AS & A Level Biology (9700)
Classification, biodiversity & conservation Banker
55 questions · about 7% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Conservation & threats · Core · Paper 4 style · 6 marks- (a) Give three reasons why it is important to maintain biodiversity. [3]
- (b) The Indus river dolphin lives only in the Indus river system and is endangered. Its range is split into sections by barrages built for irrigation. Suggest three threats to this species, other than barrages. [3]
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- (a) Any three: moral/ethical duty to other species and future generations; economic value – food, timber, crops, ecotourism; source of medicines and new genes for crop breeding; ecosystem services – pollination, nutrient recycling, flood control, oxygen; stability of food webs/ecosystems; aesthetic/cultural value.
1 mark each for three different reasons.
- (b) Any three: low water flow in the dry season because water is taken for irrigation; pollution from industry, sewage and agricultural chemicals; accidental capture in fishing nets (bycatch); dolphins stranded in irrigation canals; reduced fish prey from overfishing; noise/boat traffic; small, isolated populations losing genetic diversity.
1 mark each for any three suitable threats.
- (a) Explain how national parks such as Khunjerab National Park help to conserve endangered species. [3]
- (b) Astola Island became Pakistan's first marine protected area. Suggest two activities that might be restricted in a marine protected area and explain how one of them benefits marine species. [3]
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- (a) Habitat is protected from development, farming and logging; hunting is banned or controlled; whole ecosystems are protected, so food chains and interactions are kept; species can be monitored and studied.
1 mark each for any three of: habitat protected; hunting/poaching controlled; whole ecosystem/all species protected; monitoring/research; controlled tourism raises money.
- (b) Fishing (especially trawling or net fishing); collecting corals/shells; anchoring boats; dumping waste/pollution; development of the coastline. E.g. banning fishing lets fish populations recover and breed; this provides food for other species and fish spread to surrounding areas.
1 mark each for two restricted activities. 1 mark for a valid explanation of benefit.
- (a) Outline the roles of zoos in the conservation of endangered species such as the snow leopard. [4]
- (b) Explain why a captive population started from only a few founder animals may have problems after several generations. [3]
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- (a) Captive breeding to increase numbers; maintaining genetic diversity using studbooks and exchanging animals between zoos (avoiding inbreeding); reintroduction of animals into the wild; research into behaviour, nutrition and reproduction; education of the public and fund-raising for conservation in the wild; keeping a safe population if wild populations are lost.
1 mark each for any four of: captive breeding; studbooks/exchange to keep genetic diversity; reintroduction; research; education/raising awareness; fund-raising; safe 'reserve' population; assisted reproduction (IVF, AI).
- (b) Small gene pool / low genetic diversity; inbreeding increases the chance of offspring being homozygous for harmful recessive alleles; this reduces fertility and survival (inbreeding depression) and the population is less able to adapt to change or disease.
1 mark: low genetic diversity / small gene pool. 1 mark: inbreeding → more homozygous recessive harmful alleles expressed. 1 mark: reduced fitness/fertility or less able to adapt (e.g. to disease).
- (a) Describe how in vitro fertilisation (IVF) and embryo transfer could be used to produce offspring of a rare antelope, using a female of a common, closely related species as a surrogate mother. [4]
- (b) Explain the advantages of using a surrogate of a common species and of keeping frozen embryos and sperm in a 'frozen zoo'. [3]
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- (a) Give the rare female hormones to stimulate many eggs to mature (superovulation); collect eggs (from ovaries); collect sperm from a rare male (or use frozen sperm); mix eggs and sperm in a dish (in vitro) for fertilisation; grow embryos for a few days in culture; implant embryo into the uterus of the surrogate female of the common species, whose cycle has been synchronised with hormones.
1 mark each for any four of: hormones to stimulate egg production; eggs collected; sperm collected / thawed from frozen store; fertilisation in a dish/outside the body; embryos cultured; embryos implanted into surrogate; surrogate's cycle synchronised / prepared with hormones.
- (b) Rare females are not put at risk by pregnancy and can produce more eggs/offspring; more offspring per year from each rare female; frozen material can be stored for many years and transported between countries instead of animals; genetic diversity is preserved and can be reintroduced into future populations, even from animals that have died.
1 mark each for any three of: protects rare female / more offspring; many embryos per rare female; long-term storage; easy transport between zoos/countries; preserves genetic diversity for the future; use material from dead animals.
- (a) Describe how seeds are prepared and stored in a seed bank. [3]
- (b) Many farmers in northern Pakistan have replaced old local wheat varieties with modern high-yielding varieties. Explain why seeds of the old local varieties are being collected and stored. [3]
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- (a) Seeds are cleaned and checked; dried to a low water content (about 5 %); sealed in airtight containers; stored at low temperature (about −20 °C); samples are germinated regularly to check viability, and new seed is grown when viability falls.
1 mark each for any three of: dried to low water content; airtight containers; low temperature (e.g. −20 °C); germination tests at intervals; regrow to collect fresh seed; records kept.
- (b) Old varieties contain alleles not found in modern varieties (genetic diversity); some may give resistance to diseases, drought, cold or salinity; these alleles could be bred into future crop varieties as the climate or pathogens change; once the varieties are no longer grown they may be lost for ever.
1 mark: preserves genetic diversity / alleles. 1 mark: useful traits (e.g. drought or disease resistance). 1 mark: for future breeding / prevent loss (extinction of variety).
- (a) Outline the role of CITES in conservation. [2]
- (b) In one mountain area of Pakistan, a small number of permits to hunt old male markhor are sold each year to foreign hunters. Most of the money goes to the local community, which employs game guards. Suggest how this scheme can help the markhor population to increase. [3]
- (c) The markhor's IUCN Red List category improved from Endangered to Near Threatened. State what the IUCN Red List is. [1]
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- (a) CITES is an international agreement that regulates or bans international trade in endangered species and products made from them; species are listed in appendices according to how threatened they are (e.g. Appendix I – commercial trade banned; others need permits).
1 mark: international agreement controlling trade in endangered species / their products. 1 mark: species listed in appendices with different levels of restriction / permits needed.
- (b) Local people gain income from markhor, so they value live animals and stop poaching; game guards protect markhor from illegal hunting; only a few old males past their prime breeding age are taken, so the breeding population is hardly affected; money can also be used for habitat protection and community projects.
1 mark each for any three of: income gives incentive to conserve; reduced poaching / guards employed; only few old males removed so breeding not affected; funds habitat protection; community support for conservation.
- (c) A list that assesses and classifies species according to their risk of extinction (e.g. Critically Endangered, Endangered, Vulnerable).
1 mark.
Prosopis juliflora, a shrub from Central and South America, was introduced to Pakistan to provide fuel wood and to stabilise sand. It has since spread over large areas of grazing land and river banks in Sindh.
- (a) Suggest why Prosopis juliflora has been able to spread so rapidly in Sindh. [3]
- (b) Explain how the spread of an invasive species such as P. juliflora can reduce biodiversity. [2]
- (c) Suggest two methods that could be used to control P. juliflora and give one disadvantage of one of them. [3]
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- (a) No natural predators, herbivores or diseases that control it in its new range; it tolerates drought and salty soils better than native plants; it produces many seeds that are spread by livestock eating the pods; it outcompetes native plants for water and light.
1 mark each for any three of: lack of natural predators/pathogens; tolerance of drought/salinity; large seed production; seeds dispersed by animals; outcompetes native species (deep roots for water, shading).
- (b) It outcompetes native plants, which decline or disappear locally; animals that depend on those plants for food or habitat also decline, so species diversity (and the variety of habitats) falls.
1 mark: native plants outcompeted / lost. 1 mark: knock-on loss of animals that depend on them / habitat becomes uniform.
- (c) Mechanical removal (uprooting/cutting); herbicides; using the wood for fuel/charcoal to encourage harvesting; introducing a biological control agent (seed-eating insect) after testing; replanting native species. Disadvantage: e.g. herbicides may kill native plants or pollute water; cutting is labour-intensive and stumps resprout; a biological control agent might attack native species.
1 mark each for two methods. 1 mark for a valid disadvantage of one of them.
- (a) Explain why the elephant population in a fenced reserve may need to be controlled. [2]
- (b) Discuss the advantages and disadvantages of culling and of contraception as methods of controlling the population. [4]
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- (a) With no natural predators and no way to migrate out of the reserve, the population grows beyond the carrying capacity; elephants destroy trees and vegetation, reducing habitat and food for other species (and for themselves), lowering biodiversity.
1 mark: population exceeds carrying capacity / no predators or migration. 1 mark: overgrazing / habitat damage reduces biodiversity.
- (b) Culling: reduces numbers quickly; meat/ivory/hide may provide income; but it is unpopular/ethically controversial, disturbs social groups and causes stress to remaining animals. Contraception (e.g. immunocontraceptive vaccine darted into females): humane and reversible, keeps social groups intact; but slow to reduce numbers (elephants are long-lived), must be repeated, expensive and needs each female to be identified and treated.
1 mark each for any four, with at least one advantage and one disadvantage for each method: culling fast / income vs ethical objection / social disruption; contraception humane / reversible / groups intact vs slow effect / repeated treatment / cost / labour.
Forty captive-bred gazelles were released into a protected desert reserve in 2014. The population was counted every two years.Year: 2014, 2016, 2018, 2020, 2022, 2024Number: 40, 52, 71, 66, 94, 118
- (a) Calculate the percentage increase in population from release (2014) to 2024. [2]
- (b) Suggest reasons for the fall in numbers between 2018 and 2020. [2]
- (c) All 40 animals released came from one captive population. Discuss whether this reintroduction can be considered a long-term success. [3]
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- (a) 195 %
M1: (118 − 40) ÷ 40 × 100. A1: 195 %.
- (b) Drought/reduced food or water; disease; predation; poaching; animals moving out of the reserve.
1 mark each for any two.
- (c) Success: population has nearly tripled and is breeding in the wild. Concerns: all descended from 40 related animals, so genetic diversity is low – risk of inbreeding and less ability to adapt to disease or climate change; numbers fluctuate and the population is still small; long-term success needs continued monitoring, protection and perhaps introduction of unrelated animals.
1 mark: evidence of success (increase, breeding). 1 mark: low genetic diversity / founder effect / inbreeding risk. 1 mark: still small/fluctuating, needs further protection or new genetic stock.
- (a) List the taxonomic ranks in order from domain to species. [2]
- (b) The tiger is Panthera tigris and the lion is Panthera leo. State the lowest taxonomic rank that both share and explain how you know. [2]
- (c) State the domain and kingdom to which the tiger belongs. [1]
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- (a) Domain, kingdom, phylum, class, order, family, genus, species.
2 marks all in correct order; 1 mark with one error.
- (b) Genus; both have the same first part of the binomial name (Panthera), which is the genus.
1 mark: genus. 1 mark: first name in binomial = genus.
- (c) Eukarya; Animalia.
1 mark (both needed).
- (a) Copy and complete the table to compare the three domains.Feature | Bacteria | Archaea | Eukaryanucleus surrounded by envelope | ? | ? | ?size of ribosomes in cytoplasm | ? | ? | ?peptidoglycan in cell wall | ? | ? | ?membrane lipids | ? | ? | ? [4]
- (b) Suggest why organisms in the domain Archaea were once classified with bacteria. [1]
- (c) State one feature in which Archaea are more similar to Eukarya than to Bacteria. [1]
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- (a) nucleus: absent, absent, presentribosomes: 70S, 70S, 80Speptidoglycan: present, absent, absentmembrane lipids: unbranched fatty acids with ester bonds; branched chains with ether bonds; unbranched fatty acids with ester bonds
1 mark per correct row.
- (b) They are prokaryotic: single cells with no nucleus or membrane-bound organelles, similar in size and shape to bacteria.
1 mark.
- (c) Their RNA polymerase is more complex (similar to that of eukaryotes) / some have histone proteins associated with DNA / similar initiation of protein synthesis.
1 mark.
Descriptions of four eukaryotic organisms:W: multicellular body made of thread-like hyphae; cell walls of chitin; feeds by secreting enzymes and absorbing the products; stores glycogenX: single cell with chloroplasts and one flagellum; no cell wall; can also take in food when kept in the darkY: multicellular; cellulose cell walls; chloroplasts; stores starchZ: multicellular; no cell walls; heterotrophic, ingests food; nervous coordination
- (a) Identify the kingdom to which each organism belongs. [4]
- (b) Explain why organism X is not placed in the plant kingdom, even though it has chloroplasts. [2]
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- (a) W Fungi; X Protoctista; Y Plantae; Z Animalia.
1 mark each.
- (b) It is unicellular and has no cellulose cell wall; plants are multicellular with cellulose cell walls (and it can move by a flagellum and feed heterotrophically).
1 mark: unicellular (plants are multicellular). 1 mark: no cellulose cell wall / has flagellum.
Information about the genetic material of four viruses:Influenza virus: RNA, one strand, in eight segmentsSmallpox (variola) virus: DNA, two complementary strandsT4 bacteriophage: DNA, two complementary strandsHIV: RNA, one strand (two identical copies per particle)
- (a) Explain why viruses are not classified in any of the three domains. [2]
- (b) State the two features of the nucleic acid that are used to classify viruses. [2]
- (c) Use these features to classify each virus. [2]
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- (a) Viruses are not cells (acellular: no cytoplasm, ribosomes or cell membrane); they cannot metabolise or reproduce outside a host cell.
1 mark: not made of cells / no ribosomes or cytoplasm. 1 mark: only replicate inside host cells / no metabolism.
- (b) Whether it is DNA or RNA; whether it is single-stranded or double-stranded.
1 mark each.
- (c) Influenza virus: single-stranded RNA; smallpox virus: double-stranded DNA; T4 bacteriophage: double-stranded DNA; HIV: single-stranded RNA.
2 marks all four; 1 mark for two or three.
A segment of the gene for ribosomal RNA was sequenced in four organisms. The table shows the number of base differences between each pair of organisms in a 100-base region.J–K: 31; J–L: 33; J–M: 6; K–L: 35; K–M: 30; L–M: 32
- (a) Using the data, state which organism is most closely related to organism J and explain your answer. [2]
- (b) Organisms J and M are prokaryotes that live in hot, acidic springs. Organism K is a bacterium found in soil. Organism L is a yeast. Suggest which domain J and M belong to, giving a reason from the data. [2]
- (c) Explain why comparing sequences of genes such as ribosomal RNA genes led to the three-domain classification system. [2]
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- (a) Organism M; it has the fewest differences in its rRNA gene sequence from J (6), so they shared a common ancestor most recently.
1 mark: M. 1 mark: fewest base differences → most recent common ancestor.
- (b) Archaea; J and M differ from K (a bacterium) by about as many bases as from L (a eukaryote), so they are not closely related to bacteria even though they are prokaryotes.
1 mark: Archaea. 1 mark: as different from bacterium K as from eukaryote L (data quote: 31 and 33 differences).
- (c) rRNA genes are found in all organisms and change slowly by mutation; the number of differences reflects time since organisms shared a common ancestor; sequencing showed archaea were as different from bacteria as from eukaryotes, so three groups were needed.
1 mark: all organisms have the gene / differences accumulate over time (molecular clock). 1 mark: archaea shown to be distinct from bacteria, giving three domains.
- (a) Biodiversity can be considered at three levels. Name and describe each level. [3]
- (b) Describe how a student could place quadrats randomly in a grassland, and explain why random sampling is used. [3]
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- (a) Ecosystem (habitat) diversity – the range of different ecosystems/habitats in an area; species diversity – the number of different species and the evenness of their abundance (species richness and relative abundance); genetic diversity – the variety of alleles within a species/population.
1 mark each: ecosystem/habitat diversity; species diversity (number and relative abundance of species); genetic diversity (variation of alleles within species).
- (b) Lay two tapes at right angles along two sides of the area to form axes; use a random number generator to select pairs of coordinates; place the quadrat at each coordinate. Random sampling avoids bias in choosing where to sample, so the samples are representative of the whole area.
1 mark: grid / tapes as axes. 1 mark: random number generator / table for coordinates. 1 mark: avoids bias, representative samples (allows statistical testing).
Pitfall traps were set in a cotton field near Multan. 60 ground beetles were caught, marked with a dot of non-toxic paint on the underside and released. Two days later 75 beetles were caught, of which 12 were marked.
- (a) Calculate the estimated population size of ground beetles using the Lincoln index. [2]
- (b) State three assumptions that must be true for this estimate to be valid. [3]
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- (a) 375 beetles
M1: N = (n₁ × n₂) ÷ m = (60 × 75) ÷ 12. A1: 375.
- (b) Marked beetles mix randomly with the rest of the population; marks do not rub off; marking does not affect survival or behaviour (e.g. make beetles more visible to predators); no significant births, deaths, immigration or emigration between samples; all beetles equally likely to be caught.
1 mark each for any three.
A student counted the plants of each species in quadrats in two areas of the Margalla Hills: site A, natural forest edge, and site B, a pine plantation.Site A (species 1–6): 18, 12, 9, 7, 3, 1Site B (species 1–4): 38, 6, 4, 2
- (a) Calculate Simpson's index of diversity (D = 1 − Σ(n/N)²) for each site. Give your answers to 2 decimal places. [3]
- (b) Compare the biodiversity of the two sites and suggest a reason for the difference. [2]
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- (a) Site A: D = 0.76; site B: D = 0.40
M1: correct use of (n/N)² values for one site (e.g. A: 0.1296 + 0.0576 + 0.0324 + 0.0196 + 0.0036 + 0.0004 = 0.2432). A1: A = 0.76. A1: B = 1 − (0.5776 + 0.0144 + 0.0064 + 0.0016) = 0.40.
- (b) Site A (natural forest) has higher diversity: more species and more even numbers; site B is dominated by one species. The plantation has one tree species, so fewer niches/habitats and food sources; management (clearing undergrowth) removes other plants.
1 mark: A higher diversity – more species and more even abundance (B dominated by species 1). 1 mark: plantation = single tree species / fewer niches / management.
An ecologist recorded the percentage tree cover and the number of bird species at eight sites around Islamabad.Site: 1, 2, 3, 4, 5, 6, 7, 8Tree cover / %: 12, 25, 38, 47, 55, 63, 71, 84Bird species: 9, 14, 12, 19, 17, 24, 28, 26
- (a) State a null hypothesis for this investigation. [1]
- (b) Calculate Spearman's rank correlation coefficient, rs = 1 − 6Σd² ÷ (n(n² − 1)). Show your working. Give your answer to 2 decimal places. [3]
- (c) The critical value for n = 8 at p = 0.05 is 0.738. Explain what your result shows. [2]
- (d) Suggest one reason why a high tree cover is associated with more bird species. [1]
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- (a) There is no correlation between percentage tree cover and the number of bird species.
1 mark.
- (b) rs = 0.93
M1: ranks – tree cover 1–8 in order; bird species ranks 1, 3, 2, 5, 4, 6, 8, 7. M1: Σd² = 6. A1: 1 − 36 ÷ 504 = 0.93.
- (c) 0.93 is greater than 0.738, so reject the null hypothesis; there is a significant positive correlation between tree cover and the number of bird species (more tree cover, more bird species).
1 mark: rs > critical value, reject null. 1 mark: significant positive correlation (not chance at p < 0.05).
- (d) More trees provide more niches – nesting sites, shelter and food sources (insects, fruit) for different species.
1 mark.
A student placed 10 quadrats, each 0.5 m × 0.5 m (0.25 m²), at random on a school playing field of area 1200 m² and counted the dandelion plants in each.Numbers: 3, 5, 2, 0, 4, 6, 1, 3, 2, 4
- (a) Calculate the mean number of dandelion plants per quadrat. [1]
- (b) Calculate the mean density of dandelions per m². [1]
- (c) Estimate the total number of dandelions on the field. [1]
- (d) Suggest why this estimate may not be accurate. [1]
Show answer
- (a) 3.0
30 ÷ 10 = 3.0.
- (b) 12 per m²
3.0 ÷ 0.25 = 12 m⁻².
- (c) 14 400
12 × 1200 = 14 400.
- (d) Only 10 quadrats sampled (small sample); dandelions may be clumped/unevenly distributed; counting errors at quadrat edges.
1 mark for any one valid reason.
A survey estimated the number of mature individuals of a vulture species in a region at 3400 in 2014 and 1450 in 2024.One criterion used by the IUCN states that a population reduction over 10 years of at least 80 % qualifies a species as Critically Endangered, at least 50 % as Endangered and at least 30 % as Vulnerable.
- (a) Calculate the percentage decrease in the population between 2014 and 2024. Give your answer to 3 significant figures. [2]
- (b) Use the criterion given to state the category the species qualifies for, and explain your choice. [1]
- (c) Calculate the mean decrease in number of individuals per year. [1]
- (d) Suggest why a survey of mature individuals, rather than all individuals, is used to judge extinction risk. [1]
Show answer
- (a) 57.4 %
M1: (3400 − 1450) ÷ 3400 × 100. A1: 57.4 %.
- (b) Endangered, because the decline (57.4 %) is at least 50 % but less than 80 %.
1 mark: Endangered with reason.
- (c) 195 individuals per year
(3400 − 1450) ÷ 10 = 195 per year.
- (d) Only mature individuals can breed and produce the next generation, so their number determines whether the population can recover.
1 mark.
Students compared seaweed cover on two rocky shores near Karachi. On each shore they chose five places that looked typical, placed a 1 m² quadrat at each and estimated by eye the percentage cover of each seaweed species. One shore was sampled at low tide on a morning, the other in the afternoon a week later when the tide was coming in.
- (a) Identify three limitations of this method and suggest an improvement for each. [6]
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- (a) Percentage cover estimated by eye is subjective – use a quadrat divided into 100 small squares/point quadrat, and the same person to estimate. Quadrats placed where the student chose – use random coordinates (or regular intervals along a belt transect). Only 5 quadrats – sample more (e.g. 20) until the running mean levels off. Some species may be misidentified – use an identification key/photograph. Samples taken at different states of the tide or on different days – sample at the same time on the same day.
1 mark per limitation and 1 mark per matching improvement, for three pairs (max 6).
Students used mark–release–recapture to estimate the number of land snails in a garden. They noted three problems:1: water-based paint used to mark the shells washed off in rain2: bright paint was put on the top of the shell, where birds could see it3: the second sample was taken 20 minutes after release, close to the release point
- (a) For each problem, explain how it would affect the population estimate (too high or too low). [3]
- (b) Suggest an improvement for each problem. [3]
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- (a) 1: marked snails in the second sample are fewer than they should be, so the estimate is too high. 2: if marked snails are eaten more, fewer marked are recaptured, so the estimate is too high. 3: marked snails have not mixed, so recapturing near the release point gives too many marked snails (estimate too low) – or too few if sampling elsewhere (too high).
1 mark each: (1) too high; (2) too high; (3) explained either way – not randomly mixed, so ratio not representative.
- (b) 1: use waterproof, non-toxic paint/permanent marker and check it does not rub off in a trial. 2: mark on the underside or in an inconspicuous place / use a dull colour. 3: leave a longer time (e.g. 24 h) for mixing and sample from the whole area.
1 mark per improvement.
Belt transects were used to sample butterflies at six urban and six rural sites near Lahore. Simpson's index of diversity was calculated for each site.Urban: mean D = 0.62, 95 % confidence interval ± 0.07Rural: mean D = 0.81, 95 % confidence interval ± 0.05
- (a) A student concluded that butterfly diversity is higher in rural sites than in urban sites. Use the data to evaluate this conclusion. [3]
- (b) Suggest two reasons why butterfly diversity is lower in urban areas. [2]
Show answer
- (a) Mean D is higher in rural sites (0.81 vs 0.62). The 95 % confidence intervals (rural 0.76–0.86, urban 0.55–0.69) do not overlap, so the difference is significant and the conclusion is supported for these sites. But only 6 sites of each type, sampled in one season/year, in one city, so it may not apply generally.
1 mark: rural higher with data. 1 mark: CIs do not overlap so significant. 1 mark: limitation – few sites / one season / one region.
- (b) Fewer flowering plants/nectar sources and larval food plants; habitat loss and fragmentation by buildings and roads; pollution/pesticide use; higher temperature/light at night.
1 mark each.
Lichen diversity on tree trunks was recorded at seven distances (5 m to 200 m) from a busy road. Spearman's rank correlation between distance from the road and lichen diversity gave rs = 0.86. The critical value for n = 7 at p = 0.05 is 0.79.
- (a) A student concluded that pollution from vehicles on the road causes the decrease in lichen diversity. Evaluate this conclusion. [4]
- (b) Suggest one further measurement that would test whether vehicle pollution is the cause. [1]
Show answer
- (a) The significant positive correlation shows lichen diversity increases with distance from the road (rs = 0.86 > 0.79), which is consistent with pollution from traffic. However, correlation does not prove causation; other variables also change with distance, e.g. dust, light, humidity, trampling, tree species or bark pH; pollution was not measured; only one road and 7 points, so the conclusion is tentative.
1 mark: correlation significant and positive, supports link. 1 mark: correlation ≠ causation. 1 mark: named confounding variable(s) changing with distance. 1 mark: pollution not measured / small sample / one road.
- (b) Measure the concentration of nitrogen dioxide/sulfur dioxide (or particulates) in the air at each distance and correlate with lichen diversity.
1 mark.
- (a) Plan an investigation to compare the species richness of plants in a grazed area and an ungrazed (fenced) area of hillside. [6]
Show answer
- (a) Choose two areas of similar slope, aspect and soil; mark out each with tapes as axes; use random number coordinates to place 1 m² (or 0.25 m²) quadrats; use at least 20 quadrats in each area (or until a running mean levels off); count the number of different plant species in each quadrat, using an identification key; also record abundance/percentage cover if Simpson's index is to be calculated; sample both areas at the same time of year; calculate mean species per quadrat and compare with a t-test.
1 mark each for any six of: areas matched for other variables (slope, aspect, soil); random sampling method with coordinates; quadrat size stated; sufficient number of quadrats (≥10, ideally 20+); DV species per quadrat identified with key; same time/season; abundance recorded for Simpson's index; means calculated; t-test to compare.
- (a) Plan a field investigation into how the abundance of a fern changes with light intensity from the inside of a woodland into an open field. [6]
- (b) Name the statistical test that should be used to analyse the results and explain your choice. [1]
Show answer
- (a) Lay a tape from inside the wood, across the edge and into the field; place quadrats at regular intervals (e.g. every 2 m) along the tape – a belt or interrupted belt transect; in each quadrat record the abundance of the fern as percentage cover or number of fronds; measure light intensity at ground level in each quadrat with a light meter, at the same height and at the same time of day (or several readings averaged); repeat with two or three more transects at random positions along the wood edge; also record soil moisture/temperature as other variables; plot abundance against light intensity.
1 mark each for any six of: transect line across the edge; quadrats at regular intervals (belt/interrupted belt); abundance measure (% cover, count); light meter readings at each quadrat; same time of day / weather / height of meter; repeat transects; record other variables (soil moisture, pH, temperature); at least 10 sampling points; plot scatter graph of abundance against light.
- (b) Spearman's rank correlation – to test for a correlation between two variables (light intensity and abundance) whose data may not be normally distributed.
1 mark: Spearman's rank with reason (testing correlation between two variables).
A student wants to find out whether the number of mayfly larvae in streams flowing into Rawal Lake is related to the concentration of dissolved oxygen in the water. She will use a kick-sampling net for 1 minute at each site and measure dissolved oxygen with a meter.
- (a) Identify the independent and dependent variables. [2]
- (b) In a field study the other variables cannot be controlled. Name three variables that might affect the number of mayfly larvae and describe how the student could take account of them. [4]
Show answer
- (a) Independent: dissolved oxygen concentration (site). Dependent: number (abundance) of mayfly larvae.
1 mark each.
- (b) Any three: water temperature; flow rate; type of stream bed (stones vs mud); depth; pH; shading. Choose sites that are similar in these variables (e.g. same depth and stone size), and measure them at each site (thermometer, flow meter, pH probe) so that their effects can be considered.
1 mark each for three variables (max 3). 1 mark for: choose similar sites or measure and record them to take into account.
A wetland plant was counted in four zones of equal area around a pond. The total was 120 plants.Zone 1: 28; zone 2: 41; zone 3: 19; zone 4: 32
- (a) State the null hypothesis. [1]
- (b) Calculate the value of χ², using χ² = Σ(O − E)²/E. Give your answer to 2 decimal places. [3]
- (c) The critical value at p = 0.05 for 3 degrees of freedom is 7.82. Explain what the result shows. [2]
Show answer
- (a) The plants are distributed evenly (randomly) across the four zones; any difference from the expected numbers is due to chance.
1 mark.
- (b) χ² = 8.33
M1: E = 120 ÷ 4 = 30 for each zone. M1: (O − E)²/E = 0.133, 4.033, 4.033, 0.133. A1: 8.33.
- (c) 8.33 > 7.82, so reject the null hypothesis; the difference between observed and expected numbers is significant – the plants are not evenly distributed (more in zone 2, fewer in zone 3).
1 mark: χ² greater than critical value, reject null. 1 mark: distribution significantly different from even / not due to chance.
Students will collect invertebrates by kick-sampling in a shallow, fast-flowing stream in the hills and measure the water temperature and flow rate.
- (a) Identify three hazards of this fieldwork and suggest a precaution for each. [3]
- (b) State the overall level of risk and give a reason. [1]
Show answer
- (a) Slipping or falling into the stream/drowning – work in pairs, wear boots with grip, stay in shallow water, check water level and weather. Infection from contaminated water (e.g. bacteria from animal urine or sewage) – wear gloves, cover cuts, wash hands. Sunburn/heat stroke – hat, sunscreen, drinking water. Insect or snake bites – long sleeves, look before placing hands. Getting lost – tell someone the route and return time.
1 mark for each hazard with a matching precaution (max 3).
- (b) Medium – working in or near flowing water has a risk of serious harm, although it can be reduced by the precautions.
1 mark: level with justification (accept low with good justification of shallow water and supervision).
An ecologist plans to measure Simpson's index of diversity of plants in quadrats at 100 m intervals from 100 m to 1000 m from a cement factory, where dust from the factory settles on the vegetation.
- (a) State a testable hypothesis for this investigation. [1]
- (b) State the null hypothesis. [1]
- (c) Predict the results and explain your prediction. [2]
Show answer
- (a) Species diversity of plants increases as distance from the cement factory increases.
1 mark.
- (b) There is no correlation between distance from the factory and plant species diversity.
1 mark.
- (c) Diversity will be lowest near the factory and increase with distance, because dust settling on leaves near the factory blocks light and stomata (and changes soil pH), so only a few tolerant species survive there.
1 mark: prediction (diversity increases with distance). 1 mark: explanation (dust reduces photosynthesis / alters soil pH; only tolerant species survive).
