(b) Why is this a coherent FFL? Show by drawing a diagram using activating ( ) and / or inhibiting ( -- ) interactions between S, R and X.

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1 Solution to Homework 5 for Exercise 1. Modeling a coherent feed-forward loop (FFL). (a) Write down a reaction description (e.g. A + B C) of the following system: 1. Synthesis of R is activated by two (independent) catalysts: S and X, at rates k 1, and k 2 respectively. 2. Degradation of R and X takes place spontaneously (no catalyst) at rate k 3 and k 5 respectively. 3. S activates the synthesis of X at rate k 4. Assume 1 st order kinetics in all reactions. S S + R, rate k 1 X X + R, rate k 2 R Ø, rate k 3 S S + X, rate k 4 X Ø, rate k 5 (b) Why is this a coherent FFL? Show by drawing a diagram using activating ( ) and / or inhibiting ( -- ) interactions between S, R and X. Writing the logical diagram for activation ( ) or inhibition ( ), we see that both paths from S to R have the same property: the same number of inhibitory interactions: S X R (c) Write down the equations for the rate of change for R and X, calculate their steady-state concentrations, and determine their stability. R = k 1 S + k 2 X k 3 R X = k 4 S k 5 X Steady state for X means that X =0, resulting in Steady state for R means that R = 0, resulting in R. We determine stability of steady state by calculating Jacobian. We have two variables, R and X, which gives rise to a 2x2 Jacobian matrix: J= 0 τ = -k 3 -k 5 < 0 ; Δ=k 3 k 5 > 0

2 Thus, the steady state point is stable. (d) Draw the signal-response curve (RR as function of S). What is the behavior of R if S goes through consecutive step-wise increases? R S S 0 S 1 S The red solid line indicates the change in S, followed by thee response inn R S towards a new steady state value (dotted line). Similar graph, just now the horizontal axis is time. (e) Is this circuit still a coherent FFLL if X is not catalyzing R with rate k 2. Instead, X is catalyzing the decay of R with rate k 2. Draw a diagram of this system using the approach in (b). S X R Since X now catalyzes the decay of R, it means that the presence of an X moleculee is reducing (= inhibiting) ) the availability of R molecules.

3 (f) With the change described in (g), draw the new signal-response curve. What do you call this type of functional response? The inhibitory interaction between X and R makes this an incoherent FFL. From lecture, we know this circuit acts as a sniffer. Here s an a example for a choice of the rate constants. Exercise 2. Dynamical systems analysis Consider the classical love story of Romeo and Juliet. (a) In this version of the story, s Juliet is a fickle lover; the more m Romeoo loves her, the more she wants to run and hide. However, when this discourages Romeo, Juliet begins to find him strangely attractive. Using R as a measure of Romeo s love (R positive) or hate (R negative) for Juliet, and J vice versa, wee can write the following system: dr dj aj, br dt dt where a and b are positive numbers. Find the steady state of o this system, and classify its stability. Will they live happily ever after? (i) Steady state when R = J = 0. Only possible when R=J= =0, when Romeo and Juliet are indifferent to each other. (ii) Stability analysis by calculating the Jacobian. Note, for linear systems the Jacobian is always equal to the interaction matrix (why?). 0 Jacobian( R, J ) b linear). a is indepe endent of R and J (againn because the system iss 0

4 Eigenvalues: calculate τ and Δ. τ = 0 and Δ=ab. Since both a and b are positive, Δ is positive. Thus, the solution is stable centers (see figure in Strogatz class material ch. 5) where R and J will go in circles around the origin. If the initial point was (R,J)=(0,0), where they were indifferent to each other, they will stay so. (iii) They will not live happily ever after they will both love each other only about 1/4 of the time (when both R and J are positive). The rest of the time, at least 1 will hate the other. (b) Consider the generalized version of this system, with dr dj ar bj, br aj dt dt with the parameters a and b may have either sign. Why can a be described as a measure of cautiousness, and b a measure of responsiveness? Identify the steady state of this system. (i) We can describe a as a measure of cautiousness, since a measures the quality and the amount of response to being in love. Example Romeo. If a is positive, then it qualifies R s response to being in love telling how much more Romeo will fall in love (or hate) for already being in love (or hate, respectively). We can describe b as responsiveness, since it captures how much Romeo will fall in love based on Juliet s feelings, and vice versa. (ii) Steady states: This is a linear system, and the only solution to R =J =0 is (R,J)=(0,0). (c) Analyze the stability of the steady state in (b) if a<0 and b>0 (also called cautious lovers why?). Will they live happily ever after? Hint: find the eigenvectors corresponding to the eigenvalues, if necessary use MatLab (with appropriate choice of test values for a and b). An example of how to calculate eigenvectors is given on page 131 (Example 5.2.1) in the Strogatz text. We analyze stability by calculating the Jacobian: a b Jacobian ( R, J ) b a Now, τ = 2a and Δ=a 2 -b 2. (i) Since a<0, it means that τ < 0, and the system is stable as long as Δ >0 (with possible solutions being either stable node or stable spiral): τ 2-4Δ=4a 2-4a 2 +4b 2 =4b 2 > 0. Hence, the only stable solution is a node (spirals are impossible since λ is a real number, not complex).

5 If Δ < 0, the steady state is unstable (saddle). (ii) If Δ >0, the steady state is stable and the system WILL eventually end up in (R,J)=(0,0). However, in this point Romeo and Juliet are indifferent to each other. The only possibility for happily ever after (assuming we all agree that is both R,J > 0) is if the steady state is unstable (saddle) with the eigenvector associated with the unstable eigenvalue pointing into the positive quadrant: Δ=a 2 -b 2 < 0 => a 2 < b 2 (1) The eigenvalues are: a b. If equation (1) is satisfied, we find that a b 0 and a b 0, and the solution is a saddle where λ + is the unstable eigenvalue. The eigenvector associated with λ + is found through backsubstitution: a ( A I) x b And: (a-λ + ) x 1 + b x 2 =0 Substituting value for λ +, we find: -b x 1 +b x 2 = 0. Hence, we find that x 1 =x 2 b x a x The eigenvector corresponding to the unstable eigendirection x. This is a 1 vector that points along the diagonal into the positive quadrant, independent of the values of a and b. In conclusion: If they start out liking each other (and Δ < 0), they will end up happily ever after. Even if one of them dislikes the other, they will live happily ever after, as long as the amount of dislike is less than the amount of love. (d) Now consider the system in (b) with a>0 and b>0 (so called eager beaver ). What is the stability of the steady state? Will they live happily ever after? (i) If a>0, the steady state found in (c) is always unstable since τ>0. We can have either unstable node or a saddle as solutions. (ii) To answer the question of happily ever after, we again have to calculate the eigenvectors. As before, the two eigenvalues are: a b 1 The eigenvector associated with λ + is as before: x. This time, they ll live 1 happily ever after independent of the sign of Δ, if they have some initial love. Also

6 (same as above): Even if one of them dislikes the other, they will live happily ever after, as long as the amount of dislike is less than the amount of love.