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Unity and Diversity — Free Biology SL Practice Questions

1FoundationMCQWater as a solvent1 markPaper 1~2 min
A student adds a drop of vegetable oil to a test tube of water and shakes the mixture. The oil initially disperses into small droplets but quickly coalesces into a separate layer above the water. Which property of water best explains why oil does not dissolve in water?
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2MasteryMCQWater as a solvent1 markPaper 1~2 min
A small crystal of potassium permanganate is placed in a beaker of still distilled water at 25 °C. The purple colour spreads evenly throughout the water over time without stirring. Which property of water is most directly responsible for the permanganate ions becoming evenly distributed?
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3FoundationMCQTheories of evolution (Darwin, Lamarck)1 markPaper 1~2 min
Lamarck proposed that traits acquired during an organism's lifetime through use or disuse could be inherited by offspring. Which observation provides the strongest evidence against this hypothesis?
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4MasteryMCQTheories of evolution (Darwin, Lamarck)1 markPaper 1~2 min
A species of land snail shows shell colour variation from pale yellow to dark brown. In areas with darker soil, dark brown shells are more common; in areas with lighter soil, pale yellow shells predominate. A student proposes that individual snails darken their shells during their lifetime in response to predation pressure, and that this darkening is inherited by their offspring. Which statement correctly identifies the key difference between the student's proposal and Darwin's theory of natural selection?
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5FoundationMCQHuman impact on ecosystems and biodiversity1 markPaper 1~2 min
Selective logging removed 30% of the large canopy trees from a tropical rainforest. Five years later, bird species richness in the logged area was significantly lower than in adjacent undisturbed area. Which of the following most directly explains this decline in bird species richness?
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6MasteryMCQImportance of biodiversity1 markPaper 1~2 min
Fragment X is a forest with 12 tree species and a Simpson's Diversity Index of D=0.85D = 0.85. Fragment Y is an adjacent forest with 8 tree species and D=0.92D = 0.92. Which statement best explains the difference in overall biodiversity between the two fragments?
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7FoundationMCQStructure of DNA and RNA1 markPaper 1~2 min
In the Hershey–Chase experiment, bacteriophages were grown separately in media containing 32^{32}P or 35^{35}S. After infection of bacteria, blending and centrifugation produced a pellet of bacterial cells that was radioactive only when phages had been grown in 32^{32}P medium. Which statement correctly explains this outcome?
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8MasteryMCQStructure of DNA and RNA1 markPaper 1~2 min
One strand of a DNA double helix has the sequence 5-ATGC-35'\text{-ATGC-}3'. Which of the following correctly represents the complementary strand?
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9FoundationMCQProkaryotic vs eukaryotic cells1 markPaper 1~2 min
A newly discovered unicellular organism has a single circular chromosome located in a region of cytoplasm not enclosed by a membrane, and contains 70S ribosomes. Which statement correctly classifies this organism and identifies the most direct structural evidence for that classification?
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10MasteryMCQProkaryotic vs eukaryotic cells1 markPaper 1~2 min
In E. coli, ribosomes can begin translating an mRNA molecule before transcription of that molecule is complete. This coupled transcription–translation does not occur in human liver cells. What is the most direct structural explanation for this difference?
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11FoundationMCQEvolutionary relationships and cladistics1 markPaper 1~2 min
A cladogram shows the following topology: species A is the outgroup; B is the first ingroup lineage to diverge; a subsequent node gives rise to E one branch and, on the other branch, node X from which C and D diverge as sister taxa. Which statement about synapomorphies is correct?
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12MasteryMCQClassification systems1 markPaper 1~2 min
A cladogram shows four plant groups — moss (Bryophyta), fern (Pteridophyta), pine (Pinophyta), and rose (Magnoliophyta) — with the following branching pattern: moss diverges first, then ferns diverge from a clade containing pines and roses. Which conclusion is most consistent with this cladogram and the principles of cladistic classification?
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13MasterySAQ-SWater potential in plants5 marksPaper 2~8 min
A student places a red blood cell and an Elodea leaf cell into separate beakers of distilled water (ψ=0kPa\psi = 0\,\text{kPa}). The cells are observed under a microscope after 10 minutes. Red blood cell: ψs=800kPa\text{Red blood cell: } \psi_s = -800\,\text{kPa} Elodea leaf cell: ψs=800kPa,ψp=+600kPa\text{Elodea leaf cell: } \psi_s = -800\,\text{kPa},\quad \psi_p = +600\,\text{kPa}
(a)
Calculate the initial water potential of the Elodea leaf cell before it is placed in distilled water. [1 mark]
(b)
Describe the difference in the appearance of the red blood cell and the Elodea cell after 10 minutes in distilled water. [2 marks]
(c)
Explain, using water potential values, why the Elodea cell reaches equilibrium with the distilled water without bursting, while the red blood cell does not. [2 marks]
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14MasterySAQ-SWater’s role in temperature regulation5 marksPaper 2~8 min
A coastal city experiences a sea breeze during the day. The land surface heats to 35C35\,^{\circ}\text{C}, while the sea surface remains at 22C22\,^{\circ}\text{C}. The specific heat capacity of dry soil is 800Jkg1K1800\,\text{J\,kg}^{-1}\text{K}^{-1} and the specific heat capacity of seawater is 4200Jkg1K14200\,\text{J\,kg}^{-1}\text{K}^{-1}.
(a)
State what is meant by specific heat capacity. [1 mark]
(b)
A 1.0kg1.0\,\text{kg} sample of dry soil and a 1.0kg1.0\,\text{kg} sample of seawater each absorb 5000J5000\,\text{J} of solar energy. Calculate the temperature increase of each sample. [2 marks]
(c)
Using your answers from (b), explain how the difference in specific heat capacities between land sea leads to the formation of a sea breeze during the day. [2 marks]
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15ChallengeSAQ-LWater as a solvent7 marksPaper 2~11 min

Data

- Hydrated radius of DOX+\text{DOX}^+: 0.85nm0.85\,\text{nm} - Hydrated radius of Na+\text{Na}^+: 0.36nm0.36\,\text{nm} - Hydrated radius of Cl\text{Cl}^-: 0.33nm0.33\,\text{nm} - Concentration of –COO⁻ groups in the fully swollen hydrogel: 0.12moldm30.12\,\text{mol}\,\text{dm}^{-3} - Water absorbed by hydrogel: 15.0g15.0\,\text{g} per gram of dry polymer - Volume of one water molecule: 3.0×1029m33.0 \times 10^{-29}\,\text{m}^3 - Avogadro constant: NA=6.02×1023mol1N_A = 6.02 \times 10^{23}\,\text{mol}^{-1} - Density of water: 1.00gcm31.00\,\text{g}\,\text{cm}^{-3}
A medical researcher investigates a polymer hydrogel for controlled drug delivery. The hydrogel consists of cross-linked poly(acrylic acid) (PAA) chains. At physiological pH 7.4, the carboxyl groups (–COOH) dissociate to carboxylate ions (–COO⁻). A positively charged drug molecule, doxorubicin (DOX+\text{DOX}^+), is loaded into the hydrogel.
(a)
Calculate the number of water molecules directly associated with each –COO⁻ group in the fully swollen hydrogel. [3 marks]
(b)
Explain why the hydrogel swells significantly more in pure water than in 0.9%0.9\% (mass/volume) NaCl solution, with reference to the properties of water and the interactions between ions and the polymer network. [2 marks]
(c)
Using the hydrated radius data provided, evaluate whether the movement of DOX+\text{DOX}^+ through the hydrogel in the bloodstream would be more restricted than that of Na+\text{Na}^+ or Cl\text{Cl}^-, and assess one additional factor that limits the accuracy of this hydrogel model for predicting DOX+\text{DOX}^+ release rate in vivo. [2 marks]
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16MasterySAQ-STheories of evolution (Darwin, Lamarck)8 marksPaper 2~12 min
A population of a small mammal species lives on an island with two distinct habitats: a dark, forested area and a light, sandy beach. The mammals are preyed upon by birds that hunt by sight. A biologist recorded fur colour frequencies over several generations. Generation 1 — 45%45\% light fur, 55%55\% dark fur — 60%60\% light fur, 40%40\% dark fur Generation 10 — 82%82\% light fur, 18%18\% dark fur — 21%21\% light fur, 79%79\% dark fur - Average offspring per breeding pair, Generation 1: 4.24.2 - Average offspring per breeding pair, Generation 10: 3.83.8
(a)
State the selection pressure acting on the mammal population in both habitats. [1 mark]
(b)
Calculate the percentage change in the average number offspring per breeding pair from Generation 1 to Generation 10. [3 marks]
(c)
Explain how Darwin's theory of natural selection accounts for the observed change in fur colour frequency in the beach habitat between Generation 1 and Generation 10. [4 marks]
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17MasterySAQ-STheories of evolution (Darwin, Lamarck)6 marksPaper 2~9 min

Data

- Year 1: weed population =10000= 10\,000 plants; herbicide efficacy =95%= 95\% killed - Year 8: weed population =12000= 12\,000 plants; herbicide efficacy =15%= 15\% killed - Estimated frequency of spontaneous resistance mutation: 11 in 10001\,000 plants per generation - Average generation time of weed: 11 year
A farmer notices that a species of weed in his fields has become increasingly resistant to a common herbicide over a period of 8 years. In Year 1, the herbicide at the recommended dose killed 95%95\% of the weed population. By Year 8, the same dose killed only 15%15\% of the weed population. The farmer also observed that individual weed plants exposed to sub-lethal doses of the herbicide sometimes produced offspring that were slightly more resistant, even without direct herbicide exposure to those offspring.
(a)
Calculate the number of plants in the Year 1 population expected to carry a new resistance mutation. [2 marks]
(b)
Explain the role of mutation in the evolution of herbicide resistance according to Darwin's theory of natural selection. [2 marks]
(c)
Discuss whether the farmer's observation of increased resistance in offspring of exposed plants supports Lamarck's theory of evolution rather than Darwin's theory. [2 marks]
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18ChallengeSAQ-LTheories of evolution (Darwin, Lamarck)7 marksPaper 2~11 min
A population of Poecilia reticulata (guppies) lives in a stream in Trinidad. One section of the stream contains the predator fish Crenicichla alta; an adjacent upstream section, separated by a waterfall, is predator-free. The waterfall prevents gene flow between the two populations. Guppies in the predator-present section mature at a smaller size and produce more, smaller offspring than those in the predator-absent section. A researcher recorded the following data over one generation: Section — Average offspring per female — Survival rate to reproductive age Predator-present — 20 — 10%10\% Predator-absent — 8 — 60%60\%
(a)
Calculate the average number offspring per female that survive to reproductive age in each section. [2 marks]
(b)
Explain how Lamarck's theory of evolution would account for the higher fecundity observed in the predator-present population, referring to the mechanism by which traits are transmitted between generations. [2 marks]
(c)
Evaluate whether Darwin's theory of natural selection or Lamarck's theory provides the more scientifically valid explanation for the observed differences in guppy life-history traits, using the data provided. [3 marks]
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19MasterySAQ-SHuman impact on ecosystems and biodiversity6 marksPaper 2~9 min
A tropical rainforest was fragmented over 30 years due to agricultural expansion. The maps below show the forest distribution before and after this period.
(a)
State one way in which forest fragmentation could reduce biodiversity. [1 mark]
(b)
Explain how the isolation of patches A, B, and C could affect the genetic diversity of a species originally distributed across the whole forest. [2 marks]
(c)
The original forest area was 7500km27500\,\text{km}^2. Calculate the total area of forest remaining after 30 years and determine the percentage of the original forest that has been lost. [3 marks]
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20MasterySAQ-SHuman impact on ecosystems and biodiversity5 marksPaper 2~8 min
A study examined the impact of agricultural runoff on a lake ecosystem. The graph below shows the concentration of dissolved oxygen (mgL1)(\text{mg}\,\text{L}^{-1}) at different depths in the lake before and after a fertilizer spill.
(a)
State one difference between the two oxygen profiles shown in the graph. [1 mark]
(b)
Explain the sequence of events by which fertilizer entering the lake leads to a reduction in dissolved oxygen at depth. [2 marks]
(c)
Calculate the percentage decrease in dissolved oxygen at 15m15\,\text{m} depth after the spill. Before the spill, dissolved oxygen at 15m15\,\text{m} was 7.0mgL17.0\,\text{mg}\,\text{L}^{-1}; after the spill it was 0.5mgL10.5\,\text{mg}\,\text{L}^{-1}[2 marks]
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21ChallengeSAQ-LImportance of biodiversity9 marksPaper 2~14 min
A team of ecologists is studying the impact of deforestation ecosystem service value in a tropical rainforest reserve. The reserve has an area of 500km2500\,\text{km}^2 and currently supports a population of 12001200 jaguars, a keystone species. The annual economic value of ecosystem services (carbon sequestration, water purification, and pollination) provided by the reserve is estimated at USD 2.52.5 million per km2\text{km}^2 at the current jaguar population density. Deforestation reduces the jaguar population by 15%15\% per year. The ecosystem service value per km2\text{km}^2 decreases proportionally to jaguar population density.
(a)
Calculate the percentage decrease in the total annual ecosystem service value of the reserve after 33 years of deforestation. [3 marks]
(b)
Explain why a keystone species like the jaguar has a disproportionate effect on ecosystem service value compared to a non-keystone species. [2 marks]
(c)
Evaluate the use of economic valuation of ecosystem services a tool for biodiversity conservation policy, using the data provided as an example. [4 marks]
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22MasterySAQ-SDNA replication and transcription5 marksPaper 2~8 min
A replication fork forms during DNA replication in a bacterium. - Rate of DNA polymerase III: 10001000 nucleotides per second - Length of the bacterial chromosome: 4.6×1064.6 \times 10^{6} base pairs - Energy required to add one nucleotide to a growing strand: 2.0×1019J2.0 \times 10^{-19}\,\text{J}
(a)
State the role of DNA helicase in DNA replication. [1 mark]
(b)
Calculate the minimum time, in seconds, required to replicate the entire bacterial chromosome. Assume replication begins at a single origin of replication and proceeds bidirectionally. [2 marks]
(c)
The total energy consumed during replication of the bacterial chromosome is 1.84×1012J1.84 \times 10^{-12}\,\text{J}. Explain why the total number of nucleotides added during replication is 9.2×1069.2 \times 10^{6}, and not 4.6×1064.6 \times 10^{6}[2 marks]
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23MasterySAQ-SDNA replication and transcription5 marksPaper 2~8 min
A segment of double-stranded DNA has the following sequence: 5-ATGCCTAG-3(coding strand)5'\text{-ATGCCTAG-}3' \quad \text{(coding strand)} 3-TACGGATC-5(template strand)3'\text{-TACGGATC-}5' \quad \text{(template strand)}
(a)
State the direction in which RNA polymerase synthesizes mRNA. [1 mark]
(b)
State one way in which the base composition of mRNA differs from that of the DNA template strand. [1 mark]
(c)
Write the complete mRNA sequence transcribed from the template strand, clearly labelling the 55' and 33' ends. Show your working. [3 marks]
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24ChallengeSAQ-LGene expression regulation7 marksPaper 2~11 min
A research team investigates regulation of the lac operon in E. coli under different environmental conditions. The relative rate of transcription of the lacZ gene (encoding β\beta-galactosidase) is measured under three conditions, normalised so that Condition 1 = 0.10.1. Condition — Glucose — Lactose — Relative transcription rate 1 — present — absent — 0.10.1 2 — absent — present — 100100 3 — present — 2.02.0
(a)
State which condition produces the lowest transcription rate of lacZ and identify the molecule responsible for repressing transcription under that condition. [1 mark]
(b)
Calculate the fold-change in transcription rate between Condition 1 and Condition 2. [2 marks]
(c)
Explain the molecular mechanism that accounts for the difference in transcription rates between Condition 1 and Condition 2. [2 marks]
(d)
Evaluate the biological significance of the transcription rate observed in Condition 3 compared to Condition 2, with reference to catabolite repression and the efficiency of energy use in E. coli. [2 marks]
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25MasterySAQ-SRole of ribosomes, mitochondria, and chloroplasts7 marksPaper 2~11 min
A plant cell chloroplast carries out photosynthesis. Under high light intensity, a single chloroplast produces 5.4×1017mol5.4 \times 10^{-17}\,\text{mol} of glucose per second. The chloroplast contains 200 thylakoid stacks (grana), each with an average of 50 thylakoid discs.
(a)
State where the light-dependent reactions of photosynthesis occur in the chloroplast. [1 mark]
(b)
Explain how the products of the light-dependent reactions are used in the Calvin cycle. [2 marks]
(c)
Calculate the rate of glucose production per thylakoid disc in mols1\text{mol}\,\text{s}^{-1}. Give your answer in standard form to 2 significant figures. [2 marks]
(d)
A student claims that doubling the number of thylakoid discs per granum would double the rate of glucose production by the chloroplast. Evaluate this claim. [2 marks]
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26MasterySAQ-SRole of ribosomes, mitochondria, and chloroplasts5 marksPaper 2~8 min
A bacterial culture is grown under optimal conditions. Each ribosome synthesizes a polypeptide of 300 amino acids in 15s15\,\text{s}. The average mass of an amino acid is 110Da110\,\text{Da}, where 1Da=1.66×1024g1\,\text{Da} = 1.66 \times 10^{-24}\,\text{g}.
(a)
State the sizes of the large and small subunits of a bacterial ribosome. [1 mark]
(b)
Explain how the structure of a ribosome enables peptide bond formation during translation. [2 marks]
(c)
Calculate the rate of protein synthesis, in gs1\text{g}\,\text{s}^{-1}, for one ribosome. Give your answer in standard form to 2 significant figures. [2 marks]
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27ChallengeSAQ-LMembrane structure and fluid mosaic model7 marksPaper 2~11 min

Data

- Average width of a phospholipid bilayer: 7.5nm7.5\,\text{nm} - Diameter of a single cholesterol molecule: 0.8nm0.8\,\text{nm} - Ratio of cholesterol to phospholipid molecules in a typical mammalian cell membrane: 1:21:2 - Lateral diffusion coefficient (DD) of a phospholipid molecule at 37°C37°\text{C}: 1.0×108cm2s11.0 \times 10^{-8}\,\text{cm}^2\,\text{s}^{-1} - Mean square displacement in two dimensions: x2=4Dt\langle x^2 \rangle = 4Dt - 1μm=104cm1\,\mu\text{m} = 10^{-4}\,\text{cm}
The fluid mosaic model describes the dynamic structure of cell membranes. The below shows a representation of a plasma membrane with various components labelled A to E.
(a)
Calculate the mean time, in seconds, for a phospholipid molecule to diffuse laterally a root-mean-square distance of 2.0μm2.0\,\mu\text{m} (the length of a typical bacterial cell). Show all working. [3 marks]
(b)
Explain how cholesterol (component B), present at the ratio given, affects membrane fluidity at 37°C37°\text{C} and at 20°C20°\text{C}[2 marks]
(c)
The fluid mosaic model proposes that membrane proteins and lipids undergo free lateral diffusion. Evaluate this claim by referring to membrane rafts and explaining how their molecular composition restricts the lateral diffusion of associated proteins. [2 marks]
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28MasterySAQ-SEvolutionary relationships and cladistics6 marksPaper 2~9 min
The cladogram below shows proposed evolutionary relationships among five tetrapod groups: amphibians, reptiles, birds, marsupials, and placental mammals. Four shared derived characters (synapomorphies) are mapped onto the tree. Group — Amniotic egg — Endothermy — Mammary glands — Hair/fur Amphibians — — — — Reptiles — ✓ — — — — Birds — ✓ — — Marsupials — ✓ — — — ✓ Placental mammals — ✓ — — — ✓
(a)
State the number of tetrapod groups in the table that possess the amniotic egg. [1 mark]
(b)
The cladogram places birds as the sister group to reptiles, and marsupials as the sister group to placental mammals, with both mammal groups together forming the outgroup to reptiles + birds. Using the parsimony principle, calculate the minimum number of evolutionary origins (steps) required to explain the distribution of all four characters across the five groups. Assume each character, once gained, is never lost. Show your reasoning for each character. [3 marks]
(c)
A student proposes that endothermy evolved independently in birds and in the mammal lineage, rather than once in a common ancestor of all four non-amphibian groups. Using the data in the table, explain why the cladogram supports the student's proposal. [2 marks]
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29MasterySAQ-SEvolutionary relationships and cladistics5 marksPaper 2~8 min
Researchers studying six insect species (A–F) have identified three shared derived characters: (i) folded wings, (ii) social behaviour, (iii) modified mouthparts for piercing. The presence (+) or absence (–) of each character is shown below. Species — (i) folded wings — (ii) social behaviour — (iii) piercing mouthparts A — – — – B — + — – C — + — – D — + — + E — – — + F — – — + The outgroup (ancestral state) for all three characters is absent (–). Characters evolve only in the direction absent → present (no reversals).
(a)
State the principle of parsimony as applied to cladogram construction. [1 mark]
(b)
A cladogram is constructed in which species B, C, and D form a monophyletic clade, with A as the outgroup to all others, and E and F forming a separate clade sister to the BCD clade. Calculate the total number of character-state changes required to map all three characters onto this cladogram. [3 marks]
(c)
The distribution of character (iii) across the six species is described as an example of homoplasy. Evaluate this claim by referring to the cladogram topology and the distribution of character (iii). [1 mark]
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30ChallengeSAQ-LEvolutionary relationships and cladistics8 marksPaper 2~12 min

Data

Parsimony — the cladogram requiring the fewest independent evolutionary changes is preferred.
A research team investigates evolutionary relationships among four species of flowering plants (Silene: species P, Q, R, S) and one outgroup species T. A cladogram is constructed from three shared derived characters (synapomorphies). The table below summarises character presence (+) or absence (−). Species — Character 1: Red flowers — Character 2: Fused petals — Character 3: Nectar spur P — + — − Q — + — − R — + — + S — − — − T (outgroup) — − — −
(a)
Determine the most parsimonious cladogram for species P, Q, R, S, and T. Describe the branching order and state which synapomorphy defines each internal node. [3 marks]
(b)
Explain how the principle of parsimony is applied when choosing between two alternative cladograms in this analysis. [2 marks]
(c)
The team also sequences the chloroplast gene rbcL from all five species. Evaluate the use of morphological characters compared with molecular sequence data for determining evolutionary relationships in Silene, referring to the data above. [3 marks]
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