Textbook Question
The elongation of the leading strand during DNA synthesis
a. Progresses away from the replication fork
b. Occurs in the 3′→5′ direction
c. Produces Okazaki fragments
d. Depends on the action of DNA polymerase
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Ch. 16 - The Molecular Basis of Inheritance
Urry11th EditionCampbell BiologyISBN: 9789357423311
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Table of contents
- Ch. 1 - Evolution, the Themes of Biology, and Scientific Inquiry8
- Ch. 2 - The Chemical Context of Life8
- Ch. 3 - Water and Life6
- Ch. 4 - Carbon and the Molecular Diversity of Life9
- Ch. 5 - The Structure and Function of Large Biological Molecules9
- Ch. 6 - A Tour of the Cell6
- Ch. 7 - Membrane Structure and Function6
- Ch. 8 - An Introduction to Metabolism6
- Ch. 9 - Cellular Respiration and Fermentation10
- Ch. 10 - Photosynthesis7
- Ch. 11 - Cell Communication7
- Ch. 12 - The Cell Cycle10
- Ch. 13 - Meiosis and Sexual Life Cycles10
- Ch. 14 - Mendel and the Gene Idea14
- Ch, 15 - The Chromosomal Basis of Inheritance7
- Ch. 16 - The Molecular Basis of Inheritance9
- Ch. 17 - Gene Expression: From Gene to Protein9
- Ch. 18 - Regulation of Gene Expression10
- Ch. 19 - Viruses6
- Ch. 20 - DNA Tools and Biotechnology8
- Ch. 21 - Genomes and Their Evolution8
- Ch. 22 - Descent with Modification: A Darwininan View of Life6
- Ch. 23 - The Evolution of Populations5
- Ch. 24 - The Origin of Species6
- Ch. 25 - The History of Life on Earth6
- Ch. 26 - Phylogeny and the Tree of Life7
- Ch. 27 - Bacteria and Archaea8
- Ch. 28 - Protists7
- Ch. 29 - Plant Diversity I: How Plants Colonized Land7
- Ch. 30 - Plant Diversity II: The Evolution of Seed Plants7
- Ch. 31 - Fungi5
- Ch. 32 - An Overview of Animal Diversity4
- Ch. 33 - An introduction to Invertebrates6
- Ch. 34 - The Origin and Evolution of Vertebrates7
- Ch. 35 - Vascular Plant Structure, Growth, and Development8
- Ch. 36 - Resource Acquisition and Transport in Vascular Plants8
- Ch. 37 - Soil and Plant Nutrition10
- Ch. 38 - Angiosperm Reproduction and Biotechnology8
- Ch. 39 - Plant Responses to Internal and External Signals8
- Ch. 40 - Basic Principles of Animal Form and Function8
- Ch. 41 - Animal Nutrition7
- Ch. 42 - Circulation and Gas Exchange7
- Ch. 43 - The Immune System6
- Ch. 44 - Osmoregulation and Excretion6
- Ch. 45 - Hormones and the Endocrine System8
- Ch. 46 - Animal Reproduction8
- Ch. 47 - Animal Development8
- Ch. 48 - Neurons, Synapses, and Signaling6
- Ch. 49 - Nervous Systems8
- Ch. 50 - Sensory and Motor Mechanisms5
- Ch. 51 Animal Behavior6
- Ch. 52 - An Introduction to Ecology and the Biosphere7
- Ch. 53 - Population Ecology10
- Ch. 54 - Community Ecology9
- Ch. 55 - Ecosystems and Restoration Ecology8
- Ch. 56 - Conservation Biology and Global Change6
Chapter 16, Problem 7
A biochemist isolates, purifies, and combines in a test tube a variety of molecules needed for DNA replication. When she adds some DNA to the mixture, replication occurs, but each DNA molecule consists of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. What has she probably left out of the mixture?
a. DNA polymerase
b. DNA ligase
c. Okazaki fragments
d. Primase
Verified step by step guidance1
Understand the process of DNA replication, which involves the synthesis of a new DNA strand complementary to the template strand. This process requires several enzymes and components.
Recognize that during DNA replication, the leading strand is synthesized continuously, while the lagging strand is synthesized in short segments known as Okazaki fragments.
Identify the role of DNA ligase in DNA replication. DNA ligase is responsible for joining the Okazaki fragments on the lagging strand, creating a continuous DNA strand.
Consider the problem description: the DNA molecules consist of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. This suggests that the Okazaki fragments are not being joined together.
Conclude that the absence of DNA ligase in the mixture is likely the reason for the presence of unjoined Okazaki fragments, as DNA ligase is needed to connect these fragments into a continuous strand.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
DNA Replication
DNA replication is the process by which a cell duplicates its DNA, ensuring that each daughter cell receives an exact copy of the genetic material. It involves unwinding the double helix, synthesizing new strands using existing strands as templates, and requires various enzymes and proteins to facilitate the process.
Recommended video:
Guided course
03:37
Introduction to DNA Replication
Okazaki Fragments
Okazaki fragments are short sequences of DNA nucleotides synthesized discontinuously on the lagging strand during DNA replication. These fragments are later joined together by DNA ligase to form a continuous strand, as DNA polymerase can only synthesize DNA in the 5' to 3' direction.
Recommended video:
Guided course
05:15Leading & Lagging DNA Strands
DNA Ligase
DNA ligase is an enzyme crucial for DNA replication and repair, responsible for joining Okazaki fragments on the lagging strand. It catalyzes the formation of phosphodiester bonds between adjacent DNA fragments, ensuring the continuity and integrity of the newly synthesized DNA strand.
Recommended video:
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02:09Step 1) Create Recombinant DNA
Related Practice
Textbook Question
In a nucleosome, the DNA is wrapped around
a. Histones
b. Ribosomes
c. Polymerase molecules
d. A thymine dimer
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Textbook Question
E. coli cells grown on 15N medium are transferred to 14N medium and allowed to grow for two more generations (two rounds of DNA replication). DNA extracted from these cells is centrifuged. What density distribution of DNA would you expect in this experiment?
a. One high-density and one low-density band
b. One intermediate-density band
c. One high-density and one intermediate-density band
d. One low-density and one intermediate-density band
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Textbook Question
The spontaneous loss of amino groups from adenine in DNA results in hypoxanthine, an uncommon base, opposite thymine. What combination of proteins could repair such damage?
a. Nuclease, DNA polymerase, DNA ligase
b. Telomerase, primase, DNA polymerase
c. Telomerase, helicase, single-strand binding protein
d. DNA ligase, replication fork proteins, adenylyl cyclase
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Textbook Question
Although the proteins that cause the E. coli chromosome to coil are not histones, what property would you expect them to share with histones, given their ability to bind to DNA (see Figure 5.14)?
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