3. The table on the last page compares for each of four different proteolytic enzymes the chemical bonding structure of a classical substrate with the structures of two competitive inhibitors. For each substrate structure an arrow indicates the position of the scissile bond, i.e., the bond that is cleaved through catalytic action. For each enzyme, one of the inhibitors is a classical competitive inhibitor while the other is a transition-state inhibitor analog. While ordinary competitive inhibitors are associated with (dissociation) inhibitor equilibrium con- stants of ~10-³ to 10-6 M, transition-state analogs exhibit inhibitor constants ≤ 10-⁹ M. (a) For each enzyme draw a circle around those parts of the substrate that account for specificity of substrate recognition. (b) For each enzyme identify the transition-state inhibitor analog by drawing a circle around it and give a brief explanation of why it mimics the structure of the transition-state species. (c) Draw a "generic" Lineweaver-Burk plot that would apply to each enzyme in which there are only three straight lines that separately represent (1) initial velocity data in the absence of an inhibitor, (2) initial velocity data in the presence of the classical competitive inhibitor, and (3) initial velocity data in the presence of the transition-state inhibitor analog.

I am a bit confused about this PPT slide explanation.    Why would a high energy transition state be favored when it requires more activation energy in an enzyme-substrate binding scenario?

24. Consider the figure below, which is an alternate way to depict the energy changes occurring during a reaction from Substrate (S) to Product (P), when uncatalyzed (curve A) and when catalyzed by an enzyme (curve B). Note that curve B is not the same way we modeled an enzyme-catalyzed reaction in class; this model is a different, perhaps slightly more realistic, way of conceptualizing the energy changes over the course of a reaction than what we did in class. S and ES represent the transition states for reaction of the free substrate (S) or the enzyme-substrate complex (ES). T T activation energy for uncatalyzed reaction EST S edhosob nohtum od bluo woH abiow yedio 2 wwolaixa-y odi no vaigin oroda dapng odt ni o nfog) r gibadhoa P B ES EP progress of reaction activation energy for catalyzed reaction a) Explain briefly why, in curve B, the energy state of the enzyme-substrate complex is less than the energy state of the substrate alone. b) Suppose the enzyme in the diagram was mutated…

1.  Provide the best coenzyme(s) for each step shown below. Give a 1–2 sentence rationale for your choice. 2.  Reaction (C) can require multiple coenzymes, or the enzyme may only require one. Briefly describe how the reaction that uses only one coenzyme works. Be specific about the mechanism.

1. For enzymatic reaction, a mechanism was proposed by Michaelis and Menten as follows: ES k, and k,' ES à E and P k,. E + S a. Use steady state assumption, derive expression for the reaction rate. Where E is concentration of enzyme, S substrate, ES complex of E and S, E = E, – ES. (If you have difficulty in doing it, please consult lecture note) b. Assume K = 0.038 mol.L' at 25 °C, when the substrate concentration is 0.156 Mol.L', the rate of the reaction is 1.21 m mol/L.s. The maximum rate of conversion reaction is reached at high substrate concentrations. Calculate the maximum rate of this enzyme catalyzed reaction.

1. If a molecule is interating with its side chains of an enzymes active site but it is not the substrate of the enzyme what kind of enzyme regulation is this?  2. What is the change in thetype of bond between Ser 80 -> Arg.  3. Will this change cause the complex to be more or less stable. Explain

6. Examine the plot of the enzyme-catalyzed reaction below: Wheat-germ acid phosphatase shows Michaelis-Menten kinetics when acting on para-nitrophenol phosphate 16 14 12 10 8 4 2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 (S] / mM a) Is this reaction better characterized by Michaelis-Menten kinetics or simple mass action? b) What is the approximate value of Vmax? Km? c) Assuming that this plot was made at 100 nM enzyme, draw the curve, including axis labels, for 300 nM enzyme. d) Imagine that this enzyme is inhibited by a competitive inhibitor with Ki (for inhibitor) = Km (for substrate). Note that v = (kcat [E] tot [S]/Km) / (1 + [S]/Km + [I]/Ki). Draw the curve of reaction velocity (v) versus [S] for [I] = 1 mM. e) Now consider an experiment where you simultaneously introduce substrate and competitive inhibitor at equal concentrations. Make a plot of v versus concentration of [S] = [I], i.e. the X-axis should be the concentration of both I and S, which are equal. v/ uM min-

1. - - Substrate Enzyme - 1 2 3 BOO - Look at the above diagram and understand what is being shown. Explain AND draw a graph depicting how the rate of the reaction depicted here is changing. ☐ L 2. How will the rate of the reaction change with time after 3? How can we increase the rate of the reaction after 3? I 100 B L

7. An enzyme-catalyzed reaction proceeds by the mechanism below: E+S1ES --2E+P E+A 3 EA EA+S4→ EAS --5→ EA + P E+I6 → EI EAS +17→ EAIS -8 EIS + P A. B. C. E = enzyme, S = substrate, I = inhibitor, P = product and A = activator Rate constants (k's) for the forward reactions are: K1, K2, k3, K4, k5, k6, k7, and k8 Rate constants (k's) for the reverse reactions are: k-1, k-3, K.4, k.6, and k.7 Write the enzyme balance for this mechanism. How many total equations will result from applying the RAPID EQUILIBRIUM ASSUMPTION? Using any concentrations of species in the mechanism and any of the rate constants (k's), write ONE of the equations that would result from applying the QUASI STEADY STATE ASSUMPTION. (ONLY ONE EQUATION; ANY OF THEM ARE FINE)

1. The enzyme you are studying uses the substrate pictured at right. An inhibitor box in the freezer contains two tubes labeled Inhibitor #1 and Inhibitor #2, along with the structures of the compounds contained within them and their respective Ki. Which one is more likely to be the non-competitive inhibitor? Inhibitor #1 K₁ = 10 μM m CH₂OH OH -0, $ OH OH Inhibitor #2 K₁ = 50 μM Briefly Justify: OH CH₂OH OH OH OH Substrate