An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 SL (TZ2) IB Diploma Programme Biology paper. Not affiliated with or reproduced from IB.
卷一
Answer all 30 multiple choice questions. No calculators allowed.
30 題目 · 30 分
題目 1 · 選擇題
1 分
Antidiuretic hormone (ADH) plays a critical role in osmoregulation. Which of the following best describes the effect of a high blood solute concentration on ADH secretion and the subsequent cellular response in the collecting duct of the kidney?
A.ADH secretion increases, leading to the fusion of aquaporin-containing vesicles with the luminal membrane of collecting duct cells.
B.ADH secretion decreases, leading to the endocytosis of aquaporins from the luminal membrane of collecting duct cells.
C.ADH secretion increases, causing active transport of water molecules across the basolateral membrane.
D.ADH secretion decreases, resulting in the passive movement of water from the lumen into the interstitial fluid.
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解題
When blood solute concentration is high, osmoreceptors in the hypothalamus detect this change and stimulate the posterior pituitary gland to secrete ADH. ADH travels via the blood to the collecting ducts of the kidneys, where it binds to receptors on the basolateral membrane. This initiates a signal transduction pathway that causes vesicles containing aquaporin water channels to fuse with the luminal (apical) membrane. This increases the permeability of the collecting duct to water, allowing water to be reabsorbed passively down its concentration gradient into the hypertonic medulla.
評分準則
Award [1] mark for the correct answer (A). No partial marks.
題目 2 · 選擇題
1 分
Condensation reactions form glycosidic bonds between monosaccharides. When three molecules of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) are linked together to form the trisaccharide maltotriose, what is the molecular formula of the resulting carbohydrate?
A.\(\text{C}_{18}\text{H}_{36}\text{O}_{18}\)
B.\(\text{C}_{18}\text{H}_{34}\text{O}_{17}\)
C.\(\text{C}_{18}\text{H}_{32}\text{O}_{16}\)
D.\(\text{C}_{18}\text{H}_{30}\text{O}_{15}\)
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解題
A glucose molecule has the molecular formula \(\text{C}_6\text{H}_{12}\text{O}_6\). Three glucose molecules would contain \(3 \times 6 = 18\) carbon atoms, \(3 \times 12 = 36\) hydrogen atoms, and \(3 \times 6 = 18\) oxygen atoms (\(\text{C}_{18}\text{H}_{36}\text{O}_{18}\)). To link three monosaccharides together to form a trisaccharide, two condensation reactions must occur. Each condensation reaction releases one water molecule (\(\text{H}_2\text{O}\)). Therefore, two water molecules (\(2 \times \text{H}_2\text{O} = \text{H}_4\text{O}_2\)) must be subtracted from the total: Carbon = 18, Hydrogen = \(36 - 4 = 32\), Oxygen = \(18 - 2 = 16\). This yields the molecular formula \(\text{C}_{18}\text{H}_{32}\text{O}_{16}\).
評分準則
Award [1] mark for the correct answer (C). No partial marks.
題目 3 · 選擇題
1 分
Somatic and germline mutations have different implications for an organism and its offspring. Which statement correctly distinguishes between these two types of mutations?
A.Somatic mutations can be passed on to offspring via sexual reproduction, whereas germline mutations only affect the individual in which they occur.
B.Somatic mutations occur only in non-reproductive cells and can lead to cancer in the individual, while germline mutations occur in gamete-producing cells and can be inherited by offspring.
C.Germline mutations occur due to environmental mutagens, whereas somatic mutations are always a result of errors in DNA replication.
D.Somatic mutations affect every cell in the offspring's body, whereas germline mutations only affect a specific tissue type in the parent.
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解題
Somatic mutations occur in body (somatic) cells and are not passed on to offspring because they do not affect gametes. They can, however, lead to uncontrolled cell division (cancer) in the affected individual. Germline mutations occur in the germline cells (which produce gametes) and can be inherited by the offspring, potentially affecting every cell in the offspring's body.
評分準則
Award [1] mark for the correct answer (B). No partial marks.
題目 4 · 選擇題
1 分
An electron micrograph of a eukaryotic cell shows a high density of rough endoplasmic reticulum (rER) and a large number of Golgi vesicles near the plasma membrane. What is the most likely function of this cell?
A.Synthesis of steroid hormones such as testosterone
B.Production and secretion of extracellular enzymes
C.Rapid generation of ATP via aerobic respiration
D.Storage of glycogen and lipids for metabolic energy
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解題
A high density of rough endoplasmic reticulum (rER) indicates that the cell is actively synthesizing large quantities of proteins. The presence of numerous Golgi vesicles near the plasma membrane suggests these proteins are being packaged, processed, and secreted outside the cell via exocytosis. This is characteristic of secretory cells, such as those producing extracellular digestive enzymes.
評分準則
Award [1] mark for the correct answer (B). No partial marks.
題目 5 · 選擇題
1 分
In a population of forest-dwelling beetles, a new avian predator that detects prey primarily by sight is introduced. If the forest floor is dark brown, which of the following best describes the expected evolutionary response of the beetle population over successive generations?
A.Individual green beetles will mutate their color genes to become brown during their lifetime to avoid being eaten.
B.The proportion of brown alleles in the gene pool will increase as green beetles are preferentially preyed upon.
C.The beetles will learn to hide under leaves, causing their offspring to be born with flatter bodies.
D.The mutation rate for brown coloration will increase significantly in response to the presence of the avian predator.
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解題
In the presence of a visual predator, beetles that match the dark brown color of the forest floor are less likely to be spotted and eaten. Green beetles, being more visible, will have a higher rate of predation. Consequently, brown beetles are more likely to survive, reproduce, and pass on their alleles for brown coloration to their offspring. Over generations, the frequency of the brown allele in the gene pool will increase. Natural selection operates on existing genetic variation; individual beetles cannot deliberately mutate their colors during their lifetime to adapt, nor does the predator induce directed mutations.
評分準則
Award [1] mark for the correct answer (B). No partial marks.
題目 6 · 選擇題
1 分
An inhibitor is added to an enzyme-catalyzed reaction. It is observed that increasing the concentration of the substrate decreases the inhibition, eventually restoring the maximum rate of reaction (\(V_{\max}\)) to its original value. Which type of inhibition and change in the Michaelis constant (\(K_m\)) are occurring?
A.Non-competitive inhibition, with an increase in \(K_m\)
B.Non-competitive inhibition, with no change in \(K_m\)
C.Competitive inhibition, with an increase in \(K_m\)
D.Competitive inhibition, with no change in \(K_m\)
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解題
In competitive inhibition, the inhibitor competes with the substrate for the active site of the enzyme. By increasing the substrate concentration, the substrate outcompetes the inhibitor, and the reaction rate can reach the original maximum velocity (\(V_{\max}\)). However, because more substrate is required to achieve the same rate of reaction, the apparent affinity of the enzyme for its substrate is lower, which is represented by an increase in the Michaelis constant (\(K_m\)).
評分準則
Award [1] mark for the correct answer (C). No partial marks.
題目 7 · 選擇題
1 分
Ocean acidification is a major environmental consequence of increasing atmospheric carbon dioxide (\(\text{CO}_2\)) levels. How does this acidification affect marine calcifying organisms, such as reef-building corals?
A.Decreased dissolved \(\text{CO}_2\) levels reduce the photosynthesis of zooxanthellae, starving the coral.
B.An increase in \(\text{H}^+\) ions decreases the concentration of carbonate ions (\(\text{CO}_3^{2-}\)), making it harder for corals to secrete calcium carbonate (\(\text{CaCO}_3\)).
C.Lower pH levels directly dissolve the organic tissues of the coral polyps, leaving only the skeleton.
D.Increased hydrogen ions combine with calcium ions to form highly soluble calcium hydride, preventing skeletal formation.
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解題
When carbon dioxide dissolves in seawater, it reacts with water to form carbonic acid, which dissociates into hydrogen ions (\(\text{H}^+\)) and hydrogen carbonate ions. The increased concentration of \(\text{H}^+\) ions leads to a reaction with carbonate ions (\(\text{CO}_3^{2-}\)) to form bicarbonate (\(\text{HCO}_3^-\)). This reduces the concentration of free carbonate ions, which marine calcifying organisms like corals need to react with calcium (\(\text{Ca}^{2+}\)) to build their calcium carbonate (\(\text{CaCO}_3\)) skeletons.
評分準則
Award [1] mark for the correct answer (B). No partial marks.
題目 8 · 選擇題
1 分
During active physical exercise, ventilation rate and depth increase. Which of the following describes the muscular actions that occur during active expiration?
A.The external intercostal muscles contract, and the diaphragm contracts and flattens.
B.The internal intercostal muscles contract, and the abdominal muscles contract to push the diaphragm upwards.
C.The external intercostal muscles contract, and the abdominal muscles relax.
D.The internal intercostal muscles relax, and the diaphragm contracts.
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解題
Ventilation of the lungs requires antagonistic muscle pairs. During active expiration (such as during exercise), the process is active rather than passive. The internal intercostal muscles contract to pull the rib cage down and inward, and the abdominal muscles contract to push the abdominal organs against the diaphragm, forcing it upward. This rapidly decreases the volume of the thoracic cavity, increasing pressure and forcing air out of the lungs.
評分準則
Award [1] mark for the correct answer (B). No partial marks.
題目 9 · multiple_choice
1 分
During intensive exercise, multiple body systems must coordinate to maintain cellular respiration in skeletal muscle. Which of the following correctly describes a physiological coordination mechanism occurring during this transition?
A.An increase in blood pH stimulates the medullary respiratory center to decrease the breathing rate.
B.Epinephrine released from the adrenal glands increases cardiac output and dilates bronchioles.
C.Decreased blood \(CO_2\) concentration leads to parasympathetic stimulation of the sinoatrial node.
D.The somatic nervous system triggers constriction of arterioles supplying skeletal muscles to increase blood pressure.
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解題
During exercise, sympathetic stimulation triggers the release of epinephrine from the adrenal glands. Epinephrine binds to adrenergic receptors, which increases cardiac output (by increasing heart rate and stroke volume) and causes bronchodilation (widening of airways) to facilitate increased gas exchange.
評分準則
Award 1 mark for identifying option B as the correct physiological mechanism.
題目 10 · multiple_choice
1 分
Which of the following represents a positive feedback loop associated with global climate change?
A.Increased atmospheric carbon dioxide leads to increased terrestrial plant growth, which removes more carbon dioxide from the atmosphere.
B.Higher temperatures increase the rate of chemical weathering of silicate rocks, which sequesters carbon dioxide over geological timescales.
C.Rising temperatures melt Arctic permafrost, releasing trapped methane gas which further increases global temperatures.
D.Increased evaporation leads to more low-level cloud cover, reflecting more incoming solar radiation back into space.
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解題
A positive feedback loop is a process where the output of a system amplifies the initial stimulus. Rising global temperatures cause Arctic permafrost to melt, which releases trapped methane (a potent greenhouse gas) into the atmosphere. This increases the greenhouse effect and leads to further warming.
評分準則
Award 1 mark for identifying option C as the correct positive feedback loop.
題目 11 · multiple_choice
1 分
A person consumes a high-salt meal without drinking water. What physiological changes are expected in response to the resulting increase in blood osmolarity?
A.Osmoreceptors in the hypothalamus shrink, stimulating the release of ADH from the posterior pituitary, leading to increased aquaporins in the collecting duct.
B.Osmoreceptors in the medulla oblongata swell, inhibiting the release of ADH, leading to decreased water reabsorption in the proximal convoluted tubule.
C.Baroreceptors in the aorta stimulate the anterior pituitary to release ADH, decreasing the permeability of the collecting duct to water.
D.Osmoreceptors in the hypothalamus swell, stimulating the thyroid gland to release calcitonin, decreasing sodium reabsorption.
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解題
When blood osmolarity increases, water leaves osmoreceptor cells in the hypothalamus by osmosis, causing them to shrink. This stimulates the release of antidiuretic hormone (ADH) from the posterior pituitary. ADH travels to the kidneys, where it binds to receptors on the collecting duct cells, initiating a signaling cascade that inserts aquaporins into the apical membrane, increasing water reabsorption.
評分準則
Award 1 mark for identifying option A as the correct sequence of osmoregulatory events.
題目 12 · multiple_choice
1 分
A researcher isolates a novel single-celled organism from an alkaline hot spring. Upon analysis, the organism is found to possess a peptidoglycan cell wall, 70S ribosomes, and a circular DNA molecule not associated with histones. Which of the following features would also be expected in this organism?
A.An extensive network of rough endoplasmic reticulum for protein transport.
B.Transcription and translation occurring simultaneously in the cytoplasm.
D.Membrane-bound mitochondria with folded cristae for ATP synthesis.
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解題
The presence of a peptidoglycan cell wall, 70S ribosomes, and naked circular DNA indicates that the organism is a bacterium (prokaryote). Prokaryotes lack membrane-bound organelles such as mitochondria, endoplasmic reticulum, or a nucleus. Since there is no nuclear envelope separating the DNA from the cytoplasm, transcription and translation can occur simultaneously in the cytoplasm.
評分準則
Award 1 mark for identifying option B as the correct prokaryotic characteristic.
題目 13 · multiple_choice
1 分
An enzyme-catalyzed reaction is studied in the presence of a metabolic inhibitor. It is observed that the maximum velocity (\(V_{max}\)) of the reaction remains unchanged, but the concentration of substrate required to reach half of the maximum velocity (\(K_m\)) increases. What type of inhibition is occurring?
A.Non-competitive inhibition, where the inhibitor binds to an allosteric site and changes the shape of the active site.
B.Competitive inhibition, where the inhibitor binds to the active site and can be overcome by increasing substrate concentration.
C.End-product inhibition, where the final product of the pathway acts as an irreversible inhibitor of the first enzyme.
D.Uncompetitive inhibition, where the inhibitor binds only to the enzyme-substrate complex.
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解題
In competitive inhibition, the inhibitor competes directly with the substrate for binding to the active site. Because high concentrations of substrate can overcome the inhibitor's effects, the maximum rate of reaction (\(V_{max}\)) remains unchanged. However, more substrate is required to achieve the same rate, which increases the Michaelis constant (\(K_m\)).
評分準則
Award 1 mark for identifying option B as competitive inhibition with its corresponding kinetic parameters.
題目 14 · multiple_choice
1 分
Why are lipids more suitable than carbohydrates for long-term energy storage in highly mobile animals?
A.Lipids release about half as much energy per gram as carbohydrates, reducing metabolic heat loss.
B.Lipids are highly soluble in water, allowing rapid transport through the circulatory system during flight or running.
C.Lipids store approximately twice as much energy per gram as carbohydrates and are stored without associated water weight.
D.Lipids can be rapidly mobilized through anaerobic respiration pathways during sudden burst activities.
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解題
Lipids (triglycerides) are excellent for long-term energy storage in mobile animals because they are hydrophobic and store about twice as much energy per gram (\(38\text{ kJ/g}\)) compared to carbohydrates (\(17\text{ kJ/g}\)). Because they are hydrophobic, they do not attract or bind water, preventing extra, non-functional water weight.
評分準則
Award 1 mark for identifying option C as the reason lipids are better suited for long-term energy storage in mobile animals.
題目 15 · multiple_choice
1 分
A single nucleotide substitution occurs in the middle of an exon of a protein-coding gene. The resulting mRNA codon changes from 5'-UAC-3' to 5'-UAA-3'. What is the most likely consequence of this mutation on the synthesized polypeptide?
A.A missense mutation occurs, resulting in the substitution of a single amino acid with no change to the overall length of the protein.
B.A silent mutation occurs, and the polypeptide sequence remains completely unchanged.
C.A nonsense mutation occurs, causing premature termination of translation and resulting in a truncated polypeptide.
D.A frameshift mutation occurs, altering the reading frame for all subsequent codons down the message.
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解題
The codon 5'-UAC-3' codes for the amino acid tyrosine. The codon 5'-UAA-3' is a stop codon. A mutation that changes an amino-acid-specifying codon into a stop codon is a nonsense mutation. This results in premature termination of translation, producing a shorter (truncated) and usually non-functional polypeptide.
評分準則
Award 1 mark for identifying option C as a nonsense mutation leading to a truncated polypeptide.
題目 16 · multiple_choice
1 分
What is the primary role of the pulmonary surfactant secreted by Type II pneumocytes in the alveoli?
A.To phagocytose airborne pathogens and debris that enter the alveolar spaces.
B.To reduce the surface tension of the water film lining the alveoli, preventing alveolar collapse during expiration.
C.To act as a physical barrier that prevents carbon dioxide from diffusing too rapidly into the blood.
D.To increase the thickness of the respiratory membrane to regulate oxygen diffusion rates.
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解題
Type II pneumocytes secrete a lipoprotein mixture called pulmonary surfactant. This surfactant reduces the surface tension of the fluid lining the alveoli, which prevents the alveoli from collapsing during expiration and reduces the effort needed to expand the lungs during inspiration.
評分準則
Award 1 mark for identifying option B as the correct role of surfactant secreted by Type II pneumocytes.
題目 17 · 選擇題
1 分
In an experiment monitoring pancreatic secretion, blood glucose levels were artificially elevated above the normal set point. What is the response of the pancreatic islet cells and the target mechanism in the liver to bring blood glucose back to homeostasis?
When blood glucose levels rise above normal, beta cells in the pancreatic islets of Langerhans detect this change and increase the secretion of insulin into the bloodstream. Insulin travels to target cells, primarily in the liver and muscles, where it stimulates the uptake of glucose and its conversion into glycogen (glycogenesis), thereby lowering blood glucose levels back to the normal range.
評分準則
Award 1 mark for the correct option (C). Alternate options are incorrect because alpha cells secrete glucagon (not insulin), and insulin stimulates glycogenesis, not glycogenolysis or gluconeogenesis.
題目 18 · 選擇題
1 分
In the CRISPR-Cas9 system used for gene editing, what is the specific role of the single guide RNA (sgRNA)?
A.It cleaves the phosphodiester bonds on both strands of the target DNA.
B.It binds to the Cas9 protein and directs it to a complementary target DNA sequence.
C.It repairs the double-strand break by homologous recombination.
D.It synthesizes a new DNA template strand using the host cell's machinery.
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解題
In the CRISPR-Cas9 system, the single guide RNA (sgRNA) is a synthetic RNA sequence that binds to the Cas9 endonuclease and guides it to a specific sequence of genomic DNA through complementary base pairing, ensuring highly precise targeting. The Cas9 enzyme itself performs the double-strand cleavage.
評分準則
Award 1 mark for the correct option (B). Option A describes the role of the Cas9 enzyme itself. Option C describes the cell's natural repair mechanism. Option D is incorrect as sgRNA does not synthesize DNA.
題目 19 · 選擇題
1 分
During moderate exercise, a mammalian subject's cardiac output increased from \(6.0\text{ dm}^3\text{ min}^{-1}\) to \(18.0\text{ dm}^3\text{ min}^{-1}\). If the heart rate increased from \(60\text{ bpm}\) to \(120\text{ bpm}\), what was the change in stroke volume?
A.It increased by \(50\text{ cm}^3\).
B.It decreased by \(50\text{ cm}^3\).
C.It increased by \(150\text{ cm}^3\).
D.It remained unchanged.
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解題
Cardiac output (CO) is calculated as Heart Rate (HR) multiplied by Stroke Volume (SV). At rest, CO is \(6.0\text{ dm}^3\text{ min}^{-1}\) (6000 \(\text{cm}^3\text{ min}^{-1}\)) and HR is 60 bpm, giving a resting stroke volume of \(6000 / 60 = 100\text{ cm}^3\). During exercise, CO is \(18.0\text{ dm}^3\text{ min}^{-1}\) (18000 \(\text{cm}^3\text{ min}^{-1}\)) and HR is 120 bpm, giving an exercise stroke volume of \(18000 / 120 = 150\text{ cm}^3\). The change in stroke volume is \(150\text{ cm}^3 - 100\text{ cm}^3 = +50\text{ cm}^3\), representing an increase of \(50\text{ cm}^3\).
評分準則
Award 1 mark for the correct option (A). Calculations must correctly convert decimeters cubed to centimeters cubed and use the cardiac output formula.
題目 20 · 選擇題
1 分
Which statement correctly describes the greenhouse effect on Earth?
A.Short-wave solar radiation is absorbed by greenhouse gases in the atmosphere, warming the troposphere directly.
B.Long-wave radiation emitted from the Earth's surface is absorbed and re-emitted by greenhouse gases, trapping heat.
C.Ozone in the stratosphere absorbs ultraviolet radiation and traps it in the lower atmosphere.
D.Greenhouse gases reflect incoming short-wave radiation back into space before it can reach the surface.
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解題
Solar radiation that reaches the Earth is primarily short-wave radiation. Much of this passes through the atmosphere and warms the Earth's surface. The warmed surface then re-emits this energy as long-wave infrared radiation. Greenhouse gases in the atmosphere absorb this long-wave radiation and re-radiate it in all directions, including back to Earth, trapping heat in the atmosphere.
評分準則
Award 1 mark for the correct option (B). Solar radiation is short-wave, while emitted heat from Earth is long-wave. Stratospheric ozone absorbs UV but is not the driver of the greenhouse effect.
題目 21 · 選擇題
1 分
As a spherical cell increases in size, how do its surface area, volume, and surface-area-to-volume ratio change?
A.Surface area increases, volume increases, and surface-area-to-volume ratio increases.
B.Surface area increases, volume increases, and surface-area-to-volume ratio decreases.
C.Surface area decreases, volume increases, and surface-area-to-volume ratio decreases.
D.Surface area increases, volume decreases, and surface-area-to-volume ratio remains constant.
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解題
As a cell increases in radius, both its surface area and volume increase. However, volume increases with the cube of the radius, while surface area increases with the square of the radius. Consequently, the surface-area-to-volume ratio decreases as the cell gets larger.
評分準則
Award 1 mark for the correct option (B). Candidates must recognize that both surface area and volume increase, but the ratio of the two decreases.
題目 22 · 選擇題
1 分
An inhibitor is added to an enzyme-catalyzed reaction. Increasing the substrate concentration gradually restores the rate of reaction to its maximum value (\(V_{\max}\)). What type of inhibition is occurring, and where does the inhibitor bind?
A.Non-competitive inhibition; the inhibitor binds to an allosteric site.
B.Competitive inhibition; the inhibitor binds to the active site.
C.Non-competitive inhibition; the inhibitor binds to the active site.
D.Competitive inhibition; the inhibitor binds to an allosteric site.
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解題
In competitive inhibition, the inhibitor structurally resembles the substrate and binds to the active site. If the concentration of the substrate is significantly increased, the substrate outcompetes the inhibitor for the active site, allowing the reaction rate to reach the normal maximum (\(V_{\max}\)). Non-competitive inhibitors bind to allosteric sites and decrease the overall capacity of the enzyme, so increasing substrate concentration does not restore \(V_{\max}\).
評分準則
Award 1 mark for the correct option (B). The overcoming of inhibition by high substrate concentration is the hallmark of competitive inhibition at the active site.
題目 23 · 選擇題
1 分
Which of the following correctly pairs a polysaccharide with its monomer composition and main biological function?
A.Cellulose / \(\beta\)-glucose monomers / energy storage in plant cells
B.Glycogen / \(\beta\)-glucose monomers / energy storage in animal liver cells
C.Amylose / \(\alpha\)-glucose monomers / energy storage in plant cells
D.Amylopectin / \(\beta\)-glucose monomers / structural support in plant cell walls
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解題
Amylose is a linear form of starch made of alpha-glucose monomers; its primary function is energy storage in plants. Cellulose is made of beta-glucose and functions in structural support, not energy storage. Glycogen is made of alpha-glucose (not beta-glucose). Amylopectin is a branched starch made of alpha-glucose and functions in energy storage, not structural support.
評分準則
Award 1 mark for the correct option (C). Reject options associating beta-glucose with glycogen or amylopectin, or describing cellulose as an energy storage molecule.
題目 24 · 選擇題
1 分
Which statement describes the correct sequence of events in the development of antibiotic resistance in a population of bacteria?
A.Exposure to antibiotic \(\rightarrow\) mutations occur to resist the antibiotic \(\rightarrow\) resistant bacteria survive and reproduce \(\rightarrow\) proportion of resistant bacteria increases.
B.Random mutations create resistance alleles \(\rightarrow\) exposure to antibiotic \(\rightarrow\) non-resistant bacteria die while resistant bacteria survive and reproduce \(\rightarrow\) proportion of resistant bacteria increases.
C.Exposure to antibiotic \(\rightarrow\) bacteria adapt to tolerate the drug \(\rightarrow\) tolerance is passed down to offspring \(\rightarrow\) population becomes resistant.
D.Random mutations create resistance alleles \(\rightarrow\) non-resistant bacteria adapt by altering their cellular pathways \(\rightarrow\) all bacteria survive.
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解題
Natural selection operates on pre-existing genetic variation. Random mutations occur in a bacterial population prior to antibiotic exposure. When an antibiotic is introduced, it acts as a selective pressure, killing susceptible bacteria while allowing those with the pre-existing resistance allele to survive and reproduce. Over time, the frequency of resistance in the population increases.
評分準則
Award 1 mark for the correct option (B). Reject options suggesting that the antibiotic causes the mutation or that individual bacteria acquire resistance through adaptation during their lifetime.
題目 25 · 選擇題
1 分
During intense exercise, the human body coordinates cardiac output and ventilation rate to meet the metabolic demands of active skeletal muscles. Which of the following correctly describes how this physiological integration is mediated?
A.Increased blood pH stimulates peripheral chemoreceptors, which signal the medulla oblongata to increase both heart rate and ventilation rate.
B.Increased partial pressure of carbon dioxide in the blood is detected by central chemoreceptors, leading to increased parasympathetic stimulation of the sinoatrial node.
C.Sensing of decreased blood pH by chemoreceptors triggers the medulla oblongata to increase sympathetic stimulation to the heart and motor signals to ventilation muscles.
D.Decreased blood oxygen levels directly stimulate the sinoatrial node to depolarize faster, while the motor cortex independently increases ventilation via the phrenic nerve.
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解題
During exercise, increased cell respiration in skeletal muscles produces more carbon dioxide (\(\text{CO}_2\)), which reacts with water in the blood to form carbonic acid, thereby lowering blood pH. This decrease in pH is detected by central chemoreceptors in the medulla oblongata and peripheral chemoreceptors in the carotid and aortic bodies. The medulla oblongata coordinates the homeostatic response by increasing sympathetic stimulation to the sinoatrial (SA) node of the heart (increasing heart rate and cardiac output) and sending increased motor signals via the phrenic and intercostal nerves to the diaphragm and external intercostal muscles (increasing ventilation rate and depth). Option A is incorrect because blood pH decreases, not increases. Option B is incorrect because increased \(\text{CO}_2\) leads to sympathetic, not parasympathetic, stimulation. Option D is incorrect because the SA node is not directly stimulated by oxygen levels in this manner.
評分準則
Award 1 mark for the correct option (C). No partial marks are awarded for incorrect options.
題目 26 · 選擇題
1 分
As global temperatures rise, the melting of Arctic permafrost leads to the decomposition of previously frozen organic matter by methanogenic archaea. Which statement correctly identifies the type of feedback loop this represents and its primary effect on climate change?
A.Positive feedback loop, because the released methane decreases the greenhouse effect, cooling the planet.
B.Negative feedback loop, because the increased decomposition rate increases carbon dioxide sink capacity.
C.Positive feedback loop, because the released methane traps more infrared radiation, further increasing global temperatures.
D.Negative feedback loop, because the growth of Arctic vegetation on melted permafrost completely offsets the released greenhouse gases.
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解題
A positive feedback loop is a process in which the end products of an action cause more of that action to occur, amplifying the initial change. Rising global temperatures cause Arctic permafrost to melt, allowing methanogens to decompose organic matter and release methane (\(\text{CH}_4\)), a potent greenhouse gas. This methane traps more outgoing longwave (infrared) radiation in the atmosphere, warming the planet further and causing even more permafrost to melt. Option A is incorrect because methane increases the greenhouse effect. Options B and D are incorrect because this represents an amplifying (positive) feedback loop, not a self-regulating (negative) one, and the growth of vegetation does not completely offset the high warming potential of the released greenhouse gases.
評分準則
Award 1 mark for the correct option (C). No partial marks are awarded for incorrect options.
題目 27 · 選擇題
1 分
A runner completes a marathon on a hot day without drinking sufficient water. Which of the following correctly describes the homeostatic response of their body to restore water balance?
A.Osmoreceptors in the hypothalamus detect a decrease in blood solute concentration, stimulating the posterior pituitary to secrete more ADH, which decreases the water permeability of the collecting duct.
B.Osmoreceptors in the hypothalamus detect an increase in blood solute concentration, stimulating the posterior pituitary to secrete more ADH, which increases the water permeability of the collecting duct.
C.Osmoreceptors in the medulla oblongata detect an increase in blood solute concentration, stimulating the anterior pituitary to secrete less ADH, which increases the water permeability of the collecting duct.
D.Osmoreceptors in the kidneys detect a decrease in blood solute concentration, stimulating the adrenal cortex to secrete aldosterone, which decreases water reabsorption in the collecting duct.
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解題
Dehydration due to sweating causes a decrease in blood volume and an increase in blood solute concentration (osmolarity). Osmoreceptors in the hypothalamus detect this increase in solute concentration and trigger the release of antidiuretic hormone (ADH) from the posterior pituitary gland. ADH travels via the bloodstream to the kidneys, where it binds to receptors on the cells of the collecting ducts, increasing their permeability to water (via the translocation of aquaporins). This allows more water to be reabsorbed back into the blood, reducing blood osmolarity and producing concentrated urine. Therefore, option B is correct.
評分準則
Award 1 mark for the correct option (B). No partial marks are awarded for incorrect options.
題目 28 · 選擇題
1 分
A microbiology student isolates a single-celled organism from an alkaline hot spring. Which combination of features would confirm that this organism is a eubacterial prokaryote rather than a eukaryote?
A.The presence of a cell wall made of peptidoglycan, 70S ribosomes, and circular DNA not associated with histones.
B.The presence of a plasma membrane, 80S ribosomes, and linear DNA packaged into chromosomes.
C.The absence of a nuclear membrane, the presence of 80S ribosomes, and circular DNA associated with histone proteins.
D.The presence of membrane-bound organelles, 70S ribosomes, and double-stranded DNA in a nucleoid region.
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解題
Eubacterial prokaryotes are characterized by having a cell wall made of peptidoglycan, 70S ribosomes in the cytoplasm, and a single circular chromosome of naked DNA (not associated with histone proteins). Eukaryotes have 80S ribosomes in the cytoplasm, linear DNA wrapped around histone proteins, and do not possess peptidoglycan cell walls (plant cell walls are made of cellulose, fungal cell walls of chitin, and animal cells have no cell wall). Therefore, option A describes a combination unique to eubacterial prokaryotes. Option B describes eukaryotic features. Options C and D contain impossible or mixed combinations of prokaryotic and eukaryotic features.
評分準則
Award 1 mark for the correct option (A). No partial marks are awarded for incorrect options.
題目 29 · 選擇題
1 分
An investigator measures the rate of an enzyme-catalyzed reaction under different substrate concentrations in the presence of a constant concentration of an inhibitor. They observe that as the substrate concentration is increased to very high levels, the rate of reaction approaches the same maximum rate (\(V_{\text{max}}\)) as the uninhibited control reaction. What type of inhibitor was used, and how does it affect the enzyme?
A.A non-competitive inhibitor that binds to the active site, preventing substrate binding.
B.A competitive inhibitor that binds to an allosteric site, altering the shape of the active site.
C.A competitive inhibitor that binds to the active site and can be outcompeted by high substrate concentrations.
D.A non-competitive inhibitor that binds to an allosteric site and cannot be outcompeted by high substrate concentrations.
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解題
In the presence of a competitive inhibitor, the maximum rate of reaction (\(V_{\text{max}}\)) can still be reached if the substrate concentration is sufficiently increased. This is because competitive inhibitors have a structural similarity to the substrate and bind to the active site of the enzyme. At extremely high substrate concentrations, the substrate molecules vastly outnumber the inhibitor molecules, making it highly probable that a substrate molecule will bind to the active site rather than an inhibitor, thereby overcoming the inhibition. Non-competitive inhibitors bind to an allosteric site (an alternative site) and alter the conformation of the active site so that the substrate can no longer bind or be converted to product, which reduces the overall \(V_{\text{max}}\) regardless of how much substrate is added. Thus, option C is correct.
評分準則
Award 1 mark for the correct option (C). No partial marks are awarded for incorrect options.
題目 30 · 選擇題
1 分
Humans store chemical energy as both glycogen (a carbohydrate) and triglycerides (a lipid). Why are lipids preferred for long-term energy storage over carbohydrates?
A.Lipids are soluble in water, allowing them to be transported rapidly through the blood to respiring tissues.
B.Lipids release about twice as much energy per gram when respired compared to carbohydrates and are stored without associated water.
C.Lipids are easily mobilized because they can be converted directly into glucose under anaerobic conditions.
D.Lipids form rigid fibrous structures that provide mechanical support to the cell membrane while storing chemical energy.
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解題
Triglycerides (lipids) are superior to carbohydrates like glycogen for long-term energy storage because they release approximately double the energy per gram when metabolized (about \(37\text{ kJ/g}\) compared to \(17\text{ kJ/g}\) for carbohydrates). Additionally, lipids are hydrophobic and insoluble in water, meaning they can be stored compactly in adipose tissue without attracting water (anhydrous storage). In contrast, glycogen is hydrophilic and binds with water molecules, increasing the storage weight significantly. Option A is incorrect because lipids are insoluble in water. Option C is incorrect because lipids cannot be broken down anaerobically to yield glucose. Option D is incorrect because structural carbohydrates like cellulose form rigid fibrous structures, but lipids do not.
評分準則
Award 1 mark for the correct option (B). No partial marks are awarded for incorrect options.
卷二 甲部
Answer all questions in Section A. Calculator allowed.
5 題目 · 37 分
題目 1 · Data Analysis
13 分
The Australian desert hopping mouse (*Notomys alexis*) is highly adapted to survive in extremely arid conditions without drinking active liquid water. Researchers investigated the physiological and anatomical mechanisms that allow this species to conserve water so effectively.
In the first study, researchers measured plasma antidiuretic hormone (ADH) concentration and urine osmolarity in *Notomys alexis* over a 5-day period of complete water deprivation. The results are shown in Table 1.
**Table 1: Plasma ADH levels and urine osmolarity in *Notomys alexis* during water deprivation** | Day of water deprivation | Plasma ADH concentration (pg mL\(^{-1}\)) | Urine osmolarity (mOsmol L\(^{-1}\)) | |---|---|---| | Day 0 (Hydrated) | 1.5 | 800 | | Day 1 | 4.5 | 2100 | | Day 2 | 8.0 | 4500 | | Day 3 | 12.0 | 6500 | | Day 4 | 15.0 | 8200 | | Day 5 | 15.5 | 9100 |
In a second study, researchers compared the relative medullary thickness (RMT) of the kidney (which is proportional to the relative length of the loop of Henle) and the maximum recorded urine concentration across five different rodent species. The results are shown in Table 2.
(a)(i) State the relationship between plasma ADH concentration and urine osmolarity in *Notomys alexis* over the 5-day water deprivation period. [1]
(a)(ii) Identify the day interval during which the greatest increase in urine osmolarity occurred. [1]
(b) Calculate the percentage increase in plasma ADH concentration from Day 0 to Day 4. Show your working. [2]
(c) Explain the physiological processes that lead to the change in plasma ADH levels and urine osmolarity between Day 0 and Day 3 when water is deprived. [4]
(d) Using the data in Table 2, analyze the relationship between relative medullary thickness and maximum urine concentration across the five rodent species. [3]
(e) Based on all the provided data, evaluate the claim that *Notomys alexis* relies solely on hormonal control (ADH secretion) to survive in extremely arid desert environments. [2]
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解題
**(a)(i)** There is a strong positive correlation/relationship; as plasma ADH concentration increases, urine osmolarity also increases.
**(a)(ii)** Between Day 1 and Day 2 (increase of \(2400\text{ mOsmol L}^{-1}\)).
**(c)** - Water deprivation causes blood solute concentration/osmolarity to increase (water potential decreases). - This increase is detected by osmoreceptors in the hypothalamus. - The hypothalamus stimulates the posterior pituitary gland to secrete/release more ADH (antidiuretic hormone) into the blood. - ADH binds to receptors on the cells of the collecting ducts in the kidney nephrons. - This triggers the insertion of aquaporins (water channels) into the luminal membranes of the collecting duct cells. - This increases the permeability of the collecting ducts to water. - Water is reabsorbed by osmosis down the concentration gradient into the hypertonic medulla, producing a highly concentrated urine.
**(d)** - There is a positive correlation between relative medullary thickness (RMT) and maximum urine concentration. - Desert species (*N. alexis*, *P. hermannsburgensis*, *D. merriami*) have both high RMT values (7.2 to 8.5) and very high maximum urine concentrations (6000 to 9100 mOsmol L\(^{-1}\)). - Non-desert species/semi-arid species have lower RMT values (4.0 to 5.5) and produce less concentrated urine (1500 to 3000 mOsmol L\(^{-1}\)). - A higher RMT indicates a longer loop of Henle, which allows the kidney to generate a steeper concentration gradient in the medulla, enhancing water reabsorption.
**(e)** - **Supporting the claim:** Hormonal control via ADH is clearly vital because as ADH rises, the urine osmolarity increases (Table 1), showing immediate, dynamic physiological control of water retention. - **Refuting the claim:** Table 2 shows that anatomical adaptations (a high RMT/long loop of Henle) are also necessary to physically achieve such high concentration gradients; ADH alone cannot concentrate urine to this level without the structural medullary framework. Behavior or other physiological factors (e.g., metabolic water production, nocturnal burrowing) also contribute to survival.
**(c) [Maximum 4 marks]** - [1] Water deprivation increases blood osmolarity / decreases blood water potential. - [1] Detected by osmoreceptors in the hypothalamus, which signals the posterior pituitary to release ADH. - [1] ADH binds to target cells on the collecting duct. - [1] Increases permeability of the collecting ducts to water / stimulates insertion of aquaporins. - [1] More water is reabsorbed by osmosis into the hypertonic medulla / blood, leaving highly concentrated urine.
**(d) [Maximum 3 marks]** - [1] State the overall positive correlation (higher RMT results in higher max urine concentration). - [1] Support with specific numerical data comparing at least two species (e.g., *N. alexis* has RMT of 8.5 and urine of 9100 mOsmol L\(^{-1}\) whereas *R. norvegicus* has RMT of 4.0 and 1500 mOsmol L\(^{-1}\)). - [1] Explain the biological mechanism: larger RMT means longer loops of Henle, which generate a larger solute gradient in the medulla to allow more osmosis/water reabsorption.
**(e) [Maximum 2 marks]** - [1] Support: Table 1 shows that ADH is actively secreted/elevated dynamically during water stress to concentrate urine. - [1] Refute: Table 2 shows anatomical adaptations (RMT/long loop of Henle) are essential to establish the concentration capacity, meaning survival is a combination of anatomical and hormonal factors (accept behavioral factors like staying in burrows / avoiding heat).
題目 2 · Short Answer
6 分
During physical exertion, the cardiovascular system is regulated by both the nervous and endocrine systems.
(a) Describe how the medulla oblongata coordinates the increase in heart rate during exercise. [3] (b) Explain how the hormone epinephrine (adrenaline) interacts with the heart to support this physiological response. [3]
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解題
Part (a) During exercise, chemoreceptors detect increased levels of blood carbon dioxide and a decrease in pH. Proprioceptors in muscles also detect movement. These receptors send nerve impulses to the cardiovascular control centre in the medulla oblongata. In response, the medulla oblongata sends excitatory impulses down the sympathetic nerve to the sinoatrial (SA) node, releasing noradrenaline to increase the heart rate.
Part (b) Epinephrine (adrenaline) is secreted into the blood by the adrenal glands during stress or physical exertion. It travels through the blood and binds to beta-adrenergic receptors on the cell membranes of the sinoatrial (SA) node and cardiac muscle. This binding stimulates the SA node to depolarize faster, increasing heart rate, and enhances muscle contractility, thereby increasing stroke volume.
評分準則
Part (a): - Award [1] for detecting changes in blood pH/carbon dioxide via chemoreceptors or muscle movement via proprioceptors. - Award [1] for transmission of signals to the cardiovascular centre in the medulla oblongata. - Award [1] for sympathetic nerve pathway transmission of impulses to the sinoatrial (SA) node (releasing noradrenaline to increase heart rate).
Part (b): - Award [1] for epinephrine release from the adrenal glands into the circulatory system. - Award [1] for binding of epinephrine to specific receptors on the SA node/cardiac muscle cells. - Award [1] for increasing the rate of depolarization of the SA node (increasing heart rate) and increasing the strength of contraction (stroke volume).
題目 3 · Short Answer
6 分
Climate change leads to numerous feedback loops that accelerate global warming.
(a) Explain how the melting of Arctic permafrost acts as a positive feedback mechanism in global warming. [3] (b) Describe how ocean warming affects the ability of oceans to act as carbon sinks and the subsequent impact on marine calcifying organisms. [3]
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解題
Part (a) Permafrost holds vast reservoirs of organic carbon (dead plant and animal material frozen for thousands of years). As global temperatures rise, the permafrost melts, allowing decomposers (bacteria and fungi) to break down this organic matter. Under aerobic conditions, this releases carbon dioxide, and under anaerobic conditions, it releases methane. These greenhouse gases trap more infrared radiation in the atmosphere, warming the planet further and causing more permafrost to melt.
Part (b) Warm water has lower gas solubility, meaning that as oceans warm, their ability to dissolve and store carbon dioxide from the atmosphere decreases, reducing their efficiency as carbon sinks. Additionally, the absorption of excess carbon dioxide forms carbonic acid, reducing ocean pH (acidification). This lowers the concentration of carbonate ions, making it harder for marine calcifying organisms (like corals and molluscs) to secrete calcium carbonate shells.
評分準則
Part (a): - Award [1] for stating that permafrost contains large stores of organic carbon/matter. - Award [1] for stating that melting permafrost allows microbial decomposition, releasing carbon dioxide or methane. - Award [1] for explaining that these released greenhouse gases trap more heat in the atmosphere, causing further warming and melting (positive feedback).
Part (b): - Award [1] for stating that rising ocean temperatures decrease the solubility of carbon dioxide, reducing carbon sink capacity. - Award [1] for explaining that high carbon dioxide levels cause ocean acidification (decreased pH). - Award [1] for stating that acidification reduces the availability of carbonate ions, inhibiting calcium carbonate shell/skeleton formation in marine organisms.
題目 4 · Short Answer
6 分
The regulation of blood glucose concentration is a key homeostatic process in humans.
(a) Describe how blood glucose is lowered following a meal rich in carbohydrates. [3] (b) Explain the physiological changes that occur when blood glucose levels fall significantly below normal limits during prolonged fasting. [3]
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解題
Part (a) Following a meal, blood glucose concentration rises. This is detected by beta cells in the islets of Langerhans within the pancreas. These cells secrete insulin into the blood. Insulin binds to receptors on target cells (liver and muscle cells), stimulating them to absorb glucose from the blood. It also activates enzymes that convert glucose into glycogen (glycogenesis) for storage, reducing blood glucose levels.
Part (b) During fasting, blood glucose drops below the normal range, which is detected by alpha cells in the islets of Langerhans. These cells secrete glucagon into the blood. Glucagon targets hepatocytes (liver cells), stimulating glycogenolysis (the hydrolysis of stored glycogen back into glucose) and gluconeogenesis (the synthesis of glucose from non-carbohydrates). This glucose is then released into the blood, raising blood glucose levels back to homeostatic norms.
評分準則
Part (a): - Award [1] for beta cells in pancreatic islets of Langerhans detecting high blood glucose and secreting insulin. - Award [1] for insulin promoting the uptake of glucose by liver/muscle cells. - Award [1] for insulin stimulating glycogenesis (conversion of glucose to glycogen) or increasing metabolic use of glucose.
Part (b): - Award [1] for alpha cells in pancreatic islets of Langerhans detecting low blood glucose and secreting glucagon. - Award [1] for glucagon targeting liver cells to stimulate glycogenolysis (breakdown of glycogen to glucose). - Award [1] for glucagon stimulating gluconeogenesis (synthesis of glucose from amino acids/glycerol) to release glucose into the bloodstream.
題目 5 · Short Answer
6 分
Metabolic pathways are tightly regulated to prevent the overproduction of metabolites.
(a) Distinguish between competitive and non-competitive enzyme inhibition, referring to the active site and substrate concentration. [3] (b) Explain how end-product inhibition regulates a metabolic pathway, using the pathway that converts threonine to isoleucine as an example. [3]
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解題
Part (a) Competitive inhibitors bind directly to the enzyme's active site because they are structurally similar to the substrate, preventing substrate binding. Non-competitive inhibitors bind to an allosteric site (a site away from the active site), causing a conformational change in the active site so the substrate can no longer bind. While competitive inhibition can be overcome by increasing the substrate concentration (eventually reaching the same maximum rate, Vmax), non-competitive inhibition cannot be overcome by adding more substrate (lowering the Vmax).
Part (b) In the metabolic pathway converting threonine to isoleucine, isoleucine acts as the end-product. The first enzyme in the pathway is threonine dehydratase. When isoleucine accumulates to high concentrations, it binds allosterically to threonine dehydratase. This inhibits the enzyme, stopping the conversion of threonine to the intermediate, thus shutting down the entire pathway. When isoleucine levels drop, the pathway resumes because the inhibitor dissociates from the enzyme.
評分準則
Part (a): - Award [1] for stating that competitive inhibitors bind to the active site whereas non-competitive inhibitors bind to an allosteric site. - Award [1] for stating that competitive inhibitors are structurally similar to the substrate whereas non-competitive inhibitors are not. - Award [1] for stating that competitive inhibition can be overcome by high substrate concentrations (reaching normal Vmax), whereas non-competitive inhibition reduces the maximum rate (Vmax) regardless of substrate concentration.
Part (b): - Award [1] for identifying isoleucine as the end-product and threonine dehydratase as the first enzyme of the pathway. - Award [1] for explaining that excess isoleucine binds to the allosteric site of threonine dehydratase, altering its active site and inhibiting its activity. - Award [1] for explaining that this is a negative feedback loop where low levels of isoleucine cause the inhibitor to release, restarting the pathway.
卷二 乙部
Answer one chosen extended response question. Up to one extra mark available for quality of construction.
2 題目 · 32 分
題目 1 · extended-response
16 分
a) Explain the principle of negative feedback control in homeostasis, using the regulation of blood glucose concentration in humans as an example. [6]
b) Describe how the human kidney regulates water balance (osmoregulation) during periods of dehydration. [5]
c) Compare the roles of the nervous system and the endocrine system in maintaining homeostasis. [4]
Quality of Construction: Up to one extra mark is available for the clarity, organization, and logical flow of the entire response. [1]
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解題
a) Homeostasis relies on negative feedback, where a deviation of a physiological variable from its set point triggers a corrective response that opposes or reverses the change. Blood glucose levels are monitored continuously by the pancreas. When blood glucose is elevated (e.g., after eating), beta cells in the pancreatic islets of Langerhans detect this change and secrete insulin into the bloodstream. Insulin binds to target cell receptors, stimulating body tissues (especially muscle and adipose) and the liver to absorb glucose from the blood. It also promotes glycogenesis, the conversion of glucose to glycogen for storage in the liver and muscle cells. This lowers blood glucose back to its normal set point, reducing insulin secretion. Conversely, when blood glucose drops below the set point (e.g., during exercise or fasting), alpha cells in the islets of Langerhans detect this and secrete glucagon. Glucagon targets hepatocytes (liver cells) to stimulate glycogenolysis (the breakdown of stored glycogen into glucose) and gluconeogenesis (the synthesis of glucose from non-carbohydrate sources). The newly released glucose enters the blood, raising blood glucose levels back to the set point, which in turn reduces glucagon secretion.
b) During dehydration, blood water potential decreases (the solute concentration or osmolarity of the blood increases). This change is detected by specialized sensory receptors called osmoreceptors located in the hypothalamus. In response, the hypothalamus stimulates the posterior pituitary gland to release antidiuretic hormone (ADH) into the bloodstream. ADH is carried to the kidneys, where it binds to specific receptors on the basolateral membranes of cells lining the collecting ducts. This binding triggers a signaling cascade that causes intracellular vesicles containing aquaporins (water channel proteins) to fuse with the luminal membrane. The insertion of aquaporins significantly increases the water permeability of the collecting duct walls. As the dilute fluid flows down the collecting duct through the highly concentrated solute gradient of the renal medulla, water is rapidly drawn out of the tubule by osmosis into the interstitial fluid and back into the surrounding capillaries (vasa recta). This leads to a low volume of highly concentrated, hypertonic urine being produced, conserving water and restoring blood water potential back to normal.
c) Both the nervous and endocrine systems coordinate homeostasis, but they differ in pathways, speed, and duration: - Transmission pathway: The nervous system transmits signals via electrical impulses (action potentials) along specialized cells called neurones, whereas the endocrine system uses chemical messengers (hormones) transported globally through the circulatory system (blood plasma). - Speed of transmission: Nervous transmission is extremely rapid (occurring in milliseconds), whereas endocrine signaling is generally much slower to take effect (seconds, minutes, or even hours). - Duration of response: The effects of nervous stimulation are typically short-lived and transient, whereas endocrine responses are often long-lasting and persistent. - Target specificity: The nervous system has highly localized and specific targets (synapsing directly onto precise muscle fibers or glands), while the endocrine system can have widespread, systemic effects, acting on any tissue or organ in the body that expresses the specific receptors for that hormone.
評分準則
Part a) [6 marks max] - Negative feedback involves a change in a variable triggering a response that counteracts/reverses the initial deviation to restore a set point. [1] - High blood glucose is detected by beta cells in the pancreatic islets of Langerhans. [1] - Beta cells secrete insulin into the bloodstream. [1] - Insulin stimulates target cells (e.g., muscle/liver) to take up glucose from the blood. [1] - Insulin stimulates glycogenesis (conversion of glucose to glycogen) in the liver/muscles. [1] - Low blood glucose is detected by alpha cells in the pancreatic islets of Langerhans. [1] - Alpha cells secrete glucagon. [1] - Glucagon stimulates glycogenolysis (breakdown of glycogen to glucose) / gluconeogenesis in the liver. [1]
Part b) [5 marks max] - Dehydration causes low blood water potential / high blood osmolarity (solute concentration). [1] - Osmoreceptors in the hypothalamus detect the increase in blood osmolarity. [1] - Hypothalamus stimulates the posterior pituitary gland to secrete ADH. [1] - ADH travels through the blood and binds to cells of the collecting ducts. [1] - ADH causes the insertion of aquaporins (water channels) into the collecting duct membranes. [1] - Collecting duct permeability to water increases, allowing water to move out by osmosis. [1] - Water moves into the hypertonic medulla and is reabsorbed into blood capillaries, resulting in a small volume of concentrated urine. [1]
Part c) [4 marks max] - Nervous system utilizes electrical impulses along neurones AND endocrine system utilizes chemical hormones transported via blood. [1] - Nervous transmission is much faster/immediate AND endocrine transmission is slower. [1] - Nervous responses are short-lived/temporary AND endocrine responses are longer-lasting. [1] - Nervous system has highly localized/specific targets AND endocrine system has widespread/systemic targets (any cell with appropriate receptors). [1] - Both coordinate responses to stimuli to maintain a stable internal environment (homeostasis). [1]
Quality of Construction [1 mark] - 1 mark is awarded if the response is organized logically, presents clear explanations across all sections, and uses appropriate scientific terminology consistently. [1]
題目 2 · extended-response
16 分
a) Explain the principle of negative feedback control in homeostasis, using the regulation of blood glucose concentration in humans as an example. [6]
b) Describe how the human kidney regulates water balance (osmoregulation) during periods of dehydration. [5]
c) Compare the roles of the nervous system and the endocrine system in maintaining homeostasis. [4]
Quality of Construction: Up to one extra mark is available for the clarity, organization, and logical flow of the entire response. [1]
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解題
a) Homeostasis relies on negative feedback, where a deviation of a physiological variable from its set point triggers a corrective response that opposes or reverses the change. Blood glucose levels are monitored continuously by the pancreas. When blood glucose is elevated (e.g., after eating), beta cells in the pancreatic islets of Langerhans detect this change and secrete insulin into the bloodstream. Insulin binds to target cell receptors, stimulating body tissues (especially muscle and adipose) and the liver to absorb glucose from the blood. It also promotes glycogenesis, the conversion of glucose to glycogen for storage in the liver and muscle cells. This lowers blood glucose back to its normal set point, reducing insulin secretion. Conversely, when blood glucose drops below the set point (e.g., during exercise or fasting), alpha cells in the islets of Langerhans detect this and secrete glucagon. Glucagon targets hepatocytes (liver cells) to stimulate glycogenolysis (the breakdown of stored glycogen into glucose) and gluconeogenesis (the synthesis of glucose from non-carbohydrate sources). The newly released glucose enters the blood, raising blood glucose levels back to the set point, which in turn reduces glucagon secretion.
b) During dehydration, blood water potential decreases (the solute concentration or osmolarity of the blood increases). This change is detected by specialized sensory receptors called osmoreceptors located in the hypothalamus. In response, the hypothalamus stimulates the posterior pituitary gland to release antidiuretic hormone (ADH) into the bloodstream. ADH is carried to the kidneys, where it binds to specific receptors on the basolateral membranes of cells lining the collecting ducts. This binding triggers a signaling cascade that causes intracellular vesicles containing aquaporins (water channel proteins) to fuse with the luminal membrane. The insertion of aquaporins significantly increases the water permeability of the collecting duct walls. As the dilute fluid flows down the collecting duct through the highly concentrated solute gradient of the renal medulla, water is rapidly drawn out of the tubule by osmosis into the interstitial fluid and back into the surrounding capillaries (vasa recta). This leads to a low volume of highly concentrated, hypertonic urine being produced, conserving water and restoring blood water potential back to normal.
c) Both the nervous and endocrine systems coordinate homeostasis, but they differ in pathways, speed, and duration: - Transmission pathway: The nervous system transmits signals via electrical impulses (action potentials) along specialized cells called neurones, whereas the endocrine system uses chemical messengers (hormones) transported globally through the circulatory system (blood plasma). - Speed of transmission: Nervous transmission is extremely rapid (occurring in milliseconds), whereas endocrine signaling is generally much slower to take effect (seconds, minutes, or even hours). - Duration of response: The effects of nervous stimulation are typically short-lived and transient, whereas endocrine responses are often long-lasting and persistent. - Target specificity: The nervous system has highly localized and specific targets (synapsing directly onto precise muscle fibers or glands), while the endocrine system can have widespread, systemic effects, acting on any tissue or organ in the body that expresses the specific receptors for that hormone.
評分準則
Part a) [6 marks max] - Negative feedback involves a change in a variable triggering a response that counteracts/reverses the initial deviation to restore a set point. [1] - High blood glucose is detected by beta cells in the pancreatic islets of Langerhans. [1] - Beta cells secrete insulin into the bloodstream. [1] - Insulin stimulates target cells (e.g., muscle/liver) to take up glucose from the blood. [1] - Insulin stimulates glycogenesis (conversion of glucose to glycogen) in the liver/muscles. [1] - Low blood glucose is detected by alpha cells in the pancreatic islets of Langerhans. [1] - Alpha cells secrete glucagon. [1] - Glucagon stimulates glycogenolysis (breakdown of glycogen to glucose) / gluconeogenesis in the liver. [1]
Part b) [5 marks max] - Dehydration causes low blood water potential / high blood osmolarity (solute concentration). [1] - Osmoreceptors in the hypothalamus detect the increase in blood osmolarity. [1] - Hypothalamus stimulates the posterior pituitary gland to secrete ADH. [1] - ADH travels through the blood and binds to cells of the collecting ducts. [1] - ADH causes the insertion of aquaporins (water channels) into the collecting duct membranes. [1] - Collecting duct permeability to water increases, allowing water to move out by osmosis. [1] - Water moves into the hypertonic medulla and is reabsorbed into blood capillaries, resulting in a small volume of concentrated urine. [1]
Part c) [4 marks max] - Nervous system utilizes electrical impulses along neurones AND endocrine system utilizes chemical hormones transported via blood. [1] - Nervous transmission is much faster/immediate AND endocrine transmission is slower. [1] - Nervous responses are short-lived/temporary AND endocrine responses are longer-lasting. [1] - Nervous system has highly localized/specific targets AND endocrine system has widespread/systemic targets (any cell with appropriate receptors). [1] - Both coordinate responses to stimuli to maintain a stable internal environment (homeostasis). [1]
Quality of Construction [1 mark] - 1 mark is awarded if the response is organized logically, presents clear explanations across all sections, and uses appropriate scientific terminology consistently. [1]
Paper 3 甲部
Answer all structured experimental questions.
3 題目 · 15 分
題目 1 · Short Answer
5 分
An experiment was conducted to investigate the effect of copper sulfate concentration on the activity of the enzyme catalase, extracted from potato tubers. Catalase breaks down hydrogen peroxide into water and oxygen. The volume of oxygen gas produced in 3 minutes was measured using a gas syringe.
The data collected is shown below:
0.0 mmol dm-3 copper sulfate: 28.4 cm3 of oxygen
2.0 mmol dm-3 copper sulfate: 21.2 cm3 of oxygen
4.0 mmol dm-3 copper sulfate: 14.5 cm3 of oxygen
6.0 mmol dm-3 copper sulfate: 8.1 cm3 of oxygen
8.0 mmol dm-3 copper sulfate: 3.2 cm3 of oxygen
10.0 mmol dm-3 copper sulfate: 0.8 cm3 of oxygen
(a) Describe the relationship between the concentration of copper sulfate and the volume of oxygen gas collected. [2]
(b) Identify two variables, other than the concentration of copper sulfate, that must be controlled in this experiment. [2]
(c) Suggest how the experiment could be modified to determine if copper sulfate acts as a competitive or non-competitive inhibitor. [1]
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解題
(a) There is an inverse/negative relationship: as the concentration of copper sulfate increases, the volume of oxygen collected decreases. This suggests that copper sulfate acts as an inhibitor of catalase. The reduction is fairly steady but shows a slight non-linear trend as it approaches zero at 10.0 mmol dm-3.
(b) Control variables include: 1. Temperature of the reaction mixture (controlled using a thermostatically controlled water bath). 2. pH of the solution (using a buffer). 3. Concentration and volume of hydrogen peroxide substrate. 4. Volume and concentration of the potato catalase extract.
(c) To determine the type of inhibition, repeat the experiment with a fixed concentration of inhibitor while progressively increasing the concentration of hydrogen peroxide (substrate). If the maximum rate of reaction (Vmax) can be reached at very high substrate concentrations, the inhibition is competitive. If the rate remains suppressed regardless of substrate concentration, it is non-competitive.
評分準則
(a) [Max 2] Award 1 mark for stating that increasing copper sulfate concentration decreases the volume of oxygen collected (negative correlation). Award 1 mark for describing the trend details, such as pointing out that the reaction is almost completely inhibited at 10.0 mmol dm-3 or that the decline is non-linear.
(b) [Max 2] Award 1 mark for each valid control variable identified (up to 2). Accept: pH, temperature, substrate concentration, substrate volume, enzyme concentration, or enzyme volume. Reject: "amount of substrate" or "amount of enzyme" unless specified as volume or concentration.
(c) [Max 1] Award 1 mark for suggesting the addition of more/increasing substrate (hydrogen peroxide) concentration to test if the inhibition can be overcome (competitive) or not (non-competitive).
題目 2 · Short Answer
5 分
An investigation was carried out to study the effect of atmospheric carbon dioxide (\(\text{CO}_2\)) concentration on the stomatal density of dwarf willow (Salix herbacea) leaves. Plants were grown in specialized chambers with varying concentrations of \(\text{CO}_2\) for 8 weeks. Epidermal peels of the lower epidermis were analyzed under a light microscope.
The findings are summarized below:
380 ppm \(\text{CO}_2\): 155 stomata mm-2
500 ppm \(\text{CO}_2\): 132 stomata mm-2
620 ppm \(\text{CO}_2\): 115 stomata mm-2
740 ppm \(\text{CO}_2\): 102 stomata mm-2
860 ppm \(\text{CO}_2\): 94 stomata mm-2
(a) State the relationship between atmospheric \(\text{CO}_2\) concentration and stomatal density in Salix herbacea. [1]
(b) Outline how the student could determine the area of the microscope field of view in order to calculate the stomatal density. [2]
(c) Explain the physiological advantage to the plant of adjusting its stomatal density under elevated \(\text{CO}_2\) levels. [2]
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解題
(a) There is a clear negative correlation/inverse relationship: as the concentration of \(\text{CO}_2\) increases, the mean stomatal density of the leaves decreases.
(b) To determine the area: 1. Place a stage micrometer under the microscope at the same magnification used for counting stomata. 2. Measure the diameter of the circular field of view. 3. Divide the diameter by 2 to find the radius (\(r\)). 4. Calculate the area using the formula \(\text{Area} = \pi r^2\).
(c) Under elevated \(\text{CO}_2\), the concentration gradient of carbon dioxide between the outside air and the inside of the leaf is steeper, so sufficient \(\text{CO}_2\) can diffuse into the leaf through fewer stomatal openings to maintain optimal rates of photosynthesis. Having fewer stomata reduces the total surface area available for the loss of water vapor by transpiration, thus conserving water and improving the plant's water-use efficiency under changing environmental conditions.
評分準則
(a) [Max 1] Award 1 mark for stating that stomatal density decreases as \(\text{CO}_2\) concentration increases (inverse relationship).
(b) [Max 2] Award 1 mark for measuring the diameter of the field of view using a stage micrometer. Award 1 mark for calculating the area using \(\pi r^2\) (or showing how radius is derived from diameter and applied to the circle area formula).
(c) [Max 2] Award 1 mark for stating that elevated \(\text{CO}_2\) levels maintain high diffusion rates/photosynthesis despite fewer stomata. Award 1 mark for explaining that fewer stomata reduce water loss via transpiration / improve water conservation / improve water-use efficiency.
題目 3 · Short Answer
5 分
A student calibrated a light microscope using a stage micrometer where each small division represents \(10\ \mu\text{m}\). They observed that 10 divisions of the eyepiece graticule aligned precisely with 4 divisions of the stage micrometer.
(a) Calculate the value of one division of the eyepiece graticule under this magnification, in micrometres (\(\mu\text{m}\)). Show your working. [2]
(b) Using this calibrated microscope, the student measured the diameter of a human cheek epithelial cell to be 15 eyepiece graticule divisions. Calculate the actual diameter of this cell in micrometres (\(\mu\text{m}\)). [1]
(c) Distinguish between the resolution of a light microscope and a scanning electron microscope (SEM). [2]
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解題
(a) First, determine the physical distance of the aligned stage micrometer divisions: \(4 \text{ divisions} \times 10\ \mu\text{m/division} = 40\ \mu\text{m}\). Since 10 eyepiece graticule divisions span this exact distance, the calibration value of one eyepiece division is: \(\frac{40\ \mu\text{m}}{10\text{ divisions}} = 4\ \mu\text{m}\) per eyepiece division.
(b) The actual diameter of the cell is calculated by multiplying the number of eyepiece graticule divisions by the calibration factor: \(15 \text{ divisions} \times 4\ \mu\text{m/division} = 60\ \mu\text{m}\).
(c) 1. The scanning electron microscope (SEM) has a much higher resolution (typically around \(1\text{ to }10\text{ nm}\)) compared to a light microscope (limited to about \(200\text{ nm}\) or \(0.2\ \mu\text{m}\)). 2. This is because the wavelength of an electron beam in an SEM is much shorter than the wavelength of visible light used in a light microscope, allowing much smaller structures to be distinguished as separate entities.
評分準則
(a) [Max 2] Award 1 mark for correct working showing the conversion of stage divisions to micrometres (\(4 \times 10 = 40\ \mu\text{m}\)). Award 1 mark for the correct final value of \(4\ \mu\text{m}\) per division (must include units).
(b) [Max 1] Award 1 mark for the correct calculation: \(60\ \mu\text{m}\) (accept correct calculation based on an incorrect calibration value from part a, i.e., error carried forward).
(c) [Max 2] Award 1 mark for stating that the SEM has a much higher resolution than a light microscope (accept numerical values like 1-10 nm vs 200 nm). Award 1 mark for explaining that this difference is due to the shorter wavelength of electrons compared to visible light waves.
Paper 3 乙部
Answer all of the questions from one chosen Option (e.g. Option D - Human Physiology).
5 題目 · 20 分
題目 1 · Short Answer
4 分
The liver plays a central role in maintaining homeostatic balance in the human body. Explain how the liver regulates blood glucose levels when they are elevated, and outline the pathway by which surplus amino acids are processed for excretion.
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解題
When blood glucose levels rise above normal, the pancreas secretes insulin, signaling hepatocytes to take up glucose from the bloodstream and convert it into glycogen (glycogenesis). In the case of surplus amino acids, the liver performs deamination, which removes the amine group to yield highly toxic ammonia. The liver immediately converts this ammonia to urea via the ornithine cycle, which is then released back into the bloodstream to be filtered and excreted by the kidneys.
評分準則
Award 1 mark for each of the following up to 4 marks: [1] Insulin is released by pancreas in response to high blood glucose. [2] Hepatocytes absorb glucose and convert it to glycogen (glycogenesis). [3] Deamination of surplus amino acids removes the amine group. [4] Toxic ammonia is produced during deamination and is converted into urea (which is transported to the kidneys for excretion).
題目 2 · Short Answer
4 分
Skeletal muscle contraction requires the coordinated action of several proteins and calcium ions. Describe the precise role of calcium ions and ATP in the sliding filament theory of muscle contraction.
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解題
During muscle contraction, calcium ions are released from the sarcoplasmic reticulum and bind to troponin. This binding induces a conformational change that moves tropomyosin away from the myosin-binding sites on the actin filament. ATP then binds to the myosin head, causing it to detach from the actin filament and breaking the cross-bridge. Hydrolysis of this ATP to ADP and inorganic phosphate (Pi) provides the energy to cock the myosin head into its high-energy state, allowing it to bind to the next active site on actin.
評分準則
Award 1 mark for each of the following up to 4 marks: [1] Calcium ions bind to troponin. [2] Tropomyosin shifts to expose binding sites on the actin filament. [3] ATP binds to the myosin head to cause its detachment from actin (breaking the cross-bridge). [4] Hydrolysis of ATP (to ADP and Pi) provides energy to cock / reset the myosin head.
題目 3 · Short Answer
4 分
Explain how the pituitary gland and the kidneys interact to regulate the solute concentration of blood plasma during periods of dehydration.
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解題
Dehydration leads to an increase in blood solute concentration, which is detected by osmoreceptors in the hypothalamus. The posterior pituitary gland is stimulated to release antidiuretic hormone (ADH) into the bloodstream. ADH travels to the kidneys, where it binds to receptors on the collecting ducts, increasing their permeability to water by inducing the insertion of aquaporins. Consequently, more water is reabsorbed into the hypertonic medulla, resulting in a low volume of highly concentrated urine.
評分準則
Award 1 mark for each of the following up to 4 marks: [1] Osmoreceptors in the hypothalamus detect high blood osmolarity / low water potential. [2] Posterior pituitary releases ADH. [3] ADH increases the permeability of the collecting ducts (or distal convoluted tubules) to water by inserting aquaporins. [4] More water is reabsorbed back into the blood, resulting in a small volume of highly concentrated urine.
題目 4 · Short Answer
4 分
During intense physical exercise, active muscles undergo rapid cellular respiration. Explain how the Bohr shift facilitates the delivery of oxygen to these highly active tissues.
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解題
High rates of cellular respiration in active muscles generate large amounts of carbon dioxide and lactic acid, lowering local blood pH. The increased acidity (higher hydrogen ion concentration) and carbon dioxide levels decrease hemoglobin's affinity for oxygen. This is represented by a rightward shift of the oxygen dissociation curve, known as the Bohr shift. As a result, hemoglobin unloads/releases oxygen much more easily to the tissues that require it most.
評分準則
Award 1 mark for each of the following up to 4 marks: [1] Cellular respiration produces carbon dioxide and lactic acid, lowering blood pH. [2] Increased acidity decreases hemoglobin's affinity for oxygen. [3] This causes a rightward shift in the oxygen-hemoglobin dissociation curve (Bohr shift). [4] Hemoglobin releases/unloads more oxygen to the respiring tissues at any given partial pressure of oxygen.
題目 5 · Short Answer
4 分
Monoclonal antibodies are widely used in diagnostics and therapeutics. Outline the steps involved in the production of monoclonal antibodies using hybridoma cells.
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解題
The production process begins by injecting a specific antigen into a host animal, such as a mouse, to stimulate an immune response. Plasma B cells producing the target antibodies are harvested from the animal's spleen. These B cells are fused with immortal myeloma (cancer) cells to form hybridoma cells. The hybridoma cells are then screened to select the clone producing the desired monoclonal antibody, which is then cultured in vitro to harvest the antibodies on a large scale.
評分準則
Award 1 mark for each of the following up to 4 marks: [1] Injection of antigen into an animal to stimulate B cell production. [2] Harvesting plasma/B cells from the spleen of the immunized animal. [3] Fusion of these B cells with myeloma/cancer cells to create hybridoma cells. [4] Screening, cloning, and culturing of the specific hybridomas to harvest monoclonal antibodies.
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