Shown below are Km, and Vmax values obtained for an enzyme A which catalyze the transformation ofthe following substrates. Enzyme concentration used was 0.01 M.Km, mM0.02Vmax, mM/min5.3Substrate121.513.732.610040.1250.05621. Which substrate have the highest affinity for the enzyme? Explain.2. Which will show higher efficiency of converting the substrate to the product? Show solutions andеxplain.

Km or the Michaelis-Menten constant is the substrate concentration required to produce half-maximum…

Answered: 1. Which substrate have the highest…

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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An enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.
Consider an enzyme that follows standard Michaelis-Menten kinetics and has the following kinetic constants: %3| k2 = 1.5 x 10? s1 Еo 3D 1 х 104 М = 1 x 104 M a. What is the value of the maximum rate VM? b. Prepare a hand-drawn quantitative plot on graph paper (not a simple sketch nor an EXCEL-generated graph) of the enzymatic reaction rate versus substrate concentration (like Fig 3-3) using the kinetic parameters given above. Be sure to label the axes and include numeric values on the axes. C. Based upon your hand drawn saturation plot, at what substrate concentration is the enzymatic reaction rate 75% of Vm?
In an enzyme kinetics study, three inhibitors resulted to the following results: Inhibitor ABC Inhibitor XYZ Inhibitor PQR Without Inhibitor Vmax 40.2 mM/ sec 40.3 mM/ sec 12.32 mM/ sec 65.43 mM/ sec Km 24.3 mM 28.5 mM 24.3 mM 15.7 mM b. What type of inhibitor is Inhibitor PQR? Why do you say so?
Please note the following Lineweaver-Burk plot for the enzyme Virbraniumase reacting with a substrate: -0.4 . 1/V 5 4 3 2 -0.2 0 0.2 y = 5.2781x + 1.3338 R² = 0.9967 0.4 0.6 0.8 1/[S] 1 Based on the information provided, what is the Vmax for this reaction?
A particular enzyme-catalyzed reaction has an apparent Vmax = 9.00 nmol s-1 and α' = 3.00 when 2.00 µmol L-1 inhibitor X is present and uncompetitively inhibiting the reaction. Calculate Vmax for the uninhibited reaction in nmol s-1.
The Lineweaver - Burk plot (Figure 1) shows an enzyme-catalyzed reaction in the absence and presence of 0.1µM inhibitor (ketoconazole). O Estimate Vmax and Km in the absence and presence of the inhibito.. (ii) Determine the type of inhibition shown by the inhibitor. Explain. 0.1- 0.08 0.06 - With inhibitor 0.04 0.02 Without inhibitor 0.4 -0.2 -0.02 0.2 0.4 0.6 0.8 1.0 1/{S\(&M-1) 1/vo (pmol-11 min)
Staphylococcal nuclease has a ΔΔG‡ of -84.1 kJ mol-1 at 25.0 °C. If the uncatalyzed rate is 0.630x10-13 µmol s-1, calculate the enzyme-catalyzed rate in µmol s-1. (Use R = 8.3145 J mol-1 K-1)
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
In an enzyme kinetics study, three inhibitors resulted to the following results: Inhibitor ABC Inhibitor XYZ Inhibitor PQR Without Inhibitor Vmax 40.2 mM/ sec 40.3 mM/ sec 12.32 mM/ sec 65.43 mM/ sec Km 24.3 mM 28.5 mM 24.3 mM 15.7 mM d. Draw the estimated Michaelis Menten Curve of Inhibitor ABC and the curve without the inhibitor.
The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.
The accompanying figure shows three Lineweaver–Burk plots for enzymereactions that have been carried out in the presence, or absence, of aninhibitor. Indicate what type of inhibition is predicted based on eachLineweaver–Burk plot. For each plot indicate which line corresponds to thereaction without inhibitor and which line corresponds to the reaction withinhibitor present.
you are trying to come up with a drug to inhibit the activity of an enzyme thuogth to have a role in a liver disease. in the lab the enzyme was shown to have a Km of 1x10-6 M and Vmax of 0.1 micromoles/min.mg measruing at room temperature. you developed a mixed non-competitve inhibitor with a ki=0.4x10-6M and a Ki` pf 0/2 x 10-5 What will be the apparent Km in the presence of 1.0x10-6 M?

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1. Which substrate have the highest affinity for the enzyme? Explain. 2. Which will show higher efficiency of converting the substrate to the product? Show solutions and explain.