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chapter16

Course: BIO 311C, Spring 2008
School: University of Texas
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16 Lecture Chapter Outline The Molecular Basis of Inheritance Overview: Life's Operating Instructions In April 1953, James Watson and Francis Crick shook the scientific world with an elegant double-helical model for the structure of deoxyribonucleic acid, or DNA. Your genetic endowment is the DNA you inherited from your parents. Nucleic acids are unique in their ability to direct their own replication. The...

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16 Lecture Chapter Outline The Molecular Basis of Inheritance Overview: Life's Operating Instructions In April 1953, James Watson and Francis Crick shook the scientific world with an elegant double-helical model for the structure of deoxyribonucleic acid, or DNA. Your genetic endowment is the DNA you inherited from your parents. Nucleic acids are unique in their ability to direct their own replication. The resemblance of offspring to their parents depends on the precise replication of DNA and its transmission from one generation to the next. It is this DNA program that directs the development of your biochemical, anatomical, physiological, and (to some extent) behavioral traits. Concept 16.1 DNA is the genetic material The search for genetic material led to DNA. Once T. H. Morgan's group showed that genes are located on chromosomes, the two constituents of chromosomes--proteins and DNA--were the candidates for the genetic material. Until the 1940s, the great heterogeneity and specificity of function of proteins seemed to indicate that proteins were the genetic material. However, this was not consistent with experiments with microorganisms, such as bacteria and viruses. The discovery of the genetic role of DNA began with research by Frederick Griffith in 1928. He studied Streptococcus pneumoniae, a bacterium that causes pneumonia in mammals. One strain, the R strain, was harmless. The other strain, the S strain, was pathogenic. Griffith mixed heat-killed S strain with live R strain bacteria and injected this into a mouse. The mouse died, and he recovered the pathogenic strain from the mouse's blood. Griffith called this phenomenon transformation, a phenomenon now defined as a change in genotype and phenotype due to the assimilation of foreign DNA by a cell. 16-1 Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. For the next 14 years, scientists tried to identify the transforming substance. Finally in 1944, Oswald Avery, Maclyn McCarty, and Colin MacLeod announced that the transforming substance was DNA. Still, many biologists were skeptical. Proteins were considered better candidates for the genetic material. There was also a belief that the genes of bacteria could not be similar in composition and function to those of more complex organisms. Further evidence that DNA was the genetic material was derived from studies that tracked the infection of bacteria by viruses. Viruses consist of DNA (or sometimes RNA) enclosed by a protective coat of protein. To replicate, a virus infects a host cell and takes over the cell's metabolic machinery. Viruses that specifically attack bacteria are called bacteriophages or just phages. In 1952, Alfred Hershey and Martha Chase showed that DNA was the genetic material of the phage T2. The T2 phage, consisting almost entirely of DNA and protein, attacks Escherichia coli (E. coli), a common intestinal bacteria of mammals. This phage can quickly turn an E. coli cell into a T2-producing factory that releases phages when the cell ruptures. To determine the source of genetic material in the phage, Hershey and Chase designed an experiment in which they could label protein or DNA and then track which entered the E. coli cell during infection. They grew one batch of T2 phage in the presence of radioactive sulfur, marking the proteins but not DNA. They grew another batch in the presence of radioactive phosphorus, marking the DNA but not proteins. They allowed each batch to infect separate E. coli cultures. Shortly after the onset of infection, they spun the cultured infected cells in a blender, shaking loose any parts of the phage that remained outside the bacteria. The mixtures were spun in a centrifuge, which separated the heavier bacterial cells in the pellet from lighter free phages and parts of phage in the liquid supernatant. They then tested the pellet and supernatant of the separate treatments for the presence of radioactivity. Hershey and Chase found that when the bacteria had been infected with T2 phages that contained radiolabeled proteins, Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. 16-2 most of the radioactivity was in the supernatant that contained phage particles, not in the pellet with the bacteria. When they examined the bacterial cultures with T2 phage that had radiolabeled DNA, most of the radioactivity was in the pellet with the bacteria. Hershey and Chase concluded that the injected DNA of the phage provides the genetic information that makes the infected cells produce new viral DNA and proteins to assemble into new viruses. The fact that cells double the amount of DNA in a cell prior to mitosis and then distribute the DNA equally to each daughter cell provided some circumstantial evidence that DNA was the genetic material in eukaryotes. Similar circumstantial evidence came from the observation that diploid sets of chromosomes have twice as much DNA as the haploid sets in gametes of the same organism. By 1947, Erwin Chargaff had developed a series of rules based on a survey of DNA composition in organisms. He already knew that DNA was a polymer of nucleotides consisting of a nitrogenous base, deoxyribose, and a phosphate group. The bases could be adenine (A), thymine (T), guanine (G), or cytosine (C). Chargaff noted that the DNA composition varies from species to species. In any one species, the four bases are found in characteristic, but not necessarily equal, ratios. He also found a peculiar regularity in the ratios of nucleotide bases that are known as Chargaff's rules. In all organisms, the number of adenines was approximately equal to the number of thymines (%T = %A). The number of guanines was approximately equal to the number of cytosines (%G = %C). Human DNA is 30.9% adenine, 29.4% thymine, 19.9% guanine, and 19.8% cytosine. The basis for these rules remained unexplained until the discovery of the double helix. Watson and Crick discovered the double helix by building models to conform to X-ray data. By the beginnings of the 1950s, the race was on to move from the structure of a single DNA strand to the three-dimensional structure of DNA. Among the scientists working on the problem were Linus Pauling in California and Maurice Wilkins and Rosalind Franklin in London. 16-3 Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. Maurice Wilkins and Rosalind Franklin used X-ray crystallography to study the structure of DNA. In this technique, X-rays are diffracted as they passed through aligned fibers of purified DNA. The diffraction pattern can be used to deduce the threedimensional shape of molecules. James Watson learned from their research that DNA was helical in shape, and he deduced the width of the helix and the spacing of nitrogenous bases. The width of the helix suggested that it was made up of two strands, contrary to a three-stranded model that Linus Pauling had recently proposed. Watson and his colleague Francis Crick began to work on a model of DNA with two strands, the double helix. Using molecular models made of wire, they placed the sugarphosphate chains on the outside and the nitrogenous bases on the inside of the double helix. This arrangement put the relatively hydrophobic nitrogenous bases in the molecule's interior. The sugar-phosphate chains of each strand are like the side ropes of a rope ladder. Pairs of nitrogenous bases, one from each strand, form rungs. The ladder forms a twist every ten bases. The nitrogenous bases are paired in specific combinations: adenine with thymine and guanine with cytosine. Pairing like nucleotides did not fit the uniform diameter indicated by the X-ray data. A purine-purine pair is too wide, and a pyrimidine-pyrimidine pairing is too short. Only a pyrimidine-purine pairing produces the 2-nm diameter indicated by the X-ray data. In addition, Watson and Crick determined that chemical side groups of the nitrogenous bases would form hydrogen bonds, connecting the two strands. Based on details of their structure, adenine would form two hydrogen bonds only with thymine, and guanine would form three hydrogen bonds only with cytosine. This finding explained Chargaff's rules. The base-pairing rules dictate the combinations of nitrogenous bases that form the rungs of DNA. However, this does not restrict the sequence of nucleotides along each DNA strand. The linear sequence of the four bases can be varied in countless ways. Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. 16-4 Each gene has a unique order of nitrogenous bases. In April 1953, Watson and Crick published a succinct, one-page paper in Nature reporting their double helix model of DNA. Concept 16.2 Many proteins work together in DNA replication and repair The specific pairing of nitrogenous bases in DNA was the flash of inspiration that led Watson and Crick to the correct double helix. The possible mechanism for the next step, the accurate replication of DNA, was clear to Watson and Crick from their double helix model. During DNA replication, base pairing enables existing DNA strands to serve as templates for new complementary strands. In a second paper, Watson and Crick published their hypothesis for how DNA replicates. Essentially, because each strand is complementary to the other, each can form a template when separated. The order of bases on one strand can be used to add complementary bases and therefore duplicate the pairs of bases exactly. When a cell copies a DNA molecule, each strand serves as a template for ordering nucleotides into a new complementary strand. One at a time, nucleotides line up along the template strand according to the base-pairing rules. The nucleotides are linked to form new strands. Watson and Crick's model, semiconservative replication, predicts that when a double helix replicates, each of the daughter molecules will have one old strand and one newly made strand. Other competing models, the conservative model and the dispersive model, were also proposed. Experiments in the late 1950s by Matthew Meselson and Franklin Stahl supported the semiconservative model proposed by Watson and Crick over the other two models. In their experiments, they labeled the nucleotides of the old strands with a heavy isotope of nitrogen (15N), while any new nucleotides were indicated by a lighter isotope ( 14N). Replicated strands could be separated by density in a centrifuge. Each model--the semiconservative model, the conservative model, and the dispersive model--made specific predictions about the density of replicated DNA strands. 16-5 Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. The first replication in the 14N medium produced a band of hybrid (15N-14N) DNA, eliminating the conservative model. A second replication produced both light and hybrid DNA, eliminating the dispersive model and supporting the semiconservative model. A large team of enzymes and other proteins carries out DNA replication. It takes E. coli 25 minutes to copy each of the 5 million base pairs in its single chromosome and divide to two form identical daughter cells. A human cell can copy its 6 billion base pairs and divide into daughter cells in only a few hours. This process is remarkably accurate, with only one error per ten billion nucleotides. More than a dozen enzymes and other proteins participate in DNA replication. Much more is known about replication in bacteria than in eukaryotes. The process appears to be fundamentally similar for prokaryotes and eukaryotes. The replication of a DNA molecule begins at special sites, origins of replication. In bacteria, this is a specific sequence of nucleotides that is recognized by the replication enzymes. These enzymes separate the strands, forming a replication bubble. Replication proceeds in both directions until the entire molecule is copied. In eukaryotes, there may be hundreds or thousands of origin sites per chromosome. At the origin sites, the DNA strands separate, forming a replication bubble with replication forks at each end. The replication bubbles elongate as the DNA is replicated, and eventually fuse. DNA polymerases catalyze the elongation of new DNA at a replication fork. As nucleotides align with complementary bases along the template strand, they are added to the growing end of the new strand by the polymerase. The rate of elongation is about 500 nucleotides per second in bacteria and 50 per second in human cells. In E. coli, two different DNA polymerases are involved in replication: DNA polymerase III and DNA polymerase I. In eukaryotes, at least 11 different DNA polymerases have been identified so far. 16-6 Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. Each nucleotide that is added to a growing DNA strand is a nucleoside triphosphate. Each has a nitrogenous base, deoxyribose, and a triphosphate tail. ATP is a nucleoside triphosphate with ribose instead of deoxyribose. Like ATP, the triphosphate monomers used for DNA synthesis are chemically reactive, partly because their triphosphate tails have an unstable cluster of negative charge. As each nucleotide is added to the growing end of a DNA strand, the last two phosphate groups are hydrolyzed to form pyrophosphate. The exergonic hydrolysis of pyrophosphate to two inorganic phosphate molecules drives the polymerization of the nucleotide to the new strand. The strands in the double helix are antiparallel. The sugar-phosphate backbones run in opposite directions. Each DNA strand has a 3' end with a free hydroxyl group attached to deoxyribose and a 5' end with a free phosphate group attached to deoxyribose. The 5' 3' direction of one strand runs counter to the 3' 5' direction of the other strand. DNA polymerases can only add nucleotides to the free 3' end of a growing DNA strand. A new DNA strand can only elongate in the 5' 3' direction. Along one template strand, DNA polymerase III can synthesize a complementary strand continuously by elongating the new DNA in the mandatory 5' 3' direction. The DNA strand made by this mechanism is called the leading strand. The other parental strand (5' 3' into the fork), the lagging strand, is copied away from the fork. Unlike the leading strand, which elongates continuously, the lagging stand is synthesized as a series of short segments called Okazaki fragments. Okazaki fragments are about 1,0002,000 nucleotides long in E. coli and 100200 nucleotides long in eukaryotes. Another enzyme, DNA ligase, eventually joins the sugarphosphate backbones of the Okazaki fragments to form a single DNA strand. DNA polymerases cannot initiate synthesis of a polynucleotide. They can only add nucleotides to the 3' end of an existing chain that is base-paired with the template strand. The initial nucleotide chain is called a primer. Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. 16-7 In the initiation of the replication of cellular DNA, the primer is a short stretch of RNA with an available 3' end. The primer is 510 nucleotides long in eukaryotes. Primase, an RNA polymerase, links ribonucleotides that are complementary to the DNA template into the primer. RNA polymerases can start an RNA chain from a single template strand. After formation of the primer, DNA pol III adds a deoxyribonucleotide to the 3' end of the RNA primer and continues adding DNA nucleotides to the growing DNA strand according to the base-pairing rules. Returning to the original problem at the replication fork, the leading strand requires the formation of only a single primer as the replication fork continues to separate. For synthesis of the lagging strand, each Okazaki fragment must be primed separately. Another DNA polymerase, DNA polymerase I, replaces the RNA nucleotides of the primers with DNA versions, adding them one by one onto the 3' end of the adjacent Okazaki fragment. The primers are converted to DNA before DNA ligase joins the fragments together. In addition to primase, DNA polymerases, and DNA ligases, several other proteins have prominent roles in DNA synthesis. Helicase untwists the double helix and separates the template DNA strands at the replication fork. This untwisting causes tighter twisting ahead of the replication fork, and topoisomerase helps relieve this strain. Single-strand binding proteins keep the unpaired template strands apart during replication. To summarize, at the replication fork, the leading strand is copied continuously into the fork from a single primer. The lagging strand is copied away from the fork in short segments, each requiring a new primer. It is conventional and convenient to think of the DNA polymerase molecules as moving along a stationary DNA template. In reality, the various proteins involved in DNA replication form a single large complex, a DNA replication machine. Many protein-protein interactions facilitate the efficiency of this machine. For example, helicase works much more rapidly when it is in contact with primase. The DNA replication machine is probably stationary during the replication process. Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. 16-8 In eukaryotic cells, multiple copies of the machine may anchor to the nuclear matrix, a framework of fibers extending through the interior of the nucleus. The DNA polymerase molecules reel in the parental DNA and extrude newly made daughter DNA molecules. Enzymes proofread DNA during its replication and repair damage in existing DNA. Mistakes during the initial pairing of template nucleotides and complementary nucleotides occur at a rate of one error per 100,000 base pairs. DNA polymerase proofreads each new nucleotide against the template nucleotide as soon as it is added. If there is an incorrect pairing, the enzyme removes the wrong nucleotide and then resumes synthesis. The final error rate is only one per ten billion nucleotides. DNA molecules are constantly subject to potentially harmful chemical and physical agents. Reactive chemicals, radioactive emissions, X-rays, and ultraviolet light can change nucleotides in ways that can affect encoded genetic information. DNA bases may undergo spontaneous chemical changes under normal cellular conditions. Mismatched nucleotides that are missed by DNA polymerase or mutations that occur after DNA synthesis is completed can often be repaired. Each cell continually monitors and repairs its genetic material, with 100 repair enzymes known in E. coli and more than 130 repair enzymes identified in humans. In mismatch repair, special enzymes fix incorrectly paired nucleotides. A hereditary defect in one of these enzymes is associated with a form of colon cancer. In nucleotide excision repair, a nuclease cuts out a segment of a damaged strand. DNA polymerase and ligase fill in the gap. The importance of the proper functioning of repair enzymes is clear from the inherited disorder xeroderma pigmentosum. These individuals are hypersensitive to sunlight. Ultraviolet light can produce thymine dimers between adjacent thymine nucleotides. This buckles the DNA double helix and interferes with DNA replication. In individuals with this disorder, mutations in their skin cells are left uncorrected and cause skin cancer. 16-9 Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. The ends of DNA molecules are replicated by a special mechanism. Limitations of DNA polymerase create problems for the linear DNA of eukaryotic chromosomes. The usual replication machinery provides no way to complete the 5' ends of daughter DNA strands. Repeated rounds of replication produce shorter and shorter DNA molecules. Prokaryotes do not have this problem because they have circular DNA molecules without ends. The ends of eukaryotic chromosomal DNA molecules, the telomeres, have special nucleotide sequences. Telomeres do not contain genes. Instead, the DNA typically consists of multiple repetitions of one short nucleotide sequence. In human telomeres, this sequence is typically TTAGGG, repeated between 100 and 1,000 times. Telomeres protect genes from being eroded through multiple rounds of DNA replication. Telomeric DNA tends to be shorter in dividing somatic cells of older individuals and in cultured cells that have divided many times. It is possible that the shortening of telomeres is somehow connected with the aging process of certain tissues and perhaps to aging in general. Telomeric DNA and specific proteins associated with it also prevents the staggered ends of the daughter molecule from activating the cell's system for monitoring DNA damage. Eukaryotic cells have evolved a mechanism to restore shortened telomeres in germ cells, which give rise to gametes. If the chromosomes of germ cells became shorter with every cell cycle, essential genes would eventually be lost. An enzyme called telomerase catalyzes the lengthening of telomeres in eukaryotic germ cells, restoring their original length. Telomerase uses a short molecule of RNA as a template to extend the 3' end of the telomere. There is now room for primase and DNA polymerase to extend the 5' end. It does not repair the 3'-end overhang, but it does lengthen the telomere. Telomerase is not present in most cells of multicellular organisms. Therefore, the DNA of dividing somatic cells and cultured cells tends to become shorter. 16-10 Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. Telomere length may be a limiting factor in the life span of certain tissues and of the organism. Normal shortening of telomeres may protect organisms from cancer by limiting the number of divisions that somatic cells can undergo. Cells from large tumors often have unusually short telomeres, because they have gone through many cell divisions. Active telomerase has been found in some cancerous somatic cells. This overcomes the progressive shortening that would eventually lead to self-destruction of the cancer. Immortal strains of cultured cells are capable of unlimited cell division. Telomerase may provide a useful target for cancer diagnosis and chemotherapy. Lecture Outline for Campbell/Reece Biology, 7th Edition, Pearson Education, Inc. 16-11
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University of Texas - BIO - 311C
Chapter 15The Chromosomal Basis of InheritanceLecture OutlineOverview: Locating Genes on Chromosomes Today we know that genes-Gregor Mendel's "hereditary factors"-are located on chromosomes. A century ago, the relationship of genes and chromosom
University of Texas - BIO - 311C
Chapter 17From Gene to ProteinLecture OutlineOverview: The Flow of Genetic Information The information content of DNA is in the form of specific sequences of nucleotides along the DNA strands. The DNA inherited by an organism leads to specific t
University of Texas - BIO - 311C
Chapter 18The Genetics of Viruses and BacteriaLecture OutlineOverview: Microbial Model Systems Viruses and bacteria are the simplest biological systems- microbial models in which scientists find life's fundamental molecular mechanisms in their m
University of Texas - BIO - 311C
Chapter 20DNA Technology and GenomicsLecture OutlineOverview: Understanding and Manipulating Genomes One of the great achievements of modern science has been the sequencing of the human genome, which was largely completed by 2003. Progress began
University of Texas - BIO - 311C
Chapter 19Eukaryotic GenomesLecture OutlineOverview: How Eukaryotic Genomes Work and Evolve Two features of eukaryotic genomes present a major information-processing challenge. First, the typical multicellular eukaryotic genome is much larger th
University of Texas - BIO - 311C
Chapter 21The Genetic Basis of DevelopmentLecture OutlineOverview: From Single Cell to Multicellular Organism The application of genetic analysis and DNA technology to the study of development has brought about a revolution in our understanding
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Chapter 22 Descent with Modification: Darwinian View of LifeLecture OutlineOverview: Darwin Introduces a Revolutionary Theory On November 24, 1859, Charles Darwin published On the Origin of Species by Means of Natural Selection. Darwin's book dre
University of Texas - BIO - 311C
Chapter 23Lecture OutlineThe Evolution of PopulationsOverview: The Smallest Unit of Evolution One common misconception about evolution is that organisms evolve, in a Darwinian sense, during their lifetimes. Natural selection does act on individ
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Introduction to Electrical and Computer Engineering (EE302)Fall 2007Course:EE302; Unique #s: 16190, 16195, 16200 Lecture: MWF 9-10AM in ENS 127 Lab: 16190 meets T 11-1 PM in ACA1.108 16195 meets TH 11-1 PM in ACA1.108 16200 meets TH 3-5 PM in ACA
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EE 302, Introduction to Electrical and Computer Engineering - Honors Dr. Archie Holmes, Jr. Exam #2Name: _ EID: _Please remember. Read the entire exam before starting If you feel you need more information than is given, please ask! Show all wo
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EE 302, Introduction to Electrical and Computer Engineering - Honors Dr. Archie Holmes, Jr. Exam #2 - Fall 2004Name: _ EID: _Please remember. Read the entire exam before starting All answers must include units and an appropriate number of signif
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EE 302, Introduction to Electrical and Computer Engineering - Honors Dr. Archie Holmes, Jr. Exam #2Name: _ EID: _Please remember. Read the entire exam before starting All answers must include units and an appropriate number of significant figur
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EE 302, Introduction to Electrical and Computer Engineering - Honors Dr. Archie Holmes, Jr. Exam #3Name: _ EID: _Please remember. Read the entire exam before starting If you feel you need more information than is given, please ask! Show all wo
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EE 302, Introduction to Electrical and Computer Engineering - Honors Dr. Archie Holmes, Jr. Exam #3Name: _ EID: _Please remember. Read the entire exam before starting All answers must include units and an appropriate number of significant figur
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University of Texas - EE - 302
Practice ProblemFind Vo in the circuit.Practice ProblemFind Vx and Vo in the circuit.Practice ProblemFind vo and io in the circuit.Practice ProblemFind the currents and voltages in the circuit.Practice ProblemCalculate Rab.Practice Pro
University of Texas - EE - 302
EE302 Exam2 Review Problems1. Use the node-voltage method to find the power associated with the 2A source.24 2ADC55V3P = 40W2. Find the current Io flowing through the 10 resistor using the current analysis technique that results in the
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Exam 2 Practice ProblemsFor Problems 1-21, use the figures at the end of this document. 1. Calculate the power dissipated by each resistor in Figure 1.Resistor 2 10 5 8 3 Power 2.40W 0.100W 5.00W 6.50W 1.30W2. Calculate the power being delivered
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EE302 Practice Problems for Exam #31. Use a series of source transformations to find the current Io flowing through the 2 resistor in the circuit below. Mark the direction of the current Io.Io1A2. When a 15k resistor is connected to the termin
University of Texas - EE - 302
EE302 Practice Problems for Exam #31. Use a series of source transformations to find the current Io flowing through the 2 resistor in the circuit below. Mark the direction of the current Io.10A41 40 24A5DC10VIo1A2. When a 15k resi
University of Texas - EE - 302
University of Texas - EE - 302
Node Voltage Review ProblemsSome tips for writing KCL equations Label branch currents in all branches as I1, I2, I3. Mark current directions as going left to right or up to down. Then write KCL for every node in terms of I1, I2, I3., with curren
University of Texas - EE - 302
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University of Texas - EE - 302
EE302 A Circuit a Day Club (ACDC)Rules Solve one circuit a day from the list of problems given or from the homework problems. You may not roll over problems from one day to another. Note down the problem number in the spreadsheet given. This is
University of Texas - EE - 302
EE302 Fall '07 HomeworkHW1 HW2 HW3 HW4 HW5 HW6 HW7 HW8 HW9 HW101.6, 1.11, 1.18, 1.20, 1.28, 1.36 2.12, 2.16, 2.17, 2.18, 2.20, 2.22 2.26, 2.32, 2.34, 2.38, 2.41, 2.45 3.6, 3.10, 3.12, 3.17, 3.20, 3.22 3.36, 3.39, 3.40, 3.44, 3.50, 3.52 3.56, 3.60
University of Texas - EE - 302
Chapter 1, Problem 6. The charge entering a certain element is shown in Fig. 1.23. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 msChapter 1, Problem 11.A rechargeable flashlight battery is capable of delivering 85 mA for about 12 h. H
University of Texas - EE - 302
EE302 Homework #1 Chapter 1, Problem 6. The charge entering a certain element is shown in Fig. 1.23. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 msChapter 1, Solution 6. (a) At t = 1ms, i = (b) At t = 6ms, i =dq 80 = = 40 A dt 2dq
University of Texas - EE - 302
Chapter 1, Problem 6. The charge entering a certain element is shown in Fig. 1.23. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 msChapter 1, Problem 11.A rechargeable flashlight battery is capable of delivering 85 mA for about 12 h. H
University of Texas - EE - 302
In the circuit in Fig. 2.76, obtain v1, v2, and v3.Chapter 2, Problem 16.Determine Vo in the circuit in Fig. 2.80.62+ 9V + _ Vo _ + _ 3VObtain v1 through v3 in the circuitFind I and Vab in the circuit.Determine io in the circuitFin
University of Texas - EE - 302
In the circuit in Fig. 2.76, obtain v1, v2, and v3.Chapter 2, Problem 16.Determine Vo in the circuit in Fig. 2.80.62+9V+ _Vo+ _3V_Obtain v1 through v3 in the circuitFind I and Vab in the circuit.Determine io in the circ
University of Texas - EE - 302
Chapter 2, Problem 26. For the circuit in Fig. 2.90, io =2 A. Calculate ix and the total power dissipated by the circuit.ixio 2 4 8 16The voltage across the 8 resistor is that across the 16.A. TrueB. FalseChapter 2, Problem 32. Find i1 thr
University of Texas - EE - 302
Chapter 2, Problem 26. For the circuit in Fig. 2.90, io =2 A. Calculate ix and the total power dissipated by the circuit.ixio 2 4 8 16The voltage across the 8 resistor is that across the 16.A. TrueB. FalseChapter 2, Problem 32. Find i1 thr
University of Texas - EE - 302
Chapter 3, Problem 6. Use nodal analysis to obtain v0 in the circuit in Fig. 3.55. The current I1 is equal toA. V0/4B. 3AC. (V0 -12)/4D. None of the aboveChapter 3, Problem 10. Find i0 in the circuit in Fig. 3.59.The supernode method needs
University of Texas - EE - 302
Chapter 3, Problem 36. Rework Prob. 3.6 using mesh analysis.Chapter 3, Problem 39. Determine the mesh currents i1 and i2 in the circuit shown in Fig. 3.85.Chapter 3, Problem 40. For the bridge network in Fig. 3.86, find Io using mesh analysis.C
University of Texas - EE - 302
Chapter 3, Problem 56. Determine v1 and v2 in the circuit of Fig. 3.101.The voltages across all the resistors will be the same because the values of the resistors are the same. A. True B. FalseChapter 3, Problem 60. Calculate the power dissipated
University of Texas - EE - 302
Chapter 4, Problem 24.Use source transformation to find the voltage Vx in the circuit of Fig. 4.92.3A8 + + _ Vx 1040 V102 VxChapter 4, Problem 26.Use source transformation to find io in the circuit of Fig. 4.94.53Aio46A2
University of Texas - EE - 302
Chapter 4, Problem 40. Find the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.107.+ Vo 20 k10 k a70 V + _ b+ 4 VoChapter 4, Problem 43. Find the Thevenin equivalent looking into terminals a-b of the circuit in Fig. 4.11
University of Texas - EE - 302
Chapter 5, Problem 8. Obtain vo for each of the op amp circuits in Fig. 5.47.Figure 5.47 for Prob. 5.8Chapter 5, Problem 10. Find the gain vo/vs of the circuit in Fig. 5.49.Figure 5.49 for Prob. 5.10Chapter 5, Problem 13. Find vo and io in th
University of Texas - EE - 302
EE302 Homework #1 Chapter 1, Problem 6. The charge entering a certain element is shown in Fig. 1.23. Find the current at: (a) t = 1 ms (b) t = 6 ms (c) t = 10 msChapter 1, Solution 6. (a) At t = 1ms, i = (b) At t = 6ms, i =dq 80 = = 40 A dt 2dq
University of Texas - EE - 302
EE302 Homework #2 Chapter 2, Problem 12. In the circuit in Fig. 2.76, obtain v1, v2, and v3.Chapter 2, Solution 12. + 15v -loop 2 25v + + 20v For loop 1, For loop 2, For loop 3, + 10v + v1 + v2 + v3 -loop 1loop 3-20 -25 +10 + v1 = 0 -10 +1
University of Texas - EE - 302
EE302 Homework #3 Chapter 2, Problem 26. For the circuit in Fig. 2.90, io =2 A. Calculate ix and the total power dissipated by the circuit.ix io 2 4 8 16 Chapter 2, Solution 26. If i16= io = 2A, then v = 16x2 = 32 Vi8 =v =4A, 8i4 =v = 8 A
University of Texas - EE - 302
EE302 Homework #4 Chapter 3, Solution 6. i1 + i2 + i3 = 0v0 - 12 v0 v0 - 10 + + =0 4 6 2or v0 = 8.727 VChapter 3, Solution 10.At node 1:V - V3 V1 +4+ 1 = 0, 8 1also by Ohm's law: I 0 =V1 8 V1 V3 V3 - V1 + + =0 4 4 1At node 3: - 2 I 0 +
University of Texas - EE - 302
EE302 Homework #5 Chapter 3, Solution 36.4 i1 i2 10 V + i312 V+I16I22Applying mesh analysis gives, 12 = 10I1 6I2 -10 = -6I1 + 8I2or 6 5 - 3 I 1 - 5 = - 3 4 I 2 =5 -3 6 -3 5 6 = 11, 1 = = 9, 2 = = -7 -3 4 -5 4 -
University of Texas - EE - 302
EE302 Homework #6 Chapter 3, Solution 56. + v1 2 2i222 2+v212 V+i1i3For loop 1, 12 = 4i1 2i2 2i3 which leads to 6 = 2i1 i2 i3 For loop 2, 0 = 6i2 2i1 2 i3 which leads to 0 = -i1 + 3i2 i3 For loop 3, 0 = 6i3 2i1 2i2 whic
University of Texas - EE - 302
Introduction to Electrical and Computer EngineeringEE302 Prof. Nina Telang1Welcome to UT!What is the Mission of UT? To transform lives for the benefit of society through the core values of learning, discovery, freedom, leadership, individual
University of Texas - EE - 302
Unit 1: The Professional Engineer1ObjectivesWho is an Engineer?What is the Engineering Design Cycle?2ABET* Definition of EngineeringThe profession in which a knowledge of the mathematical and natural sciences gained by study, experien
University of Texas - EE - 302
Energy and Power1EnergyEnergy is the capacity of matter to perform work Energy is conserved. It can be neither created nor destroyed. The unit of energy is the Joule and is the same as the unit of work.2PowerPower = energy per un
University of Texas - EE - 302
Unit 2: Basic Circuit TheoryUnit 2.1 Units and Significant FiguresLearning ObjectivesWhat are engineering units and notations? How many significant figures should you use?International System of Units (SI), formerly MKSSystems of UnitsUn
University of Texas - EE - 302
Unit 2: Basic Circuit TheoryUnit 2.2: Physical Basics1Learning Objectives Charge Current and Voltage Energy and Power Ideal basic circuit element Passive sign convention2Charge Charles CoulombFundamental electric quantity Charg
University of Texas - EE - 302
Energy and Power1EnergyEnergy is the capacity of matter to perform work Energy is conserved. It can be neither created nor destroyed. The unit of energy is the Joule and is the same as the unit of work.2PowerPower = energy per un
University of Texas - EE - 302
Unit 2: Basic Circuit TheoryUnit 2.3: Circuits and Circuit Elements Chapter 21Learning Objectives Review Resistance Ohm's Law Short circuit and open circuit Nodes, branches and loops Kirchhoff's Voltage Law (KVL) Kirchhoff's Current Law
University of Texas - EE - 302
Unit 2: Basic Circuit TheoryUnit 2.5: Series and Parallel CircuitsLearning Objectives Series Circuits Voltage sources Resistors Parallel Circuits Current Sources ResistorsHow do series circuits divide voltageHow do parallel circuits divide c
University of Texas - EE - 302
Unit 4: Other Circuit Analysis TechniquesUnit 4.1 Source Transformations Unit 4.2 Thevenin's Theorem/Norton's TheoremLearning Objectives Transform voltage sources with series resistors into current sources with parallel resistors source transf
University of Texas - EE - 302
Unit 3: Node and Mesh Circuit AnalysisUnit 3.1 Node Voltage Analysis Chapter 3Learning Objectives Why is there a need for more powerful techniques to solve circuits. What is the node voltage method. How do we apply it.Practice ProblemCalcul