Advanced Protein Models again: adding regulation

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1 Advanced Protein Models again: adding regulation James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University April 1, 2014

2 Outline 1 Simple Regulations 2

3 Simple Regulations Abstract This lecture is going to talk about regulation in protein synthesis.

4 Simple Regulations Now for simplicity, let s set the leakage Λ = 0 and all the other constants in these two models to 1 except for the concentration of the transcription factor R and τ. Then we have two simple models we can look at. C = C = R 1 τ R 1 + R 1 τ C, C repression SS ( ) 1 = τ 1 + R C, C activation SS = τr 1 + R (1) (2)

5 Simple Regulations Now for simplicity, let s set the leakage Λ = 0 and all the other constants in these two models to 1 except for the concentration of the transcription factor R and τ. Then we have two simple models we can look at. C = C = R 1 τ R 1 + R 1 τ C, C repression SS ( ) 1 = τ 1 + R C, C activation SS = τr 1 + R We also a simple model where the transcription factor enhances the loss of the protein by selectively increasing production of the proteins that break down C. We will assume the transcription factor binds to C and degrades it through some process. We have the following reactions: l [C] + [TF] [TFC] where l is the rate at which C and TF combine to form the complex TFC which then immediately degrades. (1) (2)

6 Simple Regulations Hence, although we also know the complex also breaks apart at the rate l which we denote in equation form as l [TFC] [C] + [TF]. effectively, the backwards rate is l = 0 and so we only have the forward association reaction.

7 Simple Regulations Hence, although we also know the complex also breaks apart at the rate l which we denote in equation form as l [TFC] [C] + [TF]. effectively, the backwards rate is l = 0 and so we only have the forward association reaction. The rate of change of [C] is then d [C] dt = amount of [C] lost in the reaction = [TFC] = l [C] [TF] using the fact the concentration of the complex is the product of the concentrations of the components used to produce it standard chemistry!

8 Simple Regulations So replacing [C] by C and the transcription factor [TF] by R, we see this use of the transcription factor leads to a loss term of C loss = RC where for simplicity we set l = 1. We then add that to our standard protein transcription model to obtain Equation 3. C degradation SS = C = 1 R C 1 τ C (3) τ 1 + Rτ where we also set the usual term β = 1.

9 Simple Regulations So replacing [C] by C and the transcription factor [TF] by R, we see this use of the transcription factor leads to a loss term of C loss = RC where for simplicity we set l = 1. We then add that to our standard protein transcription model to obtain Equation 3. C degradation SS = C = 1 R C 1 τ C (3) τ 1 + Rτ where we also set the usual term β = 1. Now depending on the value of τ, we can see that a protein can be cleared out of the biological system more rapidly using this degradation method rather than repression.

10 Simple Regulations So replacing [C] by C and the transcription factor [TF] by R, we see this use of the transcription factor leads to a loss term of C loss = RC where for simplicity we set l = 1. We then add that to our standard protein transcription model to obtain Equation 3. C degradation SS = C = 1 R C 1 τ C (3) τ 1 + Rτ where we also set the usual term β = 1. Now depending on the value of τ, we can see that a protein can be cleared out of the biological system more rapidly using this degradation method rather than repression. We show this in two graphs. In the next figure, we show the steady state values for the degradation and repression models as a function of R for τ = 0.5. Now the response time for repression is τ ln(2), so protein production ramps down quickly for this value of τ. So we see repression is a better strategy for clearing the protein.

11 Simple Regulations

12 Simple Regulations However, it τ is larger, degradation is a better way to go. In the this figure, we show how the steady states change with R and degradation is clearer faster. The code to generate these plots is shown in the next slide.

13 Simple Regulations CSSREP tau, R) tau./(1+r) ; CSSDEG tau, R) tau. / ( 1 + tau. R) ; tau = 0. 5 ; r = l i n s p a c e ( 0, 2 0, ) ; p l o t ( r, CSSREP( tau, r ), +, r, CSSDEG( tau, r ), ˆ ) ; x l a b e l ( x = [ TF ] /K ) ; y l a b e l ( Steady S t a t e Value ) ; t i t l e ( Steady S t a t e f o r r e p r e s s i o n and d e g r a d a t i o n model w i t h tau = 0. 5 ) ; l e g e n d ( r e p r e s s i o n, d e g r a d a t i o n, l o c a t i o n, e a s t ) ; tau = 2. 0 ; p l o t ( r, CSSREP( tau, r ), +, r, CSSDEG( tau, r ), ˆ ) ; x l a b e l ( x = [ TF ] /K ) ; y l a b e l ( Steady S t a t e Value ) ; t i t l e ( Steady S t a t e f o r r e p r e s s i o n and d e g r a d a t i o n model w i t h tau = 2 ) ; l e g e n d ( r e p r e s s i o n, d e g r a d a t i o n, l o c a t i o n, e a s t ) ;

14 Simple Regulations Example Example Examine the difference between the steady state values for repression and degradation in the case τ = 1.0. Comment on the results. The plot shows the winning strategy. Solution CSSREP tau, R) tau./(1+r) ; CSSDEG tau, R) tau. / ( 1 + tau. R) ; r = l i n s p a c e ( 0, 2 0, ) ; tau = 1. 0 ; p l o t ( r, CSSREP( tau, r ), +, r, CSSDEG( tau, r ), ˆ ) ; x l a b e l ( x = [ TF ] /K ) ; y l a b e l ( Steady S t a t e Value ) ; t i t l e ( Steady S t a t e f o r r e p r e s s i o n and d e g r a d a t i o n model w i t h tau = 1 ) ; l e g e n d ( r e p r e s s i o n, d e g r a d a t i o n, l o c a t i o n, e a s t ) ;

15 Simple Regulations Example Examine the difference between the steady state values for repression and degradation in the case τ = 3.0. Comment on the results. The plot shows the winning strategy. Solution CSSREP tau, R) tau./(1+r) ; CSSDEG tau, R) tau. / ( 1 + tau. R) ; r = l i n s p a c e ( 0, 2 0, ) ; tau = 3. 0 ; p l o t ( r, CSSREP( tau, r ), +, r, CSSDEG( tau, r ), ˆ ) ; x l a b e l ( x = [ TF ] /K ) ; y l a b e l ( Steady S t a t e Value ) ; t i t l e ( Steady S t a t e f o r r e p r e s s i o n and d e g r a d a t i o n model w i t h tau = 3 ) ; l e g e n d ( r e p r e s s i o n, d e g r a d a t i o n, l o c a t i o n, e a s t ) ;

16 Simple Regulations Homework Examine the difference between the steady state values for repression and activation in the case τ = 1.5. Here you would use CSSACT in your code. Comment on the results Examine the difference between the steady state values for repression and activation in the case τ = 2.5. Here you would use CSSACT in your code. Comment on the results Examine the difference between the steady state values for repression and degradation in the case τ = 4.0. Comment on the results Examine the difference between the steady state values for repression and degradation in the case τ = 6.0. Comment on the results.

17 The final models we want to discuss involve feedback of the transcription factor signal onto itself. There are two forms: negative feedback and positive feedback and in both this can be done also with a time delay.

18 The final models we want to discuss involve feedback of the transcription factor signal onto itself. There are two forms: negative feedback and positive feedback and in both this can be done also with a time delay. For these equations, we are again setting the leakage Λ = 0 all the other constants to be 1 for simplicity. the feedback equations are seen in Equation 4 and Equation 5. R = R 1 R, negative feedback (4) τ R = 1 + R 1 + R 1 R, positive feedback (5) τ

19 The final models we want to discuss involve feedback of the transcription factor signal onto itself. There are two forms: negative feedback and positive feedback and in both this can be done also with a time delay. For these equations, we are again setting the leakage Λ = 0 all the other constants to be 1 for simplicity. the feedback equations are seen in Equation 4 and Equation 5. R = R 1 R, negative feedback (4) τ R = 1 + R 1 + R 1 R, positive feedback (5) τ These models can not be solved with our Calculus tools, so we must use numerical techniques as we have discussed.

20 Here are some sample runs to show you how to generate interesting plots. In this code, we are using the more general model rather than the simpler one in Equation 4. Hence, in the code, the C which represented the protein concentration is now being thought of as its own transcription factor. So the R in the model is variable and we get the following equations. ( 1 R = Λ + γ 1 + R b ( R R b = Λ + γ 1 + R b ) 1 τ ) 1 τ R, negative feedback (6) R, positive feedback (7)

21 f t, Lambda, gamma, b, tau, R) Lambda + gamma./(1+r. ˆ b ) R/ tau ; g t, R) f ( t, Lambda, gamma, b, tau, R) ; h =. 1 ; tau = 1 ; Lambda = 1 ; gamma =. 5 ; b = 1 ; F i n a l T i m e = ; R I n i t = 2 ; M a x I t e r s = c e i l ( F i n a l T i m e /h ) ; [ htime, Rhat ] = FixedRK ( g, 0, R I n i t, h, 4, M a x I t e r s ) ; p l o t ( htime, Rhat ) ;

22 This generates the plot

23 The positive feedback case is next. We have f t, Lambda, gamma, b, tau, R) Lambda + gamma ( R. ˆ b )./(1+R. ˆ b ) R/ tau ; g t, R) f ( t, Lambda, gamma, b, tau, R) ; h =. 1 ; tau = 1 ; Lambda = 1 ; gamma = 4 ; b = 1 ; TF = ; R I n i t = 2 ; M a x I t e r s = c e i l (TF/h ) ; [ htime, Rhat ] = FixedRK ( g, 0, R I n i t, h, 4, M a x I t e r s ) ; p l o t ( htime, Rhat ) ;

24 This generates the plot

25 If we add a delay of time d, the models are changed to Equation 8 and Equation 9. ( ) R 1 = Λ + γ 1 R, negative feedback (8) 1 + R(t d) τ ( ) R(t d) R = Λ + γ 1 R, positive feedback (9) 1 + R(t d) τ

26 If we add a delay of time d, the models are changed to Equation 8 and Equation 9. ( ) R 1 = Λ + γ 1 R, negative feedback (8) 1 + R(t d) τ ( ) R(t d) R = Λ + γ 1 R, positive feedback (9) 1 + R(t d) τ We don t know how to solve a model with delays, so we are going to show you a quick way to an estimate of the solution. We would have to be more sophisticated if we were really serious, but this will be enough for now. We will use a modified Euler approach.

27 If we add a delay of time d, the models are changed to Equation 8 and Equation 9. ( ) R 1 = Λ + γ 1 R, negative feedback (8) 1 + R(t d) τ ( ) R(t d) R = Λ + γ 1 R, positive feedback (9) 1 + R(t d) τ We don t know how to solve a model with delays, so we are going to show you a quick way to an estimate of the solution. We would have to be more sophisticated if we were really serious, but this will be enough for now. We will use a modified Euler approach. Let s introduce the code eulerdelay as seen below. The basic idea is like this. If the delay is D and the step size is h, it takes m = D/h steps to reach the time when the delayed information starts coming back.

28 So remove the delay part of the model for those time steps and for the first m time steps solve R = Λ + γ 1 τ R negative feedback R = Λ 1 τ R positive feedback using Euler or Runge-Kutta methods.

29 So remove the delay part of the model for those time steps and for the first m time steps solve R = Λ + γ 1 τ R negative feedback R = Λ 1 τ R positive feedback using Euler or Runge-Kutta methods. Then switch to the delayed models for the remainder of the time steps. So we need to define a function g for the model dynamics before the delay kicks in and a function f for the steps with a delay.

30 f u n c t i o n [ htime, Rhat ] = e u l e r d e l a y ( Lambda, gamma, b, tau, d e l a y, TF, R I n i t, h, f, g ) % % Lambda i s the leakage, % gamma i s the capacity 5 % b i s t h e H i l l s c o e f f i c i e n t, % tau i s l i k e t h e r e s p o n s e time % delay i s the time delay, TF i s the f i n a l time % h i s t h e s t e p s i z e % f i s the dynamics a f t e r the delay time has been passed 10 % g i s the dynamics before the delay time has been reached % M = d e l a y /h ; N = c e i l (TF/h ) ; htime = z e r o s (1,N ) ; 15 Rhat = z e r o s (1,N ) ; Rhat ( 1 ) = R I n i t ; htime ( 1 ) = 0 ; % s o l v e f o r steps before the delay so we use the dynamics g % which a r e not d e l a y e d 20 f o r i =1:M Rhat ( i +1) = Rhat ( i ) + g ( i h, Lambda, gamma, b, tau, Rhat ( i )) h ; htime ( i +1) = i h ; end % now we have r e a c h e d t h e d e l a y and we s w i t c h to t h e d e l a y model 25 % w i t h dynamics f f o r i = M+1:N 1 Rhat ( i +1) = Rhat ( i ) + f ( i h, Lambda, gamma, b, tau, Rhat ( i M), Rhat ( i )) h ; htime ( i +1) = i h ; end 30 end

31 Example Let s use this code on the negative feedback with delay model for τ = 0.7, a delay of 1.5, Λ = 2, γ = 1 and the Hill coefficient b = 1. We ll use a step size h =.05 with a final time of 10.0 and initialize with R = 2. Solution g t, Lambda, gamma, b, tau, R) Lambda + gamma R/ tau ; f t, Lambda, gamma, b, tau, S, R) Lambda + gamma./(1+ S. ˆ b ) R/ tau ; h =. 0 5 ; tau = 0. 7 ; d e l a y = 1. 5 ; Lambda = 2 ; gamma = 1. 0 ; b = 1 ; TF = ; R I n i t = 2 ; [ htime, Rhat ] = e u l e r d e l a y ( Lambda, gamma, b, tau, d e l a y, 3 0, R I n i t, h, f, g ) ; p l o t ( htime, Rhat ) ;

32 This generates the plot

33 Example Finally, let s use this code on the positive feedback with delay model with the same parameter values: τ = 1, a delay of 1.5, Λ = 0, γ = 4 and the Hill coefficient b = 1. We ll use a step size h =.1 with a final time of 30.0 and initialize with R = 2. Solution v t, Lambda, gamma, b, tau, R) Lambda R/ tau ; u t, Lambda, gamma, b, tau, S, R) Lambda + gamma (S. ˆ b )./(1+ S. ˆ b ) R/ tau ; h =. 1 ; tau = 1 ; d e l a y = 1. 5 ; Lambda = 0 ; gamma = 4 ; b = 1 ; TF = ; R I n i t = 2 ; [ htime, Rhat ] = e u l e r d e l a y ( Lambda, gamma, b, tau, d e l a y, 3 0, R I n i t, h, u, v ) ; p l o t ( htime, Rhat ) ;

34 This generates the plot

35 Homework Use this code on the negative feedback with delay model for τ = 2, a delay of 3.5, Λ = 1, γ = 2 and the Hill coefficient b = 2. Use a step size h =.05 with a final time of 30.0 and initialize with R = Use this code on the positive feedback with delay model for τ = 0.5, a delay of 1.5, Λ = 0, γ = 5 and the Hill coefficient b = 1. We ll use a step size h =.05 with a final time of 30.0 and initialize with R = Use this code on the negative feedback with delay model for τ = 4, a delay of 1.5, Λ = 0, γ = 3 and the Hill coefficient b = 1. We ll use a step size h =.1 with a final time of 30.0 and initialize with R = Use this code on the positive feedback with delay model for τ = 1, a delay of 3.0, Λ = 1, γ = 6 and the Hill coefficient b = 2. We ll use a step size h =.05 with a final time of 30.0 and initialize with R = 2.