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Continuity and Change — Free Biology SL Practice Questions

1FoundationMCQConcept of water potential in plants1 markPaper 1~2 min
A plant cell has a solute potential of 0.8 MPa-0.8 \text{ MPa} and a pressure potential of +0.3 MPa+0.3 \text{ MPa}. The cell is placed in a solution with a water potential of 0.4 MPa-0.4 \text{ MPa}. What is the initial direction of net water movement, and why?
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2MasteryMCQImportance in plant hydration and transport1 markPaper 1~2 min
A 10 cm segment of potato tissue is placed in a 0.5 mol dm30.5\ \text{mol dm}^{-3} sucrose solution for 30 minutes, after which it has become flaccid. The segment is then transferred to pure water. Which statement correctly explains the net movement of water into the potato cells after transfer?
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3FoundationMCQMitigation strategies and global agreements1 markPaper 1~2 min
Two graphs are presented to a student. The first shows global average temperature anomaly rising sharply after 1970; the second shows atmospheric CO₂ concentration rising sharply after 1950. The student concludes that rising CO₂ causes global warming because the two variables are positively correlated. Which statement is the most scientifically appropriate evaluation of this conclusion?
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4MasteryMCQMitigation strategies and global agreements1 markPaper 1~2 min
A climate summit is evaluating four mitigation strategies. Strategy X captures CO₂ from power plant exhaust and stores it in deep geological formations before it enters the atmosphere. Strategy Y involves planting fast-growing eucalyptus trees across large areas of degraded land. Strategy Z coats the ocean surface with reflective particles to reduce solar energy absorption. Strategy W replaces coal-fired power plants with nuclear reactors. Which strategy directly removes CO₂ already present in the atmosphere?
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5FoundationMCQAsexual vs sexual reproduction1 markPaper 1~2 min
A gardener propagates a rose by taking a stem cutting from a parent plant and growing it in soil. The new plant is genetically identical to the parent. Which statement correctly identifies the type of reproduction and its most significant advantage in this context?
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6MasteryMCQAsexual vs sexual reproduction1 markPaper 1~2 min
In the life cycle of a fern, the sporophyte generation is diploid with 2n=202n = 20 chromosomes. Spores are produced in structures called sporangia. How many chromosomes are present in the nucleus of a fern spore, and by which type of cell division are spores produced?
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7FoundationMCQMendelian inheritance1 markPaper 1~2 min
In pea plants, tall stems (T) are dominant over short stems (t). A gardener crosses two heterozygous tall plants. What proportion of the offspring is expected to have tall stems?
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8MasteryMCQMendelian inheritance1 markPaper 1~2 min
In pea plants, round seeds (R) are dominant over wrinkled seeds (r). A heterozygous round-seeded plant is crossed with a wrinkled-seeded plant. What proportion of the offspring is expected to have wrinkled seeds?
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9FoundationMCQDNA structure and replication mechanism1 markPaper 1~2 min
In the Meselson–Stahl experiment, E. coli were grown in 15^{15}N medium, then transferred to 14^{14}N medium for exactly one generation. DNA was extracted analysed by density-gradient centrifugation, producing a single band at an intermediate density. Which mechanism of DNA replication is consistent with this result?
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10MasteryMCQDNA structure and replication mechanism1 markPaper 1~2 min
In the Meselson–Stahl experiment, E. coli were grown in 15^{15}N medium until all DNA was fully labelled, then transferred to 14^{14}N medium. After two rounds of replication, the DNA was centrifuged in a caesium chloride density gradient. Which result, and which replication model, are correctly paired?
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11FoundationMCQMechanisms of temperature, pH, and water balance1 markPaper 1~2 min
In metabolic acidosis, plasma pH falls and bicarbonate concentration decreases. Which of the following correctly describes the primary mechanism by which breathing rate increases?
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12MasteryMCQRegulation of blood glucose levels1 markPaper 1~2 min
A patient with type 1 diabetes mellitus has a fasting blood glucose concentration of 18.0 mmol L118.0 \text{ mmol L}^{-1}. She self-administers an appropriate dose of insulin. Which statement best describes the immediate mechanism by which insulin lowers blood glucose concentration in her liver cells?
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13FoundationMCQTranscription and translation processes1 markPaper 1~2 min
A tRNA has the anticodon sequence 3'-AAG-5'. What is the mRNA codon recognised by this tRNA?
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14MasteryMCQTranscription and translation processes1 markPaper 1~2 min
A cell-free translation system containing all necessary components translates the synthetic mRNA 5-AUG UUU UAA-35'\text{-AUG UUU UAA-}3', producing a dipeptide. A point mutation changes this mRNA to 5-AUG UUC UAA-35'\text{-AUG UUC UAA-}3'. What is the most likely product of translating the mutated mRNA?
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15FoundationMCQMechanisms of natural selection1 markPaper 1~2 min
In a grassland, meadow voles (Microtus pennsylvanicus) occupy patches of dark soil and patches of light soil. Hawks hunt voles by sight. Over 10 generations, the allele for dark fur increases in frequency on dark soil patches and the allele for light fur increases in frequency on light soil patches, while voles with intermediate fur colour decline in both patch types. Which type of natural selection is acting on fur colour in this population?
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16MasteryMCQMechanisms of natural selection1 markPaper 1~2 min
In a meadow, grasshoppers show heritable variation in body colour: green, yellow, and brown. A visually hunting bird species colonises the meadow. After three generations, the frequency of brown grasshoppers has increased significantly while the frequencies of green and yellow grasshoppers have both decreased. Which type of natural selection best describes this change in the grasshopper population?
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17FoundationMCQRenewable resources and ecological conservation1 markPaper 1~2 min
A village depends on a nearby forest for firewood. The forest has been logged at 5 hectares per year for 20 years, while natural regrowth adds 2 hectares per year. The village council proposes to classify the forest as a renewable resource. Which condition must be met for this classification to be valid?
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18MasteryMCQSustainable agricultural practices1 markPaper 1~2 min
A study compares two adjacent fields over 10 years. Field X uses conventional tillage and synthetic fertilisers; Field Y uses no-till farming with cover crops and manure. The data are shown below. Year 1 — 12 — 5.5 — 2 — 4.8 Year 10 — 15 — 4.2 — 1.5 — 5.3 Which statement best explains why Field Y's crop yield increased over 10 years while Field X's yield decreased?
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19FoundationMCQTypes of mutations: Point, frameshift, chromosomal1 markPaper 1~2 min
The DNA template strand of a gene contains the sequence 5'-ATG GGC TAC CTA-3'. A single nucleotide substitution changes the final nucleotide of the fourth codon from A to T. Which type of mutation has occurred?
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20MasteryMCQTypes of mutations: Point, frameshift, chromosomal1 markPaper 1~2 min
A bacterial gene has the coding strand sequence 5'-TAC GCA TGG-3'. A single adenine nucleotide is inserted after the first base. Which type of mutation has occurred, and what is the immediate consequence for translation?
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21FoundationMCQMitosis vs meiosis1 markPaper 1~2 min
A researcher applies colchicine to human liver cells undergoing mitosis. Colchicine prevents tubulin polymerisation, blocking spindle fibre formation. At which stage of mitosis will most cells become arrested?
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22MasteryMCQMitosis vs meiosis1 markPaper 1~2 min
A student uses a light microscope to examine two cells. In a cell from a growing onion root tip, individual chromosomes are aligned along the cell equator. In a cell from a developing lily anther, homologous pairs of chromosomes are aligned along the cell equator. Which row correctly identifies the stage and process occurring in each cell?
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23MasterySAQ-SConcept of water potential in plants5 marksPaper 2~8 min
A student places a plant cell from a freshwater aquatic plant (Elodea canadensis) into a concentrated salt solution. The cell initially has a solute potential (Ψs\Psi_s) of 0.8MPa-0.8\,\text{MPa} and a pressure potential (Ψp\Psi_p) of +0.3MPa+0.3\,\text{MPa}. The external salt solution has a water potential (Ψw\Psi_w) of 1.4MPa-1.4\,\text{MPa}. After 10 minutes, the cell has undergone plasmolysis.
(a)
State the initial water potential (Ψw\Psi_w) of the cell. [1 mark]
(b)
Describe the direction of net water movement and the change in pressure potential (Ψp\Psi_p) as plasmolysis occurs. [2 marks]
(c)
Predict whether net water movement into or out of the cell will continue after plasmolysis complete, given that the cell's solute potential (Ψs\Psi_s) at the point of incipient plasmolysis 1.4MPa-1.4\,\text{MPa}. Justify your answer using water potential values. [2 marks]
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24MasterySAQ-SConcept of water potential in plants5 marksPaper 2~8 min
A student uses a pressure bomb to measure water potential in a branch of a maple tree (Acer saccharum). The branch is cut and placed in the pressure chamber with the cut end exposed. Pressure is applied until xylem sap just reappears at the cut surface. The pressure required is 0.8MPa0.8\,\text{MPa}. In a pressure bomb, the applied pressure equals the magnitude of the xylem water potential, and the xylem water potential approximates the leaf cell water potential.
(a)
State the water potential of the leaf cells, including its sign. [1 mark]
(b)
The solute potential (Ψs\Psi_s) of the leaf cells is 1.5MPa-1.5\,\text{MPa}. Calculate the pressure potential (Ψp\Psi_p) of the leaf cells. [2 marks]
(c)
Root cells have a water potential of 0.3MPa-0.3\,\text{MPa}. Using the values from (a) and (b), explain how water moves from roots to leaves through the xylem, and deduce whether the pressure potential calculated in (b) is consistent with water movement occurring by the cohesion-tension mechanism. [2 marks]
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25ChallengeSAQ-LConcept of water potential in plants7 marksPaper 2~11 min

Data

- Water potential of soil solution: Ψsoil=1.2MPa\Psi_{soil} = -1.2\,\text{MPa} - Balancing pressure recorded when sap just begins to exude: Pbalance=+0.8MPaP_{balance} = +0.8\,\text{MPa} - Standard formula: Ψw=Ψs+Ψp\Psi_w = \Psi_s + \Psi_p
A plant physiologist investigates the water relations of the halophyte Salicornia europaea, which grows in coastal salt marshes. Cells in the stem accumulate high concentrations of sodium chloride (NaCl) and the organic solute glycine betaine to maintain turgor despite the high salinity of the surrounding soil water. A 5.0cm5.0\,\text{cm} segment of S. europaea stem is placed in a sealed pressure chamber (Scholander bomb) containing soil solution. The stem segment is allowed to equilibrate fully with the soil solution before measurements are taken.
(a)
State the water potential (Ψw\Psi_w) of the stem cells after equilibration with the soil solution. [1 mark]
(b)
The balancing pressure recorded equals the pressure potential (Ψp\Psi_p) of the stem cells. Calculate the solute potential (Ψs\Psi_s) of the stem cells. [2 marks]
(c)
Explain how the accumulation of NaCl and glycine betaine in the vacuoles of S. europaea stem cells enables the plant to maintain a positive turgor pressure (Ψp\Psi_p) when growing in soil where Ψsoil=1.2MPa\Psi_{soil} = -1.2\,\text{MPa}. Refer to the values calculated in (a) and (b) in your answer. [2 marks]
(d)
Evaluate the use of the Scholander pressure chamber technique for measuring water potential in halophyte stem cells. In your answer, discuss one strength and one limitation, with specific reference to the assumption that the stem segment equilibrates fully with the surrounding solution before measurement. [2 marks]
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26ChallengeSAQ-LMitigation strategies and global agreements11 marksPaper 2~17 min
A proposed reforestation programme plants 500,000ha500{,}000\,\text{ha} of tropical forest and 300,000ha300{,}000\,\text{ha} of temperate forest. A mature tropical forest sequesters 2.5t C ha1yr12.5\,\text{t C ha}^{-1}\,\text{yr}^{-1} and a mature temperate forest sequesters 1.8t C ha1yr11.8\,\text{t C ha}^{-1}\,\text{yr}^{-1}. The current global deforestation rate is 10,000ha yr110{,}000\,\text{ha yr}^{-1}. The carbon content of CO2\text{CO}_2 by mass is 27.3%27.3\%, and 1t=1000kg1\,\text{t} = 1000\,\text{kg}.
(a)
(i) Calculate the total annual carbon sequestration, in tonnes of carbon, for the entire reforestation programme, assuming all trees reach maturity. [3]
(ii) The tropical forest component of the programme removes carbon from the atmosphere. Determine the mass of CO2\text{CO}_2, in kg, equivalent to the carbon sequestered annually by the tropical forest alone. [2 marks]
(b)
Explain two biological mechanisms by which reforestation contributes to climate change mitigation beyond direct carbon storage in tree biomass. [2 marks]
(c)
Using your answer from (a)(i) and the global deforestation rate, evaluate whether reforestation alone can be considered an effective long-term mitigation strategy. In your answer, consider one biological limitation and one socio-economic factor. [4 marks]
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27MasterySAQ-SCauses and effects of climate change5 marksPaper 2~8 min
The Arctic sea ice minimum extent (the smallest area covered by ice each year) has decreased significantly since satellite records began in 1979. A study published in 2020 reported that the September 2019 minimum extent was 4.154.15 million km2\text{km}^2, compared to an average of 6.856.85 million km2\text{km}^2 for the period 1981–2010.
(a)
State the term used to describe the proportion of incoming solar radiation reflected by a surface. [1 mark]
(b)
Calculate the percentage decrease in Arctic sea ice minimum extent from the 1981–2010 average to September 2019. Show your working. [2 marks]
(c)
Explain how the loss of Arctic sea ice creates a positive feedback loop that amplifies global warming. [2 marks]
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28ChallengeSAQ-LMitigation strategies and global agreements7 marksPaper 2~11 min

Data

- Power plant generates 500MW500\,\text{MW} of electricity (1MW=106Js11\,\text{MW} = 10^6\,\text{J}\,\text{s}^{-1}) - The CCS system requires 25%25\% of the plant's gross energy output to operate - The plant emits 0.8kg CO20.8\,\text{kg CO}_2 per kWh of electricity generated (before capture), based on gross output - The CCS system captures 90%90\% of the CO2\text{CO}_2 from flue gases - 1year=8760hours1\,\text{year} = 8760\,\text{hours}
A carbon capture and storage (CCS) facility is proposed adjacent to a coal-fired power plant. The system captures CO2\text{CO}_2 from flue gases and stores it in deep saline aquifers. Environmental impact assessments must consider both the energy penalty of the system and the biological consequences of potential CO2\text{CO}_2 leakage.
(a)
Calculate the net reduction in CO2\text{CO}_2 emissions, in tonnes per year, achieved by implementing this CCS system. [3 marks]
(b)
Explain the biological mechanism by which CO2\text{CO}_2 leakage from storage sites into the ocean could affect marine phytoplankton communities. [2 marks]
(c)
Using evidence, evaluate whether CCS technology alone is sufficient to limit global average temperature rise to 1.5C1.5\,^\circ\text{C} above pre-industrial levels, as targeted by the Paris Agreement. [2 marks]
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29ChallengeSAQ-LAsexual vs sexual reproduction10 marksPaper 2~15 min
A study investigated the reproductive strategies of two freshwater crustacean species, Daphnia pulex and Artemia franciscana. Both species can reproduce asexually (parthenogenesis) or sexually, depending on environmental cues. Population growth rates were measured in controlled laboratory cultures over 30 days: Species — Asexual rate (day1)(\text{day}^{-1}) — Sexual rate (day1)(\text{day}^{-1}) Daphnia pulex — 0.320.320.080.08 Artemia franciscana — 0.250.250.120.12 Genetic diversity was assed using gel electrophoresis of a polymorphic enzyme locus. The number of distinct alleles present in a population of 100 individuals after 30 days was: Species — Asexual reproduction — Sexual reproduction Daphnia pulex — 2 alleles — 8 alleles Artemia franciscana — 3 alleles — 6 alleles
(a)
(i) Calculate the percentage advantage in population growth rate of asexual over sexual reproduction for each species. [2]
(ii) State which species would benefit more from asexual reproduction in a stable, resource-rich environment, justifying your answer using your calculated values. [1 mark]
(b)
Explain the biological mechanisms by which sexual reproduction generates greater genetic diversity than asexual reproduction, and use the allele data to support your answer. [3 marks]
(c)
Evaluate the hypothesis that sexual reproduction is more advantageous than asexual reproduction in environments that change unpredictably over time, using both the data and your knowledge of the costs and benefits of each reproductive mode. Reach a reasoned, qualified conclusion. Total: 10 marks [4 marks]
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30MasterySAQ-SSexual cycles in animals and plants6 marksPaper 2~9 min
The European white water lily (Nymphaea alba) is a flowering plant that reproduces sexually. Its life cycle involves alternation of generations, with both haploid (nn) and diploid (2n2n) stages. The gametophyte generation is reduced and dependent on the sporophyte.
(a)
Describe the process of gametogenesis in the anthers of Nymphaea alba, including the change in chromosome number. [3 marks]
(b)
Calculate the number of chromosomes in a zygote of Nymphaea alba if each gamete contains 4242 chromosomes. [1 mark]
(c)
Explain one advantage of sexual reproduction over asexual reproduction for Nymphaea alba in a changing environment. [2 marks]
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31ChallengeSAQ-LAsexual vs sexual reproduction24 marksPaper 2~36 min
Hieracium pilosella (mouse-ear hawkweed) reproduces both asexually via apomixis (seed production without fertilisation) and sexually. A research team monitored two populations over three growing seasons. Population A (stable field — consistent rainfall and temperature): - Average seed set per plant: 450seeds year1450\,\text{seeds year}^{-1} - 95%95\% of seeds produced via apomixis; 5%5\% via sexual reproduction - Seedling survival rate: 72%72\% - 92%92\% of seedlings genetically identical to mother plant (microsatellite analysis) Population B (variable field — unpredictable rainfall, frequent pest outbreaks): - Average seed set per plant: 310seeds year1310\,\text{seeds year}^{-1} - 40%40\% of seeds produced via apomixis; 60%60\% via sexual reproduction - Seedling survival rate: 48%48\% - 38%38\% of seedlings genetically identical to mother plant (microsatellite analysis)
(a)
(i) Calculate the number of sexually produced seeds per plant per year for each population. [2]
(ii) State which population produces more sexually derived seeds and suggest one reason for this difference based on the environmental conditions described. [2 marks]
(b)
Explain the cellular mechanism of apomixis that results in offspring genetically identical to the mother plant, and contrast this with the mechanism of sexual reproduction that generates genetic variation. [3 marks]
(c)
Evaluate the hypothesis: "Environmental stability favours asexual reproduction, while environmental variability favours sexual reproduction." Use the data provided to support your evaluation. [5] Total: [12 marks]
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32ChallengeSAQ-LMendelian inheritance9 marksPaper 2~14 min
A geneticist is studying a family with a history of cystic fibrosis (CF), an autosomal recessive disorder caused by mutations in the CFTR gene. The pedigree below shows the inheritance of CF across three generations. - Individuals I-1 and I-2 are unaffected. - Individual II-1 is affected with CF and is the child of I-1 and I-2. - Individual II-2 is an unaffected child of I-1 and I-2. - Individual II-3 is unaffected; their brother, II-4, is affected with CF. - Individual III-1 is the unaffected child of II-2 and II-3. Use the allele symbols FF (dominant, normal) and ff (recessive, CF).
(a)
State the genotypes of individuals I-1 and I-2. [1 mark]
(b)
Calculate the probability that individual II-2 is a carrier of the CF allele. Show your reasoning. [3 marks]
(c)
Explain how the Law of Segregation is demonstrated by the inheritance of CF from generation I to generation II in this pedigree. [2 marks]
(d)
Evaluate the use of pedigree analysis a tool for predicting the risk of autosomal recessive disorders in future offspring, using evidence from this pedigree to support your answer. [3 marks]
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33MasterySAQ-SPunnett squares and genetic ratios6 marksPaper 2~9 min
In pea plants, the allele for purple flower colour (PP) is dominant over the allele for white flower colour (pp). A farmer crosses a heterozygous purple-flowered plant with a white-flowered plant.
(a)
State the expected genotypic ratio of the offspring from this cross. [1 mark]
(b)
Calculate the probability that two randomly selected offspring from this cross are both homozygous for flower colour. [2 marks]
(c)
Explain why the observed phenotypic ratio from this cross might differ from the expected 1:11:1 ratio in a small sample of 10 offspring. [3 marks]
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34ChallengeSAQ-LMendelian inheritance7 marksPaper 2~11 min
In a breeding experiment an agricultural research station, a true-breeding strain of pea plants with round, yellow seeds (RRYYRRYY) was crossed with a true-breeding strain with wrinkled, green seeds (rryyrryy). The F1F_1 generation all had round, yellow seeds. The F1F_1 plants were then self-fertilized to produce an F2F_2 generation of 320 seeds. The following results were recorded: - Round, yellow seeds: 175 - Round, green seeds: 55 - Wrinkled, yellow seeds: 65 - Wrinkled, green seeds: 25
(a)
State the expected phenotypic ratio for the F2F_2 generation, assuming the two genes assort independently. [1 mark]
(b)
Calculate the chi-squared (χ2\chi^2) value for the observed data, using the formula χ2=Σ(OE)2E\chi^2 = \Sigma\dfrac{(O - E)^2}{E} and the expected ratio from (a). [3 marks]
(c)
The critical value of χ2\chi^2 at 3 degrees of freedom (p=0.05p = 0.05) is 7.82. Deduce whether the calculated χ2\chi^2 value supports the hypothesis that the two genes assort independently. [2 marks]
(d)
Evaluate the usefulness of Mendel's Law of Independent Assortment for predicting inheritance patterns of genes located on the same chromosome. [1 mark]
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35MasterySAQ-SDNA structure and replication mechanism5 marksPaper 2~8 min
DNA replication is a highly coordinated process involving multiple enzymes. In a eukaryotic cell, replication begins at multiple origins of replication along the chromosome. The below (not to scale) shows a replication bubble with two replication forks moving in opposite directions away from a single origin.
(a)
State the role of DNA helicase in DNA replication. [1 mark]
(b)
A replication fork moves at a rate of 5050 base pairs per second. A chromosome segment of 150,000150{,}000 base pairs is replicated from a single origin located at the centre of the segment, with two forks moving in opposite directions at equal speed. Calculate the minimum time, in minutes, required to completely replicate this segment. [2 marks]
(c)
Explain why eukaryotic chromosomes use multiple origins of replication rather than a single origin. [2 marks]
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36MasterySAQ-SEnzymes involved in DNA replication8 marksPaper 2~12 min
A student analyses a of a replication fork in E. coli. The shows the following labelled components: DNA helicase, single-stranded binding proteins (SSBs), DNA primase, DNA polymerase III, and the leading and lagging strands.
(a)
Describe the sequence of events involving helicase, SSBs, and primase at the replication fork, starting from the moment the parental DNA is first opened. [3 marks]
(b)
An Okazaki fragment on the lagging strand is 12001200 nucleotides long. DNA primase synthesises an RNA primer of 1010 nucleotides, and DNA polymerase III adds nucleotides at a rate of 800nucleotides s1800\,\text{nucleotides s}^{-1}. Assume primer synthesis occurs at the same rate. Calculate the time, in seconds, required to complete synthesis of this Okazaki fragment, including primer synthesis. Show your working. [2 marks]
(c)
Explain why the lagging strand must be synthesised as discontinuous fragments, and state one consequence of this for the final structure of the replicated strand. [3 marks]
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37ChallengeSAQ-LDNA structure and replication mechanism7 marksPaper 2~11 min
The bacterium Escherichia coli has a single circular chromosome of 4.6×1064.6 \times 10^{6} base pairs. Replication proceeds bidirectionally from a single origin of replication (oriC). Each replication fork adds nucleotides at 1000nucleotidess11000\,\text{nucleotides}\,\text{s}^{-1}.
(a)
Calculate the minimum time, in seconds, required to replicate the entire E. coli chromosome. Show all working. [3 marks]
(b)
Explain, with reference to two specific features of the replication mechanism, why the actual replication time exceeds this calculated minimum. [2 marks]
(c)
The two strands of the DNA double helix run antiparallel, yet DNA polymerase III can only synthesise in the 535' \rightarrow 3' direction. Evaluate the significance of this constraint for the mechanism of replication, using structural evidence from the double helix. [2 marks]
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38MasterySAQ-SRegulation of blood glucose levels6 marksPaper 2~9 min
A research team investigates the effect of a new drug, GlucoStat, on blood glucose regulation in healthy laboratory rats. Rats are divided into a control group (injected with saline) and an experimental group (injected with GlucoStat). After 30 minutes, both groups receive an intravenous injection of glucose solution (0.5g kg10.5\,\text{g kg}^{-1} body mass). Blood glucose concentration is measured every 15 minutes for 120 minutes. - Baseline blood glucose (both groups): 4.8mmol L14.8\,\text{mmol L}^{-1} - Control group: peak at 15 min = 10.2mmol L110.2\,\text{mmol L}^{-1}; returns to baseline by 90 min - GlucoStat group: peak at 15 min = 9.1mmol L19.1\,\text{mmol L}^{-1}; returns to baseline by 60 min
(a)
Describe the mechanism by which a rise in blood glucose concentration leads to insulin secretion from pancreatic beta cells. [2 marks]
(b)
Calculate the mean rate of blood glucose clearance in the GlucoStat group, from peak to baseline. Express your answer in mmol L1min1\text{mmol L}^{-1}\text{min}^{-1}[2 marks]
(c)
Using the data, evaluate the effect of GlucoStat on blood glucose regulation. [2 marks]
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39MasterySAQ-SRegulation of blood glucose levels5 marksPaper 2~8 min
A researcher investigates insulin action using cultured human skeletal muscle cells (myocytes). Cells are incubated in medium containing 20mmolL120\,\text{mmol}\,\text{L}^{-1} glucose. One group is treated with 100nM100\,\text{nM} insulin; a control group receives no insulin. After 30 minutes, GLUT4 glucose transporter proteins are counted. Group — GLUT4 on cell surface membrane — GLUT4 internal vesicles Control — 200 per cell — 4800 per cell Insulin-treated — 4800 per cell — 200 per cell Total GLUT4 per cell (surface + internal) = 5000 in both groups.
(a)
Describe the mechanism by which insulin causes an increase in the number of GLUT4 transporters on the cell surface membrane. [2 marks]
(b)
Calculate the percentage of total GLUT4 proteins located on the cell surface membrane in the insulin-treated group. Show your working. [2 marks]
(c)
Using the data and your knowledge of membrane transport, evaluate whether the rate of glucose uptake in the insulin-treated cells would increase proportionally with the increase in surface GLUT4 number. [1 mark]
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40ChallengeSAQ-LRegulation of blood glucose levels8 marksPaper 2~12 min
A clinical study investigated blood glucose regulation in two groups of healthy adults (n=15n = 15 each) following an oral glucose tolerance test (OGTT). Participants fasted for 12 hours, then consumed 75g75\,\text{g} of glucose dissolved in water. Blood samples were taken every 30 minutes for 180 minutes. Mean blood glucose concentration (mmolL1\text{mmol}\,\text{L}^{-1}): Time (min) — Group A (control) — Group B (soluble fibre) 0 — 4.8 — 4.7 30 — 8.2 — 6.9 60 — 7.5 — 6.1 90 — 6.0 — 5.2 120 — 5.1 — 4.9 180 — 4.9 — 4.8 Group B consumed 10g10\,\text{g} of soluble fibre 30 minutes before the glucose drink.
(a)
Calculate the percentage change in blood glucose concentration from t=0t = 0 to t=30mint = 30\,\text{min} for Group A. [2 marks]
(b)
Explain how blood glucose concentration is returned to baseline after the peak at t=30mint = 30\,\text{min} in Group A, including the role of insulin and its mechanism of action target cells. [3 marks]
(c)
Evaluate the hypothesis that soluble fibre supplementation improves glucose regulation by slowing glucose absorption, using both the data provided and your knowledge of blood glucose homeostasis. [3 marks]
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41ChallengeSAQ-LTranscription and translation processes7 marksPaper 2~11 min
A study was conducted on the expression of the β\beta-globin gene in human erythroblast precursor cells. Researchers isolated mRNA from these cells and used it as a template for complementary DNA (cDNA) synthesis. The cDNA was then sequenced. The following data were obtained from the coding strand of the β\beta-globin gene and the corresponding mature mRNA transcript. - Length of β\beta-globin gene (coding strand): 1600bp1600\,\text{bp} - Length of β\beta-globin mature mRNA (after processing): 626nt626\,\text{nt} - Number of amino acids in the final β\beta-globin polypeptide: 146146 - The gene contains three exons and two introns.
(a)
Calculate the number of nucleotides in the combined introns of the β\beta-globin gene. Show your working. [2 marks]
(b)
Explain why the number of nucleotides in the mature mRNA (626nt626\,\text{nt}) is greater than three times the number of amino acids in the polypeptide. [2 marks]
(c)
Evaluate whether the data support the conclusion that all exonic sequences in the β\beta-globin gene are translated into the polypeptide. [3 marks]
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42MasterySAQ-SRole of mRNA, tRNA, and ribosomes7 marksPaper 2~11 min
The below shows a ribosome during translation in a prokaryotic cell. The mRNA sequence is 5-AUGGCUAAC-35'\text{-AUGGCUAAC-}3'. A large and small ribosomal subunit are shown with mRNA threaded between them. Three tRNA binding sites are visible: the E site (empty), the P site (occupied by a tRNA with anticodon 3-UAC-53'\text{-UAC-}5' carrying methionine and a growing peptide), and the A site (approached by a tRNA carrying alanine). The codons are 5-AUG-35'\text{-AUG-}3', 5-GCU-35'\text{-GCU-}3', and 5-AAC-35'\text{-AAC-}3'.
(a)
State the anticodon sequence, written 353' \to 5', of the tRNA that binds to the third codon AAC. [1 mark]
(b)
Describe how the structure of tRNA enables it to carry a specific amino acid and to recognise a specific mRNA codon. [2 marks]
(c)
The third codon is 5-AAC-35'\text{-AAC-}3'. Calculate the total number of hydrogen bonds formed between this codon and its complementary anticodon, using the rules A–U=2\text{A–U} = 2 bonds and G–C=3\text{G–C} = 3 bonds. [2 marks]
(d)
The wobble hypothesis states that the third base of anticodon can form non-Watson–Crick base pairs with the third base of a codon. Explain one advantage this gives to a cell during translation. [2 marks]
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43ChallengeSAQ-LTranscription and translation processes7 marksPaper 2~11 min
In an investigation of translation efficiency, researchers used an in vitro translation system containing ribosomes, tRNAs, amino acids, ATP, and GTP. They added a synthetic mRNA with the sequence 5-AUG GUA CCC UAA GCA UGG-35'\text{-AUG GUA CCC UAA GCA UGG-}3'.
(a)
Determine the amino acid sequence of the polypeptide produced from this mRNA. Use three-letter amino acid abbreviations and show your codon assignments. [3 marks]
(b)
Explain the molecular events at the ribosome that prevent the codons GCA and UGG from being translated. [2 marks]
(c)
A mutation in a tumour suppressor gene introduces a premature stop codon near the middle of the coding sequence. Evaluate whether the resulting truncated protein is likely to be functional, referring to protein structure and function. [2 marks]
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44MasterySAQ-SMechanisms of natural selection7 marksPaper 2~11 min
The peppered moth (Biston betularia) is a classic example of natural selection. Before industrialisation, most peppered moths in England had light, speckled colouration (typica morph); a dark melanic morph (carbonaria morph) was very rare. During the Industrial Revolution, soot from factories darkened tree trunks and killed light-coloured lichens. Bird predation experiments showed that on soot-darkened trees, typica moths were eaten more frequently than carbonaria moths. - Frequency of carbonaria morph in Manchester in 1848: 2%2\% - Frequency of carbonaria morph in Manchester in 1895: 98%98\% - Generation time of peppered moth: 11 year
(a)
State three necessary components of natural selection. [3 marks]
(b)
Calculate the average annual percentage-point increase in the frequency of the carbonaria morph from 1848 to 1895. Show your working. [2 marks]
(c)
After clean-air legislation in the late 20th century reduced industrial pollution, carbonaria morph frequency declined again. Evaluate whether this reversal supports or contradicts the conclusion that the original rise in carbonaria frequency was caused by natural selection rather than by a non-heritable change individual moth colouration. [2 marks]
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45MasterySAQ-SMechanisms of natural selection7 marksPaper 2~11 min
A population of cheetahs (Acinonyx jubatus) lives in a nature reserve in Namibia. A researcher measures the running speed of 200200 adult cheetahs. The fastest 10%10\% of cheetahs (those running over 110kmh1110\,\text{km\,h}^{-1}) produce, on average, 44 surviving cubs per year. The slowest 10%10\% (those running under 85kmh185\,\text{km\,h}^{-1}) produce, on average, 11 surviving cub per year. The remaining 80%80\% produce, on average, 22 surviving cubs per year. Running speed is a heritable, polygenic trait.
(a)
State the type of natural selection acting on running speed in this population. [1 mark]
(b)
Calculate the mean number of surviving cubs per year per cheetah for the whole population of 200200 individuals. [2 marks]
(c)
Explain how this selection pressure will change the mean running speed of the population over successive generations. [2 marks]
(d)
Predict the effect of this selection the genetic diversity of running speed alleles in the population over many generations, and justify your prediction. [2 marks]
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46ChallengeSAQ-LMechanisms of natural selection7 marksPaper 2~11 min
The peppered moth (Biston betularia) is a classic example of natural selection driven by industrial pollution. In 19th-century England, two phenotypes existed: light-coloured (typica) and dark-coloured (carbonaria). Before industrialisation, typica moths were camouflaged against lichen-covered bark. After pollution darkened tree trunks with soot, carbonaria increased in frequency. Observed frequencies: Period — carbonaria frequency — typica frequency Pre-industrial (1848) — 2%2\%98%98\% Post-industrial (1895) — 95%95\%5%5\% Relative fitness values (ww): Phenotype — Polluted forest — Unpolluted forest carbonaria — 0.950.950.400.40 typica — 0.500.500.900.90 Generation time of Biston betularia: 1 year.
(a)
State the selection coefficient (ss) against the typica phenotype in the polluted forest, using s=1ws = 1 - w[1 mark]
(b)
The change in frequency of a favoured allele per generation under selection can be approximated by: Δqsq(1q)wˉ\Delta q \approx \frac{s \cdot q(1-q)}{\bar{w}} where qq is the current frequency of the carbonaria allele, ss is the selection coefficient against typica, and wˉ\bar{w} is mean population fitness. Using q0=0.02q_0 = 0.02, s=0.50s = 0.50, and wˉ=1\bar{w} = 1 (approximation), calculate the predicted frequency of the carbonaria allele after one generation of selection in the polluted forest. [2 marks]
(c)
Explain how the three conditions necessary for natural selection — variation, heritability, and differential reproductive success — account for the observed increase in carbonaria frequency between 1848 and 1895. [3 marks]
(d)
The relative fitness of carbonaria in unpolluted forests is 0.400.40, compared with 0.900.90 for typica. Evaluate whether the carbonaria allele would be expected to reach fixation in a landscape containing both polluted and unpolluted forest patches, using the fitness data provided. [1 mark]
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47MasterySAQ-SSustainable agricultural practices5 marksPaper 2~8 min
A researcher studies the effect of sustainable agricultural practices on soil biodiversity in a grassland ecosystem. She compares two adjacent fields over 5 years: Field A is managed with conventional tillage and synthetic pesticides; Field B uses no-till farming, cover cropping, and integrated pest management (IPM). The Shannon-Wiener diversity index (HH) for soil invertebrates is measured each year. A higher HH value indicates greater biodiversity. Year — Field A (HH) — Field B (HH) 1 — 1.21.21.81.8 3 — 0.90.92.32.3 5 — 0.70.72.52.5
(a)
State the trend in HH for Field A and for Field B over the 5-year period. [2 marks]
(b)
Calculate the percentage change in HH for Field A from Year 1 to Year 5. [1 mark]
(c)
Explain why HH increased in Field B but decreased in Field A, with reference to the agricultural practices used in each field. [2 marks]
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48ChallengeSAQ-LRenewable resources and ecological conservation7 marksPaper 2~11 min
A field study was conducted in a temperate deciduous forest to evaluate the sustainability of harvesting wood for biofuel. Two 1ha1\,\text{ha} plots were established: Plot A (unharvested control) and Plot B (selective logging of 30%30\% of tree biomass every 5 years). Data collected after 20 years are shown below. Parameter — Plot A — Plot B Total woody biomass — 450t450\,\text{t}320t320\,\text{t} Simpson's Diversity Index (DD) — 0.870.870.720.72 Soil organic carbon — 120t120\,\text{t}95t95\,\text{t} Annual NPP — 12.5tha1yr112.5\,\text{t}\,\text{ha}^{-1}\,\text{yr}^{-1}10.8tha1yr110.8\,\text{t}\,\text{ha}^{-1}\,\text{yr}^{-1} - Harvested wood from Plot B over 20 years: 210t210\,\text{t} - Energy content of dry wood: 18.5MJkg118.5\,\text{MJ}\,\text{kg}^{-1} D=1ni(ni1)N(N1)Percentage change=neworiginaloriginal×100%D = 1 - \frac{\sum n_i(n_i - 1)}{N(N-1)} \qquad \text{Percentage change} = \frac{\text{new} - \text{original}}{\text{original}} \times 100\%
(a)
Determine the percentage change in Simpson's Diversity Index (DD) from Plot A to Plot B, and state what this value indicates about species evenness in the logged plot. [2 marks]
(b)
Explain the mechanism by which selective logging could reduce soil organic carbon levels in Plot B compared to Plot A. [3 marks]
(c)
Evaluate whether the wood harvested from Plot B represents a renewable energy source over the 20-year period. Use calculations of energy yield and total NPP energy to support your answer. [2 marks]
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49ChallengeSAQ-LRenewable resources and ecological conservation7 marksPaper 2~11 min

Data

Before construction: - Mean tidal range: 4.2m4.2\,\text{m} - Salinity in estuary: 32ppt32\,\text{ppt} - Phytoplankton primary productivity: 2.8g C m2day12.8\,\text{g C m}^{-2}\,\text{day}^{-1} - Fish species richness: 2424 species - Shorebird population (annual maximum): 12,50012{,}500 individuals - Detritus export to coastal waters: 4,500tonnes C yr14{,}500\,\text{tonnes C yr}^{-1} After construction (modelled): - Salinity behind barrage: 18ppt18\,\text{ppt} - Phytoplankton productivity: 1.1g C m2day11.1\,\text{g C m}^{-2}\,\text{day}^{-1} - Fish species richness: 99 species - Shorebird population: 1,8001{,}800 individuals - Detritus export: 800tonnes C yr1800\,\text{tonnes C yr}^{-1}
A coastal community is evaluating the installation of a tidal barrage for renewable electricity generation. The barrage would enclose a 5.0km25.0\,\text{km}^2 estuary. Environmental impact assessments collected the following baseline
(a)
Calculate the percentage reduction in shorebird population following barrage construction. State one reason why fish species richness also declines. [2 marks]
(b)
Explain the mechanism by which reduced detritus export to coastal waters could affect offshore food webs. [2 marks]
(c)
Using the data provided, evaluate whether the tidal barrage represents a sustainable solution, considering both energy generation and ecological impact. [3 marks]
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50MasterySAQ-STypes of mutations: Point, frameshift, chromosomal5 marksPaper 2~8 min
A karyotype analysis of a human patient reveals a chromosomal mutation. The patient has 46 chromosomes, but one copy of chromosome 7 has a segment that is inverted relative to the normal sequence. A different patient has 45 chromosomes formed by the fusion of two acrocentric chromosomes at their centromeres.
(a)
State the type of chromosomal mutation present in the first patient. [1 mark]
(b)
Explain why the second patient has 45 chromosomes despite retaining a full complement of genetic material. [2 marks]
(c)
Evaluate the claim that a point mutation in an exon always has a greater effect on the protein product than a point mutation in an intron of the same gene. In your answer, refer to specific mechanisms by which each mutation type can affect or fail to affect the protein. [2 marks]
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51MasterySAQ-SGene editing technologies (e.g. CRISPR)5 marksPaper 2~8 min
A researcher uses the CRISPR-Cas9 system to disrupt a gene in human cells grown in culture. She designs a single guide RNA (sgRNA) that targets a sequence near the start of the gene. After editing, she uses PCR with primers flanking the target site to amplify the region. The PCR products are separated by gel electrophoresis. Cells that were not exposed to CRISPR-Cas9 produce a single band at 500bp500\,\text{bp}. Cells exposed to CRISPR-Cas9 produce two bands: one at 500bp500\,\text{bp} and one at 320bp320\,\text{bp}.
(a)
State how the sgRNA enables the Cas9 protein to cut at one specific location in the genome. [1 mark]
(b)
Determine the size, in base pairs, of the deletion introduced by the CRISPR-Cas9 cut, and state one assumption required for this calculation. [2 marks]
(c)
Explain why the simultaneous presence of both the 500bp500\,\text{bp} band the 320bp320\,\text{bp} band indicates incomplete editing of the cell population, and predict what result would be expected if editing were 100%100\% efficient. [2 marks]
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52ChallengeSAQ-LTypes of mutations: Point, frameshift, chromosomal8 marksPaper 2~12 min
A research team studies a gene in the bacterium E. coli that codes for an enzyme involved in lactose metabolism. The following segment of the coding strand (non-template strand) of the wild-type gene is shown: 5-ATG GGC TAC CCT GAA TTC-35'\text{-ATG GGC TAC CCT GAA TTC-}3' The wild-type peptide produced from this segment is: Met–Gly–Tyr–Pro–Glu–Phe. After exposure to a chemical mutagen, two mutant strains are isolated, both unable to metabolise lactose. - Mutant 1: a single guanine (G) nucleotide is inserted between the 4th and 5th nucleotides of the coding strand shown above. - Mutant 2: a single base substitution changes the cytosine (C) at position 10 of the coding strand to adenine (A). Use the genetic code in the IB Data Booklet where required.
(a)
State the type of mutation present in Mutant 1. [1 mark]
(b)
Determine the mRNA sequence produced from the mutant 1 coding strand deduce the amino acid sequence of the resulting peptide from the point of insertion onwards. Show your working. [3 marks]
(c)
Explain why the mutation in Mutant 2 would have a different effect on protein function compared to the mutation in Mutant 1. [2 marks]
(d)
Evaluate the claim that frameshift mutations are generally more harmful to an organism than point mutations, using the specific examples from this question to support your answer. [2 marks]
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53MasterySAQ-SMitosis vs meiosis6 marksPaper 2~9 min
A single diploid cell (2n=42n = 4) undergoes nuclear division as shown below. Cell 1: Four chromosomes (two long, two short), each consisting of two sister chromatids, arranged as two homologous pairs. Cell 2: Homologous pairs have separated; one long and one short chromosome move to opposite poles. Cell 3: Two cells, each containing two chromosomes (one long, one short), each still consisting of two sister chromatids. Cell 4: Four cells, each containing two chromosomes (one long, one short) as single chromatids.
(a)
State the type of nuclear division shown. [1 mark]
(b)
State the number of chromatids per cell at the stage shown in Cell 3, and explain why this differs from the number of chromosomes per cell at Cell 4. [2 marks]
(c)
Describe one event occurring during prophase I that increases genetic variation. [1 mark]
(d)
Explain how independent assortment, acting alone, produces genetic differences between the cells in Cell 4 and the original cell in Cell 1. [2 marks]
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54MasterySAQ-SPhases of mitosis and meiosis5 marksPaper 2~8 min
A diploid cell (2n=42n = 4) undergoes meiosis. The shows the cell at metaphase I, with two homologous pairs of chromosomes arranged at the equator of the cell. Chiasmata are visible between homologous chromosomes.
(a)
State the stage of meiosis shown in the . [1 mark]
(b)
Describe the behaviour of chromosomes during metaphase I of meiosis. [2 marks]
(c)
A student examines 250 cells from anther undergoing meiosis and records that 20 cells are at metaphase I. The entire meiotic division takes 48 hours. Calculate the average time, in minutes, that a cell spends in metaphase I. Show your working. [2 marks]
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55ChallengeSAQ-LMitosis vs meiosis7 marksPaper 2~11 min
A research team investigates the effects of a novel compound, Karyostatin, on cell division in the root meristem of Allium cepa. Root tip cells are treated with Karyostatin for 24 hours, then fixed, stained, and observed under a light microscope. Cells are classified by phase. Results are shown below. Phase — Control (500 cells) — Karyostatin-treated (500 cells) Interphase — 120 — 185 Prophase — 240 — 100 Metaphase — 60 — 200 Anaphase — 50 — 10 Telophase — 30 — 5
(a)
Calculate the mitotic index for both the control group and the Karyostatin-treated group, expressing each as a percentage. [2 marks]
(b)
Using the data, explain the stage at which Karyostatin disrupts mitosis and identify the most likely cellular mechanism responsible. [3 marks]
(c)
Karyostatin is proposed as a cancer therapeutic. Evaluate the significance of the observed decrease in mitotic index in the treated group when assessing Karyostatin's potential effectiveness against rapidly dividing tumour cells. [2 marks]
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