Cambridge A Level Biology 9700 — 2010 May/June Paper 4 · Variant 1

This document consists of 19 printed pages, 3 lined pages and 2 blank pages. DC (AT/JG) 13968/5 © UCLES 2010 [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 on all the work you hand in. Write in dark blue or black pen. 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. 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. * 6 6 3 3 5 1 8 7 9 7 * BIOLOGY 9700/41 Paper 4 A2 Structured Questions May/June 2010 2 hours Candidates answer on the Question Paper. Additional Materials: Answer Paper available on request. 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/10 © UCLES 2010 For Examiner’s Use Section A Answer all the questions. 1 The flatback turtle, Natator depressus, is an endangered species that nests on northern Australian beaches. Fig. 1.1 shows a flatback turtle. Fig. 1.1 Each female lays approximately 50 eggs per nest, which is a smaller number than all other species of marine turtle. The eggs are buried in the sand and when the hatchlings emerge each has a mass of approximately 43 g. Unlike most marine turtles, flatback turtles spend most of their time in coastal waters. This is where they feed and mate. Fig. 1.2 shows the numbers of female flatback turtles nesting on a beach in northern Australia between 1993 and 2002. 0 100 200 300 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 number of female turtles nesting year Fig. 1.2

3 9700/41/M/J/10 © UCLES 2010 [Turn over For Examiner’s Use (a) Calculate the mean rate of decrease in the numbers of females nesting between 1993 and 2002. Show all the steps in your calculation. Answer … [2] (b) Suggest ways in which the flatback turtle could be protected. … … … … … … … … … … [5] [Total: 7]

4 9700/41/M/J/10 © UCLES 2010 For Examiner’s Use 2 (a) The glycoproteins CD28 and CD40 are found on the surface of T-lymphocytes (T-cells). They are binding sites for cell-signalling molecules and are essential for triggering the cloning of T-cells in an immune response. A monoclonal antibody (mAb), which could block the CD40 signalling pathway, was produced from hamsters using the hybridoma method. Outline the procedure, starting with a hamster, for producing mAbs suitable for use in another mammal, such as a mouse. … … … … … … … … [4]

5 9700/41/M/J/10 © UCLES 2010 [Turn over For Examiner’s Use (b) The ability of the mAb produced in (a) to prevent rejection of transplanted hearts in mice was compared with that of a protein, P, which blocks the CD28 signalling pathway. Four groups of mice were treated as follows: • group A – no treatment • group B – treated with protein P only • group C – treated with mAb only • group D – treated with both mAb and protein P. Fig. 2.1 shows the percentage survival of the transplanted hearts in the four groups of mice over a period of 80 days. 0 20 40 time / days 60 80 100 75 50 25 0 percentage survival of transplanted hearts D B C A Fig. 2.1 With reference to Fig. 2.1 (i) describe the effectiveness of the four different treatments … … … … … … … … [4]

6 9700/41/M/J/10 © UCLES 2010 For Examiner’s Use (ii) suggest an explanation for the differences in survival of the transplanted hearts in groups B and D. … … … … [2] (c) Examination of the transplanted hearts showed that the hearts in group A had significant damage to their coronary arteries, whereas in group D these blood vessels appeared normal. Explain the importance of the coronary arteries of the heart. … … … … … … [3] (d) State two uses of mAbs in humans, other than preventing rejection of transplanted tissue. 1. … … 2. … … [2] [Total: 15]

7 9700/41/M/J/10 © UCLES 2010 [Turn over For Examiner’s Use 3 Fig. 3.1 shows the structure of part of a seminiferous tubule of a human testis. E F G H Fig. 3.1 (a) Name cells E to H. E … F … G … H … [4] (b) Describe how cell F is produced from cell E in the process of spermatogenesis. … … … … … … … … [4] [Total: 8]

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9 9700/41/M/J/10 © UCLES 2010 [Turn over For Examiner’s Use 4 Fig. 4.1 is a photomicrograph of a transverse section through the leaf of a C4 plant. J K L Fig. 4.1 (a) (i) Identify structures J to L. J … K … L … [3] (ii) Outline how this leaf anatomy adapts the plant for high rates of carbon fixation at high temperatures. … … … … … … … … … … [4]

10 9700/41/M/J/10 © UCLES 2010 For Examiner’s Use (b) Sorghum is a C4 plant and Sorghum bicolor is a major food crop in dry tropical regions. The leaves of S. bicolor are covered with a layer of wax made up of a mixture of esters and free fatty acids, with a melting point of 77– 85 °C. Waxes from the leaves of non- tropical plants tend to have melting points lower than this. For example, wax from the bayberry, Myrica sp., has a melting point of 45 °C. Suggest how the wax on sorghum leaves helps the plant to survive in dry, tropical regions. … … … … [2] (c) An investigation was carried out into the response of sorghum to being kept at a low temperature for a short period of time. Soybean plants, which are better adapted than sorghum for growth in subtropical and temperate climates, were used for comparison. Plants of sorghum and soybean were kept at 25 °C for several weeks and then at 10 °C for three days. The temperature was then increased to 25 °C again for seven days. Day length, light intensity and carbon dioxide concentration were kept constant throughout. The uptake of carbon dioxide, as mg CO2 absorbed per gram of leaf dry mass, was measured • at 25 °C before cooling • on each of the three days at 10 °C • for seven days at 25 °C. The results are shown in Table 4.1. Table 4.1 plant carbon dioxide uptake / mg CO2 g–1 at 25 °C, before cooling at 10 °C at 25 °C (mean over days 4 to 10) day 1 day 2 day 3 sorghum 48.2 5.5 2.9 1.2 1.5 soybean 23.2 5.2 3.1 1.6 6.4

11 9700/41/M/J/10 © UCLES 2010 [Turn over For Examiner’s Use (i) Compare the changes in carbon dioxide uptake in sorghum and soybean during the three days at 10 °C. … … … … [2] (ii) During the cooling period, the ultrastructure of the sorghum chloroplasts changed. The membranes of the thylakoids moved closer together, eliminating the spaces between them. The size and number of grana became reduced. Explain how these changes could be responsible for the low rate of carbon dioxide uptake by sorghum even when returned to a temperature of 25 °C. … … … … … … … [4] [Total: 15]

12 9700/41/M/J/10 © UCLES 2010 For Examiner’s Use 5 Anole lizards are found throughout the Caribbean and the surrounding mainland. There are many species. Each species is found only on one island or a small group of islands, apart from Anolis carolinensis which is found in mainland Florida. Fig. 5.1 shows the distribution of four species of anole lizards. Florida U.S.A. Cuba Andros Acklins Anolis smaragdinus Anolis brunneus Anolis carolinensis Anolis porcatus 0 50 100 km Fig. 5.1 An investigation was carried out into the relationships between these four lizard species, using DNA analysis. The base sequences of a region of mitochondrial DNA from the four species were compared. The results are shown in Table 5.1. The smaller the number, the smaller the differences between the base sequences of the two species.

13 9700/41/M/J/10 © UCLES 2010 [Turn over For Examiner’s Use Table 5.1 A. brunneus A. brunneus A. smaragdinus A. smaragdinus 12.1 A. carolinensis A. carolinensis 16.7 15.0 A. porcatus A. porcatus 11.3 8.9 13.2 (a) With reference to Table 5.1, state the species to which A. brunneus appears to be most closely related. … [1] (b) The researchers put forward the hypothesis that the three species, A. brunneus, A. smaragdinus and A. carolinensis, have originated from three separate events in which a few individuals of A. porcatus spread directly from Cuba to three different places. Explain how the results in Table 5.1 support the researchers’ hypothesis. … … … … … … [3] (c) Explain how a population of A. porcatus that became isolated on an island could evolve into a new species. … … … … … … … … [4] [Total: 8]

14 9700/41/M/J/10 © UCLES 2010 For Examiner’s Use 6 (a) Explain what is meant by the term gene mutation. … … … … [2] (b) Rickets is a childhood disorder involving the softening and weakening of bones. It is usually caused by a lack of vitamin D, calcium ions or phosphate ions. A rare form of rickets that cannot be successfully treated with vitamin D therapy is caused by a mutant allele on the X chromosome. Fig. 6.1 shows a pedigree chart for a family that has a history of this condition. 1 3 4 5 8 9 10 11 6 7 2 key: male without rickets female without rickets male with rickets female with rickets Fig. 6.1 Using the symbols XR for the mutant allele on the X chromosome Xr for the non-mutant allele on the X chromosome state the genotypes of the following individuals. 1 … 3 … 9 … 10 … [4]

15 9700/41/M/J/10 © UCLES 2010 [Turn over For Examiner’s Use (c) The gene in which this mutation occurs codes for a protein found in the cells of the proximal convoluted tubule of the kidney. This protein is involved in phosphate ion transport across membranes. Suggest why individuals with this mutant allele show symptoms of rickets. … … … … [2] [Total: 8]

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17 9700/41/M/J/10 © UCLES 2010 [Turn over For Examiner’s Use 7 (a) The initial stages of respiration convert one molecule of glucose into two molecules of a 3C compound. State (i) the name given to these initial stages … [1] (ii) where these stages occur in cells … [1] (iii) the total number of ATP molecules formed during these stages. … [1] (b) Most of the ATP formed in respiration is produced within the mitochondria by oxidative phosphorylation. (i) State the location, in the mitochondrion, of oxidative phosphorylation. … [1] (ii) Outline the process of oxidative phosphorylation. … … … … … … … … … … … [5]

18 9700/41/M/J/10 © UCLES 2010 For Examiner’s Use (c) In an investigation, mammalian liver cells were homogenised (broken up) and the resulting homogenate centrifuged. Samples of the complete homogenate and samples containing only nuclei, only ribosomes, only mitochondria or only the remaining cytosol were incubated with: 1 glucose 2 pyruvate 3 glucose and cyanide 4 pyruvate and cyanide Cyanide inhibits oxidative phosphorylation. After incubation the presence or absence of carbon dioxide and lactate in each sample was determined. The results are summarised in Table 7.1. Table 7.1 samples of homogenate complete only nuclei only ribosomes only mitochondria only cytosol carbon dioxide lactate carbon dioxide lactate carbon dioxide lactate carbon dioxide lactate carbon dioxide lactate 1 glucose ✓ ✓ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✓ 2 pyruvate ✓ ✓ ✘ ✘ ✘ ✘ ✓ ✘ ✘ ✓ 3 glucose and cyanide ✘ ✓ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✓ 4 pyruvate and cyanide ✘ ✓ ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✓ ✘ = absent ✓ = present (i) With reference to Table 7.1, name the two organelles not involved in respiration. 1. … 2. … [1]

19 9700/41/M/J/10 © UCLES 2010 [Turn over For Examiner’s Use (ii) Explain why carbon dioxide is produced when mitochondria are incubated with pyruvate but not when they are incubated with glucose. … … … … … … [3] (iii) Explain why, in the presence of cyanide, lactate is produced but carbon dioxide is not. … … … … … … [3] [Total: 16]

20 9700/41/M/J/10 © UCLES 2010 For Examiner’s Use 8 (a) Cystic fibrosis (CF) is an inherited disease. (i) Explain briefly how two parents who do not have CF may have a child with CF. … … … … [2] (ii) Describe two ways in which CF affects the lungs. … … … … [2] (b) One potential way of treating CF is by using gene therapy. (i) Outline, with reference to CF, what is meant by gene therapy. … … … … [2] (ii) Describe one possible advantage and one possible disadvantage of using gene therapy to treat CF. advantage … … disadvantage … … [2] [Total: 8]

21 9700/41/M/J/10 © UCLES 2010 [Turn over For Examiner’s Use Section B Answer one question. 9 (a) Explain how the palisade mesophyll cells of a leaf are adapted for photosynthesis. [8] (b) Outline the light-independent stage of photosynthesis. [7] [Total: 15] 10 (a) Describe the structure of a kidney nephron and its associated blood vessels. [7] (b) Explain how glomerular filtrate is formed. [8] [Total: 15] … … … … … … … … … … … … … … … … … … …

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24 9700/41/M/J/10 © UCLES 2010 For Examiner’s Use … … … … … … … … … … … … … … … … … … … … … … … Copyright Acknowledgements: Question 1 © www.iucn-mtsg.org/species/images/flatback.jpg. Question 4 © http://delta-intkey.com/grass/images/pennis80.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.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2010 question paper for the guidance of teachers 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 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 May/June 2010 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 AS/A LEVEL – May/June 2010 9700 41 © UCLES 2010 1 (a) (275 – 90) or 185 or 1705 for 1 mark 10 10 10 18.5 ;; A 19 R 18 [2] (b) 1 avoid disturbance to, nest sites/nesting females ; R ref. to mating 2 protect, nest sites/young, from predators ; 3 avoid sea pollution ; 4 example of pollution ; e.g. do not throw rubbish into sea / avoid discharge from boats/light pollution (beaches) 5 take care when fishing (with nets) ; 6 stop hunting of adults ; A trading ban on turtle products 7 captive breeding programmes/AW ; 8 conservation areas/zoos ; 9 education/ecotourism ; [5 max] [Total: 7] 2 (a) 1 hamster injected with, antigen/CD40 ; 2 B cells/plasma cells, with ability to make antibody taken ; 3 from spleen ; 4 (B cells/plasma cells) fused with, tumour/cancer/myeloma, cell ; 5 use of, fusogen/PEG ; 6 (hybrid) cells cultured/AW ; R use of fermenter 7 check cells for mAb production ; 8 (antibody making) cells mass produced/AW ; A use of fermenter [4 max]

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9700 41 © UCLES 2010 (b) (i) accept mouse survival for heart survival 1 in A, 100% hearts survive 10 days or no heart survives 20 days ; 2 in D, 100% hearts survive, 80 days/to end of investigation ; 3 in B, 100% hearts survive 30 days or 10% hearts survive, 80 days/to end of investigation ; 4 in C, 100% hearts survive 30 days or 75% hearts survive, 80 days/to end of investigation ; penalise once for no mention of percentage in mps 2, 3 and 4 [4] (ii) 1 in D, both pathways/CD28 and CD40, blocked ; 2 so T-cells cannot be cloned/no immune response ; 3 in B, CD40 pathway is not blocked/only CD28 is blocked ; 4 so T cells can still be cloned/immune response triggered ; [2 max] (c) 1 carry blood to, cardiac/heart, muscle/tissue/cells ; 2 supply oxygen ; 3 supply, nutrient/named nutrient ; 4 for, energy release/respiration ; R produce energy [3 max] (d) two of the following: 1 diagnosis of, disease/named disease ; e.g. gonorrhoea/HIV 2 treatment of disease ; e.g. directing drugs to cancerous cells A autoimmune disease but not tissue or blood typing 3 pregnancy testing/drug testing ; 4 (passive) vaccine production ; [2 max] [Total: 15]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9700 41 © UCLES 2010 3 (a) E – spermatogonium/germinal epithelial cell ; F – secondary spermatocyte ; G – spermatid ; R spermatozoa H – Sertoli cell/nurse cell ; [4] (b) Accept identification of cells from diagram. 1 cell E mitosis ; 2 (E / spermatogonia) increases in size/AW ; 3 becomes a primary spermatocyte ; 4 (primary spermatocyte) meiosis I ; 5 forms secondary spermatocyte(s) ; 6 2n to n/diploid to haploid/halving chromosome number ; [4 max] [Total: 8] 4 (a) (i) J – epidermis/epidermal cell ; K – mesophyll (cell) ; L – bundle sheath (cell) ; [3] (ii) 1 mesophyll cells tightly packed/AW ; 2 so O2 cannot reach bundle sheath cells ; 3 light independent stage/Calvin cycle or RuBP, in bundle sheath cells ; 4 ref. malate shunt ; 5 maintains high CO2 concentration (in bundle sheath cells) ; 6 PEP carboxylase, has high optimum temperature/has higher affinity for CO2 /doesn’t accept O2 ; 7 (PEP carboxylase) not denatured ; 8 photorespiration is avoided ; [4 max] (b) 1 reduces water loss/AW ; 2 wax does not melt ; 3 shiny surface reflects radiation ; [2 max]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9700 41 © UCLES 2010 (c) (i) greater reduction in sorghum than in soybean ; use of comparative figures ; e.g. sorghum 5.5 to 1.2 or by 4.3 soybean 5.2 to 1.6 or by 3.6 [2] (ii) reject ‘no’ for all points 1 less surface area ; 2 less absorption of light ; 3 less, photophosphorylation / light dependent reaction ; 4 less chemiosmosis ; 5 (due to) smaller thylakoid space or reduced proton gradient ; 6 less ATP (produced) ; 7 less reduced NADP (produced) ; 8 light-independent reaction / Calvin cycle, slows down ; 9 less carbon dioxide, fixed / combined with PEP ; R uptake [4 max] [Total: 15] 5 (a) (A.) porcatus ; [1] (b) 1 A. brunneus, A. smaragdinus and A. carolinensis have smaller differences with A. porcatus (than with others)/AW ; 2 therefore more closely related to A. porcatus (than to each other) ; 3 use of figures ; 4 AVP ; e.g. comment about figures for A. brunneus with A. smaragdinus/ ref. different times of separation [3 max]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9700 41 © UCLES 2010 (c) 1 allopatric speciation ; 2 (lizard populations) separated by water ; 3 geographical/physical, barrier ; 4 no, breeding/gene flow, between populations ; 5 mutations occur ; 6 different selection pressures/different (environmental) conditions ; 7 genetic change ; e.g. different alleles selected for/change in allele frequency/ change in gene pool/advantageous alleles passed on ; 8 (can result in) different chromosome numbers ; 9 genetic drift ; 10 ultimately, reproductively isolated/cannot interbreed ; [4 max] [Total: 8] 6 (a) change in, DNA/base sequence ; produces different allele ; ref. different, protein/polypeptide, produced ; [2 max] (b) 1 – XrXr ; 3 – XrY ; 9 – XRXr ; 10 – XRY ; [4] (c) answers must refer to phosphate ions 1 altered shape/non-functional/no, carrier protein ; 2 less/no, reabsorption of phosphate ions (into blood) ; 3 from, glomerular filtrate/lumen of/proximal convoluted tubule ; 4 more/all, phosphate ions excreted ; 5 low phosphate ion concentration in, blood/bones ; R no phosphate ion conc [2 max] [Total: 8]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9700 41 © UCLES 2010 7 (a) (i) glycolysis ; [1] (ii) cytoplasm/cytosol ; [1] (iii) 4 ; A 4 – 2 = 2 [1] (b) (i) inner membrane/cristae/stalked particles ; [1] (ii) 1 reduced, NAD/FAD ; 2 dehydrogenase enzymes ; 3 release hydrogen ; A H R H2/H+ 4 hydrogen splits into proton and electron ; 5 electrons flow down, ETC/AW ; 6 energy released ; 7 protons pumped (across inner membrane/from matrix) ; 8 into intermembrane space ; 9 proton gradient ; 10 protons pass through, ATP synthase/stalked particle ; 11 oxygen final, hydrogen/proton, acceptor ; [5 max] (c) (i) nuclei and ribosomes ; [1] (ii) 1 glycolysis, does not occur in mitochondrion/only occurs in cytosol or cytoplasm ; 2 pyruvate produced in glycolysis ; 3 pyruvate can enter mitochondrion/glucose cannot enter mitochondrion ; 4 carbon dioxide produced/decarboxylation, in, Krebs/link reaction ; [3 max] (iii) 1 cyanide, inhibits cytochrome oxidase is a non-competitive inhibitor ; 2 reduced NAD not oxidised/AW ; 3 Krebs cycle stops ; 4 alternative H acceptor needed/pyruvate is H acceptor/pyruvate is reduced ; R H+ 5 lactate produced in cytoplasm ; 6 by anaerobic respiration ; [3 max] [Total: 16]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9700 41 © UCLES 2010 8 (a) (i) 1 parents, heterozygous/carriers ; 2 CF allele recessive ; 3 CF child homozygous recessive ; [2 max] (ii) 1 thick/sticky/dehydrated, mucus produced ; 2 mucus not moved effectively by cilia/mucus accumulates ; R mucus blocks airway 3 reduced gaseous exchange/longer diffusion pathway ; 4 difficulty in breathing/AW ; 5 infections/(mucus) traps bacteria ; 6 lungs are scarred ; [2 max] (b) (i) 1 alters genotype ; 2 insert, dominant/normal, allele ; R gene 3 into, affected/appropriate, cells ; 4 use of vector/named vector ; 5 ref. recombinant DNA ; [2 max] (ii) advantage 1 treats cause not symptoms ; 2 no, physiotherapy/antibiotics/etc, needed ; 3 less time consuming than others treatments ; max 1 disadvantage 4 effects only last for a few days (at present)/low uptake by target cells ; 5 only target lung cells (at present) ; 6 side effects ; max 1 [2 max] [Total: 8]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9700 41 © UCLES 2010 9 (a) 1 closely packed to absorb maximum light ; 2 vertical/at right angles to surface of leaf to reduce number of cross walls ; 3 large vacuole pushes chloroplasts to edge of cell ; 4 chloroplasts at edge short diffusion path for carbon dioxide ; 5 chloroplasts at edge to absorb maximum light ; 6 large number of chloroplasts to absorb maximum light ; 7 cylindrical cells or air spaces to circulate gases/provide a reservoir of CO2 ; 8 large surface area for diffusion of gases ; 9 moist cell surfaces for diffusion of gases ; 10 cell walls thin for maximum light penetration/diffusion of gases ; 11 chloroplasts can move towards light ; 12 chloroplasts can move away from high light intensity to avoid damage ; [8 max] (b) 13 Calvin cycle/stroma ; 14 carbon dioxide fixed by RuBP ; 15 rubisco ; 16 2 molecules of GP formed ; A PGA 17 (GP) forms TP ; A GALP/PGAL 18 use of ATP ; 19 use of, reduced NADP/NADPH ; 20 from light dependent stage ; 21 some TP forms, hexose/sucrose/starch/cellulose/glycerol ; 22 some TP converted to acetyl CoA ; 23 some TP used to regenerate RuBP ; 24 using ATP ; allow either mp 18 or mp 24 marks can be awarded on a diagram [7 max] [Total: 15]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9700 41 © UCLES 2010 10 (a) 1 renal/Bowman’s, capsule ; 2 ref. podocytes ; 3 (proximal convoluted tubule/distal convoluted tubule/capsule) in cortex ; 4 proximal convoluted tubule ; 5 loop of Henle ; 6 (loop) in medulla ; 7 distal convoluted tubule ; 8 afferent arteriole ; 9 glomerulus ; 10 efferent arteriole ; 11 capillary network around/proximal convoluted tubule/loop/distal convoluted tubule ; 12 collecting duct ; accept points on a labelled diagram [7 max] (b) 13 endothelium of, blood capillaries/glomerulus ; 14 more/large, gaps between endothelial cells ; 15 podocytes ; 16 large gaps between podocytes/filtration slits ; 17 basement membrane, selective barrier/acts as a filter ; 18 prevents, large protein/RMM > 68 000, passing through ; 19 no cells pass through ; 20 named molecule which is filtered ; e.g. urea/water/glucose/uric acid/creatinine/ Na+/K+/Cl- ; 21 high, blood/hydrostatic, pressure in glomerulus ; 22 afferent arteriole wider than efferent arteriole ; 23 lower pressure in, renal/Bowman’s, capsule ; 24 fluid forced into capsule/ultrafiltration ; [8 max] [Total: 15]