5 | Model reduction: conservation laws and steady states Flashcards
Consider a molecular species produced at a constant rate and degraded according to Michaelis-Menten kinetics:
dt/dx=k1− Vmax*x/(Km+x)
k1=1 nM/min
Vmax=2.5 nM/min
Km=2 nM
What is the steady-state concentration of x?
(ungraded quiz)
set ODE = 0, solve for x, enter given values
–>
1,33
A→k+B
B→k-A
with k+=3/min, k-=2/min, and xA(0)=xB(0)=1nM
What is the steady state for A?
(ungraded quiz)
ODEs, find conservation law, set ODE for A = 0 and substitute for B
–>
0,8
A→k+B
B→k-A
with k+=3/min, k-=2/min, and xA(0)=xB(0)=1nM
What is the response time for this model?
(ungraded quiz)
We obtain the single equation (
dxA/dt=−k + xA + k−(xtot−xA) = k−xtot−(k++k−)xA.
This equation has the same form as the single protein example (linear & autonomous), and accordingly, the response time is
log(2)/(k++k−) ≈ 0.1386.
Consider the ODE
dx/dt = ax(b−x), with state variable x, x(0)>0 and parameters a,b>0
Which of the following statements is true?
a. The system admits one steady state, and it is stable
b. The system admits two steady states, but only the larger one is stable
c. The system admits two steady states, but only the smaller one is stable
d. The system admits one steady state, and it is unstable
e. The system admits two steady states, and both are stable.
f. The system doesn’t admit any steady state
(ungraded quiz)
The equation 0=dx/dt=ax(b−x) admits the two solutions
xss,1=0
xss,2=b
For positive initial condition, x always approaches b (logistic growth). Therefore, xss,2 is stable, whereas xss,1 is unstable.
The correct answer is:
b. The system admits two steady states, but only the larger one is stable
Model reduction in biochemical reaction models
Aims ?
▶ reducing the number of state variables
▶ reducing the number of parameters
▶ structural insightModel reduction in biochemical reaction models
Model reduction in biochemical reaction models
Strategies for model reduction?
▶ conservation laws
▶ steady states
▶ quasi-steady state approximation (timescale separation)
▶ (and more)
Response time
Definition ?
For a reaction kinetic system approaching a (stable) steady state (= relaxing towards it),
the characteristic timescale of x(t) → xss is of interest, called response time τ , which is defined as the time by which an initial difference (x0 − xss ) is halved, i.e.,
(x(τ) - xss) = 1/2 * (x0 - xss)
Response time
What does a smaller τ mean?
Faster timescale!
The difference is halved quicker
Response time
What does a smaller τ mean?
Slower timescale!
The difference is halved slower
Response time
For the single protein model ?
∗ →ksyn S →kdeg ∗
the response time τ coincides with the half-life of S:
τ = ln(2) / kdeg .
Response time
In more general settings, for more complex systems
In more general settings, τ may depend on the (initial) state x0 of the system, i.e. τ (x0)
to find response time:
a) Write out the differential equations describing the system.
b) Linearize these equations around the steady state.
c) response time should be ln(2)/ (the rate constants for the qss0 (?)
Response time
for first order equations?
For first-order reactions or systems that can be approximated as first-order:
The response time τ is given by:
τ = 1 / (sum of rate constants)
For example, in a simple reversible reaction A ⇌ B with forward rate constant k₁ and reverse rate constant k₋₁:
τ = 1 / (k₁ + k₋₁)
Response time for the logistic differential equation
dx(t)/dt = x(t) (1- x(t))
τ(x0) = log (1 + 1/x0)
Response time
𝜏𝑞𝑠𝑠 if we have a qss:
𝑑𝐶𝑞𝑠𝑠/𝑑𝑡 = 𝑘𝑎 ∗ 𝐷 ∗ (𝑃𝑡𝑜𝑡 − 𝐶𝑞𝑠𝑠) − 𝑘𝑑 ∗ 𝐶𝑞𝑠𝑠 = 0
𝑑𝐶𝑞𝑠𝑠/𝑑𝑡 = 𝑘𝑎 ∗ 𝐷 ∗ (𝑃𝑡𝑜𝑡 − 𝐶𝑞𝑠𝑠) − 𝑘𝑑 ∗ 𝐶𝑞𝑠𝑠 = 0
⇔ 𝑘𝑎 ∗ 𝐷 ∗ 𝑃𝑡𝑜𝑡 − (𝑘𝑎 ∗ 𝐷 + 𝑘𝑑) ∗ 𝐶𝑞𝑠𝑠 = 0 ⇔ 𝐶𝑞𝑠𝑠 = ⋯
⇒ 𝜏𝑞𝑠𝑠 =ln(2) / (𝑘𝑎∗𝐷+𝑘𝑑)
Response time
𝜏𝑎𝑝𝑝𝑟 if we have a reduced ODE system eg:
𝑑𝐷𝑎𝑝𝑝𝑟/𝑑𝑡 = 0 = −𝑘𝑎 ∗ 𝐷𝑎𝑝𝑝𝑟 ∗ (𝑃𝑡𝑜𝑡 − 𝐶𝑞𝑠𝑠) + 𝑘𝑑 ∗ 𝐶𝑞𝑠𝑠 − 𝑘𝑒𝑙 ∗ 𝐷𝑎𝑝𝑝𝑟
= −(𝑘𝑎 ∗ 𝐷𝑎𝑝𝑝𝑟 ∗ (𝑃𝑡𝑜𝑡 − 𝐶𝑞𝑠𝑠) − 𝑘𝑑 ∗ 𝐶𝑞𝑠𝑠) − 𝑘𝑒𝑙 ∗ 𝐷𝑎𝑝𝑝𝑟
= −𝑑𝐶𝑞𝑠𝑠/𝑑𝑡 − 𝑘𝑒𝑙 ∗ 𝐷𝑎𝑝𝑝𝑟
= − 0 − 𝑘𝑒𝑙 ∗ 𝐷𝑎𝑝𝑝𝑟
⇒ 𝜏𝑎𝑝𝑝𝑟 = ln(2) / 𝑘𝑒l
