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2. ODEs and Result

2.1 ODEs

From what has been discussed earlier, we can list the following system of ordinary differential equations:

$$ \frac{d[A T P]}{d t}=v_{A T P}-v_{P P K 1}-v_{ {in,pst }}-v_{G D P 2 G T P}-N G A R \tag{i} $$

$$ \frac{d[A D P]}{d t}=-\frac{d[A T P]}{d t} \tag{ii} $$

$$ \frac{d\left[P i_{c u l}\right]}{d t}=-v_{m t, Pi} \cdot N \cdot V_{ {cell }} / V_{c u l} \tag{iii} $$

$$ \frac{d\left[P i_{ {sur }}\right]}{d t}=v_{m t, P i}-v_{p i t}-v_{ {in,pst }} \tag{iv} $$

$$ \frac{d\left[P i_{i n}\right]}{d t}=v_{ {pit }}+2 \cdot v_{ {in,pst }}-v_{A T P}+N G A R \tag{v} $$

$$ \frac{d[ PolyP] }{d t}=v_{PPk1}-v_{PPK2}-v_{NADK} \tag{vi} $$

$$ \frac{d\left[ { Lactate }_{ {cul }}\right]}{d t}=-v_{ {LldP }} \cdot N \cdot V_{ {cell }} / V_{ {cul }} \tag{vii} $$

$$ \frac{d\left[ { Lactate }_{ {in }}\right]}{d t}=v_{ {LldP }}-v_{ {Lac }} \tag{viii} $$

$$ \frac{d\left[N A D^{+}\right]}{d t}=-v_{N A D H}+v_{M, r 2}-v_{N A D K} \tag{ix} $$

$$ \frac{d[N A D H]}{d t}=v_{N A D H}-v_{M r 2} \tag{x} $$

$$ \frac{d\left[N A D P^{+}\right]}{d t}=v_{N A D K}-v_{N A D P 2 N A D P H} \tag{xi} $$

$$ \frac{d[N A D P H]}{d t}=v_{N A D P 2 N A D P H} \tag{xii} $$

$$ \frac{d\left[M_{o}\right]}{d t}=-v_{M, r 1}-v_{M, r 2}+v_{E E T}/n_e \tag{xiii} $$

$$ \frac{d\left[M_{r}\right]}{d t}=-\frac{d\left[M_{o}\right]}{d t} \tag{xiv} $$

$$ \frac{d[G D P]}{d t}=-v_{P P K 2}-v_{G D P 2 G T P} \tag{xv} $$

$$ \frac{d[G T P]}{d t}=-\frac{d[G D P]}{d t} \tag{xvi} $$

$$ \frac{d[N G A R]}{d t}=N G A R-n g a r \tag{xvii} $$

$$ \frac{d I}{d t}=n_{e} \cdot F \cdot v_{E E T} \cdot I E P S \cdot N \cdot V_{ {cell }} / 60-I \tag{xviii} $$

$$ \frac{d I E P S}{d t}=v_{I E P S} \tag{xix} $$

$$ \frac{d\left[m R N A_{P P K 2}\right]}{d t}=k_{0,PPK2}-d_{0,PPK2} \cdot\left[m R N A_{P P K 2}\right] \tag{xx} $$

$$ \frac{d[P P K 2]}{d t}=k_{1,PPK2} \cdot\left[m R N A_{PPK2}\right]-d_{1,PPK2} \cdot[P P K 2] \tag{xxi} $$

$$ \frac{d\left[m R N A_{N A D K}\right]}{d t}=k_{0,NADK}-d_{0,NADK} \cdot\left[m R N A_{N A D K}\right] \tag{xxii} $$

$$ \frac{d[N A D K]}{d t}=k_{1,NADK} \cdot\left[m R N A_{NADK}\right]-d_{1,NADK} \cdot[N A D K] \tag{xxiii} $$

When the cells are transferred to M9 medium to generate electricity, the expression of the enzyme gene in the cells has been induced for some time, and the enzyme concentration tends to be stable. Therefore, we assume that the variables expressed in equations $( x x \sim x x iii)$ have reached equilibrium, that is, the value of the equation is 0. So we have:

$$ \begin{gathered} {[P P K 2]_{s s}=\frac{k_{1,PPK2} \cdot k_{0,PPK2}}{d_{1,PPK2} \cdot d_{0,PPK2}}} \\ {[N A D K]_{s s}=\frac{k_{1,NADK} \cdot k_{0,NADK}}{d_{1,NADK} \cdot d_{0,NADK}}} \end{gathered} $$


2.2 Results

We used MATLAB to solve the ODEs listed above (see details in ode codes.m ), and plotted the values of each variable over time:

beaker
beaker
beaker
beaker

From the simulations of the above model, we can draw the following conclusions:

  1. The variation of current output with time was in good agreement with the variation of current measured with time in wild-type Shewanella. This indicates that our model is able to correctly capture the factors affecting Shewanella current output and reasonably explains why such a pattern is present.
  2. After the introduction of dual hydrolases, the ATP concentration of S.oneidensis was higher than that of the other groups, indicating an increase in the metabolic level. At the same time, its phosphorus aggregation capacity and electricity production efficiency are also enhanced, suggesting that introducing dual hydrolases is feasible.

Conclusion 1 gives us a better understanding of modeled process, i.e. the coupling relationship between phosphorus metabolism and electricity generation in Shewanella, while conclusion 2 proposes the theoretical support for our further engineering of Shewanella, which greatly helped us in project design.