BIOCHEMISTRY-ACHIEVE (1 TERM)
BIOCHEMISTRY-ACHIEVE (1 TERM)
9th Edition
ISBN: 9781319402853
Author: BERG
Publisher: MAC HIGHER
Question
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Chapter 8, Problem 46P
Interpretation Introduction

(a)

Interpretation:

V0 versus pH curve when the substrate concentration is much greater than the enzyme KM should be drawn.

Concept introduction:

KM is the Michaelis constant which is the substrate concentration when the reaction rate is half of the maximum rate achieved by a system. Michaelis constant is a measure of its binding affinity to a substrate. The higher the Michaelis constant, the lower affinity for its substrate. That means an enzyme with lower KM needs a higher concentration of substrate to achieve Vmax.

Interpretation Introduction

(b)

Interpretation:

V0 versus pH curve when the substrate concentration is much less than the enzyme KM should be drawn.

Concept introduction:

KM is the Michaelis constant which is the substrate concentration when the reaction rate is half of the maximum rate achieved by a system. Michaelis constant is a measure of its binding affinity to a substrate. The higher the Michaelis constant, lower affinity for its substrate. That means an enzyme with lower KM needs a higher concentration of substrate to achieve Vmax.

Interpretation Introduction

(c)

Interpretation:

The pH at which the velocity will equal half of the maximal velocity attainable under these conditions should be determined.

Concept introduction:

KM is the Michaelis constant which is the substrate concentration when the reaction rate is half of the maximum rate achieved by a system. Michaelis constant is a measure of its binding affinity to a substrate. The higher the Michaelis constant, lower affinity for its substrate. That means for an enzyme with lower KM needs a higher concentration of substrate to achieve Vmax.

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An enzyme-catalyzes the isomerization of substrate S to product P. The enzyme has a molecular weight of 120,000 g/mol. In assays using 1 μg of enzyme per assay the Km was 3 x 10^-3M and the Vmax was 2.75 μmole per minute. What would be the Kcat (turnover number or molecular activity) of the enzyme under these conditions? 2.75 min^-1? 3,300,000 min^-1? 330,000 s^-1? 19,800,000 min^-1? 5,500 s^-1?
molecule A Plot of velocity versus substrate B Lineweaver-Burk plot 1/v Km 1 Vmax (S) Vmx 1 V max 1/2Vmax 1/Vmax -1/Km Km [S] 1/[S] fppt.com molecule Exercise The following data describe an enzyme-catalyzed reaction. Plot these results using the Lineweaver-Burk method, and determine values for KM and Vinax- The symbol mM represents millimoles per liter; 1 mM = 1 × 10 3 mol L. (The concentration of the enzyme is the same in all experiments.) Velocity (mM sec-) Substrate Concentration (тм) 2.5 0.024 5.0 0.036 10.0 0.053 15.0 0.060 20.0 0.061 fppt.com
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
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