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DNA polymorphisms

Course: BIOC BIOC2603, Spring 2007
School: HKU
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Principles BIOC2603 of Molecular Genetics DNA polymorphisms Brian Wong bcwwong@hkucc.hku.hk Genetic variations in human Compare genomic DNA of any two individuals: 99.9% identical (approximately 1 difference per 1 kb) Size of human genome is 3 x 109 bp so 0.1% difference corresponds to about 3 x 106 bp difference These genetic differences result in individual differences in disease susceptibility, responses to...

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Principles BIOC2603 of Molecular Genetics DNA polymorphisms Brian Wong bcwwong@hkucc.hku.hk Genetic variations in human Compare genomic DNA of any two individuals: 99.9% identical (approximately 1 difference per 1 kb) Size of human genome is 3 x 109 bp so 0.1% difference corresponds to about 3 x 106 bp difference These genetic differences result in individual differences in disease susceptibility, responses to drug, etc Finding genetic variations associated with a disease phenotype can help to locate genes responsible for the disease and these variations can also be used for diagnostic purposes Genetic variations can also be used for personal identification and kinship testing Genetic variations resulting in individual differences CYP1A2 in liver is responsible for metabolizing caffeine. People with a certain polymorphism in this gene result in a slow metabolizer phenotype and are more likely to suffer heart attack. Those with a fast metabolizer polymorphism have a lower risk. Cornelis MC et al (2006) JAMA 295:11351141 Discontinuous and continuous variations Caused by variations in a single gene Caused by variations in many genes, and interactions between genes and the environment. From: Griffiths AJF et al (1999) Modern genetics analysis. Freeman. Polymorphisms Locus: a defined position in the genome, can be a gene, a stretch of DNA, a restriction enzyme recognition site, or a single nucleotide Alleles: alternative forms of the same gene/locus, caused by mutations Mutation: a change in the gene sequence. Occurs at a low frequency in a population. A locus is polymorphic if the least frequent allele occurs at 1% or greater in a population Polymorphisms Phenotypes: color, morphology, etc Protein polymorphism (allozymes) Differences in size and charge because of amino acid substitutions Identified using native gel electrophoresis Other hemoglobin variants Other hemoglobin variants From: Griffiths AJF et al (1999) Modern genetics analysis. Freeman. Polymorphisms DNA polymorphisms Single-nucleotide polymorphism (SNP) Variable number of tandem repeats (VNTR) polymorphism: Minisatellite polymorphism: total size 0.5 - 30kb. The unit size of the repeats is variable (15-100bp) but most share a common core sequence (GGGCAGGANG). Minisatellites are also found near telomeres (10-15kb of TTAGGG repeats). Microsatellite polymorphism: the unit size of the repeats is short (e.g. 1 to 6 bp). Also known as short tandem repeats polymorphism or simple sequence repeat polymorphism. Both SNP and VNTR polymorphism can give rise to restriction fragment length polymorphism (RFLP) Restriction fragment length polymorphism Variations in the size of DNA fragments produced after restriction enzyme digestion. Due to: Single nucleotide polymorphism: single base change abolishes or creates a restriction enzyme recognition site VNTR: a region containing variable number of tandem repeats flanked by two restriction enzyme recognition sites. To test for RFLP, DNA samples are digested with an appropriate restriction enzyme. The DNA fragments produced are separated by gel electrophoresis and transferred to a nylon membrane (Southern blotting). The fragment length polymorphism is detected by hybridization using a DNA probe near the polymorphic site. RFLP due to SNP detected by Southern blot hybridization and PCR 300 bp 100 bp P1, P2 = primers for PCR * = polymorphic site Allele A Allele B Southern Blots PCR products Gelehrter TD et al (1998) Principles of Medical Genetics. 2nd ed. RFLP due to variable number of tandem repeat polymorphism detected by Southern blot hybridization tandem repeat Probe 1 hybridizes to an unique DNA sequence outside the tandem repeats (locus-specific probe): this detects polymorphisms in a single locus. Probe 2 hybridizes to the tandem repeats. This detects polymorphisms in a multiple loci. Gelehrter TD et al (1998) Principles of Medical Genetics. 2nd ed. Single nucleotide polymorphism (SNP) Most abundant form of variations in human, occurs about 1 in 1000 bp Single base change Transition (pyrimidine pyrimidine; purine purine) C T; G A Transversion (pyrimidine purine) C A; C G; T A; T G Most are biallelic polymorphism; tri- and tetra-allelic SNPs are very rare. May result in creation or abolishment of restriction enzyme recognition site, give rise to restriction fragment length polymorphism Single nucleotide polymorphism (SNP) SNPs can be found in Coding regions (cSNPs, very few are found) Synonymous: different SNP alleles code for the same amino acid Non-synonymous: different SNP alleles code for different amino acid Non-coding regions (non-coding SNPs) Promoters 5 or 3 untranslated regions Splice sites Introns Intergenic region Shastry BS (2002) SNP alleles in human disease and evolution. J Hum Genet 47:561566 Microsatellite polymorphism Most human microsatellite loci are identified from randomly cloned genomic DNA fragments: Cloned genomic fragments that hybridized to labeled simple repetitive sequence (e.g. CATTCATTCATT) are sequenced to determine the unique DNA sequences flanking the simple repeats. These unique flanking sequences are used as primers for the polymerase chain reaction (PCR) to detect the microsatellite locus. Variation in the number of simple sequence repeats will appear as polymorphism in the sizes of the PCR products after gel electrophoresis. Microsatellites are highly polymorphic, some can have nine or more alleles. 9 repeats 8 repeats 6 repeats 5 repeats Microsatellite in the tyrosine hydroxylase (TH) gene on human chromosome 11p15.5. PCR using primers flanking the CATT repeats amplified alleles whose sizes differ by multiples of 4 bp. (9/5) (9/6) (5/6) (9/9) (9/5) (9/8) (5/6) From: Hearne CM et al (1992) Microsatellites for linkage analysis of genetic traits. Trends in Genetics 8:288294 Sources of DNA variation in a population Mutation Chemical modifications in DNA Errors in DNA replication and repair Strand slippage due to mispairing mediated by direct repeats in DNA sequences during replication give rise to VNTR polymorphism Recombination Independent assortment and crossing-over Unequal crossing-over at nonallelic site of sister or nonsister chromatids due to misalignment of nonallelic repetitive DNA may also give rise to VNTR Migration Introduction of genes from one population into another Normal replication Backward slippage causes insertion Forward slippage causes deletion Strand slippage during replication causes variation in number of short tandem repeats. From: Strachan T and Read AP (1999) Human Molecular Genetics 2. Fig 9.5 Generation of minisatellite polymorphisms: (A) Unequal crossing-over and (B) unequal sister chromatid exchange result in duplication or deletion of repeats. From: Strachan T and Read AP (1999) Human Molecular Genetics 2. Fig 9.7 How is polymorphism maintained? Neutral theory of evolution: most mutations are neutral (no advantageous nor disadvantageous phenotypic effects) and the allele frequency changed by genetic drift Balancing selection: mutation creates variations and natural selection favored different alleles at different times, in different places, in different combinations heterozygote advantage individual heterozygous for a locus is more favourable e.g. in sickle-cell anaemia, heterozygous individuals are resistant to malaria while individuals homozygous for the locus are susceptible to malaria Applications of DNA polymorphisms Serve as markers Population genetics and evolutionary studies. E.g. inferring demographic history Molecular diagnosis of diseases Mapping e.g. genetic maps of the human genome e.g. identification of disease susceptibility genes DNA fingerprinting e.g. forensics and paternity tests DNA sequence variations used in phylogenetic studies Mitochondrial DNA from 182 individuals of diverse geographic origin, including 121 native Africans, Europeans, Papua New Guineans, Asians, a native Australian, African-Americans were compared. There was a greater sequence variations among Africans than among Asian or Europeans. Since mutations accumulated over time, more mutations mean longer time to evolve variations so concluded that modern human originated in Africa. Wilson AC and Cann RL (1992) Scientific American 266(4):22-27. DNA polymorphisms and personalized medicine Identify polymorphisms (e.g. SNP) that correlate with drug response custom design a drug treatment plan for a person based on the predicted drug response Identify polymorphisms (e.g. SNP) that associated with a disease develop screening method to identify individuals susceptible to the help disease to find gene responsible for the disease different alleles of the SNP loci Roses A.D. (2000) Pharmacogenetics and the practice of medicine. Nature 405:857865 Polymorphism and drug response CYP450: CYP2D6 and CYP2C19 genes are important in the metabolism of ~25% of all prescription drugs. based on genotypes of these genes, patients can be classified as poor, intermediate, extensive, or ultra rapid metabolizer knowing the genotypes of a person at these genes may help physicians in individualizing treatment selection and dosing for drugs metabolized through these genes, e.g. antidepressants, antipsychotics, beta-blockers, and some chemotherapy drugs diagnostic test available for genotyping of CYP2D6 and CYP2C19 Use of DNA polymorphism in disease diagnosis The restriction site for MstII is destroyed in S globin allele The restriction site for MstII is present in the normal A globin allele RFLPs as diagnostic markers in diseases: The sickle cell mutation destroys an Mst II site and generates a disease-specific RFLP. Uses of DNA polymorphisms in gene mapping Serve as markers that correlate with phenotypic variations (e.g. drug response, disease susceptibility) because of its proximity to the actual underlying genetic factor, i.e. the genetic marker is close to the genetic factor such that there is little recombination between them DNA polymorphism can be used to detect linkage. A,B,C: alleles of a polymorphic marker normal disease DNA polymorphism can be used to detect linkage. A,B,C: alleles of a polymorphic marker normal disease No correspondence between the inheritance of alleles and inheritance of the disease no linkage between marker and disease gene. They segregate independently. H: dominant disease allele h: normal allele Correspondence between the inheritance of alleles and inheritance of the disease those inherited allele C always inherited the disease. The marker and disease gene are linked. Pierce BA (2005) Genetics 2nd ed. A microsatellite locus can show linkage to a disease gene M: M: M: M: 3 repeats 5 repeats 2 repeats 8 repeats Griffiths et al (2005) Introduction to Genetic analysis 8th ed. A microsatellite locus can show linkage to a disease gene 8 repeats 5 repeats 3 repeats 2 repeats All affected individuals have M allele. M allele is linked to the disease allele P. An RFLP linked to a disease gene Griffiths et al (2005) Introduction to Genetic analysis 8th ed. An RFLP linked to a disease gene The normal allele, d, segregates with RFLP allele 2. The disease allele D, segregates with RFLP allele 1 (except in child 8). So the gene responsible for the disease is on the same chromosome as the RFLP marker. D: dominant disease allele normal disease Griffiths et al (2005) Introduction to Genetic analysis 8th ed. Population association studies Linkage disequilibrium (LD): non-random association of alleles of two adjacent loci, i.e. two loci do not segregate independently. LD is high when two loci are physically close to each other (5-20kb). LD decreases as distance between 2 loci increases. Haplotype Haplotype = haploid genotype Haplotype refers to a set alleles on the same chromosomal region. Each individual has two haplotypes for any given genomic region, one from paternal chromosome, one from maternal chromosome. SNP represents a single nucleotide variation. Haplotype represents a much longer sequence of nucleotide that tends to be inherited together. Individual 1 Individual 2 Individual 3 Individual 4 (http://www.hapmap.org/) Linkage disequilibrium Ancestor Current generation A1 to J1 are alleles of genetic loci A to J (can be SNP, RFLP, mini-satellite, etc) surrounding a gene locus (N) in an ancestor. Different ancestors may have different combinations of alleles. The normal gene (N) was mutated to a disease gene (D) in this ancestor. In subsequent generations, recombination (crossing-over) between homologous chromosomes lead to exchange of alleles such that D was linked to alleles other than those that originally presented. After many generations, only the alleles of the closest loci that surround the disease gene are retained. The disease gene is in linkage disequilibrium with F1 and E1. To find the disease gene in the current generation, we identify alleles of marker loci (e.g. F1 and E1) that are more commonly found in persons with the disease phenotype. The disease gene should be close to, i.e. in linkage disequilibrium with these marker loci. The chromosome around the marker loci are then screened for genes. Using haplotypes to deduce gene position SNPs 4 and 5 are in linkage disequilibrium with the disease gene patients tend to have SNP 4 and 5. The disease gene is around SNP 4 and 5. Principle of multi-locus DNA fingerprinting. A, B, C are minisatellite loci sharing a similar core sequence in the tandem repeats (e.g. 1kb in length). A, B, and C can be on the same or different chromosomes and each locus have different number of repeats on each homologous chromosome. DNA is digested with an restriction enzyme that does not cut within the repeat region. This generate a set of DNA fragments with different sizes corresponding to variation in the number of repeats. The DNA fragments are analyzed by Southern Blotting using a probe that hybridizes to the repeat regions. The probe therefore identifies polymorphism at many loci at a time. The resulting hybridization pattern is known as DNA fingerprint. Use of DNA fingerprinting (A): paternity test M (mother), C (Child) F1, F2 (two possible fathers) Child has minisatellite repeat lengths inherited from either the mother or F1 forensics The minisatellite repeat lengths in the specimen match those of suspect 1 (B): Note: current DNA fingerprinting methods use multiple singlelocus probes (e.g. short tandem repeat polymorphisms). From: Strachan T and Read AP (1999) Human Molecular Genetics 2. Fig 17.19 Use of DNA polymorphism in forensic investigation Background: Tsar Nicholas II, Tsarina Alexandra (wife), and 4 daughters, 1 son, three servants, 1 doctor were killed on July 16, 1918 during the Russian revolution bodies were buried together in July 1991, Russian government recovered 1000 bone fragments from the burial site and assembled into 5 females and 4 males believed to be the remains of the Tsar family. analysis of the amelogenin gene allowed the determination of sex of the bones analysis of five short tandem repeat polymorphic loci in the nine skeletons suggested that 5 skeletons belonged to a single family (Tsar, Tsarina, and three children) how to confirm whether the remains were those of the Russian royal family? Use of DNA polymorphism in forensic investigation Extract mitochondrial DNA (mtDNA) from bone recovered from burial site Extract mitochondrial DNA from Prince Philip who is a maternal relative (grand nephew) of Tsarina Alexandra Comparison of the mtDNA sequences reveal the sibling status of the children and the identification of the mother The mtDNA sequence of the putative Tsar was compared to that of two relatives of maternal descent The two relatives have the same mtDNA sequence of the putative Tsar, except at position 16169: Tsar is heteroplasmic (C/T) while the relative is homoplasmic (T) Could the bone really be the remains of Tsar Nicholas II? ALEXEI Lineage of Tsarina Alexandra, showing relationship to Prince Philip Lineage of Tsar Nicholas II, showing relationship to two maternal relatives tested (from Gill P. et al (1994) Nature Genetics 6: 130135) To further confirm heteroplasmy in the Tsar's lineage, the mtDNA was extracted from the bone of Georgij Romanov, Tsar's brother. The mtDNA sequence of Georgij Romanov matched that of the putative Tsar and was also heterplasmic (C/T) at position 16169. This confirmed heteroplasmy in Tsar's lineage. The mother of Tsar Nicholas must also be heteroplasmic, and that the heteroplasmy segregated to homoplasmy in later generations. 72% C, 28% T 38% C, 62% T Lineage of Tsar Nicholas II, black symbols indicate maternal lineage Sequence of mtDNA around position 16169 from bones of putative Tsar Nicholas II, the bone of Georgij Romanov and blood of Xenia Cheremeteff-Stfiri From: Ivanov P.L. et al (1996) Nature Genetics 12:417420 References Griffiths AJF et al (2000) Introduction to Genetic Analysis. 7th ed. Freeman and Company, New York. Chapters 5, 13, and 14 or Griffiths AJF et al (1999) Modern Genetic Analysis. Freeman and Company, New York. Chapters 11 and 12. Hartwell LH (2004) Genetics: From Genes to Genomes. 2nd ed. McGraw-Hill. Chapter 11. Strachan T and Read AP (2004) Human Molecular Genetics 3. Garland Science. Chapter 7.
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HKU - BIOC - BIOC2603
Genetic Engineering and Applications (BIOC 2603)Dr. Danny Chan Department of BiochemistryEmail: chand@hkusua.hku.hk Phone: 28199842Topics in year 1 Restriction enzymes. DNA modifying enzymes. Cloning using plasmids as vectors. cDNA library. Screening
HKU - BIOC - BIOC2603
BIOC2603 (2007) BIOC2603 (2007)Genomes and GenomicsBrian Wong bcwwong@hkucc.hku.hkReferencesBrown TA (2007) Genomes 3rd ed. Chapters 7-8 TA (2007) Genomes 3rd ed Chapters (see http:/www.ncbi.nlm.nih.gov/books/ for 2nd ed.) International Human Genome S
HKU - BIOC - BIOC2603
BIOC2603 Principles of Molecular GeneticsRecombination Brian Wong bcwwong@hkucc.hku.hkGenetic recombination: reassortment of genetic material(A) Independent assortment (for genes on different chromosomes) (B) crossing over (for genes on the same chromo
HKU - BIOC - BIOC2603
BIOC2603 Principles of Molecular GeneticsRecombinationBrian Wong bcwwong@hkucc.hku.hkPart IISite-specific recombinationSite-specific recombination Reciprocal recombination between two short specific DNA sequences. No net synthesis or deletion of DNA
HKU - BIOC - BIOC2603
BIOC2603 Principles of Molecular GeneticsSplicing and RNA processing RNA editing Gene evolution: exons and introns Dr. Mai Har Sham Department of BiochemistryAdvanced ReferencesWatson et al. (2004) Molecular Biology of the Gene. 5th Ed. Pearson/ Benjam
Skyline College - PHYS - 121
12.) Which of the following are permissible amounts of charge for an object to have? Select all that are correct (and none that are incorrect) to receive credit. (You should assume the elementary charge value is exactly e=1.610-19 C.)A neutral object is
USC - BISC - 220
" \ r toN.t "I.:l 2 S pring 010 bisc220sileaders@gmail.com* " -l'3ls l t oSarah,SeanandYuree www.usc.edu/siBISC 220 SI- Exam #2 ReviewCHAPTER 8: ENERGY AND METABOLISM p 1.T he m olecule ictured n t he r ight i s A T F o ATP + H2O-+ DP + Ir A The bre
Abraham Baldwin Agricultural College - ACC - 11101
CHAPTER 7 DEDUCTIONS AND LOSSES: CERTAIN BUSINESS EXPENSES AND LOSSES SOLUTIONS TO PROBLEM MATERIALSQuestion/ Problem 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24Topic Bad debts: cash basis taxpayer Bad debts: business versus nonbusine
TAMU Kingsville - MGMT - 5337
Statistics Data Paper
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
The Petroleum Institute - PHYSICS - 191262
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
National University of Singapore - CVE - 3132
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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
Waterloo - CS - 447
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
Waterloo - CS - 447
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
Waterloo - CS - 447
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
Waterloo - CS - 447
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 ;
Waterloo - CS - 447
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
Waterloo - CS - 447
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
Waterloo - CS - 447
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
Waterloo - CS - 447
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
Waterloo - REC - 280
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
Waterloo - REC - 280
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.