3. (а)0.0050 M operate at one-quarter of its maximum rate?At what substrate concentration would an enzyme with a kcat of 30.0 s-1 and a Km of(b)trations [So]: ½ Km, 2 KM, and 10 KM.Determine the fraction of Vmax that would be obtained at the following substrate concen-(c)1 (HIV-1) has been the object of innumerable studies to developeffective chemotherapeutic agents. It has been shown that p6*known as the late assembly protein is an inhibitor of HIV protease.An assay was developed using an artificial polypeptide substratecontaining a p-nitrophenylalanine residue at the cleavage pointthat undergoes a small change in absorption at 295 nm upon bondhydrolysis that could be followed spectrophotometrically. Thecleavage of the peptide bond is shown schematically on the right.Results of the assay are given in the table below.The protease of the human immunodeficiency virus-LysNH2AlaNleArgAlaVal-Nle-NH-CH-Ċ–NH-GluCH2LysNH2AlaNO2NleArgH2Ó N HIV-1 proteaseAlaVo (nmole/min)in presence of10рМ рб* prо-VoVal-Nle-NH-CH-COO-+ NH-Glu[S](nmole/min)(µM)CH2tein104.632.70155.883.46206.944.74NO2259.266.063010.786.494012.148.065014.939.71 Construct a Lineweaver-Burk plot to determine what type of an inhibitor the p6* protein is. Extract fromthe plot Vmax, KM, and the Ki of the protein.

To study an enzyme kinetics, Michaelis-Menten kinetics equation is best known mathematical…

Answered: 3. (а) 0.0050 M operate at one-quarter…

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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5. For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s-1, k-1 = 3.1 ⅹ 104 s-1, and k2 = 3.4 ⅹ 105 s-1. a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach equilibrium or the steady state? Justify your answer. b) What is kcat for this reaction? Justify your answer. c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1, and each enzyme has two active sites. d) What substrate concentration would be required for the reaction in (c) to reach half of Vmax. Justify your answer mathematically. e) A second Michaelis-Menten enzyme has k1 = 4.2 ⅹ 107 M-1 s-1, k-1 = 6.1 ⅹ 104 s-1, and k2 = 5.3 ⅹ 102 s-1. Which enzyme is most efficient? 6. A pharmaceutical company is trying to develop a
For an enzyme that displays Michaelis-Menton kinetics, what is thereaction velocity, V (as a percentage of V max , observed at the followingvalues?[S] = K M[S] = 0.5K M[S] = 0.1K M[S] = 2K M[S] = 10K M
Assume that the experiments performed in the absence of inhibitors were conducted by adding 5 μL of a 2 mg/mL enzyme stock solution to an assay mixture with a total volume of 1 mL. Take into account that XYZase is a monomeric enzyme with a molecular mass of 45,000 Daltons. Hint: To calculate the ???? in units of per second (s−1), you must first determine the ???? in micromoles per second (μmol/sec). Please explain step by step
Calculate the mass of invertase (in mg) and concentration of invertase (in mM) contained in a 25.0mL sample of yeast extract that has 3,000 total units of activity, assuming that pure invertase has a specific activity of 1,000 units/mg with a mass of 270kD.
From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.
1) Is the ratio of the forward rate constant and reverse rate constant changed by the presence of an enzyme catalyst? 2) 4) What is the simplest mathematical relationship between substrate concentrations [S], initial velocity (Vo) of an enzyme catalyzed reaction, the maximal velocity of the reaction (Vmax), and the ½ maximal Vo (i.e. Km) ? (Hint: what is the Michaelis-Menten Equation? 3) Graphically illustrate the most useful derivation of the Michaelis-Menton equation (that darn linearized, double-reciprocal)
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. ?maxVmax(μmol min−1) ?MKM(mM) Wild‑type 100 10 Mutant 1 0.1 Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which ?−1k−1 is much larger than ?2,k2, which of the following statements are correct? 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 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. ?0=V0=
SUBSTRATE CONCENTRATION [S] µM INITIAL VELOCITY V0 s-1 10 0.13 25 0.27 50 0.45 100 0.65 150 0.77 200 0.85 300 0.94 500 1.03 (i) a) Construct an empty table with the following column headings: Substrate concentration [S] and initial velocity (Vi) where [S] has the unit µM, and Vi has the unit mM/s. (ii) The table provided is the enzyme kinetic data for your mutated enzyme, whereby Vi was expressed using the unit ∆A(405 nm)/s. Using the standard curve, express Vi with the unit mM/s rather than ∆A(405 nm)/s. Place your answer in the table above alongside the appropriate [S]. Hint: To answer this question you need to use this standard curve equation=0.0419x (The slope of the line is= 0.0419) (iii) The unmutated form of your protein has a Km of 25 µM and a Vmax of 43 mM/s. The enzyme kinetic data for your enzyme with the amino acid substitution should now be displayed in the table above. Based on these data, what is Vmax? Km? and…
List 4 major types of inhibition modes and clearly indicate the effect on Vmax and KM for each mode?2. What is the effect of each of the 4 types of inhibitors on the initial rate of an enzyme catalyzed reaction?3. A potent inhibitor effectively inhibits an enzyme catalyzed reaction. What kind of a Ki value you would expect for a potent inhibitor?4. Considering PNPP → PNP reaction, would you expect to see more intense or pale color for the reaction that contain the inhibitor? Explain.
At what substrate concentration would an enzyme with a kcat of 25.0 s-1 and a KM of 3.5 mM operate at 25% of its maximal rate? How many reactions would the enzyme catalyze in 45 seconds when it is fully saturated with substate, assuming the enzyme has one active site?
The 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"
An enzyme catalyzes a reaction at a velocity of 20 μmol/min when the concentration of substrate (S)is 0.01 M. The Km for this substrate is 1 × 10-5 M. Assuming that Michaelis-Menten kinetics arefollowed, what will the reaction velocity be when the concentration of S is 1 ×10-6 M?

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