unsaturation is larger. Assuming that an increase in width of the
distribution linearly decreases the free energy of activation, the
$un_f$ parameter must follow
-$x^{\mathrm{stdev}\left(un_\mathrm{ves}\right)}$ where $x > 1$, so a
-convenient starting base for $x$ is 2:
+$a^{\mathrm{stdev}\left(un_\mathrm{ves}\right)}$ where $a > 1$, so a
+convenient starting base for $a$ is $2$:
\begin{equation}
un_f = 2^{\mathrm{stdev}\left(un_\mathrm{ves}\right)}
\label{eq:unsaturation_forward}
\end{equation}
+The most common $\mathrm{stdev}\left(un_\mathrm{ves}\right)$ is around
+$1.5$, which leads to a $\Delta \Delta G^\ddagger$ of
+$\Sexpr{format(digits=3,to.kcal(2^1.5))}
+\frac{\mathrm{kcal}}{\mathrm{mol}}$.
+
\setkeys{Gin}{width=3.2in}
<<fig=TRUE,echo=FALSE,results=hide,width=5,height=5>>=
curve(2^x,from=0,to=sd(c(0,4)),
dependent upon the charge of the monomer and the average charge of the
vesicle. If either the vesicle or the monomer has no charge, there
should be no effect of charge upon the rate. This leads us to the
-following equation, $x^{-\left<ch_v\right> ch_m}$, where
+following equation, $a^{-\left<ch_v\right> ch_m}$, where
$\left<ch_v\right>$ is the average charge of the vesicle, and $ch_m$
is the charge of the monomer. If either $\left<ch_v\right>$ or $ch_m$
is 0, the adjustment parameter is 1 (no change), whereas it decreases
if both are positive or negative, as the product of two real numbers
-with the same sign is always positive. A convenient base for $x$ is
+with the same sign is always positive. A convenient base for $a$ is
60, resulting in the following equation:
\label{eq:charge_forward}
\end{equation}
+The most common $\left<{ch}_v\right>$ is around $-0.165$, which leads to
+a range of $\Delta \Delta G^\ddagger$ from
+$\Sexpr{format(digits=3,to.kcal(60^(-.165*-1)))}
+\frac{\mathrm{kcal}}{\mathrm{mol}}$ to $0\frac{\mathrm{kcal}}{\mathrm{mol}}$.
+
<<fig=TRUE,echo=FALSE,results=hide,width=7,height=7>>=
x <- seq(-1,0,length.out=20)
y <- seq(-1,0,length.out=20)
As in the case of unsaturation, void formation is increased by the
presence of lipids with mismatched curvature. Thus, a larger
distribution of curvature in the vesicle increases the rate of lipid
-insertion into the vesicle. However, a species with curvature $e^-1$
+insertion into the vesicle. However, a species with curvature $e^{-1}$
will cancel out a species with curvature $e$, so we have to log
transform (turning these into -1 and 1), then take the absolute value
(1 and 1), and finally measure the width of the distribution. Thus, by
using the log transform to make the range of the lipid curvature equal
-between positive and negative, taking the absolute value to cancel out
+between positive and negative, and taking the average to cancel out
exactly mismatched curvatures, we come to an equation with the shape
-$x^{\mathrm{stdev}\left|\log cu_\mathrm{vesicle}\right|}$, and a
-convenient base for $x$ is 10, yielding:
-
-{\color{red} Shouldn't a vesicle of -1,1,0 have the same activation
- energy as a vesicle of 0,0,0? It doesn't currently.}
+$a^{\left<\log cu_\mathrm{vesicle}\right>}$. A convenient base for $a$
+is $10$, yielding:
\begin{equation}
- cu_f = 10^{\mathrm{stdev}\left|\log cu_\mathrm{vesicle}\right|}
+ % cu_f = 10^{\mathrm{stdev}\left|\log cu_\mathrm{vesicle}\right|}
+ cu_f = 10^{\left<\log cu_\mathrm{vesicle} \right>}
\label{eq:curvature_forward}
\end{equation}
+The most common $\left<\log {cu}_v\right>$ is around $-0.165$, which leads to
+a range of $\Delta \Delta G^\ddagger$ from
+$\Sexpr{format(digits=3,to.kcal(60^(-.165*-1)))}
+\frac{\mathrm{kcal}}{\mathrm{mol}}$ to $0\frac{\mathrm{kcal}}{\mathrm{mol}}$.
+% 1.5 to 0.75 3 to 0.33
<<fig=TRUE,echo=FALSE,results=hide,width=7,height=5>>=
-curve(10^x,from=0,to=max(c(sd(abs(log(c(0.8,1.33)))),
- sd(abs(log(c(1,1.33)))),
- sd(abs(log(c(0.8,1)))))),
+curve(10^x,from=0,to=max(abs(c(mean(log(c(0.8,1.33))),
+ mean(log(c(1,1.33))),
+ mean(log(c(0.8,1)))))),
main="Curvature forward",
xlab="Standard Deviation of Absolute value of the Log of the Curvature of Vesicle",
ylab="Curvature Forward Adjustment")
@
<<fig=TRUE,echo=FALSE,results=hide,width=7,height=5>>=
-curve(to.kcal(10^x),from=0,to=max(c(sd(abs(log(c(0.8,1.33)))),
- sd(abs(log(c(1,1.33)))),
- sd(abs(log(c(0.8,1)))))),
+curve(to.kcal(10^(x^2)),from=0,to=max(abs(c(mean(log(c(0.8,1.33))),
+ mean(log(c(1,1.33))),
+ mean(log(c(0.8,1)))))),
main="Curvature forward",
xlab="Standard Deviation of Absolute value of the Log of the Curvature of Vesicle",
ylab="Curvature Forward Adjustment (kcal/mol)")
a membrane. If a molecule has an unsaturation level which is different
from the surrounding membrane, it will be more likely to leave the
membrane. The more different the unsaturation level is, the greater
-the propensity for the lipid molecule to leave. Therefore, an equation
-with the shape
+the propensity for the lipid molecule to leave. However, a vesicle
+with some unsaturation is more favorable for lipids with more
+unsaturation than the equivalent amount of less unsatuturation, so the
+difference in energy between unsaturation is not linear. Therefore, an
+equation with the shape
$x^{\left|y^{-\left<un_\mathrm{ves}\right>}-y^{-un_\mathrm{monomer}}\right|}$
where $\left<un_\mathrm{ves}\right>$ is the average unsaturation of
the vesicle, and $un_\mathrm{monomer}$ is the average unsaturation. In
-this equation, as the average unsaturation of the vesicle is larger,
+this equation, as the average unsaturation of the vesicle is larger,
+
+\textcolor{red}{I don't like this equation; the explanation above
+ seems really contrived. Need to discuss.}
\begin{equation}
un_b = 10^{\left|3.5^{-\left<un_\mathrm{ves}\right>}-3.5^{-un_\mathrm{monomer}}\right|}
\newpage
\subsubsection{Charge Backwards}
+As in the case of monomers entering a vesicle, monomers leaving a
+vesicle leave faster if their charge has the same sign as the average
+charge vesicle. An equation of the form $ch_b = x^{\left<ch_v\right>
+ ch_m}$ is then appropriate, and using a base of 20 for $x$ yields:
+
\begin{equation}
ch_b = 20^{\left<{ch}_v\right> {ch}_m}
\label{eq:charge_backwards}
\newpage
\subsubsection{Curvature Backwards}
+
+The less a monomer's intrinsic curvature matches the average curvature
+of the vesicle in which it is in, the greater its rate of efflux. If
+the difference is 0, $cu_f$ needs to be one. To map negative and
+positive curvature to the same range, we also need take the logarithm.
+Increasing mismatches in curvature increase the rate of efflux, but
+asymptotically. \textcolor{red}{It is this property which the
+ unsaturation backwards equation does \emph{not} satisfy, which I
+ think it should.} An equation which satisfies this critera has the
+form $cu_f = a^{1-\left(b\left(\left<\log cu_\mathrm{vesicle} \right>
+ -\log cu_\mathrm{monomer}\right)^2+1\right)^{-1}}$. An
+alternative form would use the aboslute value of the difference,
+however, this yields a cusp and sharp increase about the curvature
+equilibrium, which is decidedly non-elegant. We have chosen bases of
+$a=7$ and $b=20$.
+
\begin{equation}
- cu_f = 7^{1-\left(20\left(\log_{e} cu_\mathrm{vesicle}-\log_{e} cu_\mathrm{monomer}\right)^2+1\right)^{-1}}
+ cu_f = 7^{1-\left(20\left(\left<\log cu_\mathrm{vesicle} \right> -\log cu_\mathrm{monomer}\right)^2+1\right)^{-1}}
\label{eq:curvature_backwards}
\end{equation}
projects / ool / lipid_simulation_formalism.git / commitdiff