Cambridge A Level Biology 9700 — 2007 May/June Paper 4 · Variant 1

This document consists of 20 printed pages, 2 lined pages and 2 blank pages. SP (NF/KS) T26307/6 © UCLES 2007 [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. * 5 0 4 5 7 4 0 9 0 3 * BIOLOGY 9700/04 Paper 4 A2 Structured Questions May/June 2007 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 9700/04/M/J/07 For Examiner’s Use © UCLES 2007 Section A Answer all the questions. 1 The western lowland gorilla, Gorilla gorilla, has become an endangered species although it has no known enemies, except humans. Gorillas are herbivorous, feeding on fruit, shoots, tree bark and leaves. Fig. 1.1 shows a western lowland gorilla. Fig. 1.1 (a) Suggest three reasons why the western lowland gorilla has become an endangered species. 1. … … 2. … … 3. … …[3] (b) (i) Explain how captive breeding programmes in zoos may help in the protection of endangered species, such as the western lowland gorilla. … … … … … …[3]

3 9700/04/M/J/07 [Turn over For Examiner’s Use © UCLES 2007 (ii) State two disadvantages of captive breeding programmes. 1. … … 2. … …[2] [Total: 8]

4 9700/04/M/J/07 For Examiner’s Use © UCLES 2007 2 Aerobic respiration consists of three main processes. Fill in the table to show the major products of each process. process major products glycolysis … … … Krebs cycle … … … oxidative phosphorylation … … … [8] [Total: 8]

5 9700/04/M/J/07 [Turn over For Examiner’s Use © UCLES 2007 3 (a) Three different strains, A, B and C, of a species of bacterium were grown on nutrient agar in a divided petri dish until they formed ‘lawns’ covering the agar. Three discs of filter paper were soaked in a solution of a penicillin antibiotic and one disc placed in contact with each of the bacterial strains for 10 minutes. After 24 hours, zones of clearance, indicating bacterial cell death, were seen in the ‘lawns’ of strains A and B. The appearance of the petri dish 24 hours after addition of antibiotic is shown in Fig. 3.1. zone of clearance zone of clearance ‘lawn’ of bacteria growing on nutrient agar position where disc of filter paper soaked in antibiotic had been placed strain A strain B strain C Fig. 3.1 (i) The effectiveness of the antibiotic is proportional to the area of the zone of clearance. Measure the diameters (d) of each of the zones of clearance of bacterial strains A and B and record them to the nearest mm in Table 3.1. Using
r2, calculate the area in mm2 of the zone of clearance for each strain of bacterium and record them in Table 3.1. Calculate the ratio of the area for strain A to the area for strain B and record the ratio in Table 3.1. Table 3.1 bacterial strain A B diameter (d) of zone of clearance / mm … … area of the zone of clearance / mm2 … … area for strain A : area for strain B … [3]

6 9700/04/M/J/07 For Examiner’s Use © UCLES 2007 (ii) Explain the different effects of the antibiotic on bacterial strains A, B and C. … … … … … … …[4] (iii) Describe the role of natural selection in the spread of bacterial strains, such as A and B, when an antibiotic is used. … … … … … … …[4] (b) The β-lactam antibiotics, such as penicillin, are similar shaped molecules to the substrate of a bacterial enzyme, transpeptidase. Explain the mode of action of β-lactam antibiotics on susceptible bacteria. … … … … … … …[4] [Total: 15]

7 9700/04/M/J/07 [Turn over For Examiner’s Use © UCLES 2007 4 (a) Describe the role of insulin in the regulation of blood glucose concentration. … … … … … … …[4] Question 4 continues on page 8

8 9700/04/M/J/07 For Examiner’s Use © UCLES 2007 (b) Fig. 4.1 shows some of the steps involved in the production of bacteria capable of synthesising human insulin. mRNA for human insulin isolated DNA coding for human insulin produced DNA coding for human insulin cloned DNA coding for human insulin inserted into a plasmid vector plasmid vector inserted into bacterium Fig. 4.1 State the role of each of the following enzymes in the production of bacteria capable of synthesising human insulin, reverse transcriptase … … … DNA polymerase … … … restriction enzymes (restriction endonucleases) … … … DNA ligase. … … …[6] [Total: 10]

9 9700/04/M/J/07 [Turn over For Examiner’s Use © UCLES 2007 5 Rice, Oryza sativa, is a grass that is grown as a cereal crop in many parts of the world. In most rice-growing regimes, the rice fields are flooded with water while the rice is actively growing. Fig. 5.1 shows cultivation of rice. Fig. 5.1 (a) (i) Describe one structural feature of the tissues in the submerged stems and leaves of rice that is an adaptation for growth in water. … … … …[2] (ii) Explain the importance of the adaptation you have described in (i). … … …[2]

10 9700/04/M/J/07 For Examiner’s Use © UCLES 2007 (b) An investigation was carried out into the effect of flooding on the growth of the submerged stems of rice plants. Young rice plants were grown in a container in which the level of water was increased in 10 cm steps, over a period of seven days. The mean length of the submerged internodes (lengths of stem between two leaves) and the concentration of ethene in the rice stems was measured each day. As a control, rice plants were grown in identical conditions but the water level was kept constant throughout the seven days. The results are shown in Fig. 5.2. 0 20 40 60 7 6 5 4 3 2 1 0 0 20 40 60 80 7 6 5 4 3 2 1 0 0.0 0.5 1.0 1.5 7 6 5 4 3 2 1 0 depth of water / cm mean length of internodes / cm mean concentration of ethene / mm3 dm–3 submerged submerged not submerged not submerged time / days time / days time / days Fig. 5.2

11 9700/04/M/J/07 [Turn over For Examiner’s Use © UCLES 2007 (i) With reference to Fig. 5.2, describe the effect of increasing water level on the length of the submerged internodes. … … … …[2] (ii) Suggest advantages to the rice plants of the effect that you have described in (i). … … … …[2] (iii) With reference to Fig. 5.2, describe the effect of increasing water level on the concentration of ethene in the rice stems. … … … …[2]

12 9700/04/M/J/07 For Examiner’s Use © UCLES 2007 (c) Application of gibberellin can also affect the growth of rice plants. In a further investigation, various concentrations of gibberellin were applied to submerged rice stems. The stems were placed, for three days in closed containers, in which the air supply either contained pure air or contained ethene. Ethene is a gas that is secreted by plant tissues and acts as a plant growth regulator. The results are shown in Fig. 5.3. 0 20 40 60 80 no ethene with ethene 100 60 80 40 20 0 mean increase in length of stems / mm concentration of gibberellin / nmol dm–3 Fig. 5.3 (i) State the meaning of the term plant growth regulator. … …[1] (ii) Using your knowledge of the effects of gibberellin, and the results shown in Fig. 5.2, suggest an explanation for the results shown in Fig. 5.3. … … … … … …[3] [Total: 14]

14 9700/04/M/J/07 For Examiner’s Use © UCLES 2007 6 Fig. 6.1 is a photomicrograph of a section through the ovary of a mammal. Fig. 6.1 (a) Name A and B. A … B …[2] B A

15 9700/04/M/J/07 [Turn over For Examiner’s Use © UCLES 2007 Fig. 6.2 shows part of the sequence of processes by which female gametes are produced. cells in germinal epithelium oogonia C P Q secondary oocyte Fig. 6.2 (b) With reference to Fig. 6.2, (i) name the cell at stage C; … [1] (ii) draw a label line on Fig. 6.1 to a cell at stage C; [1] (iii) name the types of cell division that take place at P and Q. P … Q … [1] (c) Describe one way in which genetic variation between secondary oocytes is achieved during meiosis. … … … …[3] [Total: 8]

17 9700/04/M/J/07 [Turn over For Examiner’s Use © UCLES 2007 7 (a) Sometimes a gene has more than two alleles, termed multiple alleles. The ABO blood group system in humans is controlled by a gene with three alleles, IA, IB and Io. Alleles IA and IB are codominant and Io is recessive to both. The blood group AB is the result of codominance. Explain what is meant by codominance. … … … … …[3] (b) In humans, a gene that codes for the production of a protein, called factor VIII, is located on the X chromosome. The dominant allele for this gene produces factor VIII, but the recessive allele does not produce factor VIII. A person who is unable to make factor VIII has haemophilia in which the blood fails to clot properly. Explain why a man with haemophilia cannot pass haemophilia to his son but may pass haemophilia to his grandson. … … … … … …[3]

18 9700/04/M/J/07 For Examiner’s Use © UCLES 2007 (c) A gene for feather colour in chickens is carried on an autosome. This gene has two alleles, black (CB) and splashed-white (CW). When a male chicken with black feathers is mated with a female chicken with splashed-white feathers, all the offspring have blue feathers. This also occurs when a male chicken with splashed-white feathers is crossed with a female with black feathers. black feathers splashed-white feathers Fig. 7.1 Another gene may cause stripes on feathers (barred feathers). This gene is carried on the X chromosome. The allele for barred feathers (XA) is dominant to the allele for non- barred feathers (Xa). In chickens the male is homogametic and has two X chromosomes while the female is heterogametic and has one X chromosome and one Y chromosome. barred feathers Fig. 7.2

19 9700/04/M/J/07 [Turn over For Examiner’s Use © UCLES 2007 (i) A male chicken with black, non-barred feathers was crossed with a female chicken with splashed-white, barred feathers. All the offspring had blue feathers, but the males were barred and the females were non-barred. Using the symbols given above draw a genetic diagram to show this cross. parents’ male, black, female, splashed-white, phenotype non-barred feathers. barred feathers. genotype gametes offspring genotypes phenotypes male, blue, barred feathers. female, blue, non-barred feathers. [5] (ii) Explain how a farmer could use a breeding programme to find out the genotype of a male chicken with blue, barred feathers. … … … … … …[3] [Total: 14]

20 9700/04/M/J/07 For Examiner’s Use © UCLES 2007 8 (a) Explain what is meant by artificial selection. … … … … … … … …[4] (b) In a plant breeding programme, corn, Zea mays, was bred in an attempt to produce a high yield of protein in the grain. The results of this programme are shown in Fig. 8.1. 0 4 8 12 16 20 24 60 45 55 50 40 35 30 25 20 15 10 5 0 mass of protein / g per 100 g of grain generations Fig. 8.1

21 9700/04/M/J/07 [Turn over For Examiner’s Use © UCLES 2007 (i) With reference to Fig. 8.1, calculate the percentage increase in grain protein by the end of the experiment. Show your working. Answer …% [2] (ii) Suggest why the protein yield does not increase steadily in each generation. … … … …[2] [Total: 8] Section B starts on page 22

22 9700/04/M/J/07 For Examiner’s Use © UCLES 2007 Section B Answer one question. 9 (a) Describe how a nerve impulse crosses a cholinergic synapse. [9] (b) Explain the roles of synapses in the nervous system. [6] [Total: 15] 10 (a) Describe the structure of a chloroplast. [9] (b) Explain how the palisade mesophyll cells of a leaf are adapted for photosynthesis. [6] [Total: 15] … … … … … … … … … … … … … … … … … …

23 9700/04/M/J/07 For Examiner’s Use … … … … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2007 [Turn over

24 9700/04/M/J/07 © UCLES 2007 … … … … … … … … … … … … … … … … … … … … … … … … Copyright Acknowledgements: Question 1 Fig. 1.1 © TONY CAMACHO / 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. For Examiner’s Use

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2007 question paper 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. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. 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 2007 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme Syllabus Paper GCE A/AS LEVEL – May/June 2007 9700 04 © UCLES 2007 Section A 1 (a) 1. killed / hunted, qualified ; e.g. for meat / for fur / blood sport / takes human food / thought to be dangerous A poaching (unqualified) 2. war ; 3. sale of live young ; 4. habitat destruction / AW ; 5. loss of / competition for food ; 6. AVP ; e.g. disease [3 max] (b) (i) 1. fewer animals need to be caught (for zoos) ; 2. ref. becoming pregnant ; e.g. IVF / finding a mate 3. reintroduction into the wild ; 4. research easier with captive animals / AW ; 5. ref. increase in numbers ; 6. ante or postnatal care ; [3 max] (ii) 1. inbreeding / AW ; 2. gene pool too small ; 3. no fear of humans / difficulty in socialising with other gorillas ; 4. difficulty in, finding food / reproducing ; 5. ref. transfer of pathogens ; 6. ref. effects of captivity ; e.g. stress [2 max] [Total: 8] 2 process major products glycolysis ATP ; pyruvate ; reduced NAD ; Krebs cycle ATP ; reduced NAD / reduced FAD ; CO2 ; oxidative phosphorylation ATP ; water ; NAD / FAD ; [8 max] R NADP throughout [Total: 8]

Page 3 Mark Scheme Syllabus Paper GCE A/AS LEVEL – May/June 2007 9700 04 © UCLES 2007 3 (a) (i) bacterial strain A B diameter (d) / mm 24 16 ; area / mm2 452 - 453 201 – 201.2 ; A ecf ratio of area A : area B 2.25 : 1 A 9:4 A ecf [3] (ii) 1. penicillin kills more of strain A than strain B or C / AW ; 2. ref. different active or binding sites ; 3. A produces less penicillinase than B or C ; 4. C is resistant (to penicillin) ; 5. C has mutation ; 6. penicillin cannot bind to enzymes ; 7. penicillin inactivated by C / C produces much penicillinase ; 8. AVP ; e.g. B is evolving into a more resistant strain / variation in carriers across membrane [4 max] (iii) 1. antibiotic, is selective agent / provides selective pressure ; 2. resistant survive / susceptible die ; 3. ref. reproduction ; 4. resistants pass on, mutation / allele ; R gene 5. ref. vertical transmission ; 6. increases frequency of allele in population ; 7. may pass advantageous mutation to other species / ref. plasmid transfer ; 8. ref. horizontal transmission ; [4 max] accept reference to strains A, B and C in correct context for points 2, 3 and 4 (b) 1. competitive inhibitors (of transpeptidase) ; 2. binds to enzyme ; 3. blocks active site ; 4. crosslnks in peptidoglycan wall cannot form ; 5. weakens cell wall ; 6. lysis / cell bursts ; 7. ref. high internal pressure of bacterial cell ; [4 max] [Total: 15] 4 (a) 1. norm concentration of blood glucose is 80 - 120 mg 100cm-3 ; (A within range) 2. ß cells of, Islets of Langerhans / pancreas, detect increase ; 3. ref. K+ channels close / role of Ca 2+ ; 4. secrete insulin ; 5. ref. glycogenesis ; 6. increased uptake of glucose (by cells) ; 7. increased use of glucose in respiration / glucose converted to fat ; 8. ref. negative feedback / described ; [4 max]

Page 4 Mark Scheme Syllabus Paper GCE A/AS LEVEL – May/June 2007 9700 04 © UCLES 2007 (b) reverse transcriptase makes, cDNA / single strand of DNA ; from (human) mRNA ; DNA polymerase produces, second strand of DNA / double stranded DNA ; ref. links nucleotides (in context of backbone formation) ; ref. semiconservative replication / ref. complementary base pairing ; [max 2] restriction enzymes cut DNA / cut plasmid ; R cuts gene A cuts out gene at specific sites / at palindromic sites ; to give sticky ends ; A blunt ends [max 2] DNA ligase seals nicks in sugar-phosphate backbone ; forms rDNA ; by adding phosphate group ; [max 2] [6 max] [Total: 10] 5 (a) (i) air spaces (between cells) / aerenchyma ; in mesophyll / cortex ; formed by cell death ; [2 max] (ii) provides oxygen ; for aerobic respiration / because conditions are anaerobic ; ref. diffusion ; AVP ; e.g. allows escape of ethene / buoyancy / active transport [2 max] (b) (i) internode length increases as water depth increases ; use of figures ; (2 days) 2 depths + 2 lengths ignore units [2] (ii) part of plant is (always) above water ; access to light ; access to, air / oxygen / carbon dioxide ; ref. pollination / flowering ; [2 max] (iii) ethene concentration increases up to 30 or 40 cm water depth ; fluctuation / plateau between 30 or 40 cm to 60 cm water depth ; comparison between when water level is constant and when water level increases ; [2] (c) (i) substance that affects growth / development ; [1] (ii) 1. gibberellin causes increase in stem length ; 2. detail of mechanism ; e.g. cell elongation 3. gibberellin has greater effect with ethene present ; 4. more gibberellin could be secreted as water depth increases ; 5. gibberellin could remain constant but have greater effect because more ethene secreted ; 6. more gibberellin could be transported through plant as water depth increases ; 7. AVP ; [3 max] [Total: 14]

Page 5 Mark Scheme Syllabus Paper GCE A/AS LEVEL – May/June 2007 9700 04 © UCLES 2007 6 (a) A – germinal epithelium ; B – Graafian follicle ; [2] (b) (i) primary oocyte ; [1] (ii) label to primary oocyte on Fig. 6.2 ; [1] (iii) P - mitosis Q - meiosis ; both required for mark [1] (c) either independent assortment ; homologous / maternal and paternal, chromosomes position themselves either way up / AW ; on equator (of spindle) ; so segregate randomly / any combination of maternal and paternal chromosomes can end up in daughter cells ; AVP ; e.g. occurs during metaphase 1 or crossing over / chiasmata ; between, chromatids of homologous chromosomes / non-sister chromatids ; genetic material on maternal and paternal chromosomes swap places / AW ; leads to new combination of alleles ; R genes AVP ; e.g. breaking established linkage groups / occurs during prophase 1 [3 max] [Total: 8] 7 (a) both alleles, influence phenotype / are expressed ; ref. more than 2 phenotypes possible ; phenotype of heterozygote different from either homozygote ; [3] (b) son receives Y chromosome from father ; Y chromosome does not carry haemophilia allele ; father will pass haemophilia allele to daughter(s) ; daughter will be, a carrier / heterozygous / XHXh ; daughter may pass allele to, her son / his grandson ; accept on diagram [3 max] (c) (i) (male) CBCBXaXa ; x (female) CWCWXAY ; (gametes) CBXa CWXA or CWY ; CBCWXAXa ; CBCWXaY ; (male, blue, barred) (female, blue, non-barred) accept other symbols but only with key if male XY and female XX then mark gametes and offspring genotypes to max 2 if other symbols used but no key then mark to max 2 [5]

Page 6 Mark Scheme Syllabus Paper GCE A/AS LEVEL – May/June 2007 9700 04 © UCLES 2007 (ii) blue colour is heterozygous / CBCW ; test cross ; with non-barred female ; if all offspring barred, must be XAXA / homozygous ; if some offspring non-barred, must be XAXa / heterozygous ; [3 max] [Total: 14] 8 (a) 1. human ; 2. applies selection pressure ; 3. for benefit of human ; 4. choose / breed, parents with suitable trait ; 5. named example (species and characteristic) ; 6. select offspring ; 7. repeat over several generations ; 8. increased allele frequency ; [4 max] (b) (i) 140 (%) ;; 2 marks for correct answer (14/10 x 100 = 1 mark) [2] (ii) genetic variation ; ref. polygenes ; environmental variation ; AVP ; e.g. sampling / experimental, error [2 max] [Total: 8] Section B 9 (a) 1. action potential / depolarisation, reaches presynaptic membrane ; 2. calcium (ion) channels open / presynaptic membrane becomes more permeable to Ca2+ ; 3. Ca2+ flood into presynaptic neurone ; R membrane 4. this causes vesicles of (neuro)transmitter to move towards presynaptic membrane ; 5. ref. acetylcholine / ACh ; 6. vesicle fuses with presynaptic membrane / exocytosis ; 7. ACh released into synaptic cleft ; 8. ACh diffuses across (cleft) ; 9. ACh binds to receptor (proteins) / AW ; 10. on postsynaptic membrane ; R neurone 11. proteins change shape / channels open ; 12. sodium ions rush into postsynaptic neurone ; R membrane 13. postsynaptic membrane depolarised ; 14. action potential / nerve impulse ; 15. AVP ; e.g. action of acetylcholinesterase [9 max]

Page 7 Mark Scheme Syllabus Paper GCE A/AS LEVEL – May/June 2007 9700 04 © UCLES 2007 (b) 16. ensure one-way transmission ; 17. receptor (proteins) only in postsynaptic, membrane / neurone ; ora 18. vesicles only in presynaptic neurone ; ora 19. ref. adaptation ; 20. increased range of actions ; 21. due to interconnection of many nerve pathways ; 22. ref. inhibitory synapses ; 23. involved in memory / learning ; 24. due to new synapses being formed ; 25. AVP; e.g. summation / discrimination [6 max] [Total:15] 10 (a) 1. biconvex disc ; 2. 3-10 µm diameter ; 3. double, membrane / envelope ; 4. internal membrane system ; 5. flattened or fluid-filled sacs / thylakoids ; 6. arranged in stacks / grana ; 7. hold pigments / named pigment ; 8. ref. clusters of pigments / AW ; 9. (membrane of grana) hold ATP synthase ; 10. intergranal lamellae ; 11. stroma / ground substance ; 12. lipids / starch grains ; 13. contains enzymes of Calvin cycle ; 14. stroma contains ribosomes / DNA etc ; 15. AVP ; e.g. variation in shape between species [9 max] accept on labelled diagram (b) 16. closely packed -- to absorb more incident light / AW ; 17. palisade mesophyll near upper surface of leaf -- to maximize light interception ; 18. arranged at right angles to leaf surface -- to reduce number of light absorbing walls ; 19. cylindrical cells -- producing air spaces between cells ; 20. air spaces -- act as reservoir of carbon dioxide ; 21. large surface area -- for gas exchange ; 22. cell walls thin -- so short diffusion pathway ; 23. large vacuole -- pushes chloroplasts to edge of cell ; 24. chloroplasts on periphery -- to absorb light more efficiently ; 25. large number of chloroplasts -- to maximise light absorption ; 26. chloroplasts can move within cells -- towards light ; 27. chloroplasts can move away from high light intensity -- to avoid damage ; 28. AVP ; [6 max] accept chlorophyll for chloroplast for 23, 24 and 25 only [Total: 15]