Stochastic analysis of biochemical reaction networks with absolute concentration robustness

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1 Stochastic analysis of biochemical reaction networks with absolute concentration robustness David F. Anderson Department of Mathematics University of Wisconsin - Madison Boston University Probability/Statistics seminar Nov. 21st, 2013

2 Big picture Biochemical/population networks can range from simple to very complex. Example 1: A + B C. Example 2: A + B 2B B A Example 3: Gene transcription & translation: G κ 1 G + M M κ 2 M + P M κ 3 P κ 4 G + P κ 5 κ 5 B transcription translation degradation degradation Binding/unbinding of Gene Cartoon representation: 1 1 J. Paulsson, Physics of Life Reviews, 2,

3 Big picture Example 4: EnvZ/OmpR signaling system 2 2 Guy Shinar and Martin Feinberg, Structural Sources of Robustness in Biochemical Reaction Networks, Science, 2010

4 Big picture Hanahan and Weinberg, The Hallmarks of Cancer, Cell, 2000.!

5 Big picture Metabolic Pathways Roche Applied Science!

6 Big picture For complex models, simulation is often used to explore the possible dynamics (for both deterministic and stochastic models). One problem: key system parameters are oftentimes unknown, or known only up to an order of magnitude. Want an alternative approach: discover what pieces of the network architecture determine overall system behavior. Best if results do not depend upon particular choices of rate constants (which are never really known). how can we cut through the bewildering complexity of biological systems and say something sensible? Mathematics. This story is far from complete really just beginning.

7 Big picture Will tell two stories: 1. ( A., Craciun, Kurtz) Provide network conditions that guarantee both an especially stable deterministic model and an especially stable stochastic model. 2. ( A., Enciso, Johnston) Provide network conditions that guarantee deterministic model has component which is Absolutely robust, stochastic model has extinction event! This research is part of chemical reaction network theory, which is part of systems biology. A network property called deficiency will play a key role in the analysis.

8 Networks: {S, C, R} Example: A B S = {A, B}. C = {A, B}. R = {A B}. Example: A + B 2B B A S = {A, B}. C = {A + B, 2B, B, A}. R = {A + B 2B, B A}.

9 Networks: {S, C, R} Example: E + S ES E + P, E S S = {E, S, ES, P}. C = {E + S, ES, E + P, E,, S}. R = {E + S ES, ES E + S,... }.

10 Dynamics: stochastic Example: A + B α 2B B β A (R1) (R2) X(t) = ([ 0 X(0) + R 1 (t) 2 = [ 1 X(0) + R 1 (t) 1 For Markov models can take R 1 (t) = Y 1 (α R 2 (t) = Y 2 (β ] [ 1 1 ] + R 2 (t) t 0 t 0 ]) ([ 1 + R 2 (t) 0 [ 1 1 ]. ) X A (s)x B (s)ds ) X B (s)ds where Y 1, Y 2 are independent unit-rate Poisson processes. ] [ 0 1 ])

11 Dynamics: stochastic For general system, we have S = {X 1,..., X d }, with R : d y ki X i i=1 κ k d y kix i or y k y k i=1 and complexes found at either end of reaction arrow: C = {y k, y k}. The intensity (or propensity) of kth reaction is λ k : Z d 0 R. By analogy with before: X(t) = X(0) + k R k (t)(y k y k ), with X(t) = X(0) + k Y k ( t Y k are independent, unit-rate Poisson processes. 0 ) λ k (X(s))ds (y k y k ),

12 Mass-action kinetics The standard intensity function chosen is mass-action kinetics: λ k (x) = κ k ( ( ) x x i! y ik!) = κ k y k (x i y ik )!. i i Example: If S 1 anything, then λ k (x) = κ k x 1. Example: If S 1 + S 2 anything, then λ k (x) = κ k x 1 x 2. Example: If 2S 2 anything, then λ k (x) = κ k x 2 (x 2 1).

13 Other ways to understand model We can also describe the model as a continuous time Markov chain with infinitesimal generator Af (x) = k λ k (x)(f (x + ζ k ) f (x)). where ζ k = y k y k. Dynkin s formula (See Ethier and Kurtz, 1986, Ch. 1) yields Ef (X(t)) f (X 0 ) = E t 0 (Af )(X(s))ds, Letting f (y) = 1 x(y) above so that E[f (X(t))] = P{X(t) = x} = p x(t), gives Kolmogorov s forward equation (chemical master equation) p t (x) = k λ(x ζ k )p t(x ζ k ) p t(x) k λ k (x)

14 Morris-Lecar Voltage satisfying dv dt = f (V (t), N(t)) = 1 C ( Iapp g Ca m (V (t))(v (t) V Ca ) V K ) g L (V V L ) g o K N(t)(V (t) ). with number of open potassium channels following Open α(v (t)) Closed β(v (t))

15 Morris-Lecar Voltage satisfying dv dt = 1 C ( Iapp g Ca m (V (t))(v (t) V Ca ) V K ) g L (V V L ) g o K N(t)(V (t) ). with number of open potassium channels following Open α(v (t)) Closed β(v (t)) N(t) = N(0) Y close ( t 0 ) ( t ) β(v (s))n(s)ds + Y open α(v (s))(n tot N(s))ds. 0

16 Dynamics: deterministic model Assuming V is a scaling parameter (volume times Avogadro s number), X i = O(V ), and X V (t) = def X(t)/V, λ k (X(t)) V ( κ k X V (t) y k ), Then, X V (t) 1 V X 0 + k LLN for Y k says 1 V Y k(vu) u so a good approximation is solution to ( 1 t ) V Y k V κ k X V (s) y k ds (y k y k ) 0 x(t) = x(0) + k t 0 κ k x(s) y k ds (y k y k ), where is standard mass-action kinetics. u v = u v 1 1 uv d d,

17 Dynamics. Example A + B α 2B B β A (R1) (R2) Stochastic equations X(t) = X(0) + Y 1 (α t 0 ) [ 1 X A (s)x B (s)ds 1 ] ( t ) [ 1 + Y 2 β X B (s)ds 0 1 ]. Deterministic equations or or x(t) = x(0) + α t 0 [ 1 x A (s)x B (s)ds 1 ẋ(t) = αx ax b [ 1 1 ] t [ 1 + β x B (s)ds 0 1 ] + βx B [ 1 1 ẋ A (t) = αx A x B + βx B ẋ B (t) = αx ax B βx B. ] ],

18 Story 1: special stability for both models Network reversibility conditions Definition Let S = {S i }, C = {y k }, and R = {y k y k} denote the sets of species, complexes, and reactions, respectively. The triple {S, C, R} is called the chemical reaction network. Definition Let G be the directed graph with nodes given by the complexes C and directed edges given by the reactions R = {y k y k}, and let G 1,..., G l denote the connected components of G. {G j } are the linkage classes of the reaction network. Example A + B α 2B B β A (R1) (R2) Has two linkage classes.

19 Network reversibility conditions Definition A chemical reaction network, {S, C, R}, is called weakly reversible if each linkage class is strongly connected. A network is called reversible if y k y k R whenever y k y k R. Intuition for probabilists: If the network is weakly reversible, then, thinking of the complexes as states of a Markov chain, the linkage classes are the irreducible communicating equivalence classes of classical Markov chain theory. BUT, these equivalence classes do not correspond to the communicating equivalence classes of the Markov chain model of the reaction network.

20 Network properties Definition S = span {yk y k R}{y k y k } is the stoichiometric subspace of the network. For c R d we say c + S and (c + S) R d >0 are the stoichiometric compatibility classes and positive stoichiometric compatibility classes of the network, respectively. Denote dim(s) = s. Implication: Solutions bound to stoichiometric compatibility class. 1. Deterministic: x(t) x(0) c + S for all t Stochastic: communication classes contained within c + S.

21 Network properties Definition The deficiency of a chemical reaction network, {S, C, R}, is δ = C l s, where C is the number of complexes, l is the number of linkage classes of the network graph, and s is the dimension of the stoichiometric subspace of the network. Lemma (Feinberg) The deficiency of a network is nonnegative. Example A + B 2B B A (R1) (R2) n = 4, l = 2, s = 1 = δ = 1. But, A + B C B A (R1) (R2) n = 4, l = 2, s = 2 = δ = 0.

22 Deficiency zero theorem of Feinberg Theorem (The Deficiency Zero Theorem - Deterministic) Let {S, C, R} be a weakly reversible, deficiency zero chemical reaction network with mass action kinetics. Then, for any choice of rate constants κ k, within each positive stoichiometric compatibility class there is precisely one equilibrium value c, κ k c y k (y k y k ) = 0, k and that equilibrium value is locally asymptotically stable relative to its compatibility class. Actually have stronger result: for each η C, κ k c y k = k:y k =η c is said to be a complex balanced equilibrium. k:y k =η κ k c yk. (1)

23 Deficiency Zero Theorem - stochastic Theorem (A., Craciun, Kurtz, 2010) Let {S, C, R} be a chemical reaction network with rate constants κ k. Suppose that the deterministic system is complex balanced with equilibrium c R m >0. Then, for any irreducible communicating equivalence class, Γ, the stochastic system has a product form stationary measure π(x) = M where M is a normalizing constant. m i=1 c x i i, x Γ, (2) x i! Useful for stochastic averaging.

24 Enzyme kinetics Consider the possible enzyme kinetics given by E + S ES E + P, E S Not too difficult to see that Γ = Z 4 0. Therefore, in distributional equilibrium the specie numbers are independent and have Poisson distributions.

25 Non-independence Consider the simple reversible system S 1 κ 1 κ 2 S 2. Suppose that X 1 (0) + X 2 (0) = N = X 1 (t) + X 2 (t) = N for all t. δ = n l s = = 0. An equilibrium to the system that satisfies the complex balance equation is ( ) κ2 κ 1 c =,, κ 1 + κ 2 κ 1 + κ 2 and the product form stationary distribution for the system is π(x) = M cx 1 1 c x 2 2 x 1! x 2!, x Γ.

26 Non-independence Using that X 1 (t) + X 2 (t) = N for all t yields π 1 (x 1 ) = M cx 1 1 c N x 1 2 x 1! (N x 1 )! = M x 1!(N x 1 )! cx 1 1 (1 c 1) N x 1. After setting M = N!, we see that X 1 is binomially distributed. Similarly, π 2 (x 2 ) = ( N x 2 ) c x 2 2 (1 c 2) N x 2.

27 Enzyme kinetics Consider the slightly different enzyme kinetics given by E + S ES E + P, E 0 We see S + ES + P = N. In distributional equilibrium E has Poisson distribution, S, ES, P have a multinomial distribution, and E is independent from S, ES, and P.

28 Story 2: Absolute Concentration Robustness Guy Shinar and Martin Feinberg, Structural Sources of Robustness in Biochemical Reaction Networks, Science, AIM:

29 Story 2: Absolute Concentration Robustness A + B α 2B B β A (R1) (R2) ẋ A (t) = αx A (t)x B (t) + βx B (t) ẋ B (t) = αx A (t)x B (t) βx B (t) M = def x A (0) + x B (0), Solving for equilibria: x A = β/α, x B = M β/α, Network has absolute concentration robustness in species A.

30 X tot := X + XD + XT + X p + X py + XDY p Y tot := Y + X py + XDY p + Y p.

31 Differing in one species We say that two complexes, y 1, y 2, differ in species S i if for some α > 0. y 1 = y 2 + αe i Examples: 1. differ in species B. A, A + B 2. differ in species Y p. XT, XT + Y p

32 Terminal and non-terminal complexes Viewing complexes as states of a Markov chain, we say two complexes are strongly linked if they are part of the same communication class (of this Markov chain). the communication classes are called strong linkage classes. A strong linkage class is called terminal if the communication class is recurrent. Transient complexes are called non-terminal. k 1 XD X XT X p k 2 [ D] k 6 k 3 k4 X p +Y X p Y X+Y p k 7 k 8 k 9 [ T] XD+Y p XDY p XD+Y k 10 k 11 k 5

33 Conservative system We say there is a global conservation relation if there is a w R d >0 for which w (y k y k ) = 0. Example A + B 2B B A. has a conservation relation of w = [ 1 1 ]. since [ 1 1 [ 1 1 ] [ 1 1 ] [ 1 1 ] = 0 ] = 0.

34 Theorems Theorem (Marty Feinberg and Guy Shinar, Science, 2010 deterministic) Consider a deterministic mass-action system that has a deficiency of one. admits a positive steady state and has two non-terminal complexes that differ only in species S, then the system has absolute concentration robustness in S. Theorem (Anderson, Enciso, Johnston stochastic) Consider a reaction network satisfying the following: has a deficiency of one, the deterministic model admits a positive steady state, has two non-terminal complexes that differ only in species S, (new) is conservative, then for every positive recurrent state of the Markov model the intensity of each non-terminal complex is zero.

35 Do we need a conservation relation? - yes Example A + B α B β A + 2B, have n = 4, l = 2, s = 1 = δ = 1. No conservation relation (but x B x A is conserved), stable fixed point for ODE (if x B (0) > x A (0)) can never run out of B. x A = β α, x B = M + β α,

36 Flavor of proof Consider ẋ = κ k x y k (y k y k ) = Y A κ Ψ(x) k:y k y k Consider the vector space on R C with basis ω y = 1 in yth component. Define Ψ : R d 0 R C by Ψ(x) y = x y = d i=1 x y i i. Consider the mapping A κ : R C R C given by A κ(ψ) = κ k (ω y k ω yk ) Ψ y, y C k:y k =y so that A κ Ψ(x) = y C κ k x y k (ω y k ω yk ) k:y k =y Now define Y : R C R d to be linear mapping with Y (ω y) = y = Y (A κ(ψ(x))) = ẋ

37 Flavor of proof ẋ = f (x) = Y A κ Ψ(x). Note that x is an equilibrium value if Ψ(x) ker(a κ) or Ψ(x) ker(y A κ). So what is A κ? A quick (and easy) calculation shows { A ij κ j i for j i κ = l j κ j l for i = j A T κ is the infinitesimal generator for the complex space CTMC! We know all about the kernel (they correspond to stationary distributions).

38 Flavor of proof ẋ = f (x) = Y A κ Ψ(x). Lemma Let (S, C, R) denote a chemical reaction network with terminal strong linkage classes Λ 1,..., Λ t. Then, for any κ R r >0, ker(a κ) has a basis {b 1,..., b t} where supp(b θ ) = Λ θ for all θ = 1,..., t. Lemma Let (S, C, R) denote a chemical reaction network with deficiency δ and t terminal strong linkage classes. Then dim(ker(ya κ)) δ + t. (3) assumption equilibrium for the ODE model tells ker(y A κ). Not weakly reversible, so can t be linear combination of b i. Deficiency is one, so second lemma tells us that the kernel looks like where c charges all complexes. {c, b 1,..., b t},

39 Flavor of proof Recap: ker(y A κ) has basis {c, b 1,..., b t}, where c charges all complexes supp(b θ ) = Λ θ for all θ = 1,..., t. Rest of proof: consider a modified network with stochastic kinetics, prove that that system satisfies conclusion of theorem by using kernel condition, finally argue the original system does too.

40 An example A + B α 2B B β A (R1) (R2)

41 An example A + B α 2B B β A 2A B + C. (R1) (R2) (R3) Note C = 6, l = 3, s = 2 = δ = 1. w = (1, 1, 1) is conservation relation.

42 flavor of proof Only work with A + B reduced network. Call it {S, C, R }. Call process X. Suppose there is a positive recurrent state X 0 which charges a non-terminal complex. We want to reach a contradiction (with kernel condition). We denote the Nth return time to X 0 as t N. It follows that we have X (t N ) = X (0) for all N so that X (t N ) = X (0) + k ( tn ) Y k κ i (X (s)) y k ds (y k y k ) = X (0), (4) 0 and so ( tn ) Y k κ i (X (s)) y k ds (y k y k ) = 0, (5) k 0 Relatively easy to show that for each y k tn 0 (X (s)) y k ds, as N,

43 flavor of proof and ( tn ) Y k κ i (X (s)) y k ds (y k y k ) = 0, k 0 tn 0 (X (s)) y k ds, as N, multiplying and dividing by appropriate terms, 0 = [ ( 1 tn ) κ k tn Y k κ k (X (s)) y k ds 1 ] tn (X (s)) y k ds (y k κ k 0 (X (s)) y k k ds 0 t y k ). N 0 We have 1 lim N t N tn 0 (X (s)) y k ds = X I π I(X)X y k, (6) Define the vector G π R C via [G π] y := X I π I(X)X y, (7) for y C, and note that [G π] y > 0 for all remaining y. Can then argue that [G π] is in the kernel of Y A κ, which it can not be (contradiction).

44 Generalization Theorem Let (S, C, R) be a conservative chemical reaction network for which the following assumptions hold: 1. There are non-empty sets of non-terminal complexes C and C such that, if y C then there exists a y C such that y y. 2. For some choice of rate constants {k i } r i=1, the following property holds: if Ψ ker(ya k ) and Ψ y = 0 for all y C, then Ψ ȳ = 0 for all non-terminal ȳ. Then, for any choice of stoichiometrically admissible kinetics, all non-terminal complexes of the network are off at each positive recurrent state of the stochastically modeled system. Remark: The second condition in assumption 2 above may also be formulated in the following way, which may be more intuitive to some readers. 2. For some choice of rate constants {k i } r i=1, the following property holds: if Ψ ker(ya k ) has support on a non-terminal complex, then Ψ has support on some y C.

45 Generalization, example XD k 1 X k 3[T ] XT k 5 X p k 2 [D] k 4 X p + Y k 6 X py k 8 X + Y p k 7 k 9 k XD + Y p XDY p 11 XD + Y k 10 (8) Deficiency is two. k 12 k XT + Y p XTY p 14 XT + Yp. k 13 XD + Y p XD, and XT + Y p XT. Ordering non-terminal complexes as {XD, X, XT, X p + Y, X py, XD + Y p, XDY p, XT + Y p, XTY p}. (9) there are two basis vectors of YA k with non-zero support on the complexes as ordered in (9); {[2, 2, 1, 2, 1, 2, 1, 0, 0], [2, 2, 1, 2, 1, 0, 0, 2, 1]}.

46 Quasi-stationary distribution A + B α 2B B β A X A (0) + X B (0) = M, Find π Q M so that for τ absorption time and x transient states, Satisfies lim Pν(X(t) = x τ > t) = t πq M(x). πm(x) Q = P π Q (X(t) = x τ > t). M Can show that quasi-stationary distribution for A converges to Poisson πm(x) Q e (β/α) (β/α) x, as M. x!

47 Quasi-stationary distribution: EnvZ-OmpR signaling system 3 3 Guy Shinar and Martin Feinberg, Structural Sources of Robustness in Biochemical Reaction Networks, Science, 2010

48 Quasi-stationary distribution Quasi stationary probabilities X tot = 100 Y tot = 3500 X tot = 1000 Y tot = X tot = Y tot = Poisson Molecules of Y p

49 That is the story. Thanks! Collaborators: Matthew Johnston, Germán Enciso, Gheorghe Craciun, and Tom Kurtz References: 1. Guy Shinar and Martin Feinberg, Structural Sources of Robustness in Biochemical Reaction Networks, Science, David F. Anderson, Germán Enciso, and Matthew Johnston, Stochastic analysis of biochemical reaction networks with absolute concentration robustness, submitted. 3. David F. Anderson, Gheorghe Craciun, Thomas G. Kurtz, Product-form stationary distributions for deficiency zero chemical reaction networks, Bulletin of Mathematical Biology, Vol. 72, No. 8, , Funding: NSF-DMS , and NSF-DMS