Difference between revisions of "Team:Kent/Modeling"

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<p align="center"> More to come soon... </p>
 
<p align="center"> More to come soon... </p>
  
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<h3 align="left"> Contents </h3>
 
<h3 align="left"> Contents </h3>
 
<a href="#c1">Variables</a>  <br>
 
<a href="#c1">Variables</a>  <br>
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</table>
 
</table>
  
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<table>
 
<tr>
 
<tr>
 
     <th> Event </th> <th> Description </th> <th> Value </th>
 
     <th> Event </th> <th> Description </th> <th> Value </th>
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We have taken a small proportion of the cell and the bulk outside of the cell and extrapolated it to represent the whole system.
 
We have taken a small proportion of the cell and the bulk outside of the cell and extrapolated it to represent the whole system.
 
</p><br><br>
 
</p><br><br>
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<p>The typical length of an E coli cell is \[L~2 \mu m] and the typical diameter, \[d~1 \mu m]. If we consider the cell to be composed of a cylinder of length \[l=L-d] and two hemispheres at each end then the volume of the sphere is:</p>
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<p>The typical length of an E coli cell is \(L \approx 2 \mu m\) and the typical diameter, \(d \approx 1 \mu m\). If we consider the cell to be composed of a cylinder of length \(l=L-d\) and two hemispheres at each end then the volume of the sphere is:</p>
 
<br><br>
 
<br><br>
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\[V_cell = V_cylinder + V_sphere]
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\[V_{cell} = V_{cylinder} + V_{sphere}\]
 
<br>
 
<br>
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\[V_cell = \pi  (\frac{d}{2})^2 (L-d)+ \frac{4}{3} \pi  (\frac{d}{2})^3 = 1.309 (\mu m)^3]
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\[V_{cell} = \pi  \big(\frac{d}{2}\big)^2 (L-d)+ \frac{4}{3} \pi  \big(\frac{d}{2}\big)^3 = 1.309 (\mu m)^3\]
 
<br>
 
<br>
 
<p>The volume of the cell inside the observation box is</p>
 
<p>The volume of the cell inside the observation box is</p>
 
<br><br>
 
<br><br>
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\[V_ob = L_x L_y L_z = 0.08(\mu m)^3 ]
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\[V_{ob} = L_x L_y L_z = 0.08(\mu m)^3\]
 
<br><br>
 
<br><br>
 
<p>From this we can work out the proportion of the volume of cell inside the observation box</p>
 
<p>From this we can work out the proportion of the volume of cell inside the observation box</p>
 
<br><br>
 
<br><br>
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\[ \frac{V_ob}{V_cell } = 6.112% ]
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\[\frac{V_{ob}}{V_{cell} } = 6.112%\]
 
<br><br>
 
<br><br>
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<p>LIkewise fro the surface area of the cell</p>
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<p>Likewise for the surface area of the cell</p>
 
<br><br>
 
<br><br>
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\[A_cell = A_cylinder + A_sphere]
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\[A_{cell} = A_{cylinder} + A_{sphere}\]
 
<br><br>
 
<br><br>
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\[A_cell = \pi d(L-d) + 4 \pi (\frac{d}{2})^2 = 6.28 (\mu m)^2
+
\[A_cell = \pi d(L-d) + 4 \pi (\frac{d}{2})^2 = 6.28 (\mu m)^2 \]
 
<br><br>
 
<br><br>
 
<p>The proportion of the cell's surface area inside the observation box is: </p>
 
<p>The proportion of the cell's surface area inside the observation box is: </p>
 
<br><br>
 
<br><br>
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\[\frac{A_ob}{A_cell} = 2.54%]
+
\[\frac{A_{ob}}{A_{cell}} = 2.54% \]
 
<br><br>
 
<br><br>
 
<a name="c3"></a><h3 align="center"> M</h3>
 
<a name="c3"></a><h3 align="center"> M</h3>
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<p>Our Matlab code can be found here: <a href="    ">Kent15_Model.m</a></p>
 
<p>Our Matlab code can be found here: <a href="    ">Kent15_Model.m</a></p>
  
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</div>
 
</div>
  

Revision as of 17:10, 6 September 2015


iGEM Kent 2015

Modeling

Modeling is important as it allows us to describe the system mathematically. If we change some of the parameters in our system we can see how this will affect the system, this is especially important when the some of the parameters are unknown. The main aim of our model is to demonstrate the production of our nanowires in an interactive and interesting way.



More to come soon...