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Unformatted text preview: ChE 101 2012 Problem Set 5 Solutions 1. (a) The full ODE system to be solved is d [ P ] dt = k cat [ ES ] d [ ES ] dt = k f [ E ][ S ]- ( k cat + k b )[ ES ] d [ E ] dt = ( k cat + k b )[ ES ]- k f [ E ][ S ] d [ S ] dt = k b [ ES ]- k f [ E ][ S ] Under the equilibrium assumption, the simplified equations to solve are [ P ] = [ S ]- [ S ] d [ S ] dt =- k cat [ E ] [ S ] K d + [ S ] K d = k b k f Under PSSA, the equations are [ P ] = [ S ]- [ S ] d [ S ] dt =- k cat [ E ] [ S ] K m + [ S ] K m = k b + k cat k f These equations were solved using the Matlab script below. Note a few key changes from our usual ODE solution scripts: First, Matlab’s ODE solvers use a default absolute tolerance of 10- 6 . This means that variables with values below 10- 6 are treated as zero by the solver. This is problematic for our setup, since the initial enzyme concentration is already at this threshold value, so the concentrations of E and ES will not be accurately simulated without a reduction in the absolute tolerance to a suitably low value (we chose 10- 16 in the script below). Second, when high levels of accuracy are desired in the solution (as mandated by our constraint on the error tolerance), or when the rate law is computationally expensive to evaluate (also true in our case because of the order of magnitude differences between the rate constants), ode113 is typically a better solver to use than ode45. 1 clear clc close a l l set (0 , ' DefaultTextInterpreter ' , ' latex ' ) kcat = 1 e1 ; kb = 1 e3 ; kf = 1 e6 ; E0 = 1 e- 6; S0 = 1 e- 3; tf = 1 e3 ; tvals = linspace (0 , tf ,25) ; options = odeset ( ' RelTol ' , 1 e- 3, ' AbsTol ' , 1 e- 16) ; [ T1 , Y1 ] = ode113 ( @ ( t , y ) [ kcat * y (2) ; kf * y (3) * y (4)- ( kcat + kb ) * y (2) ; ( kcat + kb ) * y (2)- kf ←- * y (3) * y (4) ; kb * y (2)- kf * y (3) * y (4) ] , tvals , [ 0 ; 0 ; E0 ; S0 ] , options ) ; Kd = kb / kf ; [ T2 , Y2 ] = ode45 ( @ ( t , y )- kcat * E0 * y ./( Kd + y ) , tvals , S0 , options ) ; Km = ( kb + kcat )/ kf ; [ T3 , Y3 ] = ode45 ( @ ( t , y )- kcat * E0 * y ./( Km + y ) , tvals , S0 , options ) ; figure ( ' Position ' , [0 240 800 600]) axes ( ' FontSize ' , 14) plot ( T1 , Y1 ( : , 1 ) * 1000 , ' o-- ' , T2 , ( S0- Y2 ) * 1000 , ' s-- ' , T3 , ( S0- Y3 ) * 1000 , ' ˆ-- ' , ' ←- MarkerSize ' , 10) xlabel ( ' Time ( seconds ) ' , ' FontSize ' , 16) ylabel ( ' $ [P] $ (mM) ' , ' FontSize ' , 16) leg = legend ( ' Exact Calculation ' , ' Equilibrium Assumption ' , ' PSSA ' ) ; set ( leg , ' FontSize ' , 16) set ( leg , ' Interpreter ' , ' latex ' ) set ( leg , ' Location ' , ' Best ' ) set ( leg , ' Box ' , ' o f f ' ) The resulting solutions are shown in Figure 1. Note that under the specified conditions, all three approaches yield identical kinetics. This is not surprising for two reasons: First, we have k b k cat , which is the condition required for validity of the equilibrium assumption (i.e. the enzyme-substrate complex needs to reach equilibrium on time scales much faster than that of product formation). Second, we have [much faster than that of product formation)....
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- Winter '11
- Equilibrium, Rh, k2, Rate equation, Lineweaver–Burk plot, k2 k4