Cambridge A Level Biology 9700 — 2019 May/June Paper 4 · Variant 3

1 (a) Fig. 1.1 shows a cholinergic synapse. presynaptic knob A B D C Fig. 1.1 Complete Table 1.1, using the letters A, B, C or D from Fig. 1.1, to show the location of compounds and structures associated with a cholinergic synapse. You may use A, B, C and D once, more than once, or not at all. Table 1.1 compound or structure location acetylcholine .......................... voltage-gated channel .......................... receptor protein .......................... acetylcholinesterase .......................... [4] (b) Explain what is meant by a voltage-gated channel. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (c) Explain the role of acetylcholinesterase in a synapse. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (d) Outline the roles of synapses in the nervous system. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 11]

Mark scheme: 1(a) compound / structure location acetylcholine C and D ; voltage-gated channel A ; receptor protein B ; acetylcholinesterase C ; 4 1(b) any two from: 1 ref. to ion (transport) ; 2 open / closes, when, voltage / depolarisation / charge / electrical potential, changes ; A arrival of an action potential 3 detail ; e.g. ion specificity / hydrophilic pore / transmembrane / (made of) protein 2 1(c) any three from: 1 breaks down acetylcholine ; 2 (so) acetylcholine leaves, binding site / receptor / B ; ora 3 depolarisation stops in post-synaptic membrane / B ; 4 stops continuous action potentials (in post-synaptic membrane) ; 5 recycle ACh / described ; 3 Question Answer Marks 1(d) any two from: 1 one-way transmission ; 2 interconnection of nerve pathways ; 3 AVP ; e.g. memory / learning integration of impulses 2

2 Albinism is a condition that results from disruption of a biosynthetic pathway involving the enzyme tyrosinase. It is estimated that 1 in 17 000 people in the world has albinism. (a) A recessive mutation in the TYR gene, which codes for the enzyme tyrosinase, is one cause of albinism. Individuals with this form of albinism are homozygous recessive. Describe the phenotype of a person with albinism. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) A recessive mutation in a different gene causes a type of albinism that mainly affects the eyes (ocular albinism). A person with this condition has reduced clarity of vision and involuntary eye movements. Fig. 2.1 shows the pattern of inheritance of ocular albinism in one family. The pattern indicates sex-linked inheritance. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Key: = unaffected female = carrier female = unaffected male = male with ocular albinism Fig. 2.1 (i) Explain why Fig. 2.1 supports sex-linked inheritance of ocular albinism. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) Draw a genetic diagram to show how individuals 1 and 2 cannot have a child with ocular albinism. key to symbols .................................... .................................... parental genotypes gametes offspring genotypes offspring phenotypes [4] (iii) Ocular albinism may be caused by a base deletion mutation. This mutation results in a non-functional protein. Explain how a base deletion mutation can result in a non-functional protein. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (iv) Ocular albinism is a non-progressive disorder and clarity of vision remains stable throughout life. A female has a family history of ocular albinism but she does not have any symptoms. A test to find out if she has the mutant allele is available. Suggest one reason for taking this test and one reason against taking this test. for ...................................................................................................................................... ........................................................................................................................................... against ............................................................................................................................... ........................................................................................................................................... [2] [Total: 13]

Mark scheme: 2(a) any two from: 1 no melanin ; 2 white hair ; 3 (very) pale skin ; 4 pink / very pale / AW, eyes ; 2 Question Answer Marks 2(b)(i) any three from: 1 more males / fewer females, have ocular albinism ; 2 males with (one copy of) recessive allele have ocular albinism / females with (one copy of) recessive allele are carriers ; 3 males (with ocular albinism) receive recessive allele from mothers ; 4 fathers cannot pass on ocular albinism to their sons ; 5 (because) males pass Y chromosome to sons ; 6 males with ocular albinism pass recessive allele to daughters ; 7 detail from Fig. 2.1 ; e.g. all daughters (3, 5, 7) of male 1 are carriers / all sons (9 and 10) of male 1 do not have ocular albinism / males 11, 12, 13, get ocular albinism from mothers 3 and 5 3 2(b)(ii) symbols normal allele = A OA1 allele = a ; parental genotypes XaY x XAXA and gametes Xa Y and (all) XA ; offspring genotypes (phenotypes correctly linked) XAXa XAY ; normal / carrier, female normal male ; 4 Question Answer Marks 2(b)(iii) any two from: 1 frameshift / every triplet from site of deletion is changed ; 2 primary structure / tertiary structure / 3D shape, is changed or (protein) folded incorrectly; 3 (protein has) changed, binding / active, site; 4 premature STOP codon leads to incomplete protein ; 2 2(b)(iv) any one from: for 1 decide whether to have children ; 2 earlier treatment if child born with ocular albinism ; any one from: against 3 test may be expensive ; 4 condition, is mild / not life threatening ; 5 can’t be treated ; 2

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3 The cilia of ciliated epithelial cells show variation in length, within an individual and between different individuals. Samples of ciliated epithelial tissue were removed from the airways of healthy people and the mean cilia length for each individual was calculated. The people in the study formed two groups: • people who were exposed to a harmful environmental factor • people who were not exposed to a harmful environmental factor. The results are shown in Fig. 3.1. Key: people not exposed to a harmful environmental factor people exposed to a harmful environmental factor 25 20 15 percentage frequency 10 5 0 0 2 4 6 8 10 mean length of cilia / μm Fig. 3.1 (a) (i) Name the type of variation shown for mean cilia length in each group of people in Fig. 3.1. ..................................................................................................................................... [1] (ii) Suggest two harmful environmental factors that people might have been exposed to that could explain the difference between the two curves in Fig. 3.1. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (b) The development and final length of cilia of epithelial cells is controlled by many genes. Fig. 3.2 summarises the interactions of some of these genes. The arrows represent the genes being switched on. regulatory gene A regulatory gene B regulatory gene C 6 structural genes 2 structural genes 11 structural genes Fig. 3.2 (i) With reference to Fig. 3.2, explain how genes such as A, B and C are able to switch on other genes. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [4] (ii) Describe how microarray analysis can be used to identify the genes switched on by the product of gene C. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [5] (c) Microarray analysis has shown that ciliated epithelial cells and other epithelial cells in the airways express genes that code for receptor proteins on cell surface membranes. These receptor proteins detect specific chemicals. When the receptor proteins bind to specific chemicals, this causes coughing, constriction of the airways and an increase in cilia activity. In people who have asthma, the response can be severe and life-threatening. Suggest how knowledge of the genes for these receptor proteins could help in the treatment of asthma. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 15]

Mark scheme: 3(a)(i) continuous; 1 Question Answer Marks 3(a)(ii) any two from: 1 smoking ; 2 air pollution ; 3 workplace pollution ; 4 previous lung disease ; 2 3(b)(i) 1 ref. to transcription factors ; any three from: 2 TF / gene product / protein, bind to DNA ; 3 (TF binds to) promoter ; 4 ref. to binding of RNA polymerase ; R if operator mentioned 5 (so) mRNA is made / transcription occurs ; ora 6 detail from Fig. 3.2 ; gene A, switches on / controls, 4 genes or genes B and C and 2 other genes gene B, switches on / controls, 6 genes gene C, switches on / controls, 11 genes 4 Question Answer Marks 3(b)(ii) any five from: 1 obtain mRNA from (basal) cell (expressing gene C) ; 2 reverse transcription of mRNA to produce cDNA ; 3 add fluorescent label to (c)DNA ; I colour 4 (microarray has) ssDNA probes ; 5 each from a, different / known, gene ; 6 (c)DNA hybridises / AW, to, probes / ssDNA (on microarray) ; 7 fluorescence shows the expressed genes ; 8 AVP ; e.g. ref. to washing off excess cDNA (after hybridisation) UV light / laser scanning (to record fluorescence pattern) intensity of fluorescence gives quantitative measure 5 3(c) any three from: 1 sequence gene(s) ; 2 predict, primary structure / amino acid sequence ; 3 model receptor protein (3D) structure ; 4 find molecules, that can block receptors ; 5 without triggering effects ; 6 ref. to limiting receptor protein production ; 3

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4 Rice, Oryza sativa, is an important food crop for human consumption. Two varieties of rice are T2A-1 and Shanyou 2. Shanyou 2 has been developed in the traditional way by selective breeding (artificial selection) and T2A-1 is an example of a genetically modified (GM) organism. (a) T2A-1 rice is described as insect-resistant as it produces a chemical that is toxic to insect pests. Outline how genetic engineering gave T2A-1 rice the trait of insect resistance. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] (b) Countries vary in the extent to which they grow GM varieties instead of traditional crop varieties. • The USA, Canada and countries in South America plant GM varieties on a large proportion of their crop-growing land, reaching 94% for soybeans in the USA. • In most European countries, 0% of the crop-growing land is planted with GM varieties. Scientists used computer models to predict the effects of a global ban that reduced the cultivation of GM crops to 0% everywhere. Table 4.1 shows the results of this modelling on world crop yields and the purchase price of each crop. Table 4.1 percentage percentage change crop change in in crop yield purchase price rapeseed –0.14 +1.96 rice –0.12 +1.58 soybeans –1.40 +4.05 (i) Explain what the data in Table 4.1 show about the social implications of growing GM crops. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) A worldwide ban on growing GM crop varieties would mean that more land would be needed to grow traditional crops, such as Shanyou 2. This would involve converting forest and grassland to crop-growing land. Table 4.2 shows the predicted changes in carbon dioxide emissions associated with this change in use of land. Table 4.2 change in carbon dioxide emissions change in use of land / million kg CO2 forest to crop-growing land + 608 726 grassland to crop-growing land + 276 042 Discuss what the data in Table 4.2 indicate about the environmental implications of growing GM crops. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 8]

Mark scheme: 4(a) any four from: 1 gene(s) from, another species / variety / (soil) bacterium / Bacillus thuringensis ; 2 ref. to restriction enzyme ; 3 ref. to (Ti) plasmid / vector / Agrobacterium tumefaciens / gene gun ; 4 (DNA) ligase ; 5 ref. to recombinant, DNA / plasmid ; 6 introduced into rice, cells / genome / DNA ; 7 (new) gene, expressed / transcribed, 8 toxic protein produced ; 9 ref. to marker genes / insert promoter ; 4 4(b)(i) any two from: 1 (crop) yields would decrease if GM banned ; 2 prices would increase if GM banned ; 3 AVP ; e.g. issues involving, food security / social unrest / wars 2 4(b)(ii) growing GM crops 1 helps protect, ecosystems / habitat / biodiversity ; ora 2 helps reduce, greenhouse effect / global warming / climate change / risk of flooding ; ora 2

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5 (a) Fig. 5.1 is a photograph of a section through a kidney. Fig. 5.1 On Fig 5.1, use label lines and letters to label the regions of the kidney where the following are located: L – loop of Henlé D – distal convoluted tubule P – podocyte cells. [3] (b) In a cold environment, people urinate more frequently. This condition is called cold diuresis. In a cold environment, homeostatic mechanisms keep heat loss to a minimum. Blood is diverted away from the skin so that more blood circulates around the main body organs, including the kidney. Suggest and explain why cold diuresis occurs. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] (c) The water potential of the blood is maintained at a constant level by a mechanism called negative feedback. With reference to the maintenance of the water potential of the blood, explain what is meant by negative feedback. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [5] (d) ADH affects the number of water channel proteins, aquaporins, in the cell surface membranes of the cells of the distal convoluted tubule. State what happens to the water potential and volume of the urine if cells in the distal convoluted tubule have more aquaporins in their cell surface membranes. water potential .......................................................................................................................... volume of urine ........................................................................................................................ [1] [Total: 13]

Mark scheme: 5(a) 3 5(b) any four from: 1 blood pressure increases in kidney ; 2 (so) more ultrafiltration ; 3 (therefore) increased volume of urine ; 4 ADH concentration in blood decreases ; 5 ref. to less reabsorption of water in collecting duct ; 6 AVP ; e.g. blood pressure drop in skin sweating stops so increases volume of blood 4 L ; D ; P ; Question Answer Marks 5(c) any five from: 1 osmoreceptors detect decrease (in water potential) ; ora A change 2 (osmoreceptors) in hypothalamus ; 3 ADH released (into blood) ; 4 by posterior pituitary (gland) ; 5 ref. to increase in aquaporins in, collecting duct / DCT; ora 6 more water reabsorbed ; ora 7 water potential returns to set point ; 5 5(d) water potential – decrease and volume – decrease ; 1

6 (a) Regressive evolution is a change in a population over time that involves the loss of certain phenotypic characteristics. It is thought to be caused by either genetic drift or natural selection. An example of regressive evolution is the loss of eyes in one form of the Mexican cavefish, Astyanax mexicanus. These eyeless cavefish live in caves that are in total darkness. There are three theories to explain how the loss of eyes in the cavefish has occurred. Theory A There is no advantage to having eyes in a cave that is in total darkness, where energy sources are scarce. Having eyes is a disadvantage as there may be an energy cost. Theory B A mutation has occurred in a single gene. This mutation has two effects: • a lack of eye development • an increase in the number of chemoreceptors on the skin. Theory C Various mutations occurred in the genes responsible for eye development over a period of time. By chance, these mutations increased in frequency in small isolated populations. Eventually this produced a population of eyeless cavefish. (i) State one theory, A, B or C, which describes genetic drift as the cause of loss of eyes. ................. [1] (ii) State and explain which theory or theories are based on natural selection as the cause of loss of eyes. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [4] (b) There are several separate populations of eyeless A. mexicanus in different caves. There are populations of A. mexicanus that spend time in areas with light and the fish in these populations have eyes. The mitochondrial DNA (mtDNA) of eyeless A. mexicanus was compared to the mtDNA of A. mexicanus with eyes. (i) Suggest how DNA evidence can help find out whether the eyeless A. mexicanus and the A. mexicanus with eyes are the same species or different species. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Suggest why mtDNA is used instead of nuclear DNA when studying the closeness of the relationship between populations. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] [Total: 10]

Mark scheme: 6(a)(i) (theory) C ; 1 Question Answer Marks 6(a)(ii) 1 (theories) A and B ; any three from 2 (A) energy acts as a selection pressure / having eyes is a selective disadvantage ; ora 3 detail of mp2 ; e.g. less energy used means lower need for food 4 (B) selection for more chemoreceptors / more chemoreceptors is a selective advantage ; 5 detail of mp4 ; e.g. more chemoreceptors allows detection of more food 6 no eyes / more chemoreceptors, allows individuals to, survive / reproduce ; AW 7 pass on advantageous, mutation / alleles, to offspring / increase in allele frequency ; ora 4 6(b)(i) any two from: 1 count, base / nucleotide, sequence differences ; 2 few(er) differences in (mt)DNA sequences suggest same species ; ora 3 fewer mutations means less time has passed since, most recent common ancestor ; 2 Question Answer Marks 6(b)(ii) any three from: mitochondrial DNA accept ora throughout 1 inherited from mother alone ; 2 recombination / crossing over, doesn’t occur ; 3 mutations, occur at a constant rate / molecular clock ; 4 mtDNA mutates faster (than nuclear DNA) ; 5. many copies of mtDNA (per cell) ; 6 is smaller / has fewer genes ; 7 not associated with histones ; 8 mtDNA analysis is quicker ; 3

7 The passage below outlines one method used by a student to estimate the population size of an animal species. Complete the passage by using the most appropriate scientific term or terms. A student estimated the population size of an animal species using the mark-release- ................................... technique. This can only be used for ................................... animals such as mice. The student caught a sample of mice using humane traps. The student then marked the mice in a way that did not ................................... them. For example, the mark did not make them more visible to potential ................................... . The student released the marked mice and left enough time for them to randomly ................................... with the population. The student then caught a second sample of mice and noted the numbers of marked and unmarked mice. The student estimated the population size by multiplying the number of mice in the first sample by the number in the second sample, then dividing by the number of ................................... mice in the second sample. [6]

Mark scheme: 7 recapture ; small / mobile / AW ; harm ; predators ; mix / disperse / integrate ; marked ; 6

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8 (a) Table 8.1 shows some of the stages of glycolysis and respiration in anaerobic conditions in yeast. The stages are not listed in the correct order. Table 8.1 stage description of stage A triose phosphate is dehydrogenated by NAD B pyruvate is decarboxylated C ethanal is formed D ethanol is formed E glucose is phosphorylated by ATP F pyruvate is formed G ethanal is reduced by alcohol dehydrogenase H fructose bisphosphate is formed I ATP is made by substrate-linked phosphorylation Complete Table 8.2 to show the correct order of the stages. One stage has been completed for you. Table 8.2 correct order letter of stage 1 ......... 2 ......... 3 ......... 4 ......... 5 F 6 ......... 7 ......... 8 ......... 9 ......... [4] (b) Yeast respires glucose in aerobic and anaerobic conditions. Explain why less ATP is produced when yeast respires in anaerobic conditions compared to when yeast respires in aerobic conditions. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [5] [Total: 9]

Mark scheme: 8(a) correct order letter of stage 1 E 2 H 3 A 4 I 5 F 6 B 7 C 8 G 9 D E and H in correct position ; A and I in correct position ; B and C in correct position ; G and D in correct position ; 4 Question Answer Marks 8(b) any five from: in yeast 1 only glycolysis occurs ; 2 (net) 2 ATP produced ; 3 no oxygen as (final) electron acceptor ; 4 (so) pyruvate does not enter mitochondria ; 5 ETC / chemiosmosis / oxidative phosphorylation / Krebs cycle / link reaction, does not occur ; 6 (ETC / chemiosmosis / oxidative phosphorylation) produce most ATP ; 7 pyruvate / ethanal, converted to ethanol ; 8 ethanol still energy-rich / AW ; 5

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9 (a) Describe how the structure of a chloroplast is related to its functions. [8] (b) Describe how you would separate chloroplast pigments using chromatography. [7] [Total: 15]

Mark scheme: 9(a) any eight from: 1 ground substance / stroma ; 2 for, light independent stage / Calvin cycle ; 3 contains enzymes / named enzyme e.g. rubisco ; 4 also, sugars / lipids / starch / ribosomes / DNA ; 5 internal membrane system ; 6 for, light dependent stage ; 7 fluid-filled sacs / thylakoids ; 8 grana are stacks of thylakoids ; 9 (grana) hold (photosynthetic) pigments ; 10 (grana) have large surface area for (maximum) light absorption ; 11 (pigments are arranged in), light harvesting clusters / photosystems ; 12 primary pigment / reaction centre / chlorophyll a, surrounded by accessory pigments ; 13 (accessory pigments) pass energy to, primary pigment / reaction centre / chlorophyll a ; 14 different photosystems absorb light at different wavelengths ; 15 membranes hold, ATP synthase / electron carriers ; 16 for, photophosphorylation / chemiosmosis ; 8 Question Answer Marks 9(b) any seven from: 1 grind leaf ; 2 use, solvent / named solvent ; 3 leaf extract contains mixture of pigments ; 4 ref. to concentrate extract ; 5 further detail ; e.g. pencil line drawn / extract placed on chromatography paper / extract placed on thin layer chromatography plate 6 paper placed (vertically) in jar of (different) solvent ; 7 solvent rises up paper ; 8 each pigment travels at different speed ; 9 pigments separated as they ascend ; 10 distance moved by each pigment is unique ; 11 Rf value ; 12 two dimensional chromatography ; 13 better separation of pigments ; 7

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10 (a) Describe how ATP is synthesised by oxidative phosphorylation. [8] (b) Describe how you would carry out an investigation, using a simple respirometer, to measure the effect of temperature on the respiration rate of insect larvae. [7] [Total: 15] .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. 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Mark scheme: 10(a) any eight from: 1 reduced, NAD / FAD ; 2 passed to ETC ; 3 inner membrane / cristae ; 4 hydrogen released (from reduced, NAD / FAD) ; 5 split into electrons and protons ; 6 electrons pass along, ETC / carriers / cytochromes ; 7 energy released pumps protons into intermembrane space ; 8 proton gradient is set up ; 9 protons diffuse, (back) through membrane / down gradient ; 10 ATP synthase / stalked particles / protein channels ; 11 (ATP produced from) ADP and (inorganic) phosphate ; 12 idea of oxygen as final electron acceptor ; 13 addition of proton (to oxygen) to form water / (oxygen) reduced to water ; 8 Question Answer Marks 10(b) any seven from: 1 detail of placing larvae in respirometer ; e.g. on gauze / use spoon 2 use and role of, KOH / AW; 3 water bath / maintain temperature ; 4 choose 4 different temperatures ; 5 mark initial position of liquid drop ; 6 leave for set time ; 7 measure distance moved by drop (in set time) ; 8 (use tap to) reset respirometer ; 9 (at least 3) replicates ; 10 calculate mean values ; 11 calculate rate as distance ÷ time ; 12 plot graph of rate v temperature ; 7

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