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E = enzyme, S= substrate • The key does not fit the lock quite perfectly, but rather exercises a certain strain on it. 1. Both the substrate and the enzyme have to change their shape for binding. 2. When bound to the enzyme the bond is stretched such that it is ready to break. 3. After the bond breaks, a new bond forms somewhere else in the substrate. 4. This changes the shape of the substrate and it can easily unbind from the enzyme. Without the enzyme The interaction with the enzyme lowers the energy When bound, entropy lower Sum of contributions The enzyme has reduced ∆Gǂ , but cannot change ∆G. • • • • • This is the picture for a system with a single enzyme and a single substrate molecule. If we have many substrate molecules, the chemical potential of the reactants grows (kBT ln cS), pulling up the left side of the free energy landscape. ∆Gǂ decreases, ∆G increases. If we have many product molecules, the chemical potential of the reactants grows (kBT ln cP), pulling up the right side of the free energy landscape. ∆G decreases. • Generalize the reaction coordinate to count the substrate molecules -> cyclic machine Many enzymes can be regarded as simple cyclic machines; they work by random-walking down a one-dimensional free energy landscape. The net descent of this landscape in one forward step is the value of ∆G for the reaction S -> P. • Enzymes display saturation kinetics – the reaction velocity – d[S]/dt – is not linear in [S], but saturates at large [S]. • In the ratchet picture, large [S] -> large ∆G -> perfect ratchet -> velocity saturates • In the enzyme picture, large [S] -> the enzyme is occupied most of the time. • Another way to make the ratchet perfect (irreversible) is to lower [P] -> the kBT ln cP becomes strongly negative. • In the enzyme picture, there is no product, P, to bind to the enzyme to go back to the substrate, S. Mechanochemical motors – 2D energy landscape Macroscopic world 1:1 gear w1 = load, w2 = machine If rigid with N teeth: = + 2n/N If slippery, rubber teeth (N=3, w1 = w2 = 0): Preferred motion along the valleys. = = 2 bump along the valley slipping (from one valley to the next) What is the effect of applying a load, w1 > 0, and a driving torque, w2 > 0 (w1 < w2 )? Like in 1D, if the load is too large, the machine could get stuck at = = 2. In 2D, there is also the option of slipping to the next valley. If slipping becomes too large, the machine doesn’t work!! Microscopic world • A mechanochemical machine that catalyses a reaction S->P and has a second binding site that can attach to any point along a periodic track a ratchet on a 2D free energy landscape • Model for kinesin: catalyses ATP->ADP and moves along microtubule • the two coordinates are: the number of substrate molecules processed and the position along the microtubule. If the valleys on the energy landscape sufficiently deep motion effectively only along valleys = tight coupling effectively 1D (coordinate along the valley) Real Enzymes • In metabolism (glycolysis), a derivative of glucose, phosphoglycerate, transfers one of his phosphate groups to ADP ATP ADP ATP Phosphoglycerate kinase 1,3-Bisphosphoglycerate 3-Phosphoglycerate • If phosphoglycerate would transfer its phosphate group to water no ATP • To avoid that, phosphoglycerate kinase engulfs phosphoglycerate and ADP protecting them from water. the • Phosphoglycerate kinase consists of two domains, connected with a hinge. Some of the amino acids required for catalysis are in one domain and some in the other. When the hinge closes these amino acids come in contact with each other and the substrates, phosphoglycerate and ADP . Phosphoglycerate kinase ATP The enzyme will stretch the substrates into their transition state. The kinetics of enzymes (Michaelis-Menten) • Most often we don’t know the energy landscape • Instead, assume: 1. almost no product abrupt drop of G before products no back step 2. EP -> E + P is very rapid, neglect EP as an intermediate state 3. There is only one molecule of enzyme • The reaction is E+S cs k1 k1 k2 ES E+P • The fraction of time that the enzyme is unoccupied is PE , and occupied, PES = 1 - PE . Then d PE k1cS (1 PES ) (k1 k2 ) PES dt In steady state: k1cS PES k1 k2 k1cS The rate at which a single enzyme creates the products is k2PES. Then, if cE is the concentration of enzyme and k1 k2 KM k1 ; v vmax vmax k2cE cs K M cs Michaelis-Menten law • KM = Michaelis constant; cS = KM v = vmax/2 • linear at small cS , saturation at large cS • The Lineweaver-Burk plot (linearized plot) 1 1 v vmax KM 1 cs • E.g.: pancreatic carboxypeptidase cleaves amino acids from the end of a polypeptide. Pancreatic carboxypeptidase acting on a peptide with only two amino acids. KM = 6.4 mM, vmax = 0.085 mM s-1