Biochemistry: Concepts and Connections (2nd Edition)
2nd Edition
ISBN: 9780134641621
Author: Dean R. Appling, Spencer J. Anthony-Cahill, Christopher K. Mathews
Publisher: PEARSON
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Textbook Question
Chapter 8, Problem 6P
Would you expect an “enzyme” designed to bind to 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 a protein actually be a catalyst? Explain why or why not.
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Students have asked these similar questions
b) Enzymes accelerate reactions by facilitating the formation of the transition state. Define transition
state and activation energy. For full credit, you need to present the actual graph (for an endergonic or
exergonic reaction - make sure to specify your choice) highlighting each term?
c) Explain how an irreversible inhibitor for an enzymatic reaction differs from reversible inhibitors.
Provide specific example of an irreversible inhibitor and its target enzyme
d) Determine the Vo as a function of Vmax when the substrate concentration is equal to 10 KM or 20
KM. What does this tell you about an enzyme ability to reach 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
Consider an enzyme that catalyzes the reaction S2 P, by the following simple
reaction mechanism:
k,
E + S 2 E•S →E
kcat
+ P
Suppose the enzyme acquires a mutation that causes k1 to be 10-times smaller
than for the wild-type (non-mutant) enzyme.
Suppose you measure the initial reaction rate (vo) at several different [S] for the
mutant and the wild-type enzymes.
Under what conditions would the mutation have a greater effect on the
reaction rate (vo) of the mutant enzyme compared to the wild-type enzyme - at
very low [S], or at very high [S]? Explain briefly how you decided.
Chapter 8 Solutions
Biochemistry: Concepts and Connections (2nd Edition)
Ch. 8 - Prob. 1PCh. 8 - The enzyme urease catalyzes the hydrolysis of urea...Ch. 8 - An enzyme contains an active site aspartic acid...Ch. 8 - The folding and unfolding rate constants for a...Ch. 8 - In some reactions, in which a protein molecule is...Ch. 8 - Would you expect an “enzyme” designed to bind to...Ch. 8 - The initial rate for an enzyme-catalyzed reaction...Ch. 8 - a. If the total enzyme concentration in Problem 7...Ch. 8 - Prob. 9PCh. 8 - Prob. 10P
Ch. 8 - The following data describe the catalysis of...Ch. 8 - At 37 oC, the serine protease subtilisin has kcat...Ch. 8 - The accompanying figure shows three...Ch. 8 - The steady-state kinetics of an enzyme are studied...Ch. 8 - The same enzyme as in Problem 14 is studied in the...Ch. 8 - Enalapril is an anti-hypertension “pro-drug"...Ch. 8 - Initial rate data for an enzyme that obeys...Ch. 8 - Prob. 18PCh. 8 - Suggest the effects of each of the following...Ch. 8 - The inhibitory effect of an uncompetitive...Ch. 8 - Prob. 21PCh. 8 - Prob. 22PCh. 8 - Prob. 23PCh. 8 - In kinetics experiments, the hydrolysis of the...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biochemistry and related others by exploring similar questions and additional content below.Similar questions
- Consider the hypothetical biochemical pathway shown below. Assume that each letter (A, B, C, etc) represents a molecule and each number (1, 2, 3, etc) represents an enzyme. Draw arrows indicating all the probable feedback inhibition interactions that would be expected to regulate the activity of enzymes in this pathway.arrow_forwardAn 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 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.arrow_forwardExamine the figure below, which compares the energetics of a catalyzed and uncatalyzed reaction during the progress of the reaction from substrate (S) to product (P). The highest peak in such a diagram corresponds to the transition state, which is an unstable, high-energy arrangement of substrate atoms that is intermediate between substrate and product. The free energy required to surmount this barrier to the reaction is termed the activation energy. Enzymes function by lowering the activation energy, thereby allowing a more rapid approach to equilibrium. UNCATALYZED activation energy progress of reaction CATALYZED activation energy S ES | progress of reaction free energy free energyarrow_forward
- Would you expect an “enzyme” designed to bind to its target substrate astightly as it binds the reaction transition state to show a rate enhancementover the uncatalyzed reaction? In other words, would such a protein actuallybe a catalyst? Explain why or why not.arrow_forwardHow does the Michaelis-Menten equation explain why the rate of an enzyme-catalyzed reaction reaches a maximum value at high substrate?arrow_forwardIn enzyme catalysed reactions, the energy level of the enzyme/substrate (or ES) complex is higher (or raised) compared to the uncatalyzed reaction. List 4 factors that contribute to this raised energy level and explain how each of these factors contribute to the higher energy level of the ES complexarrow_forward
- Suppose you have 100 molecules of an enzyme (an aspartyl protease) that requires aspartate in the active site for catalysis. If the pka of Asp side chain is 4.5 and you did your experiment in the presence of the proper substrate concentration and environment and at pH 4.5. Then: O a. Repeating the experiment at pH 7.5 results in achieving Vmax O b. Vmax is decreased O c. All enzyme molecules will be active O d. Vmax can be achieved O e. Repeating the experiment at pH 1.5 results in achieving Vmaxarrow_forwardAn uncatalyzed reaction has keq=50. in the presence of an appropriate enzyme.the forward rate of the reaction increased by 20-fold.what is the equilibrium constant in the presence of the enzyme?arrow_forwardGraphically explain the term saturation of the enzyme? Why is the rate of an enzymecatalyzed reaction proportional to the amount of E.S complex?arrow_forward
- 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…arrow_forwardUnder the following conditions, fill in the blanks. Then, describe why this inhibitor is the type of inhibitor you identified it as. If you were to add 5nM of a reversible inhibitor, the Km for the measured enzyme catalyzed reaction would ______ (Increase, Decrease, Stay the same) to ______µM (choose appropriate value) and Vmax would _______ (Increase, Decrease, Stay the same) to ______µMs-1. So, this inhibitor is a ______ (Competitive, Uncompetitive, Mixed) inhibitor. Conditions: kcat = 130 s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µMarrow_forwardThe 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.arrow_forward
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Enzyme Kinetics; Author: MIT OpenCourseWare;https://www.youtube.com/watch?v=FXWZr3mscUo;License: Standard Youtube License