Cambridge A Level Biology 9700 — 2009 May/June Paper 4 · Variant 1

This document consists of 19 printed pages, 4 lined pages and 1 blank page. SPA SHW 00311 6/08 V00747/4 © 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 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. * 4 9 6 5 1 1 8 6 0 1 * BIOLOGY 9700/04 Paper 4 A2 Structured Questions May/June 2009 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

9700/04/M/J/09 © UCLES 2009 For Examiner’s Use Section A Answer all the questions. 1 (a) The respiratory quotient (RQ) is used to show what substrates are being metabolised in respiration. The RQ of a substrate may be calculated using the formula below: RQ = molecules of CO2 given out molecules of O2 taken in When the unsaturated fatty acid linoleic acid is respired aerobically the equation is: C18H32O2 + 25O2 ……… CO2 + 16H2O + energy (i) Calculate how many molecules of carbon dioxide are produced when one molecule of linoleic acid is respired aerobically. answer … [1] (ii) Calculate the RQ for linoleic acid. answer … [1] (b) Hummingbirds feed on nectar from flowers only during daylight hours. Nectar is rich in sugars. Fig. 1.1 shows a hummingbird. Fig. 1.1 2

3 9700/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use A study of aerobic respiration in captive hummingbirds was carried out. The hummingbirds were allowed to feed freely and then made to fast for four hours in constant conditions. During this time their RQ values were calculated every 40 minutes. Fig. 1.2 shows the results from this study. 0.4 0 time after feeding / min RQ 40 80 120 160 200 240 0.5 0.6 0.7 0.8 0.9 1.0 1.1 Fig. 1.2 Describe and explain the results shown in Fig. 1.2. … … … … … … … [4]

4 9700/04/M/J/09 © UCLES 2009 For Examiner’s Use (c) Hummingbirds regulate their body temperature whereas butterflies do not regulate their body temperature. Explain briefly the effect of an increase in temperature on the rate of respiration of a butterfly. … … … … [2] [Total: 8]

5 9700/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use 2 (a) The steroid hormones oestrogen and progesterone are secreted by the ovary. State precisely the sites of secretion of each. oestrogen … progesterone … [2] (b) The most effective oral contraceptives for general use are the so-called combined oral contraceptives (COCs), which contain both oestrogen and progesterone. Explain how COCs produce their effects. … … … … … … … … [4] (c) Describe two social implications of the use of contraceptives. 1 … … 2 … … [2] [Total: 8]

6 9700/04/M/J/09 © UCLES 2009 For Examiner’s Use 3 (a) The African elephant, Loxodonta africana, is a large herbivorous mammal which eats tree leaves. It has the longest gestation period of any land mammal and normally produces one offspring at a time. Its habitat is mainly savannah. Suggest how human activities have caused the African elephant to become endangered. … … … … … … [3] (b) The meerkat, Suricata suricatta, also lives in the savannah of southern Africa. It is a carnivorous mammal and feeds on insects, worms, snails and other invertebrates. It grows up to 30 cm in length and lives in large family groups in burrows. Fig. 3.1 shows a meerkat. Fig. 3.1 With reference to the information given, suggest why the meerkat is less likely than the elephant to become endangered. … … … … … … [3] [Total: 6]

7 9700/04/M/J/09 © UCLES 2009 [Turn over BLANK PAGE Question 4 starts on page 8

8 9700/04/M/J/09 © UCLES 2009 For Examiner’s Use 4 (a) Fig. 4.1 shows a section through a maize fruit. A B C D Fig. 4.1 (i) Name the parts labelled A to D. A … B … C … D … [2] (ii) Describe the function of the endosperm. … … … … … … [3]

9 9700/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use (b) The corn earworm, is the larva of a moth Helicoverpa zea, that is a serious pest of maize. Insecticides containing pyrethrum have long been used to control this insect. These act by irreversibly inhibiting the enzyme acetylcholinesterase, which normally catalyses the hydrolysis of acetylcholine. (i) Describe how an insecticide could irreversibly inhibit acetycholinesterase. … … … … [2] (ii) Suggest the effects on synapses of this irreversible inhibition of acetylcholinesterase. … … … … [2] (c) Some populations of H. zea have developed resistance to pyrethrum. This occurs as the result of a point mutation of the acetylcholinesterase gene. Many different such mutations have been identified in different populations. Explain how a point mutation in the acetylcholinesterase gene could confer resistance to pyrethrum. … … … … … … [3]

10 9700/04/M/J/09 © UCLES 2009 For Examiner’s Use (d) • A group of corn earworms was collected from a field where the farmer had reported resistance to insecticides containing pyrethrum. • Another group was collected from a field where the insects showed no resistance (were susceptible). • Some individuals from these two groups were crossed with each other to form a hybrid group. Insects from each of the three groups were then exposed to a range of concentrations of pyrethrum. The percentage of the insects that were dead after 24 hours was recorded. The results are shown in Table 4.1. Table 4.1 dose of insecticide / g per group % mortality of insects after 24 hours resistant group hybrid group susceptible group 0 0 0 0 0.1 0 0 50 0.5 0 23 63 1.5 7 45 94 2.5 12 50 100 5.0 42 89 100 10.0 80 100 100 30.0 100 100 100 (i) With reference to Table 4.1, compare the effect of the insecticide on the resistant group and on the hybrid group. … … … … … … [3]

11 9700/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use (ii) Assuming that resistance is conferred by a single point mutation in the gene for acetycholinesterase, suggest an explanation for the overall differences between all three groups of insects in Table 4.1. … … … … [2] [Total: 17]

12 9700/04/M/J/09 © UCLES 2009 For Examiner’s Use 5 The fungus Penicillium chrysogenum is grown in fermenters on an industrial scale to produce penicillin, using a batch culture system. (a) Explain why batch culture, rather than continuous culture, is used for the production of penicillin. … … … … … … [3] (b) Temperature and pH are normally controlled in the fermenter. Temperature is kept constant, while pH is held at a value of 5.5 for the first stage of the fermentation and then raised to 6.8 and kept constant for the remainder of the fermentation period. Fig. 5.1 shows how the pH and the concentration of penicillin in the culture change over time, when the pH is controlled and when the pH is not controlled. 4 5 6 7 8 9 10 11 0 40 80 120 160 200 0 5 10 15 20 25 30 35 time / hours pH penicillin concentration / arbitrary units Key pH controlled pH not controlled Key pH controlled pH not controlled Fig. 5.1

13 9700/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use With reference to Fig. 5.1, describe and explain the differences in the concentration of penicillin in the culture when the pH is controlled and when the pH is not controlled. … … … … … … … … [4] (c) Explain why penicillin affects bacteria but not viruses. … … … … [2] [Total: 9]

14 9700/04/M/J/09 © UCLES 2009 For Examiner’s Use 6 (a) Describe the role of insulin in the regulation of blood glucose concentration. … … … … … [3] (b) State two advantages of treating diabetes with insulin produced by gene technology. 1 … … 2 … … [2] (c) One of the steps in the production of bacteria capable of producing human insulin is the insertion of the gene coding for human insulin into a plasmid vector. Fig. 6.1 shows one of the artificial plasmids constructed to act as a vector. ampicillin resistance gene DNA of plasmid tetracycline resistance gene target site for the restriction enzyme BamHI G G A T C C C C T A G G Fig. 6.1

15 9700/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use (i) With reference to Fig. 6.1, explain the importance of the plasmid having a single target site for a particular restriction enzyme, such as BamHI. … … … … [2] (ii) The genes for ampicillin resistance and tetracycline resistance on the plasmid allow the genetic engineer to distinguish between bacteria that have taken up different circles of DNA. Complete the table to show whether bacteria which have taken up each different circle of DNA are, or are not resistant to ampicillin, to tetracycline or to both. Show presence of resistance with a tick (✓) and absence of resistance with a cross (✗). circle of DNA taken up by bacteria bacteria resistant to ampicillin bacteria resistant to tetracycline unaltered plasmids recombinant plasmids that have taken up the wanted gene circles of the wanted gene [3] (d) (i) Explain why genes for antibiotic resistance are now rarely used as markers in gene technology. … … … … … … [3] (ii) Describe the use of one alternative marker gene that can be used instead of an antibiotic gene. … … … … [2] [Total: 15]

16 9700/04/M/J/09 © UCLES 2009 For Examiner’s Use 7 (a) The inheritance of coat colour in horses is complex but all horses have one of two base colours, red (chestnut) or black. The base colour is controlled in a simple monohybrid way. • When chestnut stallions and mares are mated the foals are always chestnut. • When black stallions are mated with black mares, either black or chestnut foals may be produced. Draw a genetic diagram to show how two parents with black coat colour can produce a chestnut foal and the probability of such an event occurring. Choose a letter symbol to represent coat colour. [4]

17 9700/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use (b) Five other genes can modify the base coat colour. One of these genes is the C gene. There are two alleles of this gene, C and CCR. • C does not affect the base coat colour. • CCR may modify the base coat colour. • If a chestnut horse has at least one CCR allele its phenotype will be palomino, which is a light cream colour. • If a black horse has at least one CCR allele its effect will not be noticeable in the phenotype. Complete the genetic diagram below. parental genotype aaCCCR AaCC parental phenotype …………………… …………………… gametes …………… …………… …………… …………… offspring genotypes …………… …………… …………… …………… offspring phenotypes …………… …………… …………… …………… [4] [Total: 8]

18 9700/04/M/J/09 © UCLES 2009 For Examiner’s Use 8 (a) Fig. 8.1 shows a scanning electron micrograph of a section through a leaf of the Christmas rose, Helleborus niger. M { } N Fig. 8.1 Name M and N. M … N … [2] (b) Gases leave and enter the leaf through pores called stomata. Describe and explain how a stoma is opened. … … … … … … … … … … … … [6]

19 9700/04/M/J/09 © UCLES 2009 [Turn over For Examiner’s Use (c) Fig. 8.2 outlines the main reactions in the light-dependent stage of photosynthesis. increasing energy level light reduced NADP photosystem II light photosystem I H2O R S 2e– 2e– 2e– Fig. 8.2 (i) Name the process shown by the dotted arrows ( ). … [1] (ii) Describe what happens to water at R. … … … … [2] (iii) State the product formed as electrons flow along S. … [1] (iv) Explain briefly the role of reduced NADP in the light-independent stage. … … … … [2] [Total: 14]

9700/04/M/J/09 © UCLES 2009 For Examiner’s Use Section B Answer one question. 9 (a) Explain how changes in the nucleotide sequence of DNA may affect the amino acid sequence in a protein. [7] (b) Explain how natural selection may bring about evolution. [8] [Total: 15] 10 (a) Describe the part played by the proximal convoluted tubules in the functioning of the kidneys. [8] (b) Explain how the collecting ducts in the kidneys may reduce the loss of water from the body. [7] [Total: 15] … … … … … … … … … … … … … … … … … 20

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24 9700/04/M/J/09 © UCLES 2009 … … … … … … … … … … … … … … … … … … … … … … … Copyright Acknowledgements: Question 1b © White-whiskered hermit hummingbird; Science Photo Library. Question 8a © Christmas rose leaf; Eye of Science/Science Photo Library. Question 3 Figure 3.1 © www.swasafaris de swasalbum bg_namib%20meerkat 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. For Examiner’s Use

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2009 question paper for the guidance of teachers 9700 BIOLOGY 9700/04 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 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 – May/June 2009 9700 04 © UCLES 2009 1 (a) (i) 18 ; [1] (ii) 0.72 ; allow ecf from (i) [1] (b) 1 RQ value falls steeply, initially / 40–80 min ; 2 then, very little change / AW ; 3 sugar / carbohydrate, metabolised at start ; A named carbohydrate 4 then fat metabolised ; 5 (due to) fasting / carbohydrate running out ; [4 max] (c) 1 increase in rate of respiration ; 2 kinetic energy increases / more enzyme-substrate complexes / enzyme activity increases ; 3 effects of too high a rise in temperature ; e.g. denaturation of enzymes 4 AVP ; e.g. Q10 = 2 [2 max] [Total: 8] 2 (a) oestrogen follicle (cells) / granulosa (cells) / theca ; progesterone corpus luteum ; A follicle (cells) [2] (b) 1 (oestrogen / progesterone affect) hypothalamus / anterior pituitary ; 2 inhibit secretion of, FSH / LH / GnRH ; 3 follicles do not develop ; 4 no ovulation ; R ref to eggs 5 ref. negative feedback ; 6 alters cervical mucus to stop sperm ; 7 prevents implantation / effect on endometrium ; R endometrium thickens [4 max]

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 04 © UCLES 2009 (c) any two from 1 (advantage of smaller population), less poverty / less starvation / less disease ; 2 greater care for children that are born ; 3 (benefit to adult women), fitter women / more women working ; 4 more promiscuity ; 5 more, STD / breast cancer / cervical cancer ; 6 population decrease ; [2 max] [Total: 8] 3 (a) 1 loss of habitat ; A deforestation 2 building / industry / farming / localised use of wood ; ignore logging / timber production 3 difficulty in finding food ; A increased competition R no food 4 poaching / hunting ; 5 ref. ivory trade ; [3 max] (b) 1 of no use to humans ; 2 protected in burrows ; 3 variety of food ; 4 small quantity of food required ; 5 short gestation ; 6 large number of offspring ; 7 camouflaged ; 8 (sophisticated) early warning system ; [3 max] [Total: 6]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 04 © UCLES 2009 4 (a) (i) A pericarp / fruit coat B scutellum / cotyledon C plumule / embryo shoot D radicle / embryo root 0 or 1 = 0 marks, 2 or 3 = 1 mark, 4 = 2 marks ; ; [2] (ii) 1 food / starch / nutrients ; 2 for use, during germination / before photosynthesis / before leaves emerge above ground ; 3 to provide glucose for, respiration / ATP production ; ignore energy 4 to produce cellulose for cell wall production ; 5 to produce protein for, cell division / growth (of plant) ; R growth of cells [3 max] (b) (i) 1 permanently ; 2 binds with / blocks, active site ; 3 binds with, another part of enzyme / allosteric site ; 4 change (shape) of active site ; [2 max] (ii) when acetylcholinesterase is inhibited 1 acetylcholine remains attached to receptors (on post-synaptic membrane) ; 2 sodium channels on post-synaptic (membrane) remain open ; 3 membrane remains depolarised ; 4 action potentials / nerve impulses, continue to be produced ; [2 max] (c) 1 different sequence of, bases / nucleotides, causes different, sequence of amino acids / primary structure ; 2 acetylcholinesterase has a different, shape / tertiary structure ; 3 acetylcholine can still bind with, active site / acetylcholinesterase / enzyme or active site remains functional ; 4 (but) pyrethrum / inhibitor, cannot bind with, acetylcholinesterase / enzyme ; 5 inhibition is allosteric / AW ; [3 max]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 04 © UCLES 2009 (d) (i) 1 below 0.5 µg no insects killed in either group ; 2 at 0.5 µg hybrid insects killed but resistant insects survived ; 3 at 10 µg all insects killed in hybrid group but only 80% killed in resistant group ; 4 at 30 µg all insects killed in both groups ; penalise lack of units once [3 max] (ii) 1 resistant and susceptible insects are homozygous ; 2 hybrid insect is heterozygous ; 3 hybrid insect shows codominance / mutant allele and normal allele both have an effect ; allow ref to gene here [2 max] [Total: 17] 5 (a) marking points refer to batch culture 1 (penicillin) is a secondary, metabolite / product ; R Penicillium 2 more penicillin is produced (per unit time) ; A higher yield comparative statement 3 in the later stages of growth (of the culture) / after main growth phase is over ; 4 (penicillin produced when, fungus / Penicillium,) is short of nutrients ; R no nutrients left alternative points for 2 and 3 for continuous culture (ora) 2a less penicillin is produced (per unit time) ; comparative statement 3a continuous culture remains in, exponential / active growth, phase ; [3 max] (b) description when pH is controlled (blue unbroken line) 1 penicillin is produced throughout the time period ; when pH not controlled (blue dotted line) 2 penicillin production increases to a maximum and then decreases ; 3 2 penicillin figs plus 2 time figs (to support 1 or 2) ; ignore pH figs explanation 4 (pH affects) enzymes (involved in penicillin production) ; when pH is controlled 5 optimum pH for enzymes is at approx pH 7 ; when pH not controlled 6 pH, high / above 7, decreases / stops, penicillin production ; 7 (pH, high / above 7), causes change in active site of enzymes / AW ; [4 max]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 04 © UCLES 2009 (c) 1 penicillin affects (bacterial) cell wall production ; A affects cross-linkages 2 inhibits, glycoprotein peptidases / enzymes involved with constructing (bacterial) cell wall ; 3 viruses do not have cell walls ; [2 max] [Total: 9] 6 (a) 1 increases, cellular uptake of glucose (from blood) / membrane permeability to glucose ; 2 (by), liver / muscle / adipose, cells ; 3 increased, respiration / metabolism, of glucose ; A increased glycolysis 4 causes conversion of glucose to, glycogen / fat ; A inhibits glycogenolysis 5 (blood glucose concentration maintained between) 80–120 mg per 100 cm3 ; A single value between 80–120 [3 max] (b) 1 it is identical to human insulin / ora ; 2 (more) rapid response ; 3 no / fewer, rejection problems / side effects / allergic reactions ; 4 ref. to ethical / moral / religious, issues ; 5 cheaper to produce in large volume / unlimited availability ; R cheap to produce 6 less risk of, transmitting disease / infection ; 7 good for people who have developed tolerance to animal insulin ; [2 max] (c) (i) 1 single target site will be in correct resistance gene ; 2 (gene to be inserted has) complementary sticky ends to target site sticky ends ; 3 more cuts would fragment plasmid ; [2 max] (ii) circle of DNA taken up by bacteria bacteria resistant to ampicillin bacteria resistant to tetracycline unaltered plasmids

; recombinant plasmids that have taken up the wanted gene
 ; circles of the wanted gene   ; [3]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 04 © UCLES 2009 (d) (i) 1 risk spread of resistance to other bacteria ; 2 spread of resistance makes the use of antibiotics less effective / AW ; 3 via, conjugation / transformation / uptake of plasmids ; A description 4 via, ‘phage / transduction ; A description 5 ref. R plasmid multiple resistance (MDR) / extreme resistance (XDR) ; [3 max] (ii) 1 gene for fluorescent substance ; 2 source of gene ; e.g. from jellyfish 3 substance fluoresces when exposed to appropriate light ; or 4 lacZ gene / gene for β-galactosidase ; 5 splits non-blue substrate ; 6 product is blue ; [2 max] [Total: 15] 7 (a) key ; black upper case, chestnut lower case gametes ; offspring genotypes and chestnut identified ; 25% / 0.25 / ¼ / 1 in 4, (probability) ; ignore ratios [4] (b) parental genotype aaCCCR AaCC parental phenotype palomino / cream black ; gametes aC aCCR AC aC ; offspring genotypes AaCC aaCC AaCCCR aaCCCR ; any order offspring phenotypes black chestnut black palomino / cream ; order linked to genotype order ecf can be applied to offspring genotypes and phenotypes [4] [Total: 8]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 04 © UCLES 2009 8 (a) M – palisade ; N – vascular bundle / phloem and xylem / vein ; [2] (b) 1 ref. ABA absence ; 2 H+ transported out of guard cells, actively / using ATP ; 3 low H+ conc / negative charge, inside cell ; 4 K+ channels open / K+ diffuses into cell ; 5 water potential of cell falls ; A decrease in solute potential 6 water moves into cell by osmosis ; 7 volume of guard cells increase / turgor increases ; guard cells: 8 have hoops of cellulose microfibrils which ensure increase in length rather than diameter ; 9 have ends that are joined together ; 10 have, thicker inner walls / thinner outer walls ; 11 curve apart / bend, (to open stoma) ; [6 max] (c) (i) cyclic photophosphorylation ; [1] (ii) photolysis ; (water splits into) 2e–, 2H+ and (½)O2 ; enzyme is involved ; [2 max] (iii) ATP ; [1] (iv) hydrogen carrier ; GP, reduced / hydrogen added ; R H2 to, TP / 3 carbon sugar ; uses ATP ; [2 max] [Total: 14]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 04 © UCLES 2009 9 (a) 1 code is three, bases / nucleotides ; A triplet code 2 (gene) mutation ; R chromosome mutation 3 base, substitution / addition / deletion ; 4 addition / deletion, large effect (on amino acid sequence) ; 5 frame shift ; 6 completely new code after mutation / alters every 3 base sequence which follows ; 7 (substitution) often has no effect / silent mutation ; 8 different triplet but same amino acid / new amino acid in non-functional part of protein ; 9 (substitution) may have big effect (on amino acid sequence) ; 10 could produce ‘stop’ codon ; 11 sickle cell anaemia / PKU / cystic fibrosis ; 12 reference to transcription or translation in correct context ; A description 12a AVP ; e.g. protein produced, is non-functional / not produced / incomplete [7 max] (b) 13 individuals in population have great reproductive potential / AW ; 14 numbers in population remain roughly constant ; 15 variation in members of population ; 16 environmental factors / named factor (biotic or abiotic) ; linked to 17 and 18 17 (cause) many, fail to survive / die / do not reproduce ; 18 those best adapted survive / survival of the fittest ; 19 (reproduce to) pass on alleles ; R genes 20 genetic variation leads to change in phenotype ; 21 ref: changes in, gene pool / allele frequency ; 22 over time produces evolutionary change ; 23 new species arise from existing ones / speciation ; 24 directional / stabilising, selection ; [8 max] [Total: 15]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 04 © UCLES 2009 10 (a) 1 selective reabsorption ; 2 (pct cells have) villi / microvilli / large surface area ; 3 (pct cells have) many mitochondria ; 4 Na+ leave pct cells ; 5 by active transport ; 6 Na+ concentration falls in (pct) cells / Na+ concentration gradient ; 7 Na+ (diffuse) from lumen into (pct) cells ; 8 through, transporter / carrier, proteins ; ignore channel proteins 9 cotransport ; 10 of, glucose / amino acids / vitamins / chloride ions ; 11 (from pct cells) into intercellular fluid ; linked to 10 12 (then) diffusion into blood ; linked to 10 13 (normally) all glucose reabsorbed ; 14 some water reabsorbed ; 15 some urea reabsorbed ; 16 AVP ; e.g. creatinine secreted into lumen [8 max] accept sodium ions but reject sodium or Na penalise once only

Page 11 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – May/June 2009 9700 04 © UCLES 2009 (b) 17 ADH affects collecting duct ; 18 binds to receptor on membrane ; 19 increase membrane permeability (to water) / more water channels ; 20 ref. enzyme controlled reactions ; 21 produces (active) phosphorylase ; 22 (which causes) vesicles with, water channels / aquaporins ; must be linked to 23 23 to, move to / fuse with, (plasma) membrane ; 24 more water flows out of collecting duct ; 25 down / along, water potential gradient ; 26 (then) into blood ; 27 urine (more) concentrated / small volume of urine ; 28 ref. negative feedback ; 29 AVP ; e.g. role of loop of Henle in creating water potential gradient movement of urea increases water potential gradient [7 max] [Total: 15]