Thi d d i h F M k 4 0 4 T u r n i n g a M i c r o c a p i l l a r y i n t o a S e p a r a t i o n R a t c h e t by Martin Bier, September 2003 Microcapillaries are silica tubes of about 50 µm in inner diameter. Over the past decade microcapillaries have found wide application in analytical chemistry as it is possible to create so called plug like flow in these capillaries. When liquid in a tube is pushed by ordinary hydrodynamic pressure, the resulting Poiseuille flow is fastest in the center of the tube and zero at the wall. The electro-osmotic flow in a microcapillary appears comes out plug like through the following process. The inner surface of a fused silica capillary is covered with silanol groups (Si-OH), which are ionized to SiO - at ph > 2. The negatively charged surface is counterbalanced by positive H 3 O + ions from the buffer, forming the so-called electric double layer. When an electric field is applied along the length of the microcapillary the positive ions in the diffuse part of the double layer will migrate towards the cathode. For a sufficiently narrow microcapillary (and 50 µm generally suffices) the ions entrain the water which resluts in the electroosmotic plug like flow. Biomolecules like DNA strands and proteins generally carry a negative charge. At ph values of more than 3, the electrophoretic pull on a anion toward the anode is generally weaker than the drag force of the buffer s electroosmotic flow toward the cathode. Only at lower ph do the anions migrate toward the anode. Next imagine the setup sketched in Fig. 1. There is an electric field along the length of the capillary. However, interspaced by a distance s 1 there are thick strontium titanate (SrTiO 3 ) rings around the capillary of a width s 2.
Strontium titanate has a high dielectric permittivity (ε r! 300). This means that along the stretch s 2 the electric field will be sucked out of the water (ε r! 80). The ph and the electric field have to be tuned such that nions that are dissolved in the aqueous solution will be pulled by the electric field in the direction of the anode when in a domain s 1. When in the domain s 2, where the electric field should be much smaller, the net force should be with the drag of the flowing liquid into the direction of the cathode. Electroosmotic flow amounts to about 1 mm/s, so for s1! 1.0 mm and s2! 0.1 all the anions should be concentrated at the position that is indicated in Fig. 1 with a black solid dot. Next we switch the electric field off. This will immediately stop the electroosmotic flow (inertia is negligible on the microscopic scale). The anions will now start to diffuse freely along the length of the capillary. Imagine that the anions are a mixture of a single aminoacid and a dimer of two aminoacids. Another possibility would be to use a monosaccharide and a disaccharide. The monomer is smaller and will therefore diffuse faster. The off-time "T off for the electric field has to be tuned such that it is much more likely for the monomer to diffuse a distance s 2 to the left than it is for the dimer. When, after such a "T off, the electric field is turned on again, a monomer is much more likely going to end up at the next ring to the left than is a dimer. When the field is repeatedly turned on and off, the monomer will be transported towards the anode faster than the anode and the mixture will separate. Note that the analytes are separated solely on the basis of their diffusivity. The exact value of the charge is of no consequence as long as it is such that the forces in the s 1 and s 2 point in opposite directions. The energy
input from the electric oscillationand the anisotropy of the setup combine to effectively ratchet the diffusion of the analytes, i.e. diffusion is allowed and amplified in one direction and inhibited in the opposite direction. Eventually faster net transport will occur for the more diffusive analyte. The ratchet potential is the combined result of an electrical field and hydrodynamic flow. The capillary is coated with a material of higher dielectric permittivity. By making the right periodic carving pattern in this material, an electric field can be obtained that looks like in the drawing: strong for a short distance, weak for a longer distance. The pressure and the flow of the liquid is to be such that it is balanced exactly by the average electric force. This will give rise to the ratchet potential shown at the bottom of Fig. 1. When the field is on, the net forces F 1 and F 2 in the regions s 1 and s 2 should be sufficiently large to not beovercome by the Brownian motion. This means that the products F 1 s 1 and F 2 s 2 should be at least a order of magnitude larger than kt (the average amount of energy in Brownian motion). Aminoacids like glycine or alanine have a diffusion coefficient of about 10-9 m 2 /sec at 20 C. Following the diffusion equation, <x 2 > = 2Dt, we see that it takes about 10 seconds for the an aminoacid to diffuse 0.1 mm (the width i=of the ring). On and off time of between a second and a minute are very managable. Larger molecules, like long DNA strands, have smaller diffusion coefficients (D! 10-11 m 2 /sec for a ribosome) and take longer to diffuse a distance of 0.1 mm. The sorting will be faster, better and more efficient if the spatial periods are smaller. Because of the square in the diffusion formula the time inter-
val "T off can be reduced by a factor 100 when s 1 and s 2 are reduced by a factor 10. When our device is shrunk from the millimeter level to the micrometer level, separation will be a million times as fast. This will be the appropriate scale for the separation of DNA strand, the so-called DNA fingerprinting. DNA strands can be up to several thousand units long. In the appendices I provide material to prove that such strands can be quickly separated with a small device even if they only differ in length by a few units. The derivations in the appendices show that if no major engineering problems arise this method will have a speed and a resolution that far exceed the methods currently in use.
Thi d d i h F M k 4 0 4 Ring of SrTiO 3 Ring of SrTiO 3 Microcapillary Tube ANODE CATHOD + Electric Field - Electroosmotic Flow Net Force on an anion s 1 s 2
Thi d d i h F M k 4 0 4 Appendix 1 When the electric field and the and the fluid flow are off, there is no net force on the particle and there is free diffusion. A particle starting at t=0 at x=0 will, at time t will be found between x 1 and x 2 with a probability: p = x 2 x 1 P( x, t)dx where P( x, t) = 1 ---------------- exp 4πDt x 2 --------- 4Dt. P(x,t) describes a spreading Gaussian. When the potential is on, it needs to be on long enough for all particles to slide down to the energy minimum. Going down the long slope takes the longest and this will lead to a condition. F 1 = β v 1 leads to F 1 = β s 1 /t on. So we get t on > β s 1 /F 1. Here β represents the coefficient of friction which is related to the diffusion coefficient D by Einstein s Fluctuation-Dissipation Theorem, i.e. β = kt/d. As mentionded before, we get a Gaussian spreading when the potential is off. For the ratchet to operate optimally, we need the Gaussian to have a standard deviation of about the length of the short segment, i.e <s 2 2 > = 2Dt off. Notice that t off is independent of electric field, flow speed v, or any other equipment parameter. It is determined by the diffusion coefficients of the involved analytes. Generally, t off will be the factor that limits the speed of the separation. Obviously, taking smaller spatial periods will speed up the separation. One cycle of the system lasts T = t on + t off. The probability of a forward
step (i.e. ending up at the next ring to the left in Fig. 1) after one cycle is p f, and the probability of a backward step after one cycle isp b. Calculating p f and p b involves integrating the above Gaussian w.r.t. x and leads to socalled Error functions. In the course of many succesive periods an analyte travels along the array of periods as a spreading Gaussian. Appendix 2 derives the speed and increasing width of the blot in terms of p f and p b. Appendix 2, furthermore, gives an estimate for how long the device has to run and how lengthy (how many periods) the device has to be, for 2 analytes to be separated into two distinct blots. Appendix 3 is a Mathematica program. One enters the length of the short and long segment in the device. Next the diffusion coefficients of two analytes are entered. The program first calculates an order-of-magnitude estimate for t off and the associated p f and p b. Next the program is able to search for a t off that leads to the fastest separation of the two analytes. The necessary number of spatial periods is also calculated.
Thi d d i h F M k 4 0 4 Appendix 2 Consider a stochastic stepper as described in the main text, but let the stepsize be 1 and every unit of time the particle can move forward (+1 with probability p f ), backward (-1 with probability p b ) or stay in the same place (0 with probability 1-p f -p b ). We have for the drift speed of such a particle <v> = p f (+1) + p b (-1) = p f - p b. Rescaling to a stepsize L and a period T we have <v> = (p f - p b )L/T. After running the system for τ seconds, the covered distance averages: Λ =(p f - p b )L(τ/T). To calculate the variance, the procedure is somewhat more involved. Go back to the scaled system (L=1, T=1). In one unit of time the average displacement is <X> = p f - p b. For the second moment we have <X 2 > = p f (+1) 2 + p b (-1) 2 = p f + p b. For the variance we derive σ 2 = <X 2 > - <X> 2 = p f + p b - (p f - p b ) 2. After some manipulation this can be written as: 2 σ sc = p f ( 1 p f ) + p b ( 1 p b ) + 2 p f p b. To unscale, the variance must be multiplied with L 2. Variances add up and therefore we find in unscaled variables after running the system for τ seconds: σ 2 = { p f ( 1 p f ) + p b ( 1 p b ) + 2 p f p b } L 2 τ T --. The standard deviation is the square root of the variance, i.e. simply σ. Suppose we now have two analytes in our device. After running the system for τ seconds, the difference in average covered distance equals: Λ 1 - Λ 2 =
{(p f1 - p b1 ) - (p f2 - p b2 )} L (τ/t). This difference should be larger than the sum (σ 1 + σ 2 ). (Λ 1 - Λ 2 )/(σ 1 + σ 2 ) > 1 translates into an estimate for how long we have to run the device to get an obvious separation: σ sc1 + σ sc2 τ > ----------------------------------------------------------------- ( p f 1 p b1 ) ( p f 2 p b2 ) 2 T. (τ/t) gives the number of time periods necessary to bring about a separation. Subsequently, the quantity (τ/t) has to be multiplied with Max[(p f1 - p b1 ),(p f2 - p b2 )] to obtain an estimate for how many spatial periods the device needs to have to bring about a separation.
RatchetPump1.nb 1 (* First enter the length of the long and the short segment in the ratchet *) s1 = 0.0000009; s2 = 0.0000001; (* Next enter the diffusion coefficients for the two analytes *) Df1 = 31.62 10^-11; Df2 = 31.64 10^-11; (* Let the computer first give an estimate for a reasonable and recommended toff *) t1offrec = (s2)^2/(2 Df1)//N; t2offrec = (s2)^2/(2 Df2)//N; Print[StringForm["Long slope: `` Short slope `` ",s1, s2]] Print[StringForm["Diffusion coefficient for 1st analyte `` ",Df1]] Print[StringForm["Diffusion coefficient for 2nd analyte `` ",Df2]] Print[StringForm["For the 1st analyte the recommended off time is `` seconds",t1offrec]] Print[StringForm["For the 2nd analyte the recommended off time is `` seconds",t2offrec]] -7-7 Long slope: 9. 10 Short slope 1. 10-10 Diffusion coefficient for 1st analyte 3.162 10-10 Diffusion coefficient for 2nd analyte 3.164 10 For the 1st analyte the recommended off time is 0.0000158128 seconds For the 2nd analyte the recommended off time is 0.0000158028 seconds
RatchetPump1.nb 2 (* This is a little routine to search for the best possible toff, *) (* i.e., the toff that leads to the fastest separation *) (* We start with a toff that is an order of magnitude smaller than the *) (* minimum recommended toff. Next we raise toff 5% each step and scan about *) (* 4 orders of magnitude *) toff = Min[t1offrec,t2offrec]/40 periodnumber = 10^9; NN=24 Do[toff = (1.02)^m toff; pf1 = (1/2) Erfc[s2/(2 Sqrt[Df1 toff])]//n; pb1 = (1/2) Erfc[s1/(2 Sqrt[Df1 toff])]//n; pf2 = (1/2) Erfc[s2/(2 Sqrt[Df2 toff])]//n; pb2 = (1/2) Erfc[s1/(2 Sqrt[Df2 toff])]//n; var1 = pf1 (1 - pf1) + pb1 (1 - pb1) + 2 pf1 pb1; var2 = pf2 (1 - pf2) + pb2 (1 - pb2) + 2 pf2 pb2; numbofper = ((Sqrt[var1] + Sqrt[var2])/((pf1 - pb1) - (pf2 - pb2)))^2//n; duration = numbofper toff; length = Max[pf1,pf2]*numbofPer*(s1+s2); t1[m] = {duration,length}//n; t2[m] = {Log[10,duration],Log[10,length]}//N,{m,NN}] tabel1 = Table[t1[i],{i,NN}]; tabel2 = Table[t2[i],{i,NN}]; loggraph = ListPlot[tabel2,AxesLabel -> {"Log(duration)", "Log(length)"}, PlotRange -> {{2.0,7.0},{-2.0,3.0}},PlotJoined -> True] Do[If[t1[j-1][[1]] > t1[j][[1]] && t1[j+1][[1]] > t1[j][[1]], k=j],{j,2,nn-1}]; Print[StringForm["The optimum takes `` seconds, is `` meter long and has toff = `` ",t1[k][[1]],t1[k][[2]],toff (1.02)^k]] Print[StringForm["We scan the lower part of the branch to see what best fits available time and length"]] Do[kk = k-8+i; Print[StringForm[" `` seconds and [2]],toff (1.02)^kk]], {i,8}] `` meter at toff = `` ",t1[kk][[1]],t1[kk][ -7 3.9507 10 24
RatchetPump1.nb 3 Log(length) 3 2 1 0-1 3 4 5 6 7 Log(duration) -2 -Graphics- The optimum takes 295.281 seconds, is 1.18369 meter long and has toff = 0.000198211 We scan the lower part of the branch to see what best fits available time and length 58896. seconds and 0.0702664 meter at toff = 0.000172554 16153.3 seconds and 0.0952325 meter at toff = 0.000176006 5046.5 seconds and 0.133804 meter at toff = 0.000179526 1875.92 seconds and 0.194681 meter at toff = 0.000183116 854.116 seconds and 0.292875 meter at toff = 0.000186779 483.29 seconds and 0.45451 meter at toff = 0.000190514 340.298 seconds and 0.725134 meter at toff = 0.000194324 295.281 seconds and 1.18369 meter at toff = 0.000198211 (* Enter the length of time the potential is "off" and calculate the number of necessary periods*) toff = 0.36; pf1 = (1/2) Erfc[s2/(2 Sqrt[Df1 toff])]//n; pb1 = (1/2) Erfc[s1/(2 Sqrt[Df1 toff])]//n; pf2 = (1/2) Erfc[s2/(2 Sqrt[Df2 toff])]//n; pb2 = (1/2) Erfc[s1/(2 Sqrt[Df2 toff])]//n; var1 = pf1 (1 - pf1) + pb1 (1 - pb1) + 2 pf1 pb1; var2 = pf2 (1 - pf2) + pb2 (1 - pb2) + 2 pf2 pb2; numbofper = ((Sqrt[var1] + Sqrt[var2])/((pf1 - pb1) - (pf2 - pb2)))^2//n; Print[StringForm["For the 1st analyte pf1= `` and pb1= ``",pf1,pb1]] Print[StringForm["For the 2nd analyte pf2= `` and pb2= ``",pf2,pb2]] Print[StringForm["`` time periods are required to separate the analytes",numbofper]] Print[StringForm["`` spatial periods are required to separate the analytes",numbofper* Max[pf1,pf2]]] -63 For the 1st analyte pf1= 0.0312037 and pb1= 2.00526 10-33 For the 2nd analyte pf2= 0.0938162 and pb2= 9.71657 10 55.2595 time periods are required to separate the analytes 5.18424 spatial periods are required to separate the analytes