What is the Vmax of an enzyme at 1 µM with an activation free energy (AG#) of 55000 J/mol at 298K? -AG # = 6.2x10¹2 RT k cat S-1

A true transition state can bind to an enzyme active site with a KT as low as 7.00×10-26 M. This is a remarkable number, with interesting consequences. Consider a hypothetical solution of an enzyme in equilibrium with a ligand that binds with a Kp of 3.45×10-27 M. a If the concentration of free enzyme, [E], is equal to the concentration of the enzyme-ligand complex, [EL], what would [L], the concentration of free ligand, be? [L] = M

384 Enzyme Structure and Function Q6.2: The enzyme active site provides an optimized physical and chemical microenvironment to lower the activation energy of a reaction and promote catalysis. Describe three catalytic reaction mechanisms used by enzymes and provide an example of each type using specific enzymes.

Use the equation:  ln(k1/k2)=−Ea/R(1/T1−1/T2)  to calculate the activation energy k1 is 0.00316 k2 is 0.0106 R is 8.314 T2 is 298  T1 is 273

For an enzyme that follows Michaelis –Menten kinetics, k1=1X106 M-1 sec-1, k-1= 2x103 sec-1 and k2=2X 102 sec-1.                                                                                                          a.  What is the Km for the enzyme ?    If the enzyme concentration is 10 nanomoles in 1 ml, what is Vmax for the enzyme ?    What is the catalytic efficiency for this enzyme?

The peak on the blue line marked by the #2 represents which of the following? Free energy, G Transition state (‡) S 2 3 ES EP 4 AG* P uncat Reaction coordinate JAG cat The energy required for the enzymatic conversion of substrate to product. The binding energy (AGB) of the substrate binding the enzyme. Desolvation of the substrate and the active site of the enzyme.. Solvation of the product and the active site of the enzyme..