=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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3.An enzyme catalyzed reaction is studied and the following kinetic analysis is obtained: |[S), mM 0.050 0.075 v, (µM min") 0.93 1.264 1.77 2.14 3.7 0.125 0.175 0.935 a. Using Excel, make a fully labeled Lineweaver-Burk plot and determine the Km and Vmax for the enzyme b. The reactions were set up dissolving 1 mg of the enzyme (MW= 100000 Da) in 100 ml of final reaction buffer. Determine the turnover number for the enzyme assuming 1 active site exists per enzyme molecule? c. Determine the catalytic efficiency for the enzyme d. The same reactions are performed in presence of an inhibitor A and the resulting velocities determined: v plus inhibitor, (µM min) 0.272 0.37 0.518 0.626 |1.08 |[S), mM 0.050 0.075 0.125 0.175 0.935 Plot these data on the same graph as above and determine the new Km and Vmax and the type of inhibitor (competitive, non-competitive). e. Can the effects of the inhibitor be over-ridden by adding more substrate? Why?
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-
The compound below which is labelled in two positions with 13c (as represented by the shaded spheres) was prepared and used to probe the mechanism of the aldolase enzyme employed in glycolysis. OH O Predict the pattern of 13c radiolabelling observed in the products: OH HO он 20,PO O LOPO,? но, OH 20,PO но. Hక్కి OH 20,PO Но, он 20;PO, HO. LOPO,2-
explain the results of this tables
In glycolysis, fructose 1,6-bisphosphate is converted to two products with a standard free- energy change (AG) of 23.8 kJ/mol. Under what cellular conditions will the free-energy change be negative, enabling the reaction to proceed spontaneously to form products? O In a cell where the concentrations of products and reactants yield a low value of the mass-action ratio Q ○ The reaction will not go to the right spontaneously under any conditions because the AG is positive ○ The reaction will proceed spontaneously to the right if there is a high concentration of products relative to the concentration of fructose 1,6-bisphosphate ○ The reaction will proceed spontaneously to the right if there is a high concentration of enzyme to catalyze the reaction O None of these conditions is sufficient
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
An enzyme catalyzes a reaction with a Km of 6.50 mM and a Vmax of 2.00 mM s¹. Calculate the reaction velocity, Vo, for each substrate concentration. Vo : Vo: Vo : [S] = 4.00 mM [S] = 6.50 mM [S] 14.0 mM = n. mm. s-1 mm. sl mm. s-1
If the enzyme lactase has a Vo of 0.111111111111 mM per minute when [S] = 1.0 mM, and a Vo of 0.20 mM per minute when [S] = 5.0 mM, what is its Km? 0.125 mM 0.25 mM 0.50 mM 1.25 mM 5.0 mM
The enzymatic activity of an enzyme with Kg = 2 mM that converts substrate S into product P is measured at an initial substrate concentration S, of 10 µM. After 5 min, the substrate concentration is halved. What is the rate constant k, the maximal velocity vmax and the concentration of product after 12 min?
1. The optimal conditions for salivary lysozyme (hydrolyzing glycoproteins ofbacterial wall) are 37 C- temperature and pH is 5.2. Explain the decrease in this enzyme activity if the temperature will rise up to 60 °C and pH will be changed to 8.0. To answer the question: a) draw the graph of the velocity dependency on temperature and pH; b) calculate the relative enzyme activity if 10 mg of lysozyme catalyzes the formation of 5 uM of the product per 2 minutes. 2 Consider the matic reaction schee: Asnaragine + H20 Aspartate+ NH3:
You are evaluating the kinetics of an enzyme catalyzed reaction containing 5.5 μM total enzyme and 11.2 μM substrate. At this substrate concentration, you determine that the Vo = 88.6 μmol mL-¹. s-¹. If the Vmax 833.3 mM s the KM is: . == " 30.9 μΜ 10.4 μΜ Ο 124.6 μΜ 234 μΜ 94.5 μM
Table 1 shows the kinetic data that have been obtained for glucoamylase from Aspergillus niger at different temperatures in the production of glucose from maltodextrin. The rates of reaction measured are using an enzymes concentration of 0.003 μmolml-¹. Table 1: Glucoamylase Activity at different temperatures Maltodextrin Concentration (μmol/L) 0.5 1 1.5 2.2 5.5 11 Initial Reaction Velocity (μmol/L.s) T = 40°C 2.1 3.5 4.9 5.6 589 8.3 9.3 T = 60°C 5.2 9.7 13 16 23 31

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