QUESTION 17calculate the KM ANDthe total amount of enzyme present in these experiments.The following equations relate to this question. (Hint 1: some data is directly apparent without need to do calculations; Hint 2:Watch your exponents.)Kcat=Substrateconcentration(mm)12461001,000The turnover number for an enzyme is known to be 5,000 min-1. From the following set of data,250334376500500Vmax[Et]Initialvelocity(μmol/min)167G. 10 mMH. 250 uM1. 2 mMJ. 100 mMPart 1Total Enzyme amount (choose from options A-E)Part 2 (Enzyme KM: (choose from options G-J)A. Total enzyme = 10 umolB. Total enzyme = 0.1 umolOC. Total enzyme = 100 μmolD. Total enzyme = 1000 μmolE. Total enzyme = 10 μmolOFVo =Vmax[S]KM + [S]

As given in the question, Kcat (turnover number) = 5000 min-1 KM = ? Et (total enzyme concentration)…

Answered: QUESTION 17 calculate the KM AND the…

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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If the enzyme lactase has a Vo of 0.40 mM per minute when [S] = 1.25 mM, and a Vo of 1.0 mM per minute when [S] = 5.0 mM, what is its Vmax? Calculate the Km of the above enzyme. Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the above enzyme.
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Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the above enzyme (lactase). (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) 0.125 per minute 0.25 per minute 0.50 per minute 1.25 per minute 5.0 per minute
When 10 micrograms of an enzyme with a molecular mass of 80,000 Daultons (grams/mol) is added to a solution containing its substrate at a concentration 100 times the Km, it catalyzes the conversion of 65 micromoles of substrate into product in 3 minutes. What is the enzyme's turnover number (in units of min-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.
For an enzyme-catalyzed reaction, the velocity was determined at two different concentrations of the substrate. Estimate the value of Vmax. [S] (MM) 10 20 88 nmol/min 79 nmol/min 67 nmol/min V(nmol/min) 51 nmol/min 27 Not enough information is given to form a reasonable estimation. 48
The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (uM) 1.3 - Inhibitor + Inhibitor 2.50 0.62 2.6 4.00 1.42 6.5 6.30 2.65 13.0 7.60 3.12 26.0 9.00 3.58 What is the Km of the inhibited enzyme reaction?
How can you find Kcat if you are only given Vmax, Km and [E] ? I tried using Kcat=Vmax/[E] but that didn't work. How do you know if [E] is the same as [E]total, and if it isn't how do you find it from this information: The Vmax for a particular enzyme is 10 nmols/L/s. The Km for its substrate is 5 microM. If the enzyme concentration is 10 nM, what is the kcat? a.80 nmoles/L/s b.8000 nmoles/L/s c.2 nmoles/L/s d.50 nmoles/L/s
Question #2: Consider the following data for Cell G run in 1 L chemostat, with a feed containing 30 g/L substrate: Feedrate (ml/hr) (c) Calculate 27.5 24.2 22.1 18.8 Steady State Cell Mass Concentration (g/L) 12.0 14.7 15.8 16.7 Steady State Substrate Concentration (g/L) 10.0 5.56 (d) Using the calculated kinetic parameters, 3.70 (a) Determine the kinetic parameters for growth (b) At each dilution rate, calculate Yx/s. Is it constant with dilution rate and substrate concentration? 2.23 (i) the critical dilution rate, D. (ii) the dilution rate corresponding to maximum productivity, Dm (iii) the maximum cell productivity (cells/L/hr) (i) Generate a plot of X and S vs D for a feed concentration of 25 g/L. (ii) What is the critical dilution rate for this condition? (iii) What is the dilution rate corresponding to maximum productivity at this condition? (iv) What is the maximum cell productivity at this condition? (v) For a reactor operating at this maximum productivity, what would be the…
Question 6 Review Concept 8.4 Enzymes. Match the term and its description. Each term can only be used once. These are nonprotein enzyme helpers. [ Choose) These inhibitors bind to the active site of an enzyme, [ Choose ) competing with the substrate to decrease reaction > >
Which of the following statements are true about the relationships of [S], KM, and Vmax? (Choose all that are true) As the [S] is increased, vo approaches the limiting value, Vmax KM = Vmax/2 The rate of product formed, vo, is at Vmax when [S] <<< KM KM and Vmax assist in finding the rate of the enzyme catalyzed reaction only if the reaction is considered irreversible.
You obtain a calculated Vmax of 4.26uM/s and a Km of 122.5uM from a kinetics experiment performed using 0.5uM enzyme. What is the catalytic efficiency of this enzyme?

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