Cambridge A Level Biology 9700 — 2014 May/June Paper 4 · Variant 3

This document consists of 22 printed pages and 2 lined pages. DC (NF/CGW) 91959 © UCLES 2014 [Turn over Cambridge International Examinations Cambridge International Advanced Level * 5 1 4 1 8 1 0 4 4 7 * BIOLOGY 9700/43 Paper 4 A2 Structured Questions May/June 2014 2 hours Candidates answer on the Question Paper. Additional Materials: Answer Paper available on request. READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces provided at the top of this page. Write in dark blue or black pen. You may use a pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Section A Answer all questions. Section B Answer one question. 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. Electronic calculators may be used. For Examiner’s Use Section A 1 2 3 4 5 6 7 8 Section B 9 or 10 Total

2 9700/43/M/J/14 © UCLES 2014 Section A Answer all the questions. 1 (a) The unicellular green alga, Chlorella, a photosynthetic protoctist, was originally studied for its potential as a food source. Although large-scale production proved to be uneconomic, the many health benefits provided by Chlorella mean that it is now mass produced and harvested for use as a health food supplement. Fig. 1.1 shows cells of Chlorella. Chlorella Fig. 1.1 In one study into the productivity of Chlorella, carbon dioxide concentration was altered to investigate its effects on the light-independent stage of photosynthesis. • A cell suspension of Chlorella was illuminated using a bench lamp. • The suspension was supplied with carbon dioxide at a concentration of 1% for 200 seconds. • The concentration of carbon dioxide was then reduced to 0.03% for a further 200 seconds. • The concentrations of RuBP and GP (PGA) were measured at regular intervals. • Throughout the investigation the temperature of the suspension was maintained at 25 °C. The results are shown in Fig. 1.2. 3 2 50 0 100 150 200 time / s 250 300 350 400 concentration / arbitrary units 1 0 1 % CO2 0.03 % CO2 GP RuBP Fig. 1.2

3 9700/43/M/J/14 © UCLES 2014 [Turn over (i) State precisely where in the chloroplast RuBP and GP are located. …[1] (ii) Explain why the concentration of RuBP changed between 200 and 275 seconds. … … … … … …[2] (iii) Calculate the rate of decrease per second in the concentration of GP between 200 and 350 seconds. Show your working and give your answer to two decimal places. answer … arbitrary units per second [2] (b) Explain how the decrease in the concentration of GP leads to a decreased harvest for commercial suppliers of Chlorella. … … … … … …[2] [Total: 7]

4 9700/43/M/J/14 © UCLES 2014 2 Russian scientists have discovered the fruits of a flowering plant, Silene stenophylla, in the food store in a burrow of a ground squirrel in frozen sediments in Siberia. Dating techniques suggest that the fruits were stored by the ground squirrel about 32 000 years ago, shortly before the ground became permanently frozen. Tissue samples were taken from the fruits and grown in a nutrient culture medium. After treatment with plant hormones to stimulate the growth of roots and shoots, 36 complete plants were produced. These ‘regenerated’ plants, which looked identical to one another, flowered and after cross-pollination, produced seeds that were able to germinate. (a) Explain why cross-pollination produces more genetic variation among the offspring than self-pollination. … … … … … … …[3] (b) The flowers of modern-day S. stenophylla look similar, but not identical, to the flowers of the ‘regenerated’ plants. Outline how DNA sequencing could be used to compare the DNA of modern-day and ‘regenerated’ S. stenophylla. … … … … … … … … … …[4]

5 9700/43/M/J/14 © UCLES 2014 [Turn over (c) Suggest a simple experiment, using plants of modern-day and ‘regenerated’ S. stenophylla to find out whether, after 32 000 years, they are still the same species. … … … … …[2] [Total: 9]

6 9700/43/M/J/14 © UCLES 2014 3 (a) Wheat, Triticum aestivum, owes its origin to hybridisation involving three different, but related, species of grass, A, B and C. Each of these species had seven pairs of chromosomes (2n = 14). The hybridisation process is shown in Fig. 3.1. species A 2n = 14 (AA) species B 2n = 14 (BB) sterile hybrid 2n = 14 (AB) × Y fertile hybrid 2n = 28 (AABB) species C 2n = 14 (CC) A = one set of chromosomes from species A B = one set of chromosomes from species B Key: C = one set of chromosomes from species C sterile hybrid 2n = 21 (ABC) × Z fertile T. aestivum 2n = 42 (…) Fig. 3.1 (i) Using the symbols in the key, complete Fig. 3.1 by writing in the chromosome sets of T. aestivum. [1] (ii) At the points labelled Y and Z in the hybridisation process, a fertile hybrid was produced from a sterile hybrid. Explain why the hybrid (AB) is sterile and what occurred at the point labelled Y in Fig. 3.1. … … … … … … … … …[4]

7 9700/43/M/J/14 © UCLES 2014 [Turn over Question 3 continues on page 8

8 9700/43/M/J/14 © UCLES 2014 (b) In 2012, permission was granted for a field trial in the UK of genetically modified T. aestivum. The wheat carries a gene, taken from peppermint plants, that results in the wheat leaves releasing a volatile, non-toxic chemical, (E)-β-farnesene (Eβf), into the atmosphere. Eβf is not only produced by various species of plants. It is also secreted by aphids when they are disturbed by a predator. Two experiments have been performed into the effect of Eβf on the behaviour of aphids feeding on leaves in closed containers. Experiment 1 Either 10 cm3 of air from a syringe that contained plant leaves that secrete Eβf or 10 cm3 of air from a syringe with no such leaves was added to the containers of feeding aphids. Experiment 2 Either 20 cm3 of air containing 50ng of Eβf or 20 cm3 of air containing no Eβf was added to the containers of feeding aphids. In both experiments, the number of aphids that stopped feeding and moved away from the food leaves was counted. The results are shown in Table 3.1. Table 3.1 Experiment 1 Experiment 2 air added to containers of feeding aphids 10 cm3 air that had been in contact with leaves secreting Eβf 10 cm3 air that had not been in contact with leaves secreting Eβf 20 cm3 air containing 50ng Eβf 20 cm3 air containing no Eβf number of aphids in containers 99 113 132 106 number of aphids that stopped feeding and moved away from the food leaves 54 1 111 0

9 9700/43/M/J/14 © UCLES 2014 [Turn over (i) Discuss the extent to which the results of these experiments support the idea that Eβf is an alarm signal for aphids. … … … … … … … … … …[4] (ii) Other experiments show that Eβf attracts predators of aphids, such as ladybirds. Explain how growing genetically modified wheat secreting Eβf could increase the yield of wheat. … … … … … … … … …[3]

10 9700/43/M/J/14 © UCLES 2014 (iii) Suggest why growing this genetically modified wheat might be acceptable to people who object to the growth of genetically modified insect-resistant maize or cotton. … … … … … … … …[3] [Total: 15] 4 (a) Spermatogenesis takes place in the seminiferous tubules, in the testis. Fig. 4.1 is a diagram showing some of the cells in a small sector of a seminiferous tubule. spermatogonium … primary spermatocyte … secondary spermatocyte … spermatid … spermatozoan … Fig. 4.1 (i) On Fig. 4.1, state whether each of the labelled cells is haploid or diploid. Write n if the cell is haploid and 2n if the cell is diploid. [2] (ii) Spermatogenesis involves meiosis, mitosis, growth and maturation. State which of these processes is involved in each of the following steps in spermatogenesis. spermatogonium to primary spermatocyte … spermatid to spermatozoan (sperm) …[2]

11 9700/43/M/J/14 © UCLES 2014 [Turn over (iii) State one role of a Sertoli cell. … … …[1] (b) In some men, spermatogenesis does not take place successfully and the sperm that are produced are unable to fertilise an egg. A form of IVF called intra-cytoplasmic sperm injection (ICSI) may enable them to father a child with their partner. In ICSI, a sperm cell is inserted into a secondary oocyte using a very tiny needle. Outline the treatment required in order to obtain mature oocytes as part of an IVF procedure. … … … … … … … … … … …[4]

12 9700/43/M/J/14 © UCLES 2014 (c) One approach to helping an infertile man to father children is to extract immature spermatids from his testes and culture them in the laboratory, in conditions that may increase the number of them that develop into functioning sperm. These can then be used for IVF. An investigation was carried out to see if adding reproductive hormones to a culture of immature spermatids affects their development. Samples of spermatids were collected from men in whom the spermatids did not normally develop into functioning sperm. The spermatids were cultured in a suitable liquid medium, kept at 30 °C. The samples were divided into four groups. No hormones were added to one group. FSH, testosterone or both were added to the other groups. The percentage of spermatids that developed into elongated cells in each group after 24 hours and 48 hours was calculated. The results are shown in Table 4.1. Table 4.1 hormones added percentage of spermatids that developed into elongated cells after 24 hours after 48 hours none 20 21 FSH 32 31 testosterone 21 19 FSH and testosterone 39 44 (i) With reference to Table 4.1, describe the effects of adding reproductive hormones on the development of the spermatids after 48 hours. … … … … … … … … … … …[4]

13 9700/43/M/J/14 © UCLES 2014 [Turn over (ii) Suggest a reason for the apparent reduction in the percentage of elongated cells between 24 hours and 48 hours in some of the samples. … … …[1] (iii) Suggest why the culture medium was maintained at a temperature of 30 °C, and not at core body temperature (37 °C). … … … … … …[2] [Total: 16]

14 9700/43/M/J/14 © UCLES 2014 5 Deer mice, Peromyscus maniculatus, are small rodents that live in North America. Like all mammals, their blood contains haemoglobin which combines with oxygen in the lungs, and unloads its oxygen in respiring tissues. Deer mice show variation in their genotypes for the genes that code for the α-polypeptide chain of haemoglobin. In most populations of deer mice, the majority of individuals have the genotype A1A1, while a smaller number have the genotype A0A0. (a) In mice with the genotype A1A1, the amino acid at position 64 in the α-polypeptide chain is aspartic acid. In mice with the genotype A0A0, the amino acid at this position is glycine. Suggest how the change from aspartic acid to glycine in the α-polypeptide chain could have been brought about. … … … … … …[2] (b) The genotypes of deer mice from three different populations, each living at a different altitude, were analysed. Fig. 5.1 shows the relative proportions of deer mice with aspartic acid (white areas) and glycine (black areas) at position 64 in the α-polypeptide of their haemoglobin. altitude 4347 m 1158 m 620 m Fig. 5.1

15 9700/43/M/J/14 © UCLES 2014 [Turn over (i) Describe the effect of altitude on the frequency of the haemoglobin alleles in these populations of deer mice. … … … … … …[2] (ii) The partial pressure of oxygen is relatively low at high altitudes. Haemoglobin containing glycine at position 64 in the α-polypeptide chain has a higher affinity for oxygen than haemoglobin with aspartic acid at this position. Suggest how natural selection could account for the difference in allele frequency in deer mice living at high altitudes and low altitudes. … … … … … … … … … …[4] [Total: 8]

16 9700/43/M/J/14 © UCLES 2014 6 (a) The passage below outlines how sensory receptors work. Complete the passage by using the most appropriate scientific term(s). A sensory receptor cell responds to a stimulus by opening ion … in its cell surface membrane. Sodium ions flood into the cell causing the membrane to become … . This is called the … potential. If this potential is large enough to reach a … then an action potential is transmitted to the central nervous system. An increase in the strength of the stimulus will result in an increase in the … of action potentials transmitted. [5] (b) Describe how an action potential is transmitted along a sensory neurone in a mammal. … … … … … … … … … … … … … … …[5] [Total: 10]

17 9700/43/M/J/14 © UCLES 2014 [Turn over Question 7 starts on page 18

18 9700/43/M/J/14 © UCLES 2014 7 Occasionally during meiosis, homologous chromosomes fail to separate at anaphase. This is known as non-disjunction. Turner’s syndrome is the most common chromosome mutation in human females. It can occur due to non-disjunction in meiosis during gametogenesis. Some resulting gametes will be missing an X chromosome. Some forms of Turner’s syndrome occur when one of the pair of X chromosomes is not missing but has become damaged. The damaged X chromosome may have been broken and re-formed so that part of its structure is lost. Fig. 7.1 is a diagram of a normal X chromosome and two forms of ‘damaged’ X chromosomes, X1 and X2. • In X1, a section of the ‘p’ arm of the chromosome is missing. This deletion leads to reduced height of the female and abnormalities such as narrowing of the aorta. • In X2, a section of the ‘q’ arm of the chromosome is missing. This deletion leads to little or no development of the ovaries. K p q normal X X1 X2 p q p q Fig. 7.1 (a) Name structure K. …[1]

19 9700/43/M/J/14 © UCLES 2014 [Turn over (b) Explain why X1 and X2 result in different phenotypes. … … … … … …[2] (c) Mothers with the X1 form of Turner’s syndrome can pass on the chromosome mutation to their daughters but not to their sons. Complete the genetic diagram below to show how the chromosome mutation X1 may be passed on to daughters from a mother with Turner’s syndrome. parental female with normal male phenotypes Turner’s syndrome parental genotypes XX1 gametes genotypes of daughters phenotypes of daughters [4] [Total: 7]

20 9700/43/M/J/14 © UCLES 2014 8 (a) A mitochondrion contains DNA and ribosomes and is the organelle in which aerobic respiration takes place. Suggest the functions of the DNA and ribosomes in a mitochondrion. … … … … … … … …[3] (b) Oxidative phosphorylation takes place in the mitochondrion. Different stages of oxidative phosphorylation are listed below. They are not listed in the correct order. stage description of stage Q protons diffuse through the channel protein into the matrix R a proton gradient is set up across the crista S hydrogen atoms split into protons and electrons T protons combine with electrons and oxygen atoms to form water U electrons are passed from carrier to carrier V reduced NAD releases hydrogen atoms to cytochrome carriers W energy from electron transfer is used to pump protons into the intermembrane space X ATP synthase produces ATP Complete Table 8.1 to show the correct order of the stages. Two of the stages have been done for you.

21 9700/43/M/J/14 © UCLES 2014 [Turn over Table 8.1 correct order letter of stage 1 V 2 … 3 … 4 … 5 R 6 … 7 … 8 … [4] (c) ATP can be converted to ADP and inorganic phosphate by the enzyme ATPase. State the type of reaction taking place. … [1] (d) Some parasitic worms, such as tapeworms, live in a mammalian gut where there is no oxygen. Suggest how a tapeworm produces ATP in this environment. … … … … … … … … … … …[5] [Total: 13]

22 9700/43/M/J/14 © UCLES 2014 Section B Answer one question. 9 (a) It has been stated that the kingdom Protoctista can be described as a very diverse group of organisms that share only a few common features. Discuss the ways in which members of the kingdom Protoctista are similar to each other and ways in which they differ. [7] (b) With reference to any named species of plant or animal, explain why this species is considered to be endangered and outline the reasons that have caused it to become endangered. [8] [Total: 15] 10 (a) Describe the action of penicillin on bacteria. [8] (b) Outline the use of microorganisms in the extraction of heavy metals from their ores. [7] [Total: 15] … … … … … … … … … … … … … … … … …

23 9700/43/M/J/14 © UCLES 2014 [Turn over … … … … … … … … … … … … … … … … … … … … … … … … … … … …

24 9700/43/M/J/14 © UCLES 2014 … … … … … … … … … … … … … … … … … … … … … … … … … 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. 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.

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2014 series 9700 BIOLOGY 9700/43 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 43 © Cambridge International Examinations 2014 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 accepted) 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 Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 43 © Cambridge International Examinations 2014 1 (a) (i) stroma ; [1] (ii) lower CO2 concentration ; less, carbon fixation / CO2 combining with RuBP / RuBP converted to GP ; RuBP reformed from TP ; [max 2] (iii) 0.01 ;; A 0.012 or 1.8 ÷ 150 or 150 0.2 2.0 − or 200 350 0.2 2.0 − − for 1 mark [2] (b) less TP ; (so less) conversion to, (other) carbohydrates / lipids / amino acids / proteins ; A named examples, e.g. glucose / hexose / cellulose / starch AVP ; e.g. 1 – (amino acids) used to make proteins for, growth / cell division e.g. 2 – (carbohydrate / lipid) for respiration for, growth / cell division [max 2] [Total: 7] 2 (a) idea of cross-pollination involves two (parents) / self-pollination one (parent) ; ref. outbreeding / inbreeding ; (two parents) have different, genotypes / sets of alleles ; idea of new combinations of alleles in offspring ; [max 3] (b) (total) DNA / genome, cut into fragments ; by restriction enzymes ; DNA, denatured / made single stranded ; ref. primers / (modified) PCR ; ref. dideoxynucleotides / chain termination ; DNA / Taq, polymerase ; copies of different lengths produced ; electrophoresis ; A description detection, of fluorescence / by laser scanner ; sequence of, bases / nucleotides, read (by computer) ; [max 4]

Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 43 © Cambridge International Examinations 2014 (c) cross(-pollinate) them ; A description (if same species) offspring, are fertile / can themselves produce seeds; ora [2] [Total: 9] 3 (a) (i) AABBCC ; [1] (ii) meiosis unsuccessful (in, sterile hybrid / AB) ; gametes not formed ; bivalents cannot form / chromosomes cannot pair up / chromosomes are not homologous ; polyploidy occurs / chromosomes double ; A tetraploid failure of cell division / all chromosomes in one daughter cell ; A description chromosomes can now form pairs / gametes can be formed / meiosis can be completed ; [max 4] (b) (i) in presence of Eβf large number aphids, stop feeding / move ; in absence of Eβf, few / no, aphids, stop feeding / move ; air in Experiment 1, has other chemicals / not pure Eβf or air in Experiment 2 has only Eβf ; Eβf concentration in Experiment 2 may be unnaturally high or Eβf concentration unknown in Experiment 1 ; different volumes of air in Experiment 1 and Experiment 2 ; comparative data quote ; e.g. 55% versus 84% or 54 out of 99 versus 111 out of 132 54.5% versus 0.9% or 54 out of 99 versus 1 out of 113 84% versus 0% or 111 out of 132 versus 0 out of 106 [max 4]

Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 43 © Cambridge International Examinations 2014 (ii) Eβf stops aphids settling ; Eβf attracts, predators of aphids / ladybirds ; attacked aphids secrete more Eβf ; aphids not, eating / taking nutrients from, wheat ; [max 3] (iii) gene / Eβf, already in, peppermint / various plant species ; Eβf not, toxic / harmful to human health ; no new chemical added to human diet ; does not kill insects (unlike Bt maize or cotton) ; aphids still available for, predators / food web ; [max 3] [Total: 15] 4 (a) (i) spermatagonium – 2n primary spermatocyte – 2n secondary spermatocyte – n spermatids – n spermatozoan – n ;; all five correct for two marks three or four correct for one mark [2] (ii) (spermatogonium to primary spermatocyte) growth / mitosis ; (spermatid to sperm) maturation ; [2] (iii) any 1 from provide nutrients for sperm(atid) ; protect sperm from attack from immune system ; regulation of, sperm production / FSH ; AVP ; e.g. removes excess cytoplasm during sperm maturation / guides sperm to centre of tubule [max 1]

Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 43 © Cambridge International Examinations 2014 (b) FSH ; (hormone) given to stimulate follicle development ; GnRH agonists / GnRH receptor antagonists ; to prevent, LH surge / ovulation ; human chorionic gonadotrophin ; (hormone) given to stimulate maturation of oocytes ; (mature oocytes) collected from ovaries (just before ovulation) ; ref. use of, fine tube / needle / ultrasound ; [max 4] (c) (i) FSH (alone) / FSH + testosterone, increases development (of spermatids into, spermatozoa / elongated cells) ; testosterone (alone) has very little effect ; FSH + testosterone causes greatest increase of development ; use of, comparative / manipulated, figures ; [4] (ii) (reduction is very small so) may be, insignificant / random / due to chance ; (some cells) may have died ; [max 1] (iii) temperature, similar to testes / in range 30 °C to 35 °C / lower than core ; spermatozoa production, will not proceed at 37 °C / at high temperature ; [2] [Total: 16] 5 (a) random / spontaneous ; mutation ; base / nucleotide / triplet, change / substitution ; R addition / deletion [max 2] (b) (i) as altitude increases frequency of A0 increases ; ora for A1 A0 more frequent at high altitudes / A1 more frequent at low altitudes / intermediate frequency of either allele at intermediate altitude ; [2]

Page 7 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 43 © Cambridge International Examinations 2014 (ii) idea of (pre-existing) genetic variation in deer mouse population ; at high altitude mice with, glycine / A0, more likely to survive / have selective advantage ; ora mice (with A0 ) reproduce (at high altitude) ; ora and pass on the A0 allele ; ora partial pressure / concentration, of O2 acts as a selection pressure ; ref. to disadvantage of haemoglobin with very high affinity at low altitude ; as less able to unload oxygen (in respiring tissues) ; [max 4] [Total:8] 6 (a) channels ; I voltage-gated depolarised ; A positive inside receptor / generator ; threshold ; frequency ; A number per second / rate R speed [5] (b) action potential stimulates neighbouring area of membrane ; AW Na+, moves sideways / attracted to areas at resting potential ; A local circuit causes, Na+ ion channels to open / 2nd depolarisation ; (transmission) in one direction due to, hyperpolarisation / refractory period ; myelin sheath / Schwann cell ; sheath insulates, axon / dendron / neurone ; depolarisation / action potential, only at nodes of Ranvier / unmyelinated part ; ora saltatory conduction / action potential ‘jumps’ from node to node ; [max 5] [Total:10]

Page 8 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 43 © Cambridge International Examinations 2014 7 (a) centromere ; [1] (b) idea that different genes, are present / missing ; R alleles different, proteins / poypeptides, produced / missing ; [2] (c) XY ; X X1 X Y ; XX XX1 ; normal Turner’s ; [4] [Total:7] 8 (a) (DNA for) transcription / codes for mRNA ; (ribosomes for) translation ; synthesis of, respiratory enzymes / named enzyme / inner membrane proteins ; [max 3] (b) correct order letter of stage 1 V 2 S 3 U 4 W 5 R 6 Q 7 X 8 T S U W all above R ; S U W in correct order ; Q X T all below R ; Q X T in correct order ; [4]

Page 9 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 43 © Cambridge International Examinations 2014 (c) hydrolysis / dephosphorylation / exothermic / exergonic ; [1] (d) anaerobic respiration ; substrate level phosphorylation (in glycolysis); at triose phosphate pyruvate step ; (net) gain of 2ATP (per glucose) ; A 2 used and 4 produced pyruvate, reduced / gains hydrogens (from reduced NAD) ; forming lactate ; NAD regenerated / NADH2 re-oxidised ; this allows glycolysis to continue ; I ethanol pathway [max 5] [Total:13] 9 (a) similarities eukaryotic (cells) ; detail of eukaryotic cell ;; e.g. nucleus / linear DNA / chromosomes associated with histones / (named) membrane-bound organelles / 80S ribosomes differences single-celled or colonial / multicellular ; autotrophic or heterotrophic ; motile or unable to move ; cell wall or no cell wall ; vacuole or no vacuole ; different life cycles ; [max 7]

Page 10 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 43 © Cambridge International Examinations 2014 (b) fall in numbers ; danger of becoming extinct ; ref. (IUCN / International Union for Conservation of Nature) / red list ; one mark for idea, additional mark if qualified with point specific to named example e.g. habitat destruction ; detail ; climate change ; detail ; e.g. rise in temperature increase in disease ; detail ; increase in, predators / grazers ; detail ; decrease in food ; detail ; named pollutant and habitat affected ; detail ; hunting / killing / poaching / removal (plant) ; detail ; e.g. trade in animal parts, selling rare plants increased competition ; detail ; lack of human education ; detail ; disturbance to breeding sites ; detail ; [max 8] [Total:15]

Page 11 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2014 9700 43 © Cambridge International Examinations 2014 10 (a) bacteria walls made of peptidoglycans ; bacteria secrete autolysins ; make holes in cell wall / AW ; to allow wall to stretch during growth / AW ; (glycoprotein) peptidases form cross-links (between peptidoglycans) ; (penicillin) inhibits (glycoprotein) peptidases ; cross-links (between peptidoglycans) do not form ; cell wall weakened ; bacteria take in water by osmosis ; increased turgor pressure causes cell to burst ; AW AVP ; e.g. competitive inhibition [max 8] (b) ref. bioleaching ; Acidithiobacillus / A. ferrooxidans ; A Thiobacillus / T. ferrooxidans low grade ores / (mine) waste ; two metals ; e.g. copper, zinc, cobalt, uranium, lead, nickel, gold, silver I iron insoluble ore turned into soluble products ; ore piled up ; acidic conditions created / pH low(ered) / pH 1.5 – 3 ; different bacteria at different temperatures ; chemoautotrophic ; A description oxidation (reactions) ; sulfide / S 2– to sulfate / SO4 2– ; (direct oxidation of ore) Fe 2+ / ferrous
Fe 3+ / ferric ; Fe 3+ oxidise other ores ; product, drains / leaches / is washed, into pool ; metal displaced by adding scrap iron ; [max 7] [Total:15]