Cambridge A Level Biology 9700 — 2018 May/June Paper 4 · Variant 2

1 The Sumatran tiger lives on the island of Sumatra, Indonesia. Fig. 1.1 shows a Sumatran tiger. Fig. 1.1 (a) The International Union for Conservation of Nature (IUCN) is the world’s largest global environmental organisation. The IUCN Red List of Threatened Species™ evaluates the conservation status of plant and animal species. The Sumatran tiger is categorised as critically endangered on the IUCN Red List. There are approximately 400 Sumatran tigers left in the wild. Suggest reasons why the Sumatran tiger has become critically endangered. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (b) Some zoos in Europe have captive breeding programmes for the Sumatran tiger, which breeds well in captivity. Once cubs have reached maturity they may be moved to other zoos. For example, as part of a coordinated breeding programme carried out in 2013, one tiger born in Chester Zoo in England was moved to France and another tiger born in the same zoo was moved to Germany. This type of coordinated breeding programme is essential for the survival of the Sumatran tiger. Explain why it is important to the gene pool of Sumatran tigers to move tigers from one zoo to another. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (c) Some animals do not breed well in captivity. List the methods of assisted reproduction that may be used with animals that do not breed well in captivity. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 8]

Mark scheme: 1(a) any 3 of: 1 killed by humans ; 2 difficulty in finding a mate ; 3 loss of habitat / habitat fragmentation / habitat polluted / deforestation ; 4 food shortage ; 5 (new) diseases ; 3 1(b) any 3 of: 1 maintains or increases, gene pool / genetic diversity / genetic variation ; 2 maintains or increases, heterozygosity ; 3 hybrid vigour ; 4 reduces / prevents, inbreeding depression ; 5 reduce chances of harmful recessive alleles coming together ; 3 1(c) any 2 of: 1 IVF or description; 2 embryo transfer ; 3 surrogacy or description ; 4 artificial insemination ; 2

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2 (a) In the sweet pea plant, Lathyrus odoratus, one gene codes for flower colour and one gene codes for pollen grain shape. Flower colour is either purple or red. Pollen grain shape is either long or round. The inheritance of these genes is an example of autosomal linkage. • The allele F for purple flowers is dominant over the allele f for red flowers. • The allele G for long pollen grains is dominant over allele g for round pollen grains. Explain the meaning of the term autosomal linkage. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) A dihybrid cross was carried out between homozygous dominant and homozygous recessive sweet pea plant parents to produce the F1 generation. The offspring from the F1 generation were crossed to produce the F2 generation. (i) Draw a genetic diagram to show a dihybrid cross between two offspring from the F1 generation. Assume that these genes are closely linked and that there are no crossing over events. [4] (ii) The actual results of the dihybrid cross are shown in Table 2.1. Table 2.1 phenotypes of F2 offspring number of individuals purple flowers, long pollen grains 284 purple flowers, round pollen grains 21 red flowers, long pollen grains 21 red flowers, round pollen grains 55 State how the results support the fact that this is an example of autosomal linkage. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[1] (c) (i) In a test cross, an individual of known genotype is crossed with an individual that has a dominant phenotype but unknown genotype. State the genotype of the known individual in a test cross. .......................................................................................................................................[1] (ii) A test cross was carried out with sweet pea plants known to be heterozygous for both flower colour and pollen grain shape. The results of the test cross are shown in Table 2.2. Table 2.2 phenotypes of offspring of test number of cross individuals purple flowers, long pollen grains 215 purple flowers, round pollen grains 30 red flowers, long pollen grains 32 red flowers, round pollen grains 210 The result of a test cross can be used to determine a crossover value (COV). A crossover value is the percentage of the total number of offspring showing recombination. The crossover value (COV) can be calculated using the formula shown in Fig. 2.1. number of recombinants # COV = 100 total number of individuals Fig. 2.1 Calculate the COV from the results shown in Table 2.2. COV = .......................................................% [1] (iii) Suggest what information about the relative distance between the linked genes can be gained from crossover values. ........................................................................................................................................... .......................................................................................................................................[1] [Total: 10]

Mark scheme: 2(a) (autosomal) not a sex chromosome ; (linkage) genes on the same chromosome / alleles inherited together ; 2 2(b)(i) 1 F1 genotypes FG/fg or FGfg or (FG) (fg) ; A FfGg 2 gametes FG fg FG fg ; 3 F2 genotypes FGFG FGfg FGfg fgfg ; A FFGG FfGg FfGg ffgg 4 F2 phenotypes purple long purple long purple long red round ecf for mp3 from incorrect gametes (mp2) R if sex linkage 4 2(b)(ii) any 1 of: do not show 9:3:3:1 ratio ; large(r) numbers of parental phenotypes / low(er) numbers of recombinant phenotypes ; 1 2(c)(i) fgfg / homozygous recessive ; A ffgg 1 2(c)(ii) 12.7 or 13 ; 1 2(c)(iii) low, COV / crossover value, indicates genes closer together or high, COV / crossover value, indicates genes further apart ; 1

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3 Within a population, the variation for one characteristic is usually the result of genetic and environmental causes. (a) Meiosis is one source of genetic variation. (i) Describe the events that take place during prophase I of meiosis in an animal cell. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[4] (ii) Explain how independent assortment of homologous chromosomes leads to genetic variation during meiosis I. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (b) Both genetic and environmental factors contribute to variation in body mass in a bird population. One cause of variation in body mass in chickens is infection with the gut protoctist Eimeria. When fed with the same diet, infected chickens have a lower gain in body mass than healthy chickens in the same population. (i) Suggest reasons why infected chickens have a lower gain in body mass than healthy chickens. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Suggest one environmental factor that affects the growth and reproduction of Eimeria. ........................................................................................................................................... .......................................................................................................................................[1] (c) A study was carried out to investigate the effect of treating infected chickens with extracts from a plant, Bidens pilosa. B. pilosa is used in traditional medicine for the treatment of some infectious diseases. • Three populations, each containing 25 chickens infected with Eimeria were observed. • The body mass of each chicken was measured at the start of the study. • Each population was given a different treatment for 56 days: standard diet with no B. pilosa extract standard diet with low dose of B. pilosa extract standard diet with high dose of B. pilosa extract. • The body mass of each chicken was measured at the end of the study and the gain in body mass over the 56 days was calculated. • The mean gain in body mass was calculated for each population. The results are shown in Table 3.1. Table 3.1 treatment mean gain in body mass / g no B. pilosa 1773.0 ±23.9 low dose of B. pilosa 2093.1 ±34.2 high dose of B. pilosa 2033.3 ±29.9 A statistical analysis of the results of the study of these three populations confirmed that there was a significant difference in mean gain in body mass between low dose B. pilosa and high dose B. Pilosa extract treatments. (i) Name a statistical test that could be used to analyse the results of this study. .......................................................................................................................................[1] (ii) Sketch on Fig. 3.1 curves to show the pattern of variation for gain in body mass in chickens treated with low dose B. pilosa extract and in chickens treated with high dose B. pilosa extract. Label one curve ‘low dose’ and the other curve ‘high dose’. number of chickens 2063 body mass gain / g Fig. 3.1 [3] [Total: 13]

Mark scheme: 3(a)(i) any 4 of: 1 chromosomes become visible ; 2 condensation / coiling / supercoiling ; A description 3 nuclear envelope or nuclear membrane, disintegrates / disappears ; 4 nucleolus, disintegrates / disappears ; 5 centrioles migrate to (opposite) poles ; I centrosomes 6 spindle forms / microtubules assemble ; 7 synapsis / bivalents form / homologous chromosomes pair up ; 8 chiasmata formation / crossing over ; 4 3(a)(ii) any 2 of: 1 (homologous chromosomes / bivalents) line up randomly on the, equator / metaphase plate ; 2 produces (many) different chromosome combinations ; 3 number of possible combinations is 2n ; 4 new combinations of maternal and paternal alleles ; 2 3(b)(i) any 2 of: 1 protoctist uses up nutrients ; 2 loss of appetite / inability to feed ; 3 idea of impaired absorption / diarrhoea ; 4 (so) less nutrients available for, growth / gain of body mass / increase in biomass ; 5 AVP ; e.g. idea of energy used to fight infection / ref. to immune response 2 Question Answer Marks 3(b)(ii) (chicken) body temperature / competition with microbes / presence of drugs / presence of plant extract ; 1 3(c)(i) t-test ; 1 3(c)(ii) any 3 of: 1 two labelled bell-shaped curves ; 2 high dose curve shifted to left of low dose curve ; 3 high dose curve peak to left of 2063 ; 4 high dose curve, narrower / peak higher ; 3

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4 Adult Drosophila fruit flies feed on yeasts on the surface of ripe fruits. Female flies lay their eggs inside the fruit and the larvae feed inside the fruit before developing into adults. The tree Morinda citrifolia produces fruits called noni fruits. Noni fruits contain a toxin and are avoided by nearly all Drosophila species, including Drosophila yakuba in mainland Africa. One population of D. yakuba lives on the island of Mayotte off the coast of Africa. The population of D. yakuba on Mayotte has specialised to feed on noni fruit. Flies of this population produce enzymes that give them resistance to the toxin in noni fruits. Adults are able to feed on yeasts on the fruit surface and their eggs and larvae can develop and grow inside the fruit. Scientists have used bioinformatics to estimate that the unique D. yakuba population on Mayotte originated in the last 30 000 years. (a) Describe how a population of Drosophila can develop resistance to noni fruit toxin. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] (b) Explain why living on an island increases the likelihood of the noni toxin-resistant population of D. yakuba on Mayotte becoming a separate species to the mainland population of D. yakuba. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (c) Suggest how scientists can use bioinformatics to estimate the length of time that the population of D. yakuba has been living on Mayotte. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] [Total: 10]

Mark scheme: 4(a) any 4 of: 1 spontaneous / random, mutation ; 2 produces an allele coding for enzyme (that gives resistance) ; 3 toxin is selection pressure ; 4 flies or larvae, with, allele / enzyme / mutation, have selective advantage ; 5 (flies with, allele / enzyme / mutation) survive / reproduce ; ora 6 pass on this allele (to offspring) ; 7 this allele increases in frequency ; 4 Question Answer Marks 4(b) any 3 of: 1 separated by a geographical barrier / description ; 2 different, environmental conditions / selection pressures ; 3 allopatric speciation ; 4 reproductive isolation / no breeding with mainland flies / no gene flow (between two populations) ; A pre-zygotic isolation 5 new allele not, lost / diluted in frequency ; 6 founder effect / genetic drift ; 7 (more pronounced) in small population ; 3 4(c) any 3 of: 1 obtain, DNA / genome / gene / base / nucleotide, sequences (from database) ; 2 obtain, amino acid / protein, sequences (from database) ; 3 compare sequences (of two populations) ; 4 calculate percentage of, bases / amino acids, changed ; 5 mitochondrial DNA ; 6 molecular clock / mutation rate, (to estimate time) ; 3

5 (a) A severe reduction of blood flow to the brain causes cells to die. This is called a stroke. The after-effects of a stroke can range from recovery to permanent brain damage and death. A new emergency gene therapy treatment for people who are at risk of brain damage from a stroke was tested in mice. • The human granulocyte colony-stimulating factor, hG-CSF, is a protein that stimulates the production of stem cells in bone marrow. • mRNA coding for hG-CSF was obtained and used to make cDNA. • This cDNA was inserted into an adeno-associated virus (AAV) vector and given in eye drops to mice just after they experienced a stroke. (i) Explain what is meant by gene therapy. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Describe the roles of reverse transcriptase and DNA polymerase in making cDNA for hG-CSF. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (iii) The AAV vector used was unable to replicate itself within the target cells. Suggest why the researchers chose a vector that could not replicate. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (b) A study was carried out to investigate the effect of the gene therapy described in (a). Four groups of mice were used. • Group A mice had a stroke. They received eye drops containing AAV vector carrying cDNA for hG-CSF once only. • Group B mice had a stroke. They received eye drops containing AAV vector carrying cDNA for hG-CSF four times. • Group C mice had a stroke. They received eye drops containing AAV vector carrying the GFP gene coding for green fluorescent protein, instead of the cDNA for hG-CSF, once only. • Group D mice did not have a stroke. They were not given any eye drops. (i) Explain why the mice in group C were used in the study. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[3] (ii) Explain why the mice in group D were used in the study. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[1] (c) Table 5.1 summarises some results from the study and shows: • the percentage of mice surviving • the percentage of brain volume occupied by fluid-filled space • the score on a behavioural test in which normal mice score 0.5 and brain-damaged mice score nearer to 1.0. Table 5.1 mouse treatment percentage of percentage of behavioural test group mice surviving brain occupied score / arbitrary by fluid-filled units space A 63 3.6 0.67 B 100 3.0 0.67 C 25 5.2 0.90 D 100 3.0 0.50 Use the results in Table 5.1 to evaluate the benefits of gene therapy treatment, with AAV vector carrying the gene for hG-CSF, for people who have a stroke. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] [Total: 14]

Mark scheme: 5(a)(i) any 2 of: 1 to treat disease caused by, faulty / recessive, allele ; A cure 2 deliver, gene / allele / DNA (into target cells of individuals) ; 3 named example ; e.g. SCID / cystic fibrosis 2 5(a)(ii) (reverse transcriptase) uses mRNA to make (ss)DNA ; (DNA polymerase) makes DNA double-stranded / description ; 2 5(a)(iii) 1 to prevent virus spreading (throughout the body) ; 2 to limit, side effects / immune response / cancer / illness / infection / cell destruction ; 2 5(b)(i) any 3 of: 1 control ; 2 idea of to see if gene therapy has worked or to compare the effect of no gene therapy with gene therapy or to compare the effect of no hG-CSF gene with presence of hG-CSF gene ; 3 to see where the gene (carried by the vector), goes (in, body / brain) or to see if the gene (carried by the vector), enters cells or to see where cells have been transformed ; 4 to see the effects of the, vector / virus, alone ; 3 Question Answer Marks 5(b)(ii) any 1 of: 1 (control) to compare results (of mice without stroke) to mice with stroke ; 2 to establish baseline figures (in mice without a stroke) / act as a reference point ; 3 to make the study valid ; 1 Question Answer Marks 5(c) any 4 of: survival 1 increases survival ; brain damage 2 reduces (percentage of brain occupied by) fluid filled space ; behaviour test 3 improves / lowers, behavioural test score ; A score closer to normal general 4 comparative data quote to support, mp1 / mp2 / mp3 ; A processed data mp1 25% v 63% 25% v 100% 63% v 100% mp2 5.2% v 3.6% 5.2% v 3.0% 3.6% v 3.0% mp3 0.90 v 0.67 5 eye drops four times is better than eye drops once ; ora 6 if effective with mice likely to be effective with humans ; 7 ref. to unknown differences between mice and humans / study is only on mice ; 8 treatment is, non-invasive / quick ; 4

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6 Fig. 6.1 is an outline diagram showing two stages of aerobic respiration. pyruvate (3C) stage X acetyl coenzyme A (2C) step 1 oxaloacetate (4C) citrate reduced Q (6C) stage Y Q step 3 step 2 reduced R ATP ADP + Pi Q R reduced Q intermediate reduced Q compounds Q (5C) Fig. 6.1 (a) With reference to Fig. 6.1: name stage X ................................................................................................................................................... state the precise location of stage Y ................................................................................................................................................... state the numbered steps in stage Y in which decarboxylation occurs ................................................................................................................................................... name the type of reaction by which ATP is made during step 3 ................................................................................................................................................... name substances Q and R. ...............................................................................................................................................[5] (b) Uncontrolled cell division can lead to a cancerous tumour. Many cancer cells break down the amino acid glutamine and convert it to a 5-carbon intermediate compound, which is shown in Fig. 6.1. Suggest how the breakdown of glutamine can lead to the production of ATP in a cancer cell, other than that directly produced during step 3. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 7]

Mark scheme: 6(a) link reaction ; matrix of mitochondrion ; 2 and 3 ; substrate–linked phosphorylation ; A substrate-level phosphorylation (Q =) NAD and (R=) FAD ; 5 6(b) reduced, NAD / FAD, to ETC ; A description ref. to oxidative phosphorylation / chemiosmosis ; A description 2

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7 Fig. 7.1 shows part of a section through the leaf of a C4 plant such as maize. The letters A, B and C show three types of cell found in the leaf. A B C stoma Fig. 7.1 (a) Complete Table 7.1 by using the letters A, B or C from Fig. 7.1 to show the location of several compounds associated with photosynthesis in C4 plants. You may use A, B and C once, more than once, or not at all. Table 7.1 compound location PEP carboxylase .................. RuBP .................. rubisco .................. [3] (b) Explain why the cells in C form a tight ring around the cells in B. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (c) Several enzymes have a role in the light independent stage of photosynthesis. Fig. 7.2 shows the activity of one of these enzymes at different temperatures for a C3 plant, Pisum sativum, and a C4 plant, Amaranthus hypochondriacus. 100 90 80 Amaranthus 70 hypochondriacus C4 60 percentage of 50enzyme activity 40 30 Pisum sativum 20 C3 10 0 0 10 20 30 40 50 temperature / °C Fig. 7.2 Describe the differences between the two curves and suggest explanations for these differences. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] [Total: 8]

Mark scheme: 7(a) compound location PEP carboxylase C RuBP B rubisco B 3 7(b) any 2 of: 1 stop, air / oxygen, getting to cells B / rubisco / RuBP / bundle sheath cells ; 2 so rubisco does not catalyse reaction of oxygen with RuBP ; 3 reduce / stop, photorespiration ; 2 7(c) any 3 of: describe 1 (enzyme in) Amaranthus / C4, peaks, at higher temperature or (enzyme in) Amaranthus / C4, has higher optimum temperature or (enzyme in) Pisum / C3, has higher activity at low(er) temperatures ; ora 2 data quote ; e.g. C4 – 45 °C and C3 – 30 °C at 100% or C4 – 70% and C3 – 25% at 50 °C ; explain 3 enzyme in, Pisum / C3, less thermostable / denatures at lower temperature ; ora 4 AVP ; e.g. Amaranthus adapted to a hot climate / ora e.g. enzyme in Amaranthus has more S-S bonds 3

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8 (a) Describe how an action potential arriving at a presynaptic neurone can result in the release of acetylcholine into the synaptic cleft. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] (b) Lidocaine is a local anaesthetic drug which blocks voltage-gated Na+ channels in the postsynaptic membrane of a synapse. Explain the effect of lidocaine action on the postsynaptic neurone. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (c) Fig. 8.1 is a diagram of a neuromuscular junction. A C B Fig. 8.1 (i) With reference to Fig. 8.1, name structures A, B and C. A ............................................................................. B ............................................................................. C ............................................................................. [3] (ii) Outline the importance of structure A in the transmission of nerve impulses. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (iii) Explain why the structures labelled C are needed in a neuromuscular junction. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[4] [Total: 15]

Mark scheme: 8(a) any 4 of: 1 (presynaptic) membrane depolarised ; 2 calcium (ion) channels open / (presynaptic membrane becomes) more permeable to Ca2+ ; 3 Ca2+ move in (to presynaptic, neurone / knob) ; R membrane 4 causes vesicles (of ACh) to move towards (presynaptic) membrane ; 5 vesicle fuses with (presynaptic) membrane / exocytosis ; 4 8(b) any 2 of: 1 no Na+ enter (postsynaptic neurone) ; 2 no depolarisation of (postsynaptic) membrane ; 3 no, action potentials / impulses (transmitted) ; 2 8(c)(i) A – myelin sheath / Schwann cell ; B – sarcolemma ; C – mitochondria ; 3 8(c)(ii) any 2 of: 1 insulates axon / prevents movement of ions / AW ; 2 increases speed (of impulses) ; 3 saltatory conduction / described ; 2 Question Answer Marks 8(c)(iii) 1 produces / releases, ATP ; any 3 of mp2-5: 2 (for) production / recycling, of ACh ; 3 (for) production / movement, of vesicles or exocytosis ; 4 (for) contraction of sarcomere ; 5 (for) sodium potassium pumps / active transport ; 4

9 (a) Outline the characteristic features of organisms in the domain Archaea. [8] (b) Describe the role of seed banks in the protection of endangered species. [7] [Total: 15]

Mark scheme: 9(a) any 8 of: 1 no, nucleus / nuclear envelope ; 2 circular DNA / plasmid ; A circular chromosome 3 cell membrane composition different (from, bacteria / eukarya) ; 4 no membrane-bound organelles / named example ; 5 70S ribosomes ; 6 cell wall present ; 7 cell wall does not contain peptidoglycans ; 8 cells divide by, binary / multiple, fission ; 9 (mostly) unicellular ; 10 (some) have histones associated with DNA ; 11 (some) are extremophiles / description ; 8 Question Answer Marks 9(b) any 7 of: 1 variety / collection, of seeds ; 2 long-term storage ; 3 (storage) to, prevent germination / maintain dormancy ; 4 detail of storage conditions ; e.g. frozen or stored at temperatures below –13 °C dried 5 seeds regularly tested for viability / description ; 6 seeds re-stocked regularly ; 7 maintain genetic diversity / genetic material preserved / acts as a gene bank ; 8 detail of future use ; e.g. can be used after environmental disaster / new diseases / used to increase numbers 9 maintains biodiversity ; 10 research / education ; 11 AVP ; e.g. takes up less space than growing plants / ref. to cheaper to store than whole plants 7

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10 (a) Describe the process of selective reabsorption and explain how the cells of the proximal convoluted tubule are adapted to carry out selective reabsorption. [9] (b) Describe the role of the hypothalamus and the role of the posterior pituitary in osmoregulation. 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Mark scheme: 10(a) any 9 of: structure 1 microvilli ; A brush border 2 many mitochondria ; 3 tight junctions ; 4 folded, basal membrane / described ; 5 many, transport proteins / cotransporters / pumps / carrier proteins ; process 6 active transport of Na+, out of (pct) cells / into blood ; 7 Na+ ion concentration, gradient (produced) / reduced in cell ; 8 Na+ enters (pct) cells from, lumen / tubule / filtrate ; 9 by facilitated diffusion / using carrier protein ; 10 cotransport (from lumen to pct cell) ; 11 of, glucose / amino acids / ions ; only award if mp10 given 12 AVP ; e.g. water enters (pct) cells 9 Question Answer Marks 10(b) any 6 of: 1 hypothalamus detects water potential of blood ; 2 (using) osmoreceptors ; 3 (if) water potential, decreases / is low ; 4 ADH produced ; 5 ref. to neurosecretion / neurosecretory cells ; 6 which extend into posterior pituitary ; 7 ADH released from posterior pituitary ; 8 enters blood stream ; 9 water potential of blood, increases / returns to set point ; 10 negative feedback ; 6