From what has been discussed earlier, we can list the following system of ordinary differential equations:
$$ \frac{d[a t p]}{d t}=v{A T P}-v{P P K 1}-v{\text {inpst }}-v{G D P 2 G T P}-N G A R \tag{i} $$
$$ \frac{d[a d p]}{d t}=-\frac{d[a t p]}{d t} \tag{ii} $$
$$ \frac{d\left[p i_{c u l}\right]}{d t}=-v{m t P{i}} \cdot N \cdot V{\text {cell }} / V{c u l} \tag{iii} $$
$$ \frac{d\left[p i_{\text {sur }}\right]}{d t}=v{m t P i}-v{p i t}-v{\text {inpst }} \tag{iv} $$
$$ \frac{d\left[P i{i n}\right]}{d t}=v{\text {pit }}+2 \cdot v{\text {inpst }}-v{A T P}+N G A R \tag{v} $$
$$ \frac{d[\text { polyp] }}{d t}=v{p p k 1}-v{p p k 2}-v{NADK} \tag{vi} $$
$$ \frac{d\left[\text { lactate }_{\text {cul }}\right]}{d t}=-v{\text {LldP }} \cdot N \cdot V{\text {cell }} / V{\text {cul }} \tag{vii} $$
$$ \frac{d\left[\text { Lactate }{\text {in }}\right]}{d t}=v{\text {Lldp }}-v{\text {Lac }} \tag{viii} $$
$$ \frac{d\left[n a d^{+}\right]}{d t}=-v{N A D H}+v{M r 2}-v{N A D K} \tag{ix} $$
$$ \frac{d[n a d h]}{d t}=v{N A D H}-v{M r 2} \tag{x} $$
$$ \frac{d\left[n a d p^{+}\right]}{d t}=v{N A D K}-v{N A D P 2 N A D P H} \tag{xi} $$
$$ \frac{d[n a d p h]}{d t}=v{N A D P 2 N A D P H} \tag{xii} $$
$$ \frac{d\left[m_{o}\right]}{d t}=-v{M r 1}-v{M r 2}+v{E E T} \tag{xiii} $$
$$ \frac{d\left[M{r}\right]}{d t}=-\frac{d\left[m_{o}\right]}{d t} \tag{xiv} $$
$$ \frac{d[g d p]}{d t}=-v{P P K 2}-v{G D P 2 G T P} \tag{xv} $$
$$ \frac{d[g t p]}{d t}=-\frac{d[g d p]}{d t} \tag{xvi} $$
$$ \frac{d[n g a r]}{d t}=N G A R-n g a r \tag{xvii} $$
$$ \frac{d I}{d t}=n{e} \cdot F \cdot v{E E T} \cdot I E P S \cdot N \cdot V{\text {cell }} / 60-I \tag{xviii} $$
$$ \frac{d I E P S}{d t}=v{I E P S} \tag{xix} $$
$$ \frac{d\left[m r n a_{p p k 2}\right]}{d t}=k 0{p p k 2}-d 0{p p k 2} \cdot\left[m r n a_{p p k 2}\right] \tag{xx} $$
$$ \frac{d[p p k 2]}{d t}=k 1{p p k 2} \cdot\left[m R N A{p p k 2}\right]-d 1{p p k 2} \cdot[p p k 2] \tag{xxi} $$
$$ \frac{d\left[m R N A{N A D K}\right]}{d t}=k 0{\text {nadk }}-d 0{\text {nadk }} \cdot\left[m r n a{\text {nadk}}\right] \tag{xxii} $$
$$ \frac{d[n a d k]}{d t}=k 1{\text {nadk }} \cdot\left[m R N A{\text {nadk }}\right]-d 1{\text {nadk }} \cdot[n a d k] \tag{xxiii} $$
When the cells were transferred to M9 medium to generate electricity, the expression of the enzyme gene in the cells had been induced for some time, and the enzyme content tended 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:
$$ \begin{gathered} {[p p k 2]{s s}=\frac{k 1{p p k 2} \cdot k 0{p p k 2}}{d 1{p p k 2} \cdot d 0{p p k 2}}} \ {[n a d k]{s s}=\frac{k 1{\text {nadk }} \cdot k 0{\text {nadk }}}{d 1{\text {nadk }} \cdot d 0{\text {nadk }}}} \end{gathered} $$
We used MATLAB to solve the ODEs listed above (see details in ode_codes.m ), and plotted the values of each variable over time:
From the simulations of the above model, we can draw the following conclusions:
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 proposed the theoretical support for our further engineering of Shewanella, which greatly helped us in project design.