BIOCHEM-ACHIEVE(FIRST DAY DISCOUNTED)
9th Edition
ISBN: 2818000069358
Author: BERG
Publisher: MAC HIGHER
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Chapter 36, Problem 9P
Interpretation Introduction
Interpretation:
The reason behind the bacteria placed in acidic solution in the state of starvation continues moving should be determined.
Concept introduction:
The energy in the biological system is present in the form of chemicals. Different types of cells use different energy sources. The bacterial flagellum uses the proton motive force to move.
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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?
Chapter 36 Solutions
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