Biochemistry: The Molecular Basis of Life
6th Edition
ISBN: 9780190209896
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
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Chapter 15, Problem 71TQ
Summary Introduction
To review:
The prevalence of gout only in mammals.
Introduction:
Gout is caused by anabnormality of purine
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The preferred substrate is T because the enzyme half-saturates at 7.00 mM for T, but requires 28.0 mM for U, and 112 mM for S.
b
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The rate constant k 2 with substrate S is 9.60×107 sec-1, with substrate T, k 2 = 6.00×104 sec-1, and with substrate U, k 2 = 2.40×106 sec-1.
Calculate the catalytic efficiency with S, T, and U.
Catalytic efficiency with S =
Catalytic efficiency with T =
Catalytic efficiency with U =
Does enzyme A use substrate S, substrate T, or substrate U with greater catalytic efficiency?
Fumerase catalyzes the conversion of fumerate to malate.
fumerate + H2O ⇋ malate
The turnover number, kcat, for fumerase is 8.00×102 sec-1. The Km of this enzyme for fumerate is 5.00×10-3 μmol mL-1.
a
In an experiment using 2.00×10-3 μmol·mL-1, what is Vmax?
Suppose you wanted to make a buffer of exactly pH 7.00 using KH2PO4 and Na2HPO4. If the final solution was 0.18 M in KH2PO4, you would need 0.25 M Na2HPO4. Now assume you wish to make a buffer at the same pH, using the same substances, but want the total phosphate molarity ([HPO42−]+[H2PO−4]) to equal 0.20 M. What concentration of the Na2HPO4 would be required?
Chapter 15 Solutions
Biochemistry: The Molecular Basis of Life
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