=display&toolld3calendar-mybb,GmailYouTubeMapscalendar-toolHomeCoursesCA Question Completion Status:203D70100| 11凸|12D|| 1301415016170QUESTION 14A newly isolated bacterial isocitrate dehydrogenase is found to have a molecular mass of 180KDA and SDS-PAGE shows it to be a homodimer. 1mg ofenzyme produces 0.05 mmoles of product per minute per mL.What is the turnover number of this enzyme in µmol product produced per pmol of enzyme subunit per mL?O a. 4.5Ob.4500O c. 9000O d.9QUESTION 15The absorption spectrum of NAD" and its reduced form NADH are shown below:Absorption2,42,01,6Click Save and Submit to save and submit. Click Save All Answers to save all answers.Save All Ar811JAN9.tvW...LO

Enzyme kinetics is a fascinating and important topic in biochemistry. It can be used to determine…

Answered: A newly isolated bacterial isocitrate…

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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6. Examine the plot of the enzyme-catalyzed reaction below: Wheat-germ acid phosphatase shows Michaelis-Menten kinetics when acting on para-nitrophenol phosphate 16 14 12 10 8 4 2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 (S] / mM a) Is this reaction better characterized by Michaelis-Menten kinetics or simple mass action? b) What is the approximate value of Vmax? Km? c) Assuming that this plot was made at 100 nM enzyme, draw the curve, including axis labels, for 300 nM enzyme. d) Imagine that this enzyme is inhibited by a competitive inhibitor with Ki (for inhibitor) = Km (for substrate). Note that v = (kcat [E] tot [S]/Km) / (1 + [S]/Km + [I]/Ki). Draw the curve of reaction velocity (v) versus [S] for [I] = 1 mM. e) Now consider an experiment where you simultaneously introduce substrate and competitive inhibitor at equal concentrations. Make a plot of v versus concentration of [S] = [I], i.e. the X-axis should be the concentration of both I and S, which are equal. v/ uM min-
Below is kinetic data obtained for an enzyme-catalyzed reaction. The enzyme concentration is fixed at 100 nM. Using a Lineweaver-Burke plot, calculate the catalytic efficiency for this reaction. Report your answer in scientific notation to three significant figures in units of M-1s-1.
8 7- 6 2- 180 225 270 315 [Substrate] (nM) 45 90 135 360 405 450 495 540 58! What is the KM of the enzyme data shown above? rate (nM/sec) 3.
Use the graph below to determine which of the following is the optimal temperature for catalase. Catalase Activity 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 10 20 30 40 50 60 Temperature (°C) O 0.8 O 1.45 O 37 O 20 Percent Increase in Oxygen per Minute
The 2 eynzymes are affected by a substrate. Choose all correct answers.
What is the initial velocity measurement for the hydrolysis of 50 μM amoxicillin by the enzyme?
The hydrolysis of a substrate, S, by an enzyme has been studied in the lab. The following initial rates, vo, were recorded at different concentrations. [S] (M) Vo (M/min) 2.10-10-4 1.20.10-6 4.20-10-4 3.10.10-6 9.30-10-4 6.30-10-6 1.42-10-3 9.10-10-6 A. Determine the rate constants for degradation of the substrate B. What is the rate of reaction at [S] = 1.1.104 M? C. Explain, why enzymes can make reactions go faster? Does enzymes also catalyse the reverse reaction from product to substrate?
Below is kinetic data obtained for an enzyme-catalyzed reaction. The enzyme concentration is fixed at 100 nM. Using a Lineweaver-Burke plot, calculate the kcat value for this reaction. Report your answer to three significant figures in units of 1/sec.
An enzyme catalyzes the reaction M-N. The enzyme is present at a concentration of 0.0000000022 M, and the Vmax is 0.0000028 Ms1. The Km for substrate is 5.5 µM. Calculate kcat-
Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. A. 0.10 per minuteB. 0.25 per minuteC. 0.50 per minuteD. 1.0 per minuteE. 2.0 per minute
15) The graph at right shows the results of reaction rate vs. substrate concentration for a Michaelis-Menten type enzyme 16) Th a. True b. False* reaction rate substrate concentration
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

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