From Single-Cell Simulation to Emergent Systems (1)From Single-Cell Simulation to Emergent Systems (1).pdf

From Single－Cell Simulation to Emergent Systems

The question of how to shape a system of metabolic networks from a single-cell perspective and how to compute it to a macroscopic reactor system has always been a difficult problem for reductionism and emergence. Let's start with the microscopic metabolic network to consider a system.

Axiom 1: DNA can be transcribed into RNA and then translated into proteins, which are then decatalyzed by proteins or inhibited or regulated metabolic networks.

Hypothesis 1: There is an oil-producing strain with DNA-to-RNA transcription as full expression.

Axiom 2: Promoters in DNA can be represented by Boolean algebra (0 or 1, which means start or no), which can be represented by

SU (2)

Group representation Axiom 3: Enzymes are three-dimensional objects (

SO (3)

groups) whose motion (rotation, vibration, translation) is expressed in terms of quaternion (SU(2) group).

Axiom 4: The integral of the probability density function is 1 , which can be represented by

S U (2)

a group.

Defines operator a as an operator that rotates state 0 to state 1.

a = U_{a} = e^{- i θ σ_{y} / 2} = \cos (\frac{θ}{2}) I - i \sin (\frac{θ}{2}) σ_{y}

Defines

b

an operator as an operator that rotates state 1 to state 0.

b = U_{b} = e^{- i θ σ_{y} / 2} = \cos (\frac{Φ}{2}) I + i \sin (\frac{Φ}{2}) σ_{y}

Rule:

\begin{aligned} | ⟨ 1 | a | 0 ⟩ |^{2} = \sin^{2} (\frac{θ}{2}) \\ | ⟨ 1 | b | 1 ⟩ |^{2} = \cos^{2} (\frac{ϕ}{2}) \\ | ⟨ 0 | a | 0 ⟩ |^{2} = \cos^{2} (\frac{θ}{2}) \\ | ⟨ 0 | b | 1 ⟩ |^{2} = \sin^{2} (\frac{ϕ}{2}) \end{aligned}

| ⟨ 1 | a | 0 ⟩ |^{2}

Represents the probability that operator A will convert state 0 to state 1.
Therefore, the changed probability density function is:

\begin{aligned} p_{1}^{'} = p_{a} \cdot \sin^{2} (\frac{θ}{2}) + (1 - p_{a}) \cdot \cos^{2} (\frac{ϕ}{2}) \\ p_{0}^{'} = p_{a} \cdot \cos^{2} (\frac{θ}{2}) + (1 - p_{a}) \cdot \sin^{2} (\frac{ϕ}{2}) \end{aligned}

Lemma 1:

SO (3)

The group and

SU (2)

the group are homomorphic.

Axiom 5: Shannon Entropy is used to measure the uncertainty of a probability distribution by following the formula:

H = - \sum_{i} P_{i} \log P_{i}

For the promoter of the binary state, we have:

H = - P_{0} \log P_{0} - P_{1} \log P_{1}

Then: The flux J of a metabolic reaction can be expressed as:

J = P_{0} \cdot v_{0} + P_{1} \cdot v_{1}

v_{0}

is the reaction rate at which the promoter state is 0.

v_{1}

is the reaction rate at promoter state 1.

Lemma 2: At steady state, the probability density distribution is equivalent to the flux density distribution.

The Gaussian flux theorem is formulated in the two-dimensional case as:

\int_{V} (\nabla \cdot J) d A = ∮_{S} J \cdot n d s

\begin{aligned} J (x, y) & = (P_{x}, P_{y}) \\ \nabla \cdot J & = \frac{\partial P_{x}}{\partial x} + \frac{\partial P_{y}}{\partial y} = 0 \\ 在稳态下, P_{x} & = \frac{J_{x}}{J_{total}}, P_{y} = \frac{J_{y}}{J_{total}} \end{aligned}

Lemma 3: In biological systems, the system tends to reach a state of equilibrium through some kind of optimization principle. Common optimization principles include the principle of maximizing entropy and the principle of minimum free energy.

Here, we can combine the two and utilize the Lagrangian multiplier method to maximize the aromatic entropy of the system by introducing constraints, while keeping the flux J constant.

Constraints:

1. Probability Normalization:

P_{0} + P_{1} = 1

2. Flux fixation:

J = P_{0} \cdot v_{0} + P_{1} \cdot v_{1}

The Lagrangian function

L

is denoted as:

L = - P_{0} \log P_{0} - P_{1} \log P_{1} + λ (P_{0} + P_{1} - 1) + μ (J - P_{0} v_{0} - P_{1} v_{1})

where and

λ

μ

is the Lagrange multiplier.

P_{1}

Find the partial derivative of

L

about

P_{0}

and make it equal to zero:

\begin{aligned} \frac{\partial L}{\partial P_{0}} = - \log P_{0} - 1 + λ - μ v_{0} = 0 \\ \frac{\partial L}{\partial P_{1}} = - \log P_{1} - 1 + λ - μ v_{1} = 0 \end{aligned}

Subtract to get:

- \log P_{0} + \log P_{1} - μ v_{0} + μ v_{1} = 0

Easy to get:

\log (\frac{P_{1}}{P_{0}}) = μ (v_{1} - v_{0})

Take the index:

\frac{P_{1}}{P_{0}} = e^{μ (v_{1} - v_{0})}

P_{0} = \frac{1}{1 + e^{μ (v_{1} - v_{0})}}, P_{1} = \frac{e^{μ (v_{1} - v_{0})}}{1 + e^{μ (v_{1} - v_{0})}}

Axiom 5: According to the Boltzmann distribution in thermodynamics, the relationship between probability and energy is:

\frac{P_{1}}{P_{0}} = e^{- β Δ G}

Lemma 4: In summary:

μ (v_{1} - v_{0}) = - β Δ G \Rightarrow μ = - \frac{β Δ G}{v_{1} - v_{0}}, β = \frac{1}{k_{B} T}

Then the flux formula can be expressed as:

J = \frac{v_{0} + v_{1} e^{- β Δ G}}{1 + e^{- β Δ G}} = v_{0} + \frac{v_{1} - v_{0}}{1 + e^{β Δ G}}