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

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Foundations of Chemical Kinetics Lecture 23: The chemical master equation: Stationary distributions

Marc R. Roussel Department of Chemistry and Biochemistry

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The stationary distribution

◮ In ordinary chemical kinetics, a closed chemical system has a

unique equilibrium composition, usually given as a set of concentrations.

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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.

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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.

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The stationary distribution (continued)

◮ Recall the chemical master equation:

dP(N, t) dt =

r∈R

ar(N − νr)P(N − νr, t) −

r∈R

ar(N)P(N, t) =

r∈R

[ar(N − νr)P(N − νr, t) − ar(N)P(N, t)]

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The stationary distribution (continued)

◮ Recall the chemical master equation:

dP(N, t) dt =

r∈R

ar(N − νr)P(N − νr, t) −

r∈R

ar(N)P(N, t) =

r∈R

[ar(N − νr)P(N − νr, t) − ar(N)P(N, t)]

◮ 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(+)

r

be the propensity for forward reaction r, and a(−)

r

be the propensity for the corresponding reverse reaction. νr is the stoichiometry vector for the forward reaction.

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The stationary distribution (continued)

◮ Then we can write the CME as

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The stationary distribution (continued)

◮ Then we can write the CME as

dP(N, t) dt =

r∈R+
a(+)

r

(N − νr)P(N − νr, t) − a(+)

r

(N)P(N, t)

+
r∈R−
a(−)

r

(N + νr)P(N + νr, t) − a(−)

r

(N)P(N, t)

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The stationary distribution (continued)

◮ Then we can write the CME as

dP(N, t) dt =

r∈R+
a(+)

r

(N − νr)P(N − νr, t) − a(+)

r

(N)P(N, t)

+
r∈R−
a(−)

r

(N + νr)P(N + νr, t) − a(−)

r

(N)P(N, t)

=
r∈R+
a(+)

r

(N − νr)P(N − νr, t) − a(−)

r

(N)P(N, t)

+
a(−)

r

(N + νr)P(N + νr, t) − a(+)

r

(N)P(N, t)

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The stationary distribution (continued)

◮ Then we can write the CME as

dP(N, t) dt =

r∈R+
a(+)

r

(N − νr)P(N − νr, t) − a(+)

r

(N)P(N, t)

+
r∈R−
a(−)

r

(N + νr)P(N + νr, t) − a(−)

r

(N)P(N, t)

=
r∈R+
a(+)

r

(N − νr)P(N − νr, t) − a(−)

r

(N)P(N, t)

+
a(−)

r

(N + νr)P(N + νr, t) − a(+)

r

(N)P(N, t)

◮ For a detailed balanced solution, each bracketed pair of terms

would be zero.

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Example: Stationary distribution for the Michaelis-Menten mechanism

E + S

κ1

− − ⇀ ↽ − −

κ−1 C κ2

− − ⇀ ↽ − −

κ−2 E + P

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Example: Stationary distribution for the Michaelis-Menten mechanism

E + S

κ1

− − ⇀ ↽ − −

κ−1 C κ2

− − ⇀ ↽ − −

κ−2 E + P

Variable ordering: (NC, NE, NP, NS)

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Example: Stationary distribution for the Michaelis-Menten mechanism

E + S

κ1

− − ⇀ ↽ − −

κ−1 C κ2

− − ⇀ ↽ − −

κ−2 E + P

Variable ordering: (NC, NE, NP, NS) dP(N, t) dt = κ1(NE + 1)(NS + 1)P(NC − 1, NE + 1, NP, NS + 1) + κ−1(NC + 1)P(NC + 1, NE − 1, NP, NS − 1) + κ2(NC + 1)P(NC + 1, NE − 1, NP − 1, NS) + κ−2(NE + 1)(NP + 1)P(NC − 1, NE + 1, NP + 1, NS) − P(NC, NE, NP, NS) [κ1NENS + κ−1NC + κ2NC + κ−2NENP]

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Example: Stationary distribution for the Michaelis-Menten mechanism

E + S

κ1

− − ⇀ ↽ − −

κ−1 C κ2

− − ⇀ ↽ − −

κ−2 E + P

Variable ordering: (NC, NE, NP, NS) dP(N, t) dt = κ1(NE + 1)(NS + 1)P(NC − 1, NE + 1, NP, NS + 1) + κ−1(NC + 1)P(NC + 1, NE − 1, NP, NS − 1) + κ2(NC + 1)P(NC + 1, NE − 1, NP − 1, NS) + κ−2(NE + 1)(NP + 1)P(NC − 1, NE + 1, NP + 1, NS) − P(NC, NE, NP, NS) [κ1NENS + κ−1NC + κ2NC + κ−2NENP]

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Example: Stationary distribution for the Michaelis-Menten mechanism

E + S

κ1

− − ⇀ ↽ − −

κ−1 C κ2

− − ⇀ ↽ − −

κ−2 E + P

Variable ordering: (NC, NE, NP, NS) dP(N, t) dt = κ1(NE + 1)(NS + 1)P(NC − 1, NE + 1, NP, NS + 1) + κ−1(NC + 1)P(NC + 1, NE − 1, NP, NS − 1) + κ2(NC + 1)P(NC + 1, NE − 1, NP − 1, NS) + κ−2(NE + 1)(NP + 1)P(NC − 1, NE + 1, NP + 1, NS) − P(NC, NE, NP, NS) [κ1NENS + κ−1NC + κ2NC + κ−2NENP]

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Example: Stationary distribution for the Michaelis-Menten mechanism

E + S

κ1

− − ⇀ ↽ − −

κ−1 C κ2

− − ⇀ ↽ − −

κ−2 E + P

Variable ordering: (NC, NE, NP, NS) dP(N, t) dt = κ1(NE + 1)(NS + 1)P(NC − 1, NE + 1, NP, NS + 1) + κ−1(NC + 1)P(NC + 1, NE − 1, NP, NS − 1) + κ2(NC + 1)P(NC + 1, NE − 1, NP − 1, NS) + κ−2(NE + 1)(NP + 1)P(NC − 1, NE + 1, NP + 1, NS) − P(NC, NE, NP, NS) [κ1NENS + κ−1NC + κ2NC + κ−2NENP]

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Example: Stationary distribution for the Michaelis-Menten mechanism

E + S

κ1

− − ⇀ ↽ − −

κ−1 C κ2

− − ⇀ ↽ − −

κ−2 E + P

Variable ordering: (NC, NE, NP, NS) dP(N, t) dt = κ1(NE + 1)(NS + 1)P(NC − 1, NE + 1, NP, NS + 1) + κ−1(NC + 1)P(NC + 1, NE − 1, NP, NS − 1) + κ2(NC + 1)P(NC + 1, NE − 1, NP − 1, NS) + κ−2(NE + 1)(NP + 1)P(NC − 1, NE + 1, NP + 1, NS) − P(NC, NE, NP, NS) [κ1NENS + κ−1NC + κ2NC + κ−2NENP]

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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 κ1 = 10, κ−1 = 1, κ2 = 5 and κ−2 = 2 s−1.

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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 κ1 = 10, κ−1 = 1, κ2 = 5 and κ−2 = 2 s−1. dP(0, 1, 0, 3)/dt = P(1, 0, 0, 2) − 30P(0, 1, 0, 3) dP(1, 0, 0, 2)/dt = 30P(0, 1, 0, 3) + 2P(0, 1, 1, 2) − 6P(1, 0, 0, 2) dP(0, 1, 1, 2)/dt = P(1, 0, 1, 1) + 5P(1, 0, 0, 2) − 22P(0, 1, 1, 2) dP(1, 0, 1, 1)/dt = 20P(0, 1, 1, 2) + 4P(0, 1, 2, 1) − 6P(1, 0, 1, 1) dP(0, 1, 2, 1)/dt = P(1, 0, 2, 0) + 5P(1, 0, 1, 1) − 14P(0, 1, 2, 1) dP(1, 0, 2, 0)/dt = 10P(0, 1, 2, 1) + 6P(0, 1, 3, 0) − 6P(1, 0, 2, 0) dP(0, 1, 3, 0)/dt = 5P(1, 0, 2, 0) − 6P(0, 1, 3, 0)

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Example: Stationary distribution for the Michaelis-Menten mechanism (continued)

◮ Solving these equations with P(NC, NE, NP, NS) = 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

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Example: Stationary distribution for the Michaelis-Menten mechanism (continued)

◮ Solving these equations with P(NC, NE, NP, NS) = 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.

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Example: Stationary distribution for the Michaelis-Menten mechanism (continued)

◮ Does the solution obey detailed balance?

If so, we should have κ1NENSP(NC, NE, NP, NS) = κ−1(N1 + 1)P(NC + 1, NE − 1, NP, NS − 1) κ2NCP(NC, NE, NP, NS) = κ−2(NE + 1)(NP + 1)P(NC − 1, NE + 1, NP + 1, NS) for all values of (NC, NE, NP, NS) where these equations make sense.

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Example: Stationary distribution for the Michaelis-Menten mechanism (continued)

◮ Does the solution obey detailed balance?

If so, we should have κ1NENSP(NC, NE, NP, NS) = κ−1(N1 + 1)P(NC + 1, NE − 1, NP, NS − 1) κ2NCP(NC, NE, NP, NS) = κ−2(NE + 1)(NP + 1)P(NC − 1, NE + 1, NP + 1, NS) for all values of (NC, NE, NP, NS) where these equations make sense.

◮ Example: For (NC, NE, NP, NS) = (1, 0, 1, 1),

κ2NCP(NC, NE, NP, NS) = 0.198 and κ−2(NE + 1)(NP + 1)P(NC − 1, NE + 1, NP + 1, NS) = 0.198.

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Example: Stationary distribution for the Michaelis-Menten mechanism (continued)

◮ Note:

NS =

NSP(NC, NE, NP, NS)

= 3P(0, 1, 0, 3) + 2P(1, 0, 0, 2) + 2P(0, 1, 1, 2) + P(1, 0, 1, 1) + P(0, 1, 2, 1) = 0.0948 NP =

NPP(NC, NE, NP, NS) = 2.3693