Introduction

Building a model can make our experiment more efficient and fast. The experimental data can provide us with the parameters to build the model, and the model can provide guidance for the experiment and solve the difficult problems in the experiment.

Selection and Optimization of PAM Sites

Protospacer Adjacent Motif (PAM) is a short sequence adjacent to the target DNA sequence, and refers to the specific sequence required when Cas proteins are used to recognize foreign DNA sequences in the CRISPR-Cas system. In CRISPR systems, Cas proteins typically identify a target DNA sequence by identifying and binding to a PAM site and performing DNA cuts or other modifications at that location. Different types of Cas proteins have specific PAM preferences, which also limits the scope of the CRISPR-Cas system. Therefore, the correct identification and design of PAM sites is of great significance for the application of CRISPR-Cas system.

In the process of this study, in order to explore the preference of PAM sites of CRISPR-MAD7 system in Corynebacterium glutamicum, to find the PAM sites that are most suitable for the work of MAD7 nuclease complex, and improve the efficiency and accuracy of CRISPR-MAD7 system in gene editing technology in Corynebacterium glutamicum.

In this regard, based on our experimental data, we built a model of the difficulty of binding MAD7 protein to PAM sites, and set up a good template for the subsequent team to select better PAM sites and improve the efficiency of gene editing when studying CRISPR-related technologies.

The editing of the genome of Corynebacterium glutamicum by MAD7 system can be simplified to the following equation.

$$\begin{equation} \begin{split} [pre&\text{-}RNA] \xrightarrow{k1} & [gRNA] \\ & ↓d_{1} & ↓d_{2} \end{split} \tag{1.1} \end{equation}$$

$$\begin{equation} \begin{split} [M&AD7] + [g&&RNA]\xrightleftharpoons[k_{\text{off}}]{k_{\text{on}}}[MAD7\text{-}gRNA] \\ & ↓d_{3} && ↓d_{2} \end{split} \tag{1.2} \end{equation}$$

$$[MAD7\text{-}gRNA] + [PAM] \xrightarrow{k2} [Del]\text{ } \tag{1.3}$$

We can set out differential equations.

We first specify that [pre-RNA] and [gRNA] are the concentrations of RNA precursors and RNA that complete the modification, respectively.\(k_{1}\) represents the conversion rate, \(d_{1}\) and \(d_{2}\) represent the decomposition rate, respectively. Both are degraded by various factors in the cellular environment.

For reaction (1.1), the differential equation is

$$\frac{{ d[gRNA] }}{{ dt }} = k_{1}([pre\text{-}RNA] - d_{1}[pre\text{-}RNA]) - d_{2}[gRNA] \tag{1.4}$$

When the formation rate is 0, the reaction equilibrium is reached

$$[gRNA] = \frac{{ k_{1} [pre\text{-}RNA] (1-d_{1}) }}{{ d_{2 }}} \tag{1.5}$$

For reaction (1.2), [MAD7] represents the translated protein concentration, [gRNA] represents the pre-RNA that has completed the modification, [MAD7-gRNA] is a complex, kon represents the binding rate, \(k_{off}\) represents the off-target rate, and \(d_{2}\) and \(d_{3}\) represent the degradation rate. The differential equation is

\begin{equation} \begin{split} \frac{{ d[MAD7\text{-}gRNA] }}{{ dt }} & = k_{on}([MAD7] - d_{3}[MAD7])([gRNA] - d_{2}[gRNA]) \\ & - k_{off}[MAD7\text{-}gRNA]\end{split} \tag{1.6}\end{equation}

When the formation rate is 0, the reaction equilibrium is reached

$$\frac{{ k_{on}[MAD7][gRNA](1-d_{2})(1-d_{3}) }}{{ k_{off }}} = [MAD7\text{-}gRNA] \tag{1.7}$$

For reaction (1.3), [Del] is cut successfully, \(k_{-2}\) is not cut, or repaired successfully, not complete the splicing, the differential equation is

$$\frac{{ d[Del] }}{{ dt }} = k_2[MAD7\text{-}gRNA][PAM] - k_{-2}[Del] \tag{1.8}$$

When the formation rate is 0, the reaction equilibrium is reached

$$ [Del] = \frac{{ k_2 }}{{ k_{-2 }}} [MAD7\text{-}gRNA][PAM] \tag{1.9}$$

we stipulate

$$\frac{{ k_{on }}}{{ k_{off }}}=K,\text{ }\text{ }\text{ }\text{ }\text{ }\frac{{ k_{2 }}}{{ k_{-2 }}}=K_{2},\text{ }\text{ }\text{ }1-d_{1}=D_{1},\text{ }\text{ }\text{ }\text{ }1-d_{2}=D_{2}\text{ },\text{ } \text{ } 1-d_{3}=D_{3}\text{ }$$

By combining equations (1.5), (1.7), and (1.9), it can be obtained that

$$ [Del] = \frac{{ K^2 D_{1}D_{2}D_{3}k_{1}[MAD7][pre\text{-}RNA] }}{{ d_{2}[PAM] }} \tag{1.10}$$

For the genome editing of Corynebacterium glutamicum, we transferred the plasmid carrying MAD7 protein and gRNA gene, and the final equation showed the relationship between the transcribed MAD7 and pre-RNA and the PAM site under certain conditions, and finally obtained the efficiency of gene editing. To this end, we designed a related experiment, randomly selected 11 PAM sites to edit the genome of Corynebacterium glutamicum, in order to revise our model.

We try to design the different PAM loci in a total of 11, respectively is: TTTA (+), TTTC (-), TTTT (-), TTTG (-), TTTC (+), TTTA (-), TTTT (-), CTTC (+), CTTA (+), CTTG (-), CTTT (+). By comparing multiple PAM sites, we observed that the CRISPR-MAD7 system exhibited the best performance when the PAM site in Corynebacterium glutamicum is TTTC(+). This means that gene editing at this PAM site is easier to achieve and performs well in terms of success and efficiency. Based on this finding, we determined that this PAM site will be an important basis for our subsequent experiments.

Design of downstream products

In our project, the "oxidation rate" is the core point of our downstream product design. The oxidation rate of lycopene was investigated by constructing the oxidation rate model of lycopene. And given by the model, in the relevant temperature and pH conditions, for rapid detection, these actual scenarios will be simulated in the model.

Lycopene, as a highly unsaturated conjugated polyolefin, will degrade under the conditions of light, adaptation temperature and oxygen. In 1989, German scientist Paolo DiMascio experimentally believed that its antioxidant capacity was twice that of beta-carotene and 100 times that of vitamin E, and in subsequent epidemiological studies, it was found that it had a strong ability to quench singlet oxygen and clear free radicals.

Therefore, the lycopene oxidation rate model can be constructed to simulate the oxidation degree in the real environment, and make full preparation for the production of downstream relevant detection reagents.

First, we express the lycopene oxidation process using the following chemical equation:

$$Matter \xrightarrow{k1} M\cdot \tag{2.1}$$

$$M\cdot + O_{2} \xrightarrow{k2} MOO\cdot \tag{2.2}$$

$$Lycopene + MOO\cdot \xrightarrow{k3} MOOH + LOO^- \tag{2.3}$$

Reaction (2.1) describes the process by which substances absorb energy and generate free radicals under the influence of light, heat and enzymes.

Reaction (2.2) describes the process by which a free radical forms a peroxy radical after exposure to oxygen.

Reaction (2.3) describes the process of lycopene in contact with peroxy free radicals to quench free radicals.

In these reactions, we can create differential equations to simulate the system. From these equations, we can better understand the kinetic principle of lycopene oxidation.

Using a set of ordinary differential equations:

Reaction (2.1)

Before this, we specify that [Matter] is the concentration of a substance, [M·] is the concentration of free radicals produced, and we specify [\(k_{2}\)] as the forward reaction rate. Substances absorb energy to produce free radicals. This step is irreversible.

$$Matter \xrightarrow{k1} M\cdot \tag{2.1}$$

We can translate this into a differential equation, using the law of action mass, which describes the relationship between the reaction concentration and the reaction rate, that is, the reaction rate is proportional to the reaction concentration.

$$\frac{d[M\cdot]}{dt} = k_1 [Matter]$$

Reaction (2.2)

We specify that [M·] is the concentration of free radicals, [\(O_{2}\)] is the concentration of oxygen, [MOO·] is the concentration of peroxy free radicals, \(k_{2}\) is the forward reaction rate, and \(k_{-2}\) is the reverse reaction rate. Since both have high energies, the reaction is a reversible process.

$$M\cdot + O_{2} \xrightarrow{k2} MOO\cdot \tag{2.2}$$

Using the mass action law, we can derive the differential equation of (2)

$$\frac{d[MOO\cdot]}{dt} = k_2 [O_2][M\cdot]-k_{-2} [MOO\cdot]$$

Reaction (2.3)

Before this, we specify that [MOO·] is the concentration of peroxy free radicals produced, [Lycopene] is the concentration of lycopene, [LOO-] is the concentration of substances produced after lycopene quench free radicals, and \(k_{3}\) is specified as the forward reaction rate. In actual circumstances, the quenching reaction is usually faster. And the products converted to non-radical compounds rarely react with free radicals again, so as a whole tend to be irreversible. This reaction is irreversible.

$$Lycopene + MOO\cdot \xrightarrow{k3} MOOH + LOO^- \tag{2.3}$$

Using the mass action law, we can derive the differential equation of (3)

$$\frac{d[MOOH]}{dt} = k_3 [Lycopene][MOO\cdot]$$

Ideal model construction

In this regard, we simulated the oxidation rate of lycopene by setting different temperature gradients at 1 atmosphere pressure, and then predicted the oxidation of lycopene at different temperatures.

The oxidative decomposition degree of lycopene was measured by ultraviolet spectrophotometer at different temperatures and at 1 standard atmosphere. Among them, the wavelength of the light wave we detected is 450nm.

Here, we can see that the absorbance of lycopene still changes to some extent even without the peroxidation reaction. After reviewing the literature, we found that before REDOX, lycopene will absorb energy, undergo isomerization, and finally reach isomerization equilibrium, and the oxidation reaction begins. We can get the following simplified equation and derive the differential equation.

$$trans\text{-}Ly \xrightarrow{k_4} cis\text{-}Ly$$

$$\frac{d[ cis\text{-}Ly ]}{dt} = k_{-4}[cis\text{-}Ly]$$

We draw a graph of lycopene’s continuous oxidation and decomposition over time at the same temperature, and build a model based on it to predict the half-life of lycopene and its oxidation and decomposition rate at different times and temperatures.

After a series of data analysis and comparison, it was found that as the temperature increased, the isomerization rate continued to increase, and as the temperature increased, the oxidation rate also continued to increase. Based on the above data, we calculated the isomerization rate and oxidation rate constants at different temperatures.

Here, we can consider the isomerization reaction and decomposition as first-order kinetic reactions, and calculate the reaction rate constant. List integral formulas:

$$ln⁡(\frac{C_{A,0}}{C_A}) = k_A t$$

Temperature / K	Transformation reaction constant /s−
288.15	3.18371 * 10−6
293.15	7.18917 * 10−6
298.15	1.04145 * 10−5
303.15	2.90727 * 10−5
308.15	4.02419 * 10−5
313.15	4.52545 * 10−5
318.15	5.28303 * 10−5
323.15	6.94847 * 10−6

Among them, the activation energy of the isomerization reaction is \(93.1\ KJ/mol\).

We assume that the oxidation reaction follows a first-order reaction and calculate the reaction rate constant of the oxidation reaction. List integral formulas:

$$ln⁡(\frac{C_{A,0}}{C_A}) = k_A t$$

Temperature / K	Oxidation rate constant / s−
288.15	8.36403 * 10−5
293.15	1.19052 * 10−4
298.15	1.30874 * 10−4
303.15	2.48636 * 10−4
308.15	2.61558 * 10−4
313.15	3.02013 * 10−4
318.15	3.33117 * 10−4
323.15	3.88275 * 10−4

The activation energy of the oxidative degradation reaction is \(69.53\ KJ/mol\).

Based on the above reaction, we can calculate the half-life at each temperature and observe the progress of oxidation.

Temperature / K	Half-life period / s
293.15	5822
298.15	5296
303.15	2787
308.15	2650
313.15	2298
318.15	2080
323.15	1785

By consulting relevant literature, we selected the Arrhenius equation to fit the relevant data obtained.

We can obtain two related curves, one is the predicted \(ln(k)\) value and the other is the actual \(ln(k)\) value. Through fitting and data correction, we compare the two on the same chart.

Based on the above data, we can modify the Arrhenius equation and obtain the following equation.

$$ln(k) = -\frac{E_{a}}{kT} + ln(A)$$

In \(ln(A)\), after correction, A=260.91. We can compare the corrected equation with the actual values and fit them accordingly.

It can be seen that the modified parameters can better fit the actual values. In this regard, we can provide a new direction for the application of lycopene on this equation, which can predict the service life of downstream products of lycopene.Based on this equation, let’s better design downstream products and apply lycopene to a wider range of fields.