None An enzyme X is placed at 4 different pH values and the following data are gathered. Choosethe pH value which is the best for the function of enzyme X. - = no activity, += low activity,++= high activity52432345+++Give detailedSolution withexplanation needed.don't use Ai foranswering this

Step 1:Enzymes are biological molecules that catalyze or speed up a biological reaction. They are…

Answered: An enzyme X is placed at 4 different pH…

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Data from enzyme inhibition are used to determine a Kmapp and Vmax PP. Comparison of these values with assays run without inhibitor are used to understand how the inhibition is occurring. This is useful for better understanding the active site as well as the practical aspect of pharmaceutical drugs. Below are idealized Line-Weaver Burke plots of different types of inhibitors. Comnetitive Uncomnetitive Mixed +Inh +Inh 4Inh Anh Inh Anh [S] [S] [S] a. How does the value of Vmax for the enzyme compare to the Vmax PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed b. How does the value of Km for the enzyme compare to the Km PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed c. For each situation in Model 1, consider an inhibitor that is better than the one shown on the graph. Answer the following questions for each type of inhibition: i. How would the KmPP change? ii. How would the Vmax PP change?
The accompanying figure shows three Lineweaver–Burk plots for enzymereactions that have been carried out in the presence, or absence, of aninhibitor. Indicate what type of inhibition is predicted based on eachLineweaver–Burk plot. For each plot indicate which line corresponds to thereaction without inhibitor and which line corresponds to the reaction withinhibitor present.
1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…
Consider the given data for an enzyme-catalyzed reaction. Determine the Vm, Km and the type of inhibition based on the given data below Substrate concentration, uM 30 50 100 300 900 slope y-intercept Complete the table below (include correct units). Experiment A Vm Km Experiment A (Initial velocity without inhibitor, uM-min) Type of Inhibition: 10.4 14.5 22.5 33.8 40.8 Experiment B (Initial velocity with inhibitor, uM-min) 5.1 7.3 13.3 25.7 37.2 Experiment B
Briefly comment on the differences of using a fixed-time assay versus a kinetic assay to measure enzyme activity. Is it reasonable to assume that the reaction velocity obtained by measuring the amount of product after 30 minutes in a fixed-time assay is directly proportional to absorbance? How could you determine whether this was the case? Word limit 180 words including citation and reference
Part 1: Assess the following partial results section below by editing it for brevity by omitting any unnecessary parts (1 point), explain why you decided to remove certain sections (1 point): To evaluate inhibitory effects of the selected molecules, 10mM stock solutions of each molecule were prepared in DMSO. A reaction mixture (200μl) was prepared with the same formula optimized for the enzyme activity assay (0.1 M Tris-HCl ph 8, 0.1 M KCI, 25 mM NaCl, 0.25 mM ATP, and two units of inorganic yeast pyrophosphatase) with 10 µM of the sample molecule. The reaction mixture was incubated for 20 minutes at ambient temperature. Enzymatic reaction was triggered by addition of the substrate B (0.2 mM) and the absorbance of the product was monitored at 290 nm for 10 minutes. Six out of 15 sample molecules showed appreciable inhibition at 10 μM (Figure 5). Three of the molecules, A3, A6, and A7 exhibited more than 50% inhibition of the enzyme activity and were further diluted to find the minimal…
True or False Immobilization improves the stability of the enzyme. EnaLne, has a half-life of 10 days in free solution, but under identical conditions of temperature, pH, and medium composition, the measured half-life of a packed column is 30 days. The enzyme is immobilized in a porous sphere 5 mm in diameter.
The following statements refer to enzyme inhibition. Match the statement to the one of the following descriptors to which it is best associated. Descriptors: competitive inhibition; non-competitive inhibition; un-competitive; covalent inhibition. 9a. Inhibition is not reversed even after the inhibitor (1) is removed from solution by dialysis or drug metabolism/excretion. 9b. Inhibitor and substrate reversibly compete for occupancy of a common binding site 9c. The inhibitor binds reversibly only to the preformed E.S (enzyme-substrate) complex forming an inactive E.S.I. 9d. The inhibitor binds reversibly and independently of substrate to an allosteric site producing E.I or a ternary E.S.I complex which can't form product. 9f. The relative amount of inhibition decreases as [S] (the concentration of substrate) increases and S better competes for occupancy of the active site.
A с Wan WWW GHEDE MAK am2 Increasing CE a-1 (no inhibitor) Slope #KY... ...." 7-15 7 a>c²=1 [no inhibitori Slope R 101 9-19 aver-Burk plots in this figure represent the activities of enzymes in the 4-5 n. 155 B D MIN -a'/KM 0.8- 1/V >~- -1/K 0 01 m(app) d'Nmax a=2 a=1.5 1[S] per mM inhibitor a=10 Slope K
Please use the graph below to explain the differences between the 2 enzymes whose activities are plotted (enzyme 1-blue and enzyme 2-red). Using appropriate biochemical terminology, how do these differences affect the activity and function of Enzyme 1 versus Enzyme 2? Reaction velocity (vo) Vmax Vmax 2 0 0 Enzyme 1 Enzyme 2 [Substrate]
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.
Which statements are false? Initial velocities of enzyme reactions are best obtained in the absence of product because it simplifies analysis. Initial velocities refer to the velocity of the reaction right after it is initiated. The velocity of the reaction as a function of measuring time are curved just like an isothermal binding curve because of substrate binding to the enzyme. Initial velocities correspond to the pre-steady state condition for free enzyme. Initial velocities can sometimes be measured by spectroscopy such as UV/Vis spectroscopy when monitoring the production of NADH from NAD+. The velocity of the reaction will eventually go to zero. The reaction will reach equilibrium because of the presence of the enzyme. It is always better to use substrate rather than product to measure enzyme kinetics.

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