Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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Question
An uncompetitive inhibitor interacts with the enzyme•substrate complex to form a ternary
complex (ES•I). This equilibrium reaction can be described as follows:
ES + I ⇌ ESI
Modify the simplified kinetic scheme you drew for E + S ⇌ E + P to include this equilibrium
expression
Expert Solution
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Step 1
Enzyme inhibition is a process by which the activity of an enzyme is altered. Inhibitors are substances that decreases the catalytic activity of an enzyme by altering the interaction between the enzyme & substrate.
Enzyme inhibition can be classified into the following types:
- Reversible
- Irreversible
The reversible enzyme inhibition can be categorized into the following types:
- Competitive
- Non-Competitive
- Uncompetitive
Km indicates the Michaelis constant & Vmax (Vm) refers to the maximum velocity attained by the reaction. E refers to the Enzyme, S refers to the substrate, P refers to the product & I refer to the inhibitor.
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- The typical Michaelis-Menten equation mathematically describes the overall rate of the reaction as V (this is because biologists don't like math). What does V actually mean? (write the definition of V in differential equation form). V= d( )/dt Reaction rate Substrate concentration V max ·½V TURKarrow_forwardWhich of the following statements are true about the relationships of [S], KM, and Vmax? (Choose all that are true) As the [S] is increased, vo approaches the limiting value, Vmax KM = Vmax/2 The rate of product formed, vo, is at Vmax when [S] <<< KM KM and Vmax assist in finding the rate of the enzyme catalyzed reaction only if the reaction is considered irreversible.arrow_forwardThe enzyme, fumarate, has the following kinetic constants: k 1 k 2 k -1 where k 1 = 10 9 M -1 s -1 k -1 =4.4 x 10 4 s -1 k 2 = 10 3 s -1 a. What is the value of the Michaelis constant for this enzyme? b. At an enzyme concentration of 10 -6 M, what will be the initial rate of product"arrow_forward
- Some enzymes have catalytic activity only limited by diffusion. Which rate constants of an enzyme- catalyzed reaction is/are rate limiting for the enzyme? How does this line up/compare to the rate limiting step of Michaelis-Menten Enzyme Kinetics? (Please show work and correct answer)arrow_forwardYou are working on an enzyme that obeys standard Michaelis-Menten kinetics. Based on the following reaction expression, what is the Km value for this enzyme? E+SESE + P . . . k₁ = 880.8 M-¹5-1 k.₁ = 42.18 S-1 k₂ = 56.29 S-1arrow_forwardConsider 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 Oa. The reaction is now spontaneous due to the addition of enzyme b. The rate of the enzyme catalyzed reaction is faster than the uncatalyzed reaction O C. The reaction is exergonic O d. The change in free energy for the reaction is greater in the catalyzed reaction, compared to the uncatalyzed reaction e. The enzyme stabilizes the transition state for the reaction Released Energy pesarrow_forward
- A competitive inhibitor interacts with the free enzyme to form an enzyme•inhibitor complex(E•I). This equilibrium reaction can be described as follows:E + I ⇌ EIModify the simplified kinetic scheme for the reaction E + S ⇌ E + P to include this equilibriumexpressionarrow_forwardNOTE: the enzyme-inhibitor complex requires 450 joule/mol to dissociateJOULE/MOLarrow_forward
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