force is from collisions from ater molecules. If the protein ere struck head-on every second by a ater molecule that bounced straight back, the average force ould be the change in momentum of the ater molecule: < F > t = p. What are the number of collisions that must take place every second in order to obtain the drag force you estimated in part (b)? The average force exerted by each ater molecule if there is one collision per second is equal to ( p)/(1 second) = m v /(1 second), here v is the instantaneous thermal velocity of a ater molecule and m is its mass: m = 18/ N A kg 6 kg v k T B = = 6 m kg J = 640 m/s Average force = N =.8 11 pn Therefore, number of collisions per second 480 pn/(.8 11 pn/s) 1 1 collisions/second or 1 collisions/picosecond.

3 Problem The flo of heat satisfies Fick's la and the diffusion equation. The flux of heat is proportional to the thermal gradient, here the constant of proportionality is the thermal conductivity, K. The flo of heat then changes the temperature of a unit volume of solid by cρ, here c is the specifc heat and ρ is the density. For the case of heat flo in solids, the equivalent to the diffusion coefficient (ith units of m /s) is D thermal = K / cρ (a) For ater at 5 C, the thermal conductivity, K = W/m.K. What is D thermal for ater? K = W/m.K c = 1cal/gm/K = 4.18 J/kg/K ρ = 1gm/cc = kg/m Therefore, D thermal = K / cρ = / 4.18 / m / s = m / s = 0.15 mm / s (b) Ho long does it take heat to diffuse nm, roughly the distance from the center of a globular protein to the surrounding fluid? For diffusion in -dimensions, < r >= 6Dt Therefore, heat ill diffuse a distance of nm in ( m) 11 t =< r > / 6Dthermal = s ps m / s (c) Ho long does it take a ater molecule to diffuse that distance? Water molecule ill diffuse a distance of nm in ( m) t =< r > / 6D = 6.8 s 680 ps 6. m / s Problem 4 Consider a bacterium (radius ~1 µm) simming through ater at a speed of 5 µm/s. (a) Ho much force do the flagella of the bacteria have to generate to maintain that speed? For a bacterium simming through ater at a constant speed of 5 μm/s, acceleration = 0 and F app = F drag = v k drag = v 6πηR

4 = 5 µ m s (6π kg m s 6 m) = 0.47pN (b) The motor protein kinesin generates a force of 6 pn. Given that the viscosity of the cytoplasm is ~00 times that of ater (for large objects like organelles), ho fast could a single kinesin molecule move a bacterium though a cell? Express your anser in µm/s. Viscosity of the cytoplasm = 00 (viscosity of ater) = 1 Pa.s. Motor protein kinesin generates a froce of 6pN Therefore, the speed ith hich kinesin can move a bacterium through the cytoplasm Fdrag 6pN can be estimated as F app = F drag = k drag v v = = = 0. µ m/s k 6π (1Pa.s)(1µ m) drag Problem 5 A solution of gold spheres is stored on a shelf. After a eek or so, it is noticed that the spheres have settled stably in the container and that the density (as estimated from the color) decreases e-fold every 0 mm from the bottom to top. Given that the density of gold is 19. times that of ater, hat is the diameter of the gold spheres? The density (d) of gold spheres as a function of the height (z) in the container is governed by the Boltzmann relation d = d 0 exp( m' gz / k BT ) here m' = V ( ρ g ρ ) = 18. Vρ Here ρ g and ρ are the density of gold and ater, respectively, and V is the volume of each of the gold spheres. The characteristic height at hich the density of gold spheres decreases e-fold is k T / m' g = k T /18.V g = 0 mm B B ρ Solve for V k BT = m ρ g Therefore diameter of gold sphere = ( V / 4π ) 1/ 1 Å 4

5 Problem 6 A protein solution as placed in an analytic centrifuge and spun until equilibrium as reached. The measurement as then repeated. But it as realized, too late, that prior to the second run, the protein had been denatured. The average size of a denatured protein is larger than that of a native, compact protein. (a) Do you expect the distribution of molecules to be the same, at equilibrium, for the to different runs? Explain. The concentration of molecules as a function of height in a test-tube, after spinning in a centrifuge depends, is described by the Boltzmann distribution: C( z) = C0 exp( m' az / kbt ), here m' is the effective mass, and a the acceleration due to the spinning centrifuge. The distribution of molecules at equilibrium depends on the effective mass of the molecule. Since the native and the denatured protein have approximately the same effective mass, the equilibrium distribution is expected to be the same for the to runs. (b) The time that it takes to reach equilibrium depends on the sedimentation constant of the molecules. Which run ill reach equilibrium faster, the one ith the native protein, or the one ith the denatured protein? Explain. The time that it takes to reach equilibrium is inversely proportional to the drift velocity v d v d m' D given by = = S, here the diffusion constant D depends on the volume of the a kbt molecule, ith a more compact molecule having a larger diffusion constant. Therefore, for to molecules ith the same effective mass, the more compact molecule ill have a larger sedimentation constant, and hence a larger drift velocity, and ill reach equilibrium faster. In this case, the native protein is more compact and ill have a higher drift velocity. Problem 7 (for Phys 594 students) The figure shos some experimental data on Bronian motion taken by Jean Perrin. Perrin took colloidal particles of radius 0.7 µm. He atched their projections into the xy plane. He observed the location of a particle, aited 0 s, then observed again and plotted the net displacement in that time interval. He collected 508 data points in this ay, and calculated the root-mean-square displacement to be d = 7.84 µm. (a) Given the radius of the particles, predict the value of d, and compare ith the measured value. In -dimensions, < rr 4DDDD, here DD = kk BB TT 6ππππππ = m /s for 0.7 µm particle. Therefore rms displacement d = 4DDDD 8. µm for t = 0s, hich is ithin error of the measured value. 5

6 (b) The concentric circles dran on the figure have radii d/4, d/4, d/4, Ho many dots do you expect to find in each of the rings? Ho do your expectations compare ith the actual numbers? The probability that the displacement of the particle has a magnitude beteen rr and rr + dddd in time t, in -dimensions, is: 1 rr rrrrrr rr PP( rr, tt)dddd = exp (ππππππππ) = exp 4ππππππ 4DDDD DDDD 4DDDD Therefore, the number of particles expected ithin the first circle is: NN 0 dd/4 0 PP( rr, tt) dr 8, ith the total number of particles NN 0 = 508 Similarly, the number of particles ithin the next ring is: NN 0 dd/ dd/4 PP( rr, tt) dr 74 and the next ring is 99, and so on I counted ~0 dots in the innermost circle, and ~75 dots in the next ring, and gave up after that. 6