Lecture 5

Self-Avoiding Chain

• A self-avoiding chain is a freely jointed chain the does not cross over itself.
  
  • One of the first scientists to work on this problem is Flory.
• We want to quantify how the chain avoids crossing itself. This is done by examing the free energy of the chain.

Free Energy Discussion: $F=E-TS$

We need expression for the energy and the entropy of the self-avoiding chain.

• Approximate the entropy as that of the FJC.
  $$\rightarrow S = k\ln P(r) + \text{const.} = k \left(-\frac{3r^2}{2Nb^2} \right) + \text{const.}$$
• Energy term will be used to penalize crossing over.
  
  • Consider a single monomer $i$ (link) along the chain.
    
    • Assign an energy penalty for having non-neighboring monomers near the monomer $i$.
      $$E_i = kT(\text{# of other monomers around})$$
    • Let this number be described by the density of monomers $\rho$ multiplied by some small volume $V$ that encolses the monomer $i$.
      $$\Rightarrow E_i = kT(\rho V) = kT \left( \frac{3N}{4 \pi r^3} V \right) \\ \Rightarrow F \approx kTN\frac{3NV}{4\pi r^3} + kT\frac{3r^2}{2Nb^2}$$

We want to minimize $F$ with respect to $r$ to figure out the optimal size of the chain.
$$0 = \frac{\partial F}{\partial r} = \frac{3N^2V}{4\pi} \frac{-3}{r^4} + \frac{3r}{Nb^2} \\ \rightarrow r^2 \approx \left(\frac{3}{4\pi}\right)^{2/5} N^{6/5} (b^2V)^{2/5}$$
This is an important result! For the self-avoiding chain $\left<r^2\right> \propto N^{6/5}$ whereas for the other models $\left<r^2\right> \propto N$.

Aside: DNA

DNA behavior is well predicted by that of a worm-like chain. We know the following for a worm-like chain:
$$\sqrt{\left<r^2\right>} = r_{\text{RMS}} = \sqrt{Nb(2\xi)}$$
Lets examine the RMS length of DNA from different organisms.

• Lambda Phage
  
  
  • $N \approx 50,000$ ; $R \approx 30nm$
    
    • $r_{\text{RMS}} \approx 1300nm$
  • $r_{\text{RMS}} >> R$
• E Coli
  
  
  • $N \approx 4.6 \cdot 10^6$ ; $R \approx 0.5\mu m$
    
    • $r_{\text{RMS}} \approx 13\mu m$
  • $r_{\text{RMS}} >> R$
• Human Cell
  
  
  • $N \approx 3.3 \cdot 10^9$ ; $R \approx 3\mu m$ (Cell Nucleus)
    
    • $r_{\text{RMS}} \approx 300\mu m$
  • $r_{\text{RMS}} > R$

In each case the RMS length of the DNA is much bigger than the size of it’s enclosing cell structure. Single cell organisms like Lambda Phage and E Coli take care of this by having very hard cell walls to keep DNA enclosed at a high pressure. In human cells however, the nucleus walls of very weak. Tight packing of DNA into chromosomes ensures that the DNA to remains contained in the nucleus.

Concluding Remarks about Polymers

The MSD distance of a polymer always has behaves like something of the form:
$$\left<r^2\right> = Nb(2\xi)$$

• If $\xi<<Nb$ then the polymer is floppy and dense.
• If $\xi>>Nb$ then the polymer is straight and stiff.

Lipids

Lipids are molecules found in cells and one of their functions is to group together forming cell membranes. Consists of a head and a tail. The head is hydrophilic - attracted to water - while the tail is hydrophobic - repelled by water.

An example of the different kind of groupings lipids can make.

Do Lipids Always Self-Assemble into Membranes?

There is a trade off between lipids being spread apart or grouped together like in membranes.

• Free Lipids
  
  • High entropy
  • High energy
• Grouped Lipids
  
  • Low entropy
  • Low energy

Free Energy Discussion

Hypothesis: lipids transition from free to grouped when $E \approx TS$.

We can quantify the energy of a single lipid assuming that it has an approximately cylindrical shape as:
$$E_1 = \gamma (2\pi R \ l) \\ \gamma = \text{surface tension}$$
For free lipids, we can approximate the entropy of a single lipid as that of an ideal gas particle.
$$S = k \ln Z + \frac{\bar{E}}{T} \\ \bar{E} = \text{average energy} \\ Z = \frac{Z_1^N}{N!} \\ Z_1 = \frac{1}{h} \int dx \ dp \ e^{-\left(p^2/2m\right) / kT} = \frac{1}{h} L \sqrt{2\pi m k T} \\ \text{let } \lambda \equiv \frac{h}{\sqrt{2\pi m k T}} \\ \Rightarrow Z_1 = \frac{L}{\lambda} \\ $$
For an ideal gas the average energy $\bar{E} = \frac{3}{2}kTN$. Now lets put this all together to calculate $TS$, the energy at which we hypothesized free lipids will group together.
$$\ln Z = 3N \ln Z_1 - \ln N! \\ \rightarrow \ln Z = N \left[ \ln\left(\frac{L^3}{\lambda^3 N}\right) + 1 \right] \\ \text{let } \rho \equiv N/L^3 \\ \ln Z = N \left[-\ln\left(\lambda^3\rho\right) + 1 \right] \\ \Rightarrow TS = kTN \left[-\ln\left(\lambda^3\rho\right) + 5/2 \right]$$
This final expression above is the transition energy. Lets find the corresponding density $\rho$ where the transition occurs.
$$\text{let } NE_1=TS \\ \rightarrow 2\pi N\gamma R \ l = kT N \left( 5/2 - \ln \lambda^3 \rho \right) \\ \ln\lambda^3\rho = \frac{5}{2} - \frac{2\pi\gamma R \ l}{kT} \\ \text{let } l_0 \equiv \frac{kT}{2\pi\gamma R \ l} \\ \Rightarrow \rho_c = \frac{1}{\lambda^3} e^{5/2} e^{-l/l_0}$$
So the prediction is that $\rho_c \propto e^{-l/l_0}$ where $l$ is the length of the lipid’s chain. It turns out that this corresponds to very low critical densities. Therefore, even at low lipid density they easily start packing together.

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1D Diffusion Notes

1 Cell Diffusion with No State Switching

$P(x|t)$

Prediction

The probability density for cell location given a time $t$ should look like a normal distribution. As time increases this prediction becomes less accurate since more and more particles get absorbed at the right side boundary.
$$P(x|t) = \frac{2}{\sqrt{2\pi\sigma^2}} e^{-x^2/2\sigma^2} \\ \sigma^2 = 2Dt$$
The extra factor of two in $P(x|t)$ is because of the reflective boundary at the origin. In an 1D random walk so $\left<x\right> = 0$, therefore $\sigma^2 = \left<x^2\right>$, and $\left<x^2\right> = Nb^2$.
$$\Rightarrow 2Dt = Nb^2 \\ t = N \Delta t \\ \Rightarrow D = \frac{b^2}{2\Delta t}$$

Results

At small times the data is very much inline with prediction.

At large times the data starts to diverge from a normal distribution as more and more cells are absorbed.

$P(t)$

Prediction

$P(t)$ is the probability density of the first passage time. It is expressed as the following:
$$P(t) = -D \frac{\partial P(x|t)}{\partial x}$$
This equation has a solution for our set of initial and boundary conditions. In the large time limit this solution can be expressed as:
$$P(t) = \frac{2D}{L^2} \sum_{m=1}^{\infty} \beta_m e^{-\beta_m^2 D t / L^2} \sin\left(\beta_m \frac{x}{L}\right) |_{x=L}$$
In the following plots only the first term, $m=1$, from the series is used.

Results - Update

The results are plotted here on a normal scale and on a semilog scale. The prediction at large times does seems to be in agreement with the simulations. Update, I have since gone back and remade the plots as a histogram. In creating the histograms I weight the probability density in each bin by the width of the bin to ensure that the whole probability density is properly normalized.

Old Plots
I believe the thickness of the band in the data is due in part because of the large time scale plotted and that on small timescales there is lots of variation in the probability density.

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