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Revision as of 14:16, 16 October 2018
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Model
Overview
The models we built included four parts. First, we established a fluorescent protein model to screen out the most suitable fluorescent protein, the main modeling method here is grayscale analysis. Then, for the large amount of measured OD values, we drew the growth curve of yeasts and it fitted logistic model. It described the growth situation of the yeasts after plasmid introduction, and we compare it with yeasts without any foreign plasmid. The growth curve also offers the best measuring point and the best measuring interval. What’s more, we drew the degradation curve of the fluorescent protein, which helps us know different characteristics of the two chosen fluorescent proteins better. Finally, we constructed a model to illustrate the oscillation of KaiA, KaiB and KaiC protein called Mars Model, it explained the reason why the cycle reduced in yeasts nicely. Modeling work integrated with experiments tightly made our project complete and convincing.
As a lot of fluorescent proteins and luciferase existing, an evaluation model to evaluate the performance of every fluorescent protein was established to find the most suitable one.
relationship | ai=aj | ai>aj(a little) | ai>aj | ai>aj(obviously) | ai>>aj |
score | 1 | 3 | 5 | 7 | 9 |
There were five factors we should consider. To determine the weight of them, we discussed with Prof. Zhao Kun, school of chemical engineering and technology, Tianjin University and professors in Tianjin Institute of Industrial Biotechnology, Chinese Academy of Sciences, they gave us a suggested order as shown below.
Between every two standards, select numerical value 1, 3, 5, 7 to measure the relative relation between index.
Stokes | QY*Brightness | Bleaching time | ph | Maturing time | |
Stokes | 1 | 3 | 5 | 1/3 | 5 |
QY*Brightness | 1/3 | 1 | 3 | 1/5 | 3 |
Bleaching time | 1/5 | 1/3 | 1 | 1/7 | 1 |
ph | 3 | 5 | 7 | 1 | 7 |
Maturing time | 1/5 | 1/3 | 1 | 1/7 | 1 |
Construct standard matrix shown as below.
$$A = \begin{bmatrix} 1 & 3 & 5 & 1/3 & 5 \\ 1/3 & 1 & 3 & 1/5 & 3 \\ 1/5 & 1/3 & 1 & 1/7 & 1 \\ 3 & 5 & 7 & 1 & 7 \\ 1/5 & 1/3 & 1 & 1/7 & 1 \end{bmatrix}$$
Next, calculate the concordance index (CI) to judge whether the matrix is reasonable.
$$CI = {\lambda_{max}-n \over n-1} = 0.0340$$
$$\lambda_{max} = 5.1361$$
λmax -- the largest eigenvalue of the evaluation matrix and n is the dimension of the matrix.
According to the dimension of the matrix, RI value could be found from the form below.
n |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
RI |
0 |
0 |
0.58 |
0.90 |
1.12 |
1.24 |
1.32 |
1.41 |
1.45 |
In our model, because n=5, RI=1.12.
$$CR = {CI \over RI} = 0.0304$$
Because CI<0.1, the matrix passes the consistency check. Therefore, the method we used to find the weight above is suitable.
Calculate the corresponding eigenvector of λmax and normalize it, W=[0.0762 0.1607 0.3621 0.0389 0.3621].
At last, using the method mentioned in the [1], the score table is shown as below.
Score table | |||||
---|---|---|---|---|---|
score | 0 | 25 | 50 | 75 | 100 |
Stokes | 0 | 0-0.02 | 0.02-0.05 | 0.05-0.1 | 0.1- |
QY*Brightness | 0 | 0-0.01 | 0.01-0.04 | 0.04-0.06 | 0.06- |
Bleaching time | 0 | 0-0.01 | 0.01-0.10 | 0.10-0.20 | 0.20- |
pH | 0 | 1.5-2.5 8.5-9.5 | 2.5-3.5 7.5-8.5 | 3.5-4.5 6.5-7.5 | 4.5-6.5 |
Maturing time | 0 | 0-0.01 | 0.01-0.05 | 0.05-0.30 | 0.30- |
The score of every fluorescent protein can be obtained shown as Figure1.
![](https://static.igem.org/mediawiki/2018/c/cb/T--Tianjin--tutu1.png)
Figure1 Selection of report genes
![](https://static.igem.org/mediawiki/2018/f/f8/T--Tianjin--tutu2.png)
Figure2 Fluorescent proteins could be chose from
It shows that mCherry, mOrange and EYFP ranks top 3. However, after our pre-experiment, it was found that the signal of mOrange was too weak. The reason maybe was that we sequenced the part of mOrange and found that there were some mutations. Therefore, we finally chose mCherry(66.0336 ranking 12/478) and EYFP(34.9807 ranking 82/478) as our report genes.
The degradation of the fluorescent protein is also of great importance to our experiment. Therefore, the change of fluorescence intensity with time was measured, and here are the results.
![](https://static.igem.org/mediawiki/2018/0/05/T--Tianjin--tu3.png)
Figure3 EYFP Degradation Curve
![](https://static.igem.org/mediawiki/2018/f/f1/T--Tianjin--tu4.png)
Figure4 mCherry Degradation Curve
To better explain the degradation of the fluorescent protein, we consulted a large number of documents[2,3,4,5]. It shows that the degradation of the fluorescent protein is exponential.
The EYFP degradation curve can be described by function (1):
$$f(x) = a·e^{bx} + c·e^{dx} (1) $$
Coefficients (with 95% confidence bounds):
a = 133, b =-0.005066, c =-44.38, d =-0.02168
![](https://static.igem.org/mediawiki/2018/7/7a/T--Tianjin--tu5.jpg)
Figure5 Fitted EYFP Degradation Curve
Goodness of fit:
SSE: 76.77
R-square: 0.9443
Adjusted R-square: 0.9363
RMSE: 1.912
The mCherry degradation curve also can be described by function (1), but the coefficients are different.
Coefficients (with 95% confidence bounds):
a = 613.8, b = -0.0003886, c = 0.0003207, d =0.06852
![](https://static.igem.org/mediawiki/2018/4/42/T--Tianjin--tu6.jpg)
Figure6 Fitted mCherry Degradation Curve
Goodness of fit:
SSE: 82.27
R-square: 0.9482
Adjusted R-square: 0.9404
RMSE: 2.028
The fitted degradation curve of mCherry approximates a straight line, so it can be simplified as a linear function:
$$ f(x) = p_1·x + p_2 (2)$$
Coefficients (with 95% confidence bounds):
$$ p_1 = -0.2287 , p_2 = 613.6$$
![](https://static.igem.org/mediawiki/2018/3/35/T--Tianjin--tu7.jpg)
Figure7 Linear mCherry Degradation Curve
Goodness of fit:
SSE: 82.67
R-square: 0.9479
Adjusted R-square: 0.9455
RMSE: 1.938
It can be seen that the degradation rate of different fluorescent protein is very different, which made it clearer for us to know about the characteristics of the two fluorescent protein and offered important information to our experiment.
For the OD600 values we got, we did some processing and modeling work. And here are our steps and results.
There were three groups in our experiment. They were blank control group, partial control group and experimental group. After getting all the data, first, we drew a histogram and a scattergram of time and maximum OD600 values (Figure8, 9). These results were very instructive to experiments that these results told us the best measuring point and the best measuring interval.
![](https://static.igem.org/mediawiki/2018/0/0e/T--Tianjin--tu8.png)
![](https://static.igem.org/mediawiki/2018/0/0d/T--Tianjin--tu9.png)
![](https://static.igem.org/mediawiki/2018/0/0e/T--Tianjin--tu8.png)
Figure8 histogram of Time-Maximum OD Value
![](https://static.igem.org/mediawiki/2018/0/0d/T--Tianjin--tu9.png)
Figure9 Scatter gram of Time-Maximum OD Value
From the beginning to the maximum OD600 value, it fits the logistic model. The block effect of resource and environment for the growth of yeasts is reflected in the growth rate r, which makes r decrease with the increase in the number of yeasts x. Express r as a function r(x) of x, and take a simple and convenient linear reduction function r(x)=a+bx. In order to give a real meaning to the coefficients a and b in the growth rate function, we introduced two parameters:
(1)Intrinsic growth rate r : r is the growth rate when x=0 (in theory);
(2)Population capacity xm : xm is the largest yeast amount that can be accommodated by resources and the When x=xm, the quantity of yeasts is no longer increasing, that is r(xm)=r+bxm=0, then b=-r/xm.
r and xm values in our experiments are shown in the chart below.
YPD | SC | BY4741 | D-THREE | |
---|---|---|---|---|
r |
|
| 0.0164 | 0.0172 |
|
| 0.8523 | 0.8034 | |
pABaC+p1m | pABaC+p1E | pABaC+p2N | pABaC+p1F | |
r | -0.002364402 | 0.001746617 | -0.002826764 | -0.001905785 |
0.402523944 | 0.508816901 | 0.424323944 | 0.542298592 | |
pCiRbS+p1m | pCiRbS+p2N | pCiRbS+p1F | pbCiRS+p1m | |
r | -0.006367923 | -0.007098618 | -0.007176452 | -0.007853975 |
0.410507042 | 0.254873239 | 0.446169014 | 0.315098592 | |
pbCiRS+p1E | pbCiRS+p2N | pbCiRS+p1F | pABaC+pCiRbS+p1m | |
r | -0.024143608 | -0.012145451 | 0.002428334 | -0.006280764 |
0.413985915 | 0.458239437 | 0.270442254 | 0.337278873 | |
pABaC+pCiRbS+p1E | pABaC+pCiRbS+p2N | pABaC+pCiRbS+p1F | pABaC+pbCiRS+p1m | |
r | 0.002305512 | -0.00217225 | 0.002272595 | 0.002039534 |
0.33171831 | 0.293661972 | 0.303701408 | 0.289346479 | |
pABaC+pbCiRS+P1e | pABaC+pbCiRS+p2N | pABaC+pbCiRS+p1F |
| |
r | 0.001894111 | -0.003848457 | -0.007151104 |
|
0.301574648 | 0.345819718 | 0.329769014 |
The resulting growth rate function is $$r(x) = r(1 - {x \over x_m}) $$ Replacing intrinsic growth rate with r(x), get $${dx \over dt} = rx(1 - {x \over x_m}) , x(0) = x_0 (3)$$
Factor rx in the function shows the growth trend of yeast amount itself, while factor (1 - x/xm) reflects the block effects of resources and environment to the yeast quantity growth. Obviously, the bigger x is, the former factor is bigger and the latter factor is smaller. The growth of yeast amount is the result of the two factors.
Take x as the horizontal axis and dx/dt as the vertical axis, we obtained a parabola (Figure 1), when x = xm/2, dx/dt reaches the maximum. As shown in Figure 10, dx/dt changes with the increasing x, and we can do the following analysis to the curve x(t).
Setting when t = 0 x0 < xm/2, with the increase of t, dx/dt increases, so x is growing faster and faster and the curve x(t) is raised downward; dx/dt decreases when x0 > xm/2, x grows slower and slower and the curve x(t) is raised upward. x = xm/2 is the inflection point of the curve. When x → xm , dx/dt → 0, so x = xm is the asymptote of x(t). From the above analysis, we can draw the figure as shown in Figure 11.
Actually, equation (1) can be solved by separation of a variable method as
$$x(t) = {x_m \over {1+({x_m \over x_0}-1)e^{-rt}}} (4) $$
![](https://static.igem.org/mediawiki/2018/f/fe/T--Tianjin--tu1010.jpg)
Figure10 example x-dx/dt curve
![](https://static.igem.org/mediawiki/2018/0/00/T--Tianjin--tu11.jpg)
Figure11 example t-x curve
By Analyzing our data, following results were obtained similar to the above. Figure12,13 shows the t-x curve and x-dx/dt curve of our blank control group, partial control group and experimental group. To make the results clearer, we enlarged two of them.
![](https://static.igem.org/mediawiki/2018/4/46/T--Tianjin--tu12.jpg)
Figure12 t-x curve
![](https://static.igem.org/mediawiki/2018/1/15/T--Tianjin--tu13.jpg)
Figure13 x-dx/dt curve