Chapter 10: Problem 4
The J-factor connection Derive the relation between the \(J\) -factor and the free energy of cyclization described in Section \(10.3 .3 .\) In particular. construct a lattice model of the cyclization process 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 \(\varepsilon_{\mathrm{b}}\) to the binding of complementary ends, \(z_{\text {loop }}\) to the looped configurations, and \(\epsilon_{\text {?ol }}\) as the energy associated with DNA-solvent interactions when the length of the DNA corresponds to one monomer. Further, assume that the solvent energy for a dimer is \(2 x_{\text {sol }}\) and for a looped configuration is identical to that of a monomer. Use a Lagrange multiplier \(\mu\) to impose the constraint that \(N_{\mathrm{tot}}=\) \(N_{1}+2 N_{2}+N_{c}\) (b) Minimize the free energy with respect to \(N_{1}, N_{2},\) and \(N_{c}\) to find expressions for the concentrations of the three species. Then use the fact that \(J\) is the concentration of monomers at which \(N_{2}=N_{\mathrm{c}}\) 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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