Cambridge A Level Biology 9700 — 2018 Oct/Nov Paper 4 · Variant 3

1 The Bali starling, Leucopsar rothschildi, is found on the island of Bali, Indonesia. Fig. 1.1 shows a Bali starling. 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 Bali starling is categorised as critically endangered on the IUCN Red List, which means that it is nearly extinct in the wild. The Bali starling is protected by Indonesian law and trapping of the birds for sale as pets is illegal. In 2001 there were only six Bali starlings in the wild and about 1000 in captivity. (i) Suggest reasons, other than the trapping of the birds for sale as pets, for the very low numbers of Bali starlings in the wild. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (ii) Describe the roles zoos can take in the protection of the Bali starling. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (b) Explain the consequences on genetic biodiversity of having a wild population of just six birds. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] [Total: 9]

Mark scheme: 1(a)(i) any three from 1 habitat destruction ; 2 competition for food / food shortage ; 3 predation / predators ; 4 disease / infection ; 5 AVP ; e.g. pollution / pesticide use / removed from wild (to zoos) to conserve species / problems finding a mate 3 1(a)(ii) any two from 1 (captive) breeding ; 2 release into wild / insurance populations ; 3 education / increase awareness (of illegal bird trade) ; 4 research, diet / habitat / breeding / behaviour / genetic diversity ; 5 raise money / work with (Bali) government, to set up reserves ; 6 veterinary / health, care ; 2 1(b) any four from 1 low / little / less / decreased (genetic) (bio)diversity / variation ; 2 low / little / less / decreased, heterozygosity ; 3 harmful recessive alleles may come together ; 4 inbreeding depression ; 5 no / less / decreased, hybrid vigour ; 6 not / less, able to adapt to, new / changed, selection pressure / disease / environment ; 4

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2 In the fruit fly, Drosophila melanogaster, two different genes control body colour and eye colour. • G/g are alleles of the body colour gene. • G results in grey body, g results in black body. • R/r are alleles of the eye colour gene. • R results in red eyes, r results in brown eyes. Each gene is autosomal. A dihybrid cross was carried out using a fly with a grey body and red eyes crossed with a fly with a black body and brown eyes. Both parents were homozygous for both genes. The offspring from the F1 generation were crossed to obtain the F2 offspring. (a) A statistical test showed that the results of the cross were significantly different from those expected. State the name of the statistical test used and state the expected phenotypic ratio for the F2 generation. statistical test ............................................................................................................................ expected ratio ........................................................................................................................... [2] (b) A test cross can be carried out in order to identify flies from an F2 generation that are heterozygous for both genes. Draw a genetic diagram to show how a test cross between a heterozygous grey‑bodied, red‑eyed F2 fly and a fly with a black body and brown eyes can produce four different offspring phenotypes. Use the symbols G/g and R/r. [4] (c) The results of the test cross in (b) are shown in Table 2.1. These results are significantly different from the expected results. Table 2.1 phenotypes of offspring of number of test cross individuals grey body, red eyes 123 grey body, brown eyes 7 black body, red eyes 6 black body, brown eyes 132 Describe how these results are different from the expected results and explain why they are different. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [5] [Total: 11]

Mark scheme: 2(a) 9 : 3 : 3 : 1 ; 2 2(b) parents: GgRr x ggrr gametes: GR Gr gR gr x gr ; offspring: GgRr Ggrr ggRr ggrr ; phenotypes: grey (body) grey (body) black (body) black (body) red (eyes) brown (eyes) red (eyes) brown (eyes) ; 4 2(c) any five from 1 expect (ratio) / (ratio) not, 1 : 1 : 1 : 1 ; 2 get, many / more / most / majority / large numbers of, grey(body)-red(eye) and black(body)-brown(eye) (flies / offspring) ; 3 majority resemble (original) parents / have parental phenotypes ; 4 1 : 1 ratio of two parental types and 1 : 1 ratio of recombinant types ; 5 genes / loci / eye colour and body colour, are, linked / on same chromosome ; 6 (alleles) G+R linked and g+r linked (in, F1 / heterozygotes) ; 7 no, random / independent, assortment ; 8 crossing over (produces recombinants / minority phenotypes) ; 9 low numbers of recombinants show, genes / loci, close together ; 5

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3 (a) Meiosis is one process that contributes to genetic variation. (i) State precisely the stage of meiosis where single chromosomes line up on the equator. ...................................................................................................................................... [1] (ii) Outline the events taking place during anaphase I of meiosis. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (iii) Describe how crossing over during meiosis leads to genetic variation. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] Mutation also causes genetic variation. Some populations of water hemp, Amaranthus tuberculatus, have evolved herbicide resistance as a result of a mutation. This is a problem for farmers as water hemp grows in crop fields, lowering productivity. Two populations of water hemp were tested for resistance to the herbicide mesotrione. One was a population known to be resistant (control) and the other was a test population, whose resistance was unknown. • Leaves were removed and immersed in a radioactively labelled solution of mesotrione. • The leaves absorbed some mesotrione and became radioactive. • Resistant leaves are able to degrade mesotrione by metabolism. • The time for 50% of absorbed mesotrione to degrade was calculated by measuring the radioactivity of the leaves. The results are shown in Table 3.1. Table 3.1 population of water mean time for 50% of standard hemp absorbed mesotrione to deviation degrade / hours test 27.5 4.75 control 10.1 2.34 (b) (i) Explain how the results in Table 3.1 show that the two populations differ in their resistance to mesotrione. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) Explain why this example of genetic variation is important for natural selection in water hemp populations. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (iii) Farmers can send in a sample of leaves of water hemp from their fields to a laboratory to be tested for resistance to mesotrione or other herbicides. Suggest the benefit of this to a farmer. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [1] (c) The null hypothesis states there is no significant difference between the mean times for 50% of absorbed mesotrione to degrade in the two populations. A t‑test can be carried out to compare these two means. The critical value for t at the p = 0.05 significance level is 2.23. (i) Use the formula in Fig. 3.1 to calculate the value of t. Show your working. Key x 1 - x 2 x = mean t = s 21 s 22 s = standard deviation + n1 = 6 (number of readings for test population) e n 1 n 2 o n2 = 6 (number of readings for control population) Fig. 3.1 t = .......................................................... [2] (ii) Use your calculated value of t to explain whether the null hypothesis should be accepted or rejected. accept or reject ................................................. explanation ........................................................................................................................ ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] [Total: 14]

Mark scheme: 3(a)(i) metaphase, II / 2 ; 1 3(a)(ii) any two from 1 spindle (fibres) / microtubules, contract / shorten ; 2 chromosome(s), move / pulled, to pole(s) / centriole(s) ; 3 centromeres, move / lead / pulled, first ; 4 homologous chromosomes separate or pair (of chromosomes) / bivalent, pulled apart ; 2 Question Answer Marks 3(a)(iii) any two from 1 non-sister chromatids ; 2 swap / exchange, alleles / sections of DNA ; 3 chiasmata ; 4 (giving) new / different, allele combinations or recombination ; 2 3(b)(i) any two from 1 test (leaves / population / group) takes, longer / higher time or test group degrades mesotrione more slowly ; 2 so test group, less resistant / fewer have resistance ; 3 test standard deviation is, larger / higher, so (population) more variable ; 2 3(b)(ii) any two from 1 can, adapt / evolve / survive long term ; 2 resistant / mutant, plants / hemp, have selective advantage ; 3 in areas where / when, mesotrione is used ; 4 resistant / mutant, plants survive and reproduce ; 2 3(b)(iii) any one from 1 lets farmer choose, different / correct / effective / best, herbicide ; 2 less money wasted due to, wrong herbicide / reduced crop yield ; 1 3(c)(i) numerator working 17.4 ; denominator working √3.76 + 0.91 OR √4.67 OR 2.16 ; 2 3(c)(ii) any two from 1 reject because, t / calculated value / it, is, higher / greater / more, than, 2.23 / critical value ; 2 probability of result by chance is less than, 0.05 / 5% or p < 0.05 ; 3 there is a significant difference between the, means / populations / test and control ; 2

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4 The Hawaiian archipelago is a group of volcanic islands in the Pacific ocean. Fig. 4.1 shows the relative locations of five of these islands. Moloka’i O’ahu Maui Lana’i Hawai’i Fig. 4.1 Table 4.1 shows the size and age of these five islands and the total number of Mecyclothorax ground beetle species and their species density, on each island. Data for the island of Maui is shown as two distinct regions, West and Haleakalã. This is because they formed at different times from two separate volcanoes. Table 4.1 island area / km2 age of island / total number of species density / number million years Mecyclothorax species of species per km2 Hawai’i 10 433 0.4 30 0.003 Lãna’i 364 1.3 3 0.008 Maui (West) 443 1.3 27 0.061 Maui 1440 1.1 116 0.081(Haleakalã) Moloka’i 673 1.9 – 1.8 43 0.064 O’ahu 1545 3.7 – 2.6 20 .................... Fig. 4.2 shows a ground beetle of the genus Mecyclothorax. All the beetle species of this genus on the Hawaiian archipelago form a monophyletic group, descended from one original colonising species that reached Maui from Australia. Fig. 4.2 (a) (i) Complete Table 4.1 by calculating the density of Mecyclothorax beetle species on the island of O’ahu. [1] (ii) Use Table 4.1 to explain why the island of Hawai’i has the lowest density of Mecyclothorax beetle species. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (iii) Use Fig. 4.1 and Table 4.1 to suggest why O’ahu has a lower number of Mecyclothorax beetle species than Moloka’i. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (b) The rate of speciation of Mecyclothorax beetles on the slopes of the volcano Haleakalã, on Maui, is the highest recorded for any genus or location on earth. The volcano last erupted 400 years ago. This produced lava flows which cut through ancient forest and vegetation, dividing it into many separate microhabitats. Explain how the large number of Mecyclothorax beetle species on Haleakalã developed. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] [Total: 9]

Mark scheme: 4(a)(i) (O’ahu) 0.013 ; 1 4(a)(ii) any two from 1 Hawai’i is youngest island / formed only 0.4 million years ago ; 2 less / little, time for, Mecyclothorax / beetles / them, to, colonise / arrive / speciate ; 3 Hawai’i has large area so, density / number of species per km2, is low ; 4 early stage of succession / low habitat diversity ; 2 4(a)(iii) any two from 1 O’ahu is further away from Maui (than Moloka’i) ; 2 fewer, Mecyclothorax / beetles / species, reached O’ahu ; 3 O’ahu was colonised later ; 4 less time for speciation ; 2 4(b) any four from 1 allopatric speciation ; 2 geographical isolation / population became divided ; 3 different, mutations / alleles / gene pools, in different, populations ; 4 ref. to different selection pressures ; 5 no, gene flow / interbreeding, between populations ; 6 genetic differences accumulated ; 7 natural selection ; 8 genetic drift / founder effect ; 4

5 (a) Leber’s congenital amaurosis (LCA) is an autosomal recessive eye disease. LCA results in eye disorders, including severe loss of vision, at birth. LCA has been successfully treated by gene therapy, using a virus instead of a plasmid as the vector. Adeno‑associated virus (AAV) vectors containing the therapeutic allele were injected directly into the retina, the layer at the back of the eye containing the photoreceptor cells. People who had been blind from a young age were able to see again. There is a risk associated with the injection method used to deliver the vectors, as it might cause the retina to detach, damaging vision. This method of delivery was first used for LCA before being trialled on other retinal diseases that gradually reduce the vision of people as they get older. (i) Suggest the main steps involved in creating recombinant DNA for this example of gene therapy. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [4] (ii) Explain why the fact that LCA is an autosomal recessive genetic disease makes it suitable for treatment with gene therapy. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (iii) Suggest why the retinal injection method of gene therapy was used for LCA before it was trialled on other retinal diseases that gradually reduce the vision of people as they get older. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (b) Scientists tried to create an improved virus vector for gene therapy. step 1 – The scientists used a special form of the polymerase chain reaction (PCR). This form of PCR causes mutations in the DNA sequence of AAV by base substitution. step 2 – The viruses containing different base substitutions were tested. This was done by using the different viruses to deliver a new gene, the gene for green fluorescent protein (GFP), into the photoreceptor cells of mice, using the retinal injection method. step 3 – The best virus, known as 7m8, caused the photoreceptor cells in the retina of the mouse to fluoresce brightly, even when the recombinant virus was injected into the fluid inside the eye instead of into the retina itself. step 4 – The 7m8 virus was used to cure a mouse with LCA by injecting this virus containing the therapeutic allele into the fluid inside the eye of the mouse. (i) Suggest how errors occurring during PCR can cause base substitution mutations in the DNA sequence of AAV. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (ii) Explain why the photoreceptor cells of the mouse fluoresced in step 3. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (iii) Predict the impact of the 7m8 AAV on treatment for age‑related retinal diseases. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[1] [Total: 14]

Mark scheme: 5(a)(i) any four from 1 synthesise / obtain, therapeutic / correct / normal / dominant, allele / (c)DNA ; 2 from, mRNA / (cells of) healthy person / gene library ; 3 ref. to probe / electrophoresis / sequencing, for identification ; 4 ref. to PCR to amplify, gene / DNA ; 5 restriction, enzyme / endonuclease, + role ; 6 (DNA) ligase + role ; 7 ref. to add promoter ; 4 5(a)(ii) any two from 1 (can) add / insert, correct / normal / therapeutic / dominant / functional, DNA / allele ; 2 (only) need one, allele / copy, per cell ; 3 to, cure disease / correct phenotype / restore function / restore vision ; 4 to, make / synthesise, correct / functional, protein ; 5 no need to, edit / knock out / remove / delete, faulty allele (as would be the case if faulty allele was dominant) ; 2 5(a)(iii) any two from 1 LCA patients already, blind / have lost vision / cannot see ; 2 less likely to, harm / add to problem / worsen vision, for LCA patients ; 3 AVP ; e.g. other age-related diseases may involve, many genes / dominant alleles 2 5(b)(i) any three from 1 wrong base added / wrong bases pair up / base pair mismatch ; 2 with / to, template, DNA / strand(s) ; 3 Taq polymerase, inaccurate / no proofreading ; 4 in, extension / elongation, stage ; 5 mistake / mutation, replicated / copied, (many times) ; 6 AVP ; e.g. high temperature speeds up replication and increases chance of more mistakes 3 5(b)(ii) any two from 1 AAV / 7m8 / virus / vector, crossed fluid / reached photoreceptors / reached retina ; 2 (virus) delivered (new / GFP) gene / DNA, to photoreceptors / retina cells ; 3 (GFP) gene / DNA, expressed in, photoreceptors / retina cells ; 4 GFP / protein product, is fluorescent ; 2 5(b)(iii) (7m8) will decrease risk of / increase treatment by, gene therapy ; 1

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6 (a) Fig. 6.1 outlines some of the steps of glycolysis. glucose step 1 glucose 6–phosphate step 2 fructose 6–phosphate step 3 fructose 1, 6–bisphosphate step 4 triose triose phosphate phosphate step 5 pyruvate pyruvate Fig. 6.1 (i) State the precise location of glycolysis in the cell. ...................................................................................................................................... [1] (ii) With reference to Fig. 6.1: state the steps where phosphorylation occurs ........................................................................................................................................... state the step where oxidation occurs ........................................................................................................................................... name the type of reaction by which ATP is made during step 5. ........................................................................................................................................... [3] (b) Some cancer cells have different metabolic requirements from normal cells. These cancer cells obtain most of their ATP from glycolysis, even if oxygen is available. State how the glucose and oxygen requirements of these cancer cells differ from normal cells. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] [Total: 6]

Mark scheme: 6(a)(i) cytoplasm ; 1 6(a)(ii) phosphorylation: 1 (and) 3 ; oxidation: 5 ; reaction in step 5: substrate-linked phosphorylation ; 3 6(b) any two from cancer cells 1 need / use, more glucose ; 2 need / use, less oxygen ; 3 get, little energy / 2 ATP / small amount of ATP, per glucose (molecule) / from glycolysis ; 2

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7 (a) Fig. 7.1 is a transmission electron micrograph of a chloroplast. A B C Fig. 7.1 Many compounds and structures involved in photosynthesis are located in a chloroplast. Using the labels A, B or C, complete Table 7.1 to show the location of four of these compounds or structures. You may use each of the letters A, B and C once, more than once, or not at all. Table 7.1 compound or structure location ATP synthase rubisco starch grain phospholipid bilayer [3] (b) Elodea canadensis is an aquatic plant that lives submerged in freshwater. Equal‑sized plants of E. canadensis were exposed to different wavelengths of light for the same period of time. As each plant photosynthesised, the number of bubbles of oxygen leaving the plant was counted. For each wavelength, the rate of oxygen production was calculated. The results are shown in Fig. 7.2. rate of oxygen production / bubbles per minute 400 450 500 550 600 650 700 wavelength of light / nm Fig. 7.2 Describe and explain the results shown in Fig. 7.2. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (c) Chlorophyll b, carotene and xanthophyll are known as accessory pigments. Describe the role of the accessory pigments in photosynthesis. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] [Total: 8]

Mark scheme: 7(a) compound or structure location ATP synthase C rubisco B starch grain A phospholipid bilayer C ;;; 3 Question Answer Marks 7(b) any three from description 1 high(est) / peaks, at, 450 nm and 650 nm / blue and red or lowest at, 545 nm / 535–555 nm / green ; explanations for blue / red 2 Elodea / plant / chloroplasts / (named) pigments, absorb / use, this, (specified) wavelength / (named) colour of light ; 3 light / wavelengths, used for, photosynthesis / photophosphorylation ; 4 so more oxygen, produced / released, from photolysis ; 5 AVP ; e.g. ref. non-cyclic / light-dependent stage / Z-scheme 3 7(c) any two from 1 pass, energy / light / photons, to, chlorophyll a / primary pigment / reaction centre ; 2 absorb different wavelengths to, chlorophyll a / primary pigment / reaction centre ; 3 form / in, light-harvesting cluster / photosystem / antenna complex ; 2

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8 (a) Outline the roles of sensory receptor cells in the mammalian nervous system. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) Fig. 8.1 shows the changes in potential difference (p.d.) across the membrane of a receptor cell over a period of time. The membrane was stimulated at time A and at time B with stimuli of different intensities. +60 +40 +20 0 membrane CC DD –20 potential / mV –40 thresholdthreshold potentialpotential –60 E –80 0 1 2 3 4 5 time / ms stimulus stimulus A B Fig. 8.1 (i) State which of the letters C, D and E on Fig. 8.1 correspond to each of these events. You may use each of the letters C, D or E once, more than once or not at all. The Na+ / K+ pump is operating .................... The voltage‑gated Na+ channels are open .................... The voltage‑gated K+ channels are open .................... [3] (ii) Explain why stimulus A did not result in an action potential being produced whereas stimulus B did. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (iii) Describe the importance of the refractory period in the transmission of action potentials. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (iv) Describe how action potentials are transmitted along a myelinated axon. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [4] [Total: 14]

Mark scheme: 8(a) any three from 1 detect / respond to, (change in) stimulus / stimuli ; 2 two examples from – light / heat / sound / touch / pressure / pain / chemicals / taste / smell / tension ;; 3 (act as) transducers / convert stimulus energy to electrical energy ; 4 produce, generator / receptor / action, potential ; 5 passes impulse, to / along, sensory neurone ; 3 8(b)(i) Na+/K+ pump: C or D and E ; Na+ channels open: C ; K+ channels open: D ; 3 8(b)(ii) for A (ora for B) 1 generator / receptor, potential ; 2 threshold not, reached / exceeded / crossed ; 2 Question Answer Marks 8(b)(iii) any two from 1 limits / controls, (maximum) frequency of action potentials ; 2 (action potentials / impulses) travel in one direction ; 2 8(b)(iv) any four from 1 local circuit / movement of ions from positive to negative region ; 2 (causes) opening of Na+ channels ; 3 at next, node (of Ranvier) / gap in myelin sheath ; 4 causing next / new, action potential / depolarisation ; 5 saltatory conduction ; 6 one-way transmission ; 7 AVP ; e.g. myelin (sheath) / Schwann cells, insulate axon / prevent (named) ion movement / impermeable to (named) ions / speed up transmission / lengthens local circuits 4

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

Mark scheme: 9(a) any eight from 1 no, nucleus / nuclear envelope ; 2 circular DNA ; 3 histone-like / HU / HN-S / INF, proteins associated with DNA ; 4 no membrane-bound organelles ; 5 named example of absent organelle ; 6 70S / 18 nm, ribosomes ; 7 cell wall made of, peptidoglycan / murein ; 8 binary fission ; 9 unicellular ; 10 spherical / rod / spiral / comma / corkscrew, shape or may form, pairs / tetrads / strings / filaments / clusters / palisades ; 11 size detail ; 12 Gram-negative have, extra / second / outer, membrane ; 13 AVP ; (other structural feature) 14 AVP ; (other non-structural feature) 8 9(b) any seven from 1 collect / take, plants / seeds, from the wild ; 2 (from) many countries / international effort ; 3 (especially from) areas at risk from climate change ; 4 grow / cultivate / plant, seeds / plants (in botanic gardens) ; 5 increase, plant / seed, number ; 6 tissue culture / cloning ; 7 (store in) seed bank ; 8 maintain / measure, genetic diversity or ref. to gene bank ; 9 cooperate with, governments / charities / agencies / universities ; 10 for research ; 11 to conserve habitats ; 12 to restore habitats ; 13 to reintroduce species (to wild) ; 14 educate (public) / raise awareness ; 15 raise funds ; 16 AVP ; e.g. prepare suitable habitats / growing conditions 7

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10 (a) Describe how the structure of the nephron and its associated blood vessels are adapted to the process of ultrafiltration. [8] (b) Describe the effects of insulin on its main target tissues and explain how this leads to changes in blood glucose concentration. 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Mark scheme: 10(a) any eight from 1 capillary / glomerulus, endothelium ; 2 has, many / large, gaps / pores / holes / fenestrations ; 3 afferent arteriole wider than efferent arteriole ; 4 high, blood / hydrostatic, pressure in glomerulus ; 5 fluid forced, out of glomerulus / into Bowman’s capsule ; 6 podocytes ; 7 have interdigitating, processes / extensions ; 8 that form, filtration slits / slit pores ; 9 basement membrane ; 10 (mesh of) collagen / glycoprotein (fibres) ; 11 acts as (main), selective barrier / filter ; 12 large / RMM > 68 000, proteins / molecules, do not pass through ; 13 (red / white blood) cells, do not pass through ; 14 water / solutes / ions, pass through ; 8 10(b) any seven from 1 two examples from: liver / muscle / adipose ; 2 increases glucose uptake ; 3 increases permeability of cells to glucose ; 4 more, transport / GLUT (4), proteins (added to membrane) ; 5 by vesicles fusing with cell surface membrane ; 6 stimulates, glycogen synthesis / glycogenesis ; 7 glucokinase / hexokinase / glycogen synthase ; 8 lipid / triglyceride / fatty acid, synthesis stimulated ; 9 increase in respiration ; 10 (so) more (facilitated) diffusion of glucose into (liver) cells ; 11 decreases, glycogenolysis / lipolysis / gluconeogenesis ; 12 decreases blood glucose concentration ; 7