Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 1, Problem 13P
Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.
An Estimation of Minimal Genome Size for a Living Cell Studies of existing cells to determine the minimum number of genes for a living cell have suggested that 206 genes are sufficient. If the ratio of protein-coding genes to non-protein-coding genes is the same in this minimal organism as the genes of Mycoplasma genitulium, how many proteins are represented in these 206 genes. How many base pairs would be required to form the genome of this minimal organism if the genes are the same size as M genilalium genes? (Section 1.5)
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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 1 Solutions
Biochemistry
Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...
Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...Ch. 1 - Answers to all problems are at the end of this...
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