Team:UFSCar-Brasil/part1.html
Plasmolysis
How long Bug Shoo could be stored?
Fitting Data
Starting from the experimental data presented in Table 1 in the appendix, a 3D graph (Figure 1) was constructed. It was used to better system understanding, which is the relationship between absorbance, time and PEG 6000 concentration.
Figure 1: Experimental plots of absorbance versus time versus PEG concentration.
This graph was used to find best-fitted surface to experimental points. To this, parameter r² was used, because it shows the percentage of points that fit in a certain curve. Thus, we observed the following equation to explain this behaviour:
\(z = z_0+ax+by+cx²+dy²+fxy\).
This curve obtained r² = 0.95457, which means that more than 95% of data are modeled by this function. Using the MatLab software to function fitting. We also find the value of equation constants (Appendix Table 2).
The surface equation is:
\(z = 1.07+0.06x-0.044y-0.04x²+0.0009y²+0.003xy\) (Eq. 1)
Where, X means time and Y means PEG concentration. From known curve was possible to simulate a surface that models most likely behavior of the three variables in question.
Figure 2: Fitted surface obtained using the Eq. 1.
Results and Discussion
Our goal to find main points of system, meaning those points where plane tangent in surface has a null gradient value. First, it was analyzed (using a Hessian matrix) on these important points to make possible to know if these are the maximum points, minimum or saddle points of the function.
The calculation of the Hessian matrix is given by:
$$ H[z(x,y)] = \begin{bmatrix} \frac{\partial^2 z}{\partial x^2} & \frac{\partial^2 z}{\partial z,\partial y} \\ \frac{\partial^2 z}{\partial z,\partial y} & \frac{\partial^2 z}{\partial y^2} \end{bmatrix} $$
thus to find the value of the Hessian of z(x,y) we obtained the partial derivatives of the second order, whose values are below:
$$ \frac{\partial^2 z}{\partial x^2} = 2c = - 0,08774 $$
$$ \frac{\partial^2 z}{\partial y^2} = 2d = - 0,00178 $$
$$ \frac{\partial^2 z}{\partial x,\partial y} = \frac{\partial^2 z}{\partial y,\partial x} = f = 0,00319 $$
with the known partial derivatives, we have the Hessian:
$$ H[z(x,y)]=(-0,08774)(0,00178)-(0,00319)^2=-0,0001663533 $$
As H[f(x,y)] <0, we know that this function posses a saddle point and there are no minimum or maximum values (Guidorizzi, 1997). Saddle point is understood as the point where the slope is zero, but it is not a maximum or minimum value. In it there is the highest elevation in one direction and lowest in the perpendicular direction.
Now, known the point we want to find, we take the partial derivatives of the first order function and made them equal to zero. To know:
$$ \frac{dz}{dx}=a+2cx+fy=0 $$
$$ \frac{dz}{dy}=b+2dy+fx=0 $$
Therefore:
$$ -\frac{a+2cx}{f}=y=\frac{-b+fx}{2d} $$
replacing the values that are found in Table 2 of the appendix, we have that $$x=\frac{(2{da}-bf)} {(f²-4cd)}=1,5$$
And thus y=22.2 and z=0.63 . Then it is possible to concluded that for optimal absorbance point (z = 0.63), it is necessary 22.2% PEG 6000 where it gets 1.5 weeks of plasmolysis (Figure 3).
Figure 3: Extrapolated fitted surface for extreme values of PEG concentrations and time.
It is possible to observe that for very long time values the absorbance decrease to zero very fast and the time do not surpass 8 weeks. The PEG concentrations also increase the absorbance values for both extreme concentrations, when approximate of zero and pass of 30% of PEG 6000 the absorbance value explodes. Despite mathematically valid, biologically this means that near to zero there will be no control of cell division, and for values near to 30% of PEG 6000, culture reach a plateau of minimum cell division, followed by a decreasing function.
Conclusions
Through these analyzes we could first find one function that suited in r² = 95%, which is very satisfactory, and validates our whole discussion about the obtained surface. Then, using Hessian matrix, we found a saddle point, which is particularly biologically relevant. The variation of absorbance along time indicates two things to us: first, cell numbers, and second, the metabolic activity of cells, since they are active, they should divide increasing absorbance. Due to its plasmolysed state, variation rates of absorbance values should be low, but not zero, because if so, they would be dead and would not serve for our propose. As it is a saddle point, we know that it is not null. Finally, the point was found through gradient of the function and we concluded that for the minimum value of absorbance the best PEG concentration is 22,2% with a maximum time of 8 weeks of storage.
Appendix
Table 1: Experimental data obtained in microbiology lab
| Time | PEG | Absorbance |
|---|---|---|
| Weeks | % | OD |
| 24h | 0 | 1,111 |
| 1 | 0 | 1,114 |
| 2 | 0 | 1,057 |
| 24h | 5 | 0,854 |
| 1 | 5 | 0,845 |
| 2 | 5 | 0,813 |
| 24h | 10 | 0,729 |
| 1 | 10 | 0,779 |
| 2 | 10 | 0,725 |
| 24h | 15 | 0,675 |
| 1 | 15 | 0,694 |
| 2 | 15 | 0,714 |
| 24h | 20 | 0,532 |
| 1 | 20 | 0,635 |
| 2 | 20 | 0,609 |
| 24h | 25 | 0,532 |
| 1 | 25 | 0,635 |
| 2 | 25 | 0,609 |
Table 2: Constants values of the expression fitted:
| Value | Error | |
|---|---|---|
| z0 | 1,07642 | 0,03278 |
| a | 0,06261 | 0,05302 |
| b | -0,04442 | 0,00406 |
| c | -0,04387 | 0,02244 |
| d | 8,90E-04 | 1,45E-04 |
| f | 0,00319 | 0,0014 |
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