V23. The graph below is a graph of Vmax(a) Label the graph clearly with both the Vmax and the Km. Estimate the Km from thisgraph giving the correct units.v/Vmax1.00.50.00.0vs [S] for an enzyme.Enzyme Activity vs [Substrate]:0.20.4Substrate (HM):0.6(b) If the Vmax = 25 mmoles per minute per µmole of enzyme calculate the Keat andthe specificity constant."

As given in the question, V/Vmax was plotted with substrate concentration. V= rate of a reaction…

Answered: 23. The graph below is a graph of Vmax…

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?
Use the Michaelis-Menten equation to complete the enzyme kinetic data set, whenKm is known to have a value of 1 mmol L−1 [S] (mmol L−1) Vo (μmol L−1 min−1) 0.5 50 1.0 2.0 3.0 10.0
An enzyme has a rate enhancement of 1.3x106. Calculate the value of ΔΔG‡ at 25.0 °C in kJ mol-1. (Hint: be sure to pay attention to units and signs.) (R = 8.3145 J mol-1 K-1)
The steady-state kinetics of an enzyme are studied in the absence and presence of an inhibitor (inhibitor A). The initial rate is given as a function of substrate concentration in the following table: v[(mmol/L)min- [S] (mmol/L) No inhibitor Inhibitor A 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38 (a) What kind of inhibition (competitive, uncompetitive, or mixed) is involved? (b) Determine Vmax and Ky in the absence and presence of inhibitor.
Use the Michaelis-Menten equation to complete the enzyme kinetic data set, when Km is known to have a value of 1 mmol L-1
L(24 points) Explain How is the Michaelis constant defined. and what does a low or high value for Km tell you? What is the difference between the velocity and initial velocity of an enzyme reaction? What determines the efficiency of an enzyme reaction, and what terms are used to describe it? 2. (50 points) About how to obtain kinetic data experimentally Lisa decides to obtain values for the Km and Vmax of an enzyme she has just isolated from liver cells (it is now pure), using a Michaelis Menten plot. Describe in detail what kinds of measurements she would have to make, and what she would need to plot on graphs in order to estimnte the values for Km and Vmax. (Show the kinds of graphs she would have to plot, and how these will allow her to estimate Km and Vmax.) Describe how she would be able to obtain Vmax experimentally and from the Michaelis Menten plot - what conditions are needed and what would be measured). Also, describe what she would have to do to obtain the turnover number of…
If 287.9 umol of enzyme X has a Vmax = 47.8 mmol/sec, what is the value of kcat %3D sec-1? Please report answer with 1 decimal place. Please do not report units. Your Answer: Answer units MacBook Air 888 F5 F4 F3 F2 %23 %24 %24
What is the Vmax of an enzyme at 1 µM with an activation free energy (AG#) of 55000 J/mol at 298K? -AG # RT k cat = 6.2× 10¹² (Ideal gas constant R = 8.314 J K¯1 mol-¹1) 0.14 μM s-1 -1 S-1 1.4 μM s 14 μM s-1 0.14 mM s-1 1.4 mM s-1
8) plot of enzyme activity with and without an inhibitor present gave the following plot. What type of inhibitor is present? How does this inhibitor function? What changes are seen in Vmax and KM? Draw a line that approximates the result from addition of twice as much inhibitor to the reaction. 1/v (v in mM/min) 0.8 0.6 0.4 0.2 0 0.2 0.4 06 1/[s] ([S] in mM) 0.8
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
Use the Michaelis-Menten equation to complete the enzyme kinetic data set, when Km is known to have a value of 1 mmol/L. Upload your tabulated answers and show your calculations.
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

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