. One imitated the male multicellular organism, while the other was for the female (called G+ cells and G- cells respectively). In this project, the modified G+ and G- cells would be cultivated separately first, and then put together. Therefore, we had to simulate the two phases. The first part focused on gamete forming (from G+/G- to P+/P-), while the second part was forming of the “zygotes” (G cells, from P+ and P- cells) and the separation of them to G+ and G-. These processes are shown in figure 1.

1. Although signal molecule growth rate of each *E. coli* changes, considering the number of colonies of *E. coli*, we ignore the growth rate changes.

2. For a quorum sensing system, *cre* gene was activated to adapt the changes of the environment only when the amount of signal molecule reached a certain threshold. So we could assume that the concentration in each *E. coli* is nearly the same.

According to the previous assumption, the amount of signal molecule can be expressed by N

_{sign.}Where K

_{sig} is a coefficient constant that represents the amount of the signal molecule produced by one

*E. coli*. We know from the experimental data that when OD600 reaches 0.5,

So we make k_{sig} =4.82×10^{8}
When the amount of signal molecule reaches the threshold value, the *cre* gene is activated to start generating the *cre*. We let t_{1} be the point when the amount of signal molecule reaches the threshold value. We let t_{2} be the point when the amount of *cre* reaches a certain number and the

*E. coli* starts out to transform. So here is the equation to describe the *cre.*

T is the half-life. K_{c} is a growth constant.

If we let k_{c}=6.1538×10^{-11}, t_{2}=1400, then we get figure 2, 3, 4.

Fig 2 Amount of *E. coli* (k_{c}=6.1538×10^{-11}，t_{2}=1400)

Fig 3 Amount of signal molecule (k_{c}=6.1538×10^{-11}，t_{2}=1400)

Fig 4 Amount of *cre* (k_{c}=6.1538×10^{-11}，t_{2}=1400)

If we let k_{c} =6.1538×10^{-11}，t_{2}=300, then we get figure 5, 6, 7.

Fig 5 Amount of *E. coli'*** **( k_{c} =6.1538×10^{-11}，t_{2}

Fig 6 Amount of signal molecule (k_{c} =6.1538×10^{-11}，t_{2}=300）

Fig 7 Amount of *cre* （k_{c} =6.1538×10^{-11}，t_{2}=300）

If we let k_{c} be nine different values: 0.0615×10^{-8}, 0.1231×10^{-8} , 0.1846×10^{-8}, 0.2462×10^{-8}, 0.3077×10^{-8}, 0.3692×10^{-8}, 0.4308×10^{-8}, 0.4923×10^{-8}, 0.5538×10^{-8}, then we get figure 8,

Fig 8 Amount of *cre*（t_{2}=1400）

According to this figure, although we do not know the exact value of kc, we are sure that, despite of the value of k_{c}, the increasing speed of *cre* will be sharply accelerated in 600 minutes and will be slow down to zero in 1000 minutes. These figures implicate that we could test the fluorescence after about 600- minute culture.

2.2.2 The probability of transition

Then we estimate that when the concentration of enzymes is x, the probability of transition is the thing that determine the cause of G+/G- transferring to P+/P-. Using the previous results, we know that concentrations of *cre* at each time.

Based on the assumption 2, we know that concentrations of *cre* for the each *E. coli *(ξ_{1},…, ξ_{n}…) are random, independent and identically distributed.

Drawn N samples from the data obtained in our experiments, and then:

Step 1. Calculating the sample's expectations E (ξ) =μ, variance D (ξ) =σ^{2 }and standardization

Step 2. By the Lindbergh - Levi central limit theorem， we get function of concentration:

So we can get the concentration after standardized variables, and probability of unturned bacterial under this concentration is Φ(x). Then the ratio of transformation is 1-Φ(x).

Since our test has not been completed, the lacking of this part of the data led us not to reach concrete results.

**3 Mix two gametes group**

**3.1 Model hypothesis**

1) The broth and culture condition for each gamete and zygote is the same.

2) The nutrition of each plate for bacteria growth is sufficient and there is no stress from environment.

3) It takes some time for zygote to differentiate into gametes of different gender, we assume that this process lasts for a long time and the differentiation of zygote is after the combination of gametes.

**3.2 modeling**

3.2.1Gametes combination

After G+ and G- cells has been cultured separately and converted into P+ and P- , two culture systems are mixed to form a new system. Five different kinds of cells as mentioned above exist in the system, and only the G+ and G- cells have the ability to proliferate while the “zygotes”, G cells can also divide. The moving traces of G cells in the petri dish are independent of each other. Because the medium in the petri dish is homogeneous, its trajectory can be approximately viewed as Brownian motion.

Simply we can take the whole petri dish as a region "E" which has a circle boundary. The collection of locations which P+（or P-）moving through all the place called state space(That is the entire dish area). Since Brownian motion is essentially a Markov random process, its Markov property:

For any

we have

let locations of P+ and P- were n1 and n2, then P+ and P- will combine when p[x(n1)]=p[x(n2)] at the same time.

So Markov process is describing a probability of that gametes reach each state (position). In this process, two gametes become zygote through collision. Therefore when at the same time t, P+ and P- reach the same state E0 (That collide with each other form a zygote). Thus, we describe a stochastic process with complete binding process. We can calculate the gametes P+ and P- where at the states of e1 and e2 in initial moment by a Markov probability equation:

1) How long it takes for gametes to combine into a zygote?

2) The combination probability at time t.

3.2.2 Zygote division

When the entire population of gamete combine into zygotes after a period of culture, these zygotes will redivide into G+ and G- cells. This is a natural process of cell division.