please answer and write correct statement for each question and please answer all parts or better to leave it. If correct I will upvote and positive feedback thanks Multiple answers: Multiple answers are accepted for this questionSelect one or more answers and submit. For keyboard navigation. SHOW MORE VaDecreasing the free energy of activation decreases the reaction rate.b The catalytic efficiency of an enzyme cannot exceed the diffusion rate of encounters between E and S.When [S] = Km the velocity = Vmax/2.Lineweaver-Burk double reciprocal (1/V vs. 1/S) plots convert the hyperbolic Michaelis-Menten V vs. S plot into a straightd.line.eAccording to the Michaelis-Menten equation the v/Vmax ratio = 0.4 when [S] = 4 Km-If Vmax = 100 umol/mL/sec and Km = 2 mM the velocity of the reaction when [S] = 20 mM is 10 umol/mL/sec.The Michaelis-Menten equation describes the relationship between initial reaction velocity and substrate concentrationunder steady state conditions.

Enzymes are usually made of protein molecules which catalyze biochemical reactions by lowering the…

Multiple answers: Multiple answers are accepted for this question Select one or more answers and submit. For keyboard navigation. SHOW MORE V Decreasing the free energy of activation decreases the reaction rate. a b. The catalytic efficiency of an enzyme cannot exceed the diffusion rate of encounters between E and S.

Multiple answers: Multiple answers are accepted for this question Select one or more answers and submit. For keyboard navigation. SHOW MORE V Decreasing the free energy of activation decreases the reaction rate. a b. The catalytic efficiency of an enzyme cannot exceed the diffusion rate of encounters between E and S.

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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Answer TRUE or FALSE.a. According to the lock-and-key model of enzyme action, the active site of an enzyme is not flexible in shape.b. In an enzyme-catalyzed reaction, the compound that does not undergo a chemical change is called the substrate.c. The nonprotein portion of a conjugated enzyme is not the enzyme’s active sited. Simple enzymes are composed only of protein; conjugated enzymes have nonprotein cofactors.
Trend observed in graph and conclusion about the effect of temperature on enzyme activity. In your answer i) include a concise description of the trend observed in the graph shown , and ii) explain this trend using the language presented in this unit and your biochemical knowledge of enzymes and reactions. In your conclusion, provide a logical argument supported by molecular theory that would explain any change observed in enzyme activity
Use the Michaelis-Menten equation to complete the enzyme kinetic data set, when Km is known to have a value of 1 mmol/L. Upload your tabulated answers and show your calculations.
Answers should be specific and mention pathway details from the book. PLEASE ANSWER ALL. im very confused, will upvote. Briefly describe the difference between a reversible and irreversible reaction in a metabolic pathway.
Below is kinetic data obtained for an enzyme-catalyzed reaction. The enzyme concentration is fixed at 100 nM. Using a Lineweaver-Burke plot, calculate the Vmax value for this reaction. Report your answer to four significant figures in units of uM/min.
Hi :-) Does anyone know how to calculate the amount of an enzyme in mg when you're only given this data: Volume of the original sample containing the enzyme in microlitres. The change in absorbance for the enzyme reaction. The velocity equation. Molar extinction coefficient. Please let me know what formula to use. I'm confused where to begin. Thank you!
If an enzyme catalyzed reaction has a KM of 5mM and a Vmax of 60 nm/sec, the substrate concentration at 30 nM/sec is? Thank you.
The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate ?0V0 for an enzyme‑catalyzed, single‑substrate reaction E+S↽−−⇀ES⟶E+PE+S↽−−⇀ES⟶E+P. The model can be more readily understood when comparing three conditions: [S]<<?m[S]<<Km, [S]=?m[S]=Km, and [S]>>?m[S]>>Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity ?0V0 where steady state conditions are assumed. [Etotal][Etotal] refers to the total enzyme concentration and [Efree][Efree] refers to the concentration of free enzyme.
Which 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.
Reaction 25.2 Glucose (aq) + PO43- (aq) ⇄ Glucose-PO43- (aq) Reaction 25.2 is nonspontaneous and very slow. Use this information to sketch a reaction diagram for Reaction 25.2. Clearly label the x axis, y axis, products, reactants, and activation energy.
Mark any/all statements that are TRUE regarding reactions and enzyme catalysis. Group of answer choices -The role of an enzyme in catalysis is to increase the difference in free-energy between the products and reactants (increase the G for the reaction). -None of these is correct. -The free-energy of a reaction (G) determines the rate of reaction. -In thermodynamics, the term "spontaneous" means that a reaction requires energy to proceed. -The transition state of an enzyme-catalyzed reaction is destabilized in comparison to the transition state of the uncatalyzed reaction.
The Km value of an enzyme-catalyzed reaction and its Vmax is 70 mmol/min. What is the rate of the reaction of the reaction when the substrate concentration is 7 × 10−2 mmol/min? a. 35 mmol/min b. 50 mmol/min c. 60 mmol/min d. 70 mmol/min