Chapter 10: Problem 4
Derive the relation between the \(J\) -factor and the free energy of cyclization described in Section \(10.3 .3 .\) In par. ticular, construct a lattice model of the cyclization pro. cess by Imagining a box of \(\Omega\) lattice sites each with volume \(v,\) We consider three species of DNA: monomers with sticky ends that are complementary to each other, dimers which reflect two monomers sticking together, and DNA circles in which the two ends on the same molecule have stuck together. The number of molecules of each species is \(N_{1}, N_{2}\) and \(N_{c},\) respectively. (a) Use the lattice model to write down the free energy of this assembly and attribute an energy \(\&_{b}\) to the binding of complementary ends, reap to the looped configurations, and \(r_{\text {sol }}\) as the energy associated with DNA-solvent inter actions when the length of the DNA corresponds to one monomer. Further, assume that the solvent energy for a dimer is \(2 e_{\text {sal}}\) and for a looped configuration is identical to that of a monomer. Use a Lagrange multiplier \(\mu\) to impose the constraint that \(N_{\text {tot }}=N_{1}+2 N_{2}+N_{c}\) (b) Minimize the free energy with respect to \(N_{1}, \mathrm{N}_{2},\) and \(N_{c}\) to find expressions for the concentrations of the three species. Then use the fact that 1 is the concentration of monomers at which \(N_{2}=N_{c}\) to solve for the unknown Lagrange multiplier \(\mu\) and to obtain an expression for the J-factor in terms of the looping free energy.
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