Principal Component Analysis

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Squalene Epoxide and Lanosterol Synthase Binding Model

Aim

The aim of a model describing the binding between squalene epoxide and lanosterol synthase is to find the optimum concentration and binding rates that we require for visual detection of squalene epoxide in the fingermark sample from the crime scene. The more squalene epoxide and lanosterol synthase that bind the more likely it will be that squalene epoxide will be visually detected.

Results

Squalene epoxide is an intermediate in cholesterol synthesis. Squalene epoxide is degraded into lanosterol, a precursor for cholesterol by lanosterol synthase. These reactions can be described by the schematic:

$$ \large{ \ce{LS + SE<=>[K_{1}][K_{2}] PC ->[K_{3}] La}. } $$

Where \(LS\) is the concentration of lanosterol synthase, \(SE\) is the concentration of squalene epoxide, \(PC\) is the concentration of the 1st intermediate, protosterol cation, and \(La\) is the concentration of lanosterol, the full complex. \(K_{1}\), \(K_{3}\) are the forward reaction rates, and \(K_{2}\) is the reverse reaction rate.

The initial concentration of squalene epoxide was defined to be \(SE_{0}\) and two parameters of the system were defined as:

$$ \large{ \begin{equation*} \lambda=\frac{K_{1}}{K_{2}} SE_{0}, \qquad \gamma=\frac{K_{3}}{K_{2}}. \end{equation*} } $$

The initial concentration of lanosterol synthase was defined to be \(LS_{0}\) and the ratio between initial concentrations was defined as:

$$ \large{ \begin{equation*} v_{0}=\frac{LS_{0}}{SE_{0}}. \end{equation*} } $$ Sensitivity analysis was performed to find the optimum values for the two parameters, \(\gamma\) and \(\lambda\), and the ratio, \(v_{0}\) which give the highest concentration of the final complex, lanosterol. Consider \(\lambda\) and \(\gamma\) first by setting \(v_{0}\) as the suggested value from (Eq 7) and setting a range of values for \(\gamma\) and \(\lambda\). The range of values chosen has the maximum value as twice the suggested value from (Eq 10).

From Figure 1, it can be seen that there are optimal values for both parameters where increasing them has no effect on lanosterol formation. This was further investigated by looking at each parameter individually and the effect on lanosterol formation over time. Consider the effect of \(\lambda\) by setting \(v_{0}\) and \(\gamma\) as the suggested values, (Eq 7) and (Eq 10).

Figure 2 suggests that the concentration of lanosterol does not increase after \(\lambda = 0.28\). Therefore if the value of \(\lambda\) is smaller than the suggested value, \(\lambda = 0.64\), but larger than \(\lambda = 0.28\) the same concentration of lanosterol will be formed. Recall that \(\lambda = \frac{K_{1}SE_{0}}{K_{2}}\), therefore the initial concentration of squalene epoxide or the binding ratio can be less than the suggested value. The effect of \(\gamma\) on lanosterol formation can be considered by setting \(v_{0}\) and \(\lambda\) as the suggested values, (Eq 7) and (Eq 10).

Figure 3, similar to Figure 2, suggests that the concentration of lanosterol does not increase after \(\gamma = 0.4\). Therefore if the value of \(\gamma\) is smaller than the suggested value, \(\gamma = 1\), but larger than \(\gamma = 0.4\) the same concentration of lanosterol will be formed. Recall that \(\gamma = \frac{K_{3}}{K_{2}}\), therefore the binding rate ratio can be less than the suggested value. The effect of \(v_{0}\) on lanosterol formation can be considered by setting \(\gamma\) and \(\lambda\) as the suggested values, (Eq 10).

Figure 4, similar to Figure 2 and 3, suggests that the concentration of lanosterol does not increase after \(v_{0} = 1.5\). Therefore if the value of \(v_{0}\) is smaller than the suggested value, \(v_{0} = 2.56\), but larger than \(v_{0} = 1.5\) the same concentration of lanosterol will be formed. Recall that \(v_{0} = \frac{LS_{0}}{SE_{0}}\), and \(SE_{0}=0.82 \mu M\). From this the best concentration of lanosterol synthase to have in the fingerprint ageing device is:

$$ \large{ \begin{equation*} LS_{0}=1.23 \mu M. \end{equation*} } $$

The results of the model describing lanosterol formation has been passed on to the lab to be used in future decision making.

Method

Using the law of mass action (1) the binding reaction schematic was written as a system of ordinary differential equations (ODEs):

$$ \large{ \begin{eqnarray} \frac{dLS}{dt}&=&K_{2}PC - K_{1}LS\cdot SE, \nonumber \\ \frac{dSE}{dt}&=&K_{2}PC - K_{1}LS\cdot SE, \nonumber \\ \frac{dPC}{dt}&=&K_{1}LS \cdot SE - K_{2} PC- K_{3}PC,\tag{Eq 1}\\ \frac{dLa}{dt}&=&K_{3} PC. \nonumber \end{eqnarray} } $$

with initial conditions:

$$ \large{ \begin{eqnarray} LS(0)&=&LS_{0}, \quad \mu M \nonumber \\ \nonumber SE(0)&=&SE_{0}, \quad \mu M\\ PC(0)&=&0, \quad \mu M \tag{Eq 2}\\ La(0)&=&0. \quad \mu M \nonumber \end{eqnarray} } $$

The parameters were estimated by considering the steady state of the system. Setting the left hand side of (Eq 1) to zero gives:

$$ \large{ \begin{eqnarray} K_{2} PC&=&K_{1} LS \cdot SE, \nonumber \\ K_{1} LS \cdot SE&=&K_{2} PC - K_{3} PC. \tag{Eq 3} \end{eqnarray} } $$

Rearranging (Eq 3) gives:

$$ \large{ \begin{equation} \frac{PC}{LS \cdot SE}=\frac{K_{1}}{K_{2}}. \tag{Eq 4} \end{equation} } $$

Considering the first binding reaction, it was found that the total concentration of lanosterol synthase, \(LST\), will be equal to:

$$ \large{ \begin{equation} LST=LS+PC. \tag{Eq 5} \end{equation} } $$

Now using (Eq 4) and (Eq 5) it can be written that:

$$ \large{ \begin{equation} \frac{LS}{LST}=\frac{1}{\frac{K_{1}}{K_{2}} SE_{0} + 1}. \tag{Eq 6} \end{equation} } $$

It is known that the ratio between lanosterol synthase and squalene epoxide is:

$$ \large{ \begin{equation} v_{0}= 2.56, \tag{Eq 7} \end{equation} } $$

and that they bind at a 1:1 ratio (2) . Therefore the ratio of free lanosterol synthase to total lanosterol synthase will be:

$$ \large{ \begin{equation} \frac{LS}{LST}=\frac{1.56}{2.56}. \tag{Eq 8} \end{equation} } $$

By substituting (Eq 8) into equation (Eq 6) the ratio between \(K_{1}\) and \(K_{2}\) can be found:

$$ \large{ \begin{equation} \frac{K_{1}}{K_{2}}= 0.78\quad \mu M^{-1}. \tag{Eq 9} \end{equation} } $$

For (Eq 3), (Eq 6) and (Eq 8) can be used to find the ratio between \(K_{3}\) and \(K_{2}\):

$$ \large{ \begin{equation} \frac{K_{3}}{K_{2}}=1. \tag{Eq 10} \end{equation} } $$

From Goodman's 1964 paper (3) , it can be calculated that the suggested initial concentration of squalene is: \(SE_{0}=\) 0.82 \(\mu M\). It is then assumed that this will be a reasonable estimate for the initial concentration of squalene epoxide. Therefore, from (Eq 9) and (Eq 10) the estimated values for \(\lambda\) and \(\gamma\) are found to be:

$$ \large{ \begin{equation} \lambda=0.64, \qquad \gamma=1.\tag{Eq 11} \end{equation} } $$

By running the ode23 solver over one hundred different values for both parameters and the ratio \(v_{0}\), sensitivity analysis can be performed. The range of values has the mean as the estimated values, (Eq 7) and (Eq 11). The results are shown in Figure 1, where the centre of the plot represents the suggested concentration of complex formed when the suggested binding rates are used.

References

1. Guldberg CM, Waage P. Concerning chemical affinity. Erdmanns Journal fr Practische Chemie 1879; 127: 69-114.
2. Boutaud O, Dolis D, & Schuber F. Preferential cyclization of 2, 3 (S): 22 (S), 23-dioxidosqualene by mammalian 2, 3-oxidosqualene-lanosterol cyclase. Biochemical and Biophysical Research Communications 1992; 188(2): 898-904.
3. Goodman DS. Squalene in human and rat blood plasma. Journal of Clinical Investigation 1964; 43(7): 1480.