Cambridge A Level Biology 9700 — 2017 Oct/Nov Paper 4 · Variant 3

1 The St. Lawrence river in Canada has been identified as an area with very high biodiversity. (a) Explain how the term biodiversity can be considered at different levels. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) The St. Lawrence river is rich in species of aquatic mammals, especially whales. In spring, thousands of whales swim from the Atlantic ocean up the St. Lawrence river. Thirteen different species of whale have been recorded. One of these is the blue whale, Balaenoptera musculus. Fig. 1.1 is a diagram of a blue whale. dorsal fin Fig. 1.1 Photographs of blue whales are analysed by computer so that individual whales can be identified and population abundance can be estimated. Each blue whale is different in colour and in the shape of dorsal fin. Different whales also have different patterns of scarring. (i) State the two sources of phenotypic variation in the blue whale population. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[1] (ii) Suggest one reason why it is difficult to measure the actual population size of the blue whale. ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[1] (c) The St. Lawrence river runs through very busy industrial areas. It has many ports for cargo ships, a good fishing trade and many whale-watching boat trips for tourists. Decades of whale hunting has caused a large decrease in whale population sizes. Since whale hunting was banned, whale populations have not recovered. Seven of the thirteen whale species in the St. Lawrence river have been rated as endangered species. Suggest two reasons why the populations of whales have not recovered since the ban on whale hunting. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (d) Fat samples from under the skin of individual whales of several different species were taken. These were analysed and the concentrations of the toxins DDT and PCBs were measured. Suggest why whales were found to have accumulated very high concentrations of DDT and PCBs in their fatty tissues. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (e) Algal blooms sometimes occur in the area of the Atlantic ocean near the St. Lawrence river. These result from rapid population growth of unicellular algae such as Alexandrium tamarense. (i) A. tamarense produces saxitoxin, a neurotoxin that causes muscle paralysis by acting on voltage-gated sodium ion channels in neurones. Saxitoxin can kill whales. Suggest how saxitoxin results in the death of a whale. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Algae, such as A. tamarense, used to belong to the kingdom Plantae but are now classified in the kingdom Protoctista. State one reason why A. tamarense is classified in the kingdom Protoctista and not in the kingdom Plantae. ........................................................................................................................................... .......................................................................................................................................[1] [Total: 11]

Mark scheme: 1(a) two from: 1 variation in / diversity of, ecosystems / habitats ; 2 number of / (how) many / variety of / diversity of, species ; 3 the (relative) abundance of each species ; 4 genetic diversity / range of alleles, within a species ; 2 1(b)(i) genes and environment ; 1 1(b)(ii) one from: 1 whales, mobile / swim / migrate ; 2 (they inhabit) large, area / distances ; 3 live, underwater / at great depths ; 1 1(c) two from: 1 (water) pollution from, industry / boats ; 2 accidents involving / damaged by, boats / fishing gear ; 3 lack of / competition for, food / krill / prey ; 4 noise / vibration, disturbs whale, communication / behaviour / mating ; 5 illegal, whaling / hunting ; 6 reproduction rate is slow / one offspring at a time / long gestation ; 2 Question Answer Marks 1(d) two from: toxins (PCBs / DDT) are 1 present in, water / river / sea ; 2 ingested / absorbed by, producers / phytoplankton / algae ; 3 bioaccumulation or toxins, pass up / accumulate up, food chain ; 4 persistent / long-lasting / not broken down (in environment / whale) ; 5 fat / lipid, soluble ; 2 1(e)(i) two from: 1 sodium ions do not enter (neurones / nerve cells / axons) ; 2 (neurones) cannot depolarise or cannot, generate / transmit, impulses / action potentials ; 3 reason for death ; 2 1(e)(ii) one from: unicellular / not multicellular ; motile / have flagella ; 1

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2 The light dependent stage of photosynthesis in a suspension of isolated chloroplasts can be investigated using the Hill reaction. Dichlorophenolindophenol (DCPIP) can be used to follow the process. DCPIP is a blue dye which turns colourless when it is reduced by accepting hydrogen and electrons. oxidised DCPIP reduced DCPIP blue colourless (a) (i) The natural hydrogen acceptor in chloroplasts that is replaced by DCPIP in the Hill reaction is NADP. Reduced NADP is then used in the light independent stage of photosynthesis. State the role of reduced NADP in the light independent stage. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Predict and explain the effect on the concentration of RuBP in the chloroplasts if DCPIP becomes reduced instead of NADP. ........................................................................................................................................... .......................................................................................................................................[1] DCPIP can be used to investigate the effect of light wavelength on the rate of the light dependent stage of photosynthesis. One method is described here: • Dip a small glass capillary tube into a beaker containing a suspension of isolated chloroplasts. Some of the chloroplast suspension will be drawn into tube A, forming a colour standard. • Mix the rest of the chloroplast suspension with blue DCPIP solution. Dip a second capillary tube into this to draw up a sample, to form tube B. • Store the chloroplast-DCPIP mixture for future use by wrapping the beaker in foil and covering it with a removable foil lid, as shown in Fig. 2.1. foil lid beaker foil Fig. 2.1 • Place tube A and tube B on a white tile under a bench lamp placed 15 cm away and immediately cover them with a purple filter, as shown in Fig. 2.2. light source white tile far side A of filter B Fig. 2.2 • Measure the time for tube B to change colour to match that of tube A. This is the time taken for DCPIP to decolourise. • Repeat four more times using the same tube A and a fresh tube B, placing the tubes under a different coloured filter each time. (b) Explain the reasons for: the colour standard tube A ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... covering the beaker containing the chloroplast-DCPIP mixture with foil. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] (c) Table 2.1 shows the results of the experiment. Table 2.1 colour of filter wavelength of time taken to rate of light light / nm decolourise dependent stage of DCPIP / s photosynthesis / s–1 purple 425 37 27.0 blue 450 84 11.9 green 525 480 2.1 orange 625 45 ......................... red 675 50 20.0 (i) Complete Table 2.1 by calculating the rate of the light dependent stage of photosynthesis for the orange wavelength, using the formula: 1000 rate = where t = time in seconds. [1] t (ii) With reference to Table 2.1, describe and explain the effect of light wavelength on the rate of the light dependent stage of photosynthesis. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[5] [Total: 13]

Mark scheme: 2(a)(i) 1 reduces, GP / glycerate (3) phosphate ; 2 to, TP / triose phosphate ; 2 2(a)(ii) RuBP, decreases / less either because it, reacts / is used up / is converted or because it is not, replaced / regenerated ; AW 1 2(b) any four in total: tube A 1 for comparison / to compare ; 2 to see, end-point / when all DCPIP has been reduced, in B ; foil (max 3) 3 to, stop / limit, light entering (the beaker / mixture) or to stop light reaching chlorophyll ; 4 to, stop / limit, light dependent reaction occurring ; 5 to, stop / limit, DCPIP, decolourising / being reduced ; 6 so all tests start with same colour (of DCPIP–chloroplast mixture) ; 4 2(c)(i) 22.2 ; 1 Question Answer Marks 2(c)(ii) five from: description of rate of, photosynthesis / light (dependent) reaction (max 2) 1 (it is) highest / fastest / most, in purple / at 425 nm ; 2 (it is) lowest / slowest / least, in green / at 525 nm ; explanation (max 3) 3 chlorophyll absorbs purple and orange (best) but does not absorb green ; 4 accessory pigments ; 5 light, excites electrons / triggers electron transport ; 6 non-cyclic photophosphorylation ; 7 action spectrum ; 5

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3 Fig. 3.1 shows a red squirrel, Sciurus vulgaris. This species is native to the British Isles, meaning it has lived there for at least 10 000 years. In the 1800s a related but slightly larger species, the grey squirrel, Sciurus carolinensis, was introduced from North America. Fig. 3.1 A history of the interaction between red squirrels and grey squirrels includes these facts: • Grey squirrels, which occupy a similar niche to red squirrels, expanded their range rapidly after introduction. • Grey squirrels carry a virus that does not harm them, but which kills red squirrels. • Red squirrels became extinct in many parts of the British Isles. • Recently some areas have seen a reversal of this trend. There has been a decrease in numbers of grey squirrels. Red squirrels have re-colonised their former areas of habitat. • This has happened in areas where protection of a native predator species, the pine marten, Martes martes, has resulted in an increase in the number of these predators. • Analysis of pine marten faeces shows that they catch and eat many more grey squirrels than red squirrels. (a) Use the information given to identify the causes of the extinction of red squirrels in many parts of the British Isles. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (b) Describe how the level of molecular similarity between the two species, S. vulgaris and S. carolinensis, can be investigated. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] (c) Explain how the differing effect of pine marten predation on red and grey squirrel populations is due to natural selection and the separate evolution of each squirrel species. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] [Total: 8]

Mark scheme: 3(a) (interspecific) competition (with greys) ; virus / disease / infection, passed, from greys / to reds ; 2 3(b) three from: 1 DNA / base / nucleotide, sequences ; 2 mitochondrial / mt, DNA ; 3 protein / polypeptide / amino acid, sequences ; 4 genetic fingerprinting / DNA profiling ; 5 compare (sequences from reds and greys) ; 3 3(c) three from: 1 pine marten / predation, is / was, selection pressure ; 2 red squirrel better adapted (to pine marten predation) ; ora 3 detail / suggestion ; e.g. red squirrel, faster / better camouflaged ora 4 (two squirrel species arose by) allopatric speciation / AW ; 5 different, selection pressures / predators (in two places / for two species) ; 6 red squirrels and pine martens co-existed for, 10 000 years / long time ; 3

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4 There are many different strains of the soil bacterium Bacillus thuringiensis. Each produces slightly different types of Cry-proteins, which are toxic to insects. Some types of cotton, known as Bt cotton, have been genetically modified to produce one of these proteins, Cry1Ac. This protein acts specifically to kill the larvae of butterflies and moths, including the cotton bollworm, Helicoverpa zea, a serious pest of cotton crops. The genetically modified cotton contains a ‘genetic package’ that includes: • the gene coding for Cry1Ac, the Bt protein • a promoter • a herbicide resistance gene that is used as a marker. (a) (i) Suggest the advantages of using, in Bt cotton, the gene coding for Cry1Ac, rather than one of the genes coding for other types of the Cry-protein. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Explain why a promoter is included in the genetic package. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (iii) Suggest how the herbicide resistance gene can be used as a genetic marker. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[3] (b) Table 4.1 shows information about the cultivation of Bt cotton and non-GM cotton by farmers in India in 2002–2003. Table 4.1 Bt cotton non-GM cotton mean yield of cotton / kg ha–1 264 196 seed cost / rupees ha–1 629 196 insecticide cost / rupees ha–1 503 851 net income / rupees ha–1 2118 1253 (i) With reference to Table 4.1, compare the costs involved in growing Bt cotton with the costs involved in growing non-GM cotton. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[2] (ii) Table 4.1 shows that farmers who grow Bt cotton have a higher net income than those who grow non-GM cotton. Use the information in the table to suggest one reason why some farmers in India choose to grow non-GM cotton, rather than Bt cotton. ........................................................................................................................................... .......................................................................................................................................[1] (c) In one region of India, Andhra Pradesh, a severe drought in 2002–2003 meant that Bt cotton grew less well than other varieties of cotton that were better adapted for the conditions. Suggest how a variety of Bt cotton that is better adapted to dry conditions could be produced from the existing varieties of Bt cotton, without using gene technology. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[3] [Total: 13]

Mark scheme: 4(a)(i) two from: 1 (only) kills / targets / acts on, specific / some, insects / pests ; 2 does not kill, beneficial / useful, insects ; 3 (such as) pollinators / bees / predators of pests ; 4 to conserve / protect, biodiversity / food web ; ora 5 idea that other Cry proteins might not kill, right pests / bollworm ; 2 4(a)(ii) two from: 1 (so, new / foreign / inserted) gene(s) are, expressed / switched on / transcribed (and translated) ; 2 RNA polymerase binds (at promoter) ; 3 ref. to correct / template, strand ; 4 to control quantity of Cry(1Ac / protein) made ; 5 to control, where / which part(s) of plant, make Cry(1Ac / protein) ; 2 4(a)(iii) three from: 1 insert, herbicide resistance gene / it, next to, Bt / Cry(1Ac), gene ; 2 spray / add, herbicide on (transformed) plants / protoplasts / cells ; 3 survivors have, Bt / Cry(1Ac), gene ; 4 to identify, successful / GM / insect-resistant, plants ; 3 Question Answer Marks 4(b)(i) two from: 1 Bt seed costs more but insecticide costs less ; 2 total cost is more for Bt than for non-GM ; 3 manipulated figure(s) comparing both Bt and non-GM ; 2 4(b)(ii) one from: non-GM seeds are cheap(er) / (more) affordable ; non-GM / it, is cheap(er), overall / to grow ; 1 4(c) three from: 1 selective breeding / artificial selection ; 2 cross Bt cotton with a (Bt) variety that grows well in, dry / drought ; 3 select / choose, offspring with Bt (trait / gene) and grow well in, dry / drought ; 4 repeat (crossing / selection) for several generations ; 3

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5 (a) Explain the importance of the myelin sheath in the transmission of action potentials. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] (b) Acetylcholine is a neurotransmitter. Describe the events that occur at a synapse that lead to the release of acetylcholine. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[5] [Total: 9]

Mark scheme: 5(a) four from: 1 insulator / ions cannot pass through it ; 2 depolarisation / action potentials, occur at nodes of Ranvier (only) ; 3 long(er) local, circuits / currents ; 4 action potential jumps from node to node / saltatory conduction ; 5 transmission / conduction, fast(er) ; 4 5(b) five from: 1 action potential / depolarisation, at presynaptic membrane ; 2 Ca2+ channels open / increased permeablity to Ca2+ ; 3 Ca2+ enter, (presynaptic) neurone / knob / axoplasm / AW ; 4 by (facilitated) diffusion / down concentration gradient ; 5 vesicles, of acetylcholine / neurotransmitter, fuse with membrane ; 6 ACh / neurotransmitter, enters / exocytosed into, synaptic cleft ; 5

6 Chemiosmosis is the term used to describe the synthesis of ATP using a proton gradient across a membrane in a mitochondrion or chloroplast. It was first demonstrated by Peter Mitchell in 1961. (a) • In some of his experiments, Peter Mitchell used mitochondria that had been isolated from cells. • The mitochondria were kept in liquid, in glass dishes, to which ADP, Pi and other substances were added. • The temperature, pH and water potential were kept constant. • After a period of time he checked for the presence of ATP. The contents of some of the dishes are shown in the table below. Complete the table using a tick (3) if ATP was produced and a cross (7) if no ATP was produced. dish contents ATP produced mitochondria + ADP + Pi + acetyl CoA + oxygen mitochondria + ADP + Pi + acetyl CoA mitochondria + ADP + Pi + low concentration of protons (H+) mitochondria + ADP + Pi + high concentration of protons (H+) [2] (b) Explain the consequences to a mitochondrion if the water potential of the liquid in the dishes is higher than the water potential of the mitochondrial matrix. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] (c) State the specific role of oxygen in the mitochondrion. ................................................................................................................................................... ...............................................................................................................................................[1] (d) Name the enzyme used to produce ATP in chemiosmosis. ...............................................................................................................................................[1] (e) Describe the role of the inner mitochondrial membrane (crista) in chemiosmosis. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] [Total: 10]

Mark scheme: 6(a) contents of dishes ATP produced mitochondria + ADP + Pi + acetyl CoA + oxygen 9 mitochondria + ADP + Pi + acetyl CoA 8 mitochondria + ADP + Pi + low concentration of protons (H+) 8 mitochondria + ADP + Pi + high concentration of protons (H+) 9 ; ; 2 or 3 correct = 1 mark 4 correct = 2 marks 2 6(b) two from: water enters (mitochondrion / matrix) ; by osmosis / down the water potential gradient ; membranes ruptured / mitochondrion bursts ; 2 6(c) final electron (and proton) acceptor (in ETC) ; 1 6(d) ATP synth(et)ase ; 1 Question Answer Marks 6(e) four from: 1 (site of) electron transport chain ; 2 moves / pumps, protons / H+, to inter-membrane space ; 3 electrochemical / proton / H+, gradient ; 4 protons / H+, diffuse to matrix ; 5 through, stalked particles / ATP synth(et)ase ; 6 ADP + Pi Æ ATP ; 7 oxidative phosphorylation ; 4

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7 (a) Galactosaemia is a rare genetic disease in which the build-up of the monosaccharide galactose can result in an enlarged liver, kidney failure and brain damage. Galactose is produced in the body from the digestion of the sugar lactose, found in milk. Galactosaemia is caused by a recessive mutation of the GALT gene. The normal dominant allele codes for an enzyme which converts galactose to glucose. (i) Suggest how a person with galactosaemia can minimise damage to the liver, kidney and brain. ........................................................................................................................................... .......................................................................................................................................[1] (ii) Explain how a mutation in the GALT gene could result in a change in the enzyme responsible for the metabolism of galactose. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... .......................................................................................................................................[4] (b) If the phenotypes of parents are known, the probabilities of having a child with galactosaemia, an unaffected child (healthy, not a carrier) or a child who is a carrier can be calculated. Complete Table 7.1 to show the results of these calculations. Table 7.1 parent 1 parent 2 percentage percentage percentage probability probability probability of of having a of having an having a child child with unaffected child who is a carrier galactosaemia unaffected carrier 0 50 50 carrier carrier ......................... ......................... ......................... has unaffected galactosaemia ......................... ......................... ......................... has carrier 50 0 50 galactosaemia [2] (c) Testing for galactosaemia is usually carried out on newborn babies, although it is possible to carry out the test on a foetus. Explain how the presence of galactosaemia in a foetus may be determined. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[2] [Total: 9]

Mark scheme: 7(a)(i) consume, less / no, milk / lactose / (named) dairy products ; 1 7(a)(ii) four from: 1 changes / different, base / nucleotide (sequence), in, DNA / gene ; 2 changes / different, mRNA, codon / triplet ; 3 changes / different, primary structure of, polypeptide / protein / enzyme ; 4 changes / different, tertiary structure (of, polypeptide / protein / enzyme) ; 5 changes / different, allosteric / active, site ; 6 enzyme, non-functional / does not convert galactose (to glucose) ; 4 7(b) parent 1 parent 2 % prob. affected child % prob. unaffected child % prob . carrier child unaffected carrier 0 50 50 carrier carrier 25 25 50 ; unaffected affected 0 0 100 ; carrier affected 50 0 50 2 7(c) two from: genetic screening ; obtain fetal, cells / DNA ; by, amniocentesis / chorionic villus sampling ; electrophoresis + probe ; 2

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8 (a) Homeostasis in mammals involves negative feedback mechanisms. Outline how a negative feedback mechanism works. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] (b) Fig. 8.1 is a diagram summarising the role of the nervous and endocrine systems in the control of blood glucose concentration. hypothalamus neurones neurones adrenal β cells α cells glands pancreas insulin glucagon adrenaline decrease in increase in blood glucose concentration blood glucose concentration Fig. 8.1 With reference to Fig. 8.1, describe the role of the nervous system in the control of blood glucose concentration. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...............................................................................................................................................[4] (c) If the core temperature of the human body falls, the hypothalamus sends impulses to activate several physiological responses, some of which are listed below. For each one, state how it would help to bring the core temperature back to normal. vasoconstriction ........................................................................................................................ ................................................................................................................................................... ................................................................................................................................................... shivering ................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... increasing secretion of adrenaline ............................................................................................ ................................................................................................................................................... ...............................................................................................................................................[4] [Total: 12]

Mark scheme: 8(a) four from: 1 change in factor away from, the norm / set-point ; 2 detected / sensed by, receptor ; 3 hormone released or (nerve) impulse sent ; 4 (hormone / impulse) reaches, target organ / effector ; 5 (effector) performs corrective action ; 6 (factor) returns to, norm / set-point ; 4 8(b) four from: 1 hypothalamus detects change in blood glucose concentration ; 2 autonomic / motor / nerve, impulses ; 3 (so) β cells secrete insulin when blood glucose increases ; 4 (so) α cells secrete glucagon when blood glucose decreases ; 5 (so) adrenal gland secretes adrenaline either when blood glucose decreases or due to fear / shock / excitement / stress ; 6 nervous control supplements, endocrine control / control by pancreas ; 4 Question Answer Marks 8(c) four from: vasoconstriction 1 arterioles in skin get narrow(er) ; 2 less blood flow through (skin / surface) capillaries ; 3 (so) less heat lost (to surroundings) ; shivering 4 muscle contraction ; 5 releases / provides / gives, heat / thermal energy ; increasing secretion of adrenaline 6 increases, rate of respiration / metabolic rate ; 7 more heat, released / provided / given (by respiration) ; 4

9 (a) Explain how rice is adapted to grow with its roots submerged in water. [6] (b) Explain how the leaves of maize or sorghum are able to maximise carbon dioxide fixation at high temperatures. [9] [Total: 15]

Mark scheme: 9(a) six from: 1 aerenchyma ; 2 in stem and roots ; 3 help oxygen to, move / diffuse, to roots ; 4 shallow roots ; 5 air (film) trapped on underwater leaves ; 6 fast internode growth ; 7 (modified) growth regulated by, gibberellin / ethene ; 8 anaerobic respiration, underwater / when submerged ; 9 tolerant to high ethanol concentration / high tolerance to ethanol ; 10 ethanol dehydrogenase (switched on in anaerobic conditions) ; 11 AVP ; e.g. growth stops / carbohydrates conserved / quiescence, in short-term (flash) floods 6 Question Answer Marks 9(b) nine from: 1 RuBP / rubisco, in bundle sheath (cells) ; 2 away from, oxygen / air ; 3 to avoid photorespiration ; 4 carbon dioxide combines with PEP ; 5 (catalysed by) PEP carboxylase ; 6 in mesophyll (cells) ; 7 forms oxaloacetate ; 8 converted to malate ; 9 malate passes to bundle sheath (cells) ; 10 (malate) releases (high concentration of) carbon dioxide ; 11 RuBP, carboxylated / reacts with carbon dioxide ; 12 PEP carboxylase / enzyme(s), has high optimum temperature / tolerate high temperatures ; 9

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10 (a) Describe the types of gene mutation that can occur and explain how some mutations can lead to a shortened polypeptide. [6] (b) Outline the effects of the mutant sickle cell allele on the phenotype of a person with sickle cell anaemia. 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Mark scheme: 10(a) six from: 1 base / nucleotide, substitution ; 2 missense / silent, mutation ; 3 base / nucleotide, insertion / addition ; 4 base / nucleotide, deletion ; 5 may cause frameshift ; 6 alters triplets of following, base / nucleotide, sequence ; 7 (premature) stop codon gives shortened polypeptide ; 8 does not code for amino acid ; 9 nonsense mutation ; 6 Question Answer Marks 10(b) nine from: 1 homozygous for, mutant allele / HbS ; 2 altered β polypeptide in haemoglobin ; 3 haemoglobin / β-globin, less soluble ; 4 in low(er) oxygen (concentration) ; 5 (Hb) forms long fibres ; 6 red blood cells, sickle / form crescent shape ; 7 (RBCs) carry less oxygen ; 8 (RBCs) get stuck in capillaries ; 9 blocks blood flow ; 10 causes pain ; 11 sickle cell crisis ; 12 RBCs break down faster / lack of RBCs ; 13 protection against, malaria / Plasmodium infection ; 9