Study Guide with Student Solutions Manual and Problems Book for Garrett/Grisham's Biochemistry, 6th
6th Edition
ISBN: 9798214167343
Author: GARRETT
Publisher: Cengage Learning US
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Chapter 14, Problem 21P
Answers to all problems are at the end οΓthis book. Detailed solutions are available in the Student Solutions Manual. Study Guide, and Problems Book.
Superbug infections are becoming more common around the world. Many of these infections arise from the action of
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Study Guide with Student Solutions Manual and Problems Book for Garrett/Grisham's Biochemistry, 6th
Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Prob. 3PCh. 14 - Prob. 4PCh. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Prob. 6PCh. 14 - Prob. 7PCh. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Answers to all problems are at the end οf this...Ch. 14 - Answers to all problems are at the end οf this...
Ch. 14 - Answers to all problems are at the end of this...Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Prob. 14PCh. 14 - Prob. 15PCh. 14 - Prob. 16PCh. 14 - Prob. 17PCh. 14 - Prob. 18PCh. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Answers to all problems are at the end οΓthis...Ch. 14 - Prob. 22PCh. 14 - Prob. 23PCh. 14 - Prob. 24PCh. 14 - Prob. 25PCh. 14 - Prob. 26P
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- 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)arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Assessing the Cross-Overs of Flavodoxin For flavodoxin (pdb id = 5NLL), identify (he right-handed cross-overs and the left-handed cross-overs in the parallel -sheet.arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. A Simple Genome and Its Protein-Encoding Capacity The genome of the Mycoplasma genitalium consists of 523 genes, encoding 484 proteins, in just 5S0.U74 base pairs (see Table 1.6). What fraction of the M. genitalium genes encode proteins? What do you think the other genes encode? If the fraction of base pairs devoted to protein-coding genes is the same as the fraction of the total genes that they represent, what is tlie average number of base pairs per protein-coding gene? If it takes three base pairs to specify an amino acid in a protein, how many amino acids are found in the average M. genitalium protein? If each amino acid contributes, on average, 12U daltons to the mass of a protein, what is the mass of an average M. genitalium protein? (Section 1.5)arrow_forward
- 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. The Biosynthetic Capacity of Cells The nutritional requirements of Escherichia coli cells are far simpler than those of humans, yet the macromolecules found in bacteria are about as complex as those of animals. Because bacteria can make all their essential biomolecules while subsisting on a simpler diet, do you think bacteria may have more biosynthetic capacity and hence more metabolic complexity than animals? Organize your thoughts on this question, pro and con, into a rational argument. (Section 1.5)arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Consider the following peptide sequences: EANQIDEMLYNVQCS LTTLE DTVPW LG VHLDITVPL SWTWTLYVKL QQNWGGLWILTLVWFLM CNMKHGDSQCDERTYP YTREQSDGHIPKMNCDS AGPFGPDGPTIGPK Which of the preceding sequences would be likely to be found in each of the following: A parallel -sheet An antiparallel -sheet A tropocollagen molecule The helical portions of a protein found in your hairarrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. The Dimensions of Prokaryotic Cells and Their Constituents Escherichia coli cells are about 2 μm (microns) long and 0.8 μm in diameter. (Section 1.5) a. How many E. coli cells laid end to end would fit across the diameter of a pinhead? (Assume a pinhead diameter of 0.5 mm.) b. What is the volume of an E. coli cell? (Assume it is a cylinder, with the volume of a cylinder given by V=r2h , where = 3.14.) c. What is the surface area of ail E coli cell? What is the surface-to- volume ratio of an E coli cell? d. Glucose, a major energy-yielding nutrient, is present in bacterial cells at a concentration of about 1 mM. What is the concentra¬tion of glucose, expressed as mg/mL? How many glucose molecules are contained in a typical E. coli cell? (Recall that Avogadro’s number =6.0231023 .} e. A number of regulatory proteins are present in E. coli at only one or two molecules per cell. If we assume that an E. coli cell contains just one molecule of a particular protein, what is the molar concentration of this protein in the cell? If the molecular weight of this protein is 4023 , what is its concentration, expressed as mg/mL? f. An E coli cell contains about 15,000 ribosomes, which carry out protein synthesis. Assuming ribosomes are spherical and have a diameter of 20 nm (nanometers), what fraction of the E. coli cell volume is occupied by ribosomes? g. The E coli chromosome is a single DNA molecule whose mass is about 3109 daltons. This macromolecule is actually a linear array of nucleotide pairs. The average molecular weight of a nucleotide pair is 660, and each pair imparts 0.34 nm to the length of the DNA molecule. What is the total length of the E. coli chromosome? How does this length compare with the overall dimensions of an E. coli cell? How many nucleotide pairs does this DNA contain? The average E. coli protein is a linear chain of 360 amino acids. If three nucleo¬tide pairs in a gene encode one amino acid in a protein, how many different proteins can the E. coli chromosome encode? (The answer to this question is a reasonable approximation of the maximum number of different kinds of proteins that can be expected in bacteria.)arrow_forward
- 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. Evaluation of -Helices in Proteins The hem agglutinin protein in influenza virus contains a remarkably long -helix, with 53 residues. How long is this -helix (in nm)? How many turns does this helix have? The typical residue in an -helix is involved in two H bonds. How many H bonds are present in this helix?arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Preparing cDNA Libraries from Different Cells Describe an experimental protocol for the preparation of to cDNA libraries, one from anaerobically grown yeast cells and the second from aerobically grown yeast cell.arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Solving the Sequence of an Oligopeptide From Sequence Analysis Data Amino acid analysis of a decapeptide revealed the presence of the following products: The following facts were observed: Neither car boxy peptidase A nor B treatment of the- decapeptide had any effect. Trypsin treatment yielded two tetrapcptides and free Lys. Clostripain treatment yielded a tetrapcptide and a hexapeptidc. Cyanogen bromide treatment yielded an octapeptide and a dipeptide of sequence NP (using the one-letter codes). Chymotrypsin treatment yielded two tripeptides and a telrapeptide. The N-terminal chymotryptic peptide had a net charge of — 1 at neutral pi I and a net charge of —3 al pH 12. One cycle of Ed man degradation gave the PTH derivative What is the ammo acid sequence of this decapeptide?arrow_forward
- 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. (Research Problem) The Nature and Roles of Linear Motifs in Proteins In addition to domains and modules, there are other significant sequence patterns in proteins—known as linear motifs—that are associated with a particular function. Consult the biochemical literature to answer the following questions: 1. What are linear motifs? 2. How are they different from domains?. 3. What are their functions? 4. How can they be characterized? 5. There are several papers that are good starting points for this problem. Neduva, V., and Russell, R., 2005. Linear motifs: evolutionary interaction switches. FEBS Letters 579:3342-3345. Gibson, T., 2009. Cell regulation: determined to signal discrete cooperation. Trends in Biochemical Sciences 34:471-482. Diella, K. Haslam, N., Chica., C. et aL, 2009. Understanding eukaryotic linear motifs and their role in cell signaling and regulation. Frontiers of Bioscience 13:6580-6603.arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. CRISPR/Cas9: Design of a gRNA to Target the Human PVALB Gene The human PVALB gene, which encodes the Ca2+-binding protein parvalbumin, can be Targeted by CRISPR/Cas9, at the protospacer sequence - ATGCAGGAGGGTGGCGAGAGGGGCCGAGAT- followed by a -TGG-PAM trinucleotide. Give the sequence of the spacer region of a gRNA that will target the complementary DNA strand at this site. Include at the 3'-end of your gRNA sequence a region that will form a stem-loop structure with a 5'-AGCAUAGCUGUAAAAC- sequence downstream in the gRNA to create the dsRNA-binding site for Cas9.arrow_forwardAnswers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book. Separating Glycated Hb From Normal Hb (Integrates with Chapters 5 and 6.) Human hemoglobin can react with sugars in the blood (usually glucose) to form covalent adducts. The α-amino groups of N-terminal valine in the Hb -subunits react with the C-1 (aldehyde) carbons of monosaceharides to form aldimine add wets, which rearrange to form very stable ketoamine products. Quantitation of this “glycated hemoglobin is important clinically, especially for diabetic individuals. Suggest at least three methods by which glycated lib (also referred to as HbA1c) could be separated from normal Hb and quantitated.arrow_forward
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