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Course: BISC 320, Fall 2010
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RNA 10/29/10 Transcription: synthesis Prokaryotes - transcription initiation o Elongation of mRNA o Termination of transcription - eukaryotes - - - - the coding part of the gene that expresses proteins through transcription and translation (figure 1) o can be turned on or off o transcription initiation: (more RNA polymerase that attaches at a section) RNA polymerase travels on 1 strand o There are about 5000...

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RNA 10/29/10 Transcription: synthesis Prokaryotes - transcription initiation o Elongation of mRNA o Termination of transcription - eukaryotes - - - - the coding part of the gene that expresses proteins through transcription and translation (figure 1) o can be turned on or off o transcription initiation: (more RNA polymerase that attaches at a section) RNA polymerase travels on 1 strand o There are about 5000 core RNA polymerases o 1 comes on the start, goes along, and creates RNA (elongates) o ribosomes can translate at the same time that mRNA is begin made Putting together the ribosome in prokaryotes: Figure 2 o Shine Dalgarno sequence: about 6 bases: where the 30s ribosome loads on for translation, and loads and goes upstream until the AUG start codon, where a 50S is loaded This creates a 70S full ribosome Then, read codons to add amino acids o In bacteria, transcription and translation happens simultaneously Short half-life time: 3 minutes, if you stop transcription/translation This allows bacteria to adjust to new environments quickly Hows transcription initiated in prokaryotes? (figure 4) o There are about 5000 core polymerases (mostly studied in e. coli gram negative bacteria) o TATA is about 25 base pairs in length o Alpha alpha beta beta omega are subunits of core polymerases o Attaches o Sigma 70 helps locate where the TATA region is (the only transcription initiation factor) o After core starts to slide downstream o CTD tails stick off alpha, attach to UP sequence o during elongation, BPing is opening: a bubble about 14 base pairs in length of SS DNA as polymerase copies o Beta subunit is where Rifampcin attaches and freezes transcription in bacteria (transcription inhibitor) o omega is necessary to restore denatured RNA polymerase in vitro to its fully functional form - Methods of Termination: figure 6 o Rho independent transcription termination the RNA snapback handle of a tennis racket is G+C rich and the 3 end of the mRNA has 6 or >Us, hydrogen bonded to the template DNA. As soon as base pairing happens, tennis racquet handle has G-C Where RNA still made, U rich (about 6) Racquet attaches to polymerase, core falls off, all ribosomes making protein falls off the open end Hit stop translation codon Allow long pieces of mRNA uninterrupted, but pulling tennis racquet causes discrete end o Rho dependent termination utilizes a rho factor (6 copies of a 55K Dalton monomer per rho factor) +ATP (ADP) and the handle is less G+C rich; the 3 end of the mRNA has <Us. Requires ATP hydrolysis Rho protein is multi-subunit Different because less GC rich in the handle, and doesnt have 7 Us Rho protein attaches by polymerase, Racquet flies up and all falls off o 11/1/10 Eukaryotes 1. pre-mRNA by RNA polymerase II a. general transcription factors i. TF II: DABFEHS ii. Enhance and double stranded DNA sequence iii. Activator iv. Chromatin remodeling complex v. BRE TATA Int DCE DCE MTE DPE DCE ATG TAC b. RNA polymerase II makes pre-mRNA i. Travels across gene, with bubble (ss) ii. Pre-mRNA mRNA 1. In nucleus, introns are tossed out c. Poly-A tail - transcription factors o heterodimeric2 different subunits o transcription factor II: TF II: DABFEHS o initiation at D o histones wrap around, bending of DNA o chromosome remodeling complex: if transcription is turned on, then this is he tobe transcribed area of a chromosome (figure 7) o euchromatin vs. heterochromatin: figure 8 - RNA polymerase II o 12 subunits o 1 has a CTD tail (largest): figure 9 CTD: carboxene terminal domain: the loading dock One CTD tail has a set of repeating amino acids (about 52 times in - - mammals) It hangs behind the traveling pol II downstream Tyr ser (2) pro thr ser (5) pro ser (7) Serine has hydroxyl group, which can be phosphorylated by kinase to serine-phosphate (which is reversible by phosphatase) CTD tail phosphorylation controls what is being loaded (2, 5, or 7) If phosphorolation at these three serines, GTFs unloaded, and transcription elongation factors TFIIs, hsPT5 loaded de-phosphorylation unloads the elongation factors, then: o phosphorylation at Ser 2 loads splicing factors o phosphorylation at Ser 5 cap on 5 mRNA end. Loading of capping factors. o DNA proofreading (from online notes) exonucleases: XPD and XPB o Modification of pre-mRNA as it travels Pre-mRNA is full of E (extrons) and I (introns) Need RNA splicing to remove introns RNA polymerase traveling o About 100 bases/minute o 1. Guanylyl transferase o 2. Methyl transferase o if no cap, Kozac allows 40s to load next to 60s o Transcribed from DNA RNA to make figure10b termination of transcription and poly A o cut loose o add dd poly A tail o PAP attaches puts on a few As Poly-A-polymerase Uses ATP Adds 3-4 As on 3 OH end o Then PABII Poly-A binding protein adds 50 to 200 As o They load via the RNA pol II CTD tail: CPSF (cleavage polyadenylation specificity factor) and CstF (cleavage stimulation factor) o Poly A addition is via PAP first, then PABIIp adds 50 to 200 As. These two polymerases are loaded sequentially via the RNA pol II CTD tail 1. How does transcription start with poI/III? a. RNA Pol I makes rRNA b. RNA Pol III makes tRNA c. Certain part of the chromosome codes for rRNA, not translated because not protein d. Figures 12/13 2. Splicing (~60s) RNA splicing a. ~22,000 genes in human genome b. 23 chromosomes c. look for polymerase II coding regions d. but more kinds of proteins because variation from not being strict with exons/introns e. CD40 proteinchanges in splicing programmable, metastatic (travels) f. Small nuclear ribonuclear protein: snRNPs i. U1 binds to 5 SS 1. + SR protein (gets U1 to bind) a. control the binding of U1 or not b. figure 15 2. +BBP (branch point binding) ii. U2 binds to branch point iii. U4, U5, U6 all attach together. U5 binds to 3 splice site iv. U1 and U4 then off, then Lariet forms (makes 2 5 connection) 3. RNA alternative splicing a. 4 sequences (intro) control if splicing occurs in pre-mRNA or not: b. ESE or ISE i. ESE SR protein, facilitates binding of U1, binds first to the exon/intron ii. Enhances the binding of U1 c. ESS, ISS i. Binds hnRNPs ii. Represses U1 binding, thus splicing too d. SR proteins (figure 15) e. RNA editing i. Guide RNA ii. Deamination: 1. Ex: CAAUAA A=UI=C (ionosine) iii. Introns are shorter than exons 1. Splicosomes are about 60 sr from snRPs (1, 2, 3, 4, 5, 6), machinery of a splicosome 2. Controller: SR protein family (has a long repeat of serine-argenine, and serine can be phosphorylated) 3. Have 5 and 3 splice sites, free introns, and connect exons f. Alternative RNA editing i. Variations increase the types of RNA: there are 22,000 human genes, but many more proteins found (X100), to alternatively splice original premRNAmultiply products from 1 gene ii. SV40: infects kidney cells of African green monkeys, (the monkey virus) 105 copies of new virus from each cell infected. About 3 days find/destroy host cell completelyliquid debris: FIGURE 16 iii. SR protein controls T or t antigen formation iv. CD44: 1. 20 exons, 19 introns 2. mostly takes out all introns 3. exons 5, 6, 7 have variationsmetastatic cancer cells 4. not on ordinary cancel cells, tells the difference between normal/metastatic cancer cells 5. exon 27 left outmuscular dystrophy atophy 6. melanoma 7. a few more 4. RNA editing a. GA3 GUAUACCU CUAACAUAUGGA mRNA After GA3, endo cut, then UTP puts in 2 Us at the cut on top uridyl transferase, then ligase connects i. Consequence for ribosome: new Uschanges the set of codonsframeshift mutation that changes the reading frames of codons b. Deamination: CAAUAA (stop codon) i. When C loses an amino group, it becomes U ii. Deamination of A becomes I (Inosine) iii. A=UI=C 1. In naturation of tRNA, made by pol III, has adenines. tRNA converts a to I so that it binds to C instead c. In the nucleus of a eukaryotic cell, mRNA is made in the nucleus to be exported out i. The 5 end 1st through a nuclear pore ii. Proteins that escort mRNA: hnRNP, RAN protein iii. Export signals 1. Take protein in the nucleus, signal to leave is a leucine-rich stretch on protein about to leave (that came in). Binds to exportin 1 (another nuclear protein) iv. Import signals 1. A protein in the cytoplasm to go to the nucleus has this sequence: proline, lysine, lys, lys (basic), arg, lys, arg 2. Usually argine lysine rich, binds to the alpha-importin 3. Nuclear pores have many proteins associated with the double membrane (2X PLBL) 4. Some is active transport in both directions like nTPs, DNA pol, RNA pol is imported and exported - translation o A, P, E sites: Figure 18 o tRNA anticodon aminoacyl synthases: charges the tRNA class I: 2 OH on ribose class II: 3 OH on ribose charged tRNA has an amino acid aminoacyl synthetases: takes 20 kinds of these enzymes: 1 kind hooks up a particular amino acid to connect tRNA looks at secondary structure and places to A on 3 end correct amino acid adenosine has amino acids: two classes: (depending on where you hook up the amino acid) o class 1: 2 OH (on ribose of A) o class II: 3 OH (on ribose of A) anticodon on tRNA XXX a traveling ribosome base pairs with the correct anticodon (on Asite), then hooks to carboxy end of protein (on P-site), then empty dwells (at E site), then leaves - initiation in prokaryotes: * factor=protein o 3 factors: IF3: initiation factor 3 loads on 30s subunit to SD sequence IF3 + 30s S.D. 30s slides to AUG, using IF2, uses GTP for energy IF2+GTPmRNA IF1+50s+N-formyl-met-tRNA First codon is AUG, bring in proper tRNA to base pair with AUG, then N-formyl removed by deformylase 1st-3rdish amino acid (sometimes) taken off o Initiation in eukaryotes EIF1A eIF3 eIF5b+GTP eIF2 + tRNA-Met all above attach to 40s, which becomes scanning, sliding 43s , and then becomes 48s when eIF4b attaches, which activates A subunit of eIF4F to become a helicase attaches eIF4F (E, G, A subunits) eIF4b Activates A helicase (A subunit of 4F) aminoacyl synthetases that ends correct aa to CCA ending, binds to tRNA, and puts correct aa on CCA by looking at entire structure of tRNA polyribosomes: many ribosomes travel in a row eIF4F is in the cytoplasm circularizes the mRNA: figure 19 constant supply of 40s on the same strand o Elongation in prokaryotes (of protein) EF-Tu-GTP-tRNA chargedA site Requires GTP hydrolysis Bring in 1 at a time charged tRNAs into the A site In eukaryotes, use eEF1 Peptidyl transferase Latest amino acid is added at the COOH terminal end Hoods amino acids together Part of the 23s tRNA (an enzyme, not a protein) An enzyme, not a protein Translocation EF-G-GTP (AP site translocation) Swinging back and forth of the polypeptide chain between the A and P sites Double translocation: PAP, or simple directly transport AP Makes A site open so that when a ribosome goes downstream to next codon, another tRNA can attach o Termination of Translation Ribosome falls off STOP codons: UAA, UAG, UGA No anticodon for tRNA, the last amino acid is whatever the codon right before the stop codon was In both prokaryotes and eukaryotes, the 2 subunits of the ribosome comes apart, and the protein (COOH end) falls free A site encounters this and no tRNA To fall apart, release factors: Prokaryotes: o Pull apart the ribosome to 2 subunits: A site is open because no amino acid, so the factor pulls in here o RF1=UAA or UAG o RF2 = UGA or UAA RF3 stimulates RF1, RF2 Eukaryotes: eRF1 + GTP to pull apart ribosome eRF3 stimulates eRF1 - MORF o Open reading frame: a region that can be translated by ribosomes (about 10 codons long) o A sequence built into a region that allows correct finding of AUG o near 5 end of message (upstream) o a 43s eukaryotic ribosome scans looking for the correct AUG, allows 43s to bypass incorrect AUGs - IRES o Internal ribosome sites: entry IRES at positions where no cap, so allows translation anyways If ribosome starts at an unusual place on the 5 mRNA o You can have a start without a cap: o 43s binds first viable protein eIF4F at a new site further downstream. eIF4F binds and allows the 43s to scan for AUG o Found in eukaryotic viral mRNA - tmRNA o piece of mRNA about 10 codons matches a tRNA and mRNA o place where restart can happen on broken mRNA strand - - - - - ribosome cant come off at a break, release factors cant do anything o restart at the A site, trMNA comes in, and has a stop codon comes tRNA, but translates its own mRNA, so ribosome can fall off, and gives a signal for destruction of incorrectly made protein nonsense mediated decay o stop codon too early? No problem o then these proteins come in: upf1, upf2, upf3 bind to stalled ribosome o the ribosome falls off and removes the 5 cap o 5 lytic activity (because no cap, exonuclease) destroys mRNA nonstop mediated decay o ribosome should stop beforehand, at the stop o starts to translate the poly A tail on 3 eukaryotic mRNA o decay mRNA, and protein with many lysines o skI7 protein + exosome sKI7 allows exosome to bind exosome destroys mRNA (exonucleolytic) 35 decay polylysine tail nibbled away 64 codons o comma-less code: no space/comma between codons, its continuous o degenerate code: 61 codons20 amino acids, for example 4 codons for valine suppression of o nonoverlapping: each codon is discrete (doesnt share anything with the next base/codon) genetic code: reading frames o wobble allows as few as 31 kinsd of tRNA o the code, itself: m Nirenberg experiments o suppression of mutationsintragenic and intergenic 3 possible reading frames o only 1 gives biologically useful (few exceptions where short mRNA sequences read by various reading frames) wobble o a fast moving ribosome flys around and wobbles o the cell can reduce the number of essential tRNA to code for amino acids instead of 61 tRNA (how many codons there are), wobble reduces this number to about 31 If the anticodon in the tRNA is: The codon base can be (#3 position) G C, U C G A U U A, G I (a base from adenine being A, U, C deaminated) - - - - - o dont need as many tRNA genes, reduced exactness in A position o in some mitochondria (which have there own DNA, xcription, xlation), 22 tRNA UUUU o Synthetic mRNA: if add ribosome, what protein is made? o In vitro o Every three codes for phenalalanine (ala) o Shows that it has to be a triplet code (no commas) Francis Crick, Sydney Brenner o rII region of a T4 (a virus that infects e. coli, rII is the region) o rII region controls how rapidly the T4 phage can lyse (destroy) the cell infected o by using organic dyesframeshift mutations in the rII regioninsertion or deletion of rII regiondeduce that triplet commaless code Marshall Nirenberg o Cracks all 61 code 20 amino acids can be separated o can put a radioactive atom on any of the 20 amino acids (hot), and add 19 nonradioactive aa (cold) o attach ribosome. Ribosomes a scaffold. o Amino acid is collected onto ribosome. Any ribosome holding a cold amino acid is not radioactive. But the 20th is hot o Done for each amino acid separately o Built short 3-base pair long synthetic RNA o Goes into A site, and only stay if codon and anticodon match up o If codon is GUG, anticodon is CAC, can match up with which amino acid was radioactive Examples of mutations o Point mutation: a pase pair changes o Insertion/deletion o Amplification (a sequence is reiterated many times) o Rearrangement of DNA (order changes) Suppression of mutations o 2nd mutation that cancels phenotypic effect of the first mutation o intragenic point mutation: if there is a single base change, then the amino acid changes, and the 3-D shape of the protein is incorrect you can have a second suppressor mutation that suppresses the bad folding for example, to form a dimer, the second mutaiton allows formation of a dimer again o intergenic change in the anticodon in tRNA (mutation here is on another gene) this changeit puts the correct amino acid example: streptomycin binds the wrong bad protein drug resistance prokaryotic operons - operon: a set of genes that are usually between 3-12, in ds circular chromosome of bacteria, controlled as sets. In bacterial operons, if 1 gene is on in the set, they are all on (transcriptive) (only in bacteria) - polycistronic mRNA: each mRNA coded by many genes in a tandem set o example: figure 20 o coordinated set: if turn on transcription of these four genes, all at once: make a 4 gene long message (in eukaryotes, make 1 gene-length mRNA, each gene is regulated independently) o coordination of ratio of synthesis of these proteins o many biocomeical pathways depend on several proteins in fixed ratios (A B C D, each step requiring its own enzyme 1 2 or 3) in a biological cell, need a certain amount of each converted to the next, need to limit how much of B/C made o cistron: gene (at RNA level) o discovered because _______ saw genes turned on in sets learned how genes truend on/off o regenerative medicine (internal medicine) about transcription o both lac and trp are polycistronic messages: figure 21 ex: Z:Y:A=100:50:30lactose operon - Lactose (lac) operon o Z: betagalactosidase cleaves disaccharide lactose glucose+galactose o on/off switch for translation dependent on the concentration of lactose o i gene has its own non-cistronic translation o for lac repressor protein, when binds to operator site and represses ability of mRNA to go downstream, it turns off transciption of ongoing pathway o lactose allows discharge if lac repressor protein binds lactose, then no represson (i genes product cant bind to operator if lactose is present) o high lactosemake beta-galactosidase (read polycistronic message) break down to glucose and galactose lactose is inducer o the promoter is 2 parts on right, binds RNA polymerase, but needs CAP (CRP) protein to bind for RNA polymerase, only binds if cyclooadenosinemonophosphate (CAMP must bind to CAP protein) o high concentration of lactose (inducer) plus cAMP (needs low glucose)=ON o but cant make cAMP if high concentration of glucose o what if mutate DNA i gene? Cannot regulate transcription with lactose operon always turned on: lactose doesnt matter but glucose does o transcription by RNA polymerase controls how much protein is made. Each message20-300 ribosomes o High IPTG, a chemical that turns on the lactose operon (like lac) - - Its the natural inducer of lac operon. Its used because it doesnt break down from beta-galactisidase, gratuitous inducer. Causes transcription by removing the repressor o X-gal+beta-galactisidaseblue, good color marker for the presence of betagalactisidase. Is promoter indicable by a chemical. Tryptophan operon o EDCBA all enzymes part of a pathway that converts chroismic acid tryptophan o Figure 23 o Trp operon is derepressible operon o Low concentration of trp turns on the operon (derepresses transcription) Removes trp repressor protein from operon So RNA polymerase can attach to TATA on promoter (trp binds to trp repressor protein)allosteric changebinds to the operon o 1 is palindromic to 2, 2 is palindromic to 3, 3 is palindromic to 4 o 2 can make a tennis racquet with 3, or 3 with 4 o a tennis racquet causes attenuation (stop at 139, rho-independent termination at 139) o critical region number 1 has 42 bases 14 codons 2 are UGGs, 2 or trp if not enough charged tryp, stalls at this UGG, which causes 2/3 racquet, which goes past This allows region 4 to be free as transcription continues (no transcription termination at position 139) and produce an ~ 6000 base mRNA. o how good is regulation of trp? 600 times up down gene regulation repression: 70 fold: reduces amount of trp message when repressor attached attenuation: 8-10 fold: if make proteins attenuation is only found in amino acid tpe operons 14 codons, 2 stall places?? other operons: o Organic chemical I3P studies operon Turns on transcription in operon Makes sure repressor doesnt bind Good to study rate of transctioption (how fast RNA polymerase travels) All cells in culture represson is on Example: when was EG message made? How soon translated? Lambda phage virus o Bacterial phage: a virus that infects bacteria o A bacterial cell can attach itself: figure 24a o Lysis produces more viruses Lambda phage DNA replicated, and transctiption of certain genes, translationviral proteins ot make a new virus o Lysogeny Integration of lambda DNA into the e. coli DNA Homology for site on lambda circle: att site (between gal and bio) 24b Crossing over (recombination) between lambda circle of DNA lineralized and bacterial DNA Att pp and BB similar crossover: 15 B.B. () are the same as pp: 24c o need these proteins: 17:30 3 IHF protein coded by e. coli (integration host factor) 3 integrase: lambda gene protein allows crossing over each bind to 3 sides on the pp DNA o cos R and cos L: overhangs are homologous: easy to circularize after infetion (24d) o FIGURE 25 o C1 and cro code for proteins that race after infection of if cro inds to Or3 or C1 bind to Or2/Or3 (one or the other) o Crolysis, prevents prm ( a repressor) o C1lysogeny (prevents Pr) (a repressor) o CIII repressor protein stabilizes CII o Cell damage affects lysogeny recA only binds to ss DNA: figure 26 if reduce CI, Cro takes over (then lysis) o Xis (binds 2 sites on ds DNA) is exisionase that pulls lambda phage out o Fis (binds 1 site) binds to ATT siteexcision of lambda DNA o How to make a gap? Damage cells, UV light o Recap If binds Cro protein, lysis. Its a repressor that prevents C1 from binding If CI protein binds, then lysogeny RecA protein Lysis OR lysogeny Lysis productive infection, bacterial cell destroyed and many phages released Lysogeny: integration of lambda DNA to bacterial chromosome. o recA pulls out of lysogeny because its a ss DNA binding protein o RecA* causes autcleavage of CI, which has a falling concentration, so CIs destroyed, and return to lysis At top of genetic site, OR, OR2, OR3 with promoters - Eukaryotic gene regulation o Protein motifs that bind to DNA: transcription factors o DS DNA enhance + sequences like SV 40, yeast o Chromatin remodeling o Locus control regions ex/ globin proteins LCR globin genes o Signal transduction eukaryotic gene regulation o cell differentiation: stem? Embroyonic? How much and where transcription regulation? - Transcription factors are proteins that turn on genes o Protein motif (helix 100p, leucine zipper), homeodomain protein (helix-turnhelix), zinc finger, leucine zipper, helix-loop-helix o Has high specificity for a specific sequence in an enhancer o WHAT Usually binds in major groove of enhancer sequence (a sequence if the enhancer has the protein motif in its major groove, a nearby gene will be turned on o Example: SV40 FIGURE 28 The 72 base pair enhancer sequence: 72 72 21 21 21 Regulates level of early or late transcription. In it, binding sites for transcription factors. Only need one, if have one backwards, works fine. If moved somewhere else (other site of gene) still works. Its about changing how the DNA makes the bend. - DS DNA enhance+sequences (eukaryote) figure 29: chromatin remodeling - Globin: a protein in hemoglobin in RBCs o Embreyogenesis o 5 versions o LCR: lows control region o Figure o 5 different timings, by adult, only beta version, epsilon is fetal o figure 30 a shows the genes and b shows the regulation - signal transduction o signal: ligand (from outside world) binds to receptor in membrane o signal goes across cytoplasm across into nucleus to turn on gene transctiption o many copies of signal if similar o enhancer sequence, all nearby genes turned on o figure 31 o TNFalpha and IL1 is to ligandreceptortransduction o Virus infects figure 31b
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ADMS 2610 FINAL 2005Question 1.Parol Evidence Rule - rule that prevents a party from introducing evidence that would add to orcontradict the terms of a contract. Limits the kinds of evidence that may be used to prove termsof a contract. Cannot contrad
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York University - ADMS - 2000
York University - ADMS - 2000
York University - ADMS - 2000
AK/ADMS 3330.3.0MIDTERM EXAMINATION1 of 4WINTER 2011_Question 1 (22 marks)A commercial pool management company is seeking to reduce the costs of maintainingswimming pools. The maintenance cost is mostly driven by the amount of chlorine that isneed
York University - ADMS - 2000
AK/ADMS 3330 3.0FINAL EXAMINATIONWINTER 2006Multiple choice is not included.Question 1 (20 marks)For the following linear programming problem:Maxs.t.-X + 2Y6X 2 Y 3-2X + 3Y 6X+Y38X 4 Y 5X, Y 0a) Solve the LP Model using the graphical method.
York University - ADMS - 2000
AK/ADMS 3330 - M, N, P, Q, R, S, T, UQuantitative Methods II - Winter 2005MIDTERM EXAMINATIONFebruary 26th 2005Prof. M. RochonProf. M. KarakulProf. A. MarshallSection S Section R Section MTime Allowed: 3 hoursSection U Section P Section N Se
York University - ADMS - 3000
York UniversityAtkinson Faculty of Liberal and Professional StudiesFINANCEAK/ADMS 3530.03COMMON MIDTERM EXAMINATIONSaturday, October 23, 20046:00 - 9:00 p.m.Answers provided at the end of paper.Question 1 (15 marks)(a)John and Teresa have decide
York University - ADMS - 3000
AK/ADMS 3530.03 Finance Final Exam Summer 2006Solutions provided at the end of the paper.1. You are expected to pay $1,883.33 per month on a one-year loan with a principal of $20,000. What is the EAR of this loan? A) B) C) D) 13.00% 13.80% 23.19% 25.82%
York University - ADMS - 3000
AK/ADMS 3530Midterm ExamSummer 2006May 21, 2006Type A ExamThis exam consists of 28 multiple choice questions and carries a total of 100 points. Choosethe response which best answers each question. Circle your answers below, and fill in youranswers
York University - ADMS - 3000
Winter 2005-2006YORK UNIVERSITYATKINSON FACULTY OF LIBERAL ANDPROFESSIONAL STUDIESSchool of Administrative StudiesIntermediate Accounting - AK/ADMS 3585M Tuesdays 7-10PMDirector: Sandra IacobelliMidterm #1January 31, 2006Instructions:1. This tes
York University - ADMS - 3000
Name: _Section: _WINTER 2005YORK UNIVERSITYSCHOOL OF ADMINISTRATIVE STUDIESADMS 3595 Intermediate Financial Accounting IIMidterm ExamPROFESSOR SUNG KWONTypePoints AvailableI.40 @ 2 = 80_II. Bonds Payable10_III. Income Tax Accounting10_1
Humber - ACCT - 108
Assignment 1Answers for Research and Writing based questions are to be typed up in a Word 2007/2010 document.Save the document as Your First Name.docx and submit in the Assignments tool in Blackboard underAssignment 1. Each Research and Writing based q
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Chapter 6Telecommunications and NetworksChapter OutlineThe Evolution ofNetworkingData TransmissionCharacteristicsKey TermstelecommunicationsBandwidthbandwidthbits per second (bps)broadband mediumnarrowband mediumanalog signalfrequency range
Humber - ACCT - 108
Chapter 7The Internet and the World Wide WebChapter OutlineThe Internet: A GlobalNetworkCommunicationsElectronic MailChat RoomsInstant MessagingBlogsSocial NetworkingServicesMessage BoardsStreaming Video SitesTelecommutingFile Transfer Prot
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bCOURSE OUTLINEDISTANCE EDUCATIONACADEMIC YEAR 2010 - 2011It is the students responsibility to retain course outlines for possible future use in supportof applications for transfer credit to another educational institution.PROGRAM(S):AccountingCOU
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Spywares: Any software that surreptitiously gathers user information through the user's Internet connection without his or herknowledge, usually for advertising purposes is known as spyware. Spyware applications are normally packed as secret element off
Humber - ACCT - 108
Computer Technology:How computers workComponents of a Computer SystemCategories of ComputersInput Technology: The Keyboard, Touch Screens, the Mouse and Other Point-and-Click DevicesData Representation: Bits and BytesThe System Unit, The Power Suppl
Humber - ACCT - 108
ACCT 108 33Final Exam At Home PortionQuestion 1:A customer is applying for a car loan. The cost of the car will be entered in cell C5. It will be financed at a rate entered in cellC7 for a period set in cell C8. The dealer estimates that the car will
Humber - ACCT - 108
CarLoanCalculatorCarPriceFutureValueRateNumberofPaymentsMonthlyPayment$20,500.0008.15%60($417.139)
Humber - ACCT - 108
College FundFuture Value of SavingsOut of State Annual TuitionRateNumber of Monthly PaymentsPayment Amount (per period)Future Value of SavingsPercent of Tuition Saved$40,000.004.50%15$375.00$96,155.5060%
Humber - ACCT - 108
4th of July Office PicnicSaturday from 5 11pmRiverfront ParkFood and FireworksGames for all agesBring your family and we will supply:Hot dogsHamburgersChipsSaladsDesertsSoft drinks
Humber - ACCT - 108
TheWorldWideWebBy Linda EricksenThe idea of a comprehensive computer network that would allow communication among users ofvarious computers developed over time. These developments brought together the network ofnetworks known as the Internet, which in
Humber - ACCT - 108
Transylvania SoftwareMiami, FloridaTo:Mr. WhiteChairman, Transylvania SoftwareFrom:Heather BondVice President, MarketingSubject:May Sales DataThe May sales data clearly indicate that Pittsburgh is outperforming our other geographic areas.It is
Humber - ACCT - 108
Revenue By Product Category$ 600,000$ 500,000$ 400,000$ 300,000$ 200,000$ 100,000$Word ProcessingSpreadsheetsDatabaseRevenue By City$250,000$200,000Word ProcessingSpreadsheetsDatabase$150,000$100,000$50,000$-M ilwaukeeHarrisburgReven
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State University Marching Band RosterLast NameFirst NameSexTurnerHainesTownesPowellJohnsDrakefordBallardWohrleWilliamsRichmanByruchBrownRobinsonQuaison-SackeyBrennanBeySirkinPatriotisStevensNormanCostelloBethalaMorrisJohnsonPlu
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TripIDMarktingRep1 Kenny2 Frank3 Diane4 Kenny5 Kenny6 Patty7 Frank8 Peter9 Bob10 Peter11 Patty12 Frank13 Diane14 Diane15 Frank16 Frank17 Peter18 Patty19 Bob20 Frank21 Kenny22 Diane23 Patty24 Peter25 Peter26 Bob27 BobPassengers
Humber - ACCT - 108
TripIDMarktingRep1 Kenny2 Frank3 Diane4 Kenny5 Kenny6 Patty7 Frank8 Peter9 Bob10 Peter11 Patty12 Frank13 Diane14 Diane15 Frank16 Frank17 Peter18 Patty19 Bob20 Frank21 Kenny22 Diane23 Patty24 Peter25 Peter26 Bob27 BobPassengers
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IT DepartmentNameFRANKRENAHEATHERJEFFZEVARIELLARRYHARRYJEFFSARAHMARYJASONKONSTANTINEMATTRICHARDRICKMATTHEWINDEEPATRICIAMICHAELMEGANEDLINDSAYROBINKRISPHILJUANJOHNJEFFDAVIRICHARDJAMESTHOMASLINDSEYDANIELTIMROBERTLORITE
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Scholarship TotalsFreshman TotalJunior TotalSenior TotalSophomore TotalGrand TotalFootball Team StatisticsFirst Name Last NameSex Height (ft.) Height (in.) Weight (lb.)InstrumentClassScholarship AmountFreshman Total$91,410Junior Total$57,46
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The Classic Car Cruise4/23/2012MakeMustangThunderbirdChrysler 300CCorvette StingrayFord Deuce CoupeChevrolet ImpalaPackard PhaetonLincoln ContinentalPontiac GTOStutz BearcatCadillac ElDoradoOldsmobile GTOModelYear19641955195719661932
Humber - ACCT - 108
ABC234567891011DEFGHAmortization Schedule1Ent e r Loan Parame t e rs as Indicat e dPrincipal$400,000Annual Interest6.25%Date of First Payment2/1/2008Term of Loan (years)30Extra Payment Every Period$200Your Last NameSiddique
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04/23/2012Dr. Smith's Grade BookTest 1Basil, JamesBrown, JamesCoulter, SharonEdwards, MelissaEidsen, MattFegin, RichardFord, JuddGlassman, KrisJones, SamLaquer, LindaMartin, ShellyMoldof, AdamPeters, JanPons, AlexSimon, EricStutz, JoelA
Humber - ACCT - 211
Chapter19ReportingandAnalyzingCashFlowsQuestions1.The purpose of the cash flow statement is to report detailed information about the majorcash receipts (inflows) and cash payments (outflows) during a period. This includesseparately identifying the