Macroscopic containers and nanometer hole (Nanoscale systems)

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SUMMARY

This discussion focuses on the translocation of polymeric chain molecules, such as DNA or proteins, through a nanometer-sized hole in an insulating wall separating two macroscopic containers. The entropy of the segments in each container is calculated using the formula S(X) ~ -kBln(X), leading to the total entropy SC = kB•ln(m) + kB•ln(N-m). The entropic free energy G is derived as a function of the number of translocated segments m, with maximum and minimum values identified, and the free energy barrier ΔG calculated. Additionally, the relationship between the translocation time τ and the total length N is explored, demonstrating that τ is proportional to N for large values.

PREREQUISITES
  • Understanding of polymer statistics and entropy, specifically S(X) ~ -kBln(X).
  • Familiarity with thermodynamic concepts such as free energy and enthalpy.
  • Knowledge of Arrhenius equations and their application in reaction kinetics.
  • Basic proficiency in mathematical modeling and graphing functions.
NEXT STEPS
  • Study the implications of entropy in polymer physics and its applications in biophysics.
  • Learn about the Arrhenius equation and its relevance to reaction rates in chemical kinetics.
  • Explore the concept of free energy barriers in thermodynamic processes.
  • Investigate computational methods for modeling polymer translocation dynamics.
USEFUL FOR

This discussion is beneficial for physicists, biochemists, and researchers in materials science focusing on polymer dynamics, thermodynamics, and molecular biology, particularly those studying the behavior of macromolecules in confined environments.

themonk
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Homework Statement


Consider two macroscopic containers separated by an insulating wall, which however has a circular hole of a few nanometers radius and negligible thickness. The containers (1 and 2) enclose aqueous solutions of a polymeric chain molecules (e.g. a DNA or protein). These often have strong interactions between various locations on the chain, that allow them to maintain "coiled up" spherical shapes up to several tens of nanometers in diameter. Such molecules can be "uncoiled" in solution by adding a salt or by heating

A. Consider an uncoiled polymer of N monomers translocating through the hole. At some time during the process, let m segments (1 ≤ m ≤ N-1) be in container 1, whereas the remaining N-m segments have passed through the hole into container 2. From the theory of polymer statistics, one can show that the entropy S of a chain of X segments with one end anchored at a fixed point (in this case the hole), is given by S(X) ~ -kBln(X). Here kB is Boltzmann’s constant. Note that these entropies are relative to that of the “coiled” polymer, and are hence negative. What is the entropy of the m segments in container 1? What is the entropy of the N-m segments in container 2 ? Hence what is the total entropy of the chain, SC?

B.
Assuming that the enthalpies of interaction of the polymer with the solvent are the same on both sides, the enthalpic contribution to the free energy is a constant and can be ignored. But the entropic term depends on the number of segments m that have translocated. Hence, write down the entropic free energy G = -TSC as a function of m. At what values of m (1 ≤ m ≤ N-1) does the free energy have maximum and minimum values? What are these values Gmax and Gmin? Hence, what is the free energy barrier ΔG = Gmax - Gmin for translocation of the strand, given its total length N? Also plot the dependence of the normalized free energy G/kBT on the number of translocated segments m, for N = 100.

C.
If a ‘rate constant’ k for full translocation can be approximated as an Arrhenius expression k ~ koexp(-ΔG/kBT), then the translocation time is τ ~ 1/k. Find τ as a function of N. Show that for large N, the translocation time is proportional to N. How does this compare with the theory discussed in class ?

Homework Equations


The Attempt at a Solution


A.
Since the entropy is measured as S(X) ~ -kB•ln(X) where X is the length of the segment and S(X) represents a coiled polymer, making the result negative. This means that the entropy for an uncoiled polymer would be positive. S(m) ~ -(-kB• ln(m))= kB•ln(m) is the entropy of the m segments in container 1. For the entropy of an N-m polymer i.e. container 2, the equation is S(N-m) ~ -(-kB• ln(N-m))= kB•ln(N-m)
Total Entropy is kB•ln(N-m)+ kB•ln(m) = Sc

This seems too simple. I do not believe this can be the answer for A.

B.
I do not know how to approach this problem. This is the "logic" I have come up with so far.

From the Sc above
G=-T•( kB•ln(N-m)+ kB•ln(m))
The maximum value (closest to positive) of Free Energy is when m is when ln(m) and ln(N-m) are small, which is nearly half way through. N needs to be (m+1), so that the ln(N-m)= 1. The minimum is when m=1.


C.
I will attempt this after A and B
 
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Any suggestion anyone?
 

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