Biochemistry (Looseleaf)
Biochemistry (Looseleaf)
9th Edition
ISBN: 9781319114800
Author: BERG
Publisher: MAC HIGHER
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Chapter 9, Problem 11P
Interpretation Introduction

Interpretation:

The expected value for kcat at pH 6.0 is to be calculated.

Concept introduction:

The biological catalysts which lower the minimum amount of energy required by the reactants to convert into product are known as enzymes. An enzyme does not alter the potential energy of reactant and product. Carbonic anhydrase is an enzyme used in catalytic interconversion.

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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 Question Content Area 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?
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