# Biomod/2013/Sendai/calcuation

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- Fig.1 Assumed state of vesicle in the system +
Fig.1 Assumed state of vesicle in the system
- Fig.2 Assumed relation of each free energy +
Fig.2 Assumed relation of each free energy
Line 73: Line 79: - ・Assumed situation
+

- N0 surfactant molecules units exist in solution, and N units of them participate in the formation of the vesicle. One vesicle consists of n surfactant molecules units.
+ - In this time,we use DOPC (C44H84NO8P).
+ ・Assumptions
- A : Surface area of one molecule =0.6 [nm2].
+ N0 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.
- V : Volume of the water =10-4 [m3].
+ In the Lipo-Hanabi project, we adopt DOPC (C44H84NO8P)as the surfactant molecule.
+ A : Surface area of DOPC molecule =0.6 [nm2].
+ V : Volume of water =10-4 [m3].
m : Molecular weight of DOPC =785 [g/mol].
m : Molecular weight of DOPC =785 [g/mol].
- T : Temperature of the system T=300[K].
+ T : Temperature T=300[K].
- Boltzmann’s constant and Planck's constant use the follows.
+ Boltzmann’s constant and Planck's constant are as follows.
-
+

- ・Calculation result
+ ・Results
- Free energy “F” of the vesicle is given as follows*.(*Reference)
+ Free energy F of the vesicle is given as follows*.(*Reference)
-
+
- In this time,
+ where,
-
+
- (pf)bulk is a partition function of the single discrete molecule underwater, and (pf)vesicle is a partition function of the vesicle.
+ (pf)bulk is a partition function of a single discrete molecule underwater, and (pf)vesicle is a partition function of the vesicle.
- Therefore, free energy “F”,
+ Therefore, free energy F is
-
+
- In this calculation, we assume that N, T, and V are constant,
+ We assume that N, T, and V are constants:
-
+
+ Because α= 0 when N0=N
- α= 0 at the moment of N0 – N = 0, so simplify,
+
-
+ Here, we approximate,
- We do not know the from of (pf)vesicle, so this time we assume,
+
-
+ N/n=X is the number of the vesicles in the whole space. Therefore the free energy is derived as follows,
- N/n=X means the number of the vesicle of the whole system. So we calculate free energy by using this,
+
-
+ Here, assume a vesicle with radius 100μm is decomposed into 10 and 100 vesicles. Calculated free energy of each situation is:
- Here, a vesicle which radius is 100μm changes to each 10 and 100 vesicles, calculate each free energy,
+
-
+

+
Fig.3 The relation of each free energy by calculation

+
- +

- From these results, we get the follow (fig 3).
+ - + -
+ - Fig.3 The relation of each free energy by calculation
+ - + - + - From fig 3, free energy is smaller when the radius of vesicle is small.

+ ・Discussion
・Discussion
- When all molecules participate in the formation of the vesicle, the free energy becomes a one-tenth time when the number of vesicle increase 10 times.  In addition, we think that the vesicle of the small radius is more stable.

+ 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 Key DNA.

・Reference
・Reference
- Yukio Suezaki "Physics of lipid film" + Y. Suezaki, "Shishitsu maku no butsuri (Physics of lipid membrane)", Kyusyu Univ. Press, 2007(in Japanese)

## Current revision

Biomod2013 Sendai ver2.0

# Biomod2013  TeamSendai

## Calculation

### Destruction the liposome

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 Key DNA.

・Hypothesis
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.
 Fig.1 Assumed state of vesicle in the system Fig.2 Assumed relation of each free energy

・Assumptions
N0 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.
In the Lipo-Hanabi project, we adopt DOPC (C44H84NO8P)as the surfactant molecule.
A : Surface area of DOPC molecule =0.6 [nm2].
V : Volume of water =10-4 [m3].
m : Molecular weight of DOPC =785 [g/mol].
T : Temperature T=300[K].
Boltzmann’s constant and Planck's constant are as follows.

・Results
Free energy F of the vesicle is given as follows*.(*Reference)
where,
(pf)bulk is a partition function of a single discrete molecule underwater, and (pf)vesicle is a partition function of the vesicle.
Therefore, free energy F is
We assume that N, T, and V are constants:
Because α= 0 when N0=N
Here, we approximate,
N/n=X is the number of the vesicles in the whole space. Therefore the free energy is derived as follows,
Here, assume a vesicle with radius 100μm is decomposed into 10 and 100 vesicles. Calculated free energy of each situation is:

Fig.3 The relation of each free energy by calculation

・Discussion
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 Key DNA.

・Reference
Y. Suezaki, "Shishitsu maku no butsuri (Physics of lipid membrane)", Kyusyu Univ. Press, 2007(in Japanese)