LecturesPart18

LecturesPart18 - Computational Biology, Part 18 Biochemical...

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Unformatted text preview: Computational Biology, Part 18 Biochemical Kinetics II Robert F. Murphy Copyright © 1996, 1999-2006. Copyright All rights reserved. General form of ordinary differential equations s For a set of n unknown functions yi (for i=1 For =1 to n) we are given a set of n functions fi′ we that specify the derivatives of each yi with respect to some independent variable x Euler’s method s The simplest numerical integration method The is Euler’s method. It simply converts each Euler’s It differential to a difference differential s and then calculates the value of ∆ yi by and multiplying the right hand side of each differential equation by the step size ∆ x Euler’s method s We can rewrite this as a recursion formula We that allows us to calculate values of the functions yi at a series of x values. We introduce a second subscript j to indicate which x value we refer to (note that yi,j now refers to a value not a function). value function Euler’s method s Note the asymmetry of this method: the Note derivative (fi′ ) that is used to span the ∆ x is derivative calculated only for the x value at the only beginning of the interval. In regions where beginning fi′ is increasing with x, this leads to x, underestimation of ∆ y, and, in regions underestimation where fi′ is decreasing with x, to x, overestimation of ∆ y. overestimation Euler’s method s Consider dy/dx=6x+2. The analytical Consider dy/dx The solution is y=3x2+2x. s The graph shows the Euler’s method The approximation for y and the analytical solution for y (along with dy/dx). dy/dx). Euler's Method vs. Analytical Solution 35 Note how the approximation always underestimates y since dy/dx is increasing y(Euler) 30 y(Analyt) 25 y 20 18 16 14 dy/dx 12 10 8 20 15 6 4 10 5 2 0 0 0 1 2 x 3 Midpoint method s A better estimate would come from better evaluating the fi′ at the midpoint of the ∆ x midpoint interval. The problem: we know x at the midpoint but we don’t know the yi at the midpoint (yet). The solution is to use Euler’s method to estimate ∆ y and then rere estimate ∆ y using the derivatives evaluated estimate halfway along the line segment encompassing the original ∆ y. Midpoint method (2nd order Runge-Kutte) s This is called the midpoint method or the This second-order Runge-Kutte method. second-order Midpoint method (2nd order Runge-Kutte) s Again consider dy/dx=6x+2. Again dy/dx s The graph shows the midpoint The approximation for y and the analytical solution for y (along with dy/dx at x and x+0.5). along and Midpoint Method vs. Analytical Solution for dy/dx =6 x +2 35 Note that the approximation and the analytical solution are identical in this case. y(2¡RK) 30 y(Analyt) 25 y dy/dx(x) dy/dx(x+Dx/2) 20 15 10 5 0 0 0.5 1 1.5 x 2 2.5 3 Midpoint method (2nd order Runge-Kutte) s Now consider dy/dx=3y. The analytical Now dy/dx The solution is y=e3x. s The graph shows the Euler, Midpoint and The analytical solutions (along with derivatives). analytical Note that the midpoint method is better than Euler’s method but it does not give results identical to the analytical solution since dy/dx now depends on y. Three methods for dy/dx =3 y 100 300 90 y(2¡RK) 80 y(Analyt) 70 y(Euler) 60 dy/dx(x) 250 200 dy/dx(x+Dx/2) y 50 150 40 100 30 20 50 10 0 0 0 0.5 1 x 1.5 Fourth-order Runge-Kutta s The midpoint method can be extended by The considering other intermediate estimates. The most frequently used variation is the fourth-order Runge-Kutta method which considers one estimate at the initial point, two estimates at the midpoint, and one estimate at a trial endpoint. estimate Fourth-order Runge-Kutta s Here are the formulas. Interactive demonstration s (Modify enzyme kinetics model to (Modify incorporate midpoint method) incorporate k1= km1= k2= 0.1 0.005 0.1 t 1.2 1 0.8 0.6 0.4 0.2 0 0 0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 2.7 3 3.3 3.6 3.9 4.2 4.5 4.8 5.1 5.4 5.7 6 6.3 6.6 6.9 7.2 7.5 7.8 8.1 1 8.4 0.98.7 9 0.8 9.3 0.79.6 0.69.9 10.2 0.5 E deltat= timescale= C P 10 0 0 0.75355 9.74905 0.24645 0.0045 0.57750019 9.56241726 0.42249981 12 0.01508293 0.45063468 9.42059136 0.54936532 0.03004333 0.3586179 9.31044282 0.6413821 0.04817507 0.29155566 9.22293967 0.70844434 0.06861598 10 0.24250696 9.15176234 0.75749304 0.09074462 0.20654139 9.09243117 0.79345861 0.11411021 0.18012304 9.04173889 0.81987696 0.13838415 8 0.16069745 8.9973715 0.83930255 0.16332595 0.14640868 8.9576496 0.85359132 0.18875909 E 0.13590238 8.92134871 0.86409762 0.21455367 C 6 0.24061391 0.12818673 8.88757282 0.87181327 P S 0.12253343 8.85566439 0.87746657 0.26686904 0.11840648 8.82513974 0.88159352 0.29326674 0.11541065 8.79564242 0.88458935 4 0.31976822 0.11325406 8.76690933 0.88674594 0.34634473 0.11172089 8.73874597 0.88827911 0.37297492 0.11065133 8.71100841 0.88934867 2 0.39964292 0.10992694 8.68359003 0.89007306 0.42633691 0.10945986 8.65641176 0.89054014 0.4530481 0.10918482 8.62941491 0.89081518 0.47976991 0 0.10905335 8.6025559 0.89094665 12 0.50649744 2 4 6 8 10 0.10902936 8.57580233 0.89097064 0.53322703 0.10908601 8.54913009 0.89091399 0.55995592 0.1092033 8.52252125 0.8907967 0.58668205 Comparison of Euler and midpoint 0.10936632 8.49596246 0.89063368 0.61340386 0.10956395 8.46944377 0.89043605 0.64012018 0.10978787 8.44295777 0.89021213 0.6668301 0.11003189 8.41649893 0.88996811 0.69353296 0.11029138 8.39006315 0.88970862 0.72022823 0.11056286 8.36364735 0.88943714 0.74691551 0.11084374 8.33724926 0.88915626 0.77359449 0.11113209 8.31086718 0.88886791 0.80026491 0.11142644 8.28449986 0.88857356 0.82692658 C 1 S 0.3 0.90497738 En 0.85 0.64723564 0.5013201 0.39560851 0.31863629 0.26237865 0.22114699 0.19086891 0.16860672 0.15222864 0.14018058 0.13132543 0.12482875 0.1200768 0.11661727 0.11411627 0.11232699 0.11106678 0.11020031 0.10962719 0.10927292 0.10908214 0.10901369 0.10903693 0.10912905 0.10927304 0.10945622 0.10966913 0.1099047 0.11015764 0.11042402 0.11070086 0.11098597 0.1112777 0.11157479 Sn 9.85 9.63903889 9.47989967 9.3573247 9.26084049 9.183136 9.11903997 9.06485682 9.01792441 8.97631316 8.93861763 8.90381029 8.87113765 8.84004576 8.81012663 8.7810792 8.75268107 8.72476767 8.69721715 8.66993918 8.64286672 8.61595001 8.58915205 8.56244534 8.53580941 8.50922904 8.48269285 8.45619241 8.42972141 8.40327516 8.37685015 8.3504438 8.32405415 8.29767977 8.27131961 Cn 0.15 0.35276436 0.4986799 0.60439149 0.68136371 0.73762135 0.77885301 0.80913109 0.83139328 0.84777136 0.85981942 0.86867457 0.87517125 0.8799232 0.88338273 0.88588373 0.88767301 0.88893322 0.88979969 0.89037281 0.89072708 0.89091786 0.89098631 0.89096307 0.89087095 0.89072696 0.89054378 0.89033087 0.8900953 0.88984236 0.88957598 0.88929914 0.88901403 0.8887223 0.88842521 Pn 0.00819675 0.02142043 0.03828381 0.0577958 0.07924265 0.10210702 0.12601209 0.1506823 0.17591548 0.20156296 0.22751513 0.25369111 0.28003104 0.30649064 0.33303706 0.35964592 0.38629911 0.41298315 0.43968801 0.4664062 0.49313214 0.51986164 0.54659159 0.57331963 0.600044 0.62676337 0.65347672 0.68018328 0.70688248 0.73357386 0.76025707 0.78693183 0.81359792 0.84025518 0 k1= km1= k2= 0.1 0.005 0.1 t 0 =A6+deltat =A7+deltat =A8+deltat E 1 =B6+(­_k1*En*Sn+(_km1+_k2)*_Cn)*deltat =B7+(­_k1*En*Sn+(_km1+_k2)*_Cn)*deltat =B8+(­_k1*En*Sn+(_km1+_k2)*_Cn)*deltat En =E+(­_k1*E*S+(_km1+_k2)*_C)*deltat/2 =E+(­_k1*E*S+(_km1+_k2)*_C)*deltat/2 =E+(­_k1*E*S+(_km1+_k2)*_C)*deltat/2 =E+(­_k1*E*S+(_km1+_k2)*_C)*deltat/2 deltat= timescale= S 10 =C6+(­_k1*En*Sn+_km1*_Cn)*deltat =C7+(­_k1*En*Sn+_km1*_Cn)*deltat =C8+(­_k1*En*Sn+_km1*_Cn)*deltat Sn =S+(­_k1*E*S+_km1*_C)*deltat/2 =S+(­_k1*E*S+_km1*_C)*deltat/2 =S+(­_k1*E*S+_km1*_C)*deltat/2 =S+(­_k1*E*S+_km1*_C)*deltat/2 0.3 =1/(_k1*C6+_km1+_k2) C 0 =D6+(_k1*En*Sn­(_km1+_k2)*_Cn)*deltat =D7+(_k1*En*Sn­(_km1+_k2)*_Cn)*deltat =D8+(_k1*En*Sn­(_km1+_k2)*_Cn)*deltat P 0 =E6+(_k2*_Cn)*deltat =E7+(_k2*_Cn)*deltat =E8+(_k2*_Cn)*deltat Cn =_C+(_k1*E*S­(_km1+_k2)*_C)*deltat/2 =_C+(_k1*E*S­(_km1+_k2)*_C)*deltat/2 =_C+(_k1*E*S­(_km1+_k2)*_C)*deltat/2 =_C+(_k1*E*S­(_km1+_k2)*_C)*deltat/2 Pn =P+(_k2*_C)*deltat/2 =P+(_k2*_C)*deltat/2 =P+(_k2*_C)*deltat/2 =P+(_k2*_C)*deltat/2 Solving differential equations using Maple Solving a single differential equation using Maple s (Define the differential equation (Define dy/dx=6x+2 using the diff operator, diff assigning the equation to the name deq1) deq1 s (Integrate analytically using dsolve) (Integrate dsolve s (Integrate using boundary condition y(0)=5) (Integrate y(0)=5 s (Integrate using boundary condition y(0)=a) (Integrate y(0)=a s (Integrate dy/dx=bx+c analytically) (Integrate analytically) s (Integrate using boundary condition y(0)=a) (Integrate y(0)=a Solving a single differential equation using Maple s (Define values for constants a,b,c) (Define a,b,c s (Substitute for constants using subs) (Substitute subs s (Use subs to convert the solution that is in (Use the form of an equation (y(x)=...) to a the to functional form for plotting) functional s (Plot the function using plot) (Plot plot Solving a set of differential equations numerically using Maple s (Define the differential equations for the (Define enzyme catalyzed reaction discussed in Part 17) 17 Solving a set of differential equations numerically using Maple s (Plot the solutions using odeplot) (Plot odeplot s (Plot the phase planes using odeplot) (Plot odeplot Other systems for exploration s The following slides describe two additional The biochemical systems that can be modeled by simple modifications of the model already developed. developed. Biochemical System 2 s Reversable catalysis. Biochemical System 2 dE = - k1 ES - k- 2 EP + ( k- 1 + k 2 )C dt dS = - k1 ES + k- 1C dt dC = k1 ES + k- 2 EP - ( k- 1 + k 2 )C dt dP = k 2C - k- 2 EP dt Biochemical System 3 s Catalytic byproduct. Biochemical System 3 dE = - k1 ES - k- 2 EP + k- 1BC + k 2C dt dS = - k1 ES + k- 1BC dt dC = k1 ES + k- 2 EP - k- 1BC - k 2C dt dP = k 2C - k- 2 EP dt ...
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This note was uploaded on 01/13/2012 for the course BIO 101 taught by Professor Staff during the Fall '10 term at DePaul.

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