The Roman snail (Helix pomatia) is found on chalk grassland in parts of southern England. Shell colour in this species is controlled by a single gene with two alleles. The allele for brown shell (B) is dominant to the allele for pale shell (b).
(a)State the genotype of a snail with a pale shell.(1)
(b)A brown-shelled snail was crossed with a pale-shelled snail (bb). Of the 100 offspring produced, 48 were brown-shelled and 52 were pale-shelled. Use a genetic diagram to determine the genotype of the brown-shelled parent.(3)
(c)Explain why crossing an organism showing the dominant phenotype with a homozygous recessive individual (a test cross) allows its genotype to be determined.(2)
(Total for Question 1 is 6 marks)
2
A plant breeder crossed two pea plants that were both heterozygous for seed shape (Rr) and seed colour (Yy). Round shape (R) is dominant to wrinkled (r) and yellow colour (Y) is dominant to green (y). The two genes assort independently. A chi-squared test was used to test whether the 320 offspring produced from this dihybrid cross (RrYy x RrYy) fitted the expected 9:3:3:1 ratio. Observed numbers: round yellow = 160, round green = 74, wrinkled yellow = 70, wrinkled green = 16.
(a)State the null hypothesis for this chi-squared test.(1)
(b)Calculate the expected number of offspring in each of the four phenotype classes.(2)
(c)Calculate the chi-squared value for this data. Show your working.(3)
(d)The critical value of chi-squared at p=0.05 for 3 degrees of freedom is 7.82. Using this value and your answer to part (c), state and explain the conclusion that should be drawn about the offspring ratio.(2)
(Total for Question 2 is 8 marks)
3
In a species of squash, fruit colour is controlled by two genes. Gene 1 has a dominant allele (A) that suppresses colour production, giving a white fruit regardless of genotype at gene 2; the recessive allele (a) allows colour to be produced. When colour is not suppressed, gene 2 determines the colour: dominant allele (B) gives yellow fruit, recessive allele (b) gives green fruit. This is an example of epistasis.
(a)Define the term epistasis.(1)
(b)Two squash plants, both heterozygous at both genes (AaBb), were crossed. Using a Punnett square (or equivalent working), determine the phenotypic ratio of the offspring, stating how many of 16 offspring would be white, yellow and green.(4)
(c)Explain how this result differs from the phenotypic ratio expected if the two genes assorted independently with no epistatic interaction, and explain the genetic basis for this difference.(2)
(Total for Question 3 is 7 marks)
4
The ability to taste a bitter-tasting compound called PTC is controlled by a single gene, with the allele for tasting (T) dominant to the non-tasting allele (t). In a sample of a UK population, 16% of individuals were found to be non-tasters.
(a)State the Hardy-Weinberg equations that link (i) allele frequencies and (ii) genotype frequencies.(2)
(b)Calculate the frequency of the non-tasting allele (t) in this population.(2)
(c)A different town has a population of 8500 people in Hardy-Weinberg equilibrium for this gene, with the same allele frequencies as above. Calculate the number of people in this town who are heterozygous carriers of the non-tasting allele.(3)
(d)State two assumptions that must be true for a population to be in Hardy-Weinberg equilibrium.(2)
(Total for Question 4 is 9 marks)
5
Genetic drift is the random change in allele frequencies from one generation to the next, due to chance rather than natural selection. Two examples of genetic drift are the founder effect and the population bottleneck effect.
(a)Define the term genetic drift.(1)
(b)A small group of 20 finches was blown off course by a storm and colonised a remote Atlantic island, founding a new population. Explain, using the term founder effect, why the allele frequencies in the new island population might differ significantly from the allele frequencies in the mainland population from which the finches came.(3)
(c)A population of 50000 elephant seals was reduced to just 30 individuals after a period of intense hunting in the 19th century, before recovering to over 200000 today. Explain why the genetic diversity of the modern population remains low despite its large size, using the term population bottleneck.(3)
(Total for Question 5 is 7 marks)
6
Ecologists studied the beak depth of a population of finches on a small island over several generations. Before a period of drought, beak depth followed a normal distribution with a mean of 9.0 mm. During the drought, only large, hard-shelled fruits were available, and finches with deeper, stronger beaks were better able to crack these open and survive. After the drought, the mean beak depth in the surviving population had increased to 10.4 mm, with the whole distribution shifted towards larger values.
(a)Identify the type of natural selection described.(1)
(b)Explain, with reference to the data, how directional selection produced this change in the finch population.(3)
(c)A different island population of the same finch species experienced a change in food supply such that only very small, soft seeds and very large, hard seeds became available, with no intermediate-sized seeds. Predict and explain, using an appropriate named type of selection, how the distribution of beak depth in this population would change over several generations.(3)
(Total for Question 6 is 7 marks)
7
New species can arise through allopatric speciation (where populations are separated by a geographical barrier) or sympatric speciation (where populations diverge within the same geographical area).
(a)Define the biological species concept, in terms of reproductive isolation.(2)
(b)Compare how allopatric speciation and sympatric speciation can lead to the formation of a new species, referring to isolating mechanisms, genetic divergence and reproductive isolation in your answer.(6)
(Total for Question 7 is 8 marks)
8
This question is based on the required practical investigating the distribution and abundance of organisms using quadrats. A group of students investigated the population size of daisy plants in a school playing field with an area of 800 m2. They used a random number generator to obtain two random numbers, which were used as coordinates to position a 0.5 m x 0.5 m quadrat at ten random locations across the field. The number of daisy plants rooted within each quadrat was counted. Results (number of daisy plants per quadrat): 4, 6, 5, 7, 3, 5, 6, 4, 5, 5.
(a)Explain why the students used random numbers to position the quadrats, rather than placing them wherever was convenient.(2)
(b)Calculate the mean number of daisy plants per quadrat, and use this to estimate the total daisy population in the field.(3)
(c)Suggest one way the reliability of this population estimate could be improved, and explain why this would help.(2)
(Total for Question 8 is 7 marks)
9
This question is based on the required practical investigating the distribution and abundance of organisms, using the mark-release-recapture technique. Students estimated the population size of woodlice in a section of woodland. On the first day, they collected and marked 45 woodlice using a small dot of non-toxic paint, then released them back into the woodland. Two days later, they collected a second sample of 60 woodlice, of which 15 were found to be marked.
(a)State two assumptions that must be made for the mark-release-recapture method to give a valid population estimate.(2)
(b)Use the Lincoln index equation below to estimate the total population of woodlice in this section of woodland. Population estimate = (number caught in first sample x number caught in second sample) / number of marked individuals recaptured(3)
(c)The paint used to mark the woodlice was bright orange. Explain how this might cause the population estimate calculated in part (b) to be inaccurate.(2)
(Total for Question 9 is 7 marks)
10
Ecological succession is the process by which the species composition of a community changes over time.
(a)Explain what is meant by the term climax community in the context of ecological succession.(2)
(b)Describe how primary succession occurs on an area of bare sand at a newly formed coastal sand dune system, from the first colonisation by pioneer species through to the formation of a climax community. In your answer, refer to changes in the abiotic environment and to the biotic community at each stage.(6)
(Total for Question 10 is 8 marks)
11
The diagram below summarises part of the nitrogen cycle in a farmland ecosystem. Atmospheric nitrogen gas (N2) --[process X, via bacteria genus Y in root nodules]--> Ammonium ions in soil --[nitrification: Nitrosomonas then Nitrobacter]--> Nitrate ions in soil --[uptake]--> Plant proteins --[decomposition/death and feeding]--> Ammonium ions in soil --[process Z, denitrifying bacteria, under anaerobic conditions]--> Atmospheric nitrogen gas (N2).
(a)Name process X in the diagram and the genus of bacteria (Y) responsible for it.(2)
(b)Explain the roles of Nitrosomonas and Nitrobacter bacteria in the nitrogen cycle.(3)
(c)Explain why waterlogged soil reduces the availability of nitrate ions to plant roots.(3)
(Total for Question 11 is 8 marks)
12
In a UK grassland ecosystem, producers (grass) had a gross primary productivity (GPP) of 20000 kJ m-2 yr-1. The producers used 8000 kJ m-2 yr-1 of this energy in respiration.
(a)Using the equation NPP = GPP - R, calculate the net primary productivity (NPP) of the grass.(2)
(b)Primary consumers (e.g. rabbits and insects) in this grassland converted 1200 kJ m-2 yr-1 of this NPP into their own biomass. Calculate the percentage efficiency of energy transfer from producers to primary consumers.(2)
(c)Secondary consumers in this grassland (e.g. shrews and small birds) obtained 150 kJ m-2 yr-1 of energy from feeding on the primary consumers. Calculate the percentage efficiency of energy transfer from primary consumers to secondary consumers, and state which of the two trophic transfers calculated in this question was more efficient.(3)
(d)Explain why the percentage of energy transferred between trophic levels in an ecosystem is generally low.(2)
(Total for Question 12 is 9 marks)
13
Ecologists compared the biodiversity of two adjacent habitats using Simpson's Index of Diversity: D = 1 - sum[(n/N)2], where n = number of individuals of a species and N = total number of individuals of all species recorded. Habitat A (ancient woodland): 75 individual plants recorded: 15 oak seedlings, 10 birch seedlings, 20 bramble plants, 5 ferns and 25 bluebells. Habitat B (adjacent monoculture wheat field): 100 individual plants recorded: 90 wheat plants, 5 of weed species 1, 3 of weed species 2 and 2 of weed species 3.
(a)Calculate the value of Simpson's Index of Diversity (D) for Habitat A. Show your working.(3)
(b)Calculate the value of Simpson's Index of Diversity (D) for Habitat B.(2)
(c)Using your calculated values, compare the biodiversity of the two habitats and suggest one reason for the difference.(3)
(Total for Question 13 is 8 marks)
14
Comparing the DNA base sequences or amino acid sequences of homologous genes/proteins between species provides evidence for evolutionary relationships, alongside non-molecular evidence such as anatomical structures and the fossil record.
(a)Outline what is meant by a molecular clock, and state the key assumption on which it is based.(2)
(b)Evaluate the different types of evidence (molecular evidence, such as DNA or protein sequence comparison, and non-molecular evidence, such as anatomical structures and the fossil record) that can be used to establish evolutionary relationships between species, and explain why molecular evidence is generally considered more reliable than anatomical evidence alone.(6)
(Total for Question 14 is 8 marks)
Mark scheme · AB7 Genetics, Populations, Evolution and Ecosystems
Question 1
(a) B1 bb (homozygous recessive) oe
(a) Answer: bb
(b) M1 correct genetic diagram/Punnett square set up for a cross of Bb x bb (or equivalent gamete diagram)
(b) M1 gametes B and b from the unknown parent combine with b from the pale parent to give offspring genotypes Bb, Bb, bb, bb (1 Bb : 1 bb)
(b) A1 genotype of brown parent = Bb (heterozygous); the predicted 1:1 ratio matches the observed 48:52 ratio cao
(b) Answer: Bb (heterozygous)
(c) B1 the homozygous recessive parent can only contribute a recessive allele, so the phenotypes of the offspring reveal the alleles carried by the unknown parent oe
(c) B1 if all offspring show the dominant phenotype the unknown parent is homozygous dominant; if approximately half the offspring show the recessive phenotype the unknown parent is heterozygous oe
(c) Answer: A homozygous recessive test-cross partner only contributes recessive alleles, so the offspring ratio (all dominant, or 1:1 dominant:recessive) reveals whether the unknown parent is homozygous dominant or heterozygous.
Question 2
(a) B1 there is no significant difference between the observed numbers of offspring in each phenotype class and the numbers expected from a 9:3:3:1 ratio (any difference is due to chance) oe
(a) Answer: There is no significant difference between the observed and expected (9:3:3:1) numbers of offspring; any difference is due to chance.
(b) M1 correct method shown, e.g. 9/16 x 320
(b) A1 all four expected values correct: 180, 60, 60, 20 cao
(b) Answer: 180, 60, 60, 20
(c) M1 correct use of chi2 = sum[(O-E)2/E] with at least two correct (O-E)2/E terms
(c) M1 all four (O-E)2/E terms correct: 2.22, 3.27, 1.67, 0.80 (awrt 2 dp)
(c) A1 chi2 = 7.96 (awrt) cao
(c) Answer: chi2 = 7.96 (awrt)
(d) B1 calculated chi2 (7.96) is greater than the critical value (7.82) ft from (c)
(d) B1 the null hypothesis is rejected; there is a significant difference between observed and expected ratios, suggesting the two genes may not be assorting independently (e.g. could be linked) oe
(d) Answer: Since 7.96 > 7.82, the null hypothesis is rejected: there is a significant difference between observed and expected ratios, suggesting the genes may be linked rather than assorting independently.
Question 3
(a) B1 the interaction of genes at different loci, in which one gene (the epistatic gene) masks or suppresses the expression of another gene (the hypostatic gene) oe
(a) Answer: Epistasis is the interaction of genes at different loci in which one gene masks or suppresses the expression of another gene.
(b) M1 correct standard dihybrid ratio derived: 9 A_B_ : 3 Abb : 3 aaB_ : 1 aabb
(b) M1 A_B_ and Abb genotypes (9 + 3) both give white phenotype, since the dominant A allele suppresses colour regardless of the B/b genotype
(b) M1 aaB_ (3) gives yellow and aabb (1) gives green
(b) A1 final ratio stated as 12 white : 3 yellow : 1 green (out of 16) cao
(b) Answer: 12 white : 3 yellow : 1 green (out of 16 offspring)
(c) B1 without epistasis, four distinct phenotypes would be expected in a 9:3:3:1 ratio, but with epistasis only three phenotypes are seen, in a 12:3:1 ratio oe
(c) B1 this is because the dominant allele at gene 1 (A) masks the effect of gene 2, so two of the four genotype classes (A_B_ and Abb) produce the same (white) phenotype oe
(c) Answer: Without epistasis four phenotype classes (9:3:3:1) would be expected; with epistasis only three are seen (12:3:1) because the dominant A allele masks gene 2, merging the A_B_ and Abb classes into a single white phenotype.
Question 4
(a) B1 p + q = 1
(a) B1 p2 + 2pq + q2 = 1
(a) Answer: p + q = 1 and p2 + 2pq + q2 = 1
(b) M1 q2 = 0.16, so q = √0.16
(b) A1 q = 0.4 cao
(b) Answer: q = 0.4
(c) M1 p = 1 - q = 0.6 ft from (b)
(c) M1 2pq = 2 x 0.6 x 0.4 = 0.48
(c) A1 number of carriers = 0.48 x 8500 = 4080 cao ft
(c) Answer: 4080 people
(d) B1 any one of: no mutation occurs; the population is large (genetic drift is negligible); mating is random; no natural selection acts (no allele confers a survival/reproductive advantage); no gene flow/migration into or out of the population
(d) B1 any second, different assumption from the list above
(d) Answer: For example: no mutation occurs, and mating within the population is random (any two valid assumptions from the accepted list).
Question 5
(a) B1 the random change in allele frequency in a population from one generation to the next, due to chance rather than natural selection oe
(a) Answer: Genetic drift is the random change in allele frequency in a population from generation to generation, due to chance.
(b) B1 the small founding group carries only a small, random sample of the alleles present in the mainland gene pool oe
(b) B1 by chance this small sample may not be representative of the mainland allele frequencies (some alleles may be over-represented, under-represented, or absent) oe
(b) B1 because the founding population is small, this chance effect (genetic drift/founder effect) has a proportionally larger effect on allele frequencies than it would in a large population oe
(b) Answer: The founding population carries only a small, random sample of mainland alleles, which may not represent mainland frequencies; because the population is small, this chance effect has a large impact on allele frequencies.
(c) B1 only 30 individuals (and the alleles they carried) survived the population bottleneck, so much of the original genetic diversity/many alleles were lost by chance oe
(c) B1 even though the population has since grown to over 200000, all these individuals are descended from the surviving 30, so alleles lost during the bottleneck cannot be regained (without new mutation) oe
(c) B1 this reduces the gene pool/genetic diversity available for future natural selection, making the population more vulnerable to environmental change or disease oe
(c) Answer: The bottleneck (30 survivors) lost much of the original allelic diversity by chance; the recovered population descends only from these survivors, so diversity remains low and the species is more vulnerable to future change.
Question 6
(a) B1 directional selection
(a) Answer: Directional selection
(b) B1 before the drought there was a range of variation in beak depth within the population, arising from mutation and new allele combinations produced by sexual reproduction oe
(b) B1 during the drought, only finches towards the deeper-beak extreme of the distribution could feed successfully and survive to reproduce; the selection pressure favoured one extreme of the phenotypic range oe
(b) B1 alleles for greater beak depth were passed to offspring in greater numbers over successive generations, so the mean beak depth of the population increased/the distribution shifted to the right oe
(b) Answer: Existing variation in beak depth meant some finches had deeper beaks; only these survived the drought and reproduced, passing on alleles for deeper beaks, so the population mean shifted upward.
(c) B1 disruptive selection (named)
(c) B1 both extremes of the phenotypic range (small beak depth and large beak depth) are selected for, since these types can exploit the two available food sources oe
(c) B1 individuals with intermediate beak depth are selected against (cannot efficiently use either food source), so the distribution becomes bimodal (two peaks) rather than a single normal distribution oe
(c) Answer: Disruptive selection: both small- and large-beaked finches are favoured (each suited to one seed type), while intermediate beak depths are selected against, producing a bimodal distribution.
Question 7
(a) B1 a species is a group of organisms with similar morphology/genetics/characteristics oe
(a) B1 that can interbreed to produce fertile offspring oe
(a) Answer: A species is a group of organisms with similar characteristics that can interbreed to produce fertile offspring.
(b) L3 (5-6): A detailed and coherent comparison that clearly explains both allopatric and sympatric speciation, correctly identifies the isolating mechanism in each case, and links this to genetic divergence and the eventual evolution of reproductive isolation between the two populations. Scientific terminology is used accurately throughout.
(b) L2 (3-4): The answer describes both allopatric and sympatric speciation and identifies at least one relevant isolating mechanism, with some link to divergence or reproductive isolation, but detail, accuracy or the comparison between the two routes may be incomplete.
(b) L1 (1-2): Basic or isolated statements about geographical separation or speciation, with little explanation of mechanism or reproductive isolation. Little or no valid comparison is made.
(b) L0 (0): No relevant content.
(b) Indicative content:
Allopatric speciation begins when a geographical barrier (e.g. a river, mountain range or stretch of sea) physically separates a population into two groups.
This prevents gene flow (interbreeding) between the two separated populations.
Each isolated population is exposed to different environmental/selection pressures, and mutation and genetic drift occur independently in each population.
Over many generations, the allele frequencies of the two populations diverge.
Eventually the accumulated genetic differences mean that, even if the barrier is removed, the two populations can no longer interbreed to produce fertile offspring: they have become reproductively isolated, forming two separate species.
Sympatric speciation occurs without a geographical barrier; populations remain in the same physical area but become reproductively isolated by other mechanisms.
Examples of isolating mechanisms in sympatric speciation include: temporal isolation (breeding at different times), behavioural isolation (different courtship behaviours/mate preferences), ecological isolation (using different micro-habitats or food sources within the same area), or polyploidy in plants (an instant change in chromosome number preventing successful interbreeding with the parent population).
As with allopatric speciation, reduced gene flow between the diverging groups allows allele frequencies to change independently (through selection and/or drift) until interbreeding is no longer possible.
In both cases, the key requirement for speciation is a reduction or cessation of gene flow between the diverging populations, allowing genetic divergence to accumulate until reproductive isolation is complete; the routes differ mainly in whether a geographical barrier or another isolating mechanism causes this reduction in gene flow.
(b) Answer: Levels-of-response essay; see indicative content.
Question 8
(a) B1 random placement avoids sampling bias, e.g. students unconsciously choosing areas with more or fewer daisies oe
(a) B1 a randomly obtained sample is representative of the whole field, so the results can validly be used to estimate the population of the whole field oe
(a) Answer: Random placement avoids bias in quadrat positioning, so the sample obtained is representative of the whole field and results can be validly extrapolated.
(b) M1 number of quadrats needed to cover the field = 800/0.25 = 3200
(b) A1 estimated population = 5.0 x 3200 = 16000 daisy plants cao
(b) Answer: 16000 daisy plants
(c) B1 e.g. increase the number of quadrats sampled oe (accept other valid suggestions, e.g. use a larger quadrat if the species is patchily distributed)
(c) B1 with linked explanation: this reduces the effect of anomalous/unrepresentative results and gives a more reliable estimate of the true mean/reduces the effect of chance variation between samples oe
(c) Answer: Increase the number of quadrats sampled, as this reduces the effect of chance variation between samples and gives a more reliable estimate of the true mean.
Question 9
(a) B1 any one of: no individuals die, are born, immigrate or emigrate between release and recapture; the marking method does not affect survival or behaviour (e.g. does not increase visibility to predators); the mark does not fade or rub off before recapture; marked individuals mix randomly/fully with the rest of the population before recapture
(a) B1 any second, different assumption from the list above
(a) Answer: For example: no individuals die, are born, or migrate into/out of the area during the study, and marked individuals mix randomly with the rest of the population before recapture (any two valid assumptions).
(b) M1 correct substitution: N = (45 x 60)/15
(b) M1 45 x 60 = 2700
(b) A1 N = 2700/15 = 180 woodlice cao
(b) Answer: 180 woodlice
(c) B1 bright paint may make marked woodlice more visible/conspicuous to predators, so more marked individuals may be predated (removed) before the second sample is taken oe
(c) B1 this would reduce the number of marked individuals available to be recaptured, so the population estimate calculated would be an overestimate of the true population size oe
(c) Answer: Bright paint may make marked woodlice easier for predators to spot, reducing the number of marked individuals surviving to recapture; this would make the calculated population estimate too high.
Question 10
(a) B1 the final, relatively stable community reached at the end of succession oe
(a) B1 which remains in equilibrium with the prevailing environmental (abiotic) conditions as long as they do not change oe
(a) Answer: A climax community is the final, stable community formed at the end of succession, which remains in equilibrium with the prevailing environmental conditions.
(b) L3 (5-6): A detailed, well-sequenced description of primary succession that correctly identifies pioneer species and their tolerance of harsh abiotic conditions, explains how the abiotic environment (e.g. soil/humus formation, water retention) changes as a result of colonisation, and links this to a logical sequence of seral stages of increasing species richness and biomass up to a climax community. Scientific terminology is used accurately throughout.
(b) L2 (3-4): The answer describes pioneer colonisation and at least one further seral stage, with some reference to changing abiotic conditions and/or increasing species diversity, but the sequence, detail or accuracy may be incomplete.
(b) L1 (1-2): Basic or isolated statements about succession or pioneer species, with little explanation of how abiotic or biotic conditions change over time.
(b) L0 (0): No relevant content.
(b) Indicative content:
Bare sand is initially a harsh abiotic environment: little water retention, few nutrients/no soil, high salinity and exposure to wind.
Pioneer species (e.g. lichens and salt-tolerant, drought-tolerant plants such as marram grass) are able to colonise this harsh environment first, often with adaptations such as long roots or the ability to fix nitrogen.
As pioneer plants grow, photosynthesise and eventually die, decomposition by microorganisms adds organic matter (humus) to the sand.
This changes the abiotic conditions: the developing soil retains more water and has more available nutrients, and there is more shelter from wind, making conditions less hostile.
These improved conditions allow new, less tolerant species (e.g. grasses, then shrubs) to colonise and out-compete the original pioneers in some areas.
At each successive seral stage, species diversity/richness and total biomass generally increase, and the community becomes structurally more complex, creating new niches (e.g. shade, shelter) for further species.
Interspecific competition increases as more species colonise, and less well-adapted earlier species may be replaced.
Eventually a stable climax community (e.g. woodland) is reached, determined largely by the climate of the area, which persists in equilibrium with the abiotic environment unless a major disturbance occurs.
(b) Answer: Levels-of-response essay; see indicative content.
Question 11
(a) B1 nitrogen fixation
(a) B1 Rhizobium (accept free-living Azotobacter)
(a) Answer: Process X = nitrogen fixation; bacteria genus Y = Rhizobium (in root nodules of legumes; accept free-living Azotobacter).
(b) B1 Nitrosomonas oxidise ammonium ions to nitrite ions oe
(b) B1 Nitrobacter oxidise nitrite ions to nitrate ions oe
(b) B1 together these nitrifying bacteria convert ammonium into nitrate (nitrification), a form of nitrogen that can be absorbed by plant roots and used to build amino acids/proteins/nucleic acids oe
(b) Answer: Nitrosomonas oxidise ammonium to nitrite; Nitrobacter oxidise nitrite to nitrate; this nitrification produces nitrate, the form of nitrogen that plants can absorb and use to build proteins and nucleic acids.
(c) B1 waterlogged soil contains little or no oxygen (anaerobic conditions) oe
(c) B1 anaerobic conditions favour denitrifying bacteria, which convert nitrate ions into nitrogen gas (denitrification), removing nitrate from the soil oe
(c) B1 nitrifying bacteria (Nitrosomonas/Nitrobacter) require oxygen and cannot function well in anaerobic conditions, so less nitrate is produced from ammonium in the first place, further reducing nitrate availability to plant roots oe
(c) Answer: Waterlogging creates anaerobic conditions, which favour denitrifying bacteria (converting nitrate to N2 gas) and inhibit oxygen-requiring nitrifying bacteria, so nitrate is both removed and produced more slowly, reducing availability to plants.
Question 12
(a) M1 NPP = 20000 - 8000
(a) A1 NPP = 12000 kJ m-2 yr-1 cao (correct units required)
(a) Answer: 12000 kJ m-2 yr-1
(b) M1 (1200/12000) x 100
(b) A1 10% cao
(b) Answer: 10%
(c) M1 (150/1200) x 100
(c) A1 12.5% cao
(c) B1 transfer from primary consumers to secondary consumers (12.5%) was more efficient than from producers to primary consumers (10%) ft from candidate's values
(c) Answer: 12.5%; the primary-to-secondary consumer transfer was more efficient than the producer-to-primary consumer transfer.
(d) B1 any one of: not all organisms/parts of organisms at a trophic level are eaten by the next level (e.g. roots, bones, uneaten material); not all ingested material is digested/absorbed, some is egested as faeces; much absorbed energy is lost as heat through respiration (and used for movement, excretion) rather than being converted to biomass
(d) B1 any second, different point from the list above
(d) Answer: For example: much energy is lost as heat through respiration, and not all biomass at one trophic level is eaten or fully digested by the next (any two valid reasons).
Question 13
(a) M1 correct method, calculating (n/N)2 for at least three of the five species
(a) M1 sum[(n/N)2] = 0.24 (awrt 2 dp)
(a) A1 D = 1 - 0.24 = 0.76 (awrt) cao
(a) Answer: D = 0.76 (awrt)
(b) M1 sum[(n/N)2] = 0.81 (awrt 2 dp)
(b) A1 D = 1 - 0.81 = 0.19 (awrt) cao
(b) Answer: D = 0.19 (awrt)
(c) B1 Habitat A (woodland) has a much higher Simpson's Index (D approx 0.76) than Habitat B (wheat field, D approx 0.19), so Habitat A has much greater biodiversity ft from (a) and (b)
(c) B1 Habitat B is dominated by a single species (wheat, 90% of individuals recorded), reducing evenness and therefore diversity oe
(c) B1 this is because Habitat B is a managed monoculture (e.g. herbicides used to remove weed species and only one crop species deliberately planted), reducing the range and evenness of species present, whereas the woodland is unmanaged in this way oe
(c) Answer: Habitat A has much higher biodiversity (D approx 0.76) than Habitat B (D approx 0.19), because Habitat B is a managed monoculture dominated by a single crop species, reducing species evenness and diversity.
Question 14
(a) B1 the molecular clock uses the number of differences in the DNA base sequence (or amino acid sequence) between two species to estimate how long ago they diverged from a common ancestor oe
(a) B1 it assumes that mutations (and therefore sequence differences) accumulate at a broadly constant, known rate over time oe
(a) Answer: The molecular clock uses the number of DNA/protein sequence differences between species to estimate time since divergence from a common ancestor, assuming mutations accumulate at a roughly constant rate.
(b) L3 (5-6): A well-structured evaluation that accurately describes molecular, anatomical and fossil evidence for evolutionary relationships, correctly distinguishes homologous from analogous structures, identifies specific limitations of each type of evidence, and gives a clear, well-reasoned explanation of why molecular evidence is generally more reliable. Scientific terminology is used accurately throughout.
(b) L2 (3-4): The answer describes at least two types of evidence (e.g. molecular and anatomical) with some valid comparison or limitation identified, but the explanation of relative reliability may be partial, or one type of evidence may be omitted or described only briefly.
(b) L1 (1-2): Basic or isolated statements about evidence for evolution (e.g. fossils or similar structures show relatedness), with little evaluation or explanation of reliability.
(b) L0 (0): No relevant content.
(b) Indicative content:
Molecular evidence: comparing DNA base sequences or amino acid sequences of the same gene/protein (e.g. a widely conserved protein) in different species; a greater number of sequence differences indicates a more distant evolutionary relationship (longer time since divergence from a common ancestor).
Anatomical (morphological) evidence: homologous structures (the same underlying structure/bone arrangement, but adapted for a different function in different species) provide evidence of a shared common ancestor.
A key limitation of anatomical evidence is that analogous structures (similar in function/appearance but arising from unrelated ancestry) can be produced by convergent evolution, and could mistakenly be interpreted as evidence of close relatedness.
Fossil evidence: the fossil record can show transitional forms and can be dated (e.g. using the rock strata in which fossils are found), providing evidence of how species have changed over time.
A key limitation of fossil evidence is that fossilisation is a rare event, soft tissues rarely fossilise, and the fossil record is therefore incomplete, so evolutionary relationships based on fossils alone may be uncertain.
Molecular evidence is generally considered more reliable because it is quantitative and can be compared precisely across very many species and genes, is not affected by convergent evolution in the way that analogous physical structures can be, and does not depend on the chance survival of fossils.
A limitation of the molecular clock itself is that mutation rates are not identical in every gene or every lineage, so molecular clock estimates usually need to be calibrated against independent evidence, such as dated fossils.
The most robust evolutionary trees are usually built by combining molecular, anatomical and fossil evidence, since each source can help address the limitations of the others.
(b) Answer: Levels-of-response essay; see indicative content.