Question

Molecular driving forces by Dill & Bromberg. Ch 8, Qn 18.

states 18. Lattice model of igand binding to a protein. You placehave a ligand L and a protein P. When the ligand binds, lp you the energy change is favorable by SEo (Eo <0) because stant- it makes 5 contacts with the protein. As shown in foldedFigure 8.11, the protein is modeled using two-dimensional lattice particles. The protein has a rigid portion (squares containing a P) and a flexible loop (circles containing a P at one end). The ligand (rectangle containing an L) is rigid gy verand occupies two lattice sites. an ideal Figure 8.11 es C- are both (a) Counting only lateral contacts (and not diagonal contacts), find the number of conformers W of the f an ideal flexible loop. Assume (b) What are the energetic and entropic contributions to the free energy change due to ligand binding? Write 3. What is e gas into an expression for AFoinding Eo =-1 kcal mol-1 ? the proper configuration for ligand binding), what (c) For T = 300 K, what is Ahinding-hound-Funbound if from food rocess? 3. What is oom from (d) What is the dissociation temperature Tdissoc? (e) If the flexible loop were made rigid instead (and in would be the new AFbinding? (0 Does the ligand bind more tightly or more weakly otion of a e enthalpy with the rigid loop? 45

Answer 1

@ ConGimes means -h No diffrent struchurer petsbi. eie a Specific Compou nd. By Counting ooly (aheval eibleoop 1S bound to t ligd lake rall em ts decreased and iven tha! ta deexase in eneqy '* 5e. entu moec rart of pen as Cm"gurations So, entrey decreases by on % ά-the Osgrnalenhtry. Oç.ventat ε。--1 kcal/mole T: 500k , hound nbou 스Funding= -5 kcal /mol be dissociafed coken the available Thrmal enerjy Crss*» nding crer Ks Beltz m nstant aimal tneray' : ksTdisseciahian ko IT 2516 35 kelvin

Olf tf. Į1mWe loop was made 、TJ 4hen lateval Contacts wou be 6 instead of 5, So (H- H indin - lalth -the neus vigid loop th bondinq coill be more -tight as adt Compared to a Spheical molecule -IKare is more literal Cont