1. Enzyme X has a molecular weight of 35,000 Da. When this enzyme is assayed forenzyme activity, the corresponding line of the double reciprocal plot has the equation y =75,253X + 102,392. The substrate concentration was in mM and the velocity wasmeasured in mM min". From these data calculate the following:a. KM and Vmaxb. kcat if the enzyme concentration was 2 mg/mL.2. When you run enzyme X from question one on an SDS-PAGE you find that, afterCoomassie staining, there is only one band on the gel that runs just below the 20,000Dalton molecular weight marker. What does this result tell you about the nature of thequaternary structure of your enzyme?

Enzyme kinetics is used to study the rates of enzyme catalysed biochemical reactions. In this study…

Answered: 1. Enzyme X has a molecular weight of…

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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Similar questions

If the data from an enzyme experiment is plotted as a Lineweaver-Burk plot, and the Vmax is 0.02 mol/sec, and x-intercept is –2.5 mM then what is the KM value? Show yourwork/reasoning.
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=
The initial velocity data shown in the table were obtained for an enzyme. Each assay at the indicated substrate concentration was initiated by adding enzyme to a final concentration of 0.01 nM. Derive Km, Vmax, kcat, and the specificity constant. [S] (mM) Velocity (x10^7) 0.10 0.96 0.125 1.12 0.167 1.35 0.250 1.66 0.50 2.22 1.0 2.63
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…
200 ml of a 2% protein solution are available, containing an enzyme to be purified. Half of the sample is subjected to method A, consisting of fractionated precipitations, and 5 ml of final solution are obtained, with a protein concentration equal to 5 mg / ml and enzymatic activity equal to 2000 U / ml. The other half is subjected to method B, consisting of ion exchange chromatography, and a final solution of 10 ml is obtained, with a protein richness equal to 10 mg / ml, and with an enzymatic activity equal to 2000 U / ml. You want to know: a) Which of the methods has provided the purest enzyme. b) By which of the methods has the greatest amount of protein been obtained.
The turnover number for an enzyme is known to be 5000min-1. Given the following set of data, Substrate concentration(mM) 1, 2, 4, 6, 100, 1,000 Initial Rate(micromol/min) 167, 250, 334, 376, 498, 499 a) What is the Km of the enzyme for the substrate? (do this without using a calculator) b) What is the total amount of enzyme present in the assay?
1. Make a Lineweaver-Burk plot and use the plot to complete the information in the table and the following questions. a. Is it possible for the enzyme to overcome the effect of the inhibitor in question from the chart. Explain. b. What prevents this enzyme from being an even more catalytically efficient enzyme? c. What do single molecule data indicate about the validity of ensemble data?d. What is the reason that humans are insensitive to sulfa drugs?
An enzyme is found that catalyzes the reaction X ⇌ Y. Researchers find that the Km for the substrate X is 4 μM, and the kcat is 20 min−1.(a) In an experiment, [X] = 6 mM, and V0 = 480 nM min−1. What was the [Et] used in the experiment?(b) In another experiment, [Et] = 0.5 μM, and the measured V0 = 5 μM min−1. What was the [X] used in the experiment?(c) The compound Z is found to be a very strong competitive inhibitor of the enzyme, with an α of 10. In an experiment with the same [Et] as in (a), but a different [X], an amount of Z is added that reduces V0 to 240 nM min−1. What is the [X] in this experiment?(d) Based on the kinetic parameters given above, has this enzyme evolved to achieve catalytic perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic perfection.
Each plot has its own limitations. Match the plot with the correct statement on limitations. 1.Hanes-Woolf 2.Michaelis Menten 3.Lineweaver-Burk limitations: A. This plot appears to allow for more accurate calculation of these parameters, but is prone to error, as the y-axis takes the reciprocal of the rate of reaction –this increases any small errors in measurement at low substrate concentration, where they are most likely to occur. B.The advantage of this plot is that it does not not overemphasizing the data obtained at low [S]. The disadvantage is that any inaccurate measures of substrate concentration will be exaggerated as [S] is used on both axes. C.It is difficult to obtain a meaningful value of Km from this plot because of the high substrate concentrations needed to reach Vmax. Vmax has to be estimated by extrapolation- which is partly subjective and introduces error. D.The advantage of this plot is that it overcomes uneven spacing of points, and undue weight of a…
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
. Lineweaver-Burk plot is the most used linearization method of the MichaelisMenten equation, but the main drawback is a/an:a. overemphasis to low substrate concentration and less emphasis to high substrate concentration.b. overemphasis to high substrate concentration and less emphasis to low substrate concentration.c. exaggeration to both ends of the reciprocal of the substrate concentration.d. lack of independent variables to x- and y- coordinates.Kinetic data were gathered for Enzyme X, its substrate, and a known competitive inhibitor Z. Upon plotting the linearized kinetic data, one could surmise that the plot of the inhibited enzyme would have:a. a slope and y-intercept both lower in magnitude than the plot of the uninhibited enzyme using Eadie-Hofstee method.b. a constant slope and y-intercept that is lower in magnitude with the plot of the uninhibited enzyme using Eadie-Hofstee method.c. a slope and y-intercept greater in magnitude than the plot of the uninhibited enzyme using…
A research group discovers a new version of happyase, which they call happyase*, that catalyzes the chemical reaction The researchers begin to characterize the enzyme. (a) In the first experiment, with [Et] at 4 nM, they find that the Vmax is 1.6 uM/s. Based on this experiment, what is the kcat for happyase*? (Include appropriate units.) (b) In another experiment, with [Et] at 1 nM and [HAPPY] at 30 uM, the researchers find that V0 = 300 nM/s. What is the measured Km of happyase* for its substrate HAPPY? (Include appropriate units.) (c) Further research shows that the purified happyase* used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase* preparation and the two experiments repeated, the measured Vmax in (a) is increased to 4.8 uM/s, and the measured Km in (b) is now 15 uM. Based on this information, can you figure out what type of inhibitor is ANGER? (Use table 6.9 at the end of the…

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