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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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
- Examine 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_forwardHow does the Michaelis-Menten equation explain why the rate of an enzyme-catalyzed reaction reaches a maximum value at high substrate?arrow_forwardWould 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_forward
- In 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_forwardSuppose 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_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_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_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_forward
- 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 Energyarrow_forwardWhen enzyme solutions are heated, there is a progessive loss of catalytic activty over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at 45 degrees Celsius lost 50% of its activity in 12 minutes, but when incubated at 45 degrees Celsius in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 minutes. Suggest why thermal denaturation of hexokinase was retarded in the presence of one substrates.arrow_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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