Cambridge A Level Biology 9700 — 2019 May/June Paper 2 · Variant 3

1 Fig. 1.1 is a diagram showing the structure of a section of a DNA molecule. X Key hydrogen bond Fig. 1.1 (a) Draw a circle around one monomer of DNA in Fig. 1.1. [1] (b) Name the two bases forming the base pair at X in Fig. 1.1 and give a reason for your answer. bases ........................................................................................................................................ reason ....................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (c) The statements 1–5 describe events that occur during DNA replication. 1 DNA polymerase forms a phosphodiester bond 2 DNA double helix forms 3 hydrogen bonds break 4 hydrogen bonds form 5 two strands of the double helix separate Write the numbers 1 to 5 in the spaces below to show the order in which these events occur. The first one has been done for you. 3 ................ ................ ................ ................ ................ [1] (d) The telomere is a region found at the end of a chromosome. Outline the function of telomeres. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 6]

Mark scheme: 1(a) one nucleotide circled ; 1 1(b) cytosine and guanine ; three hydrogen bonds between the base pairs ; 2 1(c) (3) 5,4,1,2 ; 1 1(d) I maintains length of chromosomes / prevents chromosome shortening any two from: permit continued replication ; prevent loss of genes ; I prevents gene damage A genetic / coded, information A protein coding regions of DNA protect ends of chromosomes from being, degraded / AW ; AVP ; e.g. prevents ends of chromosomes from being attached to each other not mistaken for a break in DNA that needs repairing 2

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2 Fig. 2.1 is a photomicrograph showing a transverse section through a bronchus and the surrounding lung tissue. R PP QQ magnification x40 Fig. 2.1 (a) Calculate the actual diameter of the lumen of the bronchus using the line P–Q. Write down the formula used to make your calculation. Show your working and give your answer to the nearest micrometre (µm). formula actual diameter = ................................................... µm [2] (b) (i) In Fig. 2.1, the area labelled R shows a section through some alveoli. Name the tissue that lines the air spaces of the alveoli. ..................................................................................................................................... [1] (ii) Suggest how the alveoli are adapted for gas exchange. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (c) In the leaves of plants, both gas exchange and transpiration occur through open stomata. Suggest how the arrangement of cells in the leaf of a dicotyledonous plant contributes to the loss of water by transpiration. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (i) Describe the data shown in Fig. 2.2. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) Suggest why the transpiration rate changed between 09:00 and 10:00. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 13]

Mark scheme: 2(a) correctly stated formula ; e.g. actual diameter = image length / magnification magnification = image length / actual diameter allow a magnification triangle 1750 (µm) ; (for 70 mm) credit also measurements made to 0.5 mm A 1700 (µm) (for 68 mm) A 1725 (µm) (for 69 mm) A 1775 (µm) (for 71 mm) A 1800 (µm) (for 72 mm) 2 2(b)(i) squamous epithelial (tissue) ; A pavement epithelial (tissue) 1 Question Answer Marks 2(b)(ii) any two from: (alveoli have) thin wall / wall is one cell thick ; A wall composed of thin cells R thin membrane R thin cell wall idea of short diffusion distance ; in context of between alveolar space and, blood / capillary elastic tissue, qualified ; A elastin / elastic fibres e.g. stretch and recoil for, ventilation / inhalation and exhalation maintaining steep, diffusion / partial pressure, gradient (many) alveoli provide a large surface area (for diffusion) ; I idea that a single alveolus has a large surface area AVP ; surfactant, prevents alveolar collapse / maintains inflation I ref. to moist lining 2 2(c) any three from: 1 loosely packed spongy mesophyll cells / spongy mesophyll cells surround air spaces / air spaces between spongy mesophyll cells ; 2 large, surface / area, for evaporation (of water to air space) ; In context of spongy mesophyll cells, surfaces / cell walls 3 (cells arranged so that), air spaces above the stomata / sub-stomatal air spaces ; 4 ref. to evaporation creates saturated air spaces / water vapour collects in air spaces ; 5 idea of creates water potential gradient between air in air spaces and air in external environment ; 6 AVP ; e.g. guard cells situated in epidermis to form stomatal pore ref. to xylem, close proximity to spongy mesophyll cells for passage of water / passes on water to spongy mesophyll cells 3 Question Answer Marks 2(d)(i) accept au for arbitrary units and first hour etc. for 08.00 etc. any three from: 1 between 08.00 and 09.00, steep increase / increase from 2.5 au to 5 au ; 2 highest rate / peak, at 09.00 ; A maximum rate is at 09.00 3 fluctuates / AW, between 10.00 and 15.00 ; 4 (continued) decrease between 15.00 and 19.00 ; A steep /steepest, decrease between 15.00 and 16.00 A less steep (than 15.00 to 16.00) decrease between 16.00 and 19.00 5 comparative data quote for decrease ; e.g. 3.8 au at 15.00 to 0.5 au at 19.00 or 3.8 au at 15.00 to 2.00 au at 16.00 or 2.0 au at 16.00 to 0.5 au at 19.00 3 2(d)(ii) any two from: change in external environmental factor decrease in light intensity / AW ; A (less sunlight because it was) cloudy decrease in temperature / AW ; temperature decreased because it was cloudy is mp2 only decrease in, wind speed / air movement / AW ; increase in humidity / raining / AW ; stomata, decrease in (aperture) size / open less wide ; decreased evaporation (rate) ; water vapour diffuses out less rapidly / less steep water potential gradient ; AW AVP ; e.g. suggestion of xerophytic feature occurring in response to increased, temperature / light intensity (to 09.00) 2

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3 Cholera is an infectious disease caused by the bacterium Vibrio cholerae. The pathogen causes disease by disrupting the activity of the epithelial cells in the human intestine. (a) Complete Table 3.1 to compare the structure of V. cholerae and an epithelial cell from the human intestine. Use a tick (3) to indicate the presence of the structure. Use a cross (7) to indicate the absence of the structure. The first row has been completed for you. Table 3.1 epithelial cell from the cell structure V. cholerae human intestine 3 7 cell wall cell surface membrane ribosomes large permanent vacuole organelles surrounded by a double membrane [4] (b) Describe the methods that can be used to prevent the spread of cholera. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [5] (c) The World Health Organization has suggested that people with HIV/AIDS take a longer time to recover from cholera and are at an increased risk of death from cholera. The human immunodeficiency virus (HIV) only infects certain types of cell. These cells have CD4 receptor proteins in their cell surface membrane. Helper T-lymphocytes have CD4 receptor proteins. (i) Fig. 3.1 is a diagram of HIV showing the glycoprotein gp120. This glycoprotein is embedded in a membrane envelope which surrounds the viral protein coat. envelope gp120 protein coat genetic material Fig. 3.1 The glycoprotein gp120 is important in allowing HIV to only infect certain types of cell. Suggest the role of gp120. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) People with HIV/AIDS have a low helper T-lymphocyte count. Explain how a low helper T-lymphocyte count could reduce the body’s ability to produce antibodies against V. cholerae. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 13]

Mark scheme: 3(a) one mark each row cell structure V. cholerae epithelial cell from the human intestine cell wall 9 8 cell surface membrane 9 9 ; ribosomes 9 9 ; large permanent vacuole 8 8 ; organelles surrounded by a double membrane 8 9 ; 4 Question Answer Marks 3(b) I better sanitation unqualified I ref. to unclean water any five from : ref. to treat drinking water ; A clean / sterilised / chlorinated / safe / bottled / boiled A ref. to water treatment plants provide sewage treatment plants ; A treat waste water / containment of sewage / provide latrines (in temporary camps) ; ref. to keeping sewage and (drinking) water sources separate ; A examples e.g. avoid swimming downstream of sewage outlets avoid defaecating near rivers used for washing water treatment plants upstream of sewage disposal bury faeces ref. to vaccination / providing vaccines ; ref. to education in ways to prevent spread ; ref. to (encourage) hand washing ; ref. to food hygiene ; e.g. use of clean cooking utensils / covering food to prevent flies landing / washing food in treated water / cook food thoroughly / peel fruit and vegetable washed in (suspected) contaminated water discourage use of human faeces for fertiliser / use fertiliser other than human faeces / do not irrigate plant food with contaminated water; control breeding of, vectors / flies ; rapid diagnosis ; rapid treatment (for earlier recovery to minimise risk of spreading) ; ref. to oral rehydration, salts / therapy ; use of antibiotics ; ref. to isolation / quarantine ; 5 Question Answer Marks 3(c)(i) to, bind / attach / AW, HIV / virus / pathogen, to, host cell / T-lymphocyte / T-cell ; (because it) binds / is complementary, to, CD4 receptor proteins / AW ; ref. to binding leads to, entry of virus into cell / endocytosis / fusion of viral envelope with host cell (surface) membrane ; 2 3(c)(ii) ref. to less cytokine (released) ; in context of fewer helper T-lymphocytes any one from: (so) fewer B-lymphocytes divide by mitosis / AW ; A humoral response decreased (so) fewer B -lymphocytes stimulated to differentiate into plasma cells / AW ; (so) fewer plasma cells to, produce / secrete antibodies ; 2

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4 Phloem sieve tube elements and xylem vessel elements are cells that are involved with the transport of substances in plants. (a) Describe two differences between the structure of a phloem sieve tube element and a xylem vessel element. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Sucrose is transported in phloem sieve tubes. A student carried out a test to identify the presence of sucrose in a sample of sap taken from inside a phloem sieve tube. One of the steps in the procedure instructed the student to heat the phloem sap with hydrochloric acid. Explain why it is necessary to carry out this step in the procedure. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (c) The process of loading sucrose into a phloem sieve tube element involves a companion cell. (i) Fig. 4.1 shows a stage in the process of loading sucrose into the phloem. cell surface membrane of companion cell cytoplasm cellulose cell wall of companion cell of companion cell ATP H+ H+ ADP + Pi not to scale Fig. 4.1 Companion cells contain large numbers of mitochondria and ribosomes. Explain the roles of mitochondria and ribosomes for the stage shown in Fig. 4.1. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Describe and explain how sucrose is transported from the cell wall of the companion cell into the cytoplasm. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3]

Mark scheme: 4(a) any two from: note that some mark points must have both xylem and phloem mentioned (‘v’) and some mark points only need xylem or phloem (‘or’ – below the line) xylem phloem no cytoplasm / hollow / no contents v (peripheral / little / some) cytoplasm I protoplasm R full of cytoplasm / AW or has (a few) organelles A examples of organelles mitochondria or ER I has SER / RER / ribosomes R has nucleus ; lignified R idea of lignin within element v no lignin / (only) cellulose ; no end wall(s) / no sieve plate(s) A end walls have broken down or sieve plate(s) / perforated end wall (s) A (end walls) have sieve pores (side walls) contain pits R piths or no pits ; ; no plasmodesmata or plasmodesmata (to companion cells) ; ; thick(er), cell wall / walled or thin(ner), cell wall / walled ; thicker lignified wall = thicker wall mark only (for lignin mark need to state what phloem has) 2 Question Answer Marks 4(b) any two from: to hydrolyse / break, glycosidic bonds ; to produce, reducing sugars / fructose and glucose ; A to make reactive groups (to Benedicts) available to be able to test with Benedict’s solution to obtain a positive result ; A description of positive result sucrose, is a non-reducing sugar / has no reactive groups when tested with Benedict’s solution ; 2 4(c)(i) (large number of mitochondria) provide ATP, for active transport / for protein synthesis ; (large number of ribosomes) to synthesise the, proton pumps / carrier proteins / AW ; 2 4(c)(ii) any three from: H+, pumped / actively transported (out of companion cell into cell wall) or ref. to higher concentration of H+ (in cell wall) / H+ gradient builds up (between cell wall and cytoplasm of companion cell) ; H+ move back (into companion cell) by facilitated diffusion ; I sucrose through, co-transporter protein / cotransporter ; A symporter H+ cotransports sucrose / sucrose moves (into cell) with H+ ; sucrose is transported against the concentration gradient ; A sucrose transported by secondary active transport 3

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5 (a) Fig. 5.1 is a photomicrograph of human blood cells that have been placed in a solution of sodium chloride. Y crenated red blood cell Fig. 5.1 (i) State the function of the cell labelled Y. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) The red blood cell labelled in Fig. 5.1 is described as crenated because it has an abnormal, shrivelled appearance. Explain how this red blood cell has become crenated. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (b) Researchers studied the effect of altitude on the red blood cell count of humans. They recorded the red blood cell count of healthy adults from populations living at four different altitudes in a mountainous region. The results are shown in Table 5.1. Table 5.1 altitude of population mean red blood cell count population / m above sea level / number of cells ×106 mm−3 A 695 4.50 B 1676 5.14 C 2003 5.04 D 2118 5.11 (i) Calculate the percentage increase in mean red blood cell count between population A and population D. .......................................................% [1] (ii) Describe and suggest explanations for the data shown in Table 5.1. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [4] (c) Carbonic anhydrase is an enzyme found in red blood cells which has an important role in the transport of respiratory gases. Explain why a non-competitive inhibitor of carbonic anhydrase will reduce the supply of oxygen to actively respiring tissues. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 10]

Mark scheme: 5(a)(i) phagocytosis (of, dead cells / pathogens / AW) ; A endocytosis of pathogens 1 5(a)(ii) (sodium chloride) solution has lower water potential (than cytoplasm) ; ora A Ψ for water potential water, leaves / AW, cell by osmosis ; 2 5(b)(i) 13.56 ; A 13.6 / 14 1 Question Answer Marks 5(b)(ii) allow (red) blood cell count for mean red blood cell count allow ora any four from: 1 description of data ; e.g. lowest mean red blood cell count for, (population) A / lowest altitude (populations) B, C and D / all other altitudes, have a higher mean red blood cell count than A / 695m altitude (population) B has highest mean red blood cell count (population) B has higher mean red blood cell count than C and D physiological explanation population A compared to B/C/D 2 decrease in partial pressure of oxygen in the atmosphere with increase in altitude / AW or lower partial pressure of / less, oxygen in, inhaled air / lungs / alveoli AW 3 ref. to haemoglobin ; e.g. lower saturation of haemoglobin with oxygen (in lungs) lower affinity of haemoglobin for oxygen (in lungs) more haemoglobin required to carry sufficient oxygen to tissues 4 suggestion of increase in count at 1676 m and above ; e.g. rate of production of red blood cells increases higher erythropoietin (EPO) levels ref. to changed proportion of plasma to red blood cells / plasma volume decreases 5 little difference between B, C and D / AW ; 6 differences in altitude are not great / AW ; in context of B, C, D 7 ref. to genetic differences / adaptation to environment ; allow if general statement includes population A 8 AVP ; e.g. ref. to differences in demographics 4 Question Answer Marks 5(c) less haemoglobinic acid is formed / fewer hydrogen ions to bind to haemoglobin (to cause dissociation of oxygen) ; any one from: binds to allosteric site / site other than active site (of enzyme), qualified ; e.g. causes change in shape of active site substrate cannot bind to active site fewer / no, enzyme-substrate / ES, complex formation prevents / reduces / AW, carbonic acid / H2CO3, formation / AW ; A fewer hydrogen ions produced from carbonic acid dissociation 2

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6 Fig. 6.1 is a diagram of a vertical section of the human heart and the associated blood vessels. L P R S M T N Fig. 6.1 (a) Complete Table 6.2 by writing the letter from Fig. 6.1 that identifies each of the structures described. Table 6.2 structure in the heart letter valve preventing back flow of blood into a ventricle blood vessel carrying blood under highest pressure chamber that pumps blood to the lungs [3] (b) Describe the short-term effects of nicotine on the cardiovascular system. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] Question 6 continues on page 16 (c) Cardiac myocytes are heart muscle cells. Fetal cardiac myocytes are present before birth and divide rapidly during fetal development. Soon after birth the cell cycle in most of these fetal cardiac myocytes stops, forming adult cardiac myocytes. Most of the adult cells have completed the cell cycle but in some the cell cycle stops at the end of mitosis. (i) State the difference between adult cardiac myocytes that have completed the cell cycle and those that have stopped at the end of mitosis. ..................................................................................................................................... [1] (ii) Suggest why it could be a disadvantage that the cell cycle has stopped in most adult cardiac myocytes. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 9]

Mark scheme: 6(a) structure in the heart letter valves preventing back flow of blood into the ventricle S ; blood vessel carrying blood under highest pressure P ; chamber that pumps blood to the lungs N ; 3 Question Answer Marks 6(b) treat as neutral effects of nicotine on the brain and ref. to adrenaline any three from: increased heart rate ; I increased heart beat increased blood pressure ; increases risk of, blood clotting / thrombus formation ; A increases risk of thrombosis A increases stickiness of platelets damages the endothelium ; AVP ; e.g. stimulates vasoconstriction / reduces the diameter of blood vessels A named blood vessel types 3 6(c)(i) any one from: two nuclei vs one ; ref. to, DNA / chromosome, content ; cytokinesis has not occurred ; A cytoplasm has not divided (into two cells) 1 6(c)(ii) any two from: in context of cardiac myocyte or overall heart function unable to replace, damaged / worn out / old, cardiac myocytes ; R repair myocytes unable to repair (damaged) cardiac, muscle / tissue ; A heart tissue repair (to cardiac muscle may be) with, unspecialised cells / scar tissue ; 2

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