429-2009-HW7key

429-2009-HW7key - Systems Bioengineering III: Homework 7...

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Unformatted text preview: Systems Bioengineering III: Homework 7 Key Rahul Karnik November 9, 2009 GENERAL NOTES : 1. Unless otherwise specified in the question, you can use a single production rate β and a single decay rate α for all proteins in the model. I have solved using separate rates, but you will get full credit if you use a single rate as well. 2. The first protein in a cascade (generally X ) is assumed to be present in its inactive form and simply needs the input signal S x in order to rapidly be converted to the active form. Subsequent proteins (say Y ) in the network need to be induced and transcribed first, even if their activating signal (say S y ) is present throughout. If already transcribed, these proteins would follow the same rapid activation dynamics as the first protein. For example, if S y is present, Y = Y * and we assume that the two are essentially interchangeable. 1: Alon 3.2 We start with the approximate ODE for ( X/K ) n >> 1: dX dt = βK n X n- αX Setting the derivative to zero, we get: X n +1 = βK n α = βK n +1 αK X = K β αK 1 n +1 Now we can calculate the sensitivity: Sensitivity( X,β ) = d ln X d ln β = d 1 n +1 ln β d ln β = 1 n + 1 1 2: Alon 3.3 [The solution presented here uses the linear growth model that applies when K xx << β/α and K yy << βα . You can also use the full ODE to get delay and response times.] The ODE for X is: dX dt = β Θ( X < K xx )- αX In order for X to autoregulate itself, β/α > K xx . Since X = 0 at t = 0, there is no autoregulation until the level of X reaches K xx . So the initial regime is that while X < K xx : dX dt = β- αX As in the book, we can assume that β >> αX , and we get linear growth as long as X < K xx . X = βt Setting X = K xx , we get the timepoint beyond which X begins to autoregulate itself: t xx = K xx β Once the level of X is above K xx , we get proportional control to keep the level of X at this value. So the overall equation for X is: X = ( βt ,t < t xx K xx ,t ≥ t xx The ODE for Y is: dY dt = β Θ( X < K xy )Θ( Y < K yy )- αY We assume that K xy < K xx , otherwise X will never have an effect on Y . We also assume that K xy > K yy , otherwise Y would never autoregulate itself. Initially, X < K xy and Y < K yy are below the threshold for regulating Y . So Y follows the ODE: dY dt = β- αY As with X , we can assume that β >> αY , and we get linear growth for Y : Y = βt Once the level of Y reaches K yy , negative autoregulation provides proportional control to keep Y at this level. This is the second regime. Finally, when X reaches K xy (and stays above this level), transcription of Y is turned off and the level of Y follows the ODE: dY dt =- αY with the initial condition Y ( t XY ) = K yy , which we can solve to get: Y = K yy e- α ( t- t XY ) 2 As with X , we can calculate the critical timepoints t Y Y and t XY , which are the timepoints at which Y and X respectively begin to regulate the production of Y . Setting Y = K yy we get: t Y Y = K yy β Setting X...
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This note was uploaded on 03/30/2010 for the course SBE 580.429 taught by Professor Joelbader during the Fall '09 term at Johns Hopkins.

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429-2009-HW7key - Systems Bioengineering III: Homework 7...

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