Cambridge A Level Biology 9700 — 2009 Oct/Nov Paper 4 · Variant 2

This document consists of 18 printed pages, 3 lined pages and 3 blank pages. DC (SJH/SW) 18304/5 © 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. * 2 4 2 6 3 1 4 1 1 6 * BIOLOGY 9700/42 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 9700/42/O/N/09 For Examiner’s Use © UCLES 2009 Section A Answer all the questions. 1 All living organisms are divided into five kingdoms. The table below lists some features possessed by living organisms and some processes that they carry out. Place a tick or a cross in the table to indicate the presence or absence of the feature or process in any or all members of the kingdom. The first row has been done for you. feature or process kingdom Prokaryotae Protoctista Fungi Plantae Animalia 80s ribosomes ✗ ✓ ✓ ✓ ✓ cell walls contain chitin circular DNA endoplasmic reticulum most species unicellular autotrophic heterotrophic [6] [Total: 6]

3 9700/42/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 2 (a) A recent study of the house mouse, Mus musculus, on the island of Madeira resulted in the following observations. ● There are six distinct populations. ● The mice are associated with human settlements. ● The populations are located in different valleys separated by steep mountains. ● Each population has a different diploid number of chromosomes. As a result of these observations it has been suggested that speciation is taking place. Fig. 2.1 is a map of Madeira showing the distribution of the six populations. mountains 3 6 1 2 4 5 Fig. 2.1 Using the information in Fig. 2.1, state the likely isolating mechanism and the type of speciation taking place. isolating mechanism … type of speciation …[2]

4 9700/42/O/N/09 For Examiner’s Use © UCLES 2009 (b) Explain how speciation is occurring in the house mouse populations of Madeira. … … … … … … … … …[5] [Total: 7] www.OnlineExamHelp.com

5 9700/42/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 3 (a) Cell walls of bacteria contain peptidoglycans. Peptidoglycans are long chains of the sugars N-acetylmuranic acid (NAM) and N-acetylglucosamine (NAG) which alternate along the chain. A short peptide chain of three to five amino acids is attached to each NAM and these form cross-links with similar peptide chains from adjacent strands. Fig. 3.1 shows a diagram representing part of a peptidoglycan structure. NAM peptide chains cross-link NAG NAM NAG NAM NAG NAM NAG NAM NAG NAM NAG Fig. 3.1 (i) Name the type of reaction that takes place to assemble the peptide chains that form the cross-links. …[1] (ii) Describe the mode of action of antibiotics, such as penicillin, on bacteria. … … … … … … … …[4] (iii) Suggest the name of the type of enzyme that assembles the peptide chains that form the cross-links in peptidoglycans. …[1]

6 9700/42/O/N/09 For Examiner’s Use © UCLES 2009 (b) State why antibiotics, such as penicillin, have no effect on viruses. … …[1] (c) Bacteria may be Gram-positive or Gram-negative. Fig. 3.2 shows a diagram of part of the cell walls of both Gram-positive and Gram-negative bacteria. inner membrane peptidoglycan outer membrane inner membrane peptidoglycan periplasmic space Gram-positive Gram-negative Gram-positive bacteria cell walls have Gram-negative bacteria cell walls have a a peptidoglycan content of 50% peptidoglycan content of 10 – 20% Fig. 3.2 Suggest why Gram-positive bacteria are more susceptible to the action of penicillin than Gram-negative bacteria. … … … …[2]

7 9700/42/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 (d) There is evidence that some bacteria have developed resistance to antibiotics. One form of pneumonia, a serious lung disease, is caused by the bacterium Streptococcus pneumoniae. The Canadian Health Service has carried out a survey to show how the resistance of S. pneumoniae to penicillin has changed over the last 20 years. Fig. 3.3 shows the results of this survey. 0 1987 2 4 6 8 10 12 14 16 18 percentage of penicillin resistant S. pneumoniae year 1991 1995 1999 2003 2007 Fig. 3.3 Describe the results shown in Fig. 3.3 and explain how some strains of S. pneumoniae may have become resistant to penicillin. … … … … … … … … … …[5] [Total: 14]

8 9700/42/O/N/09 For Examiner’s Use © UCLES 2009 4 (a) In Queensland, Australia, the effect of the water-holding capacity of soil on the yield of sorghum and wheat was investigated. • Four test plots were prepared, two with high water-holding capacity (HWC) soil and two with low water-holding capacity (LWC) soil. • Sorghum seeds were sown on one plot with HWC soil and one plot with LWC soil. • Wheat seeds were sown on the second plot with HWC soil and the second plot with LWC soil. • The plots were regularly watered or irrigated throughout the growing season. • The yield of sorghum and wheat from all four plots was measured at the end of the growing season. Fig. 4.1 shows the results of this investigation. 0 sorghum LWC yield / kg per hectare sorghum HWC wheat LWC wheat HWC 1000 2000 3000 4000 5000 6000 Fig. 4.1 www.OnlineExamHelp.com

9 9700/42/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 (i) Describe and explain the results shown in Fig. 4.1. … … … … … … … …[4] (ii) State two factors, other than water, light and temperature that would have to be controlled during this investigation to ensure that the results were valid. 1 … 2 …[2] (b) Sorghum is able to carry out photosynthesis at high temperatures by preventing photorespiration. Explain how sorghum is able to prevent photorespiration. … … … … … … … …[4] [Total: 10]

10 9700/42/O/N/09 For Examiner’s Use © UCLES 2009 5 (a) Fig. 5.1 shows a section through part of a human testis. A B Fig. 5.1 Name structures A and B. A ……………………………………………………… B ……………………………………………………… [2] (b) Spermatogenesis, the production of sperm, begins in the testes of a boy around the age of 11 and can continue for the rest of his life. Fig. 5.2 outlines the sequence of events that occur during spermatogenesis. germinal epithelium spermatogonium cell division 1 stage C cell division 2 cell division 3 maturation cell D cell E cell F spermatozoan Fig. 5.2

11 9700/42/O/N/09 For Examiner’s Use © UCLES 2009 With reference to Fig. 5.2, (i) state which cell division is mitotic, …[1] (ii) state which cells are haploid, … …[2] (iii) state what is happening to the cell during stage C. …[1] (c) The middle piece of a spermatozoan contains many mitochondria. Suggest why a spermatozoan needs so many mitochondria. … … … …[2] (d) Some couples have difficulty in conceiving. This could be due to a problem with either the male or female reproductive systems. (i) Suggest reasons why a man may be infertile. … … … … … …[3] [Turn over

12 9700/42/O/N/09 For Examiner’s Use © UCLES 2009 (ii) In vitro fertilisation (IVF) is a widely used treatment for infertility. Explain what is meant by the term in vitro fertilisation. … … … …[2] (iii) At one IVF clinic, over 1000 treatment cycles were monitored. The number of live births was recorded as a percentage of the number of treatment cycles for each age group. The results are shown in Table 5.1. Table 5.1 age of women/years percentage of live births per treatment cycle under 34 27.6 34 to 36 22.3 37 to 39 18.3 40 to 42 10.0 above 42 less than 5.0 The data in Table 5.1 show that there is a decrease in the percentage of live births per treatment cycle with increasing age. Explain this trend. … … … … … …[3] [Total: 16] www.OnlineExamHelp.com

13 9700/42/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 6 (a) The pancreas acts both as an exocrine and an endocrine gland. (i) Describe the parts of the pancreas involved in its endocrine function. … … … … … …[3] (ii) State precisely the group of compounds to which the pancreatic hormone insulin belongs. …[1] (b) People with insulin-dependent (type 1) diabetes require regular injections of insulin. In the past the insulin used came from animal sources such as pigs. Diabetics now use human insulin that has been manufactured using gene technology. Describe the advantages of treating diabetics with insulin produced by gene technology. … … … … … …[3] [Total: 7]

14 9700/42/O/N/09 For Examiner’s Use © UCLES 2009 7 Pompe disease is a rare neuromuscular disease caused by an autosomal recessive allele. This allele prevents the production of an enzyme called acid alpha-glucosidase (AG), which breaks down glycogen in muscle cells. Glycogen can build up in muscle cells causing damage to the cells. This damage leads to muscle weakness which gets worse with time. (a) Explain how two parents, both of whom produce normal amounts of AG, can produce a child with Pompe disease. … … … … … …[3] (b) One form of treatment is enzyme replacement therapy where AG is given through regular injections. (i) Suggest how AG may be manufactured. … …[1] (ii) Name the hormone that stimulates the breakdown of glycogen in liver cells. …[1] (iii) State under what conditions glycogen would need to be broken down in liver or muscle cells. … …[1] (c) The MN blood group system is based on the presence of glycoproteins M and N, on the surface membrane of red blood cells, which act as antigens. State what is meant by the term antigen. … …[1]

15 9700/42/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 (d) The type of MN antigen on the surface membrane of red blood cells is controlled by a single gene with two alleles, LM and LN. The phenotypes of the MN blood group system are MM, MN and NN. Complete the genetic diagram to show how the MN blood group is inherited. parental phenotypes MN x MN parental genotypes … … gametes … offspring genotypes … offspring phenotypes … [3] (e) Allele frequencies for LM and LN vary in different human populations throughout the world. Table 7.1 shows the LM and LN allele frequencies from five populations. Table 7.1 population allele frequency / % LM LN Canadian Inuit 91 9 Egyptian 52 48 German 55 45 Chinese 57 43 Nigerian 55 45 Discuss the data shown in Table 7.1. … … … … … …[3] [Total: 13]

16 9700/42/O/N/09 For Examiner’s Use © UCLES 2009 8 (a) In the majority of plants the leaf is the main photosynthetic organ. List four ways in which the structure of a dicotyledonous leaf is adapted for gas exchange. 1 … … 2 … … 3 … … 4 … … [4] In an experiment to investigate the effect of light intensity on the rate of photosynthesis, the following procedure was carried out. ● Discs were cut, using a cork borer, from the photosynthetic tissue of the brown alga, Fucus serratus, a common seaweed of rocky shores. ● Ten discs were placed in each of four beakers containing 50 cm3 of sea water. The discs are denser than sea water and therefore initially sink to the bottom of the beaker. ● Each beaker was illuminated with a bench lamp placed at different distances, d, from the beaker. ● With time the discs began to rise to the surface of the water. ● The time, t, in minutes, at which the fifth disc from each batch reached the surface was recorded. ● The rate of photosynthesis was determined by calculating 1000 / t. A student’s set of results is shown in Table 8.1. Table 8.1 distance of beaker from lamp, d / cm light intensity 1 / d 2 time for fifth disc to reach the surface t / min rate of photosynthesis 1000 / t 5 0.04 23 43.5 10 0.01 36 27.8 15 0.004 52 19.2 20 … 88 11.4

17 9700/42/O/N/09 [Turn over For Examiner’s Use © UCLES 2009 (b) Calculate the value for light intensity when the distance between beaker and lamp was 20 cm. Record the value in the space in Table 8.1. [1] (c) Explain why the discs rise to the surface after being illuminated for a length of time. … … … … … …[3] (d) Using the data in Table 8.1, describe the relationship between light intensity and the rate of photosynthesis. … … … …[2] (e) The student found that there was no increase in the rate of photosynthesis when two lamps were placed 5 cm from the beaker. Suggest why there was no increase in the rate of photosynthesis. … … … …[2] [Total: 12] www.OnlineExamHelp.com

18 9700/42/O/N/09 For Examiner’s Use © UCLES 2009 Section B Answer one question. 9 (a) Outline the main features of the Krebs cycle. [9] (b) Explain the role of NAD in aerobic respiration. [6] [Total: 15] 10 (a) Describe how a nerve impulse crosses a cholinergic synapse. [9] (b) Explain the roles of synapses in the nervous system. [6] [Total: 15] … … … … … … … … … … … … … … … … … …

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24 9700/42/O/N/09 BLANK PAGE Copyright Acknowledgements: Question 5, Fig. 5.1 © P608/189; Human testis; 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.

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 9700/42 Paper 42 (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 42 © UCLES 2009 Section A Question Expected Answers Marks kingdom process or feature Prokaryotae Protoctista Fungi Plantae Animalia 80s ribosomes
    cell walls contain chitin



; circular DNA 



; endoplasmic reticulum     ; most species unicellular  


; autotrophic  

; heterotrophic   
 ; 1 one mark for each correct row if there are any blanks in a row then award no marks for that row [6] [Total: 6] 2 (a) isolating mechanism - geographical / mountains / physical barrier ; type of speciation – allopatric ; [2] (b) 1 2 3 4 5 6 7 8 mouse populations separated by mountains ; no, breeding / gene flow, between populations ; mutations occur ; different selection pressures / different (environmental) conditions ; genetic change ; e.g. different alleles selected for / change in allele frequency / change in gene pool / advantageous alleles passed on ; (results in) different chromosome numbers ; genetic drift ; (different populations ultimately) cannot interbreed ; R different species [5 max] [Total: 7]

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 42 © UCLES 2009 3 (a) (i) condensation ; [1] (ii) 1. autolysins ; 2. make holes in cell walls ; 3. in, growing / developing, bacteria ; 4. (antibiotic), inhibits / acts on, (another) enzyme ; 5. so peptidoglycan chains cannot link up / stops cross-links forming ; 6. cell wall becomes weaker / AW ; 7. turgor of cell not resisted (by cell wall) / AW ; 8. cell bursts ; [4 max] (iii) (glycoprotein) peptidase ; R other peptidase [1] (b) viruses have no cell wall ; [1] (c) 1 2 3 assume gram+ unless otherwise stated (gram+) penicillin can reach, cell wall / peptidoglycan, directly /AW / (gram-) ora ; (gram-) outer membrane provides protection (from penicillin) / (gram+) ora ; (gram+) more % peptidoglycan in wall (so greater effect from penicillin) / (gram-) ora ; [2 max]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 42 © UCLES 2009 (d) 1 2 3 4 5 6 7 8 9 10 accept antibiotic for penicillin and bacteria for S. pneumoniae throughout increase in resistance (throughout time period) ; paired figs + units ; overuse / misuse, of penicillin ; some S. pneumoniae survive ; mutation (in S. pneumoniae) ; resistance, gene / allele ; resistance passed to other bacteria ; e.g. plasmid transfer resistant strain, multiplies ; idea of many produced beta – lactamase produced ; breaks down penicillin ; point 7 accept vertical or horizontal transfer point 8 accept vertical transfer only [5 max] [Total: 14] 1. yield for sorghum is greater than yield for wheat (in any soil type) ; 2. yield for wheat is better in HWC soil / little difference in yield for sorgham ; 3. paired figs ; only award if linked correctly to mp 1 or mp2 4. sorghum is adapted to live in arid environment / AW ; 5. and 6. any two of the following ;; feature function extensive / deep, root system maximises water absorption curled leaves / leaves small surface area / wazy leaves / bulliform leaf cells / hinged leaf cells / reduced stomata numbers / stomata in pits reduces water loss high silica content / more sclerenchyma / more strengthening tissue reduces wilting 4 (a) (i) [4 max] www.OnlineExamHelp.com

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 42 © UCLES 2009 (ii) number of seeds sown ; density of seeds sown / area of plot ; minerals / fertilisers ; wind / shelter ; soil pH; [2 max] (b) 1. ref. bundle sheath cells; 2. light independent stage occurs / RuBP found (in bundle sheath cells) ; 3. RuBP / rubisco, kept away from, air / oxygen ; 4. by mesophyll cells ; 5. limits uptake of O2 / maintains high CO2 concentration (in bundle sheath cells) ; 6. enzymes / PEP carboxylase, have high optimum temperature ; 7. approx 450C ; 8. not denatured ; [4 max] [Total: 10] 5 (a) A – Leydig cell / interstitial cell ; B – (wall of) seminiferous tubule ; [2] (b) (i) 1 ; [1] (ii) mark first two answers E ; A secondary spermatocyte F ; A spermatid spermatozoan ; [2 max] (iii) cells grow in size / cells grow larger ; [1] (c) 1 2 3 ATP production / provides energy ; R produces energy (for) movement of flagellum ; R tail (for) production of acrosomal enzymes ; [2 max]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 42 © UCLES 2009 (d) (i) 1. infectious disease causes damage ; A mumps / Chlamydia / STDs 2. lower sperm count / absence of sperm ; 3. damaged / abnormal / immobile / lazy , sperm ; 4. blocked sperm ducts / lack of seminal fluid ; 5. named genetic condition ; e.g. CF 6. autoimmune reaction to sperm ; 7. reduced testosterone ; 8. effect of chemical damage ; e.g. chemotherapy / hormones in drinking water [3 max] (ii) (fertilisation of) oocyte by sperm ; in glass dish ; A appropriate glassware R test tube AVP ; e.g. sperm injected into oocyte [2 max] (iii) 1. ovulation less likely ; 2. (older) oocytes less likely to be fertilised / oocytes less viable ; 3. implantation less likely (in uterus of older woman) ; 4. miscarriage rate increases (with age) ; 5. (as) lower concentration of hormones / unbalanced hormones (in older woman) / start of menopause ; 6. (as) genetic defects / mutations, increase (with age) ; [3 max] [Total: 16] 6 (a) (i) ignore refs to function islets of Langerhans ; scattered throughout pancreas / AW ; alpha and beta cells ; blood supply (to carry hormones away) ; [3 max] (ii) globular protein ; [1]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 42 © UCLES 2009 (b) 1 2 3 4 5 6 7 it is identical to human insulin / fits membrane receptor on (target) cells ; (more) rapid response ; no / fewer, rejection problems / side effects / allergic reactions ; ref. to ethical / moral / religious, issues ; cheaper to produce in large volume / unlimited availability ; R cheap to produce less risk of, transmitting disease / infection ; good for people who have developed tolerance to animal insulin ; [3 max] [Total: 7] 7 (a) parents, carriers / heterozygous ; child homozygous recessive ; ¼ / 0.25 / 25%, chance ; mutation ; [3 max] (b) (i) gene technology / genetic engineering / description ; [1] (ii) glucagon ; [1] (iii) low blood glucose concentration / during or after exercise ; R sugar [1] (c) foreign / non-self / cell recognition ; stimulates immune response / AW ; [1 max] (d) parental genotypes LMLN x LMLN gametes LM or LN LM or LN ; parental genotypes and gametes for one mark offspring genotypes LMLM LMLN LMLN LNLN ; offspring phenotypes MM MN MN NN ; [3] penalise once for omission of L

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 42 © UCLES 2009 (e) Canadian Inuit, allele frequencies / LMLN ratio, different from others ; high frequency of LM / low frequency of LN , compared to other populations ; R just highest LM / lowest LN less outbreeding / more inbreeding ; AVP; e.g. LM has selective advantage in Inuit environment [3 max] [Total: 13] 8 (a) 1 2 3 4 5 6 7 stomata ; air spaces (between cells) ; thin cell walls ; moist internal walls ; thin leaf ; cylindrical palisade cells ; large surface area of, palisade / mesophyll, cells ; [4 max] (b) 0.0025 / 2.5 x 10 -3; A 0.003 only if 0.0025 in answer [1] © 1 2 3 4 photosynthesis takes place ; oxygen is produced ; collects, inside disc / on surface of disc ; disc, less dense / more buoyant ; [3 max] (d) rate of photosynthesis increases as light intensity increases ; paired data quotes from columns 2 and 4 ; [2] (e) 1 2 3 light intensity no longer limiting ; carbon dioxide, concentration / rate of diffusion, now limiting ; temperature, too high / denatures enzymes ; [2 max] [Total: 12] www.OnlineExamHelp.com

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 42 © UCLES 2009 Section B: only one question to be answered. 9 (a) 1 2 3 4 5 6 7 8 9 10 11 12 13 acetyl CoA combines with oxaloacetate ; to form citrate ; 4C to 6C ; decarboxylation / CO2 released ; dehydrogenation / oxidation / release of hydrogen ; reduced NAD produced / NAD accepts hydrogen ; reduced FAD produced / FAD accepts hydrogen ; ATP produced ; substrate level phosphorylation ; series of, steps / intermediates ; A many named steps off a diagram enzyme catalysed reactions ; oxaloacaetate regenerated ; occurs in mitochondrial matrix ; [9 max] accept diagram (b) 14 15 16 17 18 19 20 21 22 coenzyme ; for dehydrogenase ; reduced ; carries, electrons and protons / hydrogen / NAD from Krebs cycle ; and glycolysis ; to ETC / electron carrier chain / oxidation ; reoxidised / regenerated hydrogen removed ; ATP produced ; [6 max] [Total: 15]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9700 42 © UCLES 2009 10 (a) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 action potential / depolarisation, reaches presynaptic membrane ; (Ca2+) channels open in presynaptic membrane / presynaptic membrane becomes more permeable to (Ca2+) ; R calcium / Ca / Ca+ Ca2+ (flood) into presynaptic, neurone / knob ; R membrane (this causes) vesicles of, acetylcholine / ACh ; (to) move towards presynaptic membrane / (to) fuse with presynaptic membrane; ACh released into synaptic cleft / exocytosis of ACh ; ACh diffuses across (cleft) ; ACh binds to receptor (proteins) / AW ; on postsynaptic membrane ; proteins change shape / channels open ; sodium ions (rush) into postsynaptic neurone ; R membrane postsynaptic membrane depolarised ; action potential / nerve impulse ; action of acetylcholinesterase ; [9 max] (b) 15 16 17 18 19 20 21 22 23 24 ensure one-way transmission; receptor (proteins) only in postsynaptic, membrane / neurone ; ora vesicles only in presynaptic neurone ; ora adaptation / ACh amount reduces due to overuse of synapse ; wide range of responses ; due to interconnection of many nerve pathways ; inhibitory synapses affect other synapses ; involved in memory / learning ; due to new synapses being formed ; summation / discrimination ; [6 max] [Total: 15] www.OnlineExamHelp.com