D1.0.25Short answer questions, write answers on blank paper.A plot of enzyme activity with and without an inhibitor present gave the following plot.0.20.151/Vv in mM/s0.10.05000.050.1y=0.67x+0.020.15y=0.2x+0.020.20.250.31/[S]. [S] in mMWithout 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 asnecessary--answers that do not contain required units will be marked incorrect

LB plot is a double reciprocal plot which is constructed by taking inverse values of substrate…

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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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Vo 60 50 40 30 20 mm/sec 0100 0 O 0 0 50 100 150 200 250 300 Substrate (mM) Use the Michaelis-Menten plot to answer this question. What is the estimated value of Vmax of the enzyme catalyzed reaction (square data points)? (choose the one best answer) 10
Consider the given data for an enzyme-catalyzed reaction. Determine the Vm, Km and the type of inhibition based on the given data below Substrate concentration, uM 30 50 100 300 900 slope y-intercept Complete the table below (include correct units). Experiment A Vm Km Experiment A (Initial velocity without inhibitor, uM-min) Type of Inhibition: 10.4 14.5 22.5 33.8 40.8 Experiment B (Initial velocity with inhibitor, uM-min) 5.1 7.3 13.3 25.7 37.2 Experiment B
The following data were collected in the study of a new enzyme and an inhibitor of the new enzyme: Vo (nmol/sec) [S] (µM). 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 Vmax of the inhibited enzyme reaction?
An enzyme catalyzes a reaction with a Km of 9.50 mM and a Vmax of 2.10 mM · s-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 3.25 mM vo : mM · s-1 [S] = 9.50 mM mM · s-1 Vo :
Refer to the following graph when answering questions. These words may be used to fill: competitive, uncompetitive, noncompetitive, irreversible. -1 mg , μmol min-1 v-1 0.2 0.1 THA -4-2 CBA + 2 4 [S]-1, (mM-1) Word 1, The value of Km for the enzyme depicted by curve A is 2, The value of Vmax for the enzyme depicted by curve A is 3, Curve B depicts the effect of an inhibitor on the system described by curve A. This inhibitor is a inhibitor. 4, Curve C depicts the effect of a different inhibitor of the system described by curve A. This second inhibitor is a inhibitor. D
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
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
Given the line-weaver Burke plot below for Enzyme Y, identify the Vmax for Enzyme Y. -10 30 1/v 1/[S] 5 10 Enter the numerical value for Vmax only, no letters or symbols. Round answer to 2 significant figures.
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)
From the following data of an enzymatic reaction, determine the type of inhibition, the KM for the substrate and Ki for the enzyme-inhibitor complex. [S] mM 2.0 3.0 4.0 10.0 15.0 [P] mM/h (without inhibitor) 139 179 213 313 370 [P] mM/h ([I]=6mM) 88 121 149 257 313
Given the Km and Vmax values of three enzymes, A, B and C: Enzyme Km Vmax A 2.0 x 10-3 M 4.0 B 4.2 x 104M 1.8 C 1.0 x 10-3 M 2.5 Which enzyme would have the highest activity? Why?
The table below contains experimental data for the enzyme A; substrate concentration (mM) and initial velocity (mM/min). Plot a suitable graph and calculate the Km and Vmax for this enzyme. In 2. your answer, insert an image of the graph and show your calculations. Substrate concentration; [S] (mM) Initial Velocity; Vo (mM/min) 10.00 33.33 5.00 20.00 3.33 14.29 2.50 11.11

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