One way of expressing the rate at which an enzyme can catalyze a reaction is to state its turnover number. The turnover number is the maximum number of substrate molecules that can be acted on by one molecule of enzyme per unit of time. The table gives the turnover number of four representative enzymes. Enzyme Ribonuclease Fumarase Lactate dehydrogenase Urease Substrate Turnover number (per second) RNA 100 800 1000 10,000 fumarate molecules: fumarate lactate urea How many molecules of fumarate can one molecule of fumarase act on in 12.1 min?

One way of expressing the rate at which an enzyme can catalyze a reaction is to state its turnover number. The turnover number is the maximum number of substrate molecules that can be acted on by one molecule of enzyme per unit of time. The table gives the turnover number of four representative enzymes. Enzyme Ribonuclease Fumarase Lactate dehydrogenase Urease Substrate Turnover number (per second) RNA 100 800 1000 10,000 fumarate molecules: fumarate lactate urea How many molecules of fumarate can one molecule of fumarase act on in 12.1 min?

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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When the final product of a series of enzymatically-catalyzed reactions binds to the first enzyme in the pathway to limit its production, it generally uses ___ because the structure of this final product is generally not similar to that of any of the enzyme's normal substrates. Zymogen activation Covalent modification Competitive inhibition Allosteric activation Allosteric inhibition
A generalized enzyme active site is shaped like a hemisphere with a radius of 45Å. The active site holds the following amino acids in a homeostatic solution (pH = 7.38): -HAVARILKHAVARILKHAVARILK- Assuming the charge is distributed uniformly along the hemisphere, determine the force at which this active site acts upon a single ATP molecule at the center of the hemisphere.
Which of the following statements regarding enzymes and transition states is true? stabilization of the transition state must be less than stabilization of ES for catalysis to occur binding of substrate to an enzyme often causes strain, thus promoting transition state formation the transition state conformation of an enzyme catalyzed reaction is identical to the conformation seen in the uncatalyzed transition state formation of the transition state always assures that the reaction will proceed to product none of the above are true
Which of the following statements are false? Initial velocities of enzyme reactions are best obtained in the absence of product because it simplifies analysis. Initial velocities refer to the velocity of the reaction right after it is initiated. The velocity of the reaction as a function of measuring time are curved just like an isothermal binding curve because of substrate binding to the enzyme. Initial velocities correspond to the pre-steady state condition for free enzyme. Initial velocities can sometimes be measured by spectroscopy such as UV/Vis spectroscopy when monitoring the production of NADH from NAD+. The velocity of the reaction will eventually go to zero. The reaction will reach equilibrium because of the presence of the enzyme. It is always better to use substrate rather than product to measure enzyme kinetics.
Which of the following is NOT true about enzymes? Enzyme inhibitors interact reversibly or irreversibly with an enzyme to alter its Km and /or Vmax values. Enzymes typically act under milder conditions of temperature and pH than non-enzyme chemical catalysts. Zymogens are the inactive precursors of enzyme. Nucleophilic groups can catalyze reactions through the transient formation of covalent bonds with the substrate. Transition state stabilization can significantly increase the activation energy for a reaction.
Please handraw this graph with all the necessary detailed information: Imagine that I text enzyme rate for four different temperatures: 10 degrees celsius, 20 degrees celsisus, 30 degree celsius, and 40 degree celsius, in separate tubes. The enzyme appears to work faster as temperature increases, but completely ceases activity at 40 degrees celcius. Sketch a graph to show this outcome, but here you will graph product formation (nmoles/mL) vs. time (minutes). The graph should be 4 lines and HANDDRAWN. Include a legend if necessary. You do not need precise quantitivate values, but most show the correct trends on the graph.
The active (catalytic) site of an enzyme contains the side chains of amino acid residues that areconserved because they participate in the protein’s catalytic activity. The bulk of the enzyme,however, is not part of the active site. A substantial amount of energy is required to synthesizeenzymes. Why are these molecules so large?
For each of the statements below about the dUTPase enzyme, mark whether it is true or false. If it is false, change the language to make the statement true. _______If the dUTPase enzyme is the rate determining step in a larger metabolic pathway, the reaction likely lies near to equilibrium and is irreversible. _______For the dUTPase reaction, it is not likely that DG’ = DG°’ _______ dUTPase has a higher affinity for its substrate than for its product, but both of these are higher than the affinity for the transition state _______Kinetic analysis of dUTPase using Michaelis-Menten methods assumes that the formation of the E*S complex has a large negative value for DG
All of the following are ways of controlling enzyme mechanism, except A. Edman degradation B. zymogens C. phosphorylation D. allosteric
From the following data of an enzymatic reaction, determine the type of inhibition, the KM for the substrate and Ki for the enzyme-inhibitor complex. [S] mM 2.0 3.0 4.0 10.0 15.0 [P] mM/h (without inhibitor) 139 179 213 313 370 [P] mM/h ([I]=6mM) 88 121 149 257 313
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
The concept of “induced fit” refers to the fact that: a. enzyme specificity is induced by enzyme-substrate binding. b. enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction. c. enzyme-substrate binding induces movement along the reaction coordinate to the transition state. d. substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation. e. when a substrate binds to an enzyme, the enzyme induces a loss of water (desolvation) from the substrate.