Cambridge A Level Biology 9700 — 2009 Oct/Nov Paper 4 · Variant 1

This document consists of 19 printed pages, 4 lined pages and 1 blank page. DCA (SJH/SW) 12492/4 © UCLES 2009 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces provided at the top of this page. Write in dark blue or black pen. Do not use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Section A Answer all questions. Section B Answer one question. 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. * 3 7 0 3 1 3 3 2 4 1 * BIOLOGY 9700/41 Paper 4 Structured Questions A2 Core October/November 2009 2 hours Candidates answer on the Question Paper. No Additional Materials are required. For Examiner’s Use 1 2 3 4 5 6 7 8 Section B 9 or 10 Total

2 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 Section A Answer all the questions. 1 (a) The squirrel monkey, Saimiri sciureus, of Costa Rica has become an endangered species. Fig. 1.1 shows a squirrel monkey. Fig. 1.1 Explain what is meant by the term endangered species. … … … …[2] (b) Discuss possible ways in which the squirrel monkey could be protected. … … … … … … …[4] [Total: 6]

3 9700/41/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 2 (a) Asellus aquaticus is a small freshwater crustacean. 200 A. aquaticus were released into a pond where there had previously been none. The pond was favourable for their growth and reproduction. Describe and explain the expected changes in the population size of A. aquaticus over the following few months. … … … … … … … … …[5] (b) In order for natural selection to occur a population must show phenotypic variation. Explain why variation is important in natural selection. … … …[2] [Total: 7] www.OnlineExamHelp.com

4 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 3 Proteases that work in alkaline conditions are made in large quantities for use in the detergent industry. The microorganism that is generally used for this is the bacterium Bacillus subtilis. An investigation was carried out to compare three potential production methods: ● using free cells of B. subtilis ● using B. subtilis cells immobilised in cubes of agar ● using B. subtilis cells immobilised in beads of sodium alginate. To immobilise the cells in agar, the agar was dissolved and cooled. A suspension of B. subtilis was then added. The agar-bacterium mixture was poured into sterile dishes and allowed to solidify. It was then cut into cubes with sides of 2 mm. (a) (i) Explain why the agar was cooled before the suspension of B. subtilis was added. … …[1] (ii) Describe how cells of B. subtilis could be immobilised in beads of alginate. … … … … … …[3] (b) A liquid medium containing glucose, a nitrogen source and various mineral ions was made up, and 50 cm3 placed into each of three flasks. Samples of a culture of free cells of B. subtilis, agar cubes containing immobilised B. subtilis and alginate beads containing B. subtilis were placed in the three flasks. Each flask contained the same number of bacteria. All the flasks were incubated at 37 °C for 48 hours. Samples of the liquid medium in each flask were taken at six hourly intervals and the concentration of protease measured. The results are shown in Fig. 3.1.

5 9700/41/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 free cells cells in agar cubes cells in alginate beads Key: 0 0 10 20 30 40 50 60 6 12 18 24 time / hours concentration of protease / arbitrary units 30 36 42 48 Fig. 3.1 (i) With reference to Fig. 3.1, compare the results for the free cells of B. subtilis and cells immobilised in alginate beads. … … … … … … … …[4]

6 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 (ii) Suggest why lower concentrations of protease were produced by B. subtilis immobilised in agar cubes than B. subtilis immobilised in alginate beads. … … … …[2] (c) Two new cultures of immobilised B. subtilis were set up as described in (b). However, this time a repeat batch fermentation method was used, in which the liquid medium was replaced every 24 hours. This was continued until the cubes or beads had begun to disintegrate. The results are shown in Table 3.1. Table 3.1 number of batches before cubes or beads disintegrated total fermentation time / hours total protease produced / arbitrary units mean productivity of protease / arbitrary units per hour agar cubes 6 144 1792 12.44 alginate beads 9 216 3264 15.11 With reference to Table 3.1 (i) calculate the percentage increase in the total protease produced when the bacteria were immobilised in alginate rather than agar. Show your working. …[2] (ii) explain why using bacteria immobilised in alginate rather than agar would be a more cost-effective production of protease. … … … … … …[3] [Total: 15]

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8 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 4 Modern varieties of wheat have developed from numerous hybridisation events between different species of wild grasses. Fig. 4.1 shows some of the possible steps that are believed to have been involved in the development of bread wheat, Triticum aestivum. The letters A, B and C represent three different sets of seven chromosomes. einkorn Triticum urartu 14 chromosomes AA × goat grass 1 Aegilops speltoides 14 chromosomes BB hybridisation and doubling of chromosome number emmer wheat Triticum turgidum 28 chromosomes AABB × goat grass 2 Aegilops tauschii 14 chromosomes CC hybridisation and doubling of chromosome number bread wheat Triticum aestivum 42 chromosomes … Fig. 4.1 (a) Complete Fig. 4.1 by writing letters to represent the sets of chromosomes in bread wheat. Write your answer on Fig. 4.1. [1]

9 9700/41/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 (b) Explain why hybridisation between emmer wheat and goat grass 2 would have produced a sterile hybrid, if doubling of chromosome number had not occurred. … … … … … …[3] (c) With reference to Fig. 4.1, suggest why Triticum urartu and Triticum turgidum are classified as different species. … … … …[2] (d) Triticum turgidum emerged as a new species without being geographically isolated from Triticum urartu. Outline how geographical isolation may result in speciation. … … … … … …[3] [Total: 9]

10 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 5 (a) Hormones are secreted by endocrine glands. Explain what is meant by the term endocrine gland. … … … …[2] (b) Fig. 5.1 shows the changes in concentration in the blood of follicle stimulating hormone (FSH) and luteinising hormone (LH) during the first half of the menstrual cycle. 0 2 4 6 8 10 12 14 days from start of menstrual cycle FSH LH concentration of hormone in blood /arbitrary units Fig. 5.1 With reference to Fig. 5.1, describe, (i) the changes that take place in the ovary during this time, as a result of the action of FSH … … … …[2] (ii) the role of LH. … …[1]

11 9700/41/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 (c) In preparation for in-vitro fertilisation (IVF), women are injected with FSH. Explain why treatment with FSH is a necessary preparation for IVF. … … … …[2] (d) The standard treatment with FSH and clomiphene (clomifene) causes significant side-effects. Clomiphene occupies oestrogen receptors, blocking a negative feedback mechanism. (i) Explain briefly what is meant by negative feedback. … …[1] (ii) Outline the feedback mechanism that is blocked by clomiphene. … …[1] (e) Recently a so-called ‘mild’ treatment has been introduced in the hope of avoiding the side-effects of the standard treatment. This treatment does not use clomiphene. Instead, an antagonist to LH secretion is used. The days in the first half of the menstrual cycle on which injections of FSH and clomiphene are given in the two treatments are shown by asterisks (*) in Fig. 5.2. 0 2 4 6 8 10 12 14 days from start of menstrual cycle * * * * * * * * FSH antagonist to LH secretion * * * * * * * * * * * * * * * * FSH mild treatment standard treatment clomiphene Fig. 5.2 (i) With reference to the concentrations of LH shown in Fig. 5.1, show, using an asterisk on Fig. 5.2 when the antagonist to LH secretion should first be given. Put your asterisk into the grey area on Fig. 5.2. [1]

12 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 (ii) Suggest why an antagonist to LH secretion forms part of the mild treatment. … …[1] (f) The average dose of FSH given in the mild treatment is 1300 international units (IU), compared with an average dose of 1800 IU in the standard treatment. This could lead to the mild treatment being less effective. The outcomes of an investigation into the two treatments are shown in Table 5.1. Table 5.1 mild treatment standard treatment mean number of oocytes harvested per treatment cycle 6.7 8.5 mean number of embryos produced per treatment cycle 2.8 3.8 percentage of pregnancies resulting in live birth 43.4 44.7 With reference to Table 5.1, compare the effectiveness of the two treatments. … … … … … …[3] (g) FSH consists of two polypeptide chains which are encoded by genes on different chromosomes. The two genes, together with their promoters, have been inserted into bacteria to produce the hormone used in fertility treatments. Explain briefly why promoters need to be transferred into the recipient bacteria together with the two genes for the FSH polypeptides. … … …[2] [Total: 16]

13 9700/41/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 6 (a) A husband and wife who already have a child with cystic fibrosis (CF) elected to have their second child tested for the condition while still a fetus in very early pregnancy. The results of the test, a DNA banding pattern, were discussed with a genetic counsellor. The relevant DNA banding pattern produced by electrophoresis is shown in Fig. 6.1. father first child fetus mother direction of movement of DNA fragments electrophoresis gel band of DNA Fig. 6.1 With reference to Fig. 6.1, explain why, (i) the fetus will develop CF, … …[1] (ii) the positions of the bands of DNA of the first child and of the fetus indicate that the mutant allele for CF has a deletion in comparison with the normal allele. … … … …[2] (b) Explain briefly the need to discuss the result of the test with a genetic counsellor. … … … … … … … …[4] [Total: 7]

14 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 7 (a) The fruit fly, Drosophila melanogaster, feeds on sugars found in damaged fruits. A fly with normal features is called a wild type. It has a striped body and its wings are longer than its abdomen. There are mutant variations such as an ebony coloured body or vestigial wings. These three types of fly are shown in Fig. 7.1. wild type ebony body vestigial wing Fig. 7.1 Wild type features are coded for by dominant alleles, A for wild type body and B for wild type wings. Explain what is meant by the terms allele and dominant. allele … … dominant … …[2] www.OnlineExamHelp.com

15 9700/41/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 (b) Two wild type fruit flies were crossed. Each had alleles A and B and carried alleles for ebony body and vestigial wings. Draw a genetic diagram to show the possible offspring of this cross. [6]

16 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 (c) When the two heterozygous fruit flies in (b) were crossed, 384 eggs hatched and developed into adult flies. A chi-squared (χ2) test was carried out to test the significance of the differences between observed and expected results. χ2 =  (O – E)2 E where  = sum of O = observed value E = expected value (i) Complete the missing values in Table 7.1. Table 7.1 phenotypes of Drosophila melanogaster grey body long wing grey body vestigial wing ebony body long wing ebony body vestigial wing observed number (O) 207 79 68 30 expected ratio 9 3 3 1 expected number (E) 216 72 72 24 O – E -9 … - 4 6 (O – E)2 81 … 16 36 (O – E)2 E 0.38 … 0.22 1.50 [3] (ii) Calculate the value for χ2. χ2 = … [1]

17 9700/41/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 Table 7.2 relates χ2 values to probability values. As four classes of data were counted the number of degrees of freedom was 4 – 1 = 3. Table 7.2 gives values of χ2 where there are three degrees of freedom. Table 7.2 probability greater than 0.50 0.20 0.10 0.05 0.01 0.001 values for χ2 2.37 4.64 6.25 7.82 11.34 16.27 (iii) Using your value for χ2, and Table 7.2, explain whether or not the observed results were significantly different from the expected results. … … … …[2] [Total: 14]

18 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 8 (a) Fig. 8.1 shows the results from two experiments carried out to investigate the effect of light intensity and carbon dioxide concentration on the rate of photosynthesis. light intensity / arbitrary units rate of photosynthesis /arbitrary units 0 0 1 2 3 4 5 6 7 1 2 3 4 5 6 7 experiment 1 (25 °C 0.04% CO2) experiment 2 (25 °C 0.4% CO2) Fig. 8.1 (i) Describe and explain the results shown in Fig. 8.1 for experiment 1. … … … … … …[3] (ii) Describe and explain the difference between the results for experiment 1 and experiment 2. … … … … … …[3]

19 9700/41/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 (b) The optimum temperature for many plants living in temperate regions is approximately 25 °C. Explain why the rate of photosynthesis in these plants decreases at temperatures above 25 °C. … … … … … … … … …[5] [Total: 11] www.OnlineExamHelp.com

20 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 Section B Answer one question. 9 (a) Describe the process of glycolysis. [7] (b) Describe the structure and synthesis of ATP and its universal role as the energy currency in all living organisms. [8] [Total: 15] 10 (a) Describe a reflex arc and explain why such reflex arcs are important. [7] (b) Describe the structure of a myelin sheath and explain its role in the speed of transmission of a nerve impulse. [8] [Total: 15] … … … … … … … … … … … … … … … … … For Examiner’s Use [Turn over

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24 9700/41/O/N/09 For Examiner’s Use © UCLES 2009 Copyright Acknowledgements: Question 7, Fig 7.1 wild type © www.exploratorium.edu/exhibits/mutant_flies/normal.gif ebony body © www.exploratorium.edu/exhibits/mutant_flies/ebony.gif; vestigial wing © www.exploratorium.edu/exhibits/mutant_flies/short-wings.gif 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. … … … … … … … … … … … … … … … … … … … … … … … www.OnlineExamHelp.com

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2009 question paper for the guidance of teachers 9700 BIOLOGY 9700/41 Paper 41 (Theory 2), 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2009 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 41 © UCLES 2009 Section A Question Expected Answers Marks 1 (a) 1 2 3 species threatened with extinction ; numbers reduced to critical level / population too small ; such low numbers that reproduction is affected ; [2 max] (b) 1 2 3 4 5 6 7 (maintain colony) in zoo ; captive breeding (programme) ; assisted reproduction ; e.g. IVF educate public ; national parks / conservation areas ; habitat protection ; ban, hunting / poaching ; [4 max] [Total:6] 2 (a) 1 2 3 4 5 6 7 8 9 population increases slowly at first / ref. lag phase ; (because) adjusting to pond environment ; (then) steep increase / log phase / exponential increase / rapid growth or reproduction phase ; (because) abundant food source / named other factor ; stationary phase ; fall in population size / death phase / decline phase ; (due to) predation / build up of waste ; competition for named resource ; e.g. food shortage idea of further increase and fall / ref. population size may be cyclic ; [5 max] (b) variation means the presence of different characteristics ; resulting in different survival rates / AW ; (leads to) reproductive, success / failure ; [2 max] [Total: 7]

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 41 © UCLES 2009 3 (a) (i) so that, the bacteria were not killed / enzymes not denatured ; [1] (ii) 1. bacteria put into (solution of) sodium alginate ; 2. place mixture in syringe ; 3. add drops of mixture to calcium chloride solution ; 4. calcium ions replace sodium ions (to form beads) ; 5. bacteria trapped in beads ; [3 max] (b) (i) note comparison between blue line and black line ignore references to red line - agar 1. both increase up to, 18 / 24, hours ; 2. both similar, initially / up to 18 hours ; 3. biggest difference at 24 hours / rate of increase for immobilised cells greater than free cells between 18 and 24 hours ; 4. after 24 hours immobilised cells rate decreases while free cells rate continues to increase or after 39 hours free cells rate is greater than immobilised cells rate ; 5. free cells final concentration is still lower than highest value attained by immobilised cells ; 6. use of comparative figures ; [4 max] (ii) 1. (could be) less surface area (to volume ratio) in cubes than beads ; 2. (could be) a greater diffusion distance to centre of cubes than beads ; 3. agar may be less permeable (to substrate) than alginate ; 4. something in agar may inhibit bacterial enzymes ; 5. some protease adsorbed by agar ; [2 max]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 41 © UCLES 2009 (c) (i) 82.14 / 82.1 / 82 (%) ; ; allow one mark for suitable working if incorrect answer [2] (ii) 1. can use alginate (beads) many times ; 2. (reduces cost of), materials / energy / labour ; 3. fewer bacterial cultures needed / less time spent immobilising bacteria ; 4. more protease produced (per hour) (using alginate) ; 5. can run fermentation for longer time ; 6. less time wasted between fermentations ; answers must imply comparison [3 max] [Total:15] 4 (a) AABBCC ; [1] (b) 1 2 3 4 if doubling of chromosomes has not occurred chromosomes would not be able to pair ; because chromosomes in the two sets are not homologous ; during, prophase 1 / meiosis 1; (therefore) gametes cannot be produced ; [3 max] (c) 1 2 3 unable to, breed / reproduce ; to produce fertile offspring ; reproductively isolated ; [2 max] (d) 1 2 3 4 5 species split into two populations by (geographical) barrier ; different, selection pressures / (environmental) conditions, (on the two populations) ; different features, selected / advantageous ; change in, gene pools / allele frequencies ; (over time) become unable to interbreed ; [3 max] [Total: 9] www.OnlineExamHelp.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 41 © UCLES 2009 5 (a) ductless gland ; secretes (hormone) into blood ; [2] (b) (i) 1. follicle, develops / matures / grows ; 2. detail follicle ; e.g. antrum / corona / theca 3. (follicle) secretes oestrogen (and progesterone) ; [2 max] (ii) trigger ovulation / description ; [1] (c) 1 2 3 4 to produce many (mature) oocytes at same time ; superovulation ; make harvesting easier ; IVF procedure has low success rate ; [2 max] (d) (i) a change sets off events that counteract the change / AW / example described ; [1] (ii) oestrogen inhibition of, GnRH / FSH ; [1] (e) (i) day 9 ; [1] (ii) prevent ovulation / so oocytes can be harvested ; [1] (f) 1 2 3 4 very little difference in percentage of pregnancies resulting in live birth ; standard (slightly) more oocytes (per cycle) ; ora standard (slightly) more embryos (per cycle) ; ora comparative figs ; [3 max] (g) 1 2 3 (promoter needed) to ensure genes are, expressed / switched on ; to produce, correct product / correct hormone / FSH ; ref. human / eukaryote, gene in, bacteria / prokaryote ; [2 max] [Total: 16]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 41 © UCLES 2009 6 (a) (i) same band of DNA as, first / affected, child ; [1] (ii) 1. father and mother, have normal and mutant alleles / are heterozygous ; 2. mutant / CF, DNA is, shorter / lighter ; 3. therefore travels further ; [2 max] (b) 1 2 3 4 5 6 outcome of test needs explanation / counsellor gives advice on options ; already have one affected child to care for or problems / cost, of care ; ref. termination ; life expectancy increasing with improved drugs ; gene therapy, not as yet successful / likely to be temporary ; possibility of, pre-implantation genetic diagnosis (PGD) / artificial insemination by donor sperm (AID), on another occasion ; [4 max] [Total: 7] 7 (a) allele different / alternative, form of a gene ; A variety of a gene dominant (allele) that always expresses itself in the phenotype when present / (allele) which influences the phenotype even in the presence of an alternative allele / AW ; [2] (b) parental phenotype ; e.g. striped / long x striped / long A wild x wild parental genotype ; e.g. AaBb x AaBb gametes ; e.g. AB Ab aB ab offspring genotypes ;; offspring phenotypes ; must be linked to genotypes [6] accept other symbols if key used penalise once for no key but only if genetic cross works

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 41 © UCLES 2009 phenotypes of Drosophila melanogaster grey body long wing grey body vestigial wing ebony body long wing ebony body vestigial wing observed number (O) 207 79 68 30 expected ratio 9 3 3 1 expected number (E) 216 72 72 24 O – E –9 7 –4 6 (O – E)2 81 49 16 36 E E) (O 2 − 0.38 0.68 0.22 1.50 (c) (i) [3] (ii) 2.78 ; apply ecf [1] (iii) χ 2 value represents probability of > 0.05 ; no significant difference ; (probability shows) differences due to chance ; [2 max] [Total:14] www.OnlineExamHelp.com

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 41 © UCLES 2009 8 (a) (i) at low light intensity 1. rate of photosynthesis increases as light intensity increases ; 2. light intensity is limiting factor ; at higher light intensity 3. graph, levels off / forms a plateau / rate becomes constant ; 4. CO2 / some other factor, becomes limiting ; [3 max] (ii) 1. above light intensity of 1 rate is always higher for expt. 2 ; 2. plateau reached at lower light intensity for expt. 1 ; 3. maximum / plateau, rate is double for expt. 2 ; 4. expt 2 has much more CO2 (conc) (compared to expt 1) ; 5. CO2, no longer limiting after 4.2 in expt.2 / is limiting in expt. 1 up to 2.8 ; [3 max] (b) 1 2 3 4 5 6 7 8 9 10 enzymes, denatured / active site changes shape ; rubisco / enzyme in cyclic photophosphorylation ; Calvin cycle affected / description ; less photolysis ; less ATP produced ; increased rate of respiration ; respiration rate faster than photosynthesis rate / ref. compensation point ; increased rate of transpiration ; stomatal closure ; less CO2 uptake ; [5 max] [Total:11]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 41 © UCLES 2009 Section B: only one question to be answered. 9 (a) 1 2 3 4 5 6 7 8 9 10 11 (glucose) phosphorylated by ATP ; raises energy level / overcomes activation energy ; hexose bisphosphate ; lysis / splitting, of, glucose / hexose ; R sugar splitting breaks down to two TP ; A GALP / GADP / G3P / PGAL 6C → 2 x 3C ; dehydrogenation / description ; 2 NAD reduced formed (from each TP to pyruvate formed) ; 4 ATP produced / net gain of 2 ATP ; pyruvate produced ; reduced NAD → oxidative phosphorylation / redox ; accept flow diagram [7 max] (b) 12 13 14 15 16 17 18 19 20 21 22 23 nucleotide ; adenine + ribose / pentose + three phosphates ; loss of phosphate leads to energy release / hydrolysis releases 30.5 kJ ; ADP + Pi ↔ ATP (reversible reaction) ; synthesised during, glycolysis / Krebs cycle / substrate level phosphorylation ; synthesised, using electron carriers / oxidative phosphorylation / photophosphorylation ; in, mitochondria / chloroplasts ; ATP synthase / ATP synthetase ; chemiosmosis / description; used by cells as immediate energy donor ; link between energy yielding and energy requiring reactions / AW ; active transport / muscle contraction / Calvin cycle / protein synthesis ; [8 max] [Total: 15]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 41 © UCLES 2009 10 (a) 1 2 3 4 5 6 7 8 9 10 11 12 13 strong stimulus in receptor / AW ; action potential / impulses, along sensory neurone ; dorsal root of spinal nerve ; into spinal cord ; synapse with intermediate neurone ; (then) motor neurone ; action potential / impulses, to effector ; action potential / impulses, to brain ; response ; e.g. knee jerk 5 max can be on diagram fast / immediate ; stops / limits, damage / danger ; automatic / no conscious thought ; innate / stereotyped / instinctive ; [7 max] (b) 14 15 16 17 18 19 20 21 22 23 24 25 Schwann cells ; wrap around axon ; sheath mainly lipid ; (sheath) insulates axon (membrane) ; Na+ / K+, cannot pass through sheath / can only pass through membrane at nodes ; depolarisation (of axon membrane) cannot occur where there is sheath / only at nodes of Ranvier ; local circuits between nodes ; action potentials ‘jump’ between nodes ; saltatory conduction ; increases speed / reduces time, of impulse transmission ; up to 100 ms-1 ; speed in non-myelinated neurones about 0.5 ms-1 ; [8 max] [Total: 15] www.OnlineExamHelp.com