ANSWER ONLY C AND D 1. A Lineweaver-Burk Plot is shown below.3025Curve Ay = 3.1207x + 2.49782015Curve By = 1.0003x + 2.3602105-3157111/[Catechol] (mM1)With these curves, determine the following enzyme parameters. Show all pertinent solutions.a. Km of Curve A and Curve Bb. Vmax of Curve A and Curve Bc. Assuming that one of these curves corresponds to the kinetics of one enzyme and one substrate, which curverepresents the effect of an inhibitor? Why do you say so?d. What type of inhibition is exhibited by your answer in question c? Why do you say so?1/V, (units of activity 1)

If an enzyme follows Michaelis-Menten Kinetics, a plot of the reciprocal of the reaction…

Answered: 1. A Lineweaver-Burk Plot is shown…

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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Explain why each of the following data sets from a Lineweaver–Burk plot are not individually ideal for determining KM for an enzymecatalyzed reaction that follows Michaelis–Menten kinetics.
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…
1.1)the following data duscribe an enzyme-catalyzed reaction(hydrolysis of cabobenzoxyglycyl-L-tryptophan) Plot these results using a lineweaver-Burk method, and determine values for Km and Vmax. substrate concenrate(mM) Velocity(mM.sec-1) 2,5 0.024 5 0.036 10 0.053 15 0.060 20 0.061 25 0.062 1.2) If the Km of an enzyme for it's substrate remains constant as the concentration of the inhibitor icreaces, what can be said about the mode of inhibition and why? 1.3) calculate the turnover number for an enzyme, assuming Vmax is 0.5M.sec-1 and the concentration of the enzyme used is 0.002M . why is it usefull to know this? 1.4) discuss the mechanism of the bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues. make use of relavant graphs and diagrams to explain your answer.
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
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.
Using the appropriate graph and table above, explain what the N426S mutation appears to be doing to the enzyme’s function. Discuss the kinetic parameter changes and their meaning in this context, not the structure of the enzyme, which was not given to you.
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=
Lisa decides to obtain values for the Km and Vmax of an enzyme that was isolated from liver cells.. Using the Michaelis Menten plot. In what kind of measurements are needed and what would be needed to plot on a graph to estimate Km and Vmax?
Although graphical methods are available for accurate determination of the Vmax and Km of an enzyme-catalyzed reaction, sometimes these quantities can be quickly estimated by inspecting values of V0 at increasing [S]. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained:
1.1)the following data duscribe an enzyme-catalyzed reaction(hydrolysis of cabobenzoxyglycyl-L-tryptophan) Plot these results using a lineweaver-Burk method, and determine values for Km and Vmax. substrate concenrate(mM) Velocity(mM.sec-1) 2,5 0.024 5 0.036 10 0.053 15 0.060 20 0.061 25 0.062
The Keq (25C) of the reaction below is 635.67. Fructose 1,6-biphosphate <-->fructose -6-phosphate + Pi. a) What is the standard Gibbs free energy change for this reaction? b) if the concentrationof fructose 1,6 biphosphate is adjusted to 0.85 M and that of fructose 6 phosphate and phosphate adjusted to 0.055 M, what is the actual free energy change
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"

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