Use the program. Each patch on the lattice represents one cell. mRNAs are drawn as red dots close to the centre of each patch. The number of turtles on a patch represents the number of mRNAs in a cell. Transcription of mRNAs occurs at a constant rate k. In one time step, there is a probability kδt that a new mRNA is made in each cell. There is a finite rate of breakdown of mRNAs γ. In one time step, there is a probability γδt that each mRNA is destroyed. Let the total number of mRNAs be M and the number of cells be Ncells( = number of lattice sites in this model) . The program plots the mean number of mRNAs per cell (m) =M / Ncells . As M is large, we can treat it as a deterministic variable and write down an ODE: dM/dt = kNcells - γM Dividing this equation by Ncells, we also get the deterministic ODE for m . d(m) /dt= k- γ(m) From this, it can be seen that d(m) /dt =0 when (m) = k/ γ . This is the stationary (or equilibrium) value of (m) . Q1. Add the exact solution to this ODE into the model. Although the mean number of mRNAs per site follows a smooth deterministic function, the number of mRNAs in any one cell follows the stochastic dynamics of births and deaths. Use the following line of code to count the turtles in one particular patch (e. g. the patch with co- ordinates 10 10) count turtles-on patch 10 10 here is the code globals [ ltot ; lattice size atot ; total area = LxL time npop ] to setup clear-all set ltot max-pxcor + 1 set atot ltot * ltot ask patches [ set pcolor white ] reset-ticks plot-mrna end to go birth-death tick set time dt * ticks plot-mrna end to birth-death ask patches [ if random-float 1. 0 < dt * k [ sprout 1 [ set color red set shape "dot" set size 0. 5 set xcor pxcor + 0. 5 * (random-float 1. 0 - 0. 5) set ycor pycor + 0. 5 * (random-float 1. 0 - 0. 5) ] ] ] ask turtles [ if random-float 1. 0 < dt * gamma [die] ] end to plot-mrna set-current-plot "mrna" set-current-plot-pen "mrna" set npop (count turtles) / atot plotxy time npop end