Cambridge A Level Biology 9700 — 2015 Oct/Nov Paper 4 · Variant 3

This document consists of 21 printed pages, 2 blank pages and 1 lined page. DC (LK/FD) 98848/3 © UCLES 2015 [Turn over Cambridge International Examinations Cambridge International Advanced Level * 3 3 1 6 5 5 5 3 1 4 * BIOLOGY 9700/43 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 9700/43/O/N/15 © UCLES 2015 Section A Answer all the questions. 1 (a) Yeast cells sometimes carry out anaerobic respiration. Fig. 1.1 outlines the process of anaerobic respiration in yeast cells. S\UXYDWH FRPSRXQG; FRPSRXQG: FRPSRXQG< HWKDQRO FRPSRXQG= Fig. 1.1 (i) Identify compounds W, X and Y. W … X … Y … [3] (ii) State two differences between anaerobic respiration in yeast cells and anaerobic respiration in human muscle cells. … … … … … [2]

3 9700/43/O/N/15 © UCLES 2015 [Turn over (b) Dinitrophenol (DNP) is a compound used as a herbicide. DNP inhibits respiration by interfering with the formation of the proton gradient between mitochondrial membranes. When DNP was added to isolated mitochondria the following observations were made: • fewer ATP molecules were produced • more heat energy was released • the uptake of oxygen remained constant. Suggest explanations for these observations. fewer ATP molecules produced … … … more heat energy released … … … constant oxygen uptake … … …[3] [Total: 8]

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5 9700/43/O/N/15 © UCLES 2015 [Turn over 2 A vaccine, NicVAX, is being developed to help people stop smoking tobacco. Injection of NicVAX into the body causes production of antibody molecules that bind to nicotine. (a) Outline the immune response that leads to the production of these anti-nicotine antibodies. … … … … … … … … … … … …[5] (b) Mice injected with NicVAX produce B-lymphocytes that mature into cells responsible for the production of antibody (plasma cells). Outline how these B-lymphocytes can be used to produce monoclonal antibody. … … … … … … … … … …[4]

6 9700/43/O/N/15 © UCLES 2015 (c) Tobacco smoking during pregnancy has adverse side-effects on the developing fetus. An investigation was carried out to find out whether vaccinating pregnant women with NicVAX might offer some protection for the developing fetus. Two different monoclonal antibodies, produced in response to NicVAX, were used in this investigation: • Nic-IgG • Nic311. Nicotine, or nicotine plus one of the monoclonal antibodies, was injected into the maternal circulation. The concentrations of nicotine in the fetal circulation were measured at intervals. The results of the investigation are shown in Fig 2.1.              WLPHPLQXWHV QLFRWLQHFRQFHQWUDWLRQ LQIHWDOFLUFXODWLRQQJFP²     QLFRWLQH QLFRWLQH  1LF QLFRWLQH  1LFOJ* Fig. 2.1

7 9700/43/O/N/15 © UCLES 2015 [Turn over (i) With reference to Fig. 2.1, describe the results obtained for nicotine only. … … … … … …[2] (ii) Discuss the extent to which these results support the idea that vaccination with NicVAX could protect the developing fetus of a woman who smokes tobacco. … … … … … … … …[3] (d) State one medical use of monoclonal antibodies, other than their use in producing vaccines. … …[1] [Total: 15]

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9 9700/43/O/N/15 © UCLES 2015 [Turn over 3 Atlantic salmon, Salmo salar, is one of the most important fish species farmed for human consumption. Fig. 3.1 shows an Atlantic salmon. Fig. 3.1 Infectious pancreatic necrosis (IPN) is a serious viral disease currently affecting farmed salmon. (a) (i) Describe how artificial selection could be used to produce a population of salmon that is resistant to IPN. … … … … … … …[3] (ii) Suggest problems that may arise from artificial selection. … … … … … …[2]

10 9700/43/O/N/15 © UCLES 2015 (b) A laboratory investigation was carried out to compare the artificially selected farmed Atlantic salmon with farmed salmon that had not been artificially selected. Three groups of young fish were set up in carefully controlled conditions as follows: • Group A: artificially selected salmon • Group B: non-artificially selected salmon • Group C: non-artificially selected salmon. During this investigation, only groups A and B were exposed to IPN on day 0. The percentages of salmon that died (percentage mortality) were calculated and are shown in Fig. 3.2.                    WLPHGD\V JURXS% JURXS$ JURXS& SHUFHQWDJH PRUWDOLW\ Fig. 3.2

11 9700/43/O/N/15 © UCLES 2015 [Turn over (i) Describe and explain the differences in percentage mortality between groups A and B. … … … … … … … … … … … …[4] (ii) Suggest a reason for the mortality in group C. … … …[1] [Total: 10]

12 9700/43/O/N/15 © UCLES 2015 4 Haemophilia A and haemophilia B are common hereditary disorders of blood clotting. Haemophilia A is a sex-linked genetic disorder that affects approximately 1 in 20 000 males worldwide. It is caused by a recessive allele of a gene coding for a clotting factor and results in excessive bleeding. There is currently no cure, but symptoms of haemophilia can be treated with a transfusion of a clotting factor to slow down the bleeding. (a) State how genetic screening could reduce the number of cases of haemophilia. … … … … …[2] (b) (i) Some genetic disorders can be treated with gene therapy. Outline the aims of gene therapy. … … … … …[2] (ii) Suggest why haemophilia A is a suitable disorder for treatment with gene therapy. … … …[1]

13 9700/43/O/N/15 © UCLES 2015 [Turn over (c) Haemophilia A and haemophilia B are caused by mutations in different blood clotting genes, F8 and F9 respectively. Both disorders have been treated with gene therapy involving the use of a vector. (i) Table 4.1 shows the lengths, in kilobases (kb), of the F8 and F9 genes. Table 4.1 haemophilia gene gene length / kb A F8 >8 B F9 1.4 With reference to Table 4.1, suggest why gene therapy using the F9 gene has been more successful than using the F8 gene. … … … … …[2]

14 9700/43/O/N/15 © UCLES 2015 (ii) Two frequently used vectors in gene therapy are compared in Table 4.2. Table 4.2 feature vector adenovirus retrovirus genetic material of virus double-stranded DNA single-stranded RNA expression of inserted gene high gene expression gene expression in dividing cells only host immune response to virus high low With reference to Table 4.2, explain the advantages and disadvantages of using adenovirus rather than retrovirus as a vector. … … … … … … … …[3] [Total: 10]

15 9700/43/O/N/15 © UCLES 2015 [Turn over 5 (a) (i) Explain what is meant by the term biodiversity. … … … … …[2] (ii) Explain why it is important to ensure that biodiversity is maintained. … … … … … … …[3]

16 9700/43/O/N/15 © UCLES 2015 (b) To reduce the loss of global biodiversity, areas of habitat have been protected. Fig 5.1 shows the changes in the total area protected and in global biodiversity from 1965 to 2005, in terrestrial and marine habitats. 0.6 0.9 1.2 9 0 year key: global biodiversity total area protected 1965 1975 1985 1995 2005 18 terrestrial global biodiversity / arbitrary units total area protected / km2 × 106 0.6 0.9 1.2 9 0 year 1965 1975 1985 1995 2005 18 marine global biodiversity / arbitrary units total area protected / km2 × 106 Fig. 5.1 (i) With reference to Fig. 5.1, compare the relationship between total area protected and global biodiversity in terrestrial and marine habitats: • between 1970 and 1990 • between 1990 and 2005. between 1970 and 1990 … … … between 1990 and 2005 … … …[3] (ii) Suggest why a smaller area of marine habitats has been protected than of terrestrial habitats. … … … … …[2] [Total: 10]

17 9700/43/O/N/15 © UCLES 2015 [Turn over 6 (a) One important function of the kidney nephron is selective reabsorption. This involves the rapid transfer of water across cell surface membranes. The rapid transfer of water requires the presence of protein channels known as aquaporins. Fig. 6.1 is a diagram of a nephron. $ % & ' - + * ( ) Fig. 6.1 With reference to Fig. 6.1, complete the table by inserting the correct letter for each description. description of region of nephron letter region where no aquaporins are present in the tubule wall cells region where aquaporins and glucose transport proteins are present in tubule wall cells region where aquaporins are always present in the tubule wall cells but no glucose transport proteins are present region where tubule wall cells are modified to produce filtration slits [4]

18 9700/43/O/N/15 © UCLES 2015 (b) (i) The urine of people on different types of diet was analysed. • people on a low protein diet had a mean urea concentration of 2.40 g dm–3 • people on a high protein diet had a mean urea concentration of 14.76 g dm–3. Calculate the percentage increase in the concentration of urea between the low and high protein diets. Show your working. answer … % [2] (ii) Explain why an increase in the quantity of protein in the diet leads to an increase in the concentration of urea in the urine. … … … … … … …[2] [Total: 8]

19 9700/43/O/N/15 © UCLES 2015 [Turn over 7 (a) Fig. 7.1 shows the absorption spectra of chlorophyll a and chlorophyll b and a corresponding action spectrum.     ZDYHOHQJWKRIOLJKWQP SHUFHQWDJH OLJKW DEVRUSWLRQ UDWHRI SKRWRV\QWKHVLV               ; < DFWLRQVSHFWUXP FKORURSK\OOE FKORURSK\OOD Fig. 7.1 (i) Explain why peak X of the action spectrum is higher than peak Y. … … … … …[2] (ii) Explain why most plants appear green. … … … … …[2]

20 9700/43/O/N/15 © UCLES 2015 (iii) Chlorophyll b is an accessory pigment. Outline the role played by accessory pigments in the light-dependent stage of photosynthesis. … … … … …[2] (b) Describe the effects on a plant if its environmental temperature rises well above the usual temperature range. … … … … … … … … … … …[5]

21 9700/43/O/N/15 © UCLES 2015 [Turn over (c) Complete the following paragraph by using the most suitable words to fill in the gaps. A chloroplast is surrounded by two phospholipid membranes. It has an internal ground substance called the stroma which is the site of the Calvin cycle. The stroma contains enzymes such as … and also sugars, lipids and starch. A chloroplast has an internal membrane system of fluid-filled sacs called … which can be stacked to form grana. Grana membranes hold photosynthetic pigments so that the light-dependent stage of photosynthesis can take place. The stroma contains circular … which codes for some of the chloroplast proteins made by its own small … . [4] [Total: 15] 8 The fruit fly, Drosophila melanogaster, is widely used in genetic research. It has many phenotypic variants in features such as body colour, wing shape and eye colour. Two variations from the normal-winged, grey-bodied phenotype are: • vestigial (very short) wings, coded for by the recessive allele of the gene N/n • ebony (black) body colour, coded for by the recessive allele of the gene G/g. (a) Using the symbols given, state the possible genotypes of normal-winged, grey-bodied fruit flies. … …[2] (b) Describe how you would determine the genotype of a normal-winged, grey-bodied fly. … … … … … … …[3]

22 9700/43/O/N/15 © UCLES 2015 (c) One of the genes for eye colour is carried on the X chromosome. This gene has different alleles coding for: • red eyes • orange eyes • white eyes. The allele for red eyes (R) is dominant to the allele for orange eyes (o) and dominant to the allele for white eyes (w). The allele for orange eyes is dominant to that for white eyes. Using these symbols, draw a genetic diagram to show how a cross between a white-eyed male fruit fly with a red-eyed female fruit fly will produce male and female offspring that are either red-eyed or orange-eyed. [4] [Total: 9]

23 9700/43/O/N/15 © UCLES 2015 [Turn over Section B Answer one question. 9 (a) Outline oogenesis in a human female. [9] (b) Describe and explain the changes to the uterus during the menstrual cycle. [6] [Total:15] 10 (a) Outline how hybridisation leads to polyploidy in wheat and how this benefits farmers. [8] (b) Discuss the detrimental environmental and economic effects of growing genetically modified herbicide-resistant oil seed rape. [7] [Total: 15] … … … … … … … … … … … … … … … … … …

24 9700/43/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 9700/43 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 43 © 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 43 © Cambridge International Examinations 2015 1 (a) (i) W = ethanal ; A acetaldehyde / C2H4O X = carbon dioxide ; A CO2 Y = reduced NAD ; A NADH / NADH2 / NADH+ + H+ [3] (ii) in yeast cells – ora for muscle cells 1 ethanol produced as opposed to, lactate / lactic acid ; 2 irreversible ; 3 different dehydrogenases involved / reduction of ethanal instead of pyruvate / AW ; 4 two steps / two enzymes involved / decarboxylation / ref. to (pyruvate) decarboxylase / CO2 production ; [max 2] (b) fewer ATP molecules produced no / fewer, protons / H+, move through, ATP synth(et)ase / stalked particles or less steep, proton / H+, gradient ; I chemiosmosis more heat energy released H+ gradient / electron flow / ETC, energy converted to, heat / thermal energy; constant oxygen uptake ETC still works / oxygen acts as final electron acceptor ; I oxidative phosphorylation still works [3] [Total:8] 2 (a) 1 NicVAX / vaccine, recognised as, non-self / foreign ; 2 ref. to antigen presenting cells ; 3 (recognised / bound, by), specific / particular / certain, B-lymphocytes ; I correct / right 4 clonal selection ; 5 clonal expansion / mitosis / cell division, of B-lymphocytes ; 6 T-helper cells stimulate B-lymphocytes ; 7 T-helper cells release cytokine ; 8 B-lymphocytes, become / mature into, plasma cells ; 9 plasma cells, secrete / produce, antibody ; A B-cell for B-lymphocyte throughout [max 5]

Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 43 © Cambridge International Examinations 2015 (b) 1 plasma cells / B-lymphocytes, extracted from (mouse) spleen ; I blood 2 fused with, myeloma / cancerous / tumour / malignant, cells ; I ‘mixed with’ 3 use of, a fusogen / electrofusion ; A EFF–AFF / detergent 4 formation of hybridoma cells ; 5 identify hybridoma cells with, specific / anti-nicotine / relevant, antibody ; 6 large-scale culture / grow in fermenter ; 7 AVP ; e.g. detail of cell identification [max 4] (c) (i) increase from 0 to 30 mins or rapid / steep, increase from 0 to 15 mins ; (from 30 mins) decrease then, gradual / slow / gentle, increase ; I steady [2] (ii) 1 (both) antibodies reduce nicotine (concentration in the fetal circulation) ; 2 at specified time quote concentration for nicotine and either Nic-IgG or Nic311 plus units or compare maximum concentrations for nicotine = 12.5 ng cm–3 and Nic-IgG = 2.0 ng cm–3 and Nic311 = 5.5 ng cm–3 ; units need to be quoted once only 3 lower nicotine (concentration) gives fewer adverse side-effects in the fetus ; 4 Nic-IgG, is more effective / reduces the fetal nicotine (concentration) to a lower level, (than Nic311) ; 5 AVP ; e.g. do not know concentration of nicotine that is harmful to fetus / idea that nicotine still present in fetal circulation [max 3 ] (d) pregnancy testing / diagnosis of disease / treatment of disease / delivery of drugs / blood or tissue typing ; I monoclonal antibodies kill pathogens [1] [Total:15]

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 43 © Cambridge International Examinations 2015 3 (a) (i) 1 expose salmon to, IPN / virus ; 2 (humans) select / choose / identify, unaffected / resistant, individuals ; A survivors 3 breed them together ; 4 repeat for several generations ; [max 3] (ii) 1 increase in homozygosity ; 2 harmful recessive alleles may be expressed ; 3 inbreeding depression / loss of hybrid vigour ; 4 limited gene pool / decrease in genetic variation ; 5 AVP ; e.g. loss of desirable traits [max 2] (b) (i) accept ora throughout 1 comparative statement that group A, have lower percentage mortality ; 2 after 30 days no more in group A die or rise in deaths in group B, throughout / until 45 days ; 3 at specified time in days quote mortality for both A and B plus % unit ; A ‘percentage mortality’ for unit 4 (more) resistance / less susceptibility, (to IPN) in group A ; 5 ref. to resistance allele(s) ; A resistance gene R immunity / tolerance 6 infection spreads throughout / reservoir of infection in, group B ; [max 4] (ii) another, disease / pathogen, could be present ; by chance / random event ; e.g. pollution / temperature variation [max 1] [Total:10] 4 (a) 1 identify females, with the recessive allele / who are carriers ; 2 if embryo has allele can choose abortion ; 3 select unaffected IVF embryo (to implant) ; A pre-implantation genetic diagnosis 4 women can choose not to have children ; [max 2]

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 43 © Cambridge International Examinations 2015 (b) (i) 1 insert a, functional / normal / dominant / correct, allele ; R remove / replace, faulty allele I gene 2 to obtain, functional / normal / correct, protein / polypeptide ; A e.g. clotting factor 3 reduce the symptoms (of the disorder) ; 4 restore / modify / enhance, cellular functions ; A e.g. enzyme reaction / clotting process / membrane transport 5 increase, quality of life / life expectancy / survival ; A live normal life penalise germ-line therapy once only [max 2] (ii) 1 caused by a recessive allele ; 2 serious / common, disorder ; [max 1] (c) (i) 1 F9 gene is shorter ; 2 easier to insert into, plasmid / vector / adenovirus ; 3 easier to enter nucleus ; I into cell 4 easier to integrate into genome ; ora throughout for F8 gene [max 2] (ii) adenovirus advantage 1 (double-stranded) DNA so no, reverse transcription / making cDNA; I single-stranded to double-stranded step alone 2 high gene expression so produce more (therapeutic) protein ; adenovirus disadvantage 3 high immune response so adenovirus may be removed before it reaches target cells ; 4 high immune response so, allergies / side effects ; [max 3] [Total:10] 5 (a) (i) 1 diversity of, habitats / ecosystems ; 2 number of different species ; 3 genetic diversity within a species ; [max 2]

Page 7 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 43 © Cambridge International Examinations 2015 (ii) 1 maintain, food chains / food webs or maintain, stability / balance, in ecosystems ; I ecological 2 maintain, genetic diversity / genetic variation / gene pool ; 3 resources (for humans) ; e.g. biofuel / food / medicine / wood 4 aesthetic reasons / (eco)tourism ; 5 maintain, nutrient cycle / soil structure / climate stability ; [max 3] (b) (i) between 1970 and 1990 1 in terrestrial, as protected areas increase, biodiversity decreases A negative correlation / inversely proportional 2 in marine (general trend) as protected areas increase, biodiversity increases ; A positive correlation 3 exceptions ; e.g. dip, from 1980 / till 1985 (in marine) / rise, from 1970 / till 1975 (in terrestrial) ; between 1990 and 2005 4 in both habitats as total area protected increases, biodiversity decreases ; [max 3] (ii) 1 marine environments are difficult to, patrol / monitor ; 2 lack of public, awareness / interest ; 3 international ownership issues ; A example 4 difficult to, set / mark / recognise, boundaries ; 5 AVP ; e.g. problem of mobile populations [max 2] [Total:10] 6 (a) G ; C ; J ; B ; [4] (b) (i) 515 (%) ;; allow one mark for working e.g. 100) ( 2.40 2.40 14.76 × − or 100) ( 2.40 12.36 × [2]

Page 8 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 43 © Cambridge International Examinations 2015 (ii) high protein diet 1 hydrolysed / digested, to amino acids (in gut) ; A broken down 2 excess amino acids cannot be stored ; 3 deaminated (in liver) / ornithine cycle, to produce urea ; 4 more urea in blood filtered into nephron(s) ; [max 2] [Total:8] 7 (a) (i) 1 more light absorbed by chlorophyll ; 2 short / blue, wavelengths have more energy ; ora A suitable figures for X (in range 400–500nm) or for Y (in range 600–700nm) 3 (so) greater rate of photosynthesis ; [max 2] (ii) 1 contain chlorophyll ; 2 reflects / does not absorb, green light ; A reflects / does not absorb, 500–600 nm [2] (iii) 1 absorbs light, wavelengths / colours, not absorbed by, primary pigment / reaction centre / P680 / P700 ; 2 passes (light) energy to, primary pigment / reaction centre / P680 / P700 ; [2] (b) 1 decrease in rate of photosynthesis ; A photosynthesis stops 2 rubisco / enzyme, denatured ; 3 less / no, carbon dioxide, fixed / binds to RuBP ; 4 (initial) increase in transpiration ; A high transpiration 5 loss of turgor / wilting ; 6 ABA production ; 7 (eventually) stomata close ; 8 reduction in carbon dioxide uptake ; 9 photorespiration / rubisco binds to oxygen instead of carbon dioxide ; [max 5]

Page 9 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 43 © Cambridge International Examinations 2015 (c) rubisco / RuBP carboxylase-oxygenase ; thylakoids ; I membranes DNA ; ribosomes ; A 70S R 80S [4] [Total:15] 8 (a) NNGG NNGg NnGG NnGg ;; 4 correct = 2 marks 2/3 correct = 1 mark [2] (b) 1 test cross ; 2 cross fly with, vestigial wing and ebony body fly or double / homozygous, recessive fly / nngg fly ; 3 if some offspring have vestigial wing and / or ebony body genotype is heterozygous ; A if, some offspring have recessive trait / not all offspring have dominant trait, genotype is heterozygous 4 if offspring all have normal wing and / or grey body genotype is homozygous ; A if offspring all have dominant trait genotype is homozygous A short for vestigial and black for ebony throughout [max 3] (c) (white male) (red female) parental genotypes XwY × XRXo ; gametes Xw Y XR Xo ; offspring genotypes XRXw XoXw XRY XoY ; offspring phenotypes red-eyed female orange-eyed female red-eyed male orange-eyed male ; wrong symbols = 0 superscript R on Y chromosome = 0 superscripts w/o on Y chromosome = 1 (for correct line 4) no X and Y = max 2 (for correct lines 3 and 4) ecf alleles written as subscripts not superscripts = max 3 ecf superscript R written as small r = max 3 [4] [Total:9]

Page 10 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 43 © Cambridge International Examinations 2015 9 (a) 1 germinal epithelial cells form oogonia ; A primordial germ cells form oogonia 2 by mitosis ; A mitosis increases number of oogonia 3 ref. to germinal epithelial cells / oogonia, are, diploid / 2n ; 4 oogonia , grow / mature ; 5 (oogonia) start meiosis to form primary oocytes ; 6 meiosis stops at prophase 1 ; 7 stage, 1 / 2 / 3 / 4 / 5 / 6, occurs in, embryo / fetus ; 8 many primary oocytes in baby girl at birth ; 9 primary oocyte completes meiosis I ; 10 at / after, puberty ; A correct ref. to each menstrual cycle / before ovulation 11 produces secondary oocyte and (first) polar body ; 12 products (of meiosis I) are two haploid cells ; 13 secondary oocyte undergoes meiosis II at fertilisation ; 14 produces ovum and (second) polar body ; 15 AVP ; e.g. ref. to events occur in follicles correct names required for all mp except mp6, mp7, mp10, mp12 and mp15 [max 9] (b) 1 fall in concentration of progesterone ; 2 endometrium (uterine lining) breaks down ; I ‘thins’ 3 menstruation / period, occurs ; 4 follicular / granulose, cells secrete oestrogen ; I oestrogen produced 5 oestrogen concentration rises ; 6 (oestrogen) stimulates, proliferation / thickening / increase in blood vessels, of endometrium ; 7 corpus luteum secretes progesterone ; I progesterone produced 8 progesterone concentration increases ; 9 (progesterone) maintains endometrium ; I ‘thickens’ [max 6] [Total:15]

Page 11 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 43 © Cambridge International Examinations 2015 10 (a) 1 hybrids / offspring from cross between two species, infertile / sterile ; A AA × BB gives, sterile / infertile, AB 2 (normal) meiosis cannot occur ; 3 chromosomes do not pair up ; A set A chromosomes, not homologous to / do not pair with, set B 4 (spontaneous) doubling of chromosome number / formation of, tetraploid / AABB (emmer wheat) ; A chromosome doubling I doubling idea for mp 4 if context not chance occurrence but ecf for mp 6 5 non-disjunction (in mitosis) ; A in meiosis (unreduced gametes) 6 restores fertility / (AB) gametes can now form ; must be linked to mp 4 7 second hybridisation and polyploidy gives, hexaploid ; A 4n (emmer wheat) × 2n (wild goat grass) and chromosome number doubling → 6n A AABB × CC → ABC and doubling to AABBCC benefits 8 hybrid vigour ; 9 large grains ; 10 high yield ; 11 beneficial characteristic / named example, introduced by parent of hybrid ; A example e.g. shorter stems plus benefit / grain remains attached to ear more strongly plus benefit [max 8]

Page 12 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2015 9700 43 © Cambridge International Examinations 2015 (b) environmental 1 GM / genetically modified, rape may itself be, a weed / invasive ; 2 pollen transfer to / hybridisation with, wild relatives ; 3 resistant gene transfer to, non-GM crops / wild relatives ; I other plants 4 (resulting) hybrid offspring invasive ; 5 (intensive) use of herbicide selects for herbicide-resistant weeds ; 6 (intensive use of herbicide) reduces biodiversity ; economic 7 problem with competition between crops and herbicide-resistant weeds ; 8 idea of, contamination of organic farming / accidental mixing of GM crops with non-GM, financial consequences ; 9 high cost of / poor farmers cannot afford, GM, seeds / plants ; 10 cost of herbicide ; 11 cost of problems with pollution ; 12 cost of human health problems ; [max 7] [Total:15]