AS & A Level Biology (9700)
Selection & evolution Banker
50 questions · about 6% 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.
Q1Natural & artificial selection · Core · Paper 4 style · 6 marksSince 2016, many cases of typhoid in Sindh have been caused by strains of Salmonella Typhi that are resistant to several antibiotics (extensively drug-resistant typhoid).
- (a) Explain how a population of bacteria can become resistant to an antibiotic. [4]
- (b) Suggest two ways in which the spread of antibiotic-resistant bacteria can be reduced. [2]
Show answer
- (a) A random mutation produces an allele giving resistance (e.g. coding for an enzyme that breaks down the antibiotic); when the antibiotic is used it acts as a selection pressure; bacteria without the allele are killed, resistant bacteria survive and reproduce (binary fission), passing on the allele; the allele frequency increases over generations; (resistance alleles on plasmids can also pass between bacteria by horizontal gene transfer/conjugation).
1 mark each for any four of: mutation (random, already present); antibiotic is the selection pressure; resistant survive and reproduce; allele passed to offspring; allele frequency increases; plasmid transfer.
- (b) Any two of: use antibiotics only when needed/prescribed; complete the full course; avoid using antibiotics in animal feed; improve sanitation and clean drinking water; vaccination against typhoid; infection control/hand washing.
1 mark each.
- (a) Describe what is meant by stabilising, directional and disruptive selection. [3]
- (b) Human birth mass is an example of stabilising selection. Explain why. [2]
Show answer
- (a) Stabilising: individuals with intermediate phenotypes are favoured, extremes selected against; the mean stays the same and variation decreases. Directional: individuals at one extreme are favoured; the mean shifts in that direction. Disruptive: both extremes are favoured over the intermediate; can produce two distinct groups.
B1 each.
- (b) Babies of very low birth mass have a higher death rate (less able to keep warm/more infections) and very high birth mass increases complications in birth; intermediate birth masses have the highest survival, so the mean is maintained and variation is reduced.
B1 extremes have higher mortality; B1 intermediate favoured/mean maintained.
Sahiwal cattle are a dairy breed from Punjab, valued for tolerating heat and resisting ticks. Breeders want to increase their milk yield.
- (a) Describe how breeders could use artificial selection to increase the milk yield of Sahiwal cattle over many generations. [4]
- (b) Suggest why breeders should also keep selecting for heat tolerance while selecting for milk yield. [1]
- (c) State one difference between artificial selection and natural selection. [1]
Show answer
- (a) Measure milk yield of cows and select those with the highest yield; bulls cannot be measured directly, so select bulls whose mothers/daughters (progeny testing) have high yields; breed selected cows with selected bulls (e.g. by artificial insemination); select the highest-yielding offspring and breed them; repeat over many generations so the frequency of alleles for high yield increases.
1 mark each for any four of the points.
- (b) Selecting only for yield may reduce/lose alleles for heat tolerance or tick resistance, which are needed in Pakistan's climate.
B1.
- (c) In artificial selection humans choose which individuals breed, for characteristics useful to humans; in natural selection the environment/selection pressures decide which survive and reproduce.
B1.
Maize farmers in Khyber Pakhtunkhwa buy new hybrid maize seed every year. The seed is produced by crossing two inbred (homozygous) lines.
- (a) Explain why the inbred lines are often small and produce low yields. [2]
- (b) Explain why the F1 hybrid plants are vigorous and uniform. [2]
- (c) Explain why farmers do not save seed from their hybrid crop to sow the next year. [2]
Show answer
- (a) Repeated inbreeding increases homozygosity; harmful recessive alleles are more likely to be expressed in the homozygous state (inbreeding depression).
B1 homozygous; B1 harmful recessive alleles expressed.
- (b) Hybrids are heterozygous at many loci, so harmful recessive alleles are masked (hybrid vigour); both parents are homozygous, so all F1 plants have the same genotype and are uniform.
B1 heterozygous/hybrid vigour; B1 identical genotypes, uniform.
- (c) Seed from F1 × F1 is the F2 generation; alleles segregate, producing many different genotypes, so plants are not uniform and many have lower yield.
B1 segregation/variety of genotypes in F2; B1 not uniform/lower yield.
Stem rust is a fungal disease that can destroy wheat crops. Wheat breeders in Pakistan cross local high-yielding varieties with varieties that carry a resistance allele.
- (a) Outline how breeders could produce a high-yielding variety that is resistant to stem rust. [3]
- (b) New races of the stem rust fungus sometimes appear that can infect previously resistant wheat. Explain how this happens. [3]
Show answer
- (a) Cross a high-yielding variety with a resistant variety; grow offspring and expose them to the fungus, selecting plants that are resistant and have high yield; breed these together/backcross to the high-yielding parent for several generations, selecting each time, until the variety is true-breeding for both features.
B1 cross the two varieties; B1 select resistant, high-yield offspring (test by infecting); B1 repeat for many generations/backcrossing.
- (b) Mutation in the fungus produces a new allele allowing it to infect resistant wheat; widely planted resistant wheat is a selection pressure; fungi with the new allele reproduce, so the new race spreads/allele frequency increases.
B1 mutation in fungus; B1 resistant wheat selects for it; B1 the new race survives, reproduces and spreads.
Rock pocket mice live in deserts in North America. Most mice living on pale sandy rock have pale fur; most mice living on dark lava flows have dark fur. Dark fur is caused by a dominant allele of the Mc1r gene. Owls, which hunt by sight, are their main predators.
- (a) Explain how natural selection has led to most mice on lava flows having dark fur. [4]
- (b) Suggest why a new dominant mutation for dark fur could spread through a population faster than a new recessive mutation would. [2]
Show answer
- (a) There is variation in fur colour in the population, caused by different alleles; owls see and catch pale mice more easily on dark rock (predation is the selection pressure); dark mice are camouflaged, so more survive and reproduce; they pass on the dark allele, so its frequency increases over generations.
1 mark each for: variation/alleles; predation as selection pressure; dark mice survive and reproduce; allele passed on/frequency increases.
- (b) A dominant allele is expressed in heterozygotes, so it shows in the phenotype immediately and is selected for at once; a recessive allele is hidden in heterozygotes and only selected when homozygous, which is rare at first.
B1 dominant expressed in heterozygote/visible to selection; B1 recessive hidden until homozygous.
Mosquitoes of the genus Anopheles transmit malaria. In one district, the same insecticide was sprayed in houses every year. The percentage of mosquitoes surviving a standard dose was: year 1 – 2 %; year 3 – 9 %; year 5 – 31 %; year 7 – 68 %. Spraying then stopped. In year 10 the survival was 40 %.
- (a) Explain the change in survival between years 1 and 7. [4]
- (b) Suggest why survival fell between year 7 and year 10. [2]
Show answer
- (a) Some mosquitoes already carried a resistance allele (from mutation) at a low frequency (2 % survival); the insecticide killed non-resistant mosquitoes – a selection pressure; resistant mosquitoes survived and reproduced, passing on the allele; the frequency of the resistance allele increased each generation (directional selection), so survival rose to 68 %.
1 mark each for any four of: pre-existing allele/mutation; selection pressure; survivors reproduce; allele passed on; allele frequency increases; directional selection; use of data.
- (b) Without insecticide there is no advantage to resistance; the resistance allele may carry a cost (e.g. lower fertility or slower development), so non-resistant mosquitoes are selected for/out-reproduce resistant ones and the allele frequency falls.
B1 no selection for resistance; B1 resistant individuals at a disadvantage without insecticide.
Sickle cell anaemia is caused by the allele Hbˢ. People with genotype HbˢHbˢ have sickle cell anaemia; heterozygotes (HbᴬHbˢ) are mostly healthy and are less likely to die from severe malaria than HbᴬHbᴬ people. The Hbˢ allele is much more common in parts of Africa where malaria is common than in places without malaria.
- (a) Explain why the Hbˢ allele remains at a high frequency in areas where malaria is common. [4]
- (b) Suggest why the frequency of Hbˢ is falling slowly among descendants of people who moved from these areas to countries without malaria. [2]
Show answer
- (a) HbˢHbˢ people often die young/have fewer children, so Hbˢ alleles are lost; HbᴬHbᴬ people are more likely to die of malaria, so Hbᴬ alleles are lost; heterozygotes have a selective advantage – they survive malaria and reproduce, passing on both alleles; so both alleles are kept in the population (balanced polymorphism/heterozygote advantage).
1 mark each for: selection against HbˢHbˢ; selection against HbᴬHbᴬ by malaria; heterozygote advantage; both alleles maintained.
- (b) Malaria is no longer a selection pressure, so heterozygotes have no advantage; HbˢHbˢ individuals are still selected against, so the allele frequency falls slowly (slowly because most Hbˢ alleles are carried in heterozygotes, where they are not selected against).
B1 no heterozygote advantage without malaria; B1 selection against homozygotes only, so slow decline.
On an island, a severe drought killed most plants that produce small, soft seeds, leaving mainly large, hard seeds. The mean beak depth of a population of seed-eating finches was 9.2 mm before the drought and 9.7 mm in the generation hatched after it.
- (a) Calculate the percentage increase in mean beak depth. Give your answer to 3 significant figures. [2]
- (b) Name the type of selection shown and explain how the change happened. [4]
- (c) Explain why the change in mean beak depth would not have occurred if beak depth were controlled only by the environment. [1]
Show answer
- (a) 5.43 %
M1 (9.7 − 9.2) ÷ 9.2 × 100; A1 5.43 %.
- (b) Directional selection; beak depth varies and is partly genetic (polygenic); during the drought only birds with deeper beaks could crack the large, hard seeds; so they survived and bred, while birds with shallow beaks starved; alleles for deep beaks were passed on, so the mean in the next generation increased.
B1 directional; B1 variation with genetic basis; B1 deep-beaked birds survived/reproduced (food as selection pressure); B1 alleles passed on/mean shifted.
- (c) Characteristics caused only by the environment are not inherited, so offspring would not show the change.
B1.
- (a) Define the term 'species' using the biological species concept. [2]
- (b) State the difference between allopatric speciation and sympatric speciation. [2]
Show answer
- (a) A group of organisms with similar features that can interbreed to produce fertile offspring, and are reproductively isolated from other such groups.
B1 interbreed to produce fertile offspring; B1 reproductively isolated from other species.
- (b) Allopatric: populations are separated by a geographical barrier. Sympatric: a new species forms in the same area without geographical separation (e.g. by polyploidy or behavioural isolation).
B1 each.
- (a) State two differences between genetic drift and natural selection. [2]
- (b) Explain why both processes change the gene pool of a population. [2]
Show answer
- (a) Drift is random/due to chance, selection is non-random (depends on phenotype and environment); drift has most effect in small populations, selection acts in any size population; selection leads to adaptation, drift does not.
1 mark each for any two.
- (b) Both change the frequency of alleles in the population from one generation to the next; alleles can be lost or become fixed.
B1 change allele frequencies; B1 alleles lost/fixed.
A species of freshwater fish lived in one large lake. Over thousands of years the water level fell and the lake split into two smaller lakes, one with clear, rocky water and one with muddy water and dense weed.
- (a) Explain how two separate species of fish could evolve from the original population. [5]
Show answer
- (a) The two populations are geographically isolated, so no gene flow/interbreeding between them; different mutations arise in each population; different selection pressures in each lake (e.g. food, predators, water clarity) favour different alleles; genetic drift may also change allele frequencies (especially in small populations); over many generations the gene pools become so different that the two populations could no longer interbreed to produce fertile offspring even if they met (reproductive isolation) – allopatric speciation.
1 mark each for any five of: geographical isolation; no gene flow; different mutations; different selection pressures (with example); drift; allele frequencies change; reproductive isolation/cannot produce fertile offspring.
Bread wheat (Triticum aestivum) is hexaploid with 2n = 42. It is thought to have arisen when tetraploid emmer wheat (2n = 28) hybridised with a wild grass, Aegilops tauschii (2n = 14).
- (a) Calculate the number of chromosomes in the hybrid formed from one gamete of each parent. [1]
- (b) Explain why this hybrid was sterile. [2]
- (c) Explain how a fertile hexaploid species could have formed from the hybrid in the same place as its parents, and why this is an example of sympatric speciation. [3]
Show answer
- (a) 21
B1 14 + 7 = 21.
- (b) The chromosomes from the two parents are not homologous, so they cannot pair up in prophase I of meiosis; viable gametes cannot be formed.
B1 chromosomes not homologous/cannot pair; B1 meiosis fails/no functional gametes.
- (c) The chromosome number of the hybrid doubled (e.g. a failure of cell division/non-disjunction), giving 42; every chromosome now has a homologous partner, so meiosis works and fertile gametes form; the new plant cannot produce fertile offspring with either parent, so it is reproductively isolated without any geographical barrier.
B1 chromosome doubling to 42; B1 homologues can pair/fertile; B1 reproductively isolated in same area.
- (a) Explain what is meant by genetic drift. [2]
- (b) A storm carried six birds of one species to a remote island where none of that species lived. Over 200 years the island population grew to several thousand. Explain why the island population may show less genetic diversity and different allele frequencies from the mainland population (the founder effect). [3]
- (c) Suggest why a population with low genetic diversity may be at greater risk of extinction. [1]
Show answer
- (a) A random (chance) change in allele frequencies in a population from one generation to the next, not caused by selection; it has a larger effect in small populations.
B1 random/chance change in allele frequency; B1 greater effect in small populations.
- (b) The six founders carried only a small sample of the alleles in the mainland gene pool; by chance some alleles were absent and others at a higher frequency than on the mainland; the population grew from these few, so it has lower diversity; drift in the small early population changed frequencies further.
B1 founders carry a small/non-representative sample of alleles; B1 some alleles lost/frequencies different by chance; B1 low diversity as population grows from few individuals/drift.
- (c) Fewer alleles means less variation for natural selection to act on if the environment changes (e.g. a new disease).
B1.
The amino acid sequence of a respiratory protein, 104 amino acids long, was compared between humans and four other species. Number of amino acid differences from the human sequence: species K – 0; species L – 1; species M – 12; species N – 21.
- (a) Calculate the percentage of amino acids in species M's protein that are identical to the human sequence. Give your answer to 1 decimal place. [1]
- (b) Explain how these data provide evidence about the evolutionary relationships of the species. [3]
- (c) Suggest why comparing DNA base sequences of the gene gives more information than comparing amino acid sequences. [2]
Show answer
- (a) 88.5 %
B1 (104 − 12) ÷ 104 × 100 = 88.5 %.
- (b) Differences in amino acid sequence result from mutations in DNA that build up over time; the fewer differences, the more recently the species shared a common ancestor; so K and L are most closely related to humans, and N is the most distantly related.
B1 mutations accumulate over time; B1 fewer differences = more recent common ancestor; B1 correct order using data.
- (c) The genetic code is degenerate, so some base changes do not change the amino acid; these silent mutations (and changes in introns/non-coding DNA) are detected only by DNA comparison.
B1 degenerate code/silent mutations; B1 non-coding DNA also compared.
Scientists compared a 30-base section of mitochondrial DNA from three populations of a mountain goat (markhor) from different valleys.Population X: ATGCTAGCCTTAGGACATCCGTTAGCAATCPopulation Y: ATGCTGGCCTTAGGATATCCGCTAGCAATTPopulation Z: ATACTGGCTTTAGGATATCCACTGGCGATT
- (a) Count the number of base differences between X and Y, X and Z, and Y and Z. [2]
- (b) Use your answers to state which two populations are most closely related and explain your reasoning. [2]
- (c) Suggest why mitochondrial DNA is useful for studying relationships between populations of the same species. [2]
Show answer
- (a) X–Y: 4; X–Z: 9; Y–Z: 5
B2 all three correct (B1 for two).
- (b) X and Y; they have the fewest base differences, so fewest mutations have built up since they shared a common ancestor/they separated most recently.
B1 X and Y; B1 reasoning.
- (c) It is inherited only from the mother, so it does not undergo recombination/crossing over; it mutates relatively quickly, so differences build up even between closely related populations.
B1 maternal inheritance/no recombination; B1 fast mutation rate.
A horse (2n = 64) and a donkey (2n = 62) can mate to produce a mule. Mules are strong and healthy but almost always sterile.
- (a) Calculate the number of chromosomes in a body cell of a mule. [1]
- (b) Explain why mules are sterile. [3]
- (c) Horses and donkeys are separate species even though they can produce offspring. Explain why, and name this type of reproductive isolating mechanism. [2]
Show answer
- (a) 63
B1 32 + 31 = 63.
- (b) The mule has an odd number of chromosomes and the horse and donkey chromosomes are not all homologous; they cannot pair correctly in prophase I of meiosis; so gametes have unbalanced/incorrect chromosome sets and are not viable.
B1 odd number/non-homologous chromosomes; B1 cannot pair in meiosis I; B1 viable gametes not formed.
- (c) Their offspring are sterile, so genes cannot pass between the two gene pools; postzygotic isolation (hybrid sterility).
B1 offspring not fertile/no gene flow; B1 postzygotic/hybrid sterility.
- (a) State the difference between continuous variation and discontinuous variation, giving one example of each in humans. [3]
- (b) Explain why continuous variation is usually affected by both genes and the environment. [2]
Show answer
- (a) Continuous: a range of values with no distinct categories (e.g. height, mass); discontinuous: distinct categories with no intermediates (e.g. ABO blood group).
B1 continuous description; B1 discontinuous description; B1 correct examples of both.
- (b) It is controlled by many genes (polygenes) each having a small additive effect; the environment (e.g. diet) also affects how the genes are expressed, adding to the range of phenotypes.
B1 polygenic/many genes with additive effect; B1 environment modifies phenotype.
The lengths (cm) of eight leaflets taken from one neem tree were: 6.2, 5.8, 7.1, 6.5, 6.9, 5.5, 6.0, 6.8.Standard deviation s = √(Σ(x − x̄)² ÷ (n − 1)).
- (a) Calculate the mean and standard deviation of the leaflet lengths. Give the standard deviation to 2 decimal places. [3]
- (b) A second tree had leaflets with mean length 6.40 cm and standard deviation 1.20 cm. Explain what the two standard deviations show. [2]
- (c) Suggest why leaflets on one tree vary in length even though they all have the same genotype. [1]
Show answer
- (a) mean 6.35 cm; s = 0.57 cm
B1 mean; M1 Σ(x − x̄)² = 2.26; A1 0.57.
- (b) The means are similar but the second tree has a much larger spread of leaflet lengths/more variation around the mean; the first tree's leaflets are more uniform.
B1 means similar; B1 larger SD = more spread/variation.
- (c) Environmental factors differ for each leaflet, e.g. light (shade), position, age.
B1.
A researcher compared the mass at birth of lambs from two breeds of sheep. She calculated t = 1.45 with 18 degrees of freedom. The critical value of t at p = 0.05 for 18 df is 2.10.
- (a) State the null hypothesis that was tested. [1]
- (b) State what the researcher can conclude, and explain your answer. [2]
- (c) State two conditions that the data should meet for a t-test to be suitable. [2]
Show answer
- (a) There is no significant difference between the mean birth mass of lambs of the two breeds.
B1.
- (b) 1.45 < 2.10, so the probability that the difference is due to chance is greater than 0.05; accept the null hypothesis – there is no significant difference in mean birth mass.
B1 t below critical value; B1 accept null/no significant difference.
- (c) Any two of: data are continuous measurements; data are approximately normally distributed; two independent samples being compared; samples should have similar variances; sample size less than about 30 per group.
1 mark each for any two.
Shell lengths (mm) of a species of land snail were measured on an exposed hillside and in a sheltered valley.Hillside: 18.2, 19.5, 17.8, 20.1, 18.9, 19.4, 18.6, 19.9 (mean 19.05, s = 0.816)Valley: 16.9, 17.5, 16.2, 18.0, 17.1, 16.6, 17.8, 16.4 (mean 17.06, s = 0.659)t = |x̄₁ − x̄₂| ÷ √(s₁²/n₁ + s₂²/n₂). Degrees of freedom = n₁ + n₂ − 2.Critical values of t at p = 0.05: 12 df – 2.18; 14 df – 2.14; 16 df – 2.12.
- (a) Calculate the value of t. Give your answer to 2 decimal places. [2]
- (b) State the number of degrees of freedom and what the student can conclude. [2]
- (c) Suggest one reason why the result does not show that the difference is genetic. [1]
Show answer
- (a) 5.36
M1 1.99 ÷ √(0.666/8 + 0.434/8); A1 5.36 (accept 5.35–5.37).
- (b) 14 df; 5.36 > 2.14, so the difference between the mean shell lengths is significant (p < 0.05) – hillside snails have longer shells; reject the null hypothesis.
B1 14 df and comparison with 2.14; B1 significant difference/reject null.
- (c) The difference could be caused by environmental factors (e.g. food or calcium availability, temperature) rather than different alleles.
B1.
Albinism is caused by a recessive allele. In one population, 1 in every 2500 people has albinism. The Hardy–Weinberg equations are p + q = 1 and p² + 2pq + q² = 1.
- (a) Calculate the frequency of the recessive allele, q, and of the dominant allele, p. [2]
- (b) Calculate the percentage of the population who are heterozygous carriers. Give your answer to 2 significant figures. [2]
- (c) State two conditions that must apply for the Hardy–Weinberg principle to give reliable estimates. [2]
Show answer
- (a) q = 0.02; p = 0.98
M1 q² = 1/2500 = 0.0004; A1 q = 0.02 and p = 0.98.
- (b) 3.9 %
M1 2pq = 2 × 0.98 × 0.02 = 0.0392; A1 3.9 %.
- (c) Any two of: large population; random mating; no mutation; no migration in or out; no selection for or against any allele.
1 mark each for any two.
A gene in a population of 800 wild rats has two alleles, A and a. A survey found 410 AA, 300 Aa and 90 aa individuals.
- (a) Calculate the frequencies of alleles A and a in this population. [2]
- (b) Use the Hardy–Weinberg equation to calculate the expected numbers of AA, Aa and aa rats. [2]
- (c) Compare the observed and expected numbers, and suggest one explanation for the difference. [2]
Show answer
- (a) A (p) = 0.70; a (q) = 0.30
M1 A alleles = 2 × 410 + 300 = 1120 out of 1600; A1 p = 0.70, q = 0.30.
- (b) AA 392; Aa 336; aa 72
M1 p² × 800, 2pq × 800, q² × 800; A1 all three.
- (c) Fewer heterozygotes and more homozygotes than expected; this could be due to non-random mating/inbreeding (e.g. a small, isolated colony), or selection against heterozygotes.
B1 comparison; B1 valid explanation.
The tail length of field mice was measured on two islands.Island A: mean 42.6 mm, standard deviation 3.2 mm, n = 25Island B: mean 45.1 mm, standard deviation 3.6 mm, n = 36Standard error SE = s ÷ √n. The 95 % confidence interval is approximately mean ± 2 × SE.
- (a) Calculate the standard error and the 95 % confidence interval for each island. [3]
- (b) Explain what the confidence intervals suggest about the difference between the two populations. [2]
Show answer
- (a) A: SE = 0.64, 95 % CI 41.32–43.88 mm; B: SE = 0.60, 95 % CI 43.90–46.30 mm
B1 both SE; M1 mean ± 2SE; A1 both intervals.
- (b) The intervals do not overlap, so the difference between the mean tail lengths is likely to be significant (real, not due to chance); island B mice have longer tails. (A statistical test would confirm this.)
B1 no overlap; B1 likely significant difference.
A student measured the wing length (mm) of house sparrows caught and released in a rural area and in the centre of Karachi.Rural: 76.2, 77.5, 75.8, 78.1, 76.9, 77.3City: 74.9, 75.6, 74.1, 76.0, 75.2, 74.6t = |x̄₁ − x̄₂| ÷ √(s₁²/n₁ + s₂²/n₂).
- (a) Calculate the mean and standard deviation (n − 1) of each sample. Give means to 2 decimal places and standard deviations to 3 significant figures. [3]
- (b) Calculate t to 2 decimal places. [2]
- (c) The critical value of t at p = 0.05 for 10 df is 2.23. Explain what the result shows. [2]
Show answer
- (a) Rural: 76.97, s = 0.852; City: 75.07, s = 0.686
B1 means; M1 method for s; A1 both s values.
- (b) 4.25
M1 substitution; A1 4.25.
- (c) 4.25 > 2.23, so the difference between the means is significant; there is less than a 5 % probability that it is due to chance; rural sparrows have longer wings.
B1 comparison with 2.23; B1 significant, rural longer.
A hospital laboratory recorded the percentage of Salmonella Typhi samples resistant to the antibiotic ciprofloxacin, and the number of ciprofloxacin prescriptions in the city (thousands per year).Year 1: 12 %, 140; Year 2: 18 %, 165; Year 3: 27 %, 190; Year 4: 41 %, 230; Year 5: 55 %, 260.
- (a) Calculate the percentage increase in the proportion of resistant samples between year 1 and year 5. Give your answer to the nearest whole number. [2]
- (b) A student concludes that 'increasing use of ciprofloxacin caused the increase in resistance'. Evaluate this conclusion. [3]
Show answer
- (a) 358 %
M1 (55 − 12) ÷ 12 × 100; A1 358 %.
- (b) Support: both increase together (positive correlation) and greater use means stronger selection pressure for resistant bacteria. Against: correlation does not prove cause; no statistical test done; only five years of data from one city; other factors (e.g. resistant strains arriving with travellers, spread through contaminated water, which samples were sent for testing) could be involved.
B1 positive correlation/consistent with selection; B1 correlation not causation; B1 one other valid limitation.
A species of land snail has yellow-shelled and brown-shelled forms. Thrushes eat the snails, smashing the shells on stones. Students counted live snails and broken shells near thrush stones in two habitats.Grassland – live: 160 yellow, 40 brown; broken shells: 22 yellow, 28 brown.Woodland (dark leaf litter) – live: 45 yellow, 155 brown; broken shells: 31 yellow, 19 brown.
- (a) For each habitat, calculate the percentage of live snails that are yellow and the percentage of broken shells that are yellow. [2]
- (b) Use these results to explain the distribution of the two forms. [3]
- (c) Suggest one reason why the results from the broken shells might not fully represent the snails eaten by thrushes. [1]
Show answer
- (a) Grassland: live 80 % yellow, broken 44 % yellow; woodland: live 22.5 % yellow, broken 62 % yellow
B2 all four (B1 for two).
- (b) In each habitat, the less common form is over-represented among broken shells (eaten more than expected): brown in grassland, yellow in woodland; thrushes hunt by sight, so the form that contrasts with the background is seen and eaten more; the camouflaged form survives and reproduces, so its frequency is higher in that habitat (natural selection by predation).
B1 use of data showing selective predation; B1 conspicuous form eaten more; B1 camouflaged form survives/reproduces – higher frequency.
- (c) Some snails are eaten elsewhere/not at stones; other predators take snails; old shells may remain from earlier seasons; brown shells may be harder to find.
B1.
To investigate selection by birds, students made 100 green and 100 brown 'caterpillars' from modelling clay and placed them on the grass of a school field in one afternoon. The next morning they counted those with beak marks: green 14, brown 31. They concluded that birds select against brown caterpillars on grass.
- (a) Identify three limitations of this investigation that reduce confidence in the conclusion. [3]
- (b) Suggest three improvements. [3]
Show answer
- (a) Any three of: only one trial/one day; done on one field only; marks may be made by other animals (e.g. rodents, insects) not birds; clay models do not behave or smell like real caterpillars; models may have been moved or lost; identification of beak marks is subjective; models may not have been placed randomly; no statistical test.
1 mark each for any three.
- (b) Any three matching improvements: repeat on several days/sites; place models at random positions (random coordinates); use a standard key/photographs to identify bird beak marks, or have the same person check all; place an equal number on another background (e.g. bare soil) as a comparison; count marks after a fixed time; use a χ² test to see if the difference is significant.
1 mark each for any three.
A student wants to find out whether bacteria from a culture become more resistant to an antibacterial chemical after repeated exposure. She has a non-pathogenic bacterium, nutrient agar plates, sterile paper discs, the antibacterial chemical at 10 % concentration, sterile distilled water and normal microbiology apparatus.
- (a) Describe a method she could use. Your method should be safe and allow valid conclusions to be drawn. [7]
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- (a) Points: aseptic technique (flame loop/neck of bottle, work near Bunsen, lid lifted only slightly); spread a standard volume of culture evenly on agar (lawn); place discs soaked in a series of concentrations (dilute 10 % stock, e.g. 0.5, 1, 2, 5, 10 %) and a control disc with sterile water; incubate at 25 °C for a set time (e.g. 48 h), lids taped not sealed; measure diameter of clear zone of inhibition with a ruler (two diameters at right angles, mean); take bacteria from the edge of the clear zone around a disc, grow them, and repeat the exposure for several 'generations'; compare inhibition zone size at each round (smaller zone = more resistance); keep agar depth/volume, temperature, disc size and soaking time constant; at least three plates per round, calculate means; dispose of plates by autoclaving.
1 mark each for any seven points; credit specific detail.
Students compare how two non-pathogenic laboratory strains of a bacterium (strain P and strain Q) respond to the same antibiotic, using paper discs on agar plates and measuring the diameter of the clear zone around each disc.
- (a) State the independent and dependent variables. [2]
- (b) State three variables that must be standardised. [3]
- (c) Describe a suitable control and explain its purpose. [2]
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- (a) Independent: strain of bacterium (P or Q); dependent: diameter of the clear zone (zone of inhibition).
B1 each.
- (b) Any three of: concentration of antibiotic; size of paper disc/volume of antibiotic on disc; volume and density of bacteria spread; depth/type of agar; incubation temperature; incubation time.
1 mark each for any three.
- (c) A disc soaked in sterile distilled water (no antibiotic) placed on each plate; shows that any clear zone is caused by the antibiotic and not by the disc or handling.
B1 control; B1 purpose.
