Team:Dundee/Modelling2

How effective is endolysin at killing C. difficile?

Different Endolysin concentrations can have different effects on the survival of C.difficile cells. It is important therefore to understand how much endolysin is required to control C.difficile population and how can this concentration be obtained from our newly engineered E.coli. In this section we present mathematical models and quantative analyses that quantify the effect of Endolysin on a C.difficile population, Models were constructed using data that have been collected over summer, but we have also exploited data from the literature.

First, we used data from “Molecular Characterization of a Clostridium difficile Bacteriophage and Its Cloned Biologically Active Endolysin” where the size of the C.difficile population is measured (optical density, OD) as a function of the Endolysin concentration, E(μg/ml) and the time, t(mins) during which C.difficile is exposed to Endolysin.

Table 1. Level of C.diff at different time points when cultured with three different masses of endolysin

Table 1. presents the data extracted from Molecular Characterization of a Clostridium difficile Bacteriophage and Its Cloned Biologically Active Endolysin. The data shows the response of C.diff to endolysin is delayed by approximately 10 minutes. Similar data in which the C.diff was flash-frozen before the endolysin was added showed no lag phase. It is likely that freezing weakens bacteria defences and, in the later case, the action of endolysin is seen immediately after application. In the following we choose to focus on this stage where lysis is effective.

Can we formulate a relationship between endolysin and the amount of C.diff it kills?

To find a formula that relates the two variables (E and cell number), we have used curve fitting functions (Curve Fitting Tool, MATLAB). We found that exponential functions best describe the data, see Figures 3-5.

The best fit of this data led to an expression for optical density (and hence cell density) as follows: (1) Where OD(t) is the Optical Density and time t, OD12 is the optical density at time t=12 minutes and μ is the death rate (parameter). Equation 1 can then be linearised to estimate μ. Plotting ln(ODt/OD12) against time t, we will get three straight lines, of which we can derive 3 values for μE and therefore μ. The values of the slopes of these lines μE was found to be 0.04149, 0.03293 and 0.02377 for respectively 10.5, 3.5 and 0.7 μg of endolysin. Using these values, we determine the death rate parameter. μA = 1.577x104 In a second stage, we establish a relationship between optical density and cell number. To our suprise there seems to be no universal function that relates optical denisty to cell number. However, using data from OD to Cell Number Chart and the MATLAB curve fitting tool, we derive an exponential model as follows: (2) A plot of this function is given in Figure 6 below Fig. 6 Cell Number as a Function of Optical Density Combining equations (1) and (2), we can formulate a model for the number of C.diff cells as a function of endolysin concentration and time: As an example we can perform a calculation, to find out how many molecules of endolysin we need to reduce of a C.diff population by half after one hour. After some basic arithmetic, we derive the following equation: Taking OD12 = 0.8, we have been able to calculate E: E = 8.754x10-7g The endolysin our team is using has the molar mass 20869.9gmol-1. From this we can deduce the number of molecules, n, required to lyse half of a C.diff population. n is therefore the solution of the following equation: Where NA = 6.0224x1023 is Avogadro's number. We find that the number of molecules required is: n = 2.53x1013 Molecules We can use this number with other data to estimate the number of engineered E.coli we need to repress a given C.diff population. Suppose we have an initial C.diff population wit an optical density 0.8 (this equates to 1x107 bacteria). Now we have just established that 2.5x1013 molecules of endolysin can reduce this population to half its starting size (i.e. reduce by 5x106 cells) in one hour. Dividing the number of molecules by the number of cells killed, gives approximately 5x106. This is the number of molecules of endolysin (or its equivalent concentration) needed to kill one cell of C.diff in one hour. We then looked at how much endolysin can be produced by the Type VI secretion systems engineered into E.coli. As the endolysin is fused to a protein at the tip of the “needle”, our lab results show that only one molecule will be released by one needle. Other studies such as those in Type VI secretion requires a dynamic contractile phage tail-like structure , suggest that around 5 needles per cell can be produced at once and that each one takes roughly 300s to construct. This means that in 5 minutes, 5 needles can be constructed and therefore 5 molecules of endolysin produced. In one hour then, 60 molecules of endolysin can be produced per E.coli bacterium. We therefore need ~8.5x104 E.coli bacterium to kill 5x107 C.diff cells in one hour. This supports our ordinary differential equation model (Fig. 7 and see Work Package 1) which shows that it is very important to administer enough E.coli bacteria into the gut. Fig. 7 Population dynamics model for E.coli vs. C.diff A large initial amount of E.coli is required to overcome the incumbent C.diff population. Conclusions Relating optical density to cell number was an important part of this analysis. Once this relationship and that of cell-kill in response to varying quantities of endolysin were formulated, we have been able to provide a rough estimate for the number of E.coli required to overcome an incumbent C.diff population. Our calculation reveals this relationship to be: E.coli Inoculum ≥ 2% C.diff Population Of course this is a very rough estimate and again takes no account of cell division or removal processes. These will be covered in the next section. References 1. “OD to Cell Number Chart” Bloom J. Bloom Lab 2. “Molecular Characterization of a Clostridium difficile Bacteriophage and Its Cloned Biologically Active Endolysin” Mayer M.J., Narbad A., Gasson J. Institute of Food Research 2008 3. “Type VI secretion requires a dynamic contractile phage tail-like structure” Basler M., Pilhofer M., Henderson G.P., Jensen G.J., Mekalanos J.J. Nature 2012 Appendix Endolysin vs. C.diff Derivation of equations for calculating how much endolysin is needed to kill population of C.Diff We have already found an expression for optical density with an input of endolysin: This equation has been linearised by dividing each side by OD12 and taking the natural log of each side, to give: The value for μ was found from averaging values taken from the slopes of the graphs, for each endolysin value. Combining the exponential equation above, with our conversion formula for converting optical density with cell number, we get: The calculations below show how we derived the equation in the example in 'Endolysin vs. C.diff' where we found out how many molecules of endolysin we need to reduce of a C.diff population by half after one hour: