Biochemistry, The Molecular Basis of Life, 6th Edition
6th Edition
ISBN: 9780190259204
Author: Trudy McKee, James R. McKee
Publisher: Oxford University Press
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 6, Problem 27RQ
Summary Introduction
To review:
The phenomenon of compartmentation within eukaryotic cells and the problems solved by this phenomenon.
Introduction:
An important reasonto regulate biochemical reactions is due to the infrastructure of cells. Physically separating the cells into compartments makes various cellular processes easy.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Given the values for the changes in enthalpy (AH) and entropy (AS), which processes can take place at 298 K without
violating the Second Law of Thermodynamics?
(a) AH =
(b) AH = -84 kJ mol¹ (−20 kcal mol¹), AS = -125 J mol-¹ K-¹ (-30 cal mol-¹ K-¹)
(c) AH = +84 kJ mol¹ (+20 kcal mol¯¹), AS = +125 J mol¯¹ K−¹ (+30 cal mol-¹ K-¹)
(d) AH = +84 kJ mol-1 (+20 kcal mol-¹), AS = -125 J mol¯¹ K-¹ (-30 cal mol-¹ K-¹)
-84 kJ mol¹ (-20 kcal mol¹), AS = +125 J mol¯¹ K-¹ (+30 cal mol-¹ K-¹)
What chemical interactions are primarily responsible for the fact that oil and water do not mix easily?
ionic interactions
covalent bonds
hydrogen bonding
Ohydrophobic effect
The carbon-nitrogen peptide bond is rigid, but rotation can take place
about the bond between the nitrogen and α-carbon atoms and the bond
between the α-carbon and the carbonyl carbon atom. These bond angles
define the conformation of the peptide chain.
Consider the Ramachandran plot shown to answer both questions.
(degrees)
180
120
60
60
0
Which observation most likely describes part of an observed a helix?
p=-59 degrees; = +150 degrees; many Ser residues
cp-140 degrees; w = +130 degrees; many Val residues
op-57 degrees; w = -47 degrees; many Pro residues
op-60 degrees; y = -47 degrees; many Ala residues
p = +60 degrees; y = +40 degrees; many Gly residues
-60
-120
-180
-180
0
➜ (degrees)
180
Chapter 6 Solutions
Biochemistry, The Molecular Basis of Life, 6th Edition
Ch. 6 - Prob. 1QCh. 6 - Prob. 2QCh. 6 - Prob. 3QCh. 6 - Prob. 4QCh. 6 - Prob. 5QCh. 6 - Prob. 6QCh. 6 - Prob. 7QCh. 6 - Prob. 8QCh. 6 - Prob. 9QCh. 6 - Prob. 1RQ
Ch. 6 - Prob. 2RQCh. 6 - Prob. 3RQCh. 6 - Prob. 4RQCh. 6 - Prob. 5RQCh. 6 - Prob. 6RQCh. 6 - Prob. 7RQCh. 6 - Prob. 8RQCh. 6 - Prob. 9RQCh. 6 - Prob. 10RQCh. 6 - Prob. 11RQCh. 6 - Prob. 12RQCh. 6 - Prob. 13RQCh. 6 - Prob. 14RQCh. 6 - Prob. 15RQCh. 6 - Prob. 16RQCh. 6 - Prob. 17RQCh. 6 - Prob. 18RQCh. 6 - Prob. 19RQCh. 6 - Prob. 20RQCh. 6 - Prob. 21RQCh. 6 - Prob. 22RQCh. 6 - Prob. 23RQCh. 6 - Prob. 24RQCh. 6 - Prob. 25RQCh. 6 - Prob. 26RQCh. 6 - Prob. 27RQCh. 6 - Prob. 28RQCh. 6 - Prob. 29RQCh. 6 - Prob. 30RQCh. 6 - Prob. 31RQCh. 6 - Prob. 32RQCh. 6 - Prob. 33RQCh. 6 - Prob. 34RQCh. 6 - Prob. 35RQCh. 6 - Prob. 36RQCh. 6 - Prob. 37RQCh. 6 - Prob. 38RQCh. 6 - Prob. 39RQCh. 6 - Prob. 40RQCh. 6 - Prob. 41RQCh. 6 - Prob. 42RQCh. 6 - Prob. 43FBCh. 6 - Prob. 44FBCh. 6 - Prob. 45FBCh. 6 - Prob. 46FBCh. 6 - Prob. 47FBCh. 6 - Prob. 48FBCh. 6 - Prob. 49FBCh. 6 - Prob. 50FBCh. 6 - Prob. 51FBCh. 6 - Prob. 52FBCh. 6 - Prob. 53SACh. 6 - Prob. 54SACh. 6 - Prob. 55SACh. 6 - Prob. 56SACh. 6 - Prob. 57SACh. 6 - Prob. 58TQCh. 6 - Prob. 59TQCh. 6 - Prob. 60TQCh. 6 - Prob. 61TQCh. 6 - Prob. 62TQCh. 6 - Prob. 63TQCh. 6 - Prob. 64TQCh. 6 - Prob. 65TQCh. 6 - Prob. 66TQCh. 6 - Prob. 67TQCh. 6 - Prob. 68TQCh. 6 - Prob. 69TQCh. 6 - Prob. 70TQCh. 6 - Prob. 71TQCh. 6 - Prob. 72TQCh. 6 - Prob. 73TQCh. 6 - Prob. 74TQCh. 6 - Prob. 75TQCh. 6 - Prob. 76TQCh. 6 - Prob. 77TQCh. 6 - Prob. 78TQCh. 6 - Prob. 79TQCh. 6 - Prob. 80TQCh. 6 - Prob. 81TQ
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biochemistry and related others by exploring similar questions and additional content below.Similar questions
- The α and ẞ subunits of hemoglobin bear a remarkable structural similarity to myoglobin. However, certain residues that are hydrophilic in myoglobin are hydrophobic in the subunits of hemoglobin. Why might this be the case? Hemoglobin forms long, extended structures that feature repeated sequences, whereas myoglobin forms globular structures. Hydrophobic residues on the surface of hemoglobin subunits interact with similar regions on the other subunits through van der Waals interactions. Hydrophilic residues on the surface of myoglobin form ionic interactions with similar regions on other myoglobin molecules. Myoglobin is a water-soluble protein, whereas hemoglobin is found in the hydrophobic environment of membranes.arrow_forwardProtein disulfide isomerase (PDI) is a protein folding enzyme that catalyzes disulfide-sulfhydryl exchange reactions. Ribonuclease is an enzyme with two ẞ-sheets, three α-helices, and four disulfide bridges critical to the stability of its structure. PDI rapidly converts inactive, scrambled ribonuclease to enzymatically active ribonuclease. This process is driven by the decrease in free energy as the scrambled conformations are converted into the stable, active conformation of the enzyme. In contrast, PDI rapidly inactivates insulin. What does PDI's effect on insulin imply about the structure of insulin? The three-dimensional shape of insulin does not contain any disulfide bonds. The active conformation of insulin has more ẞ-sheets than α-helices. The active conformation of insulin is not the most thermodynamically preferred form. The amino acid sequence of insulin is similar to the amino acid sequence of ribonuclease.arrow_forwardIn the 1950s, Christian Anfinsen demonstrated the renaturation of the protein ribonuclease (RNase) in vitro. After reduction and the addition of urea, the protein was in an unfolded state. After removing the urea and then the reducing agent, the protein oxidized and refolded, with greater than 90% activity. If reducing agent removal occurs before removing the urea, the protein showed less than 5% activity. Why does RNase refold incorrectly if the reducing agent is removed before urea removal? Contaminants in the RNase preparation would form covalent bonds with the protein, preventing reactivation. Urea would participate in weak bonding interactions with RNase, preventing oxidation of cysteine. Disulfide bonds are not positioned correctly unless weak bonding interactions are present. The protein would not fully denature.arrow_forward
- A hospital blood processing unit uses apheresis to concentrate red blood cells (RBCs) fromwhole blood donations. Whole blood is drawn from a donor at a rate of 100 g/min and passedthrough a centrifuge-based apheresis system. The whole blood contains 45% RBC and 1%WBCs and platelets by mass, with the remainder consisting of plasma. The separated plasma andnon-RBC components are returned to the donor at a rate of 90 g/min.The goal of the system is to collect a concentrated RBC fraction containing 80% red blood cells.Assume that red blood cells do not break down during processing and that no fluid loss occurs inthe system.(a) How long will it take to collect 100g of the concentrated RBC fraction?arrow_forwardPlease give an intuitive explanation of the blood-brain barrier. Specifically, what is it, what is its structure and function, what types of cells compose it, etc.,..?arrow_forwardObtain the sequence for the 5-HT receptor HTR1A and generate a hydropathy plot usingthe ExPASY tool ProtScale, the appropriate window, and the Kyte-Doolittle weightingalgorithm. How many transmembrane domains are present in this receptor?arrow_forward
- From the reaction data below, determine whether the reaction is first order or second order and calculate the rate constant. Time (s) 0 Reactant (mM) 5.4 1 4.6 2 3.9 3 3.2 4 2.7 5 2.3 Only a plot of In[reactant] versus t gives a straight line, so the reaction is first order . The negative of the slope, k, is 0.171arrow_forwardHair grows at a rate of about 20 cm/yr. All this growth is concentrated at the base of the hair fiber, where a-keratin filaments are synthesized inside living epidermal cells and assembled into ropelike structures. Two-chan 14 Protofilament 20-30 A Two-chain Intermediate flament -Protob Protofilament Cross section of a hair The fundamental structural element of a keratin is the a helix, which has 3.6 amino acid residues per turn and a rise of 5.4 A perlum. 54A (36) Amino terminus Carbon Hydrogen Oxygen Nitrogen group Carboxyl terminus Assuming that the biosynthesis of a helical keratin chains is the rate-limiting factor in the growth of hair, calculate the rate at which peptide bonds of a-keratin chains must be synthesized (peptide bonds per second) to account for the observed yearly growth of hair. 0422 rate of peptide bond formation: Income bonds/sarrow_forwardSpecific rotation is a measure of a solution's capacity to rotate circularly polarized light. The unfolding of the a helix of a polypeptide to a random conformation is accompanied by a large decrease in specific rotation. Polyglutamate, a polypeptide made up of only 1-Glu residues, has the a helix conformation at pH 3. When researchers raise the pH to 7, there is a large decrease in the specific rotation of the solution. Similarly, polylysine (1.-Lys residues) is an a helix at pH 10, but when researchers lower the pH to 7 the specific rotation also decreases, as shown in the graph. a Helix Specific rotation Poly(Glu) a Helix Random conformation Poly(Lys) Random conformation T + ° 2 4 6 В 10 12 14 PH Complete the statements about the molecular mechanism for these changes in specific rotation. Increasing the pH of a polyglutamate solution from 6 to 7 causes the carboxyl group of each glutamate residue Comed Artwer lose a proton. The negatively charged groups in each glutamate residue…arrow_forward
- From the reaction data below, determine whether the reaction is first order or second order and calculate the rate constant. Time (s) 0 Reactant (mM) 6.2 1 3.1 2 2.1 3 1.6 4 1.3 5 1.1 Only a plot of 1/[reactant] versus t gives a straight line, so the reaction is 0.150 mM-1 s-1 . S second order . The slope, k, isarrow_forwardFrom the reaction data below, determine whether the reaction is first order or second order and calculate the rate constant. Time (s) 0 Reactant (mM) 5.4 1 4.6 2 3.9 3 3.2 4 2.7 5 2.3 Only a plot of In[reactant] versus t gives a straight line, so the reaction is s-1. . The negative of the slope, k, isarrow_forwardA protein has a molecular mass of 400 kDa when measured by size-exclusion chromatography. When subjected to gel electrophoresis in the presence of sodium dodecyl sulfate (SDS), the protein gives three bands with molecular masses of 180, 160, and 60 kDa. When electrophoresis is carried out in the presence of SDS and dithiothreitol (DTT), three bands again form, this time with molecular masses of 160, 90, and 60 kDa. How many subunits does the protein have, and what is the molecular mass of each? four subunits: 180, 160, 90, and 60 kDa three subunits: 180, 160, and 60 kDa three subunits: 160, 90, and 60 kDa four subunits: 160, 90, 90, and 60 kDa Correct Answerarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles Of Radiographic Imaging: An Art And A ...Health & NutritionISBN:9781337711067Author:Richard R. Carlton, Arlene M. Adler, Vesna BalacPublisher:Cengage Learning
Case Studies In Health Information ManagementBiologyISBN:9781337676908Author:SCHNERINGPublisher:Cengage
Medical Terminology for Health Professions, Spira...Health & NutritionISBN:9781305634350Author:Ann Ehrlich, Carol L. Schroeder, Laura Ehrlich, Katrina A. SchroederPublisher:Cengage Learning
Principles Of Radiographic Imaging: An Art And A ...
Health & Nutrition
ISBN:9781337711067
Author:Richard R. Carlton, Arlene M. Adler, Vesna Balac
Publisher:Cengage Learning
Case Studies In Health Information Management
Biology
ISBN:9781337676908
Author:SCHNERING
Publisher:Cengage
Medical Terminology for Health Professions, Spira...
Health & Nutrition
ISBN:9781305634350
Author:Ann Ehrlich, Carol L. Schroeder, Laura Ehrlich, Katrina A. Schroeder
Publisher:Cengage Learning