Model 1: Decomposition rate of wheat straw #

Decomposition rate is usu­ally es­ti­mated by ex­po­nen­tial de­cay model or lin­ear de­cay model. We will fol­low the ex­po­nen­tial de­cay model.

Exponential de­cay #

For the ex­po­nen­tial de­cay, let us take the sim­plest model equa­tion: $$\frac{x_t}{x_0}=e^{-kt}$$

Where, $x_0$ is the ini­tial dry mat­ter mass of the lit­ter (g), $x_t$ is the resid­ual dry mat­ter mass af­ter a pe­riod of de­com­po­si­tion time t and k is the av­er­age de­com­po­si­tion rate of straws.

Let ($x_i$ , $y_i$ ), ($i$ = 1, 2, 3, … n) be the ex­per­i­men­tal data and ($x_k$, $y_k$) to con­firm the ex­per­i­men­tal model:

$$y=a(1-e^{-kx})$$

Where a and k are the pa­ra­me­ters to be de­ter­mined, in or­der to de­ter­mine the es­ti­mated value of k, we will use the non-lin­ear least squares cri­te­rion, and the resid­ual sum of squares [Q (a,k)] is cal­cu­lated us­ing the for­mula:

$$Q(a,k)=\sum _{i=1}^n(y_i-a(1-e^{-k{x}_i}))$$

The al­go­rithm to reach the min­i­mum tar­get op­ti­miza­tion pa­ra­me­ter in­cludes the Gauss-Newton Method, Marquart d op­ti­miza­tion, and the DUD method.

The value of a and k is to be de­ter­mined ex­per­i­men­tally. Using a lit­er­a­ture Search [LI Yun-Le et al., 2007], the value of ‘$a$’ and ‘$k$’ can be de­ter­mined as:

Treatments	a	k	Regression Squares sum	Residual Squares sum	Index of Correlation
Original Soil	116.17	0.0096	14504.78	88.80	0.993915

Using Mathematica, we can get the sim­u­la­tion re­sult for our pro­ject:

From the above ma­nip­u­la­tion plot, we get the value of ‘$k$’ as 0.09 and the value of ‘$a$’ as 106, re­spec­tively, which must be tar­geted through wet lab ex­per­i­ments. However, these con­stants are sub­jected to change based on Wet Lab Experiments.

Model 2: Decomposition of la­bile and re­frac­tory com­po­nents in the wheat straw #

Following the two-com­part­ment model #

Consider the fol­low­ing: Ini­tial Mass = $A$ Labile Component $(M_L)$, a com­po­nent that can be de­graded Re­frac­tory Component $(M_R)$, a com­po­nent that can­not be de­graded Now, we are pre­sent­ing a model which is crop-de­pen­dent. As we know, dif­fer­ent crops will have dif­fer­ent com­po­si­tions of lignin, cel­lu­lose, pectin, and xy­lan, and we have en­gi­neered the bac­te­ria with roles of de­grad­ing spe­cific com­po­nents of the straw. So, the crop choice will de­cide the con­cen­tra­tion of bac­te­ria with spe­cific roles. Similarly, the crop choice will de­cide the rate con­stants of la­bile and re­frac­tory com­po­nents and re­frac­tory frac­tions of ini­tial mass $A$.

So, by choos­ing the ini­tial mass, re­frac­tory frac­tion ‘$a$’ of ini­tial mass $A$ and the time in which we want to achieve the de­com­po­si­tion will give us the re­quired rate con­stant of the re­frac­tory com­po­nent ($k_r$ = x) and la­bile com­po­nent ($k_l$ = y), which will help us in de­cid­ing the con­cen­tra­tion of bac­te­ria with spe­cific roles to be taken in the bac­te­r­ial con­sor­tium.

Following first-or­der Differential Equations, we have,

For Labile Component: $$\frac{d{M}_L}{dt}=-K_L{M}_L$$

For Refractory Component: $$\frac{d{M}_R}{dt}=-K_R{M}_L$$

Normalized par­al­lel first-or­der dif­fer­en­tial fol­lows: $$\frac{M}{M_0}=a{e}^{-K_{R}t}+(1-a)e^{-K_{L}t}$$

Where ‘$a$’ is the re­frac­tory frac­tion of ini­tial mass $M_0$. From equa­tion $13$, we have the tem­per­a­ture and mois­ture con­trol func­tion as: $$f(T,M(T))={1.78}^{\frac{T-23}{10}}\times 0.411\times {10}^{\frac{1.66\times {10}^6+T(79674.4+T(17.4419+7.5T))}{-2.53\times {10}^6+T(10623.3+T(2.32558+T))}}$$

As an ex­am­ple, on sim­u­lat­ing with Wolfram Mathematica, We have

In this ex­am­ple, Let us take the ini­tial mass of straw taken to be 70kg and as­sume that the straw con­tains 38% Refractory com­po­nent and rest 62% as the la­bile com­po­nent, which means 62% of the straw will be de­graded, and the rest 38% will re­main un­de­graded by the bac­te­ria. Let us as­sume that we want the straw to be de­graded in 30 days. Then, the rate con­stant for re­frac­tory frac­tion will be 0.125 day ${}^{-1}$ , and the rate con­stant for la­bile frac­tion will be 1.95 day ${}^{-1}$.

Now, Adding the fac­tor of mois­ture and tem­per­a­ture con­trol (Check Model 3 be­low) of the air to the above de­com­po­si­tion model. To do this, let ‘$k$’ be the over­all rate con­stant of the re­ac­tion, then check the de­com­po­si­tion with re­spect to Temperature and mois­ture.

Simulating this sit­u­a­tion on Mathematica, we have:

In this ex­am­ple, If we keep the Temperature at $30\mathrm{°}C$ and want to achieve the over­all de­com­po­si­tion in 25 days, then the re­quired rate con­stant will be 0.108 day ${}^{-1}$

Effect of burn­ing shown through sim­u­la­tion #

From the above sim­u­la­tion, we find that as the Temperature and the over­all rate con­stant ($k$) in­creases, the de­com­po­si­tion in­creases, and at very high tem­per­a­tures or very high rate con­stant, the graph blows off, in­di­cat­ing the burn­ing of stub­ble oc­curs!

Model 3: Calculation of Humidity and Temperature Control #

Specific hu­mid­ity (or mois­ture con­tent) is the ra­tio of the mass of wa­ter va­por to the to­tal mass of the air par­cel. Now, Mathematically, Specific Humidity or Moisture con­tent in the air is given by:

$${\text{Moisture Content}}_{\text{air}}=6.11\times 10237.3+\text{Dew Point}$$

The dew point is the Temperature to which air must be cooled to be­come sat­u­rated with wa­ter va­por, as­sum­ing con­stant air pres­sure and wa­ter con­tent. Mathematically, Dew Point is:

$$\text{Dew Point}=T_d=T-\frac{(100-\text{Relative Humidity)}}{5}$$

Now, to get the re­la­tion be­tween Dew Point and Relative hu­mid­ity, let us first con­sider the re­la­tion be­tween Relative Humidity and Temperature. Temperature ($\mathrm{°}C$)

Temperature (°C)	0	5	10	15	20	25	30	35	40	45	50
Rel. Humidity	4.8	6.8	9.4	12.8	17.3	23	30.4	39.6	51.5	65.4	83

Plotting the above table of tem­per­a­ture v/​s Relative hu­mid­ity, we have:

So, the curve fit­ted cu­bic equa­tion is: $$4.669+0.432T+0.001T^2+0.00043T^3$$

Overlapping with the tem­per­a­ture v/​s Relative hu­mid­ity plot:

Hence, the Equation sat­is­fy­ing the re­la­tion be­tween the Dew Point and the Temperature is Given by:
$$4.669+0.432T+0.001T^2+0.00043T^3$$ Placing this Equation in place of Relative Humidity in Equation $5$, we get,

$$\text{Dew Point}=(-8.6\times {10}^{-5})T^3-0.0002T^2-0.9136-19.0662$$

Plotting this, we have:

Hence, Moisture Content is given as:

$${\text{Moisture Content}}_{\text{air}}=6.11\times {10}^{\frac{7.5(-19.0662-0.9136T-0.0002T^2-0.000086T^3)}{237.3-19.0662-0.9136T-0.0002T^2-0.000086T^3}}$$

Normalizing and Plotting this re­la­tion, we have:

To in­cor­po­rate the ef­fects of fluc­tu­at­ing Air tem­per­a­tures and Moisture on de­com­po­si­tion rates in the re­gres­sion mod­els with the in­de­pen­dent vari­able, time as:

$$f(T,M(T))={1.78}^{\frac{T-23}{10}}\times 0.411\times {10}^{\frac{1.66\times {10}^6+T(79674.4+T(17.4419+7.5T))}{-2.53\times {10}^6+T(10623.3+T(2.32558+T))}}$$

Hence Control Function will be:

The above plot will con­trol the Moisture and Air Temperature fluc­tu­a­tions in the en­vi­ron­ment, whose value varies be­tween 0 and 1. Where mois­ture is con­trolled by Temperature as it is writ­ten as a func­tion of Temperature to re­duce the num­ber of in­de­pen­dent vari­ables.

Model 4: Four Component Model #

Let us in­clude two more com­po­nents, ac­tive ($M_A$) com­po­nent and sta­bi­lized ($M_S$) com­po­nent of de­com­po­si­tion prod­ucts. The ac­tive com­po­nent was as­sumed to in­clude mi­cro­bial bio­mass and non-sta­bi­lized de­com­po­si­tion prod­ucts. All trans­for­ma­tions will fol­low the first-or­der ki­net­ics. Product for­ma­tion and mass loss as $C{O}_2$ will be pro­por­tional de­pend­ing on a yield ef­fi­ciency fac­tor $ϵ$. However, only a por­tion $\gamma$ of the ac­tive com­po­nent turnover was as­sumed to be­come sta­bi­lized.

Let us look at the Active Pool of the straw. The ac­tive pool was as­sumed to in­clude mi­cro­bial bio­mass and non-sta­bi­lized de­com­po­si­tion prod­ucts. It fol­lows the First Order Kinetics as fol­lows:

$$\frac{dM[t]}{dt}=[ϵ(l\times L+r\times R+(1-\gamma )k\times M[t])-k\times M[t]]$$

Where, ‘$M[t]$’ is the mass of the ac­tive com­po­nent of straw at the time ‘$t$‘, ‘$l$’ is the rate con­stant for the la­bile com­po­nent, ‘$L$’ is the mass of the la­bile com­po­nent, ‘$r$’ is the rate con­stant for the re­frac­tory com­po­nent, ‘$R$’ is the mass of the Refractory com­po­nent, ‘$k$’ is the rate con­stant of the ac­tive com­po­nent of the straw.

Plotting the ac­tive com­po­nent, we have

Now, Let us look at the sta­bi­lized com­po­nent of the straw as fol­lows from the first-or­der ki­net­ics (without con­sid­er­ing Temperature and mois­ture ef­fects):

$$\frac{dM[t]}{dt}=(ϵ\gamma )a\times -k\times M[t]$$

Where ‘$a$’ is the rate con­stant of the ac­tive com­po­nent, $A$ is the mass of the ac­tive com­po­nent, $k$ is the rate con­stant of sta­bi­lized pools, and $M[t]$ is the mass of sta­bi­lized com­po­nent of the straw.

For com­par­i­son, by the lit­er­a­ture search [Andrén et al., 1987], let us take, $a$ = 0.0293, $A$ = 10 kg, $e$ = 0.36, $\gamma$ = 0.47

The above two mod­els for sta­bi­lized com­po­nents and Refractory com­po­nents will help us de­velop the model to es­ti­mate the pro­duc­tion of Carbon Dioxide re­leased in the air us­ing Acitve com­po­nents and Stabilized com­po­nents. It fol­lows the zero-or­der ki­net­ics:

$$\frac{d{\text{CO}}_{\text{2}}}{dt}=(1-ϵ)(l\times L+r\times R+a\times A+s\times S)$$

Where the sym­bols rep­re­sent the usual mean­ings as de­scribed above.

For com­par­i­son, by the lit­er­a­ture search [Andrén et al., 1987], let us take,

$ϵ=0.36$

$l=0.367{\text{ day}}^{-1}$

$r=0.00364{\text{ day}}^{-1}$

$a=0.0293$

$s=0.0005{\text{ day}}^{-1}$

Plotting equa­tion $22$ along with the com­par­i­son graph, we get,

Model 5: Bioplastic Decomposition Model #

Simulating the Observations through Mathematica: As we know from the above mod­els, the degra­da­tion curve usu­ally fol­lows ex­po­nen­tial de­cay. Hence, con­sid­er­ing our ansatz as:

$$dm=a{e}^{-kt}$$

Where ‘$dm$’ is the change in mass with re­spect to time, ‘a’ and ‘k’ are the con­stants, and ‘t’ is the time.

Experimental Data:

Mass (g)	0.1001	0.0904	0.0847	0.0768	0.0725
Time (hours)	0	24	48	72	96

Using Wolfram Mathematica, we fit the ex­per­i­men­tal data us­ing our ansatz from equa­tion $1$ we get:

Fitted Exponential Equation: $$0.099307e^{-0.00340519x}$$

Plotting this equa­tion:

From this graph, we have es­ti­mated that within 2500 hours or 105 days (Around 3.5 Months), 99.9801% of the bio­plas­tic will be de­com­posed.

According to BBC Science Focus, biodegrad­able plas­tics take only three to six months to fully de­com­pose, far quicker than tra­di­tional plas­tic, which can take hun­dreds of years.

Model 6: Enzyme Kinetic Model #

Xylanase #

1) xynA #

Enzyme se­creted: endo-1,4-beta-xy­lanase ( ex­tra­cel­lu­lar)

Reaction: $a[(1\to 4)-\beta -D-xylan]+n{H}_{2}O\to na[(1->4)-\beta -D-xylanoligosaccharide]$

KINETIC DIFFERENTIAL EQUATION: Assuming rate con­stants of the re­ac­tion to be kA

Code $$\frac{d[(1,4)-\beta -\text{D}-\text{xylan}]}{dt}=-\frac{V_{xynA}\times [(1,4)-\beta -\text{D}-\text{xylan}]}{K{m}_{xynA}+[(1,4)-\beta -\text{D}-\text{xylan}]}-(K_A\times [(1,4)-\beta -\text{D}-\text{xylan}])$$

Function: xy­lan degra­da­tion

E.C. 3.2.1.8

Protein fam­ily:glycosyl hy­dro­lase 11 (cellulase G) fam­ily (single mem­ber, ac­cord­ing to UniProt)

Domains: GH 11 do­main (aa 29-213) (according to UniProt)

Localization: ex­tra­cel­lu­lar (signal pep­tide)

Structure

Alphafold

• Subtiwiki

2) xynB #

Enzyme: xy­lan beta-1,4-xy­losi­dase

Reaction: $a(1->4)-\beta -D-xylanoligosaccharide+n{H}_{2}O⟶n\beta -D-xylopyranose$

KINETIC DIFFERENTIAL EQUATION: Assuming rate con­stants of the re­ac­tion to be k_B

Code $$\frac{d[(1,4)-\beta -\text{D}-xylanoligosaccharide]}{dt}=-\frac{V_{xynB}\times [(1,4)-\beta -\text{D}-[xylanoligosaccharide]}{K{m}_{xynB}+[(1,4)-\beta -\text{D}-xylanoligosaccharide]}-(K_B\times [(1,4)-\beta -\text{D}-xylanoligosaccharide])$$

Pathway:(1,4)-β-D-xylan degra­da­tion

Functioning: Catalysis of the hy­drol­y­sis of (1->4)-beta-D-xylans so as to re­move suc­ces­sive D-xylose residues from the non-re­duc­ing ter­mini.

E.C. 3.2.1.37

Protein fam­ily: gly­co­syl hy­dro­lase 43 fam­ily Localization:cell mem­brane (according to Swiss-Prot)

STRUCTURE

P94489 (AlphaFold)

Subtiwiki

3) xynC #

Enzyme: endo-xy­lanase

Function: xy­lan degra­da­tion

Product: endo-xy­lanase

E.C: 3.2.1.136

Catalyzed re­ac­tion/ bi­o­log­i­cal ac­tiv­ity: Endohydrolysis of (1->4)-beta-D-xylosyl links in some glu­curonoara­bi­noxy­lans (according to UniProt)

$$a\text{[glucuronoarabinoxylan]}+n{H}_{2}O⟶na\text{[glucuronoarabinoxylan oligosac­cha­ride]}$$

$$a [glucuronoxylan] + n H2O \longrightarrow n a [glucuronoxylan\textrm{ }oligosaccharide]$$

Protein fam­ily:glycosyl hy­dro­lase 30 fam­ily (single mem­ber, ac­cord­ing to UniProt)

Localization:extracellular (signal pep­tide)

Structure

3GTN (PDB)

Subtiwiki

4) xynD #

Enzyme: ara­bi­noxy­lan ara­bi­no­fu­ra­nohy­dro­lase

Function: ara­bi­noxy­lan degra­da­tion

E.C.: 3.2.1.55

Catalyzed re­ac­tion/ bi­o­log­i­cal ac­tiv­ity: Hydrolysis of ter­mi­nal non-re­duc­ing al­pha-L-ara­bi­no­fu­ra­noside residues in al­pha-L-ara­bi­no­sides (according to UniProt)

Protein fam­ily: gly­co­syl hy­dro­lase 43 fam­ily (according to UniProt)

Domains: CBM6 do­main (aa 382-511) (according to UniProt)

Localization: ex­tra­cel­lu­lar (signal pep­tide)

Structure

3C7O (PDB)

Subtiwiki

Degradation ki­net­ics #

bio­cyc.org

(1→4)-β-D-xylan → (1->4)-β-D-xylan oligosac­cha­ride → β-D-xy­lopy­ra­nose

Let A be (1→4)-β-D-xylan

B be (1->4)-β-D-xylan oligosac­cha­ride

C be β-D-xy­lopy­ra­nose

Let $k_1$ and $k_2$ be the rate con­stants for the re­spec­tive re­ac­tions as shown.

Now mul­ti­ply equa­tion (4) by an in­te­grat­ing fac­tor $e^{-k_{1}t}$, on both sides

Now in or­der to find [C], we will sub­sti­tute equa­tion (vi) in equa­tion (iii),

Now we will find the time when [B]{i.e, [(1->4)-β-D-xylan oligosac­cha­ride] } con­cen­tra­tion of be­comes max­i­mum

Ligninase #

Functioning: Dye-decolorizing per­ox­i­dases (DyPs) are a fam­ily of per­ox­i­dases that cat­alyze H2O2-dependent ox­i­da­tion of var­i­ous mol­e­cules. They are re­spon­si­ble for lignin degra­da­tion in lig­no­cel­lu­losic bio­mass.

Structure #

GENE: BsDyP #

PDB ID

GENE: DyP1B(Pseudomonas flu­o­rescens) #

Uniprot

Pectinase #

Gene: Pme (from Ralstonia solanacearum)

EC: 3.1.1.11

Reaction:

STRUCTURE

Uniprot

Cellulase #

Cellulase en­zyme: cglT

EC: 3.2.1.21

Function: Hydrolysis of ter­mi­nal, non-re­duc­ing beta-D-glu­co­syl residues with re­lease of beta-D-glu­cose.

STRUCTURE

link - uniprot

FACTORS AFFECTING ACTIVITY OF LIGNOCELLULOSIC ENZYMES

Several fac­tors as­so­ci­ated with the na­ture of the cel­lu­lase en­zyme sys­tem are in­flu­en­tial dur­ing the hy­drol­y­sis process. These in­clude:

1. Enzyme con­cen­tra­tion
2. Adsorption
3. Synergism
4. End-product in­hi­bi­tion
5. Mechanical de­ac­ti­va­tion (fluid shear stress or gas-liq­uid in­ter­face)
6. Thermal in­ac­ti­va­tion
7. Irreversible (non-productive) bind­ing to lignin.

Andersen, N. (2007).Enzymatic Hydrolysis of Cellulose: Experimental and Modeling Studies.

The rate of en­zy­matic hy­drol­y­sis of lig­no­cel­lu­lose is pro­foundly af­fected by the struc­tural fea­tures of cel­lu­lose :

1. Crystallinity of cel­lu­lose

2. Degree of poly­mer­iza­tion (DP), i.e. mol­e­c­u­lar weight of cel­lu­lose

3. Available/accessible sur­face area

4. Structural or­ga­ni­za­tion, i.e. macro-struc­ture (fiber) and mi­crostruc­ture (el­e­men­tary mi­crofib­ril) and par­ti­cle size

5. Presence of as­so­ci­ated ma­te­ri­als such as hemi­cel­lu­lose and lignin.

   The typ­i­cal time course of the en­zy­matic hy­drol­y­sis of the lig­no­cel­lu­losic ma­te­r­ial is char­ac­ter­ized by the rapid ini­tial rate of hy­drol­y­sis fol­lowed by slower and in­com­plete hy­drol­y­sis. This is due to the rapid hy­drol­y­sis of more eas­ily avail­able amor­phous cel­lu­lose, with con­se­quent in­crease of in­her­ent de­gree of crys­tallinity, as the hy­drol­y­sis pro­ceeds.

Andersen, N. (2007). Enzymatic Hydrolysis of Cellulose: Experimental and Modeling Studies.

Hydrolysis of cel­lu­lose dif­fers from most other en­zy­matic re­ac­tions by the fact that sub­strate is in­sol­u­ble; thus to en­sure the re­ac­tion the phys­i­cal con­tact, i.e. ad­sorp­tion of the en­zymes to the sub­strate, is pre­req­ui­site for cel­lu­lose hy­drol­y­sis.

The ef­fi­ciency of cel­lu­lases ad­sorp­tion on the sur­face of the cel­lu­lose can be char­ac­ter­ized by the par­ti­tion co­ef­fi­cient Kp [L/g] of the en­zyme be­tween the sub­strate sur­face and the wa­ter phase.

Substrate up­take and ki­netic mod­el­ing #

• Rate of nu­tri­ent con­sump­tion (in this case stub­ble degra­da­tion) of a pop­u­la­tion of bac­te­ria is greatly af­fected by pres­ence of lim­ited/​op­ti­mal/​ex­cess nu­tri­ent con­cen­tra­tions.
• So a quan­tity nu­tri­ent flux (ФN) can be mod­eled as a func­tion of the con­cen­tra­tion of bac­te­ria (B) and the con­cen­tra­tion of nu­tri­ent (N):

where μmax is the max­i­mum growth rate of the bac­terium us­ing a par­tic­u­lar nu­tri­ent and K_S is the con­cen­tra­tion of nu­tri­ent at which the growth rate is one-half of the max­i­mum.

• At high nu­tri­ent con­cen­tra­tions (N >> $K_S$), nu­tri­ent up­take ap­proaches its up­per limit re­sult­ing in ex­po­nen­tial growth at the max­i­mum rate for a par­tic­u­lar nu­tri­ent com­po­si­tion.
• At low nu­tri­ent con­cen­tra­tions (N << $K_S$), the nu­tri­ent flux equa­tion be­comes the fol­low­ing:

and hence we get

• Let’s de­fine γ as the frac­tion of nu­tri­ent flux de­voted to se­creted en­zymes (P) such as

• At low cell den­sity, γ is pro­por­tional to B. At high cell den­sity, the frac­tional in­vest­ment in P lev­els off at a max­i­mum value (γmax). The cell den­sity at which the cul­ture de­votes one-half of the max­i­mal amount of re­sources to se­creted pro­tein is Kγ.

● The rate of change of sub­strate with re­spect to time can be mod­eled by Michaelis–Menten ki­net­ics of degra­da­tion of the sub­strate by se­creted en­zymes:

$$\frac{dS}{dt}=\hspace{0.17em}-V_{max}\hspace{0.17em}•\hspace{0.17em}P\hspace{0.17em}•\hspace{0.17em}\frac{S}{(Km+S)}$$

● The rate of change of nu­tri­ent con­cen­tra­tion is the neg­a­tive of the rate of degra­da­tion of sub­strate mi­nus the rate of nu­tri­ent flux into cells:

$$\frac{dN}{dt}=-\frac{dS}{dt}-\varphi N=\frac{V_{max}\cdot P\cdot S}{K_m+S}-\frac{{\mu }_{max}\cdot N}{K_s+N}$$2

References #

1. https://​bio­cyc.org/​BSUB/​NEW-IM­AGE?type=PATH­WAY&ob­ject=PWY-6717

2. https://​bio­cyc.org/​BSUB/​NEW-IM­AGE?type=PATH­WAY&ob­ject=PWY-6717

3. https://​www.ncbi.nlm.nih.gov/​pmc/​ar­ti­cles/​PM­C3086598/

4. https://​www.uniprot.org/​unipro­tkb/​Q60026/​en­try#struc­ture

5. https://​chem­istry-eu­rope.on­lineli­brary.wi­ley.com/​doi/​10.1002/​cbic.

6. https://​www.sci­encedi­rect.com/​sci­ence/​ar­ti­cle/​pii/​S0141022918305106

7. https://​www.uniprot.org/​unipro­tkb/​A0A0S4WW32/

8. http://​sub­ti­wiki.uni-goet­tin­gen.de/​v4/​gene?id=55D7F05F8592A53A58A032C78B847782E598B268

9. LI Yun-Le, QIAO Yu-Hui, SUN Zhen-Jun, ZHANG Lu-Da, ZHANG Rui-Qing. Mathematical model of wheat straw de­com­po­si­tion rate in farm­land with dif­fer­ent lev­els of fer­til­iza­tion[J]. Chinese Journal of Eco-Agriculture, 2007, 15(3): 61-63.

10. Andrén, Olof & Paustian, O.. (1987). Barley Straw Decomposition in the Field: A Comparison of Models. Ecology. 68. 1190-1200. 10.2307/1939203.

11. Humidity. (2022, September 24) - https://​en.wikipedia.org/​wiki/​Hu­mid­ity

Bioplastic degra­da­tion model

Stubble Degradation model