Cambridge A Level Biology 9700 — 2013 May/June Paper 4 · Variant 1

This document consists of 21 printed pages, 2 lined pages and 1 blank page. DC (NF/SW) 62591/3 © UCLES 2013 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level * 4 1 5 3 4 2 7 7 2 2 * BIOLOGY 9700/41 Paper 4 A2 Structured Questions May/June 2013 2 hours Candidates answer on the Question Paper. Additional Materials: Answer Paper available on request. READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black ink. You may use a pencil for any diagrams, graphs, or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all questions in Section A and one question from Section B. Electronic calculators may be used. Circle the number of the Section B question you have answered in the grid below. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 2 3 4 5 6 7 8 Section B 9 or 10 Total

2 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use Section A Answer all the questions. 1 (a) Fig. 1.1 shows a transverse section through a dicotyledonous leaf. A B C Fig. 1.1 Name A, B and C. A … B … C … [3] (b) The leaf is the main photosynthetic organ in most plants. For the light-independent stage of photosynthesis to occur, carbon dioxide must be present. (i) Describe how carbon dioxide enters the leaf. … … … … [2]

3 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (ii) Name the compound that combines with carbon dioxide in the light-independent stage in a C3 plant. … [1] (iii) Outline the role of reduced NADP in the light-independent stage. … … … … [2] [Total: 8]

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5 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 2 A number of diseases, such as dengue fever, are spread by mosquitoes. The incidence of this disease has increased dramatically in recent years and this has been linked with the spread of the mosquito, Aedes aegypti. In an attempt to reduce the numbers of A. aegypti, genetically modified (GM) male mosquitoes were produced. One of the genes added to these mosquitoes, when switched on, results in the production of a protein which is toxic to mosquitoes. In 2010, in the Cayman Islands and in Malaysia, GM male mosquitoes were released into the wild to mate with females. All the resulting offspring died in the larval stage. (a) About 3 million GM male mosquitoes were released in the Cayman Islands. Suggest why releasing such large numbers of male mosquitoes did not immediately increase the risk of transmission of dengue fever. … … [1] (b) In Malaysia, both GM male and non-GM male mosquitoes were released in order to compare their dispersal and life span in the wild. The GM mosquitoes could be identified because they also carried a gene for green fluorescent protein (GFP). Explain why, in many examples of gene technology, fluorescent markers are used in preference to antibiotic resistance genes. … … … … [2]

6 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use (c) In addition to the gene for GFP, the DNA that has been added to the GM mosquitoes consists of • a promoter • a gene coding for a toxic protein, tTA • a binding site for tTA. When a GM mosquito larva hatches from an egg, the promoter induces the production of only a small amount of tTA, so that the larva does not die immediately. In a process of positive feedback, the tTA produced binds to the DNA as shown in Fig. 2.1. This increases the expression of the gene until the increased concentration of tTA kills the larva. binding site for tTA gene coding for tTA tTA produced promoter Fig. 2.1 (i) Suggest why this process is called positive feedback. … … [1] (ii) Explain why, in gene technology, a promoter needs to be transferred along with the desired gene. … … … … … … [3]

7 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (iii) Switching on the gene coding for tTA in the mosquito larvae, rather than in the eggs, increases the effectiveness of this method of controlling mosquito numbers. Suggest why this is so. … … … … [2] (d) GM mosquitoes carrying the tTA gene can live and reproduce normally when fed on a diet containing an added chemical, A. With reference to Fig. 2.1: (i) suggest how A could prevent death of the GM mosquitoes … … … … [2] (ii) suggest how large numbers of adult GM male mosquitoes can be produced for release into the wild, from an original stock of GM males … … … … [2] (iii) suggest why there is little danger of the gene carried by these GM mosquitoes being passed to other organisms from GM mosquitoes which escape or are released into the wild. … … … … [2] [Total: 15]

8 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use 3 The filamentous fungus, Fusarium venenatum, can be grown in a fermenter and harvested as mycoprotein. It is sold as a food in a number of different countries. The fungus is grown in continuous culture in 150 000 dm3 airlift fermenters, in which the introduction of bubbles of compressed air both oxygenates and stirs the contents. The fungus grows as narrow, branched filaments, giving the harvested mycoprotein a naturally chewy, fibrous texture. Approximately 300 kg of fungus can be harvested per hour. (a) Explain what is meant by the term continuous culture. … … … … [2] (b) After about six weeks, mutants may appear in the fungal population, for example, a more highly-branched form of the fungus. The fermenter is emptied, cleaned and repopulated with the original strain of F. venenatum every six weeks. Explain why the fermentation process should be stopped before mutants appear. … … … … … … … … [4]

9 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (c) Approximately 12% of the harvested fungus is protein. Calculate the approximate mass of protein harvested in one day during continuous culture. Show your working. answer … [1] [Total:7]

10 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use 4 (a) An experiment was carried out to investigate the effect of temperature on the rate of oxygen consumption of the lizard, Sauromalus hispidus. The body temperature of a lizard varies with environmental temperature. Several lizards were fitted with small, airtight masks that covered their heads. Air was supplied inside the mask through one tube, and collected through another. The differences between oxygen concentrations in the air supplied for inhalation and the exhaled air enabled the researchers to measure the rate of oxygen consumption of the lizards. The rate of oxygen consumption of each lizard was measured when it was at rest and when it was running. Measurements were made at different temperatures ranging from 15 °C to 40 °C. Fig. 4.1 shows the results. 15 0.01 0.10 1.00 20 25 30 temperature / °C O2 consumption / cm3 g–1 hour–1 (log scale) 35 40 at rest running Fig. 4.1

11 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (i) Running requires rapid use of ATP by muscle cells in the legs and heart of a lizard. With reference to the events occurring inside a mitochondrion, explain why a faster use of ATP requires a greater rate of oxygen consumption. … … … … … … … … … … [4] (ii) Explain the effect of temperature on the rate of oxygen consumption in Sauromalus when at rest. … … … … … … … … [3]

12 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use (b) The researchers also measured the oxygen debt that was built up when a lizard was running. They measured this for two species of lizard, Sauromalus hispidus and Varanus gouldi, at six different temperatures. The results are shown in Table 4.1. Table 4.1 temperature / °C 15 20 25 30 35 40 Sauromalus oxygen debt / cm3 O2 kg−1 70.3 81.3 93.0 102.0 118.0 154.0 Varanus oxygen debt / cm3 O2 kg−1 62.0 72.2 78.5 87.9 96.7 102.0 (i) The oxygen debts were found by using the masks described in (a). Suggest what measurements were taken, and how these measurements were used to calculate the oxygen debt. … … … … [2] (ii) Compare the oxygen debt built up by a running Varanus with that of a running Sauromalus. … … … … … … [3]

13 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (iii) Varanus is a fast-moving carnivore. Sauromalus is a slow-moving herbivore. Explain how the results in Table 4.1 indicate that Varanus is well-adapted for its mode of life. … … … … … … [3] (iv) Most lizards, including Sauromalus, have very simple lungs with no alveoli. Varanus, however, has lungs that are more like those of mammals, containing large numbers of air sacs similar to the alveoli of human lungs. Suggest how this difference could account for the differences in the oxygen debts of Sauromalus and Varanus shown in Table 4.1. … … … … [2] [Total: 17]

14 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use 5 (a) As part of the technique of In-vitro fertilisation (IVF), several oocytes are collected from a woman who is undergoing treatment. Each oocyte is checked under a microscope. Explain why oocytes that have a first polar body are used in the fertilisation process. … … … … [2] (b) It is possible to freeze embryos that are produced by IVF, using a solution containing sucrose and various salts. The embryos can later be thawed and implanted. A trial was carried out to compare the success rates of freezing oocytes in • solution A, the same solution as is used for freezing embryos • solution B, a different solution containing different concentrations of sucrose and salts. Oocytes were placed into either solution A or solution B. They were then frozen and stored at a temperature of −33 °C. Later, the oocytes were thawed and then fertilised, using intracytoplasmic sperm injection (ICSI). (i) When the oocytes were placed into solution A or solution B, they quickly reduced in size. Explain why this happened. … … … … [2]

15 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (ii) Table 5.1 shows the results of the trial. Table 5.1 number of oocytes frozen number of oocytes thawed number of oocytes that survived after thawing number of oocytes that underwent ICSI number of oocytes that were successfully fertilised solution A 60 49 6 6 3 solution B 90 90 67 66 39 With reference to Table 5.1, explain which solution is the better solution to use. … … … … [2] (iii) Suggest one advantage of being able to freeze and thaw oocytes as part of the IVF procedure. … … [1] [Total: 7]

16 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use 6 (a) Fig. 6.1 outlines how a cholinergic synapse works. A A A B ACh wave of depolarisation A C B B acetylcholinesterase Fig. 6.1 With reference to Fig. 6.1: (i) name A and B A … B … [2] (ii) name the process occurring at C … [1] (iii) state the effect of B entering the post-synaptic neurone … [1]

17 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (iv) explain the role of acetylcholinesterase in the synapse. … … … … … … [3] (b) Some synapses in the brain use the neurotransmitter dopamine. After the postsynaptic membrane has been depolarised, dopamine leaves the receptor proteins and moves back into the presynaptic neurone through specific transporter proteins. Schizophrenia is a condition in which there is a higher than usual concentration of dopamine in certain areas of the brain. The drug phenothiazine has a similar shape to dopamine and is used to treat schizophrenia. Suggest and explain what occurs at the synapse when phenothiazine is used in the treatment of schizophrenia. … … … … [2]

18 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use (c) DRD4 is a dopamine receptor in humans. The gene coding for the DRD4 receptor has a large number of alleles, of which an individual can have only two. Three alleles of the DRD4 receptor gene have the following mutations: • a single base substitution • a 21 base-pair deletion • a 13 base-pair deletion. The 13 base-pair deletion has the most serious consequences for the structure of the DRD4 receptor protein. Suggest why this is so. … … … … … … … … [3] (d) One allele of the DRD4 gene has been found more frequently amongst individuals whose personality is described as ‘impulsive and exploratory’. Describe the mechanism whereby an allele such as this could have become common in the human population. … … … … … … … … [3] [Total: 15]

19 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 7 Coat colour in cats is determined by a sex-linked gene with two alleles coding for black and orange. When black cats are mated with orange cats: • the female offspring are always tortoiseshell (black and orange patches) • the male offspring are always the same colour as their mother. (a) Explain what is meant by a sex-linked gene. sex-linked … … gene … … [2] (b) Using the symbols B for the allele for black coat and O for the allele for orange coat, complete the genetic diagram below. parental phenotypes tortoiseshell female parental genotypes gametes offspring genotypes offspring phenotypes black male [4] (c) Explain why a male cat cannot have a tortoiseshell coat. … … … … … [2] [Total: 8]

20 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use 8 Sarawak is an area of south-east Asia that is largely covered by tropical rainforest. Logging has been allowed in large parts of the forest. A study was carried out to estimate the population size of different species of mammals living in the rainforest: • before logging • immediately after logging • two years after logging • four years after logging. Table 8.1 shows the results of the study for six species of mammal. Where numbers were too small to measure the population density, the species were recorded as “present”. Table 8.1 mean number of animals km−2 mammal before logging immediately after logging two years after logging four years after logging marbled cat present 0 0 0 small-clawed otter present 0 0 0 giant squirrel 5 1 4 1 treeshrew 10 5 10 38 small squirrel 16 24 104 19 barking deer 3 1 10 present (a) Calculate the percentage rise in the small squirrel population from before logging to two years after logging. Show your working. answer … % [2]

21 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (b) Suggest why populations, such as that of the small squirrel, do not increase in size indefinitely. … … … … … … … … [4] (c) Suggest why marbled cats and small-clawed otters became extinct in this area but the other mammals did not. … … … … [2] [Total: 8]

22 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use Section B Answer one question. 9 (a) Explain the need to maintain biodiversity in an ecosystem such as a tropical rainforest. [7] (b) Discuss the advantages and the disadvantages of captive breeding programmes for mammals. [8] [Total: 15] 10 (a) Explain how the physiology of the leaves of a C4 plant, such as maize, is adapted for efficient carbon fixation at high temperatures. [7] (b) Describe how, in photosynthesis, light energy is converted into chemical energy, in the form of ATP. [8] [Total: 15] … … … … … … … … … … … … … … … … …

23 9700/41/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use … … … … … … … … … … … … … … … … … … … … … … … … … … … …

24 9700/41/M/J/13 © UCLES 2013 For Examiner’s Use … … … … … … … … … … … … … … … … … … … … … … … Copyright Acknowledgements: Fig. 1.1 © DR KEITH WHEELER/SCIENCE PHOTO LIBRARY. Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2013 series 9700 BIOLOGY 9700/41 Paper 4 (A2 Structured Questions), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 Mark scheme abbreviations ; separates marking points / alternative answers for the same point R reject A accept (for answers correctly cued by the question, or by extra guidance) AW alternative wording (where responses vary more than usual) underline actual word given must be used by candidate (grammatical variants excepted) max indicates the maximum number of marks that can be given ora or reverse argument mp marking point (with relevant number) ecf error carried forward I ignore AVP Alternative valid point (examples given as guidance)

Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 Question Expected Answers 1 (a) A – palisade, mesophyll/cell/tissue/layer; B – guard cell; C – (sub-stomatal) air space; [3] (b) (i) 1. through the stoma(ta); 2. by diffusion/description; 3. from the, atmosphere/air; [max 2] (ii) ribulose bisphosphate; I RuBP [1] (iii) reduces/donates hydrogen ; A H/hydrogen atoms/H+ AND e– R H+ / H2 GP to TP ; A PGA to PGAL [2] [Total: 8] 2 (a) male mosquitoes do not, bite humans/feed on blood transmit disease OR only females, bite humans/feed on blood/transmit disease; I GM male mosquitoes are not infected with the disease [1] (b) 1. easier to, identify/screen; 2. more economical/time saving/labour saving; 3. resistance gene(s) can be passed to other bacteria; 4. idea of antibiotics no longer effective OR requiring development of new, antibiotics/treatments; [max 2]

Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 (c) (i) production of tTA causes production of more tTA/AW; [1] (ii) 1. promoter, initiates transcription/switches on gene/causes gene expression/AW; 2. ref. binding of, RNA polymerase/transcription factors; 3. otherwise gene has to be inserted near an existing promoter; 4. this is difficult to do/this may disrupt expression of existing gene; 5. in eukaryotes precise position of promoter important; [max 3] (iii) 1. GM larvae do not die immediately; 2. so gives longer time for tTA, production/build up; 3. so tTA gets into environment (when GM larvae die) and kills non-GM larvae; 4. so (longer-lived larvae) compete with non-GM larvae (for, food/space); R ref. to larvae breeding [max 2] (d) (i) 1. chemical A has, similar shape to tTA/complementary shape to binding site; 2. so chemical A binds to, DNA/binding site, AND prevents tTA from binding; 3. chemical A, binds to/changes shape of, tTA AND so prevents tTA binding to, DNA/binding site; 4. stops positive feedback/small quantity of tTA does not kill; 5. chemical A, binds to/changes shape of/breaks down, tTA, so no longer toxic; [max 2] (ii) 1. GM males, mated/bred; R with GM females 2. mosquitoes fed chemical A; 3. males, identified/separated; 4. ref. cloning; [max 2]

Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 (iii) 1. GM males die if they cannot get chemical A; 2. (if males mate), their offspring die; 3. only mate with, other A. aegypti/their own species; [max 2] [Total: 15] 3 (a) 1. nutrients added and product removed at a steady rate/AW; 2. (so) volume kept constant; 3. organism kept at, exponential/log, phase of growth; [max 2] (b) 1. (branched fungus tangles together in clumps so) too heavy for bubbles to, lift/stir OR ref. to blocking; 2. difficult to, harvest/get desired texture; 3. mutant may be, harmful when eaten/toxic/allergenic; 4. mutant may produce, distasteful/coloured, substance; 5. mutant may be less productive; 6. mutant may have high concentration of RNA (which is difficult to lower); 7. approval for sale only applies to original strain; [max 4] (c) 864 kg; [1] [Total: 7]

Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 4 (a) (i) 1. ATP is made, in the electron transport chain/by oxidative phosphorylation; 2. oxygen is the final electron acceptor; 3. in the, inner membrane of the mitochondrion/cristae; 4. transfer of electron (between electron carriers) provides energy; 5. energy used to pump hydrogen ions (into intermembrane space); 6. creates proton gradient; 7. diffusion of hydrogen ions down their electrochemical gradient causes ATP to be synthesised; 8. ref. chemiosmosis/ATP synthase/stalked particles; 9. idea that if less oxygen (consumed/available) then fewer electrons transferred along the chain; [max 4] (ii) 1. at high temperatures, reactions/enzyme activity/metabolism, faster; 2. because, molecules/enzymes/substrates, have more kinetic energy; 3. more frequent collisions; 4. therefore, respiration/Krebs cycle/electron transport chain/production of reduced NAD, take place at a faster rate; 5. idea of increase in rate of anabolic reactions (requiring more ATP); [max 3] (b) (i) 1. oxygen consumed = oxygen inhaled – oxygen exhaled; 2. measure oxygen consumption at rest (x) and after exercise stops (y); 3. extra oxygen consumed/oxygen debt = y – x; 4. measure mass of lizard; [max 2] (ii) 1. less (oxygen debt )(for Varanus); ora 2. difference is greater at higher temperatures; 3. any two comparative figures at one temperature including units; A 102.0 cm3 O2 kg-1 at 30oC and 40oC [3]

Page 7 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 (iii) 1. Varanus uses, less anaerobic/more aerobic, respiration (when running); 2. more ATP produced per glucose molecule; 3. able to run for long time; 4. good chance of catching prey; [max 3] (iv) assume Varanus throughout 1. larger surface area, in lungs/for gas exchange; 2. more oxygen absorbed into blood (per unit time)/faster rate of gas exchange; 3. more oxygen supplied to muscles ( so oxygen debt lower); [max 2] [Total: 17] 5 (a) (indicates that they) have undergone meiosis I; so are, haploid/n ; A 23 chromosomes [2] (b) (i) water moved out of cells; down water potential gradient/into a more concentrated solution/into a lower water potential; (by) osmosis; [max 2] (ii) (B) has, higher survival of oocytes after thawing/more successful fertilisations; supporting figures; these should compare columns 1 or 2 with column 3 or 5 for both A and B raw or manipulated data can be given [2] (iii) idea of deferring, fertilisation/implantation; idea of preserving oocytes from a woman who may lose her fertility due to medical treatment; idea of fewer rounds of, hormone treatment/oocyte retrieval; [max 1] [Total: 7]

Page 8 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 6 (a) (i) A – calcium ions ; A Ca2+ R calcium/Ca/Ca+ B – sodium ions ; A Na+ R sodium/Na [2] (ii) exocytosis; [1] (iii) depolarisation (of post-synaptic membrane)/action potential ; [1] (iv) 1. splits ACh; 2. into acetate and choline; 3. stops continuous depolarisation of postsynaptic membrane/AW; 4. choline recycled (into presynaptic neurone); [max 3] (b) binds to/blocks, dopamine receptors (on postsynaptic membrane); prevents depolarization (of postsynaptic membrane); reduces effect of dopamine; R reduces amount of dopamine [max 2] (c) ref 13 base deletion frame shift/alters reading frame (after mutation); (so) all amino acids different after mutation; 3-D shape/tertiary structure, of protein changed; (whereas) 21 base-pair deletion, loses 7 amino acids/no frame shift; (whereas) substitution, may change only one amino acid/may be silent; [max 3] (d) increased chances of, survival/breeding/mating; provides a selective advantage; allele passed on (to next generation); allele increases in frequency over time; natural selection; [max 3] [Total: 15]

Page 9 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 7 (a) sex-linked (gene) carried on, one sex chromosome/X, and not on, the other/Y; gene section of DNA/sequence of nucleotides/sequence of bases, that codes for a (particular) polypeptide; [2] (b) parental phenotypes tortoiseshell female black male parental genotypes XBXO XBY; gametes XB XO XB Y; offspring genotypes XBXB XBY XBXO XOY; offspring phenotypes black female black male tortoiseshell female orange male; [4] (c) tortoiseshell is heterozygous; males, heterogametic/only one X chromosome; (therefore) only one copy of gene/only black or orange allele present; [max 2] [Total: 8] 8 (a) 550(%);; allow one mark for 104 – 16 ( x 100) 16 [2] (b) 1. limiting/density dependent, factors or described; 2. reached carrying capacity/AW; 3. competition/AW; 4. for, food/nesting sites/resources; 5. large population attracts predators; 6. large population spreads disease more easily; [max 4]

Page 10 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 (c) 1. not many to begin with; 2. are carnivorous; 3. prey numbers fell; 4. slower reproductive rate; 5. more likely to migrate (to other areas); [max 2] [Total: 8] 9 (a) 1. cultural/aesthetic / leisure, reasons; 2. moral/ethical, reasons ; e.g. right to exist/prevent extinction; 3. resource material ; e.g. wood (for building)/fibres for clothes/food for humans/(herbal) medicine 4. (eco)tourism; 5. economic benefits; 6. ref. resource / species, may have use in future/AW; e.g. medical use 7. maintains, food webs / food chains; A description 8. nutrient cycling; 9. protection against erosion; 10. climate stability; 11. maintains, (large) gene pool/genetic variation; 12. scientific research; [max 7]

Page 11 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 (b) advantages (max 5) 13. can monitor health of mother; 14. can monitor development of foetus; 15. storage of, sperm/eggs/gametes; 16. artificial insemination; 17. IVF; 18. ref. surrogate mothers; 19. international cooperation; 20. genetic records kept; 21. can prevent extinction/extend range of a species/used in restoring ecosystem; disadvantages (max 5) 22. unnatural environment; 23. stress in captivity; 24. behavioural changes; 25. reproductive cycles disrupted; 26. may reject selected mate; 27. examples of problems with release ;; 28. difficulty in finding food may not integrate into groups more susceptible to disease very little natural habitat left to release animals into [max 8] [Total: 15]

Page 12 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 10 (a) 1. in C3 plants at high temperature rubisco combines with oxygen; 2. less rubisco to combine with CO2; 3. in C4 plant such as maize idea of spatial separation of light-dependent stage from carbon fixation; 4. rubisco/RuBP, in bundle sheath cells; 5. kept away from, oxygen/air; 6. mesophyll cells, absorb CO2; 7. CO2 released to combine with RuBP; 8. avoid/reduce, photorespiration; 9. high optimum temperatures of enzymes involved; 10. Calvin cycle can continue; 11. AVP ; e.g. CO2 reacts with PEP PEP carboxylase [max 7]

Page 13 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9700 41 © Cambridge International Examinations 2013 (b) 12. light energy absorbed by chlorophyll; A photosystems/pigments 13. electron, excited/raised to higher energy level; 14. (electron) emitted by chlorophyll; A photosystems/pigments 15. passes to electron, acceptor/carrier; 16. passes along, chain of electron carriers/ETC/Electron Transfer Chain; 17. energy released used to pump protons; I ATP production here 18. into thylakoid space; 19. thylakoid membrane impermeable to protons; 20. proton gradient forms; 21. protons move down gradient; 22. through/using, ATP synthase/ATP synthetase; R ATPase 23. enzyme rotates; 24. ATP produced from ADP and Pi; [max 8] [Total: 15]