University of Washington Department of Chemistry Chemistry 355 Spring Quarter 2005

Transcription

1 University of Washington Department of Chemistry Chemistry 55 Spring Quarter 5 Homework Assignment 7: Due no later than 5: pm on 5/14/4 8., 8.6, ) According to text equation 16, after N jumps the probability of a displacement m N! places from the origin is Pm b g. Then the relative probability bn + mg /! bn mg /! of observing after 8 total jumps four displacements from the origin versus two displacements is Pbg 4 8! b8+ g/! b8 g/! P 8+ 4 /! 8 4 /! 8! b g b g b g b g b g b g b g 8+ /! 8 /! 5!! 8+ 4 /! 8 4 /! 6!! 1 8.6) Consider the fact that for spherical molecules the frictional coefficient f is proportional to the radius R (Stokes Law). Consider also that the volume is proportional to the mass and the volume of a sphere is proportional to the radius cubed. Then f is proportional to the cube root of volume and thus is also proportional to the cube root of mass. Take that proportionality (i.e. f proportional to the cube root of mass) and determine how s and D depend on mass when the molecule is spherical. Solution: Suppose the molecule is spherical. The easiest case to study is if the molecule is unhydrated in solution i.e. has no waters of hydration attached. Then the volume of one F 4 mole of spherical molecules is V M V N H G I R K J π where N is Avagadro s number. F MV Then R H G 1 / I N K J. k T RT From the statement D we get 6πηR 6πηNR 1 / RT RT F N 1 D D 1 6 NR 6 N H G I / πη πη MV KJ M 1 / mc1 Vρh M c1 V ρh F N From the statement s s M R N H G I / 6πη 6πη MV KJ 8.1) Calculate the molecular radii for myoglobin and hemoglobin. The ratio of the radii equals the cube root of the ratio of the volumes.

2 k T k T D R 6πηR k T c18. 1 J / Khb9Kg Myoglobin: R 11 6π. 1kg / m s m / s b gc h c18. 1 J / Khb9Kg 11 πb. kg / m sgc. m / sh kt Hemoglobin: R rhb rmb 186. If Hb were a tetramer, its volume would be four times that of Mb: r Hb F VHb VMb H G I K J F 1 / H G 1 / 4 I 1 / K J rmb VMb theoretical ratio V Mb b g 186. nm 5. nm The experimental ratio is close to the ) A base pair DNA may be modeled as a rigid rod that is 6.8x1-8 m in length and has a radius of 1-9 m. a) Calculate the axial ratio and the volume of a base pair DNA molecule. Solution: L L 68 P 4. d R V R L m 68 1 m m π b gc h c h b) Calculate the radius of a sphere which has an equal volume to the rod-shaped DNA in part a. Solution: 1 / 4 F 14 1 m I 9 V πr 14 1 m R 7. 1 m HG KJ c) Assuming the base pair DNA may be modeled as a rigid rod, calculate the frictional coefficient f. Assume the solvent viscosity is.1 gm cm -1 s / / 1 / / b/ g P b/ g P f f 6πηR ln bpg. ln bpg. 1 / / / 4 b6 gb. 1gm / cm sgc. 7 1 cmh b g b g π ln 68. F I HG K J b g g / s g/ s 9.

3 d) Calculate the diffusion coefficient of a base pair DNA. Assume T98K. D k T 16 c18. 1 ergs / Khb9Kg cm / s f g / s ) A base pair fragment of DNA is 68 Angstroms long and Angstroms in diameter. It has a molecular weight of 15, grams/mole. a) Calculate the frictional coefficient of this DNA fragment, assuming it can be treated as a rod of length L68 Angstoms and diameter d Angstroms. Solution: From Lecture 6 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 1 / / b/ g P the frictional coefficient is defined as f f. ln( P). 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 Pa/b. F R is ab ab determined as follows: Vsphere Vrod R ab R R H G 1 / I π K J So Pa/b68/4 Note 1 Angstrom is 1-1 m. Then a68 Ang. And b Ang. Then 1/ / 1/ / ( /) P ( /) P f f ( 6πηR) ln( P). ln( P). 1/ 1 1 6π (.1Pa s) ( 4 1 m)( 1 1 m) 1/ / ( /) ( 4) 1/ (.188Pa s)( 51 1 m ) (.8) ln(68) kg / s g / s b) Calculate the sedimentation coefficient of the DNA assuming the gram specific volume of DNA is.51 ml/gram. Solution: m V M V s f N f ( 1 ρ) ( 1 ρ) ( ) ( 6. 1 / mole)( g / s) ( 15, gm / mole) 1 (.51 ml / gm)( 1 gm / ml) sec 6.66 A S

4 c) For the purpose of calculating the frictional coefficient f, a rod-like, linear polymer can be more accurately treated as a string of N touching beads, each spherical bead has a diameter δ. The rod is therefore of length Nδ. πηn δ For such a polymer the frictional coefficient is f where η is the ln N viscosity of the solvent, assumed here to be water. Calculate the bead diameter and number of beads required to have the same total volume and length as the DNA described in part a. Solution: The length of the string of beads is 68 AngstromsNδ. The volume of the DNA cylinder is 1 9 V aπ b m 1 m.14 1 ( )( )( )( ) m 5 4 Te volume of the string of beads is the volume of the bead δ πδ π times 6 Nπδ the number of beads N or. So we have two unknowns N and d but two 6 equations 1 Nδ 68 1 m Nπδ m 6 1 Nπδ ( Nδ) πδ ( 68 1 m) πδ m ( 6 )(.14 1 m ) 18 9 δ m δ.45 1 m m π ( ) m N m Comment: The rise between adjacent base pairs is about 4 Angstroms in form DNA, so these beads are of dimension much larger than a single base pair.it means that for the purposes of calculating friction coefficients, the minimal unit in DNA is about 5-6 base pairs at least according to this model. d) Calculate the frictional coefficient and the sedimentation coefficient of the DNA described in part c.

5 1 ( )(.14)(.1Pa s)( 8)( m) πηnδ f ln N ln ( 8) kg / s. e) kg / s / ) Chromatin is the complex of DNA and proteins (mostly histones) found in all eukaryotic cells. The fundamental repeating unit of chromatin is the nucleosome particle. The properties of the DNA and the protein in nucleosome particles were determined using a combination of velocity sedimentation, dynamic light scattering, and gel electrophoresis. Dynamic light scattering studies determined the diffusion coefficient of the nucleosome particle in solution at 9K to be 4.7x1-7 cm /s. In the same solution velocity sedimentation performed at 18,1 revolutions per minute, obtained the following data: g s Time (minutes) oundary Position r (cm) Gel electrophoresis showed that the DNA molecule associated with a single nucleosome protein complex is base pairs in length. a) What is the molecular weight of the nucleosome particle? Assume the solution has a density of 1.g/cm and the specific volume of the nucleosome particle is.66cm /g. mc1 V D M V D s ρ h c1 ρh kt N A kt Obtain s from a plot of lnr versus time

6 Nucleosome Sedimentation Data Time (sec) Ln(R) Series1 The slope is 6 1 c57. 1 sec hb6sec/ ming sω sec s sec 158. S 1 π 18, 1 min Then M sk T N A c1 VρhD c6. 1 mole hc18. 1 ergs / Khc sechb9kg c1 b. 66gb1. ghc cm / sh c h g / mole b) Assuming the nucleosome particle is spherical, calculate its Stokes radius. Assume the viscosity of the solution is.1 gm cm -1 s -1. c hb g 16 k T k T ergs / K 9K D R cm 6πηR 6πb1. g / cm sgc47. 1 cm / sh c) Assuming each base pair in DNA is separated from the adjacent base pairs by about.4x1-1 m, approximately how long is a piece of DNA base pairs in length? ased on this result, and the result from part b, comment on how tightly packed the DNA is in the nucleosome. b gc h 8 Length of DNA base pairs. 4 1 cm / base pair 68 1 cm ecause the length of the DNA is much greater than the diameter of the nucleosome particle we assume the DNA must be tightly folded in the nucleosome. d) Other evidence suggests that the protein component of the nucleosome is composed of a complex of eight protein molecules. Assuming that a single DNA base pair weighs 66g/mole, estimate the total weight of the protein in the nucleosome and the weight of each component protein.

7 Weight Protein Complex Weight Nucleosome Weight DNA7,g/mole- (66g/mole)()18,g/mole Weight Protein Monomer18,/817,5g/mole e) How are the proteins and DNA packed in the nucleosome? To answer this question, assume the eight proteins form a unhydrated, spherical complex with specific volume.74 cm /g. Calculate the radius of this hypothetical protein sphere. Assume the base pair DNA behaves as a random coil polymer. Calculate the root mean square (rms) end-to-end distance. Comparing the protein sphere radius with the rms end-toend distance for the DNA, is most of the DNA packed outside or inside the protein core of the nucleosome? Explain. R protein F MV g moleof complex cc g H G 1 / I F I N A K J H G b, / gb. / g 1 c6. 1 mole h K 4 1 / c. hbg R N cm cm DNA J 5. 1 Note because RDNA >> Rprotein, the DNA occupies a larger volume than the protein and is thus largely located outside the protein core particle. cm