The end problem in eukaryotic DNA replicationa. is solved by t.pdf

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The \"end problem\" in eukaryotic DNA replication a. is solved by telomerase b. refers to loss of chromosome ends during replication c. is solved by an RNA dependent DNA polymerase d. A and B e. all of the above Solution Answer : The correct answer is E ( All of the above) Explanation : Telomerases are the ribonucleoprotein complexes containing a small RNA that serves as a template for addition of new six nucleotide repeats. Or Telomerase, the enzyme that synthesizes the telomere ends of linear chromosomes, is a specialized reverse transcriptase that contains an internal RNA template. Telomeres undergo cycles of shortening of the lagging strands due to the inability to complete synthesis. Most normal cells lose telomerase activity with increaseing age of the organism and repeated generations of the cell division. In other words, after certain number of divisions normal cells reaches the Hayflicks limit beyond which cells will no longer devide instead they die. In germ-line cells, which contain telomerase activity, telomere lengths are maintained; for example, telomerase is highly expressed only in cells that need to divide regularly especially in male sperm cells. Whereas somatic cells lack telomerase activity due to which they die after certain number of generations. In cancer cells/tumor cells, end replication problems observed in normal cells will be taken care by the telomerase enzyme which makes the cells to bypass the Hayflick limit and become immortal, thus avoiding cell death Many cancer cells are considered \'immortal\' because telomerase activity supports them to live longer than somatic cells. 90% of the cancer cells retain the ability to express telomerase and acquires the ability to proloferate indefinitely..

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Practice using layouts Anderson, Franceschi import javax..pdf

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Please, I need a correct answer and clear explanation. open image and zoom for a clear view. Thanks, Using the West Jet Financial statements, answer the following questions R1 2-4 p.106-109) b. Why do you think West Jet does not report a cost of goods sold c. Explain what “Advanced ticket sales” in the current liabilities section represents e. How much did West Jet use in 2013 to acquire new aircraft? Which statement did you find this information on? RI2-4 (Determination of items from a Canadian company\'s financial statements) Calgary-based WestJet Airlines Ltd. provides services throughout Canada and to some international destinations. Excerpts from its 2013 financial statements are in Exhibits 2-19A to 2- 19C. Exhibit 2-19A WESTJET AIRLINES LTD.\'S 2013 CONSOLIDATED STATEMENT OF EARNINGS Consolidated Statement of Earnings For the years ended December 31 (Stated in thousands of Canadian dollars, except per share amounts) Note 2013 2012 Revenue: Guest 3,337,569 3,133,492 Other 324,628 293,917 3,662,197 3,427,409 Operating expenses: Aircraft fuel 1,039,448 992,787 Airport operations 459,465 424,911 Flight operations and navigational charges 410,052 376,050 Sales and distribution 356,988 333,106 Marketing, general and administration 222,567 202,398 Depreciation and amortization 200,840 185,401 Inflight 176,907 162,633 Aircraft leasing 175,646 173,412 Maintenance 169,197 154,406 Employee profit share 51,577 46,585 3,262,687 3,051,689 Earnings from operations 399,510 375,720 Non-operating income (expense): Finance income 15 17,848 18,391 Finance cost 15 (43,447) (48,900) Gain on foreign exchange 1,136 1,061 Gain (loss) on disposal of property and equipment (2,962) 469 Loss on fuel derivatives — (6,512) (27,425) (35,491) Earnings before income tax 372,085 340,229 Income tax expense (recovery): Current 154,964 66,230 Deferred (51,601) 31,607 11 103,363 97,837 Net earnings 268,722 242,392 Exhibit 2-19B WESTJET AIRLINES LTD.\'S 2013 CONSOLIDATED STATEMENT OF FINANCIAL POSITION Consolidated Statement of Financial Position At December 31 (Stated in thousands of Canadian dollars) Note 2013 2012 Assets Current assets: Cash and cash equivalents 5 1,256,005 1,408,199 Restricted cash 6 58,106 51,623 Accounts receivable 19 42,164 37,576 Prepaid expenses, deposits and other 19 133,263 101,802 Inventory 19 36,722 35,595 1,526,260 1,634,795 Non-current assets: Property and equipment 7 2,487,734 1,985,599 Intangible assets 8 58,691 50,808 Other assets 19 70,778 75,413 Total assets 4,143,463 3,746,615 Liabilities and Shareholders\' Equity Current liabilities: Accounts payable and accrued liabilities 19 543,167 460,003 Advance ticket sales 19 551,022 480,947 Non-refundable guest credits 19 46,975 47,859 Current portion of maintenance provisions 9 76,105 34,135 Current portion of long-term debt 10 189,191 164,909 1,406,460 1,187,853 Non-current liabilities: Maintenance provisions 9 142,411 145,656 Long-term debt 10 689,204 574,139 Other liabilities 19 8,834 9,914 .

Please, I need a correct answer and clear explanation. open image an.pdf

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Patient 2 Patient #2: Jan Johnson Next, Dr. Gupta sees Jan Johnson. She is a 12-year-old female. She is in the 7th grade and has been on the school swim team for the past four years. She is a good student and is generally a happy child. Lately, Jan\'s mother has been worried about her daughter. Jan has been very irritable the past few months and at first, her mother thought it was related to the onset of puberty. In addition to mood changes, Mrs. Johnson has also noted a few other changes in Jan she is often too tired to attend swimming practice and has complained of feeling too weak to swim and of having headaches. Despite her decreased physical activity, Jan has been eating almost non-stop but has lost about 5 lbs, she is also drinking more than usual and has been vomiting frequently. Jan saw her primary care physician, and after running some tests, has sent Jan to follow up with Dr. Gupta. Jan\'s blood work and patient information from his chart are listed below. Solution 1.Jan is easily fatigued, eating too much (polyphagia) drinking too much water (polydipsia), sudden weight loss and nausea are the symptoms. All these symptoms relate to the metabolic disorder called Diabetes. 2.jan is having high blood glucose level than what is considered to be normal. This indicates the insulin receptors have become resistant to insulin which is type 2 diabetes or adequate amounts of insulin is not being secreted by the pancreas which is type 1 diabetes. If untreated, diabetes can be life threatening in the long term. It leads to diabetic retinopathy, kidney parenchymal changes, coronary artet disease, diabetic foot etc. 3. I must collect information about her diet, frequency of urination, recent injuries, family history of diabetes, any blurred vision,..

Patient 2 Patient #2 Jan Johnson Next, Dr. Gupta sees Jan Johnson. .pdf

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Non-math and physics question, but engineering orientated. Identify and discuss what you know are the significant milestones in materials used for commercial aircraft since the end of the Second World War? Include the manufacturing, maintenance and reliability aspects if relevant. Limit your answer to three principal types of material. Provide essay answers with a maximum of 500 words. Use current APA format for all references. (minimum of 3 references) Thanks! Solution On December 17, 1903, Orville and Wilbur Wright capped four years of research and design efforts with a 120-foot, 12-second flight at Kitty Hawk, North Carolina - the first powered flight in a heavier-than-air machine. Prior to that, people had flown only in balloons and gliders. The first person to fly as a passenger was Leon Delagrange, who rode with French pilot Henri Farman from a meadow outside of Paris in 1908. Charles Furnas became the first American airplane passenger when he flew with Orville Wright at Kitty Hawk later that year. Aircraft Innovations For the airlines to attract passengers away from the railroads, they needed both larger and faster airplanes. They also needed safer airplanes. Accidents, such as the one in 1931 that killed Notre Dame Football Coach Knute Rockne along with six others, kept people from flying Aircraft manufacturers responded to the challenge. There were so many improvements to aircraft in the 1930s that many believe it was the most innovative period in aviation history. Air-cooled engines replaced water-cooled engines, reducing weight and making larger and faster planes possible. Cockpit instruments also improved, with better altimeters, airspeed indicators, rate-of- climb indicators, compasses, and the introduction of artificial horizon, which showed pilots the attitude of the aircraft relative to the ground - important for flying in reduced visibility. designers already knew that as an aircraft approaches the speed of sound (Mach 1), in the transonic region, shock waves begin forming, causing a large increase in drag. Wings, already thin, had to become thinner and finer. Fineness is a measure of how thin the wing is compared to its front-to-back chord. A small, highly loaded wing has less drag and so some early types used this type, including the Bell X-1 rocket plane and the Lockheed F-104 Starfighter. But these craft had high takeoff speeds, the Starfighter causing significant pilot deaths during takeoff, and small wings fell out of use. An approach pioneered by German designers during the war was to sweep the wing at an angle, delaying the buildup of shock waves. But this made the wing structure longer and more flexible, making the aircraft more likely to suffer from bending or aeroelasticity and even causing a reversal in the action of the flight controls. Stall behaviour of the swept wing was also poorly understood and could be extremely sharp. Other problems included divergent oscillations which could build up lethal forces. In re.

Non-math and physics question, but engineering orientated.Identify.pdf

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Name three properties of the waveform that can be changed with the m.pdf

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Inventory Valuation: FIFO, LIFO, and Average The company reported the following inventory data for the year: Practice 9-7 Units Cost per Unit 300 $1750 Beginning inventory Purchases: 900 1,200 400 18.00 18.25 March 23. Units remaining at year-end. . . . . . .. Compute (1) cost of goods sold and (2) ending inventory assuming (a) FIFO inventory valuation, (b)LIFOin- ventory valuation, and (c) average cost inventory valuation. The company uses a periodic inventory system Practice 9-8 Inventory Valuation: Complications with a Perpetual System Refer to Practice 9-7. Assume that the sales occurred as follows: Units Sold 100 600 1,300 July 15 November 1 Total Compute (1) cost of goods sold and (2) ending inventory assuming (a) FIFO inventory valuation, (b) LIFO inventory valuation, and (c) average cost inventory valuation, The company uses a perpetual inventory system Solution Answer Units sold = 2400 – 400 = 2000 FIFO Cost of Goods available for sale Cost of Goods Sold Endging Inventory Units Cost/unit COG for sale Units sold Cost/unit COGS Units Cost/unit Ending inventory Beginning Inventory 300 17.5 5250 300 17.5 5250 0 17.5 0 Purchases: Mar-23 900 18 16200 900 18 16200 0 18 0 Sep-16 1200 18.25 21900 800 18.25 14600 400 18.25 7300 TOTAL 2400 43350 2000 36050 400 7300 LIFO Cost of Goods available for sale Cost of Goods Sold Endging Inventory Units Cost/unit COG for sale Units sold Cost/unit COGS Units Cost/unit Ending inventory Beginning Inventory 300 17.5 5250 0 17.5 0 300 17.5 5250 Purchases: Mar-23 900 18 16200 800 18 14400 100 18 1800 Sep-16 1200 18.25 21900 1200 18.25 21900 0 18.25 0 TOTAL 2400 43350 2000 36300 400 7050 ---Average Cost Valuation Cost of opening units + Purchases = $43350 Opening units + Purchased units = 2400 Average cost = 43350 / 2400 = $18.0625 per unit Cost of Goods Sold = 2000 units sold x $18.0625 = $36125 Value of ending Inventory = 400 units x $18.0625 = $7225 FIFO Opening + Purchases Cost of Goods Sold Closing Inventory Opening units 300 17.50 5250 300 17.50 5250.00 Jan 16 Sale 100 17.50 1750 200 17.50 3500.00 Mar 23 Purchase 900 18.00 16200 200 17.50 3500.00 900 18.00 16200.00 July 15 Sale 200 17.50 3500 500 18.00 9000.00 400 18.00 7200 Sept 16 Purchase 1200 18.25 21900 500 18.00 9000.00 1200 18.25 21900.00 Nov 1 Sale 500 18.00 9000 400 18.25 7300.00 800 18.25 14600 TOTAL 2400 43350 2000 36050 400 7300.00 Units Value Cost of Goods Sold 2000 36050 Closing Inventory 400 7300 LIFO Opening + Purchases Cost of Goods Sold Closing Inventory Opening units 300 17.50 5250 300 17.50 5250.00 Jan 16 Sale 100 17.50 1750 200 17.50 3500.00 Mar 23 Purchase 900 18.00 16200 200 17.50 3500.00 900 18.00 16200.00 July 15 Sale 600 18.00 10800 200 17.50 3500.00 0 0.00 0 300 18.00 5400.00 Sept 16 Purchase 1200 18.25 21900 200 17.50 3500.00 300 18.00 5400.00 1200 18.25 21900.00 Nov 1 Sale 1200 18.25 21900 200 17.50 3500.00 100 18.00 1800 200 18.00 3600.00 TOTAL 2400 43350.00 2000 36250.00 400 7100 Units Value Cost of Goods Sold 2000 36250 .

Inventory Valuation FIFO, LIFO, and Average The company reported the.pdf

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In December 2009, a 45-year-old female presented to the emergency department (ED) 2 days following abrupt onset of sore throat, nonproductive cough, chills, and mild fever. A chest radiograph was performed, which was normal. She was diagnosed with bronchitis and asked to follow up with her primary care physician, who subsequently started her on levofloxacin and albuterol. Four days later she presented again to the ED with worsening cough, dispread, fever (38.3 degree C; 101 degree F), and generalized lethargy. Additionally, she reported new symptoms including a global headache, dizziness, myalgias, and arthralgias. She had no abdominal pain, but reported nausea and anorexia. Her chest radiograph showed diffuse reticulonodular opacities throughout the left lung, which were not present on her visit 4 days previously. The patient was admitted for further evaluation. Questioning revealed the following: she had a history of diabetes and hypertension. she smoked an average of a pack of cigarettes daily, and she had received the seasonal influenza vaccine. Her husband was recently ill with cough, but no order symptoms. On day 2 of hospitalization the patient\'s respiratory rate increased from 22 to 46 and her oxygen saturation dropped while on oxygen administered by nasal canola. She was transferred to an intensive care unit, where her respiratory status quickly deteriorated, necessitating emergency intubation. A new chest radiograph showed bilateral involvement, and she was begun on vancomycin, anreonam, and azithromycin. Blood cultures drawn on her admission were negative, and an expectorated sputum sample taken at the same time was not processed due to poor specimen quality. A PCR test performed on a nasopharyngeal swab was positive for a viral agent, revealing the etiology of her infection (Fig. 10.1). What is the agent causing her infection? What are the key virulence factors of this agent? How does this virus change over time? What made this virus unique in 2009? Solution Answer= 1] The agent causing her infection is influenza virus Human influenza virus can evolve rapidly by 2 mechanisms 1] antigenic shift = genetic reassortment between a human & non human virus in non human host 2] antigenic drift = accumulation of mutations that facilitate evasion of host immune response New influenza viruses can be constantly emeges from environment 2] influenza constantly change by 2 different ways antigenic drift = These are small changes in genes of influenza viruses that happen continuously over time as the virus replicates Antigenic shift = is an abrupt major changes resulting in hemagglutinin & neuraminidase proteins in influenza viruses that infect humans Such a shift occured in 2009 when H1N1 virus with a new combination of genes emerged to infect people & quickly spread causing a pandemic when shift happens most people has no protection against H1N1.

In December 2009, a 45-year-old female presented to the emergency dep.pdf

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In a longitudinal study of attitude toward statistics, a doctoral student surveyed 10 doctoral students at four times during their program: (a) before taking any statistics courses; (b) after having taken one course; (c) after all required statistics courses; and (d) one semester after completing their preliminary examination. In each instance, alternate forms of the Attitude Towards Statistics instrument were used. The data were: 1. Analyze the data using an alpha of .05 in SPSS (Free trial of SPSS here: http://www- 01.ibm.com/software/analytics/spss/products/statistics/downloads.html) 2. Writing the result in APA format. Solution SPSS out put GLM v1 v2 v3 v4 /WSFACTOR=factor1 4 Polynomial /METHOD=SSTYPE(3) /EMMEANS=TABLES(factor1) COMPARE ADJ(BONFERRONI) /PRINT=DESCRIPTIVE /CRITERIA=ALPHA(.05) /WSDESIGN=factor1. Descriptive Statistics Mean Std. Deviation N v1 40.50 11.008 10 v2 42.40 11.806 10 v3 43.00 11.126 10 v4 36.80 10.390 10 Multivariate Testsa Effect Value F Hypothesis df Error df Sig. factor1 Pillai\'s Trace .768 7.725b 3.000 7.000 .013 Wilks\' Lambda .232 7.725b 3.000 7.000 .013 Hotelling\'s Trace 3.311 7.725b 3.000 7.000 .013 Roy\'s Largest Root 3.311 7.725b 3.000 7.000 .013 a. Design: Intercept Within Subjects Design: factor1 b. Exact statistic Mauchly\'s Test of Sphericitya Measure: MEASURE_1 Within Subjects Effect Mauchly\'s W Approx. Chi-Square df Sig. Epsilonb Greenhouse-Geisser Huynh-Feldt Lower-bound factor1 .509 5.209 5 .394 .688 .896 .333 Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix. a. Design: Intercept Within Subjects Design: factor1 b. May be used to adjust the degrees of freedom for the averaged tests of significance. Corrected tests are displayed in the Tests of Within-Subjects Effects table. Tests of Within-Subjects Effects Measure: MEASURE_1 Source Type III Sum of Squares df Mean Square F Sig. factor1 Sphericity Assumed 234.275 3 78.092 15.916 .000 Greenhouse-Geisser 234.275 2.064 113.520 15.916 .000 Huynh-Feldt 234.275 2.687 87.190 15.916 .000 Lower-bound 234.275 1.000 234.275 15.916 .003 Error(factor1) Sphericity Assumed 132.475 27 4.906 Greenhouse-Geisser 132.475 18.574 7.132 Huynh-Feldt 132.475 24.182 5.478 Lower-bound 132.475 9.000 14.719 Tests of Within-Subjects Contrasts Measure: MEASURE_1 Source factor1 Type III Sum of Squares df Mean Square F Sig. factor1 Linear 55.125 1 55.125 20.066 .002 Quadratic 164.025 1 164.025 18.401 .002 Cubic 15.125 1 15.125 4.946 .053 Error(factor1) Linear 24.725 9 2.747 Quadratic 80.225 9 8.914 Cubic 27.525 9 3.058 Tests of Between-Subjects Effects Measure: MEASURE_1 Transformed Variable: Average Source Type III Sum of Squares df Mean Square F Sig. Intercept 66178.225 1 66178.225 138.576 .000 Error 4298.025 9 477.558 Estimated Marginal Means factor1 Estimates Measure: MEASURE_1 factor1 Mean Std. Error 95% Confidence Interval Lower Bound Upper Bound 1 40.500 3.481 32.626 48..

In a longitudinal study of attitude toward statistics, a doctoral st.pdf

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