PES 3950/PHYS 6950: Homework Assignment 6

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1 PES 3950/PHYS 6950: Homewok Assignment 6 Handed out: Monday Apil 7 Due in: Wednesday May 6, at the stat of class at 3:05 pm shap Show all woking and easoning to eceive full points. Question 1 [5 points] A system contains a set of stationay eceptos with a concentation of feely diffusing ligands. Hee [R]... concentation of eceptos [L]... concentation of ligand and suppose you have the eaction [LR] <==> [L] + [R] a) Wite down the expession fo K D b) Find the atio of bound eceptos to total eceptos in tems of K D and [L]. c) K D is 1 nm. What pecentage of eceptos ae occupied when thee is a nm concentation of unbound ligand? Question [10 points] Hee, we con- Assume that a potein can be thought of as a chaged sphee in wate. side the effect of salt on its electical enegy. a) Compute the enegy of a chaged spheical potein of adius R, in wate, and in the pesence of monovalent salt at concentation c 0. Assume that the chaged esidues ae unifomly distibuted on the suface of the potein, i.e. Q e = 4πR π, whee = 5 angstoms is the typical adius o an amino acid. b) Plot the electical enegy of the potein as a function of its adius R fo the following fou diffeent salt concentations c 0 : 1 mm, 4 mm, 5 mm, and 100 mm. What conclusion do you daw about the effect of salt on the chaged state of a potein? Let z = 1.

2 Question 3 [15 points] In this poblem we study what would happen if we pulled on a one-dimensional co-polyme by attaching a mass m to the polyme. (Causing a foce mg.) So, ou basic system is a combination of two segments of diffeent lengths. As usual in one-dimension an individual segment can be oiented eithe up o down. So one possible state fo the system is whee both segments ae oiented down. This state is illustated below. The potential enegy of this state is mg(a 1 + a ). a) Illustate the 3 othe possible state and find the enegy of each state. b) Find the patition function. c) Find an expession fo the aveage length of the polyme at a tempeatue T fo a given foce of mg. (Assume thee ae N double segments.) Question 4 [0 points] GRADUATE STUDENTS ONLY As we have noted, some viuses ae cylindical in shape. Also, shot sections of DNA could be chaged and cylindical. In this poblem you will find the potential V () of a chaged cylinde in an ionic solution. a) Wite the thee sets of equations that goven the system - Poisson s equation, equations fo the concentation based on the Boltzmann facto, and an equation elating the chage density to the concentations. Let z = 1 to educe the witing. b) Obtain the new Poisson s Equation by combining all the equations. Assume V () is small and show that you get ( ) Co e V () = V () ε o ε kt c) Expand in the appopiate coodinate system and obtain the diffeential equation which govens V ().

3 d) The solution to this is difficult to find diectly. Fist look at an appoximation fo the egion whee is lage. Find the solution in this case - you only need to get the functional fom of the solution; you do not need to apply bounday conditions at the suface of the cylinde, but use the one at = infinity. e) Find the complete solution using Mathematica o anothe softwae. You can ask Mathematica to solve the diffeential equation d V d + 1 dv d = AV () by using the command DSolve[V []+(1/)*V []-A*V[]==0,V,]. Then plot both ou appoximation of exponential decay, found ealie, and the coect answe and comment on you solution - how does it elate to pevious esults fo spheical viuses o to esults elating to sceening of a chaged plate?

4 Homewok 6 Question 1 Solution A system contains a set of stationay eceptos with a concentation of feely diffusing ligands. Hee [R]... concentation of eceptos [L]... concentation of ligand and suppose you have the eaction [LR] <==> [L] + [R] a) Wite down the expession fo K d : answe: K d [ L][ R] [ LR] b) Find the atio of bound eceptos to total eceptos. answe: [ LR] [ L][ LR] [ L] bound LR [ R] [ R][ L] Kd total [ R] [ LR] [ LR] [ L][ LR] [ L] [ R] [ R][ L] K d c) K d is 1 nm. What pecentage of eceptos ae occupied when thee is a nm concentation of unbound ligand? [ L] Kd [ L] answe: atio = / 3 [ L] 1 Kd [ L] 1 Kd The atio is /3 and bound pecentage is 66 pecent.

5 Answe Question pat a) Simply use answe fom class (see lectue 8) U(R) = with 4eR Q 4 16e R Q 4 1 QV(R) Q 8 (R ) o R to obtain e R 1 e R V ( R) o ( R ) R o ( R ) with o (ze) kt C o pat b) 8 x c 0 = 1mM c 0 = 4mM c 0 = 5mM c 0 = 100mM 5 U (J) R (m) x 10-9

6 We plot the enegy U fo salt concentations c0 = 1; 4; 5; and 100mM, fo which the Debye length is 97; 49; 19; and 9.7 Angstom, espectively. We see that the enegy deceases as the salt concentation is inceased. This is to be expected since moe salt sceens the electical inteactions moe effectively. Matlab code: close all clea all clc %define all constants e0 = 8.85e-1; e = 80; =5e-10; z=1; k=1.38e-3; T=93; %oom temp NA=6.0*10^3; e=1.6e-19; c = [1,4,5,100]; %efeence concentation c0 R = linspace(0,3e-9,100); Q = 4*e*R.^/^; lambda = sqt(e0*e*k*t./(*(z*e)^*c*na)); %sceening lenght in m %%% fo c = 1; 4; 5; and 100mM, % the Debye length is 97; 49; 19; and 9:7_Angstom, espectively. U1 = lambda(1)./(8*pi*e0*e*(r+lambda(1))).*(q.^./r); U = lambda()./(8*pi*e0*e*(r+lambda())).*(q.^./r); U3 = lambda(3)./(8*pi*e0*e*(r+lambda(3))).*(q.^./r); U4 = lambda(4)./(8*pi*e0*e*(r+lambda(4))).*(q.^./r); figue(1) plot(r,u1,'k','linewidth',) hold on plot(r,u,'-','linewidth',) plot(r,u3,'bo','linewidth',) plot(r,u4,'g^','linewidth',) hold off xlabel('r (m)','fontsize',16) ylabel('u (J)','FontSize',16) legend('c_0 = 1mM', 'c_0 = 4mM','c_0 = 5mM','c_0 = 100mM')

7 Question 3 a) Illustate the 3 othe possible state and find the enegy of each state Hee ae the 4 states with the enegy of each state. a 1 a 1 - a a 1 + a m a E 1 = mg(a 1 + a ). E = mg(a 1 - a ). a 1 + a a 1 - a E 3 = +mg(a 1 - a ). E 4 = +mg(a 1 + a ). Basically the fou states coespond to 1) down/down ) down/up 3) up/down 4) up/up b) Find the patition function Since we have all the enegies, this is easy Z e E1 E E3 E4 e e e

8 c) Find an expession fo the aveage length of the polyme at a tempeatue T fo a given foce of mg. (Assume thee ae N double segments) This is easy. The aveage length of a double segment is given by <double segment length> = ( length in state s)(pobabilit y to be in state s) state s The aveage length of the whole chain will be the value above multiplied by N, the numbe of double segments. Fom the pictues above we see Length in state 1 L 1 = a 1 + a Length in state L = a 1 - a (we ae measuing positive length downwad in the diection of the foce) Length in state 3 L 3 = -(a 1 - a ) Length in state 4 L 4 = -(a 1 + a ) Using the Boltzmann pobability facto we get L E e s N (Ls ) Z state s L mg(a a ) mg(a a ) (a 1 a )e 1 (a1 a )e 1 N Z mg(a a ) mg(a a ) (a1 a 1 )e (a1 a 1 )e While this is ugly, it does solve the poblem.

9 Question 4 Solution As we have noted, some viuses ae cylindical in shape. In this poblem you will find the potential V() of a chaged cylinde in an ionic solution. a) Wite the thee sets of equations that goven the system - Poisson's Equation, Equations fo the concentation based on the Boltzmann facto, and an equation elating the chage density to the concentations. 1) ) () V() (1) o C C ev / kt () C o e () ev / kt () C o e 3) ( ) ec () ec () (3) b) Obtain the new Poisson's Equation by combining all the equations. Assume V() is small and show that you get C e V () o V () o kt (4) We substitute Eq () into Eq (3), and then put the esult of that in Eq (1). This gives C e o ev / kt ev / kt V() e e (5) o If V() is small compaed to kt we can expand the exponentials e ev / kt 1 ev / kt (6) Using this last esult in the pevious equation we obtain C e 1 V () o V () V () kt (7) o c) Expand in the appopiate coodinate system and obtain the diffeential equation which govens V() In cylindical coodinates the Laplacian is given by

10 Since nothing depends on o z, the deivatives on these vaiables ae zeo. We get O 1 V 1 V() (9) d V 1 dv 1 V() (10) d d d) The solution to this is difficult to find diectly. Fist look at an appoximation fo the egion whee is lage. Find the solution in this case If is lage, the equation educes to This has d V 1 V() d (11) / V () Ae (1) (8) as a solution. (We have eliminated the solution finite at = infinity.) / V () Ae so that the potential is

11 e) Comment on you solution - how does it elate to pevious esults fo spheical viuses. The solution in this limit is simila to that fom an infinite plate. Actually it is easonably easy to find the complete solution using Mathematica. If you ask Mathematica to solve the diffeential equation d V 1 dv AV() (13) d d by using the command DSolve[V''[]+(1/)*V'[]-A*V[]==0,V,] Then you get V() C J (i A) C Y ( i A) (14) 1 o o Hee J o is the Bessel function of the fist kind (with an imaginay agument) and Y o is the Bessel function of the second kind (also with an imaginay agument. We can use the bounday condition at infinity to eliminate one of these solutions. If we plot the J o function, we see that it looks like (Use A = 3 as an example) Jo Clealy, it diveges as goes to infinity and we should set C 1 = 0 to eliminate this tem. We ae left with V() C Y ( i A) (15) o

12 If we plot the eal pat of the Y o function with C = 1 and A=3, we get eal(v()) So fo this to look like an exponential, we need to take C = eal(v()) This has the expected behavio in that the potential deceases as we get away fom the oigin whee the chaged cylinde will be. If we plot both ou appoximation of exponential decay, found ealie with A = 3 and λ =1, and the coect answe, we get 3.5 Y 0 solution exponential solution eal(v())

13 We see that the two solutions ae quite close in behavio, especially as inceases, as expected. In fact, it is clumsy to use the Y o solution with an imaginay agument. It tuns out that the modified Bessel function of the second kind is also a solution to Eq (13). So ou potential (outside the cylinde) is given by. The constant C would be obtained by the bounday conditions at the suface of the cylinde V() CK o ( A) One can see this esult in a numbe of eseach papes A PHYSICAL METHOD FOR DERIVING THE ELECTROSTATIC INTERACTION BETWEEN ROD- LIKE POLYIONS AT ALL MUTUAL ANGLES STEPHEN L. BRENNER and V. ADRIAN PARSEGIAN BioPhysical Jounal VOLUME 14 (1974) p 37 Moe sophisticated teatments can also be found - See "The Electostatic Potential of B- DNA" by Jayaam et al in Biopolymes Vol (1989)

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