Identifying separated time-scales in stochastic models of reaction - PowerPoint PPT Presentation

Identifying separated time-scales in stochastic models of reaction networks • Formulating Markov models • General balance condition • Reaction networks • Heat shock model • Multiple scales • References • Example: Model of a viral in- fection • Abstract • Selecting scaling exponents An outrageous claim with Hye-Won Kang • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 1

Poisson processes Suppose Y is a unit Poisson process. Then P { Y ( t + ∆ t ) − Y ( t ) > 0 |F Y t } ≈ ∆ t . Let Y λ ( t ) = Y ( λt ). The Y λ is a Poisson process with parameter λ . If F Y λ represents the information obtained by observing Y λ ( s ), for s ≤ t , t t } = P { Y λ ( t +∆ t ) − Y λ ( t ) > 0 } = 1 − e − λ ∆ t ≈ λ ∆ t P { Y λ ( t +∆ t ) − Y λ ( t ) > 0 |F Y λ More generally t } = P { Y λ ( t +∆ t ) − Y λ ( t ) = k } = e − λ ∆ t ( λ ∆ t ) k P { Y λ ( t +∆ t ) − Y λ ( t ) = k |F Y λ k ! Law of large numbers � � � � Y ( Nu ) � � − u N →∞ sup lim � = 0 � N u ≤ u 0 • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 2

Time-change representation for a Markov chain Let Y 1 , . . . , Y m be independent unit Poisson processes. � t m � X ( t ) = X (0) + Y k ( λ k ( X ( s )) ds ) ζ k 0 k =1 (Solve from one jump until the next.) P { X ( t + ∆ t ) − X ( t ) = ζ k |F X t } ≈ λ k ( X ( t ))∆ t p t ( x ) = P { X ( t ) = x } satisfies master (forward) equation m m � � d dtp t ( x ) = λ k ( x − ζ k ) p t ( x − ζ k ) − λ k ( x ) p t ( x ) k =1 k =1 • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 3

Reaction networks We consider a network of reactions involving s 0 chemical species, S 1 , . . . , S s 0 . s 0 s 0 � � ν ′ ν ik S i ⇀ ik S i i =1 i =1 where the ν ik and ν ′ ik are nonnegative integers. the vector whose i th element is ν ik . ν k ν ′ the vector whose i th element is ν ′ ik . k ζ k = ν ′ k − ν k . • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 4

Markov chain models X ( t ) number of molecules of each species in the system at time t . ν k number of molecules of each chemical species consumed in the k th reaction. ν ′ k number of molecules of each species created by the k th reaction. λ k ( x ) rate at which the k th reaction occurs. (The propensity/intensity.) If the k th reaction occurs at time t , the new state becomes X ( t ) = X ( t − ) + ν ′ k − ν k = X ( t − ) + ζ k . The number of times that the k th reaction occurs by time t is given by the counting process satisfying � t R k ( t ) = Y k ( λ k ( X ( s )) ds ) , 0 where the Y k are independent unit Poisson processes. • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 5

Equations for the system state The state of the system satisfies � R k ( t )( ν ′ k − ν k ) X ( t ) = X (0) + k � t � = X (0) + Y k ( λ k ( X ( s )) ds ) ζ k 0 k For a binary reaction S 1 + S 2 ⇀ S 3 or S 1 + S 2 ⇀ S 3 + S 4 λ k ( x ) = κ ′ k x 1 x 2 For S 1 ⇀ S 2 or S 1 ⇀ S 2 + S 3 , λ k ( x ) = κ ′ k x 1 . For 2 S 1 ⇀ S 2 , λ k ( x ) = κ ′ k x 1 ( x 1 − 1) . • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 6

Multiple scales Fix N 0 >> 1. For each species i , define the normalized abundances (or simply, the abundances) by Z i ( t ) = N − α i X i ( t ) , 0 where α i ≥ 0 should be selected so that Z i = O (1). Note that the abun- dance may be the species number ( α i = 0) or the species concentration or something else. The rate constants may also vary over several orders of magnitude so k = κ k N β k scale the rate constants κ ′ 0 . Then k x i x j = N β k + α i + α j κ ′ κ ′ k x i ( x i − 1) = N β k +2 α i κ k z i ( z i − N − α i ) κ k z i z j 0 0 0 k x i = κ k N β k + α i κ ′ z i 0 Note that the exponent on N 0 is ρ k = β k + α · ν k . • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 7

A parameterized family of models Then, noting that ν k · α = � i ν ik α i , � t � N − α i N β k + ν k · α λ k ( Z ( s )) ds )( ν ′ Z i ( t ) = Z i (0) + Y k ( ik − ν ik ) . 0 0 0 k Let � t � Z N N − α i Y k ( N β k + ν k · α λ k ( Z N ( s )) ds )( ν ′ i ( t ) = Z i (0) + ik − ν ik ) . 0 k Then the “true” model is Z = Z N 0 . • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 8

Time-scale parameter Let Z N,γ ( t ) ≡ Z N i ( tN γ ) i � t � N − α i Y k ( N γ + β k + ν k · α λ k ( Z N,γ ( s )) ds ) ζ ik . = Z i (0) + 0 k Equation is “balanced” if max { β k + ν k · α : ζ ik > 0 } = max { β k + ν k · α : ζ ik < 0 } If the equation is not balanced then we need γ + β k + ν k · α ≤ α i (1) for all i such that ζ ik � = 0. The time-scale of species i : γ i = α i − max { β k + ν k · α : ζ ik � = 0 } • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 9

Example: Model of a viral infection Srivastava, You, Summers, and Yin (2002), Haseltine and Rawlings (2002), Ball, Kurtz, Popovic, and Rempala (2006) Three time-varying species, the viral template, the viral genome, and the viral structural protein (indexed, 1, 2, 3 respectively). The model involves six reactions, κ ′ S 1 + stuff ⇀ S 1 + S 2 1 κ ′ 2 S 2 ⇀ S 1 κ ′ 3 S 1 + stuff ⇀ S 1 + S 3 κ ′ ⇀ ∅ 4 S 1 κ ′ ⇀ ∅ 5 S 3 κ ′ 6 S 2 + S 3 ⇀ S 4 • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 10

Stochastic system � t � t κ ′ κ ′ 2 X 2 ( s ) ds ) − Y 4 ( X 1 ( t ) = X 1 (0) + Y 2 ( 4 X 1 ( s ) ds ) 0 0 � t � t κ ′ κ ′ X 2 ( t ) = X 2 (0) + Y 1 ( 1 X 1 ( s ) ds ) − Y 2 ( 2 X 2 ( s ) ds ) 0 0 � t κ ′ − Y 6 ( 6 X 2 ( s ) X 3 ( s ) ds ) 0 � t � t κ ′ κ ′ 3 X 1 ( s ) ds ) − Y 5 ( X 3 ( t ) = X 3 (0) + Y 3 ( 5 X 3 ( s ) ds ) 0 0 � t κ ′ − Y 6 ( 6 X 2 ( s ) X 3 ( s ) ds ) 0 κ ′ κ ′ 1 0 . 25 1 4 κ ′ κ ′ 0 . 025 2 2 5 κ ′ κ ′ 7 . 5 × 10 − 6 1000 3 6 • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 11

Figure 1: Simulation (Haseltine and Rawlings 2002) • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 12

Balance equations for the viral model � t � t Z N Z 1 (0) + N − α 1 Y 2 ( κ 2 N β 2 + α 2 Z N 2 ( s ) ds ) − N − α 1 Y 4 ( κ 4 N β 4 + α 1 Z N 1 ( t ) = 1 ( s ) ds ) 0 0 � t � t Z N Z 2 (0) + N − α 2 Y 1 ( κ 1 N β 1 + α 1 Z N 1 ( s ) ds ) − N − α 2 Y 2 ( κ 2 N β 2 + α 2 Z N 2 ( t ) = 2 ( s ) ds ) 0 0 � t − N − α 2 Y 6 ( κ 6 N β 6 + α 2 + α 3 Z N 2 ( s ) Z N 3 ( s ) ds ) 0 � t � t Z N Z 3 (0) + N − α 3 Y 3 ( κ 3 N β 3 + α 1 Z N 1 ( s ) ds ) − N − α 3 Y 5 ( κ 5 N β 5 + α 3 Z N 3 ( t ) = 3 ( s ) ds ) 0 0 � t − N − α 3 Y 6 ( κ 6 N β 6 + α 2 + α 3 Z N 2 ( s ) Z N 3 ( s ) ds ) 0 β 2 + α 2 = β 4 + α 1 β 1 + α 1 = ( β 2 + α 2 ) ∨ ( β 6 + α 2 + α 3 ) β 3 + α 1 = ( β 5 + α 3 ) ∨ ( β 6 + α 2 + α 3 ) β 3 ≥ β 5 ≥ β 1 ≥ β 4 ≥ β 2 ≥ β 6 • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 13

An example β 2 + α 2 = β 4 + α 1 β 1 + α 1 = ( β 2 + α 2 ) ∨ ( β 6 + α 2 + α 3 ) β 3 + α 1 = ( β 5 + α 3 ) ∨ ( β 6 + α 2 + α 3 ) β 3 ≥ β 5 ≥ β 1 ≥ β 4 ≥ β 2 ≥ β 6 β 1 0 α 1 0 − 2 2 β 2 α 2 3 3 β 3 1 α 3 1 β 4 0 γ 1 0 2 0 β 5 γ 2 3 − 5 0 β 6 γ 3 3 • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 14

Scaling parameters Ball et al. (2006) Each X i is scaled according to its abundance in the system. 0 ), X 2 = O ( N 2 / 3 For N 0 = 1000, X 1 = O ( N 0 ), and X 3 = O ( N 0 ) and we 0 take Z 1 = X 1 , Z 2 = X 2 N − 2 / 3 , and Z 3 = X 3 N − 1 0 . 0 Expressing the rate constants in terms of N 0 = 1000 κ ′ 1 1 1 2 . 5 N − 2 / 3 κ ′ 0.025 2 0 κ ′ 1000 N 0 3 κ ′ 0.25 . 25 4 κ ′ 2 2 5 . 75 N − 5 / 3 κ ′ 7 . 5 × 10 − 6 6 0 • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 15

Normalized system With the scaled rate constants and abundances, we have � t � t Z N 1 ( t ) = Z N 2 . 5 Z N . 25 Z N 1 (0) + Y 2 ( 2 ( s ) ds ) − Y 4 ( 1 ( s ) ds ) 0 0 � t � t Z N 2 ( t ) = Z N 2 (0) + N − 2 / 3 Y 1 ( Z N 1 ( s ) ds ) − N − 2 / 3 Y 2 ( 2 . 5 Z N 2 ( s ) ds ) 0 0 � t − N − 2 / 3 Y 6 ( . 75 Z N 2 ( s ) Z N 3 ( s ) ds ) 0 � t � t 3 (0) + N − 1 Y 3 ( 1 ( s ) ds ) − N − 1 Y 5 ( Z N 3 ( t ) = Z N NZ N 2 NZ N 3 ( s ) ds ) 0 0 � t − N − 1 Y 6 ( . 75 Z N 2 ( s ) Z N 3 ( s ) ds ) , 0 • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 16

Limiting system Passing to the limit, we have � t � t Z 1 ( t ) = Z 1 (0) + Y 2 ( 2 . 5 Z 2 ( s ) ds ) − Y 4 ( . 25 Z 1 ( s ) ds ) 0 0 Z 2 ( t ) = Z 2 (0) � t � t Z 3 ( t ) = Z 3 (0) + Z 1 ( s ) ds − 2 Z 3 ( s ) ds 0 0 • First • Prev • Next • Go To • Go Back • Full Screen • Close • Quit 17