Cambridge A Level Biology 9700 — 2012 May/June Paper 2 · Variant 1

This document consists of 15 printed pages and 5 blank pages. DC (AC/SW) 47762/5 © UCLES 2012 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level * 1 0 8 1 1 3 9 4 1 2 * BIOLOGY 9700/21 Paper 2 Structured Questions AS May/June 2012 1 hour 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces provided at the top of the page. Write in dark blue or black ink. You may use a soft pencil for any diagrams, graphs or rough working. Do not use red ink, staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all questions. 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. For Examiner’s Use 1 2 3 4 5 6 Total

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3 9700/21/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use 1 Name as precisely as you can the structure described in each of the following statements. (a) The blood vessel that transports deoxygenated blood from the heart. …[1] (b) The cell that ingests and digests cell debris and bacteria in the lungs. …[1] (c) The cell that secretes antibodies. …[1] (d) The epithelial cell that secretes mucus in the trachea. …[1] (e) The tissue that prevents the collapse of the trachea during inhalation. …[1] [Total: 5]

4 9700/21/M/J/12 © UCLES 2012 For Examiner’s Use 2 Fig. 2.1 shows a drawing made from an electron micrograph of two adjacent cells in a leaf. A B C D E F G vacuole Fig. 2.1 (a) Structures A and B are both visible using the light microscope, but the internal detail of these organelles shown in Fig. 2.1 is only visible using the electron microscope. Explain why the internal details of structures A and B are only visible when using the electron microscope and not when using the light microscope. … … … … … …[3]

5 9700/21/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use (b) Name in full the structures labelled C, D and E. C … D … E … [3] (c) State one role of vacuoles in plant cells. … [1] (d) Structures F and G have very different permeability properties. Explain how the composition of structures F and G determines the permeability properties of these structures. … … … … … … … … … … [4] (e) Fig. 2.1 shows two plasmodesmata connecting the adjacent cells. Describe the roles of plasmodesmata in transport in plants. … … … … … … [3] [Total: 14]

6 9700/21/M/J/12 © UCLES 2012 For Examiner’s Use 3 Haemoglobinopathies are inherited conditions linked to the structure and function of haemoglobin. Sickle cell anaemia is one of these conditions in which the transport and delivery of oxygen to tissues is less than normal. An investigation was carried out to discover the effect of sickle cell anaemia on the ability of blood to carry oxygen. Blood samples were taken from two people: • person L without sickle cell anaemia • person M with sickle cell anaemia. The percentage saturation of haemoglobin with oxygen was determined over a range of partial pressures of oxygen. Fig. 3.1 shows oxygen haemoglobin dissociation curves for the two blood samples. 00 1 2 3 4 5 6 7 partial pressure of oxygen / kPa percentage saturation of haemoglobin with oxygen 8 9 10 11 12 13 14 10 20 30 40 50 60 70 80 90 100 M L Fig. 3.1 (a) P50 is the partial pressure of oxygen at which haemoglobin is 50% saturated with oxygen. It is taken as a measurement of the affinity of haemoglobin for oxygen. (i) State the P50 for the two blood samples, L and M. L … M …[1]

7 9700/21/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use (ii) With reference to Fig. 3.1, describe how the dissociation curve for person M differs from the dissociation curve for person L. … … … … … … [3] (b) Explain the advantage of the position of the dissociation curve for people with sickle cell anaemia. … … … … … … [3] (c) The partial pressure of oxygen in the lungs at sea level is about 13.5 kPa. At an altitude of 3000 metres the partial pressure of oxygen in the lungs is about 7.5 kPa. When people move from sea level to high altitude they become adapted to the low partial pressure of oxygen. Describe and explain how humans become adapted to the low partial pressure of oxygen at high altitude. … … … … … … … … … …[4]

8 9700/21/M/J/12 © UCLES 2012 For Examiner’s Use (d) Vaccination is used to control the spread of diseases, such as measles. Explain why vaccination cannot be used to prevent sickle cell anaemia. … … … …[2] [Total: 13]

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10 9700/21/M/J/12 © UCLES 2012 For Examiner’s Use 4 Penicillin is an antibiotic that interferes with the synthesis of cell walls in bacteria. Even before penicillin became widely available in the 1940s, the enzyme penicillinase which breaks down penicillin had been isolated. This enzyme is now found in many bacteria and gives them resistance to penicillin. Fig. 4.1 is a ribbon model of the structure of the enzyme penicillinase. The arrow indicates the active site of the enzyme. Fig. 4.1 (a) Explain why the shape of the active site of an enzyme, such as penicillinase, is important. … … … … … …[3]

11 9700/21/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use (b) With reference to Fig. 4.1, identify the aspects of protein structure that are shown and those that are not shown. aspects of protein structure shown … … … … aspects of protein structure not shown … … … …[3] Fig. 4.2 shows the changes in energy during the progress of an uncatalysed reaction. progress of the reaction energy Fig. 4.2 (c) (i) Draw on Fig. 4.2 a curve to show changes in energy during the progress of the same reaction when catalysed by an enzyme. [2] (ii) State the term given to the energy level that must be overcome before a reaction can progress. …[1]

12 9700/21/M/J/12 © UCLES 2012 For Examiner’s Use (d) Antibiotic resistance is a serious worldwide problem. Suggest how antibiotics can be used effectively to avoid the development of widespread resistance in bacteria. … … … …[2] [Total: 11]

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14 9700/21/M/J/12 © UCLES 2012 For Examiner’s Use 5 The slime mould, Dictyostelium discoideum, is a eukaryote and a decomposer of protein-rich material. Fig. 5.1 shows the life cycle of D. discoideum. X Y spores (n) spores (n) amoeboid cells fusion of amoeboid cells aggregation of amoeboid cells Fig. 5.1 (a) State the type of nuclear division that occurs at X and at Y. X … Y …[1] (b) State what is meant by the term reduction division and explain why this division is necessary in a life cycle, such as that shown in Fig. 5.1. … … … … …[2]

15 9700/21/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use (c) The amoeboid cells of D. discoideum feed on protein-rich material and break it down to form ammonium ions (NH4 +). The cell membranes of D. discoideum have transporter proteins that are responsible for the excretion of ammonium ions. Describe what happens to the ammonium ions excreted by D. discoideum into the soil. … … … … … …[3] (d) Suggest why a transporter protein is required for the removal of ammonium ions from D. discoideum. … … … … …[2] [Total: 8]

16 9700/21/M/J/12 © UCLES 2012 For Examiner’s Use 6 Fig. 6.1 shows part of a DNA molecule. O O O P O O O O P O O O O P O O O P O O O O ¯ ¯ ¯ O¯ ¯ ¯ O P O O O O P O O O O O N N N N N H H H H3C N N N H H H H H O O O O O O O N N N A C G T A = adenine C = cytosine G = guanine T = thymine Key: N N N N Fig. 6.1 (a) (i) Complete Fig. 6.1 by drawing on the hydrogen bonds between the two base pairs shown. [2] (ii) State the importance of hydrogen bonding in DNA structure. … … … … …[2]

17 9700/21/M/J/12 © UCLES 2012 [Turn over For Examiner’s Use In the 1950s, Erwin Chargaff determined the relative quantities of the four bases in DNA in different organisms. His results provided important evidence for the model of DNA proposed by James Watson and Francis Crick in 1953. Some of Chargaff’s data is shown in Table 6.1. Table 6.1 organism percentage of adenine percentage of thymine percentage of guanine percentage of cytosine Escherichia coli (bacterium) 24.7 23.6 26.0 25.7 a yeast 31.3 32.9 18.7 17.1 wheat 27.3 27.1 22.7 22.8 octopus 33.2 31.6 17.6 17.6 sea urchin 32.8 32.1 17.7 17.3 chicken 28.0 28.4 22.0 21.6 human 29.3 30.0 20.7 20.0 (b) With reference to Fig. 6.1, explain how the data in Table 6.1 helps to confirm the arrangement of bases in DNA. … … … … … … … …[3]

18 9700/21/M/J/12 © UCLES 2012 For Examiner’s Use (c) Table 6.2 shows Chargaff’s data for a virus. Table 6.2 organism percentage of adenine percentage of thymine percentage of guanine percentage of cytosine a virus 24.0 31.2 23.3 21.5 (i) State how the result for the virus differs from the results for all the organisms given in Table 6.1. … … …[1] (ii) Suggest why the results for the virus are different from all the other organisms. … …[1] [Total: 9]

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2012 question paper for the guidance of teachers 9700 BIOLOGY 9700/21 Paper 2 (AS Structured Questions), maximum raw mark 60 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. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2012 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 2012 9700 21 © University of Cambridge International Examinations 2012 Mark scheme abbreviations: ; separates marking points / alternative answers for the same point R reject A accept (for answers correctly cued by the question, or by extra guidance) AW alternative wording (where responses vary more than usual) underline actual word given must be used by candidate (grammatical variants excepted) max indicates the maximum number of marks that can be given ora or reverse argument mp marking point (with relevant number) ecf error carried forward I ignore

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9700 21 © University of Cambridge International Examinations 2012 1 (a) pulmonary artery ; A pulmonary arteries [1] (b) phagocyte / macrophage ; A neutrophil / polymorphonuclear leucocyte R PMN R leucocyte / white blood cell unqualified R any incorrect qualification [1] (c) B-lymphocyte / (effector) B (cell) / plasma (cell) ; R lymphocyte alone R effector cell unqualified [1] (d) goblet (cell) ; [1] (e) cartilage ; ignore plates / rings [1] [Total: 5] 2 (a) 1 electron microscope has, higher / AW, resolution (than LM) / ora ; 2 explanation of resolution as ability to differentiate between two points (close together) ; 3 ref. to (internal) membranes (of A and B) which cannot be seen in LM ; A named membranes e.g. cristae, grana 4 AVP ; e.g. (resolution of) EM is 0.5 nm (0.0005µm) and LM is 200 nm (0.2µm) A 0.5 to 1 nm (0.001µm) resolution is equal to half the wavelength (of medium used) ref. to shorter / AW, wavelength (of electrons) / ora (must have a comparison) ref. to, width of membranes / distance apart of membranes, e.g. width of membranes in A and B is 7 nm (+/– 1) [max 3]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9700 21 © University of Cambridge International Examinations 2012 (b) C – rough endoplasmic reticulum ; penalise once only for ER instead of endoplasmic reticulum D – ribosome ; A ribosomes ignore 70S E – smooth endoplasmic reticulum ; A smooth ER if full term used for C award one mark if E = rough endoplasmic reticulum and C = smooth endoplasmic reticulum [3] (c) any one relevant e.g. store of / holds, cell sap ; R if contains organelles store of / holds, water / ions / named ion(s) / minerals / salts / pigments / (named) sugars ; R substances / molecules R storage unqualified pushes chloroplast to edge of cell ; gives, turgidity / turgor pressure / hydrostatic pressure / support / AW ; A makes, firm / rigid A controls / maintains, turgidity R gives shape / strengthen store of / holds, waste (products) R reactions occur in vacuole, unqualified [1] (d) no marks for identifying F and G if only F or G described max 3 if F and G incorrectly identified, accept mark points correctly linked to membrane and wall to max 3 1 F partially permeable A selectively permeable and G (fully / freely / AW), permeable / porous ; F is partially permeable cell surface membrane 2 phospholipid (bilayer); 3 permeable to, lipid-soluble molecules / oxygen ; A other terms for lipid-soluble treat reference to water as neutral 4 impermeable to, water-soluble / AW, molecules / ions / AW ; A other terms for water-soluble treat reference to water as neutral 5 aquaporins / proteins, provide (increased) permeability to water ; 6 transport proteins provide permeability to, ions / polar molecules ; A channel / pore / carrier, proteins G is permeable cell wall 7 cellulose ; 8 fibres ; 9 ref. to, spaces / gaps / holes / pores, (between, fibres / other cell wall components) ; [max 4]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9700 21 © University of Cambridge International Examinations 2012 (e) 1 allows transport of, water / sucrose / amino acids / organic substances / ions /minerals / salts / lipids / hormones / ATP, (from cell to cell / between cells) ; R if linked to an incorrect transport mechanism e.g. sucrose moves by osmosis 2 without crossing, membranes / walls ; A without going through protein channels 3 this is movement through the symplast ; 4 any e.g. ; companion cell to (phloem) sieve tube (element / cell) between mesophyll cells mesophyll cell to companion cell cortical cell to cortical cell / across cells of the cortex cortical cell to endodermal cell endodermal cell to, pericycle cell / xylem / phloem ignore between sieve tube elements 5 allows, communication / signalling, between cells ; [max 3] [Total: 14] 3 (a) (i) no mark if no units used at all L – 3.6 kPa ; award the mark if units only used once M – 4.5 kPa ; A in range 4.45 to 4.55 [1] (ii) ignore any similarities 1 to the right / lower (affinity) / qualified ; e.g. lower percentage saturation 2 at, higher / lower, partial pressures, small(er) difference in percentage saturation (than others) ; A ora 3 comparative data quote ; must refer to L and M allow ecf from (i) [3] (b) 1 at partial pressures in the tissues ; where oxygen is unloaded from Hb 2 haemoglobin is less saturated (than L) ; 3 because, haemoglobin / Hb, dissociates more readily ; A idea of unloading oxygen more readily even if Hb not mentioned 4 to compensate for, fewer / less effective, red blood cells / Hb ; [max 3]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9700 21 © University of Cambridge International Examinations 2012 (c) 1 haemoglobin less well saturated (in lungs at high altitude) ; 2 data quote from Fig. 3.1 ; A 80–90% saturated at ‘about 7.5 kPa’ 3 produce more red blood cells / increase in number of RBCs ; 4 more haemoglobin ; 5 idea of compensates for, smaller volume of oxygen absorbed / lower saturation (of haemoglobin) ; also accept the following adaptations 6 increase in haematocrit / AW / decrease in plasma volume ; A increase in RBCs per unit volume R decrease in blood volume 7 increase in, breathing rate / tidal volume / heart rate / stroke volume ; 8 increase in, capillary density / number of mitochondria / myoglobin / respiratory enzymes, in muscle ; 9 ref. to (increased) secretion of, erythropoietin / EPO ; 10 increase in (2,3), BPG / DPG, in red blood cells ; A rightward shift in curve [max 4] (d) 1 not caused by (named type of) pathogen / non-infectious / non-transmissible / non- communicable / AW ; 2 genetic / inherited / AW, disease ; A caused by a mutation / AW A ‘passed down from parent(s)’ R idea of congenital diseases R ‘you get it from your mother’ 3 ref. to, no immune response / no antigen(s) ; 4 affects all red blood cells so vaccine would lead to their destruction ; [max 2] [Total: 13]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9700 21 © University of Cambridge International Examinations 2012 4 (a) this can be answered in the context of penicillinase 1 complementary shape ; 2 substrate, fits into / enters / binds to / with, active site ; A enzyme-substrate complex / ESC 3 ref. to specificity ; 4 lock and key / induced fit ; A description of induced fit 5 ref. to temporary bonds form with, active site / R groups (of amino acid residues) ; [max 3] (b) shown to max 2 secondary structure ; α / alpha, helix ; R ‘helix’ / helical structure unqualified by alpha β pleated sheet ; tertiary structure / folding ; ignore 3D shape or structure globular ; not shown to max 2 amino acids / primary structure / sequence of amino acids ; (types of) R groups ; bonds / named bonds ; A peptide quaternary structure ; prosthetic group ; [max 3] (c) (i) one lower peak inside line than uncatalysed ; start and finish at, dotted lines / same energy levels as uncatalysed ; [2] (ii) activation (energy) / (energy of) activation ; [1] (d) 1 do not prescribe for viral diseases ; 2 only use when necessary / do not overprescribe ; 3 only available on prescription / not available ‘over the counter’ ; 4 people must, complete the course / take as instructed ; R take a long course 5 test to find out which is most appropriate antibiotic to use ; A use most, appropriate / effective, antibiotics A use narrow-spectrum antibiotics 6 details of sensitivity test ; 7 rotate / AW, antibiotics / use in combination ; R use many antibiotics 8 do not use same antibiotics for humans and animals ; [max 2] [Total: 11]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9700 21 © University of Cambridge International Examinations 2012 5 (a) ignore Y X = mitosis ; Y = meiosis / mitosis [1] (b) 1 chromosome number is halved / 2n
n / diploid
haploid ; A 2 sets of chromosomes
1 set of chromosomes explanation to max 1 2 restore diploid number on fusion ; R restore full set if not qualified 3 avoids number doubling with each generation ; 4 allows expression of (recessive) alleles / AW ; 5 allows variation / new combinations of chromosomes ; [2] (c) if only use formulae, these must be correct – otherwise ignore 1 nitrification / nitrifying / oxidation ; 2 ammonium ions to nitrite ions ; 3 nitrite ions to nitrate ions ; A one mark for ammonium to nitrate 4 one named microorganism in correct context Nitrosomonas / Nitrobacter ; R Rhizobium 5 ammonium / nitrate / AW, absorbed by plants / leached / AW ; R used by plant [max 3] (d) 1 ammonium ions are (positively) charged ; A hydrophilic / polar / water-soluble 2 cannot pass through, phospholipid bilayer / membrane ; either 3 active transport ; 4 moved against concentration gradient ; or 3 facilitated diffusion ; 4 moves down its concentration gradient ; [max 2] [Total: 8]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9700 21 © University of Cambridge International Examinations 2012 6 (a) (i) hydrogen bonds drawn onto Fig. 6.1 lines must go between O-H, N-H as follows two lines between A and T H –– O and N –– H ; three lines between C and G H –– O and N –– H and O –– H ; [2] (ii) 1 hydrogen bonds hold (two), polynucleotides / strands / chains, together ; A hold, (complementary) nucleotides / base pairs, together A ora e.g. prevents, unwinding / strand separation 2 (many hydrogen bonds) give stability / DNA is stable molecule / DNA is long lasting / AW ; ignore ref. to strength 3 can be broken for, transcription / replication ; 4 ref. to (double) helix ; [2] (b) 1 (named) base / nucleotide, pairing ; 2 purine – pyrimidine ; 3 percentage of A = percentage of T ; A very similar 4 percentage of C = percentage of G ; A very similar 5 data quote in support ; [max 3] (c) (i) idea that percentages of, A and T / C and G, are not the same / three percentages are similar; [1] (ii) single-stranded DNA / not double-stranded / not a double helix ; A may be other bases ; [1] [Total: 9]