Biomod/2013/Sendai/calcuation: Difference between revisions

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</br></br>
</br></br>


・Assumed situation</br>
・Assumptions</br>
N<sub>0</sub> surfactant molecule units exist in a solution, and N units of them participate in the formation of the vesicle. One vesicle consists of n surfactant molecule units.</br>
N<sub>0</sub> units of surfactant molecule exist in a solution. We assume that N units of them participate in the formation of the vesicle. One vesicle consists of n units of surfactant molecule.</br>
Here we have been working with DOPC (C<sub>44</sub>H<sub>84</sub>NO<sub>8</sub>P). </br>
In the Lipo-Hanabi project, we adopt DOPC (C<sub>44</sub>H<sub>84</sub>NO<sub>8</sub>P)as the surfactant molecule. </br>
A : Surface area of one molecule =0.6 [nm<sup>2</sup>].</br>  
A : Surface area of DOPC molecule =0.6 [nm<sup>2</sup>].</br>  
V : Volume of the water =10<sup>-4</sup> [m<sup>3</sup>]. </br>
V : Volume of water =10<sup>-4</sup> [m<sup>3</sup>]. </br>
m : Molecular weight of DOPC =785 [g/mol]. </br>
m : Molecular weight of DOPC =785 [g/mol]. </br>
T : Temperature of the system T=300[K].</br>
T : Temperature T=300[K].</br>
Boltzmann’s constant and Planck's constant use the following values.</br>
Boltzmann’s constant and Planck's constant are as follows.</br>




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</br>
</br>


・Calculation result</br>
・Results</br>
Free energy “F” of the vesicle is given as follows*.(*Reference)</br>
Free energy <i>F</i> of the vesicle is given as follows*.(*Reference)</br>




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In this time,</br>
where,</br>




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(pf)<sub>bulk</sub> is a partition function of the single discrete molecule underwater, and (pf)<sub>vesicle</sub> is a partition function of the vesicle.</br>
(pf)<sub>bulk</sub> is a partition function of a single discrete molecule underwater, and (pf)<sub>vesicle</sub> is a partition function of the vesicle.</br>
Therefore, free energy “F”is defined as,</br>
Therefore, free energy F is</br>




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In this calculation, we assume that N, T, and V are constant,</br>
We assume that N, T, and V are constants:</br>




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Because of α= 0 at the moment of N<sub>0</sub> – N = 0</br>
Because α= 0 when N<sub>0</sub>=N</br>




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This time we approximate,</br>
Here, we approximate,</br>




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N/n=X means the number of the vesicles in whole of this system. Therefore we calculate free energy by using this,</br>
N/n=X is the number of the vesicles in the whole space. Therefore the free energy is derived as follows,</br>




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Here, a vesicle which radius is 100μm changes to each 10 and 100 vesicles, calculate each free energy from,</br>
Here, assume a vesicle with radius 100μm is decomposed into 10 and 100 vesicles. Calculated free energy of each situation is:</br>




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Based on these results, we get the following figure 3 (fig 3).</br>


<img src="http://openwetware.org/images/c/c1/Fig.3.jpg" width="400"></br>
<img src="http://openwetware.org/images/c/c1/Fig.3.jpg" width="400"></br>
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</br>
</br>


From fig 3, we can conclude free energy is smaller when the radius of vesicle is small.</br></br>
</br></br>


・Discussion</br>
・Discussion</br>

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        <h2>Calculation</h2>

<h3>Destruction the liposome</h3>


</br>

Following mathematical formula reveals proves that smaller liposomes are more stable. From this calculation result, we think it is possible to collapse liposomes if we provide with energy by trigger. </br>

</br>

・Hypothesis</br> We assume that a vesicle of large radius existing underwater changes to the plural vesicles of small radius (fig 1). The free energy of each states F1: large radius, and F2: small radius, probably have the following relationship (fig 2).
 In our case, the vesicle is basically small state because a small radius is more stable.The decisive factor is, however, that there is the energy gap “δ” that must exceed to change the size of vesicle.
In our calculation we have not had the possibility to calculate δ, but to calculate the free energy in each radius. With this values of energy and radius the energy gap "δ" can be deducted.</br>

<table> <tr> <td> <img src="http://openwetware.org/images/7/73/Fig.1.jpg" width="400"> </td> <td> <img src="http://openwetware.org/images/2/2e/Fig.2.jpg" width="400"> </td> </tr> <tr> <td> Fig.1 Assumed state of vesicle in the system </td> <td> Fig.2 Assumed relation of each free energy </td> </tr> </table>


</br></br>

・Assumptions</br> N<sub>0</sub> units of surfactant molecule exist in a solution. We assume that N units of them participate in the formation of the vesicle. One vesicle consists of n units of surfactant molecule.</br> In the Lipo-Hanabi project, we adopt DOPC (C<sub>44</sub>H<sub>84</sub>NO<sub>8</sub>P)as the surfactant molecule. </br> A : Surface area of DOPC molecule =0.6 [nm<sup>2</sup>].</br> V : Volume of water =10<sup>-4</sup> [m<sup>3</sup>]. </br> m : Molecular weight of DOPC =785 [g/mol]. </br> T : Temperature T=300[K].</br> Boltzmann’s constant and Planck's constant are as follows.</br>


<div align="center"><img src="http://openwetware.org/images/2/29/Cal-00.png" width="293" height="102"></div>


</br>

・Results</br> Free energy <i>F</i> of the vesicle is given as follows*.(*Reference)</br>


<div align="center"><img src="http://openwetware.org/images/c/c8/Cal-01.png" width="200" height="70"></div>


where,</br>


<div align="center"><img src="http://openwetware.org/images/b/bf/Cal-02.png" width="424" height="139"></div>


(pf)<sub>bulk</sub> is a partition function of a single discrete molecule underwater, and (pf)<sub>vesicle</sub> is a partition function of the vesicle.</br> Therefore, free energy F is</br>



<div align="center"><img src="http://openwetware.org/images/6/6f/Cal-03.png" width="529" height="98"></div>



We assume that N, T, and V are constants:</br>



<div align="center"><img src="http://openwetware.org/images/3/33/Cal-ima-04-03.png" width="400" height="81"></div>


<div align="center"><img src="http://openwetware.org/images/d/d8/Cal-ima-04-02.png" width="490" height="85"></div>



Because α= 0 when N<sub>0</sub>=N</br>




<div align="center"><img src="http://openwetware.org/images/6/6a/Cal-05.png" width="337" height="93"></div>



Here, we approximate,</br>


<div align="center"><img src="http://openwetware.org/images/7/7c/Cal-06.png" width="254" height="42"></div>


N/n=X is the number of the vesicles in the whole space. Therefore the free energy is derived as follows,</br>



<div align="center"><img src="http://openwetware.org/images/7/77/Cal-07.png" width="400" height="42"></div>


Here, assume a vesicle with radius 100μm is decomposed into 10 and 100 vesicles. Calculated free energy of each situation is:</br>



<div align="center"><img src="http://openwetware.org/images/6/6c/Cal-08.png" width="342" height="107"></div>



<img src="http://openwetware.org/images/c/c1/Fig.3.jpg" width="400"></br> Fig.3 The relation of each free energy by calculation<br> </br>

</br></br>

・Discussion</br> When all molecules participate in the formation of the vesicle, the free energy becomes a one-tenth time and the number of vesicle increase 10 times. 
Smaller radius vesicles are more stable.Therefore, it is possible to destroy vesicles using trigger. <br><br>

・Reference</br> Yukio Suezaki "Physics of lipid film"



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