Lecture /4/ University of Washington Department of Chemistry Chemistry 4 Winter Quarter A. Polymer Properties of DNA When a linear polymer like DNA becomes long enough it can no longer be treated as a prolate ellipsoid. DNA must be viewed as a random-coil polymer. For the purposes of calculating the the frictional coefficient f, a very long random-coil polymer may be treated as a spherical molecule, with radius R H, called the hydrodynamic radius. The hydrodynamic radius is related to the radius of gyration R G by the equation: RH =.66RG (.). The radius of gyration for a very long DNA can be calculated in turn: 8 PN (.4 cm) RG = (.) 6 where N is the number of base pairs and P is a property of a polymer called the persistence length. The persistence length measures the length over which a polymer points in the same direction as its first bond. The persistence length for DNA is P=x -8 8 cm, equivalent to about 4 base pairs. The number.4 cm is the approximate distance between successive base pairs in DNA. Example : Calculate the radius of gyration R G for a DNA which is base pairs in length. Also, calculate the hydrodynamic radius R H. c h b gc hc h. 8 8 8 PN 4. cm cm. 4 cm : RG = = 6 6 6 6 Then R = 7. cm. R = 66. R = cm. G H G = 7. Example : Treating the base pair DNA as a sphere with radius R H, calculate the frictional coefficient f. Assume the viscosity of the solvent η is. poise. 6 7 Solution: f = 6πη R = 8. 84. g/ cm s cm= 9. 4 g/ s H Example : Using the frictional coefficient calculated above, determine the diffusion coefficient for base pair long DNA at T=9K. kt (.8 J / K)(.9 K) B Solution: D = = = 4.9 m / s f 9.4 kg/ s Example 4: What is the root-mean-squared displacement of a base pair DNA at T=9K after second? Assume properties for DNA from examples -. cm
Solution: The root mean squared displacement is defined for a three dimensions as R = R = x + y + z = Dt + Dt + Dt = 6Dt rms 6 Rrms = 6 4.9 m / s s =.7 m Note that unlike Newtonian mechanics where the displacement of a particle R moving at constant velocity v is proportional to time, i.e. R = v t, the rms displacement of a particle undergoing diffusive motion goes like the square root of time. B. Integrals of Gaussian functions Here we give general guidelines and rules for handling integrals that involve functions of n x r the form xe where r is a constant and n =,,, The function f ( x) = e xr is called a bell-shaped curve or a Gaussian function. A common form for integrals of importance in this class are: x e dx where n=,,, We also use integrals of the type The integral x e dx. x e dx has standard form, which means you can find it in a table of standard integrals. The answer you get depends upon whether n is an even integer, and odd integer or zero. Case : n=. The integral is e xr dx = Case : n is even. We can guarantee that n is even by making the substitution k x r ( k ) n=k, where k=,, Then we get x e dx = π. The k+ k+ r commonest cases we have used in Chemistry 4 are xr π k= or xe dx=. r 4 4 xr k= or xe dx= π 8r Case : n is odd. We can guarantee that n is even by making the substitution k+ x r k! n=k+, where k=,,, Then we get x e dx =. The k + r commonest cases we have used in Chemistry 4 are xr k= or xe dx =. r k= or xr xe dx= r π r
C. Even and Odd Functions In Chemistry 4 we encounter integrals of the form x e dx where n=,, Note the integral differs from the standard forms discussed above in that the limits are from to instead of from to. In such cases the standard forms may or may not be applicable depending on certain properties of the integrand. The property in question is whether the function in the integrand is even or odd. o An even function has the property f ( x) f ( x) =. If you change the sign of the variable the value of the function does not change. Consider the function f ( x) = e xr ( x) r xr. We find f ( x) = e = e = f ( x). Therefore f(x) is even. Similarly m mu /kbt u = uf u du = ue du = π kt B xr x r f x = x e = x e = f x, so all functions of the form k x r x e xr xe o An odd function has the property x k+ x r e k=,,, are odd. is also even. In fact k=,,, are even. f x = f x. Functions of the form D. Integrals of Even and Odd Functions An important type of integral that we encounter in Chemistry 4 is an integral + a over a symmetric interval f ( x) dx. These integrals have the property: + a + a + a + a = + = + ( ) f x dx f x dx f x dx f x dx f x dx. f(x) is even i.e. f ( x) = f ( x). Then + a + a + a + a + a = + = + ( ) = f x dx f x dx f x dx f x dx f x dx f x dx. f(x) is odd i.e. f(x)=-f(-x). Then + a + a + a + a + a + a f x dx = f x dx + f x dx = f x dx + f x dx = f x dx f x dx = The integral of an odd function over a symmetric interval is zero. An application of these principles is the evaluation of integrals of Gaussian functions.
k. For even functions of the form x e k=,,, x r k x r k x r x e dx = x e dx. For odd functions of the form x k+ x r e k=,,, k+ x r x e dx = E. Physical Examples: Kinetic Theory of Gases:. For a hypothetical one dimensional gas the velocity distribution function is m mu /kbt f u = e. The average velocity in one dimension is π kt B m mu /kbt u = uf u du = ue du π kt =, a result that is due to the B mu fact that an odd function /k B T ue is being integrated over a symmetric interval. The diffusion equation dc k T d C D d C B = = has the solution dt f dx dx D=.*^(-)m^/s 6 4 Series t= C t=. t= - 7 9 d= to.x^-m C bx, tg= CD t e x / 4 D t, where D is the diffusion coefficient. 4π The plot above is a graph of C (x,t) for a diffusion coefficient of D=.x - m /s. Note C =. The function C(x,t) can be used to calculate the average displacement of a solute particle x rms
z z z x xc x t dx xcd t e dx xc Dt e x / 4 Dt x / 4 Dt dx rms = (, ) = = = 4π 4π Average displacement.. x*c(x,t). -. 7 9 Series -. -. d= -.*^-m This result can be understood if we plot the function xc(x,t) versus x The area under this curve(i.e. the integral of the function xc(x,t) is zero which indicates that the average displacement is zero. That is z xrmx = x C( x, t) dx =. This is reasonable because for a bell-shaped distribution for every molecule that travels a distance +x, another molecule travels x. The average is thus zero. z However, the mean squared displacement is given by x = x C( x, t) = Dt This quantity is clearly non-zero, a fact which is shown by the graph below
Root Mean Square Displacement x^*c(x,t).6.4...8.6.4. Series 7 9 distance -.x^(-)m Where the area under the curve is the root mean square displacement z / x x x C( x, t) Dt rms = = L N M O Q P =