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C274/SQP368 Biology Time: 2 hours 30 minutes Advanced Higher (Revised) Specimen Question Paper for use in and after 2012 NATIONAL QUALIFICATIONS 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 additional 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 11 contains a choice. [C274/SQP368] 1 © Read carefully 1 Check that the answer sheet provided is for Biology Advanced Higher (Section A). 2 For this section of the examination you must use an HB pencil and, where necessary, an eraser. 3 Check that the answer sheet you have been given has your name, date of birth, SCN (Scottish Candidate Number) and Centre Name printed on it. Do not change any of these details. 4 If any of this information is wrong, tell the Invigilator immediately. 5 If this information is correct, print your name and seat number in the boxes provided. 6 The answer to each question is either A, B, C or D. Decide what your answer is, then, using your pencil, put a horizontal line in the space provided (see sample question below). 7 There is only one correct answer to each question. 8 Any rough working should be done on the question paper or the rough working sheet, not on your answer sheet. 9 At the end of the examination, put the answer sheet for Section A inside the front cover of the answer book. Sample Question Which of the following molecules contains six carbon atoms? AGlucose B Pyruvic acid C Ribulose bisphosphate D Acetyl coenzyme A The correct answer is A—Glucose. horizontal line (see below). A B C The answer A has been clearly marked in pencil with a D Changing an answer If you decide to change your answer, carefully erase your first answer and using your pencil, fill in the answer you want. The answer below has been changed to D. A [C274/SQP368] 2 B C D Page two SECTION A All questions in this section should be attempted. Answers should be given on the separate answer sheet provided. 1. Which of the following diagrams illustrates a peptide bond? A B O C C H N H C D H C H H C C C H N N H H H Positively charged H C Table 1 H N 4. Table 1 shows the charge of amino acids at a certain pH. Negatively charged arginine tyrosine lysine cysteine histidine glutamate aspartate Table 2 shows the number of each amino acid in a protein. H Table 2 2. A hydrophobic amino acid has an R group that is Amino acid Number arginine 13 lysine 19 Cnot polar histidine 2 Dpolar. tyrosine 7 cysteine 2 glutamate 20 aspartate 9 Anegatively charged Bpositively charged 3.A buffered solution of four amino acids was applied to the midline of a strip of electrophoresis gel. The result of running the gel is shown below. _ + 1 2 3 4 Assuming that each amino acid carries a single positive or negative charge, what is the protein’s net charge at this pH? A+34 B+4 C−4 Which of the amino acids was at its isoelectric point? A1 B2 C3 D4 [C274/SQP368] 3 Page three D−38 5. The diagram below shows how phosphate is used to modify the conformation of an enzyme, phosphorylase, and so change its activity. ATP 7. The sodium-potassium pump spans the plasma membrane. Various processes involved in the active transport of sodium and potassium ions take place either inside the cell (intracellular) or outside the cell (extracellular). ADP Which line in the table correctly applies to the binding of potassium ions? Y X Binding location of potassium ions Conformation of transport protein A intracellular not phosphorylated B extracellular phosphorylated C intracellular phosphorylated D extracellular not phosphorylated Z O H O Pi W H2O Pi Which line in the table correctly identifies the labels? Kinase Phosphatase Phosphorylase A Y Z W B W Y Z C X Y W D Y W Z 8. The Na,K–ATPase moves ions in the ratio 3 sodium : 2 potassium. 5000 of these ions are pumped across the membrane every ten seconds. The number of potassium ions moved across in one second is A200 B500 C2000 D3000. 6. The diagram below shows the distribution of protein molecules in a cell membrane. Protein Protein phospholipid bilayer Protein Protein 9. The contribution of aquaporins (AQPs) to osmosis was studied by measuring the rate of movement of radioactive water across a plasma membrane. Rates were measured in either isotonic or hypertonic external solution when the pores were either open or closed. Results are shown in the table. External solution Open AQPs Closed AQPs Isotonic 2·5 1·0 Hypertonic 20·0 1·8 Protein Which line in the table correctly identifies a peripheral and an integral membrane protein? Peripheral membrane protein Integral membrane protein A 1 5 B 2 1 C 3 4 D 5 2 [C274/SQP368] 4 Rate of water movement (units s−1) Which of the following is the dependent variable in the experiment? A External solution B Radioactivity of water C Rate of water movement DAquaporins Page four 10. To which group of signalling molecules does the sex hormone testosterone belong? A Extracellular hydrophilic B Extracellular hydrophobic C Peptide hormones 12. Which of the following diagrams correctly represents the sequence of phases in the cell cycle? A B G2 M G1 G1 S M S DNeurotransmitters 11. The diagram below represents a well in an immunoassay kit testing a blood sample from a person who may have been exposed to a virus. The substrate has been broken down to form a coloured product, so the result is positive. C G2 D S G2 M G2 G1 M S G1 1 2 5 3 6 13. Animal cells growing in culture are found to spend 20% of their time in the G2 phase of the cell cycle. G2 lasts for 4 hours. If cells spend 12% of their time in the M phase, how long does this last? 4 plastic well Which line in the table correctly identifies the roles of numbered components? 4 3 2 1 B 5 2 3 6 C 4 6 5 2 D 6 3 2 5 [C274/SQP368] 5 2 hours 4 minutes B 2 hours 12 minutes C 2 hours 24 minutes D 2 hours 40 minutes 14.Which type of ion is pumped membranes by bacteriorhodopsin? Antigen Antibody Enzyme Substrate A A across ASodium BPotassium CChloride DHydrogen 15. Which of the following would be true if a population’s gene pool remained unaltered for many generations? Page five A Genetic drift had occurred. B Mating was random. C Migration was common. D Certain alleles had a selective advantage. 16. Which line in the table correctly describes cells in meiosis? Stage Chromosome complement Number of cells A Meiosis I haploid 4 B Meiosis I diploid 2 C Meiosis II haploid 4 D Meiosis II diploid 2 20. A quadrat with sides 50 cm long was used to estimate the density of a plant species in two areas X and Y. Five random samples were taken in each of the two areas and the results are given below. 17. The crossing over that generates new allele combinations in meiosis I occurs between Asister chromatids chromosomes of Number of plants Quadrat number Area X Area Y 1 27 15 2 19 16 3 39 42 4 19 31 5 11 16 homologous B non-sister chromatids of homologous chromosomes The mean density per square metre in each of the two areas is Area X Area Y C sister chromatids of non‑homologous chromosomes A 23 24 D B 46 48 C 92 96 D 115 120 non-sister chromatids of non‑homologous chromosomes. 18. Which of the following is not a source of DNA during horizontal gene transfer in bacteria? AVirus BPlasmids C Bacterial cells DGametes 19. In birds, females are heterogametic. The gene for feather-barring in chickens is sexlinked and the allele for barred feathers is dominant to the allele for non-barred feathers. What ratio of offspring would be expected when a non-barred male is crossed with a barred female? A 1 barred male : 1 non-barred female B 1 non-barred male : 1 non-barred female C 1 barred female : 1 barred male D 1 non-barred male : 1 barred female [C274/SQP368] 6 Page six 21. Anolis lizards are found on Caribbean islands. They feed on prey of various sizes. 22.All viruses surrounding Histogram 1 shows the range of prey length eaten by Anolis marmoratus on the island of Jarabacoa, where there are five other Anolis species. Histogram 2 shows the range of prey length eaten by Anolis marmoratus on the island of Marie Galante, where it is the only Anolis species. Percentage of diet Histogram 1: Jarabacoa Island A DNA or RNA B DNA and RNA C DNA only D RNA only. of a protein coat 23. Reverse transcriptase catalyses the production of A DNA from DNA B mRNA from DNA C tRNA from mRNA D DNA from RNA. 26–30 21–25 16–20 11–15 6–10 0–5 24. Which line in the table correctly describes imprinting? Within critical period Reversible Learning process A Yes No Rapid B No Yes Slow C No No Rapid D Yes Yes Slow Prey length (mm) 46–50 41–45 36–40 31–35 26–30 21–25 16–20 11–15 6–10 Percentage of diet Histogram 2: Marie Galante Island 0–5 consist 25. The formula N = MC/R is used to estimate population size using mark and recapture data. N = population estimate Prey length (mm) M=number first captured, marked and released Which of the following statements could explain the different range of prey sizes eaten by Anolis marmoratus on the two islands? A Larger numbers of prey are found on Marie Galante. B Anolis marmoratus occupies fundamental niche on Jarabacoa. C Anolis marmoratus occupies its realised niche on Marie Galante. its D Resource partitioning takes place on Jarabacoa. C = total number in second capture R = number marked in second capture In a survey to estimate a woodlouse population, the following data were obtained: Woodlice captured, marked and released = 80 Marked woodlice in second capture = 24 Unmarked woodlice in second capture = 96 The estimated population of the woodlice was A200 B320 C400 D [C274/SQP368] 7 Page seven 3 840. [END OF SECTION A] Candidates are reminded that the answer sheet MUST be returned INSIDE the front cover of the answer book. [C274/SQP368] 8 Page eight SECTION B All questions in this section should be attempted. All answers must be written clearly and legibly in ink. 1. Bovine spongiform encephalopathy (BSE) and variant Creutzfeldt-Jakob disease (vCJD) are examples of fatal brain disease that can pass from one species to another. The nature of the infectious agent is as yet unidentified but, in both diseases, a protein known as PrPSC accumulates in brain tissue. It has been shown that PrPSC is an altered form of the normal membrane protein PrPC. Both molecules have the same primary structure (PrP) but they differ in how the PrP protein folds. Molecules of PrPSC have a lower proportion of α-helix and a higher proportion of β-sheets. Proteins are normally broken down after a certain length of time by intracellular enzymes. However, the increased β-sheet content makes PrPSC more resistant to enzymatic breakdown, which leads to its accumulation. PrP normal folding abnormal folding PrPC PrPSC can be broken down by enzymes resistant to breakdown A substance capable of breaking β-sheets (β-breaker) was tested to find out if it could make PrPSC more susceptible to the intracellular enzymes. PrPSC samples from mice and humans were each incubated for 48 hours with different concentrations of β-breaker and the percentage of PrPSC remaining after digestion was determined. The results are shown in the Figure on Page ten. In the same study, mouse PrPSC was further analysed to determine if any change in the proportions of α-helix and β-sheet had occurred. The results are shown in Table 1. In a second study, mice were treated with infectious material containing (i) PrPSC and (ii) a 1:1 mixture of PrPSC and β-breaker. Quantities of PrPSC were equivalent in both treatments. Table 2 shows the mean time to onset of symptoms of brain disease in the two groups. Different concentrations of PrPSC were prepared by diluting stock solutions. [Question 1 continues on Page ten [C274/SQP368] 9 Page nine Question 1 (continued) Figure: The effect of β-breaker on mouse and human PrPSC % PrPSC present after 48 hours 150 100 Mouse PrPSC Human PrPSC 50 0 0:1 1:1 10:1 100:1 Ratio of β-breaker to PrPSC 1000:1 Table 1: Proportions of secondary structure in mouse PrPSC before and after 48 hours incubation with β-breaker. Proportions of secondary structure (%) Secondary structure Before incubation After incubation α-helix 18 27 β-sheet 36 9 Table 2: T ime to onset of symptoms for mice treated with PrPSC with and without β-breaker. Time to onset of symptoms (days) PrPSC concentration (relative to stock solution) PrPSC PrPSC + β-breaker 1 × 10−2 129 143 1 × 10−3 145 159 1 × 10−4 173 185 [C274/SQP368] 10 Page ten Marks Question 1 (continued) (a) (i)PrPC and PrPSC have the same primary structure. Why should they be expected to fold in an identical way? (ii) Hydrogen bonds are important in the folding of proteins. Name two other types of interaction involved in protein folding. 1 1 (b) Refer to the data in the Figure. The error bars represent confidence intervals. (i) What do the confidence intervals show about the data collected? 1 (ii) How are confidence intervals used in the analysis of results? 1 (iii) Draw two conclusions about the effect of β-breaker on human and mouse PrPC and PrPSC. 2 (c) Refer to the information in Table 1. (i) Explain why the PrPSC protein should be more susceptible to breakdown by intracellular enzymes after incubation. 2 (ii) How would a control showing the proportions of secondary structure in PrPC help in judging the success of the β-breaker on PrPSC? 1 (d) Refer to the information in Table 2. (i) Describe how to produce a 1 × 10−4 concentration of PrPSC by serial dilution from the 1 × 10−2 solution. 1 (ii) What is the effect of changing the concentration on the onset of brain disease? 1 (iii) Calculate the greatest percentage improvement achieved in this experiment by using β-breaker. 1 (iv) Suggest one alteration to the treatment that might improve the performance of the β-breaker. 1 (13) [C274/SQP368] 11 Page eleven Marks 2. Gamma-aminobutyric acid (GABA) is a neurotransmitter that functions as a signalling molecule in the central nervous system. GABA binds to a receptor protein located in the plasma membrane of target cells as shown in Figure 1. Binding of a GABA molecule opens a channel that allows chloride ions (Cl−) to enter the cell. Figure 1 Figure 2 − Cl modulatory site } Drug present plasma membrane of target cell Chlorine ion movement through channel (units) GABA binding site Drug absent GABA concentration (log units) Cl− Benzodiazepines are sedative drugs that bind to the receptor protein and increase its affinity for GABA. These drugs act as allosteric modulators by binding at a site that is distinct from the GABA-binding site. Figure 2, above, shows the movement of chloride ions through the channel as GABA is increased with and without the drug being present. (a) Use the information provided to explain why the GABA receptor is described as a ligand-gated channel. 2 (b) What term describes the action of a membrane receptor in which signal binding brings about an effect in the cytoplasm? 1 (c) (i) How does the information in Figure 2 show that the affinity of the receptor for GABA has been increased by the benzodiazepine? 1 (ii) Why does the affinity of the receptor for GABA increase when the drug binds to the modulatory site? 1 (iii) What effect will chloride ion influx have on the membrane potential of the nerve cell? 1 3. Discuss the advantages of a pilot study in the development of a biological investigation. [C274/SQP368] 12 Page twelve (6) (4) Marks 4. When insulin attaches to its receptor in the plasma membrane of fat cells and muscle cells, GLUT 4 glucose transporter proteins in the cytoplasm are recruited into the membrane to take in glucose. Type 2 diabetes is associated with insulin resistance in which cells are less able to respond to insulin in this way. GLUT 4 concentration relative to control UT leg (%) A recent study concluded that moderate strength training increases the GLUT 4 content of muscle tissue in those with type 2 diabetes. Individuals taking part all did strength training on one leg (T leg) for six weeks while the other leg was left untrained (UT leg). The subjects either had type 2 diabetes or did not. At the end of the training, muscle biopsies (samples) were taken from the trained and untrained legs and compared for GLUT 4 protein content. The results are shown in the Graph below. 140 * 120 100 80 * shows significant difference UT leg 60 T leg 40 20 0 Non diabetic Diabetic Diabetic status (a) In this investigation, what would be a suitable null hypothesis? (b) What comparisons did the researchers make in order to conclude that moderate strength training increases the GLUT 4 content of muscle tissue only in those with type 2 diabetes? 1 2 (3) [C274/SQP368] 13 Page thirteen Marks 5. Thyroxine is a hormone that acts as a regulator of metabolic rate in most tissues. Thyroxine causes an increase in metabolic rate by binding to specific receptors located within the nucleus of a target cell. Hyperthyroidism is a condition caused by overproduction of thyroxine. The figure shows the average change in metabolic rate of individuals with hyperthyroidism who were treated over a twenty week period with a drug (carbimazole). The drug decreases the synthesis of thyroxine from the thyroid gland. 40 Change in metabolic rate (%) 30 20 10 0 −10 −20 0 4 8 12 16 20 Time (weeks) (a) What property of thyroxine allows it to cross the membrane of cells? 1 (b) Describe how thyroxine binding to its receptor affects transcription of genes that influence metabolic rate. 2 (c) (i) How many weeks of drug treatment were required to reach normal metabolic rate? 1 (ii) How do the data support the conclusion that the thyroid gland has large stores of thyroxine? 1 (iii) Why have changes in metabolic rate been presented as percentages? 1 (6) [C274/SQP368] 14 Page fourteen 6. Rod cells and cone cells are photoreceptors in vertebrate eyes. Membranes in these cells contain rhodopsin, a protein molecule that has a light-absorbing component. Rhodopsin generates a nerve impulse when light is absorbed. Marks light-absorbing component protein membrane (a) Name the light-absorbing component of rhodopsin. 1 (b) Explain how the absorption of a photon by rhodopsin leads to the generation of a nerve impulse. 2 (c) Give one feature of the photoreceptor system in rods that allows these cells to function in low light intensity. 1 (d) Cone cells are sensitive to different wavelengths of light. Name the component of rhodopsin that varies in cone cells. 1 (5) 3 7.(a) Describe the features of chromosomes in homologous pairs. (b)In Drosophila, the genes for wing length (W), eye colour (E), body colour (B) and presence of bristles (P) are linked. The table below gives the frequency of recombination obtained in crosses involving different pairs of the linked genes. Gene pair in cross Wing length × Eye colour 12% Wing length × Body colour 18% Wing length × Presence of bristles 15% Eye colour 6% × Body colour Body colour × Presence of bristles Frequency of recombination 3% Using the letters W, E, B, and P to identify them, give the sequence in which the genes would be arranged on the chromosome. 1 (4) [C274/SQP368] 15 Page fifteen Marks 8. The capercaillie (Tetrao urogallus) is the largest member of the grouse family. It is a lekking species found in some old Caledonian pine forests. Males are larger and noisier than females, particularly in spring when their displays make them very conspicuous. (a) What term is used to describe the condition, shown in the capercaillie, where there are distinct differences between males and females? 1 (b) State one function of the male displays in spring. 1 (c) How do females benefit from being inconspicuous? 1 (d) Explain what is meant by “lekking species”. 2 (5) [C274/SQP368] 16 Page sixteen Marks 9. The figure shows the life cycle of the macroparasitic flatworm called Schistosoma japonicum. The flatworm can live for many years within a host. In humans, if untreated, it causes the disease schistosomiasis (bilharzia) that can be fatal. Figure: Life cycle of Schistosoma japonicum PRIMARY HOST Within the body, larvae migrate to the liver as they mature Adult parasites pass eggs into large intestine Eggs hatch in water Asexual stage in snail First free-living stage Second free-living stage can penetrate skin SECONDARY HOST (a) (i) Explain why the snail may not be described as a vector. 1 (ii) Suggest a feature of the parasite’s life cycle that can lead to a higher rate of transmission. 1 (iii) Name the phylum to which parasitic flatworms such as Schistosoma belong. 1 (b) Parasites living inside a host will be exposed to attack by the host’s immune system. (i) Describe one way in which parasites may overcome the immune response of their hosts. 1 (ii) Predict, in terms of the Red Queen hypothesis, how a host species would be expected to respond to the parasite adaptation. 2 (6) [C274/SQP368] 17 Page seventeen Marks 10. In 1953, the viral disease myxomatosis was introduced to the UK to control rabbit populations. The virus is spread by the rabbit flea. In successive years from 1952, young rabbits removed from the wild population were tested against the virus and their response noted, as shown in the Table. Year in which Number of rabbits were epidemics selected from previously wild population suffered by for investigation the population Response to the virus (% of population) Death Moderate symptoms Mild symptoms 1952 0 93 5 2 1953 2 95 5 0 1954 3 93 5 2 1955 4 61 26 13 1956 5 75 14 11 1957 6 54 16 30 (a) Use the data to show that rabbits were developing resistance over the period of the study. 2 (b) Why was it important always to test the rabbits against the original strain of virus? 1 (3) 11.Answer either A or B A. Discuss the concept of niche and how it applies to parasites. (10) OR B. Discuss some of the challenges involved in the treatment and control of parasites. [END OF SPECIMEN QUESTION PAPER] [C274/SQP368] 18 Page eighteen (10) C274/SQP368 Biology Advanced Higher (Revised) Specimen Marking Instructions for use in and after 2012 [C274/SQP368] 19 NATIONAL QUALIFICATIONS © Section A 1A 2C 3B 4C 5D 6B 7B 8A 9C 10B 11D 12B 13C 14D 15B 16C 17B 18D 19A 20C 21D 22A 23D 24A 25C [C274/SQP368] 20 Page two Section B Marks Question 1 (a) (i) They have the same sequence/order of amino acids OR Idea showing understanding of primary structure. 1K (ii) Disulfide (bridges), ionic (bonds), Van der Waals (attractions), hydrophobic interactions. 1K (b) (i) Variability around mean. 1K (ii) To judge if result is due to chance. 1K (iii) As (ratio of) breaker to PrPSC increases and – the % PrPSC decreases (in mice) –there is a bigger % PrPSC decrease in mice than in humans OR it works better/has a greater effect in mice than in humans –(in mice the %) decrease gets less/the breaker gets less effective at higher concentrations. Any two 2P (i) Beta sheets resist breakdown. 1 (c) (Results show) a decrease in beta sheet content (36 to 9%). 1 (ii) Would show how close PrPSC is to normal structure after treatment. (d) (i) Add one (unit) of 10−2 to 9 (units) of diluent to get 10−3; 2P 1P and repeat with 1 of this (+ 9). OR Add 1 unit of 1 x 10−2 (concentration) to 99 units of diluent. 1K (ii) The lower the concentration (of PrPSC) the greater the delay in the onset of symptoms. 1P (iii) 11% (for 10−2). 1P (iv) Use more breaker. Incubate PrPSC with breaker (before injection)/incubate longer/use different lengths of incubation. Reduce the concentration of PrPSC. Any one 1P (13) U1 3K 7P U3 2K 1P U3 b(i) and c(ii) and d(iv) Prac links d(i) [C274/SQP368] 21 Page three Marks Question 2 (a) GABA is a ligand/substance that can bind to protein. The channel is a protein that opens/gates in response to GABA/ligand binding. Chloride passes through the protein when GABA is bound. Any two 2K (b)Transduction. 1K (c) (i) Chloride movement is (generally) greater at any GABA conc. if drug present. 1P (ii) (Binding at the modulatory site) changes the conformation of the GABA site. 1K (iii) (Make the cell more negative inside so) increase the membrane potential. 1P (6) Discuss the advantages of building a pilot study into the development of a biological investigation. (4) U1 4K 2P Question 3 1. Can be used to develop a new protocol. 2. Allows practice of established protocol. 3. Ensures appropriate range of values for independent variable. 4. Avoids results for dependent variable going “off the scale”. 5. Allows the number of repeat measurements required to be estimated. 6. Checks whether results can be produced in suitable time frame. U3 4K Question 4 1P (a) The training will have no effect on GLUT 4 content of muscle. (b) ND UT is baseline GLUT 4 (100%) and training does not produce significant increase. 2 (1K 1P) D UT is (significantly) lower GLUT 4 than baseline and exercise generates significant increase. OR 1 mark for noting controls are ND v D subjects and T v UT legs. 3 U3 (1K 2P) [C274/SQP368] 22 Page four (3) Marks Question 5 1K (a)Hydrophobic. (b) Thyroxine receptor protein is blocking transcription. OR Thyroxine binding removes repression of genes. 1 More NaKATPase in membrane so more energy expenditure/higher metabolic rate. 1 2K (c) (i) 8 weeks. 1P (ii) People need to be treated for several weeks before metabolic rate reaches normal. 1P (iii) Starting metabolic rate is different for each individual. 1P U1 3K 2P U3 1P (6) Question 6 (a)Retinal. 1K (b) Excited rhodopsin activates G protein which in turn activates many enzyme molecules. Enzyme molecules cause closure of ion channels/catalyse the removal of molecules that keep channels open. Inward leakage of positive ions / Na+ and Ca+ is halted so membrane potential increases. Hyperpolarisation/increasing charge stimulates nerve impulse. Any two 2K (c) Wide range of wavelengths absorbed OR High degree of amplification from single photon. 1K (d)Opsin. 1K U1 5K (5) Question 7 (a) Same size. Same genes in same loci. Same position of centromeres. Inherited from different parents. Alleles for a gene may be different on the two chromosomes. 5 = 3; 3 or 4 = 2; 1 or 2 =1 3K (b) WEPB or BPEW 1P U2 3K 1P [C274/SQP368] 23 Page five (4) Marks Question 8 (a) Sexual dimorphism. 1K (b) To attract females/let females select “best” male. 1K (c) Camouflage decreases chances of predation. OR Improves survival chances of young. 1K (d) In breeding season, males gather in leks/communal display areas. 1 Display enables female choice. 1 2K (5) U2 5K Question 9 (a) (i) Does not itself actively transmit parasite to another species. 1K (ii) Waterborne dispersal stage. 1K (iii)Platyhelminthes. 1K (b) (i) Mimic host antigens to evade detection. OR Modify host immune response to reduce chances of destruction. OR Antigenic variation allows rapid evolution to overcome host immune cell clonal selection. 2K (ii) Red Queen hypothesis = coevolution/“arms race”. 1 1K Host adaptation to counter parasite. 1 (6) U2 6K Question 10 (a) Trend: Death rate decreases as the prior exposure to epidemics increases. Supporting data: % deaths drop from 93 to 54% after 6 epidemics. Trend: proportion with moderate/mild symptoms increases with exposure. Supporting data: 5 26% after 3 epidemics. OR Proportion with mild symptoms increases with exposure 2 30%. 2P (b) To be sure that the response was not to a mutated virus. OR To be sure that the rabbit response was not due to other factors. U2 3P [C274/SQP368] 24 1P Page six (3) Marks Question 11 Answer either A or B A Discuss the concept of niche and how it applies to parasites. 1. Define as multi-dimensional summary of tolerances and requirements. 2. Fundamental defined. 3. Realised defined. 4. Interspecific competition when niches overlap. 5. Interspecific competition can lead to competitive exclusion/local extinction. 6. When realized niches differ, species can co-exist by resource partitioning. 7. Symbionts explained. 8. Idea of cost/benefit of the relationship between parasite and host. 9. Parasites have narrow niche – explained in relation to host specificity. 10. Degenerate idea in relation to adaptations, ie loss rather than gain. 11. Endo- and ectoparasite distinction. 12. Definitive host defined. 13. Intermediate host. 14. Vector’s role in lifecycle of parasites. 15.Example. 16.Example. (10) Any ten B Discuss some of the challenges involved in the treatment and control of parasites. 1. Parasites are difficult to culture in the laboratory. 2. Rapid antigen change. 3. Should be reflected in the design of vaccines. 4. Host and parasite metabolism are similar. 5. Makes it difficult to develop drugs . . . 6. . . . that only target the parasite/do not harm hosts. 7. Limited number of control strategies. 8. Improvement of sanitation. 9. Vector control. 10. Parasites spread most rapidly in areas where control is difficult. 11. Example – eg overcrowded refugee camps, sea lice in salmon farming. 12. Parasites more abundant in tropical climates/developing countries. 13. Improvements in parasite control reduce child mortality. 14. Result in improvements in child development and intelligence. 15. Individuals have more resources for growth and development. (10) Any ten U2 10K [END OF SPECIMEN MARKING INSTRUCTIONS] [C274/SQP368] 25 Page seven