5. THE STATISTICAL MECHANICS OF DNA

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1 INTRODUCTORY BIOPHYSICS A. Y THE STATISTICAL MECHANICS OF DNA Edoardo Milotti Dipartimento di Fisica, Università di Trieste

2 From C. R. Calladine, H. R. Drew, B. F. Luisi, and A. A. Travers, Understanding DNA: The Molecule & How It Works, 3 rd ed. (Elsevier, 2004).

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4 B-DNA major groove minor groove Stacked base pairs, distance nm Helix pitch, 3.32 nm (10 base pairs) Diameter, 2 nm

5 The structure of DNA From R. P. Wagner, Understanding Inheritance, Los Alamos Science, n. 20 (1992)

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10 From R. P. Wagner, Understanding Inheritance, Los Alamos Science, n. 20 (1992)

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12 Nucleotides datp (deoxyadenosine triphosphate) dgtp (deoxyguanosine triphosphate) dctp (deoxycytidine triphosphate) dttp (deoxythymidine triphosphate)

13 Nucleosides and nucleotides nucleoside nucleotide

14 From C. R. Calladine, H. R. Drew, B. F. Luisi, and A. A. Travers, Understanding DNA: The Molecule & How It Works, 3 rd ed. (Elsevier, 2004).

15 The nucleosome

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18 Photographs of human chromosomes in duplicate, as isolated just before cell division (at metaphase) and then sorted by length into pairs. Each number identifies the two chromosomes of a homologous pair. X and Y are nonhomologous chromosomes that determine a person s sex as female (XX) or male (XY). The two duplicate copies of any individual chromosome form an X shape because they have not yet separated entirely. Scale: chromosomes 3 are approximately 10-5 m or 10 µm long. Taken from C. R. Calladine, H. R. Drew, B. F. Luisi, and A. A. Travers, Understanding DNA: The Molecule & How It Works, 3 rd ed. (Elsevier, 2004).

19 number of base pairs in human DNA total length: nm x m diploid cells, therefore length is doubled to 2 m 46 chromosomes, therefore length of unwound DNA is about 4.3 cm/chromosome

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21 The cell nucleus euchromatin is loosely bound and actively expressed heterochromatin is tightly bound and mostly not expressed the nucleolus is a site of protein synthesis inside the nucleus

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23 From F. Gentile et al., Direct Imaging of DNA Fibers: The Visage of Double Helix, Nano Lett. 12 (2012) 6453

24 Schematics and simulation of electron transmission/diffraction through fibers A-DNAs superstructures used for image simulations. (a) The smallest bundle consisting of A-DNAs (a central one and a shell of 6), the bundle was turned slightly out of the low-index zone axis to reduce the contrast in the image due to the coherence created by the superstructure periodicity. The incident parallel electron beam is sketched by the yellow arrows. (b) The bundle viewed along the y direction and (c) the corresponding simulated TEM image at about 1 µm defocus with aligned chains. From F. Gentile et al., Direct Imaging of DNA Fibers: The Visage of Double Helix, Nano Lett. 12 (2012) 6453

25 Molecular dynamics simulations of DNA double helices aggregation. (a) Conformations of the paired DNA molecules at the beginning of the simulation and after 5 ns at different hydration conditions; the numbers (Å) refer to the thickness of the hydration layer. (b) Time-evolution of the distance between the centers of mass of the dodecamers along the simulations at different degrees of hydration. (c) Interaction energy (potential of mean force) between the DNA molecules in the 0.3 nm solvation layer as a function of their relative vertical displacement. High and low energy values are highlighted with red circles, and the corresponding conformations are shown. From F. Gentile et al., Direct Imaging of DNA Fibers: The Visage of Double Helix, Nano Lett. 12 (2012) 6453

26 TEM image with intensity profile and corresponding FFT pitch calculation of λ-dna fibers. (a) DNA fiber TEM image. The inset shows higher magnification DNA fiber details; the red arrows point out the 2.7 nm pitch of A double helix. The scale bar corresponds to a length of 20 nm. In panel b, a white rectangle is superimposed, showing where the intensity profile was measured. The peaks in plot c correspond to the alternation of bright and dark bands in the original image (b): plot c displays a two-dimensional graph where the Y-axis reports the pixel intensity integrated along the height of the rectangle and the X-axis represents the distance measured on the rectangle. Plot d shows the FFT of the signal displayed in plot c: a well-defined maximum is observed at 0.37 ± /nm, corresponding to a frequency of 2.7 ± 0.2 nm. From F.EDOARDO GentileMILOTTI et al.,- INTRODUCTORY Direct Imaging of DNA Fibers: The Visage of Double Helix, Nano Lett. 12 (2012) 6453 BIOPHYSICS - A.Y

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28 DNA is not a static structure, it has a complex and varied dynamics. This dynamics can only be described in statistical terms. Here we recall first some basic concepts of statistical mechanics, then we move to more specific issues of DNA thermodynamics.

29 An extremely short review of statistical mechanics 1. Boltzmann factor Thermal reservoir at temperature T Heat exchange Physical system, total energy E Probability of finding system with energy E is proportional to exp E k B T

30 Multilevel statistical system Consider a macrostate defined by N 1 particles at energy level E 1 with degeneracy g 1 N 2 particles at energy level E 2 with degeneracy g 2... N i particles at energy level E i with degeneracy g i... then the number of ways in which we can arrange the identical particles in the M levels is N! N 1! N M! and when we also include degeneracy, we find that the number of different ways to obtain the macrostate (thermodynamic probability) is Ω = N! g N 1 N 1 g 2 2 N 1! N 2! g N i i N i!

31 We use Stirling s approximation and we find lnn! nlnn n ( ) + N i ln g i N i ln N i + N i lnω = N ln N N i N i i ( ) = N ln N + N i ln g i N = N i N ln g i N N = N i Now the problem is finding the distribution {N i } that maximizes the thermodynamic probability (this is the distribution that is observed with the highest probability) i i N i

32 Maximization must be carried out constraining both the number of particles and the total energy lnω = N N = U = i i N i i E i N i N i N ln N i g i N expression of thermodynamic probability total number of particles is fixed total energy is fixed For constrained maximization we use the method of Lagrange multipliers and maximize the auxiliary function lnω + λn βu = N i ln g i N + λ N β E i i N i i N i i i

33 lnω + λn βu = N ln N + N i ln g i + λ N i i N i i β E i N i i N l ( lnω + λn βu ) = ln g l 1+ λ βe l = 0 N l N l = g l e λ 1 βe l

34 N i = g i e λ 1 βe i N = g l e λ 1 βe l = e λ 1 g l e βe l e λ 1 = i i i N g l e βe l = N Z N i = N Z g l e βe l Z = i g l e βe l Partition function (Zustandsumme) U = i E i N i = N Z i E i g l e βe l = N Z β i g l e βe l = N Z Z ln Z = N β β the partition function is used to determine many other thermodynamical functions

35 The entropy is simply related to the thermodynamic probability N i S = k B lnω = k B N N ln N = = k B N ln Z + k B βu i N i g i = k B N ln Z + U T 1 T = S U

36 The forces that keep the DNA together covalent bonds along the sugar-phosphate backbone hydrogen bonds between complementary bases stacking interactions between adjacent base pairs (mostly Van der Waals interactions, about 0.5 kcal/ mole, approximately 10 times weaker than hydrogen bonds in water)

37 In general, the strength of the hydrogen bonds (about 5 kcal/mole) is intermediate between Van der Waals interactions (about 0.3 kcal/mole), and covalent chemical bonds (about 100 kcal/mole). Note that 5 kcal/mole 0.2 ev/molecule This must be compared with the thermal energy at room temperature 3/2 kt 0.04 ev/molecule Therefore thermal agitation can destroy the hydrogen bonds that pair bases in the DNA molecule.

38 From M. Peyrard, Nonlinear dynamics and statistical physics of DNA, Nonlinearity 17 (2004) R1

39 Rotational degrees of freedom of DNA bases, from

40 From C. R. Calladine, H. R. Drew, B. F. Luisi, and A. A. Travers, Understanding DNA: The Molecule & How It Works, 3 rd ed. (Elsevier, 2004).

41 Molecular zipper model of DNA (Kittel, 1969) last link stays fixed N links There are N states, characterized by p = 0,..., p = N-1 open links

42 Energy to open one link: Energy to open p links: Degeneracy (rotational): Degeneracy of p open links: ε pε g g p Partition function Z = N 1 p=0 g p exp pε k B T ( )

43 ... it is easy to find a closed expression for the partition function Z = N 1 g p exp pε N 1 k B T = x p p=0 p=0 = 1 x N 1 x x = gexp ε k B T

44 then, the mean number of open links is s = 1 Z N 1 p=0 pg p exp pε k B T = 1 Z N 1 p=0 px p x = gexp ε k B T = 1 Z x d dx N 1 x p p=0 = x d ln Z dx = x d dx ( ) ln( N 1 x ) ln 1 x = x Nx N 1 1 x N x = Nx N x N 1 + x 1 x = NgN e Nε k BT g N e Nε k BT 1 + ge ε kbt 1 ge ε k BT

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46 The first-order Taylor expansion is s = Nx N x N 1 + (prove it!), and it shows that x 1 x 1 ( 2 N 1 ) + 1 ( 12 N 2 1) ( x 1) s N = 1 ( 2N N 1 ) + 1 ( 12N N 2 1) ( x 1) N 12 x 1 N 1 ( ) and the transition at x = 1 becomes infinitely sharp as N grows indefinitely

47 The transition point corresponds to x = gexp ε k B T = 1 g = exp ε k B T > 1 i.e., there can be a transition from the bound to the open state at finite temperature only if the degeneracy g is greater than 1, i.e., if each link has a true rotational freedom.

48 Notice also that for a given g, the transition temperature is determined by x = gexp ε k B T = 1 ln g = ε k B T T = ε k B ln g

49 The probability that all links are open (except, obviously, the last, fixed one) is P = 1 ( Z gn 1 exp N 1 )ε k B T = x N 1 Z = x N 1 1 x 1 x N 1 N N Δx 1 N Δx 2 close to the the transition N >> 1 and at the transition point the probability P = 1/N is very small when N >> 1.

50 The mean energy is U = N Z N 1 p=0 pεg p exp pε k B T = Nε Z N 1 p=0 px p = Nε s = Nε Nx N x N 1 + x 1 x = Nε Ng N e Nε kbt g N e Nε k BT 1 + ge ε kbt 1 ge ε k BT and therefore the corresponding heat capacity is C = du dt = dx dt du dx = k B x ε ln g ln x ( ) 2 Nε d dx Nx N x N 1 + x 1 x

51 The helix-to-coil transition When DNA is heated, hydrogen bonds break, and DNA melts, (thermal DNA denaturation) turning the linear chains into tangles (coils). This is different from the unzipping described earlier. T

52 Optical measurements can be used to detect the melting transition G. L. Baker & M. E. Alden, Optical Rotation and the DNA Helix-To- Coil Transition, J. Chem. Educ. 51 (1974) 591

53 The measurements can be translated into a fraction of intact bonds G. L. Baker, DNA helix-to-coil transition: A simplified model, Am. J. Phys. 44 (1976) 599

54 We start from the fraction of intact DNA base pairs f = 1 ( 2 1+ σ i ) = 1 ( 2 1+ σ ) i where σ i is a spin-like variable, such that σ i = +1 pair intact 1 pair broken and where J is the interaction strength between the intact and the broken links.

55 The partition function for a single base pair is Z i = exp 1 2 σ J i σ i = 1 + exp 1 2 σ J i σ i =+1 = 2cosh 1 2 J and the partition function for the system is (neglecting correlations) Z = i Z i The average value for the pseudo-spin variable is σ i = 1 σ i exp 1 Z i 2 σ ij σ i = 1 + σ i exp 1 2 σ ij σ i =+1 = tahn 1 2 J

56 and then the fraction of intact DNA base pairs is ( ) = tahn 1 2 J f = σ i = exp 1 2 J sech 1 2 J We still do not have a temperature dependence, but here we note that critical temperature J = J ( T ); f ( T C ) = 1 2 ( ) = J T a T T C ( ) + ( ) 2 + b T T C J ( T C ) = 0 expansion about the critical temperature

57 Then, keeping only the first order terms ( ) f = exp a T T C Close to the transition we can also write ( ) sech a T T C Gibbs free energy J ( T ) ΔG RT = ΔH T ΔS RT J ( T C ) = 0 ΔH T C ΔS J ( T ) ΔH T ΔH 2 RT C RT C = ΔH RT C 2 T T C ( ); a = ΔH RT C 2

58 ΔH 7.9 kcal mole (This is in line with various types of calorimetric measurements) a 0.03 K 1

59 Helicase chemical dynamics n i = number of systems with i intact bonds dn i dt = ( α + β )n i + αn i 1 + βn i+1

60 Equilibrium solution (at temperature T) dn i dt = ( α + β )n i + αn i 1 + βn i+1 = 0 α + β ( ) + α n i 1 n i + β n i+1 n i = 0

61 Since at equilibrium (see the previous discussion) n l g N l exp ( N l)ε k B T n l 1 n l = g N l+1 exp ( N l +1)ε k B T g N l exp ( N l)ε k B T = gexp ε k B T = exp ε k T ln g B k B T

62 ε k B T ln g = ε T ΔS = ΔG n l 1 n l = exp ΔG k B T ( α + β ) + α n i 1 + β n i+1 = 0 ( α + β ) + αe n i n i ΔG ΔG k T B k + βe T B = 0

63 Therefore ( ) + αe α + β ΔG ΔG k T B k + βe T B = 0 α e ΔG ΔG k T B 1 = β 1 e k B T ΔG = βe k B T k T B 1 e ΔG α β = e ΔG k B T

64 α β = e ΔG k B T means that unzipping is faster than zipping if β = αe ΔG k T B > α i.e., only if ΔG < 0 ε k B T ln g < 0 T > T M = ε k B ln g melting temperature, in agreement with the previous analysis based on statistical mechanics

65 Helicase, an unzipping enzyme

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67 from D. Goodsell, The Molecular Perspective: DNA Polymerase, The Oncologist 9 (2004) 108 The DNA Polymerase complex

68 Kary B. Mullis Born: 28 December 1944, Lenoir, NC, USA Nobel Prize in Chemistry in 1993 "for his invention of the polymerase chain reaction (PCR) method

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71 References C. R. Calladine, H. R. Drew, B. F. Luisi, and A. A. Travers, Understanding DNA: The Molecule & How It Works, 3 rd ed. (Elsevier, 2004). M. Peyrard, Nonlinear dynamics and statistical physics of DNA, Nonlinearity 17 (2004) R1 C. Kittel, Phase Transition of a Molecular Zipper, Am. J. Phys. 37 (1969) 917 G. L. Baker, DNA helix-to-coil transition: A simplified model, Am. J. Phys. 44 (1976) 599 S. G. J. Mochrie, The Boltzmann factor, DNA melting, and Brownian ratchets: Topics in an introductory physics sequence for biology and premedical students, Am. J. Phys. 79 (2011) 1121 R. P. Wagner, Understanding Inheritance, Los Alamos Science, n. 20 (1992)