Cambridge A Level Biology 9700 — 2020 May/June Paper 4 · Variant 1

1 (a) In mammals, the hormone ACTH is released from the anterior pituitary gland. ACTH stimulates the cells of the adrenal glands to synthesise the hormone cortisol. This is an example of cell signalling. The main stages of cell signalling involving ACTH are similar to the main stages of cell signalling by adrenaline. Describe the main stages of cell signalling involving ACTH. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [5] (b) If the concentration of cortisol in the blood increases above its set point, it will inhibit the production of ACTH by the anterior pituitary gland. Name this type of homeostatic mechanism. ............................................................................................................................................. [1] (c) The domestic dog, Canis lupus familiaris, is a mammal. A dog may develop a condition called Cushing’s disease, which is a result of too much cortisol being secreted by the adrenal glands. The most common cause of this is an increased secretion of ACTH because of a tumour in the anterior pituitary gland. Explain how a tumour develops. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (d) One symptom of Cushing’s disease in a dog is that the dog will want to drink much more than normal. Suggest one other reason why a dog would want to drink much more than normal. ................................................................................................................................................... ............................................................................................................................................. [1] [Total: 9]

Mark scheme: 1(a) any five from: 1 (ACTH) binds to receptor on cell surface membrane (of adrenal gland cells) ; 2 G protein activated ; 3 adenylyl cyclase activated ; 4 cyclic AMP made from ATP ; 5 (cyclic AMP) second messenger ; 6 (cyclic AMP) binds to kinase (enzyme) ; 7 ref. to enzyme cascade / activation of enzymes by phosphorylation ; 8 amplifies signal ; 9 (amplification of signal) results in synthesis of cortisol ; 1(b) negative feedback ; 1 1(c) any two from: 1 uncontrolled mitosis / continuous cell cycle / cell cycle checkpoints not controlled ; 2 abnormal mass of cells formed ; 3 no programmed cell death / no apoptosis / cells immortal ; 4 AVP ; e.g. mutation of, tumour suppressor gene / (proto)oncogene 2 Question Answer Marks 1(d) any one from: 1 decrease in water potential of the blood ; 2 eating food with high salt content ; 3 excessive panting (because of increased body temperature) ; 4 AVP ; e.g. diabetes 1

2 The patty pan squash plant, Cucurbita pepo, produces edible fruits that vary in colour. (a) The colour of the fruits is controlled by two genes, A/a and B/b, that occur on different chromosomes. • Allele A produces a white fruit colour. • Allele a does not produce a colour by itself but allows the colours coded by gene B/b to show in the phenotype. • Allele B produces a yellow fruit colour. • Allele b produces a green fruit colour. In a dihybrid cross, an AABB plant was crossed with an aabb plant. All the resulting F1 plants produced white fruits. The F1 plants were then crossed with each other to obtain the F2 generation. (i) Complete Fig. 2.1 with the gametes produced by the F1 parents, the F2 genotypes and the F2 phenotypes. State the ratio of fruit colours in the F2 offspring. F1 parents: AaBb × AaBb Fig. 2.1 ratio of fruit colours in F2 offspring .................................................................................... [4] (ii) Test crosses were carried out on two white-fruited plants, P and Q, from the F2 generation. Each of these plants had its female flowers pollinated with pollen from a green-fruited plant. For plant P, half of the offspring were white and half were yellow. For plant Q, half of the offspring were white and half were green. Deduce the genotypes of plants P and Q. plant P ....................................................... plant Q ....................................................... [2] (iii) Plants P and Q show genetic variation with respect to fruit colour alleles. Identify the process that occurred during meiosis in the F1 parents that produced this variation and the stage of meiosis at which it occurred. process .......................................................................... stage of meiosis ............................................................. [2] (b) In 1994, two new varieties of squash were grown in the USA. Both varieties showed resistance to two viral diseases of squash plants. • The variety ‘Tigress’ was developed by selective breeding. • The variety ‘Freedom II’ was developed by adding genes for viral coat proteins to the squash genome. (i) Identify, with reasons, the variety of squash that can be described as recombinant. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Outline one social advantage and one social disadvantage of the recombinant squash variety compared to non-resistant squash varieties. advantage ......................................................................................................................... ........................................................................................................................................... disadvantage ..................................................................................................................... ........................................................................................................................................... [2] [Total: 12]

Mark scheme: 2(a)(i) gametes correctly entered ; genotypes correct ; A ecf phenotypes match genotypes ; A ecf ratio of fruit colours: 12 white : 3 yellow : 1 green ; A ecf AB Ab aB ab AB AABB white AABb white AaBB white AaBb white Ab AABb white AAbb white AaBb white Aabb white aB AaBB white AaBb white aaBB yellow aaBb yellow ab AaBb white Aabb white aaBb yellow aabb green 4 Question Answer Marks 2(a)(ii) plant P: AaBB ; plant Q: Aabb ; 2 2(a)(iii) random / independent, assortment ; metaphase I ; 2 2(b)(i) Freedom II because: 1 contains, viral / foreign, DNA / genes or has DNA from 2 different, sources / organisms ; 2 ref. to genetic engineering / GMO / genetic modification / transgenic ; 2 2(b)(ii) advantage: improved yield / plants do not die of disease / crop not lost ; disadvantage: viral proteins might cause allergies / may be social resistance to GM crops ; 2

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3 Cats are members of the Felidae family. Two genera of Felidae are Leopardus and Panthera. (a) The genus Leopardus consists of species of wild cats that are small and spotted. In 2013, biologists investigated the evolution of Leopardus tigrinus in South America. Fig. 3.1 shows the locations in South America of two populations of. L. tigrinus, population A and population B. population A population B Fig. 3.1 • Population A lives in grassland and desert habitats and population B lives in forest habitats. • The cats in population A have a lighter coat colour and a different pattern of spots from the cats in population B. • Genetic analysis shows that population A is genetically distinct from population B. • Population B has now been reclassified as a new species, L. guttulus. (i) Define the term species. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Explain how the two species, L. tigrinus and L. guttulus, have evolved from one original population in South America. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [5] (b) Illegal trade threatens the survival of endangered species. Forensic tests can identify animal species from the DNA of their body parts, for example bones. Cat species differ in the number of repeats of bases C and A (CA repeats) within one particular region of DNA, known as Ple46. Table 3.1 shows the number of repeats in Ple46 for three endangered cat species. Table 3.1 number of CA repeats cat species in Ple46 Asiatic lion 22 Bengal tiger 7–8 leopard 14–15 (i) Gel electrophoresis is used to estimate the length of Ple46 in a sample of DNA to check whether the sample comes from one of the endangered cat species in Table 3.1. Outline how gel electrophoresis is carried out to confirm whether the sample comes from one of these endangered cat species. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) Customs officials are responsible for checking whether animal body parts come from species on the CITES list of endangered species. Officials can now use a small DNA barcoding kit linked to a computer database to identify a species. Suggest two advantages of using this method to identify a species. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 12]

Mark scheme: 3(a)(i) any two from: 1 group of (morphologically) similar organisms ; 2 that interbreed to produce fertile offspring ; 3 occupy same niche ; 4 AVP ; e.g. detail of molecular similarity Question Answer Marks 3(a)(ii) any five from: 1 ref. to geographical isolation ; 2 no, gene flow / interbreeding, between, A and B / populations ; 3 different, selection pressures / environmental conditions (for A and B populations) ; 4 different mutations occur (in A and B populations) ; 5 some mutations make individuals better adapted ; 6 those individuals, survive / reproduce ; 7 pass on advantageous alleles ; 8 ref. to many generations ; 9 reproductive isolation ; 10 allopatric speciation ; 5 3(b)(i) any two from mp1-4 and one for mp5 any two from: 1 DNA sample placed in well (at end of gel) ; 2 ref. to electric field / current ; 3 (negatively charged) DNA attracted to, anode / positive electrode ; 4 shorter fragments move further in unit time / ora ; one from: 5 compare band position to identify cat species ; 3 Question Answer Marks 3(b)(ii) any two from: 1 quicker test ; 2 cheaper / good for low income countries ; 3 portable / no special lab needed / no need to send off sample to be tested ; 4 many sequences can be stored / ref. to bioinformatics ; 5 AVP ; e.g. more successful prosecutions deters others form illegal trading / new enforcement measures available 2

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4 Mitochondrial complex I is a large enzyme complex that forms part of the electron transport chain. The enzyme is composed of many different polypeptides. The genes coding for these polypeptides are located either in mitochondrial DNA (mtDNA) or in nuclear DNA. Mutations in these genes can lead to the production of an enzyme that does not function efficiently. This results in a disease known as mitochondrial complex I deficiency. If severe, this can lead to death in early childhood. (a) Explain why people with mitochondrial complex I deficiency may have muscle weakness and difficulty with nervous coordination of movement. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) When mitochondrial complex I deficiency is caused by mutation in mtDNA: • a cell in an ovary produces gametes with different proportions of normal mitochondria and mitochondria that contain the mtDNA mutation (mutant mitochondria) • a person has disease symptoms when the proportion of mutant mitochondria in their cells exceeds a certain threshold • the severity of disease symptoms, and the age at which they appear, can vary greatly in the children of one woman. In a family with a history of mitochondrial complex I deficiency that is caused by a mutation in a nuclear gene, the probability of a child inheriting the mutation can be predicted. Suggest why, in families where mitochondrial complex I deficiency is caused by mtDNA mutation, it is not possible to predict the probability of a child inheriting the mutation. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (c) Genetic screening can be carried out on people with symptoms of mitochondrial complex I deficiency. Previously 7 mtDNA genes and 37 nuclear genes were sequenced. Some of the people tested did not have mutations in any of these genes. As a result, another gene was sequenced in these people and was found to be mutated. This led to the suggestion that genetic screening should sequence a larger proportion of the genome for people suspected of having this disease. Discuss the ethical reasons for and against sequencing a larger proportion of the genome for people suspected of having mitochondrial complex I deficiency. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (d) One mutation linked to mitochondrial complex I deficiency is a base substitution. It causes the amino acid glycine to be replaced by the amino acid valine in a region of α helix in a protein that is important for the formation of mitochondrial complex I. Glycine is a small amino acid with an R group of one hydrogen atom whereas valine has a larger and branched R group. Predict how the change in amino acids would affect the structure of the protein. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 10]

Mark scheme: 4(a) 1 less / no, ATP made ; 2 (so) poor muscle contraction / AW or (so) reduced impulse transmission ; 4(b) any three from: 1 different children have different amounts of mutant, mtDNA / mitochondria ; 2 only mother / siblings, above threshold of mutant, mtDNA / mitochondria, will show symptoms ; 3 ref. to not showing typical patterns of inheritance ; 4 (mutant) mtDNA / mitochondria, inherited maternally or males do not pass on (mutant), mtDNA / mitochondria ; 3 Question Answer Marks 4(c) any two from: arguments for 1 more diagnosis, means better management of those with the deficiency ; 2 can prevent losing further children to the deficiency / can decide to not have children / can choose to have children after embryo testing ; 3 ref. to possibility of using enucleated donor egg for mutated mtDNA ; any two from: arguments against 4 different opportunities for people to access the test ; 5 no cure ; 6 some people person may still lead a normal life ; 7 many mutations still unknown so may still not lead to positive diagnosis ; 3 4(d) any two from: 1 change in, formation of α helix / secondary structure ; 2 change in, folding of / bonds in, tertiary structure ; 3 change in, 3D specific shape ; 2

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5 Yeast cells are unicellular eukaryotes. Yeast cells respond to changes in the concentrations of the sugars galactose and glucose by switching on genes using transcription factors. Fig. 5.1 shows the events that occur when the sugar galactose is present and glucose is absent in the external environment. Msn2 transcription factor proteins bind to promoters of genes A and B gene A switched on gene B switched on enzyme A synthesised protein B synthesised galactose converted to glucose Fig. 5.1 (a) (i) Explain why enzyme A is described as an inducible enzyme. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) Protein B has a similar role to the protein coded for by gene Y of the lac operon. Suggest a possible role for protein B that would allow yeast cells to make use of galactose. ........................................................................................................................................... ..................................................................................................................................... [1] Scientists produced genetically engineered yeast cells. The gene coding for Msn2 transcription factor proteins and the marker gene coding for green fluorescent protein (GFP) are transcribed together to produce a single mRNA molecule. The resulting Msn2 transcription factor proteins contain a GFP region as part of their structure and are called tagged Msn2 molecules. These tagged Msn2 molecules show up as green fluorescent spots when viewed using a microscope with a very high resolution. An investigation was carried out to compare the distribution of tagged Msn2 molecules in yeast cells, when glucose is absent and when glucose is present. The results are shown in Table 5.1. Table 5.1 mean number of tagged Msn2 molecules present glucose availability cytoplasm nucleus total glucose absent 3 263 2 012 5 275 glucose present 1 755 632 2 387 (b) (i) Calculate the percentage of Msn2 molecules in the cell that are located inside the nucleus, when glucose is absent. Show your working and write your answer to two significant figures. ...................................................... % [2] (ii) When glucose is present, 26% of Msn2 molecules in the cell are located inside the nucleus. Suggest why this figure is different from your answer to (i). ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (c) Another method of tagging molecules with a fluorescent colour is to use monoclonal antibodies that have GFP attached. These monoclonal antibodies bind specifically to the molecule of interest. This is called immunofluorescent tagging. Suggest reasons why immunofluorescent tagging is not a suitable choice for this investigation. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 11]

Mark scheme: 5(a)(i) any three from: 1 not, made all the time / constitutive; 2 gene switched on / protein made, (only) when needed ; 3 triggered by absence of glucose ; 4 concentration increases when galactose present ; 5(a)(ii) transporter / channel, for galactose or to, allow galactose into cell / increase permeability to galactose ; 1 5(b)(i) 2012 × 100 ; 5275 38(%) ; 2 5(b)(ii) either any two from mp1, mp2 or mp5 or any two from mp3, mp4 or mp5 either 1 (Msn2 in nucleus) higher when glucose is absent ; 2 (as) Msn2 / transcription factors, bind to DNA in nucleus ; or 3 (Msn2 in nucleus) lower when glucose is present ; 4 (as) Msn2 is not, needed in nucleus / bound to DNA ; plus 5 AVP ; e.g. ref. to diffusion / movement, of tagged Msn2 / unbound Msn2 distributed equally in nucleus and cytoplasm ; 2 Question Answer Marks 5(c) any three from: 1 antibodies are too big to enter cells ; 2 making monoclonal antibodies is, expensive / time consuming ; 3 making monoclonal antibodies harms, mice / animals ; 4 immunofluorescence is only used on dead cells ; 3

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6 (a) At the end of glycolysis in mammalian muscle tissue, pyruvate is formed. Pyruvate can be further metabolised along two different pathways, pathway 1 and pathway 2. Fig. 6.1 outlines the two pathways. pyruvate pathway 1 pathway 2 NAD NAD reduced reduced NAD NAD X acetyl CoA lactate Fig. 6.1 (i) State the condition under which pyruvate would be metabolised along pathway 2. ..................................................................................................................................... [1] (ii) Name compound X. ..................................................................................................................................... [1] (iii) Name the two types of reaction that are involved in pathway 1. ........................................................................................................................................... ..................................................................................................................................... [2] (iv) Suggest a cell in a mammal that cannot carry out pathway 1. Give a reason for your answer. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (v) ATP is synthesised during glycolysis. Name the process by which ATP is synthesised. ..................................................................................................................................... [1] (b) The rufous hummingbird, Selasphorus rufus, feeds on the nectar produced by flowers. Nectar is a liquid containing sugars. Hummingbirds hover to maintain a position next to a flower while they feed. This requires a lot of energy. Fig. 6.2 is a rufous hummingbird. Fig. 6.2 Hummingbirds that have not fed for some time have a respiratory quotient (RQ) value of 0.7. After feeding their RQ value is 1.0. (i) Define the term respiratory quotient. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Explain why a hummingbird that has not fed for some time has an RQ value of 0.7, while a hummingbird that has fed has an RQ value of 1.0. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (iii) When animals exercise they can respire substrates obtained directly from their diet, as well as respiring substrates from energy stores. An investigation worked out the percentage of respiration that used substrates obtained directly from the diet of humans and hummingbirds as they exercised. Fig. 6.3 shows the results of this investigation. 100 90 80 70 percentage of respiration that used 60 substrates obtained directly from the diet 50 40 30 20 10 0 human hummingbird animal Fig. 6.3 Fig. 6.3 shows that hummingbirds have a higher percentage of respiration that uses substrates obtained directly from the diet. Calculate how many times greater this figure is for hummingbirds compared to humans. Show your working and write your answer to two decimal places. answer = ......................................................... [2] (iv) Suggest the benefits to a hummingbird of obtaining most of its respiratory substrates from its diet while exercising. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 15]

Mark scheme: 6(a)(i) anaerobic conditions / no oxygen available / little oxygen available ; 1 6(a)(ii) carbon dioxide ; 1 6(a)(iii) dehydrogenation ; decarboxylation ; 2 6(a)(iv) red blood cell / erythrocyte ; (so) no mitochondria ; 2 6(a)(v) substrate-linked phosphorylation ; 1 6(b)(i) volume of carbon dioxide produced ÷ volume of oxygen consumed ; per unit time ; 2 6(b)(ii) lipids / fatty acids, metabolised (not having fed) ; sugar / carbohydrates, metabolised (after feeding) ; 2 Question Answer Marks 6(b)(iii) 95 or 95 ÷ 32 ; 32 2.97 ; 2 6(b)(iv) any two from: 1 substrate immediately available to muscle cells ; 2 no need to break down, glycogen / energy stores ; 3 glycogen still available as substrate between feeds ; 4 AVP ; e.g. cannot carry stores as need to be light to fly 2

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7 (a) Structures, processes and compounds involved in photosynthesis in C4 plants include: A – carbon dioxide H – mesophyll cells B – phosphorylation I – guard cells C – palisade cells J – photorespiration D – PEP carboxylase K – photolysis E – rubisco L – oxaloacetate F – malate M – ATP synthase G – RuBP N – chlorophyll Complete Table 7.1 by matching each description with one letter chosen from A to N to show the correct structure, process or compound. You may use each number once, more than once or not at all. Table 7.1 description letter enzyme with a high optimum temperature enzyme in bundle sheath cells process that slows down the rate of photosynthesis compound that releases carbon dioxide into bundle sheath cells cells that stop oxygen reaching bundle sheath cells [5] (b) Some plants, such as rice, are adapted to grow when partially submerged in water. Describe and explain the adaptations that allow rice plants to survive in waterlogged fields. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 8]

Mark scheme: 7(a) description letter enzyme with a high optimum temperature D ; enzyme in bundle sheath cells E ; process that slows down the rate of photosynthesis J ; compound that releases carbon dioxide into bundle sheath cells F ; cells that stop oxygen reaching bundle sheath cells H ; 5 Question Answer Marks 7(b) any three from: 1 aerenchyma so oxygen can reach submerged, parts / roots ; 2 ridges on underwater leaves to hold air close to, leaf surface / stomata ; 3 grow taller to keep leaves and flowers above water level ; 4 high tolerance to ethanol so roots can respire anaerobically ; 5 produce more ethanol dehydrogenase to breakdown ethanol from anaerobic respiration ; 3

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8 (a) Sampling is used to find out the variety of species in an ecosystem and the size of the population of each species. A study was carried out to investigate the biodiversity of two fields, A and B. Field A had not been used for growing crops for 10 years. Field B had been used for growing crops until one year before the study. Random sampling was carried out. Describe how random sampling could be carried out on plant species in the two fields. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] (b) Table 8.1 shows the results of the sampling of field A. Table 8.1 number of species individuals (n) Scabiosa columbaria 13 Centaurea centaurium 15 Primula veris 26 Trifolium pratense 36 Leucanthemum vulgare 11 Silybum marianum 5 Anacamptis morio 8 total number of individuals (N) 114 Simpson’s Index of Diversity (D) is a value of the species diversity in an area. The formula for this is: n 2 D = 1 / -f d N n p ! = sum of n = number of individuals of a species N = number of individuals of all species (i) Complete Table 8.2 for Primula veris. Table 8.2 n n 2 species n N <N F Scabiosa columbaria 13 0.11 0.01 Centaurea centaurium 15 0.13 0.02 Primula veris 26 .................... .................... Trifolium pratense 36 0.32 0.10 Leucanthemum vulgare 11 0.10 0.01 Silybum marianum 5 0.04 0.00 Anacamptis morio 8 0.07 0.00 [1] (ii) Use the data in Table 8.2 to calculate Simpson’s Index of Diversity (D) for field A. Show your working and write your answer to two decimal places. D = ......................................................... [2] (iii) The value of D for field B was 0.45. Comment on the values of D for both fields. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [1] [Total: 8]

Mark scheme: 8(a) any four from: 1 ref. to use of quadrats ; 2 ref. to size (of quadrat) ; 3 grid marked out ; 4 use of random number generator 5 measure species frequency ; 6 use of key or Braun Blanquet / other named, scale ; 7 measure, species abundance / percentage cover ; 8 repeat sampling ; 9 both sites treated the same ; 8(b)(i) 0.23 and 0.05 ; 1 8(b)(ii) Σ (n÷N)2 = 0.19 ; D = 0.81 ; 2 8(b)(iii) field A has a higher species diversity than field B ; 1

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9 (a) Describe how the polymerase chain reaction is used to clone and amplify DNA. [8] (b) Explain the advantages of using recombinant DNA techniques to produce human proteins, such as factor VIII or adenosine deaminase. [7] [Total: 15]

Mark scheme: 9(a) any eight from: 1 DNA denatured / ssDNA produced ; 2 by heating to, 94 / 95 °C ; 3 add primer (DNA) ; 4 complementary base pairing (with sample DNA) ; 5 at 55–65 °C ; 6 ref. to annealing ; 7 DNA polymerase builds new strands / AW ; 8 by adding free nucleotides ; 9 at 70–75 °C ; 10 ref. to Taq polymerase thermostable ; 11 does not need replacing ; 12 new strand denatured and process repeated ; 13 process is automated ; Question Answer Marks 9(b) any seven from: 1 can produce large(r) quantities ; 2 use bacterial host / hamster cell host / insect larva cell ; 3 product exactly the same as human protein ; 4 (as) product has same amino acid sequence 5 no immune response ; 6 no side effects ; 7 no risk of transfer of disease ; 8 easier to obtain purified product ; 7

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10 (a) Describe the ultrastructure of a striated muscle fibre. [8] (b) Explain how auxin causes plant cells to elongate. 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Mark scheme: 10(a) any eight from: 1 fibres are multinucleate ; 2 cell surface membrane is sarcolemma ; 3 sarcoplasm has many mitochondria ; 4 sarcoplasmic reticulum membranes have protein pumps ; 5 transverse system tubules / T-system ; 6 ref. to myofibrils ; 7 thick filament / myosin, attached to M line ; 8 thin filament / actin, attached to Z line ; 9 interdigitation of filaments causes striated appearance ; 10 description of one of, A / H / I, bands ; 11 sarcomere is the distance between M lines ; 12 myosin is a fibrous protein with globular protein head ; 13 actin is a chain of globular protein molecules ; 14 tropomyosin / troponin, attached to actin ; allow marks from an annotated diagram Question Answer Marks 10(b) any seven from: 1 acid-growth (hypothesis) ; 2 auxin stimulates proton pumps ; 3 (in) cell surface membrane ; 4 protons / H+, pumped into cell wall ; 5 using energy / by active transport ; 6 pH of cell wall decreases / cell wall becomes (more) acidic ; 7 pH-dependent enzymes activated ; 8 ref. to expansins ; 9 bonds between cellulose microfibrils broken ; 10 idea that cell wall, ‘loosens’ / becomes more elastic / able to stretch ; 11 (more) water enters cell / turgor pressure increases ; 12 (so) cell (wall) expands ; 7