9. For the following aspartate reaction in the presence of inhibitor, Km = 0.00065 M. Determine Vmax in bothreactions and in the reaction without inhibitor, the Km. Identify whether the inhibition is competitive, non-competitiveor uncompetitive.1101001/V7060-40-30-20-10-10. For your answer in #9, explain briefly:a. 1. how / and S bind to the E as shown by the Lineweaver Burk plot.b. the significance of the following obtained values for Km and Vmax.c. effect in slope and x-interceptF000F0000Fo008F0009F000Fo00

Enzymes are biocatalyst that increases the speed of reaction by lowering the activation energy.…

Answered: 9. For the following aspartate reaction…

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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Km value should be more than 10% apart, use non-competitive inhibitor
4
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?
From the chart below, comment on the temperature dependence of this enzyme catalyzed reaction.
1. Start with the Michaelis-Menten equation and convert it to a double-reciprocal equation. Show how you would make a linear Lineweaver-Burk plot using this equation. 2. Based on th chart make a Lineweaver-Burk plot and later use the plot to complete the information in the table.
4. Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated from the bacterium Staphylococcus aureus. The enzyme is involved in carbohydrate metabolism and is a potential biocatalyst for the large-scale production of rare sugars. (8) mg V₂ μmol s1 1.50- 1.25- 1.00- 0.75- 0.50- 0.25- 0.00 0 100 200 300 [s], mM 400 500 WT (a) Estimate the KM and Vmax for the wild-type and mutant enzyme from the graph. V105A (b) Calculate the keat and keat/KM for the wild-type and mutant enzyme based on your estimated values in (a) if the total enzyme concentration is 0.5 µmol/mg. (c) Is the mutant enzyme a more or less efficient catalyst than the wild-type enzyme? Briefly explain.
make a schematic diagram of BChE/AChE inhibitory determination Butyrylthiocholine iodide (BChI), as well as acetylthiocholine iodide(AChI) molecules were utilized as substrate compounds of the reaction. Inthis regard, 5,5′‐dithio‐bis(2‐nitro‐benzoic acid) compound (DTNB) wasused to estimate BChE/AChE activity. In summary, 100 μL Tris buffersolution (HCl, 1.0M, pH 8.0, Tris) and variety of concentrations of thesample solution (50‐200 mL) were added in deionized water and also20 μL of BChE/AChE enzymes solutions added. The mixture was thenincubated for 10 minutes at 20°C. Finally, 50 μL of DTNB (0.5 mM and25 mL) BChI/AChI were added to the incubator mixture. The reactionwas also started by adding 50 μL BChI/AChI. The activity of theseenzymes was evaluated using spectrophotometric spectra with awavelength of 412 nm.
CHEM151/251 Biochemistry 1 a. Determine Km 1. The following data were obtained for a competitive inhibition study in which the [I] = 3 µM for each determination of vo in the presence of inhibitor. The Vmax = 200 μM P/min for both data sets. 200 Vo (UM P/min) 8 8 8 8 8 8 8 8 8 180 160 140 120 100 40 Name (Print)/ID #: the absence of inhibitor. Participation Question # 10 No Inhibitor 50 +Inhibitor 100 [Substrate] (UM) 150 b. Determine Km, app for the data obtained in presence of inhibitor. 200 c. Calculate the value for Ki. Note: a = 1 + [I]/Ki and a¹ = 1 + [I]/Ki*. for the data obtained in
D 1. 0.25 Short answer questions, write answers on blank paper. A plot of enzyme activity with and without an inhibitor present gave the following plot. 0.2 0.15 1/V v in mM/s 0.1 0.05 0 0 0.05 0.1 y=0.67x+0.02 0.15 y=0.2x+0.02 0.2 0.25 0.3 1/[S]. [S] in mM Without the inhibitor, what are the values of Vmax and Km? In the presence of the inhibitor, which of these items has changed and what is (are) the new value(s)? What type of inhibitor is present (explain why you have arrived at this conclusion)? Be sure to include units as necessary--answers that do not contain required units will be marked incorrect
4. a. Calculate the KM (Michaelis constant) and the vmax (the maximum initial rate) for both substrates (sphingosine and ATP). Show your work, and be careful about units. b. threo-dihydrosphingosine, a stereoisomer of sphingosine, is an inhibitor of sphingosine kinase. What kind of inhibitor (competitive, uncompetitive, noncompetitive) is threo-dihydrosphingosine? Citing information from the Lineweaver-Burke plots, explain how you can tell.
4.2
4. a. Use the data in the graph above to estimate a KM value for the enzyme in the presence of these metabolites, and enter them into the table below. b. Classify these metabolites as either activators or inhibitors, and explain your rationale below.

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