Phosphorus and energy cycles of human-bacteria-plants in space flight systems

Our project aims to use Shewanella bacteria for phosphorus accumulation and electricity generation in space. Our current idea involves three cycles:

Human-Bacteria Cycle: Human feces are fed to the bacteria, and the bacteria produce electricity for humans.
Bacteria-Plant Cycle: Plant waste is fed to the bacteria, which generate electricity to support plant growth and provide phosphorus to nourish the plants.
Human-Plant Cycle: Plants provide food for humans, and humans cultivate the plants.

In the following discussion, we will quantitatively calculate the flow of phosphorus elements and corresponding energy in each path, and finally draw a conclusion that introducing Shevanella into space system can save 70.58% of phosphate fertilizer in space phosphorus cycling.

1. Shewanella Bacteria Power Generation Efficiency

1.1 Assumptions

(a) The bacterial metabolic rate is calculated based on the amount of substrate consumed per hour or per minute.

(b) The metabolic rate of Shewanella bacteria is similar to that of common heterotrophic bacteria.

1.2 Theoretical Electron Transfer Numbers

Suppose we have:

X: Number of lactate molecules entering the NADH-dependent pathway.
1-X: Number of lactate molecules entering the formate-dependent pathway.

Then we get:

Total ATP produced: $\frac{11}{3}+5X$
Total electrons transferred: $4+8X$

In summary, each lactate molecule can produce a maximum of 12 electrons and a minimum of 4 electrons.

The detailed derivation process can be found in ODE MODEL 1.2.3.

1.3 Theoretical Lactate Consumption Rate

Each bacterium consumes approximately 10^-11 moles of lactate per hour.

Each Shewanella bacterium consumes about 10⁻¹¹ moles of lactate per hour. Converted to a per-minute consumption rate:

$$ \frac{\mathrm{1}\mathrm{0}^{\mathrm{-11}}\mathrm{mol/h} }{\mathrm{60min{/}h}}\mathrm{=1.67}\times\mathrm{1}\mathrm{0}^{\mathrm{-13}}\mathrm{mol/min} \tag{1.1} $$

1. Total bacterial number: Assume the total number of bacteria is 108 based on experimental data.

2. Total lactate consumption

$$ \mathrm{1.67}\times\mathrm{1}\mathrm{0}^{\mathrm{-13}}\mathrm{mol/min}{\times}\mathrm{10^8}=1.67*10^{-5}\mathrm{mol/min} \tag{1.2} $$

Under this assumption, $10^8$ Shewanella bacteria consume approximately $1.67*10^{-5}$ moles of lactate per minute, which is about $0.167$ micromoles of lactate.

1.4 Coulombic Efficiency

Coulombic efficiency represents the efficiency with which microorganisms convert the chemical energy in substrates into electrical energy. Coulombic efficiency formula is:

$$ CE=\frac{I\cdot t}{n\cdot F\cdot\Delta C} \tag{1.3} $$

$I$: Total current (Amps)
$t$: Time (seconds)
$n$: Number of electrons produced per mole of lactate
$F$: Faraday constant
$ΔC$: Lactate consumption (moles)

(a). Calculation of Theoretical Charge Equivalent of Electrons: Based on the previous calculation (1.2), assume that consuming one mole of lactate produces 4 moles of electrons. Thus, the electron equivalent of lactate is:

$$ n=1.67\times\mathrm{1}\mathrm{0}^{\mathrm{-5}}\mathrm{mol}\times\mathrm{4=6.68}\times\mathrm{1}\mathrm{0}^{\mathrm{-5}}\mathrm{mol\ }\mathrm{e}^- \tag{1.4} $$

Then, using the Faraday constant to convert the number of moles of electrons into charge:

$$ electrons\ into\ charge =n\times F\mathrm{=6.68}\times\mathrm{1}\mathrm{0}^{\mathrm{-5}}\mathrm{mol}\times\mathrm{96485}C/\mathrm{mol=6.44}C \tag{1.5} $$

(b). Calculation of Actual Charge: Actual charge comes from the measured current. According to our team’s experimental results, the actual current is $I = 1500 μA/cm²$. Assuming an electrode area of 2 cm², the total current is $300 μA$.

$$ actual\mathrm{\ c}harge=I\times\mathrm{t=3000}\times\mathrm{1}\mathrm{0}^{\mathrm{-6}}A\times\mathrm{60s=1.8}\times\mathrm{1}\mathrm{0}^{\mathrm{-1}}C \tag{1.6} $$

(c). Calculation of Coulombic Efficiency: $$ CE=\frac{1.8\times10^{-1}C}{6.44C}\times 100\% =2.8\% \tag{1.7} $$

Under these conditions, the Coulombic efficiency of the electricity-generating bacteria is approximately $2.8\% $. This means that only $2.8\% $ of the lactate metabolic energy is converted into electrical energy, with the rest likely used for bacterial growth and other metabolic activities.

2. Shewanella Bacteria Phosphorus Recycling Efficiency

In a space station cycle using potato cultivation as an example, every $100g$ of potatoes contains $40mg$ of phosphorus, while an adult needs 700mg of phosphorus per day ($70\%$ can be absorbed, the rest excreted in feces). This means an intake of $1g$ of phosphorus is needed, requiring about $2500g$ of potatoes, or ten potatoes.

According to our team's experimental results, the phosphorus recovery efficiency of Shewanella is about $30\% $, meaning that about $30\% $ of the phosphorus in astronauts' urine and feces can be recycled.

For potato cultivation, each $100g$ of tuber requires $0.2g $ of phosphate fertilizer, of which phosphorus constitutes about $8.5\% $. Therefore, every $100g$ of potatoes requires $17mg$ of phosphorus (most of the phosphorus comes from the original soil).

For a typical space station with three people, $3g$ of phosphorus is needed daily (equivalent to $7500g$ of potatoes), of which Shewanella bacteria can recover about $0.9g$ of phosphorus. Growing $7500g$ of potatoes requires $1.275g$ of phosphorus aside from what is provided by the soil.

$$ \frac{0.9g}{1.275g}\times 100\% =70.58\% \tag{2.1} $$

In conclusion, the phosphorus recovery efficiency of Shewanella bacteria is about $30\% $, which can save $70.58\% $ of phosphate fertilizer in space phosphorus cycling.

Furthermore, according to experimental results, $10^8$ bacteria can accumulate $0.2mg$ of phosphorus, so to handle the daily waste of three astronauts, about $15$ liters of bacterial solution ( at $10^8 cells/ml$ ) would be required.

References

[1]: Kouzuma A: Molecular mechanisms regulating the catabolic and electrochemical activities of Shewanella oneidensis MR-1. Biosci Biotechnol Biochem 2021, 85:1572-1581.