53 Pages

# ch01

Course Number: ACCOUNTING 1234, Spring 2010

College/University: Abu Dhabi University

Word Count: 12555

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C HAPTER 1 ACCOUNTING IN ACTION SUMMARY OF QUESTIONS BY STUDY OBJECTIVES AND BLOOM’S TAXONOMY Item 1. 2. 3. 4. 5. 6. 7. 8. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. a 55. a 56. a 57. a 58. 59. 60. 61. 62. 63. 64. 65. 164. 165. 166. sg st a SO 1 1 1 2 2 2 2 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2 9 9 9 9 2 2 2 3 3 4 4 2 6 6 BT K K C K K C K C K K K C K K K K C C C C C C K K K C K K C K C K K C K K Item 9....

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Abu Dhabi University - ACCOUNTING - 1234
A PPENDIX CTIME VALUE OF MONEYSUMMARY OF QUESTIONS BY STUDY OBJECTIVES AND BLOOMS TAXONOMYItem 1. 2. 11. 12. 13. 14. 15. 16. 17. 18. 48. 49. 50. 62. SO 1 1 2 2 2 2 2 3 3 3 3 3 3 1 BT K K K C C AP AP AP AP C AP AP AP K Item 3. 4. 19. 20. 21. 22. 23. 24.
UCLA - LS - ls 4
1/5/10 50% 40% 30% 20% 10% 0%In this imaginary graph, red are batting averages for baseball players using a wooden bat, green are for players using a metal bat. Would you recommend that a new player use a wood or metal bat. 1. Wood 2. Metal 3. Dont know
UCLA - LS 4 - ls 4
Chromosomes - rst reported 1882 (after Mendel) # chromosomes/nucleus in a species is constant. human, 46; corn, 10; fruit y, 8 2 pairs of arms - members of a pair are equal in size longitudinal - 2 identical units, called chromatidsIn the fruit y8 chrom
UCLA - LS 4 - ls 4
Drosophila melanogaster Short life cycle - 12 days Many progeny Easy to grow in laboratoriesCalvin Bridges, an undergraduate dishwasher, discovered a white-eyed fly (normal flies have red eyes) in one of the fly bottles he was about to washKaryotypesca
UCLA - LS 4 - ls 4
1/13/10Sex Determination 1. Sex chromosomes humans; silkworms - presence of Y chromosome Drosophila; nematodes - X to autosome ratio 2. Ploidy (in bees, males are haploid, females diploid) 3. Environment - Bonellia; a sea worm. Larvae are sexually undiff
UCLA - LS 4 - ls 4
1/18/10Hemophilia X-linked disease1. 2. 3. 4.X-linked traits appear more frequently in males. Mutation and trait never pass from father to son. Affected male always passes mutation to daughters, who serve as carriers who pass the trait to 1/2 of their
UCLA - LS 4 - ls 4
1/20/10Hypothesis. You have two plants that are both heterozygous for a single gene displaying a simple dominant/recessive relationship. To test this hypothesis you cross the two plants together and count progeny. Imagine 4 different outcomes, which expe
UCLA - LS 4 - ls 4
1/25/10Tet resistance Auxotroph revertant Bacteriophage resistancePetri dish 5 bacteria 12 hours107 bacteria12 hours107 bacteria12 hoursE. Coli (107 tets and 7 tetr)Media + tet11/25/10107 bacteriaarg- E. coliComplete mediaMinimal media107 b
UCLA - LS 4 - ls 4
1/27/1011/27/10Hfr met+ thi+ pur+ strs X F- met- thi- pur- strr Interrupted matings indicate that met enters the recipient last. Plate progeny on media to select for met+ recombinants. What media would you plate the mated bacteria on? 1. minimal + meth
UCLA - LS 4 - ls 4
2/1/10ra- X rb- 94% large 6% small1Host Range Mutants h+ (wild type) can grow on E. colis but not on E. colir. h can grow on both E. colis and E. colir.212/1/10Recombination in phage +- +r h Xr hPhenotype Clear and small Cloudy and large Cloudy a
UCLA - LS 4 - ls 4
2/3/10Lwoff concludes that; 1. lysogenic bacteria contain a noninfective structure, the prophage. 2. the prophage can turn into an infective phage and that no new phage particle is required for this. 3. In a small percentage of lysogenic bacteria, the pr
UCLA - LS 4 - ls 4
2/8/10AI J KBCDEFGHWhich rII deletion can recombine with which rII point mutation to give WT? Which rII deletion can recombine with which rII deletion to give WT?123456 789+ + +The 5 deletion mutations are used to map a point mutation.A
UCLA - LS 4 - ls 4
2/10/10Fig. 8.3DNARNAPROTEINTRANSCRIPTION RNA differs from DNA: 1. Single stranded 2. Ribose sugar 3. U instead of a TAGGCTAGCTAG TCCGATCGATCAGGCUAGCUAGTranscription goes 5 to 3.3 MAJOR TYPES OF RNA. 1. mRNA: code for protein 2. rRNA: ribosomes 3
UCLA - LS 4 - ls 4
2/14/10UUAUGUCUACUUUAGUUCCUCAUCGUGGGGCAUAGUAC MetSerThrLeuValProHisArgGlyAlaSTOP Missense Mutation UUAUGUCUACUUUAGUUACUCAUCGUGGGGCAUAGUAC MetSerThrLeuValThrHisArgGlyAlaSTOP UUAUGUCUACUUUAGUUCCUCAUCGUCGGGCAUAGUAC MetSerThrLeuValProHisArgArgAlaSTOP Nonsens
UCLA - LS 4 - ls 4
2/17/10IPTG: -galactoside isopropyl-d-thiogalactosideCan bind to Lac repressor =&gt; can induce lac operon Can not be cleaved by -galactosidase =&gt; constant concentration of inducer Commonly used inducer in the lab.-gal Protein Bacterial Protein12/17/10
UCLA - LS 4 - ls 4
Restric(on/Modica(on100%100%K12(P1) 0.002%K12100%K12(P1)encodesenzymesthatmethylateDNAandcutnon methylatedDNA. HOSTCONTROLLEDRESTRICTIONMODIFICATIONOF.CH3cytosine5methylcytosineRestriction EndonucleasesOver 200 restriction enzymes have been iso
UCLA - LS 4 - ls 4
MstII site is CCTNAGGprobeIn this case, the RFLP does not cause the D mutation, but is linked to the D locus.Hemophilia - X-linked diseaseWhat are the genotypes of the family members?Hemophilia - X-linked diseaseH YH hH -H Yh YH -7 kb 5 kbSou
UCLA - LS 4 - ls 4
REVERSE GENETICSTRANSGENIC MICEGrowth Hormone- made in small amounts in specialized cells located in the anterior pituitary gland. The following constructs were injected: metallothionein promoter-rat growth hormone. M.P. RGH protein rat growth hormone p
UCLA - LS 4 - ls 4
MN blood group frequencies MM MN NN Total 119 76 13 208 p(M) = [(119 + 119 + 76]/ 2 x 208 = .755 q(N) = [(13 + 13 + 76]/2 x 208 = .245 If in HWE expect: MM p2 (.755)2 x (208) = 118.6 MN 2pq 2(.755)(.245) x (208) = 76.9 NN q2 (.245)2 x (208) = 12.5X-linke
UCLA - LS 4 - ls 4
UCLA - LS 4 - ls 4
CancerCancerDeaths Es-matednumberofnewcancercasesanddeathsintheUnitedStatesin1997.Cancer: A cell growth disorder resulting from an alteration in genes (mutations) that lead to deregulated (transformed) cell growth. CANCER IS A GENETIC DISEASE Transform
UCLA - LS 4 - ls 4
1 Question 1. Note that the female is the heterogametic sex.22. PedigreeThe following pedigree concerns a very rare human disorder that causes premature puberty in males.1 24567891011121314151617Which mode(s) of inheritance can explain
UCLA - LS 4 - ls 4
1LS 4 Example Midterm 1 questions INTRODUCTORY GENETICS21. Mitosis/MeiosisWe consider an organism with three pairs of chromosomes: the sex chromosomes X and Y, and the pairs of autosomal chromosomes 1 and 2. The female is the heterogametic sex. The st
UCLA - LS 4 - ls 4
Name_LS 4 INTRODUCTORY GENETICS MIDTERM II practice problemsName_ 1 The following cross is done in the Pea plant. P F1 AAbb x AAbb AaBbThese F1 progeny are then testcrossed as follows: AaBb x aabb The A and B alleles are dominant and the a and b allele
UCLA - LS 4 - ls 4
Name_LS 4 INTRODUCTORY GENETICS MIDTERM II practice problemsName_ 1 The following cross is done in the Pea plant. P F1 AAbb x AAbb AaBbThese F1 progeny are then testcrossed as follows: AaBb x aabb The A and B alleles are dominant and the a and b allele
UCLA - LS 4 - ls 4
Q. 1Correct answers in bold . coli ( wild type) . coli ( ts cI) . coli ( ts cII) . coli ( ts N) . coli ( ts cro) . coli ( ts cI, ts cII) . coli ( ts cI, ts N) . coli ( ts cII; ts N) grow at 40C lyse at 40C grow at 40C grow at 40C grow at 40C lyse at 40C
UCLA - LS 4 - ls 4
NAMELS 4 INTRODUCTORY GENETICS MIDTERM III Practice ProblemsNAMEQ. 1 (14 points)Below are listed E. coli strains that are lysogenic for bacteriophage . Each strain has a prophage that contains a ts mutation (or two ts mutations). All strains of E. col
U. Houston - CHEE - 2331
CHEE 2331 Chemical Processes Energy and Energy Balances Chapter 7, pages 313-333Department of Chemical and Biomolecular Engineering EngineeringWhat is energy?The ability to do workEnergy units (F-L)Mechanical Metric Engineering S.I. erg ft-lbf J Ther
U. Houston - BIOL - 1362
1QuestionsonCH1720ToBeCoveredDuringLectureorinExam2Review1. Carryoutthefollowingtranscriptionandtranslationproblem: 2. 3. 4. Puttheeventsthatentaileukaryotictranscriptionandtranslationinchronologicalorder: A)PremRNAisprocessedandspecificsequencesareremo
U. Houston - BIOL - 1362
1B ? 1ASamespeciesordifferentspecies?2A ? 2BSpeciationPopulationof aspeciesMicroevolutionary ForcesPopulationevolves; adaptation Howdo SpeciesForm? Speciation(processofaspeciessplittinginto2more species)bridgesmicroevolutionwithmacroevolution. Bio
U. Houston - BIOL - 1362
Microevolution Microevolution: heritablegeneticchangesthatoccurina population. Naturalselectionactsonindividuals,butpopulationsarethesmallest unitsthatcanevolve. Instudyingmicroevolution,oneanalyzesvariationinnatural populations&amp;determineshow/whytheseva
U. Houston - BIOL - 1362
Onymacris unguicularis, the head standing beetle Namibia Atlantic ocean 1of350,000beetlespecies Hastypicalbeetletraits:pairofwings;6legs;hardexterior Possessesitsownuniquebehavior Priorexample illustratesthefeaturesofalllife:sharedtraits; variation;feat
U. Houston - BIOL - 1362
U. Houston - BIOL - 1362
AnswerstoLectureQuestionsfromWeekof12510 I.Whataretheproductsofmitosis? 1. Howmanycellsareproducedattheendofasinglemitoticdivision? Twocellsareproducedattheendofasinglemitoticdivision. 2. Howmanydifferentkindsofcellsareproducedattheendofasinglemitoticdiv
U. Houston - BIOL - 1362
PhylogenyAsnakeorlizardspecies? Understandingofevolutionaryrelationships(phylogeny)allows onetocomparetraitsofrelated speciestoanswersuchquestions. Constructingphylogeniesoccurs viasystematics classifying organisms&amp;determining evolutionaryrelationships.
U. Houston - BIOL - 1362
BIOTECHNOLOGY PROBLEM1Using the following mammalian DNA sequence and the restriction enzyme, Taq I, find the restriction sites and indicate where the cuts occur. Taq I : 5 T*CGA 3 3 A GC*T 5ABCDECT*CGATTAT*CGAACCTTCAT*CGATT*CGAGCCTTCG GAGC*TAATAGC
U. Houston - BIOL - 1362
U. Houston - BIOL - 1362
Protein Synthesis WorksheetDetermine the amino acid sequence for a polypeptide coded for by the following mRNA transcript:3-T A G G C T A C G G A C T G A A A T T C A T C T T A G A A A T - 5 5-A U C C G A U G C C U G A C U U U A A G U A G A A U C U U U A
U. Houston - BIOL - 1362
AnswerstoRemainingQuestionsfromdocumentQuestionsonCH1720ToBe CoveredDuringLecture 14. Ifaresearcherwaslookingtoclonethegeneforaspecificliverenzyme,itwouldbemostefficient andlogicaltofirstsearchin: A. alivercellandpickitoutwithapairoftweezers. B. ahumanli
U. Houston - BIOL - 1362
1)OnecaninferfromtheCentralDogma,thatdifferentRNAmoleculesarenecessaryinordertosynthesizeDNA fromasequenceofaminoacids(i.e.,protein). A)True B)False Answer:B 2)Inagene,thepresenceofapromoterisrequired A)fortranscriptiontooccur. B)fortranslationtooccur. C)
U. Houston - BIOL - 1362
MboI (2 sites) *GATCThe MboI enzyme will not cut at the ends even though the *GATC is present at both ends. The MboI enzyme cannot properly bind when the target sequence is at these locations. I would not expect you to know this and I would not give you
U. Houston - BIOL - 1362
U. Houston - BIOL - 1362
1Answer Key: Replication, Transcription, Translation WorksheetExercise A. DNA structure You were given this DNA sequence: 3 TACGGCATT 5; the complete double stranded sequence would look like this (use complimentary base pairing, remember to label ends c
U. Houston - BIOL - 1362
EXAM 1 REVIEW (CH 12 16; Sadava Text: CH 9-11) ALSO: Be familiar with Key Terms for CH 12-16 and go over document titled CH 12-16 Questions to Think About CHAPTER 12 (The Cell Cycle) 1) Cell division &amp; its roles in different organisms; Prokaryote v. Eukar
U. Houston - BIOL - 1362
U. Houston - BIOL - 1362
U. Houston - BIOL - 1362
BASIC QUESTIONS: GENE EXPRESSIONAnswers are not posted to questions 1-19. This is intended more as a study tool; as you review your notes, write out the answers to these questions below. 1. How does information flow in a cell ? 2. What contribution did B
U. Houston - BIOL - 1362
U. Houston - BIOL - 1362
BIOTECHNOLOGY PROBLEM1Using the following mammalian DNA sequence and the restriction enzyme, Taq I, find the restriction sites and indicate where the cuts occur. Taq I : 5 T*CGA 3 3 A GC*T 55 3 CT CGATTATCGAACCTTCAT CGATTCGAGCCTTCG GAGCTAATAGCTTGGAAGTA
U. Houston - BIOL - 1362
U. Houston - BIOL - 1362
1 Assume that a bacterial plasmid carries the gene for resistance to the antibiotic kanamycin. Using the restriction enzyme A, you open the plasmid and insert a segment of a biologically important gene isolated from a mouse. The gene was excised from the
U. Houston - BIOL - 1362
ProkaryoticGenes&amp;RegulationReCapRegulatory genePromoterOperatorABCRepressorproteinBindstooperatorblocking transcription.Proteins(enzymes)of pathwayOperon:Type Functionofgenes inoperon RepressorActive (notranscription) RepressorInactive (transcr
U. Houston - BIOL - 1362
U. Houston - BIOL - 1362
U. Houston - BIOL - 1362
1362Exam1Distribution&amp;Curve300 250 200 150 100 50 0 &lt;30 30to 40to 50to 60to 70to 80to 90to 100+ 49 79 39 59 69 89 99AVG=69MED =70 &gt;86.45=A &gt;74.49=B &gt;62.52=C &gt;50.56=D
U. Houston - BIOL - 1362
The following is an exam given in 2006; it is intended to be an example of the type &amp; style of questions given on an exam. Using this as your primary study material for the Spring 2009 exam 1 would be a mistake. BIOL 1362 Exam 1 from Spring 2006 CH 11 1.
U. Houston - BIOL - 1362
Key Terms: CH 22 26 (Sadava text: CH 21-23 &amp; 25) Evolution Natural theology Scala naturae Theory of Use &amp; Dis-use Adaptation Population Species Artificial selection Decent with modification Natural selection Differential reproductive success Heritable var
U. Houston - BIOL - 1362
Name _Mitosis and MeiosisBefore you begin, save this Lab Report Template on your computer as LastNameAPBIOMitosis For this lab you will need to access the Mitosis Virtual Microscope (http:/www.jdenuno.com/PDFfiles/Mitosis.pdf) and the Meiosis Virtual Mi
U. Houston - BIOL - 1362
1Population Genetics 1. What are the genotypic frequencies in a population composed of the following 312 AA, 361 Aa, 242 aa individuals? Total # of individuals in population = 312 + 361 + 242 = 915 Frequency of AA = 312/915 = 0.34 Frequency of Aa = 361/9
U. Houston - BIOL - 1362
1 UNIT THREE: GENETICS C hapter Chapter Sixteen: The Molecular Basis of Inheritance (Text from Biology, 6th Edition, by Campbell and Reece)The Molecular Basis of Inheritance (Chapter Sixteen)DNA AS THE GENETIC MATERIAL The The Search for the Genetic Mat
U. Houston - BIOL - 1362
1 UNIT THREE: GENETICS Chapter Fifteen: The Chromosomal Basis of Inheritance C hapter (Text from Biology, 6th Edition, by Campbell and Reece)The Chromosomal Basis of Inheritance (Chapter Fifteen)RELATING MENDELISM TO CHROMOSOMES Mendelian Mendelian Inhe