Q18: Select the graph that correctly illustrates the effect of a negative modifier (effector) on the velocity curve of an allostericenzyme. Place the correct graph in the set of axes. The solid blue curve represents the unmodified enzyme. The dashed greencurve represents the enzyme in the presence of the effector.Reaction velocitySubstrate concentrationR is the highly active form of the enzyme and T is the less active form of the enzyme. Assume that this is a positivelycooperative enzyme, meaning that the affinity for substrate increases with increasing substrate concentration.What would happen to [T]/[R] if substrate concentration is decreased?[T]/[R] would remain the same.Answer BankO [T]/[R] would increase.O [T]/[R] would decrease.

Negative modifiers are species that decrease the activity of enzymes. This leads to a decrease in…

Answered: Select the graph that correctly…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Select the graph that correctly illustrates the effect of a positive modifier (effector) on the velocity curve of an allosteric enzyme. Place the correct graph in the set of axes. The solid blue curve represents the unmodified enzyme. The dashed green curve represents the enzyme in the presence of the effector. R is the highly active form of the enzyme and T is the less active form of the enzyme. Assume that this is a positively cooperative enzyme, meaning that the affinity for substrate increases with increasing substrate concentration.
An enzymes catalyzed reaction is studied in the presence and absence of an inhibitor. The following data was obtained in the image provided. Plot 1/[S] as abscissa and 1/V as ordinate for both catalyzed reactions and reaction with inhibitor. Use the same graph for both plots Michaelis–Menten kinetics Lineweaver–Burk plot Calculate the following: Km of enzyme in the reaction without inhibitor Km' of the enzyme in the reation with inhibitor Vmax of the uninhibited reaction Vmax of the inhibited reaction What kind of inhibitor was added to the enzyme catalyzed reaction? Explain your answer in terms of changes in Km and Vmax.
In the scheme below which represents the mechanism of action for a large number of enzymes: A+B⟺AB⟶C The steady state approximation is reached when: d[AB]/dt≈0 k2≫k1 k−1≫k1 k−1=k1
Sketch the complete reaction free energy diagram for an enzyme-catalyzed conversion of a single substrate (S) into product (P), where the reaction is spontaneous in the forward direction Overlay the free energy diagram for the uncatalyzed reaction and indicate  delta delta G〒 on your sketch: Chemical step is rate limiting
On the figure below are shown three Lineweaver-Burk plots for enzyme reactions that have been carried out in the presence, or absence, of an inhibi- tor. Indicate what type of inhibition is predicted based on each Lineweaver- Burk plot. For each plot indicate which line corresponds to the reaction without inhibitor and which line corresponds to the reaction with inhibitor present. 1 1 1 [S] [S] [S]
By using Excel or GoogleSheets, graph the Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available. What are the Km and V values for the inhibited and uninhibited reactions? Is the inhibitor competitive or max noncompetitive? [S] (mM) V, No Inhibitor (mmol min-') V, Inhibitor Present (mmol min-') 1× 10-4 5 × 10-4 1.5 × 10-3 2.5 × 10-3 5 × 10-3 0.026 0.092 0.136 0.150 0.010 0.040 0.086 0.120 0.165 0.142 Activate
The following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.
The figure displays the relationship between initial rate of product formation and reactant concentration in an enzyme-catalyzed reaction with a fixed amount of enzyme. Which of the following statements best explains the shape of the rate curve at high reactant concentration?
Consider the following free energy diagram for an uncatalyzed and enzyme-catalyzed reaction. Select all the statements that are true. Without enzyme With enzyme A+B Time AB O a. The rate of the enzyme catalyzed reaction is faster than the uncatalyzed reaction O b. The change in free energy for the reaction is greater in the catalyzed reaction, compared to the uncatalyzed reaction O c. The enzyme stabilizes the transition state for the reaction Od. The reaction is exergonic е. The reaction is now spontaneous due to the addition of enzyme Released Energy
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?
You have obtained experimental kinetic data for two versions of the same enzyme, a wild‑type and a mutant differing from the wild‑type at a single amino acid. The data are given in the table. Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which ?−1 is much larger than ?2, which of the following statements are correct? The mutant version has a higher affinity for the substrate. The wild‑type version requires a greater concentration of substrate to achieve ?maxVmax. The wild‑type version has a higher affinity for the substrate. The mutant version requires a greater concentration of substrate to achieve ?maxVmax. Calculate the initial velocity of the reaction catalyzed by the wild‑type enzyme when the substrate concentration is 10 mM. The reaction equilibrium is reached once there is no net change in the concentration of the substrate or the product. Based on the data table and your initial…
Calculate 1/[S] and 1/V to complete the table. Use this data to draw a Lineweaver-Burke plot, with lines for ‘no inhibitor’ and ‘with inhibitor’. Be sure to label your axes and lines. What kind of inhibitor is it? Estimate Vmax and Km for the uninhibited reaction.

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