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Matlab Experiment of Holliday Junction

Matlab Experiment of Holliday Junction. Kenny and Ryan. Matlab Function. % ODE's for holiday junctions, A, B, C, E %where A-B and C-E are much slower than B-C function dydt = HolJunct(t,y) k1f=.0000001; % A to B rate k2f=.032; % B to C rate

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Matlab Experiment of Holliday Junction

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  1. Matlab Experiment of Holliday Junction Kenny and Ryan

  2. Matlab Function %ODE's for holiday junctions, A, B, C, E %where A-B and C-E are much slower than B-C function dydt = HolJunct(t,y) k1f=.0000001; % A to B rate k2f=.032; % B to C rate k3f=2.2*(10^-8); % C to E rate k1r=2.8*(10^-8); % B to A rate k2r=.042; % C to B rate dydt(1,1)=-k1f*y(1)+k1r*y(2); % dA/dt dydt(2,1)=-k1r*y(2)+k1f*y(1)-k2f*y(2)+k2r*y(3); % dB/dt dydt(3,1)=k2f*y(2)-k2r*y(3)-k3f*y(3); % dC/dt dydt(4,1)=k3f*y(3); % dE/dt end

  3. Matlab Function Explanation • The function is given a title with the return value of dydt • The reaction constants, k1r, k1f, etc are initialized at the top • The 4 dimension differential equation system is represented using dydt(1,1)dydt(4,1), where (1,1) represents the first row and first column of a matrix we call dydt. • The matrix dydt is thus a column vector (4 x 1) – (rows x columns) that consists of the 4 differential equations dA/dt  dE/dt

  4. Solving the System • Now that the system has been created in the function title HolJunct, we need to call and solve it within the Matlab Command Window (the function is created in a script file) • We solve/call using the command • [t,y] = ode23(@HolJunct, [0,2000000], [0 1 0.95 0]); • The above ode23 is a built-in Matlab function that solves Ordinary Differential Equations (ODE), where the function being used needs to be defined (@), the time span or interval needs to be set ([0 to 2million]), and the initial conditions for the original functions A,B,C,E need to be inputted ([0 for A, 1 for B, 0.95 for C, 0 for E]) • Type help ode23 for more information

  5. Output/Results • After typing the command for different initial conditions, the following graphs were obtained: A – B – C – E – Note the ratio between B and C remains constant since this represents short term

  6. Results Continued • After changing k3f and k1r to larger numbers, we can see a “sped-up” version of the reaction A – B – C – E –

  7. END

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