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<p> | <p> | ||
Here, each $ k_ {i \leftarrow j}, k ^ * _ {i \leftarrow j} $ is the rate constant of the state transition to $ i \leftarrow j $, (subscript $*$ is the first order rate constant),$[1],[2],[3],[4]$are the concentration of each state, $ [S ^ o], [S ^ i] $ are the concentrations of transporting substances of each region. This system can be regarded as Non-equilibrium steady state(For each state X, $d[X]/dt = 0$) finally,Non-equilibrium steady state solution of the concentration of each state ( can be completely solved as a solution to the quaternary linear equation which combines three formulas selected from eq (2.1.1) and conservation law concerning the enzyme concentration.However, the number of parameters is too large to handle, so this time we describe the model by two general approximations.First, in the non-equilibrium steady state, it is assumed that the rate-determining step in the state transition is the structural transformation of the transporter and the equilibrium state is established for the bond with the substrate.That is introducing constant $K_d^o,K_d^i$ about Fig 2.1.1. | Here, each $ k_ {i \leftarrow j}, k ^ * _ {i \leftarrow j} $ is the rate constant of the state transition to $ i \leftarrow j $, (subscript $*$ is the first order rate constant),$[1],[2],[3],[4]$are the concentration of each state, $ [S ^ o], [S ^ i] $ are the concentrations of transporting substances of each region. This system can be regarded as Non-equilibrium steady state(For each state X, $d[X]/dt = 0$) finally,Non-equilibrium steady state solution of the concentration of each state ( can be completely solved as a solution to the quaternary linear equation which combines three formulas selected from eq (2.1.1) and conservation law concerning the enzyme concentration.However, the number of parameters is too large to handle, so this time we describe the model by two general approximations.First, in the non-equilibrium steady state, it is assumed that the rate-determining step in the state transition is the structural transformation of the transporter and the equilibrium state is established for the bond with the substrate.That is introducing constant $K_d^o,K_d^i$ about Fig 2.1.1. | ||
+ | \[ | ||
+ | \begin{split} | ||
+ | K_d^o&=\dfrac{[1][S^o]}{[2]}\\ | ||
+ | K_d^i&=\dfrac{[3][S^i]}{[4]} | ||
+ | \end{split} | ||
+ | \tag{2.1.2} | ||
+ | \] | ||
</p> | </p> | ||
− | |||
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Revision as of 03:19, 2 December 2018
1)Abstract
It is difficult to apply our system to real condition with optimizing initial yeast amount to targeted salt reduction because of non-obviousness of growth of initial yeast and Na+ influx which depends on medium Na+ concentration which is changing. So we made model which gives initial value for targeted salt reduction about SseNHXS1 yeast cell which had the best performance in our assay by describing the salt absorption kinetics of yeast cells constitutively. We calculated some parameters of kinetics by our result of assay and we simulated salt reduction expected to grow the initial input yeast.As a result, it turned out to be inefficient for large Na+ absorption even when initial yeast amount took the value of upper of growth population size.From this modeling, it is suggested that Need is introducing population which had size exceeding the upper limit by preculture for our design, and model of adding by larger population reproduced the assay results.By using the results of this modeling it is possible to estimate the optimal initial yeast input when using the device of this study.
2)Transporter Kinetics Model
We have to describe influx of AVP1 which is vacuolar PPase on vacuole membrane and NHX1 which is Na+/H+ Antiporter SseNHXS1 yeast cell which had the best performance in our assay.In this model, these transporter was dealed with state transition model and number of parameters was decreased by Rapid Equilibrium and structural symmetry.[1]
2.1)4 state transition model:NHX1
One substrate Uniporter and Two substrate Antiporter which has assumption of simultaneous binding of internal and external substrates are described by 4 state transition model.(Fig 2.1.1)
Fig 2.1.1: 4 state transition model of Uniporter. $S^o$ is outer medium, $S^i$ is inner medium transporting substrate S. E is Uniporter and transitions between four different states of it are illustrated.When the cycle progresses clockwise, S is transported from outer medium into the cell.
Here, differential equations expressing the time change of each state are
\[ \begin{split} \frac{d[1]}{dt}&=-k^*_{2\leftarrow1}[S^o][1]-k_{4\leftarrow1}[1]+k_{1\leftarrow2}[2]+k_{1\leftarrow4}[4]\\ \frac{d[2]}{dt}&=-k_{1\leftarrow2}[2]-k_{3\leftarrow2}[2]+k^*_{2\leftarrow1}[S^o][1]+k_{2\leftarrow3}[3]\\ \frac{d[3]}{dt}&=-k_{2\leftarrow3}[3]-k_{4\leftarrow3}[3]+k_{3\leftarrow2}[2]+k^*_{3\leftarrow4}[S^i][4]\\ \frac{d[4]}{dt}&=-k_{1\leftarrow4}[4]-k^*_{3\leftarrow4}[S^i][4]+k_{4\leftarrow1}[1]+k_{4\leftarrow3}[3] \end{split} \tag{2.1.1} \]
Here, each $ k_ {i \leftarrow j}, k ^ * _ {i \leftarrow j} $ is the rate constant of the state transition to $ i \leftarrow j $, (subscript $*$ is the first order rate constant),$[1],[2],[3],[4]$are the concentration of each state, $ [S ^ o], [S ^ i] $ are the concentrations of transporting substances of each region. This system can be regarded as Non-equilibrium steady state(For each state X, $d[X]/dt = 0$) finally,Non-equilibrium steady state solution of the concentration of each state ( can be completely solved as a solution to the quaternary linear equation which combines three formulas selected from eq (2.1.1) and conservation law concerning the enzyme concentration.However, the number of parameters is too large to handle, so this time we describe the model by two general approximations.First, in the non-equilibrium steady state, it is assumed that the rate-determining step in the state transition is the structural transformation of the transporter and the equilibrium state is established for the bond with the substrate.That is introducing constant $K_d^o,K_d^i$ about Fig 2.1.1. \[ \begin{split} K_d^o&=\dfrac{[1][S^o]}{[2]}\\ K_d^i&=\dfrac{[3][S^i]}{[4]} \end{split} \tag{2.1.2} \]