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Form and Function — Free Biology SL Practice Questions

1FoundationMCQCompartmentalization of processes in eukaryotic cells1 markPaper 1~2 min
Cell X contains a large central vacuole, chloroplasts, and other membrane-bound organelles. Cell Y is smaller, lacks a nucleus, and has no internal membrane-bound structures. A student claims Cell Y cannot carry out oxidative phosphorylation in the same way as Cell X. Which statement best explains why?
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2MasteryMCQCompartmentalization of processes in eukaryotic cells1 markPaper 1~2 min
A fluorescent dye binds specifically to the inner mitochondrial membrane of a living eukaryotic cell. Which process does this dye most directly reveal the compartmentalization of?
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3FoundationMCQStem cells and differentiation1 markPaper 1~2 min
A researcher isolates inner cell mass cells from a human blastocyst to obtain embryonic stem cells (ESCs). She also reprograms adult skin cells from the same patient by introducing specific transcription factors to produce induced pluripotent stem cells (iPSCs). Which statement correctly compares the differentiation potential of ESCs and iPSCs?
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4MasteryMCQStem cells and differentiation1 markPaper 1~2 min
A researcher isolates three cell populations to investigate repair of damaged cardiac muscle:
(1) cells from a 5-day-old human embryo that can differentiate into any cell type, including extra-embryonic tissues such as the placenta;
(2) cells from the bone marrow of an adult human; and
(3) adult skin cells reprogrammed in the laboratory by introducing specific transcription factors. Which of the following correctly classifies cell populations 1, 2, and 3?
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5FoundationMCQRespiratory structures in animals and plants1 markPaper 1~2 min
A leaf is sealed in a transparent container fitted with a CO₂ sensor. In bright light, the CO₂ concentration inside the container decreases at a steady rate. Which statement best explains this observation?
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6MasteryMCQRespiratory structures in animals and plants1 markPaper 1~2 min
In Cyprinus carpio, the pO2\text{pO}_2 of water entering the gills is 16 kPa16 \text{ kPa} and the pO2\text{pO}_2 of blood entering the gill capillaries is 4 kPa4 \text{ kPa}. Which statement best explains why countercurrent flow maintains a more favourable diffusion gradient for oxygen uptake than concurrent flow?
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7FoundationMCQMetabolism of glucose1 markPaper 1~2 min
In glycolysis, glucose is converted to pyruvate through the following sequence: glucose → glucose-6-phosphate → fructose-1,6-bisphosphate → glyceraldehyde-3-phosphate → 1,3-bisphoglycerate → 3-phosphoglycerate → 2-phosphoglycerate → phosphoenolpyruvate → pyruvate. Which statement correctly describes an energy-related feature of this pathway?
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8MasteryMCQStructure and function of carbohydrates and lipids1 markPaper 1~2 min
A student heats a solution containing glucose with Benedict's reagent. Which observation correctly identifies both the result and its cause?
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9FoundationMCQCirculatory systems: Open vs closed1 markPaper 1~2 min
In a grasshopper, the dorsal blood vessel pumps haemolymph into the haemocoel, where it bathes organs directly. Haemolymph re-enters the dorsal vessel through valved openings called ostia during relaxation. How does the pressure in the dorsal vessel compare to the pressure in the haemocoel at different phases of one contraction cycle?
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10MasteryMCQCirculatory systems: Open vs closed1 markPaper 1~2 min
A grasshopper has an open circulatory system and a fish has a closed circulatory system. Measurements show that the hydrostatic pressure in the main vessel leaving the heart is approximately equal in both animals. Which of the following best explains why the rate of oxygen delivery to the leg muscles is lower in the grasshopper than in the fish?
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11FoundationMCQEnzymes as biological catalysts1 markPaper 1~2 min
In the metabolic pathway AEnzyme 1BEnzyme 2CA \xrightarrow{\text{Enzyme 1}} B \xrightarrow{\text{Enzyme 2}} C, a non-competitive inhibitor binds to Enzyme 2. Which molecule shows the most immediate increase in concentration?
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12MasteryMCQStructure of proteins: Primary to quaternary structure1 markPaper 1~2 min
A purified enzyme from a thermophilic bacterium is treated with a reducing agent that cleaves disulfide bonds between cysteine residues within a single polypeptide chain. The amino acid sequence remains unchanged, but the enzyme loses its three-dimensional shape and catalytic activity. Which level of protein structure is most directly disrupted?
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13FoundationMCQEcological niche concept1 markPaper 1~2 min
A grasshopper feeds on grass leaves during the day and a field mouse feeds on grass seeds at night; both live in the same grassland field. Which statement best describes the relationship between their ecological niches?
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14MasteryMCQEcological niche concept1 markPaper 1~2 min
Two species of anole lizard, Anolis evermanni and Anolis gundlachi, co-exist in the same tropical forest. A. evermanni perches on tree trunks and branches 1–3 m above the ground, while A. gundlachi occupies the forest floor among leaf litter. Both species prey on small insects. The spatial separation of these two species into distinct microhabitats is best described as:
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15MasterySAQ-SCompartmentalization of processes in eukaryotic cells5 marksPaper 2~8 min
A group of students investigated the effect of a drug on protein secretion from pancreatic acinar cells. They measured the concentration of a digestive enzyme in the extracellular medium over 3030 minutes under two conditions: normal cells and cells treated with a drug that disrupts the Golgi apparatus. All cells were maintained at 37°C37°\text{C} in an isotonic buffer. Condition — Enzyme concentration at t=0mint = 0\,\text{min} — Enzyme concentration at t=30mint = 30\,\text{min} Normal cells — 0.0mg mL10.0\,\text{mg mL}^{-1}2.4mg mL12.4\,\text{mg mL}^{-1} Golgi-disrupted cells — 0.0mg mL10.0\,\text{mg mL}^{-1}0.3mg mL10.3\,\text{mg mL}^{-1}
(a)
Describe the role of compartmentalization in the secretion of digestive enzymes from pancreatic acinar cells. [2 marks]
(b)
Calculate the percentage reduction in the mean rate of enzyme secretion caused by Golgi disruption. Show your working. [3 marks]
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16MasterySAQ-SCompartmentalization of processes in eukaryotic cells5 marksPaper 2~8 min
A researcher studies a eukaryotic cell genetically modified so that its lysosomal membrane is permeable to protons (H+\text{H}^+ ions). The cell is placed in culture medium at pH 7.2\text{pH}\ 7.2.
(a)
(i) Describe how compartmentalization normally allows lysosomes to maintain an internal pH of approximately 5.05.0. [1]
(ii) Explain why this acidic internal pH is important for lysosomal function. [1 mark]
(b)
The researcher measures the activity of a lysosomal protease in the modified cells and compares it to unmodified cells. Predict and explain the difference in protease activity observed in the modified cells. [3 marks]
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17ChallengeSAQ-LCompartmentalization of processes in eukaryotic cells16 marksPaper 2~24 min
Hepatocytes (liver cells) are highly active in endocytosis and phagocytosis, internalising materials into phagosomes that fuse with lysosomes for digestion. A research team investigates the effect of a novel drug, EndoX, on lysosomal function in hepatocytes. They measure the activity of the lysosomal hydrolase Cathepsin D and the intralysosomal pH under two conditions. Condition — Cathepsin D activity / AU — Intralysosomal pH A — untreated — 85 — 4.8 B — EndoX-treated — 22 — 6.9 - Optimum pH for Cathepsin D (determined in vitro): 4.54.5 to 5.05.0 - Cytosolic pH of hepatocytes: 7.27.2
(a)
Calculate the percentage decrease in Cathepsin D activity in EndoX-treated cells compared to untreated cells. [2 marks]
(b)
Explain the mechanism by which the change intralysosomal pH in EndoX-treated cells leads to the observed reduction in Cathepsin D activity. [3 marks]
(c)
Evaluate the importance of compartmentalisation of hydrolytic enzymes within lysosomes for the normal functioning of a eukaryotic cell, using the data provided and your knowledge of cellular biology. [3] Total: [8 marks]
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18MasterySAQ-SStem cells and differentiation5 marksPaper 2~8 min
A research group investigates the potential of stem cells to treat spinal cord injuries. They obtain two types of cells from a patient: multipotent stem cells from the patient's bone marrow, and induced pluripotent stem cells (iPSCs) generated from the patient's skin fibroblasts.
(a)
State the difference in potency between multipotent stem cells and iPSCs. [1 mark]
(b)
The researchers culture iPSCs in a medium containing specific growth factors to direct differentiation into oligodendrocytes. After 14 days, the culture contains 5.0×1055.0 \times 10^{5} cells in total, of which 3.5×1053.5 \times 10^{5} have differentiated into oligodendrocytes. The culture dish has a surface area of 25cm225\,\text{cm}^{2}. Calculate the percentage of cells that successfully differentiated into oligodendrocytes, and determine the density of differentiated oligodendrocytes per cm2\text{cm}^{2}[2 marks]
(c)
The differentiated oligodendrocytes are transplanted into the injured spinal cord of the same patient. Evaluate one potential advantage and one potential risk of using the patient's own iPSCs rather than donor-derived neural stem cells for this treatment. [2 marks]
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19MasterySAQ-SStem cells and differentiation5 marksPaper 2~8 min
A patient with type 1 diabetes has lost the function of insulin-producing beta cells in the pancreas. Scientists propose to generate new beta cells from stem cells for transplantation.
(a)
State one property of stem cells that makes them suitable for generating replacement beta cells. [1 mark]
(b)
The scientists compare two approaches: using embryonic stem cells (ESCs) from a donated embryo, and using induced pluripotent stem cells (iPSCs) derived from the patient's own skin cells. Explain one ethical advantage of using iPSCs over ESCs for this patient. [2 marks]
(c)
The scientists successfully differentiate iPSCs into insulin-producing cells and measure insulin release in response to glucose. The results are shown below. Glucose concentration — Insulin secretion 5mM5\,\text{mM}2.1ng insulin106cellsh12.1\,\text{ng insulin} \cdot 10^{-6}\,\text{cells} \cdot \text{h}^{-1} 20mM20\,\text{mM}9.8ng insulin106cellsh19.8\,\text{ng insulin} \cdot 10^{-6}\,\text{cells} \cdot \text{h}^{-1} Calculate the fold increase insulin secretion when glucose concentration is raised from 5mM5\,\text{mM} to 20mM20\,\text{mM}[2 marks]
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20ChallengeSAQ-LStem cells and differentiation7 marksPaper 2~11 min
A research team investigates the potential of induced pluripotent stem cells (iPSCs) to treat a patient with type 1 diabetes mellitus. The patient's own skin fibroblasts are reprogrammed into iPSCs using the Yamanaka factors (Oct4, Sox2, Klf4, c-Myc). These iPSCs are then differentiated into pancreatic beta cells. Data collected after 14 days of differentiation: - Untreated iPSCs: 15%15\% expressed the beta-cell marker PDX1 - Embryonic stem cells (ESCs) under the same protocol: 72%72\% expressed PDX1 - iPSCs pre-treated for 48 hours with a histone deacetylase inhibitor (HDACi): 68%68\% expressed PDX1 - Genome analysis of iPSCs before and after HDACi treatment: no change in DNA sequence; 40%40\% increase in acetylation of histone H3 at the PDX1 gene promoter
(a)
State the percentage-point increase in PDX1 expression when iPSCs are pre-treated with HDACi compared to untreated iPSCs. [1 mark]
(b)
Calculate the fold change in PDX1 expression between HDACi-treated iPSCs and ESCs. [2 marks]
(c)
Explain the molecular mechanism by which HDACi treatment increased PDX1 expression in the iPSCs. [2 marks]
(d)
Evaluate whether iPSCs or ESCs would be the more appropriate cell source for developing a patient-specific therapy for type 1 diabetes, using evidence from the data and your knowledge of stem cell biology. [2 marks]
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21MasterySAQ-SRespiratory structures in animals and plants5 marksPaper 2~8 min
A student investigates gas exchange in the leaves of a terrestrial plant, Zebrina pendula. She examines a cross-section under a light microscope and identifies the following layers: upper epidermis, palisade mesophyll, spongy mesophyll, and lower epidermis containing stomata surrounded by guard cells. Large air spaces are visible between the spongy mesophyll cells.
(a)
State one structural feature of the spongy mesophyll that facilitates gas exchange. [1 mark]
(b)
The student measures transpiration rate under two conditions. When stomata are fully open (aperture 10μm10\,\mu\text{m}), the transpiration rate is 5.2mm3min15.2\,\text{mm}^3\,\text{min}^{-1}. Transpiration rate is proportional to the square of the stomatal aperture. Calculate the transpiration rate when the aperture is 4μm4\,\mu\text{m}. Show your working. [2 marks]
(c)
A second plant of the same species is grown in arid environment with high light intensity. Evaluate how the conflicting demands of photosynthesis and water conservation affect stomatal regulation in this plant. [2 marks]
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22MasterySAQ-SRespiratory structures in animals and plants5 marksPaper 2~8 min
A student compares gas exchange in a flatworm (Planaria) and a woody stem (Tilia europaea). The flatworm lacks specialised respiratory organs. The woody stem is covered by bark containing lenticels.
(a)
State one structural feature of Planaria that allows gas exchange by diffusion alone. [1 mark]
(b)
The body thickness of Planaria is 0.5mm0.5\,\text{mm}. The oxygen flux through its body surface is 4.0×109molmm2s14.0 \times 10^{-9}\,\text{mol}\,\text{mm}^{-2}\,\text{s}^{-1}. Its oxygen consumption rate is 1.0×109molmm2s11.0 \times 10^{-9}\,\text{mol}\,\text{mm}^{-2}\,\text{s}^{-1}. Calculate whether diffusion alone can supply sufficient oxygen. Show your working. [2 marks]
(c)
Explain why lenticels are necessary for gas exchange in woody stems. [2 marks]
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23ChallengeSAQ-LRespiratory structures in animals and plants7 marksPaper 2~11 min
A group of students investigated the effect of body mass on the tracheal system of insects. They measured the total cross-sectional area of the main tracheal trunks entering the thorax of three species of grasshopper and recorded the data below. They also calculated the rate of oxygen consumption (VO2\text{VO}_2) for each species at rest. Species — Body mass / g — Total tracheal cross-sectional area / mm2\text{mm}^2VO2\text{VO}_2 / mL O2g1h1\text{mL O}_2\,\text{g}^{-1}\,\text{h}^{-1} Schistocerca gregaria — 2.02.00.400.400.180.18 Locusta migratoria — 1.51.50.300.300.180.18 Chorthippus brunneus — 0.80.80.160.160.180.18
(a)
Calculate the ratio of total tracheal cross-sectional area to body mass for each species, and state the relationship between body mass and total tracheal cross-sectional area in these grasshoppers. [2 marks]
(b)
Explain how the structure of the tracheal system insects allows efficient gas exchange without a circulatory system for oxygen transport. [3 marks]
(c)
A grasshopper of body mass 10.0g10.0\,\text{g} is proposed. Using the relationship from (a), calculate the predicted total tracheal cross-sectional area for this grasshopper. Evaluate whether a grasshopper of this mass could survive using only a tracheal system, with reference to how oxygen demand diffusion path length scale with body size. [2 marks]
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24MasterySAQ-SStructure and function of carbohydrates and lipids10 marksPaper 2~15 min
A student investigates the effect of temperature on the rate of starch digestion by amylase. She adds amylase to a starch solution at 37°C37\,°\text{C} and at 60°C60\,°\text{C}, then measures the concentration of maltose produced every 2 minutes for 10 minutes. Her results are shown below. Time (min) — Maltose at 37°C37\,°\text{C} (gL1\text{g\,L}^{-1}) — Maltose at 60°C60\,°\text{C} (gL1\text{g\,L}^{-1}) 0.0 — 0.0 2 — 1.2 — 0.3 4 — 2.4 — 0.6 6 — 3.6 — 0.9 8 — 4.8 — 1.2 10 — 6.0 — 1.5
(a)
State the structural relationship between starch and maltose. [1 mark]
(b)
Calculate the rate of maltose production at 37°C37\,°\text{C} between 2 and 6 minutes. Give your answer in gL1min1\text{g\,L}^{-1}\,\text{min}^{-1}[2 marks]
(c)
Describe how the structure of amylase allows it to break down starch into maltose. [2 marks]
(d)
Explain why the rate of maltose production is lower at 60°C60\,°\text{C} than at 37°C37\,°\text{C}. [2] (e) The student repeats the experiment at 60°C60\,°\text{C} but adds fresh amylase every 2 minutes throughout the 10-minute period. Predict and justify whether the final maltose concentration at 10 minutes would be higher than, lower than, or equal to 6.0gL16.0\,\text{g\,L}^{-1}[3 marks]
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25MasterySAQ-SStructure and function of carbohydrates and lipids7 marksPaper 2~11 min
A biology student sets up an experiment using dialysis tubing to model the digestion of starch. She fills one bag with a 2%2\% starch solution and a second bag with a 2%2\% starch solution mixed with amylase. Both bags are placed in separate beakers of distilled water at 37C37\,^\circ\text{C}. After 30 minutes, she tests the water outside each bag for the presence of starch (using iodine solution) and for reducing sugars (using Benedict's reagent). Results: - Bag 1 (starch only): iodine test — water remains yellow-brown; Benedict's test — water remains blue. - Bag 2 (starch + amylase): iodine test — water remains yellow-brown; Benedict's test — water turns orange-red.
(a)
State the property of starch that prevents it from passing through the dialysis tubing. [1 mark]
(b)
Describe the structural change that occurs to starch in Bag 2 that allows a product to pass through the tubing. [2 marks]
(c)
Determine the mass of reducing sugar that diffused into the beaker water outside Bag 2, given that the concentration of reducing sugar in that beaker is 0.5gL10.5\,\text{g}\,\text{L}^{-1} and the beaker contains 200mL200\,\text{mL} of water. [1 mark]
(d)
The student claims that Bag 1 acts a control for this experiment. Evaluate this claim, identifying what Bag 1 controls for and stating one limitation of the experimental design. [3 marks]
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26ChallengeSAQ-LStructure and function of carbohydrates and lipids8 marksPaper 2~12 min
A study investigated the effect of acarbose, an amylase inhibitor, on starch digestion in humans. Eight healthy volunteers each consumed a standard meal containing 50g50\,\text{g} of starch. Blood glucose concentration was measured at 30-minute intervals for 180 minutes after the meal. The experiment was repeated one week later with the same volunteers; each volunteer took 100mg100\,\text{mg} of acarbose immediately before the meal. Time after meal (min)(\text{min}) — Mean blood glucose without acarbose (mmolL1)(\text{mmol}\,\text{L}^{-1}) — Mean blood glucose with acarbose (mmolL1)(\text{mmol}\,\text{L}^{-1}) 0 — 4.8 30 — 7.2 — 6.1 60 — 8.5 — 6.8 90 — 7.9 — 6.4 120 — 6.5 — 5.7 150 — 5.6 — 5.2 180 — 5.0 — 4.9
(a)
Calculate the percentage decrease in peak blood glucose concentration caused by acarbose. [2 marks]
(b)
Explain the mechanism by which acarbose causes the observed reduction in peak blood glucose concentration. [3 marks]
(c)
Evaluate the use of acarbose as a treatment for type 2 diabetes mellitus, referring to both its advantages and its limitations. [3 marks]
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27MasterySAQ-SBlood circulation in mammals (cardiovascular system)6 marksPaper 2~9 min
A patient is admitted to the emergency department with a suspected pulmonary embolism. A blood clot has lodged in a branch of the pulmonary artery, obstructing blood flow to a portion of the right lung.
(a)
State the type of blood vessel that carries oxygenated blood from the lungs to the heart, and name the heart chamber it empties into. [1 mark]
(b)
Calculate the stroke volume of the patient's heart. The cardiac output is 4.2dm3min14.2\,\text{dm}^3\,\text{min}^{-1} and the heart rate is 70beatsmin170\,\text{beats}\,\text{min}^{-1}[2 marks]
(c)
Explain why the oxygen concentration of blood arriving at the left atrium is lower than normal as a result of the pulmonary embolism. [3 marks]
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28MasterySAQ-SBlood circulation in mammals (cardiovascular system)5 marksPaper 2~8 min
A researcher measures blood pressure and blood velocity in different blood vessels of a mammalian circulatory system. Vessel — Systolic / diastolic pressure (mmHg) — Mean blood velocity (cm s1\text{cm s}^{-1}) Aorta — 120 / 80 — 40 Arteriole — 60 / 40 — 10 Capillary — 30 / 20 — 0.5 Venule — 20 / 15 Vena cava — 10 / 5 — 25
(a)
Describe the relationship between blood velocity and vessel type, and explain this pattern in terms of total cross-sectional area. [2 marks]
(b)
Calculate the pulse pressure in the aorta. [1 mark]
(c)
Explain how the structural features of capillary walls allow efficient exchange of substances between blood and tissues, and explain why a reduction in blood velocity at capillaries important for this exchange. [2 marks]
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29ChallengeSAQ-LRole of the heart and blood vessels7 marksPaper 2~11 min
A medical researcher is investigating a newly discovered mutation in the gene encoding connexin-43, a protein that forms gap junctions in the sinoatrial (SA) node and atrioventricular (AV) node of the human heart. The mutation causes a 40%40\% reduction in the rate of ion flow through these gap junctions. A patient with this mutation has a resting heart rate of 72bpm72\,\text{bpm} and a stroke volume of 70mL70\,\text{mL}. Reference values: - Normal PR interval: 0.120.120.20s0.20\,\text{s} - Normal QRS duration: <0.12s< 0.12\,\text{s} - Normal corrected QT interval: <0.44s< 0.44\,\text{s} - CO=HR×SV\text{CO} = \text{HR} \times \text{SV}
(a)
Calculate the cardiac output of this patient at rest. [1 mark]
(b)
Explain the likely effect of the mutation the PR interval seen in the patient's ECG. [2 marks]
(c)
During moderate exercise the patient's heart rate rises to 140bpm140\,\text{bpm} and stroke volume increases by 15%15\% from its resting value. In a healthy individual, stroke volume would increase by 40%40\% from the same resting value at the same heart rate. Determine the patient's cardiac output during this exercise and the cardiac output a healthy individual would achieve under the same conditions. [2 marks]
(d)
Using your answer to (c) and your knowledge of cardiovascular physiology, evaluate whether the mutation is likely to limit the patient's maximum exercise capacity. [2 marks]
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30MasterySAQ-SStructure of proteins: Primary to quaternary structure5 marksPaper 2~8 min
A point mutation in the β\beta-globin gene changes the sixth codon from GAG to GTG, substituting glutamic acid (hydrophilic, negatively charged) with valine (hydrophobic, non-polar) in the β\beta-globin polypeptide. Individuals homozygous for this allele develop sickle cell anaemia; heterozygous individuals carry one normal and one mutant allele.
(a)
State the level of protein structure directly altered by this amino acid substitution. [1 mark]
(b)
Describe how the substitution of glutamic acid with valine affects the secondary and tertiary structures of the β\beta-globin polypeptide. [2 marks]
(c)
Explain how the altered tertiary structure causes haemoglobin molecules to aggregate under low-oxygen conditions, and evaluate why heterozygous individuals are largely protected from the severe symptoms of sickle cell anaemia. [2 marks]
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31MasterySAQ-SStructure of proteins: Primary to quaternary structure5 marksPaper 2~8 min
The enzyme lysozyme is a globular protein that hydrolyses bacterial cell walls. Its structure includes regions of α\alpha-helix and β\beta-pleated sheet, stabilised by hydrogen bonds, disulfide bridges, and hydrophobic interactions. A researcher treats lysozyme with a reducing agent that breaks all disulfide bridges (—S—S— bonds) but leaves peptide bonds intact.
(a)
State the level of protein structure that is directly stabilised by disulfide bridges. [1 mark]
(b)
Describe how the loss of disulfide bridges affects the α\alpha-helices and β\beta-pleated sheets of lysozyme. [2 marks]
(c)
Evaluate why breaking disulfide bridges, but not peptide bonds, is sufficient to abolish lysozyme's ability to hydrolyse bacterial cell walls. [2 marks]
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32ChallengeSAQ-LStructure of proteins: Primary to quaternary structure9 marksPaper 2~14 min
Bovine pancreatic trypsin inhibitor (BPTI) is a protein of 58 amino acids. In its native state it has a well-defined tertiary structure stabilised by three disulfide bridges (S-\text{S}-) between cysteine residues at positions 5–55, 14–38, and 30–51, and it is a potent inhibitor of the enzyme trypsin. A research team treats native BPTI with Compound X, which specifically reduces disulfide bridges to free thiol groups (SH-\text{SH}). After treatment, trypsin-inhibitory activity falls to 5%5\% of the original value. The treated BPTI is then allowed to refold in the presence of oxygen (which can re-oxidise thiol groups to disulfide bridges); after 24 hours, activity recovers to only 12%12\% of the original value. In a separate control experiment, native BPTI is denatured with 8M8\,\text{M} urea — which disrupts hydrogen bonds and hydrophobic interactions but leaves disulfide bridges intact — and then allowed to refold by dialysis to remove the urea. Activity recovers to 98%98\% of the original value.
(a)
State the two levels of protein structure that are directly stabilised by disulfide bridges, and identify which level is primarily disrupted when Compound X is applied. [2 marks]
(b)
Explain why reduction of the disulfide bridges by Compound X causes a 95%95\% loss of trypsin-inhibitory activity in BPTI. [3 marks]
(c)
Evaluate the hypothesis: 'The primary structure of BPTI alone is sufficient to determine its correct tertiary structure and full biological function.' Use all three experimental outcomes in your answer. [4 marks]
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33MasterySAQ-SEcological niche concept5 marksPaper 2~8 min
The intertidal zone of a rocky shore in the Pacific Northwest is home to two species of barnacle: Balanus glandula (acorn barnacle) and Chthamalus dalli (thatched barnacle). Balanus grows faster and outcompetes Chthamalus in the lower intertidal zone but is less tolerant of desiccation during low tide. Chthamalus can survive higher on the shore where air exposure is longer. Researchers measured the fundamental niche of each species in the laboratory by growing them across a gradient of submersion times. - Balanus fundamental niche: 882424 hours submersion per day - Chthamalus fundamental niche: 221818 hours submersion per day Researchers then cleared a section of the mid-intertidal zone (average 1212 hours submersion per day) of all barnacles and monitored recolonisation over two years. After two years, only Balanus was present in the cleared zone.
(a)
State the fundamental niche of Chthamalus dalli in terms of submersion time. [1 mark]
(b)
Calculate the overlap between the two fundamental niches a percentage of the Chthamalus fundamental niche range. [2 marks]
(c)
Explain why the realised niche of Chthamalus in the mid-intertidal zone differs from its fundamental niche, using the recolonisation data as evidence. [2 marks]
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34MasterySAQ-SEcological niche concept5 marksPaper 2~8 min
In a freshwater lake in Uganda, two species of cichlid fish (Haplochromis nyererei and Haplochromis pyrrhocephalus) coexist. Both feed on zooplankton but have different mouth morphologies allowing prey capture at different depths. Researchers measured stomach contents of 50 individuals of each species across four depth zones. The table shows the percentage of each species' total feeding time spent in each zone. Depth zone — H. nyererei — H. pyrrhocephalus 02m0\text{–}2\,\text{m}10%10\%45%45\% 24m2\text{–}4\,\text{m}25%25\%35%35\% 46m4\text{–}6\,\text{m}40%40\%15%15\% 68m6\text{–}8\,\text{m}25%25\%5%5\%
(a)
State the depth zone where H. nyererei spends the highest percentage of its feeding time. [1 mark]
(b)
Calculate the combined percentage of feeding time that H. pyrrhocephalus spends in the two shallowest zones (02m0\text{–}2\,\text{m} and 24m2\text{–}4\,\text{m}), and compare this with the equivalent combined percentage for H. nyererei. [2 marks]
(c)
Evaluate, using the data, the extent to which niche differentiation between the two species would reduce the likelihood of competitive exclusion. [2 marks]
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35ChallengeSAQ-LEcological niche concept8 marksPaper 2~12 min
A study was conducted in a temperate deciduous forest in Vermont, USA, to investigate the ecological niches of three sympatric warbler species: the Blackburnian Warbler (Setophaga fusca), the Black-throated Green Warbler (Setophaga virens), and the Yellow-rumped Warbler (Setophaga coronata). Researchers recorded foraging height and the percentage of time spent foraging on different tree parts over three breeding seasons. Table 1: Foraging data for three warbler species Species — Mean foraging height ±\pm SD (m) — Needles/leaves (%) — Branches (%) — Trunks (%) Blackburnian — 14.2±1.814.2 \pm 1.8 — 78 — 18 — 4 Black-throated Green — 9.7±2.19.7 \pm 2.1 — 62 — 30 — 8 Yellow-rumped — 5.3±2.55.3 \pm 2.5 — 45 — 38 — 17
(a)
State which two adjacent warbler species have the greatest overlap in vertical foraging range, using the SD values to justify your answer. [1 mark]
(b)
Calculate the percentage of foraging time that the Yellow-rumped Warbler spends on non-leaf substrates (branches and trunks combined), and compare this value to the equivalent figure for the Blackburnian Warbler. [2 marks]
(c)
Explain how the observed patterns of resource partitioning among these three warbler species reduce interspecific competition. [2 marks]
(d)
Evaluate the hypothesis that the niche differences observed are a result of character displacement driven by past interspecific competition. [3 marks]
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