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homologous_recombination

Course: BIOC BIOC2603, Spring 2007
School: HKU
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Principles BIOC2603 of Molecular Genetics Recombination Brian Wong bcwwong@hkucc.hku.hk Genetic recombination: reassortment of genetic material (A) Independent assortment (for genes on different chromosomes) (B) crossing over (for genes on the same chromosome) Crossover is the breakage of the two DNA molecules at the same position and their rejoining in two reciprocal recombinations. Independent assortment...

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Principles BIOC2603 of Molecular Genetics Recombination Brian Wong bcwwong@hkucc.hku.hk Genetic recombination: reassortment of genetic material (A) Independent assortment (for genes on different chromosomes) (B) crossing over (for genes on the same chromosome) Crossover is the breakage of the two DNA molecules at the same position and their rejoining in two reciprocal recombinations. Independent assortment produces 50 percent recombinants For linked genes, recombinant frequencies are less than 50 percent Four types of recombination 1. Homologous recombination (general recombination) Requires extensive sequence similarity between the two recombining DNA Takes place anywhere along the DNA A X B X C a A a b b B c C c 2. Site-specific recombination Occurs between two very limited and specific DNA sequences on both of the recombining DNA The two parental recombination sites are conserved in a simple reciprocal recombination event CD X A B A D C B Four types of recombination 3. Transposition Involves interaction of the two ends of a transposable element with a random site on the same or another DNA molecule Results in a new position and/or proliferation of the transposable element + 4. Illegitimate recombination Processes that require neither sequence homology or site specificity e.g. non-homologous end joining Biological roles of recombination To generate genetic diversity For repair and replication To control gene expression (genetic switches) Genome evolution Introduces mutations and may result in diseases Part I Homologous Recombination Homologous recombination involves breakage and reunion of genetic material A B X recombination by breakage and rejoining: involve the physical exchange of genetic material between the two recombining DNA recombination by copy choice: recombination occurs during replication by copying partly from one parent and partly from the other a A a b b B A B a A b b Homologous recombination involves breakage and reunion of genetic material phage 1 h 15N, 13C phage 2 cI X h+ 15N, 13C cI+ labeled labeled (Heavy) (Heavy) Infect bacteria in 14N, 12C (light) medium. Separate phage progenies based on the densities of DNA by using CsCl density centrifugation. h h+ + cI+ cI Heavy recombinants were identified consistent with the breakage and reunion hypothesis. The copy-choice hypothesis predicted all recombinants were labled with 14N and 12C (light). Meselson M (1964) J. Mol. Biol. 9:734-745. Homologous recombination involves breakage and reunion of genetic material knob extra piece of DNA Corn with unusual chromosome C, colored; c, colorless Wx, waxy; wx, starchy Recombination between C/c and Wx/wx Recombinants Cytological crossing-over accompanied by genetical crossing-over Homologous recombination involves breakage and reunion of genetic material one strand contains BrdU, appears dark both strands contain BrdU, appears light mitosis Grow cells in the presence of BrdU. BrdU incorporated into DNA. BrdU Stain with Hoechst 33258 and Giemsa, sister chromatid with one DNA strand containing BrdU appear dark, sister chromatid with 2 DNA strands containing BrdU appear light. no sister chromatid exchange x homologous recombination between sister chromatids Stages of a Neurospora cross In Neurospora, meiosis is followed immediately by mitosis so a total of 8 haploid spores are observed. The linear meiosis of Neurospora In Neurospora, meiosis is followed immediately by mitosis so a total of 8 haploid spores are observed. Each spore gives the genetic information of individual DNA strand of the 4 sister chromatids before meiosis. Analysing the segregation patterns of the spores help to understand the molecular mechanism of homologous recombination. chromosome First meiotic division Second meiotic division Postmeiotic mitosis Crossing-over occurs after DNA duplication Crossing-over occurs after DNA duplication Recombination in fungi In a cross between a wild-type (A) and a mutant (a) allele, majority of the spores segregate in normal 4:4 ratio (no recombination); reciprocal recombination is also observed. A A A A a a a a A A A A a a a a mitosis A A A A a a a a A A a a = a a A A 4:4 ratio: no recombination between centromere and the marker locus. Product of A allele Product of a allele A A A A A a a A A a a X A a a a a mitosis A A a a A A a a a a A A = A A a a a a A A = a a A A One recombination event between centromere and the marker locus. Recombination in fungi In addition to asci with 4:4 (A:a) ratio, non-Mendalian ratios 5:3, 3:5, 6:2, 2:6, and aberrant 4:4 segregation patterns were also observed. Gene conversions often associated with crossing-over of closely linked flanking markers. Explanations? A A A a A a a a A A A a a a a a A A A A A a a a A A a a A a a a A A A a A A a a A A a a a a a a A A A A A A a a aberrant 4:4 3:5 5:3 aberrant 5:3 2:6 6:2 Gene conversion: unidirectional transfer of genetic information Models of homologous Recombination Holliday model (1964) Meselson-Radding model (1975) Double-strand break repair model (1983) Holliday model of homologous recombination Recombination is initiated by nicking of two homologous duplex DNA at the same site on strands with the same polarity Broken strands separated from the original duplex DNA Reciprocal strand invasion: each nicked strand pairs with the complementary strands in the homologous duplex DNA Holliday model of homologous recombination Ligation creates a structure in which 4 strands of two duplex DNA are linked via a crossed structure (Holliday junction) Holliday model of homologous recombination Branch migration: movement of the Holliday junction generates heteroduplex (symmetric heteroduplex, i.e. heteroduplexes on both recombining DNA molecule) Holliday model of homologous recombination Isomerization of the Holliday junction to give a branch structure with 4 sectors of single-stranded DNA Holliday model of homologous recombination Holliday model of homologous recombination Isomerization of the Holliday junction to give a branch structure with 4 sectors of single-stranded DNA Holliday model of homologous recombination Resolution of the Holliday junction to give non-crossover (flanking markers in parental configuration) and crossover (flanking markers exchanged) recombinant configurations. Mismatch repair of heteroduplexes: either strands can be used as template GATCCTATATATCTA CTAGGATATATAGAT X CTAGGATCTATAGAT GATCCTAGATATCTA GATCCTATATATCTA CTAGGATATATAGAT or GATCCTATATATCTA GATCCTAGATATCTA CTAGGATCTATAGAT CTAGGATCTATAGAT Mismatch repair CTAGGATATATAGAT GATCCTAGATATCTA CTAGGATATATAGAT GATCCTATATATCTA Heteroduplexes or CTAGGATCTATAGAT GATCCTAGATATCTA Holliday model of homologous recombination A A A a A a a a A A A a A a a a According to the Holliday model, reciprocal recombination results in the formation of two heteroduplexes Mitosis before mismatch repair Mismatch repair in one of the heteroduplexes before mitosis Mismatch repair in both heteroduplexes before mitosis A A A a A a a a A A A a a a a a A A A A A a a a A A a a A a a a A A A a A A a a A A a a a a a a A A A A A A a a aberrant 4:4 3:5 5:3 aberrant 5:3 2:6 6:2 Holliday model needs to be modified genetic experiments suggested that heteroduplex DNA often formed on only one chromatid, i.e. asymmetric heteroduplex (non-reciprocal recombination) was observed more often: aberrant 4:4 segregations are diagnostic of symmetric heteroduplexes; they are rare as compared to 5:3 2 types of 5:3 segregation: normal 5:3 and aberrant 5:3. If there were symmetric heteroduplexes, one should expect equal proportions of the 2 types of 5:3 segregations but experiments showed that aberrant 5:3 are rare compared to normal 5:3. Meselson-Radding model of recombination a) b) Recombination initiated by a single-stranded nick on one duplex DNA. The 3 end of the nick serves as a primer for DNA synthesis, displacing a single-stranded DNA. The displaced strand then invades a homologous duplex DNA, forming a displacement loop (D-loop). The D-loop is degraded and the invading strand ligated to the recipient DNA. DNA synthesis on the donor duplex and degradation of the recipient duplex result in the extension of heteroduplex (asymmetric heteroduplex since only one duplex contains heteroduplex). c) d) Meselson-Radding model of recombination e) Ligation of the 5 and 3 ends results in the formation of a Holliday junction. The resulting Holliday junction can move by branch migration to form symmetric heteroduplex. Resolution of the Holliday junction gives either crossover or noncrossover configuration. f) Meselson-Radding model of recombination has an asymmetric and a symmetric heteroduplex phase. A A A A A A a a a A A A A a a a If the marker is on the asymmetric heteroduplex region: Mitosis before mismatch repair of heteroduplex The asymmetric phase explains why there were more 6:2 segregations and normal 5:3 segregations observed in recombinations. A A A A A a a a Mismatch repair in heteroduplex before mitosis A A A A A A a a 5:3 6:2 Meselson-Radding model of recombination has an asymmetric and a symmetric heteroduplex phase. A A A A a A a a a A A a A a a a When the marker is on the symmetric heteroduplex region, its essentially the same as the Holliday model: mitosis before mismatch repair: aberrant 4:4 mismatch repair in one of the heteroduplex: 5:3, 3:5 mismatch repair in both heteroduplex: 2:6, 6:2 Meselson-Radding model of recombination The models has an asymmetric heteroduplex and a symmetric heteroduplex phase. If markers are within the asymmetric heteroduplex region, then mismatch repair in the heteroduplex gives a 6:2 segregation pattern (gene conversion) or normal 5:3 segregation pattern if DNA replication occurs before mismatch repair. The asymmetric phase explains why there were more 6:2 segregation and 5:3 normal segregation observed. If markers are within the symmetric heteroduplex region, the aberrant 4:4, aberrant 5:3 and 6:2 segregation are explained by mismatch repair. Homologous recombination A A B C C Homologous recombination A A B B C C A B C A B C Double-strand break repair model of recombination 1. Recombination initiated by a double-strand break in one duplex. The break is enlarged to form a gap with 3 strand-stranded ends by exonucleases. 2. One of the free 3 end then invade a homologous duplex DNA, displacing a DNA strand and forms a D-loop 3. DNA repair synthesis from the 3 end of the invading strand enlarge the Dloop until the displaced strand anneals to the other free 3 end of the gapped duplex, forming asymmetric heteroduplex. 4. DNA synthesis from the second 3 end proceeds. Ligation of DNA result in the formation of 2 Holliday junctions. Branch migration can lead to formation of symmetric heteroduplex. 5. Resolution of the junctions give rise to recombinants with crossover or noncrossover configuration. Double-strand break repair model of recombination Double-strand break repair model of recombination Double-strand break repair model of recombination Double-strand break repair model A A a a a a a a A A A a a a a a marker within the gap gap repair mismatch repair before mitosis marker within asymmetric heteroduplex Mitosis before mismatch repair A A a a a a a a A A A a a a a a 2:6 3:5 Double-strand break repair model The gap are repaired by two rounds of single-strand repair synthesis and flanked by regions of heteroduplex. Two Holliday junctions are formed, branch migration of the Holliday junctions leads to formation of symmetric heteroduplex. Gene conversion (6:2 or 2:6 segregations) is the result of i) gap repair if the marker falls within the gap; ii) mismatch repair of heteroduplex if the marker falls within flanking asymmetric heteroduplex regions. 3:5 or 5:3 segregations results if mismatch in heteroduplex is not repaired. Comparing the three models of recombination Holliday model Meselson-Radding model single-strand break in one duplex asymmetric and symmetric 1 required Double-strand break repair model double-stranded break in one duplex asymmetric and symmetric 2 required initiation duplex is the recipient of information gap repair mismatch repair Initiation single-strand breaks in both duplex symmetric 1 not required Heteroduplex No. of Holliday junctions DNA synthesis Exchange of sequence information Gene conversion reciprocal exchange of initiation duplex is the information donor of information mismatch repair asymmetric strand transfer mismatch repair The last step of homologous recombination is the resolution of Holliday junctions Holliday junction is a crossover structure that links two recombining DNA Movement of the junction (branch migration) extend the length of heteroduplex. In E. coli, resolution of Holliday junction is carried out by RuvC resolvase. crossover non-crossover Recombination proteins Recombinase RecA (E. coli) Rad51 (yeast) Branch migration RuvAB Holliday Junction resolvase RuvC (E. coli) Endonucleases (to produce DNA strand breaks for initiation of recombination) RecBCD (E. coli) Spo11 (yeast) I-SceI (yeast mitochondrial endonuclease encoded by intron of 21S rRNA gene) Recombination Proteins in E. coli RecA - promotes homologous pairing, strand exchange and formation of Holliday junctions. Binds to single-strand DNA to form a nucleoprotein filament that bind to double-strand DNA to search for homologous sequences. RuvAB - RuvA (22 kDa) binds to Holliday junctions and promote binding of RuvB to DNA; RuvB (37 kDa) catalyses branch migration; promotes branch migration of Holliday junctions in the presence of ATP, forming heteroduplex RuvC (19 kDa) - endonuclease that resolves Holliday junctions by introducing nicks into two DNA strands with the same polarity at an identical site; binds DNA substrate as a dimer in a sequence nonspecific manner but seems to cut DNA at preferred sites (5'-A/T TT G/C). RecBCD Pathway in E. coli RecBCD is made up of products from recB, recC, and recD genes, important for recombination and DNA repair RecBCD possess helicase activity, ssDNA exo- and endonuclease activity and dsDNA exonuclease activity, as well as ATPase activity Double-stranded DNA breaks (from ionization radiation, replication block etc) serve as initiation sites for homologous recombination RecBCD bind to end of a double-stranded break, the helicase activity then unwinds dsDNA and the ssDNA produced is cleaved , preferentially from 3 terminal RecBCD Pathway in E. coli When the DNA sequence (5-GCTGGTGG-3) is encountered, the 3 to 5 nuclease activity is attenuated while a 5 to 3 nuclease is activated. Therefore generating a ssDNA with the DNA sequence at the 3 end RecA protein load onto the single-stranded DNA, forming a nucleoprotein filament which direct the search of homologous sequences and strand invasion, resulting in the formation of Holliday junction. RuvAB is responsible for branch migration after the formation of Holliday junction: RuvA recognizes the Holliday junction structure and directs RuvB to the junction to catalyze branch migration RuvC is responsible for resolution of Holliday junction Early steps in homologous recombination in E. coli From: Kowalczykowski SC (2000) Trends in Biochemical Sciences 25:156-165. Double-strand breaks initiate meiotic recombination Recombination is initiated when Spo II enzyme makes double-stranded DNA cuts. Processes involving homologous recombination Chromosome segregation If crossing-over is prevented, chromosome fail to segregate properly. Genome evolution Recombination-dependent DNA replication Recombinational repair of double-strand breaks Repair damages induced by e.g. ionizing radiation Homing of homing endonuclease genes Programmed DNA rearrangement resulting in changes of gene expression e.g. mating type switching in yeast Exon and gene duplication via unequal crossing over Exon shuffling via recombination between interspersed repeats Exon and gene duplication via unequal crossing over Exon and gene duplication via unequal crossing over Recombination-dependent Replication in Phage T4 Genome of phage T4 is a linear doublestranded DNA that are circularly permutated and terminally redundant. T4 replicates its DNA using both replication and recombination Origin-dependent replication of T4 chromosome produces daughter molecules with single-stranded 3' ends. The single-stranded end then invades other daughter molecule by homologous recombination in the presence of T4 uvsX gene product. The invading 3' end serves as primer for DNA synthesis. The displaced strand also serves as template for DNA synthesis. T4 genome. DNA sequence represented by letters. Model for recombinationdependent replication. Kreuzer KN (2000) Recombination-dependent DNA replication in phage T4. TIBS 25:165-173. Recombinational Repair of a Collapsed Replication Fork Recombinational Repair of a Collapsed Replication Fork Homing Endonuclease Genes (HEGs) HEGs encode sequence-specific endonucleases. The recognition sequence (RS) is 15-30 bp long and usually occurs once in the genome. HEGs are located in the middle of their own recognition sequences. Most HEGs are encoded by self-splicing introns (group I & II) and inteins. Inteins are internal protein fragments produced from protein splicing and usually contain endonuclease and splicing activities. The allele without the HEGs are cleaved by the homing endonuclease and the double-strand break are repaired by homologous recombination (gene conversion) using the allele containing HEGs as template. Both chromosomes will contain the HEGs after repair. From: Burt A and Koufopanou V (2004) Current Opinion in Genetics & Development 14:609-615. Practical applications of homologous recombination Genetic mapping Determine the order of genes in a chromosome Recombination frequency as a measure of distance between two markers Gene targeting Modify a chromosomal gene with a cloned copy of the gene Recombination mapping In a testcross (a heterozygote cross with a homozygous recessive), the recombinant frequency is proportional to the physical distance between two gene loci: the chance of crossing-over between two gene loci to produce recombinants is proportional to the physical distance between the two loci. A recombinant frequency of 1% is defined as 1 recombination map unit (m.u.) or centimorgan (cM). In human, 1 cM approximately equal to 1 million base pairs. A three-point testcross involving three loci can reveal the order of the loci and the distance between the loci. e.g. (# of progeny) 252 243 250 248 recombinant frequency = 498/993 = 50% e.g. (# of progeny) 236 322 10 10 recombinant frequency = 20/578 = 3.5% = 3.5 m.u. Recombination between unlinked genes by independent assortment, producing recombinant frequency of 50%. Recombination between linked genes by crossing over, producing recombinant frequency of less than 50%. In general, recombinant frequency between two gene loci indicates the physical distance between the two gene loci. 1 genetic map unit (m.u.) or centimorgan (cM) = 1% recombinant frequency. Gene targeting by homologous recombination. A: insertion vector method. B: Replacement vector method. Strachan T and Read AP (1999) Human Molecular Genetics 2. BIOS Scientific Publishers Ltd. London. Reference Griffiths A.J.F. et al (2000) Introduction to Genetic Analysis. 7th ed. Freeman and Company, New York. Chapters 5,19. or Griffiths A.J.F. et al (2002) Modern Genetic Analysis: Integrating Genes and Genomes. 2nd ed. Freeman and Company, New York. Chapter 6. Watson J.D. (2004) Molecular Biology of the Gene. 5th ed. Pearson/Benjamin Cummings, San Francisco. Chapter 10. Leach D.R.F. (1996) Genetic Recombination. Blackwell Science, Oxford.
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The Petroleum Institute - PHYSICS - 191262
80. (a) Using Eq. 7-32, the work becomes W = 2 x2 x3 (SI units understood). The plot is shown below:9(b) We see from the graph that its peak value occurs at x = 3.00 m. This can be verified by taking the derivative of W and setting equal to zero, or sim
The Petroleum Institute - PHYSICS - 191262
51. (a) The objects displacement is d = d f di = (8.00 m) + (6.00 m) + (2.00 m) k . i j Thus, Eq. 7-8 gives W = F d = (3.00 N)(8.00 m) + (7.00 N)(6.00 m) + (7.00 N)(2.00 m) = 32.0 J. (b) The average power is given by Eq. 7-42: Pavg = W 32.0 = = 8.00 W. t
The Petroleum Institute - PHYSICS - 191262
52. According to the problem statement, the power of the car is P= dW d 1 2 dv = mv = mv = constant. dt dt 2 dtThe condition implies dt = mvdv / P , which can be integrated to giveT0dt =vT0mvdv PT=2 mvT 2Pwhere vT is the speed of the car at t =
The Petroleum Institute - PHYSICS - 191262
53. (a) We set up the ratio50 km E = 1 km 1 megaton and find E = 503 1 105 megatons of TNT.FG HIJ K1/ 3(b) We note that 15 kilotons is equivalent to 0.015 megatons. Dividing the result from part (a) by 0.013 yields about ten million bombs.
The Petroleum Institute - PHYSICS - 191262
54. (a) The compression of the spring is d = 0.12 m. The work done by the force of gravity (acting on the block) is, by Eq. 7-12, W1 = mgd = (0.25 kg) 9.8 m / s2 (0.12 m) = 0.29 J. (b) The work done by the spring is, by Eq. 7-26, 1 1 W2 = kd 2 = (250 N /
The Petroleum Institute - PHYSICS - 191262
55. One approach is to assume a path from ri to rf and do the line-integral accordingly. Another approach is to simply use Eq. 7-36, which we demonstrate: W=xf xiFx dx +yf yiFy dy =4 2(2x)dx +3 3(3) dywith SI units understood. Thus, we obtain W =
The Petroleum Institute - PHYSICS - 191262
56. (a) The force of the worker on the crate is constant, so the work it does is given by WF = F d = Fd cos , where F is the force, d is the displacement of the crate, and is the angle between the force and the displacement. Here F = 210 N, d = 3.0 m, and
The Petroleum Institute - PHYSICS - 191262
57. There is no acceleration, so the lifting force is equal to the weight of the object. We note that the persons pull F is equal (in magnitude) to the tension in the cord. (a) As indicated in the hint, tension contributes twice to the lifting of the cani
The Petroleum Institute - PHYSICS - 191262
58. With SI units understood, Eq. 7-8 leads to W = (4.0)(3.0) c(2.0) = 12 2c. (a) If W = 0, then c = 6.0 N. (b) If W = 17 J, then c = 2.5 N. (c) If W = 18 J, then c = 15 N.
The Petroleum Institute - PHYSICS - 191262
59. Using Eq. 7-8, we find W = F d = ( F cos sin (x + y = Fx cos + Fy sin i+F j) i j) where x = 2.0 m, y = 4.0 m, F = 10 N, and = 150 . Thus, we obtain W = 37 J. Note that the given mass value (2.0 kg) is not used in the computation.
The Petroleum Institute - PHYSICS - 191262
60. The acceleration is constant, so we may use the equations in Table 2-1. We choose the direction of motion as +x and note that the displacement is the same as the distance traveled, in this problem. We designate the force (assumed singular) along the x
The Petroleum Institute - PHYSICS - 191262
61. The total weight is (100)(660 N) = 6.60 104 N, and the words raises at constant speed imply zero acceleration, so the lift-force is equal to the total weight. Thus P = Fv = (6.60 104)(150 m/60.0 s) = 1.65 105 W.
The Petroleum Institute - PHYSICS - 191262
62. (a) The force F of the incline is a combination of normal and friction force which is serving to cancel the tendency of the box to fall downward (due to its 19.6 N weight). Thus, F = mg upward. In this part of the problem, the angle between the belt a
The Petroleum Institute - PHYSICS - 191262
63. (a) In 10 min the cart moves d = 6.0 so that Eq. 7-7 yields W = Fdcos = (40 lb)(5280 ft) cos 30 = 1.8 105 ft lb. (b) The average power is given by Eq. 7-42, and the conversion to horsepower (hp) can be found on the inside back cover. We note that 10 m
The Petroleum Institute - PHYSICS - 191262
64. Using Eq. 7-7, we have W = Fd cos = 1504 J . Then, by the work-kinetic energy theorem, we find the kinetic energy Kf = Ki + W = 0 + 1504 J. The answer is therefore 1 .5 kJ .
The Petroleum Institute - PHYSICS - 191262
65. (a) To hold the crate at equilibrium in the final situation, F must have the same magnitude as the horizontal component of the ropes tension T sin , where is the angle between the rope (in the final position) and vertical: = sin 1FG 4.00IJ = 19.5 .
The Petroleum Institute - PHYSICS - 191262
66. From Eq. 7-32, we see that the area in the graph is equivalent to the work done. We 1 find the area in terms of rectangular [length width] and triangular [ 2 base height] areas and use the work-kinetic energy theorem appropriately. The initial point i
The Petroleum Institute - PHYSICS - 191262
67. (a) Noting that the x component of the third force is F3x = (4.00 N)cos(60), we apply Eq. 7-8 to the problem: W = [5.00 N 1.00 N + (4.00 N)cos 60](0.20 m) = 1.20 J. (b) Eq. 7-10 (along with Eq. 7-1) then yields v = 2W/m = 1.10 m/s.
The Petroleum Institute - PHYSICS - 191262
68. (a) In the work-kinetic energy theorem, we include both the work due to an applied force Wa and work done by gravity Wg in order to find the latter quantity. K = Wa + Wg leading to Wg = 2.1 102 J . (b) The value of Wg obtained in part (a) still applie
The Petroleum Institute - PHYSICS - 191262
69. (a) Eq. 7-6 gives Wa = Fd = (209 N)(1.50 m) 314 J. (b) Eq. 7-12 leads to Wg = (25.0 kg)(9.80 m/s2)(1.50 m)cos(115) 155 J. (c) The angle between the normal force and the direction of motion remains 90 at all times, so the work it does is zero. (d) The
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70. After converting the speed to meters-per-second, we find K = 2 mv2 = 667 kJ.1
The Petroleum Institute - PHYSICS - 191262
71. (a) Hookes law and the work done by a spring is discussed in the chapter. Taking absolute values, and writing that law in terms of differences F and x , we analyze the first two pictures as follows: | F | = k | x| 240 N 110 N = k (60 mm 40 mm) which y
The Petroleum Institute - PHYSICS - 191262
72. (a) Using Eq. 7-8 and SI units, we find W = F d = (2 4 (8 + c = 16 4c i j) i j) which, if equal zero, implies c = 16/4 = 4 m. (b) If W > 0 then 16 > 4c, which implies c < 4 m. (c) If W < 0 then 16 < 4c, which implies c > 4 m.
The Petroleum Institute - PHYSICS - 191262
73. A convenient approach is provided by Eq. 7-48. P = F v = (1800 kg + 4500 kg)(9.8 m/s2)(3.80 m/s) = 235 kW. Note that we have set the applied force equal to the weight in order to maintain constant velocity (zero acceleration).
The Petroleum Institute - PHYSICS - 191262
74. (a) The component of the force of gravity exerted on the ice block (of mass m) along the incline is mg sin , where = sin 1 0.91 15 gives the angle of inclination for the . inclined plane. Since the ice block slides down with uniform velocity, the work
The Petroleum Institute - PHYSICS - 191262
75. (a) The plot of the function (with SI units understood) is shown below.Estimating the area under the curve allows for a range of answers. Estimates from 11 J to 14 J are typical. (b) Evaluating the work analytically (using Eq. 7-32), we have W=2 01
The Petroleum Institute - PHYSICS - 191262
76. (a) Eq. 7-10 (along with Eq. 7-1 and Eq. 7-7) leads to vf = (2 m F cos )1/2= (cos )1/2, where we have substituted F = 2.0 N, m = 4.0 kg and d = 1.0 m. (b) With vi = 1, those same steps lead to vf = (1 + cos )1/2. (c) Replacing with 180 , and still usi
The Petroleum Institute - PHYSICS - 191262
which (by taking two derivatives) we find the acceleration to be a = 0.20 m/s2. The (constant) force is therefore F = ma = 0.40 N, with a corresponding work given by W = 2 Fx = 50 t(t 10). It also follows from the x expression that vo = 1.0 m/s. This mean
The Petroleum Institute - PHYSICS - 191262
78. The problem indicates that SI units are understood, so the result (of Eq. 7-23) is in Joules. Done numerically, using features available on many modern calculators, the result is roughly 0.47 J. For the interested student it might be worthwhile to quo
The Petroleum Institute - PHYSICS - 191262
79. (a) To estimate the area under the curve between x = 1 m and x = 3 m (which should yield the value for the work done), one can try counting squares (or half-squares or thirds of squares) between the curve and the axis. Estimates between 5 J and 8 J ar
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Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 9 January 22, 2010Patrick Lam version 1So far, weve seen a number of coverage criteria for graphs, but Ive been vague about how to actually construct graphs. For the most part, its
Waterloo - CS - 447
Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 8 January 20, 2010Patrick Lam version 1Building on the notion of a def-clear path: Denition 1 A du-path with respect to v is a simple path that is def-clear with respect to v from a
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Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 7 January 18, 2010Patrick Lam version 2Input: Directed graph G Output: List of prime paths in G, primePaths nonextendablePaths ; primePaths ; worklist all paths of length 0, i.e. no
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Software Testing, Quality Assurance & Maintenance (ECE453/CS447/SE465): Assignment 3 v1Patrick LamDue: March 3, 2010You may discuss the assignment with others, but I expect each of you to do the assignment independently. I will follow UWs Policy 71 if
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/ d d d d d d d d d dWWWdddddddddd W WW//h/hWapplyFactorToTextSize, since there only exist one big nested if-statement instead of many if statements. The choice for paths are limited, thus to cover ADC, AUC and ADUPC are simple(because you cant ha
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Software Testing, Quality Assurance & Maintenance (ECE453/CS447/SE465): Assignment 2 SolutionsPatrick LamQuestion 2 (12 points)Predicate 1 a (b c) Determination analysis: a determines p i [true (b c)] [false (b c)] (b c) false (b c) i.e. b : false, c :
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Software Testing, Quality Assurance & Maintenance (ECE453/CS447/CS647/SE465): Midterm SolutionsFebruary 10, 2009This open-book midterm has 5 questions and 90 points. Answer the questions in your answer book. You may consult any printed material (books,
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Software Testing, Quality Assurance & Maintenance (ECE453/CS447/CS647/SE465): Midterm Practice QuestionsHere are a couple of midterm practice questions. These questions may be slightly ambiguous; Ive taken more care in drafting the actual midterm questio
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Software Testing, Quality Assurance & Maintenance (ECE453/CS447/SE465): Assignment 1 SolutionsPatrick LamIve excluded solutions for questions 2, 3 and 7. Ill try to make them available later, but they wont help you with midterm preparation anyway. The C
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Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 12 January 29, 2010Patrick Lam version 2Graph Coverage for SpecicationsWell move further up the abstraction chain now and talk about testing based on specications. Specication-base
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Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 11 January 27, 2010Patrick Lam version 1Data Flow Graph Coverage for Design ElementsThe structural coverage criteria for design elements were not very satisfying: basically we only
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Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 10 January 25, 2010Patrick Lam version 1Dataow Graph Coverage for Source CodeLast time, we saw how to construct graphs which summarized a control-ow graphs structure. Lets enrich o
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Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 6 January 15, 2010Patrick Lam version 3Prime Path Coverage versus Complete Path Coverage. n0 n1 n3 n2 Prime paths: path(t1 ) = path(t2 ) = T1 = cfw_t1 , t2 satises both PPC and CP
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Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 5 January 13, 2010Patrick Lam version 2Lets consider an example of a test set which satises node coverage on D, the double-diamond graph from last time. Start with a test case t1 ;
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Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 4 January 11, 2010Patrick Lam version 2Some binary distinctionsLets digress for a bit and dene some older terms which we wont use much in this course, but which we should discuss b
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Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 3 January 8, 2010Patrick Lam version 2SubsumptionSometimes one coverage criterion is strictly more powerful than another one: any test set that satises C1 might automatically satis
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Software Testing, Quality Assurance and MaintenanceWinter 2010Lecture 2 January 6, 2010Patrick Lam version 1RIP Fault ModelRecall that a fault is something thats latent or hiding, while a failure is visible (e.g. EPIC FAIL). To get from a fault to a
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Software Testing, Quality Assurance & MaintenanceLecture 1Patrick Lam University of WaterlooJanuary 4, 2010The Testing CourseCourse mechanicsTextbook: I will be (somewhat loosely) following Ammann and Offutt. Website: http:/patricklam.ca/stqam Grace
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The Impacts of Tourism Paper REC 280 Introduction to Tourism Percent of Final Grade: 25% Due Date: Wednesday March 17, 2010 (Hard-Copy Submissions Only)Assignment Description: Imagine that you will be traveling in the near future. Describe the place(s) t
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REC 280 Impacts of Tourism Rubric (W2010) CONTENT: Responses to Assignment QuestionsExcellent Discussed in detail all the aspects of their trip (transport, accommodations, environment, geography, culture and economy) Identification of motivation for tra
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Assignment 1 Test Questions Grading Weight 5% of final course grade Due at the beginning of class on Wednesday, February 3, 2010 Question Criteria: Create 5 multiple choice questions using your notes from class to guide your development of the questions.
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Basic Tips for MCQs (July 20, 2006)Basic Tips for Writing Effective Multiple Choice Questions (MCQs): A Compilation of the Most Useful AdviceJennifer Murdock, Department of Economics, U of T To define terms, here is a sample multiple choice question whe
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Ecotourism EcotourismEcological Tourism, Economic Ecological Tourism, or Both? Tourism,http:/www.youtube.com/watch?v=2DKqivqiu_ACommonly Accepted Definition of Ecotourism Definition Nature based tourism that involved the Nature interpretation, educati
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TheEconomicsofTourism TheEconomicsofTourism Takeaminuteandlist5positiveeconomicimpactsoftourismonacommunity(or person)and5negativeeconomicimpactsof tourismonacommunity(orperson)Introduction Introduction Whatistheunderlyinggoaloftourism? The19thcentur