Cambridge A Level Biology 9700 — 2017 May/June Paper 4 · Variant 1

1 (a) The mammalian kidney is an organ involved in homeostasis. Explain what is meant by the term homeostasis. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [1] (b) Fig. 1.1 shows a section through a kidney. A B Fig. 1.1 (i) With reference to Fig. 1.1, name structures A and B. A ........................................................................... B ........................................................................... [2] (ii) On Fig. 1.1, use label lines and letters to label where: U – ultrafiltration occurs L – the loop of Henle is found C – blood urea concentration is low. [3] (c) Describe the roles of the hypothalamus and the posterior pituitary in osmoregulation. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [5] [Total: 11]

Mark scheme: 1(a) maintain / keep / restore, constant / stable / set-point / within narrow limits, internal environment / in body ; 1 1(b)(i) A – pelvis ; note if labelled medulla as affects ecf in part (ii) B – ureter ; 2 1(b)(ii) A full labels instead of letters if region A (pelvis) was mislabelled as medulla in (i) can apply: ecf for L placed in pelvis ecf U placed in medulla only if word cortex also written by U / ultrafiltration U – pointing to the cortex ; L – pointing to the medulla ; C – pointing to the renal vein ; 3 Question Answer Marks 1(c) max 5 of: 1 hypothalamus detects (changes in) water potential (of the blood) ; 2 osmoreceptors shrink when, low / less, water in blood ; ora 3 ADH, produced / made, in hypothalamus ; 4 if low, water / Ψ, ADH secreted from posterior pituitary ; ora R ADH produced in posterior pituitary 5 ref. to neurosecretory cells or impulse / ADH transported, from hypothalamus to posterior pituitary ; 6 aquaporins ; 7 ADH increases permeability of, distal convoluted tubule / collecting duct ; ora 8 ADH causes, more water reabsorption / smaller volume of urine / more concentrated urine ; ora A both with and without ADH compared 5

2 Corals grow in shallow seawater. Corals consist of colonies of small animals called polyps. These polyps have photosynthetic protoctists called algae inside their cells, which is advantageous both to the coral polyps and to the algae. The algae that live within the cells of the polyps can also live independently as free-living algae. (a) The rate of photosynthesis of algae that live within the cells of coral polyps is higher than that of free-living algae. Suggest and explain how living inside the cells of coral polyps increases the rate of photosynthesis in these algae compared to free-living algae. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) The relative abundance of five different chloroplast pigments in the algae of corals was determined. The results are shown in Table 2.1. Table 2.1 chloroplast percentage of pigment total chlorophyll a 39 peridinin 39 chlorophyll c2 13 dinoxanthin 7 β-carotene 2 Outline the method you would use to separate and identify the pigments present in an extract of these algae. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] (c) Table 2.2 shows the light wavelengths at which each algal chloroplast pigment shows its two largest peaks of light absorption. Table 2.2 chloroplast pigment peak 1 wavelength peak 2 wavelength / nm / nm chlorophyll a 430 662 peridinin 456 485 chlorophyll c2 450 396 dinoxanthin 442 471 β-carotene 454 480 Corals kept in tanks are often illuminated by lamps radiating mostly violet and blue light with wavelengths in the range of 400–490 nm. With reference to Table 2.1 and Table 2.2, suggest why lamps radiating mostly violet and blue light are expected to increase coral growth. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3]

Mark scheme: 2(a) 1 2 (from) coral / polyp, respiration / metabolism or for algae, Calvin cycle / light independent reactions ; A correct use of mineral ions 2 2(b) max 4 of: 1 (paper / thin layer) chromatography / chromatogram ; 2 place spot of, extract / pigments, on pencil mark / at base of, paper / TLC plate ; 3 dry and repeat (to concentrate spot) ; 4 dip, paper / chromatogram, in solvent / so solvent travels up paper ; A named organic solvent (I water) R if spot submerged 5 measure distance travelled by solvent (front) and pigment (spot) ; 6 (calculate) Rf value distance travelled by pigment = distance travelled by solvent (front) ; 7 look up / compare results with, known Rf values (to identify pigments) ; 4 2(c) max 3 of: 1 pigments absorb, violet-blue / 400–490 nm / lamp colours, well / best / most / at 8 out of 10 peaks ; 2 rate of photosynthesis of algae increases with more light absorbed ; 3 coral growth (increases) with more (algal) photosynthesis ; R products respond to give growth 4 chlorophyll a and peridinin are, most abundant pigments / most important ; 5 AVP ; e.g. violet-blue / 400–490 nm, predominate at the depths where corals live 3

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3 Oil seed rape (canola), Brassica napus, has been genetically modified to be resistant to herbicides containing glufosinate ammonium. The genetically modified (GM) oil seed rape contains the bar gene, obtained from a soil bacterium. This gene codes for an enzyme that converts glufosinate ammonium into a non-toxic compound. (a) Outline the advantages to farmers of growing glufosinate-resistant oil seed rape. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (b) The bar gene was introduced into the oil seed rape using plasmids. The plasmids also contained a promoter taken from thale cress, Arabidopsis thaliana. (i) Outline the structure of a plasmid. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) Explain how the properties of plasmids make them suitable for use during genetic modification programmes. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (iii) Describe the role of a promoter in gene expression. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (c) The pollen of oil seed rape is transferred from one flower to another by insects. After pollination, fertilisation and seed formation can occur. One of the potential problems of growing glufosinate–resistant oil seed rape is that pollen from these plants could be transferred to the flowers of wild relatives, such as wild radish, Raphanus raphanistrum. This could result in genetic changes in these wild species. An experiment was carried out to investigate whether glufosinate–resistant hybrids between GM oil seed rape and wild radish plants are likely to compete successfully with non-hybrid or non-resistant plants in the natural environment. • Type 1 hybrids were produced by transferring pollen from wild radish (diploid number 18) to glufosinate–resistant oil seed rape (diploid number 38). • Type 2 hybrids were produced by transferring pollen from glufosinate–resistant oil seed rape to wild radish. • Each hybrid was then crossed with wild radish over several generations. • The resulting offspring were then grown in field trials, together with normal wild radish. • The height of the plants and number of seeds each produced were measured. Then the plants were tested for the bar gene. Table 3.1 shows the results. Table 3.1 type of plant number of mean height presence of seeds per / cm bar gene plant offspring from 265 22.3 absent type 1 hybrid and wild radish 99 28.3 present offspring from 3958 88.7 absent type 2 hybrid and wild radish 2047 95.0 present wild radishes 3515 76.5 absent (i) Predict the diploid number of chromosomes in a hybrid between oil seed rape and wild radish. ...................................................................................................................................... [1] (ii) Suggest how the researchers could have determined whether or not the bar gene was present in the plants. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [1] (iii) Many varieties of GM oil seed rape are male sterile, meaning that they do not produce pollen. With reference to Table 3.1, suggest the advantages to the environment of growing male sterile varieties of GM oil seed rape, rather than GM varieties that produce pollen. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] [Total: 14]

Mark scheme: 3(a) max 2 of: 1 can, kill / control, weeds ; R kill, insects / pests 2 reduce competition / increase yield (of rape) ; 3 AVP ; e.g. manual weeding / hoeing, difficult / expensive ref. to glufosinate converted to non-toxic compound 2 3(b)(i) circle of / circular, DNA ; I loop R single-stranded small / supplementary ; 2 3(b)(ii) max 3 of: 1 small so can be taken up by, cells / bacteria ; 2 replicate, independently / fast ; A have ori / origin of replication / high copy number 3 (DNA) has restriction site(s) / can be cut by restriction enzymes ; A have polylinker 4 have, marker genes / genes for resistance (for screening) ; 5 AVP ; e.g. circular so, increased stability / reduced host cell degradation 3 3(b)(iii) max 2 of: 1 RNA polymerase binds ; 2 so, transcription / mRNA synthesis, begins / occurs / allowed ; 3 AVP ; e.g. correct / template, strand is transcribed ref. to tissue-specific / inducible, expression 2 3(c)(i) 28 ; 1 Question Answer Marks 3(c)(ii) max 1 of: spray with herbicide and, those that die did not have the bar gene / those that survive did have the bar gene ; add gene for fluorescence with bar gene and test plants under UV / use PCR with primer complementary to bar gene / use (gene) probe (on Southern blot) of electrophoresis gel ; 1 3(c)(iii) max 3 of: advantage of male sterile GM variety 1 avoid transferring, bar / resistance, gene to wild, radish / relations ; ora 2 avoid superweeds ; ora 3 avoid type 2 hybrids ; ora disadvantage of type 2 hybrids (from GM variety that produces pollen) 4 taller (than wild radish) ; A very tall / 88 cm / 95 cm 5 produce, more / many, seeds (than wild radish) ; A 3958 / 443 more 6 may (out)compete, wild radish / crops ; 3

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4 (a) ATP is used or produced at different stages in the respiration of glucose in aerobic conditions. Complete the table to show whether ATP is used or produced at each stage of respiration. Write either YES or NO in each box. stage of respiration ATP used ATP produced glycolysis link reaction Krebs cycle oxidative phosphorylation [2] (b) An experiment was carried out to investigate the effect of epicatechin on mitochondrial respiration in mice. Epicatechin is a naturally occurring compound in cocoa beans and so is present in chocolate. Two groups of mice, group A and group B, were used in this experiment. • Group A was given water containing epicatechin, twice a day for 15 days. • Group B was given water without epicatechin, twice a day for 15 days. After 15 days, the structure of mitochondria from striated muscle cells in both groups of mice was examined. The surface area of the inner membrane of the mitochondria was divided by the surface area of the outer membrane to obtain a ratio for each mouse. Table 4.1 shows the mean ratios for the two groups of mice. Table 4.1 group mean ratio A 2.0 : 1 B 1.7 : 1 The mice in group A were able to exercise longer than the mice in group B. With reference to Table 4.1, explain why the mice in group A were able to exercise for longer than the mice in group B. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [5] [Total: 7]

Mark scheme: 4(a) stage of respiration ATP used ATP produced glycolysis yes yes link reaction no no Krebs cycle no yes oxidative phosphorylation no yes ;; 4 correct = 2 marks, 2 or 3 rows correct = 1 mark If ticks and crosses used need all 4 correct for maximum 1 mark 2 Question Answer Marks 4(b) max 5 of: group A (accept ora for group B throughout) accept ‘they’ = group A 1 higher ratio ; 2 larger / more, inner membrane / cristae (than B) ; 3 more, ETCs / cytochromes / ATP synth(et)ase / stalked particles ; I ATPase 4 oxidative phosphorylation ; 5 more ATP produced ; 6 muscles can contract for, longer / more time / without getting tired ; I exercise longer I muscles contract faster 7 AVP ; e.g. chemiosmosis or detail thereof: H+ move, down gradient / through ATP synth(et)ase I ATPase If B and A switched round penalise once only 5

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5 The red poppy, Papaver rhoeas, and several species of daisy of the family Compositae often co-exist as weeds of wheat fields. Fig. 5.1 shows changes in the percentage frequency of red poppies and daisies in an area of wheat fields over a six year period from 1998 to 2003. From 1985, the herbicide metsulfuron-methyl was used to control weeds in this area of wheat fields. This practice continued throughout the six year period. 1998 showed the first occurrence of a red poppy known as biotype X. This red poppy had a specific mutation not present in normal red poppies. daisies normal red poppy biotype X red poppy 100 80 60 percentage frequency 40 20 0 1998 1999 2000 2001 2002 2003 year Fig. 5.1 (a) Describe how the percentage frequencies of daisies and red poppies changed over the six year period. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] (b) Metsulfuron-methyl acts by inhibiting an enzyme called acetolactate synthetase, which is needed for the daisies and red poppies to synthesise three amino acids essential for growth. The specific mutation carried by the red poppies of biotype X occurred within the gene coding for this enzyme. The mutation changed amino acid 197 of acetolactate synthetase from proline to leucine. (i) Suggest the effect of this mutation on the structure and activity of the acetolactate synthetase enzyme of biotype X red poppies. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (ii) Suggest the effect of this mutation on the biotype X red poppies in the presence of metsulfuron-methyl. ...................................................................................................................................... [1] (iii) With reference to Fig. 5.1, predict and explain the effect of biotype X red poppies on the relative proportions of weeds and wheat in the area of wheat fields in 2003 compared to 1998. ........................................................................................................................................... . .......................................................................................................................................... . .......................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (c) Suggest how stopping the use of the herbicide metsulfuron-methyl and replacing it with a herbicide that inhibits a different target enzyme in weeds would affect the abundance of red poppies of biotype X. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] [Total: 14]

Mark scheme: 5(a) I increase or decrease ‘by x%’ when difference from start time to end calculated max 4 of: 1 decrease in daisies and normal poppies, overall / in 6 years / after 1999–2000 ; 2 (decrease in, daisies / normal poppies) from 50% to 15% ; 3 increase in poppy biotype X from 1% to 70% ; 4 increase in, total / combined, poppies from 52% to 85% ; 5 daisies and normal red poppies are always equal in % frequency ; A remain equal 6 steep / huge / dramatic, decrease in daisies and normal red poppy after 2001 or increase in X, is steeper after 2001 ; 4 5(b)(i) max 3 of: 1 change in primary structure ; 2 change in, tertiary / 3D / globular, structure ; 3 active site, binds substrate / forms ESC ; 4 metsulfuron-methyl does not, inhibit / bind to, enzyme ; 5 enzyme, functions / forms amino acids ; 3 Question Answer Marks 5(b)(iii) max 3 of: in 2003 (compared to 1998) 1 more, weeds / poppies, and less wheat / higher proportion of weeds in wheat ; I wheat yield 2 most weeds are now, poppy biotype X / resistant to herbicide ; 3 poppy biotype X, not killed by / resistant to herbicide ; 4 wheat have, more competition for / less access to, space / light / water / minerals ; I nutrients 3 5(c) max 3 of: 1 (biotype X) poppies, die / do not survive / do not breed ; 2 their, numbers / abundance. would decrease ; 3 selection pressure, removed / changed / new ; 4 (biotype X) mutant / resistance, allele no longer, advantageous / selected for / passed on ; 5 possibility of beneficial mutation in gene for different enzyme or could adapt / evolve resistance, to new herbicide ; R if new herbicide causes mutation 3

6 (a) Describe how tropomyosin and myosin are each involved in the sliding filament model of muscle contraction. (i) tropomyosin ....................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) myosin ............................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [4] (b) Striated muscle is made up of many specialised muscle cells known as muscle fibres or myocytes. There are two different types of muscle fibre in striated muscle: • fast twitch muscle fibres that contract quickly, but rapidly fatigue (get tired) • slow twitch muscle fibres that contract slowly and continue to contract for a long time. Table 6.1 shows some features of fast twitch and slow twitch muscle fibres. Table 6.1 feature fast twitch fibre slow twitch fibre respiration mainly anaerobic mainly aerobic glycogen concentration high low capillaries few many Use the information in Table 6.1 to suggest and explain one advantage of: (i) the high glycogen concentration in fast twitch fibres ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) many capillaries supplying slow twitch fibres. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 10]

Mark scheme: 6(a)(i) max 2 of: 1 tropomyosin / it, covers / uncovers, myosin binding sites on actin ; R inhibits R active site 2 when calcium ions bind to troponin, tropomyosin / it, moves / changes shape ; 3 allows myosin to, bind to actin / form cross-bridges ; ora 2 Question Answer Marks 6(a)(ii) max 4 of: 1 ATP hydrolysis / ATP → ADP + Pi ; 2 (causes myosin) head to, pivot / rotate / tilt / stand up ; 3 myosin / head, binds to actin / forms cross-bridges with actin ; R active site 4 ADP and Pi detach ; 5 (myosin) head, swings back / returns to previous position ; 6 actin is moved / power stroke occurs ; 7 (new) ATP binds ; 8 myosin / head, detaches from actin / cross-bridges break ; A mps in any order apart from 1, 4 and 7 which must be linked to correct action 4 6(b)(i) max 2 of: 1 to, supply / provide, (enough / plenty of) glucose ; 2 for glycolysis ; 3 as little ATP is produced by anaerobic respiration ; 4 as few capillaries are present (to supply glucose directly) ; 2 6(b)(ii) max 2 of: 1 to, supply / provide, (enough / plenty of) oxygen ; 2 aerobic respiration / oxidative phosphorylation ; 3 to remove, carbon dioxide / lactate ; A lactic acid 4 to, avoid fatigue or promote, stamina / endurance (for exercise / work) ; 2

7 (a) The stems of raspberry plants have spines. Fig. 7.1 shows part of a raspberry plant. spines Fig. 7.1 The colour of the spines is controlled by two genes, A/a and B/b. The two genes are on different pairs of chromosomes. • Allele A produces a pink anthocyanin pigment in the spines. • Allele B has no effect by itself, but increases the colour produced by allele A to give red spines. • Alleles a and b have no effect on colour. • In the absence of anthocyanin, the spines are green. State the colour of the spines of raspberry plants with the genotypes Aabb and aaBB. Aabb ......................................................................................................................................... aaBB ................................................................................................................................... [2] (b) Plants with the genotype AaBb were crossed with plants with the genotype aabb. The resulting seeds were sown and the seedlings grown until their stems developed spines. Use a genetic diagram to show the outcome of this cross, including the ratio of offspring phenotypes. [5] (c) Suggest why the ratio you have given in your genetic diagram would be different if the genes A/a and B/b were on the same homologous pair of chromosomes. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] [Total: 9]

Mark scheme: 7(a) Aabb – pink ; aaBB – green ; 2 7(b) 1 parents phenotypes red green ; 2 gametes AB Ab aB ab × ab ; 3 offspring genotypes AaBb Aabb aaBb aabb ; 4 offspring phenotypes (must be linked) red spines pink spines green spines green spines ; } 5 ratio 1 : 1 : 2 ; ecf mp 3 derived from incorrect 2 mp 4 matching incorrect 3 mp 5 matching incorrect 4 5 7(c) max 2 of: 1 genes would be, linked / inherited together ; 2 no independent assortment ; 3 ratio 1:1 / only two classes (of phenotypes) ; A red and green or pink and green 4 rare cross-over events / recombination (gives small numbers of third phenotype) ; 2

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8 There is considerable variation in the ecosystems that occur in the continent of North America. These include coniferous forest, prairie grassland, scrub and desert. Large areas of land that once contained natural ecosystems are now used for agriculture. (a) Explain how the variation in ecosystems in North America contributes to biodiversity. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] (b) The diversity of some beetle species that feed on animal dung (faeces) was investigated at two types of grassland site in North America. The first type of grassland site was grazed by cattle and the second type of site was not grazed. Dung beetles were collected, identified and counted from two areas of the same total size. Some of the results are shown in Table 8.1. Table 8.1 beetle species number of dung number of dung beetles on beetles on grassland grassland grazed not grazed by cattle Onthophagus pennsylvanicus 4267 6641 Canthon ebenus 2005 774 Canthon pilularius 353 108 Onthophagus hecate 218 85 total 6843 7608 (i) State the null hypothesis for a statistical test comparing the data from the two types of site. ........................................................................................................................................... ...................................................................................................................................... [1] (ii) State how many genera and how many species of beetle are shown in Table 8.1. genera ................................................. species ................................................. [1] (iii) Simpson’s Index of Diversity for the beetles on the grassland grazed by cattle was calculated as 0.521 using the formula: n D = 1 – Σ N2 n = number of individuals of each species present in the sample N = the total number of all individuals of all species Calculate Simpson’s Index of Diversity for the beetles on the grassland that was not grazed. Complete Table 8.2 to show your working. Show all working to three decimal places. Write your final answer on the dotted line. Table 8.2 species number on n n 2 grassland N  N not grazed Onthophagus pennsylvanicus 6641 Canthon ebenus 774 Canthon pilularius 108 Onthophagus hecate 85 total 7608 Simpson’s Index of Diversity = ............................................................ [3] (iv) Describe what the results in Table 8.1 and both figures for Simpson’s Index of Diversity show about the effect of grazing on the diversity of dung beetles. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2]

Mark scheme: 8(a) max 4 of: 1 different habitats ; 2 different niches ; 3 many (different) species / large variety of species ; 4 ref. to (much) genetic diversity within a species ; 5 different selection pressures ; 6 ref. to adaptation ; 7 different, climate / rainfall / temperature / soil / topography / conditions ; 4 8(b)(i) both sites are the same / no (significant) difference between two sites ; 1 8(b)(ii) genera 2 and species 4 ; 1 Question Answer Marks 8(b)(iii) all figures to 3 d.p. to score but only penalise extra d.p. or rounding error associated with extra d.p. once species number on grassland not grazed n / N (n / N)2 Onthophagus pennsylvanicus 6641 0.873 0.762 Canthon ebenus 774 0.102 0.010 Canthon pilularius 108 0.014 0.000 Onthophagus hecate 85 0.011 0.000 total 7608 0.772 n / N figures correct / numbers of each species divided by total ; (n / N)2 calculated and added up ; ecf from incorrect column 1 including figures with fewer / more than 3 d.p. 0.228 ; ecf total figure subtracted from 1 3 8(b)(iv) greater species evenness on grazed grassland ; ora A mostly, one species / O. pennsylvanicus, on not grazed grazing increases (dung beetle species) (bio)diversity ; ora if opposite conclusion reached check answer for (iii) and apply ecf for mp2 if D > 0.521 2

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9 (a) Explain how dip sticks function to test for glucose in a sample of urine. [8] (b) Outline how a high blood glucose concentration returns to normal in a healthy person. [7] [Total: 15]

Mark scheme: 9(a) max 8 of: 1 stick has, pad containing / immobilised, enzymes ; 2 glucose oxidase ; 3 peroxidase ; 4 stick dipped in urine ; A person, urinates / AW, on stick 5 glucose reacts to give hydrogen peroxide ; 6 (hydrogen peroxide reacts with) colourless substance / chromogen ; R dye / pigment 7 to give, colour change / coloured substance ; A change to any named colour 8 compare with colour chart ; 9 more glucose gives darker colour ; 10 specific / only detects glucose ; 11 AVP ; e.g. does not give current blood glucose concentration not numerical 8 Question Answer Marks 9(b) max 7 of: high blood glucose concentration 1 detected by β cells ; I alpha cells I receptors 2 in, islets of Langerhans / pancreas ; 3 (more) insulin secreted ; I produced 4 into blood ; 5 increases glucose absorption in liver (by phosphorylating glucose) ; 6 increases permeability to glucose in, muscle / fat, cells or adds GLUT 4 proteins to cell surface membranes of, muscle / fat, cells ; 7 increases (rate of) respiration of glucose ; 8 conversion of glucose to glycogen / glycogenesis ; 9 inhibits secretion of glucagon / decreases gluconeogenesis ; 10 negative feedback ; 7

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10 (a) Describe the structure of a motor neurone. [6] (b) Describe and explain the transmission of an action potential in a myelinated neurone. 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Mark scheme: 10(a) max 6 of: 1 dendrites (lead to cell body) ; R at both ends 2 nucleus in, cell body / soma ; R if cell body not at one end 3 many mitochondria (in cell body) ; 4 much RER / Nissl’s granules (in cell body) ; 5 long / one, axon ; A an axon 6 synaptic, knobs / termini / boutons, at end furthest from cell body ; 7 Schwann cells / myelin ; 8 nodes of Ranvier ; accept points on labelled diagram 6 Question Answer Marks 10(b) max 9 of: 1 Na+ / sodium ion, channels open ; I ligand or voltage gated 2 Na+ enters, cell / axon ; A Na ions / sodium ions 3 inside / p.d., becomes, less negative / positive / +40 mV or causes depolarisation (in correct context) ; 4 Na+ / sodium ion, channels close ; ecf from mp1 I ligand or voltage-gated 5 K+ / potassium ion, channels open ; ecf from mp1 I ligand or voltage-gated 6 K+ moves out (of cell) ; A K ions / potassium ions 7 inside / p.d., becomes negative / A negative figure or causes repolarisation (in correct context) ; 8 local circuits ; 9 myelin (sheath) / Schwann cells, insulate / prevent ion movement ; 10 action potential / depolarisation, only at, nodes (of Ranvier) ; 11 saltatory conduction / action potential jumps from node to node ; A impulse for AP 12 one-way / unidirectional, transmission ; 13 AVP ; e.g. hyperpolarisation / refractory period 9