Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Chapter 1, Problem 15P
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.
Intracellular Transport of Proteins The endoplasmic reticulum (HR) is a site of protein synthesis. Proteins made by ribosomes associated with the ER may pass into the EH membrane or enter the lumen of the HR. Devise a pathway by which:
- a plasma membrane protein may reach the plasma membrane.
- a secreted protein may be deposited outside the cell. (Section 1.5)
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Chapter 1 Solutions
Biochemistry
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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. The Dimensions of Eukaryotic Cells and Their ConstituentsAssume that liver cells are cuboidal in shape, 20 m oil a side. (Section 1.5) How many liver cells laid end to end would fit across the diameter of a pinhead? (Assume a pinhead diameter of 0.5 mm) What is the volume of a liver cell? (Assume it is a cube.) What is the surface area of a liver cell? What is the surface-to- volume ratio of a liver cell? How does this compare to the surface-to-volume ratio of an E. coli cell (compare this answer with that of problem 3c)? What problems must cells with low surface-to-volume ratios confront that do not occur in cells with high surface-to-volume ratios? A human liver cell contains two sets of 23 chromosomes, each set being roughly equivalent in information content. The total mass of DNA contained in these 46 enormous DNA molecules is 4 1012 Because each nucleotide pair contributes 660 daltons to the mass of DNA and 0.34 run to the length of DNA, what is the total number of nucleotide pairs and the complete length of the DNA in a liver cell? How does this length compare with the overall dimensions of a liver cell? The maximal information in each set of liver cell chromosomes should be related to the number of nucleotide pairs in the chromosome set's DNA. This number can be obtained by dividing the total number of nucleotide pairs just calculated by 2. What is this value? If this information is expressed in proteins that average 400 amino acids in length and three nucleotide pairs encode one amino acid in a protein, how many different kinds of proteins might a liver cell be able to produce? (In reality, liver cell DNA encodes approximately 20,000 different proteins. Thus a large discrepancy exists between the theoretical information content of DNA in liver cells and the amount of information actually expressed.)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. Understanding the Densities of Membrane Components Sucrose gradients for separation of membrane proteins must be able to separate proteins and protein—lipid complexes with a wide range of densities, typically 1.00 to 1.35 g/mL. a. Consult reference books (such as the CRC Handbook of Biochemistry ) and plot the density of sucrose solutions versus percent sucrose by weight (g sucrose per 100 g solution), and versus percent by volume (g sucrose per 100 mL solution). Why is one pint linear and the other plot curved? b. What would be a suitable range of sucrose concentrations for Separation of three membrane-derived protein—lipid examples with densities of 1.03, 1.07, and 1.08 g/mL?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. Comparing Membrane Barrel Structures Compare the porin proteins, which have transmembrane pores constructed from -barrels, with the Wza protein, which has a transmembrane pore constructed from a ring of -helices. How many amino acids are required to form the -barrel of a porin? How many would be required to form the same-sized pore from -helices?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. Understanding Diffusion of Proteins in Membranes Protein lateral motion is much slower than that of lipids because proteins are larger than lipids. Also, some membrane proteins can diffuse freely through the membrane, whereas others art bound or anchored to other protein structures, in the membrane. The diffusion constant for the membrane protein fibronectin is approximately 0.7 X 10-12 cm/sec, whereas that for rhodopsin is about 3 X10-9 cm/sec. Calculate the distance traversed by each of these proteins in 10 msec. What could you surmise about the interactions of these proteins with other membrane components?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. Cells as Steady-State Systems Describe what is meant by the phrase "cells tire steady-state systems." (Section 1.4)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. 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. 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. 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. Determining the Length of a Keratin Molecule The central rod domain of a keratin protein is approximately 312 residues in length. What is the length (in A) of the keratin rod domain? If this same peptide segment were a true -helix how long would it be? If the same segment were a -sheet, what would its length be?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. Cell Structure Without consulting the figures in this chapter, sketch the characteristic prokaryotic and eukaryotic cell types and label their pertinent organelle and membrane systems. (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. Solving the Sequence of an Oligopeptide From Sequence Analysis Data Amino acid analysis of ail oligopeptide seven residues long gave The following fads were observed: a. Trypsin treatment had no apparent effect. b. The phenylthiohydantoin released by Lid mini degradation was c. Brief chymotrypsin treatment yielded several products, including a dipeptide and a tetrapeptide. The amino acid composition of the tetrapeptide was Leu, Lyi. and Met. d. Cyanogen bromide treatment yielded a dipeptide, a tetrapeptide, and free Lys. What is the amino acid sequence of this heptapeptide?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. Phosphorylation of Proteins Introduces New Properties Phosphoproteins are formed when a phosphate group is esterified to an —OH group df a Ser, Thr, or Tyr side chain. At typical cellular pH values, this phosphate group bears two negative charges (— OPO32:_). Compare this side-chain modification to the 2U side chains of the Common amino acids found in proteins and comment on the novel properties that it introduces into side-chain possibilities.arrow_forward
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