fix up CF1 explanation and length issues; remove complaint about unsaturation

[ool/lipid_simulation_formalism.git] / kinetic_formalism.Rnw

diff --git a/kinetic_formalism.Rnw b/kinetic_formalism.Rnw
index 8b387e28d27c3ae07b3de4e4235f084dab454267..afd5984a9bc79ceee84952b0f991e12ac80d50ee 100644 (file)
--- a/kinetic_formalism.Rnw
+++ b/kinetic_formalism.Rnw
@@ -148,6 +148,11 @@ $1.5$, which leads to a $\Delta \Delta G^\ddagger$ of
 $\Sexpr{format(digits=3,to.kcal(2^1.5))}
 \frac{\mathrm{kcal}}{\mathrm{mol}}$.
 
 $\Sexpr{format(digits=3,to.kcal(2^1.5))}
 \frac{\mathrm{kcal}}{\mathrm{mol}}$.
 

+It is not clear that the unsaturation of the inserted monomer will
+affect the rate of the insertion positively or negatively, so we do
+not include a term for it in this formalism.
+
+

 \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)),
 \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)),


@@ -316,6 +321,13 @@ a range of $\Delta \Delta G^\ddagger$ of
 $\Sexpr{format(digits=3,to.kcal(2^(3.4)))}
 \frac{\mathrm{kcal}}{\mathrm{mol}}$.
 
 $\Sexpr{format(digits=3,to.kcal(2^(3.4)))}
 \frac{\mathrm{kcal}}{\mathrm{mol}}$.
 

+While it could be argued that increased length of the monomer could
+affect the rate of insertion into the membrane, it's not clear whether
+it would increase (by decreasing the number of available hydrogen
+bonds, for example) or decrease (by increasing the time taken to fully
+insert the acyl chain, for example) the rate of insertion or to what
+degree, so we do not take it into account in this formalism.
+

 
 <<fig=TRUE,echo=FALSE,results=hide,width=7,height=5>>=
 curve(2^x,from=0,to=sd(c(12,24)),
 
 <<fig=TRUE,echo=FALSE,results=hide,width=7,height=5>>=
 curve(2^x,from=0,to=sd(c(12,24)),


@@ -371,62 +383,6 @@ 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,
 
 the vesicle, and $un_\mathrm{monomer}$ is the average unsaturation. In
 this equation, as the average unsaturation of the vesicle is larger,
 

-\begin{equation}
-  un_b = 10^{\left(2^{- \left< un_\mathrm{ves} \right> }
-      -2^{-un_\mathrm{monomer}}\right)^2}
-  \label{eq:unsaturation_backward}
-\end{equation}
-
-The most common $\left<un_\mathrm{ves}\right>$ is around $1.7$, which leads to
-a range of $\Delta \Delta G^\ddagger$ from
-$\Sexpr{format(digits=3,to.kcal(10^((2^-1.7-2^-0)^2)))}
-\frac{\mathrm{kcal}}{\mathrm{mol}}$ for monomers with 0 unsaturation
-to
-$\Sexpr{format(digits=3,to.kcal(10^((2^-1.7-2^-4)^2)))}\frac{\mathrm{kcal}}{\mathrm{mol}}$
-for monomers with 4 unsaturations.
-
-
-<<fig=TRUE,echo=FALSE,results=hide,width=7,height=7>>=
-grid <- expand.grid(x=seq(0,4,length.out=20),
-                    y=seq(0,4,length.out=20))
-grid$z <- 10^((2^-grid$x-2^-grid$y)^2)
-print(wireframe(z~x*y,grid,cuts=50,
-          drape=TRUE,
-          scales=list(arrows=FALSE),
-          xlab=list("Average Vesicle Unsaturation",rot=30),
-          ylab=list("Monomer Unsaturation",rot=-35),
-          zlab=list("Unsaturation Backward",rot=93)))
-rm(grid)
-@ 
-<<fig=TRUE,echo=FALSE,results=hide,width=7,height=7>>=
-grid <- expand.grid(x=seq(0,4,length.out=20),
-                    y=seq(0,4,length.out=20))
-grid$z <- to.kcal(10^((2^-grid$x-2^-grid$y)^2))
-print(wireframe(z~x*y,grid,cuts=50,
-          drape=TRUE,
-          scales=list(arrows=FALSE),
-          xlab=list("Average Vesicle Unsaturation",rot=30),
-          ylab=list("Monomer Unsaturation",rot=-35),
-          zlab=list("Unsaturation Backward (kcal/mol)",rot=93)))
-rm(grid)
-@ 
-
-\subsubsection{Unsaturation Backward II}
-
-Unsaturation also influences the ability of a lipid molecule to leave
-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. 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,
-

 \begin{equation}
   un_b = 7^{1-\left(20\left(2^{-\left<un_\mathrm{vesicle} \right>} - 2^{-un_\mathrm{monomer}} \right)^2+1\right)^{-1}}
   \label{eq:unsaturation_backward}
 \begin{equation}
   un_b = 7^{1-\left(20\left(2^{-\left<un_\mathrm{vesicle} \right>} - 2^{-un_\mathrm{monomer}} \right)^2+1\right)^{-1}}
   \label{eq:unsaturation_backward}


@@ -523,9 +479,7 @@ 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
 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
+asymptotically.  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,
 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,


@@ -627,6 +581,20 @@ rm(grid)
 \newpage
 \subsubsection{Complex Formation Backward}
 
 \newpage
 \subsubsection{Complex Formation Backward}
 

+Complex formation describes the interaction between CHOL and PC or SM,
+where PC or SM protects the hydroxyl group of CHOL from interactions
+with water, the ``Umbrella Model''. PC ($CF1=2$) can interact with two
+CHOL, and SM ($CF1=3$) with three CHOL ($CF1=-1$). If the average of
+$CF1$ is positive (excess of SM and PC with regards to complex
+formation), species with negative $CF1$ (CHOL) will be retained. If
+average $CF1$ is negative, species with positive $CF1$ are retained.
+An equation which has this property is
+$CF1_b=a^{\left<CF1_\mathrm{ves}\right>
+  CF1_\mathrm{monomer}-\left|\left<CF1_\mathrm{ves}\right>
+    CF1_\mathrm{monomer}\right|}$, where difference of the exponent is
+zero if the average $CF1$ and the $CF1$ of the specie have the same
+sign, or double the product if the signs are different. A convenient
+base for $a$ is $1.5$.

 
 
 \begin{equation}
 
 
 \begin{equation}


@@ -634,13 +602,15 @@ rm(grid)
   \label{eq:complex_formation_backward}
 \end{equation}
 
   \label{eq:complex_formation_backward}
 \end{equation}
 

-The most common $\left<CF1_\mathrm{ves}\right>$ is around $0.925$, which leads to
-a range of $\Delta \Delta G^\ddagger$ from
+The most common $\left<CF1_\mathrm{ves}\right>$ is around $0.925$,
+which leads to a range of $\Delta \Delta G^\ddagger$ from

 $\Sexpr{format(digits=3,to.kcal(1.5^(0.925*-1-abs(0.925*-1))))}
 $\Sexpr{format(digits=3,to.kcal(1.5^(0.925*-1-abs(0.925*-1))))}

-\frac{\mathrm{kcal}}{\mathrm{mol}}$ for monomers with complex formation $-1$
-to
+\frac{\mathrm{kcal}}{\mathrm{mol}}$ for monomers with complex
+formation $-1$ to

 $\Sexpr{format(digits=3,to.kcal(1.5^(0.925*2-abs(0.925*2))))}\frac{\mathrm{kcal}}{\mathrm{mol}}$
 $\Sexpr{format(digits=3,to.kcal(1.5^(0.925*2-abs(0.925*2))))}\frac{\mathrm{kcal}}{\mathrm{mol}}$

-for monomers with length $2$ to $0\frac{\mathrm{kcal}}{\mathrm{mol}}$ for monomers with complex formation $0$.
+for monomers with complex formation $2$ to
+$0\frac{\mathrm{kcal}}{\mathrm{mol}}$ for monomers with complex
+formation $0$.

 
 
 <<fig=TRUE,echo=FALSE,results=hide,width=7,height=7>>=
 
 
 <<fig=TRUE,echo=FALSE,results=hide,width=7,height=7>>=