Week 5 Discussion Worksheet Answer Key

CHEM153A (Michael’s sections) Name: _M__Z% Week 5 Discussion Worksheet Enzyme kinetics and carbohydrates 1. The graph below shows the rate of an enzyme-catalyzed reaction relative to substrate concentration. 1) Label Viax, ¥2 Vimax, and Kw. 1) Point out which region of the graph follows first-order enzyme kinetics and which region follows zero-order enzyme kinetics. 1) Noncompetitive inhibition is a specific type of enzyme inhibition characterized by an inhibitor binding to an allosteric site resulting in decreased efficacy of the enzyme. Draw a new line to show how this might affect the rate of reaction relative to substrate concentration. (©) ax /&74» [51, nate of Acaeton w ot plipenobint o (s1; nate 4~ @ Vorovx . Thas A4 76&0-0& ko nat cd. Vimovx (D ok Lows (s, et ot caofion ul anewases Lunzonty o/ [s]. Thw @ : - - - - - =" ;_"VMM oy \,/Mf’OLaCLp Jeine hed . Rate of reaction K 15} Em (m('//mm?/_ol. Howerete, Umes 2. Given that AG’® for the reaction S<P is negative in the direction of S5P, reaction equilibrium favors the formation of which substance? p { B. The reaction does not proceed in either direction C. The reaction proceeds equally in both directions D. S 3. How would the addition of a catalyst to the reaction S=P affect the difference between the free " energies of S and P in their ground states (AG"®)? > A. AG’® would become more negative Fst B. The change in AG’® cannot be predicted without more information Q° "_'l AG? C. AG’® would become less negative > 2 @) AG’° would not change 4. When deriving the Michaelis-Menten equation, what assumptions are made? Circle all that apply. A. [E]>>>[S] wo, achuall (51577 [E{]‘ /@ The amount of P converted back to S is negligible Catalysis is rate limiting step, not substrate binding D. d[ES)/dtz0 V5, Mfual’l«xj_ ALEs )/ elt = O

5. For the enzyme-catalyzed reaction E +S < ES < E + P, what equation defines the rate at which ES is formed? ([Et] = total enzyme concentration, [ES] = enzyme-substrate complex concentration, [$] = substrate concentration, [P]= product concentration, k; = rate constant for ES formation from E and S, k1 = reverse reaction rate constant) . B ket A K.[S) E +5 E’_" E5Z:*£’+P [f{-}" [E] t LES] K1([Ee) = [ES])[S] A o\ B K.1[S]([ES]-[E:]) S ) - . S ES fowmadien = k. [EI[s] = k (L€~ Ces1)LS] 6. If the enzyme-catalyzed reaction E + S < ES < E + P is proceeding at or near the Vmax of E, what can be deduced about the relative concentrations of S and ES? o ok Kot €57 A. S is abundant, [ES] is vanishingly low ' -y ot ¢ B. [S] is vanishingly low, [ES] is vanishingly low (<1 E+S <= ES S is abundant, [ES] is at its highest point . ‘ ol Iack €5] D. [S] is vanishingly low, [ES] is at its highest point Zr ' Chotlieos prant e / 5 FMK Sl 6 hon #‘lwt .f—d’) ez 7. What relative values of Ky and ket would describe an enzyme with a high catalytic efficiency? ! A. High Km, low ke, gN Km, IOW Kcat ,C B. Low K, low Kcat CL/+ \& .(‘ . %L = e L, ngh Km, hlgh Kcat C ) J ‘ a KM Low Kwm, high kcat 8. Provide information for the given molecule below— A. Is the molecule an aldose or ketose? (O ([ (ode CW A-«A—ZLM) | B. Does this molecule have L or D stereochemistry? D = OH ?w p o / ’( chinak E:l- C. What is the name of this molecule? "af\,éacfn-d-& fw o D Show the mechanism in which the nmblecule below becomes a six-member cyclic ring. H (Cor" \‘</ CSLO‘f\ \ H—— OH o S \O' HO ——H 4 oH K HO ~—t—H n N2}/ w H= —8Hf\ W OH B