BIOCHEMISTRY (LOOSELEAF)-W/ACCESS
9th Edition
ISBN: 9781319425784
Author: BERG
Publisher: Macmillan Higher Education
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Chapter 27, Problem 51P
Interpretation Introduction
Interpretation:
The
Concept introduction:
The homeostasis is the capability of the body to cope up with the external environment and keep the internal part of the body stable. The most common example of homeostasis is the average temperature of the human body, which is maintained at 98.6 degrees.
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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?
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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 27 Solutions
BIOCHEMISTRY (LOOSELEAF)-W/ACCESS
Ch. 27 - Prob. 1PCh. 27 - Prob. 2PCh. 27 - Prob. 3PCh. 27 - Prob. 4PCh. 27 - Prob. 5PCh. 27 - Prob. 6PCh. 27 - Prob. 7PCh. 27 - Prob. 8PCh. 27 - Prob. 9PCh. 27 - Prob. 10P
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