In the free energy diagram below, molecule A (with energy x) is converted to molecule B (with energy z). The peak of the energy curve has energy y. The values x, y, and z represent some numerical value of free energy. A Free Energy--> B Reaction Pathway --> Which of the following represents: Z
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- Based on the definition of kcat, substitute a value that can be measured and yet still represents the value associated with the original concentration of the R. What would the rate or velocity of the reaction be equal to under these circumstances? How can cells increase Vmax? What variable that we could change would directly impact Vmax? Would the value of KM be affected by the ways you determined that Vma,x could be increased? What does this indicate about KM? Thinking about how catalysts work, about the Michaelis-Menten Equation, and the definition of kcat, what specifically does the enzyme change in the reaction mechanism to increase the rate? If an enzyme follows the 2 step mechanism proposed by Michaelis-Menten, what do you know about this enzyme? Be very specific and comprehensive. Please answer very soon will give rating surelyShown below are reduction potentials for four half-reactions. Which of the coupled reactions is favorable? (Note that for Cytochrome c you must multiply the reduction potential by 2 for each coupled reaction because only one electron is involved) a) 2 Cytochrome c (Fe3+) + H2O ó 2 Cytochrome c (Fe2+) + O2 b) NADH + Succinate- ó NAD+ + Fumarate- c) Fumarate- + H2O ó Succinate- + O2 d) All of the aboveWhich of the following is true under the following conditions: an enzyme displaying Michaelis-Menten kinetics where the enzyme concentration is 10 nM, the substrate concentration is 45 mM, and the Km is 50 µM? a) The enzyme has low catalytic efficiency for the substrate. b)The rate of catalysis is near half-maximal velocity. c)The enzymatic reaction is near maximal velocity. d)Halving the substrate concentration has little effect on the catalytic rate. e) There is not enough information provided.
- For an enzyme that obeys Michaelis-Menten kinetics, in order for the reaction velocity (v) to be 80% of the maximal velocity (Vmax), the substrate concentration must be…. Choice 1 of 5:½ Km Choice 2 of 5:2 Km Choice 3 of 5:4 Km Choice 4 of 5:8 Km Choice 5 of 5:not enough information is givenConsider the analogy of the jiggling box containing coins that was described on page 85. The reaction, the flipping of coins that either face heads up (h) or tails up (T), is described by the equation h ↔ T, where the rate of the forward reaction equals the rate of the reverse reaction.a. What are ΔG and ΔG° in this analogy? b. What corresponds to the temperature at which the reaction proceeds? What corresponds to the activation energy of the reaction? assume you have an “enzyme,” called jigglase, which catalyzes this reaction. What would the effect of jigglase be and what, mechanically, might jigglase do in this analogy?Would you expect an “enzyme” designed to bind its target substrate as tightly as it binds the reaction transition state to show a rate enhancement over the uncatalyzed reaction? In other words, would such protein be a catalyst? Use a reaction energy diagram to explain why or why not.
- Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false? a. As [S] increases, the initial velocity of reaction V0 also increases. b. At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km. c. Km is the [S] at which V0 = 1/2 Vmax. d. The shape of the curve is a hyperbola. e. The y-axis is a rate term with units of μm/min.The following reaction coordinate diagram charts the energy of a substrate molecule (S) as it passes through a transition state (X‡) on its way to becoming a stable product (P) alone or in the presence of one of two different enzymes (E1 and E2). How does the addition of either enzyme affect the change in Gibbs free energy (ΔG) for the reaction? Which of the two enzymes binds with greater affinity to the substrate? Which enzyme better stabilizes the transition state? Which enzyme functions as a better catalyst?At what substrate concentration would an enzyme with a kcat of 25.0 s-1 and a KM of 3.5 mM operate at 25% of its maximal rate? How many reactions would the enzyme catalyze in 45 seconds when it is fully saturated with substate, assuming the enzyme has one active site?
- In a Lineweaver-Burk graph, the lines representing the uninhibited and inhibited enzyme catalyzed reaction meet each other on the x-axis. The type of inhibition which is occurring is: a) competitive b) noncompetitive c) uncompetitive d) allosteric CO2 exerts direct activity upon hemoglobin by: a) blocking oxygen from binding to the heme group b) displacing BPG from the central cavity c) oxidizing Fe+2 to Fe+3 which does not bind oxygen d) forming an N-terminal carbamate which favors the T-state The dominant motif found in hemoglobin and myoglobin is: a) helix-turn-helix b) twisted beta sheet c) beta barrel d) random coil Which of these is an ketohexose? a) fructose b) glucose c) ribose d) erythrose Which of these is a constitutional isomer of d-glucose? a) fructose b) galactose c) l-glucose d) ribose Which of these is an enantiomer of d-glucose? a) d-fructose b) d- galactose c) l-glucose d) d-ribose Which of these is a diastereomer of…An enzyme catalyzes a reaction in which substrate A is cleaved into two products, P and Q. In the catalytic mechanism, the enzyme converts A to an covalently-bound reaction intermediate X and product P, P then desorbs from the enzyme, and in a second chemical step, the enzyme converts the intermediate X in the EX complex to the final product Q (in EQ), which then desorbs from the enzyme E. You discover two inhibitors of this enzyme, I and J. I is a competitive inhibitor of the substrate A, and has nearly double the molecular weight of J. On the other hand, J is a mixed inhibitor of enzyme E, and its inhibitory effect on Km / Vmax (the slope effect from the double reciprocal plot) is greater than that of 1 / Vmax (the intercept effect in a double reciprocal plot). That is Kis < Kii . At low pH, the conversion of EX to EQ is greatly slowed, kcat is decreased, and the intercept effect of inhibitor J is elevated, that is, the value of Kii is diminished. When a high, fixed concentration…Enzyme X and enzyme Y catalyze the same reaction and exhibit the νo versus [S] curves shown below. Which enzyme is more effi cient at low [S]? Which is more effi cient at high [S]?