Cambridge A Level Biology 9700 — 2022 May/June Paper 4 · Variant 2

1 The jaguar, Panthera onca, is a large cat that lives mainly in South America. The majority of jaguars have light brown fur with black spots, as shown in Fig. 1.1. Some jaguars have completely black fur, as shown in Fig. 1.2. Fig. 1.1 Fig. 1.2 (a) The pigments involved in fur colour are produced as a result of biochemical pathways that take place in cells called melanocytes. These pathways are similar to those that occur in human melanocytes. The melanocortin 1 receptor (MC1R) is located on the cell surface membrane of melanocytes and is coded for by the MC1R gene. Fig. 1.3 outlines the processes that occur in jaguar melanocytes. phaeomelanin (light brown) MC1R not activated tyrosine substance X MC1R activated eumelanin MC1R (black) cell surface membrane Fig. 1.3 (i) Name the substance represented by X. ..................................................................................................................................... [1] (ii) When MC1R is activated a second messenger is produced in the cell. Give an example of a second messenger. ..................................................................................................................................... [1] (iii) Substance X is also produced in humans, but a mutation of the TYR gene can result in substance X not being produced. Describe the phenotype of a person with this mutation. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (b) The MC1R gene has two alleles and is located on an autosome. • When two jaguars with light brown fur mate all the offspring have light brown fur. • When two jaguars with black fur mate either all the offspring will have black fur or some offspring will have black fur and some will have light brown fur. Using symbols, construct a genetic diagram to show how two jaguars with black fur can produce some offspring with black fur and some offspring with light brown fur. [4] [Total: 8]

Mark scheme: 1(a)(i) DOPA / dopaquinone / melanin ; 1 1(a)(ii) cyclic AMP / cAMP ; A calcium ions 1 1(a)(iii) any two from: 1 albino / albinism ; 2 no pigment or melanin in skin / pale skin or white hair ; Ignore fair, skin / hair 3 red / pink, eyes / irises ; A blue 4 poor vision / described or eyes sensitive to light or jerky eye movements ; Ignore blindness 2 1(b) symbols ; e.g. B allele for black fur b allele for (light) brown fur parental genotypes Bb x Bb ; gametes B b B b offspring genotypes BB Bb (Bb) bb ; offspring phenotypes black black (black) (light) brown ; accept gametes mark from a Punnett square 4

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2 (a) Photosynthesis is affected by many environmental factors. (i) Explain why light intensity can be a limiting factor in photosynthesis. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) The concentration of carbon dioxide (CO2) can also be a limiting factor. It has an effect on the Calvin cycle in the light-independent stage of photosynthesis. Ribulose bisphosphate (RuBP), triose phosphate (TP) and glycerate 3-phosphate (GP) are three important molecules in the Calvin cycle. Fig. 2.1 shows how the concentration of GP changes when the concentration of CO2 is reduced from 0.04% (atmospheric) to 0.008%. Complete Fig. 2.1 by sketching the lines for RuBP and TP when the concentration of CO2 is reduced from 0.04% to 0.008%. 0.04% 0.008% atmospheric CO2 reduced CO2 concentration GP TP relative concentration of Calvin RuBP cycle molecules GP time Fig. 2.1 [2] (iii) In very dry conditions, CO2 concentration can become the main limiting factor of photosynthesis in plants. Explain how very dry conditions cause CO2 concentration to become the main limiting factor of photosynthesis in plants. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (b) A factor that can limit the rate of photosynthesis is the rate of regeneration of RuBP. Sedoheptulose-1,7-bisphosphatase (SBPase) is an enzyme in the Calvin cycle that controls the rate of regeneration of RuBP. SBPase is coded for by the gene SBPase. In an experiment, wheat plants were genetically modified to make more SBPase by introducing the SBPase gene from another grass species, Brachypodium distachyon. The resulting GM wheat plants were named Sox4. • Wild type plants (not GM) and Sox4 plants were grown in a greenhouse. • Light intensity, CO2 concentration and temperature were kept constant. • Mature plants were removed and dried to measure the biomass. Fig. 2.2 shows the mean plant biomass for the wild type plants and GM Sox4 plants. 35 30 25 20 mean plant biomass / g 15 10 5 0 wild type Sox4 wheat plants Fig. 2.2 (i) Calculate the percentage change in mean plant biomass when Sox4 plants are grown compared to wild type plants. Show your working. percentage change = ..................................................... % [2] (ii) Suggest and explain why Sox4 plants have a different mean plant biomass than wild type plants. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [4] (iii) Some soils may be deficient in nitrates. Suggest how nitrate deficiency could limit the quantity of SBPase made by Sox4 plants. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] [Total: 15]

Mark scheme: 2(a)(i) any two from: 1 ref. to light energy / photons ; 2 for, light-dependent stage / photophosphorylation / photolysis / photoexcitation / photoactivation ; 3 to make, reduced NADP / ATP ; 4 to open stomata (for CO2 to enter) ; 2(a)(ii) correctly showing that TP reduces to a plateau below GP ; correctly showing that RuBP increases to a plateau ; 2 Question Answer Marks 2(a)(iii) 1. decrease in / no, CO2, diffusing / moving, to, air spaces / palisade cells / mesophyll cells / chloroplasts / stroma ; any two from: 2 stress response / AW ; 3 ref. to abscisic acid / ABA ; 4 ref. to action of Ca2+ as a second messenger ; 5 stomata close ; 3 2(b)(i) 28 20.5 20.5   100 or 7.5 20.5  100 ; A 20.4 or 20.6 A 37.3 or 35.9 answer to 1 decimal place 36.6 (%) ; apply ecf 2 Question Answer Marks 2(b)(ii) any four from: no mention of increased or more – ecf and max 3 1 Sox4 plants have an extra copy of SBPase gene ; 2 (overall) increased, expression / transcription, of SBPase gene (so more SBPase) ; 3 increased (rate of) / more, regeneration / production, of RuBP ; 4 (so) increased / more, carbon fixation / Calvin cycles / light independent reaction / TP / GP ; 5 more, glucose / sucrose, for respiration / ATP production or more, starch / lipid, for storage or more cellulose for cell walls ; 6 more, amino acids / proteins, for growth ; 7 more, cell division / cell elongation / mitosis ; 8 AVP ; e.g.(new) SBPase / SBPase gene, more efficient 4 2(b)(iii) nitrogen needed to produce: nucleotides for, transcription (of SBPase gene) / expression (of SBPase gene) / mRNA production ; amino acids for, protein synthesis / SBPase production / enzyme production ; 2

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3 Aerobic respiration occurs when oxygen is available. A much greater energy yield is obtained from aerobic respiration than from respiration in anaerobic conditions. (a) Explain why the link reaction only occurs when oxygen is available. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] (b) Outline the stage of aerobic respiration that occurs in the cytoplasm of eukaryotic cells. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [5] [Total: 9]

Mark scheme: 3(a) any four from: accept ora 1 pyruvate converted to, acetyl group / 2C group / acetyl CoA (in link reaction) ; 2 NAD needed (in link reaction) ; 3 (where) NAD becomes reduced ; 4 reduced NAD goes to ETC ; 5 ETC / oxidative phosphorylation, works if oxygen is present or oxygen is the final electron acceptor ; 6 (so) reduced NAD is oxidised / NAD is regenerated ; 3(b) any five from: 1 glycolysis ; 2 phosphorylation of glucose ; 3 splitting of fructose (1,6) bi(s)phosphate / AW ; 4 (into) two, triose phosphate / TP ; 5 (TP) oxidised / dehydrogenated, to pyruvate ; 6 net 2 ATP / 4 ATP, produced ; 7 (2) reduced NAD produced ; 5

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4 In 1973, a technique for genetic engineering was used for the first time. Recombinant DNA was made using a plasmid and this was successfully transferred into an organism. In 2012, a new technique for genetic engineering, called gene editing, was developed. (a) Table 4.1 lists some statements about the two genetic engineering techniques. Complete Table 4.1 to compare the original genetic engineering technique using a plasmid vector with the newer technique of gene editing. For each row, place a tick (3) in the correct column if the statement applies and leave a blank if the statement does not apply. Table 4.1 statement genetic engineering gene editing using a plasmid It may cause the organism to produce a different protein. It may cause a single base pair in a gene to be changed. The success of the technique can be evaluated using marker genes. It may use the CRISPR system. It uses DNA ligase. [5] (b) Cassava plants, Manihot esculenta, produce roots that have a high starch content. These roots are an important food source in tropical regions. The growth of cassava plants is reduced by competition from weeds. Scientists used gene editing to develop two types of cassava plant with different mutations (changes to the DNA). The gene edited cassava plants showed resistance to the herbicide glyphosate. In susceptible plants, glyphosate prevents synthesis of three amino acids from a precursor molecule called shikimate. Fig. 4.1 shows the concentration of shikimate in the wild type (not gene edited) and the two types of gene edited cassava plant after they were exposed to three different concentrations of glyphosate. 30 Key: glyphosate / μmol dm–3 0 50 20 shikimate 500 concentration / arbitrary units 10 0 A B C type of cassava plant Fig. 4.1 (i) Identify the letter in Fig. 4.1 that represents the wild type cassava plant. ..................................................................................................................................... [1] (ii) Explain the social benefit of this example of gene editing. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] [Total: 9]

Mark scheme: 4(a) statement genetic engineering using a plasmid gene editing It may cause the organism to produce a different protein.   ; It may cause a single base pair in a gene to be changed.  ; The success of the technique can be evaluated using marker genes.  ; It may use the CRISPR system.  ; It uses DNA ligase.  ; 5 4(b)(i) C ; 1 Question Answer Marks 4(b)(ii) any three from: 1 increased / high, (crop) yield ; 2 idea of less food shortages / more food production / helps solve global demand for food ; 3 less need to remove weeds (by hand) ; 4 more income for, growers / farmers or ref. to economic benefit for, country / region ; 5 cheaper / lower cost, to buy ; 6 AVP ; e.g. less need for deforestation of other land 3

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5 Bison are a type of large wild cattle. Ancestors of modern bison appeared in Asia 2.5 million years ago. This ancestral bison species increased its range into Europe and North America. While the ancestral species is now extinct, its descendants include Bison bonasus, the European bison, and Bison bison, the American bison. Fig. 5.1 shows an American bison. Fig. 5.1 (a) After the end of the last ice age, populations of the ancestral bison were separated by sea and by forests that were not suitable as habitats. The separation resulted in the evolution of the European bison and the American bison. Explain how this separation resulted in the evolution of the two bison species. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] (b) Table 5.1 compares features of European bison and American bison. Table 5.1 feature European bison, American bison, B. bonasus B. bison type of food grass and higher vegetation grass and low vegetation such as leaves of bushes and trees height / m 2.1 2.0 maximum mass / kg 1000 1270 Assess the relative importance of natural selection and genetic drift in producing the different heights and masses of the two species of bison. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (c) The European bison has a nuclear genome that is very similar to that of the American bison. The European bison has a mitochondrial genome that is more similar to that of wild cattle of the genus Bos than to the American bison. Discuss what this implies about the evolutionary history of the European bison. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (d) Outline how practical techniques could be used to test the hypothesis that farmed cattle are closely related to European bison. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] [Total: 14]

Mark scheme: 5(a) any four from: 1 geographical isolation / barrier qualified ; Ignore separation 2 no, interbreeding / gene flow (between European and American populations) ; Ignore reproductive isolation here (needs to be late on in process) 3 different, environmental (conditions) / selection pressures ; 4 random / different, mutations ; 5 different, alleles selected for / gene pool / changes in allele frequency ; 6 populations have different, morphological / physiological / behavioural, features ; 7 after, a long time / many generations, reproductive isolation occurs / cannot interbreed to produce fertile offspring ; 8 allopatric (speciation) ; Question Answer Marks 5(b) any three from: 1 not possible / difficult, to come to a (reliable) conclusion ; A AW 2 small difference in height / heights are similar (so natural selection not more important than genetic drift) ; 3 taller / lighter, bison could be an advantage in reaching higher vegetation (so natural selection may be important) ; 4 difference in mass, may not be due to genetic factors / may be due to nutrient value of diet / may be due to quantity of food, (so natural selection may not be important) ; 5 genetic drift (in ancestral population) may have led to, European bison being, taller / lighter or American bison being, shorter / heavier ; 3 5(c) 1 European bison has a, close / recent / common, ancestor with that of, cattle / Bos or European bison is more closely related to, cattle / Bos, (than to American bison) ; 2 European bison may have interbred with, cattle / Bos ; 3 mitochondria are inherited from, mother / female ; 3 Question Answer Marks 5(d) any four from: 1 obtain / use, blood / cells / tissue / DNA ; 2 from, both species / cattle and European bison ; 3 use PCR to, replicate / amplify, DNA ; 4 use, electrophoresis (described) / DNA profiling / DNA fingerprinting, to see genetic differences / compare the banding (patterns) ; 5 sequence DNA and, count / identify, (percentage) difference / similarity or sequence DNA and compare the, nucleotide / base, sequence ; 6 use, genome / DNA, microarray ; R for gene expression 7 ref. to bioinformatics / database / (computer) software ; 8 similarity indicates, close relationship / more recent split ; 9 if they can breed to produce (viable) offspring (they are closely related) ; 4

6 (a) Gibberellin is a plant hormone that has an important role in seed germination. Fig. 6.1 is a diagram of a section through a barley seed (grain). aleurone layer endosperm embryo Fig. 6.1 On Fig. 6.1: • Draw one arrow, labelled G, to show the movement of gibberellin during germination. • Draw one arrow, labelled A, to show the movement of amylase during germination. [2] (b) Outline the role of amylase in seed germination. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] Describe the results shown in Fig. 6.2. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] [Total: 9]

Mark scheme: 6(a) G – embryo to aleurone layer ; A – aleurone layer to endosperm ; 2 6(b) any three from: 1 hydrolyses / breaks down / digests / converts, starch / amylose ; 2 in endosperm ; 3 to, maltose / glucose ; 4 (glucose) used by embryo ; 5 (glucose used) for, respiration / growth ; 3 Question Answer Marks 6(c) any four from: accept distilled water for A, 500mg dm-3 for B and 1000mg dm-3 for C. 1 seeds of B and C germinated before seeds of A or seeds soaked in gibberellin germinated before seeds soaked in distilled water ; ora 2 rate of (increase in percentage) germination was, highest in C or higher in C than, B / A or higher as the concentration of gibberellin increases ; ora 3 final / day 10 / maximum (percentage) germination was, highest in C or higher in C than, B / A or higher as the concentration of gibberellin increases ; ora 4 data quote between two groups ; mp1 – A by day 6 and B and C by day 2 mp1 – A germinates four days after B and C ora mp2 – A – 2.2 B – 5.2 C – 9.2, percentage germination day-1 mp3 – A – 22% B – 52% C – 92% mp3 – A to C = 70% more, A to B = 30% more, B to C = 40% more 5 percentage germination in C levels off after 9 days ; 6 AVP ; e.g. no seeds germinate by day 1 4

7 The bacterium, Escherichia coli, can use glucose or disaccharides, such as lactose, in its metabolism. Lactose needs to be hydrolysed by the enzyme β-galactosidase to form glucose and galactose, which can then be used by E. coli. The production of β-galactosidase is controlled by a length of DNA called the lac operon. (a) Fig. 7.1 shows the lac operon when lactose is absent. RNA polymerase repressor P O lacZ lacY lacA Fig. 7.1 On Fig. 7.2, draw the positions of RNA polymerase and the repressor molecule when lactose is present. P O lacZ lacY lacA Fig. 7.2 [2] (b) The protein coded for by lacY is not involved in the control of gene expression. (i) Name the type of gene represented by lacY. ..................................................................................................................................... [1] (ii) Name the protein product coded for by lacY and state the precise role of this protein. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (c) In an investigation into the growth of E. coli, a sample of the bacterium was grown in a medium that contained limited concentrations of glucose and lactose. The population size of E. coli was measured at regular intervals. Fig. 7.3 shows the population growth curve obtained for this investigation. E. coli population size time Fig. 7.3 Describe and suggest explanations for the population growth curve shown in Fig. 7.3. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] [Total: 9]

Mark scheme: 7(a) RNA polymerase attached to promoter ; repressor detached from operator ; 2 7(b)(i) structural (gene) ; 1 7(b)(ii) lactose permease ; A beta-galactoside permease movement / transport, of lactose into cell or makes (cell surface) membrane (more) permeable to lactose ; 2 7(c) 1 population increases, then levels off, increases, then levels off / AW ; any three from: 2 (first increase) glucose, metabolised / respiration ; 3 (first levelling off) glucose has run out ; 4 (first levelling off) lactose, hydrolysed / broken down, into, glucose / galactose ; 5 (second increase) glucose / galactose, metabolised / respired ; 6 (second levelling off) lactose / glucose / galactose, has run out ; 7 AVP ; e.g. idea of time delay (for, lactose permease / beta galactosidase, to work) 4

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8 The role of sensory receptor cells in mammals is to detect stimuli and generate action potentials in sensory neurones. Human taste buds on the tongue contain chemoreceptor cells. Different chemoreceptor cells respond to different chemical stimuli. Fig. 8.1 is a diagram of chemoreceptor cells in a taste bud. Y tight junction support cell chemoreceptor cell A sensory neurone B Fig. 8.1 (a) Name the structures in the region Y. ............................................................................................................................................. [1] (b) Suggest a reason for the tight junctions between the chemoreceptor cells. ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [1] (c) Chemoreceptor cell A responds to sodium ions (Na+) in salt. Describe how the contact of cell A with Na+ can result in an action potential in sensory neurone B. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [7] [Total: 9]

Mark scheme: 8(a) microvilli ; A brush border 1 8(b) no movement of, chemicals / substances / ions, between the, chemoreceptor / cells / (cells) A ; Ignore food A named dissolved substance 1 Question Answer Marks 8(c) any seven from: 1. Na+ ions enter, (chemo)receptor cell / A, through, microvilli / Y / channel (proteins) ; 2 cell surface membrane depolarised ; 3 receptor potential ; 4 ref. to threshold ; 5 (voltage-gated) Ca2+ / calcium, channels open ; 6 Ca2+ ions enter, cytoplasm / cell / A ; A presynaptic knob 7 vesicles (of neurotransmitter), move towards / fuse with, cell surface membrane / presynaptic membrane ; 8 exocytosis (described) / secretion, of (named) neurotransmitter ; 9 neurotransmitter binds to receptor on, postsynaptic / (sensory) neurone / B, membrane ; 10 Na+ / sodium, channels open or Na+ enter (sensory) neurone / B ; 11 postsynaptic membrane / (sensory) neurone membrane, depolarised ; 12 ref. to threshold ; allow either mp4 or mp12 7

9 (a) Ultrafiltration in the kidney takes place between the glomerulus and the Bowman’s capsule. The afferent blood vessel carrying blood to the glomerulus has a wider lumen than the efferent blood vessel. Explain why the lumen of the afferent blood vessel needs to be wider than the lumen of the efferent blood vessel. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Fig. 9.1 is a diagram of part of the glomerulus and Bowman’s capsule. A B blood capillary lumen of Bowman’s capsule Fig. 9.1 (i) Name A and B. A ........................................................................................................................................ B ........................................................................................................................................ [2] (ii) Describe the roles of A and B in the formation of the glomerular filtrate. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] [Total: 7]

Mark scheme: 9(a) to generate a high, blood pressure / hydrostatic pressure ; Ignore to increase the blood pressure to force, plasma / fluid / water, through basement membrane / into the Bowman’s capsule / into lumen ; R osmosis / diffuse Ignore named substance e.g. glucose 9(b)(i) A – basement membrane ; B – podocyte ; 2 9(b)(ii) any three from: A basement membrane 1 acts as a, filter / filtration barrier ; 2 only, named molecules / ions / molecules less than 68 000–70 000 (R)MM, can pass through or red blood cells / large proteins, cannot pass through ; B podocyte 3 has, gaps / slits / projections / pores ; 4 allows, fluid / water / filtrate, to pass into the, lumen / Bowman’s capsule ; 5 produces, A / basement membrane ; 3

10 The African penguin, Spheniscus demersus, lives along the coast of South Africa. Fig. 10.1 shows two African penguins. Fig. 10.1 (a) One way of estimating the size of a population of African penguins is to use the mark-release- recapture method. Suggest three assumptions that must be made for the mark-release-recapture method to be valid. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) The International Union for the Conservation of Nature (IUCN) Red List of Threatened Species has categorised the African penguin as endangered and could become extinct. Suggest reasons why the African penguin has become endangered and could become extinct. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (c) Fig. 10.2 shows the estimated numbers of African penguins in the years of 1800, 1900 and 2000. 4 000 000 3 000 000 estimated 2 000 000 population size 1 000 000 0 1800 1900 2000 year Fig. 10.2 Calculate the mean yearly decrease in population size of the African penguin between 1900 and 2000. answer = .................................................. yr –1 [1] (d) The African penguin is a member of the kingdom Animalia. Outline the characteristic features of the kingdom Animalia. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4]

Mark scheme: 10(a) any three from: 1 marking, not harmful (described) / cannot be removed (described) ; 2 constant population size ; 3 no, births / deaths (described) or no, immigration / emigration / migration ; 4 (sufficient time for) marked individuals to mix with rest of population ; 5 (penguins are) mobile ; 10(b) any three from: 1 loss of habitat / example ; 2 climate change ; 3 predation ; 4 competition for, food / resources / breeding sites / example ; 5 new disease ; 6 pollution / example ; 7 numbers get so low that population can’t recover / AW ; 3 10(c) 13 000 / 1.3  104 ; 1 Question Answer Marks 10(d) any four from: 1 multicellular ; 2 eukaryote / eukaryotic / (cells) contain a nucleus ; 3 specialised cells / tissues / organs ; 4 heterotrophic (nutrition) / described ; 5 nervous system ; 6 cilia / flagella ; 7 mobile / motile ; 4

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