Cambridge A Level Biology 9700 — 2011 May/June Paper 4 · Variant 1

This document consists of 23 printed pages, 3 lined pages and 2 blank pages. DC (AC/DJ) 34054/5 © UCLES 2011 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black ink. You may use a pencil for any diagrams, graphs, or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all questions in Section A and one question from Section B. Circle the number of the Section B question you have answered in the grid below. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. * 6 9 6 5 2 7 2 0 3 6 * BIOLOGY 9700/41 Paper 4 A2 Structured Questions May/June 2011 2 hours Candidates answer on the Question Paper. Additional Materials: Answer Paper available on request. For Examiner’s Use Section A 1 2 3 4 5 6 7 8 Section B 9 or 10 Total

2 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use Section A Answer all the questions. 1 The polar bear, Ursus maritimus, lives in the Arctic regions of the USA, Canada, Norway and Russia. Polar bears move across the Arctic ice sheet to hunt prey such as seals. Fig. 1.1 shows a polar bear. Fig. 1.1 The area over which the Arctic ice sheet extends varies throughout the year. Fig. 1.2 shows the variation in the extent of the Arctic ice sheet for the months of July to November for the years 1979 and 2009. 12 July Aug Sept month Oct Nov 10 8 6 4 2 0 extent of Arctic ice sheet / km2 × 106 Key 1979 2009 Fig. 1.2

3 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use (a) Calculate the percentage reduction in the area over which the ice sheet extends between 1979 and 2009 for the month of September. Give your answer to the nearest whole number. Show your working. answer … % [2] (b) In 2008 the government of the USA classified U. maritimus as an endangered species because it is under threat of extinction. Suggest what has caused U. maritimus to have become endangered. … … … … … … [3] (c) U. maritimus is a eukaryote. Beneficial bacteria, which are prokaryotic cells, live in the gut of U. maritimus. State three differences between the cells of U. maritimus and its gut bacteria. 1. … 2. … 3. … [3] [Total: 8]

4 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use 2 When gold is associated with mineral ores such as iron sulfide, the sulfides must be oxidised to release the gold particles. Since the mid 1990s, gold has been extracted from such ores by bioleaching. Suitable bacteria oxidise iron sulfide to soluble iron sulfate, releasing Fe3+ and SO4 2– ions. The reaction releases heat energy and temperatures within a heap of ore that is being bioleached (a bioheap) can reach 70 °C or higher. Examples of bacteria used in this bioleaching are shown in Table 2.1. Table 2.1 example of bacterium temperature range for growth / °C activity natural habitat Acidithiobacillus ferrooxidans 35 – 45 oxidise iron and sulfur compounds acid springs Sulfobacillus thermosulfidooxidans 45 – 65 Sulfolobus metallicus 65 – 95 (a) With reference to Table 2.1, suggest (i) a natural habitat for organisms such as S. thermosulfidooxidans and S. metallicus … … [1] (ii) why all three species of bacteria, rather than just one species, are mixed with ore in a bioheap. … … … … … … [3]

5 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use (b) The rate of oxidation of the iron in iron sulfide ore was compared in the presence and absence of A. ferrooxidans at pH 2.0. The results are shown in Fig. 2.1. in presence of A. ferrooxidans in absence of A. ferrooxidans 2.0 concentration of Fe3+ ions / arbitrary units 1.0 1.5 0.5 0 0 5 10 15 20 time / days 25 Fig. 2.1 (i) With reference to Fig. 2.1, describe the effect of A. ferrooxidans on the oxidation of the ore. … … … … … … [3]

6 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use (ii) Explain why bioleaching is now used on a large scale throughout the world. … … … … … … [3] (c) Gold-bearing sulfide ores often contain arsenic, which is potentially toxic to the bacteria used in bioleaching. However, arsenic-resistant strains of A. ferrooxidans have been found in some mines. The activity of two strains of the bacterium, in the presence and absence of arsenic ions, is shown in Table 2.2. Table 2.2 oxidation rate of iron in the ore / mg dm–3 h–1 strain of A. ferrooxidans arsenic ions absent arsenic ions present 1 16 15 2 48 47 Describe the results shown in Table 2.2 and explain the role of natural selection in the evolution of arsenic-resistant bacteria. … … … … … … … … [4] [Total: 14]

7 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 3 (a) Outline the technique of in-vitro fertilisation (IVF). … … … … … … … … [4] (b) For IVF to be successful, a sperm must have an undamaged plasma (cell surface) membrane, an intact acrosome (a sperm’s large lysosome) and be capable of producing ATP for movement. One method of assessing the quality of a sample of sperm is to mix it with three chemical probes that bind to specific components of the sperm. The probes fluoresce when the sperm are examined with a microscope using ultra-violet (UV) light, allowing their uptake to be determined. The three probes fluoresce with different colours. • Probe 1 combines with DNA and fluoresces red, but can enter a sperm only when its plasma membrane is damaged. • Probe 2 combines with sugars in the acrosome and fluoresces yellow, but can enter the acrosome only when the acrosome membrane is damaged. • Probe 3 combines with mitochondria and fluoresces bright green in sperm with active mitochondria and less brightly when the mitochondria are less active. A sample of sperm was mixed with all three probes and examined using UV light. Complete Table 3.1 by placing ticks (3) in the appropriate boxes to describe the appearance of sperm that would be suitable for use in IVF. Table 3.1 appearance of sperm suitable for use in IVF target of probe red yellow green colourless DNA acrosome mitochondria [3]

8 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use (c) The technique of intracytoplasmic sperm injection (ICSI) involves injecting a single, chosen sperm into an oocyte. This technique is often used when standard IVF has failed. Researchers in Hawaii think that the acrosome of the sperm should be removed before the sperm is injected into the oocyte. Suggest one reason why it might improve the success rate of ICSI to remove the acrosome before injecting a sperm into an oocyte. … … [1] [Total: 8]

9 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 4 Almost 40% of adults with cystic fibrosis (CF) develop a form of diabetes known as cystic fibrosis-related diabetes (CFRD). This is thought to happen because the build-up of thick secretions in the pancreas destroys β cells. (a) Explain how the destruction of β cells causes diabetes. … … … … … … … … [4] (b) The bacterium Pseudomonas aeruginosa can cause chronic (long-lasting) lung infections. A person with CFRD is likely to have poorer lung function and a greater likelihood of having a chronic lung infection than a person who has CF but does not have CFRD. An investigation was carried out to find out if the severity of damage to lung function in a person with CFRD is affected by • their gender • whether or not they have a chronic P. aeruginosa infection. The investigators measured lung function by recording the maximum volume of air that can be expelled from the lungs in the first one second of a forced expiration. This is known as FEV1. The lower the median FEV1, the poorer the lung function.

10 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use Table 4.1 summarises the results of this investigation. All the 812 people in the study had cystic fibrosis. Table 4.1 without chronic P. aeruginosa infection with chronic P. aeruginosa infection male female male female with CFRD without CFRD with CFRD without CFRD with CFRD without CFRD with CRFD without CFRD number of people 44 110 52 93 106 166 121 120 FEV1 71.1 71.4 53.6 73.6 49.0 59.0 42.0 61.0 With reference to Table 4.1 (i) discuss whether or not there appears to be a positive correlation between having a chronic P. aeruginosa infection and having CFRD … … … … … … [2] (ii) calculate the percentage difference between the FEV1 of males and females without CFRD and without P. aeruginosa infection. Show your working answer … % [2]

11 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use (iii) outline the conclusions that can be drawn concerning the relationship between gender and the severity of lung damage in a person with CFRD and with P. aeruginosa infection. … … … … … … … … [3] (c) In a person with CF, damage to lung function and the increased likelihood of chronic infections are the result of the secretion of thick mucus. Explain why thick mucus is secreted in the lungs of a person with CF. … … … … … … … … [4] [Total: 15]

12 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use 5 Both sorghum and maize are important food crops in dry regions of the world, but sorghum is able to produce higher yields than maize in very dry conditions. This is partly because sorghum plants have a smaller leaf area than maize, and also because sorghum leaves have rows of motor cells along the midrib of the upper surface of the leaf, allowing the leaves to roll up. (a) Explain how these two features adapt sorghum plants for growth in very dry conditions. … … … … … … [3] (b) Sorghum is a staple food in Africa, but the major storage protein that it contains, kaffirin, is not easily digested by protease enzymes. The main cause of this is cross-linking between kaffirin molecules. The digestibility of the protein in five varieties of sorghum was measured when raw, and after cooking. Digestibility was measured as the percentage of the protein that would be broken down to amino acids during digestion. The results are shown in Fig. 5.1. HDI Macia Seredo variety of sorghum NK8828 Sudan 100 80 60 40 20 0 digestibility (percentage of proteins broken down to amino acids) Key raw cooked Fig. 5.1

13 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use With reference to Fig. 5.1 (i) compare the digestibility of raw and cooked sorghum protein … … … … [2] (ii) using your knowledge of protein structure and enzyme activity, suggest reasons for the differences you have described in (i). … … … … … … [3] [Total: 8]

14 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use 6 Canavan disease is a non sex-linked inherited condition that causes progressive damage to neurones of the brain. Symptoms of the condition include a loss of motor skills and mental retardation. The symptoms appear in early infancy and many children with this condition die by the age of four years. People with Canavan disease lack an enzyme called aspartoacylase which breaks down N-acetyl aspartate. The build up of N-acetyl aspartate can interfere with the formation of the myelin sheath, particularly in neurones of the brain. (a) Enzymes such as aspartoacylase display specificity. Outline what is meant by specificity of an enzyme. … … … … [2] (b) Complete the genetic diagram below to show how an unaffected man and an unaffected woman could produce a child with Canavan disease. key to symbols … … parental phenotypes unaffected man unaffected woman parental genotypes gametes offspring genotypes offspring phenotypes [3]

15 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use (c) Explain the importance of the myelin sheath in the functioning of a neurone. … … … … … … … … [3] [Total: 8]

16 9700/41/M/J/11 © UCLES 2011 BLANK PAGE Question 7 starts on page 17

17 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use 7 (a) Complete the following passage about ATP by writing in the missing words. All living organisms use energy. The most common immediate source of energy is adenosine triphosphate (ATP) which is used in every cell for the movement of ions against a concentration gradient, known as … . ATP is known as the universal currency of energy. ATP is a phosphorylated nucleotide which is known as a ‘high energy’ molecule. It is made of an organic base, adenine, a 5 carbon sugar named … and three phosphate groups. ATP is very soluble in … and easily transported within the cell. The removal of the outer phosphate group by the process of … releases energy. The energy released as a result of this reaction can be channelled directly into other reactions in the cell. A certain proportion of this energy is lost as … . ATP is continually broken down and is reformed at a fast rate by the process of respiration. [5]

18 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use (b) During a sporting event an athlete may have to carry out anaerobic respiration in addition to aerobic respiration to produce sufficient ATP. Fig. 7.1 outlines both processes in a muscle cell and shows how a liver cell is linked to these processes. Liver cell Muscle cell glucose additional oxygen blood lactate blood glucose glucose lactate glycolysis ATP pyruvate link reaction acetyl coA Krebs cycle carbon dioxide oxidative phosphorylation ATP water Fig. 7.1

19 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use You may refer to Fig. 7.1 in answering questions (i) to (v) below. (i) Glucose produced in the liver cell can be released into the blood to maintain blood glucose concentration. State one use of glucose within the liver cell. … … [1] (ii) Suggest why anaerobic respiration is said to be less efficient than aerobic respiration. … … … … [2] (iii) Complete the table to indicate, within the muscle cell, the precise locations of glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation. process precise location glycolysis link reaction Krebs cycle oxidative phosphorylation [4]

20 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use (iv) Glucose is phosphorylated at the start of glycolysis in the muscle cell. Suggest why this phosphorylated glucose does not diffuse out of the cell into the surrounding tissue fluid. … … … … [2] (v) Additional oxygen is required in the metabolic pathways involved in the conversion of lactate to glucose. State the term given to this additional oxygen. … [1] [Total: 15]

21 9700/41/M/J/11 © UCLES 2011 [Turn over BLANK PAGE Question 8 starts on page 22

22 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use 8 The hedgehog, Erinaceus europaeus, is a small carnivorous mammal native to Northern Europe. Fig. 8.1 shows a hedgehog. Fig. 8.1 Hedgehogs were introduced onto a small group of islands off the west coast of Scotland in 1974. The hedgehog population has increased so that there are now over 5 000 breeding pairs. These hedgehogs have no natural predators on these islands and their diet consists mainly of bird’s eggs. Fig. 8.2 shows the hedgehog population density in the year 2000. north island south island 10 km Key high / medium density low density absent Fig. 8.2

23 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use Table 8.1 shows the changes in the populations of the species of birds from 1983 to 2000. Table 8.1 breeding pairs in 1983 breeding pairs in 2000 % change in population north island oystercatcher 928 1122 +21 lapwing 1104 1364 +24 redshank 486 733 +51 south island oystercatcher 907 1403 +55 lapwing 1869 1287 –31 redshank 1288 760 –41 (a) Using Fig. 8.2 and Table 8.1, describe the relationship between the hedgehog population density and the changes in the populations of lapwings and redshanks. … … … … … … … … [3] (b) Suggest an explanation for the increase in the oystercatcher population on the south island, despite the increase in the hedgehog population. … … … … [2]

24 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use (c) Explain why the population of hedgehogs on one of these islands may eventually become a different species from that on mainland Scotland. … … … … … … … … [4] [Total: 9]

25 9700/41/M/J/11 © UCLES 2011 [Turn over For Examiner’s Use Section B Answer one question. 9 (a) Outline the ways in which the endocrine and nervous systems carry out their roles in control and coordination in animals. [8] (b) Describe the part played by auxins in apical dominance in a plant shoot. [7] [Total: 15] 10 (a) Describe how non-cyclic photophosphorylation produces ATP and reduced NADP. [9] (b) Outline the steps of the Calvin cycle. [6] [Total: 15] … … … … … … … … … … … … … … … … … …

26 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use … … … … … … … … … … … … … … … … … … … … … … … … … … … …

27 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use … … … … … … … … … … … … … … … … … … … … … … … … … … … …

28 9700/41/M/J/11 © UCLES 2011 For Examiner’s Use … … … … … … … … … … … … … … … … … … … … … … … Copyright Acknowledgements: Fig. 1.1 Stephen J. Krasemann/Science Photo Library. Fig. 8.1 Ian Gowland/Science Photo Library. Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2011 question paper for the guidance of teachers 9700 BIOLOGY 9700/41 Paper 4 (A2 Structured Questions), maximum raw mark 100 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9700 41 © University of Cambridge International Examinations 2011 Mark scheme abbreviations: ; separates marking points / alternative answers for the same point R reject A accept (for answers correctly cued by the question, or by extra guidance) AW alternative wording (where responses vary more than usual) underline actual word given must be used by candidate (grammatical variants excepted) max indicates the maximum number of marks that can be given ora or reverse argument mp marking point (with relevant number) ecf error carried forward I ignore AVP Alternative valid point (examples given as guidance)

Page 3 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9700 41 © University of Cambridge International Examinations 2011 1 (a) 36 ;; allow one mark for number not rounded up i.e. 35.7 or allow working of 7 X × 100 [2] (b) 1. reduction in extent of ice sheet ; 2. reduction in number of, seals / prey / food or increased competition for food ; 3. idea of increased distance to travel to find food ; 4. loss / destruction, of breeding sites ; 5. result of named human activity ; e.g. mining / drilling / killing / building / pollution 6. disease ; [3 max] (c) applies to U. maritimus but accept ora 1. DNA linear ; 2. DNA in nucleus or has, nuclear membrane / nucleus ; 3. DNA, associated with protein / in chromosomes ; 4. ribosome, 22 nm diameter / 80s ; 5. membrane bound organelles / named organelle ; 6. no cell wall ; 7. size up to 40µm ; [3 max] [Total: 8] 2 (a) (i) any one from ; hot springs sulphur springs geysers geothermals marine vent volcanic area hot desert [1] (ii) 1. each bacterium grows at a different temperature (range) ; 2. (the heap) heats up ; 3. idea of when temperature kills one species of bacterium others are still active or as temperature increases process can continue ; 4. increased oxidation of heap ; 5. more productive / enables increased yield of gold ; [3 max]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9700 41 © University of Cambridge International Examinations 2011 (b) (i) 1. A. ferrooxidans increases, oxidation of the ore / production of Fe3+ ; 2. little difference in effect 0–5 days ; 3. greatest effect after 15 days ; 4. comparative figs for with and without A. ferrooxidans on a single day ; [3 max] (ii) 1. cheaper (than other methods) ; 2. does not require energy input ; 3. does not require other chemicals to be purchased ; 4. does not require specialist equipment ; 5. can be done in situ ; 6. less labour needed ; 7. bacteria are self-replicating / AW ; 8. more environmentally friendly than other methods / no harmful emissions / AW ; 9. useful for extraction from, low grade ores / waste ; [3 max] (c) must have at least one D mark to score 4 marks D1 both strains give similar rate with and without arsenic ions ; D2 both strains are arsenic-resistant ; D3 strain 2, more active / higher oxidation rate, (than strain 1) ; E4 arsenic acts as a selective, agent / pressure ; E5 mutation / AW, produces resistant bacteria ; E6 resistant bacteria survive / ora ; E7 resistant allele passed on ; E8 frequency of allele increases (in population) ; [4 max] [Total: 14] 3 (a) 1. ref. hormone treatment ; 2. results in, superovulation / many oocytes / many follicles, maturing at same time ; 3. oocytes harvested ; 4. detail of harvesting ; 5. mixed with sample of sperm ; 6. in special medium ; 7. idea of, waiting for three days / wait until 6–8 cell stage ; 8. embryos placed in uterus ; 9. ref. maintenance of endometrium ; 10. sperm / sperm nucleus / sperm DNA, may be injected into oocyte ; [4 max] (b) one mark for a
in the correct box more than one
in a row = no mark ignore crosses DNA – colourless ; acrosome – colourless ; mitochondria – green ; [3] (c) 1. (hydrolytic) enzymes may damage oocyte ; 2. (acrosome contents) affect development of fertilised oocyte ; [1 max] [Total: 8]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9700 41 © University of Cambridge International Examinations 2011 4 (a) 1. β cells detect glucose levels or no detection of blood glucose conc. ; 2. β cells secrete insulin or no insulin released ; 3. when blood glucose concentration rises or when blood glucose concentration rises ; 4. (insulin causes) muscle cells / adipose tissue / liver cells or muscle cells / adipose tissue / liver cells ; 5. to increase uptake of glucose from blood / increased membrane permeability to glucose or do not take up excess glucose ; 6. (insulin causes liver cells) to convert glucose to glycogen or glucose not converted to glycogen (by liver cells) ; 7. (insulin causes liver cells) to increase respiration of glucose or rate of respiration of glucose does not increase ; 8. (if no β cells) no control of blood glucose levels / AW or no control of blood glucose levels / AW ; [4 max] (b) (i) 1. (yes) more people with infection have CFRD than those without infection ; 2. use of ‘with CFRD’ comparative figs ; either using number of people – 44 / 52 / 96 (no infection) against 106 / 121 / 227 (with infection) or using FEV1 values – 71.1 / 53.6 / 124.7 (no infection) against 49.0 / 42.0 / 91.0 (with infection) or 28.5% males against 35.8% females (no infection) or 38.9% males against 50.05% females (with infection) 3. AVP ; e.g. we do not know how the sample was chosen (so this may not be a valid conclusion) [2 max] (ii) 100 71.4 2.2 × ; = 3.08 / 3.1 ; or 100 73.6 2.2 × ; = 2.99 / 3.0 ; [2] (iii) 1. more lung damage in females (with CFRD) than in males ; 2. females (with CFRD) have lower FEV1 than males ; 3. use of figures ; e.g. males FEV1 49 whereas female FEV1 42 or female FEV1 1.16 times lower than male FEV1 [3] (c) 1. CFTR protein acts as chloride channel (in cell membranes) ; with CF 2. faulty (CFTR) gene ; 3. faulty / non-functional, (CFTR) protein produced ; 4. chloride ions not able to move out (of cell) ; 5. by active transport ; 6. so less water passes out (of cell) ; 7. down water potential gradient ; A by osmosis 8. mucus secreted contains less water ; [4 max] [Total: 15]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9700 41 © University of Cambridge International Examinations 2011 5 (a) 1. (either feature) reduces water loss by, transpiration / evaporation ; 2. reduction in, number of stomata / surface area, (for, transpiration / evaporation) ; 3. rolling leaves traps moist air ; 4. idea of reduced, diffusion / water potential, gradient (between leaf and trapped air) ; [3 max] (b) (i) cooked protein more digestible than raw protein ; use of figures ; accept any named comparison between cooked and raw [2] (ii) cooked 1. cooking breaks cross-links (in kaffirin) ; A bonds 2. ref. to named bond ; e.g. hydrogen / ionic / disulphide / covalent 3. tertiary / 3D / quaternary, structure disrupted / AW ; 4. protease can now bind, more / easier, with polypeptides ; 5. enzyme-substrate complexes can form ; 6. so more protein is digested to amino acids ; [3 max] [Total: 8] 6 (a) enzyme acts on only one substrate ; shape of active site is complementary to substrate ; AVP ; e.g. substrate held by temporary bonds / ES complex [2 max] (b) symbols (must be of same letter) ; parental genotypes and gametes ; offspring genotypes and phenotypes linked ; [3] (c) 1. insulates axon (membrane) ; 2. depolarisation occurs only at nodes (of Ranvier) / AW ; 3. local circuits ; 4. saltatory conduction / AW ; 5. speeds transmission of, action potential / impulse ; 6. AVP ; e.g. speed increases up to 50 times / 100ms–1 [3 max] [Total: 8]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9700 41 © University of Cambridge International Examinations 2011 7 (a) active transport ; ribose ; water ; hydrolysis ; A dephosphorylation heat ; [5] (b) (i) (converted to) glycogen / lipid ; (used in) glycolysis / respiration ; [1 max] (ii) anaerobic 1. less ATP / only 2 ATP ; 2. per mol glucose ; 3. lactate still contains energy / only glycolysis involved / stages other than glycolysis not involved ; 4. not sustainable / cannot go on indefinitely / AW ; [2 max] (iii) process precise location glycolysis cytoplasm / cytosol ; link reaction mitochondrial matrix ; Krebs cycle mitochondrial matrix ; oxidative phosphorylation inner mitochondrial membrane / cristae ; [4] (iv) 1. cannot pass through phospholipid bilayer ; 2. too big to fit through (glucose’s) protein channel ; 3. no specific transport protein ; 4. AVP ; e.g. used up as soon as it is made [2 max] (v) oxygen debt ; [1] [Total:15] 8 (a) north island 1. fewer / less abundant, hedgehogs allow increase (in both lapwing and redshank) ; 2. breeding pair figs for either bird for 1983 and 2000 or % change in population over that time for either bird ; south island 3. presence of hedgehogs causes decrease (in both lapwing and redshank) ; 4. breeding pair figs for either bird for 1983 and 2000 or % change in population over that time for either bird ; [3 max] (b) 1. (oystercatchers have) less competition ; 2. hedgehogs mostly eat lapwing and redshank eggs / hedgehogs don’t eat oystercatcher eggs ; 3. (oystercatcher) eggs are, too large / camouflaged / inaccessible / distasteful or oystercatchers defend their, nests / eggs ; [2 max]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9700 41 © University of Cambridge International Examinations 2011 (c) 1. idea of geographical isolation ; 2. no interbreeding / gene flow, between populations ; 3. mutations occur ; 4. different, selection pressures / environmental conditions ; 5. genetic change / AW ; 6. genetic drift ; 7. (eventually) reproductive isolation ; 8. allopatric speciation ; [4 max] [Total: 9] 9 (a) endocrine 1. hormones ; 2. chemical messengers ; A chemicals that transfer information 3. ductless glands / (released) into blood ; 4. target, organs / cells ; 5. ref. receptors on cell membranes ; 6. example of named hormone and effect ; nervous 7. impulses/ action potentials ; R electrical, signals / current 8. along, axon / neurones / nerve fibres ; R nerves R across 9. synapse (with target) / neuromuscular junction ; 10. ref. receptor / sensory neurones ; 11. ref. effector / motor neurones ; differences – endocrine 12. slow effect / ora ; 13. long lasting effect / ora ; 14. widespread effect / ora ; 15. AVP ; e.g. extra detail of synapse / hormone changes triggered within cells [8 max] (b) 16. IAA / plant growth regulator ; R plant hormone 17. synthesised in, growing tips / apical buds / meristems ; R root tip 18. moves by diffusion ; 19. moves by active transport ; 20. from cell to cell ; 21. also, mass flow / in phloem ; 22. stimulates cell elongation ; R cell enlargement 23. inhibits, side / lateral, buds / growth ; A inhibits branching 24. plant grows, upwards / taller ; A stem elongates 25. auxin not solely responsible or interaction between auxin and other plant growth regulators ; 26. AVP ; e.g. role of ABA and lateral bud inhibition 27. AVP ; e.g. cytokinins antagonistic to IAA / gibberellins enhance IAA [7 max] [Total: 15]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9700 41 © University of Cambridge International Examinations 2011 10 (a) 1. photosystem I (PI) and photosystem II (PII) involved ; 2. light harvesting clusters ; 3. light absorbed by accessory pigments ; 4. primary pigment is chlorophyll a ; 5. energy passed to, primary pigment / chlorophyll a ; 6. electrons, excited / raised to higher energy level ; 7. (electrons) taken up by electron acceptor ; 8. (electrons) pass down electron carrier chain (to produce ATP) ; 9. PII has (water splitting) enzyme ; 10. water split into protons, electrons and oxygen ; A equation 11. photolysis ; 12. electrons from PII pass to PI / electrons from water pass to PII ; 13. to replace those lost ; give either in relation to PI or PII 14. protons and electrons combine with NADP (to produce reduced NADP) ; can award these marking points from a diagram [9 max] (b) 15. RuBP combines with carbon dioxide ; 16. rubisco ; 17. forms unstable 6C compound ; 18. produces two molecules of, GP / PGA ; 19. GP / PGA, converted to TP ; 20. by reduced NADP and ATP ; 21. from light dependent stage ; 22. TP used to regenerate RuBP ; 23. using ATP ; 24. TP can form, hexose / fatty acids / acetyl CoA [6 max] [Total: 15]