TOMA given enzyme is able to utilize five different substrates. The Km's havemeasured for each substrate. Which substrate is most likely the biochemirelevant one in the cell? In other words, which one is most likely the predvivo substrate for this enzyme? (Choose the one best answer).Substrate 1 with KM= 4.7 × 10-1 M.Substrate 2 with Km = 1.5 × 10-2 M.Substrate 3 with KM = 1.5 x 10-8 M.×Substrate 4 with Km= 4.7 × 10-³ MSubstrate 5 with KM = 4. 7 × 10-5 M What is an order of the movement of electrons through the electron transport chain?Choose the one most correct answer.NADH->Complex II->Ubiquinone->Complex III -> Cyt C -> Complex IVFADH2 -> Complex II -> Ubiquinone -> Complex III -> Cyt C -> Complex IVFADH2->Ubiquinone->Complex I->Cyt C->Complex II->Complex IVFADH2 -> Complex I -> Cyt C -> Complex III -> Ubiquinone -> Complex IVE7

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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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An enzyme reversibly binds a substrate. The rate constant (kcat) for catalytic conversion of enzyme-substrate complex into product and free enzyme is 1.0 s-1. The rate constant for dissociation of enzyme-substrate complex (k-1) is 9.0s^-1. 1) For each substrate binding event, what is the probability that substrate is converted into product rather than dissociated without conversion? 2) What is the probability that 20 consecutive binding events result in no product formation? 3) If you evaluated 1,000 different sets of 20 consecutive binding events, what would be the average number of product molecules formed (per 20 binding events)? 4) What is the probability that 20 consecutive binding events result in a number of product molecules at least equal to this average?
The equil ibrium constant for the attachment of a substrate to the active site of an enzyme was measured as 200.In a separate experiment, the rate constant for the secondorder attachment was found to be 1.5 x 108 dm3 mol-1 s- 1.What is the rate constant for the loss of the unreacted substrate from the active site?
You design an enzyme assay and choose a substrate concentration equal to the Km of the enzymatic reaction. You measure the rate of the reaction at this substrate concentration to be 75.0 μmol/min. Calculate the Vmax of this reaction in μmol/min. Express your answer with one decimal place. Your previous attempt suffered from low signal-to-noise ratio when you used a substrate concentration equal to the Km. So you decide to increase the substrate concentration to 0.45 mM. You remeasure the rate of the reaction at this new substrate concentration and find it to be 135.0 μmol/min. Using the same Vmax value found above, calculate the Km value of this enzymatic reaction. Express your answer in μM with one decimal place. Considering the previous two questions, what would be the rate of the reaction at a substrate concentration of 0.01 mM? Express your answer in μmol/min with one decimal place.
Enzymes are influenced by physical and chemical conditions affecting kinetic characteristics. Rubisco function is no different, and changes in temperature have profound influences on its activity based on changes in __Blank A___ and __Blank B___ Options for Blank A: - Vmax - Kineton Ratio - denaturization - H-bond linkage - Substrate availability Options for Blank B: - susceptibility to feed-forward interactions - cellular localization - mechanism of regulation - membrane binding characteristics - kinetics of catalytic function
For an enzyme with a substrate Km of 0.001 M. a) What substrate concentration for an initial rate equal to one-half maximaly velocity? and b) for an initial rate equal to 0.9 maximal velocity> * a) [S} = 0.001 M, b) [S] = 0.009 M a) [S} = Km, b) [S] = 0.09 M a) [S] = 0.01 M, b) [S] = 0.9 M a) [S] = 1 M, b) [S] = 9 M
A hypothetical enzyme that follows Michaelis-Menten kinetics functions has a substrate concentration 50mM at 50% of Vmax. Find the substrate concentration when the velocity of reaction is equal to 75% of Vmax.
(a) At what substrate concentration would an enzyme with a kcat of 30.0s−1 and a Km of 0.0050 M operate at one-quarter of its maximum rate?(b) Determine the fraction of Vmaxthat would be obtained at the following substratecon centrations [S]: ½Km, 2Km, and 10Km.(c) An enzyme that catalyzes the reaction X ⇌ Y is isolated from two bacterial species.The enzymes have the same Vmax but different Km values for the substrate X. Enzyme A has a Km of 2.0 μM, and enzyme B has a Km of 0.5 μM. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 μM. Which curve corresponds to which enzyme?
One substrate name is X, and this substrate have two enzymes.(E1and E2) E1and E2, are enzymed and they related in the synthesis of an amino acid and an antibiotic, respectively.Reaction rates of enzymes , at an optional concentration of E1and E2 were determine at different concentrations of substrate X. Which data is suitable labeled for E1 and why ?
An enzyme is mutated such that the KM of the mutated enzyme is lower than the KM of the original enzyme. Which of the following statements is true (select all correct answers)? Question 2 options: If the original enzyme and the mutated enzyme are mixed with the same concentration of substrate, the mutated enzyme will have v0 closer to Vmax (hint: consider the Michaelis-Menten equation) The effect of a particular competitive inhibitor will be less on the mutated enzyme than it is on the original enzyme (hint: consider the reaction scheme for a competitive inhibitor) The effect of a particular uncompetitive inhibitor will be less on the mutated enzyme than it is on the original enzyme (hint: consider the reaction scheme for an uncompetitive inhibitor) The mutated enzyme has higher affinity for the substrate than the original enzyme does
When k2 >> k-1, KM approximates the affinity of the enzyme•substrate complex.The (circle one) higher / lower the KM, the more tightly the enzyme binds to its substrate
An enzyme that follows the MWC (concerted) model for allostery has a T/R ratio of 300 in the absence of substrate. Suppose that a mutation were to reverse that ratio (T/R = 1/300 = 3.3 x 10–3 in the absence of any substrate). How would this mutation affect the relation between the rate of the reaction and substrate concentration, i.e., what would a Vo vs. [S] plot look like, and why?
The covalent catalytic mechanism of an enzyme depends on a single active site Cys whose pK is 8. A mutation in a nearby residue alters the micro environment so that this pK increases to 10. Would the mutation cause the reaction rate to increase or decrease? Explain.

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