Key ideas about dynamic models

Transcription

1 GSTDMB 202: DYNAMICAL MODELLING FOR BIOLOGY AND MEDICINE Lecture 2 Introduction to modelling with differential equations Markus Owen ey ideas about dynamic models Dynamic models describe the change in the state of a system with time. Solution/trajectory: the set of future states given a particular initial state. Steady state: a solution which is steady/constant/not changing in time. Periodic solution/limit cycle: an oscillatory solution, i.e. one that repeats eactly the same values at an interval known as the period. The stability of a steady state describes what happens to the system if it starts close to that steady state: : if we start close to the steady state, the system converges to that steady state un: if we start close to the steady state, the system diverges from that steady state Bifurcation: the number or stability of steady states (or periodic solutions) changes as a parameter varies. Qualitative analysis: determines information about qualitative properties of solutions and bifurcations. Steady states and stability are important here. Quantitative analysis: determines numerical values for solutions, bifurcations, etc, usually via computer simulation (ecept for special cases). 2 Stability: eamples Consider a ball rolling on a smooth landscape. The ball can be placed at the top of a hill and will stay there for all time - this is a steady state. In practice, any small disturbance will lead to the ball rolling down one side or the other. This is an eample of an un steady state. Another steady state is at the bottom of a valley. After any small disturbance the ball will roll back to the bottom. This is an eample of a steady state. un Another eample is a rigid pendulum: 3

2 st order Ordinary Differential Equations Dynamic models for the dependence of single state variable, on the independent variable t. The solution is (t) Describe the rate of change of, written d dt In general, this may depend on itself and on time, t: (e.g. time dependent parameters in circadian models) d = f (, t) dt We will consider the autonomous case, when the rate of change of does not depend on t, but only on the state variable itself. d f () In this case qualitative analysis is reasonably straightforward Steady states are where f()=0. The phase-line diagram shows us where is increasing, where it is decreasing, and where any steady states lie. Relies on being able to sketch or plot the graph of f() 4 Phase-line diagrams () Enable qualitative analysis of st order autonomous ODEs. Given d/dt=f(), sketch f() Remember that f() is the rate of change of f() d/f() > 0, increasing d/f() < 0, decreasing 5 Phase-line diagrams (2) Steady states where f() crosses the horizontal ais (d/dt=f()=0) Stable if f() crosses from positive to negative Un if f() crosses from negative to positive f() un un 6

3 Phase-line diagrams (3) Also tells you how fast is increasing or decreasing Easiest to indicate graphically with arrows Arrows to right (left) for increasing (decreasing) This reinforces our understanding of (in)stability f() un un 7 Sketching the graph of a function f() f(0) - where the graph crosses the vertical ais f (0) - slope at =0 (f is shorthand for df/d, the gradient of f()) anywhere else obvious where f()=0? is f() polynomial (a n + b n- + c n d + e)? quadratic (highest power is 2), a 2 + b + c - always one ma or min cubic (highest power is 3) - one inflection, or one ma and one min In general, at most n- turning points what about f() as get very large positive (negative)? Does it go up, down, or become flat (i.e. approach a horizontal asymptote)? Any vertical asymptotes? 8 Sketching the graph of a function f() Eample: f() = r(-/) = r - r 2 /. f(0)=0 2. f ()=r-2r/, so f (0)=r, f() is increasing through the origin 3. clearly f()=r(-)=r(0)=0 so crosses at = 4. f() is quadratic, so one turning point. f() = r(-/) 4 quiz Q

4 Sketching the graph of a function f() Eample: f() = r(-/)(-a). f(0)=0, f () is harder to work out 2. clearly f()=r(-)(-a)=(0)(-a)=0 so crosses at = 3. clearly f(a)=ra(-a/)(a-a)=ra(-a/)(0)=0 so crosses at =a 4. As gets very large, f() gets very large negative... r.(large).(-large/).(large) = -r(large) 3 / could also see this by epanding brackets f() is cubic, so one ma and one min (or one inflection). f() = r(-/)(-a) quiz Q a Sketching the graph of a function f() Eample: f() = 2 /(+ 2 ). f(0)=0 2. f ()=2/(+^2) 2, so f (0)=0, and as gets large, f () goes to zero (horizontal asymptote). Another way to see f (0)=0, for small, f() looks like 2 ). 3. clearly f(0) 0 4. As gets large, f() approaches one (the on the bottom is insignificant). If you re not convinced, consider =0, 00, /(+0 2 ) = 00/0 = /(+00 2 ) = 0000/000 = /2 2 3 quiz Q2 Eponential population growth Eamples from Population growth illustrate many key points d r is the number of individuals r is a parameter, the rate of growth per capita. It has dimensions of /(time) Steady state: any steady solution has d/0 Hence r =0. Assuming r >0, this must mean =0 Can see this, and more, from phase-line diagram: f() = r un for small values of, grows slowly as increases, its rate of growth increases this gives characteristic eponential growth: t 2

5 Eponential population growth? d r Solution is (t) = (0)e rt (quiz Q6). Log of population data should be straight line... Here we show global human population data Take logs of both sides: ln((t)) = rt +(0) Population (millions) Population log(population growth rate (millions)) (millions/year)) t (years) t (years) Growth rate is slowing (after a period of acceleration) 3 Logistic growth d r r measures the maimum rate of growth. It has dimensions of /(time). measures the carrying capacity for the population. It has dimensions of number of individuals. Effectively, we have replaced a constant per capita growth rate r, with a rate that decreases as the population size increases. This models the depletion of resources as a population grows. Steady states are where d/0, which is where =0 or = f() = r(-/) quiz Q9 un representative solutions t 4 Logistic growth? d r Consider eperimental data on the growth of yeast. Dynamics look a bit like logistic growth... Yeast growth (Saccharomyces cerevisiae CEN.P3) Optical density Ept Ept 2 Ept 3 Ept Logistic model Ept 2 Ept 3 Gratuitous transition time (min) But logistic growth is too slow at first, and too fast later. 5

6 Allee effect d r ( a) Many species ehibit lower or even negative growth rates at low numbers. Here, the per capita growth rate is: r ( a) This is negative if <a, i.e. if the population is too small (NB: 0<a<) f() = r(-/)(-a) quiz Q0 a un a representative solutions t Bistability: two steady states. Final state depends on initial value, (0). 6 Production & degradation models In each of the preceding cases, the form of f() (straight line, quadratic, cubic) makes it easy to sketch the phase-line diagram (and there is no qualitative dependence on parameter values) Net we will consider models for simple feedback loops, such as may arise with transcriptional autoregulation Rate of change of = production decay P() TF () δ TF () d P() P() represents the effect of a Transcription Factor on its own synthesis We will consider various common forms for P() What are we interested in? Steady states: production and turnover of are balanced How fast are steady states reached? Any bifurcations (ideal for ep tal validation)? 7 Constant production d A P() =A, a constant. This could model constitutive transcription As usual, steady states satisfy d/0, hence A = δ So the steady state TF level is = A/δ Does this fit with our biological understanding and intuition? d/ A-δ A No bifurcations A/δ A/δ representative solutions t Larger δ faster decay to steady state 8

7 Saturating positive feedback P() = A h + Transcription rate increases with TF,, but saturates to a maimum rate A e.g. TF binds to its own promoter This is an eample of a Hill function: P() = An h n + n Half maimal response at = h A A/2 h quiz Q A/(h+) Model equation becomes: d A h + How to sketch the phase-line diagram and find steady states? 9 Saturating positive feedback P() = d A h + It turns out there are two qualitatively different phase-line diagrams Algebra: steady states satisfy A/(h+) = δ One obvious solution is =0 the other has A/(h+) = δ, hence A = δh + δ, hence δ = A - δh, hence = (A-δh)/δ If A < δh then this steady state is negative and not biologically relevant Interpretation: if TF turnover rate too large, TF level decays to zero There is a bifurcation at A = δh, (i.e. change in number or stability of steady states) A h + Graphically, d/dt is the difference between the curve P() and the line δ 20 d Saturating positive feedback A h + P() = A h + Graphically, d/dt is the difference between the curve P() and the line δ It is easy to see the effect of increasing δ which is the slope of the line δ Increasing δ δ A/(h+) A/(h+) (A-δh)/δ un The difference is always negative: is always decreasing to zero 2

8 Saturating positive feedback P() = Graphically, d/dt is the difference between the curve P() and the line δ It is easy to see the effect of increasing δ which is the slope of the line Increasing δ A h + A/(h+) - δ (A-δh)/δ A/(h+) - δ un Increasing δ effectively pulls the curve of d/dt down through the ais 22 Saturating positive feedback P() = A h + We see that as δ increases, the steady state TF level decreases, until it reaches zero. Beyond this point TF production cannot be sustained. We can summarise this information in a Bifurcation diagram, which shows steady states and their stability as a parameter varies Solid lines indicate steady states, dashed lines un steady states steady state TF level, bifurcation point = A h 0 A/h TF turnover rate, δ Could test this structure against eperiment but bifurcation will be at TF levels below detection threshold. 23 Sigmoidal positive feedback P() = An h n + n Transcription rate increases with TF,, but saturates to a maimum rate A A Hill function with order > is an eample of a sigmoid curve Half maimal response at = h A A/2 h quiz Q2 Model equation becomes: d An h n + n How to sketch the phase-line diagram and find steady states? 24

9 Sigmoidal positive feedback P() = An h n + n d An h n + n Again there are two qualitatively different phase-line diagrams Algebra: steady states satisfy A n /(h n + n ) = δ One obvious solution is =0 the other has A n- /(h n + n ) = δ hard to solve in general. We resort to graphical analysis. 25 Sigmoidal positive feedback P() = An h n + n d An h n + n Graphically, d/dt is the difference between the curve P() and the line δ It is easy to see the effect of increasing δ which is the slope of the line A n /(h n + n ) δ Increasing δ δ A n /(h n + n ) un Three steady states - BISTABILITY The difference is always negative: is always decreasing to zero 26 Sigmoidal positive feedback P() = An h n + n We see that the zero steady state is always. As δ increases, the nonzero steady state TF level decreases, until it disappears in a bifurcation. Beyond this point TF production cannot be sustained. We can summarise this information in a Bifurcation diagram, which shows steady states and their stability as a parameter varies. Solid lines indicate steady states, dashed lines un steady states. steady state TF level bifurcation point TF turnover rate, δ Could test this structure against eperiment at bifurcation TF levels could be above detection threshold. 27

10 Negative feedback P() = Ah h + Transcription rate decreases with TF,. Maimum rate A, minimum zero A decreasing Hill function: P() = Ahn h n + n Half maimal response at = h A A/2 h Ah/(h+) Model equation becomes: d Ah h + How to sketch the phase-line diagram and find steady states? 28 Negative feedback P() = Ah h + Algebra: steady states satisfy Ah/(h+) = δ No obvious solutions - need to cross multiply and solve a quadratic... Graphically, d/dt is the difference between the curve P() and the line δ δ d/ A/(h+) - δ Ah/(h+) The pictures are qualitatively the same whatever the parameters Always just one steady state TF level As δ increases the TF level falls 29 Discussion We have introduced simple ordinary differential equation (ODE) models for single state variables. Steady states and their stability are crucial determinant of system dynamics. Changes in number or stability of steady states are called bifurcations. For st order autonomous ODEs, the phase-line diagram can tell us most of the qualitative information we d like to know about the system dynamics: if you can sketch the graph, you can sketch the dynamics... steady states, stability AND qualitative solution behaviour (fast, slow, increasing, decreasing, etc), bifurcations. solutions cannot oscillate For st order non-autonomous ODEs (e.g. circadian models with time dependent parameters) solutions can oscillate (driven by e.g. day-night cycle) Net: Using CellDesigner to build and simulate single variable models models with > state variable - more comple dynamics possible, analysis more difficult, often resort to computer simulation 30