When 10 micrograms of an enzyme with a molecular mass of 80,000 Daultons (grams/mol) is added toa solution containing its substrate at a concentration 100 times the Km, it catalyzes the conversion of65 micromoles of substrate into product in 3 minutes. What is the enzyme's turnover number (in unitsof min-1)?

Before going into the calculations for finding the turnover number (kcat), we need to convert…

Answered: When 10 micrograms of an enzyme with a…

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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The turnover number for an enzyme is known to be 5000min-1. Given the following set of data, Substrate concentration(mM) 1, 2, 4, 6, 100, 1,000 Initial Rate(micromol/min) 167, 250, 334, 376, 498, 499 a) What is the Km of the enzyme for the substrate? (do this without using a calculator) b) What is the total amount of enzyme present in the assay?
The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1
The protein catalase is an enzyme that catalyzes the decomposition of hydrogen peroxide:2 H2O2 (aq) → 2 H2O (l) + O2 (g)and has a Michaelis-Menten constant of 25 × 10-3 mol·dm-3 and a turnover number of 4.0×107s-1.The total enzyme concentration is 0.016×10-6 mol·dm-3 and the initial substrate concentration is4.32×10-6 mol·dm-3 Calculate the maximum reaction rate (????) for this enzyme, and the initial rateof this reaction. Note that catalase has a single active site.
(a) At what substrate concentration would an enzyme with a kcat of 30.0s−1 and a Km of 0.0050 M operate at one-quarter of its maximum rate?(b) Determine the fraction of Vmaxthat would be obtained at the following substratecon centrations [S]: ½Km, 2Km, and 10Km.(c) An enzyme that catalyzes the reaction X ⇌ Y is isolated from two bacterial species.The enzymes have the same Vmax but different Km values for the substrate X. Enzyme A has a Km of 2.0 μM, and enzyme B has a Km of 0.5 μM. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 μM. Which curve corresponds to which enzyme?
What is the concentration of enzyme (in mM) needed to achieve a Vmax of 8.00 mM/s if the enzyme has a turnover number of 1.50s-1?
when saturated with substrate, an enzyme has a maximum initial rate of 110mumoles of substrate converted to product per second. At a substrate concentration of 100mu M, the same enzyme converts substrate to product at a rate of 0.010mmoles/ sec. Assuming that Michaelis - Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2x10^-3M.
Urease enzyme hydrolysed urea at [S]= 0.03 mmol/L with a Km value of around 0.06 mmol/L. The initial velocity observed was 1.5x103 mmol/L.min1. Calculate the initial velocity of the enzymatic reaction when using [S]= 0.12 mmol/L.
An enzyme-catalyzed reaction is allowed to proceed for 3.4 minutes under steady-state conditions (with large excess of substrate). If the enzyme concentration is 7.9 nM, and the concentration of product formed is 6.3 µM, what is the kcat of this enzyme? Please give your answer to two significant figures in s1.
At what substrate concentration would an enzyme with a kcat of 25.0 s-1 and a KM of 3.5 mM operate at 25% of its maximal rate? How many reactions would the enzyme catalyze in 45 seconds when it is fully saturated with substate, assuming the enzyme has one active site?
Assume that an enzyme-catalyzed reaction follows the scheme shown: E + S ES E + P k₁ = 1 x 10%/M-s k-1 = 2.5 x 10%/s k2= 3.4 x 107s What is the dissociation constant for the enzyme-substrate, Ks? What is the Michaelis constant, Km, for this enzyme? What is the turnover number, Kcat, for this enzyme? What is the catalytic efficiency for the enzyme? If the initial Et concentration is 0.25mM, what is Vmax?
You will perform the protocol below for the calf intestinal alkaline phosphatase (CIP) provided. For each reaction, your final enzyme concentration should be 10 nM CIP. Note: Enzymes purchased are typically labelled with their “units of activity” (U), as this relates to how much enzyme is needed to catalyze a reaction. The 100 nM CIP provided has approximately 3 U/mL and was diluted 1 in 1,000 from a 500 U/mL purchased enzyme. 1) Create a table (similar to the one below) to help you determine and keep track of what to add to each of the cuvettes in which your reactions will be measured. The five different concentrations of PNPP should be: 25, 50, 100, 200, 300 μM. Each reaction will be in a final volume of 1 mL and contain 10 nM alkaline phosphatase. Concentrations of stock solutions: 1.0 mM PNPP, 100 nM calf intestinal phosphatase
a) Determine kcat (in units of sec-1) for a particular enzyme, given the following information: Vo = 144 mmol/min; [S] = 2 mM; Km = 0.5 mM; Enzyme Molecular weight = 40,000 mg/mmole; 8 mg of enzyme used in assay generating this data. b) In general, explain how the total enzyme concentration affects turnover number and Vmax?

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