Cambridge A Level Biology 9700 — 2018 Oct/Nov Paper 2 · Variant 3

1 (a) Aphids are small insects which feed directly on phloem sap. The salivary glands of aphids have secretory cells that make and release a variety of proteins that assist in feeding. Fig. 1.1 is a transmission electron micrograph of a small area of a salivary gland cell of an aphid. mitochondrion Golgi bodies Fig. 1.1 Describe the role of Golgi bodies in secretory cells, such as the salivary gland cells of aphids. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] (b) (i) Explain why secretory cells have large numbers of mitochondria. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (ii) Mitochondria are partly controlled by the nucleus, but can also function independently. Suggest the features of mitochondria that allow them to function independently of the nucleus. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (c) Aphids are important vectors of plant viral diseases. (i) Describe the structure of a typical virus. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (ii) Suggest how viruses are able to pass from one plant cell to the next without crossing membranes. ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [1] [Total: 11]

Mark scheme: 1(a) any three from 1 modification / process / described, of, proteins / polypeptides ; 2 further detail of modification ; examples of modification folding of polypeptides / protein folding assembly of polypeptides to form quaternary structure addition of (named) prosthetic group addition of sugars / addition of carbohydrate / glycosylation cutting of polypeptides removal of, methionine / first amino acid making proteins functional 3 packaging into vesicles / formation of Golgi vesicles / formation of secretory vesicles ; A ‘budding off’ / ‘breaking off’ / transport proteins in Golgi vesicles 4 formation of (primary) lysosomes ; 5 AVP ; e.g. modification of lipids synthesis of, phospholipids / glycolipids synthesis of cell wall polysaccharides 3 1(b)(i) (mitochondria) provide / make / produce, ATP ; A needs a lot of, ATP / energy R ‘produce / make, energy’ (ATP / energy required for) protein synthesis / movement of (secretory / Golgi) vesicles / exocytosis ; 2 Question Answer Marks 1(b)(ii) any two from 1 mitochondria have (circular) DNA ; 2 mitochondria, have / make, (70S) ribosomes (for, protein synthesis / translation) ; R if 80S ribosomes 3 mitochondria can divide ; R by mitosis 4 AVP ; 5 AVP ; e.g. (mitochondrial / mt) DNA codes for (some mitochondrial) proteins mRNA transcribed from mtDNA mitochondria produce their own tRNA can replicate DNA I mitochondria have a double membrane 2 1(c)(i) any three from protein coat / capsid ; A protein layer R ‘cell wall of protein’ nucleic acid (core) ; R if ‘in a nucleus’ DNA or RNA ; acellular / non-cellular / ‘not a cell’ ; size – accept within range 15 nm to 1000 nm ; AVP ; e.g. (protein coat made of) capsomeres surrounded by, membrane / envelope / (phospho)lipid bilayer I antigens / enzymes 3 1(c)(ii) any one from (viruses) pass through plasmodesmata ; A ‘cytoplasmic strands, through cell walls / between cells’ via symplast pathway ; 1

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2 (a) Proteins are macromolecules composed of many amino acids. (i) Two amino acids are represented in the diagram in Fig. 2.1. Complete the diagram to show how the two amino acids react together to form a dipeptide. H R1 O H R2 O N C C + N C C H OH H OH H H Fig. 2.1 [3] (ii) State what is represented by R1 and R2 in Fig. 2.1. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [2] (b) Amylose and cellulose are polysaccharides. Fig. 2.2 shows the structure of part of a cellulose molecule. H OH CH2OH H OH CH2OH H H O H H O O O O O O H O H H O H CH2OH H OH CH2OH H OH Fig. 2.2 With reference to Fig. 2.2, state how the structure of a cellulose molecule differs from the structure of an amylose molecule. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (c) Cellulose is the main component of plant cell walls. Explain why cellulose is suitable as a component of plant cell walls. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] [Total: 11]

Mark scheme: 2(a)(i) water shown formed from –OH and –H ; A H2O formed without indicating where from peptide bond shown correctly between C of carboxylic acid and N of amino group ; complete dipeptide drawn ; A hydrogen ‘up or down’ 3 2(a)(ii) R group / side chain / variable group / residual group / functional group ; indicate, different amino acids / type of amino acid ; A specific to each amino acid A the two amino acids are different A two examples of R groups, e.g. –H and –CH3 both amino acids have different R groups = 2 marks 2 2(b) any two from straight chain / linear, v helix / helical ; I coil(ed) R branched / branching β-glucose, not α-glucose ; A β-1:4 glycosidic bond v α-1:4 glycosidic bond (β-)glucose / monomers / residues, are arranged, rotated 180° to each other / AW ; AVP ; e.g. more hydrogen bonds 2 Question Answer Marks 2(c) any four from 1 (molecules) form fibrils and fibres ; 2 hydrogen bonding between (cellulose) molecules ; 3 (cellulose molecule) is straight / linear ; 4 (straight chain allows) molecules lie parallel to each other ; mp4 dependent on mp3 5 gives strength (to cell wall) to, prevent cell bursting / withstand (turgor) pressure / AW ; R if only in context of a cellulose molecule 6 ref. to fibres at angles / criss-cross / AW ; 7 idea of many gaps, in wall / between fibres, allowing passage of water / named) substances / making cell wall permeable ; I plasmodesmata R partially / AW, permeable 8 AVP ; e.g. cellulose is insoluble many –OH groups (for hydrogen bonding) 4

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3 (a) Fig. 3.1 is a diagram of a monomer of the nucleic acid, messenger RNA. D F E Fig. 3.1 (i) Name D, E and F in Fig. 3.1. D ........................................................................................................................................ E ........................................................................................................................................ F ........................................................................................................................................ [3] (ii) State one way in which the structure of DNA differs from the structure of messenger RNA. ........................................................................................................................................... ...................................................................................................................................... [1] (b) Telomeres are repeating sequences of bases located at the ends of DNA molecules. These repeating sequences do not code for proteins. The enzyme telomerase ensures that telomeres do not shorten each time DNA is replicated. Fig. 3.2A shows the end of a DNA molecule during replication. DNA polymerase cannot attach to the region labelled X, so it cannot complete the synthesis of the new strand without the action of telomerase. Telomerase synthesises additional lengths of DNA that are added to the telomere. These additional lengths are used by DNA polymerase to complete the process of replication. Fig. 3.2B is an enlarged view of region X to show the action of the enzyme telomerase. A X 5’ 3’ T T A G G G T T A G G G T T A G G G T T A G G G T T A G A A T C C C A A T C C C A A T C C 3’ 5’ B G C A free nucleotides A T C T telomerase G T T A G C A A U C C C A A U C template RNA Fig. 3.2 Telomerase contains a short length of RNA that acts as a template for the synthesis of DNA as shown in Fig. 3.2B. Explain how a molecule of telomerase synthesises additional lengths of DNA. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] (c) Telomerase is not present in prokaryotic cells. Suggest why prokaryotes do not have telomerase. ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [1] (d) One of the ways to diagnose lung cancer is to determine the concentration of telomerase in cells from the lining of the bronchus. Explain why determining the activity of telomerase may be useful in the diagnosis of lung cancer. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] [Total: 11]

Mark scheme: 3(a)(i) D (nitrogenous organic / nucleotide) base ; A pyrimidine / uracil / cytosine R purine E ribose ; I pentose F phosphate (group) ; 3 3(a)(ii) any one from two, strands / polynucleotides (not one) ; A two chains R polypeptides deoxyribose (not ribose) ; A correct ref. to –OH (on C2) instead of –H group on the pentose / sugar thymine (not uracil) ; R thiamine have base pairs ; (double) helix / helical (not straight chain) ; longer ; 1 3(b) any four from 1 template RNA binds to (part of) region X ; 2 in active site of, telomerase / enzyme ; 3 DNA / free, nucleotides, pair with / bind to / align with, RNA ; 4 ref. to complementary, bases / base pairs ; 5 A–T and C–G and U–A ; 6 phosphodiester bonds form, between (DNA) nucleotides / AW ; treat as neutral any refs. to likely enzymes, e.g. ligase, etc. 7 telomerase moves, in the direction of the arrow / to the right ; 8 AVP ; e.g. hydrogen bonds between bases telomerase acts as a reverse transcriptase 4 Question Answer Marks 3(c) prokaryotes have, circular DNA ; A ‘loop of DNA’ I plasmid(s) do not have telomeres ; do not have chromosomes ; AVP ; e.g. ‘DNA has no ends’ 1 3(d) any two from cancer cells divide, uncontrollable / continuously ; ORA A cancer cells divide but do not undergo cell death telomerase is required to ensure, chromosomes, do not shorten ; telomerase is, synthesised / produced / found, in cancer cells ; ORA high levels of telomerase indicate cells that are cancerous / AW ; AVP ; e.g. idea that gene for telomerase switched on 2

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4 (a) Describe the roles of the sinoatrial node (SAN) and the atrioventricular node (AVN) in the initiation and control of the cardiac cycle. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] (b) The Purkyne fibres pass down the septum and extend to the cardiac muscle at the base (apex) of the heart. Explain why it is important that the Purkyne fibres extend to the base of the heart. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [2] (c) The activity of the SAN is controlled by the nervous system. Noradrenaline is released by nerve cells in the SAN. Fig. 4.1 shows the role of noradrenaline in causing calcium ions (Ca2+) to enter a cell in the SAN. noradrenaline phospholipid bilayer Ca2+ Ca2+ Ca2+ Ca2+ small gap cell surface membrane of SAN cell Ca2+ ATP cAMP (second messenger) active enzyme opens channel protein inactive enzyme active enzyme Fig. 4.1 With reference to Fig. 4.1, outline the process of cell signalling. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [4] [Total: 10]

Mark scheme: 4(a) any four from I signals R nerve impulse first time SAN 1 acts as a pacemaker / initiates heart beat / initiates cardiac cycle ; A regulates heartbeat A described, e.g. as rhythm / emits impulses at regular intervals 2 releases / AW, waves of excitation / (electrical) impulses ; A ref. to, action potentials / depolarisation R nerve impulses 3 spread across / AW, atria / atrial walls or leads to atrial, systole / contraction(s) ; AVN 4 allows a (short) delay / ~0.1 s ; 5 passes the impulse / wave of excitation, to the Purkyne fibres / down the septum ; A Bundle of His R nerve impulse 6 detail ; e.g. so atria contract before ventricles allows ventricles to fill / allow atria to empty completely so atria have, emptied / contracted, before ventricular contraction begins so atria and ventricles don’t contract at the same time 4 4(b) any two from contraction / ventricular systole, begins at the base of the ventricles ; both ventricles contract at the same time ; blood is forced, upwards / into arteries / through (named) semi-lunar valves ; A pushed idea that so ventricles, empty / pump out most of the blood ; AVP ; e.g. so impulses / waves of excitation, travel upwards from the base 2 Question Answer Marks 4(c) any four from 1 noradrenaline is, signalling, molecule / compound ; A noradrenaline is the signal I noradrenaline is a (neuro)transmitter 2 SAN cell is the target cell ; 3 noradrenaline / signalling molecule, binds to (cell surface) receptors (of target cell) ; A ‘fits’ R ‘receptor cells’ 4 receptors are specific to noradrenaline ; A ref. to complementary (shapes) 5 ref. to formation of cAMP / second messenger or activating enzyme ; 6 response is the opening of channel proteins (so calcium ions enter) ; 4

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5 The bacteria that cause tuberculosis (TB) can be found in many parts of the body including the lungs. (a) (i) State the name of the bacterium that causes TB. ...................................................................................................................................... [1] (ii) The presence of the pathogen in the lungs attracts phagocytes to the area of infection. The phagocytes release elastase, which digests elastin. Many people with TB feel tired all the time. Suggest and explain how the effect of phagocytes on tissues in the lungs leads to people feeling tired all the time. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...................................................................................................................................... [3] (b) Discuss the biological factors and social factors that make TB a difficult disease to control. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [5] [Total: 9] Question 6 starts on page 14

Mark scheme: 5(a)(i) Mycobacterium, tuberculosis / bovis ; 1 5(a)(ii) any three from 1 phagocytes / macrophages / neutrophils / elastase, breaks down / weakens, alveolar walls ; 2 (reduced elastin so) alveoli cannot stretch and recoil / alveoli overstretch ; 3 alveoli burst / one large air sac formed ; 4 surface area (for gas exchange) is reduced ; 5 less oxygen absorbed by blood / reduced oxygen supply (to tissues) / not enough oxygen for (aerobic) respiration / AW ; 3 5(b) R cost any five from 1 Transmission ; e.g. aerosol / droplet, infection I airborne A lack of pasteurised milk overcrowded, housing / living conditions / AW migration / travel, from areas with high rates of TB reservoir of infection in people not, diagnosed / not treated 2 HIV / immunodeficiency ; e.g. infection with HIV may activate (dormant) TB (pathogen) immunodeficiency / AW, makes people more susceptible to TB intravenous drug taking increases risk of, immunodeficiency / HIV 3 Diagnosis is difficult ; e.g. TB (pathogen) remains dormant / symptomless carriers limited access to health care for diagnosis / AW idea that contact tracing is difficult 5 Question Answer Marks 5(b) 4 Treatment ; e.g. treatment is long term / treatment may not start early enough people may not finish the course of treatment limited access to health care for treatment 5 Acquired resistance ; e.g. bacterium / TB (pathogen), develops resistance to, antibiotic(s) / drug(s) multi-drug resistance / MDR-TB / XDR-TB 6 Vaccination ; e.g. vaccine may not always be effective in providing protection difficult to achieve herd immunity concerns / misconceptions, about vaccination limited access to vaccination (programmes) poor (immune) response in people who are malnourished poor thermostability of vaccine 7 Education ; e.g. limited education / low awareness, about prevention for general population limited education / low awareness, for health care professionals 5

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6 (a) As part of a study of the mitotic cell cycle, a student made stained sections of a root tip of onion, Allium cepa, and observed them with a light microscope. The student made drawings of six of the cells, A to F, using the high power of the microscope, as shown in Fig. 6.1. A D B E C F Fig. 6.1 (i) Complete Table 6.1 to show the sequence of stages in the mitotic cell cycle, using the letters, A to F, as shown in Fig. 6.1. Table 6.1 sequence of stages cell 1 A 2 3 4 5 6 [1] (ii) Table 6.2 shows some events that occur during the mitotic cell cycle in A. cepa. Complete Table 6.2 by naming the stage of the cell cycle when each event occurs. Table 6.2 event in the cell cycle name of the stage in the cell cycle DNA replication division of centromeres condensation of chromatin contraction of spindle fibres organisation of metaphase chromosomes at the equator [4] (b) Explain the importance of mitosis in the immune response. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... .............................................................................................................................................. [3] [Total: 8]

Mark scheme: 6(a)(i) [A] C E D B F ; 1 6(a)(ii) I early and late except where indicated event in the cell cycle name of the stage in the cell cycle DNA replication interphase / S phase ; division of centromeres anaphase ; R late anaphase condensation of chromatin prophase ; contraction of spindle fibres anaphase ; organisation of chromosomes at the equator metaphase 4 Question Answer Marks 6(b) I antibodies have receptors any three from 1 clones / AW, of B/T cells consist of very few cells ; AW = types of / specific / groups of 2 (mitosis occurs) during clonal expansion ; 3 many, B cells / plasma cells, to make antibodies ; 4 many, T-helper / Th, cells to secrete, cytokines / lymphokines / interleukins ; 5 many, T-killer cells / T-cytotoxic cells, to kill, infected cells / cancer cells ; 6 memory cells are produced for, secondary response / faster response (when antigen encountered again) ; A during a secondary infection (implying of the same pathogen) 7 need to make genetically identical cells ; 3

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