Cambridge A Level Biology 9700 — 2017 Feb/March Paper 4 · Variant 2

1 (a) Fig. 1.1 shows diagrams of nephrons from the kidneys of three different mammals, P, Q and R. These are all drawn to the same scale. P Q R urine with high urine with low urine with very water potential water potential low water potential Fig. 1.1 (i) State the relationship between the length of the loop of Henle and the water potential of the urine of the three mammals. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [1] (ii) The nephron of a camel is similar to that of mammal R. Suggest why it is important that the camel produces urine with a very low water potential. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (b) The epithelial cells of the proximal convoluted tubule are adapted to carry out selective reabsorption. Table 1.1 lists three features of epithelial cells of the proximal convoluted tubule. For each feature, explain how it helps the process of selective reabsorption. Table 1.1 feature explanation ......................................................................................................... ......................................................................................................... microvilli ......................................................................................................... ......................................................................................................... ......................................................................................................... ......................................................................................................... ......................................................................................................... many mitochondria ......................................................................................................... ......................................................................................................... ......................................................................................................... ......................................................................................................... ......................................................................................................... tight junctions between cells ......................................................................................................... ......................................................................................................... ......................................................................................................... [6] [Total: 9]

Mark scheme: 1(a)(i) the longer the loop of Henle the lower the water potential of the urine ; 1 1(a)(ii) 1 dry environment / AW ; 2 need to conserve water / AW ; 2 1(b) microvilli: 1 large(r) surface area ; 2 for absorption of, Na+ / glucose / amino acids ; many mitochondria: 3 provide, energy / ATP ; R produce energy 4 for, Na+ / K+, pumps ; OR for active transport of, Na+ / K+ ; tight junctions between cells: 5 hold adjacent cells together ; 6 fluid cannot pass between cells / substances must pass through cells ; 6

2 (a) During a sporting event, an athlete carries out respiration in aerobic conditions. (i) Complete Table 2.1 to state the precise locations within a muscle cell of glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation. Table 2.1 process precise location glycolysis link reaction Krebs cycle oxidative phosphorylation [2] (ii) In a muscle cell, molecules of glucose are phosphorylated at the start of glycolysis. Suggest why the phosphorylated glucose molecules cannot diffuse out of the cell. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (b) Sometimes an athlete will need to carry out respiration in anaerobic conditions to produce ATP. Explain why the respiration of glucose in anaerobic conditions produces less ATP than in aerobic conditions. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [5] (c) For a short time after exercise, a person continues to breathe more heavily than at rest to take in more oxygen than normal. Explain the use of this extra oxygen. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] [Total: 12]

Mark scheme: 2(a)(i) 4 correct = 2 marks 2 / 3 correct = 1 mark glycolysis: 1 cytoplasm / cytosol / sarcoplasm ; link reaction: 2 mitochondrial matrix ; Krebs cycle: 3 mitochondrial matrix ; oxidative phosphorylation: 4 inner (mitochondrial) membrane / cristae ; 2 2(a)(ii) two from: 1 too big to pass through (membrane / glucose’s protein channel) ; 2 polar / AW ; 3 no specific, transport / carrier / channel, protein (for phosphorylated glucose) ; 2 max Question Answer Marks 2(b) five from: in anaerobic conditions: 1 only, glycolysis / conversion of glucose into pyruvate, occurs ; 2 (only) produces 2 molecules of ATP (net) ; 3 (only) substrate-linked phosphorylation (occurs) ; 4 pyruvate converted to lactate ; 5 lactate is energy-rich / AW ; 6 oxygen not available as final electron acceptor ; 7 electron transport chain / chemiosmosis / oxidative phosphorylation, does not occur ; 8 most ATP is produced (in aerobic conditions) in, electron transport chain / chemiosmosis / oxidative phosphorylation ; 5 max 2(c) three from: 1 reference oxygen debt ; 2 converts lactate to, pyruvate / glucose ; 3 in liver (cells) ; 4 re-oxygenate, haemoglobin / myoglobin ; 5 meet demands of continued increased metabolic rate / AW ; 3 max

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3 One example of chemical control and co-ordination in plants is in the breakdown of food reserves during the germination of cereal grains, such as wheat. (a) Fig. 3.1 is a diagram of a grain of wheat. The numbered arrows indicate the sequence of events that occurs after the uptake of water by the grain of wheat. 2 A B 1 3 C Fig. 3.1 (i) Name structures A, B and C, shown in Fig. 3.1. A ........................................................................................................................................ B ........................................................................................................................................ C ........................................................................................................................................ [3] (ii) With reference to the three numbered arrows in Fig. 3.1, describe the sequence of events that follows the uptake of water by the grain of wheat. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [6] (b) Growth in plants occurs in areas such as shoot and root tips. Growth occurs by cell division and cell elongation. Fig. 3.2 shows the effect of the plant growth regulator auxin on mean cell length in the shoot tips of wheat seedlings. 240 200 160 mean cell length 120 / μm 80 40 0 0 10 20 30 40 50 time / minutes auxin applied Fig. 3.2 (i) With reference to Fig. 3.2, calculate the overall rate of increase in mean cell length over the time period shown. overall rate of increase ............................................................... [2] (ii) The pH of the cell walls decreased after the application of auxin. Outline the events that occurred to cause this decrease in pH. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (iii) Explain the increase in the length of the cells following the decrease in pH of the cell walls. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] [Total: 16]

Mark scheme: 3(a)(i) A – aleurone layer ; B – endosperm ; C – embryo ; 3 Question Answer Marks 3(a)(ii) six from: 1 embryo / C, produces / releases, gibberellin ; 2 (arrow 1) gibberellin moves into, aleurone layer / A ; 3 gibberellin stimulates production of amylase ; 4 (arrow 2) amylase moves into, endosperm / B 5 hydrolyses / breakdown, starch to maltose ; 6 (arrow 3) maltose / glucose, moves into, embryo / C ; 7 for respiration / to release energy / for ATP production ; 8 for, germination / growth ; 6 max 3(b)(i) 3.6 ; µm min–1 OR µm / min ; 2 3(b)(ii) two from: 1 auxin binds with receptor (on cell surface membrane) ; 2 proton pumps activated ; 3 H+ pumped into cell wall ; 2 max Question Answer Marks 3(b)(iii) three from: 1 expansins activated / AW ; A optimum pH for expansins 2 loosens bonds in cellulose microfibrils ; 3 K+ ions enter cells (lower water potential) ; 4 (so) cells take in water by osmosis / AW ; 5 (increase in turgor causes) cell walls to stretch ; 3 max

4 (a) An investigation was carried out to measure the rate of photosynthesis at different concentrations of carbon dioxide. Two different plants, barley and sugar cane, were tested at two different temperatures, 10 °C and 25 °C. The results are shown in Fig. 4.1. sugar cane at 25 °C 40 35 30 barley at 25 °C 25 rate of photosynthesis 20 / arbitrary units barley at 10 °C 15 10 sugar cane at 10 °C 5 0 0 100 200 300 400 500 600 carbon dioxide concentration / arbitrary units Fig. 4.1 (i) Suggest why, in all four experiments, the rate of photosynthesis became constant as the carbon dioxide concentration increased. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) With reference to Fig. 4.1, describe the difference in the rate of photosynthesis, at 10 °C, between barley and sugar cane. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (iii) Sugar cane is a C4 plant. Barley is not a C4 plant. Explain why, at 25 °C, sugar cane had a higher rate of photosynthesis than barley. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [4] (b) Carbon dioxide for photosynthesis enters the leaves through open stomata. Stomata do not stay open all the time. Changes in environmental conditions can cause stomata to close. Describe these conditions and explain how stomatal closure benefits the plant. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] [Total: 13]

Mark scheme: 4(a)(i) three from: 1 reference to limiting factors ; 2 (limiting factor) not carbon dioxide ; 3 (limiting factor possibly) light intensity / temperature ; 2 max 4(a)(ii) three from: sugar cane: ORA barley 1 (rate of photosynthesis) higher rate, at lower concentrations of CO2 / initially ; 2 levels off / becomes constant, at lower rate of photosynthesis ; 3 levels off / becomes constant, at a lower carbon dioxide concentration ; 4 data quote to support mp2 or mp3 ; e.g. mp2 – sugar cane at 7–7.5 au and barley at 14 au OR mp3 – sugar cane at 60–70 au and barley at 500 au 3 max Question Answer Marks 4(a)(iii) four from: 1 sugar cane / C4 plants, can reduce photorespiration ; 2 correct ref. to adaptation to reduce photorespiration ; e.g. RuBP and rubisco, in bundle sheath cells / kept away from air OR tightly packed mesophyll cells 3 enzymes in, sugar cane / C4 plants, have high(er) optimum temperature ; 4 carbon dioxide absorbed by mesophyll cells ; 5 carbon dioxide, fixed by PEP carboxylase / combines with PEP ; 6 PEP carboxylase has higher affinity for carbon dioxide than rubisco ; 4 max 4(b) conditions (max two): 1 low light intensity ; A at night / in the dark 2 dry conditions ; 3 high temperatures ; 4 high light intensity ; 5 high wind speed / AW ; benefits (max two): 6 reduce transpiration (rate) ; 7 (so) conserves water ; 8 retains turgidity of cells ; 9 (physical) support of plant / prevents wilting ; 4 max

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5 The foxglove, Digitalis purpurea, is a common plant in many parts of the world. Fig. 5.1 shows a foxglove. Fig. 5.1 Flower colour in foxgloves is controlled by two genes that interact with each other. • Dominant allele M codes for an enzyme involved in the production of a purple pigment. • Recessive allele m codes for a non-functioning enzyme so no purple pigment is produced, resulting in a white colour. • Dominant allele D interacts with allele M to produce dark purple flowers. • Recessive allele d does not interact with allele M. • Neither allele D nor allele d interact with allele m. A double homozygous foxglove with dark purple flowers was crossed with a double homozygous recessive foxglove with white flowers. All the offspring had dark purple flowers. (a) Explain what is meant by the term homozygous. ................................................................................................................................................... .............................................................................................................................................. [1] (b) Using the symbols above, state the genotype of the offspring with dark purple flowers. .............................................................................................................................................. [1] (c) Two of these offspring with dark purple flowers were crossed. This cross produced a mixture of plants with three different flower colours: • dark purple • purple • white. Draw a Punnett square to show the possible genotypes and phenotypes of the mixture of plants from this cross. Write down the expected ratio of plants with each flower colour. ratio ............................................................... [6] [Total: 8]

Mark scheme: 5(a) having identical alleles (of a gene) ; 1 5(b) MmDd ; 1 5(c) gametes MD Md mD md MD MMDD dark purple MMDd dark purple MmDD dark purple MmDd dark purple Md MMDd dark purple MMdd purple MmDd dark purple Mmdd purple mD MmDD dark purple MmDd dark purple mmDD white mmDd white md MmDd dark purple Mmdd purple mmDd white mmdd white order D/d followed by M/m, or M/m followed by D/d are both acceptable one mark for gametes ; two marks for genotypes ; ; two marks for phenotypes linked to genotypes ; ; ratio = 9 dark purple : 3 purple : 4 white ; 6

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6 The Hardy-Weinberg principle may be used to calculate allele and genotype frequencies for a gene, within a population. The Hardy-Weinberg principle uses these equations: p2 + 2pq + q2 = 1 p + q = 1 (a) Within a population of butterflies, the allele for brown wing colour, B, is dominant to the allele for white wing colour, b. 40% of the butterflies in this population are white. Use the Hardy-Weinberg principle to calculate the percentage of butterflies in the population that are heterozygous for the gene controlling wing colour. Show your working. answer ........................................................... % [3] (b) Describe four situations where the Hardy-Weinberg principle does not apply. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] [Total: 7] Question 7 starts on page 16

Mark scheme: 6(a) 46.5 – 47 ;;; if answer incorrect allow marks for working q2 = 0.4 ; p = 0.368 OR p = 0.37 ; 3 6(b) four from: 1 mutation ; 2 migration (into, or out of, the population) ; 3 non-random mating occurs ; 4 the population is small ; 5 selective pressure occurs against one of the, alleles / genotypes ; 6 reproduction is asexual ; 7 organism is haploid ; 4 max

7 (a) Most reflex arcs pass through the spinal cord and involve different types of neurones. Name and state the functions of the three types of neurone in a spinal reflex arc. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) Some poisons affect the functioning of synapses between neurones. Listed below are four organisms, A, B, C and D, each of which produces a different poison that affects the functioning of synapses. A Hapalochlaena lunulata, the blue-ringed octopus B Conus textile, the textile cone sea snail C Clostridium botulinum, a bacterium D Physostigma venenosum, the calabar bean plant Table 7.1 summarises the action of the poisons from organisms A, B, C and D. Table 7.1 source of action of poison at synapse poison A blocks sodium channels B blocks calcium channels C prevents exocytosis D inhibits acetylcholinesterase Suggest and explain how poison from each of these organisms affects the functioning of synapses. A ............................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... B ............................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... C ............................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... D ............................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... [6] [Total: 9]

Mark scheme: 7(a) sensory neurone – receives, input / impulses, from receptor ; relay / intermediate / internuncial, neurone – passes impulses on to motor neurone ; motor neurone – sends impulses / output, to the effector ; 3 7(b) six from: A: 1 Na+ cannot enter post-synaptic neurone ; 2 no, depolarisation / action potential, (in post-synaptic neurone) ; B 3 Ca2+ cannot enter pre-synaptic neurone ; 4 vesicles cannot, move towards / fuse with, pre-synaptic membrane ; C 5 ACh cannot be released ; 6 into synaptic cleft ; D 7 ACh not broken down ; 8 continuous depolarisation / action potential, of post-synaptic neurone ; 6 max

8 Palm oil is an edible vegetable oil that comes from the fruit of oil palm trees. • Palm oil is cheap. • It is the world’s most widely used vegetable oil. • The yield of oil per hectare from oil palm trees is thirty times more than that of oil from maize. • Around 85% of the world’s palm oil is now produced in Indonesia. Table 8.1 shows the change in forest cover on the three main Indonesian islands between 2000 and 2010. Table 8.1 island Sumatra Borneo Java area of island 42 72 12 / million hectares forest cover in 2000 15 33 3 / million hectares forest cover in 2010 12 27 1 / million hectares loss of forest cover between 2000 and 2010 3 2 / million hectares .................... percentage loss of forest cover between 20 67 2000 and 2010 .................... (a) Complete Table 8.1 to show the loss of forest cover and the percentage loss of forest cover for Borneo between 2000 and 2010. [2] (b) The Sumatran forest is the natural habitat for the Sumatran orangutan, Pongo abelii. Fig. 8.1 shows a Sumatran orangutan. Fig. 8.1 The Sumatran orangutan is classified as critically endangered on the IUCN Red List of Threatened Species. Loss of forest cover can separate family groups from other groups. (i) Suggest the genetic consequences of the separation of family groups. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (ii) Suggest ways in which the Sumatran orangutan can be protected in its natural environment. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (iii) Describe the role of zoos in the protection of endangered species such as the Sumatran orangutan. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] [Total: 11]

Mark scheme: 8(a) 6 ; 18 ; 2 8(b)(i) three from: 1 inbreeding depression / lack of hybrid vigour ; 2 more chance that harmful recessive alleles may be expressed ; 3 decrease in heterozygosity / increase in homozygosity ; 4 less genetic variation ; 3 max 8(b)(ii) three from: 1 use sustainable palm oil plantations ; 2 create / leave, corridors between family groups in different parts of the forest ; 3 ban hunting ; 4 create national parks ; 5 educate local people ; 6 re-locate orangutans ; 7 reforestation ; 3 max Question Answer Marks 8(b)(iii) three from: 1 captive breeding ; 2 detail e.g. IVF / ICSI / sperm banks ; 3 education ; 4 release back into the wild ; 5 research ; 6 health monitoring ; 3 max

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9 (a) Explain what is meant by the terms ecosystem and niche. [7] (b) Describe the sampling techniques that could be used to measure the distribution and abundance of organisms in a habitat. [8] [Total: 15]

Mark scheme: 9(a) seven from: ecosystem: 1 self-sustaining unit ; 2 self-contained / defined, area ; 3 community of organisms ; 4 living and non-living / biotic and abiotic ; 5 reference to, interactions / interdependence ; 6 reference to, energy flow / food webs ; niche: 7 role of organism / how it fits in, (in an ecosystem) ; 8 (including) where it lives ; 9 how it obtains its energy / reference trophic level ; 7 max Question Answer Marks 9(b) eight from: 1 random (sampling) ; 2 (frame) quadrat ; 3 use cover scale ; 4 estimate % cover ; 5 species frequency ; 6 systematic sampling ; 7 line / belt, transect ; 8 sample at set distances ; 9 mark – release – recapture ; 10 method of, capture / marking ; 11 returned to habitat and left ; 12 population estimate = no of individuals in first sample no of individuals in second sample no of individuals marked in second sample × ; 13 appropriate mathematical / statistical technique ; 8 max

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10 (a) Describe the principles of the polymerase chain reaction (PCR). [9] (b) Explain why plasmids are frequently used in gene technology. 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Mark scheme: 10(a) nine from: 1 production of a large number of copies of a length of DNA / amplification of DNA ; 2 rapid ; 3 only small sample of DNA needed ; 4 DNA, denatured / separated into two strands, by heat / at 95 °C ; 5 primer (DNA) added ; 6 reference to annealing at, 60–65, °C ; 7 reference complementary base pairing ; 8 DNA / Taq, polymerase ; 9 replicates (template) strand at, 70–75, °C ; 10 heated again to separate strands / process repeated ; 11 Taq polymerase, is heat stable / has high optimum temperature ; 12 does not need replacing each cycle ; 13 efficient process ; 9 max Question Answer Marks 10(b) six from: 1 small / circular, piece of (double-stranded) DNA ; 2 replicate independently / multiple origins of replication ; 3 high copy number ; 4 easy to extract from bacteria ; 5 can be cut using restriction, enzyme / endonuclease ; 6 gene / DNA, can be inserted ; 7 can be taken up by bacteria ; 8 may contain genes for antibiotic resistance / can carry marker genes ; 9 helps in identifying transformed bacteria ; 10 acts as a vector ; 11 may carry promoter ; 6 max

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