Cambridge A Level Biology 9700 — 2015 Oct/Nov Paper 4 · Variant 1

This document consists of 21 printed pages, 1 blank page and 2 lined pages. DC (LK/FD) 98847/2 © UCLES 2015 [Turn over Cambridge International Examinations Cambridge International Advanced Level * 1 4 3 7 9 3 3 8 6 8 * BIOLOGY 9700/41 Paper 4 A2 Structured Questions October/November 2015 2 hours Candidates answer on the Question Paper. Additional Materials: Answer paper available on request. READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces provided at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Section A Answer all questions. Section B Answer one question. Circle the number of the Section B question you have answered in the grid below. You may lose marks if you do not show your working or if you do not use appropriate units. Electronic calculators may be used. 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 Section A 1 2 3 4 5 6 7 8 Section B 9 or 10 Total

2 9700/41/O/N/15 © UCLES 2015 Section A Answer all the questions. 1 (a) The molecules listed below are all associated with respiration. ATP synthase glucose ATP NAD oxaloacetate pyruvate citrate oxygen From these molecules identify: a phosphorylated nucleotide … a 3-carbon compound … a coenzyme … an enzyme … [4] (b) A sample of tree sap, rich in sugars, was found to be contaminated with yeast. This sample was tested for the concentration of ethanol at regular intervals. The results are shown in Fig. 1.1. 0 0 15 30 45 60 1 2 3 ethanol concentration / arbitrary units time / hours 4 5 6 Fig. 1.1

3 9700/41/O/N/15 © UCLES 2015 [Turn over (i) Calculate the percentage increase in ethanol concentration between 15 and 45 hours. Show your working. answer … % [2] (ii) Suggest why the concentration of ethanol decreased after 45 hours. … … …[1] [Total: 7]

4 9700/41/O/N/15 © UCLES 2015 BLANK PAGE

5 9700/41/O/N/15 © UCLES 2015 [Turn over 2 Gold ions (Au3+) are toxic to most microorganisms. However, the bacterium Delftia acidovorans is frequently found in sticky layers, called biofilms, that form on the surface of gold deposits. D. acidovorans produces a peptide synthase that catalyses the synthesis of a small peptide called delftibactin. When isolated, delftibactin can precipitate Au3+ ions as small particles of metallic gold. Delftibactin is a secondary metabolite. (a) Name another example of a secondary metabolite and explain what is meant by the term. example … explanation … … … …[3] (b) A mutant strain of D. acidovorans has been identified in which the gene coding for peptide synthase is inactive. The wild-type (normal) and mutant D. acidovorans were grown on agar plates and then flooded with gold chloride solution, which contains Au3+ ions. The appearance of such a plate after this treatment is shown in Fig. 2.1. mutant 'DFLGRYRUDQV wild-type 'DFLGRYRUDQV halo of small particles of gold Fig. 2.1 With reference to Fig. 2.1, suggest how delftibactin protects D. acidovorans from toxic Au3+ ions. … … … … … … …[3]

6 9700/41/O/N/15 © UCLES 2015 (c) Wild-type and mutant D. acidovorans were grown in standardised conditions: • wild-type and mutant bacteria were grown in the absence of Au3+ ions • wild type and mutant bacteria were grown in the presence of Au3+ ions • mutant bacteria were grown in the presence of Au3+ ions and of delftibactin. The results are shown in Fig. 2.2. 108 number of bacteria (log scale) in the absence of Au3+ ions wild-type bacteria mutant bacteria in the presence of Au3+ ions in the presence of Au3+ ions and delftibactin Key 109 1010 1011 Fig. 2.2 Explain whether or not the results shown in Fig. 2.2 support the idea that delftibactin is protective. … … … … … … … … …[4]

7 9700/41/O/N/15 © UCLES 2015 [Turn over (d) The secondary metabolite, delftibactin, could be used to remove the toxic Au3+ ions that are present in the waste produced by gold mining. Describe how delftibactin could be produced on a large scale. … … … … … … … … … … … …[5] [Total: 15]

8 9700/41/O/N/15 © UCLES 2015 3 The blackcap, Sylvia atricapilla, is a small song bird. It is a summer visitor to parts of northern Europe, where it breeds. Many blackcaps spend the winter (overwinter) in southern Europe, particularly in Spain. As a result of many people putting out food for birds in their gardens, some birds can survive the winter in the UK. Scientists measured the genetic variation between blackcaps from two forest sites in Germany, 800 km apart. Both sites included birds that had overwintered in Spain and in the UK. The measurements were made shortly after the birds returned from their winter feeding grounds. (a) Explain how DNA sequencing can be used to measure the genetic variation of birds. … … … … … … … … …[4] (b) The measurements of genetic variation showed that: • birds that overwinter in the same country (Spain or the UK) shared many alleles, even though they were living 800 km apart in Germany in the summer • birds that overwintered in different countries (Spain or the UK) shared fewer alleles, even though they were living in the same forest in Germany in the summer • the genetic differences between the birds that overwinter in Spain suggest that they no longer breed with those that overwinter in the UK.

9 9700/41/O/N/15 © UCLES 2015 [Turn over Explain how these blackcaps could evolve into two distinct species. … … … … … … … … …[4] [Total: 8]

10 9700/41/O/N/15 © UCLES 2015 4 Bread wheat, Triticum aestivum, is a hexaploid that has developed from diploid wild grasses. (a) Outline the process by which T. aestivum has developed from wild grasses. … … … … … … … … …[4] (b) Wheat seeds begin to germinate when they are in warm conditions and can take up water. Fig. 4.1 shows a germinating wheat seedling. root developing vascular bundles shoot apex coleoptile pericarp and testa aleurone layer endosperm scutellum seed Fig. 4.1

11 9700/41/O/N/15 © UCLES 2015 [Turn over The endosperm contains starch stores. There are also small quantities of sucrose stored in the aleurone layer. Water uptake stimulates the production of a plant growth regulator in the seed, which in turn activates the synthesis of enzymes in the aleurone layer. These enzymes hydrolyse starch to maltose and glucose. Name the plant growth regulator involved in the activation of the synthesis of the enzymes. …[1] (c) An investigation was carried out into the role of a gene, TaSUT1, which codes for a sucrose transporter protein, in the germination of wheat seeds. • Wheat seeds were germinated and left to grow for 3, 7 or 10 days. • Samples of tissues from the roots, seeds and shoots of the seedlings were tested for the presence of mRNA transcribed from TaSUT1. • The extracted mRNA was mixed with a probe, and then placed on agarose gel across which a voltage was applied. The results are shown in Fig. 4.2. root 3 7 10 3 7 10 3 7 10 seed days shoot Fig. 4.2 (i) Suggest why the researchers looked for mRNA transcribed from the TaSUT1 gene, rather than for the gene itself. … … … …[2] (ii) Explain what the results in Fig. 4.2 indicate about the sequence of activity of TaSUT1, from day 3 to day 10, in the root, seed and shoot of a seedling. … … … … …[2]

12 9700/41/O/N/15 © UCLES 2015 (d) TaSUT1 codes for the sucrose transporter protein, SUT. This protein transports only sucrose. To investigate where this protein was present in a germinating wheat seedling, a fluorescent antibody for SUT was added to sections of tissues from the seedling. (i) Suggest how this enabled the researchers to determine the areas where SUT was located. … … … … …[2] (ii) Immediately after germination began, SUT was found in the membranes of cells in the aleurone layer. It was also determined that the most common sugar in the endosperm in the first hours after germination was sucrose. Explain how these results support the hypothesis that the first source of sugar for the embryo during germination is sucrose from the aleurone layer and not sugars produced by the hydrolysis of starch. … … … … …[2]

13 9700/41/O/N/15 © UCLES 2015 [Turn over (iii) SUT appeared in the developing phloem tissue within three days of the start of germination. Outline how sucrose is transported in phloem. … … … … … … … …[3] [Total: 16]

14 9700/41/O/N/15 © UCLES 2015 5 (a) Explain the meaning of the term biodiversity. … … … … …[2] (b) The Javan gibbon, Hylobates moloch, is an endangered species. Fig. 5.1 shows a female Javan gibbon with an infant. Fig. 5.1 Javan gibbons live in fragmented patches of undisturbed forest in western Java, Indonesia. Habitat loss has reduced the population of wild gibbons to around 4500 individuals. (i) Suggest why the separation of their habitat into small fragments, rather than a single large area, poses a threat to the long-term survival of this species. … … … … … … …[3]

15 9700/41/O/N/15 © UCLES 2015 [Turn over (ii) Several zoos in Java keep Javan gibbons, but few of these are involved in breeding programmes. Suggest how, other than through captive breeding programmes, zoos in Java could contribute to the conservation of the Javan gibbon. … … … … … … …[3] [Total: 8]

16 9700/41/O/N/15 © UCLES 2015 6 Scorpions are predatory arthropods. They have a pair of grasping claws at the front of their bodies and a tail with a stinger. The stinger is used to inject venom into their prey to cause paralysis. Fig. 6.1 shows a scorpion. Fig. 6.1 (a) Scorpion venom contains two active components: • a toxin that affects ion channels at synapses of the nervous system of their prey • an inhibitor of an enzyme found at these synapses. For each component of the venom, suggest and explain one way in which it may stop the correct functioning of the synapse. toxin … … … … … inhibitor … … … … …[4]

17 9700/41/O/N/15 © UCLES 2015 [Turn over (b) Scorpions stand very still on the sand. Moving prey will disturb grains of sand, and scorpions are able to detect this movement using sensory organs, known as slit hairs, at the tips of their legs. Some of the cells of a slit hair act as sensory receptors. (i) State the role of a sensory receptor. … … …[1] (ii) When a slit hair is bent by the movement of the sand the potential difference across the cell surface membranes of the slit hair cells becomes more positive inside compared to the outside. State the name given to the initial change in potential difference that may lead to an action potential. …[1] (iii) Action potentials may then be sent by the cells in the slit hairs to the central nervous system (CNS) of the scorpion. Explain how the scorpion is able to distinguish between a small and a large movement of sand. … … … … … … …[2] [Total: 8]

18 9700/41/O/N/15 © UCLES 2015 7 (a) One way to estimate the rate of photosynthesis is to measure the rate of uptake of carbon dioxide. Fig. 7.1 shows the relationship between light intensity and relative carbon dioxide uptake and production in a dicotyledonous plant. carbon dioxide uptake < ; carbon dioxide produced light intensity 0 0 Fig. 7.1 (i) State one physical factor that may limit the rate of photosynthesis at Y. …[1] (ii) State two features of a dicotyledonous leaf that can affect the rate of photosynthesis. … …[2] (iii) Explain the shape of the curve as light intensity increases from 0 to X. … … … … …[2]

19 9700/41/O/N/15 © UCLES 2015 [Turn over (b) The uptake of radioactively-labelled carbon dioxide in chloroplasts was investigated. Three tubes, each containing different components of chloroplasts, were exposed to light. The results of the investigation are shown in Table 7.1. Table 7.1 tube contents uptake of radioactively- labelled carbon dioxide / counts per minute A stroma and grana 96 000 B stroma, ATP and reduced NADP 97 000 C stroma 4 000 (i) Name the substance that combines with carbon dioxide in a chloroplast. …[1] (ii) Explain why the results in tube B are similar to those in tube A. … … … … …[2] (iii) Explain why the uptake in tube C was less than the uptake in tube B. … … … … …[2]

20 9700/41/O/N/15 © UCLES 2015 (c) Complete the following paragraph by using the most suitable words to fill in the gaps. In a photosystem, several hundred accessory pigment molecules surround a primary pigment molecule, called … , in the … membrane. The position of the primary pigment is also called the … . Light energy is absorbed by the accessory pigments and passed on to the primary pigment. Electrons are excited to a higher energy level. They are emitted from the primary pigment and are captured by electron acceptors and eventually pass along the … , producing ATP. [4] [Total: 14]

21 9700/41/O/N/15 © UCLES 2015 [Turn over 8 (a) In shorthorn cattle, coat colour is controlled by a codominant pair of alleles. Coat colour can be red, white or roan, which is a mixture of red and white. The presence of horns is controlled by a separate pair of alleles. The allele coding for horns is recessive to the allele coding for hornless cattle. (i) Choose suitable symbols for the following: allele for red coat colour … allele for white coat colour … allele for horns … allele for no horns … [2] (ii) Using the symbols you have chosen, write down the genotypes of: white, hornless cattle … roan, horned cattle …[3] (b) Some fast-growing breeds of cattle have been produced by artificial selection. Outline the ways in which artificial selection differs from natural selection. … … … … … … … … … …[4] [Total: 9]

22 9700/41/O/N/15 © UCLES 2015 Section B Answer one question. 9 (a) Outline how the oestrogen/progesterone contraceptive pill works to prevent pregnancy. [6] (b) Discuss the biological, ethical and social implications of using this contraceptive pill. [9] [Total: 15] 10 (a) Describe and explain the structural features of a wind-pollinated plant. [9] (b) Discuss the benefits of cross-pollination. [6] [Total: 15] … … … … … … … … … … … … … … … … … …

23 9700/41/O/N/15 © UCLES 2015 [Turn over … … … … … … … … … … … … … … … … … … … … … … … … … … … …

24 9700/41/O/N/15 © UCLES 2015 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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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. … … … … … … … … … … … … … … … … … … … … … … … …

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the October/November 2015 series 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2015 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 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 accepted) 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 AVP alternative valid point (examples given as guidance)

Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 1 (a) ATP ; pyruvate ; NAD ; ATP synthase ; [4] (b) (i) 1980 ; ; Allow one mark for 0.25 0.25 5.2 − (× 100) or 0.25 4.95 (× 100) [2] (ii) ethanol evaporated ; other microorganism metabolises ethanol ; [max 1] [Total:7] 2 (a) example: penicillin / other named antibiotic ; explanation any two from: substance made by a microorganism during stationary phase / AW ; A growth of microorganism has almost stopped produced, when there is a shortage of nutrients / when population is under stress ; not needed for normal metabolism (of microorganism) ; [max 3] (b) 1 wild-type bacteria, secretes / releases, delftibactin ; I produces 2 delftibactin makes soluble gold ions into insoluble gold ; A precipitates gold 3 insoluble gold is not toxic ; ora 4 insoluble gold, stays outside the bacteria / not in bacterial cytoplasm ; 5 (so) no / fewer, soluble gold ions enter bacterium (from solution) ; A D. acidovorans for wild-type A Au / metallic gold / solid / gold particles / gold precipitate for insoluble gold A Au3+ / gold ions / ions for soluble gold [max 3]

Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 (c) 1 without Au3+ numbers of wild-type and mutants similar / AW ; support 2 with Au3+ fewer mutants than wild-type ; ora 3 with Au3+ fewer mutants than without Au3+ ; ora 4 with Au3+ and (added) delftibactin more mutants than with Au3+ alone ; ora does not support 5 with Au3+ and (added) delftibactin fewer mutants than without Au3+ ; ora 6 only one set of data / no statistical analysis ; A no repeats [max 4] (d) 1 grow the wild-type, bacterium / D. acidovorans ; 2 in fermenter ; 3 ref. to (fed) batch culture ; 4 ref. to sterilised ; 5 nutrients at start (batch) / nutrients at intervals (fed) ; 6 carbon / nitrogen, sources ; 7 ref. to aeration / provide oxygen ; 8 ref. to constant temperature / water jacket ; A environmental conditions kept constant 9 details of fermenter ; e.g. paddles / stirrers 10 harvest delftibactin / downstream processing ; [max 5] [Total:15] 3 (a) 1 DNA, denatured / strands separated ; 2 ref. to adding primer ; 3 copies of genes / pieces of DNA, of different lengths produced ; 4 ref. to use of DNA polymerase ; A PCR 5 ref. to fluorescent dyes / radioactive probes ; 6 ref. to electrophoresis / detail ; 7 DNA / base, sequence, read / visualised ; 8 (DNA / base sequence), can be compared ; [max 4]

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 (b) 1 allopatric speciation ; 2 (due to) geographic isolation ; 3 different (winter), selection pressures / environments ; 4 sympatric speciation ; 5 (two) populations have different, features / behaviours ; 6 (two) populations do not interbreed / mates within same population ; 7 ref. reproductive isolation ; 8 (over time populations) cannot breed (as different species) ; 9 AVP ; e.g. different mating calls / mutation [max 4] [Total:8] 4 (a) 1 cross between, two wild grasses / einkorn and goat grass ; 2 hybrid / offspring, sterile ; 3 chromosome doubling ; 4 due to nondisjunction ; 5 formation of, tetraploid / 4n / polyploid ; 6 diploid / 2n, gametes now formed ; 7 new cross with a, diploid / 2n, wild grass ; 8 hybrid / offspring, sterile ; 9 hybrid / offspring, triploid / 3n ; 10 chromosome doubling ; allow mp2 or mp8 not both allow mp3 or mp10 not both [max 4]

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 (b) gibberellin / gibberellic acid ; [1] (c) (i) 1 gene would be present in every cell ; 2 (when gene expressed) mRNA is in large amounts ; 3 difficult to, isolate / identify / extract, gene ;ora for mRNA [max 2] (ii) (from day 3 to day 10) 1 activity / SUT production, in seed decreases because darkness (of bands) decreases ; 2 activity / SUT production, in shoot increases because darkness (of bands) increases ; 3 activity / SUT production, remains constant in root because darkness (of bands) stays the same; [max 2] (d) (i) 1 (fluorescent) antibody binds with, SUT / the sucrose transporter protein ; 2 view / photograph, tissues / sections, with a microscope ; 3 fluorescent areas indicate presence of SUT ; [max 2] (ii) 1 presence of SUT in aleurone layer indicates sucrose moves (from aleurone layer to endosperm) ; 2 hydrolysis of starch, produces glucose or maltose / does not produce sucrose ; [2] (iii) 1 active transport / pumping, of hydrogen ions out of companion cells ; 2 (at source sucrose) loaded, by cotransport / with hydrogen ions (into companion cells) ; 3 water moves into, companion cell / sieve tube (element) ; 4 by osmosis ; 5 idea of a hydrostatic pressure gradient ; 6 mass flow ; [max 3] [Total:16]

Page 7 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 5 (a) 1 diversity of, habitats / ecosystems ; 2 number of different species ; 3 genetic diversity within a species ; [max 2] (b) (i) 1 less choice of mates ; 2 could lead to inbreeding ; 3 inbreeding depression / decrease in hybrid vigour ; 4 decrease in, genetic variation / heterozygosity ; A smaller gene pool 5 ref. to possible difficulties in finding enough food ; 6 idea that small areas are more vulnerable to damage than larger ones ; 7 more easily exposed to danger outside area ; [max 3] (ii) 1 educate people about the gibbons ; 2 can research gibbons to find about their, behaviour / habitat requirements; 3 ref. to health care ; 4 adequate food ; 5 AVP ; e.g. fundraising for conservation projects in the wild / protection from predators or hunters [max 3] [Total:8]

Page 8 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 6 (a) toxin (max 2) 1 calcium ions do not enter presynaptic, neurone / knob ; 2 ACh / neurotransmitter, not released into synaptic cleft ; 3 sodium ions do not enter, neurone / axon ; 4 no depolarisation of (postsynaptic) membrane / no action potentials in (postsynaptic) neurone ; inhibitor (max 2) 5 blocks / binds to, acetylcholinesterase ; 6 ACh / neurotransmitter, remains attached to receptors ; 7 continuous, depolarisation of postsynaptic membrane / action potentials in postsynaptic neurone ; 8 stops recycling of ACh / neurotransmitter / AW ; [max 4] (b) (i) convert / transduce, stimulus into a, nerve / electrical, impulse ; A named stimulus [1] (ii) receptor / generator, potential ; [1] (iii) 1 idea that the larger the intensity of stimulus the greater the frequency of action potentials ; 2 further detail ; e.g. ref. to all-or-nothing law / all action potentials have same p.d. 3 may involve more, receptors / neurones ; [max 2] [Total:8]

Page 9 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 7 (a) (i) carbon dioxide concentration / temperature ; [1] (ii) 1 stomata, number / size ; 2 number / size, of chloroplasts ; 3 leaf surface area / thinness of lamina ; 4 number / size, of intercellular airspaces ; 5 rubisco concentration ; 6 age / senescence ; [max 2] (iii) 1 respiration (rate) greater than photosynthesis (rate) ; 2 (so) overall there is a net production of carbon dioxide / AW ; 3 at X, idea that photosynthesis = respiration / compensation point ; [max 2] (b) (i) RuBP / ribulose bisphosphate ; [1] (ii) 1 grana site of light-dependent stage ; [2] 2 ATP and reduced NADP produced ; (iii) (without ATP and reduced NADP) 1 less / no, GP converted to TP ; 2 less / no, RuBP / ribulose bisphosphate, can be regenerated ; 3 light-independent stage / Calvin cycle, cannot occur (as much) ; [max 2] (c) chlorophyll a ; thylakoid ; I grana / granum reaction centre ; electron transport chain ; A ETC [4] [Total:14]

Page 10 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 8 (a) (i) CR CW ; must have same upper case letter plus different superscript (upper or lower) h H ; accept any lower case followed by upper case of same letter [2] (ii) CWCWHH ; CWCWHh ; CRCWhh ; [3] (b) artificial selection – accept ora for natural selection. 1 humans, act as selection pressure / choose parents ; 2 reduced genetic variation / smaller gene pool ; 3 inbreeding depression ; 4 loss of hybrid vigour ; 5 faster ; 6 for benefit of humans / not for benefit of animals ; 7 increased homozygosity / decreased heterozygosity ; 8 increased chance that harmful recessive alleles will, come together / be expressed ; [max 4] [Total:9]

Page 11 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 9 (a) 1 synthetic hormones used ; 2 as they do not get broken down quickly / act for longer ; 3 oestrogen / progesterone, concentrations remain high ; 4 inhibits secretion of, FSH / LH / GnRH ; I stops 5 from anterior pituitary gland ; 6 ref. to negative feedback ; 7 inhibits ovulation / no ovulation ; 8 alters cervical mucus to stop sperm ; 9 prevents implantation / effect on endometrium ; 10 AVP ; e.g. taken daily for 21 days / stops for 7 days to allow menstruation (or) taken daily throughout month. [max 6] (b) biological – negative 1 rise in blood pressure / increased chance of blood clots ; 2 nausea / headaches ; 3 increased risk of breast cancer ; 4 increase in STDs ; biological – positive 5 regular / no menstruation ; 6 reduced risk of developing, ovarian cysts / ovarian cancer / uterine cancer ; 7 reduced risk of uterine infections ; A pelvic social / ethical – negative 8 (sexual freedom has led to) more marriage breakdowns ; 9 (so more) single parent families ; 10 increase in promiscuity ; 11 religious / cultural, objection ; social / ethical – positive 12 reduction in, unwanted pregnancies / abortions ; 13 women have control over their fertility ; A ‘bodies’ 14 ref. to population control ; [max 9] [Total:15]

Page 12 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 41 © Cambridge International Examinations 2015 10 (a) 1 lots of pollen grains made ; 2 pollen grains, small / light ; 3 pollen grains, smooth / aerodynamic ; 4 (so), easily carried by the wind / more chance of pollination ; 5 anthers are, versatile / loosely attached / attached at one point (to filaments) ; 6 anthers / stamens / androecium, on long filaments / hang out (of flower) / exposed ; 7 to release pollen (into, wind / air) ; 8 stigmas hang out (of flower) / exposed ; 9 stigmas, large surface area / hairy / feathery / branched ; 10 to catch pollen ; 11 no / small, petals / corolla / calyx / perianth / sepals ; 12 no, nectar / scent, produced ; 13 so no energy wasted ; [max 9] (b) 1 ref. to outbreeding ; 2 increased genetic variation / increased genetic diversity / larger gene pool ; 3 increased heterozygosity / decreased homozygosity ; 4 less likely that harmful recessive alleles will, come together / be expressed ; 5 (increased) hybrid vigour ; 6 decreased / no, inbreeding depression ; 7 ability to, adapt to / survive in, changing (environmental) conditions ; 8 reduced susceptibility to, disease / pests ; 9 AVP ; e.g. positive effect on insects [max 6] [Total:15]