Introduction

According to the comparison of the blank group and the experimental group data in Tables 1, 2, 3 and 4, we found that gene editing has a positive or negative effect on the growth of Escherichia coli. The purpose of our project is to use the gene-edited Escherichia coli to convert as much creatinine and urea in the human body into harmless or even beneficial polyglutamic acid as possible. Therefore, we established mathematical models to find out the conditions for producing the most polyglutamic acid and to ensure that the gene-edited Escherichia coli can survive normally in the intestinal environment under these conditions.

Table 1 Creatinine-Creatinine Urea Nitrogen (CUN) Reference Table
Experimental group 1
Component creatinine (mg/L) OD urea (mg/L)
Time/Group 1 2 3 1 2 3 1 2 3
Oh 113 113 113 0 0 0 0 0 0
48h 100.83 98.35 106.99 11.54 10.74 11.1 4.34 4.56 5.11
Blank group 1
Component creatinine (mg/L) OD urea (mg/L)
Time/Group 1 2 3 1 2 3 1 2 3
Oh 113 113 113 0 0 0 0 0 0
48h 112.34 111.98 112.78 12.34 13.03 12.56 0.003 0.002 0.003
Table 2 Urea-Ammonia Test Control Table
Experimental group 2
Component Urea (mg/L) OD
Time/Group 1 2 3 1 2 3
Oh 10000 10000 10000 0 0 0
48h 9980.34 9976.25 9969.56 10.34 9.66 10.09
Blank group 2
Component urea (mg/L) OD
Time/Group 1 2 3 1 2 3
Oh 10000 10000 10000 0 0 0
48h 9998.21 9999.03 9999.23 11.24 12.02 12.21
Table 3 L-Glutamic Acid - Polyglutamic Acid (M9) Experimental Control Table
Experimental group 3
Component Polyglutamic Acid (mg/L) OD
Time/Group 1 2 3 1 2 3
Oh 0 0 0 0 0 0
48h 79.45 90.23 86.24 11.67 12.45 13.26
Blank group 3
Component Polyglutamic Acid (mg/L) OD
Time/Group 1 2 3 1 2 3
Oh 0 0 0 0 0 0
48h 0.002 0.003 0.003 13.45 14.23 12.34
Table 4 Creatinine + Urea - Glutamine (No Nitrogen Source M9) Experiment Control Table
Experimental group 4
Component Creatinine (mg/L) Urea (mg/L) Polyglutamic Acid (mg/L) OD
Time/Group 1 2 3 1 2 3 1 2 3 1 2 3
Oh 113 113 113 10000 10000 10000 0 0 0 0 0 0
48h 105.23 107.56 102.34 9986.46 19986.45 9988.34 1.52 0.97 1.05 8.45 7.98 8.05
Blank group 4
Component Creatinine (mg/L) Urea (mg/L) Polyglutamic Acid (mg/L) OD
Time/Group 1 2 3 1 2 3 1 2 3 1 2 3
Oh 113 113 113 10000 10000 10000 0 0 0 0 0 0
48h 112.45 112.09 112.66 9999.02 9998.78 9996.45 0.002 0.004 0.003 12.45 13.14 12.98

Based on this, we established two models, a product synthesis model and an engineering bacterial growth model. Due to the fact that our project divided the entire metabolic pathway into three reaction stages, namely, creatinine-->creatine+urea, urea-->r-PGA, and urea-->ammonia, which included intermediate products such as urea and ammonia, and then conducted experiments on the entire metabolic pathway, namely, creatinine+urea-polyglutamic acid (nitrogen-free M9), the product synthesis model was further divided into three single substrate synthesis models (stage synthesis models) and one double substrate synthesis model (overall synthesis model). For the engineering bacterial growth model, we first established a universal Escherichia coli growth curve, then compared the difference between the gene-edited Escherichia coli and the wild-type Escherichia coli, and according to the competition model, adjusted the parameters of the gene-edited Escherichia coli to obtain the competition curve of Escherichia coli. Finally, we simulated the actual intestinal environment and put our gene-edited Escherichia coli into it to confirm that it can survive normally in the human intestine.

Product Synthesis Model

Synthesis Model

According to the three metabolic pathways, three different substrates are separated out to form a single substrate synthesis model. By plotting the average rate and time relationship, the substrate concentration at the maximum average rate is found, and then combined with the double substrate synthesis model of the complete pathway to obtain the product yield at the optimal substrate concentration, that is, the maximum production rate of polyglutamic acid, which should be noted that the production rate here is for the whole process, not for a certain period of time.

(1) Single Substrate Synthesis Model

For the single substrate synthesis model, first calculate the average value of three parallel experiments, that is, the formula:

$\overline{X}=\cfrac{X_1+X_2+X_3+...+X_N}{N}$

$\overline{X}$ is the average value of three parallel experiments, unit is mg/L;

$X_1...X_N$ is the substrate concentration at each time node of each group of experiments, unit is mg/L;

$N$ is the number of parallel experiment groups.

Then calculate the average concentration of each time period (the time interval here is 12h), that is, the formula:

$\overline{Y}_{(i)}=\cfrac{X_{i+1}+X_i}{2}$

$\overline{Y}_{(i)}$is the average concentration of the i-th time period, unit is mg/L;

$X_i$ is the average concentration of the i time node, unit is mg/L.

Then obtain the average rate of each time period, that is, the formula:

$V_i=\cfrac{|X_i-X_{i+1}|}{t}$

$V_i$ is the average rate of the i time period, unit is $mg/(L·h)$;

$X_i$ is the average concentration of the i time period, unit is $mg/L$;

$t$ is the time interval, here is 12h, that is, t=12, unit is h.

According to the average rates obtained in different time periods, fit the corresponding rate-time equation, the form is as follows:

$V(t)=p_n t^n+p_{n-1} t^{n-1} +···+p_1 t^1+C$

$V(t)$ is the reaction rate at t time point, unit is $mg/(L·h)$;

$p_n$ is the coefficient term corresponding to t to the n times power;

$C$ is the constant term, since the reaction rate is 0 when the time is 0, so the intercept term is set to 0, that is, $C=0$.

Finally, in order to obtain the substrate concentration at the maximum rate, it is necessary to take the time as 0 and the time when the maximum rate is reached as the upper and lower limits, and to integrate the average rate, that is, the formula:

$X_{res}=X_{tot}-X_{con}=X_{tot}-∫_0^TV(t)dt$

$X_{res}$ is the substrate concentration at the maximum rate, unit is mg/L;

$X_{tot}$ is the total substrate concentration before the reaction starts, unit is mg/L;

$X_{con}$ is the substrate concentration consumed when the maximum rate is reached, unit is mg/L;

$V(t)$ is the reaction rate at t time point, unit is $mg/(L·h)$;

$t$ is the time when the maximum rate is reached, unit is h.

(2) Double Substrate Synthesis Model

Establish a linear regression model as follows:

$V_{PA}=R_1·X_{ct}+R_2·X_{urea}$

$V_{PA}$ is the average rate of polyglutamic acid production, unit is $mg/(L·h)$;

$R_1,R_2$ are regression coefficients;

$X_{ct}$ is the average concentration of creatinine, unit is mg/L;

$X_{urea}$ is the average concentration of urea, unit is mg/L.

Use the least squares method to fit the linear model and find the regression coefficients that best fit the given data set to minimize the square error between the predicted value and the actual observation value, that is, first calculate the slope:

$β=\cfrac{∑(x_i-\overline{x})(y_i-\overline{y})}{∑(X_i-X)^2}$

Then calculate the intercept:

$β_0=\overline{y}−β\overline{x}$

Finally, substitute the original equation.

Data analysis

We took the average of three parallel experiments to obtain the concentrations of substrates and products at different times, and took the average every 12 hours to obtain the average concentration of the product in this time, and then divided the amount of consumed product in each 12 hours by the time to obtain the average reaction rate in this period of time, as shown in the following four tables:

Table 5 Creatinine-Creatine + Urea Data Processing Table
Component Creatinine Urea
Time/Group 1 2 3 Average value(mg/L) Average concentration(mg/L) Average rate(mg/(L-h)) 1 2 3 Average value(mg/L) Average concentration(mg/L) Average rate(mg/(L-h))
Oh 113 113 113 113.00 112.01 0.17 0 0 0 0.00 0.33 0.06
12h 11.45 110.35 111.23 111.01 109.72 0.22 0.67 0.55 0.76 0.66 1.24 0.10
24h 108.26 106.23 110.78 108.42 105.75 0.45 1.89 2.04 1.55 1.83 2.69 0.14
36h 101.43 100.45 107.34 103.07 102.57 0.08 3.45 3.24 3.99 3.56 4.12 0.09
48h 100.83 98.35 106.99 102.06 4.34 4.56 5.11 4.67
Table 6 Urea-Ammonia Data Processing Table
Component Urea
Time/Group 1 2 3 Average value (mg/L) Average concentration(mg/L) Average rate mg/(L•h)
Oh 10000 10000 10000 10000.00 9998.23 0.30
12h 9996.78 9995.34 9997.24 9996.45 9994.28 0.36
24h 9993.35 9992.78 9991.45 9992.12 9985.84 1.05
36h 9985.12 9980.35 9973.24 9979.57 9977.48 0.35
48h 9980.34 9976.25 9969.56 9975.38
Table 7 L-Glutamine - Polyglutamine (M9) Data Processing Table
Component Polyglutamic Acid Glutamate
Time/Group 1 2 3 Average value(mg/L) Average concentration(mg/L) Average rate(mg/(L-h) 1 2 3 Average value(mg/L) Average concentration(mg/L) Average rate(mg/(L-h)
Oh 113 113 113 0.00 7.05 1.17 500 500 500 500.00 398.32 16.95
12h 10.23 13.23 18.83 14.10 22.59 1.42 268.34 301.44 320.15 296.64 225.20 11.91
24h 29.45 32.55 31.27 31.09 52.32 3.54 156.88 170.24 134.12 153.75 128.10 4.28
36h 70.23 78.73 71.67 73.54 79.43 0.98 100.23 98.12 108.99 102.45 79.52 3.82
48h 79.45 90.23 86.24 85.31 85.31 55.23 35.66 78.9 56.60
Table 8 Creatinine + Urea-Polyglutamine (No Nitrogen Source M9) Data Processing Table
Component Creatine Urea Polyglutamic acid
Time/Group 1 2 3 Average value(mg/L) Average concentration(mg/L) Average rate(mg/(L-h)) 1 2 3 Average value(mg/L) Average concentration(mg/L) Average rate(mg/(L-h)) 1 2 3 Average value(mg/L) Average concentration(mg/L) Average rate(mg/(L-h))
Oh 113 113 113 113.00 111.88 0.19 10000 10000 10000 10000.00 9998.68 0.22 0 0 0 0.00 0.13 0.02111
12h 111.67 110.34 110.26 110.76 110.08 0.11 9998.35 9996.34 9997.36 9997.35 9994.75 0.43 0.19 0.34 0.23 0.25 0.43 0.03000
24h 109.56 109.23 109.44 109.41 108.23 0.20 9990.34 9992.85 9993.28 9992.16 9990.22 0.32 0.67 0.61 0.56 0.61 0.75 0.02194
36h 106.34 108.34 106.45 107.04 106.04 0.17 9987.23 9989.34 9988.56 9988.29 9987.68 0.10 1.08 0.77 0.78 0.88 1.03 0.02528
48h 105.23 107.56 102.34 105.04 9986.46 9986.45 9988.34 9987.08 1.52 0.97 1.05 1.18

Data Analysis and Parameter Calculation

1.A metabolic pathway involving the conversion of Creatinine to Creatine and Urea.

According to Table 5, the following are the graphs of creatinine consumption rate and urea generation rate:

Result: the maximum rate of creatine consumption is 0.7139 $mg/(L·h)$, which is reached at 36.76h with creatine concentration of 102.3456 mg/L; the maximum rate of urea production is 0.1451 mg/(L·h), which is reached at 33.53 hours with urea concentration of 3.1470 mg/L.

2. Amino acid metabolism pathway

According to Table 6, the graph of urea consumption rate is as follows:

Result: he maximum rate of urea consumption is 1.1918 $mg/(L·h)$, which is reached at 34.34h with urea concentration of 9982.1353 mg/L.

3. L-Glutamic Acid to Polyglutamate Metabolic Pathway

According to Table 7, the L-Glutamic Acid consumption rate graph and the Polyglutamic Acid generation rate graph are as follows:

Result: the maximum rate of L-glutamine consumption is 18.5927 $mg/(L·h)$, which is reached at 9.12h with L-glutamine concentration of 381.6251 mg/L; the maximum rate of polyglutamine production is 3.9297 $mg/(L·h)$, which is reached at 34.02 hours with polyglutamine concentration of 63.2261 mg/L.

4.Creatinine + Urea - Polyglutamic Acid (Nitrogen-free M9) Metabolic Pathway

According to Table 8, the regression coefficients R_1 and R_2 in the relationship equation between the average rate of polyglutamine production and the average concentrations of creatine and urea are -0.000231 and 0.00000498, respectively. Substituting these values into the equation:

$V_{PA}=R_1·X_{ct}+R_2·X_{urea}$

we obtain:

$V_{PA}= -0.000231·X_{ct}+0.00000498·X_{urea}$

$V_{PA}$ is the average rate of polyglutamine production, in units of $mg/(L·h)$;

$R_1$ and $R_2$ are regression coefficients;

$X_{ct}$ is the average concentration of creatine, in units of mg/L;

$X_{urea}$ is the average concentration of urea, in units of mg/L.

According to the consumption rate graph of creatine and urea, when the creatine concentration is 102.3456 mg/L and the urea concentration is 9982.1353 mg/L, the maximum rate of polyglutamine production is obtained. Substituting these data into the equation, we obtain that the average rate of polyglutamine production at this time is 0.026 $mg/(L·h)$, that is, $V_{max}$=0.026 $mg/(L·h)$.

Engineered Escherichia coli Growth Model

Ideal Escherichia coli Growth Model

The growth process of a biological organism can be described graphically as an S-curve, which varies in various ways depending on environmental factors. The most well-known quantitative descriptions of the growth process of organisms are the Linear, Logistic, Richards model and Gompertz equation. Due to their fixed inflection points, they can only accurately describe one particular shape of the S-curve, or a specific part of the complete S-curve. Firstly, the modeling of the growth of Escherichia coli was carried out.

Under ideal conditions, the population exhibits exponential growth, i.e. the formula:

$\cfrac{dN}{dt}= rN$

can also be written as:

$N_t=N_0e^{rt}$

$r$ is the innate growth rate of the population;

$N$ is the population size.

Taking into account the problems of food environment competition, the model was modified to obtain the Verhulst model, resulting in the Logistic equation:

$\cfrac{dN}{dt}= rN (1 –\cfrac{N}{K})$

$K$ is called the environmental carrying capacity;

$(1-\cfrac{N}{K})$ represents the environmental resistance.

Since this growth curve is "S" shaped, it can be transformed into:

$N=\cfrac{K}{1+be^{-rt}}$

The mathematical model used to represent microorganisms is:

$\log \cfrac{N_t}{N_0} = \cfrac{a}{1+be^{-rt}}$

In addition to the traditional Logistic equation, we also consider using the Gompertz equation to reliably analyze the S-shaped growth of Escherichia coli. It is quoted from the time series and its characteristics are: slow growth at the beginning, gradually accelerating in the middle; after a certain point, the growth rate gradually slows down. Because the Gompertz model contains three unknown parameters, it has strong adaptability and can fit the reliability growth test data of many bacteria, so it is widely used.

However, there are certain limitations to the application of the Gompertz equation. It requires the test data to be divided into three equal time periods for parameter estimation. For the growth test data of many bacteria, the model parameter estimation is not particularly accurate. Therefore, an optimized fitting method is proposed to improve the Gompertz equation, i.e. the modified Gompertz equation:

$\log\cfrac{N_t}{N_0}=ae^{-e^{b-ct} }$

Considering the special conditions required by the Logistic equation and the Gompertz equation, we use the Richards equation established on the basis of the Bertalanffy growth theory. The Bertalanffy growth theory analyzed the growth of animals and found that during the growth period of animals, the growth rate of body weight was the difference between the assimilation rate and the consumption rate, and the latter two were proportional to the size of the assimilative organs and the body weight of the animals, i.e.

$\cfrac{dW}{dt}=R_a-R_t=αF-γW$

$F$ is the size of the assimilative organs;

$W$ is the weight;

$R_a$ is the assimilation rate;

$R_t$ is the consumption rate.

According to the relative growth relationship, we get the formula:

$F=βW^m$

Substituting into the original equation gives:

$\cfrac{dW}{dt}=βW^m–γW$

Integrating both sides of the equation gives:

$W=a(1-be^{-kt})^{\frac{1}{1-m}}$
$a=\cfrac{β}{γ}(1-m)^{-1}$
$b=[1-\cfrac{γ}{β}*W_0^{1-m}]$
$ k=-(1-m)* γ$

$W_0$ is the initial value of W;

When m=2, it is the Logistic equation

$$W= \cfrac{a}{1+b’e^{-kt}}$$.

when m→1, it is the Gompertz equation

$$W= a*e^{-e^{b-kx} }$$.

With the application of computer simulation technology, the research on growth models is becoming more and more in-depth. The Richards growth model is widely used for its reasonable practical meaning of parameters and its ability to describe diverse growth processes. Its expression is:

$\log\cfrac{N_t}{N_0}=\cfrac{a}{(1+e^{b-c*t})^{\frac{1}{d}}}$

In this project, we use the above three models as the basis to simulate the optimal fitting curve of Escherichia coli growth, and according to the ideal Escherichia coli growth data, we do the fitting, the fitting results are shown in the following figure, the fitting effect is very good.

The LOGISTIC model equation is as follows:

$f(t) = \cfrac{6.4894}{1 + e^{-0.3171 * (t - 8.6426)}}$

The GOMPERTZ model equation is as follows:

$f(t) = 7.1647 * e^{-e^{-0.1841 * (t - 7.1790)}}$

The RICHARDS model equation is as follows:

$f(t) = \cfrac{6.9396}{1 + e^{-0.2123 * (t - 0.0000)^{4.9718}}}$

Genetically Edited E. coli Growth Model

In order to obtain the growth model of the genetically edited E. coli, it is necessary to compare it with the native E. coli. First, the experimental data is organized to obtain Table 9 as follows:

Table 9 Growth of E. coli before and after gene editing
Time/group Experimental group OD Blank group4 OD
1 2 3 Average value 1 2 3 Average value
Oh 0 0 0 0 0 0 0 0
48h 8.45 7.98 8.05 8.16 12.45 13.14 12.98 12.85666667

According to this, the growth rate of the native E. coli is approximately 1.57 times that of the genetically edited E. coli. After substituting the ideal E. coli growth data into the equation, three equations are obtained as follows:

LOGISTIC model equation:

$f(t) = \cfrac{4.1334}{1 + e^{-0.3171 * (t - 8.6426)}}$

GOMPERTZ model equation:

$f(t) = 4.5635 * e^{-e^{-0.1841 * (t - 7.1790)}}$

RICHARDS model equation:

$f(t) = \cfrac{4.4202}{1 + e^{-0.2123 * (t - 0.0000)^{4.9718}}}$

Then, the three models are used to fit the growth data of the genetically edited E. coli, and the fitting results are shown in the following figure:

Using the Richards model to draw the growth curves of the two E. coli for comparison, the comparison chart is as follows:

Competitive Prediction Model of Genetically Edited Escherichia coli

Using the classic Lotka-Volterra competition model, a model for predicting competition between engineered bacteria and other bacterial species in the intestinal environment is described, with its general form written as:

$\cfrac{dx}{dt} =x[a_1+b_1x+c_1y]$
$\cfrac{dy}{dt} =y[a_2+b_2x+c_2y]$

Among them, the coefficients are all constants. 6 and c2 respectively reflect the density factors of the two groups, called intraspecific interaction coefficients; G1 and 6: reflecting the factors of interaction between the two groups, called interspecific interaction coefficients; a and a2 respectively represent the innate growth rates of the two groups.

$b_i < 0, c_i <0, i=1,2$

It is easy to see that there are generally four balance positions). They are: $Q(0,0)$

The intersection point $Q(0,-\cfrac{a_2}{c_2})$ between the line $x=0$ and the line $ly$

$a_2+b_2x+c_2y=0;$

The intersection point $P(-\cfrac{a_1}{b_1},0)$ between the line $y=0$ and the line $lx$:

$a_1+b_1x+c_1y=0;$

The intersection between $lx$ and $ly$ $M(x^*,y^*)$

Note that the slopes of 1: and 1: are both negative, and their relative positions are only four possibilities as shown in Figure 2-2. Let $a_1>0,a_2>0$. The following will be discussed separately.

$-\cfrac{a_1}{b_1}<-\cfrac{a_2}{b_2},-\cfrac{a_2}{c_2}<-\cfrac{a_1}{c_1}$
$\cfrac{c_1}{c_2}<\cfrac{a_1}{a_2}<\cfrac{b_1}{b_2}$

It is easy to verify that there exists a stable equilibrium point M. Due to $\cfrac{dx}{dt}=0$ on the line $lx$, it is a vertical isocline. It divides the first quadrant of the phase plane into two parts. Above the line $lx$, $\cfrac{dx}{dt} < 0$; below the line $lx$, $\cfrac{dx}{dt} > 0$.

The other three cases will not have a static equilibrium state M, so only when the following equation is satisfied, the species can survive normally:

$\cfrac{c_1}{c_2} <\cfrac{a_1}{a_2} <\cfrac{b_1}{b_2} $

We use the following differential equations to describe the growth of engineered bacteria and other species in the intestinal environment:

For species 1 (other species):

$$\cfrac{dN_1}{dt}=r_1*N_1*(1-\cfrac{N_1+\alpha *N_2}{K_1})$$

For species 2 (engineered bacteria):

$$\cfrac{dN_2}{dt}=r_2*N_2*(1-\cfrac{N_2+\beta*N_1}{K_2})$$

$\cfrac{dN_1}{dt}$and $\cfrac{dN_2}{dt}$ represent the rate of change in the population size of species 1 and species 2 over time;

$r_1$ and $r_2$ are the intrinsic growth rates of each species;

$α$ and $β$ are the competition coefficients of species 1 to species 2 and species 2 to species 1, respectively;

$K_1$ and $K_2$ are the environmental carrying capacities of species 1 and species 2, respectively,K1 and K2 are the environmental carrying capacities of species 1 and species 2, respectively,

$N_1$ and $N_2$ are the population sizes of species 1 and species 2, respectively;

$t$ is time.

In order to confirm that the engineered bacteria can be settled in the intestine environment, it must meet the following equation:

$\cfrac{r_1 K_2}{r_2 K_1}<\cfrac{r_1}{r_2} <\cfrac{r_1 αK_2}{r_2 βK_1}$

Which can be simplified to

$1<\cfrac{K_1}{K_2} <\cfrac{α}{β}$

Considering that the growth rate of the native E. coli is 1.57 times that of the engineered bacteria, we set the intrinsic growth rate r1 of species 1 to 0.157, and the intrinsic growth rate r2 of species 2 to 0.1. Since the proportion of E. coli edited by gene editing in the intestine is only a very small part, we set the initial population size N1_0 of other species to 500, and the initial population size N2_0 of the engineered bacteria to 1. The influence of the engineered bacteria on other species in the intestine is much smaller than that of other species on it, so the competition coefficient alpha of other species to the engineered bacteria is set to 0.1, which means it has a greater influence, while the competition coefficient beta of the engineered bacteria to other species is set to 0.01, which means it has almost no influence.

Since they are all in the intestinal environment, the difference in their environmental carrying capacity is not large, but at the same time, K1 is greater than K2, K1 = 1000, K2 = 900, and the time range is from 0 to 200 hours.

After inputting the parameters, the following equations are obtained:

For other species:

$$\cfrac{dN_1}{dt} = 0.157 * N_1 * (1-\cfrac{N_1 + 0.1 * N_2}{1000})$$

For engineered bacteria:

$$\cfrac{dN_2}{dt} = 0.1 * N_2 * (1-\cfrac{N_2 + 0.01 * N_1}{900})$$

The equation meets the condition $1<\cfrac{K_1}{K_2} =\cfrac{10}{9}<\cfrac{α}{β}=10$

Using Euler's method to numerically solve, to simulate the population dynamics of engineered bacteria and other species in the intestine, the graph is as follows:

After a certain period of time, both the engineered bacteria and other species reached a steady state, indicating that the engineered bacteria can coexist with other species in the intestinal environment and can be settled in the intestine.

After confirming that the engineered bacteria can survive normally in the intestine, we established a simple rule (that is, the next state of each cell depends on the state of its surrounding cells and its own state: if its own state is 1, the next state is 0; if its own state is 0, the next state is the sum of the states of the surrounding cells) to simulate the growth of a simple engineered bacteria in the intestinal environment, as shown in the following figure:

References

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