X274/13/02NATIONALQUALIFICATIONS2014FRIDAY, 16 MAY1.00 PM – 3.30 PMBIOLOGYADVANCED HIGHER(REVISED)SECTION A—Questions 1–25 (25 marks)Instructions for completion of Section A are given on Page two.SECTION B (65 marks)The answer to each question should be written in ink in the answer book provided. Any additionalpaper (if used) should be placed inside the front cover of the answer book.Rough work should be scored through.All questions should be attempted. Candidates should note that Question 10 contains a choice.Question 1 is on Pages 10, 11 and 12. Question 2 is on Page 13. Pages 12 and 13 are fold-outpages.SA*X274/13/02*
Read carefully1Check that the answer sheet provided is for Biology Advanced Higher Revised (Section A).2For this section of the examination you must use an HB pencil and, where necessary, an eraser.3Check that the answer sheet you have been given has your name, date of birth, SCN (ScottishCandidate Number) and Centre Name printed on it.Do not change any of these details.4If any of this information is wrong, tell the Invigilator immediately.5If this information is correct, print your name and seat number in the boxes provided.6The answer to each question is either A, B, C or D. Decide what your answer is, then, usingyour pencil, put a horizontal line in the space provided (see sample question below).7There is only one correct answer to each question.8Any rough working should be done on the question paper or the rough working sheet, not onyour answer sheet.9At the end of the examination, put the answer sheet for Section A inside the front cover ofthe answer book.Sample QuestionWhich of the following molecules contains six carbon atoms?AGlucoseBPyruvic acidCRibulose bisphosphateDAcetyl coenzyme AThe correct answer is A—Glucose.horizontal line (see below).ABCThe answer A has been clearly marked in pencil with aDChanging an answerIf you decide to change your answer, carefully erase your first answer and using your pencil, fill in theanswer you want. The answer below has been changed to D.A[X274/13/02]BCDPage two
SECTION AAll questions in this section should be attempted.Answers should be given on the separate answer sheet provided.1. The diagram below shows a haemocytometercounting chamber containing animal cells.The depth of the chamber is 0·01 cm.3. Which line in the table below correctlyrepresents an allosteric enzyme binding with apositive modulator?0·1 cmModulatorbinding siteactivesite0·1 cmAsecondarysite C increaseddecreased BDAffinity of enzymefor substrate centralsquareanimalcell4. The stages of muscle contraction are listedbelow.A432·0 10 cells per cm1 Phosphate ion released from myosin head.2 ATP binds to myosin head and causes it todetach from actin filament.3 Myosin head swings forward and attachesto actin filament.4 Myosin head drags along actin filament.B2·0 105 cells per cm3The sequence in which these stages occur isThe concentration of animal cells, based onthe cell count from the central square, is63C2·0 10 cells per cmD2·0 107 cells per cm3.2. Which line in the table below correctlydescribes the charges on the two componentsof nucleosomes?DNAHistone epositiveDnegativepositive[X274/13/02]Page threeA2, 1, 3, 4B2, 3, 1, 4C3, 2, 1, 4D3, 2, 4, 1.[Turn over
5. The diagram below shows cotransport(symport) of sodium ions (Na ) and glucoseinto a cell lining the gut.7. Cortisol is a steroid hormone.Which letter in the diagram below shows thefirst stage in cell signalling for this hormone?outside cellANa GlucoseNa-bindingsiteBCDproteinoutside cellGlucose-bindingsiteinside cellNa Glucosephospholipid8. The figure below shows the relative DNAcontent of cells from a culture.inside cellWhich line in the table below represents therelative concentrations of glucose and Na onthe two sides of the plasma membrane whencotransport occurs?Number of owhighDlowhighhighlowZY012Relative amount of DNA per cell(arbitrary units)116. A typical cell is estimated to possess 4 10potassium ions. Only 107 of these are used inestablishing membrane potential.In which two phases of the cell cycle are cellsin region Z?What fraction of the total potassium ions isinvolved in this AG1 and SBS and G2CG2 and MDM and G19. Which of the following would not be asubstrate for caspases?Page fourADNABactinChistoneDtubulin
10. The graph below shows the effect of carbondioxide concentration on the affinityof haemoglobin for oxygen at differentconcentrations of oxygen.Which of the following diagrams representshow the protein is altered to let the cycleprogress?120Percentage saturation of haemoglobinwith oxygen11. Retinoblastoma protein (Rb) has a rolein regulating progress through the cellcycle. It can be phosphorylated (Rb-P) orunphosphorylated (Rb).100A80Rb-P60RbSG140G2M20002468101214BRbOxygen concentration (kPa)carbon dioxide concentration 2.7 kPaRb-PSG1carbon dioxide concentration 5.3 kPacarbon dioxide concentration 10.7 kPaG2MA list of possible conclusions is given below.1Increasing the concentration of carbondioxide decreases the affinity ofhaemoglobin for oxygen.2 Increasingtheconcentrationofcarbon dioxide increases the affinity ofhaemoglobin for oxygen.3 Increasing the concentration of oxygendecreases the affinity of haemoglobin foroxygen.4 Increasing the concentration of oxygenincreases the affinity of haemoglobin foroxygen.Which conclusions are valid for the datashown in the graph between oxygenconcentrations of 2 kPa and 10 kPa?A1 and 3B1 and 4C2 and 3D2 and 4CRb-PRbSG1G2MDRbG1MRb-PSG2[Turn over[X274/13/02]Page five
12. Which line in the table below correctlydescribes processes underpinning evolution?RandomNon-randomAMutationGenetic driftBGenetic driftMutationCGenetic driftSexual selectionDNatural selectionGenetic drift13. In evolutionary theory, fitness candescribed in absolute or relative terms.15. The figure below shows a nucleus in the earlystages of meiosis I. Paternal chromosomes areshaded, maternal chromosomes are unshaded.beAbsolute fitness is the ratio ofAsurviving offspring of one phenotypecompared with other phenotypesBsurviving offspring of one genotypecompared with other genotypesCfrequencies of a particular phenotypefrom one generation to the nextgenerationDHow many different gametes wouldbe produced as a result of independentassortment?A 2B 6frequencies of a particular genotype fromone generation to the next generation.C 8D14. Chiasmata form when the chromosomes arearrangedAas pairs during meiosis IBindividually during meiosis ICas pairs during meiosis IIDindividually during meiosis II.1216. In mammals, some genes are present on the Ychromosome but not on the X chromosome.An allele of one such gene causes deafness.What is the chance of a male with deafnesscaused in this way having a child who inheritshis condition?A0%B25%C50%D100%17. In some species of bird the females are largerthan the males. This is described as[X274/13/02]Page sixAlekkingBreversed sexual dimorphismCsexual dimorphismDfemale choice.
18. Eggs from leopard geckos kept in breedingcages were collected and incubated at twotemperatures over five breeding seasons. Wheneach new gecko hatched, its gender was noted.The graph below shows how temperatureaffected gender in the population.Percentage of males inthe population10021. The virulence of an infectious organism isdefined as the case fatality risk (CFR). CFRcan be represented as the percentage ofinfections that result in death. The tablebelow shows the numbers of people infectedby the “bird flu” virus (H5N1) and thenumbers who died from it over a five otalinfectionsof H5N146981158844Numberdying fromH5N1infection324379593360402001234Breeding Season5How many females would be present in apopulation of 500 leopard geckos after fourseasons at 32·5 C?A150B200C300D35019. The beef tape worm (Taenia saginata) liveswithin the small intestine of humans. For partof its life cycle, it does not have a digestivesystem. Therefore, the parasite is said toAbe degenerateBbe a micro-parasiteCoccupy its fundamental nicheDco-exist by resource partitioning.20. Which line in the table below correctly describesthe ecological niche of a parasite?NicheHost X274/13/02]Page sevenIn which year was H5N1 most virulent?A2004B2006C2007D2008[Turn over
22. The diagram below shows a response toforeign antigens.24. Rabbits can be infested with nematodeparasites in their gut.Infestation by nematodes may be a majorfactor controlling population density ofrabbits.antigensWhich of the following procedures would testthis hypothesis?Comparing rabbit population densities invariety of B lymphocytescell proliferation/multiplicationsecretion ofantibodies intocirculationmemorycells madeWhich of the following correctly identifies thiscellular response?AapoptosisBphagocytosisCclonal selectionDinflammatory responseAinfested populations from areas withdifferent food suppliesBrabbits infested with nematodes or withother parasitesCareas with and without competition fromother herbivoresDinfestedandpopulations.The protection of non-vaccinated, vulnerablechildren is an example ofN MC/Rpopulation estimatenumber first captured, markedand releasedtotal number in second capturenumber marked in second captureIn a survey to estimate a rabbit population,the following data were obtained:Rabbits captured, marked and released 240Marked rabbits in second capture 80Unmarked rabbits in second capture 280The estimated population of the rabbits wasA560AepidemiologyB600Bherd immunityC840Cimmune surveillanceD1080.Dimmunological memory.[END OF SECTION A]Candidates are reminded that the answer sheet MUST be returned INSIDE thefront cover of the answer book.[X274/13/02]rabbit25. The following formula can be used to estimatepopulation size from mark and recapture data.Where N M C R 23. NHS Scotland vaccinates as many children aspossible with measles vaccine. This helps toreduce the spread of measles, and provides alevel of protection to those vulnerable childrenwho do not get the vaccine.non-infestedPage eight
[Turn over for SECTION B on Page ten[X274/13/02]Page nine
SECTION BAll questions in this section should be attempted.All answers must be written clearly and legibly in ink.1.Two types of muscle, red and white, can be distinguished by their colour in samples of freshtissue and can be easily separated. Red muscle cells obtain energy mainly using aerobicrespiration: they have many large mitochondria and a good supply of oxygen. White musclecells obtain energy mainly by anaerobic respiration: they have fewer mitochondria and apoorer oxygen supply. In both muscle types, glucose is the substrate for respiration. It iswidely thought that the mechanism of glucose transport into these cells is the step that limitstheir ability to use glucose, and it is considered that red muscle cells have a greater capacity forglucose transport than white muscle cells.Glucose diffuses into cells through glucose transporters (GLUTs), which are proteinmolecules embedded in cell membranes. There are several types of GLUT. GLUT1 isresponsible for glucose uptake in all cells; the membranes of muscle and fat cells also containGLUT4.The study below investigated the contribution of these two GLUTs to glucose uptake in redand white muscle cells, before and after exposure to insulin. Figure 1 shows the effect ofinsulin on glucose transport in the two types of muscle.An extract of membranes from the muscle cells was centrifuged to separate it into twoportions, plasma membrane (PM) and the internal membranes (IM) from the cytoplasm.The protein components of the membranes were separated by gel electrophoresis and blotted.The blots were exposed to radioactively-labelled antibodies specific for each of the twoGLUT proteins, to allow identification and quantification.Figure 2 shows the percentage change in total GLUT level in the two membrane fractionsfollowing the insulin treatment. In Figure 3, the blots indicate the changing abundances ofthe two GLUTs. Figure 4 shows the relative amount of GLUT4 in the two muscle types inresponse to insulin. Error bars show standard error.10·010.08·08.0Red muscleWhite ulinTreatment[X274/13/02]Figure 2: Effect of insulin on totalGLUT levels% Changes with insulinGlucose transport (µmol/g/h)Figure 1: Glucose transport with andwithout insulinPage tenRed muscleWhite muscle1001000–100-100PMIMMembrane fraction
Question 1 (continued)Figure 3: Blots showing the effect of insulin on the distribution of GLUTs 1 and 4Red MusclePMWhite MuscleIMIMPMC controlCICICICI12345678I insulinGLUT1GLUT4Figure 4: Relative amounts of GLUT4 quantified from several blotsControlInsulin1·5Red muscleGLUT4 relative to control (units)1·00·50·02·0White muscle1·51·00·50·0PMIMMembrane fraction[Question 1 continues on Page twelve[X274/13/02]Page eleven
MarksQuestion 1 (continued)(a) Use data from Figure 1 to support the statement that “red muscle cells have a greatercapacity for glucose transport than white muscle cells”.2(b) Figure 2 shows that GLUT increases in the PM fraction and decreases in the IMfraction after insulin treatment. It was concluded that both muscle types have thesame underlying GLUT response to insulin.Explain how the error bars confirm this conclusion is valid.1(c) Refer to Figure 3.(i)Describe the distribution of GLUT1 in muscle cells before insulin treatment.1(ii)Give one conclusion about the effect of insulin treatment on GLUT1.1(iii)What evidence is there that the effect of insulin on the distribution of GLUT4is the same in both types of muscle?1(d) It was hypothesised that insulin triggers the transport of additional GLUT4 tothe plasma membrane from storage on membranes in t
paper (if used) should be placed inside the front cover of the answer book. Rough work should be scored through. All questions should be attempted. Candidates should note that Question 10 contains a choice. Question 1 is on Pages 10, 11 and 12. Question 2 is on Page 13. Pages 12 and 13 are fold-out pages. NATIONAL QUALIFICATIONS 2014 FRIDAY, 16 MAY 1.00 PM – 3.30 PM [X274/13/02] Page two .
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