Microbiology: An Introduction
Microbiology: An Introduction
12th Edition
ISBN: 9780321929150
Author: Gerard J. Tortora, Berdell R. Funke, Christine L. Case
Publisher: PEARSON
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Chapter 5, Problem 2A

The following graph shows the normal rate of reaction of an enzyme and its substrate (blue) and the rate when an excess of competitive inhibitor is present (red). Explain why the graph appears as it does.

Chapter 5, Problem 2A, The following graph shows the normal rate of reaction of an enzyme and its substrate (blue) and the

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Please handraw this graph with all the necessary detailed information:   Imagine that I text enzyme rate for four different temperatures: 10 degrees celsius, 20 degrees celsisus, 30 degree celsius, and 40 degree celsius, in separate tubes. The enzyme appears to work faster as temperature increases, but completely ceases activity at 40 degrees celcius. Sketch a graph to show this outcome, but here you will graph product formation (nmoles/mL) vs. time (minutes). The graph should be 4 lines and HANDDRAWN. Include a legend if necessary. You do not need precise quantitivate values, but most show the correct trends on the graph.
In the scheme below which represents the mechanism of action for a large number of enzymes: A+B⟺AB⟶C The steady state approximation is reached when: d[AB]/dt≈0 k2≫k1 k−1≫k1 k−1=k1
The fractional saturation of an enzyme is 80% at a substrate concentration of 20 mM. In the presence of 5 mM of a competitive inhibitor the fractional saturation at the same substrate concentration is 40%. The Km for this reaction is mM, and the apparent Km is mM. The value of alpha is while the Ki of the inhibitor is mM.

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Microbiology: An Introduction

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