Foundations of Chemical Kinetics Lecture 23: The chemical master - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 23: The chemical master equation: Stationary distributions Marc R. Roussel Department of Chemistry and Biochemistry

The stationary distribution ◮ In ordinary chemical kinetics, a closed chemical system has a unique equilibrium composition, usually given as a set of concentrations.

The stationary distribution ◮ In ordinary chemical kinetics, a closed chemical system has a unique equilibrium composition, usually given as a set of concentrations. ◮ In stochastic kinetics, we have a stationary (i.e. equilibrium) probability distribution.

The stationary distribution ◮ In ordinary chemical kinetics, a closed chemical system has a unique equilibrium composition, usually given as a set of concentrations. ◮ In stochastic kinetics, we have a stationary (i.e. equilibrium) probability distribution. ◮ The stationary distribution is calculated from the chemical master equation by setting dP ( N ) / dt = 0 and requiring � P ( N ) = 1.

The stationary distribution (continued) ◮ Recall the chemical master equation: dP ( N , t ) � � = a r ( N − ν r ) P ( N − ν r , t ) − a r ( N ) P ( N , t ) dt r ∈R r ∈R � [ a r ( N − ν r ) P ( N − ν r , t ) − a r ( N ) P ( N , t )] = r ∈R

The stationary distribution (continued) ◮ Recall the chemical master equation: dP ( N , t ) � � = a r ( N − ν r ) P ( N − ν r , t ) − a r ( N ) P ( N , t ) dt r ∈R r ∈R � [ a r ( N − ν r ) P ( N − ν r , t ) − a r ( N ) P ( N , t )] = r ∈R ◮ Suppose that we split R into two sets: the set of “forward” reactions R + , and the set of “reverse” reactions R − , and that the reverse of every reaction in R + is included in R − . Let a (+) be the propensity for forward reaction r , and a ( − ) be r r the propensity for the corresponding reverse reaction. ν r is the stoichiometry vector for the forward reaction.

The stationary distribution (continued) ◮ Then we can write the CME as

The stationary distribution (continued) ◮ Then we can write the CME as dP ( N , t ) � � a (+) ( N − ν r ) P ( N − ν r , t ) − a (+) � = ( N ) P ( N , t ) r r dt r ∈R + � a ( − ) ( N + ν r ) P ( N + ν r , t ) − a ( − ) � � + ( N ) P ( N , t ) r r r ∈R −

The stationary distribution (continued) ◮ Then we can write the CME as dP ( N , t ) � � a (+) ( N − ν r ) P ( N − ν r , t ) − a (+) � = ( N ) P ( N , t ) r r dt r ∈R + � a ( − ) ( N + ν r ) P ( N + ν r , t ) − a ( − ) � � + ( N ) P ( N , t ) r r r ∈R − �� a (+) ( N − ν r ) P ( N − ν r , t ) − a ( − ) � � = ( N ) P ( N , t ) r r r ∈R + � �� a ( − ) ( N + ν r ) P ( N + ν r , t ) − a (+) + ( N ) P ( N , t ) r r

The stationary distribution (continued) ◮ Then we can write the CME as dP ( N , t ) � � a (+) ( N − ν r ) P ( N − ν r , t ) − a (+) � = ( N ) P ( N , t ) r r dt r ∈R + � a ( − ) ( N + ν r ) P ( N + ν r , t ) − a ( − ) � � + ( N ) P ( N , t ) r r r ∈R − �� a (+) ( N − ν r ) P ( N − ν r , t ) − a ( − ) � � = ( N ) P ( N , t ) r r r ∈R + � �� a ( − ) ( N + ν r ) P ( N + ν r , t ) − a (+) + ( N ) P ( N , t ) r r ◮ For a detailed balanced solution, each bracketed pair of terms would be zero.

Example: Stationary distribution for the Michaelis-Menten mechanism κ 1 κ 2 − − ⇀ − − ⇀ E + S ↽ κ − 1 C − − ↽ κ − 2 E + P − −

Example: Stationary distribution for the Michaelis-Menten mechanism κ 1 κ 2 − − ⇀ − − ⇀ E + S ↽ κ − 1 C − − ↽ κ − 2 E + P − − Variable ordering: ( N C , N E , N P , N S )

Example: Stationary distribution for the Michaelis-Menten mechanism κ 1 κ 2 − − ⇀ − − ⇀ E + S ↽ κ − 1 C − − ↽ κ − 2 E + P − − Variable ordering: ( N C , N E , N P , N S ) dP ( N , t ) = κ 1 ( N E + 1)( N S + 1) P ( N C − 1 , N E + 1 , N P , N S + 1) dt + κ − 1 ( N C + 1) P ( N C + 1 , N E − 1 , N P , N S − 1) + κ 2 ( N C + 1) P ( N C + 1 , N E − 1 , N P − 1 , N S ) + κ − 2 ( N E + 1)( N P + 1) P ( N C − 1 , N E + 1 , N P + 1 , N S ) − P ( N C , N E , N P , N S ) [ κ 1 N E N S + κ − 1 N C + κ 2 N C + κ − 2 N E N P ]

Example: Stationary distribution for the Michaelis-Menten mechanism κ 1 κ 2 − − ⇀ − − ⇀ E + S ↽ κ − 1 C − − ↽ κ − 2 E + P − − Variable ordering: ( N C , N E , N P , N S ) dP ( N , t ) = κ 1 ( N E + 1)( N S + 1) P ( N C − 1 , N E + 1 , N P , N S + 1) dt + κ − 1 ( N C + 1) P ( N C + 1 , N E − 1 , N P , N S − 1) + κ 2 ( N C + 1) P ( N C + 1 , N E − 1 , N P − 1 , N S ) + κ − 2 ( N E + 1)( N P + 1) P ( N C − 1 , N E + 1 , N P + 1 , N S ) − P ( N C , N E , N P , N S ) [ κ 1 N E N S + κ − 1 N C + κ 2 N C + κ − 2 N E N P ]

Example: Stationary distribution for the Michaelis-Menten mechanism κ 1 κ 2 − − ⇀ − − ⇀ E + S ↽ κ − 1 C − − ↽ κ − 2 E + P − − Variable ordering: ( N C , N E , N P , N S ) dP ( N , t ) = κ 1 ( N E + 1)( N S + 1) P ( N C − 1 , N E + 1 , N P , N S + 1) dt + κ − 1 ( N C + 1) P ( N C + 1 , N E − 1 , N P , N S − 1) + κ 2 ( N C + 1) P ( N C + 1 , N E − 1 , N P − 1 , N S ) + κ − 2 ( N E + 1)( N P + 1) P ( N C − 1 , N E + 1 , N P + 1 , N S ) − P ( N C , N E , N P , N S ) [ κ 1 N E N S + κ − 1 N C + κ 2 N C + κ − 2 N E N P ]

Example: Stationary distribution for the Michaelis-Menten mechanism κ 1 κ 2 − − ⇀ − − ⇀ E + S ↽ κ − 1 C − − ↽ κ − 2 E + P − − Variable ordering: ( N C , N E , N P , N S ) dP ( N , t ) = κ 1 ( N E + 1)( N S + 1) P ( N C − 1 , N E + 1 , N P , N S + 1) dt + κ − 1 ( N C + 1) P ( N C + 1 , N E − 1 , N P , N S − 1) + κ 2 ( N C + 1) P ( N C + 1 , N E − 1 , N P − 1 , N S ) + κ − 2 ( N E + 1)( N P + 1) P ( N C − 1 , N E + 1 , N P + 1 , N S ) − P ( N C , N E , N P , N S ) [ κ 1 N E N S + κ − 1 N C + κ 2 N C + κ − 2 N E N P ]

Example: Stationary distribution for the Michaelis-Menten mechanism κ 1 κ 2 − − ⇀ − − ⇀ E + S ↽ κ − 1 C − − ↽ κ − 2 E + P − − Variable ordering: ( N C , N E , N P , N S ) dP ( N , t ) = κ 1 ( N E + 1)( N S + 1) P ( N C − 1 , N E + 1 , N P , N S + 1) dt + κ − 1 ( N C + 1) P ( N C + 1 , N E − 1 , N P , N S − 1) + κ 2 ( N C + 1) P ( N C + 1 , N E − 1 , N P − 1 , N S ) + κ − 2 ( N E + 1)( N P + 1) P ( N C − 1 , N E + 1 , N P + 1 , N S ) − P ( N C , N E , N P , N S ) [ κ 1 N E N S + κ − 1 N C + κ 2 N C + κ − 2 N E N P ]

Example: Stationary distribution for the Michaelis-Menten mechanism (continued) ◮ For example, suppose that we start out with one enzyme molecule and three substrate molecules, and take κ 1 = 10, κ − 1 = 1, κ 2 = 5 and κ − 2 = 2 s − 1 .

Example: Stationary distribution for the Michaelis-Menten mechanism (continued) ◮ For example, suppose that we start out with one enzyme molecule and three substrate molecules, and take κ 1 = 10, κ − 1 = 1, κ 2 = 5 and κ − 2 = 2 s − 1 . dP (0 , 1 , 0 , 3) / dt = P (1 , 0 , 0 , 2) − 30 P (0 , 1 , 0 , 3) dP (1 , 0 , 0 , 2) / dt = 30 P (0 , 1 , 0 , 3) + 2 P (0 , 1 , 1 , 2) − 6 P (1 , 0 , 0 , 2) dP (0 , 1 , 1 , 2) / dt = P (1 , 0 , 1 , 1) + 5 P (1 , 0 , 0 , 2) − 22 P (0 , 1 , 1 , 2) dP (1 , 0 , 1 , 1) / dt = 20 P (0 , 1 , 1 , 2) + 4 P (0 , 1 , 2 , 1) − 6 P (1 , 0 , 1 , 1) dP (0 , 1 , 2 , 1) / dt = P (1 , 0 , 2 , 0) + 5 P (1 , 0 , 1 , 1) − 14 P (0 , 1 , 2 , 1) dP (1 , 0 , 2 , 0) / dt = 10 P (0 , 1 , 2 , 1) + 6 P (0 , 1 , 3 , 0) − 6 P (1 , 0 , 2 , 0) dP (0 , 1 , 3 , 0) / dt = 5 P (1 , 0 , 2 , 0) − 6 P (0 , 1 , 3 , 0)

Example: Stationary distribution for the Michaelis-Menten mechanism (continued) ◮ Solving these equations with � P ( N C , N E , N P , N S ) = 1, we get P (0 , 1 , 0 , 3) = 3 × 10 − 5 P (0 , 1 , 2 , 1) = 0 . 0495 P (1 , 0 , 0 , 2) = 8 × 10 − 4 P (1 , 0 , 2 , 0) = 0 . 4953 P (0 , 1 , 1 , 2) = 2 . 0 × 10 − 3 P (0 , 1 , 3 , 0) = 0 . 4127 P (1 , 0 , 1 , 1) = 0 . 0396

Example: Stationary distribution for the Michaelis-Menten mechanism (continued) ◮ Solving these equations with � P ( N C , N E , N P , N S ) = 1, we get P (0 , 1 , 0 , 3) = 3 × 10 − 5 P (0 , 1 , 2 , 1) = 0 . 0495 P (1 , 0 , 0 , 2) = 8 × 10 − 4 P (1 , 0 , 2 , 0) = 0 . 4953 P (0 , 1 , 1 , 2) = 2 . 0 × 10 − 3 P (0 , 1 , 3 , 0) = 0 . 4127 P (1 , 0 , 1 , 1) = 0 . 0396 Note: There is not a unique steady-state composition.

Example: Stationary distribution for the Michaelis-Menten mechanism (continued) ◮ Does the solution obey detailed balance? If so, we should have κ 1 N E N S P ( N C , N E , N P , N S ) = κ − 1 ( N 1 + 1) P ( N C + 1 , N E − 1 , N P , N S − 1) κ 2 N C P ( N C , N E , N P , N S ) = κ − 2 ( N E + 1)( N P + 1) P ( N C − 1 , N E + 1 , N P + 1 , N S ) for all values of ( N C , N E , N P , N S ) where these equations make sense.